Table of Contents
Chapter 0
1
Chapter 1
35
Chapter 2
54
Chapter 3
89
Chapter 4
132
Chapter 5
160
Chapter 6
177
Chapter 7
231
Chapter 8
295
Chapter 9
333
Chapter 10
357
Chapter 11
378
Chapter 12
423
Chapter 13
469
Chapter 14
539
Chapter 15
614
Chapter 16
658
Chapter 17
670
Chapter 0
Problems 0.1
7. True;
1. True; –13 is a negative integer.
⎛ b ⎞ ab
8. True, because a ⎜ ⎟ = .
⎝c⎠ c
2. True, because −2 and 7 are integers and 7 ≠ 0.
3. False, because the natural numbers are 1, 2, 3,
and so on.
9. False; the left side is 5xy, but the right side is
5 x 2 y.
0
4. False, because 0 = .
1
10. True; by the associative and commutative
properties, x(4y) = (x ⋅ 4)y = (4 ⋅ x)y = 4xy.
5
5. True, because 5 = .
1
11. distributive
12. commutative
6. False, since a rational number cannot have
7
is not a number
denominator of zero. In fact,
0
at all because we cannot divide by 0.
7. False, because
integer.
8. True;
13. associative
14. definition of division
25 = 5, which is a positive
15. commutative and distributive
16. associative
2 is an irrational real number.
17. definition of subtraction
9. False; we cannot divide by 0.
18. commutative
10. False, because the natural numbers are 1, 2, 3,
and so on, and 3 lies between 1 and 2.
19. distributive
20. distributive
11. True
21. 2x(y − 7) = (2x)y − (2x)7 = 2xy − (7)(2x)
= 2xy − (7 · 2)x = 2xy − 14x
12. False, since the integer 0 is neither positive nor
negative.
22. (a − b) + c = [a + (−b)] + c = a + (−b + c)
= a + [c + (−b)] = a + (c − b)
Problems 0.2
1. False, because 0 does not have a reciprocal.
2. True, because
x+2 x 2 x
= + = + 1.
2
2 2 2
23. (x + y)(2) = 2(x + y) = 2x + 2y
7 3 21
⋅ =
= 1.
3 7 21
24. 2[27 + (x + y)] = 2[27 + (y + x)] = 2[(27 + y) + x]
= 2[(y + 27) + x]
25. x[(2y + 1) + 3] = x[2y + (1 + 3)] = x[2y + 4]
= x(2y) + x(4) = (x · 2)y + 4x = (2x)y + 4x
= 2xy + 4x
3. False; the negative of 7 is −7 because
7 + (−7) = 0.
4. False; 2(3 · 4) = 2(12) = 24, but
(2 · 3)(2 · 4) = 6 · 8 = 48.
26. (1 + a)(b + c) = 1(b + c) + a(b + c)
= 1(b) + 1(c) + a(b) + a(c) = b + c + ab + ac
5. False; –x + y = y + (–x) = y – x.
6. True; (x + 2)(4) = (x)(4) + (2)(4) = 4x + 8.
1
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
27. x(y − z + w) = x[(y − z) + w] = x(y − z) + x(w)
= x[y + (−z)] + xw = x(y) + x(−z) + xw
= xy − xz + xw
51. X(1) = X
28. –2 + (–4) = –6
53. 4(5 + x) = 4(5) + 4(x) = 20 + 4x
29. –6 + 2 = –4
54. –(x – 2) = –x + 2
30. 6 + (–4) = 2
55. 0(–x) = 0
31. 7 – 2 = 5
⎛ 1 ⎞ 8 ⋅1 8
=
56. 8 ⎜ ⎟ =
⎝ 11 ⎠ 11 11
52. 3(x – 4) = 3(x) – 3(4) = 3x – 12
32. 7 – (–4) = 7 + 4 = 11
33. −5 − (−13) = −5 + 13 = 8
57.
5
=5
1
58.
14 x 2 ⋅ 7 ⋅ x 2 x
=
=
21 y 3 ⋅ 7 ⋅ y 3 y
59.
3
3
3
=
=−
−2 x −(2 x)
2x
60.
2 1 2 ⋅1 2
⋅ =
=
3 x 3 ⋅ x 3x
61.
a
a(3b) 3ab
(3b) =
=
c
c
c
34. −a − (−b) = −a + b
35. (–2)(9) = –(2 · 9) = –18
36. 7(–9) = –(7 · 9) = –63
37. (–2)(–12) = 2(12) = 24
38. 19(−1) = (−1)19 = −(1 · 19) = −19
39.
−1
⎛ 9⎞
= −1⎜ − ⎟ = 9
1
−9
⎝ 1⎠
40. –(–6 + x) = –(–6) – x = 6 – x
⎛ 7 ⎞
62. (5a ) ⎜ ⎟ = 7
⎝ 5a ⎠
41. –7(x) = –(7x) = –7x
42. –12(x – y) = (–12)x – (–12)(y) = –12x + 12y
(or 12y – 12x)
63.
−aby −a ⋅ by by
=
=
−ax
−a ⋅ x
x
−3
3
1⋅ 3
1
=− =−
=−
44. −3 ÷ 15 =
15
15
5⋅3
5
64.
7 1 7 ⋅1 7
⋅ =
=
y x y ⋅ x xy
45. −9 ÷ (−27) =
−9
9 9 ⋅1 1
=
=
=
−27 27 9 ⋅ 3 3
65.
2 5 2 ⋅ 5 10
⋅ =
=
x y x ⋅ y xy
46. (−a ) ÷ (−b) =
−a a
=
−b b
66.
1 1 3 2 3+ 2 5
+ = + =
=
2 3 6 6
6
6
67.
5 3 5 9 5 + 9 14 2 ⋅ 7 7
+ = + =
=
=
=
12 4 12 12
12
12 2 ⋅ 6 6
68.
3 7
9 14 9 − 14 −5
5 ⋅1
1
− =
−
=
=
=−
=−
10 15 30 30
30
30
5⋅6
6
43. –[–6 + (–y)] = –(–6) – (–y) = 6 + y
47. 2(–6 + 2) = 2(–4) = –8
48. 3[–2(3) + 6(2)] = 3[–6 + 12] = 3[6] = 18
49. (–2)(–4)(–1) = 8(–1) = –8
50. (−12)(−12) = (12)(12) = 144
2
ISM: Introductory Mathematical Analysis
69.
70.
4 6 4 + 6 10
+ =
=
=2
5 5
5
5
X
5
−
Y
5
=
2 3 16 15 16 − 15 1
− =
−
=
=
5 8 40 40
40
40
74.
= 6÷
l
3
m
=
a3⋅7
=
(b 4 )5
b 4⋅5
75.
−x
y2
z
xy
76.
7
is not defined (we cannot divide by 0).
0
2
=
11.
w4 s 6
y4
x9
x5
= x 9 −5 = x 4
⎛ 2a 4
12. ⎜
⎜ 7b5
⎝
z
x xy
x2
=−
÷
=−
⋅ =−
yz
y 2 xy
y2 z
x
6
⎞
(2a 4 )6
⎟ =
⎟
(7b5 )6
⎠
26 ( a 4 ) 6
=
76 (b5 )6
=
0
77.
=0
7
=
0
78.
is not defined (we cannot divide by 0).
0
13.
( x3 )6
3
=
x( x )
79. 0 · 0 = 0
Problems 0.3
14.
1. (23 )(22 ) = 23+ 2 = 25 (= 32)
64a 4⋅6
117, 649b5⋅6
64a 24
117, 649b30
x3⋅6
1+3
( x 2 )3 ( x 3 ) 2
( x3 ) 4
3. w4 w8 = w4+8 = w12
16.
4
81 = 3
4. z 3 zz 2 = z 3+1+ 2 = z 6
17.
7
−128 = −2
9 5
y y
=
x
y
9+5
12⋅4
6. ( x ) = x
12 4
=
x
8
19.
48
0.04 = 0.2
3
4
4
= x18− 4 = x14
x 2⋅3 x3⋅2
25 = 5
18.
y14
=x
=
x18
x
x12−12 = x 0 = 1
15.
x x
=
x
2. x6 x9 = x6+9 = x15
5.
b 20
⎛ w2 s 3 ⎞
( w2 s 3 ) 2 ( w2 ) 2 ( s3 ) 2 w2⋅2 s3⋅2
=
=
=
10. ⎜
⎟
⎜ y2 ⎟
( y 2 )2
y 2⋅2
y4
⎝
⎠
x
y 6y
= 6⋅ =
y
x
x
3+ 5
a 21
9. (2 x 2 y 3 )3 = 23 ( x 2 )3 ( y 3 )3 = 8 x 2⋅3 y 3⋅3 = 8 x6 y9
l m l 1
l
÷ = ⋅ =
3 1 3 m 3m
3 5
=
5
72.
x
y
(a3 )7
⎛ x2 ⎞
( x 2 )5 x 2⋅5 x10
8. ⎜ ⎟ =
=
=
⎜ y3 ⎟
( y 3 )5 y 3⋅5 y15
⎝ ⎠
5
3 1 1 18 3 2 18 − 3 + 2 17
− + = − + =
=
2 4 6 12 12 12
12
12
6
7.
X −Y
71.
73.
Section 0.3
4
1
1 1
=
=
4
16
16 2
x3⋅4
=
x6 x6
x12
=
x12
x12
Chapter 0: Review of Algebra
20.
–8
=
27
3
3
–8
3
27
=
ISM: Introductory Mathematical Analysis
−2
2
=−
3
3
21. (49)1/ 2 = 49 = 7
37. (9 z 4 )1/ 2 = 9 z 4 = 32 ( z 2 )2 = 32 ( z 2 ) 2
= 3z 2
22. (64)1/ 3 = 3 64 = 4
3/ 2
23. 9
=
3
3 13
39
39
39
=
⋅ =
=
=
2
13
13 13
13
13
132
36.
3
( 9)
3
3
38. (16 y8 )3 / 4 = ⎡ 4 16 y8 ⎤ = ⎡ 4 (2 y 2 )4 ⎤ = (2 y 2 )3
⎢⎣
⎣⎢
⎦⎥
⎦⎥
= (3) = 27
3
= 8y 6
1
24. (9) −5 / 2 =
(9)
1
25. (32) –2 / 5 =
(32)
26. (0.09)
=
1
3
10
–1/ 2
=
⎛ 1 ⎞
27. ⎜ ⎟
⎝ 32 ⎠
=
5/ 2
=
2/5
1
( 9)
5
=
1
=
5
3
1
( 5 32 )
1
=
2
1
243
1
=
(2)
2
=
⎛ 27t 3 ⎞
39. ⎜
⎟
⎜ 8 ⎟
⎝
⎠
1
4
1
=
=
0.3
0.09
10
3
41.
4/5
⎛ 64 ⎞
28. ⎜ − ⎟
⎝ 27 ⎠
4
⎛ 1 ⎞
1
⎛1⎞
= ⎜⎜ 5
⎟⎟ = ⎜ ⎟ =
16
⎝2⎠
⎝ 32 ⎠
2/3
42.
−3 / 4
2/3
c2
= a5 ⋅ b −3 ⋅
1
c2
2
9t 2
⎡ 3t ⎤
=⎢ ⎥ =
4
⎣2⎦
−3 / 4
= a5 ⋅
⎡4⎤
=⎢ ⎥
⎣ x3 ⎦
1
⋅
1
b3 c 2
= x 2 / 5 y 3 / 5 z –10 / 5 =
5 2 3 –10
x y z
−3
=
=
4−3
( x3 )−3
a5
b3 c 2
x2 / 5 y3 / 5
z2
2
2
⎛
64 ⎞
16
⎛ 4⎞
= ⎜⎜ 3 − ⎟⎟ = ⎜ − ⎟ =
27
3
9
⎝
⎠
⎝
⎠
30.
3
54 = 3 27 ⋅ 2 = 3 27
31.
3
2 x3 = 3 2
3 3
x =x
3
2 = 33 2
3
2
32.
4x = 4 x = 2 x
33.
16 x = 16 x = 4 x
4
a5b −3
4
43. 5m−2 m−7 = 5m −2+ ( −7) = 5m −9 =
50 = 25 ⋅ 2 = 25 ⋅ 2 = 5 2
29.
⎛ ⎡ 3t ⎤ 3 ⎞
= ⎜⎢ ⎥ ⎟
⎜⎣ 2 ⎦ ⎟
⎝
⎠
⎛ ⎡ 4 ⎤4 ⎞
= ⎜⎢ ⎥ ⎟
⎜ ⎣ x3 ⎦ ⎟
⎝
⎠
−3
9
9
4
x
x
=
=
=
−9
3
64
x
4
⎛ 256 ⎞
40. ⎜
⎟
⎝ x12 ⎠
1
(0.09)1/ 2
2/3
4
44. x + y –1 = x +
1
45. (3t ) –2 =
(3t )
46. (3 − z ) –4 =
2
5
m9
1
y
=
1
9t 2
1
(3 − z )4
2
47.
4
4
x
x
x
34. 4
=
=
4
16
2
16
5
5 x 2 = (5 x 2 )1/ 5 = 51/ 5 ( x 2 )1/ 5 = 51/ 5 x 2 / 5
48. ( X 3Y −3 )−3 = ( X 3 )−3 (Y −3 )−3
= X −9Y 9
35. 2 8 − 5 27 + 3 128 = 2 4 ⋅ 2 − 5 9 ⋅ 3 + 3 64 ⋅ 2
=
= 2 ⋅ 2 2 − 5 ⋅ 3 3 + 43 2
= 4 2 − 15 3 + 4 3 2
4
Y9
X9
ISM: Introductory Mathematical Analysis
x − y = x1/ 2 − y1/ 2
49.
50.
Section 0.3
u −2 v −6 w3
vw−5
w3−( −5)
=
=
u 2 v1−( −6)
w8
x
9/ 4 3/ 4
1/ 2
y
52.
= a −3 / 4b −1/ 2 a5b −4
63.
= a17 / 4 b −9 / 2
=
a17 / 4
b
=
9/ 2
53. (2a − b + c) 2 / 3 = 3 (2a − b + c)2
64.
4 3 6 9
54. (ab 2 c3 )3 / 4 = 4 (ab 2 c3 )3 = a b c
55. x
=
–4 / 5
1
=
x4 / 5
1
3
1
−
=
x 2 /15
59.
6
60.
3
5
4
]
8
=
=
=
= [x
−4 / 5 1/ 6
]
=x
–4 / 30
51/ 2
3
81/ 4
(3 x)1/ 3
3x
3
2
=
33 y 2
3y2 / 3
66.
18
= 9 =3
2
=x
5
2
=
(2 y )
y 2y
2y
=
1(3 x) 2 / 3
=
(3x)1/ 3 (3 x)2 / 3
3
(3x) 2
3x
=
2 ⋅ y1/ 3
3 y 2 / 3 ⋅ y1/ 3
=
2 y1/ 3 2 3 y
=
3y
3y
5
=
68.
6 ⋅ 51/ 2
51/ 2 ⋅ 51/ 2
3 ⋅ 21/ 4
81/ 4 ⋅ 21/ 4
=
=
6 5
5
34 2
4
16
=
34 2
2
2
5
=
(24 a10b15 )1/ 20
=
ab
2
=
2
31/ 3
=
20
31/ 3 ⋅ 32 / 3
3
3
=
(2334 )1/ 6 6 648
=
3
3
5
3
u 5 / 2 v1/ 2
=
16a10 b15
ab
21/ 2 ⋅ 32 / 3
69. 2 x 2 y –3 x 4 = 2 x6 y –3 =
70.
2 ⋅ a1/ 2 b3 / 4
2 / 4 1/ 4
a1/ 2b1/ 4 ⋅ a1/ 2 b3 / 4
a b a b
21/ 5 a1/ 2b3 / 4 24 / 20 a10 / 20b15 / 20
=
=
ab
ab
4 2
=
15 2
=
(2 y )
1/ 2
12
= 4=2
3
=
3
1
=
1/ 2
9 x2
3x
12
–2 /15
x
6
1
=
65.
67.
w3 / 5 (3w)3 / 5
3
1
3
1
=
−
=
−
5 3
5 3
5
3
5
w
(3w)
w
27 w3
1
1/ 2
x
57. 3w−3 / 5 − (3w) −3 / 5 =
58. [( x )
(2 x)
y (2 y )1/ 2
=
(2 y )
2
5 4
56. 2 x1/ 2 − (2 y )1/ 2 = 2 x − 2 y
–4 1/ 5 1/ 6
y
=
1
3
(2 x)
4 2x
2x
=
1/ 2
2y
2
=
a −3b −2 a5b −4 = (a −3b −2 )1/ 4 a5b −4
4
1/ 2
2 2x
x
2y
z
1/ 2
4(2 x)1/ 2
=
(2 x)
y
62.
4
=
2x
=
u 2 v7
51. x 2 4 xy –2 z 3 = x 2 ( xy –2 z 3 )1/ 4 = x 2 x1/ 4 y –2 / 4 z 3 / 4
=
4
61.
=
23 / 634 / 6
3
2 x6
y3
3 ⋅ u1/ 2 v1/ 2
u 5 / 2 v1/ 2 ⋅ u1/ 2 v1/ 2
=
3u1/ 2 v1/ 2
u 3v
Chapter 0: Review of Algebra
243
71.
243
= 81 = 9
3
=
3
ISM: Introductory Mathematical Analysis
= [(5k 2 ) 2 3]1/ 2 = 5k 2 31/ 2
72. {[(3a3 ) 2 ]−5 }−2 = {[32 a 6 ]−5 }−2
= {3−10 a −30 }−2
83.
= 320 a 60
73.
20
=
64 y ⋅ x
6
x
74.
s
3/ 2
5
=
3 2
s
75.
76.
=
(2 –2 x1/ 2 y –2 )3
1/ 2
⋅x
s
5/ 2
s
2/3
1
2 –6 x3 / 2 y –6
x3 / 2
85.
64 y x
x
=
=
84.
26 y 6
6 1/ 2
=
1/ 2
15 / 6
s
s
4/6
75k 4 = (75k 4 )1/ 2 = [(25k 4 )(3)]1/ 2
82.
(ab −3c)8
(a −1c 2 ) −3
3
= s11/ 6
= x2 ÷
= (31/ 4 )8 = 38 / 4 = 32 = 9
87. −
77. 32 (32) −2 / 5 = 32 (25 ) −2 / 5
78. ⎛⎜ 5 x 2 y ⎞⎟
⎝
⎠
2/5
b 24
2
x6
= x 2 ÷ x6−12 = x 2 ÷ x −6
12
x
1
x6
= x 2 ⋅ x 6 = x8
(–6)(–6) = 36 = 6
86.
8s –2
2s
88.
a5 c14
⎡ x3 ⎤
x6 ( x3 ) 2
÷⎢
=
÷
⎥
3 2
x4
x 4 ( x6 )2
⎣⎢ ( x ) ⎦⎥
= x2 ÷
= 32 (2−2 )
1
= 32 ⋅
22
9
=
4
=
3
(–6)2 ≠ −6 since –6 < 0.
Note that
8
a 3 c −6
( x 2 )3
3 2
(4 3)
a8b −24 c8
= 7 ⋅ 7 2 = 73 = 7
3 7(49)
2
x yz 3 3 xy 2 = 3 ( x 2 yz 3 )( xy 2 ) = 3 x3 y 3 z 3
= xyz
=
3
4
=−
3 2
=−
s s
( a5b−3 c )
3
4
s5
= (a5 )3 (b −3 )3 (c1/ 2 )3
= a15b −9 c3 / 2
=
= [( x 2 y )1/ 5 ]2 / 5 = ( x 2 y )2 / 25
a15 c3 / 2
b9
⎛ 3 x3 y 2 ⎞
89. (3x3 y 2 ÷ 2 y 2 z −3 ) 4 = ⎜
⎟
⎜ 2 y 2 z −3 ⎟
⎝
⎠
= x 4 / 25 y 2 / 25
4
4
79. (2 x –1 y 2 )2 = 22 x –2 y 4 =
80.
3
3
y4 x
=
3
y1/ 3 x1/ 4
=
4y
⎛ 3 x3 z 3 ⎞
=⎜
⎟
⎜ 2 ⎟
⎝
⎠
3 3 4
(3x z )
=
(2)4
4
x2
3 ⋅ y 2 / 3 x3 / 4
y1/ 3 x1/ 4 ⋅ y 2 / 3 x3 / 4
=
24
81x12 z12
=
16
3 x3 / 4 y 2 / 3
=
xy
81.
x x 2 y3 xy 2 = x1/ 2 ( x 2 y 3 )1/ 2 ( xy 2 )1/ 2
=x
1/ 2
( xy
3/ 2
1/ 2
)( x
34 x12 z12
90.
y) = x y
2 5/ 2
6
(
1
–2
2x
16 x3
)
2
=
1
( )
1/ 2 2
2
=
–2 2
(x )
(161/ 2 )2 ( x3 ) 2
1
–4
2x
16 x6
=
1
1
8 x10
= 8 x10
ISM: Introductory Mathematical Analysis
Section 0.4
18. −{−6a − 6b + 6 + 10a + 15b − a[2b + 10]}
= −{4a + 9b + 6 − 2ab − 10a}
= −{−6a + 9b + 6 − 2ab}
= 6a − 9b − 6 + 2ab
Problems 0.4
1. 8x – 4y + 2 + 3x + 2y – 5 = 11x – 2y – 3
2. 6 x 2 − 10 xy + 2 + 2 z − xy + 4
19. x 2 + (4 + 5) x + 4(5) = x 2 + 9 x + 20
= 6 x 2 − 11xy + 2 z + 6
20. u 2 + (5 + 2)u + 2(5) = u 2 + 7u + 10
3. 8t 2 − 6s 2 + 4 s 2 − 2t 2 + 6 = 6t 2 − 2 s 2 + 6
4.
5.
x +2 x + x +3 x = 7 x
21. ( w + 2)( w − 5) = w2 + (−5 + 2) x + 2(−5)
= w2 − 3w − 10
a + 2 3b − c + 3 3b
= a + 5 3b − c
22. z 2 + (–7 − 3) z + (–7)(–3) = z 2 − 10 z + 21
6. 3a + 7b − 9 − 5a − 9b − 21 = −2a − 2b − 30
23. (2 x)(5 x ) + [(2)(2) + (3)(5)]x + 3(2)
= 10 x 2 + 19 x + 6
7. 6 x 2 –10 xy + 2 − 2 z + xy − 4
= 6 x 2 − 9 xy − 2 z + 2 − 4
24. (t)(2t) + [(1)(7) + (−5)(2)]t + (−5)(7)
= 2t 2 − 3t − 35
8.
x +2 x − x −3 x = − x
9.
x + 2 y − x − 3z = 2 y − 3z
25. X 2 + 2( X )(2Y ) + (2Y )2 = X 2 + 4 XY + 4Y 2
26. (2 x)2 − 2(2 x)(1) + 12 = 4 x 2 − 4 x + 1
10. 8z – 4w – 3w + 6z = 14z – 7w
11. 9x + 9y – 21 – 24x + 6y – 6 = –15x + 15y – 27
27. x 2 − 2(5) x + 52 = x 2 − 10 x + 25
12. u − 3v − 5u − 4v + u − 3 = −3u − 7v − 3
13. 5 x − 5 y + xy − 3 x − 8 xy − 28 y
2
2
2
28. (1 ⋅ 2)
2
2
+ [(1)(5) + (–1)(2)] x + (–1)(5)
= 2x + 3 x − 5
= 2 x 2 − 33 y 2 − 7 xy
29.
14. 2 – [3 + 4s – 12] = 2 – [4s – 9] = 2 – 4s + 9
= 11 – 4s
(
3x
)
2
+2
(
)
3 x (5) + (5)2
= 3x + 10 3x + 25
15. 2{3[3x 2 + 6 − 2 x 2 + 10]} = 2{3[ x 2 + 16]}
30.
= 2{3x + 48} = 6 x + 96
2
( x)
2
( y)
2
− 32 = y − 9
31. (2 s )2 − 12 = 4 s 2 − 1
16. 4{3t + 15 – t[1 – t – 1]} = 4{3t + 15 – t[–t]}
= 4{3t + 15 + t 2 } = 4t 2 + 12t + 60
32. ( z 2 )2 − (3w)2 = z 4 − 9w2
17. −5(8 x3 + 8 x 2 − 2( x 2 − 5 + 2 x))
33. x 2 ( x + 4) − 3( x + 4)
= −5(8 x3 + 8 x 2 − 2 x 2 + 10 − 4 x)
= x3 + 4 x 2 − 3x − 12
= −5(8 x3 + 6 x 2 − 4 x + 10)
= −40 x3 − 30 x 2 + 20 x − 50
34. x( x 2 + x + 3) + 1( x 2 + x + 3)
= x3 + x 2 + 3 x + x 2 + x + 3
= x3 + 2 x 2 + 4 x + 3
7
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
35. x 2 (3 x 2 + 2 x − 1) − 4(3x 2 + 2 x − 1)
46.
= 3x 4 + 2 x3 − x 2 − 12 x 2 − 8 x + 4
= 3x 4 + 2 x3 − 13x 2 − 8 x + 4
47.
36. 3 y (4 y + 2 y − 3 y ) − 2(4 y + 2 y − 3 y )
3
2
3
2
2 x3 7 x 4
4
−
+ = 2 x2 − 7 +
x
x x
x
6 x5
2x
= 12 y 4 + 6 y 3 − 9 y 2 − 8 y 3 − 4 y 2 + 6 y
48.
= 12 y 4 − 2 y 3 − 13 y 2 + 6 y
37. x{2( x 2 − 2 x − 35) + 4[2 x 2 − 12 x]}
= x{2 x 2 − 4 x − 70 + 8 x 2 − 48 x}
= x{10 x 2 − 52 x − 70}
= 10 x3 − 52 x 2 − 70 x
2
+
4 x3
2x
2
1
−
2x
2
= 3 x3 + 2 x −
3y − 4 − 9 y − 5
3y
−6 y − 9
=
3y
−6 y 9
=
−
3y 3y
3
= −2 −
y
38. [(2 z )2 − 12 ](4 z 2 + 1) = [4 z 2 − 1](4 z 2 + 1)
x
= (4 z 2 ) 2 − 12 = 16 z 4 − 1
49. x + 5 x 2 + 5 x − 3
x2 + 5x
39. x(3x + 2y – 4) + y(3x + 2y – 4) + 2(3x + 2y – 4)
−3
= 3x 2 + 2 xy − 4 x + 3xy + 2 y 2 − 4 y + 6 x + 4 y − 8
= 3x 2 + 2 y 2 + 5 xy + 2 x − 8
Answer: x +
40. [ x 2 + ( x + 1)]2
= ( x ) + 2 x ( x + 1) + ( x + 1)
2 2
2
−3
x+5
x −1
50. x − 4 x 2 − 5 x + 4
2
x2 − 4 x
–x + 4
–x + 4
0
Answer: x – 1
= x 4 + 2 x3 + 2 x 2 + x 2 + 2 x + 1
= x 4 + 2 x3 + 3 x 2 + 2 x + 1
41. (2a )3 + 3(2a )2 (3) + 3(2a )(3) 2 + (3)3
= 8a3 + 36a 2 + 54a + 27
3 x 2 − 8 x + 17
51. x + 2 3x3 − 2 x 2 + x − 3
42. (3 y )3 − 3(3 y )2 (2) + 3(3 y )(2)2 − (2)3
= 27 y3 − 54 y 2 + 36 y − 8
3 x3 + 6 x 2
–8 x 2 + x
43. (2 x)3 − 3(2 x)2 (3) + 3(2 x)(3)2 − 33
–8 x 2 − 16 x
17 x − 3
17 x + 34
– 37
= 8 x3 − 36 x 2 + 54 x − 27
44. x3 + 3x 2 (2 y ) + 3 x(2 y ) 2 + (2 y )3
= x3 + 6 x 2 y + 12 xy 2 + 8 y 3
Answer: 3x 2 − 8 x + 17 +
45.
z 2 18 z
−
= z − 18
z
z
8
–37
x+2
1
2x2
ISM: Introductory Mathematical Analysis
Section 0.5
z+2
56. z 2 − z + 1 z 3 + z 2 + z
x3 + x 2 + 3 x + 3
52. x − 1 x 4 + 0 x3 + 2 x 2 + 0 x + 1
z3 − z 2 + z
x 4 − x3
x3 + 2 x 2
2z2
x3 – x 2
2z2 − 2z + 2
2z − 2
3x 2 + 0 x
3x 2 − 3x
3x + 1
3x − 3
4
Answer: x3 + x 2 + 3 x + 3 +
Answer: z + 2 +
z − z +1
Problems 0.5
1. 2(ax + b)
4
x −1
2. 2y(3y – 2)
3. 5x(2y + z)
x2 − 2 x + 4
53. x + 2 x + 0 x + 0 x + 0
3
2z − 2
2
2
4. 3x 2 y (1 − 3xy 2 )
x3 + 2 x 2
−2 x 2 + 0
5. 4bc(2a3 − 3ab 2 d + b3cd 2 )
−2 x − 4 x
4x + 0
4x + 8
−8
2
Answer: x 2 − 2 x + 4 −
6. 6u 2 v(uv 2 + 3w4 − 12v 2 )
7. z 2 − 7 2 = ( z + 7)( z − 7)
8
x+2
8. (x + 2)(x − 3)
9. ( p + 3)( p + 1)
3 x − 12
54. 2 x + 3 6 x 2 + 8 x + 1
10. (s – 4)(s – 2)
6x + 9x
−x +1
− x − 32
2
11. (4 x)2 − 32 = (4 x + 3)(4 x − 3)
12. (x + 6)(x – 4)
5
2
13. (a + 7)(a + 5)
5
1
Answer: 3x − + 2
2 2x + 3
14. (2t )2 − (3s )2 = (2t + 3s )(2t − 3s )
x−2
55. 3x + 2 3 x 2 − 4 x + 3
15. x 2 + 2(3)( x) + 32 = ( x + 3) 2
16. (y – 10)(y – 5)
3x2 + 2 x
–6 x + 3
–6 x − 4
7
Answer: x − 2 +
17. 5( x 2 + 5 x + 6)
= 5( x + 3)( x + 2)
7
3x + 2
18. 3(t 2 + 4t − 5)
= 3(t − 1)(t + 5)
9
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
35. y 2 ( y 2 + 8 y + 16) − ( y 2 + 8 y + 16)
19. 3( x 2 − 12 ) = 3( x + 1)( x − 1)
= ( y 2 + 8 y + 16)( y 2 − 1)
20. (3y − 4)(3y − 2)
= ( y + 4)2 ( y + 1)( y − 1)
21. 6 y 2 + 13 y + 2 = (6 y + 1)( y + 2)
36. xy ( x 2 − 4) + z 2 ( x 2 − 4) = ( x 2 − 4)( xy + z 2 )
22. (4x + 3)(x – 1)
= ( x + 2)( x − 2)( xy + z 2 )
23. 2s (6 s 2 + 5s − 4) = 2 s (3s + 4)(2s − 1)
37. b3 + 43 = (b + 4)(b 2 − 4(b) + 42 )
= (b + 4)(b 2 − 4b + 16)
24. (3z ) 2 + 2(3 z )(5) + 52 = (3 z + 5) 2
38. x3 − 13 = ( x − 1)[ x 2 + 1( x) + 12 ]
25. u 3 / 5 v(u 2 − 4v 2 ) = u 3 / 5 v(u + 2v)(u − 2v)
= ( x − 1)( x 2 + x + 1)
26. (3x
) − 1 = (3 x
2/7 2
2
2/7
+ 1)(3x
2/7
− 1)
39. ( x3 )2 − 12 = ( x3 + 1)( x3 − 1)
27. 2 x( x + x − 6) = 2 x( x + 3)( x − 2)
2
= ( x + 1)( x 2 − x + 1)( x − 1)( x 2 + x + 1)
28. ( xy ) 2 − 2( xy )(2) + 22 = ( xy − 2)2
40. 33 + (2 x)3 = (3 + 2 x)[32 − 3(2 x) + (2 x)2 ]
= (3 + 2 x)(9 − 6 x + 4 x 2 )
29. [2(2 x + 1)]2 = 22 (2 x + 1)2
= 4(2 x + 1)2
41. ( x + 3) 2 ( x − 1)[( x + 3) + ( x − 1)]
= ( x + 3)2 ( x − 1)[2 x + 2]
30. 2 x 2 [2 x(1 − 2 x)]2
= 2 x 2 (2 x)2 (1 − 2 x) 2
= ( x + 3)2 ( x − 1)[2( x + 1)]
= 2 x 2 (4 x 2 )(1 − 2 x)2
= 2( x + 3)2 ( x − 1)( x + 1)
= 8 x 4 (1 − 2 x)2
42. (a + 5)2 (a + 1) 2 [(a + 5) + (a + 1)]
31. x( x 2 y 2 − 14 xy + 49) = x[( xy )2 − 2( xy )(7) + 7 2 ]
= (a + 5)2 (a + 1)2 (2a + 6)
= 2(a + 5)2 (a + 1) 2 (a + 3)
= x( xy − 7)2
43. [P(1 + r)] + [P(1 + r)]r = [P(1 + r)](1 + r)
32. x(5x + 2) + 2(5x + 2) = (5x + 2)(x + 2)
= P (1 + r )2
33. x( x − 4) + 2(4 − x )
2
2
44. (3 X + 5 I )[( X − 3I ) − ( X + 2 I )] = (3 X + 5I )(−5I )
= −5I (3 X + 5I )
= x( x 2 − 4) − 2( x 2 − 4)
= ( x 2 − 4)( x − 2)
= (x + 2)(x – 2)(x – 2)
= ( x + 2)( x − 2)
45. ( x 2 )2 − 42 = ( x 2 + 4)( x 2 − 4)
2
= ( x 2 + 4)( x + 2)( x − 2)
34. (x + 1)(x – 1) + (x – 2)(x + 1)
= (x + 1)[(x – 1) + (x – 2)]
= (x + 1)(2x – 3)
46. (9 x 2 )2 − ( y 2 )2 = (9 x 2 + y 2 )(9 x 2 − y 2 )
= (9 x 2 + y 2 )(3 x + y )(3 x − y )
10
ISM: Introductory Mathematical Analysis
Section 0.6
47. ( y 4 ) 2 − 12 = ( y 4 + 1)( y 4 − 1)
8.
= ( y 4 + 1)( y 2 + 1)( y 2 − 1)
= ( y 4 + 1)( y 2 + 1)( y + 1)( y − 1)
9.
48. (t 2 ) 2 − 22 = (t 2 + 2)(t 2 − 2)
⎡
= (t 2 + 2) ⎢t 2 −
⎣
(
= (t 2 + 2) t +
( 2 ) ⎤⎥⎦
2 )( t − 2 )
2
10.
49. ( X + 5)( X − 1) = ( X + 5)( X + 1)( X − 1)
2
2
2
(t + 3)(t − 3) t 2
t (t + 3)(t − 3)
11.
51. y ( x 4 − 2 x 2 + 1) = y ( x 2 − 1)2 = y[( x + 1)( x − 1)]2
= y ( x + 1)2 ( x − 1) 2
52. 2 x(2 x 2 − 3 x − 2) = 2 x(2 x + 1)( x − 2)
2.
3.
4.
5.
6.
7.
a 2 − 3a
=
(a − 3)(a + 3) a + 3
=
a(a − 3)
a
x − 3x − 10
2
x −4
2
x 2 − 9 x + 20
x + x − 20
2
=
=
( x − 5)( x − 4) x − 5
=
( x + 5)( x − 4) x + 5
2 x − 16 x + 14 x
3( x − 8)
=
2 x( x − 7)
3
2
6 x2 + x − 2
2 x + 3x − 2
2
=
12 x − 19 x + 4
2
6 x − 17 x + 12
2
12.
( x + 2)( x − 5) x − 5
=
( x + 2)( x − 2) x − 2
3 x 2 − 27 x + 24
=
3( x − 8)( x − 1)
2 x( x − 7)( x − 1)
(3x + 2)(2 x − 1) 3x + 2
=
( x + 2)(2 x − 1)
x+2
=
t
t −3
(ax − b)(c − x) (ax − b)(−1)( x − c)
=
( x − c)(ax + b)
( x − c)(ax + b)
(ax − b)(−1)
=
ax + b
b − ax
=
ax + b
( x + y )( x − y )( x + y )2 ( x − y )( x + y ) 2
=
( x + y )( y − x)
(–1)( x − y )
2( x − 1)
( x + 4)( x + 1)
⋅
( x − 4)( x + 2) ( x + 1)( x − 1)
=
2( x − 1)( x + 4)( x + 1)
( x − 4)( x + 2)( x + 1)( x − 1)
=
2( x + 4)
( x − 4)( x + 2)
Problems 0.6
a2 − 9
=
= −( x + y ) 2
50. ( x 2 − 9)( x 2 − 1) = ( x + 3)( x − 3)( x + 1)( x − 1)
1.
2
(4 x − 1)(3x − 4) 4 x − 1
=
(2 x − 3)(3x − 4) 2 x − 3
=
x( x + 2)( x − 2)2
3( x − 4)( x − 2)( x − 3)( x + 2)
=
x( x − 2)
3( x − 4)( x − 3)
13.
X 2 4 4X 2 X
⋅ =
=
8 X
8X
2
14.
3x 2 14 3 x 14 3(14) x
⋅ = ⋅ =
=6
7x x
7 x
7x
15.
2m n3 2mn3 n
⋅
=
=
n 2 6m 6mn 2 3
16.
c + d 2c
2c(c + d ) 2(c + d )
⋅
=
=
c c−d
c (c − d )
c−d
17.
4x
4x 1 4x 2
÷ 2x =
⋅
=
=
3
3 2x 6x 3
18.
4 x 2 x 4 x(2 x) 8 x 2
⋅
=
=
1 3
3
3
y 2 (–1)
y2
=−
( y − 3)( y + 2)
( y − 3)( y + 2)
11
x ( x + 2)
( x − 2)2
⋅
3( x − 4)( x − 2) ( x − 3)( x + 2)
Chapter 0: Review of Algebra
19.
ISM: Introductory Mathematical Analysis
–9 x3 3 –27 x3
⋅ =
= −27 x 2
1 x
x
27.
−12Y 4
−12Y 3 1 −12Y 3
÷4 =
⋅ =
= −3Y 3
20.
Y
1
4
4
21.
22.
23.
24.
25.
26.
x −3
x−4
x −3 1
x−3
⋅
=
⋅
=
=1
1 ( x − 3)( x − 4)
1 x−3 x−3
28.
(2 x + 3)(2 x − 3)(1 + x)(1 − x)
( x + 4)( x − 1)(2 x − 3)
=
(2 x + 3)(1 + x)(–1)( x − 1)
( x + 4)( x − 1)
2
=
( x − 3)( x + 2) ( x + 3)( x − 1)
⋅
( x + 3)( x − 3) ( x + 2)( x − 2)
x + 2 ( x + 3)( x − 1)
=
⋅
x + 3 ( x + 2)( x − 2)
( x + 2)( x + 3)( x − 1)
=
( x + 3)( x + 2)( x − 2)
x −1
=
x−2
(2 x + 3)(1 + x)
x+4
y (6 x 2 + 7 x − 3) x( y − 1) + 4( y − 1)
⋅
x( y – 1) + 5( y − 1)
x 2 y ( x + 4)
=
10 x
x +1
10 x ( x + 1)
2x
⋅
=
=
( x + 1)( x − 1) 5 x
5 x( x + 1)( x − 1) x − 1
3
=
=−
( x + 3) 2
( x + 3)2 1
÷ ( x + 3) =
⋅
x
x
x+3
2
( x + 3)
x+3
=
=
x ( x + 3)
x
3
(2 x + 3)(2 x − 3) (1 + x)(1 − x)
⋅
( x + 4)( x − 1)
2x − 3
y (3 x − 1)(2 x + 3)( y − 1)( x + 4)
( y − 1)( x + 5) x 2 y ( x + 4)
(3 x − 1)(2 x + 3)
x 2 ( x + 5)
29.
x 2 + 5 x + 6 ( x + 3)( x + 2)
=
= x+2
x+3
x+3
30.
2+ x x+2
=
=1
x+2 x+2
31. LCD = 3t
2 1 6 1 6 +1 7
+ = + =
=
t 3t 3t 3t
3t
3t
( x + 2)( x + 5) ( x − 4)( x + 1)
⋅
( x + 5)( x + 1) ( x − 4)( x + 2)
x + 2 x +1
=
⋅
x +1 x + 2
( x + 2)( x + 1)
=
( x + 1)( x + 2)
=1
32. LCD = X 3
9
1
9
X
9− X
−
=
−
=
3
2
3
3
X
X
X
X
X3
33. LCD = x3 − 1
1−
( x + 3) 2 (3 + 4 x)(3 − 4 x)
⋅
4x − 3
7( x + 3)
x3
x3 − 1
=
=
( x + 3)2 (3 + 4 x)(3 − 4 x)
=
7(4 x − 3)( x + 3)
( x + 3)(3 + 4 x)(−1)(4 x − 3)
=
7(4 x − 3)
( x + 3)(3 + 4 x)
=−
7
=
=
x3 − 1
−
x3
x3 − 1 x3 − 1
x 3 − 1 − x3
x3 − 1
−1
x3 − 1
1
1 − x3
34. LCD = s + 4
4
4
s ( s + 4) 4 + s ( s + 4)
+s=
+
=
s+4
s+4
s+4
s+4
=
12
s 2 + 4s + 4 ( s + 2)2
=
s+4
s+4
ISM: Introductory Mathematical Analysis
Section 0.6
35. LCD = (2x – 1)(x + 3)
4
x
4( x + 3)
x(2 x − 1)
+
=
+
2 x − 1 x + 3 (2 x − 1)( x + 3) ( x + 3)(2 x − 1)
=
39. LCD = (x – 1)(x + 5)
−3x 2
4
−3+
x −1
−( x − 1)( x + 5)
4( x + 3) + x(2 x − 1) 2 x 2 + 3x + 12
=
(2 x − 1)( x + 3)
(2 x − 1)( x + 3)
36. LCD = (x – 1)(x + 1)
x + 1 x − 1 ( x + 1)( x + 1) ( x − 1)( x − 1)
−
=
−
x − 1 x + 1 ( x − 1)( x + 1) ( x − 1)( x + 1)
=
( x + 1) − ( x − 1)
( x + 1)( x − 1)
2
4( x + 5)
3( x − 1)( x + 5)
3x 2
−
+
( x − 1)( x + 5) ( x − 1)( x + 5) ( x − 1)( x + 5)
=
4 x + 20 − 3( x 2 + 4 x − 5) + 3x 2
( x − 1)( x + 5)
=
35 − 8 x
( x − 1)( x + 5)
2
40. LCD = (2x – 1)(x + 6)(3x – 2)
2x − 3
3x + 1
1
−
+
(2 x − 1)( x + 6) (3x − 2)( x + 6) 3 x − 2
x 2 + 2 x + 1 − ( x 2 − 2 x + 1)
4x
=
=
( x + 1)( x − 1)
( x + 1)( x − 1)
37. LCD = ( x − 3)( x + 1)( x + 3)
1
1
+
( x − 3)( x + 1) ( x + 3)( x − 3)
x+3
x +1
=
+
( x − 3)( x + 1)( x + 3) ( x − 3)( x + 1)( x + 3)
( x + 3) + ( x + 1)
=
( x − 3)( x + 1)( x + 3)
2x + 4
=
( x − 3)( x + 1)( x + 3)
2( x + 2)
=
( x − 3)( x + 1)( x + 3)
=
(2 x − 3)(3 x − 2) − (3 x + 1)(2 x − 1) + (2 x − 1)( x + 6)
(2 x − 1)( x + 6)(3 x − 2)
=
6 x 2 − 13x + 6 − (6 x 2 − x − 1) + 2 x 2 + 11x − 6
(2 x − 1)( x + 6)(3 x − 2)
=
2x2 − x + 1
(2 x − 1)( x + 6)(3 x − 2)
2
2
2
x2 + 2 x + 1
⎛ 1⎞
⎛ x 1⎞
⎛ x +1⎞
41. ⎜1 + ⎟ = ⎜ + ⎟ = ⎜
⎟ =
⎝ x⎠
⎝ x x⎠
⎝ x ⎠
x2
2
2
2
–1
⎛ 1 − xy ⎞
=⎜
⎟
⎝ x ⎠
⎛1 1⎞
⎛ y
⎛ y+x⎞
x ⎞
42. ⎜ + ⎟ = ⎜ + ⎟ = ⎜
⎟
⎝x y⎠
⎝ xy xy ⎠
⎝ xy ⎠
38. LCD = (x − 4)(2x + 1)(2x − 1)
4
x
−
( x − 4)(2 x + 1) ( x − 4)(2 x − 1)
4(2 x − 1)
x(2 x + 1)
=
−
( x − 4)(2 x + 1)(2 x − 1) ( x − 4)(2 x + 1)(2 x − 1)
4(2 x − 1) − x(2 x + 1)
=
( x − 4)(2 x + 1)(2 x − 1)
=
=
=
y 2 + 2 xy + x 2
x2 y 2
–1
⎛1
⎞
⎛ 1 xy ⎞
43. ⎜ − y ⎟ `= ⎜ − ⎟
⎝x
⎠
⎝x x ⎠
−2 x 2 + 7 x − 4
( x − 4)(2 x + 1)(2 x − 1)
2
2
1⎞
⎛
⎛ ab 1 ⎞
⎛ ab + 1 ⎞
44. ⎜ a + ⎟ = ⎜ + ⎟ = ⎜
⎟
b⎠
⎝
⎝ b b⎠
⎝ b ⎠
a 2b 2 + 2ab + 1
=
b2
–1
=
x
1 − xy
2
45. Multiplying the numerator and denominator of
7x +1
the given fraction by x gives
.
5x
46. Multiplying numerator and denominator by x
x+3
x+3
1
=
=
gives
.
2
+
−
−3
(
x
3)(
x
3)
x
x −9
13
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
47. Multiplying numerator and denominator by
2x(x + 2) gives
3(2 x)( x + 2) − 1( x + 2) ( x + 2)[3(2 x) − 1]
=
x(2 x)( x + 2) + x(2 x)
2 x 2 [( x + 2) + 1]
=
( x + 2)(6 x − 1)
2 x 2 ( x + 3)
.
56.
=
=
−12
12
=
=−
.
( x + 3)( x + 2)[9 + ( x − 7)]
( x + 3)( x + 2)2
3
x+h
3
3
=
x
=
33 x
5+ a
1
+
a
=
=
51.
1
2− 3
⋅
2+ 3 2− 3
57.
x+h x
x+h x
3
3
3 x − x+h
3
3
(
3
3
)
x + h3 x
3
50. LCD = 5 + a a
a a
33 x + h
−
a a
58.
( a ) + 1(
5+a a
a + 5+a
5+a
)
59.
5+ a a
2
a 5+a
=
2+ 3
⋅
=
2+ 3
2 2
(
2+ 3
)
2−3
4+2 6
= −4 − 2 6
−1
2 5
3+ 7
⋅
3− 7 3+ 7
2 5 3+ 7
(
2
)
3−7
15 + 35
(
)
−4
15 + 35
=−
2
49. LCD = 3 x + h ⋅ 3 x
−
2− 3
=
48. Multiplying numerator and denominator by
3(x + 3)(x + 2) gives
3( x − 1) − 1(3)( x + 3)
3(3)( x + 3)( x + 2) + ( x − 7)( x + 3)( x + 2)
3
2 2
55.
2− 3
= 2− 3
4−3
3
⋅
t− 7
t+ 7 t− 7
( x − 3) + 4
x −1
(
3t − 3 7
t2 − 7
x +1
=
5 2− 3
=
x −1
)
x +1
⋅
x +1
=
( x + 1)
)
x +1
x −1
(
4 1+ 2
−
(
)
( 2 + 3 )( 2 − 3 ) (1 − 2 )(1 + 2 )
5 ( 2 − 3 ) 4 (1 + 2 )
=
−
=
(
4−3
5 2− 3
(
1
1− 2
) − 4 (1 + 2 )
) (
−1
)
= 5 2 − 3 + 4 1 + 2 = 4 2 − 5 3 + 14
1+ 2
1+ 2 1+ 2
52.
⋅
=
=
= −1 − 2
1− 2
−1
1− 2 1+ 2
1
60.
2
53.
3− 6
2
=
(
6+ 7
=
3+ 6
3−6
5
(
⋅
)=
6 + 12
6+2 3
=−
−3
3
6− 7
6− 7
6− 7
−1
3
3+ 6
3+ 6
5
54.
⋅
) =5
(
=
5
(
6− 7
)
6−7
7− 6
)
14
(
4 x2
x +2
=
) 3(
4x2
(
x +2
x −2
)(
)
x −2
)
=
4 x2
(
x −2
3( x − 4)
)
ISM: Introductory Mathematical Analysis
Section 0.7
Set x = –4:
Problems 0.7
2(–4) + (–4)2 − 8 ⱨ 0
–8 + 16 – 8 ⱨ 0
0=0
Thus, 2 and –4 satisfy the equation.
2
1. 9 x − x = 0
Set x = 1:
9(1) − (1) 2 ⱨ 0
9 −1 ⱨ 0
8≠0
Set x = 0:
5. x(6 + x) – 2(x + 1) – 5x = 4
Set x = –2:
(–2)(6 – 2) – 2(–2 + 1) – 5(–2) ⱨ 4
–2(4) – 2(–1) + 10 ⱨ 4
–8 + 2 + 10 ⱨ 4
4=4
Set x = 0:
0(6) – 2(1) – 5(0) ⱨ 4
–2 ≠ 4
Thus, –2 satisfies the equation, but 0 does not.
9(0) − (0)2 ⱨ 0
0–0ⱨ0
0=0
Thus, 0 satisfies the equation, but 1 does not.
2. 12 − 7 x = − x 2 ; 4, 3
Set x = 4:
12 − 7(4) ⱨ − (4) 2
12 − 28 ⱨ − 16
−16 = −16
Set x = 3:
6. x( x + 1)2 ( x + 2) = 0
Set x = 0:
0(1) 2 (2) ⱨ 0
0=0
Set x = –1:
12 − 7(3) ⱨ − (3)2
12 − 21 ⱨ − 9
−9 = −9
Thus, 4 and 3 satisfy the equation.
3. z + 3( z − 4) = 5;
(–1)(0)2 (1) ⱨ 0
0=0
Set x = 2:
2(3)2 (4) ⱨ 0
72 ≠ 0
Thus, 0 and –1 satisfy the equation, but 2 does
not.
17
,4
4
17
:
4
17
⎛ 17
⎞
+ 3⎜ − 4 ⎟ ⱨ 5
4
⎝ 4
⎠
17 51
+ − 12 ⱨ 5
4 4
5=5
Set z = 4:
4 + 3(4 − 4) ⱨ 5
4+0ⱨ5
4≠5
17
Thus,
satisfies the equation, but 4 does not.
4
Set z =
7. Adding 5 to both sides; equivalence guaranteed
8. Dividing both sides by 8; equivalence
guaranteed
9. Raising both sides to the third power;
equivalence not guaranteed.
10. Dividing both sides by 2; equivalence
guaranteed
11. Dividing both sides by x; equivalence not
guaranteed
4. 2 x + x 2 − 8 = 0
Set x = 2:
12. Multiplying both sides by x – 2; equivalence not
guaranteed
2 ⋅ 2 + 22 − 8 ⱨ 0
4+4–8ⱨ0
0=0
13. Multiplying both sides by x – 1; equivalence not
guaranteed
14. Dividing both sides by (x + 3); equivalence not
guaranteed.
15
Chapter 0: Review of Algebra
15. Multiplying both sides by
ISM: Introductory Mathematical Analysis
26. 4s + 3s − 1 = 41
7 s − 1 = 41
7 s = 42
42
s=
=6
7
2x − 3
; equivalence
2x
not guaranteed
16. Adding 9 – x to both sides and then dividing
both sides by 2; equivalence guaranteed
27. 5( p − 7) − 2(3 p − 4) = 3 p
5 p − 35 − 6 p + 8 = 3 p
− p − 27 = 3 p
−27 = 4 p
27
p=−
4
17. 4x = 10
10 5
x=
=
4 2
18. 0.2x = 7
7
x=
= 35
0.2
28. t = 2 – 2[2t – 3(1 – t)]
t = 2 – 2[2t – 3 + 3t]
t = 2 – 2[5t – 3]
t = 2 – 10t + 6
11t = 8
8
t=
11
19. 3y = 0
0
y= =0
3
20. 2x – 4x = –5
–2x = –5
–5 5
x=
=
–2 2
29.
21. −8 x = 12 − 20
−8 x = −8
−8
x=
=1
−8
22. 4 − 7 x = 3
−7 x = −1
−1 1
=
x=
−7 7
30.
23. 5x – 3 = 9
5x = 12
12
x=
5
24.
x
= 2x − 6
5
x = 5(2x – 6)
x = 10x – 30
30 = 9x
30 10
x=
=
9
3
5y 6
− = 2 − 4y
7 7
5y – 6 = 14 – 28y
33y = 20
20
y=
33
4x x
=
9 2
Multiplying both sides by 9 · 2 gives
9 · 2 · 7 + 2(4x) = 9(x)
126 + 8x = 9x
x = 126
31. 7 +
2x + 3 = 8
2x = 5
5 ⎛ 5 2⎞
x=
⎜ or
⎟
2 ⎟⎠
2 ⎜⎝
32.
25. 7x + 7 = 2(x + 1)
7x + 7 = 2x + 2
5x + 7 = 2
5x = –5
–5
x=
= −1
5
16
x
x
−4 =
3
5
5x – 60 = 3x
2x = 60
x = 30
ISM: Introductory Mathematical Analysis
Section 0.7
w w w
+ −
= 120
2 6 24
Multiplying both sides by 24 gives
24 w − 12 w + 4 w − w = 2880
15w = 2880
2880
w=
= 192
15
4
r −5
3
Multiplying both sides by 3 gives
3r = 4r − 15
−r = −15
r = 15
39. w −
33. r =
34.
3x 5 x
+
=9
5
3
9 x + 25 x = 135
34 x = 135
135
x=
34
40.
x
1
− 5 = + 5x
5
5
Multiplying both sides by 5 gives
15x + x – 25 = 1 + 25x
16x – 25 = 1 + 25x
–9x = 26
26
x=−
9
35. 3x +
41.
y y y y
+ − =
2 3 4 5
60y – 30y + 20y – 15y = 12y
35y = 12y
23y = 0
y=0
36. y −
37.
38.
42.
2y − 3 6y + 7
=
4
3
Multiplying both sides by 12 gives
3(2y – 3) = 4(6y + 7)
6y – 9 = 24y + 28
–18y = 37
37
y=−
18
43.
t 5
7
+ t = (t − 1)
4 3
2
Multiplying both sides by 12 gives
3t + 20t = 42(t − 1)
23t = 42t − 42
42 = 19t
42
t=
19
44.
17
7 + 2( x + 1) 6 x
=
3
5
35 + 10(x + 1) = 18x
35 + 10x + 10 = 18x
45 = 8x
45
x=
8
x+2 2− x
−
= x−2
3
6
Multiplying both sides by 6 gives
2(x + 2) – (2 – x) = 6(x – 2)
2x + 4 – 2 + x = 6x – 12
3x + 2 = 6x – 12
2 = 3x – 12
14 = 3x
14
x=
3
x 2( x − 4)
+
=7
5
10
2x + 2(x – 4) = 70
2x + 2x – 8 = 70
4x = 78
78 39
x=
=
4
2
9
3
(3 − x) = ( x − 3)
5
4
Multiplying both sides by 20 gives
36(3 – x) = 15(x – 3)
108 – 36x = 15x – 45
153 = 51x
x=3
2 y − 7 8y − 9 3y − 5
+
=
3
14
21
14(2y – 7) + 3(8y – 9) = 2(3y – 5)
28y – 98 + 24y – 27 = 6y – 10
46y = 115
115 5
=
y=
46 2
Chapter 0: Review of Algebra
45.
46.
ISM: Introductory Mathematical Analysis
4
(5 x − 2) = 7[ x − (5 x − 2)]
3
4(5 x − 2) = 21( x − 5 x + 2)
20 x − 8 = −84 x + 42
104 x = 50
50 25
x=
=
104 52
52.
53.
(2 x − 5)2 + (3x − 3) 2 = 13x 2 − 5 x + 7
4 x 2 − 20 x + 25 + 9 x 2 − 18 x + 9 = 13x 2 − 5 x + 7
13x 2 − 38 x + 34 = 13x 2 − 5 x + 7
−33 x = −27
−27 9
x=
=
−33 11
47.
48.
54.
5
= 25
x
Multiplying both sides by x gives
5 = 25x
5
x=
25
1
x=
5
55.
4
=2
x −1
4 = 2(x – 1)
4 = 2x – 2
6 = 2x
x=3
56.
49. Multiplying both sides by 3 – x gives 7 = 0,
which is false. Thus there is no solution, so the
solution set is ∅.
50.
57.
3x − 5
=0
x −3
3x − 5 = 0
3x = 5
5
x=
3
58.
3
7
51.
=
5 − 2x 2
3(2) = 7(5 − 2 x )
6 = 35 − 14 x
14 x = 29
29
x=
14
18
x+3 2
=
x
5
5(x + 3) = 2x
5x + 15 = 2x
3x = –15
x = –5
q
1
=
5q − 4 3
3q = 5q – 4
–2q = –4
q=2
4p
=1
7− p
4p = 7 – p
5p = 7
7
p=
5
1
2
=
p −1 p − 2
p – 2 = 2(p – 1)
p – 2 = 2p – 2
p=0
2x − 3
=6
4x − 5
2x – 3 = 24x – 30
27 = 22x
27
x=
22
1 1 3
+ =
x 7 7
1 3 1
= −
x 7 7
1 2
=
x 7
7
x=
2
2
3
=
x −1 x − 2
2( x − 2) = 3( x − 1)
2 x − 4 = 3x − 3
−x = 1
x = −1
ISM: Introductory Mathematical Analysis
59.
Section 0.7
3x − 2 3x − 1
=
2x + 3 2x +1
(3x – 2)(2x + 1) = (3x – 1)(2x + 3)
64.
6 x2 − x − 2 = 6 x2 + 7 x − 3
1 = 8x
1
x=
8
60.
−2 x 2 + 5 x − 2 − 3(–2 x 2 + 7 x − 3) = 4( x 2 − 5 x + 6)
4 x 2 − 16 x + 7 = 4 x 2 − 20 x + 24
4x = 17
17
x=
4
x + 2 x +1
+
=0
x −1 3 − x
(x + 2)(3 – x) + (x + 1)(x – 1) = 0
65.
3x − x 2 + 6 − 2 x + x 2 − 1 = 0
x+5=0
x = –5
61.
y−6 6 y+6
− =
y
y y−6
Multiplying both sides by y(y − 6) gives
66.
( y − 6)2 − 6( y − 6) = y ( y + 6)
y 2 − 12 y + 36 − 6 y + 36 = y 2 + 6 y
y−2 y−2
=
y+2 y+3
x+5 = 4
(
2
y + y−6 = y −4
y=2
63.
x
x
3x − 4
−
=
x + 3 x − 3 x2 − 9
x(x – 3) – x(x + 3) = 3x – 4
67.
(y − 2)(y + 3) = (y − 2)(y + 2)
2
9
3x
=
x −3 x −3
9 = 3x
x=3
But the given equation is not defined for x = 3,
so there is no solution. The solution set is ∅.
x 2 − 3x − x 2 − 3x = 3x − 4
–6x = 3x – 4
–9x = –4
4
x=
9
y 2 − 18 y + 72 = y 2 + 6 y
72 = 24y
y=3
62.
1
3
4
−
=
x − 3 x − 2 1− 2x
(x – 2)(1 – 2x) – 3(x – 3)(1 – 2x)
= 4(x – 3)(x – 2)
x+5
)
2
= 42
x + 5 = 16
x = 11
−5
7
11
=
+
2 x − 3 3 − 2 x 3x + 5
Multiplying both sides by (2x − 3)(3x + 5) gives
−5(3 x + 5) = −7(3 x + 5) + 11(2 x − 3)
−15 x − 25 = −21x − 35 + 22 x − 33
−15 x − 25 = x − 68
−16 x = −43
43
x=
16
z−2 =3
68.
(
z−2
)
2
= 32
z–2=9
z = 11
3x − 4 − 8 = 0
3x − 4 = 8
69.
(
3x − 4
)
2
3x − 4 = 64
3x = 68
68
x=
3
19
= (8)2
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
70. 4 − 3x + 1 = 0
4 = 3x + 1
2
4 =
(
3x + 1
)
2
⎛ y 2 − 9 ⎞ = (9 − y ) 2
⎜
⎟
⎝
⎠
2
16 = 3 x + 1
15 = 3 x
x=5
y 2 − 9 = 81 − 18 y + y 2
18y = 90
90
y=
=5
18
x
2
+1 =
2
3
71.
2
⎛ x ⎞
⎛2⎞
+ 1 ⎟⎟ = ⎜ ⎟
⎜⎜
⎝3⎠
⎝ 2 ⎠
x
4
+1 =
2
9
x
5
=−
2
9
10
⎛ 5⎞
x = 2⎜ − ⎟ = −
9
⎝ 9⎠
y + 2 = 3− y
(
(6 y )
(
) ( )
x
78.
4 + 3x
= 72
x − x +1 = 1
2
)
x +1 +1
2
x = x +1+ 2 x +1 +1
2
−2 = 2 x + 1
−1 = x + 1 , which is impossible because
a ≥ 0 for all a. Thus there is no solution.
The solution set is ∅.
) =(
2
2x + 5
)
79.
2
z2 + 2z = 3 + z
2
⎛ z 2 + 2 z ⎞ = (3 + z )2
⎜
⎟
⎝
⎠
4 + 3x = 2 x + 5
x =1
75.
2
( x) = (
4 + 3x = 2 x + 5
(
2
x = x +1 +1
4x – 6 = x
3x = 6
x=2
74.
2
36y = 49
49
y=
36
4x − 6 = x
=
) = (3 − y )
6 y =7
[( x + 6)1/ 2 ]2 = 7 2
x + 6 = 49
x = 43
2
y+2
y+2 = 9−6 y + y
72. ( x + 6)1/ 2 = 7
4x − 6
y + y+2 =3
77.
2
73.
y2 − 9 = 9 − y
76.
z2 + 2z = 9 + 6z + z2
–9 = 4z
9
z=−
4
( x − 5)3 / 4 = 27
[( x − 5)3 / 4 ]4 / 3 = 27 4 / 3
x − 5 = 81
x = 86
20
ISM: Introductory Mathematical Analysis
80.
Section 0.7
1
2
−
=0
w
5w − 2
87. A =
1
2
=
w
5w − 2
R=
2
⎛ 1⎞
⎛
2 ⎞
⎜⎜
⎟⎟ = ⎜⎜
⎟⎟
⎝ w⎠
⎝ 5w − 2 ⎠
1
2
=
w 5w − 2
5w – 2 = 2w
3w = 2
2
w=
3
P=
1 − (1 + i )− n
2
88. S =
R[(1 + i )n − 1]
i
Si = R[(1 + i ) n − 1]
R=
Si
(1 + i ) n − 1
d
1 − dt
r(1 – dt) = d
r – rdt = d
–rdt = –r + d
d −r r −d
t=−
=
rd
rd
89. r =
81. I = Prt
I
r=
Pt
p ⎞
⎛
82. P ⎜ 1 +
⎟−R = 0
⎝ 100 ⎠
p ⎞
⎛
P ⎜1 +
⎟=R
⎝ 100 ⎠
R[1 − (1 + i ) − n ]
i
Ai
90.
R
p
1 + 100
83. p = 8q – 1
p + 1 = 8q
p +1
q=
8
x −a x −b
=
b−x a−x
Multiplying both sides by (b – x)(a – x) gives
(x – a)(a – x) = (x – b)(b – x)
(x – a)(a – x)(–1) = (x – b)(b – x)(–1)
(x – a)(x – a) = (x – b)(x – b)
x 2 − 2ax + a 2 = x 2 − 2bx + b 2
a 2 − b 2 = 2ax − 2bx
(a + b)(a – b) = 2x(a – b)
a + b = 2x (for a ≠ b)
a+b
=x
2
84. p = –3q + 6
p – 6 = –3q
p−6 6− p
q=
=
−3
3
91. r =
85. S = P(1 + rt)
S = P + Prt
S – P = r(Pt)
S–P
r=
Pt
2mI
B (n + 1)
2mI
B
2mI
n +1 =
rB
2mI
−1
n=
rB
r (n + 1) =
2mI
B (n + 1)
r[ B (n + 1)]
=I
2m
rB (n + 1)
I=
2m
86. r =
21
Chapter 0: Review of Algebra
92.
ISM: Introductory Mathematical Analysis
1 1 1
+ =
p q f
98.
1 1 1
= −
q f p
1 p− f
=
q
pf
q=
pf
p− f
93. P = 2l + 2 w
660 = 2l + 2(160)
660 = 2l + 320
340 = 2l
340
l=
= 170
2
The length of the rectangle is 170 m.
94.
vf
334.8
v(2500)
495 =
334.8
165, 726 = 2500v
165, 726
= 66.2904
v=
2500
Since the car is traveling at 66.2904 mi/h on a
65 mi/h highway, the officer can claim that you
were speeding.
F=
99. Bronwyn’s weekly salary for working h hours is
27h + 18. Steve’s weekly salary for working
h hours is 35h.
1
(27h + 18 + 35h) = 550
5
62h + 18 = 2750
62h = 2732
2732
h=
≈ 44.1
62
They must each work 44 hours each week.
V = πr 2 h
355 = π(2) 2 h
355 = 4πh
355
h=
4π
The height of the can is
355
≈ 28.25 centimeters.
4π
100. y = a(1 – by)x
y = ax(1 – by)
y = ax – abxy
y + abxy = ax
y(1 + abx) = ax
ax
y=
1 + abx
95. c = x + 0.0825x = 1.0825x
1.4 x
1 + 0.09 x
With y = 10 the equation is
1.4 x
10 =
1 + 0.09 x
10(1 + 0.09x) = 1.4x
10 + 0.9x = 1.4x
10 = 0.5x
x = 20
The prey density should be 20.
101. y =
96. Revenue equals cost when 450x = 380x + 3500.
450x = 380x + 3500
70x = 3500
x = 50
50 toddlers need to be enrolled.
n⎞
⎛
97. V = C ⎜ 1 − ⎟
⎝ N⎠
⎛ n⎞
2000 = 3200 ⎜ 1 − ⎟
⎝ 8⎠
2000 = 3200 – 400n
400n = 1200
n=3
The furniture will have a value of $2000
after 3 years.
102. Let x = the maximum number of customers.
8
10
=
x − 92 x − 46
8(x − 46) = 10(x − 92)
8x − 368 = 10x − 920
552 = 2x
x = 276
The maximum number of customers is 276.
22
ISM: Introductory Mathematical Analysis
Section 0.8
d
r −c
t(r – c) = d
tr – tc = d
tr − d = tc
tr − d
d
c=
=r−
t
t
109. −
103. t =
110.
Problems 0.8
1. x 2 − 4 x + 4 = 0
theorem, x 2 + 1002 = ( x + 1)2 .
( x − 2)2 = 0
x–2=0
x=2
x 2 + 10, 000 = x 2 + 2 x + 1
10, 000 = 2 x + 1
9999 = 2 x
9999
x=
= 4999.5
2
The distance from the top of the tower to the
house is x + 1 = 4999.5 + 1 = 5000.5 meters.
45 = 24d
(
24d
)
2
2025 = 24d
2025 675
3
d=
=
= 84 ≈ 84 ft
24
8
8
106. Let P be the amount in the account one year ago.
Then the interest earned is 0.073P and
P + 0.073P = 1257.
1.073P = 1257
1257
P=
≈ 1171.48
1.073
The amount in the account one year ago was
$1171.48, and the interest earned is
$1171.48(0.073) = $85.52.
or t + 2 = 0
or t = –2
t 2 − 8t + 15 = 0
(t − 3)(t − 5) = 0
t−3=0
t=3
or t − 5 = 0
or t = 5
4. (x – 2)(x + 5) = 0
x–2=0
x=2
or x + 5 = 0
or x = –5
5. x 2 − 2 x − 3 = 0
(x – 3)(x + 1) = 0
x–3=0
x=3
or x + 1 = 0
or x = –1
6. (x – 4)(x + 4) = 0
x–4=0
x=4
or x + 4 = 0
or x = –4
7. u 2 − 13u = −36
u 2 − 13u + 36 = 0
(u − 4)(u − 9) = 0
u–4=0
u=4
107. Let e be Tom’s expenses in Nova Scotia before
the HST tax. Then the HST tax is 0.15e and the
total receipts are e + 0.15e = 1.15e. The
percentage of the total that is HST is
0.15e 0.15 15
3
=
=
=
or approximately
1.15e 1.15 115 23
13%.
108.
2. (t + 1)(t + 2) = 0
t+1=0
t = –1
3.
105. s = 30 fd
Set s = 45 and (for dry concrete) f = 0.8.
45 = 30(0.8)d
(45) =
14
is a root.
61
111. 0 is a root.
104. Let x = the horizontal distance from the base of
the tower to the house. By the Pythagorean
2
1
is a root.
2
8. 3( w2 − 4 w + 4) = 0
3( w − 2)2 = 0
w–2=0
w=2
1
1
and −
are roots.
8
14
23
or u – 9 = 0
or u = 9
Chapter 0: Review of Algebra
9. x 2 − 4 = 0
(x – 2)(x + 2) = 0
x–2=0
x=2
17. − x 2 + 3 x + 10 = 0
or x + 2 = 0
or x = –2
10. 3u (u − 2) = 0
u=0
u=0
or u − 2 = 0
or u = 2
11. t 2 − 5t = 0
t (t − 5) = 0
t=0
t=0
or t – 5 = 0
or t = 5
12. x 2 + 9 x + 14 = 0
(x + 7)(x + 2) = 0
x+7=0
x = –7
ISM: Introductory Mathematical Analysis
18.
x 2 − 3x − 10 = 0
(x – 5)(x + 2) = 0
x–5=0
x=5
or x + 2 = 0
or x = –2
1 2 3
y − y=0
7
7
1
y ( y − 3) = 0
7
y=0
y=0
or y – 3 = 0
or y = 3
19. 2 p 2 = 3 p
2 p2 − 3 p = 0
p(2p – 3) = 0
p=0
or x + 2 = 0
or x = –2
2
13. 4 x + 1 = 4 x
p=0
2
4x − 4x + 1 = 0
(2 x − 1)2 = 0
2x – 1 = 0
1
x=
2
20. r 2 + r − 12 = 0
(r – 3)(r + 4) = 0
r–3=0
r=3
22. ( w − 3)2 ( w + 1)2 = 0
w−3=0
or w + 1 = 0
w=3
or w = −1
15. v(3v − 5) = −2
3v 2 − 5v = −2
23.
2
16.
or r + 4 = 0
or r = –4
21. x(x + 4)(x – 1) = 0
x = 0 or x + 4 = 0 or x – 1 = 0
x = 0 or x = –4
or x = 1
14. 2 z 2 + 9 z − 5 = 0
(2z – 1)(z + 5) = 0
2z – 1 = 0
or z + 5 = 0
1
z=
or z = –5
2
3v − 5v + 2 = 0
(3v − 2)(v − 1) = 0
3v – 2 = 0
2
v=
3
or 2p – 3 = 0
3
or p =
2
t 3 − 49t = 0
t (t 2 − 49) = 0
t (t + 7)(t − 7) = 0
t = 0 or t + 7 = 0
or t − 7 = 0
t = 0 or
t = −7 or
t=7
or v − 1 = 0
or v = 1
24. x( x 2 − 4 x − 5) = 0
x(x – 5)(x + 1) = 0
x = 0 or x – 5 = 0 or x + 1 = 0
x = 0 or x = 5
or x = –1
−6 x 2 + x + 2 = 0
6 x2 − x − 2 = 0
(2 x + 1)(3x − 2) = 0
2x +1 = 0
or 3x − 2 = 0
1
2
x=−
or
x=
2
3
25. 6 x3 + 5 x 2 − 4 x = 0
x(6 x 2 + 5 x − 4) = 0
x(2x – 1)(3x + 4) = 0
24
ISM: Introductory Mathematical Analysis
Section 0.8
x = 0 or 2x – 1 = 0 or 3x + 4 = 0
4
1
or x = −
x = 0 or x =
3
2
32. x 2 − 2 x − 15 = 0
a = 1, b = –2, c = –15
x=
26. x 2 + 2 x + 1 − 5 x + 1 = 0
x2 − 3x + 2 = 0
(x – 1)(x – 2) = 0
x–1=0
x=1
27.
=
or x – 2 = 0
or x = 2
( x − 3)( x 2 − 4) = 0
( x − 3)( x − 2)( x + 2) = 0
x − 3 = 0 or
x−2=0
x=3
or
x=2
or
or
28. 5(x + 4)(x − 3)(x − 8) = 0
x + 4 = 0 or x − 3 = 0 or
x = −4 or
x = 3 or
–(–2) ± 4 − 4(1)(–15)
2(1)
2 ± 64
2
2±8
=
2
2+8
x=
=5
2
=
x+2=0
x = −2
x −8 = 0
x=8
29. p( p − 3) 2 − 4( p − 3)3 = 0
=
( p − 3) 2 [ p − 4( p − 3)] = 0
( p − 3) 2 (12 − 3 p ) = 0
)( x − 2 ) = 0
−(−12) ± 144 − 4(4)(9)
2(4)
−b ± b 2 − 4ac
2a
5 ± 25 − 4(1)(0)
=
2(1)
5 ± 25
=
2
5±5
=
2
5+5
q=
=5
or
2
q=
31. x 2 + 2 x − 24 = 0
a = 1, b = 2, c = –24
=
−b ± b 2 − 4ac
2a
34. q 2 − 5q = 0
a = 1, b = −5, c = 0
x + 1 = 0 or x − 1 = 0
or x + 2 = 0 or x − 2 = 0
x = −1 or x = 1
or x = − 2 or x = 2
x=
2−8
= −3
2
12 ± 0
8
12 ± 0
=
8
3
=
2
3( p − 3) (4 − p) = 0
p–3=0
or 4 – p = 0
p=3
or p = 4
(
x=
=
2
( x + 1)( x − 1) x + 2
or
33. 4 x 2 − 12 x + 9 = 0
a = 4, b = –12, c = 9
x=
30. ( x 2 − 1)( x 2 − 2) = 0
–b ± b 2 − 4ac
2a
–b ± b 2 − 4ac
2a
–2 ± 4 − 4(1)(–24)
2(1)
–2 ± 100
2
−2 ± 10
=
2
−2 + 10
−2 − 10
x=
= 4 or x =
= −6
2
2
=
25
q=
5−5
=0
2
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
35. p 2 − 2 p − 7 = 0
39. 4 x 2 + 5 x − 2 = 0
a = 4, b = 5, c = –2
a = 1, b = −2, c = −7
−b ± b 2 − 4ac
2a
−5 ± 25 − 4(4)(−2)
=
2(4)
−5 ± 57
=
8
−5 + 57
−5 − 57
x=
or x =
8
8
−b ± b 2 − 4ac
p=
2a
x=
−(−2) ± (−2)2 − 4(1)(−7)
2(1)
2 ± 32
=
2
= 1± 2 2
=
p = 1+ 2 2
or
p = 1− 2 2
40. w2 − 2w + 1 = 0
a = 1, b = −2, c = 1
36. 2 − 2 x + x 2 = 0
x2 − 2 x + 2 = 0
a = 1, b = –2, c = 2
–(–2) ± 4 − 4(1)(2)
x=
2(1)
w=
−(−2) ± (−2) 2 − 4(1)(1)
2(1)
2± 0
=
2
=1
=
2 ± −4
2
no real roots
=
37. 4 − 2n + n 2 = 0
41. 0.02w2 − 0.3w = 20
n 2 − 2n + 4 = 0
a = 1, b = –2, c = 4
n=
=
0.02 w2 − 0.3w − 20 = 0
a = 0.02, b = –0.3, c = –20
–b ± b 2 − 4ac
2a
w=
−(−2) ± 4 − 4(1)(4)
2(1)
=
2 ± −12
2
no real roots
=
−b ± b 2 − 4ac
2a
−1 ± 1 − 4(2)(–5)
2(2)
−1 ± 41
4
−1 + 41
x=
4
−(–0.3) ± 0.09 − 4(0.02)(–20)
2(0.02)
0.3 ± 1.69
0.04
0.3 ± 1.3
=
0.04
0.3 + 1.3
w=
=
0.04
0.3 − 1.3
w=
=
0.04
2x2 + x − 5 = 0
a = 2, b = 1, c = –5
=
−b ± b 2 − 4ac
2a
=
38. 2 x 2 + x = 5
x=
−b ± b 2 − 4ac
2a
=
or
x=
−1 − 41
4
26
1.6
= 40 or
0.04
–1.0
= −25
0.04
ISM: Introductory Mathematical Analysis
Section 0.8
42. 0.01x 2 + 0.2 x − 0.6 = 0
a = 0.01, b = 0.2, c = –0.6
x=
44. −2 x 2 − 6 x + 5 = 0
a = –2, b = –6, c = 5
−b ± b 2 − 4ac
2a
x=
=
–0.2 ± 0.04 − 4(0.01)(–0.6)
2(0.01)
=
=
–0.2 ± 0.064
0.02
=
=
–0.2 ± (0.0064)(10)
0.02
=
–0.2 ± 0.08 10
0.02
x = −10 + 4 10 or x = −10 − 4 10
45. ( x 2 )2 − 5( x 2 ) + 6 = 0
43. 2 x 2 + 4 x = 5
Let w = x 2 . Then
2x2 + 4 x − 5 = 0
a = 2, b = 4, c = –5
w2 − 5 w + 6 = 0
(w – 3)(w – 2) = 0
w = 3, 2
–b ± b 2 − 4ac
2a
Thus x 2 = 3 or x 2 = 2, so x = ± 3, ± 2 .
–4 ± 16 − 4(2)(–5)
=
2(2)
–4 ± 56
4
–4 ± 2 14
=
4
–2 ± 14
=
2
−2 + 14
x=
2
–(–6) ± 36 − 4(–2)(5)
2(−2)
6 ± 76
−4
6 ± 2 19
=
−4
−3 ± 19
=
2
−3 + 19
−3 − 19
x=
or x =
2
2
= −10 ± 4 10
x=
−b ± b 2 − 4ac
2a
46. ( X 2 ) 2 − 3( X ) 2 − 10 = 0
Let w = X 2 . Then
=
w2 − 3w − 10 = 0
( w − 5)( w + 2) = 0
w = 5, −2
Thus X 2 = 5 or X 2 = −2, so the real solutions
or
x=
are X = ± 5.
−2 − 14
2
2
⎛1⎞
⎛1⎞
47. 3 ⎜ ⎟ − 7 ⎜ ⎟ + 2 = 0
⎝ x⎠
⎝ x⎠
1
Let w = . Then
x
3w2 − 7 w + 2 = 0
(3w − 1)( w − 2) = 0
1
w= , 2
3
1
Thus, x = 3, .
2
27
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
48. ( x −1 )2 + x −1 − 12 = 0
2
⎛ 1 ⎞
⎛ 1 ⎞
53. ⎜
⎟ − 12 ⎜
⎟ + 35 = 0
⎝ x−2⎠
⎝ x−2⎠
1
, then
Let w =
x−2
Let w = x –1. Then
w2 + w − 12 = 0
(w + 4)(w – 3) = 0
w = –4, 3
1 1
Thus, x = − , .
4 3
w2 − 12w + 35 = 0
(w – 7)(w – 5) = 0
w = 7, 5
1
1
= 5.
= 7 or
Thus,
x−2
x−2
15 11
x= , .
7 5
49. ( x –2 )2 − 9( x –2 ) + 20 = 0
Let w = x –2 . Then
w2 − 9 w + 20 = 0
(w – 5)(w – 4) = 0
w = 5, 4
1
1
1
1
= 5 or
= 4, so x 2 = or x 2 = .
Thus,
2
2
5
4
x
x
5
1
x=±
,± .
5
2
2
⎛ 1 ⎞
⎛ 1 ⎞
54. 2 ⎜
⎟ + 7⎜
⎟+3= 0
⎝ x+4⎠
⎝ x+4⎠
1
Let w =
. Then
x+4
2 w2 + 7 w + 3 = 0
(2w + 1)(w + 3) = 0
1
w = − , −3
2
1
1
1
= −3 .
= − or
Thus,
x+4
2
x+4
13
x = −6, −
3
2
⎛ 1 ⎞
⎛ 1 ⎞
50. ⎜ ⎟ − 9 ⎜ ⎟ + 8 = 0
2
⎝x ⎠
⎝ x2 ⎠
1
. Then
Let w =
x2
w2 − 9 w + 8 = 0
(w – 8)(w – 1) = 0
w = 8, 1
1
1
1
= 8 or
= 1, so x 2 = or x 2 = 1.
Thus,
2
2
8
x
x
2
x=±
, ± 1.
4
55. x 2 =
x+3
2
2 x2 = x + 3
2x2 − x − 3 = 0
(2x – 3)(x + 1) = 0
3
Thus, x = , − 1.
2
51. ( X − 5)2 + 7( X − 5) + 10 = 0
Let w = X − 5. Then
w2 + 7 w + 10 = 0
( w + 2)( w + 5) = 0
w = −2, −5
Thus, X − 5 = −2 or X − 5 = −5, so X = 3, 0.
56.
x 7 5
= −
2 x 2
Multiplying both sides by the LCD, 2x, gives
x 2 = 14 − 5 x
x 2 + 5 x − 14 = 0
(x – 2)(x + 7) = 0
Thus, x = 2, –7.
52. (3x + 2) 2 − 5(3x + 2) = 0
Let w = 3x + 2. Then
w2 − 5 w = 0
w( w − 5) = 0
w = 0, 5
2
Thus 3x + 2 = 0 or 3x + 2 = 5, so x = − , 1.
3
28
ISM: Introductory Mathematical Analysis
57.
58.
Section 0.8
3
x −3
+
=2
x−4
x
Multiplying both sides by the LCD, x(x – 4),
gives
3x + (x – 3)(x – 4) = 2x(x – 4)
61.
3x + x 2 − 7 x + 12 = 2 x 2 − 8 x
2r + 8 − (r 2 − r − 2) = 0
x 2 − 4 x + 12 = 2 x 2 − 8 x
−r 2 + 3r + 10 = 0
0 = x 2 − 4 x − 12
0 = (x – 6)(x + 2)
Thus, x = 6, –2.
r 2 − 3r − 10 = 0
(r – 5)(r + 2) = 0
Thus, r = 5, –2.
2
6
−
=5
2x +1 x −1
Multiplying both sides by the LCD,
(2x + 1)(x − 1), gives
2( x − 1) − 6(2 x + 1) = 5(2 x + 1)( x − 1)
62.
2x − 3
2x
+
=1
2 x + 5 3x + 1
Multiplying both sides by the LCD,
(2x + 5)(3x + 1), gives
(2x – 3)(3x + 1) + 2x(2x + 5) = (2x + 5)(3x + 1)
−10 x − 8 = 10 x 2 − 5 x − 5
6 x 2 − 7 x − 3 + 4 x 2 + 10 x = 6 x 2 + 17 x + 5
0 = 10 x 2 + 5 x + 3
a = 10, b = 5, c = 3
10 x 2 + 3 x − 3 = 6 x 2 + 17 x + 5
4 x 2 − 14 x − 8 = 0
2
b − 4ac = 25 − 4(10)(3) = −95 < 0, thus there
are no real roots.
59.
2
r +1
−
=0
r−2 r+4
Multiplying both sides by the LCD,
(r – 2)(r + 4), gives
2(r + 4) – (r – 2)(r + 1) = 0
2x2 − 7 x − 4 = 0
(2x + 1)(x – 4) = 0
1
Thus, x = − , 4.
2
3x + 2 2 x + 1
−
=1
x +1
2x
Multiplying both sides by the LCD, 2x(x + 1),
gives
2 x(3 x + 2) − (2 x + 1)( x + 1) = 2 x( x + 1)
63.
6 x 2 + 4 x − (2 x 2 + 3x + 1) = 2 x 2 + 2 x
4 x2 + x − 1 = 2 x2 + 2 x
2 x2 − x − 1 = 0
(2 x + 1)( x − 1) = 0
t +1 t + 3
t +5
+
=
t + 2 t + 4 t 2 + 6t + 8
Multiplying both sides by the LCD,
(t + 2)(t + 4), gives
(t + 1)(t + 4) + (t + 3)(t + 2) = t + 5
t 2 + 5t + 4 + t 2 + 5t + 6 = t + 5
2t 2 + 10t + 10 = t + 5
1
Thus, x = − , 1.
2
2t 2 + 9t + 5 = 0
a = 2, b = 9, c = 5
−b ± b 2 − 4ac
2a
−9 ± 81 − 4(2)(5)
=
2(2)
−9 ± 41
=
4
−9 + 41 −9 − 41
Thus t =
,
.
4
4
w
6( w + 1)
+
=3
60.
2−w
w −1
Multiplying both sides by the LCD,
(2 – w)(w – 1), gives
6(w + 1)(w – 1) + w(2 – w) = 3(2 – w)(w – 1)
t=
6( w2 − 1) + 2 w − w2 = 3(– w2 + 3w − 2)
5w2 + 2 w − 6 = −3w2 + 9 w − 6
8 w2 − 7 w = 0
w(8w – 7) = 0
7
Thus, w = 0, .
8
29
Chapter 0: Review of Algebra
64.
ISM: Introductory Mathematical Analysis
2
3
4
+ =
x +1 x x + 2
Multiplying both sides by the LCD,
x(x + 1)(x + 2), gives
2 x( x + 2) + 3( x + 1)( x + 2) = 4 x( x + 1)
68.
x2 + 9 x + 6 = 0
a = 1, b = 9, c = 6
70.
( x)
w=
+2
( x)−5 = 0
−b ± b 2 − 4ac
2a
−2 ± 4 − 4(1)(−5)
2(1)
−2 ± 24
=
2
−2 ± 2 6
=
2
= −1 ± 6
w=
2 x 2 − x ( x + 1) = 2( x + 1)( x − 1)
2 x2 − x2 − x = 2 x2 − 2
x2 − x = 2 x2 − 2
0 = x2 + x − 2
0 = (x + 2)(x − 1)
x = –2 or x = 1
But x = 1 does not check. The solution is –2.
Since w = x and −1 − 6 < 0, w = −1 − 6
does not check. Thus w = −1 + 6, so
(
x = −1 + 6
3 1− x
=
.
x
x
Multiplying both sides by x gives
5x – 3 = 1 – x
6x = 4
2
x=
3
66. If x ≠ –3, the equation is 5 −
2
2
Let w = x , then w2 + 2 w − 5 = 0
a = 1, b = 2, c = −5
x 2 ( x + 1)( x − 1), gives
)
2
q 2 − 12q + 32 = 0
(q – 4)(q – 8) = 0
Thus, q = 4, 8.
−
2x − 3
)
q 2 + 4q + 4 = 16q − 28
1
2
=
2
x − 1 x( x − 1) x 2
Multiplying both sides by the LCD,
(
= ( x − 6)2
(
−9 ± 92 − 4(1)(6)
=
2(1)
−9 ± 57
=
2
−9 + 57 −9 − 57
Thus, x =
,
.
2
2
67.
2
69. (q + 2) 2 = 2 4q − 7
−b ± b 2 − 4ac
2a
2
)
0 = x 2 − 21x
0 = x(x – 21)
x = 0 or x = 21
Only x = 21 checks.
5 x 2 + 13 x + 6 = 4 x 2 + 4 x
65.
x+4
9 x + 36 = x 2 − 12 x + 36
2 x2 + 4 x + 3x2 + 9 x + 6 = 4 x2 + 4 x
x=
(3
)
2
= 7 − 2 6.
z + 3 = 3z + 1
71.
(
z +3
) =(
2
)
3z + 1
2
z + 3 = 3z + 2 3z + 1
−2 z + 2 = 2 3 z
− z + 1 = 3z
= ( x − 3)2
(− z + 1) 2 =
2 x − 3 = x2 − 6 x + 9
(
3z
)
z 2 − 2 z + 1 = 3z
2
z 2 − 5z + 1 = 0
a = 1, b = −5, c = 1
0 = x − 8 x + 12
0 = (x – 6)(x – 2)
x = 6 or x = 2
Only x = 6 checks.
30
2
ISM: Introductory Mathematical Analysis
z=
Section 0.8
−b ± b 2 − 4ac
2a
75.
(
(
x −2
) =(
2x − 8
)
x+3
76.
= ( x − 12)2
(
t +2
2
2x + 1
)
2
t=
= x2
0 = x2 − 4x
0 = x(x – 4)
Thus, x = 0, 4.
(
y−2 +2
) =(
2
)
2
2
2
= (3t − 1)2
2y + 3
−b ± b 2 − 4ac
2a
)
–(−2.7) ± (−2.7) 2 − 4(0.04)(8.6)
2(0.04)
≈ 64.15 or 3.35
2
4 y − 2 = y +1
y−2
)
77. x =
y − 2 + 4 y − 2 + 4 = 2y + 3
(4
3t + 1
−(−7) ± (−7)2 − 4(9)(1)
2(9)
7 ± 13
=
18
7 + 13
Only
checks.
18
4x = x 2
74.
(
=
2 x=x
2
=
0 = 9t 2 − 7t + 1
a = 9, b = −7, c = 1
x + 2 x + 1 = 2x + 1
(2 x )
2
t = 9t 2 − 6t + 1
x +1 = 2x + 1
) =(
)
( t)
0 = x 2 − 40 x + 144
0 = (x – 4)(x – 36)
x = 4 or x = 36
Only x = 4 checks.
x +1
= (4 x − 2) 2
t + 2 = 3t + 1
t = 3t − 1
16 x = x − 24 x + 144
(
2
1
or x = 1
16
Only x = 1 checks.
2
73.
)
x=
2
x − 4 x + 4 = 2x − 8
( −4 x )
2
0 = 16 x 2 − 17 x + 1
0 = (16 x − 1)( x − 1)
−4 x = x − 12
2
2
x + 3 = 16 x 2 − 16 x + 4
x − 2 = 2x − 8
2
) = (3 x )
x + 3 +1
x + 3 + 2 x + 3 + 1 = 9x
2 x + 3 = 8x − 4
x + 3 = 4x − 2
−(−5) ± (−5)2 − 4(1)(1)
=
2(1)
5 ± 21
=
2
5 − 21
Only z =
checks.
2
72.
(
–0.2 ± (0.2) 2 − 4(0.01)(–0.6)
2(0.01)
≈ 2.65 or –22.65
78. x =
= ( y + 1) 2
16 y − 32 = y 2 + 2 y + 1
0 = y 2 − 14 y + 33
0 = (y – 11)(y – 3)
Thus, y = 11, 3.
31
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
79. Let l be the length of the picture, then its width is
l – 2.
l(l – 2) = 48
83.
l 2 − 2l − 48 = 0
(l – 8)(l + 6) = 0
l–8=0
or l + 6 = 0
l=8
or l = –6
Since length cannot be negative, l = 8. The width
of the picture is l – 2 = 8 – 2 = 6 inches.
The dimensions of the picture are 6 inches by
8 inches.
24 A = A2 + 13 A + 12
0 = A2 − 11A + 12
From the quadratic formula,
11 ± 121 − 48 11 ± 73
A=
=
.
2
2
11 + 73
11 – 73
≈ 10 or A =
A=
≈ 1.
2
2
The doses are the same at 1 year and 10 years.
A +1
c = d in Cowling’s rule when
= 1, which
24
occurs when A = 23. Thus, adulthood is achieved
at age 23 according to Cowling’s rule.
A
= 1, which is
c = d in Young’s rule when
A + 12
never true. Thus, adulthood is never reached
according to Young’s rule.
80. The amount that the temperature has risen over
the X days is
(X degrees per day)(X days) = X 2 degrees.
X 2 + 15 = 51
X 2 = 36
X = ± 36
X = 6 or X = –6
The temperature has been rising 6 degrees per
day for 6 days.
81. M =
A
A +1
d=
d.
A + 12
24
Dividing both sides by d and then multiplying
both sides by 24(A + 12) gives
24A = (A + 12)(A + 1)
Q(Q + 10)
44
1
44 M = Q 2 + 10Q
0 = Q 2 + 10Q – 44 M
From the quadratic formula with a = 1, b = 10,
c = −44 M ,
Q=
=
0
0
25
Young’s rule prescribes less than Cowling’s for
ages less than one year and greater than 10 years.
Cowling’s rule prescribes less for ages between
1 and 10.
–10 ± 100 – 4(1)(–44M )
2(1)
84. a.
–10 + 2 25 + 44M
2
= −5 ± 25 + 44M
Thus, −5 + 25 + 44M is a root.
82. g = −200 P 2 + 200 P + 20
Set g = 60.
60 = −200 P 2 + 200 P + 20
(2n − 1)v 2 − 2nv + 1 = 0
From the quadratic formula with a = 2n – 1,
b = –2n, c = 1,
v=
–(–2n) ± 4n 2 − 4(2n − 1)(1)
2(2n − 1)
v=
2n ± 4n 2 − 8n + 4
2(2n − 1)
2
2n ± 2 n 2 − 2n + 1 n ± (n − 1)
=
2(2n − 1)
2n − 1
Because of the condition that n ≥ 1, it
follows that n – 1 is nonnegative. Thus,
v=
200 P 2 − 200 P + 40 = 0
5P 2 − 5P + 1 = 0
From the quadratic formula with a = 5, b = −5,
c = 1,
5 ± 25 − 4(5)(1) 5 ± 5
P=
=
2(5)
10
(n − 1)2 = n − 1 and we have
v=
P ≈ 0.28 or P ≈ 0.72
28% and 72% of yeast gave an average weight
gain of 60 grams.
n ± (n − 1)
.
2n − 1
v = 1 or v =
32
1
.
2n − 1
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 0
87. By a program, roots are 1.5 and 0.75.
Algebraically:
nv 2 − (2n + 1)v + 1 = 0
From the quadratic formula with a = n,
b = –(2n + 1), and c = 1,
b.
v=
−[−(2n + 1)] ± [−(2n + 1)]2 − 4(n)(1)
2n
v=
2n + 1 ± 4n 2 + 1
2n
8 x 2 − 18 x + 9 = 0
(2x – 3)(4x – 3) = 0
Thus, 2x – 3 = 0 or 4x – 3 = 0.
3
3
So x = = 1.5 or x = = 0.75.
2
4
88. By a program, roots are −0.762 and 0.262.
Because 4n 2 + 1 is greater than 2n,
choosing the plus sign gives a numerator
greater than 2n + 1 + 2n, or 4n + 1, so v is
4n + 1
1
= 2 + . Thus v is
greater than
2n
2n
greater than 2. This contradicts the
restriction on v. On the other hand, because
89. By a program, there are no real roots.
90.
4n 2 + 1 is greater than 1, choosing the
minus sign gives a numerator less than 2n,
2n
= 1. This meets the
so v is less than
2n
condition on v. Thus we choose
v=
85. a.
9 2
z
z − 6.3 = (1.1 − 7 z )
2
3
9 2
1.1
7
z − 6.3 =
z − z2
2
3
3
⎛ 9 7 ⎞ 2 1.1
z − 6.3 = 0
⎜ + ⎟z −
3
⎝2 3⎠
Roots: 0.987, –0.934
91. (πt − 4)2 = 4.1t − 3
π2t 2 − 8πt + 16 = 4.1t − 3
2n + 1 − 4n 2 + 1
.
2n
π2t 2 + (–8π − 4.1)t + 19 = 0
Roots: 1.999, 0.963
When the object strikes the ground, h must
be 0, so
Mathematical Snapshot Chapter 0
0 = 39.2t − 4.9t 2 = 4.9t (8 − t )
t = 0 or t = 8
The object will strike the ground 8 s after
being thrown.
1.
b. Setting h = 68.2 gives
68.2 = 39.2t − 4.9t 2
4.9t 2 − 39.2t + 68.2 = 0
2. The procedure works because multiplying a list
by a number is the same as multiplying each
element in the list by the number, adding a
number to a list has the effect of adding the
number to each element of the list, and
subtracting one list from another is the same as
subtracting corresponding elements. The plots
match.
2
39.2 ± ( −39.2) − 4(4.9)(68.2)
2(4.9)
39.2 ± 14.1
≈
9.8
t ≈ 5.4 s or t ≈ 2.6 s.
t=
86. By a program, roots are 4.5 and –3.
Algebraically:
3.
2
2 x − 3 x − 27 = 0
(2x – 9)(x + 3) = 0
Thus, 2x – 9 = 0 or x + 3 = 0
9
So x = = 4.5 or x = –3.
2
The results agree.
33
Chapter 0: Review of Algebra
ISM: Introductory Mathematical Analysis
4. The smaller quadratic residuals indicate a better
fit. The fairly random pattern suggests that the
model cannot be improved any further. The
slight deviations from the quadratic model are
presumably due to random measurement errors.
34
Chapter 1
5. Let n = number of ounces in each part. Then we
have
2n + 1n = 16
3n = 16
16
n=
3
Thus the turpentine needed is
16
1
(1)n =
= 5 ounces.
3
3
Problems 1.1
1. Let w be the width and 2w be the length of the
plot.
2w
w
w
2w
Then area = 800.
(2w)w = 800
6. Let w = width (in miles) of strip to be cut. Then
the remaining forest has dimensions 2 – 2w by
1 – 2w.
2w2 = 800
w2 = 400
w = 20 ft
Thus the length is 40 ft, so the amount of fencing
needed is 2(40) + 2(20) = 120 ft.
2 – 2w
w
1 – 2w w
1
w
2. Let w be the width and 2w be the length.
2
2w
w
w
Considering the area of the remaining forest, we
have
3
(2 − 2 w)(1 − 2 w) =
4
3
2 − 6 w + 4 w2 =
4
w
2w
The perimeter P = 2w + 2l = 2w + 2(2w) = 6w.
Thus 6w = 300.
300
w=
= 50 ft
6
Thus the length is 2(50) = 100 ft.
The dimensions are 50 ft by 100 ft.
8 − 24w + 16 w2 = 3
16w2 − 24w + 5 = 0
(4w – 1)(4w – 5) = 0
1 5
5
Hence w = , . But w = is impossible since
4 4
4
one dimension of original forest is 1 mi. Thus
1
mi.
the width of the strip should be
4
3. Let n = number of ounces in each part. Then we
have
4n + 5n = 145
9n = 145
1
n = 16
9
4
⎛ 1⎞
Thus there should be 4 ⎜ 16 ⎟ = 64 ounces of
9
⎝ 9⎠
1
5
⎛
⎞
A and 5 ⎜ 16 ⎟ = 80 ounces of B.
9
⎝ 9⎠
7. Let w = width (in meters) of pavement. The
remaining plot for flowers has dimensions
8 – 2w by 4 – 2w.
8 – 2w
w
4 – 2w w
w
w
4. Let n = number of cubic feet in each part.
Then we have
1n + 3n + 5n = 765
9n = 765
n = 85
Thus he needs 1n = 1(85) = 85 ft3 of portland
cement, 3n = 3(85) = 255 ft3 of sand, and
5n = 5(85) = 425 ft3 of crushed stone.
8
Thus
(8 – 2w)(4 – 2w) = 12
32 − 24w + 4 w2 = 12
4w2 − 24 w + 20 = 0
w2 − 6 w + 5 = 0
(w – 1)(w – 5) = 0
35
4
Chapter 1: Applications and More Algebra
ISM: Introductory Mathematical Analysis
Hence w = 1, 5. But w = 5 is impossible since
one dimension of the original plot is 4 m. Thus
the width of the pavement should be 1 m.
13. Let p = selling price. Then profit = 0.2p.
selling price = cost + profit
p = 3.40 + 0.2p
0.8p = 3.40
3.40
p=
= $4.25
0.8
8. Since diameter of circular end is 140 mm, the
radius is 70 mm. Area of circular end is
π(radius)2 = π(70)2 . Area of square end is x 2 .
14. Following the procedure in Example 6 we obtain
the total value at the end of the second year to be
Equating areas, we have x 2 = π(70)2 .
Thus x = ± π(70)2 = ±70 π . Since x must be
1, 000, 000(1 + r ) 2 .
So at the end of the third year, the accumulated
positive, x = 70 π ≈ 124 mm.
amount will be 1, 000, 000(1 + r ) 2 plus the
9. Let q = number of tons for $560,000 profit.
Profit = Total Revenue – Total Cost
560, 000 = 134q − (82q + 120, 000)
560, 000 = 52q − 120, 000
680, 000 = 52q
680, 000
=q
52
q ≈ 13, 076.9 ≈ 13, 077 tons.
interest on this, which is 1, 000, 000(1 + r )2 r.
Thus the total value at the end of the third year
will be 1, 000, 000(1 + r ) 2 + 1, 000, 000(1 + r )2 r
= 1, 000, 000(1 + r )3 .
This must equal $1,125,800.
1, 000, 000(1 + r )3 = 1,125,800
1,125,800
(1 + r )3 =
= 1.1258
1, 000, 000
1 + r ≈ 1.04029
r ≈ 0.04029
Thus r ≈ 0.04029 ≈ 4%.
10. Let q = required number of units.
Profit = Total Revenue – Total Cost
150, 000 = 50q − (25q + 500, 000)
150, 000 = 25q − 500, 000
650, 000 = 25q, from which
q = 26, 000
15. Following the procedure in Example 6 we obtain
3, 000, 000(1 + r ) 2 = 3, 245, 000
649
(1 + r ) 2 =
600
649
1+ r = ±
600
649
r = −1 ±
600
r ≈ −2.04 or 0.04
We choose r ≈ 0.04 = 4%.
11. Let x = amount at 6% and
1
20,000 – x = amount at 7 %.
2
x(0.06) + (20,000 – x)(0.075) = 1440
–0.015x + 1500 = 1440
–0.015x = –60
x = 4000, so 20,000 – x = 16,000. Thus the
investment should be $4000 at 6% and $16,000
1
at 7 %.
2
16. Total revenue = variable cost + fixed cost
100 q = 2q + 1200
12. Let x = amount at 6% and
20,000 – x = amount at 7%.
x(0.06) + (20,000 – x)(0.07) = 20,000(0.0675)
–0.01x + 1400 = 1350
–0.01x = –50, so x = 5000
The investment consisted of $5000 at 6% and
$15,000 at 7%.
50 q = q + 600
2500q = q 2 + 1200q + 360, 000
0 = q 2 − 1300q + 360, 000
0 = (q – 400)(q – 900)
q = 400 or q = 900
36
ISM: Introductory Mathematical Analysis
Section 1.1
23. Let v = total annual vision-care expenses (in
dollars) covered by program. Then
35 + 0.80(v – 35) = 100
0.80v + 7 = 100
0.80v = 93
v = $116.25
17. Let n = number of room applications sent out.
0.95n = 76
76
n=
= 80
0.95
18. Let n = number of people polled.
0.20p = 700
700
p=
= 3500
0.20
24. a.
0.031c
b. c – 0.031c = 600,000,000
0.969c = 600,000,000
c ≈ 619,195, 046
Approximately 619,195,046 bars will have
to be made.
19. Let s = monthly salary of deputy sheriff.
0.30s = 200
200
s=
0.30
⎛ 200 ⎞
Yearly salary = 12s = 12 ⎜
⎟ = $8000
⎝ 0.30 ⎠
25. Revenue = (number of units sold)(price per unit)
Thus
⎡ 80 − q ⎤
400 = q ⎢
⎥
⎣ 4 ⎦
20. Yearly salary before strike = (7.50)(8)(260)
= $15,600
Lost wages = (7.50)(8)(46) = $2760
Let P be the required percentage increase (as a
decimal).
P(15,600) = 2760
2760
P=
≈ 0.177 = 17.7%
15, 600
1600 = 80q − q 2
q 2 − 80q + 1600 = 0
(q − 40)2 = 0
q = 40 units
26. If I = interest, P = principal, r = rate, and
t = time, then I = Prt. To triple an investment of
P at the end of t years, the interest earned during
that time must equal 2P. Thus
2P = P(0.045)t
2 = 0.045t
2
t=
≈ 44.4 years
0.045
21. Let q = number of cartridges sold to break even.
total revenue = total cost
21.95q = 14.92q + 8500
7.03q = 8500
q ≈ 1209.10
1209 cartridges must be sold to approximately
break even.
27. Let q = required number of units. We equate
incomes under both proposals.
2000 + 0.50q = 25,000
0.50q = 23,000
q = 46,000 units
22. Let n = number of shares of stock to be bought.
total investment = 4000 + 15n
total yield (goal) = 6% of total investment
= 0.06(4000 + 15n)
total yield = bond yield + stock yield
= 0.07(4000) + 0.60n
Thus,
0.06(4000 + 15n) = 0.07(4000) + 0.60n
240 + 0.9n = 280 + 0.6n
0.3n = 40
1
n = 133
3
28. Let w = width of strip. The original area is
80(120) and the new area is (120 + w)(80 + w).
w
120
80
80 + w
w
120 + w
37
Chapter 1: Applications and More Algebra
ISM: Introductory Mathematical Analysis
Thus
(120 + w)(80 + w) = 2(80)(120)
31. 10, 000 = 800 p − 7 p 2
9600 + 200w + w2 = 19, 200
7 p 2 − 800 p + 10, 000 = 0
w2 + 200 w − 9600 = 0
(w + 240)(w – 40) = 0
w = –240 or w = 40
We choose w = 40 ft.
800 ± 640, 000 − 280, 000
14
800 ± 360, 000 800 ± 600
=
=
14
14
800 + 600
For p > 50 we choose p =
= $100.
14
p=
29. Let n = number of $20 increases. Then at the
rental charge of 400 + 20n dollars per unit, the
number of units that can be rented is 50 – 2n.
The total of all monthly rents is
(400 + 20n)(50 – 2n), which must equal 20,240.
20,240 = (400 + 20n)(50 – 2n)
32. Let p be the percentage increase in market value.
Then
⎛ P ⎞ (1 + p) P
1.1⎜ ⎟ =
⎝ E ⎠ (1.2) E
20, 240 = 20, 000 + 200n − 40n 2
40n 2 − 200n + 240 = 0
1+ p
1.2
1.32 = 1 + p
p = 0.32 = 32%
1.1 =
2
n − 5n + 6 = 0
(n – 2)(n – 3) = 0
n = 2, 3
Thus the rent should be either
$400 + 2($20) = $440 or $400 + 3($20) = $460.
33. To have supply = demand,
2 p − 10 = 200 − 3 p
5 p = 210
p = 42
30. Let x = original value of the blue-chip
investment, then 3,100,000 – x is the original
value of the glamour stocks. Then the current
1
11
x.
value of the blue-chip stock is x + x, or
10
10
For the glamour stocks the current value is
1
(3,100, 000 − x) − (3,100, 000 − x ), which
10
9
simplifies to
(3,100, 000 − x).
10
Thus for the current value of the portfolio,
11
9
x + (3,100, 000 − x) = 3, 240, 000
10
10
11x + 27,900,000 – 9x = 32,400,000
2x = 4,500,000
x = 2,250,000
Thus the current value of the blue chip
11
(2, 250, 000) or $2,475,000.
investment is
10
2 p 2 − 3 p = 20 − p 2
34.
2
3 p − 3 p − 20 = 0
a = 3, b = −3, c = −20
p=
−b ± b 2 − 4ac
2a
−(−3) ± (−3) 2 − 4(3)(−20)
2(3)
3 ± 249
=
6
p ≈ 3.130 or p ≈ −2.130
The equilibrium price is p ≈ 3.13.
=
35. Let w = width (in ft) of enclosed area. Then
length of enclosed area is
300 – w – w = 300 – 2w.
38
ISM: Introductory Mathematical Analysis
Section 1.1
300 – 2w
w
(10 – x)(5 – x)2 = 72
(10 – x)(5 – x) = 36
x 2 − 15 x + 50 = 36
w
AREA
x 2 − 15 x + 14 = 0
(x – 1)(x – 14) = 0
x = 1 or 14
Because of the length and width of the original
bar, we reject x = 14 and choose x = 1. The new
bar has length 10 – x = 10 – 1 = 9 cm and width
is 5 – x = 5 – 1 = 4 cm.
PLANT
38. Volume of old style candy
150
Thus
w(300 – 2w) = 11,200
2w(150 – w) = 11,200
w(150 – w) = 5600
= π(7.1)2 (2.1) − π(2)2 (2.1)
= 97.461π mm3
Let r = inner radius (in millimeters) of new style
candy. Considering the volume of the new style
candy, we have
0 = w2 − 150w + 5600
0 = (w – 80)(w – 70)
Hence w = 80, 70. If w = 70, then length is
300 – 2w = 300 – 2(70) = 160. Since the
building has length of only 150 ft, we reject w =
70. If
w = 80, then length is
300 – 2w = 300 – 2(80) = 140. Thus the
dimensions are 80 ft by 140 ft.
π(7.1)2 (2.1) − πr 2 (2.1) = 0.78(97.461π)
29.84142π = 2.1πr 2
14.2102 = r 2
r ≈ ±3.7696
Since r is a radius, we choose r = 3.77 mm.
39. Let x = amount of loan. Then the amount
actually received is x − 0.16x. Hence,
x − 0.16 x = 195, 000
0.84 x = 195, 000
x ≈ 232,142.86
To the nearest thousand, the loan amount is
$232,000. In the general case, the amount
received from a loan of L with a compensating
p
L.
balance of p% is L −
100
p
L−
L=E
100
100 − p
L=E
100
100 E
L=
100 − p
36. Let s = length in inches of side of original
square.
s
3
3
3
3
s–6
s
3
3
3
3
s–6
Considering the volume of the box, we have
(length)(width)(height) = volume
(s – 4)(s – 4)(2) = 50
( s − 4) 2 = 25
s − 4 = ± 25 = ±5
s=4±5
Hence s = –1, 9. We reject s = –1 and choose
s = 9. The dimensions are 9 in. by 9 in.
40. Let n = number of machines sold over 600. Then
the commission on each of 600 + n machines is
40 + 0.04n. Equating total commissions to
30,800 we obtain
(600 + n)(40 + 0.04n) = 30,800
37. Original volume = (10)(5)(2) = 100 cm3
Volume cut from bar = 0.28(100) = 28 cm3
Volume of new bar = 100 – 28 = 72 cm3
Let x = number of centimeters that the length
and width are each reduced. Then
24, 000 + 24n + 40n + 0.04n 2 = 30,800
0.02n 2 + 32n − 3400 = 0
n=
39
–32 ± 1024 + 272 −32 ± 36
=
0.04
0.04
Chapter 1: Applications and More Algebra
ISM: Introductory Mathematical Analysis
43. Let q = number of units of B and
q + 25 = number of units of A produced.
1000
, and each unit of A
Each unit of B costs
q
–32 + 36
= 100. Thus the
0.04
number of machines that must be sold is
600 + 100 = 700.
We choose n =
costs
41. Let n = number of acres sold. Then n + 20 acres
7200
were originally purchased at a cost of
n + 20
each. The price of each acre sold was
⎡ 7200 ⎤
30 + ⎢
⎥ . Since the revenue from selling n
⎣ n + 20 ⎦
acres is $7200 (the original cost of the parcel),
we have
7200 ⎤
⎡
n ⎢30 +
= 7200
n + 20 ⎥⎦
⎣
1500 1000
=
+2
q + 25
q
1500q = 1000(q + 25) + 2(q)(q + 25)
0 = 2q 2 − 450q + 25, 000
0 = q 2 − 225q + 12,500
0 = (q – 100)(q – 125)
q = 100 or q = 125
If q = 100, then q + 25 = 125; if q = 125,
q + 25 = 150. Thus the company produces either
125 units of A and 100 units of B, or 150 units of
A and 125 units of B.
⎡ 30n + 600 + 7200 ⎤
n⎢
⎥ = 7200
n + 20
⎣
⎦
n(30n + 600 + 7200) = 7200(n + 20)
Principles in Practice 1.2
30n 2 + 7800n = 7200n + 144, 000
30n 2 + 600n − 144, 000 = 0
1. 200 + 0.8S ≥ 4500
0.8S ≥ 4300
S ≥ 5375
He must sell at least 5375 products per month.
2
n + 20n − 4800 = 0
(n + 80)(n – 60) = 0
n = 60 acres (since n > 0), so 60 acres were sold.
2. Since x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, and x4 ≥ 0, we
have the inequalities
150 − x4 ≥ 0
3x4 − 210 ≥ 0
x4 + 60 ≥ 0
x4 ≥ 0
42. Let q = number of units of product sold last year
and q + 2000 = the number sold this year. Then
the revenue last year was 3q and this year it is
3.5(q + 2000). By the definition of margin of
profit, it follows that
7140
4500
=
+ 0.02
3.5(q + 2000)
3q
2040
1500
=
+ 0.02
q + 2000
q
2040q = 1500(q + 2000) + 0.02q(q + 2000)
Problems 1.2
1. 3x > 12
12
x>
3
x>4
(4, ∞)
2040q = 1500q + 3, 000, 000 + 0.02q 2 + 40q
0 = 0.02q 2 − 500q + 3, 000, 000
q=
1500
. Therefore,
q + 25
500 ± 250, 000 − 240, 000
0.04
4
500 ± 10, 000
0.04
500 ± 100
=
0.04
= 10,000 or 15,000
So that the margin of profit this year is not
greater than 0.15, we choose q = 15,000. Thus
15,000 units were sold last year and 17,000 this
year.
=
2. 4x < –2
–2
x<
4
1
x<−
2
40
ISM: Introductory Mathematical Analysis
Section 1.2
8. 4s – 1 < –5
4s < –4
s < –1
(–∞, –1)
1⎞
⎛
⎜ −∞, − 2 ⎟
⎝
⎠
–1
2
–1
3. 5 x − 11 ≤ 9
5 x ≤ 20
x≤4
(−∞, 4]
9. 3 < 2y + 3
0 < 2y
0<y
y>0
(0, ∞)
4
0
4. 5 x ≤ 0
0
x≤
5
x≤0
(−∞, 0]
10.
0
5. –4x ≥ 2
2
x≤
−4
1
x≤−
2
1⎤
⎛
⎜ – ∞, − 2 ⎥
⎝
⎦
4 ≤ 3− 2y
1 ≤ −2 y
1
− ≥y
2
1
y≤−
2
1⎤
⎛
⎜ – ∞, − 2 ⎥
⎝
⎦
–1
2
11. x + 5 ≤ 3 + 2 x
− x ≤ −2
x≥2
[2, ∞)
–1
2
2
6. 2y + 1 > 0
2y > –1
1
y>−
2
⎛ 1
⎞
⎜− 2 , ∞⎟
⎝
⎠
12. –3 ≥ 8(2 – x)
–3 ≥ 16 – 8x
8x ≥ 19
19
x≥
8
⎡ 19
⎞
⎢ 8 , ∞⎟
⎣
⎠
–1
2
19
8
7. 5 – 7s > 3
–7s > –2
2
s<
7
2⎞
⎛
⎜ −∞, 7 ⎟
⎝
⎠
13. 3(2 – 3x) > 4(1 – 4x)
6 – 9x > 4 – 16x
7x > –2
2
x>−
7
⎛ 2
⎞
⎜− 7 , ∞⎟
⎝
⎠
2
7
–2
7
41
Chapter 1: Applications and More Algebra
ISM: Introductory Mathematical Analysis
14. 8(x + 1) + 1 < 3(2x) + 1
8x + 9 < 6x + 1
2x < –8
x < –4
(–∞, –4)
2
x>6
3
–x > 9
x < –9
(–∞, –9)
20. −
–4
–9
15. 2(4 x − 2) > 4(2 x + 1)
8x − 4 > 8x + 4
−4 > 4, which is false for all x.
Thus the solution set is ∅.
21.
16. 4 − ( x + 3) ≤ 3(3 − x )
1 − x ≤ 9 − 3x
2x ≤ 8
x≤4
(−∞, 4]
–5
22.
4
17. x + 2 < 3 − x
2x < 3 − 2
x<
⎛
⎜⎜ – ∞,
⎝
3 −2
2
23. −3x + 1 ≤ −3( x − 2) + 1
−3x + 1 ≤ −3x + 7
1 ≤ 7, which is true for all x. The solution is
–∞ < x < ∞.
(–∞, ∞)
2 ( x + 2) > 8(3 − x)
2 ( x + 2) > 2 2 (3 − x)
x + 2 > 2(3 – x)
x + 2 > 6 – 2x
3x > 4
4
x>
3
⎛4
⎞
⎜ 3 , ∞⎟
⎝
⎠
24. 0x ≤ 0
0 ≤ 0, which is true for all x. The solution is
–∞ < x < ∞.
(–∞, ∞)
25.
4
3
19.
3y − 2 1
≥
3
4
12 y − 8 ≥ 3
12 y ≥ 11
11
y≥
12
11
⎡
⎞
⎢ , ∞⎟
⎣ 12
⎠
11
12
3 −2⎞
⎟
2 ⎟⎠
3−2
2
18.
9 y +1
≤ 2 y −1
4
9y + 1 ≤ 8y – 4
y ≤ –5
(–∞, –5]
5
x < 40
6
5x < 240
x < 48
(–∞, 48)
1 − t 3t − 7
<
2
3
3(1 − t ) < 2(3t − 7)
3 − 3t < 6t − 14
−9t < −17
17
t>
9
⎛ 17
⎞
⎜ , ∞⎟
⎝ 9
⎠
17
9
48
42
ISM: Introductory Mathematical Analysis
26.
Section 1.2
3(2t − 2) 6t − 3 t
>
+
2
5
10
15(2t – 2) > 2(6t – 3) + t
30t – 30 > 13t – 6
17t > 24
24
t>
7
⎛ 24
⎞
⎜ 7 , ∞⎟
⎝
⎠
31.
0
2 − 0.01x
0.2
1.8 – 0.02x ≤ 2 – 0.01x
–0.01x ≤ 0.2
x ≥ –20
[–20, ∞)
24
17
32. 9 − 0.1x ≤
1
x−7
3
6 x + 39 ≥ x − 21
5 x ≥ −60
x ≥ −12
(−12, ∞)
27. 2 x + 13 ≥
–20
33. 0.1(0.03x + 4) ≥ 0.02x + 0.434
0.003x + 0.4 ≥ 0.02x + 0.434
–0.017x ≥ 0.034
x ≤ –2
(–∞, –2]
–12
28.
1 5
3x − ≤ x
3 2
18 x − 2 ≤ 15 x
3x ≤ 2
2
x≤
3
2⎤
⎛
⎜ −∞, ⎥
3⎦
⎝
–2
34.
2
3
29.
2
5
r< r
3
6
4r < 5r
0<r
r>0
(0, ∞)
3 y − 1 5( y + 1)
<
−3
−3
3y −1 > 5 y + 5
−6 > 2 y
−3 > y
y < −3
(−∞, −3)
–3
35. 12(50) < S < 12(150)
600 < S < 1800
36. 2
0
30.
y y
y
+ > y+
2 3
5
15y + 10y > 30y + 6y
25y > 36y
0 > 11y
0>y
y<0
(–∞, 0)
1
≤x≤4
2
37. The measures of the acute angles of a right
triangle sum to 90°. If x is the measure of one
acute angle, the other angle has measure 90 – x.
x < 3(90 – x) + 10
x < 270 – 3x + 10
4x < 280
x < 70
The measure of the angle is less than 70°.
7
8
t>− t
4
3
21t > –32t
53t > 0
t>0
(0, ∞)
0
43
Chapter 1: Applications and More Algebra
ISM: Introductory Mathematical Analysis
38. Let d be the number of disks. The stereo plus
d disks will cost 219 + 18.95d.
219 + 18.95d ≤ 360
18.95d ≤ 141
141
d≤
≈ 7.44
18.95
The student can buy at most 7 disks.
Problems 1.3
1. Let q = number of units sold.
Profit > 0
Total revenue – Total cost > 0
20q – (15q + 600,000) > 0
5q – 600,000 > 0
5q > 600,000
q > 120,000
Thus at least 120,001 units must be sold.
2. Let q = number of units sold.
Total revenue – Total cost = Profit
We want Profit > 0.
7.40q – [(2.50 + 4)q + 5000] > 0
0.9q – 5000 > 0
0.9q > 5000
5000
5
q>
= 5555
0.9
9
Thus at least 5556 units must be sold.
3. Let x = number of miles driven per year.
If the auto is leased, the annual cost is
12(420) + 0.06x.
If the auto is purchased, the annual cost is
4700 + 0.08x. We want Rental cost ≤ Purchase cost.
12(420) + 0.06x ≤ 4700 + 0.08x
5040 + 0.06x ≤ 4700 + 0.08x
340 ≤ 0.02x
17,000 ≤ x
The number of miles driven per year must be at least 17,000.
4. Let N = required number of shirts. Then
Total revenue = 3.5N and
Total cost = 1.3N + 0.4N + 6500.
Profit > 0
3.5 N − (1.3 N + 0.4 N + 6500) > 0
1.8 N − 6500 > 0
1.8 N > 6500
N > 3611.1
At least 3612 shirts must be sold.
5. Let q be the number of magazines printed. Then the cost of publication is 0.55q. The number of magazines sold is
0.90q. The revenue from dealers is (0.60)(0.90q). If fewer than 30,000 magazines are sold, the only revenue is
from the sales to dealers, while if more than 30,000 are sold, there are advertising revenues of
0.10(0.60)(0.90q − 30,000). Thus,
44
ISM: Introductory Mathematical Analysis
Section 1.3
if 0.9q ≤ 30, 000
⎧0.6(0.9)q
Revenue = ⎨
⎩0.6(0.9)q + 0.1(0.6)(0.9q − 30, 000) if 0.9q > 30, 000
q ≤ 33,333
⎧0.54q
=⎨
−
> 33,333
q
q
0.594
1800
⎩
Profit = Revenue − Cost
q ≤ 33,333
⎧0.54q − 0.55q
=⎨
⎩0.594q − 1800 − 0.55q q > 33,333
q ≤ 33,333
⎧−0.01q
=⎨
0.044
q
−
1800
q
> 33,333
⎩
Clearly, the profit is negative if fewer than 33,334 magazines are sold.
0.044q − 1800 ≥ 0
0.044q ≥ 1800
q ≥ 40,910
Thus, at least 40,910 magazines must be printed in order to avoid a loss.
6. Let q = number of clocks produced during regular work week, so 11,000 – q = number produced in overtime.
Then
2q + 3(11,000 – q) ≤ 25,000
–q + 33,000 ≤ 25,000
8000 ≤ q
At least 8000 clocks must be produced during the regular workweek.
3
7. Let x = amount at 6 % and 30,000 – x = amount at 5%. Then
4
3
1
interest at 6 % + interest at 5% ≥ interest at 6 %
4
2
x(0.0675) + (30,000 – x)(0.05) ≥ (0.065)(30,000)
0.0175x + 1500 ≥ 1950
0.0175x ≥ 450
x ≥ 25,714.29
3
Thus at least $25,714.29 must be invested at 6 %.
4
8. Let L be current liabilities. Then
current assets
Current ratio =
current liabilities
570, 000
3.8 =
L
3.8L = 570,000
L = $150,000
Let x = amount of money they can borrow, where x ≥ 0.
570, 000 + x
≥ 2.6
150, 000 + x
570,000 + x ≥ 390,000 + 2.6x
180,000 ≥ 1.6x
112,500 ≥ x
Thus current liabilities are $150,000 and the maximum amount they can borrow is $112,500.
45
Chapter 1: Applications and More Algebra
ISM: Introductory Mathematical Analysis
9. Let q be the number of units sold this month at
$4.00 each. Then 2500 – q will be sold at $4.50
each. Then
Total revenue ≥ 10,750
4q + 4.5(2500 – q) ≥ 10,750
–0.5q + 11,250 ≥ 10,750
500 ≥ 0.5q
1000 ≥ q
The maximum number of units that can be sold
this month is 1000.
2.
4.
−4 − 6
−10
=
= −5 = 5
2
2
5.
⎛ 7⎞
2 ⎜ − ⎟ = −7 = 7
⎝ 2⎠
6. |3 − 5| − |5 − 3| = |−2| − |2| = 2 − 2 = 0
100 + q > 5000
q > 4900
At least 4901 units must be sold.
11. For t < 40, we want
income on hourly basis
> income on per-job basis
9t > 320 + 3(40 − t )
9t > 440 − 3t
12t > 440
t > 36.7 hr
7.
x < 4 , –4 < x < 4
8.
x < 10, –10 < x < 10
9. Because 2 − 5 < 0 ,
(
)
2− 5 = − 2− 5 = 5 −2.
10. Because
12. Let s = yearly sales. With the first method, the
salary is 35,000 + 0.03s, and with the second
method it is 0.05s.
35, 000 + 0.03s > 0.05s
35, 000 > 0.02 s
1, 750, 000 > s
The first method is better for yearly sales less
than $1,750,000.
5 − 2 > 0,
11. a.
x−7 < 3
b.
x−2 <3
c.
x−7 ≤ 5
d.
x−7 = 4
e.
x+4 < 2
f.
x <3
g.
x >6
h.
x − 105 < 3
i.
x − 850 < 100
13. Let x = accounts receivable. Then
450, 000 + x
Acid test ratio =
398, 000
450, 000 + x
1.3 ≤
398, 000
517,400 ≤ 450,000 + x
x ≥ 67, 400
The company must have at least $67,400 in
accounts receivable.
Principles in Practice 1.4
12. |f(x) − L| < ε
w − 22 ≤ 0.3
13.
Problems 1.4
1.
1 1
=
2 2
3. 8 − 2 = 6 = 6
10. Revenue = (no. of units)(price per unit)
⎛ 100 ⎞
+ 1⎟ > 5000
q⎜
⎝ q
⎠
1.
2 –1 =
−13 = 13
46
p1 − p2 ≤ 9
5 − 2 = 5 − 2.
ISM: Introductory Mathematical Analysis
14.
Section 1.4
x − µ ≤ 2σ
23. 7 − 4 x = 5
7 – 4x = ±5
–4x = –7 ± 5
–4x = –2 or –12
1
x = or x = 3
2
–2σ ≤ x – µ ≤ 2σ
µ – 2 σ ≤ x ≤ µ + 2σ
15.
x =7
x = ±7
16.
−x = 2
–x = 2 or –2
x = ±2
17.
18.
19.
20.
24. 5 − 3 x = 2
5 − 3 x = ±2
−3 x = −5 ± 2
−3 x = −3 or − 7
7
x = 1 or x =
3
x
=7
5
x
= ±7
5
x = ±35
25.
x <M
−M < x < M
(–M, M)
Note that M > 0 is required.
5
= 12
x
5
= ±12
x
5
x=±
12
26.
−x < 3
x <3
–3 < x < 3
(–3, 3)
x −5 = 8
x – 5 = ±8
x=5±8
x = 13 or x = –3
27.
x
>2
4
x
x
< −2
or
>2
4
4
x < –8
or x > 8, so the solution is
(–∞, –8) ∪ (8, ∞).
4 + 3x = 6
4 + 3x = ±6
3x = –4 ± 6
3x = –10 or 2
10
2
or x =
x= −
3
3
28.
x 1
>
3 2
x
1
<−
or
3
2
3
x<−
or
2
3⎞ ⎛3
⎛
⎞
⎜ −∞, − 2 ⎟ ∪ ⎜ 2 , ∞ ⎟ .
⎝
⎠ ⎝
⎠
21. 5 x − 2 = 0
5x – 2 = 0
2
x=
5
22. 7 x + 3 = x
Here we must have x ≥ 0.
7x + 3 = x
or –(7x + 3) = x
6x = –3
–7x – 3 = x
3
1
x=− <0
x=− <0
8
2
There is no solution.
29.
x 1
>
3 2
3
x > , so the solution is
2
x+9 < 5
−5 < x + 9 < 5
−14 < x < −4
(−14, −4)
30. |2x − 17| < −4
Because –4 < 0, the solution set is ∅.
47
Chapter 1: Applications and More Algebra
31.
x−
ISM: Introductory Mathematical Analysis
1 1
>
2 2
1
1
<−
2
2
x<0
(–∞, 0) ∪ (1, ∞)
x−
32. 1 − 3x > 2
1 – 3x > 2
–3x > 1
1
x<–
3
36.
1 1
>
2 2
or x > 1
or
x−
37. |d − 35.2 m| ≤ 20 cm or |d − 35.2| ≤ 0.20
or 1 – 3x < –2
or –3x < –3
38. Let T1 and T2 be the temperatures of the two
chemicals.
5 ≤ T1 – T2 ≤ 10
or x > 1
1⎞
⎛
The solution is ⎜ −∞, − ⎟ ∪ (1, ∞).
3⎠
⎝
39.
40.
1. The bounds of summation are 12 and 17; the
index of summation is t.
2. The bounds of summation are 3 and 450; the
index of summation is m.
4 x − 1 ≥ 0 is true for all x because a ≥ 0 for all
3.
7
∑ 6i
i =1
3x − 8
≥4
2
3x − 8
≤ −4
2
3x – 8 ≤ –8
3x ≤ 0
x − 0.01 ≤ 0.005
Problems 1.5
a. Thus –∞ < x < ∞, or (–∞,∞).
35.
x − µ > hσ
Either x – µ < –hσ, or x – µ > hσ. Thus either
x < µ – hσ or x > µ + hσ, so the solution is
(–∞, µ – hσ) ∪ (µ + hσ, ∞).
33. 5 − 8 x ≤ 1
–1 ≤ 5 – 8x ≤ 1
–6 ≤ –8x ≤ –4
3
1
≥ x ≥ , which may be rewritten as
4
2
1
3
≤x≤ .
2
4
⎡1 3⎤
The solution is ⎢ , ⎥ .
⎣2 4⎦
34.
x−7
≤5
3
x−7
−5 ≤
≤5
3
−15 ≤ x − 7 ≤ 15
−8 ≤ x ≤ 22
[−8, 22]
= 6(1) + 6(2) + 6(3) + 6(4) + 6(5) + 6(6) + 6(7)
= 6 + 12 + 18 + 24 + 30 + 36 + 42
= 168
3x − 8
≥4
2
or 3x – 8 ≥ 8
or 3x ≥ 16
16
x≤0
or x ≥
3
⎡ 16
⎞
The solution is (– ∞, 0] ∪ ⎢ , ∞ ⎟ .
⎣3
⎠
or
4.
4
∑ 10 p = 10(0) + 10(1) + 10(2) + 10(3) + 10(4)
p =0
= 0 + 10 + 20 + 30 + 40
= 100
48
ISM: Introductory Mathematical Analysis
5.
9
∑ (10k + 16) = [10(3) + 16] + [10(4) + 16] + [10(5) + 16] + [10(6) + 16] + [10(7) + 16] + [10(8) + 16] + [10(9) + 16]
k =3
6.
Section 1.5
= 46 + 56 + 66 + 76 + 86 + 96 + 106
= 532
11
∑ (2n − 3) = [2(7) − 3] + [2(8) − 3] + [2(9) − 3] + [2(10) − 3] + [2(11) − 3]
n =7
= 11 + 13 + 15 + 17 + 19
= 75
7. 36 + 37 + 38 + 39 + " + 60 =
60
∑i
i =36
8. 1 + 4 + 9 + 16 + 25 =
5
∑ k2
k =1
8
9. 53 + 54 + 55 + 56 + 57 + 58 = ∑ 5i
i=3
16
10. 11 + 15 + 19 + 23 + " + 71 = ∑ (7 + 4i )
i =1
11. 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 =
8
∑ 2i
i=1
8
12. 10 + 100 + 1000 + " + 100, 000, 000 = ∑ 10 j
j=1
13.
14.
15.
43
43
k =1
k =1
135
135
101
k =35
k =35
i =1
∑ 10 = 10 ∑ 1 = 10(43) = 430
∑ 2 = 2 ∑ 1 = 2∑1 = 2(101) = 202
n
⎛
1⎞
⎛
1⎞
k =1
16.
n
⎛
1⎞
∑ ⎜⎝ 5 ⋅ n ⎟⎠ = ⎜⎝ 5 ⋅ n ⎟⎠ ∑ 1 = ⎜⎝ 5 ⋅ n ⎟⎠ (n) = 5
k =1
200
200
200
k =1
k =1
k =1
∑ (k − 100) =
∑ k − 100 ∑ 1 =
200(201)
− 100(200) = 20,100 − 20, 000 = 100
2
49
Chapter 1: Applications and More Algebra
17.
100
ISM: Introductory Mathematical Analysis
50
∑ 10k = 10∑ (i + 50)
k =51
i =1
50
23.
50
= 10∑ i + (10)(50)∑1
i =1
50(51)
+ 500(50) = 12, 750 + 25, 000
2
= 37,750
n
n
19.
k =1
n n(n + 1)(2n + 1)
=
⋅
6
n +1
n 2 (2n + 1)
=
6
20
20
24.
k =1
k =1
20(21)(41)
20(21)
+3
6
2
= 5(2870) + 3(210) = 14,980
21.
3k 2 – 200k
3 100 2 200 100
=
∑ 101
∑ k − 101 ∑ k
101 k =1
k =1
k =1
3 100(101)(201) 200 100 ⋅101
=
⋅
−
⋅
101
6
101
2
= 10,050 – 10,000 = 50
50
k =51
50
i =1
25.
50
∑ k 2 = ∑ (i + 50)2 = ∑ (i 2 + 100i + 2500)
i =1
50
50
i =1
50
i =1
50
∑ (k + 50)2 = ∑ (k 2 + 100k + 2500)
k =1
50
=
k =1
50
50
k =1
k =1
26.
=
100
1
1 ⎛ 1 ⎞ 100 2
(4) ∑ 1 − ⎜
⎟∑k
50 k =1 50 ⎝ 2500 ⎠ k =1
⎧⎪ ⎡ ⎛ 3 ⎞2 ⎤ 3 ⎫⎪
∑ ⎨⎢⎢5 − ⎜⎝ n ⋅ k ⎟⎠ ⎥⎥ n ⎬
k =1 ⎪
⎦ ⎪⎭
⎩⎣
n
3 ⎛
9 2⎞
k ⎟
= ∑ ⎜5 −
n k =1 ⎝
n2 ⎠
n
k2
n
=
50(51)(101)
50(51)
+ 100
+ 2500(50)
6
2
= 42,925 + 127,500 + 125,000 = 295,425
=
50
1
n
∑ (n + 1)(2n + 1) = (n + 1)(2n + 1) ∑ k 2
k =1
∑ k 2 + 100 ∑ k + 2500 ∑ 1
k =1
1 100 ⎛
1
⎞
k2 ⎟
∑
⎜4−
50 k =1 ⎝
2500 ⎠
=
50(51)(101)
50(51)
=
+ 100
+ 2500(50)
6
2
= 42,925 + 127,500 + 125,000 = 295,425
22.
=
n
3
3⎛ 9 ⎞ n
(5) ∑ 1 − ⎜ ⎟ ∑ k 2
n k =1 n ⎝ n 2 ⎠ k =1
15
27 n(n + 1)(2n + 1)
= ( n) − ⋅
6
n
n3
9(n + 1)(2n + 1)
= 15 −
2n 2
= ∑ i 2 + 100∑ i + 2500∑ 1
i =1
⎛ 2 ⎞ ⎫⎪
⎥ ⎜
⎟⎬
⎥⎦ ⎝ 100 ⎠ ⎭⎪
2⎤
=
100
100
⎪
⎛ 2 ⎞
∑ ⎨⎢⎢ 4 − ⎜⎝ 100 k ⎟⎠
⎪⎣
k =1 ⎩
2
1
100(101)(201)
(100) −
⋅
25
125, 000
6
1
6767
= 8−
⋅ 338,350 = 8 −
125, 000
2500
13, 233
733
=
=5
2500
2500
= 5⋅
20.
100 ⎧ ⎡
20
∑ (5k 2 + 3k ) = 5 ∑ k 2 + 3∑ k
k =1
k =1 ⎪
⎩⎣
4
1 10(11)(21)
1
(10) −
⋅
= 8−
⋅ 385
5
125
6
125
77 123
23
= 8−
=
=4
25 25
25
∑ n +1 k 2 = n +1 ∑ k 2
k =1
2
10
1
⎛ 2k ⎞ ⎤ ⎛ 2 ⎞ ⎫⎪ 1 ⎛
⎞
⎥ ⎜ ⎟⎬ = ∑ ⎜ 4 − k 2 ⎟
10
5
25
⎠
⎥⎦ ⎝ ⎠ ⎪⎭
k =1 ⎝
=
n
n
⎧⎪ ⎡
10
1
1 ⎛ 1 ⎞ 10
= (4) ∑ 1 − ⎜ ⎟ ∑ k 2
5 k =1 5 ⎝ 25 ⎠ k =1
i =1
= 10 ⋅
18.
10
∑ ⎨⎢⎢4 − ⎜⎝ 10 ⎟⎠
k =1
1
n(n + 1)(2n + 1) n
⋅
=
(n + 1)(2n + 1)
6
6
ISM: Introductory Mathematical Analysis
Chapter 1 Review
Chapter 1 Review Problems
7.
1. 5 x − 2 ≥ 2( x − 7)
5 x − 2 ≥ 2 x − 14
3 x ≥ −12
x ≥ −4
[−4, ∞)
2. 2x – (7 + x) ≤ x
2x – 7 – x ≤ x
–7 ≤ 0, which is true for all x, so –∞ < x < ∞, or
(–∞, ∞).
8.
3. –(5x + 2) < –(2x + 4)
–5x – 2 < –2x – 4
–3x < –2
2
x>
3
⎛2
⎞
⎜ 3 , ∞⎟
⎝
⎠
9.
4. –2(x + 6) > x + 4
–2x – 12 > x + 4
–3x > 16
16
x<−
3
16 ⎞
⎛
⎜ −∞, − 3 ⎟
⎝
⎠
10.
5. 3 p (1 − p ) > 3(2 + p ) − 3 p 2
3 p − 3 p2 > 6 + 3 p − 3 p2
0 > 6, which is false for all x. The solution set is
∅.
x+5 1
− ≤2
3
2
2(x + 5) – 3(1) ≤ 6(2)
2x + 10 – 3 ≤ 12
2x ≤ 5
5
x≤
2
5⎤
⎛
⎜ −∞, 2 ⎥
⎝
⎦
x x x
− >
3 4 5
20 x − 15 x > 12 x
5 x > 12 x
0 > 7x
0>x
(−∞, 0)
1
1
s − 3 ≤ (3 + 2 s )
4
8
2s – 24 ≤ 3 + 2s
0 ≤ 27, which is true for all s. Thus
–∞ < s < ∞, or (–∞,∞).
1
1
(t + 2) ≥
3
4
4(t + 2) ≥ 3t + 48
4t + 8 ≥ 3t + 48
t ≥ 40
[40, ∞)
11. 3 − 2 x = 7
3 – 2x = 7
–2x = 4
x = –2
7 ⎞
⎛
6. 3 ⎜ 5 − q ⎟ < 9
3 ⎠
⎝
15 − 7q < 9
−7q < −6
6
q>
7
6
⎛
⎞
⎜ 7 , ∞⎟
⎝
⎠
12.
5x − 6
=0
13
5x − 6
=0
13
5x – 6 = 0
6
x=
5
13. |2z − 3| < 5
−5 < 2z − 3 < 5
−2 < 2z < 8
−1 < z < 4
(−1, 4)
51
or 3 – 2x = –7
or –2x = –10
or x = 5
Chapter 1: Applications and More Algebra
14. 4 <
ISM: Introductory Mathematical Analysis
19. Let x be the number of issues with a decline, and
x + 48 be the number of issues with an increase.
Then
x + (x + 48) = 1132
2x = 1084
x = 542
2
x+5
3
2
x + 5 < −4
3
2
x < −9
3
27
x<−
2
2
x+5 > 4
3
2
or
x > −1
3
3
or x > −
2
27
⎛
⎞ ⎛ 3
⎞
The solution is ⎜ −∞, − ⎟ ∪ ⎜ − , ∞ ⎟ .
2 ⎠ ⎝ 2
⎝
⎠
or
20. Let x = purchase amount excluding tax.
x + 0.065 x = 3039.29
1.065 x = 3039.29
x = 2853.79
Thus tax is 3039.29 − 2853.79 = $185.50.
21. Let q units be produced at A and 10,000 – q at
B.
Cost at A + Cost at B ≤ 117,000
[5q + 30,000] + [5.50(10,000 – q) + 35,000]
≤ 117,000
–0.5q + 120,000 ≤ 117,000
–0.5q ≤ –3000
q ≥ 6000
Thus at least 6000 units must be produced at
plant A.
15. 3 − 2 x ≥ 4
3 – 2x ≥ 4
–2x ≥ 1
1
x≤−
2
or 3 – 2x ≤ –4
or –2x ≤ –7
7
or x ≥
2
1⎤ ⎡7
⎛
⎞
The solution is ⎜ – ∞, − ⎥ ∪ ⎢ , ∞ ⎟ .
2⎦ ⎣2
⎝
⎠
16.
5
5
i =1
i =1
5
∑ (i + 2)3 = ∑ (i3 + 6i 2 + 12i + 8)
22. Total volume of old tanks
5
5
5
i =1
i =1
i =1
= π(10)2 (25) + π(20)2 (25)
= 2500π + 10, 000π
= ∑ i3 + 6∑ i 2 + 12∑ i + 8∑ 1
i =1
2
2
= 12,500π ft 3
Let r be the radius (in feet) of the new tank.
Then
4 3
πr = 12,500π
3
r 3 = 9375
5 (6)
5(6)(11)
5(6)
+6
+ 12
+ 8(5)
4
6
2
= 225 + 330 + 180 + 40
= 775
=
17.
7
7
2
i =3
i =1
2
i =1
r = 3 9375 ≈ 21.0858
The radius is approximately 21.0858 feet.
∑ i3 = ∑ i3 − ∑ i3
7 (8)2 22 (3)2
=
−
4
4
= 784 − 9
= 775
This uses Equation (1.9). By Equation (1.8),
7
5
i =3
i =1
23. Let c = operating costs
c
< 0.90
236, 460
c < $212,814
∑ i3 = ∑ (i + 2)3 .
Mathematical Snapshot Chapter 1
1
1. Here m = 120 and M = 2 (60) = 150. For LP,
2
t
r = 2, so the first t minutes take up
of the 120
2
available minutes. For SP, r = 1, so the
150 − t
of the
remaining 150 − t minutes take up
1
120 available.
18. Let p = selling price, c = cost. Then
p – 0.40p = c
0.6p = c
c
5c
⎛2⎞
p=
=
= c + ⎜ ⎟c
0.6 3
⎝3⎠
Thus the profit is
2
2
, or 66 %, of the cost.
3
3
52
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 1
5.
t 150 − t
+
= 120
2
1
t + 300 − 2t = 240
−t = −60
t = 60
Switch after 1 hour.
x = 600
1
2. Here m = 120 and M = 2 (60) = 150. For EP,
2
t
of the
r = 3, so the first t minutes will take up
3
120 available minutes. For SP, r = 1, so the
150 − t
of the
remaining 150 − t minutes take up
1
120 available.
t 150 − t
+
= 120
3
1
t + 450 − 3t = 360
−2t = −90
t = 45
Switch after 45 minutes.
x = 310
6. Both equations represent audio being written
onto 74-minute CDs. In the first equation, 18
hours (1080 minutes) are being written to a CD
using a combination of 12-to-1 and 20-to-1
compression ratios. Here, x gives the maximum
amount of audio (600 minutes or 10 hours) that
can be written using the 12-to-1 compression
ratio. In the second equation, 26.5 hours (1590
minutes) is being written using 15-to-1 and 24to-1 compression ratios. A maximum of 310
minutes can be written at 15-to-1.
3. Use the reasoning in Exercise 1, with M
unknown and m = 120.
t M −t
+
= 120
2
1
t + 2 M − 2t = 240
−t = 240 − 2 M
t = 2 M − 240
The switch should be made after
2M − 240 minutes.
t
of the m available
R
M −t
minutes, the remaining M − t minutes use
r
of the m available.
t M −t
+
=m
R
r
t M t
+
− =m
R r r
M
⎛ 1 1⎞
t⎜ − ⎟ = m−
r
⎝R r⎠
⎛ r − R ⎞ mr − M
t⎜
⎟=
r
⎝ rR ⎠
R (mr − M )
t=
r−R
7. The first t minutes use
4. Use the reasoning in Exercise 2, with M
unknown and m = 120.
t M −t
+
= 120
3
1
t + 3M − 3t = 360
−2t = 360 − 3M
1
t = (3M − 360)
2
The switch should be made after
1
(3M − 360) minutes.
2
53
Chapter 2
b. If 200 large pizzas are being sold each
week, q = 200.
200
p = 26 −
40
p = 26 – 5
p = 21
The price is $21 per pizza if 200 large
pizzas are being sold each week.
Principles in Practice 2.1
1. a.
The formula for the area of a circle is πr 2 ,
where r is the radius.
a(r ) = πr 2
b. The domain of a(r) is all real numbers.
c.
2. a.
Since a radius cannot be negative or zero,
the domain for the function, in context, is
r > 0.
c.
The formula relating distance, time, and
speed is d = rt where d is the distance, r is
the speed, and t is the time. This can also be
d
written as t = . When d = 300, we have
r
300
t (r ) =
.
r
Problems 2.1
1. The functions are not equal because f(x) ≥ 0 for
all values of x, while g(x) can be less than 0. For
b. The domain of t(r) is all real numbers
except 0.
c.
example, f (−2) = (−2) 2 = 4 = 2 and
Since speed is not negative, the domain for
the function, in context, is r > 0.
g(−2) = −2, thus f(−2) ≠ g(−2).
2. The functions are different because they have
different domains. The domain of G(x) is [−1, ∞)
(all real numbers ≥ −1) because you can only
take the square root of a non-negative number,
while the domain of H(x) is all real numbers.
300
.
x
x
⎛ x ⎞ 300 600
.
Replacing r by : t ⎜ ⎟ =
=
x
2
x
⎝2⎠
2
d. Replacing r by x: t ( x) =
Replacing r by
e.
3. a.
x
:
4
To double the number of large pizzas sold,
use q = 400.
400
p = 26 −
40
p = 26 – 10
p = 16
To sell 400 large pizzas each week, the price
should be $16 per pizza.
3. The functions are not equal because they have
different domains. h(x) is defined for all nonzero real numbers, while k(x) is defined for all
real numbers.
⎛ x ⎞ 300 1200
t⎜ ⎟ =
.
=
x
x
⎝4⎠
4
When the speed is reduced (divided) by a
constant, the time is scaled (multiplied) by
⎛ r ⎞ 300c
.
the same constant; t ⎜ ⎟ =
r
⎝c⎠
4. The functions are equal. For x = 3 we have
f(3) = 2 and g(3) = 3 − 1 = 2, hence f(3) = g(3).
For x ≠ 3, we have
x 2 − 4 x + 3 ( x − 3)( x − 1)
=
= x − 1.
f ( x) =
x−3
x−3
Note that we can cancel the x − 3 because we are
assuming x ≠ 3 and so x − 3 ≠ 0. Thus for
x ≠ 3 f(x) = x − 1 = g(x).
f(x) = g(x) for all real numbers and they have the
same domains, thus the functions are equal.
If the price is $18.50 per large pizza,
p = 18.5.
q
18.5 = 26 −
40
q
−7.5 = −
40
300 = q
At a price of $18.50 per large pizza, 300
pizzas are sold each week.
5. The denominator is zero when x = 0. Any other
real number can be used for x.
Answer: all real numbers except 0
54
ISM: Introductory Mathematical Analysis
Section 2.1
6. Any real number can be used for x.
Answer: all real numbers
16. r 2 + 1 is never 0.
Answer: all real numbers
7. For x − 3 to be real, x – 3 ≥ 0, so x ≥ 3.
Answer: all real numbers ≥ 3
17. f(x) = 2x + 1
f(0) = 2(0) + 1 = 1
f(3) = 2(3) + 1 = 7
f(–4) = 2(–4) + 1 = –7
8. For z − 1 to be real, z − 1 ≥ 0, so z ≥ 1. We
exclude values of z for which z − 1 = 0, so
z − 1 = 0, thus z = 1.
Answer: all real numbers > 1
18. H ( s ) = 5s 2 − 3
H (4) = 5(4)2 − 3 = 80 − 3 = 77
9. Any real number can be used for z.
Answer: all real numbers
H
10. We exclude values of x for which
x+8=0
x = –8
Answer: all real numbers except –8
2
− 3 = 10 − 3 = 7
2
20
7
⎛2⎞
⎛2⎞
H ⎜ ⎟ = 5⎜ ⎟ − 3 =
−3 = −
3
3
9
9
⎝ ⎠
⎝ ⎠
19. G ( x) = 2 − x 2
11. We exclude values of x where
2x + 7 = 0
2x = –7
7
x=−
2
G (–8) = 2 − (−8)2 = 2 − 64 = −62
G (u ) = 2 − u 2
G (u 2 ) = 2 − (u 2 )2 = 2 − u 4
7
Answer: all real numbers except −
2
20. F(x) = −5x
F(s) = −5s
F(t + 1) = −5(t + 1) = −5t − 5
F(x + 3) = −5(x + 3) = −5x − 15
12. For 4 x + 3 to be real,
4x + 3 ≥ 0
4x ≥ –3
3
x≥−
4
Answer: all real numbers ≥ −
( 2 ) = 5( 2 )
21. γ (u ) = 2u 2 − u
3
4
γ (−2) = 2(−2) 2 − (−2) = 8 + 2 = 10
γ (2v) = 2(2v) 2 − (2v) = 8v 2 − 2v
13. We exclude values of y for which
γ ( x + a ) = 2( x + a) 2 − ( x + a)
y 2 − 4 y + 4 = 0. y 2 − 4 y + 4 = ( y − 2) 2 , so we
= 2 x 2 + 4ax + 2a 2 − x − a
exclude values of y for which y − 2 = 0, thus
y = 2.
Answer: all real numbers except 2.
22. h(v) =
14. We exclude values of x for which
2
x + x−6 = 0
( x + 3)( x − 2) = 0
x = −3, 2
Answer: all real numbers except −3 and 2
h(16) =
v
1
⎛1⎞
h⎜ ⎟ =
⎝4⎠
15. We exclude all values of s for which
h(1 − x) =
2s 2 − 7 s − 4 = 0
(s – 4)(2s + 1) = 0
1
s = 4, −
2
Answer: all real numbers except 4 and −
1
1
2
55
16
1
1
4
=
=
1
4
1
1
2
1
1− x
=2
Chapter 2: Functions and Graphs
23.
ISM: Introductory Mathematical Analysis
f ( x) = x 2 + 2 x + 1
28. g ( x) = x 2 / 5
( 5 32 ) = (2)2 = 4
2
g (−64) = (−64)2 / 5 = ( 5 −64 )
2
2
= ( 5 −32 5 2 ) = ( –2 5 2 ) = 4 5 4
f (1) = 12 + 2(1) + 1 = 1 + 2 + 1 = 4
g (32) = 322 / 5 =
f (–1) = (–1) 2 + 2(–1) + 1 = 1 − 2 + 1 = 0
f ( x + h) = ( x + h) 2 + 2( x + h) + 1
= x 2 + 2 xh + h 2 + 2 x + 2h + 1
24. H ( x) = ( x + 4) 2
g (t10 ) = (t10 )2 / 5 = t 4
H (0) = (0 + 4) 2 = 16
29. f(x) = 4x – 5
H (2) = (2 + 4)2 = 62 = 36
a.
H (t − 4) = [(t − 4) + 4]2 = t 2
25. k ( x) =
k (5) =
b.
x−7
x2 + 2
5−7
2
=−
2
27
5 +2
3x − 7
3x − 7
k (3 x) =
=
2
(3x) + 2 9 x 2 + 2
( x + h) − 7
x+h−7
k ( x + h) =
=
2
2
( x + h) + 2 x + 2 xh + h 2 + 2
30.
26. k ( x) = x − 3
k (4) = 4 − 3 = 1 = 1
k (3) = 3 − 3 = 0 = 0
31.
k ( x + 1) − k ( x) = ( x + 1) − 3 − x − 3
= x −2 − x −3
27.
f(x + h) = 4(x + h) – 5 = 4x + 4h – 5
f ( x + h) − f ( x )
h
(4 x + 4h − 5) − (4 x − 5) 4h
=
=
=4
h
h
f ( x) =
x
2
x+h
2
a.
f ( x + h) =
b.
f ( x + h) − f ( x )
=
h
x+h
2
− 2x
h
=
h
2
h
=
1
2
f ( x) = x 2 + 2 x
a.
f ( x + h) = ( x + h) 2 + 2( x + h)
= x 2 + 2 xh + h 2 + 2 x + 2h
f ( x) = x 4 / 3
f (0) = 04 / 3 = 0
( )
f (64) = 644 / 3 = 3 64
⎛1⎞ ⎛1⎞
f ⎜ ⎟=⎜ ⎟
⎝8⎠ ⎝8⎠
2
4/3
4
b.
= (4) 4 = 256
4
4
⎛ 1⎞
1
⎛1⎞
= ⎜3 ⎟ = ⎜ ⎟ =
⎜ 8⎟
16
⎝2⎠
⎝
⎠
32.
f ( x + h) − f ( x )
h
=
( x 2 + 2 xh + h 2 + 2 x + 2h) − ( x 2 + 2 x)
h
=
2 xh + h 2 + 2h
= 2x + h + 2
h
f ( x) = 3x 2 − 2 x − 1
a.
f ( x + h) = 3( x + h) 2 − 2( x + h) − 1
= 3( x 2 + 2 xh + h 2 ) − 2 x − 2h − 1
= 3 x 2 + 6 xh + 3h 2 − 2 x − 2h − 1
56
ISM: Introductory Mathematical Analysis
b.
33.
Section 2.1
f ( x + h) − f ( x) (3x 2 + 6 xh + 3h 2 − 2 x − 2h − 1) − (3 x 2 − 2 x − 1)
=
h
h
6 xh + 3h 2 − 2h
=
h
= 6 x + 3h − 2
f ( x) = 3 − 2 x + 4 x 2
a.
f ( x + h) = 3 − 2( x + h) + 4( x + h)2
= 3 − 2 x − 2h + 4( x 2 + 2 xh + h 2 )
b.
34.
35.
f ( x + h) − f ( x) 3 − 2 x − 2h + 4 x 2 + 8 xh + 4h 2 − (3 − 2 x + 4 x 2 )
=
h
h
−2h + 8 xh + 4h 2
=
h
= −2 + 8 x + 4h
f ( x ) = x3
a.
f ( x + h) = ( x + h)3 = x3 + 3 x 2 h + 3 xh 2 + h3
b.
f ( x + h) − f ( x) ( x3 + 3 x 2 h + 3xh 2 + h3 ) − x3 3x 2 h + 3 xh 2 + h3
=
=
= 3x 2 + 3xh + h 2
h
h
h
f ( x) =
1
x
1
x+h
a.
f ( x + h) =
b.
1
1
−h
f ( x + h) − f ( x ) x + h − x
1
x( x + h)
=
=
=
=−
h
h
h
x ( x + h) h
x ( x + h)
x −( x + h )
36.
f ( x) =
x +8
x
( x + h) + 8 x + h + 8
=
x+h
x+h
a.
f ( x + h) =
b.
x + h + 8 − x +8
x( x + h) x +x +h h+8 −
f ( x + h) − f ( x )
x
= x+ h
=
h
h
x ( x + h) h
(
=
2
2
x +8
x
) = x( x + h + 8) − ( x + h)( x + 8)
−8h
x + xh + 8 x − x − hx − 8 x − 8h
8
=
=−
x ( x + h) h
x ( x + h)
x ( x + h) h
57
x ( x + h) h
Chapter 2: Functions and Graphs
37.
38.
ISM: Introductory Mathematical Analysis
f (3 + h) − f (3) [5(3 + h) + 3] − [5(3) + 3]
=
h
h
[15 + 5h + 3] − [15 + 3]
=
h
18 + 5h − 18
=
h
5h
=
h
=5
42. x 2 + y 2 = 1
Solving for y we have y = ± 1 − x 2 . If x = 0,
then y = ±1, so y is not a function of x. Solving
for x gives x = ± 1 − y 2 . If y = 0, then x = ±1,
so x is not a function of y.
43. Yes, because corresponding to each input r there
is exactly one output, πr 2 .
f ( x) − f (2) 2 x 2 − x + 1 − (8 − 2 + 1)
=
x−2
x−2
2
2x − x + 1 − 7
=
x−2
2x2 − x − 6
=
x−2
= 2x + 3
44. a.
b.
f ( a ) = a 2 a 3 + a 3 a 2 = a 5 + a 5 = 2a 5
f (ab) = a 2 (ab)3 + a3 (ab) 2
= a 2 a 3b 3 + a 3 a 2 b 2
= a 5 b3 + a 5 b 2
= a5b 2 (b + 1)
45. Weekly excess of income over expenses is
6500 − 4800 = 1700.
After t weeks the excess accumulates to 1700t.
Thus the value of V of the business at the end of
t weeks is given by V = f(t) = 25,000 + 1700t.
39. 9y – 3x – 4 = 0
3x + 4
shows that for
The equivalent form y =
9
3x + 4
each input x there is exactly one output,
.
9
Thus y is a function of x. Solving for x gives
9y − 4
x=
. This shows that for each input y
3
9y − 4
. Thus x is a
there is exactly one output,
3
function of y.
46. Depreciation at the end of t years is
0.02t(30,000), so value V of machine is
V = f(t) = 30,000 – 0.02t(30,000), or
V = f(t) = 30,000(1 – 0.02t).
47. Yes; for each input q there corresponds exactly
one output, 1.25q, so P is a function of q. The
dependent variable is P and the independent
variable is q.
40. x 2 + y = 0
48. Charging $600,000 per film corresponds to
p = 600,000.
1, 200, 000
600,000 =
q
q=2
The form y = − x 2 shows that for each input x
there is exactly one output, − x 2 . Thus y is a
function of x. Solving for x gives x = ± − y . If,
for example, y = –1, then x = ±1, so x is not a
function of y.
The actor will star in 2 films per year. To star in
4 films per year the actor should charge
1, 200, 000
p=
= $300, 000 per film.
4
2
41. y = 7 x
For each input x, there is exactly one output
7 x 2 . Thus y is a function of x. Solving for x
49. The function can be written as q = 48p.
At $8.39 per pound, the coffee house will supply
q = 48(8.39) = 402.72 pounds per week.
At $19.49 per pound, the coffee house will
supply q = 48(19.49) = 935.52 pounds per week.
The amount the coffee house supplies increases
as the price increases.
y
. If, for example, y = 7, then
7
x = ±1, so x is not a function of y.
gives x = ±
58
ISM: Introductory Mathematical Analysis
50. a.
f(0) = 1 – 1 = 0
3
b.
c.
Section 2.2
1
1
1
52. P (1) = 1 − (1 − 0.344)0 = 1 − (1) =
2
2
2
1
1
P (2) = 1 − (1 − 0.344)1 = 1 − (0.656) = 0.672
2
2
3
27
⎛ 300 ⎞
⎛3⎞
f (100) = 1 − ⎜
= 1− ⎜ ⎟ = 1−
⎟
64
⎝ 400 ⎠
⎝4⎠
37
=
64
⎛ 300 ⎞
f (900) = 1 − ⎜
⎟
⎝ 1200 ⎠
⎛1⎞
= 1− ⎜ ⎟
⎝4⎠
1
= 1−
64
63
=
64
53. a.
3
b. Domain: 10, 12, 17, 20
g(10) = 3000, g(17) = 2300
3
54. a.
c.
55. a.
3
c.
56. a.
3
⎛ 300 ⎞
⎜ 300 + t ⎟ = 0.5
⎝
⎠
300
= 3 0.5
300 + t
300 = 300 3 0.5 + t 3 0.5
t=
300 − 300 3 0.5
3
0.5
)
4
c.
57. a.
f (1000) =
(
3
1000
2500
1,997,723.57
2,964,247.40
7.89
1.21
Principles in Practice 2.2
=
104
10, 000
=
=4
2500
2500
(
)
⎡ 3 1000(2) ⎤
10 3 2
⎦ =
f (2000) = ⎣
2500
2500
1. a.
4
Let n = the number of visits and p(n) be the
premium amount.
p(n) = 125
b. The premiums do not change regardless of
the number of doctor visits.
c.
3
10, 000 24
3
= 4 23 ⋅ 2 = 83 2
2500
f (2 I 0 ) =
–17.43
b. 63.85
≈ 77.98
c.
4
c.
–5.13
b. 1,287,532.35
78 days
=
–18.51
b. 2.64
⎛ 300 ⎞
0.500 = 1 − ⎜
⎟
⎝ 300 + t ⎠
b.
–18.97
b. –581.77
d. We solve
51. a.
Domain: 3000, 2900, 2300, 2000
f(2900) = 12, f(3000) = 10
2. a.
(2 I 0 ) 4 / 3 24 / 3 I 04 / 3
=
2500
2500
This is a constant function.
d (t ) = 3t 2 is a quadratic function.
b. The degree of d (t ) = 3t 2 is 2.
⎡ I4/3 ⎤
= 23 2 ⎢ 0 ⎥ = 23 2 f ( I0 )
⎢⎣ 2500 ⎥⎦
c.
The leading coefficient of d (t ) = 3t 2 is 3.
3. The price for n pairs of socks is given by
⎧ 3.5n 0 ≤ n ≤ 5
⎪
c(n) = ⎨ 3n
5 < n ≤ 10 .
⎪2.75n
10 < n
⎩
Thus f (2 I 0 ) = 2 3 2 f ( I 0 ) , which means
that doubling the intensity increases the
response by a factor of 2 3 2.
59
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
4. Think of the bookshelf having 7 slots, from left
to right. You have a choice of 7 books for the
first slot. Once a book has been put in the first
slot, you have 6 choices for which book to put in
the second slot, etc. The number of arrangements
is 7 · 6 · 5 · 4 · 3 · 2 · 1 = 7! = 5040.
18. g ( x) = x − 3
g (10) = 10 − 3 = 7 = 7
g (3) = 3 – 3 = 0 = 0
g (–3) = –3 − 3 = −6 = 6
19. F(10) = 1
Problems 2.2
(
)
F − 3 = −1
1. yes
F(0) = 0
⎛ 18 ⎞
F ⎜ − ⎟ = −1
⎝ 5⎠
x3 + 7 x − 3 1 3 7
= x + x − 1, which is a
2. f ( x) =
3
3
3
polynomial function.
20. f(3) = 4
f(–4) = 3
f(0) = 4
3. no
4. yes
21. G(8) = 8 − 1 = 7
G(3) = 3 − 1 = 2
5. yes
6. yes
G (−1) = 3 − (−1)2 = 2
7. no
G (1) = 3 − (1)2 = 2
8. g ( x) = 4 x –4 =
4
x4
22. F (3) = 32 − 3(3) + 1 = 1
, which is a rational function.
9. all real numbers
F(−3) = 2(−3) − 5 = −11
F(2) is not defined.
10. all real numbers
23. 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720
11. all real numbers
24. 0! = 1
12. all x such that 1 ≤ x ≤ 3
25. (4 – 2)! = 2! = 2 · 1 = 2
13. a.
26. 6! ⋅ 2! = (6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1)(2 ⋅1)
= (720)(2)
= 1440
3
b. 7
14. a.
1
27.
b. 7
15. a.
7
28.
b. 1
16. a.
0
b. 9
17. f(x) = 8
f(2) = 8
f(t + 8) = 8
(
)
f − 17 = 8
60
n!
n ⋅ (n − 1)!
=
=n
(n − 1)!
(n − 1)!
8!
8!
=
5!(8 − 5)! 5! ⋅ 3!
8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
=
(5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1)(3 ⋅ 2 ⋅1)
8⋅7⋅6
=
3 ⋅ 2 ⋅1
= 8⋅7
= 56
ISM: Introductory Mathematical Analysis
Section 2.3
29. Let i = the passenger’s income and
c(i) = the cost for the ticket.
c(i) = 4.5
This is a constant function.
b.
30. Let w = the width of the prism, then
w + 3 = the length of the prism, and
2w – 1 = the height of the prism. The formula for
the volume of a rectangular prism is
V = length · width · height.
38. a.
V ( w) = ( w + 3)( w)(2w − 1) = 2w3 + 5w2 − 3w
This is a cubic function.
31. a.
b.
1600 = 850 + 3q
750 = 3q
q = 250
36. P (5) =
37. a.
1
2!(1!)
( 14 ) ( 34 )
5!
5
5!(0!)
39. a.
1182.74
40. a.
=
0
=
6
( 161 )( 43 ) =
2(1)
c.
41. a.
5!(1)
19.12
57.69
2.21
b. 9.98
c.
–14.52
Principles in Practice 2.3
1. The customer’s price is
(c D s )( x) = c( s ( x)) = c( x + 3) = 2( x + 3)
= 2x + 6
2. g ( x) = ( x + 3) 2 can be written as
g ( x) = a(l ( x)) = (a D l )( x) where a( x) = x 2 and
l(x) = x + 3. Then l(x) represents the length of
the sides of the square, while a(x) is the area of a
square with side of length x.
9
64
1 (1)
( 1024
) =
5!
252.15
b. –62.94
34. For a committee of four, there are 4 choices for
who will be member A. For each choice of
member A, there are 3 choices for member G.
Once members A and G have been chosen, there
are two choices for member M, then one choice
for member S once members A, G, and M have
been chosen. Thus, there are
4 ⋅ 3 ⋅ 2 ⋅ 1 = 4! = 24 ways to label the members.
Similarly, a committee of five can label itself
with five labels in
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5! = 120 ways.
35. P (2) =
1218.60
c.
33. The cost for buying n tickets is
⎧ 9.5n 0 ≤ n < 12
c ( n) = ⎨
12 ≤ n
⎩8.75n
2
c.
b. 4985.27
32. The interest is Prt, so principal and interest
amount to f(t) = P + Prt, or f(t) = P(1 + rt). Since
f(t) = at + b where a (= Pr) and b (= P) are
constants, f is a linear function of t.
( 14 ) ( 43 )
742.50
b. −20.28
C = 850 + 3q
3!
1
11 5 11 16
(30) + = + =
=4
24
4 4 4
4
1
11 6 11 17
(36) + = + =
f (36) =
24
4 4 4
4
4
175
175 33
f (39) = (39) −
= 52 −
=
3
4
4
4
f (30) =
Problems 2.3
1. f(x) = x + 3, g(x) = x + 5
1
1024
a.
all T such that 30 ≤ T ≤ 39
( f + g )( x) = f ( x) + g ( x)
= ( x + 3) + ( x + 5)
= 2x + 8
b. (f + g)(0) = 2(0) + 8 = 8
61
Chapter 2: Functions and Graphs
c.
( f − g )( x) = f ( x) − g ( x)
= ( x + 3) − ( x + 5)
= −2
d.
( fg )( x) = f ( x) g ( x)
= ( x + 3)( x + 5)
ISM: Introductory Mathematical Analysis
= x 2 + 8 x + 15
e.
g.
( fg )(–2) = (−2) + 8(−2) + 15 = 3
f
f ( x) x + 3
( x) =
=
g
g ( x) x + 5
h.
( g D f )( x) = g ( f ( x)) = g (2 x) = 6 + 2 x
i.
( g D f )(2) = 6 + 2(2) = 6 + 4 = 10
( f D g )(3) = 3 + 8 = 11
i.
( g D f )( x) = g ( f ( x))
= g ( x + 3)
= ( x + 3) + 5
= x+8
f ( x) = x 2 + 1, g ( x) = x 2 − x
a.
= 2x2 − x + 1
b.
b.
c.
c.
1
1
⎛ 1⎞
( f − g) ⎜ − ⎟ = − +1 =
2
2
2
⎝
⎠
d.
( fg )( x) = f ( x) g ( x)
= ( x 2 + 1)( x 2 − x)
( g D f )(3) = 3 + 8 = 11
= x 4 − x3 + x 2 − x
( f + g )( x) = f ( x) + g ( x)
= (2 x) + (6 + x)
= 3x + 6
e.
f
f ( x) x 2 + 1
( x) =
=
g
g ( x) x 2 − x
f.
− 12 + 1
f ⎛ 1⎞
5
−
=
=
2
g ⎜⎝ 2 ⎟⎠
3
− 12 − − 12
g.
( f D g )( x) = f ( g ( x))
= f ( x 2 − x)
(f – g)(4) = (4) – 6 = –2
( fg )( x) = f ( x) g ( x) = 2 x(6 + x) = 12 x + 2 x
e.
f
f ( x)
2x
=
( x) =
g
g ( x) 6 + x
f.
f
2(2) 4 1
(2) =
= =
g
6+2 8 2
( )
( ) ( )
2
( f − g )( x) = f ( x) − g ( x)
= (2 x) − (6 + x )
= x−6
d.
( f − g )( x) = f ( x) − g ( x)
= ( x 2 + 1) − ( x 2 − x)
= x +1
2. f(x) = 2x, g(x) = 6 + x
a.
( f + g )( x) = f ( x) + g ( x)
= ( x 2 + 1) + ( x 2 − x )
( f D g )( x) = f ( g ( x))
= f ( x + 5)
= ( x + 5) + 3
= x+8
h.
j.
( f D g )( x) = f ( g ( x))
= f (6 + x)
= 2(6 + x)
= 12 + 2 x
2
3.
f.
g.
= ( x 2 − x) 2 + 1
2
= x 4 − 2 x3 + x 2 + 1
h.
( g D f )( x) = g ( f ( x))
= g ( x 2 + 1)
= ( x 2 + 1) 2 − ( x 2 + 1)
= x4 + x2
i.
62
( g D f )(−3) = (−3) 4 + (−3)2 = 90
ISM: Introductory Mathematical Analysis
4.
Section 2.3
7. ( F D G )(t ) = F (G (t ))
⎛ 2 ⎞
= F⎜
⎟
⎝ t −1 ⎠
f ( x) = x 2 + 1, g ( x) = 5
a.
( f + g )( x) = f ( x) + g ( x)
2
= ( x 2 + 1) + 5
⎛ 2 ⎞
⎛ 2 ⎞
=⎜
⎟ + 7⎜ t –1 ⎟ +1
⎝ t –1⎠
⎝
⎠
4
14
=
+
+1
(t − 1) 2 t − 1
(G D F )(t ) = G ( F (t ))
= x2 + 6
2
b.
c.
58
⎛2⎞ ⎛2⎞
( f + g) ⎜ ⎟ = ⎜ ⎟ + 6 =
3
3
9
⎝ ⎠ ⎝ ⎠
= G (t 2 + 7t + 1)
2
=
2
(t + 7t + 1) –1
2
=
2
t + 7t
( f − g )( x) = f ( x) − g ( x)
= ( x 2 + 1) – 5
= x2 – 4
d.
( fg )( x) = f ( x) g ( x)
= ( x 2 + 1)(5)
8. ( F D G )(t ) = F (G (t ))
2
= 5x + 5
= F (3t 2 + 4t + 2)
= 3t 2 + 4t + 2
(G D F )(t ) = G ( F (t ))
e.
( fg )(7) = 5(7 2 ) + 5 = 245 + 5 = 250
f.
f
f ( x) x 2 + 1
( x) =
=
g
g ( x)
5
=G
g.
( f D g )( x) = f ( g ( x)) = f (5) = 52 + 1 = 26
= 3t + 4 t + 2
h.
( f D g )(12, 003) = 26
i.
( g D f )( x) = g ( f ( x)) = g ( x 2 + 1) = 5
( t)
2
= 3( t ) + 4 ( t ) + 2
9. ( f D g )(v) = f ( g (v))
= f
=
5. f(g(2)) = f(4 – 4) = f(0) = 0 + 6 = 6
g(f(2)) = g(12 + 6) = g(18) = 4 – 36 = –32
v+2
)
1
(
v+2
)
2
+1
1
v + 2 +1
1
=
v+3
( g D f )(v) = g ( f (v))
⎛ 1 ⎞
= g⎜
⎟
⎝ v2 + 1 ⎠
1
=
+2
2
v +1
=
6. ( f D g )( p) = f ( g ( p ))
⎛ p–2⎞
= f⎜
⎟
⎝ 3 ⎠
4
=
p –2
3
=
(
12
p–2
4
⎛ 4 ⎞ p – 2 4 – 2p
( g D f )( p) = g ( f ( p )) = g ⎜ ⎟ =
=
3
3p
⎝ p⎠
=
=
63
1 + 2(v 2 + 1)
v2 + 1
2v 2 + 3
v2 + 1
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
19. ( g D f )(m) = g ( f (m))
⎛ 40m – m2 ⎞
= g⎜
⎟
⎜
⎟
4
⎝
⎠
⎛ 40m – m 2 ⎞
= 40 ⎜
⎟
⎜
⎟
4
⎝
⎠
10. ( f D f )( x) = f ( f ( x))
= f ( x 2 + 2 x − 1)
= ( x 2 + 2 x − 1)2 + 2( x 2 + 2 x − 1) − 1
= x 4 + 4 x3 + 4 x 2 − 2
11. Let g(x) = 11x and f(x) = x − 7. Then
h(x) = g(x) − 7 = f(g(x))
= 10(40m − m2 )
= 400m − 10m2
This represents the total revenue received when
the total output of m employees is sold.
12. Let g(x) = x 2 – 2 and f ( x) = x . Then
h( x) = x 2 – 2 = g ( x) = f ( g ( x))
20. ( f D g )( E ) = f ( g ( E ))
1
13. Let g ( x) = x – 2 and f ( x) = . Then
x
1
1
h( x ) =
=
= f ( g ( x))
2
g
(
x)
x –2
= f (7202 + 0.29 E 3.68 )
2
= 0.45(7202 + 0.29 E 3.68 – 1000)0.53
= 0.45(6202 + 0.29 E 3.68 )0.53
This represents status based on years of
education.
14. Let g ( x) = 9 x3 – 5 x and f ( x) = x3 – x 2 + 11.
21. a.
Then h( x) = [ g ( x)]3 – [ g ( x)]2 + 11 = f ( g ( x))
b. 1169.64
x2 − 1
15. Let g ( x) =
and f ( x) = 4 x .
x+3
22. a.
Then h( x) = 4 g ( x) = f ( g ( x)).
16. Let g(x) = 3x − 5 and f ( x) =
h( x ) =
17. a.
2 − (3x − 5)
(3x − 5) 2 + 2
2− x
x2 + 2
−0.13
b. 18.85
23. a.
. Then
194.47
b. 0.29
= f ( g ( x)).
24. a.
0.45
b. 1.61
The revenue is $9.75 per pound of coffee
sold, so r(x) = 9.75x.
Problems 2.4
b. The expenses are e(x) = 4500 + 4.25x.
c.
14.05
f −1 ( x) =
x 7
−
3 3
2. g −1 ( x) =
x 1
−
2 2
1.
Profit = revenue – expenses.
(r – e)(x) = 9.75x – (4500 + 4.25x)
= 5.5x – 4500.
18. v( x) = (4 x – 2)3 can be written as
3. F −1 ( x) = 2 x + 14
v( x) = f (l ( x)) = ( f D l )( x) where f ( x) = x3 and
l(x) = 4x – 2. Then l(x) represents the length of
the sides of the cube, while f(x) is the volume of
a cube with sides of length x.
4.
5. r ( A) =
64
x 5
+
4 4
f −1 ( x) =
A
π
ISM: Introductory Mathematical Analysis
6. r (V ) = 3
Section 2.5
⎛ 1, 200, 000 ⎞
p(q( p )) = p ⎜
⎟
p
⎝
⎠
1, 200, 000
=
3V
4π
7. f(x) = 5x + 12 is one-to-one, for if
f ( x1 ) = f ( x2 ) then 5 x1 + 12 = 5 x2 + 12, so
5 x1 = 5 x2 and thus x1 = x2 .
1,200,000
p
= 1, 200, 000 ⋅
=p
Similarly, q(p(q)) = q, so the functions are
inverses.
8. g ( x) = (5 x + 12) 2 is not one-to-one, because
g ( x1 ) = g ( x2 ) does not imply x1 = x2 . For
⎛ 11 ⎞
example, g ⎜ − ⎟ =
⎝ 5⎠
⎛ 13 ⎞
g ⎜ − ⎟ = 1.
⎝ 5⎠
9. h( x) = (5 x + 12)2 , for x ≥ −
q
, we get q = 48p. Since q > 0, p is
48
also greater than 0, so q as a function of p is
q = q(p) = 48p, p > 0.
q
⎛ q ⎞
q( p (q )) = q ⎜ ⎟ = 48 ⋅
=q
48
⎝ 48 ⎠
48 p
p(q ( p )) = p (48 p ) =
=p
48
Thus, p(q) and q(p) are inverses.
14. From p =
5
, is one-to-one.
12
If h( x1 ) = h( x2 ) then (5 x1 + 12)2 = (5 x2 + 12)2 .
Since x ≥ −
5
we have 5x + 12 ≥ 0, and thus
12
(5 x1 + 12)2 = (5 x2 + 12)2 only if
5 x1 + 12 = 5 x2 + 12, and hence x1 = x2 .
Principles in Practice 2.5
10. F ( x) = x − 9 is not one-to-one, because
1. Let y = the amount of money in the account.
Then, after one month,
y = 7250 – (1 · 600) = $6650, and after two
months y = 7250 – (2 · 600) = $6050. Thus, in
general, if we let x = the number of months
during which Rachel spends from this account,
y = 7250 – 600x. To identify the x-intercept, we
set y = 0 and solve for x.
y = 7250 – 600x
0 = 7250 – 600x
600x = 7250
1
x = 12
12
⎛ 1 ⎞
The x-intercept is ⎜12 , 0 ⎟ .
⎝ 12 ⎠
Therefore, after 12 months and approximately
2.5 days Rachel will deplete her savings. To
identify the y-intercept, we set x = 0 and solve
for y.
y = 7250 – 600x
y = 7250 – 600(0)
y = 7250
The y-intercept is (0, 7250).
Therefore, before any months have gone by,
Rachel has $7250 in her account.
F ( x1 ) = F ( x2 ) does not imply x1 = x2 . For
example, F(8) = F(10) = 1.
11. The inverse of f ( x) = (4 x − 5) 2 for x ≥
f −1 ( x ) =
5
is
4
x 5
+ , so to find the solution, we
4 4
find f −1 (23).
f −1 (23) =
23 5
+
4
4
The solution is x =
23 5
+ .
4
4
12. The inverse of V (r ) =
4 3
3V
πr is r (V ) = 3
, so
3
4π
the solution is r (100) = 3
p
1, 200, 000
3(100)
.
4π
1, 200, 000
1, 200, 000
, we get q =
.
q
p
Since q > 0, p is also greater than 0, so q as a
1, 200, 000
, p > 0.
function of p is q = q ( p) =
p
13. From p =
65
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
Problems 2.5
2. Let y = the cost to the customer and let
x = the number of rides he or she takes. Since the
cost does not change, regardless of the number
of rides taken, the equation y = 24.95 represents
this situation. The graph of y = 24.95 is a
horizontal line whose y-intercept is (0, 24.95).
Since the line is parallel to the x-axis, there is no
x-intercept.
1.
0
1
2
2.5
3
4
5
y
0
12
24
30
24
12
0
(
2.
Q. II
3. a.
Miles
(2.5, 30)
Q. I
(1, 1)
x
10
f(0) = 1, f(2) = 2, f(4) = 3, f(–2) = 0
Range: all real numbers
d. f(x) = 0 for x = –2. So a real zero is –2.
(5, 0)
1 2 3 4 5
4. a.
x
hours
f(0) = 2, f(2) = 0
b. Domain: all x ≥ 0
c.
Range: all y ≥ 2
d. f(x) = 0 for x = 2. So a real zero is 2.
5. a.
f(0) = 0, f(1) = 1, f(–1) = 1
b. Domain: all real numbers
c.
0
10
30
50
70
80
90
100
x
0
5.3
15.9
26.5
37.1
44.5
51.9
59.3
y
Cost (dollars)
y
b. Domain: all real numbers
x
40
10
(0, –6)
4. The monthly cost of x therms of gas is
⎧0.53x, if 0 ≤ x ≤ 70
y=⎨
⎩0.53(70) + 0.74( x – 70), if x >70
or
if 0 ≤ x ≤ 70
⎧0.53x,
y=⎨
⎩0.74 x − 14.7, if x > 70
60
(8, –3)
(3, 0)
24
(0, 0)
x
10
Q. IV
(–4, 5)
c.
12
(
– 1 , –2
2
Q. III
y
36
y
(2, 7)
Q. I
(0, 0)
3. The formula relating distance, time, and speed is
d = rt, where d is the distance, r is the speed, and
t is the time. Let x = the time spent biking (in
hours). Then, 12x = the distance traveled. Brett
bikes 12 · 2.5 = 30 miles and then turns around
and bikes the same distance back to the rental
shop. Therefore, we can represent the distance
from the turn-around point at any time x as
30 – 12x . Similarly, the distance from the rental
shop at any time x can be represented by the
function y = 30 – 30 – 12 x .
x
10
(100, 59.3)
Range: all nonnegative real numbers
d. f(x) = 0 for x = 0. So a real zero is 0.
6. a.
f(0) = 0, f(2) = 1, f(3) = 3, f(4) = 2
b. Domain: all x such that 0 ≤ x ≤ 4
(70, 37.1)
c.
Range: all y such that 0 ≤ y ≤ 3
20
(0, 0)
20 40 60 80 100
d. f(x) = 0 for x = 0. So a real zero is 0.
x
therms
66
ISM: Introductory Mathematical Analysis
Section 2.5
7. y = 2x
If y = 0, then x = 0. If x = 0, then y = 0.
Intercept: (0, 0)
y is a function of x. One-to-one.
Domain: all real numbers
Range: all real numbers
5
10. y = 3 – 2x
If y = 0, then 0 = 3 – 2x, x =
3
.
2
⎛3 ⎞
If x = 0, then, y = 3. Intercepts: ⎜ , 0 ⎟ , (0, 3)
⎝2 ⎠
y is a function of x. One-to-one.
Domain: all real numbers
Range: all real numbers
y
y
x
5
3
3
2
x
5
8. y = x + 1
If y = 0, then x = –1.
If x = 0, then y = 1.
Intercepts: (–1, 0), (0, 1)
y is a function of x. One-to-one.
Domain: all real numbers
Range: all real numbers
5
11. y = x 4
If y = 0, then 0 = x 4 , x = 0. If x = 0, then y = 0.
Intercept: (0, 0)
y is a function of x. Not one-to-one.
Domain: all real numbers
Range: all real numbers ≥ 0
y
1
x
–1
y
5
5
x
5
9. y = 3x – 5
If y = 0, then 0 = 3x – 5, x =
5
.
3
⎛5 ⎞
If x = 0, then y = –5. Intercepts: ⎜ , 0 ⎟ , (0, –5)
⎝3 ⎠
y is a function of x. One-to-one.
Domain: all real numbers
Range: all real numbers
10
12. y =
2
x2
If y = 0, then 0 =
, which has no solution.
x2
Thus there is no x-intercept. Because x ≠ 0 ,
Not one-to-one.
Domain: all real numbers except 0
Range: all real numbers > 0
y
x
–5
5
3
2
5
y
10
(–1, 2)
(–2, 12 )
(1, 2)
(2, 12 )
x
5
67
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
16. x = −9
If y = 0 then x = −9. Because x cannot be 0, there
is no y-intercept. Intercept: (−9, 0).
y is not a function of x.
13. x = 0
If y = 0, then x = 0. If x = 0, then y can be any
real number. Intercepts: every point on y-axis
y is not a function of x.
5
y
y
10
x
5
17. x = −|y|
If y = 0, then x = 0. If x = 0, then 0 = −|y|, y = 0.
Intercept: (0, 0)
y is not a function of x.
14. y = 4 x 2 – 16
If y = 0, then 0 = 4 x 2 – 16 = 4( x 2 − 4) ,
0 = 4( x + 2)( x − 2) , x = ±2.
If x = 0, then y = –16.
Intercepts: (±2, 0), (0, –16)
y is a function of x. Not one-to-one.
Domain: all real numbers
Range: all real numbers ≥ –16
20
x
10
ñ 10
5
x
5
y
18. x 2 = y 2
x
2
–2
5
If y = 0, then x 2 = 0, x = 0 . If x = 0, then
0 = y 2 , y = 0 . Intercept: (0, 0)
y is not a function of x.
–16
5
15. y = x3
If y = 0, then 0 = x3 , x = 0. If x = 0, then y = 0.
Intercept: (0, 0). y is a function of x. One-to-one.
Domain: all real numbers
Range: all real numbers
5
y
y
x
5
y
19. 2x + y – 2 = 0
If y = 0, then 2x – 2 = 0, x = 1. If x = 0, then
y – 2 = 0, y = 2. Intercepts: (1, 0), (0, 2)
Note that y = 2 – 2x. y is a function of x.
One-to-one.
Domain: all real numbers
Range: all real numbers
x
5
68
ISM: Introductory Mathematical Analysis
5
Section 2.5
y
⎛
10 ⎞
Intercepts: ⎜ ±
, 0 ⎟ , (0,5)
⎜
⎟
2
⎝
⎠
Domain: all real numbers
Range: all real numbers ≤ 5
2
x
1
5
5
f(x)
– √10
√10
2
2
20. x + y = 1
If y = 0, then x = 1. If x = 0, then y = 1.
Intercepts: (1, 0), (0, 1)
Note that y = 1 – x.
y is a function of x. One-to-one.
Domain: all real numbers
Range: all real numbers
5
23. y = h(x) = 3
Because y cannot be 0, there is no x-intercept. If
x = 0, then y = 3. Intercept: (0, 3)
Domain: all real numbers
Range: 3
y
1
x
1
x
5
5
y
5
3
x
5
21. s = f (t ) = 4 – t 2
If s = 0, then 0 = 4 – t 2
0 = (2 + t)(2 – t)
t = ±2. If t = 0, then s = 4.
Intercepts: (2, 0), (–2, 0), (0, 4)
Domain: all real numbers
Range: all real numbers ≤ 4
5
24. g(s) = –17
Because g(s) cannot be 0, there is no s-intercept.
If s = 0, then g(s) = –17.
Intercept: (0, –17)
Domain: all real numbers
Range: –17
s
20
4
y
t
–2
2
5
x
20
–20
–20
22.
f ( x) = 5 – 2 x 2 . If f(x) = 0, then 0 = 5 – 2x 2
25. y = h( x) = x 2 – 4 x + 1
2
2x = 5
5
x2 =
2
5
10
.
x=±
=±
2
2
If x = 0, then f(x) = 5.
If y = 0, then 0 = x 2 – 4 x + 1 , and by the
4 ± 12
= 2 ± 3 . If
2
x = 0, then y = 1. Intercepts: (2 ± 3, 0), (0,1)
Domain: all real numbers
Range: all real numbers ≥ –3
quadratic formula, x =
69
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
y
5
2–
1
3
1
x
(2, – 3)
2+
3
t 2 − 9 to be a real number,
Note that for
t 2 − 9 ≥ 0, so t 2 ≥ 9, and |t| ≥ 3. If s = 0, then
0 = t 2 − 9, 0 = t 2 − 9, or t = ±3. Because
|t| ≥ 3, we know t ≠ 0, so no s-intercept exists.
Intercepts: (−3, 0), (3, 0)
Domain: all real numbers t ≤ −3 and ≥ 3
Range: all real numbers ≥ 0
y
x
10
10
y
–8
(1, –9)
–3
27.
5
29. s = f (t ) = t 2 − 9
If y = 0, then 0 = x 2 + 2 x – 8
0 = (x + 4)(x – 2), so x = –4, 2.
If x = 0, then y = –8.
Intercepts: (–4, 0), (2, 0), (0, –8).
Domain: all real numbers
Range: all real numbers ≥ –9
–4
q
–1
26. y = f ( x) = x 2 + 2 x – 8
10
p
3
x
10
f (t ) = – t 3
If f(t) = 0, then 0 = – t 3 , t = 0 .
If t = 0, then f(t) = 0. Intercept: (0, 0)
Domain: all real numbers
Range: all real number
5
30. F (r ) = –
f(t)
1
r
1
, which has no solution.
r
Because r ≠ 0, there is no vertical-axis
intercept. Intercept: none.
Domain: all real numbers ≠ 0
Range: all real numbers ≠ 0
If F(r) = 0, then 0 = –
t
5
5
F(r)
28. p = h(q) = 1 + 2q + q 2
r
2
2
5
If p = 0, then 1 + 2q + q = 0, (1 + q) = 0, so
q = −1. If q = 0 then p = 1.
Intercepts: (−1, 0), (0, 1)
Domain: all real numbers
Range: all real numbers ≥ 0
31.
f ( x) = 2 x –1
If f(x) = 0, then 0 = 2 x –1 , 2 x –1 = 0, so
1
.
2
If x = 0, then f ( x) = –1 = 1 .
x=
70
ISM: Introductory Mathematical Analysis
Section 2.5
2
x–4
Note that the denominator is 0 when x = 4. Thus
2
, which has no
x ≠ 4. If y = 0, then 0 =
x–4
1
solution. If x = 0, then y = – .
2
1
⎛
⎞
Intercept: ⎜ 0, – ⎟
2⎠
⎝
Domain: all real numbers except 4
Range: all real numbers except 0
⎛1 ⎞
Intercepts: ⎜ , 0 ⎟ , (0,1)
⎝2 ⎠
Domain: all real numbers
Range: all real numbers ≥ 0
5
34. y = f ( x) =
f(x)
1
x
1
2
5
10
32. v = H (u ) = u – 3
y
If v = 0, then 0 = u – 3 , u – 3 = 0, so u = 3.
If u = 0, then v = –3 = 3 .
Intercepts: (3, 0), (0, 3).
Domain: all real numbers
Range: all real numbers ≥ 0
10
v
x
10
4
–1
2
35. Domain: all real numbers ≥ 0
Range: all real numbers 1 ≤ c < 8
3
3
10
8
u
10
c
p
6
33. F (t ) =
10
16
t2
If F(t) = 0, then 0 =
36. Domain: all real numbers ≥ –1
Range: all real numbers ≤ 11
16
, which has no solution.
t2
Because t ≠ 0, there is no vertical-axis intercept.
No intercepts
Domain: all nonzero real numbers
Range: all positive real numbers
10
14
␾(x)
F(t)
x
10
t
10
37. Domain: all real numbers
Range: all real numbers ≥ 0
10
9
g(x)
x
3
71
5
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
y
38. Domain: all positive real numbers
Range: all real numbers > 1
20
f(x)
Cost (dollars)
10
4
1
16
12
8
4
10 12 2 4 6 8 10
A.M.
P.M.
x
3
5
10
x
43. As price increases, quantity supplied increases; p
is a function of q.
39. From the vertical-line test, the graphs that
represent functions of x are (a), (b), and (d).
p
50
40. From the horizontal line test, the graphs which
represent one-to-one functions of x are (c) and
(d).
41. Let y = the amount that is owed and let
x = the number of monthly payments made.
Then, the amount Tara owes is represented by
the equation y = 2400 − 275x.
To determine the x-intercept, we set y = 0 and
solve for x.
y = 2400 − 275 x
0 = 2400 − 275 x
275 x = 2400
8
x=8
11
⎛ 8 ⎞
The x-intercept is ⎜ 8 , 0 ⎟ . Therefore, Tara will
⎝ 11 ⎠
have paid off her debt after 9 months.
To determine the y-intercept, we set x = 0 and
solve for y.
y = 2400 − 275x
y = 2400 − 275(0)
y = 2400
The y-intercept is (0, 2400). Therefore, before
any payments are made, Tara owes $2400.
10
q
30
210
44. As price decreases, quantity increases; p is a
function of q.
p
25
5
q
25
5
45.
1000
y
300
42. The cost of an item as a function of the time of
day, x is
⎧9, if 10:30 A.M. ≤ x < 2 : 30 P.M.
⎪8, if 2:30 P.M. ≤ x < 4 : 30 P.M.
⎪
y = ⎨13, if 4:30 P.M. ≤ x < 6 : 00 P.M.
⎪18, if 6:00 P.M. ≤ x < 8 : 00 P.M.
⎪13, if 8:00 P.M. ≤ x ≤ 10:00 P.M.
⎩
7
46.
14
x
21
y
4
4 5
47. 0.39
48. −0.50, 0.57
72
x
12
ISM: Introductory Mathematical Analysis
Section 2.5
range: (– ∞, ∞)
49. –0.61, –0.04
a.
50. 0.62, 1.73, 4.65
b. intercepts: (–1.73, 0), (0, 4)
51. –1.12
35
59.
52. No real zeros
53. −1.70, 0
54. –0.49, 0.52, 1.25
–5
25
55.
a.
maximum value of f(x): 28
b. range: (−∞, 28]
4
1
c.
–15
a.
5
–5
5
60.
maximum value of f(x): 19.60
real zeros: −4.02, 0.60
b. minimum value of f(x): –10.86
5
–5
56.
4
–5
1
–1
a.
b. intercepts: (0, 0.29), (−1.03, 0)
–2
a.
c.
maximum value of f(x): 3.94
61.
b. minimum value of f(x): –1.94
57.
range: (−∞, ∞)
real zero: −1.03
35
6
2
3
3
a.
a.
5
c.
range: [18.68, 34.21]
d. no intercept
10
5
–5
maximum value of f(x): 34.21
b. minimum value of f(x): 18.68
maximum value of f(x): 5
b. minimum value of f(x): 4
58.
5
15
–5
73
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
line y = x: (a, b) on graph, then
Problems 2.6
2a 2 + b 2 a 4 = 8 − b, but
1. y = 5x
Intercepts: If y = 0, then 5x = 0, or x = 0; if x = 0,
then y = 5 · 0 = 0.
Testing for symmetry gives:
x-axis:
–y = 5x
y = –5x
y-axis:
y = 5(–x) = –5x
origin:
–y = 5(–x)
y = 5x
line y = x: (a, b) on graph, then b = 5a, and
1
a = b ≠ 5b for all b, so (b, a) is not
5
on the graph.
2b 2 + a 2b 4 = 8 − a will not
necessarily be true, so (b, a) is not on
the graph.
Answer: (±2, 0), (0, 8); symmetry about y-axis
4. x = y 3
Intercepts: If y = 0, then x = 0; if x = 0, then
0 = y3 , so y = 0.
Testing for symmetry gives:
x-axis:
x = (– y )3 = – y 3
y-axis:
– x = y3
Answer: (0, 0); symmetry about origin
x = – y3
– x = (– y )3
origin:
2. y = f ( x) = x 2 – 4
Intercepts: If y = 0, then
x = y3
2
0 = x – 4 = ( x + 2)( x – 2) , or x = ±2; if x = 0,
line y = x: (a, b) on graph, then a = b3 , and
then y = 02 – 4 = –4 .
Testing for symmetry gives:
b = 3 a ≠ a3 for all a, so (b, a) is not
on the graph.
Answer: (0, 0); symmetry about origin
x-axis:
– y = x2 – 4
y = – x2 + 4
5. 16 x 2 − 9 y 2 = 25
y-axis:
y = (– x)2 – 4 = x 2 – 4
origin:
– y = (– x)2 – 4
Intercepts: If y = 0, then 16 x 2 = 25, x 2 =
5
so x = ± ;
4
y = – x2 + 4
line y = x: (a, b) on graph, then b = a 2 − 4, and
if x = 0, then −9 y 2 = 25, y 2 = −
2
a = ± b + 4 ≠ b − 4 for all b, so
(b, a) is not on the graph.
Answer: (±2, 0), (0, –4); symmetry about y-axis
2
x-axis:
2 4
y-axis:
16(− x) 2 − 9 y 2 = 25
16 x 2 − 9 y 2 = 25
origin:
Since the graph has symmetry about
x- and y-axes, there is also symmetry
about the origin.
line y = x: (a, b) on graph, then
2 x 2 + (– y ) 2 x 4 = 8 – (– y )
2 x2 + y2 x4 = 8 + y
16a 2 − 9b 2 = 25, and
1
a 2 = (9b 2 + 25). (b, a) on graph,
16
2(– x) 2 + y 2 (– x)4 = 8 – y
2 x2 + y2 x4 = 8 − y
origin:
16 x 2 − 9(− y ) 2 = 25
16 x 2 − 9 y 2 = 25
2 x 2 = 8, x 2 = 4, or x = ±2;
if x = 0, then 0 = 8 – y, so y = 8.
Testing for symmetry gives:
y-axis:
25
, which has
9
no real root.
Testing for symmetry gives:
3. 2 x + y x = 8 – y
Intercepts: If y = 0, then
x-axis:
25
,
16
2(– x) 2 + (– y )2 (– x) 4 = 8 – (– y )
then 16b 2 − 9a 2 = 25 and
1
1
a 2 = (16b 2 − 25) ≠ (9b 2 + 25)
9
16
2 x2 + y2 x4 = 8 + y
74
ISM: Introductory Mathematical Analysis
Section 2.6
for all b, so (b, a) and (a, b) are not
always both on the graph.
⎛ 5 ⎞
Answer: ⎜ ± , 0 ⎟ ; symmetry about x-axis,
⎝ 4 ⎠
y-axis, and origin.
9. x = – y –4
Intercepts: Because y ≠ 0, there is no
1
, which has
x-intercept; if x = 0, then 0 = –
y4
no solution.
Testing for symmetry gives:
6. y = 57
Intercepts: Because y ≠ 0, there is no
x-intercept; if x = 0, then y = 57.
Testing for symmetry gives:
x-axis:
(–y) = 57
y = –57
y-axis:
y = 57
origin:
(–y) = 57
y = –57
line y = x: (a, b) on graph, then b = 57, but a can
be any value, so (b, a) = (57, a) is not
necessarily on the graph.
Answer: (0, 57); symmetry about y–axis
x-axis:
x = – y –4
y-axis:
origin:
– x = –(– y ) –4
x = y –4
line y = x: (a, b) on graph, then a = −b −4 and
b = (−a )−1/ 4 ≠ −a −4 for all a, so
(b, a) is not on the graph.
Answer: no intercepts; symmetry about x-axis
10. y = x 2 – 25
Intercepts: If y = 0, then
x 2 – 25 = 0,
x 2 – 25 = 0, x 2 = 25, so x = ±5;
if x = 0, then y = –25 , which has no real root.
Testing for symmetry gives:
x-axis:
– y = x 2 – 25
y = – x 2 – 25
y-axis:
8. y = 2 x – 2
y = (– x)2 – 25
y = x 2 – 25
Intercepts: If y = 0, then 2 x = 2, 2 x = 2,
x = 1, so x = ±1; if x = 0, then y = –2.
Testing for symmetry gives:
– y = 2x – 2
x-axis:
origin:
– y = (– x)2 – 25
y = – x 2 – 25 .
line y = x: (a, b) on graph, then b = a 2 − 25 or
y = – 2x + 2
b 2 = a 2 − 25 and
y = 2(− x) – 2
a 2 = b 2 + 25 ≠ b 2 − 25 for all b, so
(b, a) is not on the graph.
Answer: (±5, 0); symmetry about y-axis
y = 2x – 2
origin:
– x = – y –4
x = y –4
7. x = –2
Intercepts: If y = 0, then x = –2; because x ≠ 0,
there is no y-intercept.
Testing for symmetry gives:
x-axis:
x = –2
y-axis:
–x = –2
x=2
origin:
–x = –2
x=2
line y = x: (a, b) on graph, then a = −2, but b can
be any value, so (b, a) = (b, −2) is not
necessarily on the graph.
Answer: (–2, 0); symmetry about x-axis
y-axis:
x = –(– y ) –4
– y = 2(– x) – 2
y = – 2x + 2
11. x – 4 y – y 2 + 21 = 0
Intercepts: If y = 0, then x + 21 = 0, so x = –21;
line y = x: (a, b) on graph, then b = 2a − 2 and
b+2
≠ 2b − 2 for all b, so
2
(b, a) is not on the graph.
Answer: (±1, 0), (0, –2); symmetry about y-axis
a=±
if x = 0, then –4 y – y 2 + 21 = 0,
y 2 + 4 y – 21 = 0, (y + 7)(y – 3) = 0, so y = –7 or
y = 3.
75
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
Testing for symmetry gives:
x-axis:
y=
y-axis:
2
x – 4(– y ) – (– y ) + 21 = 0
2
x + 4 y – y + 21 = 0
y-axis:
y=
(– x) – 4 y – y 2 + 21 = 0
– x – 4 y – y 2 + 21 = 0
origin:
− x3 − 2 x 2 − x
x2 + 1
(– x) – 4(– y ) – (– y ) 2 + 21 = 0
y=
a = b 2 + 4b − 21, but
b = a 2 + 4a − 21 will not necessarily
be true, so (b, a) is not on the graph.
Answer: (–21, 0), (0, –7), (0, 3); no symmetry
a=
Intercepts: If y = 0, then x 2 = 0, so x = 0;
x3 + 2 x 2 + x
b3 − 2b 2 + b
14. x 2 + xy + y 2 = 0
if x = 0, then y 3 = 0, so y = 0.
Testing for symmetry gives:
Intercepts: If y = 0, then x 2 = 0, so x = 0;
x-axis: x 2 + x(− y ) + (− y )3 = 0
if x = 0, then y 2 = 0, so y = 0.
Testing for symmetry gives:
3
x − xy − y = 0
2
y-axis: (− x) + (− x) y + y 3 = 0
x 2 + x(– y ) + (– y ) 2 = 0
x-axis:
x 2 − xy + y 3 = 0
x 2 – xy + y 2 = 0
(− x) 2 + (− x)(− y ) + (− y )3 = 0
2
(− x)2 + 1
is not necessarily
b2 + 1
true, so (b, a) is not on the graph.
Answer: (1, 0), (0, 0); no symmetry of the given
types
12. x 2 + xy + y3 = 0
2
(− x)3 − 2(− x) 2 + (− x)
x2 + 1
line y = x: (a, b) on graph, then
a 3 − 2a 2 + a
b=
, but
a2 + 1
a − 4b − b 2 + 21 = 0 and
(– x)2 + (– x) y + y 2 = 0
y-axis:
3
x + xy − y = 0
line y = x: (a, b) on graph, then
x 2 – xy + y 2 = 0
(– x) 2 + (– x)(– y ) + (– y ) 2 = 0
origin:
a 2 + ab + b3 = 0, but
x 2 + xy + y 2 = 0
b 2 + ab + a3 = 0 will not necessarily
be true, so (b, a) is not on the graph.
Answer: (0, 0); no symmetry.
13. y = f ( x) =
(− x) 2 + 1
−y =
origin:
– x + 4 y – y 2 + 21 = 0
line y = x: (a, b) on graph, then
origin:
(− x)3 − 2(− x) 2 + (− x)
line y = x: (a, b) on graph, then a 2 + ab + b 2 = 0
and b 2 + ba + a 2 = 0, so (b, a) is on
the graph.
Answer: (0, 0); symmetry about origin,
symmetry about y = x
x3 − 2 x 2 + x
x2 + 1
Intercepts: If y = 0, then
x3 − 2 x 2 + x x( x − 1) 2
=
= 0, so x = 0, 1;
x2 + 1
x2 + 1
if x = 0, then y = 0.
Testing for symmetry gives:
x-axis:
Because f is not the zero function,
there is no x-axis symmetry
15. y =
3
3
x +8
3
= 0 , which has
x +8
3
no solution; if x = 0, then y = .
8
Testing for symmetry gives:
Intercepts: If y = 0, then
76
3
ISM: Introductory Mathematical Analysis
x-axis:
–y =
y=–
y-axis:
y=
3
line y = x: (a, b) on graph, then b =
x3 + 8
3
a4
b4
, but a + b =
will not
b
a
necessarily be true, so (b, a) is not on
the graph.
Answer: no intercepts; symmetry about origin
x +8
3
3
3
17. 3x + y 2 = 9
Intercepts: If y = 0, then 3x = 9, so x = 3;
–x + 8
3
–y =
(– x )3 + 8
–y =
y=
if x = 0, then y 2 = 9, so y = ±3.
Testing for symmetry gives:
3
3
–x + 8
3
3x + y 2 = 9
x3 – 8
3
a3 + 8
−3 x + y 2 = 9
and
−3 x + y 2 = 9
line y = x: (a, b) on graph, then 3a + b 2 = 9 and
1
1
a = (9 − b 2 ), but b = (9 − a 2 ) will
3
3
not necessarily be true, so (b, a) is not
on the graph.
Answer: (3, 0), (0, ±3); symmetry about x-axis
5
4
x
= 0 , which has no
x
0
solution; if x = 0, then y = , which has no
y
solution.
Testing for symmetry gives:
x4
x-axis:
–y =
x + (– y )
y=
(– x )4
(– x) + y
x
3
18. x − 1 = y 4 + y 2 or x = y 4 + y 2 + 1
Intercepts: If y = 0, then x = 1; if x = 0, then
y 4 + y 2 = −1, so no y-intercept
Testing for symmetry gives:
x-axis:
x − 1 = (– y )4 + (− y )2
x −1 = y4 + y2
y-axis:
− x = y4 + y2 + 1
x = − y4 − y2 −1
(– x) 4
–y =
(– x) + (– y )
y=
5
–3
x4
y=
–x + y
origin:
y
3
Intercepts: If y = 0, then
y-axis:
3(− x) + (− y ) 2 = 9
origin:
x4
x+ y
x4
–x + y
3(− x) + y 2 = 9
y-axis:
3
3
a = 3 −8 ≠
for all b, so
3
b
b +8
(b, a) is not on the graph.
⎛ 3⎞
Answer: ⎜0, ⎟ ; no symmetry of the given types
⎝ 8⎠
y=
3x + (− y )2 = 9
x-axis:
line y = x: (a, b) on graph, then b =
16. y =
a4
, and
a+b
a+b =
3
(– x)3 + 8
y=
origin:
Section 2.6
origin:
− x = ( − y ) 4 + (− y ) 2 + 1
x = − y4 − y2 − 1
x4
x+ y
77
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
line y = x: (a, b) on graph, then a = b 4 + b 2 + 1
and b ≠ a 4 + a 2 + 1 for all a so (b, a)
is not on the graph.
Answer: (1, 0); symmetry about x-axis.
5
3 y = 5(– x) – (– x)3
y-axis:
3 y = –5 x + x3
3(– y ) = 5(– x) – (– x)3
origin:
3 y = 5 x – x3 .
y
line y = x: (a, b) on graph, then 3b = 5a − a3 ,
but 3a = 5b − b3 will not necessarily
be true so (b, a) is not on the graph.
x
5
(
)
Answer: (0, 0), ± 5, 0 ; symmetry about
origin
5
19. y = f ( x) = x3 – 4 x
Intercepts: If y = 0, then x3 – 4 x = 0,
x(x + 2)(x – 2) = 0, so x = 0 or x = ±2; if x = 0,
then y = 0.
Testing for symmetry gives:
x-axis:
Because f is not the zero function,
there is no x-axis symmetry.
x
–5
y = (– x) – 4(– x)
21.
y = – x3 + 4 x
– y = (– x ) – 4(– x)
x = 0, then – y = 0, so y = 0.
Testing for symmetry gives:
x – –y = 0
x-axis:
3
y = x – 4x
line y = x: (a, b) on graph, then b = a3 − 4a, but
x − y =0
a = b3 − 4b will not necessarily be
true, so (b, a) is not on the graph.
Answer: (0, 0), (±2, 0); symmetry about origin.
5
–x – y = 0
y-axis:
x – y =0
y
origin:
Since there is symmetry about the
x- and y-axes, symmetry about origin
exists.
line y = x: (a, b) on graph, then a − b = 0, thus
x
–2
x – y =0
Intercepts: If y = 0, then x = 0, so x = 0; if
3
origin:
5
–5
3
y-axis:
y
2
5
a = b , and b − a = 0, so (b, a) is
on the graph.
Answer: (0, 0); symmetry about x-axis, y-axis,
origin, line y = x.
20. 3 y = 5 x – x3
5
y
Intercepts: If y = 0, then 5 x – x3 = 0,
x
(
5+x
)(
)
5 − x = 0, so x = 0 or x = ± 5; if
x
x = 0, then y = 0.
Testing for symmetry gives:
x-axis:
5
3(– y ) = 5 x – x3
3 y = –5 x + x3
78
ISM: Introductory Mathematical Analysis
Section 2.6
graph.
5⎞
⎛ 5 ⎞ ⎛
Answer: ⎜ ± , 0 ⎟ , ⎜ 0, ± ⎟ ; symmetry about
2⎠
⎝ 3 ⎠ ⎝
x-axis, y-axis, origin
22. x 2 + y 2 = 16
Intercepts: If y = 0, then x 2 = 16, so x = ±4;
if x = 0, then y 4 = 16, so y = ±4.
Testing for symmetry gives:
2
y
5
2
x + (– y ) = 16
x-axis:
x 2 + y 2 = 16
y-axis:
(– x )2 + y 2 = 16
origin:
x 2 + y 2 = 16
Since there is symmetry about
x- and y-axes, symmetry about origin
exists.
x
5
24. x 2 − y 2 = 4
line y = x: (a, b) on graph, then a 2 + b 2 = 16
Intercepts: If y = 0, then x 2 = 4, so x = ±2;
and b 2 + a 2 = 16, so (b, a) is on the
graph.
Answer: (±4, 0), (0, ±4); symmetry about x-axis,
y-axis, origin, line y = x.
5
if x = 0, then − y 2 = 4, y 2 = −4, which has no
real roots.
Testing for symmetry gives:
y
x 2 − (− y )2 = 4
x-axis:
x2 − y 2 = 4
x
y-axis:
(− x) 2 − y 2 = 4
origin:
x2 − y 2 = 4
Since there is symmetry about x-and
y-axes, symmetry about origin exists.
5
line y = x: (a, b) on graph, then a 2 − b 2 = 4 and
a 2 = 4 + b 2 ≠ b 2 − 4 for all b, so
(b, a) is not on the graph.
Answer: (±2, 0); symmetry about x-axis, y-axis,
origin.
23. 9 x 2 + 4 y 2 = 25
25
, so
9
5
5
x = ± ; if x = 0, then 4 y 2 = 25, so y = ± .
3
2
Testing for symmetry gives:
Intercepts: If y = 0, then 9 x 2 = 25, x 2 =
x-axis:
5
y
9 x 2 + 4(− y )2 = 25
2
x
9 x + 4 y = 25
y-axis:
9(− x)2 + 4 y 2 = 25
origin:
9 x 2 + 4 y 2 = 25
Since there is symmetry about
x- and y-axes, symmetry about origin
exists.
2
−2
2
25.
6
–6
line y = x: (a, b) on graph, then 9a 2 + 4b 2 = 25
1
and b 2 = (25 − 9a 2 ). (b, a) on
4
5
6
–6
y = f ( x) = 5 − 1.96 x 2 − πx 4 . Replacing x by –x
graph, then 9b 2 + 4a 2 = 25 and
1
b 2 = (25 − 4a 2 ), so (a, b) and
9
(b, a) are not always both on the
gives y = 5 − 1.96(− x)2 − π(− x) 4 or
y = 5 − 1.96 x 2 − πx 4 , which is equivalent to
79
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
original equation. Thus the graph is symmetric
about the y-axis.
a.
2.
5
y
f(x) = x2
Intercepts: (±0.99, 0), (0, 5)
x
b. Maximum value of f(x): 5
c.
5
y = –x 2
Range: (–∞, 5]
26.
8
3.
10
y
f(x) = 1
x
4
–4
x
10
–3
y= 1
x–2
4
2
y = f ( x) = 2 x – 7 x + 5 . Replacing x by –x
gives y = 2(– x)4 – 7(– x) 2 + 5 or
4.
10
y
y = 2 x 4 – 7 x 2 + 5 , which is equivalent to
original equation. Thus the graph is symmetric
about y-axis.
Real zeros of f: ±1, ±1.58
27.
5
y = √x + 2
x
10
f(x) = √x
y
y
5
5.
x
2
f(x) = 1
x
–11
1
–1
Problems 2.7
1.
–2
5
y= 2
3x
y
y = x3 – 1
6.
x
f(x) = x3
x
5
10
y
f(x) = |x|
y = |x| – 2
x
10
80
ISM: Introductory Mathematical Analysis
7.
Section 2.7
y
10
13. Translate 3 units to the left, stretch vertically
away from the x-axis by a factor of 2, reflect
about the x-axis, and move 2 units upward.
f(x) = |x|
x
10
y = |x + 1| – 2
14. Translate 3 units to the left and 4 units
downward.
15. Reflect about the y-axis and translate 5 units
downward.
8.
5
y
f(x) =
x
16. Shrink horizontally toward the y-axis by a factor
of 3.
5x
17.
5
y = – 13 x
5
–5
9.
10
y
–5
f(x) = x2
Compared to the graph for k = 0, the graphs for
k = 1, 2, and 3 are vertical shifts upward of 1, 2,
and 3 units, respectively. The graphs for
k = –1, –2, and –3 are vertical shifts downward
of 1, 2, and 3 units, respectively.
x
10
y = 1 – (x – 1)2
3
18.
y
10.
y = (x – 1)2 + 1
f(x) = x 2
5
–5
x
10
–3
11.
10
Compared to the graph for k = 0, the graphs for
k = 1, 2, and 3 are horizontal shifts to the left of
1, 2, and 3 units, respectively. The graphs for
k = –1, –2, and –3 are horizontal shifts to the
right of 1, 2, and 3 units, respectively.
y
f(x) = √x
y = √–x
x
19.
10
5
–5
12.
10
5
y
f(x) = 1
–5
x
Compared to the graph for k = 1, the graphs for
k = 2 and 3 are vertical stretches away from the
x-axis by factors of 2 and 3, respectively. The
1
graph for k = is a vertical shrinking toward
2
the x-axis.
x
10
y= 1
2–x
81
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
Chapter 2 Review Problems
5–3 2
=
5+ 4 9
( x + 3) – 3
x
=
F ( x + 3) =
( x + 3) + 4 x + 7
F (5) =
1. Denominator is 0 when
x2 − 6 x + 5 = 0
( x − 1)( x − 5) = 0
x = 1, 5
Domain: all real numbers except 1 and 5.
2. all real numbers
5+ 4
9 3
=
=
5
5
5
h(5) =
3. all real numbers
4. all real numbers
h(–4) =
5. For x to be real, x must be nonnegative. For
the denominator x – 1 to be different from 0, x
cannot be 1. Both conditions are satisfied by all
nonnegative numbers except 1.
Domain: all nonnegative real numbers except 1.
h( x ) =
–4 + 4
0
=
=0
–4
–4
x+4
x
h(u – 4) =
6. s – 5 ≥ 0
s≥5
Domain: all real numbers s such that s ≥ 5.
7.
u+4
u
11. h(u ) =
12. H ( s ) =
f (0) = 3(0) 2 – 4(0) + 7 = 7
H (7) =
f (–3) = 3(–3) 2 – 4(–3) + 7 = 27 + 12 + 7 = 46
u–4
=
u
u–4
( s – 4) 2
3
H (−2) =
f ( x) = 3x 2 – 4 x + 7
(u – 4) + 4
(–2 – 4)2 36
=
= 12
3
3
(7 – 4)2 9
= =3
3
3
( )
2
2
⎡ 1
⎤
49
− 72
49
⎛ 1 ⎞ ⎣ 2 − 4⎦
=
= 4 =
H⎜ ⎟=
2
3
3
3
12
⎝ ⎠
2
f (5) = 3(5) – 4(5) + 7 = 75 – 20 + 7 = 62
f (t ) = 3t 2 – 4t + 7
( )
H x2 =
8. h(x) = 7; all function values are 7.
Answer: 7, 7, 7, 7
( )
( x 2 – 4)2 x 4 – 8 x 2 + 16
=
3
3
13. f(4) = 4 + 16 = 20
f(−2) = −3
f(0) = −3
f(1) is not defined.
4
9. G ( x) = x − 3
G (3) = 4 3 − 3 = 4 0 = 0
G (19) = 4 19 − 3 = 4 16 = 2
14.
G (t + 1) = 4 (t + 1) − 3 = 4 t − 2
4
G ( x3 ) = x3 − 3
1
3
⎛ 1⎞
⎛ 1⎞
f ⎜ − ⎟ = −⎜ − ⎟ +1 = +1 =
2
2
2
2
⎝
⎠
⎝
⎠
f (0) = 02 + 1 = 1
2
1
5
⎛1⎞ ⎛1⎞
f ⎜ ⎟ = ⎜ ⎟ +1 = +1 =
4
4
⎝2⎠ ⎝2⎠
x–3
10. F ( x) =
x+4
–1 – 3
4
F (–1) =
=–
–1 + 4
3
0−3
3
F (0) =
=−
0+4
4
f (5) = 53 − 99 = 125 − 99 = 26
f (6) = 63 − 99 = 216 − 99 = 117
15. a.
82
f(x + h) = 3 – 7(x + h) = 3 – 7x – 7h
ISM: Introductory Mathematical Analysis
b.
16. a.
Chapter 2 Review
f ( x + h) – f ( x) (3 − 7 x − 7 h) − (3 − 7 x) –7 h
=
=
= –7
h
h
h
f ( x + h) = 11( x + h) 2 + 4
= 11x 2 + 22 xh + 11h 2 + 4
b.
17. a.
b.
18. a.
b.
f ( x + h) – f ( x) (11x 2 + 22 xh + 11h 2 + 4) – (11x 2 + 4) 22 xh + 11h 2
=
=
= 22 x + 11h
h
h
h
f ( x + h) = 4( x + h)2 + 2( x + h) – 5 = 4 x 2 + 8 xh + 4h 2 + 2 x + 2h – 5
f ( x + h) – f ( x) (4 x 2 + 8 xh + 4h 2 + 2 x + 2h – 5) – (4 x 2 + 2 x – 5)
=
h
h
2
8 xh + 4h + 2h
=
h
= 8 x + 4h + 2
f ( x + h) =
7
7
=
( x + h) + 1 x + h + 1
f ( x + h) – f ( x )
=
h
7
x + h +1
–
7
x +1
h
=
7( x +1)–7( x + h +1)
( x + h +1)( x +1)
h
=
−7 h
−7
=
( x + h + 1)( x + 1)h ( x + h + 1)( x + 1)
19. f(x) = 3x – 1, g(x) = 2x + 3
a.
(f + g)(x) = f(x) + g(x) = (3x – 1) + (2x + 3) = 5x + 2
b. (f + g)(4) = 5(4) + 2 = 22
20.
c.
(f – g)(x) = f(x) – g(x) = (3x – 1) – (2x + 3) = x – 4
d.
( fg )( x) = f ( x) g ( x) = (3 x –1)(2 x + 3) = 6 x 2 + 7 x – 3
e.
( fg )(1) = 6(1)2 + 7(1) – 3 = 10
f.
f
f ( x) 3 x –1
( x) =
=
g
g ( x) 2 x + 3
g.
( f D g )( x) = f ( g ( x)) = f (2 x + 3) = 3(2 x + 3) –1 = 6 x + 8
h.
( f D g )(5) = 6(5) + 8 = 38
i.
( g D f )( x) = g ( f ( x)) = g (3x –1) = 2(3 x –1) + 3 = 6 x + 1
f ( x) = − x 2 , g(x) = 3x − 2
a.
( f + g )( x) = f ( x) + g ( x) = − x 2 + 3 x − 2
b.
( f − g )( x) = f ( x) − g ( x) = − x 2 − (3x − 2) = − x 2 − 3x + 2
83
Chapter 2: Functions and Graphs
ISM: Introductory Mathematical Analysis
c.
( f − g )(−3) = −(−3) 2 − 3(−3) + 2 = 2
d.
( fg )( x) = f ( x) g ( x)
23. f(x) =
( f D g )( x) = f ( g ( x)) = f ( x3 ) = x3 + 2
= (− x 2 )(3x − 2)
( g D f )( x) = g ( f ( x)) = g
= −3 x3 + 2 x 2
e.
f
f ( x)
− x2
( x) =
=
g
g ( x) 3x − 2
f.
f
−(2) 2
(2) =
= −1
g
3(2) − 2
g.
x-axis:
y-axis:
= −3 x 2 − 2
( g D f )(−4) = −3(−4) 2 − 2 = −48 − 2 = −50
origin:
, g(x) = x + 1
line y = x: (a, b) on graph, then b = 3a − a3 , but
1
a = 3b − b3 is not necessarily true, so
(b, a) is not on the graph.
( x + 1)2
(
⎞ 1
1 + x2
=
+
=
1
⎟
x2
⎠ x2
origin
26.
( x) =
x +1
4
x +1
⎛ x +1 ⎞
( g D f )( x) = g ( f ( x)) = g ⎜
⎟=
4
4
⎝
⎠
=
)
Answer: (0, 0), ± 3, 0 ; symmetry about
x +1
, g ( x) = x
4
( f D g )( x) = f ( g ( x)) = f
− y = 3(− x) − (− x)3
y = 3x − x3 , which is the original
equation.
⎛ 1
( g D f )( x) = g ( f ( x)) = g ⎜
⎝ x2
f ( x) =
y = 3(− x) − (− x)3
y = −3 x + x3
( f D g )( x) = f ( g ( x)) = f ( x + 1) =
22.
− y = 3 x − x3
y = −3 x + x3 , which is not the
original equation.
= 3(− x 2 ) − 2
x2
3
x(3 − x 2 ) = 0, x = 0, ± 3.
If x = 0, then y = 0.
Testing for symmetry gives:
( g D f )( x) = g ( f ( x))
f ( x) =
)
Intercepts: If y = 0, then 0 = 3 x − x3 ,
= g (− x 2 )
21.
x+2
25. y = 3x − x3
( f D g )( x) = f ( g ( x))
= f (3 x − 2)
1
) (
x+2 =
24. f(x) = 2, g(x) = 3
( f D g )( x) = f ( g ( x)) = f (3) = 2
( g D f )( x) = g ( f ( x)) = g (2) = 3
= −9 x 2 + 12 x − 4
i.
(
= ( x + 2)3 / 2
= −(3x − 2)2
h.
x + 2, g ( x) = x3
x +1
2
x2 y2
=4
x2 + y2 + 1
Intercepts: If y = 0, then 0 = 4, which is
impossible; if x = 0, then 0 = 4, which is
impossible.
Testing for symmetry gives:
x 2 (– y ) 2
=4
x-axis:
x 2 + (− y )2 + 1
x2 y2
x2 + y2 + 1
equation.
84
= 4 , which is the original
ISM: Introductory Mathematical Analysis
y-axis:
(– x)2 y 2
(– x) 2 + y 2 + 1
Chapter 2 Review
10
=4
y
x2 y2
origin:
= 4 , which is the
x2 + y2 + 1
original equation.
(− x)2 (− y )2
=4
( − x) 2 + (− y ) 2 + 1
x2 y2
x
10
28. y = 3x – 7
= 4, which is the
x2 + y 2 + 1
original equation.
line y = x: (a, b) on graph, then
and b 2 =
then
b2 =
4(a 2 + 1)
a2 − 4
b2 a 2
b2 + a 2 + 1
a 2b2
a 2 + b2 + 1
Intercepts: If y = 0, then 0 = 3x – 7, or x =
If x = 0, then y = –7.
Testing for symmetry gives:
x-axis:
–y = 3x – 7
y = –3x + 7, which is not the original
equation.
y-axis:
y = 3(–x) – 7
y = –3x – 7, which is not the original
equation.
origin:
–y = 3(–x) – 7
y = 3x + 7, which is not the original
equation.
line y = x: (a, b) on graph, then b = 3a − 7 and
1
a = (b + 7) ≠ 3b − 7 for all b, so
3
(b, a) is not on the graph.
⎛7 ⎞
Answer: (0, –7), ⎜ , 0 ⎟ ; no symmetry of the
⎝3 ⎠
given types
=4
. (b, a) on graph,
= 4 and
4(a 2 + 1)
, so (a, b) and (b, a)
a2 − 4
are both on the graph.
Answer: no intercepts; symmetry about x-axis,
y-axis, origin, and y = x.
27. y = 9 – x 2
Intercepts: If y = 0, then
0 = 9 – x 2 = (3 + x)(3 – x ) , or x = ±3
If x = 0, then y = 9.
Testing for symmetry gives:
x-axis:
10
y
– y = 9 – x2
y = –9 + x 2 , which is not the original
equation.
y-axis:
7
3
x
10
y = 9 – (– x)2
y = 9 – x 2 , which is the original
equation.
origin:
7
.
3
− y = 9 − (− x)
29. G (u ) = u + 4
2
If G(u) = 0, then 0 = u + 4 .
0 = u + 4,
u = –4
If u = 0, then G(u) = 4 = 2 .
Intercepts: (0, 2), (–4, 0)
Domain: all real numbers u such that u ≥ –4
Range: all real numbers ≥ 0
2
y = −9 + x , which is not the
original equation.
line y = x: (a, b) on graph, then b = 9 − a 2 and
a = ± 9 − b ≠ 9 − b 2 for all b, so
(b, a) is not on the graph.
Answer: (0, 9), (±3, 0); symmetry about y-axis.
85
Chapter 2: Functions and Graphs
10
ISM: Introductory Mathematical Analysis
G(u)
32. h(u ) = −5u
If h(u) = 0, then 0 = −5u ,
u = 0.
If u = 0, h(u) = 0.
Intercept: (0, 0)
Domain: all reals ≤ 0
Range: all reals ≥ 0
u
10
h(u)
30.
8
f ( x) = x + 1
If f(x) = 0, then 0 = x + 1.
x = –1 , which has no solution.
If x = 0, then f(x) = 1.
Intercept: (0, 1)
Domain: all real numbers
Range: all real numbers ≥ 1
10
u
–8
33. Domain: all real numbers.
Range: all real numbers ≤ 2
f(x)
5
y
x
10
x
5
31. y = g (t ) =
2
t –4
If y = 0, then 0 =
34.
y
2
, which has no solution.
t –4
f(x) = √x
2 1
If t = 0, then y = = .
4 2
⎛ 1⎞
Intercept: ⎜ 0, ⎟
⎝ 2⎠
Domain: all real numbers t such that t ≠ 4
Range: all real numbers > 0
10
8
x
y = √x – 2 – 1
35.
10
8
y
f(x) = x2
g(t)
x
10
y = – 1 x2 + 2
t
2
10
36. For 2006, t = 5. Hence
S = 150,000 + 3000(5) = $165,000.
S is a function of t.
86
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 2
37. From the vertical-line test, the graphs that
represent functions of x are (a) and (c).
38. a.
a.
b. (1.92, 0), (0, 7)
729
20
44.
b. 359.43
39.
(–∞,∞)
8
–2
8
–8
2
–20
a.
–8
b. all real numbers ≥ −9.03
–0.67; 0.34, 1.73
40.
−9.03
c.
90
−5, ±2.
45. k = 0, 2, 4
2
6
–3
–3
3
–30
–1.38, 4.68
41.
5
k = 1, 3
–2
2
2
–2
3
–3
–5
–2
–1.50, –0.88, –0.11, 1.09, 1.40
42.
a.
20
0, 2, 4
b. none
Mathematical Snapshot Chapter 2
8
–8
1.
f (23, 000) = 1510 + 0.15(23, 000 − 15,100)
= 2695
The tax on $23,000 is $2695.
2.
f (85, 000) = 8440 + 0.25(85, 000 − 61,300)
= 14,365
The tax on $85,000 is $14,365.
3.
f (290, 000) = 42,170 + 0.33(290, 000 − 188, 450)
= 75, 681.5
The tax on $290,000 is $75,681.50.
–20
(–∞,∞)
43.
20
8
–8
–20
87
Chapter 2: Functions and Graphs
4.
ISM: Introductory Mathematical Analysis
f (462, 000) = 91, 043 + 0.35(462, 000 − 336,550)
= 134,950.5
The tax on $462,000 is $134,950.50.
5. Answers may vary.
6. There should be no jump in tax as one moves from one tax bracket to the next, since it would be unfair for two
couples whose incomes differ by only $0.01 to pay substantially different taxes.
7. g ( x) = x − f ( x)
if 0 ≤ x ≤ 15,100
⎧ x − 0.10 x
⎪ x − [1510 + 0.15( x − 15,100)]
if 15,100 < x ≤ 61,300
⎪
if 61,300 < x ≤ 123, 700
⎪ x − [8440 + 0.25( x − 61,300)]
=⎨
−
+
−
x
x
[24,
040
0.28(
123,
700)]
if
123, 700 < x ≤ 188, 450
⎪
⎪ x − [42,170 + 0.33( x − 188, 450)] if 188, 450 < x ≤ 336,550
⎪⎩ x − [91, 043 + 0.35( x − 336,550)] if x > 336,550
if 0 ≤ x ≤ 15,100
⎧0.90 x
⎪0.85 x + 755
if 15,100 < x ≤ 61,300
⎪
if 61,300 < x ≤ 123, 700
⎪0.75 x + 6885
=⎨
if 123, 700 < x ≤ 188, 450
⎪0.72 x + 10,596
+
x
0.67
20,
018.50
if 188, 450 < x ≤ 336,550
⎪
⎪⎩0.65 x + 26, 749.50 if x > 336,550
8.
400,000
g(x)
200,000
0
200,000
x
400,000
88
Chapter 3
F − F1 = m ( C − C1 )
Principles in Practice 3.1
9
(C − 5)
5
9
F − 41 = C − 9
5
9
F = C + 32
5
1. Let x = the time (in years) and let y = the selling
price. Then,
In 1991: x1 = 1991 and y1 = 32, 000
In 1994: x2 = 1994 and y2 = 26, 000
The slope is
y −y
m= 2 1
x2 − x1
F − 41 =
4. To find the slope and y-intercept, let a = 1000,
then write the equation in slope-intercept form.
1
y=
(t + 1)a
24
1
y=
(t + 1)1000
24
1000 1000
y=
t+
24
24
125 125
y=
t+
3
3
125
Thus the slope, m, is
and the y-intercept, b,
3
125
is
.
3
26, 000 − 32, 000
1994 − 1991
−6000
=
3
= –2000
The car depreciated $2000 per year.
Price
(in thousands of dollars)
=
y (price)
40
30
20
10
x (time)
1990 1991 1992 1993 1994
Year
2. An equation relating the growth in enrollment to
the number of years can be found by using the
point-slope form of an equation of a line. If
S = the number of students enrolled, and
T = the number of years, then the point-slope
form can be written as
S − S1 = m (T − T1 )
5.
Let m = 14, S1 = 50 , and T1 = 3 .
S – 50 = 14(T – 3)
S – 50 = 14T – 42
S = 14T + 8
9
C + 32
5
⎛9
⎞
5( F ) = 5 ⎜ C + 32 ⎟
5
⎝
⎠
5 F = 9C + 160
0 = 9C − 5F + 160
Thus, 9C – 5F + 160 = 0 is a general linear form
9
of F = C + 32 .
5
F=
6.
3. A linear function relating Fahrenheit temperature
to Celsius temperature can be found by using the
point-slope form of an equation of a line.
F −F
77 − 41 36 9
=
=
m= 2 1 =
25 − 5
20 5
C2 − C1
F
100
–100
100
C
–100
To convert Celsius to Fahrenheit, locate the
Celsius temperature on the horizontal axis, move
vertically to the line, then move horizontally to
read the Fahrenheit temperature of the vertical
axis.
89
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
7. Right angles are formed by perpendicular lines.
The slopes of the sides of the triangle are:
0−0 0
⎧
AB ⎨m =
= =0
6−0 6
⎩
9.
7−0 7
⎧
BC ⎨m =
= =7
7−6 1
⎩
y − 7 = −5[ x − (−1)]
y − 7 = −5( x + 1)
y − 7 = −5 x − 5
5x + y − 2 = 0
10
y
7−0 7
⎧
AC ⎨m =
= =1
7−0 7
⎩
Since none of the slopes are negative reciprocals
of each other, there are no perpendicular lines.
Therefore, the points do not define a right
triangle.
x
5
10.
Problems 3.1
1. m =
10 − 1 9
= =3
7−4 3
2. m =
10 − 3
7
=
= −1
−2 − 5 −7
3. m =
−3 − (−2) −1
1
=
=−
8−6
2
2
4. m =
−4 − (−4) 0
= =0
3− 2
1
y
100
x
10
11.
5. The difference in the x-coordinates is 5 – 5 = 0,
so the slope is undefined.
6. m =
0 − (−6) 6
= =2
3−0
3
1
y − 5 = − [ x − (−2)]
4
4( y − 5) = −( x + 2)
4 y − 20 = − x − 2
x + 4 y − 18 = 0
10
y
9
2
−2 − (−2) 0
7. m =
=
=0
4−5
−1
8. m =
y − 0 = 75( x − 0)
y = 75 x
75 x − y = 0
x
10
0 − (−7) 7
=
9 −1
8
90
ISM: Introductory Mathematical Analysis
12.
Section 3.1
−8 − (−4) −4
=
= −4
−2 − (−3)
1
y − (−4) = −4[ x − (−3)]
y + 4 = −4 x − 12
4 x + y + 16 = 0
1 ⎡ ⎛ 5 ⎞⎤
⎢x − − ⎥
3 ⎣ ⎜⎝ 2 ⎟⎠ ⎦
5⎤
⎡
6( y − 5) = 2 ⎢ x + ⎥
2⎦
⎣
6 y − 30 = 2 x + 5
2 x − 6 y + 35 = 0
15. m =
y −5 =
y
20
y
10
35
6
x
x
10
20
3−0 3
=
2−0 2
3
y − 0 = ( x − 0)
2
3
y= x
2
2 y = 3x
3x − 2 y = 0
16. m =
4 −1
3
13. m =
=
1 − (−6) 7
3
( x − 1)
7
7( y − 4) = 3( x − 1)
7 y − 28 = 3x − 3
3x − 7 y + 25 = 0
y−4=
y
10
10
y
x
10
x
10
17.
2 − (−4) 6
=
= −6
5−6
−1
y − 2 = −6( x − 5)
y − 2 = −6 x + 30
6 x + y − 32 = 0
14. m =
y = 2x + 4
2x – y + 4 = 0
10
y
4
y
–2
50
x
10
91
x
10
Chapter 3: Lines, Parabolas, and Systems
18.
ISM: Introductory Mathematical Analysis
y = 5x – 7
5x – y – 7 = 0
10
21. A horizontal line has the form y = b. Thus
y = –3, or y + 3 = 0.
y
5
y
x
10
7
5
x
5
–7
19.
22. A vertical line has the form x = a. Thus x = –1,
or x + 1 = 0.
1
x−3
2
⎛ 1
⎞
2 y = 2 ⎜ − x − 3⎟
⎝ 2
⎠
2 y = −x − 6
x + 2y + 6 = 0
y=−
5
y
5
x
(–1, –1)
y
x
–6
23. A vertical line has the form x = a. Thus x = 2, or
x − 2 = 0.
3
5
–3
20.
y = 0x −
1
2
2 y = −1
2y +1 = 0
5
y
x
1
2
5
(2, –3)
y=−
5
24. A horizontal line has the form y = b. Thus y = 0.
y
5
y
5x
x
(0, 0)
–1
2
92
5
ISM: Introductory Mathematical Analysis
Section 3.1
25. y = 4x – 6 has the form y = mx + b, where m = 4
and b = –6.
10
29. x = –5 is a vertical line. Thus the slope is
undefined. There is no y-intercept.
y
x
x
10
–5
5
30. x − 9 = 5 y + 3
5 y = x − 12
1
12
y = x−
5
5
1
12
m= ,b= −
5
5
26. x − 2 = 6 or x = 8, is a vertical line. Thus the
slope is undefined. There is no y-intercept.
5
y
5
y
x
8
5
y
x
27. 3x + 5 y − 9 = 0
5 y = −3 x + 9
3
9
y = − x+
5
5
3
9
m=− ,b=
5
5
5
20
– 12
5
31. y = 3x
y = 3x + 0
m = 3, b = 0
y
5
y
9
5
x
x
5
5
28. y + 4 = 7
y=3
y = 0x + 3
m = 0, b = 3
5
32. y − 7 = 3( x − 4)
y − 7 = 3 x − 12
y = 3x − 5
m = 3, b = –5
y
3
y
3
x
x
5
5
–5
93
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
33. y = 3
y = 0x + 3
m = 0, b = 3
39.
y
5
3
x
5
1
x + 8 is in slope-intercept form.
300
⎛ 1
⎞
300 y = 300 ⎜
x + 8⎟
300
⎝
⎠
300 y = x + 2400
x − 300 y + 2400 = 0 (general form)
40. y =
34. 6 y − 24 = 0
y=4
y = 0x + 4
m = 0, b = 4
5
x 2y
3
− +
= −4
2 3
4
⎛ x 2y ⎞
⎛ 19 ⎞
12 ⎜ − +
⎟ = 12 ⎜ − ⎟
⎝ 2 3 ⎠
⎝ 4⎠
−6 x + 8 y = −57
6 x − 8 y − 57 = 0 (general form)
−8 y = −6 x + 57
3
57
y = x−
(slope-intercept form)
4
8
y
41. The lines y = 7x + 2 and y = 7x – 3 have the
same slope, 7. Thus they are parallel.
x
5
42. The lines y = 4x + 3 and y = 5 + 4x (or
y = 4x + 5) have the same slope, 4. Thus they are
parallel.
43. The lines y = 5x + 2 and –5x + y – 3 = 0 (or
y = 5x + 3) have the same slope, 5. Thus they are
parallel.
35. 2x = 5 – 3y, or 2x + 3y – 5 = 0 (general form)
2
5
3y = –2x + 5, or y = − x + (slope-intercept
3
3
form)
44. The line y = x has slope m1 = 1 and the line
y = −x has slope m2 = −1 . m1 = −
36. 3x + 2y = 6, or 3x + 2y – 6 = 0 (general form)
3
2y = –3x + 6, or y = − x + 3 (slope-intercept
2
form)
1
so the
m2
lines are perpendicular.
1
5⎞
⎛
45. The line x + 3y + 5 = 0 ⎜ or y = − x − ⎟ has
3
3⎠
⎝
1
slope m1 = − and the line y = −3x has slope
3
1
m2 = −3. Since m1 ≠ m2 and m1 ≠ −
, the
m2
lines are neither parallel nor perpendicular.
37. 4x + 9y – 5 = 0 is a general form.
4
5
9y = –4x + 5, or y = − x + (slope-intercept
9
9
form)
38. 3( x − 4) − 7( y + 1) = 2
3x − 12 − 7 y − 7 = 2
3x − 7 y − 21 = 0 (general form)
1 ⎞
⎛
46. The line x + 3y = 0 ⎜ or y = − x ⎟ has slope
3 ⎠
⎝
1
m1 = − and the line x + 6y − 4 = 0 (or
3
1
1
2⎞
y = − x + ⎟ has slope m2 = − . Since
6
6
3⎠
3
−7y = −3x + 21, or y = x − 3 (slope-intercept
7
form)
94
ISM: Introductory Mathematical Analysis
Section 3.1
56. y = –4 is a horizontal line. The perpendicular line
must be vertical and has an equation of the form
x = a. Since that line passes through (1, 1), its
equation is x = 1.
1
, the lines are neither
m2
parallel nor perpendicular.
m1 ≠ m2 and m1 ≠ −
47. The line y = 3 is horizontal and the line x = −
1
3
57. y = −3 is a horizontal line, so the perpendicular
line must be vertical with equation of the form
x = a. Since that line passes through (5, 2), its
equation is x = 5.
is vertical, so the lines are perpendicular.
48. Both lines are vertical and thus parallel.
58. The line 3 y = −
49. The line 3x + y = 4 (or y = –3x + 4) has slope
m1 = –3, and the line x – 3y + 1 = 0
1
1
1⎞
⎛
⎜ or y = 3 x + 3 ⎟ has slope m2 = 3 . Since
⎝
⎠
m2 = −
2x
2x ⎞
⎛
+ 1 has
+ 3 ⎜ or y = −
5
15 ⎟⎠
⎝
2
, so the slope of a line perpendicular
15
15
. An equation of the
to it must have slope
2
15
desired line is y − (−5) = ( x − 4) or
2
15
y = x − 35.
2
slope −
1
, the lines are perpendicular.
m1
50. The line x − 2 = 3 (or x = 5) is vertical and the
line y = 2 is horizontal, so the lines are
perpendicular.
2
59. The line 2x + 3y + 6 = 0 has slope − , so the
3
2
slope of a line parallel to it must also be − . An
3
equation of the desired line is
2
2
29
y − (−5) = − [ x − (−7)], or y = − x − .
3
3
3
x
1
51. The slope of y = − − 2 is − , so the slope of
4
4
1
a line parallel to it must also be − . An
4
1
equation of the desired line is y − 1 = − ( x − 1)
4
1
5
or y = − x + .
4
4
60. The y-axis is vertical. A parallel line is also
vertical and has an equation of the form x = a.
Since it passes through (–4, 10), its equation is
x = –4.
52. x = –4 is a vertical line. A line parallel to x = –4
has the form x = a. Since the line must pass
through (2, –8), its equation is x = 2.
61. (1, 2), (–3, 8)
8−2
6
3
=
=−
m=
−3 − 1 −4
2
53. y = 2 is a horizontal line. A line parallel to it has
the form y = b. Since the line must pass through
(2, 1) its equation is y = 1.
3
Point-slope form: y − 2 = − ( x − 1) . When the
2
x-coordinate is 5,
3
y − 2 = − (5 − 1)
2
3
y − 2 = − (4)
2
y − 2 = −6
y = −4
Thus the point is (5, –4).
54. The slope of y = 3 + 2x is 2, so the slope of a line
parallel to it must also be 2. An equation of the
desired line is y − (−4) = 2(x − 3), or y = 2x − 10.
55. The slope of y = 3x − 5 is 3, so the slope of a line
1
perpendicular to it must have slope − . An
3
1
equation of the desired line is y − 4 = − ( x − 3),
3
1
or y = − x + 5.
3
95
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
Using the point-slope form with m = 28,000 and
( x1 , y1 ) = (0, − 100, 000) gives
62. m = 3, b = 1
Slope-intercept form: y = 3x + 1. The point
(−1, −2) lies on the line if its coordinates satisfy
the equation. If x = −1 and y = −2, then
−2 = 3(−1) + 1 or −2 = −2, which is true. Thus
(−1, −2) lies on the line.
y − y1 = m ( x − x1 )
y − (−100, 000) = 28, 000( x − 0)
y + 100, 000 = 28, 000 x
y = 28, 000 x − 100, 000
Price per share (dollars)
63. Let x = the time (in years) and
y = the price per share. Then,
In 1988: x1 = 1988 and y1 = 37
In 1998: x2 = 1998 and y2 = 8
The slope is
−29
8 − 37
=
= −2.9
m=
10
1998 − 1988
The stock price dropped an average of $2.90 per
year.
66. Solve the equation for t.
L = 1.53t – 6.7
L + 6.7 = 1.53t
( L + 6.7)
=t
1.53
0.65L + 4.38 = t
The slope is approximately 0.65 and the
y-intercept is approximately 4.38.
40
67. A general linear form of d = 184 + t is
–t + d – 184 = 0.
30
68. a.
y (price)
20
10
x (time)
0
1988
1993
Year
1998
64. The number of home runs hit increased as a
function of time (in months). The given points
are ( x1 , y1 ) = (3, 14) and ( x2 , y2 ) = (5, 20) .
b. Using the points (0.5, 0.5) and (−1, −2.5)
−2.5 − 0.5
−3
=
= 2.
gives a slope of m =
−1 − 0.5
−1.5
An equation is y − 0.5 = 2(x − 0.5) or
1
y = 2x − .
2
y −y
20 − 14 6
= =3
m= 2 1 =
5−3
2
x2 − x1
Using the point-slope form with m = 3 and
( x1 , y1 ) = (3, 14) gives
These two paths are not perpendicular to each
other because the slopes are not negative
reciprocals of each other.
y − y1 = m ( x − x1 )
y − 14 = 3( x − 3)
y − 14 = 3x − 9
y = 3x + 5
69. The slopes of the sides of the figure are:
4−0 4
⎧
AB ⎨m =
= = undefined (vertical)
−0 0
0
⎩
65. The owner’s profits increased as a function of
time. Let x = the time (in years) and let
y = the profit (in dollars). The given points are
( x1 , y1 ) = (0, − 100, 000) and
( x2 ,
Using the points (3.5, −1.5) and (0.5, 0.5)
2
−1.5 − 0.5
=− .
gives a slope of m =
3.5 − 0.5
3
2
An equation is y − 0.5 = − ( x − 0.5) or
3
2
5
y = − x+ .
3
6
7−3 4
⎧
CD ⎨m =
= = undefined (vertical)
−2 0
2
⎩
y2 ) = (5, 40, 000) .
3−0 3
⎧
AC ⎨m =
=
2−0 2
⎩
y −y
40, 000 − (−100, 000) 140, 000
=
m= 2 1 =
5−0
5
x2 − x1
= 28,000
7−4 3
⎧
BD ⎨m =
=
2−0 2
⎩
Since AB is parallel to CD and AC is parallel
to BC , ABCD is a parallelogram.
96
ISM: Introductory Mathematical Analysis
Section 3.2
70. Let x = the distance traveled and let
y = the altitude. The path of descent is a straight
line with a slope of –1 and y-intercept of 3600.
Therefore, using the slope-intercept form with
m = –1 and b = 3600 gives
y = mx + b
y = (–1)x + 3600
y = –x + 3600
10
76.
10
–10
–10
10
4000
–15
–1000
–10
If the airport is located 3800 feet from where the
plane begins its landing approach, the plane will
crash 200 feet short of the airport.
The slope of the first line is
0.1875
= 0.625 , and the slope of the
m1 =
0.3
0.32
second line is m2 = −
= −1.6 . Since
0.2
1
m1 = −
, the lines are perpendicular.
m2
71. The line has slope 59.82 and passes through
(6, 1128.50). Thus C – 1128.50 = 59.82(T – 6)
or
C = 59.82T + 769.58.
72. The line has slope 50,000 and passes through
(5, 330,000). Thus R – 330,000 = 50,000(T – 5)
or R = 50,000T + 80,000.
Principles in Practice 3.2
1. Let x = the number of skis that are produced and
let y = the number of boots that are produced.
Then, the equation 8x + 14y = 1000 describes all
possible production levels of the two products.
10
73.
10
–10
2. The quantity and price are linearly related such
that p = 575 when q = 1200, and p = 725 when
q = 800. Thus ( q1 , p1 ) = (1200, 575) and
–10
( q2 ,
The graph of the equation y = −0.9x − 7.3 shows
that when x = 0, y = 7.3. Thus, the y-intercept is
7.3.
p2 ) = (800, 725) . The slope is
725 − 575
3
=− .
800 − 1200
8
An equation of the line is
m=
10
74.
15
4000
–500
p − p1 = m ( q − q1 )
3
p − 575 = − (q − 1200)
8
3
p − 575 = − q + 450
8
3
p = − q + 1025
8
10
–10
–10
The lines are parallel, which is expected because
they have the same slope, 1.5.
75. The slope is 7.1.
97
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
3. Answers may vary, but two possible points are
(0, 60) and (2, 140).
f(x) = 40x + 60
x
f(x)
0
60
2
140
1000
Problems 3.2
1. y = f(x) = –4x = –4x + 0 has the form
f(x) = ax + b where a = –4 (the slope) and b = 0
(the vertical-axis intercept).
5
y
x
f(x)
5
500
0
2. y = f(x) = x + 1 has the form f(x) = ax + b where
a = 1 (the slope) and b = 1 (the vertical-axis
intercept).
x
20
5
4. If t = the age of the child, then f(t) = the height
of the child at any age t. The height and age are
linearly related such that f(8) = 50.6. Since f(t) is
a linear function it has the form f(t) = at + b.
Since the height changes by 2.3 inches per year,
a = 2.3. Then,
f(t) = at + b
f(8) = 2.3(8) + b
50.6 = 18.4 + b
32.2 = b
Thus, f(t) = 2.3t + 32.2 is a function that
describes the height of the child at age t.
y
1
x
5
3. h(t) = 5t − 7 has the form h(t) = at + b with a = 5
(the slope) and b = −7 (the vertical-axis intercept).
10
h(t)
5. Let y = f(x) = a linear function that describes the
value of the necklace after x years. The problem
states that f(3) = 360 and f(7) = 640. Thus,
( x1 , y1 ) = (3, 360) and ( x2 , y2 ) = (7, 640) . The
t
10
y2 − y1 640 − 360 280
=
=
= 70
x2 − x1
7−3
4
Using the point-slope form with m = 70 and
( x1 , y1 ) = (3, 360) gives
slope is m =
4. f(s) = 3(5 − 2s) = 15 − 6s has the form
f(s) = as + b where a = –6 (slope) and b = 15 (the
vertical-axis intercept).
y − y1 = m ( x − x1 )
y – 360 = 70(x – 3)
y = f(x) = 70x + 150
f(s)
16
s
5
98
ISM: Introductory Mathematical Analysis
Section 3.2
2−q 2 1
= − q has the form
7
7 7
1
h(q) = aq + b where a = − (the slope) and
7
2
b = (the vertical-axis intercept).
7
⎛2⎞
−7 = −2 ⎜ ⎟ + b
⎝5⎠
4
31
b = −7 + = −
5
5
31
so f ( x) = −2 x − .
5
5. h(q) =
5
h(q)
11.
2
7
f ( x) = ax + b = −
2
2
⎛ 2⎞
x + b . Since f ⎜ − ⎟ = − ,
3
3
⎝ 3⎠
we have
2
2⎛ 2⎞
− = − ⎜− ⎟+b
3
3⎝ 3⎠
q
5
2 4
10
− =− ,
3 9
9
2
10
so f ( x) = − x −
.
3
9
b=−
6. h(q) = 0.5q + 0.25 has the form h(q) = aq + b
with a = 0.5 (the slope) and b = 0.25 (the
vertical-axis intercept).
1
h(q)
0.25
–0.5
12. Let y = f(x). The points (1, 1) and (2, 2) lie on
2 −1
=1.
the graph of f. m =
2 −1
Thus y – 1 = 1(x – 1) ⇒ y = x, so f(x) = x.
q
1
13. Let y = f(x). The points (–2, –1) and (–4, –3) lie
−3 + 1
= 1 . Thus
on the graph of f. m =
−4 + 2
y + 1 = 1(x + 2), so y = x + 1 ⇒ f(x) = x + 1.
7. f(x) = ax + b = 4x + b. Since f(2) = 8, 8 = 4(2) + b,
8 = 8 + b, b = 0 ⇒ f(x) = 4x.
14. f(x) = ax + b = 0.01x + b. Since f(0.1) = 0.01, we
have 0.01 = (0.01)(0.1) + b ⇒ b = 0.009
⇒ f(x) = 0.01x + 0.009.
8. Let y = f(x). The points (0, 3) and (4, –5) lie on
−5 − 3
= −2 . Thus
the graph of f. m =
4−0
y – 3 = –2(x – 0), so
y = –2x + 3 ⇒ f ( x) = –2x + 3.
15. The points (40, 12.75) and (25, 18.75) lie on the
graph of the equation, which is a line.
18.75 − 12.75
2
= − . Hence an equation of
m=
25 − 40
5
2
the line is p − 12.75 = − (q − 40) , which can be
5
2
written p = − q + 28.75. When q = 37, then
5
2
p = − (37) + 28.75 = $13.95.
5
9. Let y = f(x). The points (1, 2) and (–2, 8) lie on
8−2
= −2 . Thus
the graph of f. m =
−2 − 1
y – 2 = –2(x – 1), so
y = –2x + 4 ⇒ f ( x) = –2x + 4.
10. f(x) = ax + b = −2x + b.
⎛2⎞
Since f ⎜ ⎟ = −7, we have
⎝5⎠
16. The line passes through (26,000, 12) and
(10,000, 18), so
18 − 12
= −0.000375. Then
m=
10, 000 − 26, 000
p – 18 = –0.000375(q – 10,000) or
p = –0.000375q + 21.75.
99
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
17. The line passes through (3000, 940) and
740 − 940
= 0.25. Then
(2200, 740), so m =
2200 − 3000
p – 740 = 0.25(q – 2200) or p = 0.25q + 190.
20
18. The points (50, 35) and (35, 30) lie on the graph
of the equation, which is a line.
30 − 35 −5 1
=
= . Hence an equation of
m=
35 − 50 −15 3
the line is
1
p − 35 = (q − 50)
3
1
55
p = q+
3
3
r
d
0
20
23. Each year the value decreases by 0.10(1800).
After t years the total decrease is 0.10(1800)t.
Thus
v = 1800 – 0.10(1800)t
v = –180t + 1800
The slope is –180.
v
19. The line passing through (10, 40) and (20, 70)
70 − 40
= 3 , so an equation for the
has slope
20 − 10
line is
c – 40 = 3(q – 10)
c = 3q + 10
If q = 35, then c = 3(35) + 10 = 105 + 10 = $115.
1800
0
20. The line passing through (100, 79) and (400, 88)
88 − 79
= 0.03, so an equation for
has slope
400 − 100
the line is
c – 79 = 0.03(x – 100)
c = 0.03x + 76
t
10
24. The line has slope –120 and passes through
(4, 340). Thus y – 340 = –120(x – 4) or
y = f(x) = –120x + 820.
25. The line has slope 45,000 and passes through
(5, 960,000). Thus
y – 960,000 = 45,000(x – 5) or
y = f(x) = 45,000x + 735,000.
21. If x = the number of kilowatt hours used in a
month, then f ( x) = the total monthly charges
for x kilowatt hours of electricity. If f ( x) is a
linear function it has the form f ( x) = ax + b.
The problem states that f(380) = 51.65. Since
12.5 cents are charged per kilowatt hour used,
a = 0.125.
f(x) = ax + b
51.65 = 0.125(380) + b
51.65 = 47.5 + b
4.15 = b
Hence, f ( x) = 0.125x + 4.15 is a linear function
that describes the total monthly charges for any
number of kilowatt hours x.
245, 000 49, 000
=
and
15
3
y-intercept 245,000. So
49, 000
y = f ( x) =
x + 245, 000.
3
26. The line has slope
27. If x = the number of hours of service, then
f(x) = the price of x hours of service. Let y = f(x).
f(1) = 159 and f(3) = 287, so (1, 159) and
(3, 287) lie on the graph of f which has slope
287 − 159
= 64. Using (1, 159), we get
a=
3 −1
y − 159 = 64(x − 1) or y = 64x + 95, so
f(x) = 64x + 95.
22. The number of curative units from d cubic
centimeters of the drug is 210d, and the number
of curative units from r minutes of radiation is
305r. Thus
210d + 305r = 2410
42d + 61r = 482
100
ISM: Introductory Mathematical Analysis
28. a.
Section 3.2
Suppose r = respiratory rate,
l = wool length, and (l, r) lies on the graph,
which is a line. The points (2, 160) and
(4, 125) are on the line, so its slope is
125 − 160
35
= − . Thus
4−2
2
35
r − 160 = − (l − 2)
2
35
r = − l + 195
2
b. If l = 1, then r = −
100 − 65 35
=
100 − 56 44
35
y − 100 =
( x − 100)
44
35
3500
y=
x−
+ 100
44
44
35
225
y=
x+
44
11
m=
31. a.
35
225
x+
44
11
35
225
x = 62 −
44
11
1828
x=
≈ 52.2
35
52.2 is the lowest passing score on original
scale.
62 =
b.
35
(1) + 195 = 177.5
2
29. At $200/ton, x tons cost 200x, and at $2000/acre,
y acres cost 2000y. Hence the required equation
is 200x + 2000y = 20,000, which can be written
as x + 10y = 100.
32. R = 38N + 397 is a linear equation. Slope = 38.
30. P = 4x + 6y where x, y ≥ 0.
a. 240 = 4x + 6y
100
R
587
y
549
511
473
40
435
0
60
x
100
0
33. p = f(t) = at + b, f(5) = 0.32, a = slope = 0.059.
b. Since the equation can be written
2
2
y = − x + 40 , slope = − .
3
3
c.
N
5
a.
p = f(t) = 0.059t + b. Since f(5) = 0.32,
0.32 = 0.059(5) + b, 0.32 = 0.295 + b, so
b = 0.025. Thus p = 0.059t + 0.025.
b. When t = 9, then
p = 0.059(9) + 0.025 = 0.556.
600 = 4x + 6y. Since the equation can be
2
written y = − x + 100,
3
2
slope = − .
3
34. w = f(d) = ad + b, f(0) = 21,
6.3
= 0.63. Thus
a = slope =
10
w = f(d) = 0.63d + b. Since f(0) = 21, we have
20 = 0.63(0) + b, so b = 21. Hence
w = 0.63d + 21.
When d = 55, then
w = 0.63(55) + 21 = 34.65 + 21 = 55.65 kg.
d. Solving P = 4x + 6y for y gives
2
P
y = − x + . Thus any isoprofit line has
3
6
2
slope − , and lines with the same slope are
3
parallel. Hence isoprofit lines are parallel.
101
Chapter 3: Lines, Parabolas, and Systems
35. a.
ISM: Introductory Mathematical Analysis
( )
t −t
80 − 68
12 1
=
= .
m= 2 1 =
c2 − c1 172 − 124 48 4
h(1) = −16 12 + 32(1) + 8 = 24 . Thus, the vertex
is (1, 24). Since c = 8, the y-intercept is (0, 8).
To find the x-intercepts we set y = h(t) = 0.
1
1
t − 68 = (c − 124) , t − 68 = c − 31 , or
4
4
1
t = c + 37 .
4
0 = −16t 2 + 32t + 8
t=
b. Since c is the number of chirps per minute,
1
1
then c is the number of chirps in
4
4
minute or 15 seconds. Thus from part (a), to
estimate temperature add 37 to the number
of chirps in 15 seconds.
=
−32 ± 322 − 4(−16)(8)
−b ± b 2 − 4ac
=
2a
2(−16)
−32 ± 1536 −32 ± 16 6
6
=
= 1±
−32
−32
2
⎛
6 ⎞
Thus, the x-intercepts are ⎜1 +
, 0 ⎟ and
⎜
⎟
2
⎝
⎠
⎛
6 ⎞
, 0⎟ .
⎜⎜1 −
⎟
2
⎝
⎠
Principles in Practice 3.3
1. In the quadratic function
y = P ( x) = − x 2 + 2 x + 399 , a = –1, b = 2,
c = 399. Since a < 0, the parabola opens
downward. The x-coordinate of the vertex is
b
2
−
=−
=1.
2a
2(−1)
The y-coordinate of the vertex is
30
–5
( )
P (1) = − 12 + 2(1) + 399 = 400 . Thus, the vertex
–20
is (1, 400). Since c = 399, the y-intercept is
(0, 399). To find the x-intercepts we set
y = p(x) = 0.
0 = − x 2 + 2 x + 399
(
0 = − x 2 − 2 x − 399
3. If we express the revenue r as a function of the
quantity produced q, we obtain
r = pq
r = (6 – 0.003q)q
)
r = 6q − 0.003q 2
We note that this is a quadratic function with
a = –0.003, b = 6, and c = 0. Since a < 0, the
graph of the function is a parabola that opens
downward, and r is maximum at the vertex
(q, r).
b
6
=−
= 1000
q=−
2a
2(−0.003)
0 = –(x + 19)(x – 21)
Thus, the x-intercepts are (–19, 0) and (21, 0).
y
400
100
–25
25
5
r = 6(1000) − 0.003(1000)2 = 3000
Thus, the maximum revenue that the
manufacturer can receive is $3000, which occurs
at a production level of 1000 units.
x
If the model is correct, this is not a good
business, since it will lose money if more than
21 minivans are sold.
Problems 3.3
2
2. In the quadratic function h(t ) = −16t + 32t + 8 ,
a = –16, b = 32, and c = 8. Since a < 0, the
parabola opens downward. The x -coordinate of
b
32
=−
= 1 . The
the vertex is −
2a
2(−16)
y-coordinate of the vertex is
1.
f ( x) = 5 x 2 has the form f ( x) = ax 2 + bx + c
where a = 5, b = 0, and c = 0 ⇒ quadratic.
102
ISM: Introductory Mathematical Analysis
1
2. g ( x) =
2 x2 − 4
Section 3.3
10. y = f ( x) = 8 x 2 + 4 x − 1
a = 8, b = 4, c = –1
cannot be put in the form
g ( x) = ax 2 + bx + c where
a.
a ≠ 0 ⇒ not quadratic.
2
g ( x) = ax 2 + bx + c where
3⎞
⎛ 1
Vertex: ⎜ − , − ⎟
2⎠
⎝ 4
a ≠ 0 ⇒ not quadratic.
4. k (v) = 3v 2 (v 2 + 2) = 3v 4 + 6v 2 cannot be put in
b. a = 8 > 0, so the vertex corresponds to the
lowest point.
the form k (v) = av 2 + bv + c where
a ≠ 0 ⇒ not quadratic.
11. y = x 2 + x − 6
2
5. h(q) = (3 − q ) = 9 − 6q + q has form
a = 1, b = 1, c = −6
a. c = –6. Thus the y-intercept is –6.
2
h(q) = aq + bq + c where a = 1, b = −6, and
c = 9 ⇒ quadratic.
6.
b.
f (t ) = 2t (3 − t ) + 4t = −2t 2 + 10t has the form
x 2 + x − 6 = ( x − 2)( x + 3) = 0, so x = 2, −3.
x-intercepts: 2, −3
2
f (t ) = at + bt + c where a = –2, b = 10, and
c.
c = 0 ⇒ quadratic.
(
)
2
−
b
1
=−
2a
2
2
1
25
⎛ 1⎞ ⎛ 1⎞
f ⎜− ⎟ = ⎜− ⎟ − −6 = −
2
4
⎝ 2⎠ ⎝ 2⎠
s2 − 9 1 2 9
= s − has the form
7. f ( s ) =
2
2
2
1
2
f ( s ) = as + bs + c where a = , b = 0, and
2
9
c = – ⇒ quadratic.
2
8. g (t ) = t 2 − 1
b
4
1
=−
=−
2a
2 ⋅8
4
3
⎛ 1⎞
⎛ 1⎞
⎛ 1⎞
f ⎜ − ⎟ = 8⎜ − ⎟ + 4 ⎜ − ⎟ −1 = −
4
4
4
2
⎝
⎠
⎝
⎠
⎝
⎠
3. g(x) = 7 – 6x cannot be put in the form
2
−
25 ⎞
⎛ 1
Vertex: ⎜ − , − ⎟
4 ⎠
⎝ 2
12. y = f ( x) = 5 − x − 3x 2
a = –3, b = −1, c = 5
a. c = 5. Thus the y-intercept is 5.
= t 4 − 2t 2 + 1 cannot be put in
the form g (t ) = at 2 + bt + c where
a ≠ 0 ⇒ not quadratic.
b.
Vertex occurs when x = −
−b ± b 2 − 4ac
2a
−(−1) ± (−1)2 − 4(−3)(5)
2(−3)
1 ± 61
=
−6
−1 ± 61
=
6
−1 + 61 −1 − 61
x-intercepts:
,
6
6
9. y = f ( x) = −4 x 2 + 8 x + 7
a = –4, b = 8, c = 7
a.
x=
=
b
8
=−
=1.
2a
2(−4)
When x = 1, then
y = f (1) = −4(1)2 + 8(1) + 7 = 11 .
Vertex: (1, 11)
b. a = –4 < 0, so the vertex corresponds to the
highest point.
103
Chapter 3: Lines, Parabolas, and Systems
c.
−
ISM: Introductory Mathematical Analysis
b
−1
1
=−
=−
2a
2(−3)
6
15. y = g ( x) = −2 x 2 − 6 x
a = –2, b = –6, c = 0
b
−6
6
3
=−
=− =−
Vertex: −
2a
2(−2)
4
2
2
61
⎛ 1⎞
⎛ 1⎞ ⎛ 1⎞
f ⎜ − ⎟ = 5 − ⎜ − ⎟ − 3⎜ − ⎟ =
12
⎝ 6⎠
⎝ 6⎠ ⎝ 6⎠
2
9
⎛ −3 ⎞
⎛ −3 ⎞
⎛ −3 ⎞ −9
f ⎜ ⎟ = −2 ⎜ ⎟ − 6 ⎜ ⎟ =
+9 =
2
2
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠
⎛ 1 61 ⎞
Vertex: ⎜ − ,
⎟
⎝ 6 12 ⎠
⎛ 3 9⎞
Vertex: ⎜ − , ⎟
⎝ 2 2⎠
y-intercept: c = 0
2
13. y = f ( x) = x − 6 x + 5
a = 1, b = –6, c = 5
b
−6
=−
=3
Vertex: −
2a
2 ⋅1
x-intercepts: −2 x 2 − 6 x = −2 x( x + 3) = 0 , so
x = 0, –3.
9
Range: all y ≤
2
f (3) = 32 − 6(3) + 5 = −4
Vertex = (3, –4)
y-intercept: c = 5
y
2
x-intercepts: x − 6 x + 5 = (x – 1)(x – 5) = 0, so
x = 1, 5.
Range: all y ≥ –4
9
2
–3
x
–3
2
y
5
16. y = f ( x) = x 2 − 4
a = 1, b = 0, c = –4
b
0
=−
=0
Vertex: −
2a
2 ⋅1
x
1
5
(3, –4)
f (0) = 02 − 4 = −4
Vertex = (0, –4)
y-intercept: c = –4
14. y = f ( x) = –4 x 2
a = –4, b = 0, c = 0
b
0
=−
=0
Vertex: −
2a
2(−4)
x-intercepts: x 2 − 4 = ( x + 2)( x − 2) = 0 , so
x = –2, 2.
Range: all y ≥ 4
f (0) = −4(0) 2 = 0
Vertex = (0, 0)
y-intercept: c = 0
5
x-intercepts: −4 x 2 = 0 , so x = 0.
Range: all y ≤ 0
5
y
x
–2
y
2
–4
x
5
104
5
ISM: Introductory Mathematical Analysis
Section 3.3
17. s = h(t ) = t 2 + 6t + 9
a = 1, b = 6, c = 9
b
6
=−
= −3
Vertex: −
2a
2 ⋅1
19. y = f ( x) = −9 + 8 x − 2 x 2
a = –2, b = 8, c = –9
b
8
=−
=2
Vertex: −
2a
2(−2)
h(−3) = (−3) 2 + 6(−3) + 9 = 0
Vertex = (–3, 0)
s-intercept: c = 9
f (2) = −9 + 8(2) − 2(2)2 = −1
Vertex = (2, –1)
y-intercept: c = –9
x-intercepts: Because the parabola opens
downward (a < 0) and the vertex is below the
x-axis, there is no x-intercept.
Range: y ≤ –1
t-intercepts: t 2 + 6t + 9 = (t + 3)2 = 0 , so t = –3.
Range: all s ≥ 0
10
s
2
y
x
(2, –1)
t
–3
–9
3
18. s = h(t ) = 2t 2 + 3t − 2
a = 2, b = 3, c = –2
b
3
3
=−
=−
Vertex: −
2a
2⋅2
4
20. y = H ( x) = 1 − x − x 2
a = –1, b = –1, c = 1
b
−1
1
=−
=−
Vertex: −
2a
2(−1)
2
2
⎛ 3⎞
⎛ 3⎞
⎛ 3⎞
h ⎜ − ⎟ = 2 ⎜ − ⎟ + 3⎜ − ⎟ − 2
⎝ 4⎠
⎝ 4⎠
⎝ 4⎠
9 9
25
= − −2 = −
8 4
8
25 ⎞
⎛ 3
Vertex = ⎜ − , − ⎟
8 ⎠
⎝ 4
s-intercept: c = –2
2
5
⎛ 1⎞
⎛ 1⎞ ⎛ 1⎞
f ⎜ − ⎟ = 1− ⎜ − ⎟ − ⎜ − ⎟ =
4
⎝ 2⎠
⎝ 2⎠ ⎝ 2⎠
⎛ 1 5⎞
Vertex = ⎜ − , ⎟
⎝ 2 4⎠
y-intercept: c = 1
x-intercepts: Solving 1 − x − x 2 = 0 by the
quadratic formula gives
t-intercepts: 2t 2 + 3t − 2 = (2t − 1)(t + 2) = 0 , so
1
t = , –2.
2
x=
Range: all s ≥ −
5
25
8
=
y
2(−1)
−1 ± 5
2
Range: all y ≤
1
2
–2
(– 34 , – 258)
−(−1) ± (−1) 2 − 4(−1)(1)
5
x
5
(– 12 , 54 )
–2
–1 – 5
2
105
y
5
4
–1 + 5
2
x
5
=
1± 5
−2
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
21. t = f ( s ) = s 2 − 8s + 14
a = 1, b = –8, c = 14
b
−8
Vertex: −
=−
=4
2a
2 ⋅1
23.
f (4) = 42 − 8(4) + 14 = −2
Vertex = (4, –2)
t-intercept: c = 14
2
808
⎛ 5 ⎞
⎛ 5 ⎞
⎛ 5 ⎞
f ⎜ ⎟ = 49 ⎜ ⎟ − 10 ⎜ ⎟ + 17 =
.
49
49
49
49
⎝ ⎠
⎝ ⎠
⎝ ⎠
s-intercepts: Solving s 2 − 8s + 14 = 0 by the
quadratic formula:
s=
f ( x) = 49 x 2 − 10 x + 17
Since a = 49 > 0, the parabola opens upward and
f(x) has a minimum value that occurs when
5
b
−10
. The minimum value is
x=−
=−
=
2a
2 ⋅ 49 49
24.
−(−8) ± (−8)2 − 4(1)(14)
2(1)
8± 8 8± 2 2
=
= 4± 2
2
2
Range: all t ≥ –2
=
f ( x) = −3 x 2 − 18 x + 7
Since a = –3 < 0, the parabola opens downward
and f(x) has a maximum value that occurs when
b
−18
=−
= −3
x=−
2a
2(−3)
The maximum value is
f (−3) = −3(−3)2 − 18(−3) + 7 = 34.
t
25.
14
4–
2
4+
s
2
f (20) = 4(20) − 50 − 0.1(20)2 = −10 .
(4, –2)
26.
22. t = f ( s ) = s 2 + 6s + 11
a = 1, b = 6, c = 11
6
b
Vertex: −
=−
= −3
2a
2 ⋅1
f (−3) = (−3)2 + 6(−3) + 11 = 2
Vertex: (–3, 2)
t-intercept: c = 11
s-intercepts: Because the parabola opens upward
(a > 0) and the vertex is above the s-axis, there is
no s-intercept.
Range: all t ≥ 2
16
f ( x) = 4 x − 50 − 0.1x 2
Since a = –0.1 < 0, the parabola opens
downward and f(x) has a maximum value that
4
b
occurs when x = −
=−
= 20 . The
2a
2(−0.1)
maximum value is
f ( x) = x( x + 3) − 12 = x 2 + 3 x − 12
Because a = 1 > 0, the parabola opens upward
and f(x) has a minimum value that occurs when
b
3
3
=−
= − . The minimum value is
x=−
2a
2 ⋅1
2
2
57
⎛ 3⎞ ⎛ 3⎞
⎛ 3⎞
f ⎜ − ⎟ = ⎜ − ⎟ + 3 ⎜ − ⎟ − 12 = −
4
⎝ 2⎠ ⎝ 2⎠
⎝ 2⎠
27.
f ( x) = x 2 − 2 x + 4
a = 1, b = −2, c = 4
b
−2
=−
=1
v=−
2a
2(1)
t
The restricted function is g ( x) = x 2 − 2 x + 4,
11
x ≥ 1. From the quadratic formula applied to
(–3, 2)
x 2 − 2 x + 4 − y = 0, we get
s
x=
10
2 ± 4 − 4(1)(4 − y )
= 1 ± 1 − (4 − y )
2(1)
So the inverse of g(x) is g −1 ( x) = 1 + x − 3,
x ≥ 3.
106
ISM: Introductory Mathematical Analysis
8
Section 3.3
y
30. If we express the revenue r as a function of the
quantity produced q, we obtain
r = pq
r = (0.85 − 0.00045q)q
g(x)
g –1(x)
r = 0.85q − 0.00045q 2
This is a quadratic function with a = −0.00045,
b = 0.85, and c = 0. Since a < 0, the graph of the
function is a parabola that opens downward, and
r is a maximum at the vertex (q, r).
b
0.85
8500
q=−
=−
=
≈ 944
2a
2(−0.00045)
9
x
8
28.
f ( x) = − x 2 + 4 x − 3
a = −1, b = 4, c = −3
b
4
v=−
=−
=2
2a
2(−1)
r = 0.85(944) − 0.00045(944) 2 = 401.39
Thus, the maximum revenue that the
manufacturer can receive is $401.39, which
occurs at a production level of 944 units.
The restricted function is g ( x) = − x 2 + 4 x − 3,
x ≥ 2. From the quadratic formula applied to
− x 2 + 4 x − 3 − y = 0, we get
31. If we express the revenue r as a function of the
quantity produced q, we obtain
r = pq
r = (2400 – 6q)q
−4 ± 16 − 4(−1)(−3 − y )
2(−1)
= 2 ± (−1) 4 + (−3 − y )
x=
r = 2400q − 6q 2
This is a quadratic function with a = –6,
b = 2400, and c = 0. Since a < 0, the graph of the
function is a parabola that opens downward, and
r is maximum at the vertex (q, r).
b
2400
q=−
=−
= 200
2a
2(−6)
So the inverse of g(x) is g −1 ( x) = 2 + 1 − x ,
x ≤ 1.
5
y
g –1(x)
x
r = 2400(200) − 6(200)2 = 240, 000
Thus, the maximum revenue that the
manufacturer can receive is $240,000, which
occurs at a production level of 200 units.
5
g(x)
29. If we express the revenue r as a function of the
quantity produced q, we obtain
r = pq
r = (200 − 5q)q
32.
r = 200q − 5q 2
This is a quadratic function with a = –5,
b = 200, and c = 0. Since a < 0, the graph of the
function is a parabola that opens downward, and
r is maximum at the vertex (q, r).
b
200
q=−
=−
= 20
2a
2(−5)
10
40
10
n(12 − n) =
n − n 2 , where
9
3
9
10
0 ≤ n ≤ 12. Since a = − < 0 , f(n) has a
9
maximum value that occurs at the vertex.
f ( n) =
40
−
b
3
=−
=6
2a
2 − 10
9
( )
The maximum value of f(n) is
40
10
f (6) =
(6) − (6) 2 = 80 − 40 = 40 , which
3
9
corresponds to 40,000 households.
r = 200(20) − 5(20)2 = 2000
Thus, the maximum revenue that the
manufacturer can receive is $2000, which occurs
at a production level of 20 units.
33. In the quadratic function
P ( x ) = − x 2 + 18 x + 144,
a = –1, b = 18, and c = 144. Since a < 0, the
graph of the function is a parabola that opens
downward. The x-coordinate of the vertex
107
Chapter 3: Lines, Parabolas, and Systems
is −
ISM: Introductory Mathematical Analysis
f(P) is
18
b
=−
= 9 . The y-coordinate of the
2a
2(−1)
f (50) =
( )
vertex is P (9) = − 92 + 18(9) + 144 = 225 .
36. s = −4.9t 2 + 62.3t + 1.8
Since a = –4.9 < 0, s has a maximum value that
occurs at the vertex where
b
62.3
62.3 89
t=−
=−
=
=
≈ 6.36 sec.
2a
2(−4.9) 9.8 14
Thus, the vertex is (9, 225). Since c = 144, the
y-intercept is (0, 144). To find the x-intercepts,
let y = P(x) = 0.
0 = − x 2 + 18 x + 144
(
0 = − x 2 − 18 x − 144
−1
(50)2 + 2(50) + 20 = 70 grams.
50
)
When t =
0 = –(x – 24)(x + 6)
Thus, the x-intercepts are (24, 0) and (–6, 0).
89
, then
14
2
⎛ 89 ⎞
⎛ 89 ⎞
s = −4.9 ⎜ ⎟ + 62.3 ⎜ ⎟ + 1.8
⎝ 14 ⎠
⎝ 14 ⎠
= 199.825 meters.
P(x)
400
37. h(t ) = −16t 2 + 85t + 22
–20
Since a = −16 < 0, h(t) has a maximum value
that occurs at the vertex where
b
85
t=−
=−
≈ 2.7 sec. When t = 2.7,
2a
2(−16)
then
x
30
34. If k = 2, then
y = kx 2
h(t ) = −16(2.7) 2 + 85(2.7) + 22 = 134.86 feet.
y = 2 x2
This is a quadratic equation with a = 2, b = 0 and
c = 0. Since a > 0, the graph of the function is a
parabola that opens upward. The x-coordinate of
b
0
=−
=0.
the vertex is −
2a
2(2)
The y-coordinate is
38. h(t ) = −16t 2 + 16t + 4
Since a = –16 < 0, h(t) has a maximum value
that occurs at the vertex where
b
16
1
1
t=−
=−
= sec. When t = ,
2a
2(−16) 2
2
2
⎛1⎞
⎛1⎞
then, h(t ) = −16 ⎜ ⎟ + 16 ⎜ ⎟ + 4 = 8 feet.
⎝2⎠
⎝2⎠
y = 2(0) 2 = 0
Thus, the vertex is (0, 0).
8
y
39. In the quadratic function h(t ) = −16t 2 + 80t + 16 ,
a = –16, b = 80, and c = 16. Since a < 0, the
graph of the function is a parabola that opens
downward. The x-coordinate of the vertex is
b
80
5
−
=−
= .
2a
2(−16) 2
The y-coordinate of the vertex is
x
5
2
⎛5⎞
⎛5⎞
⎛5⎞
h ⎜ ⎟ = −16 ⎜ ⎟ + 80 ⎜ ⎟ + 16 = 116
⎝2⎠
⎝2⎠
⎝2⎠
1
35. f ( P ) = − P 2 + 2 P + 20 , where 0 ≤ P ≤ 100.
50
1
Because a = −
< 0 , f(P) has a maximum
50
value that occurs at the vertex.
2
b
−
=−
= 50 . The maximum value of
2a
2 − 1
(
50
⎛5
⎞
Thus, the vertex is ⎜ , 116 ⎟ . Since c = 16, the
⎝2
⎠
y-intercept is (0, 16). To find the x-intercepts, we
let y = h(t) = 0.
)
0 = −16t 2 + 80t + 16
108
ISM: Introductory Mathematical Analysis
t=
=
Section 3.4
y = 78 − x = 78 − 39 = 39. Thus, two numbers
whose sum is 78 and whose product is a
maximum are 39 and 39.
−b ± b 2 − 4ac
2a
−80 ± 802 − 4(−16)(16)
43. (1.11, 2.88)
2(−16)
−80 ± 7424 5 ± 29
=
=
−32
2
⎛ 5 + 29 ⎞
Thus, the x-intercepts are ⎜
, 0 ⎟ and
⎜
⎟
2
⎝
⎠
44. –1.61, 3.73
45. a.
none
b. one
⎛ 5 − 29 ⎞
, 0⎟ .
⎜⎜
⎟
2
⎝
⎠
c.
two
46. 14.18
h(t)
47. 4.89
160
Principles in Practice 3.4
–10
10
1. Let x = the number invested at 9% and let
y = the amount invested at 8%. Then, the
problem states
⎧ x + y = 200, 000,
⎨
⎩0.09 x + 0.08 y = 17, 200.
We eliminate x by multiplying the first equation
by –0.09 and then adding
⎧−0.09 x − 0.09 y = −18, 000,
⎨
⎩0.09 x + 0.08 y = 17, 200.
−0.01 y = −800,
y = 80, 000.
Therefore,
⎧ x = 120, 000,
⎨
⎩ y = 80, 000.
Thus, $120,000 is invested at 9% and $80,000 is
invested at 8%.
x
40. A = x(11 − x) = 11x − x 2 , so A is a quadratic
function of x where a = –1 < 0. A has maximum
value at the vertex where
b
11
11
x=−
=−
= .
2a
2(−1) 2
41. Since the total length of fencing is 500, the side
opposite the highway has length 500 – 2x. The
area A is given by
A = x(500 − 2 x) = 500 x − 2 x 2 ,
which is quadratic with a = –2 < 0. Thus A is
500
maximum when x = −
= 125. Then the
2(−2)
side opposite the highway is
500 – 2x = 500 – 2(125) = 250. Thus the
dimensions are 125 ft by 250 ft.
2. Let A = the number of deer of species A, and let
B = the number of deer of species B. Then, the
number of pounds of food pellets that will be
consumed is 4A + 2B = 4000. The number of
pounds of hay that will be consumed is
5A + 7B = 9500. Then, we have
⎧4 A + 2 B = 4000,
⎨
⎩5 A + 7 B = 9500.
If we solve the first equation for B, we obtain
⎧ B = 2000 − 2 A
⎨5 A + 7 B = 9500.
⎩
Substituting 2000 – 2A for B in the second
equation gives
5A + 7(2000 – 2A) = 9500
A = 500
Highway
x
x
500 – 2x
42. Let x, y be two numbers whose sum is 78. Thus
x + y = 78 and y = 78 − x. Their product is then
p( x) = x(78 − x) = 78 x − x 2 . Since a = −1 < 0,
p(x) has a maximum value that occurs at the
b
78
vertex where x = −
=−
= 39 and
2a
2(−1)
109
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
Thus
⎧ B = 2000 − 2 A,
⎨
⎩ A = 500.
and
⎧ B = 1000,
⎨
⎩ A = 500.
The food will support 500 of species A and 1000
of species B.
1
⎧
⎪A = 6 ,
⎪
1
⎪
⎨C = ,
2
⎪
1
⎪
⎪B = 3 .
⎩
Thus, the final mixture will consist of
3. Let A = the number of fish of species A, and let
B = the number of fish of species B. Then, the
number of milligrams of the first supplement
that will be consumed is 15A + 20B = 100,000.
The number of milligrams of the second
supplement that will be consumed is 30A + 40B
= 200,000.
⎧15 A + 20 B = 100, 000,
⎨
⎩30 A + 40 B = 200, 000.
A,
1
lb of
6
1
1
lb of B, and
lb of C.
3
2
Problems 3.4
⎧ x + 4 y = 3,
1. ⎨
⎩3 x − 2 y = −5.
(1)
(2)
From Eq. (1), x = 3 – 4y. Substituting in Eq. (2)
gives
3(3 – 4y) – 2y = –5
9 – 12y – 2y = –5
–14y = –14,
or y = 1 ⇒ x = 3 – 4y = 3 – 4(1) = –1.
Thus x = –1, y = 1.
1
We multiply the second equation by − and
2
then add.
⎧15 A + 20 B = 100, 000,
⎨
⎩−15 A − 20 B = −100, 000,
0=0
Thus, there are infinitely many solutions of the
20, 000 4
− r , B = r, where
form A =
3
3
0 ≤ r ≤ 5000.
⎧4 x + 2 y = 9, (1)
2. ⎨
⎩5 y − 4 x = 5. (2)
Rewriting the system gives
⎧4 x + 2 y = 9,
⎨
⎩−4 x + 5 y = 5.
Adding the equations gives
7y = 14
y=2
From Eq. (1) we have
4x + 2(2) = 9
4x = 5
5
x=
4
5
Thus x = , y = 2.
4
4. Let A = the amount of type A used, let
B = the amount of type B used, and let
C = the amount of type C used. If the final blend
will sell for $8.50 per pound, then
12A + 9B + 7C = 8.50, and A + B + C = 1.
Furthermore, since the amount of type B is to be
twice the amount of type A, B = 2A. Thus, the
system of equations is
⎧12 A + 9 B + 7C = 8.50,
⎪
⎨ A + B + C = 1,
⎪ B = 2 A.
⎩
Simplifying gives
⎧30 A + 7C = 8.50,
⎪
⎨3 A + C = 1,
⎪ B = 2 A.
⎩
⎧3x − 4 y = 13,
3. ⎨
⎩2 x + 3 y = 3.
(1)
(2)
Multiplying Eq. (1) by 3 and Eq. (2) by 4 gives
⎧9 x − 12 y = 39,
⎨
⎩8 x + 12 y = 12.
110
ISM: Introductory Mathematical Analysis
Section 3.4
Adding gives
17x = 51
x=3
From Eq. (2) we have
2(3) + 3y = 3
3y = –3
y = –1
Thus x = 3, y = –1.
⎧2 x − y = 1,
4. ⎨
⎩− x + 2 y = 7.
⎧3x + 5 y = 7,
8. ⎨
⎩5 x + 9 y = 7.
Multiplying Eq. (1) by 5 and Eq. (2) by –3 gives
⎧15 x + 25 y = 35,
⎨
⎩−15 x − 27 y = −21.
Adding gives –2y = 14, or y = –7. From Eq. (2)
we have
5x + 9(–7) = 7
5x = 70
x = 14
Thus x = 14, y = –7.
(1)
(2)
From Eq. (1), y = 2x – 1. Substituting in Eq. (2)
gives
–x + 2(2x – 1) = 7
3x = 9
x = 3 ⇒ y = 2x – 1 = 2(3) – 1 = 5.
Thus x = 3, y = 5.
⎧4 x − 3 y − 2 = 3 x − 7 y,
9. ⎨
⎩ x + 5 y − 2 = y + 4.
Simplifying, we have
⎧ x + 4 y = 2,
⎨
⎩ x + 4 y = 6.
Subtracting the second equation from the first
gives 0 = –4, which is never true. Thus there is
no solution.
⎧u + v = 5
5. ⎨
⎩u − v = 7
From the first equation, v = 5 − u. Substituting in
the second equation gives
u − (5 − u ) = 7
2u − 5 = 7
2u = 12
or u = 6 so v = 5 − u = 5 − 6 = −1.
Thus, u = 6, v = −1.
⎧2 p + q = 16,
6. ⎨
⎩3 p + 3q = 33.
⎧5 x + 7 y + 2 = 9 y − 4 x + 6,
⎪
10. ⎨ 21
4
11 3
2
5
⎪ 2 x− 3 y− 4 = 2 x+ 3 y+ 4.
⎩
(1)
(2)
By simplifying, we have
⎧9 x − 2 y = 4,
⎨
⎩9 x − 2 y = 4.
Both equations represent the same line, so we
have infinitely many solutions. Let y = r. Then
2
4
9 x − 2r = 4 ⇒ x = r + . Thus a parametric
9
9
2
4
solution is x = r + , y = r, where r is any real
9
9
number.
From Eq. (1), q = 16 − 2p. Substituting in Eq.
(2) gives
3 p + 3(16 − 2 p ) = 33
−3 p = −15
p = 5 ⇒ q = 16 − 2p = 16 − 10 = 6.
Thus, p = 5, q = 6.
⎧ x − 2 y = –7,
7. ⎨
⎩5 x + 3 y = –9.
(1)
(2)
(1)
(2)
1
⎧2
⎪⎪ 3 x + 2 y = 2,
11. ⎨
⎪ 3 x + 5 y = − 11 .
⎪⎩ 8
6
2
From Eq. (1), x = 2y – 7. Substituting in Eq. (2)
gives
5(2y – 7) + 3y = –9
13y = 26
y = 2 ⇒ x = 2y – 7 = 2(2) – 7 = –3.
Thus x = –3, y = 2.
Clearing fractions gives the system
⎧4 x + 3 y = 12,
⎨
⎩9 x + 20 y = −132.
Multiplying the first equation by 9 and the
second equation by –4 gives
111
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
Multiplying the first equation by 5 and the
second equation by –3 gives
⎧15 x + 50 z = 20,
⎨
⎩−15 x − 18 z = −12.
⎧36 x + 27 y = 108,
⎨
⎩−36 x − 80 y = 528.
Adding gives
–53y = 636
y = –12
From 4x + 3y = 12, we have
4x + 3(–12) = 12
4x = 48 ⇒ x = 12. Thus x = 12, y = –12.
Adding gives 32z = 8, or z =
3x + 10z = 4, we have
⎛1⎞
3x + 10 ⎜ ⎟ = 4
⎝4⎠
⎧⎪ 1 z − 1 w = 1
4
6
12. ⎨ 12
1w= 1
z
+
⎪⎩ 2
4
6
Multiplying both equations by 12 gives
⎧6 z − 3w = 2
⎨ 6 z + 3w = 2
⎩
3
2
1
x=
2
From 2x + y + 6z = 3, we have
⎛1⎞
⎛1⎞
2⎜ ⎟ + y + 6⎜ ⎟ = 3
2
⎝ ⎠
⎝4⎠
3x =
1
Adding gives 12z = 4 and so z = .
3
⎛1⎞
From the first equation we have 6 ⎜ ⎟ − 3w = 2,
⎝3⎠
1
from which w = 0. Thus z = , w = 0.
3
⎧5 p + 11q = 7,
13. ⎨
⎩10 p + 22q = 33.
y=
1
1
1
, y= , z= .
2
2
4
⎧ x + y + z = −1,
⎪
16. ⎨3x + y + z = 1,
⎪4 x − 2 y + 2 z = 0.
⎩
(1)
(2)
(1)
(2)
(3)
Subtracting Eq. (2) from Eq. (1) gives –2x = –2,
or x = 1. Substituting x = 1 in Eqs. (2) and (3)
and simplifying gives
⎧ y + z = −2,
⎨
⎩−2 y + 2 z = −4.
Multiplying the first equation by 2 gives
⎧2 y + 2 z = −4,
⎨
⎩−2 y + 2 z = −4.
By adding, we have
4z = –8
z = –2
From y + z = –2, we have
y + (–2) = –2
y=0
Thus x = 1, y = 0, z = –2.
(1)
(2)
Multiplying Eq. (1) by 2 gives
⎧10 x − 6 y = 4,
⎨
⎩−10 x + 6 y = 4.
Adding gives 0 = 8, which is never true, so the
system has no solution.
⎧2 x + y + 6 z = 3,
⎪
15. ⎨ x − y + 4 z = 1,
⎪3x + 2 y − 2 z = 2.
⎩
1
2
Therefore x =
Multiplying Eq. (1) by –2 gives
⎧−10 p − 22q = −14,
⎨
⎩ 10 p + 22q = 33.
Adding gives 0 = 19, which is never true, so the
system has no solution.
⎧5 x − 3 y = 2,
14. ⎨
⎩−10 x + 6 y = 4.
1
. From
4
(1)
(2)
(3)
⎧ x + 4 y + 3 z = 10
⎪
17. ⎨4 x + 2 y − 2 z = −2
⎪⎩ 3x − y + z = 11
From the third equation, y = 3x + z − 11.
Substituting in the first two equations gives
⎧ x + 4(3 x + z − 11) + 3 z = 10
⎨4 x + 2(3 x + z − 11) − 2 z = −2
⎩
Adding Eq. (1) and (2), and adding 2 times Eq.
(2) to Eq. (3) gives
⎧3 x + 10 z = 4,
⎨
⎩5 x + 6 z = 4.
112
ISM: Introductory Mathematical Analysis
Section 3.4
or
⎧13x + 7 z = 54
⎨10 x = 20
⎩
From the last equation we have x = 2.
Thus 13(2) + 7z = 54, and 7z = 28, hence z = 4.
Substitute these two values to solve for y:
y = 3(2) + 4 − 11 = −1
Therefore, x = 2, y = −1, z = 4.
(1)
⎧ x − y + 2 z = 0,
⎪
21. ⎨2 x + y − z = 0
(2)
⎪ x + 2 y − 3z = 0
(3)
⎩
Adding Eq. (1) to Eq. (3) gives
⎧ x − y + 2 z = 0,
⎪
⎨2 x + y − z = 0
⎪2 x + y − z = 0
⎩
We can ignore the third equation because the
second equation can be used to reduce it to
0 = 0. We have
⎧ x − y + 2 z = 0,
⎨
⎩2 x + y − z = 0.
Adding the first equation to the second gives
3x + z = 0
1
x=− z
3
Substituting in the first equation we have
1
− z − y + 2z = 0
3
5
y= z
3
Letting z = r gives the parametric solution
1
5
x = − r , y = r , z = r, where r is any real
3
3
number.
⎧ x + y + z = 18 (1)
⎪
18. ⎨ x − y − z = 12 (2)
⎪⎩3x + y + 4 z = 4 (3)
Adding Eq. (2) to both Eq. (1) and Eq. (3) gives
2 x = 30
⎧
⎨4 x + 3 z = 16
⎩
From the first equation, x = 15. Substituting in
the second equation gives
4(15) + 3 z = 16
3 z = −44
44
z=−
3
From x + y + z = 18
44
= 18
15 + y −
3
53
y=
3
53
44
Thus, x = 15, y = , z = − .
3
3
(1)
⎧ x − 2 y − z = 0,
⎪
22. ⎨2 x − 4 y − 2 z = 0
(2)
⎪− x + 2 y + z = 0
(3)
⎩
Adding Eq. (1) to Eq. (3) gives
⎧ x − 2 y − z = 0,
⎪
⎨2 x − 4 y − 2 z = 0
⎪0 = 0
⎩
We can ignore the third equation, so we have
⎧ x − 2 y − z = 0,
⎨
⎩2 x − 4 y − 2 z = 0.
Multiplying the first equation by –2 gives
⎧−2 x + 4 y + 2 z = 0,
⎨
⎩2 x − 4 y − 2 z = 0.
Adding the first equation to the second, we have
⎧−2 x + 4 y + 2 z = 0,
⎨
⎩0 = 0.
From the first equation, x = 2y + z. Setting y = r
and z = s gives the parametric solution x = 2r + s,
y = r, z = s, where r and s are any real numbers.
(1)
⎧ x − 2 z = 1,
19. ⎨
(2)
⎩ y + z = 3.
From Eq. (1), x = 1 + 2z; from Eq. (2), y = 3 – z.
Setting z = r gives the parametric solution
x = 1 + 2r, y = 3 – r, z = r, where r is any real
number.
⎧2 y + 3 z = 1,
20. ⎨
⎩3x − 4 z = 0.
From Eq. (1), y =
(1)
(2)
1 3
− z ; from Eq. (2),
2 2
4
z . Setting z = r gives the parametric
3
4
1 3
solution x = r , y = − r , z = r, where r is
3
2 2
any real number.
x=
113
Chapter 3: Lines, Parabolas, and Systems
⎧2 x + 2 y − z = 3,
23. ⎨
⎩4 x + 4 y − 2 z = 6.
ISM: Introductory Mathematical Analysis
⎧0.03 x + 0.11 y = 1.8,
⎨
⎩ y = 20 − x.
By substituting 20 – x for y in the first equation,
and then simplifying, we obtain
⎧ x = 5,
⎨
⎩ y = 15.
Thus, the final mixture should contain 5 lb of
3% nitrogen fertilizer, and 15 lb of 11% nitrogen
fertilizer.
(1)
(2)
Multiplying Eq. (2) by −
1
gives
2
⎧2 x + 2 y − z = 3,
⎨
⎩−2 x − 2 y + z = −3.
Adding the first equation to the second equation
gives
⎧2 x + 2 y − z = 3,
⎨
⎩0 = 0.
Solving the first equation for x, we have
3
1
x = − y + z . Letting y = r and z = s gives the
2
2
3
1
parametric solution x = − r + s , y = r, z = s,
2
2
where r and s are any real numbers.
27. Let C = the number of pounds of cotton, let
P = the number of pounds of polyester, and let
N = the number of pounds of nylon. If the final
blend will cost $3.25 per pound to make, then
4C + 3P + 2N = 3.25. Furthermore, if we use the
same amount of nylon as polyester to prepare,
say, 1 pound of fabric, then N = P and
C + P + N = 1. Thus, the system of equations is
⎧4C + 3P + 2 N = 3.25,
⎪
⎨C + P + N = 1,
⎪ N = P.
⎩
Simplifying gives
⎧4C + 5 N = 3.25,
⎪
⎨C + 2 N = 1,
⎪ N = P.
⎩
⎧5 x + y + z = 17
24. ⎨
⎩4 x + y + z = 14
Subtracting the second equation from the first
gives x = 3.
From the first equation we have
y + z = 17 − 5x = 17 − 5(3) = 2
Letting z = r we have the parametric solution
x = 3, y = 2 − r, z = r, where r is any real
number.
⎧ N = 0.25,
⎪
⎨C = 0.5,
⎪ P = 0.25.
⎩
Thus, each pound of the final fabric will contain
0.25 lb each of nylon and polyester, and 0.5 lb of
cotton.
25. Let x = number of gallons of 20% solution and
y = number of gallons of 35% solution. Then
(1)
⎧ x + y = 800,
⎨
(2)
⎩0.20 x + 0.35 y = 0.25(800).
From Eq. (1), y = 800 – x. Substituting in Eq. (2)
gives
0.20x + 0.35(800 – x) = 0.25(800)
–0.15x + 280 = 200
–0.15x = −80
1600
x=
≈ 533.3
3
1600 800
y = 800 − x = 800 −
=
≈ 266.7. Thus
3
3
533.3 gal of 20% solution and 266.7 gal of 35%
solution must be mixed.
28. Let F = federal tax and S = state tax. Now solve
the system
⎧ F = 0.25(312, 000 − S ),
⎨ S = 0.10(312, 000 − F ),
⎩
which is equivalent to
⎧ 4 F + S = 312, 000
⎨ F + 10 S = 312, 000,
⎩
and has solution
⎧ F = 72, 000,
⎨ S = 24, 000.
⎩
Federal tax is $72,000 and state tax is $24,000.
26. Let x = the number of pounds of 3% nitrogen
fertilizer, and let y = the number of pounds of
11% nitrogen fertilizer. Then
⎧0.03 x + 0.11 y = 0.09(20),
⎨
⎩ x + y = 20.
29. Let p = speed of airplane in still air and w = wind
speed. Now convert the time into minutes and
solve the system
114
ISM: Introductory Mathematical Analysis
Section 3.4
(1)
⎧ x = 1.20 y,
⎨
+
=
250
x
350
y
130,
000.
(2)
⎩
Substituting 1.20y for x in Eq. (2) gives
250(1.20y) + 350y = 130,000
300y + 350y = 130,000
650y = 130,000
y = 200
Thus x = 1.20y = 1.20(200) = 240. Therefore
240 units of early American and 200 units of
Contemporary must be sold.
900
⎧
⎪⎪ p + w = 175
⎨
⎪ p − w = 900 ,
⎪⎩
206
900 900 36 450
+
=
+
Thus 2 p =
175 206 7 103
3429
miles per minute
p=
721
279
miles per minute
w=
721
Multiplying by 60 to get miles per hour we have
p ≈ 285 and w ≈ 23.2
Plane speed in still air is about 285 mph and
wind speed is about 23.2 mph.
32. Let x = number of favorable comments,
y = number of unfavorable comments, and
z = number of no comments. Then
(1)
⎧ x + y + z = 250,
⎪
(2)
⎨ x = 1.625 y,
⎪ z = 0.16(250).
(3)
⎩
From Eq. (3), z = 40. Substituting for x and z in
Eq. (1), we obtain
(1.625y) + y + (40) = 250
2.625y = 210
y = 80
Thus x = 1.625y = 1.625(80) = 130. Therefore
130 liked, 80 disliked, and 40 had no comment.
30. Let r = speed of raft in still water and c = speed
of current. Then rate of raft downstream is r + c,
and rate upstream is r – c. Since
(rate)(time) = distance, we have
⎧
⎛1⎞
⎪(r + c) ⎜ ⎟ = 10,
⎪
⎝2⎠
⎨
⎪(r − c) ⎛ 3 ⎞ = 10,
⎜4⎟
⎪⎩
⎝ ⎠
or, more simply,
⎧r + c = 20,
⎪
40
⎨
⎪r − c = 3 .
⎩
Adding the equations gives
100
2r =
3
50
r=
3
10
Since r + c = 20, we have c = . Thus the
3
2
speed of the raft in still water is 16 mi/h;
3
1
speed of the current is 3 mi/h.
3
33. Let x = number of calculators produced at Exton,
and y = number of calculators produced at
Whyton. The total cost of Exton is 7.50x + 7000,
and the total cost at Whyton is 6.00y + 8800.
Thus 7.50x + 7000 = 6.00y + 8800. Also,
x + y = 1500. This gives the system
(1)
⎧ x + y = 1500,
⎨
(2)
⎩7.50 x + 7000 = 6.00 y + 8800.
From Eq. (1), y = 1500 – x. Substituting in Eq.
(2) gives
7.50x + 7000 = 6.00(1500 – x) + 8800
7.50x + 7000 = 9000 – 6x + 8800
13.5x = 10,800
x = 800
Thus y = 1500 – x = 1500 – 800 = 700.
Therefore 800 calculators must be made at the
Exton plant and 700 calculators at the Whyton
plant.
31. Let x = number of early American units and
y = number of Contemporary units. The fact that
20% more of early American styles are sold than
Contemporary styles means that
x = y + 0.20y
x = 1.20y
An analysis of profit gives
250x + 350y = 130,000. Thus we have the
system
34. Let x, y, and z be the amounts of 2.20, 2.30, and
2.60 dollars/lb coffee, respectively. Then
(1)
⎧ x + y + z = 100,
⎪
(2)
⎨2.20 x + 2.30 y + 2.60 z = 2.40(100),
⎪ y = z.
(3)
⎩
From Eq. (3), y = z. Substituting for y in Eqs. (1)
and (2) gives
115
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
20 x + 17.6 x = 3600
37.6 x = 3600
x ≈ 95.74
y = x ≈ 95.74
Thus, 95 boxes will be loose-filled and
8(95) = 760 clam-shells will be used, for a total
of 190 boxes.
⎧ x + z + z = 100,
⎨
⎩2.20 x + 2.30 z + 2.60 z = 240.
or, by simplifying,
⎧ x + 2 z = 100,
⎨
⎩2.20 x + 4.90 z = 240.
From the first equation, x = 100 – 2z.
Substituting in the second equation gives
2.20(100 – 2z) + 4.90z = 240
0.50z = 20
z = 40
From x = 100 – 2z, x = 100 – 2(40) = 20. From
y = z, y = 40. Thus, 20, 40, and 40 lb of $2.20,
$2.30, and $2.60 per lb coffee must be used,
respectively.
38. Let p1 and p2 be the amounts of the two
investments, respectively. Then the total amount
invested was p1 + p2 , and from the statement of
the problem we can write
3
( p1 + p2 ) + 600 = p1 . The return on the
10
second investment was 1120 – 384 = 736. Since
the percentage return on each was the same, and
interest
, we can write
since rate =
amt. invested
384 736
=
. This can also be written as
p1
p2
35. Let x = rate on first $100,000 and
y = rate on sales over $100,000. Then
(1)
⎧100, 000 x + 75, 000 y = 8500,
⎨
(2)
⎩100, 000 x + 180, 000 y = 14,800.
Subtracting Eq. (1) from Eq. (2) gives
105,000y = 6300
y = 0.06
Substituting in Eq. (1) gives
100,000x + 75,000(0.06) = 8500
100,000x + 4500 = 8500, 100,000x = 4000, or
x = 0.04. Thus the rate is 4% on the first
$100,000 and 6% on the remainder.
p1
p
= 2 . Hence we have the system
384 736
⎧3
⎪⎪10 ( p1 + p2 ) + 600 = p1 ,
⎨
⎪ p1 = p2 .
⎪⎩ 384 736
Simplifying, we have
3
⎧ 7
⎪⎪− 10 p1 + 10 p2 = −600,
⎨
⎪ p = 12 p .
⎪⎩ 1 23 2
12
Substituting p1 =
p2 in first equation gives
23
7 ⎛ 12
⎞ 3
− ⎜
p2 +
p2 = −600
10 ⎝ 23 ⎟⎠ 10
36. A system that describes the situation is
⎧T = L + 25, 000, 000
⎨T = L + 0.30 L
⎩
We can rewrite this as
⎧T = L + 25, 000, 000
⎨T = 1.30 L
⎩
Thus T = 1.3L and we can substitute this in the
first equation:
1.3L = L + 25, 000, 000. Solving for L
0.3L = 25, 000, 000
L = 83,333,333
T = 1.3L = 1.3(83,333,333) = 108,333,333 thus
T = $108,333,333 and L = $83,333,333.
3
p2 = −600
46
p2 = 9200
−
12
12
p2 =
(9200) = 4800 . The total
23
23
amount invested was
p1 + p2 = 4800 + 9200 = $14, 000 .
37. Let x = number of loose-filled boxes and
y = number of boxes of clam-shells that will be
filled. Then 8y clam-shells will be used. This
will take 20x + 2.2(8y) pounds of peaches.
(1)
⎧x = y
⎨20 x + 17.6 y = 3600 (2)
⎩
Substitute x = y in Eq. (2).
Thus p1 =
39. Let c = number of chairs company makes,
r = number of rockers, and l = number of chaise
lounges.
Wood used: (1)c + (1)r + (1)l = 400
Plastic used: (1)c + (1)r + (2)l = 600
116
ISM: Introductory Mathematical Analysis
Section 3.4
Aluminum used: (2)c + (3)r + (5)l = 1500
Thus we have the system
(1)
⎧c + r + l = 400,
⎪
(2)
⎨c + r + 2l = 600,
⎪2c + 3r + 5l = 1500.
(3)
⎩
Subtracting Eq. (1) from Eq. (2) gives l = 200.
Adding –2 times Eq. (1) to Eq. (3) gives
r + 3l = 700, from which
r + 3(200) = 700,
r = 100
From Eq. (1) we have c + 100 + 200 = 400, or
c = 100. Thus 100 chairs, 100 rockers and
200 chaise lounges should be made.
second gives:
5 x = 25
x=5
So y = 2x = 10
z = 70 − 3x = 70 − 15 = 55
The company should hire 5 skilled workers, 10
semiskilled workers, and 55 shipping clerks.
42. Method 1. Let a = number of minutes that pump
for tank A operates, and b = number of minutes
that pump for tank B operates. Then b = a + 5.
25a gallons are pumped from tank A and 35b
from tank B.
(1)
⎧b = a + 5,
⎨
(2)
⎩25a + 35b = 10, 000.
Since b = a + 5, substituting in Eq. (2) gives
25a + 35(a + 5) = 10,000
60a = 9825
a = 163.75
b = a + 5, b = 163.75 + 5 = 168.75. Thus
25(163.75) = 4093.75 gallons are pumped from
A, and 35(168.75) = 5906.25 gallons are pumped
from B.
Method 2. Let a = number of gallons from A,
and let b = number of gallons from B. Then
a + b = 10,000. The number of minutes the
a
. For the pump on B,
pump on A operates is
25
b
it is
. Thus
35
b
⎧a
(1)
⎪ +5 =
35
⎨ 25
⎪a + b = 10, 000.
(2)
⎩
40. Let x, y, and z, be the amounts originally
invested at 7%, 8%, and 9%, respectively. Then
(1)
⎧ x + y + z = 35, 000,
⎪
(2)
⎨0.07 x + 0.08 y + 0.09 z = 2830,
⎪0.07 x + 0.08 y + 0.10 z = 2960.
(3)
⎩
Subtracting Eq. (2) from Eq. (3) gives
0.01z = 130
z = 13,000
Subtracting 0.07 times Eq. (1) from Eq. (2)
gives
0.01y + 0.02z = 380. Letting z = 13,000, we
have 0.01y + 0.02(13,000) = 380
0.01y = 120
y = 12,000
From Eq. (1),
x + 12,000 + 13,000 = 35,000
x = 10,000
The investments are $10,000 at 7%, $12,000 at
8%, $13,000 at 9% (later 10%).
41. Let x = number of skilled workers employed,
y = number of semiskilled workers employed,
z = number of shipping clerks employed.
Then we have the system
(1)
⎧number of workers: x + y + z = 70,
⎪
wages:
16
x
+
9.5
y
+
10
z
=
725
(2)
⎨
⎪semiskilled:
y = 2x
(3)
⎩
From the last equation, y = 2x so substitute into
the first two equations:
x + 2 x + z = 70
⎧
⎨16 x + 9.5(2 x) + 10 z = 725
⎩
or
⎧ 3x + z = 70
⎨35 x + 10 z = 725
⎩
Adding −10 times the first equation to the
117
From Eq. (2), a = 10,000 – b. Substituting in
Eq. (1) gives
10, 000 − b
b
+5 =
25
35
b
b
400 − + 5 =
25
35
12b
405 =
175
5906.25 = b
Thus
a = 10,000 – b = 10,000 – 5906.25 = 4093.75.
45. x = 3, y = 2
46. x = 1.33, y = 0.67
47. x = 8.3, y = 14.0
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
Problems 3.5
4. From Eq. (2), y = x – 14. Substituting in Eq. (1)
gives
In the following solutions, any reference to Eq. (1) or
Eq. (2) refers to the first or second equation,
respectively, in the given system.
( x − 14)2 − x 2 = 28
–28x + 196 = 28
–28x = –168
x=6
If x = 6, then y = x – 14 = 6 – 14 = –8. The only
solution is x = 6, y = –8.
1. From Eq. (2), y = 3 − 2x. Substituting in Eq. (1)
gives
3 − 2x = x2 − 9
0 = x 2 + 2 x − 12
5. Substituting y = x 2 into x = y 2 gives x = x 4 ,
−b ± b 2 − 4ac
x=
2a
x4 − x = 0
(
)
x x3 − 1 = 0
−2 ± 22 − 4(1)(−12)
2(1)
−2 ± 52
=
2
= −1 ± 13
=
Thus x = 0, 1. From y = x 2 , if x = 0, then
y = 02 = 0 ; x = 1, then y = 12 = 1 . There are
two solutions: x = 0, y = 0; x = 1, y = 1.
From y = 3 − 2x, if x = −1 + 13, then
⎧ 2
6. ⎨ p − q + 1 = 0
⎩5q − 3 p − 2 = 0
y = 5 − 2 13; if x = −1 − 13, then
y = 5 + 2 13.
There are two solutions:
x = −1 + 13, y = 5 − 2 13;
From the first equation q = p 2 + 1. Substituting
into the second equation gives
5( p 2 + 1) − 3 p − 2 = 0
x = −1 − 13, y = 5 + 2 13.
5 p2 − 3 p + 3 = 0
2. From Eq. (2), y = x. Substituting in Eq. (1) gives
p=
x = x3
x − x3 = 0
(
=
)
−b ± b 2 − 4ac
2a
3 ± (−3) 2 − 4(5)(3)
2(5)
3 ± −51
=
10
Since −51 is not a real number, there are no
real solutions.
x 1 − x2 = 0
x(1 + x)(1 – x) = 0. Thus x = 0, ±1. From y = x, if
x = 0, then y = 0; if x = 1, then y = 1; if x = –1,
then y = –1. There are three solutions: x = 0,
y = 0; x = 1, y = 1; x = –1, y = –1.
3. From Eq. (2), q = p – 1. Substituting in Eq. (1)
gives
7. Substituting y = x 2 − 2 x in Eq. (1) gives
x2 − 2 x = 4 x − x2 + 8
p 2 = 5 − ( p − 1)
2 x2 − 6 x − 8 = 0
p2 + p − 6 = 0
(p + 3)(p – 2) = 0
Thus p = –3, 2. From q = p – 1, if p = –3, we
have q = –3 – 1 = –4; if p = 2, then q = 2 – 1 = 1.
There are two solutions: p = –3, q = –4;
p = 2, q = 1.
x2 − 3x − 4 = 0
(x – 4)(x + 1) = 0
Thus x = 4, –1. From y = x 2 − 2 x , if x = 4, then
we have y = 42 − 2(4) = 8 ; if x = –1, then
y = (−1)2 − 2(−1) = 3 . There are two solutions:
x = 4, y = 8; x = –1, y = 3.
118
ISM: Introductory Mathematical Analysis
Section 3.5
12. From Eq. (2), y = 3x – 5. Substituting in Eq. (1)
gives
8. From Eq. (1), y = x 2 + 4 x + 4. Substituting in
Eq. (2) gives
x 2 + (3 x − 5) 2 − 2 x(3 x − 5) = 1
x2 + 4 x + 4 − x2 − 4 x + 3 = 0
7=0
Since this is never true, the system has no
solution.
9. Substituting p = q in Eq. (2) gives
Squaring both sides gives
4 x 2 − 20 x + 24 = 0
x2 − 5x + 6 = 0
(x – 3)(x – 2) = 0
Thus x = 3, 2. From y = 3x – 5, if x = 3, then
y = 3(3) – 5 = 4; if x = 2, then y = 3(2) – 5 = 1.
Thus there are two solutions: x = 3, y = 4; x = 2,
y = 1.
q = q2 .
q = q4
13. From Eq. (1), y = x − 1. Substituting in Eq. (2)
gives
x −1 = 2 x + 2
q4 − q = 0
(
)
q q3 − 1 = 0
( x − 1) 2 = 4( x + 2)
Thus q = 0, 1. From p = q , if q = 0, then
x2 − 2 x + 1 = 4 x + 8
p = 0 = 0 ; if q = 1, then p = 1 = 1 . There are
two solutions: p = 0, q = 0; p = 1, q = 1.
10. Substituting z =
x2 − 6 x − 7 = 0
( x + 1)( x − 7) = 0
4
in Eq. (2) gives
w
Thus x = −1 or 7.
From y = x − 1, if x = −1, then y = −2; if
x = 7, then y = 6. However, from Eq. (2), y ≥ 0.
The only solution is x = 7, y = 6.
⎛4⎞
3 ⎜ ⎟ = 2w + 2
⎝ w⎠
12 = 2w2 + 2 w
14. Substituting y =
2
w + w−6 = 0
(w + 3)(w – 2) = 0
x2
1
=
+1
x −1 x −1
4
, if w = –3, then
w
4
4
z = − ; if w = 2, then z = = 2 . There are two
3
2
4
solutions: w = –3, z = − ; w = 2, z = 2.
3
Thus w = –3, 2. From z =
1 = x 2 + ( x − 1)
x2 + x − 2 = 0
(x + 2)(x – 1) = 0
Thus x = –2, 1. But x cannot equal 1 in either of
the original equations (division by zero). From
1
1
1
, if x = –2, then y =
y=
= − . The
x −1
−2 − 1
3
1
solution is x = –2, y = − .
3
11. Replacing x 2 by y 2 + 13 in Eq. (2) gives
(
1
in Eq. (1) gives
x −1
)
y = y 2 + 13 − 15
2
y − y−2 = 0
(y – 2)(y + 1) = 0
Thus y = 2, –1. If y = 2, then
15. We can write the following system of equations.
⎧⎪ y = 0.01x 2 + 0.01x + 7,
⎨
⎪⎩ y = 0.01x + 8.0.
By substituting 0.01x + 8.0 for y in the first
equation and simplifying, we obtain
x 2 = y 2 + 13 = 22 + 13 = 17 , so x = ± 17 .
If y = –1, then x 2 = y 2 + 13 = (−1)2 + 13 = 14 ,
so x = ± 14 . The system has four solutions:
x = 17 , y = 2; x = − 17 , y = 2; x = 14 ,
0.01x + 8.0 = 0.01x 2 + 0.01x + 7
0 = 0.01x 2 − 1
0 = (0.1x + 1)(0.1x – 1)
x = −10 or x = 10
If x = –10 then y = 7.9, and if x = 10 then y = 8.1.
y = –1; x = − 14 , y = –1.
119
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
The rope touches the streamer twice,
10 feet away from center on each side at
(–10, 7.9) and (10, 8.1).
20
16. We can write the following system of equations.
⎪⎧ y = 0.06 x 2 + 0.012 x + 8,
⎨
⎪⎩ y = 0.912 x + 5.
By substituting 0.912x + 5 for y in the first
equation and then simplifying, we obtain
(100, 7)
q
0
0.912 x + 5 = 0.06 x 2 + 0.012 x + 8
(
0 = 0.06 x 2 − 15 x + 50
200
2. Equating p-values gives
1
1
q+4= −
q+9
1500
2000
7
q=5
6000
30, 000
5
= 4285 ≈ 4285.71
q=
7
7
5
When q = 4285 , then
7
1
1 ⎛
5⎞
6
p=
q+4=
⎜ 4285 ⎟ + 4 = 6 ≈ 6.86
1500
1500 ⎝
7⎠
7
5 6⎞
⎛
The equilibrium point is ⎜ 4285 , 6 ⎟ .
7
7⎠
⎝
2
0 = 0.06 x − 0.9 x + 3
p
)
0 = 0.06(x – 10)(x – 5)
x = 10 or x = 5
If x = 10 then y = 14.12, and if x = 5 then
y = 9.56. The two holes are located at (10, 14.12)
and (5, 9.56).
17. The system has 3 solutions.
18. x = 2, y = 4
19. x = –1.3, y = 5.1
p
10
20. x = −1.9, y = −3.6; x = −0.3, y = 1.2;
x = 2.1, y = 8.3
6 6⎞
⎛
⎜ 4285 , 6 ⎟
7 7⎠
⎝
21. x = 1.76
5
22. x = 2.81
23. x = –1.46
5000
Problems 3.6
q
10,000
⎧35q − 2 p + 250 = 0,
3. ⎨
⎩65q + p − 537.5 = 0.
1. Equating p-values gives
4
6
+ 13
q+3= −
100
100
10
q = 10
100
q = 100
(1)
(2)
Multiplying Eq. (2) by 2 and adding equations
gives
165q – 825 = 0
q=5
From Eq. (2),
65(5) + p – 537.5 = 0
p = 212.50
Thus the equilibrium point is (5, 212.50).
4
(100) + 3 = 7
100
Thus, the equilibrium point is (100, 7).
p=
120
ISM: Introductory Mathematical Analysis
Section 3.6
(1)
⎧246 p − 3.25q − 2460 = 0,
4. ⎨
410
p
+
3
q
−
14,
452.5
=
0.
(2)
⎩
Multiplying Eq. (1) by 3 and Eq. (2) by 3.25
gives
⎧738 p − 9.75q − 7380 = 0,
⎨
⎩1332.5 p + 9.75q − 46,970.625 = 0.
Adding gives
2070.5p – 54,350.625 = 0
54,350.625
p=
= 26.25
2070.5
From Eq. (2) in original system,
14, 452.5 − 410 p 14, 452.5 − 410(26.25)
q=
=
3
3
14, 452.5 − 10, 762.5 3690
=
=
= 1230
3
3
The equilibrium point is (1230, 26.25).
8. Equating p-values gives
q
2240
+6 =
4
q+2
(q + 24)(q + 2) = 2240(4)
5. Equating p-values:
9. Letting yTR = yTC gives 4q = 2q + 5000, or
q = 2500 units.
2q + 20 = 200 − 2q
q 2 + 26q + 48 = 8960
q 2 + 26q − 8912 = 0
q=
=
−b ± b 2 − 4ac
2a
−26 ± (26) 2 − 4(1)(−8912)
2(1)
q ≈ 82.29 or −108.29
q ≥ 0 so choose q ≈ 82.29.
82.29
+ 6 ≈ 26.57.
Then p ≈
4
The equilibrium point is (82.29, 26.57).
2
TR
p
2q 2 + 2q − 180 = 0
TC
15,000
q 2 + q − 90 = 0
(q + 10)(q – 9) = 0
Thus q = –10, 9. Since q ≥ 0, choose q = 9.
Then p = 2q + 20 = 2(9) + 20 = 38. The
equilibrium point is (9, 38).
(2500, 10,000)
q
6. Equating p-values gives
5000
(q + 10)2 = 388 − 16q − q 2
10. Letting yTR = yTC gives
2q 2 + 36q − 288 = 0
14q =
q 2 + 18q − 144 = 0
(q + 24) (q – 6) = 0
Thus q = –24, 6. Since q ≥ 0, choose q = 6. Then
40
q + 1200
3
2
q = 1200
3
q = 1800 units
p = (q + 10)2 = (6 + 10)2 = 162 = 256 . The
equilibrium point is (6, 256).
30,000
y
(1800, 25,200)
7. Equating p-values gives 20 − q = q + 10 .
Squaring both sides gives
1800 units
400 − 40q + q 2 = q + 10
TC
q 2 − 41q + 390 = 0
(q – 26)(q – 15) = 0
Thus q = 26, 15. If q = 26, then
p = 20 – q = 20 – 26 = –6. But p cannot be
negative. If q = 15, then p = 20 – q = 20 – 15 = 5.
The equilibrium point is (15, 5).
TR
0
q
1000
2000
11. Letting yTR = yTC gives
0.05q = 0.85q + 600
–0.80q = 600
q = –750, which is negative. Thus one cannot
break even at any level of production.
121
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
12. Letting yTR = yTC gives
0.25q = 0.16q + 360
0.09q = 360
q = 4000 units
3
q + 9 + 0.27
200
3
p=
q + 9.27
200
This equation can be written
–3q + 200p – 1854 = 0, and the new system
to solve is
⎧−3q + 200 p − 1854 = 0,
⎨
⎩3q + 100 p − 1800 = 0.
Adding gives
3654
300 p − 3654 = 0 ⇒ p =
= $12.18 .
300
p=
900
= 1.1q + 37.3
q+3
90(q + 3) − 900 = (1.1q + 37.3)(q + 3)
13. Letting yTR = yTC gives 90 −
90q + 270 − 900 = 1.1q 2 + 40.6q + 111.9
1.1q 2 − 49.4q + 741.9 = 0
q=
=
−b ± b 2 − 4ac
2a
49.4 ± (−49.4) 2 − 4(1.1)(741.9)
16. a.
2(1.1)
49.4 ± −824
2.2
There are no real solutions, therefore one cannot
break even at any level of production.
Letting yTR = yTC gives 7q = 6q + 800, or
q = 800 units.
6000
=
3000
14. Letting yTR = yTC gives
p
(800, 5600)
TC
TR
2
0.1q + 9q = 3q + 400
2
0
0.1q + 6q − 400 = 0
q 2 + 60q − 4000 = 0
(q + 100)(q − 40) = 0
Thus q = –100, 40. Since q ≥ 0, choose
q = 40 units.
⎧3q − 200 p + 1800 = 0,
15. ⎨
⎩3q + 100 p − 1800 = 0.
a.
(1)
(2)
17. Since profit = total revenue – total cost, then
4600 = 8.35q – (2116 + 7.20q). Solving gives
4600 = 1.15q – 2116
1.15q = 6716
6716
q=
= 5840 units
1.15
For a loss (negative profit) of $1150, we solve
−1150 = 8.35q – (2116 + 7.20q). Thus
–1150 = 1.15q – 2116
1.15q = 966
q = 840 units
To break even, we have yTR = yTC , or
8.35q = 2116 + 7.20q
1.15q = 2116
q = 1840 units
p
S
10
D
0
500
q
1000
b. The new total cost equation is
yTC = 1.05(6q + 800)
yTC = 6.3q + 840
Letting yTR = yTC gives
7q = 6.3q + 840
0.7q = 840
q = 1200 units
Subtracting Eq. (2) from Eq. (1) gives
–300p + 3600 = 0
p = $12
20
500
q
1000
b. Before the tax, the supply equation is
3q – 200p + 1800 = 0
–200p = –3q – 1800
3
p=
q+9
200
After the tax, the supply equation is
122
ISM: Introductory Mathematical Analysis
Section 3.6
18. For the supply equation we fit the points (0, 1)
and (13,500, 4.50) to a straight line. We have
2500
⎛ 1250 ⎞
q+⎜
⎟
3
⎝ 3 ⎠
1300
1,562,500
q2 +
q+
=0
3
9
Using the quadratic formula,
400q = q 2 +
7
4.50 − 1
3.5
7
,
=
= 2 =
13,500 − 0 13,500 13,500 27, 000
so the line is
7
p −1 =
(q − 0)
27, 000
27,000(p – 1) = 7q
7q – 27,000p + 27,000 = 0
For the demand equation, we fit the points
(0, 20) and (13,500, 4.50) to a straight line. We
have
m=
2
2
1300
⎛ 1300 ⎞
⎛ 1,562,500 ⎞
± ⎜
⎟ − 4(1) ⎜
⎟
3
9
⎝ 3 ⎠
⎝
⎠
q=
,
2
which is not real. Thus total cost always exceeds
total revenue; there is no break-even point.
−
22. p =
31
4.50 − 20
15.5
m=
=−
=− 2
13,500 − 0
13,500
13,500
31
, so the line is
=−
27, 000
31
p − 20 = −
(q − 0)
27, 000
27,000(p – 20) = –31q
31q + 27,000p – 540,000 = 0
19. Let q = break-even quantity. Since total revenue
is 5q, we have 5q = 200,000, which yields
q = 40,000. Let c be the variable cost per unit.
Then at the break even point,
Tot. Rev. = Tot. Cost
= Variable Cost + Fixed Cost.
Thus
200,000 = 40,000c + 40,000
160,000 = 40,000c
c = $4.
1000
q
a.
4=
1000
1000
= 250 units
gives q =
q
4
b.
2=
1000
1000
= 500 units
gives q =
q
2
c.
0.50 =
1000
1000
= 2000 units
gives q =
0.50
q
⎛ 1000 ⎞
The revenue is qp = q ⎜
⎟ = 1000 , so
⎝ q ⎠
revenue of $1000 is received regardless of price.
23. After the subsidy the supply equation is
⎡ 8
⎤
p=⎢
q + 50 ⎥ − 1.50
⎣ 100
⎦
8
q + 48.50
100
The system to consider is
8
⎧
⎪⎪ p = 100 q + 48.50,
⎨
⎪ p = − 7 q + 65.
⎪⎩
100
Equating p-values gives
8
7
q + 48.50 = −
q + 65
100
100
15
q = 16.5
100
q = 110
When q = 110, then
8
8
p=
q + 48.50 =
(110) + 48.50
100
100
= 8.8 + 48.50 = 57.30 .
Thus the original equilibrium price decreases by
$0.70.
p=
20. Let q = number of pairs sold.
Total Revenue = 2.63q
Total Cost = 0.85q + 0.96q + 0.32q + 70,500
At the break-even point,
Total Revenue = Total cost, or
2.63q = 0.85q + 0.96q + 0.32q + 70,500
Solving for q gives
2.63q = 2.13q + 70,500 or 0.5q = 70,500
q = 141,000
21. yTC = 3q + 1250 : yTR = 60 q . Letting
yTR = yTC gives
60 q = 3q + 1250
1250
3
Squaring gives
20 q = q +
123
Chapter 3: Lines, Parabolas, and Systems
24. a.
ISM: Introductory Mathematical Analysis
Profit = Total Revenue – Total Cost
= 280,000(2.00) – [110,000 + 280,000(1.75)]
= 560,000 – 600,000 = –40,000.
There is a net loss of $40,000.
1
( x − 10)
2
1
y −4 = x−5
2
1
y = x − 1 , which is slope-intercept form.
2
Clearing fractions, we have
⎛1
⎞
2 y = 2 ⎜ x − 1⎟
⎝2
⎠
2y = x – 2
x – 2y – 2 = 0, which is a general form.
5. y − 4 =
b. Let q = unit sales volume. Then
40,000 = 2.00q – [110,000 + 1.75q]
150,000 = 0.25q
q = 600,000 units
25. Equating qA -values gives
7 − pA + pB = −3 + 4 pA − 2 pB
10 = 5 pA − 3 pB
Equating qB -values gives
21 + pA − pB = −5 − 2 pA + 4 pB
26 = −3 pA + 5 pB
Now we solve
⎧10 = 5 pA − 3 pB
⎨26 = −3 p + 5 p
A
B
⎩
Adding 3 times the first equation to 5 times the
second equation gives
160 = 16 pB
pB = 10
From 5 pA − 3 pB = 10, 5 pA − 3(10) = 10 or
pA = 8.
Thus pA = 8 and pB = 10.
6. Slope of a vertical line is undefined, so slopeintercept form does not exist. An equation of the
vertical line is x = 3. General form: x – 3 = 0.
7. Slope of a horizontal line is 0. Thus
y – 4 = 0[x – (–2)]
y – 4 = 0,
so slope-intercept form is y = 4. A general form
is y – 4 = 0.
5
5
7⎞
⎛
8. –3y + 5x = 7 ⎜ or y = x − ⎟ has slope .
3
3
3
⎝
⎠
Thus the line perpendicular to it has slope −
3
and its equation is y − 2 = − ( x − 1) , or
5
3
13
y = − x + . A general form is 3x + 5y – 13 =
5
5
0.
26. $17.80; 2.6 thousand units
27. 2.4 and 11.3
Chapter 3 Review Problems
1. Solving
5
⎛
⎞
9. The line 2y + 5x = 2 ⎜ or y = − x + 1⎟ has slope
2
⎝
⎠
2
5
− , so the line perpendicular to it has slope .
5
2
Since the y-intercept is −3, the equation is
2
y = x − 3. A general form is 2x − 5y − 15 = 0.
5
k −5
= 4 gives k – 5 = 4, k = 9.
3− 2
2. The equation
3
5
4−4
= 0 is true for any real
5−k
number k ≠ 5.
3. (−2, 3) and (0, −1) lie on the line, so
−1 − 3
m=
= −2. Slope-intercept form:
0 − (−2)
8−2
6
= = 3, so an
1 − (−1) 2
equation of the line is y – 8 = 3(x – 1). If x = 3,
then
y – 8 = 3(3 − 1)
y–8=6
y = 14
Thus (3, 13) does not lie on the line.
10. The line has slope
y = mx + b ⇒ y = –2x − 1. A general form:
2x + y + 1 = 0.
4. Slope of y = 3x – 4 is m = 3, so slope of parallel
line is also m = 3. Thus
y – (–1) = 3[x – (–1)]
y + 1 = 3x + 3,
Slope-intercept form: y = 3x + 2. General form:
3x – y + 2 = 0.
124
ISM: Introductory Mathematical Analysis
Chapter 3 Review
In Problems 11–16, m1 = slope of first line, and
m2 = slope of second line.
5
1
1⎞
⎛
11. x + 4y + 2 = 0 ⎜ or y = − x − ⎟ has slope
4
2⎠
⎝
1
m1 = − and 8x – 2y – 2 = 0 (or y = 4x – 1) has
4
1
slope m2 = 4 . Since m1 = −
, the lines are
m2
perpendicular to each other.
12. y – 2 = 2(x – 1) (or y = 2x) has slope m1 = 2 , and
y
x
4
3
–2
5
18. x = –3y + 4
3y = –x + 4
1
4
y = − x+
3
3
1
m=−
3
1
3⎞
⎛
2x + 4y – 3 = 0 ⎜ or y = − x + ⎟ has slope
2
4⎠
⎝
1
1
, the lines are
m2 = − . Since m1 = −
m2
2
perpendicular.
5
y
4
3
1
11 ⎞
⎛
13. x – 3 = 2(y + 4) ⎜ or y = x − ⎟ has slope
2
2⎠
⎝
x
4
1
, and y = 4x + 2 has slope m2 = 4 . Since
2
1
m1 ≠ m2 and m1 ≠ −
, the lines are neither
m2
parallel nor perpendicular to each other.
2
4⎞
⎛
14. 2x + 7y − 4 = 0 ⎜ or y = − x + ⎟ has slope
7
7⎠
⎝
2
m1 = − , and 6x + 21y = 90
7
2
2
30 ⎞
⎛
⎜ or y = − x + ⎟ has slope m2 = − . Since
7
7
7
⎝
⎠
m1 = m2 , the lines are parallel.
m1 =
19. 4 – 3y = 0
–3y = –4
4
y=
3
m=0
5
y
4
3
x
5
15. y = 3x + 5 has slope 3, and 6x − 2y = 7
7⎞
⎛
⎜ or y = 3x − 2 ⎟ has slope 3. Since m1 = m2 , the
⎝
⎠
lines are parallel.
16. y = 7x has slope m1 = 7 , and y = 7 has slope
20. y = 2x
m=2
5
1
, the
m2
lines are neither parallel nor perpendicular.
17. 3x – 2y = 4
–2y = –3x + 4
3
y = x−2
2
3
m=
2
y
m2 = 0 . Since m1 ≠ m2 and m1 ≠ −
x
5
125
Chapter 3: Lines, Parabolas, and Systems
ISM: Introductory Mathematical Analysis
21. y = f(x) = 17 − 5x has the linear form
f(x) = ax + b, where a = −5 and b = 17.
Slope = −5; y-intercept (0, 17).
10
y
(0, 9)
y
25
x
–3
x
3
5
5
24. y = f(x) = 3x – 7 has the linear form f(x) = ax + b,
where a = 3, b = –7.
Slope = 3; y-intercept (0, –7)
22. s = g (t ) = 5 − 3t + t 2 has the quadratic form
2
y
x
g (t ) = at 2 + bt + c , where a = 1, b = –3, c = 5.
Vertex: −
5
b
−3 3
=−
=
2a
2(1) 2
2
11
⎛3⎞
⎛3⎞ ⎛3⎞
g ⎜ ⎟ = 5 − 3⎜ ⎟ + ⎜ ⎟ =
2
2
2
4
⎝ ⎠
⎝ ⎠ ⎝ ⎠
⎛ 3 11 ⎞
⇒ Vertex = ⎜ , ⎟
⎝2 4 ⎠
s-intercept: c = 5
t-intercepts: Because the parabola opens upward
(a > 0) and the vertex is above the t-axis, there is
no t-intercept.
s
–7
25. y = h(t ) = t 2 − 4t − 5 has the quadratic form
h(t ) = at 2 + bt + c , where a = 1, b = –4, and
c = –5.
b
−4
=−
=2
Vertex: −
2a
2 ⋅1
8
h(2) = 22 − 4(2) − 5 = −9
⇒ Vertex = (2, –9)
y-intercept: c = –5
t -intercepts: t 2 − 4t − 5 = (t − 5)(t + 1) = 0
t
⇒ t = 5, –1
6
y
23. y = f ( x) = 9 − x 2 has the quadratic form
–1
2
f ( x) = ax + bx + c , where a = –1, b = 0 and
c = 9.
b
0
=−
=0
Vertex: −
2a
2(−1)
–5
–9
f (0) = 9 − 02 = 9
⇒ Vertex = (0, 9)
y-intercept: c = 9
x-intercepts: 9 − x 2 = (3 − x)(3 + x) = 0 , so
x = 3, –3.
126
2
5
t
ISM: Introductory Mathematical Analysis
Chapter 3 Review
(
26. y = k(t) = −3 − 3t has the linear form
k(t) = at + b, where a = −3, b = −3.
Slope = −3, y-intercept (0, −3)
5
)
29. y = F ( x) = − x 2 + 2 x + 3 = − x 2 − 2 x − 3 has
the quadratic form F ( x ) = ax 2 + bx + c , where
a = –1, b = –2, and c = –3
b
−2
=−
= −1
Vertex: −
2a
2(−1)
y
t
F (−1) = − ⎡(−1)2 + 2(−1) + 3⎤ = −2
⎣
⎦
⇒ Vertex = (–1, –2)
y-intercept: c = –3
x-intercepts: Because the parabola opens
downward (a < 0) and the vertex is below the
x-axis, there is no x-intercept.
5
27. p = g(t) = −7t has the linear form g(t) = at + b,
where a = −7 and b = 0.
Slope = −7; p-intercept (0, 0)
2
y
x
p
–1
10
–2
–3
5
t
5
x
1
− 2 = x − 2 has the linear form
3
3
1
f(x) = ax + b, where a = , b = –2.
3
1
Slope = ; y-intercept (0, –2)
3
30. y = f ( x) =
28. y = F ( x) = (2 x − 1) 2 = 4 x 2 − 4 x + 1 has the
quadratic form F ( x ) = ax 2 + bx + c , where
a = 4, b = –4, c = 1.
b
−4 1
=−
=
Vertex: −
2a
2⋅4 2
5
y
2
⎛1⎞ ⎡ ⎛1⎞ ⎤
F ⎜ ⎟ = ⎢ 2 ⎜ ⎟ − 1⎥ = 0
⎝2⎠ ⎣ ⎝2⎠ ⎦
x
⎛1 ⎞
⇒ Vertex = ⎜ , 0 ⎟
⎝2 ⎠
y-intercept: c = 1
–2
x-intercepts: (2 x − 1)2 = 0 , so x =
y
1
1
2
1
2
⎧2 x − y = 6,
31. ⎨
⎩3x + 2 y = 5.
8
(1)
(2)
From Eq. (1), y = 2x – 6. Substituting in Eq. (2)
gives
3x + 2(2x – 6) = 5
7x – 12 = 5, 7x = 17
17
17
8
⇒ y = 2x − 6 = 2 ⋅ − 6 = − .
x=
7
7
7
17
8
Thus x =
, y=− .
7
7
x
5
127
Chapter 3: Lines, Parabolas, and Systems
⎧8 x − 4 y = 7,
32. ⎨
⎩ y = 2 x − 4.
ISM: Introductory Mathematical Analysis
(1)
1
1
⎧1
⎪⎪ 3 x − 4 y = 12 ,
36. ⎨
⎪ 4 x + 3y = 5 .
⎪⎩ 3
3
(2)
Replacing y by 2x – 4 in Eq. (1) gives
8x – 4(2x – 4) = 7
16 = 7, which is never true.
There is no solution.
(2)
Multiplying Eq. (1) by –4 gives
1
⎧ 4
⎪⎪− 3 x + y = − 3 ,
⎨
⎪ 4 x + 3y = 5 .
⎪⎩ 3
3
4
1
Adding gives 4 y = ⇒ y = . From Eq. (2),
3
3
4
⎛1⎞ 5
x + 3⎜ ⎟ =
3
⎝3⎠ 3
⎧7 x + 5 y = 5
33. ⎨
⎩6 x + 5 y = 3
Subtracting the second equation from the first
equation gives x = 2. Then 7(2) + 5y = 5, or
9
9
5y = −9, so y = − . Thus x = 2, y = − .
5
5
⎧2 x + 4 y = 8 (1)
34. ⎨
⎩ 3x + 6 y = 12 (2)
Multiplying Eq. (1) by 3 and Eq. (2) by −2 gives
⎧ 6 x + 12 y = 24
⎨−6 x − 12 y = −24.
⎩
Adding gives 0 = 0. Thus, the equations are
equivalent. From EQ. (1), x = −2y + 4. Letting
y = r gives the parametric solution x = −2r + 4,
y = r, where r is any real number.
3
⎧1
⎪⎪ 4 x − 2 y = −4,
35. ⎨
⎪ 3 x + 1 y = 8.
⎪⎩ 4
2
(1)
4
2
x=
3
3
1
x=
2
Thus x =
1
1
, y= .
3
2
⎧3x − 2 y + z = −2,
⎪
37. ⎨2 x + y + z = 1,
⎪ x + 3 y − z = 3.
⎩
(1)
(2)
(1)
(2)
(3)
Subtracting Eq. (2) from Eq. (1) and adding Eq.
(2) to Eq. (3) gives
⎧ x − 3 y = −3,
⎨
⎩3 x + 4 y = 4.
Multiplying the first equation by –3 gives
⎧−3x + 9 y = 9,
⎨
⎩3 x + 4 y = 4.
Adding the first equation to the second gives
13y = 13
y=1
From the equation x – 3y = –3, we get
x – 3(1) = –3
x=0
From 3x – 2y + z = –2, we get
3(0) – 2(1) + z = –2
z=0
Thus x = 0, y = 1, z = 0.
Multiplying Eq. (2) by 3 gives
3
⎧1
⎪⎪ 4 x − 2 y = −4,
⎨
⎪ 9 x + 3 y = 24.
⎪⎩ 4
2
Adding the first equation to the second gives
5
x = 20
2
x=8
From Eq. (1),
1
3
(8) − y = −4
4
2
3
− y = −6
2
y=4
Thus
x = 8, y = 4.
128
ISM: Introductory Mathematical Analysis
Chapter 3 Review
⎧2 x + 3 y + x = 9
⎪
3
38. ⎨
5x+2 y
⎪⎩ y + 4 = 7
simplifies to
⎧7 x + 3 y = 27 (1)
⎨5 x + 6 y = 28 (2)
⎩
x=
−5 − 65
−21 − 5 65
, y=
.
4
8
(1)
(2)
From Eq. (2), y = x + 7. Substituting in Eq. (1)
we have
18
x+7 =
x+4
(x + 7)(x + 4) = 18
x 2 + 11x + 28 = 18
x 2 + 11x + 10 = 0
(x + 1)(x + 10) = 0
Thus x = –1, –10. From y = x + 7, if x = –1, then
y = –1 + 7 = 6; if x = –10, then y = –10 + 7 = –3.
Thus the two solutions are x = –1, y = 6, and
x = –10, y = –3.
(1)
⎧ x + 2 z = −2,
41. ⎨
(2)
⎩ x + y + z = 5.
From Eq. (1) we have x = –2 – 2z. Substituting
in Eq. (2) gives –2 – 2z + y + z = 5, so y = 7 + z.
Letting z = r gives the parametric solution
x = –2 – 2r, y = 7 + r, z = r, where r is any real
number.
(1)
(2)
From Eq. (2), y = 3 − x 2 . Substituting in Eq. (1)
gives
x 2 − (3 − x 2 ) + 5 x = 2
⎧ x + y + z = 0,
⎪
42. ⎨ x − y + z = 0,
⎪ x + z = 0.
⎩
2
2 x + 5x − 5 = 0
x=
−5 + 65
−21 + 5 65
, y=
, and
4
8
18
⎧
,
⎪y =
40. ⎨
x+4
⎪ x − y + 7 = 0.
⎩
Multiplying Eq. (1) by −2 gives
⎧−14 x − 6 y = −54
⎨ 5 x + 6 y = 28
⎩
Adding the equations gives
−9 x = −26
26
x=
9
Multiplying Eq. (1) by −5 and Eq. (2) by 7 gives
⎧−35 x − 15 y = −135
⎨ 35 x + 42 y = 196
⎩
Adding the equations gives
27 y = 61
61
y=
27
26
61
, y=
.
Thus, x =
9
27
⎧⎪ x 2 − y + 5 x = 2,
39. ⎨
2
⎪⎩ x + y = 3.
x=
2
−b ± b − 4ac
2a
(1)
(2)
(3)
Subtracting Eq. (3) from both Eqs. (1) and (2)
gives
⎧ y = 0,
⎪
⎨− y = 0,
⎪ x + z = 0.
⎩
The first two equations state that y = 0, and the
third implies that x = –z. Letting z = r gives the
parametric solution x = –r, y = 0, z = r, where r
is any real number.
−5 ± 52 − 4(2)(−5)
=
2(2)
−5 ± 65
=
4
−5 + 65
, then
4
−21 + 5 65
−5 − 65
y=
; if x =
, then
8
4
Since y = 3 − x 2 , if x =
−21 − 5 65
.
8
Thus, the two solutions are
y=
129
Chapter 3: Lines, Parabolas, and Systems
⎧ x − y − z = 0,
43. ⎨
⎩2 x − 2 y + 3 z = 0.
ISM: Introductory Mathematical Analysis
(1)
47. Slope is
(2)
f(1) = 5,
4
5 = − (1) + b
3
19
b=
3
Multiplying Eq. (1) by –2 gives
⎧−2 x + 2 y + 2 z = 0,
⎨
⎩2 x − 2 y + 3 z = 0.
Adding the first equation to the second gives
⎧−2 x + 2 y + 2 z = 0,
⎨
⎩5 z = 0.
From the second equation, z = 0. Substituting in
Eq. (1) gives x – y – 0 = 0, so x = y. Letting y = r
gives the parametric solution x = r, y = r, z = 0,
where r is any real number.
⎧2 x − 5 y + 6 z = 1,
44. ⎨
⎩4 x − 10 y + 12 z = 2.
Thus f ( x) = −
4
19
x+
.
3
3
5−8
−3
=
= −1 . Thus
2 − (−1)
3
f(x) = ax + b = –x + b. Since f(2) = 5,
5 = –2 + b
b=7
Thus f(x) = –x + 7.
48. The slope of f is
(1)
(2)
49. r = pq = (200 − 2q )q = 200q − 2q 2 , which is a
quadratic function with a = –2, b = 200, c = 0.
Since a < 0, r has a maximum value when
b
200
q=−
=−
= 50 units. If q = 50, then
2a
−4
r = [200 – 2(50)](50) = $5000.
Multiplying Eq. (1) by –2 gives
⎧−4 x + 10 y − 12 z = −2,
⎨
⎩4 x − 10 y + 12 z = 2.
Adding the first equation to the second gives
⎧−4 x + 10 y − 12 z = −2,
⎨
⎩0 = 0.
Solving the first equation for x, we have
1 5
x = + y − 3 z . Letting y = r and z = s gives
2 2
1 5
the parametric solution x = + r − 3s , y = r,
2 2
z = s, where r and s are any real numbers.
50. Let p1 and p2 be the prices (in dollars) of the
two items, respectively, before the tax. At the
time the difference in prices is p1 − p2 = 3.5.
After the tax, the prices are 1.05 p1 and 1.05 p2 ,
so their difference is 1.05 p1 − 1.05 p2 , or 4.1.
This gives the system
p1 − p2 = 3.5
⎧
⎨1.05 p − 1.05 p = 4.1
1
2
⎩
45. a = 1 when b = 2; a = 5 when b = 3, so
a −a
5 −1 4
m= 2 1 =
= = 4.
b2 − b1 3 − 2 1
Thus an equation relating a and b is
a − 1 = 4(b − 2)
a − 1 = 4b − 8
a − 4b = −7
When b = 5, then a = 4b − 7 = 4(5) − 7 = 13.
46. a.
−4
4
⇒ f ( x) = ax + b = − x + b . Since
3
3
Adding −1.05 times the first equation to the
second equation gives 0 = 0.425, which indicates
that the system does not have a solution. Thus
this scenario is not possible.
⎧120 p − q − 240 = 0,
51. ⎨
⎩100 p + q − 1200 = 0.
r = 206 when T = 36; r = 122 when T = 30.
r −r
122 − 206 −84
=
= 14
Thus m = 2 1 =
T2 − T1
30 − 36
−6
r − 206 = 14(T − 36)
r = 14T − 298
Adding gives 220p − 1440 = 0, or
1440
p=
≈ 6.55.
220
b. If T = 27, then
r = 14T − 298 = 14(27) − 298 = 80.
130
ISM: Introductory Mathematical Analysis
52. a.
Mathematical Snapshot Chapter 3
R = aL + b. If L = 0, then R = 1310. Thus we
have 1310 = 0 · L + b, or b = 1310. So
R = aL + 1310. Since R = 1460 when L = 2,
1460 = a(2) + 1310
150 = 2a
a = 75
Thus R = 75L + 1310.
58. x = 3.02, y = 0.14
59. x = 0.75, y = 1.43
60. x = 2.68
Mathematical Snapshot Chapter 3
1. P1 (6000) = 39.99 + 0.45(6000 − 450)
= 2537.49
P6 (6000) = 199.99
b. If L = 1, then
R = 75(1) + 1310 = 1385 milliseconds.
c.
He loses $2537.49 − $199.99 = $2337.50 by
using P1.
Since R = 75L + 1310, the slope is 75. The
slope gives the change in R for each 1-unit
increase in L. Thus the time necessary to
travel from one level to the next level is 75
milliseconds.
2. The graph shows that P2 and P3 intersect when
the second branch of P2 crosses the first branch
of P3 . Thus
59.99 + 0.40(t − 900) = 79.99
t = 950
P2 is best for usage between 494.44 and 950
minutes.
53. yTR = 16q ; yTC = 8q + 10, 000 . Letting
yTR = yTC gives
16q = 8q + 10,000
8q = 10,000
q = 1250
If q = 1250, then yTR = 16(1250) = 20, 000 .
Thus the break-even point is (1250, 20,000) or
1250 units, $20,000.
3. The graph shows that P3 and P4 intersect when
the second branch of P3 crosses the first branch
of P4 Thus
79.99 + 0.35(t − 1350) = 99.99
t ≈ 1407.14
P3 is best for usage between 950 and
1407.14 minutes.
54. C = aF + b. The points (32, 0) and (212, 100) lie
on the graph of the function. Thus its slope is
100 − 0 100 5
5
=
= , so C = F + b . Since
212 − 32 180 9
9
5
C = 0 when F = 32, 0 = (32) + b , so
9
160
5
160
. Thus C = F −
or
b=−
9
9
9
5
C = ( F − 32) . When
9
5
5
F = 50, then C = (50 − 32) = (18) = 10 .
9
9
4. The graph shows that P4 and P5 intersect when
the second branch of P4 crosses the first branch
of P5 Thus
99.99 + 0.25(t − 2000) = 149.99
t = 2200
P4 is best for usage between 1407.14 and
2200 minutes.
5. The graph shows that P5 and P6 intersect when
the second branch of P5 crosses the first branch
of P6 Thus
149.99 + 0.25(t − 4000) = 199.99
t = 4200
P5 is best for usage between 2200 and 4200
minutes.
55. Equating L-values gives
0.0042
0.0378
= 0.0005 +
0.0183 −
p
p
0.042
0.0178 =
p
0.0178 p = 0.042
p ≈ 2.36
The equilibrium pollution level is about 2.36
tons per square kilometer.
6. P6 is best for usage of greater than
4200 minutes.
56. x = 12, y = –4
7. No; answers may vary.
57. x = 7.29, y = −0.78
131
Chapter 4
Principles in Practice 4.1
second year is (1 − r )2 = (1 − 0.15)2 = 0.72 . This
pattern will continue as shown in the table.
1. The shapes of the graphs are the same. The value
of A scales the value of any point by A.
Year
Multiplicative
Decrease
Expression
0
1
0.850
1
0.85
0.851
2
0.72
0.852
3
0.61
0.853
2. If P = the amount of money invested and
r = the annual rate at which P increases, then
after 1 year, the investment has grown from P to
P + Pr = P(1 + r). Since r = 0.10, the factor by
which P increases for the first year is
1 + r = 1 + 0.1 = 1.1. Similarly, during the
second year the investment grows from P(1 + r)
to P (1 + r ) + r[ P (1 + r )] = P (1 + r )2 . Again, since
r = 0.10, the multiplicative increase for the
second year is (1 + 0.10)2 = (1.1) 2 = 1.21. This
pattern will continue as shown in the table.
Year
Thus, the depreciation is exponential with a
base of 1 – r = 1 – 0.15 = 0.85. If we graph the
multiplicative decrease as a function of years, we
obtain the following.
Multiplicative
Increase
Expression
0
1
1.10
2
1
1.1
1.11
1
2
1.21
1.12
3
1.33
1.13
1.46
4
4
y
1 2 3 4 5
4. Let t = the time at which George’s sister began
saving, then since George is 3 years behind,
t – 3 = the time when George began saving.
1.1
Therefore, if y = 1.08t represents the
multiplicative increase in George’s sister’s
Thus, the growth of the initial investment is
exponential with a base of 1 + r = 1 + 0.1 = 1.1.
If we graph the multiplicative increase as a
function of years we obtain the following.
account y = 1.08t −3 represents the multiplicative
increase in George’s account. A graph showing
the projected increase in George’s money will
have the same shape as the graph of the
projected increase in his sister’s account, but will
be shifted 3 units to the right.
y
2
1
1 2 3 4 5
x
years
5. S = P(1 + r ) n
x
years
S = 2000(1 + 0.13)5 = 2000(1.13)5 ≈ 3684.87
The value of the investment after 5 years will be
$3684.87. The interest earned over the first 5
years is 3684.87 – 2000 = $1684.87.
3. If V = the value of the car and
r = the annual rate at which V depreciates, then
after 1 year the value of the car is
V – rV = V(1 – r). Since r = 0.15, the factor by
which V decreases for the first year is
1 – r = 1 – 0.15 = 0.85. Similarly, after the
second year the value of the car is
6. Let N(t) = the number of employees at time t,
where t is in years. Then,
N (4) = 5(1 + 1.2) 4 = 5(2.2) 4 = 117.128
Thus, there will be 117 employees at the end of
4 years.
V (1 − r ) − r[V (1 − r )] = V (1 − r ) 2 . Again, since
r = 0.15, the multiplicative decrease for the
132
ISM: Introductory Mathematical Analysis
Section 4.1
3.
0.06t
⎛1⎞
7. P = e−0.06t = ⎜ ⎟
⎝e⎠
1
Since 0 < < 1 , the graph is that of an
e
exponential function falling from left to right.
x
y
0
1
2
0.89
4
0.79
6
0.70
8
0.62
10
0.55
8
y
x
5
4.
y
60
x
10
P
5.
y
8
1
10
20
t
years
x
5
Problems 4.1
1.
8
6.
y
8
x
x
5
5
2.
8
y
7.
y
y
9
x
x
5
5
133
Chapter 4: Exponential and Logarithmic Functions
8.
8
ISM: Introductory Mathematical Analysis
y
14. y = 0.4 x has base b = 0.4 and 0 < b < 1, so its
graph falls from left to right. Thus the graph is
A.
15. For 2015 we have t = 20, so
20
P = 125, 000(1.11) 20 = 125, 000(1.11)1
= 138, 750 .
x
5
9.
8
16. a.
y
For 1999, t = 1 and
P = 1,527, 000(1.015)1 = 1,549,905
b. For 2000, t = 2 and
P = 1,527, 000(1.015) 2 ≈ 1,573,154
x
5
10.
8
1
1⎛1⎞
17. With c = , P = 1 − ⎜ ⎟
2⎝2⎠
2
y
n −1
1
1 1
⎛1⎞
n = 1: P = 1 − ⎜ ⎟ = 1 − =
2
2 2
⎝ ⎠
2
1 3
⎛1⎞
n = 2: P = 1 − ⎜ ⎟ = 1 − =
4 4
⎝2⎠
3
x
1 7
⎛1⎞
n = 3: P = 1 − ⎜ ⎟ = 1 − =
8 8
⎝2⎠
5
11.
8
( )
y
18. y = 23 x = 23
19. a.
5
20. a.
8
= 8 x . Thus y = 8 x .
4000(1.06)7 ≈ $6014.52
b. 6014.52 – 4000 = $2014.52
x
12.
x
y
5000(1.05)20 ≈ $13, 266.49
b. 13,266.49 – 5000 = $8266.49
21. a.
700(1.035)30 ≈ $1964.76
b. 1964.76 – 700 = $1264.76
x
10
22. a.
4000(1.0375)24 ≈ $9677.75
b. 9677.75 – 4000 = $5677.75
13. For the curves, the bases involved are 0.4, 2, and
5. For base 5, the curve rises from left to right,
and in the first quadrant it rises faster than the
23. a.
curve for base 2. Thus the graph of y = 5 x is B.
⎛ 0.0875 ⎞
3000 ⎜1 +
4 ⎟⎠
⎝
64
≈ 11,983.37
b. 11,983.37 − 3000 = $8983.37
134
n
⎛1⎞
= 1− ⎜ ⎟ .
⎝2⎠
ISM: Introductory Mathematical Analysis
24. a.
⎛ 0.07 ⎞
2000 ⎜1 +
⎟
4 ⎠
⎝
Section 4.1
48
≈ $4599.20
b. 4599.20 − 2000 = $2599.20
25. a.
Hours
Bacteria
Expression
0
100,000
⎛ 9 ⎞
100, 000 ⎜ ⎟
⎝ 10 ⎠
1
90,000
⎛ 9 ⎞
100, 000 ⎜ ⎟
⎝ 10 ⎠
81,000
⎛ 9 ⎞
100, 000 ⎜ ⎟
⎝ 10 ⎠
2
2
72,900
⎛ 9 ⎞
100, 000 ⎜ ⎟
⎝ 10 ⎠
3
3
65,610
⎛ 9 ⎞
100, 000 ⎜ ⎟
⎝ 10 ⎠
4
4
1
5000(1.0075)30 ≈ $6256.36
b. 6256.36 – 5000 = $1256.36
10
26. a.
⎛ 0.11 ⎞
500 ⎜ 1 +
2 ⎟⎠
⎝
≈ $854.07
b. 854.07 – 500 = $354.07
27. a.
⎛ 0.0625 ⎞
8000 ⎜ 1 +
365 ⎟⎠
⎝
0
3(365)
≈ $9649.69
b. 9649.69 – 8000 = $1649.69
28. a.
b.
10
900(1.0225)
30. a.
≈ $1124.28
900(1.045)5 ≈ $1121.56
⎛ 0.04 ⎞
29. 6500 ⎜ 1 +
⎟
4 ⎠
⎝
t
⎛ 9 ⎞
after t hours is given by N (t ) = 100, 000 ⎜ ⎟ .
⎝ 10 ⎠
24
≈ $8253.28
33. Let P = the amount of plastic recycled and let
r = the rate at which P increases each year. Then
after the first year, the amount of plastic
recycled, increases from P to P + rP = P(1 + r),
since r = 0.3, the factor by which P increases for
the first year, is 1 + r = 1 + 0.3 = 1.3. Similarly,
during the second year, the amount of plastic
recycled increases from P(1 + r) to
P = 5000(1.03)t
N = 400(1.05)t
b. When t = 1, then N = 400(1.05)1 = 420.
c.
t
Thus, in general, the number of bacteria present
b. When t = 3, then P = 5000(1.03)3 ≈ 5464.
31. a.
⎛ 9 ⎞
100, 000 ⎜ ⎟
⎝ 10 ⎠
t
P(1 + r) + r[ P (1 + r )] = P (1 + r )2 . Again, since
r = 0.3, the multiplicative increase for the second
When t = 4, then N = 400(1.05)4 ≈ 486.
year is (1 + r ) 2 = (1 + 0.3)2 = (1.3) 2 = 1.69 . This
pattern will continue as shown in the table.
32. If N = N(t) = the number of bacteria present at
any time t, where t is in hours, and if
r = the rate at which the bacteria are reduced,
then, after the first hour, the number of bacteria
remaining is
N – rN = N(1 – r) = 100,000(1 – 0.1)
= 100,000(0.9) = 90,000.
Similarly, after the second hour, the number of
bacteria remaining is
N(1 – r) – r[N(1 – r)] = N (1 − r )
2
= 100, 000(1 − 0.1) 2 = 100, 000(0.9)2 = 81, 000
This pattern will continue as shown in the table.
135
Year
Multiplicative
Increase
Expression
0
1
1.30
1
1.3
1.31
2
1.69
1.32
3
2.20
1.33
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
Thus, the increase in recycling is exponential
with a base = 1 + r = 1 + 0.3 = 1.3. If we graph
the multiplicative increase as function of years,
we obtaining the following.
42.
5
y
x
5
y
3
2
1
1 2 3 4 5
x
years
43. For x = 3, P =
From the graph it appears that recycling will
triple after about 4 years.
44. f(0) ≈ 0.399; f(–1) = f(1) ≈ 0.242
( )
34. Population of city A after 5 years:
45. e kt = e k
5
70, 000(1.04) .
Population of city B after 5 years:
60, 000(1.05)5 .
Difference in populations:
46.
70, 000(1.05)5 − 60, 000(1.05)5 ≈ 8589 .
t
= bt , where b = e k
x
1
⎛1⎞
= ⎜ ⎟ = b x , where b =
x
e
⎝e⎠
e
1
47. a.
35. P = 350, 000(1 − 0.015)t = 350, 000(0.985)t ,
where P is the population after t years.
When t = 0, N = 12e −0.031(0) = 12 ⋅1 = 12.
b. When t = 10,
N = 12e −0.031(10) = 12e−0.31 = 8.8.
When t = 3, P = 350, 000(0.985)3 ≈ 334, 485.
c.
When t = 44,
N = 12e −0.031(44) = 12e−1.364 ≈ 3.1.
36. E = 14, 000(1 − 0.03)t = 14, 000(0.97)t , where E
is the enrollment after t years. When t = 12,
1
of the
4
initial amount remains. Because
1 ⎛ 1 ⎞⎛ 1 ⎞
=
, 44 hours corresponds to 2
4 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
half-lives. Thus the half-life is
approximately 22 hours.
d. After 44 hours, approximately
E = 14, 000(0.97)12 ≈ 9714.
37. 4.4817
38. 29.9641
39. 0.4966
40. 0.5134
41.
e−3 33
≈ 0.2240
3!
48. N = 75e −0.045(10) ≈ 48
y
2
49. After one half-life,
x
5
1
gram remains. After two
2
2
1 1 ⎛1⎞
1
⋅ = ⎜ ⎟ = gram remains.
2 2 ⎝2⎠
4
Continuing in this manner, after n half-lives,
half-lives,
n
4
1 ⎛1⎞
⎛1⎞
⎜ 2 ⎟ gram remains. Because 16 = ⎜ 2 ⎟ , after
⎝ ⎠
⎝ ⎠
1
4 half-lives,
gram remains. This corresponds
16
to 4 · 8 = 32 years.
136
ISM: Introductory Mathematical Analysis
50.
51.
f ( x) =
e −0.5 (0.5) x
x!
f (2) =
e−0.5 (0.5) 2
≈ 0.0758
2!
f ( x) =
e −4 4 x
x!
f (2) =
e −4 42
≈ 0.1465
2!
Section 4.2
58. a.
When p = 10, then
q = 10, 000(0.95123)10 ≈ 6065 .
b. Using a graphics calculator, 0.95123 = e − x
when x ≈ 0.05. Thus, 0.95123 ≈ e −0.05 .
(
)
p
q = 10, 000(0.95123) p ≈ 10, 000 e−0.05 .
= 10, 000e−0.05 p
q = 10, 000e −0.05(10) ≈ 6065 .
c.
5
52.
59. The first integer t for which the graph of
P = 2500(1.043)t lies on or above the horizontal
line P = 5000 is 17.
2
–2
Principles in Practice 4.2
–5
The intersection point is (0, 1).
53.
1. If 16 = 2t is the exponential form then
t = log 2 16 is the logarithmic form, where t
represents the number of times the bacteria have
doubled.
8
–5
⎛ I ⎞
2. If 8.3 = log10 ⎜ ⎟ is the logarithmic form, then
⎝ I0 ⎠
I
= 108.3 is the exponential form.
I0
5
–2
If f ( x) = 2 x , then
3. Let R = the amount of material recycled every
year. If the amount being recycled increases by
50% every year, then the amount recycled at the
end of y years is
y = 2a ⋅ 2 x = 2 x + a = f ( x + a). Thus, the graph
of y = 2a ⋅ 2 x is the graph of y = 2 x shifted
a units to the left.
54. 0.71
R (1 + r ) y = R (1 + 0.5) y = R (1.5) y Thus, the
multiplicative increase in recycling at the end of
55. 3.17
y years is (1.5) y . If we let
56. The first integer t for which the graph of
x = the multiplicative increase, then x = (1.5) y
and, in logarithmic form, log1.5 x = y .
P = 1000(1.07)t lies on or above the horizontal
line P = 3000 is 17.
⎛4⎞
57. 300 ⎜ ⎟
⎝3⎠
y
6
4.1
≈ 976
y = log1.5x
3
4.2
⎛4⎞
300 ⎜ ⎟ ≈ 1004
⎝3⎠
4.2 minutes
5
137
10
x
multiplicative
increase
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
4. Let V = the value of the boat. If the value
depreciates by 20% every year, then at the end of
y years the value of the boat is
4. log8 4 =
V (1 − r ) y = V (1.02) y = V (0.8) y . Thus, the
multiplicative decrease in value at the end of y
2
3
5. ln 20.0855 = 3
years is (0.8) y . If we let
6. ln 1.4 = 0.33647
x = the multiplicative decrease, then x = (0.8) y
and, in logarithmic form, log 0.8 x = y
7. e1.09861 = 3
8. 100.6990 = 5
y
9.
8
4
5
y
y = log0.8x
x
1
x
multiplicative
decrease
5. The equation t (r ) =
5
ln 4
can be rewritten as
r
10.
ln 4
. When this equation is graphed we find
t (r )
that the annual rate r needed to quadruple the
investment in 10 years is approximately 13.9%.
Alternatively, we can solve for r by setting
t(r) = 10.
ln(4)
r=
t (r )
r=
r=
5
y
5x
11.
ln(4)
≈ 0.139 or ≈ 13.9%
10
6. Since m = e rt , then ln m = rt.
ln m = rt
ln m
=r
t
Let m = 3 and t = 12.
ln 3
=r
12
0.092 = r
Thus, to triple your investment in 12 years,
invest at an annual percentage rate of 9.2%.
5
y
x
5
12.
5
y
5x
Problems 4.2
1. log 10,000 = 4
2. (12)2 = 144
3. 26 = 64
138
ISM: Introductory Mathematical Analysis
13.
5
Section 4.2
23. Because 10−2 = 0.01, log 0.01 = −2
y
1
24. Because 21/ 3 = 3 2, log 2 3 2 = .
3
x
8
25. Because 50 = 1, log5 1 = 0
14.
5
26. Because 5−2 =
y
1
1
27. Because 2−3 = , log 2 = −3
8
8
x
5
15.
5
1
1
= −2
, log5
25
25
1
28. Because 41/ 5 = 5 4, log 4 5 4 = .
5
29. 34 = x
x = 81
y
30. 28 = x
x = 256
x
5
31. 53 = x
x = 125
16.
5
32. 40 = x
x=1
y
33. 10−1 = x
1
x=
10
x
5
34. e1 = x
x=e
17. Because 62 = 36 , log 6 36 = 2
35. e −3 = x
18. Because 26 = 64, log 2 64 = 6.
36. x 2 = 25
Since x > 0, we choose x = 5.
19. Because 33 = 27, log3 27 = 3
37. x3 = 8
x=2
1
20. Because 161/ 2 = 4, log16 4 =
2
21. Because 71 = 7, log 7 7 = 1
38. x1/ 2 = 3
x=9
22. Because 104 = 10, 000, log10, 000 = 4
39. x −1 =
x=6
139
1
6
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
40. y = x1
x=y
49. e3 x = 2
3x = ln 2
ln 2
x=
3
41. 3−3 = x
1
x=
27
50. 0.1e0.1x = 0.5
e0.1x = 5
0.1x = ln 5
x = 10 ln 5
1
42. x = 2 x − 3
x=3
43. x 2 = 6 − x
51. e2 x −5 + 1 = 4
2
e 2 x −5 = 3
2x – 5 = ln 3
5 + ln 3
x=
2
x + x−6 = 0
(x + 3)(x – 2) = 0
The roots of this equation are –3 and 2. But since
x > 0, we choose x = 2.
44. log8 64 = x − 1
52. 6e2 x − 1 =
8 x −1 = 64
x–1=2
x=3
6e 2 x =
45. 2 + log 2 4 = 3x − 1
2 + 2 = 3x – 1
5 = 3x
5
x=
3
e2 x =
1
2
3
2
1
4
1
4
1 1
x = ln
2 4
2 x = ln
46. 3−2 = x + 2
1
= x+2
9
17
x=−
9
53. 1.60944
54. 1.45161
55. 2.00013
56. 2.30058
47. x 2 = 2 x + 8
57. If V = the value of the antique. If the value
appreciates by 10% every year, then at the end of
y years the value of the antique is
x2 − 2 x − 8 = 0
(x – 4)(x + 2) = 0
The roots of this equation are 4 and –2. But since
x > 0, we choose x = 4.
V (1 + r ) y = V (1 + 0.10) y = V (1.10) y . Thus, the
multiplicative increase in value at the end of
48. x 2 = 6 + 4 x − x 2
y years is (1.10) y . If we let
2
2x − 4x − 6 = 0
x = the multiplicative increase, then x = (1.10) y ,
and, in logarithm form, log1.10 x = y .
x2 − 2x − 3 = 0
( x − 3)( x + 1) = 0
The roots of the equation are 3 and −1. But since
x > 0, we choose x = 3.
140
ISM: Introductory Mathematical Analysis
Section 4.2
Years
y
8
7
6
5
4
3
2
1
63. T =
ln 2
≈ 36.1 minutes
0.01920
64. T =
ln 2
≈ 21.7 years
0.03194
x
0
1
2
Multiplicative increase
65. From log y x = 3 , y 3 = x ; from log z x = 2 ,
3
z 2 = x . Thus z 2 = y 3 or z = y 2 .
58. c = 3(6) ln 6 + 12 ≈ 44.25
66. x + 3e2 y − 8 = 0
1980 ⎤
⎡
59. p = log ⎢10 +
= log[10 + 990] = log1000
2 ⎥⎦
⎣
=3
⎛
⎞
E
60. 1.5M = log ⎜
11 ⎟
⎝ 2.5 × 10 ⎠
E
101.5M =
2.5 × 1011
(
)(
E = 2.5 × 1011 101.5M
3e2 y = 8 − x
8− x
e2 y =
3
⎡8 − x ⎤
ln[e2 y ] = ln ⎢
⎥
⎣ 3 ⎦
⎡8 − x ⎤
2 y = ln ⎢
⎥
⎣ 3 ⎦
1 ⎡8 − x ⎤
y = ln ⎢
2 ⎣ 3 ⎥⎦
)
E = 2.5 × 1011+1.5 M
61. a.
( ) = 2N
1
If t = k, then N = N 0 2
67.
b. From part (a), N = 2 N 0 when t = k. Thus k
is the time it takes for the population to
double.
c.
3
0
–1
4
–2
t
N1 = N 0 2 k
a.
t
N1
= 2k
N0
b. [−0.37, ∞)
N
t
= log 2 1
k
N0
t = k log 2
68.
–1
x22
2
x1 = e
4
–1
x2
u0 − 2 = A ln ( x1 )
2
ln ( x1 ) =
4
N1
N0
62. u0 = A ln ( x1 ) +
u0 −
(0, 1)
(1, 0)
x22
2
A
( )
u0 − x 2 2
2
A
141
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
R1 − R2 = log(900, 000) − log 9000
69. For y = e x , if y = 3, then 3 = e x or x = ln 3.
10
= log
–10
=2
Thus, the two earthquakes differ by 2 on the
Richter scale.
10
2. The magnitude (Richter Scale) of an earthquake
⎛ I ⎞
is given by R = log ⎜ ⎟ where I is the intensity
⎝ I0 ⎠
–10
From the graph of y = e x , when y = 3, then
x = ln 3 ≈ 1.10.
of the earthquake and I 0 is the intensity of a
I
= how
I0
many times greater the earthquake is than a zeroI
= 10, 000 , then
level earthquake. Thus, if
I0
70. For y = ln x, when y = 2, then 2 = ln x or x = e2 .
zero-level reference earthquake.
5
0
900, 000
= log 100 = log102 = 2 log 10
9000
10
R = log 10,000 = log104 = 4 log 10 = 4
The earthquake measures 4 on the Richter scale.
–5
From the graph of y = ln x, when y = 2, then
Problems 4.3
2
x = e ≈ 7.39 .
71.
1. log 30 = log(2 ⋅ 3 ⋅ 5)
= log 2 + log 3 + log 5
= a+b+c
4
2. log16 = log 24 = 4 log 2 = 4a
0
5
3. log
2
= log 2 − log 3 = a − b
3
4. log
5
= log 5 − log 2 = c − a
2
–1
1.41, 3.06
Principles in Practice 4.3
8
= log 8 − log 3 = log 23 − log 3
3
= 3 log 2 – log 3 = 3a – b
1. The magnitude (Richter Scale) of an earthquake
⎛ I ⎞
is given by R = log ⎜ ⎟ where I is the intensity
⎝ I0 ⎠
5. log
of the earthquake and I 0 is the intensity of a
6. log
I
= how
I0
many times greater the earthquake is than a zeroI
= 900, 000,
level earthquake. Thus, when
I0
7. log 36 = log(2 ⋅ 3)2 = 2 log(2 ⋅ 3)
= 2(log 2 + log 3) = 2(a + b)
zero-level reference earthquake.
R1 = log(900, 000)
When
I
= 9000
I0
R2 = log(9000)
142
6
2⋅3
= log
25
52
= log 2 + log 3 − 2 log 5
= a + b − 2c
ISM: Introductory Mathematical Analysis
Section 4.3
8. log 0.00003 = log(3 ⋅10−5 )
24. ln[ x( x + 1)]3 = 3ln[ x( x + 1)] = 3[ln x + ln( x + 1)]
= log 3 + log10−5
= log 3 − 5log10
= log 3 − 5log(2 ⋅ 5)
= log 3 − 5(log 2 + log 5)
= b − 5(a + c)
= −5a + b − 5c
9. log 2 3 =
log10 3 log 3 b
=
=
log10 2 log 2 a
10. log3 5 =
log10 5 log 5 c
=
=
log10 3 log 3 b
11. log 7 7
48
4
x +1
⎛ x +1 ⎞
25. ln ⎜
= 4[ln( x + 1) − ln( x + 2)]
⎟ = 4 ln
x
+
2
x
+2
⎝
⎠
26. ln x( x + 1)( x + 2) = ln[ x( x + 1)( x + 2)]1/ 2
1
= [ln x( x + 1)( x + 2)]
2
1
= [ln x + ln( x + 1) ln( x + 2)]
2
27. ln
= ln x − [ln( x + 1) + ln( x + 2)]
= ln x − ln( x + 1) − ln( x + 2)
= 48
( )
12. log5 5 5
5
5
15
15
⎛ 3⎞
= log5 ⎜ 5 2 ⎟ = log5 5 2 =
2
⎝ ⎠
28. ln
14. 10log 3.4 = 10log10 3.4 = 3.4
e
= − ln e2 = − log e e2 = −2
18. log3 81 = log3 34 = 4
19. log
30. ln
1
1
+ ln e3 = log10 + log e e3 = −1 + 3 = 2
10
10
21. ln ⎡ x( x + 1)2 ⎤ = ln x + ln( x + 1) 2
⎣
⎦
= ln x + 2 ln( x + 1)
2
= ln
x5
1
( x + 2)( x + 1) 5
2
1
⎡
⎤
= ln x 5 − ln ⎢ ( x + 2)( x + 1) 5 ⎥
⎣
⎦
1
2
⎡
⎤
= ln x − ⎢ ln( x + 2) + ln( x + 1) 5 ⎥
5
⎣
⎦
2
1
= ln x − ln( x + 2) − ln( x + 1)
5
5
1
x
1
= ln x 2 − ln( x + 1) = ln x − ln( x + 1)
x +1
2
x2
= ln x 2 − ln( x + 1)3
( x + 1)3
= 2 ln x − 3ln( x + 1)
23. ln
x
= ln x − [ln( x + 1) + ln( x + 2)]
( x + 1)( x + 2)
= ln x − ln( x + 1) − ln( x + 2)
1
⎡
⎤
⎡ 1
2 ⎤
⎛ x2 ⎞ 5 ⎥
x
1
⎢
31. ln ⎢
5
⎥ = ln
⎢ x + 2 ⎜⎜ x + 1 ⎟⎟ ⎥
⎢ x + 2 x +1 ⎥
⎝
⎠ ⎥
⎢⎣
⎣
⎦
⎦
20. eln π = elog e π = π
22. ln
1
= ln x 2 − ln ⎡( x + 1)2 ( x + 2)3 ⎤
⎣
⎦
( x + 1)2 ( x + 2)3
1
= ln x − ⎡ln( x + 1) 2 + ln( x + 2)3 ⎤
⎣
⎦
2
1
= ln x − [2 ln( x + 1) + 3ln( x + 2)]
2
1
= ln x − 2 ln( x + 1) − 3ln( x + 2)
2
16. ln e = log e e = 1
2
x
29. ln
15. ln e5.01 = log e e5.01 = 5.01
1
x 2 ( x + 1)
= ln ⎡ x 2 ( x + 1) ⎤ − ln( x + 2)
⎣
⎦
x+2
= ln x 2 + ln( x + 1) − ln( x + 2)
= 2 ln x + ln( x + 1) − ln( x + 2)
13. log 0.0000001 = log10−7 = −7
17. ln
x
= ln x − ln[( x + 1)( x + 2)]
( x + 1)( x + 2)
143
Chapter 4: Exponential and Logarithmic Functions
32. ln 3
x3 ( x + 2)2
ISM: Introductory Mathematical Analysis
⎡
⎤
42. log 2 ⎢ ln ⎛⎜ 5 + e2 + 5 ⎞⎟ + ln ⎛⎜ 5 + e2 − 5 ⎞⎟ ⎥
⎝
⎠
⎝
⎠
⎣
⎦
⎡ ⎛
⎤
⎞
⎛
⎞
2
2
= log 2 ⎢ ln ⎜ 5 + e + 5 ⎟ ⎜ 5 + e − 5 ⎟ ⎥
⎠⎝
⎠⎦
⎣ ⎝
2
= log 2 [ln(5 + e − 5)]
1 x3 ( x + 2) 2
= ln
3
( x + 1)3
1
= ln[ x3 ( x + 2)2 ] − ln( x + 1)3
3
1
= [ln x3 + ln( x + 2)2 − ln( x + 1)3 ]
3
1
= [3ln x + 2 ln( x + 2) − 3ln( x + 1)]
3
2
= ln x + ln( x + 2) − ln( x + 1)
3
( x + 1)3
{
}
= log 2 [ln e2 ]
= log 2 (2)
=1
43. log 6 54 − log 6 9 = log 6
33. log (6 · 4) = log 24
44. log3 3 + log 2 3 2 − log5 4 5
⎛ 10 ⎞
34. log3 ⎜ ⎟ = log3 2
⎝ 5⎠
35. log 2
= log3 31/ 2 + log 2 21/ 3 − log5 51/ 4
1 1 1
= + −
2 3 4
7
=
12
2x
x +1
36. log x 2 − log x − 2 = log
x2
45. eln(2 x ) = 5
2x = 5
5
x=
2
x−2
37. 5log 2 10 + 2 log 2 13 = log 2 105 + log 2 132
= log 2 (105 ⋅132 )
46. 4log 4 ( x ) + log 4 (2) = 3
38. 5(log x 2 + log y 3 − log z 2 )
⎛ x2 y3 ⎞
= 5log ⎜
⎟
⎜ z2 ⎟
⎝
⎠
⎡ ⎛ 2 3 ⎞5 ⎤
x y
= log ⎢⎜
⎟ ⎥
⎢⎜ z 2 ⎟ ⎥
⎠ ⎦⎥
⎣⎢⎝
4log 4 (2 x ) = 3
2x = 3
3
x=
2
2
47. 10log x = 4
x2 = 4
x=±2
39. log100 + log(1.05)10 = log ⎡100(1.05)10 ⎤
⎣
⎦
48. e3ln x = 8
( )
3
215 68
1
1
8
3
40.
log 215 + log 6 − log169 = log
2
2
1693
(
= log
41. e
)
eln x = 8
x3 = 8
x=2
215(6)8
49. From the change of base formula with b = 2,
m = 2x + 1, and a = e, we have
log e (2 x + 1) ln(2 x + 1)
log 2 (2 x + 1) =
=
log e 2
ln 2
1693
4 ln 3−3ln 4
=e
54
= log 6 6 = 1
9
ln 34 − ln 43
4
ln ⎛⎜ 33 ⎞⎟
4 ⎠
⎝
=e
=
34
3
4
=
81
64
144
ISM: Introductory Mathematical Analysis
Section 4.3
b. Given M1 = log ( A1 ) + 3 , let
50. From the change of base formula with b = 3,
m = x 2 + 2 x + 2 and a = e,
M = log (10 A1 ) + 3
log e ( x 2 + 2 x + 2)
log3 ( x 2 + 2 x + 2) =
log e 3
=
M = log10 + log ( A1 ) + 3
M = 1 + ⎡⎣ log ( A1 ) + 3⎤⎦
ln( x 2 + 2 x + 2)
ln 3
M = 1 + M1
51. From the change of base formula with b = 3,
57. y = log 6 x =
m = x 2 + 1 , and a = e, we have
(
)
log3 x 2 + 1 =
(
) = ln ( x + 1) .
log e x 2 + 1
2
log e 3
ln 3
ln x
ln 6
2
0
10
52. From the change of base formula with b = 5,
m = 9 − x 2 , and a = e, we have
(
log5 9 − x
2
)=
(
log e 9 − x 2
log3 5
) = ln (9 − x )
–2
2
ln 5
58. y = log 4 ( x + 2) =
53. eln z = 7e y
ln( x + 2)
ln 4
4
z = 7e y
z
= ey
7
z
y = ln
7
–3
10
–4
54. y = ab x so
ln x
.
ln10
ln x
Thus the graphs of y = log x and y =
are
ln10
identical.
59. By the change of base formula, log x =
x
log y = log(ab )
= log a + log b x
= log a + x log b.
This is a linear expression because it is in the
form Ax + B, where A = log b and B = log a.
60.
4
55. C = B + E
⎛ E⎞
C = B ⎜1 + ⎟
B⎠
⎝
⎡ ⎛ E ⎞⎤
ln C = ln ⎢ B ⎜ 1 + ⎟ ⎥
B ⎠⎦
⎣ ⎝
0
–1
⎛ E⎞
ln C = ln B + ln ⎜ 1 + ⎟
B⎠
⎝
y = ln(4x) = ln 4 + ln x. If f(x) = ln x, then
y = ln(4x) = f(x) + ln 4. Thus the graph of
y = ln(4x) is the graph of y = ln x shifted
ln 4 units upward.
56. M = log(A) + 3
a.
5
M = log(10) + 3 = 1 + 3 = 4
145
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
8
61.
–2
3. The magnitude (Richter Scale) of an earthquake
⎛ I ⎞
is given by R = log ⎜ ⎟ where I is the intensity
⎝ I0 ⎠
of the earthquake and I 0 is the intensity of a
8
I
= how
I0
many times greater the earthquake is than a zerolevel earthquake.
R1 = log(675, 000)
zero-level reference earthquake.
–2
ln(6x) = ln(3 ⋅ 2x) = ln 3 + ln(2x).
If f(x) = ln(2x), then y = ln(6x) = f(x) + ln 3.
Thus, the graph of y = ln(6x) is the graph of
y = ln(2x) shifted ln 3 units upward.
⎛ I ⎞
R2 = log ⎜ ⎟
⎝ I0 ⎠
Since R1 − 4 = R2
Principles in Practice 4.4
1. Let x = the number and let
y = the unknown exponent. Then
⎛ I ⎞
log(675, 000) − 4 = log ⎜ ⎟
⎝ I0 ⎠
⎛ I ⎞
log 6.75 × 105 − 4 = log ⎜ ⎟
⎝ I0 ⎠
⎛ I ⎞
log 6.75 + 5log10 − 4 = log ⎜ ⎟
⎝ I0 ⎠
⎛ I ⎞
1.829 = log ⎜ ⎟
⎝ I0 ⎠
I
101.829 =
I0
x ⋅ 32 y = x ⋅ 4(3 y −9)
32 y = 4(3 y −9)
(
log 32 y = log 4(3 y −9)
y log 32 = (3y – 9) log 4
y log 32 = 3y log 4 – 9 log 4
y(log 32 – 3 log 4) = –9 log 4
−9 log 4 −18log 2 −18log 2
y=
=
=
− log 2
log 1
log 323
2
4
y = 18
Thus, Greg used 32 to the power of 18.
2. Let S = 450.
⎛4⎞
S = 800 ⎜ ⎟
⎝3⎠
450 ⎛ 4 ⎞
=
800 ⎜⎝ 3 ⎟⎠
log
I
I0
Thus, the other earthquake is 67.5 times as
intense as a zero-level earthquake.
67.5 =
−0.1d
⎛4⎞
450 = 800 ⎜ ⎟
⎝3⎠
−0.1d
Problems 4.4
−0.1d
1. log(3x + 2) = log(2 x + 5)
3x + 2 = 2 x + 5
x=3
450
⎛4⎞
= −0.1d log ⎜ ⎟
800
⎝3⎠
450
log 800
( )
−0.1log 43
)
2. log x − log 5 = log 7
log x = log 5 + log 7
log x = log 35
x = 35
=d
20 = d
Thus, he should start the new campaign 20 days
after the last one ends.
146
ISM: Introductory Mathematical Analysis
Section 4.4
3. log 7 – log(x – 1) = log 4
7
log
= log 4
x −1
7
=4
x −1
7 = 4x – 4
4x = 11
11
x=
= 2.75
4
4. log 2 x + log 2 23 = log 2
log 2 (8 x) = log 2
8x =
8. (e3 x − 2 )3 = e3
e3(3 x − 2) = e3
3(3 x − 2) = 3
3x − 2 = 1
3x = 3
x =1
9. (81) 4 x = 9
(34 ) 4 x = 32
316 x = 32
16 x = 2
2 1
x=
= = 0.125
16 8
2
x
2
x
2
x
10. (27) 2 x +1 = 3−1
(3 )
3
8x2 = 2
1
x2 =
4
1
x = = 0.5 since x > 0
2
(
5. ln(− x) = ln x 2 − 6
= 3−1
36 x +3 = 3−1
6x + 3 = –1
6x = –4
2
x = − ≈ −0.667
3
)
11. e 2 x = 9
− x = x2 − 6
(e x ) 2 = 32
x2 + x − 6 = 0
(x + 3)(x – 2) = 0
x = –3 or x = 2
However, x = –3 is the only value that satisfies
the original equation.
x = −3
ex = 3
x = ln 3 ≈ 1.099
12. e4 x =
6. ln(4 – x) + ln 2 = 2 ln x
ln[(4 − x)2] = ln x
2 x+1
3
4
4 x = ln
2
(4 − x)2 = x 2
x=
x2 + 2 x − 8 = 0
(x + 4)(x – 2) = 0
x = –4 or x = 2
However, x = 2 is the only value that satisfies the
original equation.
x=2
ln
3
4
( 34 ) ≈ −0.072
4
13. 2e5 x + 2 = 17
17
e5 x + 2 =
2
⎛ 17 ⎞
5 x + 2 = ln ⎜ ⎟
⎝ 2 ⎠
⎛ 17 ⎞
5 x = ln ⎜ ⎟ − 2
⎝ 2 ⎠
1 ⎡ ⎛ 17 ⎞ ⎤
x = ⎢ ln ⎜ ⎟ − 2 ⎥ ≈ 0.028
5⎣ ⎝ 2 ⎠ ⎦
7. e 2 x e5 x = e14
e7 x = e14
7x = 14
x=2
147
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
14. 5e2 x −1 − 2 = 23
19. 2 x = 5
5e2 x −1 = 25
ln 2 x = ln 5
x ln 2 = ln 5
ln 5
x=
≈ 2.322
ln 2
e2 x −1 = 5
2 x − 1 = ln 5
1 + ln 5
x=
≈ 1.305
2
ln(7 2 x +3 ) = ln 9
(2 x + 3) ln 7 = ln 9
ln 9
2x + 3 =
ln 7
ln 9
2x =
−3
ln 7
1 ⎛ ln 9
⎞
− 3 ⎟ ≈ −0.935
x= ⎜
2 ⎝ ln 7
⎠
4
15. 10 x = 6
4
= log 6
x
4
x=
≈ 5.140
log 6
16.
4(10)0.2 x
=3
5
15
(10)0.2 x =
4
15
0.2 x = log
4
x=
17.
( ) ≈ 2.870
=7
5
7
102 x =
2 x = log
x=
ln 73 x − 2 = ln 5
(3x − 2) ln 7 = ln 5
ln 5
3x − 2 =
ln 7
ln 5
+2
3x =
ln 7
ln 5 + 2
x = ln 7
≈ 0.942
3
0.2
102 x
log
73 x − 2 = 5
21.
log 15
4
5
72 x +3 = 9
20.
x
22. 4 2 = 20
x
5
7
ln 4 2 = ln 20
x
ln 4 = ln 20
2
x ln 20
=
2 ln 4
2 ln 20
x=
≈ 4.322
ln 4
( 75 ) ≈ −0.073
2
18. 2(10) x + (10) x +1 = 4
2(10) x + 10(10) x = 4
12(10) x = 4
23. 2
1
3
1
x = log ≈ −0.477
3
(10) x =
− 23x
ln 2
−
=
− 23x
4
5
= ln
4
5
2x
4
ln 2 = ln
3
5
( )
4
2 x ln 5
−
=
3
ln 2
x=−
148
3ln
( 54 ) ≈ 0.483
2 ln 2
ISM: Introductory Mathematical Analysis
(
Section 4.4
)
29. log 4 (9 x − 4) = 2
24. 5 3x − 6 = 10
42 = 9 x − 4
3x − 6 = 2
9 x = 42 + 4
3x = 8
x=
ln 3x = ln 8
x ln 3 = ln 8
ln 8
x=
≈ 1.893
ln 3
30. log 4 (2 x + 4) − 3 = log 4 3
log 4 (2 x + 4) − log 4 3 = 3
2x + 4
=3
3
2x + 4
43 =
3
25. (4)53− x − 7 = 2
53− x =
9
4
(3 − x) ln 5 = ln
3− x =
x = 3−
26.
log 4
9
4
ln 53− x = ln
ln
( 94 )
2 x + 4 = 3 ⋅ 43
9
4
x=
3 ⋅ 43 − 4 188
=
= 94
2
2
31. log(3x – 1) – log(x – 3) = 2
3x − 1
log
=2
x−3
3x − 1
102 =
x −3
100(x – 3) = 3x – 1
97x = 299
299
x=
≈ 3.082
97
ln 5
ln
42 + 4 20
=
≈ 2.222
9
9
( 94 ) ≈ 2.496
ln 5
7
= 13
3x
7
= 3x
13
⎛7⎞
ln ⎜ ⎟ = ln(3x )
⎝ 13 ⎠
⎛7⎞
ln ⎜ ⎟ = x ln 3
⎝ 13 ⎠
7
ln 13
x=
≈ −0.563
ln 3
32. log( x − 3) + log( x − 5) = 1
log[( x − 3)( x − 5)] = 1
x 2 − 8 x + 15 = 10
x2 − 8x + 5 = 0
( )
x=
27. log(x – 3) = 3
8 ± (−8)2 − 4(1)(5)
2(1)
= 4 ± 11
However, x = 4 + 11 ≈ 7.317 is the only value
that satisfies the original equation.
x ≈ 7.317
103 = x − 3
x = 103 + 3 = 1003
33.
28. log 2 ( x + 1) = 4
24 = x + 1
x = 24 − 1 = 15
log 2 (5 x + 1) = 4 − log 2 (3x − 2)
log 2 (5 x + 1) + log 2 (3x − 2) = 4
log[(5 x + 1)(3 x − 2)] = 4
(5 x + 1)(3x − 2) = 24
15 x 2 − 7 x − 2 = 16
15 x 2 − 7 x − 18 = 0
x ≈ 1.353 or x ≈ −0.887
However, x ≈ 1.353 is the only value that
satisfies the original equation.
x ≈ 1.353
149
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
34. log( x + 2)2 = 2
2 log(x + 2) = 2
log(x + 2) = 1
101 = x + 2
x=8
⎛2⎞
35. log 2 ⎜ ⎟ = 3 + log 2 x
⎝x⎠
⎛2⎞
log 2 ⎜ ⎟ − log 2 x = 3
⎝x⎠
log 2
log 2
23 =
2
x
x
2
=3
x2
2
=3
x2
1
x2 =
4
1
x=±
2
However, x =
x=
36.
1
is the only value that satisfies the original equation.
2
1
= 0.5
2
ln( x − 2) = ln(2 x − 1) + 3
ln( x − 2) − ln(2 x − 1) = 3
⎛ x−2 ⎞
ln ⎜
⎟=3
⎝ 2x −1 ⎠
x−2
= e3
2x −1
e3 (2 x − 1) = x − 2
2e3 x − e3 = x − 2
x(2e3 − 1) = −2 + e3
x=
−2 + e3
≈ 0.462
2 e3 − 1
However, this value does not satisfy the original equation. The equation has no solution.
37. log S = log 12.4 + 0.26 log A
log S = log12.4 + log A0.26
log S = log ⎡12.4 A0.26 ⎤
⎣
⎦
S = 12.4 A0.26
150
ISM: Introductory Mathematical Analysis
Section 4.4
38. log T = 1.7 + 0.2068log P − 0.1334(log P )2
log T = log 50 + 0.2068 log P – 0.1334(log P)(log P)
log T = log 50 + 0.2068 log P + [–0.1334 log P] log P
log T = log 50 + log P 0.2068 + log P[ −0.1334 log P ]
(
log T = log ⎡ (50) P 0.2068
⎣⎢
)( P
−0.1334 log P
)⎤⎦⎥
T = 50 P 0.2068−(0.1334 log P )
(logb x)2 = (logb x)(logb x) = logb ( x logb x )
39. a.
When t = 0, Q = 100e −0.035(0) = 100e0 = 100 ⋅1 = 100 .
b. If Q = 20, then 20 = 100e −0.035t . Solving for t gives
20
= e −0.035t
100
1
= e−0.035t
5
1
ln = −0.035t
5
–ln 5 = –0.035t
ln 5
t=
≈ 46
0.035
−
N
40. 100 = 225e 225
N
225 9
e 225 =
=
100 4
N
9
= ln
225
4
9
N = 225ln ≈ 182
4
41. If P = 1,500,000, then 1,500, 000 = 1, 000, 000(1.02)t . Solving for t gives
1,500, 000
= (1.02)t
1, 000, 000
1.5 = (1.02)t
ln1.5 = ln(1.02)t
ln 1.5 = t ln 1.02
ln1.5
t=
≈ 20.5
ln1.02
151
Chapter 4: Exponential and Logarithmic Functions
42. If F(0) = 0, then 0 =
ISM: Introductory Mathematical Analysis
Thus
q − pe−C ( p + q )
. Thus
q ⎡1 + eC ( p + q ) ⎤
⎣
⎦
0.8t =
q − pe−C ( p + q ) = 0
− pe
−C ( p + q )
e −C ( p + q ) =
⎛ 3 − log q ⎞
t log(0.8) = log ⎜
⎟.
⎝ log 2 ⎠
= −q
q
p
t=
q
−C ( p + q ) = ln
p
3− log q
log 2
)
log(0.8)
x
x
log y = log A + a x log b
log y − log A = a x log b
log y − log A
ax =
log b
⎛
log y − log A ⎞
log a x = log ⎜
⎟
log b
⎝
⎠
⎛ log y − log A ⎞
x log a = log ⎜
⎟
log b
⎝
⎠
2 p = 80 − q
log 2 p = log(80 − q )
p log 2 = log(80 − q)
log(80 − q)
log 2
log 20
When q = 60, then p =
≈ 4.32 .
log 2
x=
log
(
log y − log A
log b
)
log a
The previous solution was the special case y = q,
1
A = 1000, b = , a = 0.8, and x = t.
2
44. The investment doubles when A = 2P.
Thus 2 P = P (1.105)t , or 2 = (1.105)t .
Solving for t gives
(
ln 2 = ln(1.105)t
ln 2 = t ln1.105
ln 2
≈7
t=
ln1.105
46. q = 500 1 − e−0.2t
a.
)
(
)
If t = 1, then q = 500 1 − e−0.2 ≈ 91 .
(
0.8t
⎛1⎞
log q = log1000 + log ⎜ ⎟
⎝2⎠
1
log q = 3 + 0.8t log
2
(
log y = log A + log b a
43. q = 80 − 2 p
⎛1⎞
45. q = 1000 ⎜ ⎟
⎝2⎠
log
y = Ab a
1
q
C=−
ln .
p+q p
p=
log(q) − 3 3 − log q
=
− log 2
log 2
)
b. If t = 10, then q = 500 1 − e−2 ≈ 432 .
c.
0.8t
We solve the equation
(
400 = 500 1 − e−0.2t
4
= 1 − e−0.2t
5
1
e−0.2t =
5
1
−0.2t = ln = − ln 5
5
ln 5
≈8
t=
0.2
log q = 3 + 0.8t (− log 2)
log(q) − 3 = 0.8t (− log 2)
152
)
ISM: Introductory Mathematical Analysis
Chapter 4 Review
47. log 2 x = 5 − log 2 ( x + 4) is equivalent to
0 = 5 − log 2 ( x + 4) − log 2 x , or
y=
ln( x + 4) ln x
. Thus the solutions of the
0 = 5−
−
ln 2
ln 2
original equation are the zeros of the function
ln( x + 4) ln x
.
y = 5−
−
ln 2
ln 2
ln
( 4 x3+5 ) .
ln 2
8
–2
5
8
–2
0
Chapter 4 Review Problems
10
1. log3 243 = 5
2. 54 = 625
–5
From the graph of this function, the only zero is
x = 4. Thus 4 is the only solution of the original
equation.
48.
1
3. 814 = 3
4. log 100,000 = 5
20
5. ln 54.598 = 4
6. 91 = 9
0
7. Because 53 = 125 , log5 125 = 3
0
2
8. Because 42 = 16 , log 4 16 = 2
1.20
49.
9. Because 3−4 =
10
1
1
= −4
, log3
81
81
3
1
1
⎛1⎞
10. Because ⎜ ⎟ = , log 1
= 3.
4 64
64
⎝4⎠
10
–10
⎛1⎞
11. Because ⎜ ⎟
⎝3⎠
–10
−2
= 32 = 9 , log 1 9 = −2
3.33
1
12. Because 4 2 = 2 , log 4 2 =
y
50. (3)2 − 4 x = 5
(3)2 y = 4 x + 5
4x + 5
2y =
3
⎛ 4x + 5 ⎞
y
ln 2 = ln ⎜
⎟
⎝ 3 ⎠
⎛ 4x + 5 ⎞
y ln 2 = ln ⎜
⎟
⎝ 3 ⎠
ln 4 x3+5
y=
ln 2
The graph of the original equation is the graph of
(
13. 5 x = 625
x=4
1
= −4
81
1
x −4 =
81
1
1
=
4
81
x
14. log x
)
x 4 = 81
x=3
153
3
1
2
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
15. 2−5 = x
1
1
=
x=
5
32
2
25.
1
log 2 x + 2 log 2 x 2 − 3log 2 ( x + 1) − 4 log 2 ( x + 2)
2
1
= e −1
e
x = –1
= log 2
0
e = 2x + 3
1 = 2x + 3
2 x = −2
x = −1
( x + 1)3 ( x + 2) 4
= log x 4 + log y 2 − 3log zw
= log x 4 + log y 2 − log( zw)3
= log x 4 y 2 − log z 3 w3
= log
19. log 8000 = log(2 ⋅10)3 = 3log(2 ⋅10)
= 3(log 2 + log10) = 3(a + 1)
2
x2
26. 4 log x + 2 log y − 3(log z + log w)
18. Because eln( x + 4) = x + 4 ,
x+4=7
x=3
= log
− ⎡ log 2 ( x + 1)3 + log 2 ( x + 2) 4 ⎤
⎣
⎦
9
17. ln(2 x + 3) = 0
9
2
⎛ 1 ⎞
= log 2 ⎜ x 2 x 4 ⎟ − log 2 ⎡ ( x + 1)3 ( x + 2)4 ⎤
⎣
⎦
⎝
⎠
16. e x =
20. log
( )
1
= log 2 x 2 + log 2 x 2
32
1
27. ln
x4 y 2
z 3 w3
x3 y 2
= ln x3 y 2 − ln z −5
z −5
= ln x3 + ln y 2 − ln z −5
= 3ln x + 2 ln y + 5ln z
1
= log 32 − log 2 2
22
1
1
a
= 2 log 3 − log 2 = 2b − a = 2b −
2
2
2
21. 3log 7 − 2 log 5 = log 73 − log 52 = log
28. ln
73
=
52
= ln( x5 y 2 z )
23. 2 ln x + ln y − 3ln z = ln x 2 + ln y − ln z 3
x2 y
1
ln x − 2(ln y + ln z )
2
4
z3
(
(
= 4 ln x + ln y 3 − ln z 2
9⎤
= log 6 2 − ⎡ log 6 4 + log 6 3
⎣
⎦
)
( yz )2
⎡ xy 3 ⎤
xy 3
30. ln ⎢
= 4 ln
= 4 ln xy 3 − ln z 2
⎥
2
2
z
⎢⎣ z ⎥⎦
24. log 6 2 − log 6 4 − 9 log 6 3
(
1
= ln x − ln( yz )2 = ln x 2 − 2 ln( yz )
1
1
29. ln 3 xyz = ln( xyz ) 3 = ln( xyz )
3
1
= (ln x + ln y + ln z )
3
22. 5ln x + 2 ln y + ln z = ln x5 + ln y 2 + ln z
= ln x 2 y − ln z 3 = ln
x
= log 6 2 − log 6 4 ⋅ 39 = log 6
)
)
= 4(ln x + 3ln y − 2 ln z )
2
9
4⋅3
= log 6
1
39,366
⎡1
31. ln ⎢
⎣⎢ x
()
y⎤
⎥ = ln
y
z
1/ 2
1
⎛ y ⎞2
= ln ⎜ ⎟ − ln x
⎝z⎠
z ⎥⎦
x
1 y
1
= ln − ln x = (ln y − ln z ) − ln x
2 z
2
154
ISM: Introductory Mathematical Analysis
Chapter 4 Review
42.
⎡⎛ x ⎞ 2 x 3 ⎤
x5
⎛ ⎞
⎢
32. ln ⎜ ⎟ ⎜ ⎟ ⎥ = ln
= ln x5 − ln y 2 z 3
2 3
⎢⎝ y ⎠ ⎝ z ⎠ ⎥
y
z
⎣
⎦
(
5
y
y = 3x
y = log 3 x
)
x
= ln x5 − ln y 2 + ln z 3 = 5ln x − 2 ln y − 3ln z
33. log3 ( x + 5) =
5
log e ( x + 5) ln( x + 5)
=
log e 3
ln 3
43.
log10 (7 x3 + 5)
34. log 2 (7 x + 5) =
log10 2
9
3
=
35. log5 19 =
log(7 x3 + 5)
log 2
x
log 2 19 4.2479
=
≈ 1.8295
log 2 5 2.3219
5
44.
ln 5
≈ 1.1610
36. log 4 5 =
ln 4
(
)
= 2y +
1
x
2
37. ln 16 3 = ln 42 + ln 3 = 2 ln 4 +
38. log
y
5
x
1
ln 3
2
5
45. log(5 x + 1) = log(4 x + 6)
5x + 1 = 4 x + 6
x=5
x3 3 x + 1
5 2
x +2
5
= log x3 3 x + 1 − log x 2 + 2
46. log 3x + log 3 = 2
log 9 x = 2
5
= log x3 + log 3 x + 1 − log x 2 + 2
1
1
= 3log x + log( x + 1) − log( x 2 + 2)
3
5
9 x = 102
9 x = 100
100
x=
9
39. 10log x + log10 x + log10 = x + x + 1 = 2 x + 1
( )
40. log102 + log(1000) − 5 = log102 + log 103 − 5
47. 34 x = 9 x +1
( )
=2+3–5=0
41. In exponential form, y = e x
34 x = 32
2
y
+2
x +1
34 x = 32( x +1)
4x = 2(x + 1)
4x = 2x + 2
2x = 2
x=1
.
155
Chapter 4: Exponential and Logarithmic Functions
48. 43− x =
ISM: Introductory Mathematical Analysis
3x
1
16
54. 10 2 = 5
3x
= log 5
2
2
x = log 5 ≈ 0.466
3
43− x = 4−2
3 – x = –2
x=5
49. log x + log(10x) = 3
log x + log 10 + log x = 3
2 log(x) + 1 = 3
2 log(x) = 2
log x = 1
(
10 x + 4 − 3 = 3
10 x + 4 = 6
x + 4 = log 6
x = log(6) – 4 ≈ –3.222
x = 101 = 10
50.
log 2 ( x + 4) = log 2 ( x − 2) + 3
⎛ x+4⎞
log 2 ⎜
⎟=3
⎝ x−2⎠
x+4
= 23 = 8
x−2
x + 4 = 8( x − 2) = 8 x − 16
20 = 7 x
20
x=
7
56. 7e3 x −1 − 2 = 1
7e3 x −1 = 3
3
e3 x −1 =
7
3x − 1 = ln
3x = ln
51. ln(log x 3) = 2
log x 3 = e
2
x=
2
xe = 3
2
2
( x e ) −e = 3−e
2
52. log 2 x + log 4 x = 3
log 2 x
=3
log 2 4
log 2 x +
log 2 x
=3
2
3
+1
7
ln 73 + 1
3
≈ 0.051
ln 4 x +3 = ln 7
(x + 3)ln 4 = ln 7
ln 7
x+3=
ln 4
ln 7
x=
− 3 ≈ −1.596
ln 4
2
2
2
x e ⋅−e = 3− e
log 2 x +
3
7
57. 4 x +3 = 7
2
x1 = 3−e
1
x= 2
3e
)
55. 3 10 x + 4 − 3 = 9
58.
3
log 2 x = 3
2
log 2 x = 2
35 / x = 2
5
ln 3 = ln 2
x
5ln 3
x=
≈ 7.925
ln 2
59. Quarterly rate =
x = 22
x=4
6
a.
53. e3 x = 14
3x = ln 14
ln14
x=
≈ 0.880
3
0.06
= 0.015
4
1
yr = 26 quarters
2
2600(1.015) 26 ≈ $3829.04
b. 3829.04 – 2600 = $1229.04
156
ISM: Introductory Mathematical Analysis
60. Monthly rate =
5 yr = 60 mo.
⎛ 0.11 ⎞
4000 ⎜1 +
12 ⎟⎠
⎝
Chapter 4 Review
0.11
12
3
66. Because
60
for
≈ $6915.66
62. a.
a.
b.
N = 600(1.05)t
t
− 40
If t = 20, R = 10e
5 = 10e
t
− 40
,
− 20
40
1
=e
2
t
− 40
= 10e
t
1
= ln = − ln 2
40
2
t = 40 ln 2 ≈ 28.
68. Let d = depth in centimeters.
When t = 5, N = 600(1.05)5 ≈ 766.
d
63. a.
(0.9) 20 = 0.0017
t
P = 6000[1 + (−0.005)] or
P = 6000(0.995)
d
ln(0.9) 20 = ln 0.0017
t
d
ln 0.9 = ln 0.0017
20
20 ln 0.0017
d=
≈ 1210 cm
ln 0.9
b. When t = 10, then
P = 6000(0.995)10 ≈ 5707.
64. If t = 2, R = 200, 000e −0.4 ≈ 134, 064
69. Tt − Te = (Tt − Te )o e − at
If t = 3, R = 200, 000e −0.6 ≈ 109, 762
e − at =
65. N = 10e −0.41t
a.
When t = 0, then N = 10e0 = 10 ⋅1 = 10 mg
−at = ln
b. When t = 2, then N = 10e−0.82 ≈ 4.4 mg
c.
When t = 10, then N = 10e −4.1 ≈ 0.2 mg
d.
ln 2
≈ 1.7
0.41
e.
Tt − Te
(Tt − Te )o
Tt − Te
(Tt − Te )o
T −T
1
a = − ln t e
t (Tt − Te )o
1 (Tt − Te )o
a = ln
t
Tt − Te
If N = 1, then 1 = 10e −0.41t . Solving for t
gives
1
= e−0.41t
10
1
−0.41t = ln = − ln10
10
ln10
t=
≈ 5.6
0.41
157
− 12
. Thus
−
b. When t = 1, N = 600(1.05)1 = 630.
c.
1
of the initial amount to be present.
8
67. R = 10e
⎛ 1 ⎞
61. 12 ⎜ 1 % ⎟ = 14%
⎝ 6 ⎠
1 ⎛1⎞
, it will take 3 · 10 = 30 days
=
8 ⎜⎝ 2 ⎟⎠
≈6.
Chapter 4: Exponential and Logarithmic Functions
ISM: Introductory Mathematical Analysis
70. For double-declining balance depreciation, the
73.
n
10
2⎞
⎛
equation is V = C ⎜ 1 − ⎟ .
⎝ N⎠
2 ⎞
⎛
700 = 1800 ⎜ 1 − ⎟
⎝ 48 ⎠
n
700 ⎛ 46 ⎞
=⎜ ⎟
1800 ⎝ 48 ⎠
7 ⎛ 23 ⎞
=⎜ ⎟
18 ⎝ 24 ⎠
10
–10
n
–10
2.53
n
20
74.
n
⎛7⎞
⎛ 23 ⎞
ln ⎜ ⎟ = ln ⎜ ⎟
⎝ 18 ⎠
⎝ 24 ⎠
⎛7⎞
⎛ 23 ⎞
ln ⎜ ⎟ = n ln ⎜ ⎟
⎝8⎠
⎝ 24 ⎠
7
ln 18
n=
≈ 22
23
ln 24
( )
( )
–5
0.37
The value drops below $700 at about 22 months.
(
1
71.
)
2
75. y = log 2 x + 1 =
0
10
5
(
)
ln x 2 + 1
ln 2
8
–2
–2
–10
(−∞, 0.37]
72.
5
0
76. (6)5 y + x = 2
10
2− x
6
2− x
ln 5 y = ln
6
2− x
y ln 5 = ln
6
5y =
–10
10
–10
10
y=
–10
ln 2−6 x
ln 5
1
10
–10
3
–10
(–1.96, –3.17), (2.93, 1.60)
–3
158
ISM: Introductory Mathematical Analysis
77.
Mathematical Snapshot Chapter 4
2. From the text, the half-life H is given by
ln 2
ln 2
or, equivalently, k =
. If H = I,
H=
k
H
ln 2
then k =
. Thus
I
− d ⋅ lnI 2 ⋅ I ⎞
⎛
P ⎜1 − e
⎟
P 1 − e− dkI
⎠
T=
= ⎝ ln 2
kI
⋅
I
e −1
e I −1
8
–5
5
(
–2
y=
3x 3x
=
= 3x − 2 .
9 32
x
If f ( x) = 3 , then we have y = 3
x −2
−d ⎞
⎛
P ⎜ 1 − ⎡ eln 2 ⎤ ⎟ P 1 − 2− d
⎣
⎦
⎠ =
= ⎝
ln 2
2 −1
e −1
⎛
1 ⎞
= P 1 − 2− d = ⎜ 1 −
⎟P .
⎝ 2d ⎠
(
= f ( x − 2) .
x
(
3
is the graph of y = 3x
9
shifted 2 units to the right.
Thus the graph of y =
1. T =
a.
(
(
T (e
T e
kI
kI
1− e
b.
)
e kI − 1
a.
) (
)
− 1)
T (e
= P or P =
−1 = P 1− e
− dkI
− dkI
(
kI
1− e
)
)
(
)
b.
)
(
) ⎤⎥
(
(
)
4.
⎥
⎥⎦
e
− 32
(
ln 2 ⋅4
8
−1
⎞
3
⎟⎟ 100 ⎜⎛1 − 2− 2 ⎟⎞
⎠=
⎝
⎠ ≈ 156
1
2
2 −1
)
)
⎞
⎟ . Thus
⎠
5
) ⎤⎥
kI
⎡
1 ⎢ P − T e −1
d = − ln
kI ⎢
P
⎣⎢
⎡
1 ⎢
P
d = ln
⎢
kI
kI
⎢⎣ P − T e − 1
−1
R = P 1 − e− dkI . From part (a),
(
⎡ P − T ekI − 1
−dkI = ln ⎢
⎢
P
⎢⎣
)
−3
⎛
P 1 − e− dkI = 100 ⎜ 1 − 2 2
⎝
− 32 ⎞
⎛
R = 100 ⎜ 1 − 2 ⎟ ≈ 65.
⎝
⎠
P − T e kI − 1
P
e
kI
ln 2 ln 2
=
H
8
−3⋅ ln 2 ⋅4 ⎞
⎛
100 ⎜1 − e 8 ⎟
⎝
⎠
=
⎡ eln 2 ⎤ 2 − 1
⎣
⎦
Pe− dkI = P − T e kI − 1
e − dkI =
(
P 1− e
1
− dkI
T ekI − 1 = P − Pe − dkI
(
T=
− dkI
⎛
100 ⎜ 1 − ⎡eln 2 ⎤
⎜ ⎣
⎦
⎝
=
−1
)
)
3. P = 100, I = 4, d = 3, H = 8, k =
Mathematical Snapshot Chapter 4
P 1 − e − dkI
)
⎥
⎦⎥
0
⎤
⎥
⎥
⎥⎦
0
10
As d changes, some of the coefficients need to
change from P to Y1 or vice versa.
159
Chapter 5
Principles in Practice 5.1
By graphing re as a function of r, we find that,
when the nominal rate r = 0.077208 or 7.7208%,
the effective rate re = 0.08 or 8%.
1. Let P = 518 and let n = 3(365) = 1095.
S = P(1 + r ) n
0.1
1095
r ⎞
⎛
S = 518 ⎜ 1 +
⎟
⎝ 365 ⎠
By graphing S as a function of the nominal rate
r, we find that when r = 0.049, S = 600. Thus, at
the nominal rate of 4.9% compounded daily, the
initial amount of $518 will grow to $600 after
3 years.
0
1000
0.1
0
4. The respective effective rates of interest are
n
⎛ r⎞
found using the formula re = ⎜1 + ⎟ − 1 .
⎝ n⎠
Let n = 12 when r = 0.11:
12
0
⎛ 0.11 ⎞
re = ⎜1 +
− 1 ≈ 0.1157 . Hence, when the
12 ⎟⎠
⎝
nominal rate r = 11% is compounded monthly,
the effective rate is re = 11.57% . When
0.2
400
2. Let P = 520 and let r = 0.052.
S = P(1 + r ) n
4
⎛ 0.052 ⎞
S = 520 ⎜1 +
365 ⎟⎠
⎝
⎛ 0.1125 ⎞
r = 0.1125: re = ⎜1 +
− 1 ≈ 0.1173 .
4 ⎟⎠
⎝
Hence in the second case when the nominal rate
r = 11.25% is compounded quarterly, the
effective rate is re = 11.73% . This is the better
effective rate of interest. To find the better
investment, compare the compound amounts, S
at the end of n years. With P = 10,000 and
re = 0.1157 ,
n
n
⎛ 365.052 ⎞
S = 520 ⎜
⎟
⎝ 365 ⎠
By graphing S as a function of n, we find that
when n = 2571, S = 750. Thus, it will take
2571
≈ 7.044 years, or 7 years and 16 days for
365
$520 to grow to $750 at the nominal rate of
5.2% compounded daily.
S1 = P (1 + r )n = 10, 000(1 + 0.1157)n , and, in the
second case, when P = 9700 and re = 0.1173
800
S2 = P (1 + r )n = 9700(1 + 0.1173)n .
S1 (20) = 10, 000(1.1157)20 ≈ 89,319.99
0
S2 (20) = 9700(1.1173)20 ≈ 89,159.52
The $10,000 investment is slightly better over
20 years.
3000
500
Problems 5.1
3. Let n = 12.
1. a.
n
⎛ r⎞
re = ⎜1 + ⎟ − 1
⎝ n⎠
b. 11,105.58 – 6000 = $5105.58
12
r ⎞
⎛
re = ⎜1 + ⎟
⎝ 12 ⎠
6000(1.08)8 ≈ $11,105.58
−1
160
ISM: Introductory Mathematical Analysis
2. a.
Section 5.1
750(1.07) = $802.50
c.
b. 802.5 – 750 = $52.50
⎛ 0.07 ⎞
(i) 1000 ⎜1 +
52 ⎟⎠
⎝
⎛ 0.07 ⎞
(ii) ⎜1 +
52 ⎟⎠
⎝
3. (1.015) 2 − 1 ≈ 0.030225 or 3.023%
52(5)
− 1000 ≈ $418.73
52
− 1 ≈ 0.07246 or 7.246%
4
⎛ 0.05 ⎞
4
4. ⎜1 +
⎟ − 1 = (1.0125) − 1 ≈ 0.05095 or
4 ⎠
⎝
5.095%
⎛ 0.04 ⎞
5. ⎜1 +
365 ⎟⎠
⎝
365
⎛ 0.06 ⎞
6. ⎜1 +
365 ⎟⎠
⎝
365
7. a.
⎛ 0.07 ⎞
d. (i) 1000 ⎜ 1 +
365 ⎟⎠
⎝
⎛ 0.07 ⎞
(ii) ⎜1 +
365 ⎟⎠
⎝
− 1 ≈ 0.04081 or 4.081%
365(5)
− 1000 ≈ $419.02
365
− 1 ≈ 0.07250 or 7.250%
9. Let re be the effective rate. Then
2000 (1 + re ) = 2950
5
− 1 ≈ 0.06183 or 6.183%
(1 + re )5 =
A nominal rate compounded yearly is the
same as the effective rate, so the effective
rate is 10%.
1 + re = 5
2
b.
⎛ 0.10 ⎞
⎜1 + 2 ⎟ − 1 = 0.1025 or 10.25%
⎝
⎠
c.
⎛ 0.10 ⎞
⎜1 + 4 ⎟ − 1 ≈ 0.10381 or 10.381%
⎝
⎠
2950
2000
2950
2000
2950
−1
2000
re ≈ 0.0808 or 8.08%.
re = 5
4
10. Let r be the monthly rate. Then
(1 + r )84 = 1835
1835
(1 + r )84 =
1000
1835
1 + r = 84
1000
1835
−1
r = 84
1000
r = 0.0072529
This gives a nominal rate of approximately
12(0.0072529) = 0.0870 ≈ 8.70% compounded
monthly.
12
d.
⎛ 0.10 ⎞
⎜1 + 12 ⎟
⎝
⎠
e.
⎛ 0.10 ⎞
⎜1 + 365 ⎟
⎝
⎠
8. a.
− 1 ≈ 0.10471 or 10.471%
365
− 1 ≈ 0.10516 or 10.516%
⎛ 0.07 ⎞
(i) 1000 ⎜ 1 +
4 ⎟⎠
⎝
4(5)
− 1000 ≈ $414.78
4
⎛ 0.07 ⎞
(ii) ⎜1 +
− 1 ≈ 0.07186 or 7.186%
4 ⎟⎠
⎝
11. From Example 6, the number of years, n, is
ln 2
≈ 8.0 years.
given by n =
ln(1.09)
12(5)
⎛ 0.07 ⎞
b. (i) 1000 ⎜1 +
12 ⎟⎠
⎝
− 1000 ≈ $417.63
12. From Example 6, the number of years, n, is
ln 2
given by n =
≈ 14.2 years.
ln(1.05)
12
⎛ 0.07 ⎞
(ii) ⎜1 +
12 ⎟⎠
⎝
− 1 ≈ 0.07229 or 7.229%
13. 6000(1.08)7 ≈ $10, 282.95
161
Chapter 5: Mathematics of Finance
ISM: Introductory Mathematical Analysis
22. Let r be the required nominal rate.
14. 3P = P (1 + r ) n
12
r ⎞
⎛
⎜1 + 12 ⎟
⎝
⎠
3 = (1 + r )n
ln 3 = n ln(1 + r)
ln 3
n=
ln(1 + r )
10
15. 21,500(1.06)
17. a.
b.
12
r ⎞
⎛
⎜1 + 12 ⎟
⎝
⎠
1+
≈ $38,503.23
⎛ 0.02 ⎞
16. 21,500 ⎜ 1 +
4 ⎟⎠
⎝
− 1 = 0.045
40
≈ $26, 247.08
= 1.045
r 12
= 1.045
12
r 12
= 1.045 − 1
12
r = 12 ⎡12 1.045 − 1⎤ ≈ 0.0441
⎣
⎦
or 4.41%.
(0.015)(12) = 0.18 or 18%
(1.015)12 − 1 ≈ 0.1956 or 19.56%
18. 2 P = P (1.01)n
2 = (1.01)n
ln 2 = n ln(1.01)
ln 2
n=
≈ 70 months
ln(1.01)
⎛ 0.0475 ⎞
⎜1 + 360 ⎟
⎝
⎠
365
23. a.
⎛ 0.0475 ⎞
⎜1 + 365 ⎟
⎝
⎠
365
b.
− 1 ≈ 0.0493 or 4.93%
− 1 ≈ 0.0486 or 4.86%
24. Let r be the nominal rate.
8
⎛ r⎞
801.06 = 700 ⎜1 + ⎟
⎝ 4⎠
19. The compound amount after the first four years
1+
is 2000(1.06)4 . After the next four years the
compound amount is
⎡ 2000(1.06) 4 ⎤ (1.03)8 ≈ $3198.54 .
⎣
⎦
r 8 801.06
=
4
700
⎛ 801.06
⎞
r = 4⎜ 8
− 1⎟ ≈ 0.0680 or 6.80%
⎜
⎟
700
⎝
⎠
20. 700 = 500(1.02) n
25. Let re = effective rate.
300, 000 = 100, 000 (1 + re )
1.4 = (1.02)
ln(1.4) = n ln(1.02)
ln(1.4)
n=
≈ 17 quarters or 4 years, 3 months
ln(1.02)
n
10
(1 + re )10 = 3
1 + re = 10 3
re = 10 3 − 1 ≈ 0.1161 or 11.61%.
21. 7.8% compounded semiannually is equivalent to
an effective rate of (1.039)2 − 1 = 0.079521 or
7.9521%. Thus 8% compounded annually,
which is the effective rate, is the better rate.
162
ISM: Introductory Mathematical Analysis
Section 5.2
26. Let P = average price of such a good,
n = number of days.
⎛ 0.0725 ⎞
2 P = P ⎜1 +
365 ⎟⎠
⎝
⎛ 0.0725 ⎞
2 = ⎜1 +
365 ⎟⎠
⎝
⎛ 0.0875 ⎞
8. 500 ⎜ 1 +
4 ⎟⎠
⎝
n
⎛ 0.135 ⎞
10. 1250 ⎜1 +
52 ⎟⎠
⎝
⎛ 0.071 ⎞
12. 12, 000 ⎜ 1 +
2 ⎟⎠
⎝
⎛ r⎞
⎜1 + 2 ⎟
⎝
⎠
1+
28
=
= 1000
−12
≈ $11,381.89
−2
≈ $11,191.31
15. Let x be the payment 2 years from now. The
equation of value at year 2 is
⎡ 50 ⎤
r = 2 ⎢ 28
− 1⎥ ≈ 0.0629 or 6.29%
⎣⎢ 21 ⎥⎦
x = 600(1.04) −2 + 800(1.04)−4
x ≈ $1238.58
28. 1000(1 − 0.01)20 = 1000(0.99) 20 ≈ $817.91
Problems 5.2
1. 6000(1.05) −20 ≈ $2261.34
2. 3500(1.06)−8 ≈ $2195.94
3. 4000(1.035)−24 ≈ $1751.83
4. 1740(1.015) −24 = $1217.21
⎛ 0.10 ⎞
7. 8000 ⎜ 1 +
12 ⎟⎠
⎝
≈ $1021.13
14. 550(1.025) −16 + 550(1.025)−20 ≈ $706.14
r 28 50
=
2
21
−22
≈ $5821.55
⎛ 0.10 ⎞
6. 6000 ⎜ 1 +
2 ⎟⎠
⎝
≈ $6838.95
13. 27, 000(1.03) −22 ≈ $14, 091.10
1000 50
=
420 21
⎛ 0.08 ⎞
5. 9000 ⎜1 +
4 ⎟⎠
⎝
−4(365)
−78
⎛ 0.053 ⎞
11. 12, 000 ⎜ 1 +
12 ⎟⎠
⎝
27. Let r = the required nominal rate.
28
≈ $385.65
⎛ 0.095 ⎞
9. 10, 000 ⎜1 +
365 ⎟⎠
⎝
n
⎛ 0.0725 ⎞
ln 2 = n ln ⎜1 +
365 ⎟⎠
⎝
ln 2
n=
≈ 3489.98 days
⎛ 0.0725 ⎞
ln ⎜ 1 +
365 ⎟⎠
⎝
or ≈ 9.56 years
⎛ r⎞
420 ⎜ 1 + ⎟
⎝ 2⎠
−12
−13
≈ $3181.93
−60
≈ $4862.31
163
Chapter 5: Mathematics of Finance
ISM: Introductory Mathematical Analysis
16. Let x be the payment at the end of 5 years. The equation of value at year 5 is
⎛ 0.08 ⎞
3000 ⎜ 1 +
12 ⎟⎠
⎝
60
+ x = 7000
⎛ 0.08 ⎞
x = 7000 − 3000 ⎜ 1 +
12 ⎟⎠
⎝
x ≈ $2530.46
60
17. Let x be the payment at the end of 6 years. The equation of value at year 6 is
2000(1.025) 4 + 4000(1.025) 2 + x = 5000(1.025) + 5000(1.025) −4
x = 5000(1.025) + 5000(1.025)−4 − 2000(1.025) 4 − 4000(1.025) 2
x ≈ $3244.63.
18. Let x be the amount of each of the equal payments. The equation of value at year 3 is
1500(1.07)3 + x(1.07) 2 + x(1.07) + x = 3500(1.07) −1 + 5000(1.07) −3
x[(1.07)2 + 1.07 + 1] = 3500(1.07)−1 + 5000(1.07)−3 − 1500(1.07)3
x=
3500(1.07)−1 + 5000(1.07)−3 − 1500(1.07)3
(1.07)2 + 2.07
x ≈ $1715.44
19. a.
NPV = 8000(1.025) −6 + 10, 000(1.025) −8 + 14, 000(1.025)−12 − 25, 000 ≈ $515.62
b. Since NPV > 0, the investment is profitable.
20. a.
NPV = 8000(1.03) −6 + 10, 000(1.03) −8 + 14, 000(1.03) −12 − 25, 000 ≈ −$586.72
b. Since NPV < 0, the investment is not profitable.
21. We consider the value of each investment at the end of eight years. The savings account has a value of
10, 000(1.03)16 ≈ $16, 047.06.
The business investment has a value of $16,000. Thus the better choice is the savings account.
22. The payments due B are 1000(1.07)5 at year 5 and 2000(1.04)14 at year 7. Let x be the payment at the end of 6
years. The equation of value at year 6 is x = 1000(1.07)5 (1.015)4 + 2000(1.04)14 (1.015) −4 x ≈ $4751.73
⎛ 0.075 ⎞
23. 1000 ⎜1 +
4 ⎟⎠
⎝
−80
⎛ 0.058 ⎞
24. 6500 ⎜ 1 +
360 ⎟⎠
⎝
≈ $226.25
−1460
≈ $5137.67
164
ISM: Introductory Mathematical Analysis
Section 5.3
25. Let r be the nominal discount rate, compounded
quarterly. Then
⎛ r⎞
4700 = 10, 000 ⎜ 1 + ⎟
⎝ 4⎠
10, 000
4700 =
32
1 + 4r
(
⎛ r⎞
⎜1 + 4 ⎟
⎝
⎠
1+
b.
−32
14. With option (a), after 18 months they have
50, 000(1 + 0.0125)6 ≈ $53,869.16
with option (b), they have
)
32
=
P = 59, 081e − (0.045)(25) ≈ $19,181
50, 000e(0.045)(1.5) ≈ $53, 491.51 .
15. Effective rate = e r − 1 . Thus 0.05 = e r − 1 ,
10, 000 100
=
4700
47
e r = 1.05 , r = ln 1.05 ≈ 0.0488.
Answer: 4.88%
r 32 100
=
4
47
16. If r is the annual rate compounded continuously,
then at the end of 1 year the compound amount
⎡ 100 ⎤
− 1⎥ ≈ 0.0955 or 9.55%
r = 4 ⎢32
⎢⎣ 47
⎥⎦
of a principal of P dollars is Per (1) = Per . This
amount must equal the compound amount of P
dollars at a nominal rate of 6% compounded
Problems 5.3
semiannually, which is P (1.03)2 . Thus
1. S = 4000e0.0625(6) ≈ $5819.97
5819.97 − 4000 = $1819.97
Per = P (1.03) 2
e r = (1.03)2
r = ln(1.03) 2
r = 2 ln 1.03 ≈ 0.0591
Answer: 5.91%
2. S = 4000e0.09(6) ≈ $6864.03
6864.03 – 4000 = $2864.03
3. P = 2500e−0.0675(8) ≈ $1456.87
17. 3P = Pe0.07t
4. P = 2500e−0.08(8) ≈ $1318.23
3 = e0.07t
0.07t = ln 3
ln 3
≈ 16
t=
0.07
Answer: 16 years
5. e0.04 − 1 ≈ 0.0408
Answer: 4.08%
6. e0.08 − 1 ≈ 0.0833
Answer: 8.33%
18.
4 P = Per (30)
4 = e30r
30r = ln 4
ln 4
r=
≈ 0.046
30
Answer: 5%
7. e0.03 − 1 ≈ 0.0305
Answer: 3.05%
8. e0.11 − 1 = 0.1163
Answer: 11.63%
19. The accumulated amounts under each option are:
9. S = 100e0.045(2) ≈ $109.42
10. S = 1000e0.03(8) ≈ $1271.25
11. P = 1, 000, 000e −0.05(5) ≈ $778,800.78
12. P = 50, 000e −0.06(30) ≈ $8264.94
13. a.
25, 000(1 + 0.035) 25 = $59, 081
165
a.
1000e(0.035)(2) ≈ $1072.51
b.
1020(1.0175) 4 ≈ $1093.30
c.
500e(0.035)(2) + 500(1.0175) 4
≈ 536.25 + 535.93 = $1072.18
Chapter 5: Mathematics of Finance
20. a.
ISM: Introductory Mathematical Analysis
4. The amount of profit earned in the first two
years is the sum of the monthly profits.
Let a = 2000,
r = 1.1, and n = 24.
On Nov. 1, 2006 the accumulated amount is
10, 000e
≈ $14,918.25 .
On Nov. 1, 2011 the accumulated amount is
(0.04)(10)
14,918.25(1.05)5 ≈ $19, 039.89 .
b.
21. a.
s=
(
2000 1 − (1.1) 24
) ≈ 176,994.65
1 − 1.1
Thus, the company earned $176,994.65 in the
first two years.
10, 000(1.045)15 ≈ $19,352.82 , which is
$312.93 more than the amount in part (a).
9000(1.0125) 4 ≈ $9458.51
5. Let R = 500 and let n = 72. Then, the present
value A of the annuity is given by
⎛ 1 − (1 + r ) − n ⎞
⎛ 1 − (1 + r ) −72 ⎞
A = R⎜
⎟ = 500 ⎜
⎟
⎜
⎟
⎜
⎟
r
r
⎝
⎠
⎝
⎠
By graphing A as a function of r, we find that
when r ≈ 0.005167, A = 30,000. Thus, if the
present value of the annuity is $30,000, the
monthly interest rate is 0.5167%, and the
nominal rate is 12(0.005167) = 0.062 or 6.2%.
b. After one year the accumulated amount of
the investment is
10, 000e0.055 ≈ $10,565.41 . The payoff
for the loan (including interest) is
1000 + 1000(0.08) = $1080. The net return
is 10,565.41 – 1080 = $9485.41.
Thus, this strategy is better by
9485.41 – 9458.51 = $26.90.
50,000
Principles in Practice 5.4
1. Let a = 64 and let r =
3
. Then, the next five
4
2
⎛3⎞
⎛3⎞
heights of the ball are 64 ⎜ ⎟ , 64 ⎜ ⎟ ,
4
⎝ ⎠
⎝4⎠
3
4
0
10,000
5
⎛3⎞
⎛3⎞
⎛3⎞
64 ⎜ ⎟ , 64 ⎜ ⎟ , 64 ⎜ ⎟ , or 48 ft, 36 ft,
⎝4⎠
⎝4⎠
⎝4⎠
3
1
27 ft, 20 ft, and 15
ft.
16
4
6. Since the man pays $2000 for 6 years and $3500
for 8 years, we can consider the payments to be
an annuity of $3500 for 14 years minus an
annuity of $1500 for 6 years so that the first 24
payments are $2000 each. Thus, the present
value is
3500a56 0.015 − 1500a24 0.015
2. Let a = 500 and let r = 1.5. Then, the number of
bacteria at the end of each minute for the first six
minutes is 500(1.5), 500(1.5)2 , 500(1.5)3 ,
4
5
≈ 3500(37.705879) − 1500(20.030405)
= 101,924.97
Thus, the present value of the payments is
$101,925. Since the man made an initial down
payment of $20,000, list price was
101,925 + 20,000 = $121,925.
6
500(1.5) , 500(1.5) , 500(1.5) , or 750, 1125,
1688, 2531, 3797, 5695.
3. The total vertical distance traveled in the air after
n bounces is equal to 2 times the sum of heights.
2
If a = 6 and r = , then when the ball hits the
3
ground for the twelfth time, n = 12 and the
distance traveled in the air is
12 ⎞ ⎤
⎡ ⎛
2
⎡ a 1− rn ⎤
⎢ 6 ⎜1 − 3 ⎟ ⎥
⎠⎥
⎥ = 2⎢ ⎝
2s = 2 ⎢
⎢ 1− r ⎥
⎢
⎥
1 − 23
⎣⎢
⎦⎥
⎢
⎥
⎣
⎦
≈ 35.72 meters
(
)
0.05
0.048
= 0.012, and n = 24.
4
⎛ 1 − (1 + r ) − n ⎞
A = R⎜
⎟
⎜
⎟
r
⎝
⎠
⎛ 1 − (1 + 0.012) −24 ⎞
⎛ 1 − (1.012)−24 ⎞
A = R⎜
⎟ = R⎜
⎟
⎜
⎟
⎜
⎟
0.012
0.012
⎝
⎠
⎝
⎠
By Graphing A as a function of R, we find that
when R = 723.03, A = 15,000. Thus the monthly
payment is $723.03 if the present value of the
7. Let r =
( )
166
ISM: Introductory Mathematical Analysis
Section 5.4
annuity is $15,000.
Problems 5.4
20,000
1. 64
⎛1⎞
64 ⎜ ⎟ = 32
⎝2⎠
2
0
0
⎛1⎞
64 ⎜ ⎟ = 16
⎝2⎠
1000
3
⎛1⎞
64 ⎜ ⎟ = 8
⎝2⎠
8. Find the annuity due. The man makes an initial
payment of $1200 followed by an ordinary
annuity of $1200 for 11 months. Thus, let
0.068
. The present
R = 1200, n = 11, and r =
12
value of the annuity due is
⎛
⎞
1200 ⎜1 + a 0.068 ⎟ ≈ 1200(1 + 10.635005)
11
⎜
⎟
12 ⎠
⎝
≈ 13,962.01
Thus, he should pay $13,962.01.
4
⎛1⎞
64 ⎜ ⎟ = 4
⎝2⎠
2. 2
2(–3) = –6
2(−3) 2 = 18
2(−3)3 = −54
3. 100
100(1.02) = 102
9. Let R = 2000 and let r = 0.057. Then, the value
of the IRA at the end of 15 years, when n = 15,
is given by
⎛ (1 + r )n − 1 ⎞
S = R⎜
⎟
⎜
⎟
r
⎝
⎠
⎛ (1 + 0.057)15 − 1 ⎞
S = 2000 ⎜
⎟ ≈ 45,502.06
⎜
⎟
0.057
⎝
⎠
Thus, at the end of 15 years the IRA will be
worth $45,502.06.
100(1.02) 2 = 104.04
4. 81
⎛1⎞
81⎜ ⎟ = 27
⎝3⎠
2
⎛1⎞
81⎜ ⎟ = 9
⎝3⎠
3
⎛1⎞
81⎜ ⎟ = 3
⎝3⎠
10. Let R = 2000 and let r = 0.057. Since the
deposits are made at the beginning of each year,
the value of the IRA at the end of 15 years is
given by
⎛ (1 + r )n +1 − 1 ⎞
S = R⎜
⎟−R.
⎜
⎟
r
⎝
⎠
Let n = 15.
⎛ (1 + 0.057)16 − 1 ⎞
S = 2000 ⎜
⎟ − 2000 ≈ 48, 095.67
⎜
⎟
0.057
⎝
⎠
Thus, the IRA is worth $48,095.67 at the end of
15 years.
5. s =
4
7
( )
5
⎡
4 ⎤
⎢1 − 7 ⎥
⎣
⎦
1 − 74
=
4 ⎡ 15,783 ⎤
7 ⎣ 16,807 ⎦
3
7
=
21, 044
16,807
()
7⎤
⎡
1 ⎢1 − 15 ⎥
⎦=
6. s = ⎣
1 − 15
167
78,124
78,125
4
5
=
19,531
15, 625
Chapter 5: Mathematics of Finance
ISM: Introductory Mathematical Analysis
⎛
23. 1200 ⎜ s13
⎝
1 ⎡1 − (0.1)6 ⎤
⎦ = 1.11111
7. s = ⎣
1 − 0.1
8. Observe that (1.1) −1 =
( )
10 ⎡1 − 10
11 ⎢
11
⎣
1 − 10
11
9. a35
0.04
10. a15
0.07
6⎤
25. 175a
6⎤
⎥
⎡
⎦ = 10 ⎢1 − ⎛ 10 ⎞ ⎥ ≈ 4.355
⎜ ⎟
⎢⎣ ⎝ 11 ⎠ ⎥⎦
32
⎞
− 1⎟ ≈ 1200(21.495297 − 1)
⎠
≈ $24,594.36
0.025
⎞
− 1⎟ ≈ 600(46.000271 − 1)
⎠
≈ $27, 000.16
0.04
12
− 25a 0.04
8
12
≈ 175(30.304595) − 25(7.881321)
≈ $5106.27
≈ 18.664613
26. 1500 + 1500a5
0.0075
12. s11
0.0125
≈ 11.713937
0.06
14. 1000a8
27. R =
≈ 8.213180
0.0075
0.05
≈ 1500 + 1500(4.889440)
≈ $8834.16
≈ 9.107914
11. s8
13. 600a6
⎛
24. 600 ⎜ s31
⎝
10
.
11
and r =
s=
1 10
10
=
. Thus a =
1.1 11
11
0.08
5000
5000
≈
≈ $458.40
a12 0.015 10.907505
28. 3000 + 250a12
0.04
≈ 3000 + 250(9.385074)
≈ $5346.27
≈ 600(4.917324) ≈ $2950.39
29. a.
≈ 1000(6.463213) ≈ 6463.21
⎛
⎞
24
⎜ 50s48 0.005 ⎟ (1.005)
⎝
⎠
≈ 50(54.097832)(1.005)24
15. 2000a18
16. 1500a15
0.02
≈ 2000(14.992031) ≈ $29,984.06
0.0075
17. 800 + 800a11
≈ $3048.85
b. 3048.85 – 48(50) = $648.85
≈ 1500(14.136995) ≈ $21, 205.49
0.035
30. Let R be the yearly payment.
275, 000 = R + Ra9 0.035
≈ 800 + 800(9.001551)
⎛
⎞
275, 000 = R ⎜ 1 + a9 0.035 ⎟
⎝
⎠
275,000 ≈ R(8.607687),
R ≈ $31,948.19
≈ $8001.24
18. 150 + 150a
59
19. 2000s36
0.07
12
0.0125
≈ 150 + 150(49.796588)
≈ $7619.49
31. R =
≈ 2000(45.115505)
≈ $90, 231.01
20. 600s16
0.02
≈ 600(18.639285) ≈ $11,183.57
21. 5000s20
0.07
22. 2000s20
0.03
≈ 5000(40.995492) ≈ $204,977.46
≈ 2000(26.870374) ≈ 53, 740.75
168
48, 000
48, 000
≈
≈ $3474.12
s10 0.07 13.816448
ISM: Introductory Mathematical Analysis
Section 5.4
32. Let x be the purchase price. In the same manner
as in Example 12,
[50, 000 − 0.08x] s10 0.06 = x
50, 000 − 0.08 x =
50, 000 = 0.08 x +
⎛
x = (1.08)6 ⎜ 5000 − 1000a5
⎝
≈ (1.08)6 [5000 − 1000(3.992710)] ≈ $1598.44
x
s10
35. s60
0.06
=
0.017
x
s10
36. a9
0.06
⎛
⎞
1
⎟
50, 000 = x ⎜ 0.08 +
⎜⎜
s10 0.06 ⎟⎟
⎝
⎠
50, 000
50, 000
x=
≈
≈ $320,800 .
1
0.08 + s 1
0.08 + 13.180795
=
0.052
37. 750a480
25, 000
. After
s10 0.06
six years the value of the fund is
25, 000
s
.
s10 0.06 6 0.06
39. R =
This accumulates to
⎡
⎤
⎢ 25, 000 s
⎥ (1.07) 4 .
6 0.06 ⎥
⎢s
⎢⎣ 10 0.06
⎥⎦
Let x be the amount of the new payment.
⎡
⎤
25, 000
4⎥
⎢
xs4 0.07 = 25, 000 −
s6 0.06 (1.07)
⎢s
⎥
⎢⎣ 10 0.06
⎥⎦
40. R =
42. a.
b.
0.01375
3000(0.01375)
(1.01375)20 − 1
≈ $131.34
( 0.1 )
)
25, 000 12
25, 000
=
≈ $531.18
a60 0.1 1 − 1 + 0.1 −60
12
12
(
0.10
$650(12)(15) = $117,000
650a
180
(
0.055
12
⎡
0.055
⎢ 1 − 1 + 12
= 650 ⎢
0.055
⎢
12
⎣
≈ $79,551.24
)
−180
⎤
⎥
⎥
⎥
⎦
43. For the first situation, the compound amount is
⎡
⎛
⎞⎤
30
⎢ 2000 ⎜ s11 0.07 − 1⎟ ⎥ (1.07)
⎝
⎠⎦
⎣
34. Let x be the final payment.
0.08
=
⎡ 1 − (1.10) −19 ⎤
= 200, 000 + 200, 000 ⎢
⎥
0.10
⎢⎣
⎥⎦
≈ $1,872,984.02
25,000
25, 000 − ⎡ 13.180795
(6.975319)(1.07)4 ⎤
⎣
⎦
x≈
4.439943
x ≈ $1725
5000 − 1000a5
3000
s20
⎡ 1 − (1.0135) −480 ⎤
= 750 ⎢
⎥
0.0135
⎢⎣
⎥⎦
≈ 55, 466.57
41. 200, 000 + 200, 000a19
⎡
⎤
25, 000 − ⎢ s25,000 s6 0.06 (1.07)4 ⎥
⎣ 10 0.06
⎦
x=
s4 0.07
0.08
1 − (1.052) −9
≈ 7.04494
0.052
⎡ (1.01)120 − 1 ⎤
=
1000
⎢
⎥
0.01
0.01
⎢⎣
⎥⎦
≈ 230, 038.69
38. 1000 s120
33. The original annual payment is
(1.017)60 − 1
≈ 102.91305
0.017
0.0135
10 0.06
5000 = 1000a5
⎞
⎠
0.08 ⎟
+ x(1.08) −6
⎡ (1.07)11 − 1 ⎤
= 2000 ⎢
− 1⎥ (1.07)30
⎢⎣ 0.07
⎥⎦
≈ $225,073,
so the net earnings are
225,073 – 20,000 = $205,073.
= x(1.08)−6
Thus
169
Chapter 5: Mathematics of Finance
ISM: Introductory Mathematical Analysis
For the second situation, the compound amount is
⎡ (1.07)31 − 1 ⎤
⎛
⎞
2000 ⎜ s31 0.07 − 1⎟ = 2000 ⎢
− 1⎥
⎝
⎠
⎢⎣ 0.07
⎥⎦
≈ $202,146,
so the net earnings are 202,146 – 60,000 = $142,146.
44. 100
1 − e−0.05(20)
≈ $1264
0.05
45. 40, 000
1 − e−0.04(5)
≈ $181, 269.25
0.04
Problems 5.5
1. R =
8000
8000
≈
≈ $273.42
a36 0.14 29.258904
12
2. A = 50a36
3. R =
0.01
≈ 50(30.107505) ≈ $1505.38
8000
8000
≈
≈ $236.19
a36 0.04 33.870766
12
Finance charge = 36(236.19) – 8000 = $502.84
4. a.
R=
500
a12
≈
0.0125
500
≈ $45.13
11.079312
b. 12(45.13) – 500 = $41.56
5. a.
R=
7500
7500
≈
≈ $221.43
a36 0.04 33.870766
12
b. 7500
c.
6. a.
0.04
= $25
12
221.43 – 25 = $196.43
R=
35, 000
35, 000
≈
≈ $851.17
a48 0.078 41.119856
12
0.078
= $227.50
12
b.
35, 000
c.
851.17 − 227.50 = $623.67
170
ISM: Introductory Mathematical Analysis
7. R =
Section 5.5
5000
5000
≈
≈ $1476.14
a4 0.07 3.387211
The interest for the first period is (0.07)(5000) = $350, so the principal repaid at the end of that period is
1476.14 – 350 = $1126.14. The principal outstanding at the beginning of period 2 is 5000 – 1126.14 = $3873.86.
The interest for period 2 is (0.07)(3873.86) = $271.17, so the principal repaid at the end of that period is
1476.14 – 271.17 = $1204.97. The principal outstanding at beginning of period 3 is
3873.86 – 1204.97 = $2668.89. Continuing in this manner, we construct the following amortization schedule.
Period
Prin. Outs.
at Beginning
Int. for
Period
Prin. Repaid
at End
1
5000.00
350.00
1476.14
1126.14
2
3873.86
271.17
1476.14
1204.97
3
2668.89
186.82
1476.14
1289.32
4
1379.57
96.57
1476.14
1379.57
904.56
5904.56
5000.00
Total
8. R =
Pmt. at
End
9000
a8
0.0475
≈
9000
≈ $1378.46
6.529036
The interest for the first period is (0.0475)(9000) = $427.50, so the principal repaid at the end of that period is
1378.46 − 427.50 = $950.96. The principal outstanding at the beginning of period 2 is 9000 − 950.96 = $8049.04.
The interest for period 2 is (0.0475)(8049.04) = $382.33, so the principal repaid at the end of that period is
1378.46 − 382.33 = $996.13. The principal outstanding at beginning of period 3 is 8049.04 − 996.13 = $7052.91.
Continuing in this manner, we construct the following amortization schedule. Note the adjustment in the final
payment.
Period
Prin. Outs.
at Beginning
Int. for
Period
Pmt. at
End
Prin. Repaid
at End
1
9000.00
427.50
1378.46
950.96
2
8049.04
382.33
1378.46
996.13
3
7052.91
335.01
1378.46
1043.45
4
6009.46
285.45
1378.46
1093.01
5
4916.45
233.53
1378.46
1144.93
6
3771.52
179.15
1378.46
1199.31
7
2572.21
122.18
1378.46
1256.28
8
1315.93
62.51
1378.44
1315.93
2027.66
11,027.66
9000.00
Total
171
Chapter 5: Mathematics of Finance
9. R =
ISM: Introductory Mathematical Analysis
900
900
≈
≈ $193.72
a5 0.025 4.645828
The interest for period 1 is (0.025)(900) = $22.50, so the principal repaid at the end of that period is
193.72 – 22.50 = $171.22. The principal outstanding at the beginning of period 2 is 900 – 171.22 = $728.78. The
interest for that period is (0.025)(728.78) = $18.22, so the principal repaid at the end of that period is
193.72 – 18.22 = $175.50. The principal outstanding at the beginning of period 3 is 728.78 – 175.50 = $553.28.
Continuing in this manner, we obtain the following amortization schedule. Note the adjustment in the final
payment.
Period
Prin. Outs.
at Beginning
Int. for
Period
Prin. Repaid
at End
1
900.00
22.50
193.72
171.22
2
728.78
18.22
193.72
175.50
3
553.28
13.83
193.72
179.89
4
313.39
9.33
193.72
184.39
5
189.00
4.73
193.73
189.00
68.61
968.61
900.00
Total
10. R =
Pmt. at
End
10, 000
10, 000
≈
≈ $2045.22
a5 0.0075 4.889440
The interest for period 1 is (0.0075)(10,000) = $75, so the principal repaid at the end of that period is
2045.22 – 75 = $1970.22. The principal outstanding at the beginning of period 2 is 10,000 – 1970.22 = $8029.78.
The interest for period 2 is (0.0075)(8029.78) = $60.22, so the principal repaid at the end of that period is
2045.22 – 60.22 = $1985. The principal outstanding at the beginning of period 3 is 8029.78 – 1985 = $6044.78.
Continuing in this manner, we construct the following amortization schedule. Note the adjustment in the final
payment.
Period
Prin. Outs.
at Beginning
Int. for
Period
Pmt. at
End
1
10,000.00
75.00
2045.22
1970.22
2
8029.78
60.22
2045.22
1985.00
3
6044.78
45.34
2045.22
1999.88
4
4044.90
30.34
2045.22
2014.88
5
2030.02
15.23
2045.25
2030.02
226.13
10,226.13
10,000.00
Total
11. From Eq. (1),
100
⎤
ln ⎡⎢ 100−1000(0.02)
⎣
⎦⎥ ≈ 11.268 .
n=
ln(1.02)
Thus the number of full payments is 11.
12. a.
Prin. Repaid
at End
2000
2000
≈
≈ $52.67
a48 0.01 37.973959
172
ISM: Introductory Mathematical Analysis
b.
52.67 a13
0.01
Section 5.5
≈ 52.67(12.133740)
b. 48(228.88) – 8500 = $2486.24
≈ $639.08
c.
100
⎤
ln ⎡⎢ 100− 2000(0.015)
⎥⎦
⎣
≈ 23.956. Thus the
17. n =
ln1.015
number of full payments is 23.
(639.08)(0.01) ≈ $6.39
d. 52.67 – 6.39 = $46.28
e.
48(52.67) – 2000 = $528.16
18. R =
18, 000
13. Each of the original payments is
.
a15 0.035
9500
a60
0.0077
⎡
⎤
0.0077
= 9500 ⎢
⎥
−60
⎣⎢ 1 − (1.0077)
⎦⎥
≈ $198.31
After two years the value of the remaining
⎡
⎤
18, 000 ⎥
a
. Thus the new
payments is ⎢
⎢a
⎥ 11 0.035
15
0.035
⎣⎢
⎦⎥
semi-annual payment is
18, 000a11 0.035
1
⋅
a15 0.035
a11 0.04
19. Present value of mortgage payments is
−360 ⎤
⎡
0.076
⎢ 1 − 1 + 12
⎥
600a360 0.076 = 600 ⎢
⎥
0.076
12
⎢
⎥
12
⎣
⎦
≈ $84,976.84
This amount is 75% of the purchase price x.
0.75x = 84,976.84
x = $113,302.45 ≈ $113,302
18, 000(9.001551)
1
⋅
11.517411
8.760477
≈ $1606.
20. For the 15-year mortgage, the monthly payment
is
⎡
⎤
240, 000
0.005
⎥
= 240, 000 ⎢
⎢ 1 − (1 + 0.005 )−180 ⎥
a180 0.005
⎣
⎦
(
=
14. R =
15. a.
2000
2000(0.014)
=
≈ $49.49
a60 0.014 1 − (1.014) −60
Monthly interest rate is
Monthly payment is
≈ $2025.26
The finance charge is
180(2025.26) – 240,000 = $124,546.80
For the 25-year mortgage, the monthly payment
is
⎡
⎤
240, 000
0.005
= 240, 000 ⎢
⎥
−
300
a300 0.005
⎣⎢ 1 − (1 + 0.005)
⎦⎥
0.092
.
12
⎡
0.092
245, 000
⎢
12
= 245, 000 ⎢
a300 0.092
0.092
⎢ 1 − 1 + 12
12
⎣
≈ $2089.69
(
)
⎤
⎥
−300 ⎥
⎥
⎦
≈ $1546.32
The finance charge is
300(1546.32) − 240,000 = 223,896.00
Thus the savings is
223,896.00 − 124,546.80 = $99,349.20
⎛ 0.092 ⎞
b. 245,000 ⎜
⎟ = $1878.33
⎝ 12 ⎠
c.
21.
2089.69 – 1878.33 = $211.36
d. 300(2089.69) – 245,000 = $381,907
16. a.
Monthly interest rate is
)
25, 000
a60
0.0125
−
25, 000
a60 0.01
⎡
⎤
1
1
⎢
⎥
= 25, 000
−
⎢a
⎥
a
60 0.01 ⎥
⎢⎣ 60 0.0125
⎦
⎡
⎤
0.0125
0.01
= 25, 000 ⎢
−
⎥
−60
1 − (1.01) −60 ⎥⎦
⎢⎣ 1 − (1.0125)
≈ $38.64
0.132
= 0.011 .
12
Monthly payment is
⎡
⎤
8500
0.011
= 8500 ⎢
⎥
−
48
a48 0.011
⎢⎣ 1 − (1.011)
⎥⎦
≈ $228.88
173
Chapter 5: Mathematics of Finance
ISM: Introductory Mathematical Analysis
22. The government’s payment is
( y − x)a60 0.0925
7. a.
A = 200a13
0.04
≈ $1997.13
12
⎡
5000
5000
=⎢
−
⎢ a 0.0925 a 0.04
60 12
⎣⎢ 60 12
⎡ a 0.0925
60 12
= 5000 ⎢1 −
⎢
a60 0.04
⎢⎣
12
⎤
⎥a
⎥ 60
⎦⎥
b.
0.0925
12
S = 200s13
0.04
≈ 200(16.626838)
≈ $3325.37
⎤
⎥
⎥
⎥⎦
8. 150 s14
0.04
− 150 = 150(18.291911) − 150
≈ 2593.79
⎡ 1−(1+ 0.0925 )−60
12
⎢
0.0925
⎢
12
= 5000 ⎢1 −
−60
1−(1+ 0.04
12 )
⎢
0.04
12
⎣⎢
(
⎤
⎥
⎥
⎥
⎥
⎦⎥
9. 200 s13
10. 250a20
)
−60
⎡
⎤
0.0925
0.04 ⎥
⎢ 1 − 1 + 12
= 5000 ⎢1 −
⋅
−60
0.0925 ⎥
1 − 1 + 0.04
⎢
⎥
12
⎣
⎦
≈ $589.89
(
≈ 200(9.985648)
11.
)
− 200 ≈ 200(13.532926) − 200
0.025
≈ $2506.59
≈ 250(15.589162) ≈ $3897.29
5000
5000
≈
≈ $886.98
s5 0.06 5.637093
12. a.
Chapter 5 Review Problems
0.08
12
7000
7000
≈
≈ $206.67
a36 0.04 33.870766
12
5
2
⎛2⎞
1. s = 3 + 2 + 2 ⋅ + " + 3 ⎜ ⎟
3
⎝3⎠
6⎤
⎡
3 ⎢1 − 23 ⎥ 3 ⎡ 665 ⎤
⎦ = ⎣ 729 ⎦ = 665
= ⎣
2
1
81
1− 3
3
b. 36(206.67) – 7000 = $440.12
( )
13. Let x be the first payment. The equation of value
now is
x + 2 x(1.07) −3 = 500(1.05) −3 + 500(1.03) −8
x ⎡1 + 2(1.07) −3 ⎤ = 500(1.05)−3 + 500(1.03) −8
⎣
⎦
12
⎛ 0.05 ⎞
2. ⎜1 +
⎟
12 ⎠
⎝
− 1 ≈ 0.0512 or 5.12%
x=
1 + 2(1.07)−3
x ≈ $314.00
3. 8.2% compounded semiannually corresponds to
an effective rate of (1.041)2 − 1 = 0.083681 or
8.37%. Thus the better choice is 8.5%
compounded annually.
⎡
0.01375 ⎤
= 3500 ⎢
⎥
−3
a3 0.01375
14.
⎣⎢ 1 − (1.01375) ⎦⎥
≈ $1198.90
The interest for the first period is
(0.01375)(3500) = $48.13, so the principal
repaid at the end of that period is
1198.90 − 48.13 = $1150.77. The principal
outstanding at the beginning of period 2 is
3500 − 1150.77 = $2349.23. The interest for that
period is (0.01375)(2349.23) = $32.30. The
principal repaid at the end of that period is
1198.90 − 32.30 = $1166.60. The principal
outstanding at the beginning of period 3 is
2349.23 − 1166.60 = $1182.63. Continuing, we
R=
4. NPV = 3400(1.035) −4 + 3500(1.035) −8 − 7000
≈ −$1379.16
5. Let x be the payment at the end of 2 years. The
equation of value at the end of year 2 is
1000(1.04) 4 + x = 1200(1.04) −4 + 1000(1.04) −8
x = 1200(1.04)−4 + 1000(1.04)−8 − 1000(1.04)4
≈ $586.60
6. 250a48
0.005
500(1.05)−3 + 500(1.03) −8
≈ 250(42.580318) ≈ $10, 645.08
174
3500
ISM: Introductory Mathematical Analysis
Chapter 5 Review
obtain the following amortization schedule. Note the adjustment in the final payment.
Period
Int. for
Period
Prin. Outs.
at Beginning
Prin. Repaid
at End
1
3500.00
48.13
1198.90
1150.77
2
2349.23
32.30
1198.90
1166.60
3
1182.63
16.26
1198.89
1182.63
96.69
3596.69
3500.00
Total
15. R =
Pmt. at
End
15, 000
15, 000
≈
≈ $3067.84
a5 0.0075 4.889440
The interest for period 1 is (0.0075)(15,000) = $112.50, so the principal repaid at the end of that period is
3067.84 – 112.50 = $2955.34. The principal outstanding at beginning of period 2 is
15,000 – 2955.34 = $12,044.66. The interest for period 2 is 0.0075(12,044.66) = $90.33, so the principal repaid at
the end of that period is 3067.84 – 90.33 = $2977.51. Principal outstanding at the beginning of period 3 is
12,044.66 – 2977.51 = $9067.15. Continuing, we obtain the following amortization schedule. Note the
adjustment in the final payment.
Period
Int. for
Period
Prin. Outs.
at Beginning
Pmt. at
End
Prin. Repaid
at End
1
15,000
112.50
3067.84
2955.34
2
12,044.66
90.33
3067.84
2977.51
3
9067.15
68.00
3067.84
2999.84
4
6067.31
45.50
3067.84
3022.34
5
3044.97
22.84
3067.81
3044.97
339.17
15,339.17
15,000.00
Total
16. 540a84
(
0.10
12
)
−84 ⎤
⎡
0.10
⎢ 1 − 1 + 12
⎥
= 540 ⎢
⎥ ≈ $32,527.80
0.10
⎢
⎥
12
⎣
⎦
17. The monthly payment is
⎡
⎤
0.055
11, 000
⎢
⎥
12
= 11, 000 ⎢
⎥ ≈ $255.82
48
−
a48 0.055
0.055
−
+
1
1
⎢
⎥
12
12
⎣
⎦
The finance charge is 48(255.82) – 11,000 = $1279.36
(
)
175
Chapter 5: Mathematics of Finance
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 5
1.
0.085
= 0.0425, thus R = 0.0425(25,000) = 1062.50.
2
1 − (1.0825) −25
P = 25, 000(1.0825)−25 + 1062.50 ⋅
1.0825 − 1
≈ $26,102.13
2.
0.065
= 0.0325 , thus R = 0.0325(10,000) = 325.
2
On a graphics calculator, let Y1 = 10,389 and Y2 = 10,000(1 + x)^ – 7 + 325(1 – (1 + x)^ – 7)/
(
)
(1 + x) − 1 .
The curves intersect at 0.0590. The yield is 5.9%.
3. The normal yield curve assumes a stable economic climate. By contrast, if investors are expecting a drop in
interest rates, and with it a drop in yields from future investments, they will gladly give up liquidity for long-term
investment at current, more favorable, interest rates. T-bills, which force the investor to find a new investment in
a short time, are correspondingly less attractive, and so prices drop and yields rise.
176
Chapter 6
Principles in Practice 6.1
1. There are 3 rows, one for each source. There are two columns, one for each raw material. Thus, the size of the
matrix is 3 × 2. Alternatively, she could use a 2 × 3 matrix.
2. The first column consists of 1’s each representing the 1 hour needed for each phase of project 1. The second
column consists of 2’s for each phase of project 2 and so on. In general the nth column will consist of 2n ’s, each
representing the 2n hours needed for each phase of project n. The time-analysis matrix is as follows.
⎡1 2 4 8 16 ⎤
⎢1 2 4 8 16 ⎥
⎢
⎥
⎢⎣1 2 4 8 16 ⎥⎦
Problems 6.1
1. a.
The size is the number of rows by the columns. Thus A is 2 × 3, B is 3 × 3, C is 3 × 2, D is 2 × 2, E is 4 × 4,
F is 1 × 2, G is 3 × 1, H is 3 × 3, and J is 1 × 1.
b. A square matrix has the same number of rows as columns. Thus the square matrices are B, D, E, H, and J.
c.
An upper triangular matrix is a square matrix where all entries below the main diagonal are zeros. Thus H
and J are upper triangular. A lower triangular matrix is a square matrix where all entries above the main
diagonal are zeros. Thus D and J are lower triangular.
d. A row vector (or row matrix) has only one row. Thus F and J are row vectors.
e.
A column vector (or column matrix) has only one column. Thus G and J are column vectors.
2. A has 4 rows and 4 columns. Thus the order of A is 4.
3. a21 is the entry in the 2nd row and 1st column, namely 6.
4. a14 is the entry in the 1st row and 4th column, namely 6.
5. a32 is the entry in the 3rd row and 2nd column, namely 4.
6. a34 is the entry in the 3rd row and 4th column, namely 0.
7. a44 is the entry in the 4th row and 4th column, namely 0.
8. a55 is the entry in the 5th row and 5th column. But A has only 4 rows and 4 columns. Thus a55 does not exist.
9. The main diagonal entries are the entries on the diagonal extending from the upper left corner to the lower right
corner. Thus the main diagonal entries are 7, 2, 1, 0.
10. ⎡ 2
⎢0
⎢
⎢0
⎢
⎣⎢ 0
3 4 5⎤
4 5 6 ⎥⎥
0 6 7⎥
⎥
0 0 8 ⎦⎥
177
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
11. ⎡ −2 ⋅1 + 3 ⋅1 −2 ⋅1 + 3 ⋅ 2 −2 ⋅1 + 3 ⋅ 3 −2 ⋅1 + 3 ⋅ 4 −2 ⋅ 1 + 3 ⋅ 5 ⎤ ⎡ 1 4 7 10 13⎤
⎢ −2 ⋅ 2 + 3 ⋅1 −2 ⋅ 2 + 3 ⋅ 2 −2 ⋅ 2 + 3 ⋅ 3 −2 ⋅ 2 + 3 ⋅ 4 −2 ⋅ 2 + 3 ⋅ 5⎥ = ⎢ −1 2 5 8 11⎥
⎢
⎥ ⎢
⎥
⎣ −2 ⋅ 3 + 3 ⋅1 −2 ⋅ 3 + 3 ⋅ 2 −2 ⋅ 3 + 3 ⋅ 3 −2 ⋅ 3 + 3 ⋅ 4 −2 ⋅ 3 + 3 ⋅ 5 ⎦ ⎣ −3 0 3 6 9 ⎦
(
(
1+1 2
⎡
1 + 12
12. ⎢ (−1)
⎢
2 +1 2
2 + 12
⎢⎣ (−1)
)
)
(
(
)
)
(−1)1+ 2 12 + 22 ⎤ ⎡ 2 −5⎤
⎥
=⎢
2+ 2 2
2 ⎥ ⎣ −5
8⎥⎦
(−1)
2 +2 ⎥
⎦
13. 12 · 10 = 120, so A has 120 entries. For a33 , i = 3 = j, so a33 = 1. Since 5 ≠ 2, a52 = 0. For a10, 10 , i = 10 = j,
so a10, 10 = 1. Since 12 ≠ 10, a12, 10 = 0.
14. The main diagonal is the diagonal extending from the upper left corner to the lower right corner.
a.
1, 0, –5, 2
b. x, y, z
15. A zero matrix is a matrix in which all entries are zeros.
a.
⎡0
⎢0
⎢
⎢0
⎢
⎢⎣ 0
0 0 0⎤
0 0 0 ⎥⎥
0 0 0⎥
⎥
0 0 0 ⎥⎦
b.
⎡0
⎢0
⎢
⎢0
⎢
⎢0
⎢0
⎢
⎢⎣ 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0⎤
0 ⎥⎥
0⎥
⎥
0⎥
0⎥
⎥
0 ⎥⎦
16. If A is 7 × 9, then A T is 9 × 7.
T
⎡ 6 −3⎤
⎡ 6 2⎤
=⎢
17. A T = ⎢
⎥
⎥
⎣2 4⎦
⎣ −3 4 ⎦
18. A T = [ 2 4 6 8]
T
⎡ 2⎤
⎢ 4⎥
=⎢ ⎥
⎢6⎥
⎢ ⎥
⎣⎢ 8 ⎦⎥
3 −4 ⎤
T ⎡1
⎡ 1 3 7 3⎤
⎢3 2
5⎥⎥
19. A T = ⎢⎢ 3 2 −2 0 ⎥⎥ = ⎢
⎢ 7 −2 0 ⎥
⎢
⎥
⎣⎢ −4 5 0 1⎦⎥
1⎥⎦
⎢⎣ 3 0
178
ISM: Introductory Mathematical Analysis
Section 6.1
b. From F, the entry in row 2 (deluxe) and
column 3 (blue) is 3. Thus in February,
3 blue deluxe models were sold.
T
⎡ 2 −1 0 ⎤
⎡ 2 −1 0 ⎤
⎢
⎥
T
20. A = ⎢ −1 5 1⎥ = ⎢⎢ −1 5 1⎥⎥
⎢⎣ 0 1 3⎥⎦
⎢⎣ 0 1 3⎥⎦
21. a.
c.
A and C are diagonal matrices.
b. All are them are triangular matrices.
T
⎡ 2 −1 0 ⎤
⎡ 2 −1 0 ⎤
⎢
⎥
22. A = −1 5 1 = ⎢ −1 5 1⎥
⎢
⎥
⎢
⎥
⎣ 0 1 3⎦
⎣ 0 1 3⎦
d. In both January and February, the deluxe
blue models (row 2, column 3) sold the
same number of units (3).
T
Since A T = A, the matrix of Problem 20 is
symmetric.
e.
In January a total of 0 + 1 + 3 + 5 = 9
deluxe models were sold. In February a total
of
2 + 3 + 3 + 2 = 10 deluxe models were sold.
Thus more deluxe models were sold in
February.
f.
In January a total of 2 + 0 + 2 = 4 red
widgets were sold, while in February a total
of 0 + 2 + 4 = 6 red widgets were sold. Thus
more red widgets were sold in February.
g.
Adding all entries in matrix J yields that a
total of 38 widgets were sold in January.
⎡ 1 7⎤
T
⎡ 1 0 −1⎤
⎢ 0 0⎥
23. A T = ⎢
=
⎢
⎥
⎣7 0 9 ⎥⎦
⎣ −1 9 ⎦
T
⎡ 1 7⎤
⎡ 1 0 −1⎤
⎢
(A ) = 0 0⎥ = ⎢
=A
⎢
⎥
⎣ 7 0 9 ⎥⎦
⎣ −1 9 ⎦
T T
24. Equating corresponding entries gives 2x = 4,
y = 6, z = 0, and 3w = 7. Thus x = 2, y = 6, z = 0,
7
w= .
3
30. The sums of the entries in the columns are 680,
710, 1510, and 6690. The sum of the entries in
the rows are 680, 710, 1510, and 6690. The
amount an industry consumes is equal to the
amount of its output. Industry B has to increase
output by (0.20)(90) = 18 units and industry C
has to increase output by (0.20)(120) = 24 units.
All other producers have to increase it by
(0.20)(420) = 84 units.
25. Equating corresponding entries gives 6 = 6,
2 = 2, x = 6, 7 = 7, 3y = 2, and 2z = 7. Thus
2
7
x = 6, y = , z = .
3
2
26. Equating entries in the 3rd row and 3rd column
gives 7 = 8, which is never true, so there is no
solution.
31. By equating entries we find that x must satisfy
x 2 + 2000 x = 2001 and x 2 = − x .
The second equation implies that x < 0. From the
27. Equating corresponding entries gives 2x = y,
7 = 7, 7 = 7, and 2y = y. Now 2y = y yields y = 0.
Thus from 2x = y we get 2x = 0, so x = 0. The
solution is x = 0, y = 0.
28. [125 275
⎡ 0.95⎤
⎢1.03 ⎥
⎢
⎥
⎣1.25 ⎦
29. a.
The entries in row 1 (regular) and column
4 (purple) give the number of purple regular
models sold. For J the entry is 2 and for F
the entry is 4. Thus more purple regular
models were sold in February.
first equation, x 2 + 2000 x − 2001 = 0 ,
(x + 2001)(x – 1) = 0, so x = –2001.
400]
⎡ 3 −2 ⎤
32. ⎢
1⎥⎥
⎢ −4
⎢⎣ 5 6 ⎥⎦
⎡3
33. ⎢
⎢1
⎢4
⎢
⎣⎢ 2
From J, the entry in row 3 (super-duper)
and column 2 (white) is 7. Thus in January,
7 white super-duper models were sold.
179
1
7
3
6
1⎤
4 ⎥⎥
1⎥
⎥
2 ⎦⎥
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
Principles in Practice 6.2
⎡120 80 ⎤ ⎡110 140 ⎤
1. T = J + F = ⎢
⎥+⎢
⎥
⎣105 130 ⎦ ⎣ 85 125⎦
⎡120 + 110 80 + 140 ⎤ ⎡ 230 220 ⎤
=⎢
⎥=⎢
⎥
⎣ 105 + 85 130 + 125⎦ ⎣190 255⎦
⎡ x1 ⎤ ⎡ 40 ⎤
⎡ 248⎤
⎢
⎥
⎢
⎥
2. 0.8 ⎢ x2 ⎥ − ⎢ 30 ⎥ = 2 ⎢⎢319 ⎥⎥
⎢⎣ x3 ⎥⎦ ⎢⎣ 60 ⎥⎦
⎢⎣532 ⎥⎦
⎡ 0.8 x1 ⎤ ⎡ 40 ⎤ ⎡ 496 ⎤
⎢ 0.8 x ⎥ − ⎢ 30 ⎥ = ⎢ 638⎥
2⎥ ⎢ ⎥ ⎢
⎢
⎥
⎢⎣ 0.8 x3 ⎥⎦ ⎢⎣ 60 ⎥⎦ ⎢⎣1064 ⎥⎦
⎡ 0.8 x1 − 40 ⎤ ⎡ 496 ⎤
⎢ 0.8 x − 30 ⎥ = ⎢ 638⎥
2
⎢
⎥ ⎢
⎥
⎢⎣ 0.8 x3 − 60 ⎥⎦ ⎢⎣1064 ⎥⎦
Solve 0.8 x1 − 40 = 496 to get x1 = 670 .
Solve 0.8 x2 − 30 = 638 to get x2 = 835 .
Solve 0.8 x3 − 60 = 1064 to get x3 = 1405 .
Problems 6.2
−3 + 4 ⎤ ⎡ 4 −3 1⎤
2 + 2 0 + (−3)
⎡ 2 0 −3⎤ ⎡ 2 −3 4 ⎤ ⎡
4+6
0 + 5⎥⎥ = ⎢⎢ −2 10 5⎥⎥
1. ⎢⎢ −1 4 0 ⎥⎥ + ⎢⎢ −1 6 5⎥⎥ = ⎢⎢ −1 + (−1)
⎢⎣ 1 −6 5⎥⎦ ⎢⎣ 9 11 −2 ⎥⎦ ⎢⎣
1 + 9 −6 + 11 5 + (−2) ⎥⎦ ⎢⎣ 10 5 3⎥⎦
2 + 7 + 2 −7 + (−4) + 7 ⎤ ⎡11 −4 ⎤
⎡ 2 −7 ⎤ ⎡ 7 −4 ⎤ ⎡ 2 7 ⎤ ⎡
2. ⎢
=
⎥ + ⎢ −2
⎥ + ⎢ 7 2 ⎥ = ⎢ −6 + (−2) + 7
6
4
1
4 + 1 + 2 ⎥⎦ ⎢⎣ −1 7 ⎥⎦
−
⎣
⎦ ⎣
⎦ ⎣
⎦ ⎣
⎡ 1 4 ⎤ ⎡ 6 −1⎤ ⎡ 1 − 6 4 − (−1) ⎤ ⎡ −5 5⎤
7 − 2⎥⎥ = ⎢⎢ −9 5⎥⎥
3. ⎢⎢ −2 7 ⎥⎥ − ⎢⎢ 7 2 ⎥⎥ = ⎢⎢ −2 − 7
⎢⎣ 6 9 ⎥⎦ ⎢⎣ 1 0 ⎥⎦ ⎢⎣ 6 − 1
9 − 0⎥⎦ ⎢⎣ 5 9⎥⎦
⎡1
6⎤ ⎢ 2 ⋅ 4
⎡ 4 −2
1⎢
4.
2 10 −12 ⎥⎥ = ⎢ 12 ⋅ 2
⎢
2⎢
⎢⎣ 0 0
8⎥⎦ ⎢ 1 ⋅ 0
⎣2
1 (−2)
2
1 ⋅10
2
1 ⋅0
2
⎤
⎥ ⎡ 2 −1 3⎤
1 ( −12) ⎥ = ⎢ 1
5 −6 ⎥⎥
2
⎥ ⎢
0 4 ⎥⎦
1 ⋅ 8 ⎥ ⎢⎣ 0
2
⎦
1 ⋅6
2
5. 2[2 −1 3] + 4[−2 0 1] − 0[2 3 1]
= [4 −2 6] + [−8 0 4] − [0 0 0]
= [4 − 8 − 0 −2 + 0 − 0 6 + 4 − 0]
= [−4 −2 10]
6.
[7
7 ] is a matrix and 66 is a number, so the sum is not defined.
180
ISM: Introductory Mathematical Analysis
⎡1 2 ⎤
7. ⎢
⎥ has size 2 × 2, and
⎣3 4 ⎦
Section 6.2
⎡7 ⎤
⎢ 2 ⎥ has size 2 × 1. Thus the sum is not defined.
⎣ ⎦
⎡ 2 −1⎤
⎡ 0 0 ⎤ ⎡ 2 −1⎤ ⎡ 0 0 ⎤ ⎡ 2 −1⎤
8. ⎢
+ 3⎢
⎥
⎥=⎢
⎥+⎢
⎥=⎢
⎥
⎣7 4⎦
⎣0 0⎦ ⎣7 4 ⎦ ⎣ 0 0⎦ ⎣7 4⎦
1⎤ ⎡ −6 ⋅ 2 −6(−6) −6 ⋅ 7
−6 ⋅ 1⎤ ⎡ −12 36 −42 −6 ⎤
⎡ 2 −6 7
9. −6 ⎢
=
=⎢
⎥
1 6 −2 ⎦ ⎣ −6 ⋅ 7
−6 ⋅1 –6 ⋅ 6 −6(−2) ⎥⎦ ⎢⎣ −42 −6 −36 12 ⎥⎦
⎣7
⎡ 1 −1⎤
⎡ −6 9 ⎤ ⎡ 1 −1⎤ ⎡ −18
⎢2 0⎥
⎢
⎥ ⎢
⎥ ⎢
⎥ − 3 ⎢ 2 6⎥ = ⎢ 2 0⎥ − ⎢ 6
10. ⎢
⎢ 3 −6 ⎥
⎢ 1 −2 ⎥ ⎢ 3 −6 ⎥ ⎢ 3
⎢
⎥
⎢
⎥ ⎢
⎥ ⎢
⎣⎢ 4 9 ⎦⎥
⎣⎢ 4 5⎥⎦ ⎣⎢ 4 9 ⎦⎥ ⎣⎢ 12
27 ⎤ ⎡ 19 −28⎤
18⎥⎥ ⎢⎢ −4 −18⎥⎥
=
−6 ⎥ ⎢ 0
0⎥
⎥ ⎢
⎥
15⎦⎥ ⎣⎢ −8 −6 ⎦⎥
⎡ 1 −5 0 ⎤
⎡10 0 30 ⎤ ⎡ 1 −5 0 ⎤ ⎡ 2 0 6 ⎤ ⎡ 3 −5 6 ⎤
1⎢
⎢
⎥
11. ⎢ −2 7 0 ⎥ + ⎢ 0 5 0 ⎥⎥ = ⎢⎢ −2 7 0 ⎥⎥ + ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ −2 8 0 ⎥⎥
5
⎣⎢ 4 6 10 ⎦⎥
⎣⎢ 5 20 25⎥⎦ ⎢⎣ 4 6 10 ⎦⎥ ⎣⎢ 1 4 5 ⎦⎥ ⎣⎢ 5 10 15⎦⎥
⎡ 1 0 0 ⎤ ⎛ ⎡ 1 2 0 ⎤ ⎡ 4 −2 2 ⎤ ⎞ ⎡ 3
12. 3 ⎢ 0 1 0 ⎥ − 3 ⎜ ⎢0 −2 1⎥ − ⎢ −3 21 −9 ⎥ ⎟ = ⎢ 0
⎢
⎥ ⎜⎜ ⎢
⎥ ⎢
⎥⎟ ⎢
1 0 ⎦ ⎠⎟ ⎣ 0
⎣ 0 0 1⎦ ⎝ ⎣0 0 1⎦ ⎣ 0
⎡3
= ⎢0
⎢
⎣0
= ⎡ 12
⎢ −9
⎢
⎣ 0
0 0⎤
⎡ −3
3 0⎥ − 3 ⎢ 3
⎥
⎢
0 3⎦
⎣ 0
0 0 ⎤ ⎡ −9
3 0⎥ − ⎢ 9
⎥ ⎢
0 3⎦ ⎣ 0
6⎤
−12
72 −30 ⎥
⎥
3
0⎦
4 −2 ⎤
−23 10 ⎥
⎥
−1 1⎦
12 −6 ⎤
−69 30 ⎥
⎥
−3 3⎦
⎡ −6 −5⎤
⎡ −6 −5⎤ ⎡ −1(−6) −1(−5) ⎤ ⎡ 6 5⎤
= (−1) ⎢
13. −B = − ⎢
⎥
⎥=⎢
⎥=⎢
⎥
⎣ 2 −3⎦
⎣ 2 −3⎦ ⎣ −1(2) −1(−3) ⎦ ⎣ −2 3⎦
1 − (−5) ⎤
⎡ 2 − (−6)
⎡8 6 ⎤ ⎡ −8 −6 ⎤
14. −( A − B) = − ⎢
⎥ = − ⎢1 0 ⎥ = ⎢ −1 0 ⎥
3
−
2
−
3
−
(
−
3)
⎣
⎦
⎣
⎦ ⎣
⎦
⎡0 0⎤ ⎡ 2 ⋅ 0 2 ⋅ 0⎤ ⎡0 0⎤
15. 2O = 2 ⎢
⎥=⎢
⎥=⎢
⎥=O
⎣0 0⎦ ⎣ 2 ⋅ 0 2 ⋅ 0⎦ ⎣0 0⎦
⎡ 2 − (−6) + (−2) 1 − (−5) + (−1) ⎤ ⎡ 6 5⎤
16. A − B + C = ⎢
⎥=⎢
⎥
⎣ 3 − 2 + (−3) −3 − (−3) + 3⎦ ⎣ −2 3⎦
⎧⎪ ⎡ 4 2 ⎤ ⎡ −18 −15⎤ ⎫⎪
1⎤
⎡ −6 −5⎤ ⎪⎫
⎡ 22 17 ⎤ ⎡ 66 51⎤
⎪⎧ ⎡ 2
17. 3(2A – 3B) = 3 ⎨2 ⎢
⎥ − 3 ⎢ 2 −3⎥ ⎬ = 3 ⎨ ⎢ 6 −6 ⎥ − ⎢ 6 −9 ⎥ ⎬ = 3 ⎢ 0 3⎥ = ⎢ 0 9⎥
−
3
3
⎪⎩ ⎣
⎦
⎣
⎦ ⎪⎭
⎦ ⎣
⎦ ⎭⎪
⎣
⎦ ⎣
⎦
⎩⎪ ⎣
⎡ −4 −4 ⎤ ⎡ 0 0 ⎤
18. 0(A + B) = 0 ⎢
⎥=⎢
⎥=O
⎣ 5 −6 ⎦ ⎣ 0 0 ⎦
181
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
19. 3(A – C) is a 2 × 2 matrix and 6 is a number. Therefore 3(A – C) + 6 is not defined.
1⎤ ⎡ –2 + (−6) −1 + (−5) ⎤ ⎡ 2 + (−8) 1 + (−6) ⎤ ⎡ −6 −5⎤
⎡2
=
=
20. A + (C + B) = ⎢
⎥ + ⎢ −3 + 2
−
3
3
3 + (−3) ⎥⎦ ⎢⎣3 + (−1) −3 + 0 ⎥⎦ ⎢⎣ 2 −3⎥⎦
⎣
⎦ ⎣
1⎤
⎡ −6 −5⎤
⎡2
⎡ −2 −1⎤
− 3⎢
+ 2⎢
21. 2B − 3A + 2C = 2 ⎢
⎥
⎥
⎥
⎣ 2 −3⎦
⎣ 3 −3⎦
⎣ −3 3⎦
3⎤ ⎡ −4 −2 ⎤
⎡ −12 −10 ⎤ ⎡ 6
=⎢
−⎢
⎥
⎥+⎢
⎥
⎣ 4 −6 ⎦ ⎣ 9 −9 ⎦ ⎣ −6 6 ⎦
⎡ −18 −13⎤ ⎡ −4 −2 ⎤ ⎡ −22 −15⎤
=⎢
+
=
3⎥⎦ ⎢⎣ −6 6 ⎥⎦ ⎢⎣ −11
9 ⎥⎦
⎣ −5
⎡ −6 −3⎤ ⎡ −12 −10 ⎤ ⎡ 6 7 ⎤
22. 3C − 2B = ⎢
⎥−⎢
⎥=⎢
⎥
⎣ −9 9 ⎦ ⎣ 4 −6 ⎦ ⎣ −13 15⎦
23.
⎧⎪ ⎡ −6 −5⎤
1⎤
⎡ −2 −1⎤ ⎫⎪
1
1 ⎡2
A − 2(B + 2C) = ⎢
⎥ − 2 ⎨ ⎢ 2 −3⎥ + 2 ⎢ −3 3⎥ ⎬
−
3
3
2
2⎣
⎦
⎦
⎣
⎦ ⎪⎭
⎩⎪ ⎣
1
⎡1
⎤
⎪⎧ ⎡ −6 −5⎤ ⎡ −4 −2 ⎤ ⎪⎫
2⎥
=⎢
− 2 ⎨⎢
⎥+⎢
⎥⎬
3
3
⎢
⎥
⎪⎩ ⎣ 2 −3⎦ ⎣ −6 6 ⎦ ⎪⎭
⎣2 − 2⎦
29 ⎤
1⎤
1⎤
⎡ 21
⎡1
⎡ −10 −7 ⎤ ⎡⎢ 1
2⎥
2 ⎥ ⎡ −20 −14 ⎤
2 ⎥
⎢
=⎢
−2⎢
=
−
=
⎥ ⎢3
⎢ −8
⎥ ⎢ 19
3
15
⎢3 − 3⎥
−
4
3
⎥
6
⎣
⎦ ⎣2 − 2⎦ ⎣
⎦ ⎣ 2 − 2 ⎥⎦
2⎦
⎣2
24.
1⎤
1⎤
⎡ 41 61 ⎤
⎡1
⎡ −8 −6 ⎤ ⎡⎢ 1
1
2⎥
2 ⎥ ⎡ 40 30 ⎤ ⎢
2⎥
−5⎢
=
+
=
A − 5(B + C) = ⎢
⎥ ⎢3
⎢ 5 0 ⎥ ⎢ 13
3
1
0
⎢3 − 3⎥
−
⎥
2
⎣
⎦ ⎣2 − 2⎦ ⎣
⎦ ⎣ 2 − 32 ⎥⎦
2⎦
⎣2
⎡ −4 −4 ⎤ ⎡ −12 −12 ⎤
25. 3( A + B) = 3 ⎢
⎥=⎢
⎥
⎣ 5 −6 ⎦ ⎣ 15 −18⎦
3⎤ ⎡ −18 −15⎤ ⎡ −12 −12 ⎤
⎡6
3A + 3B = ⎢
⎥+⎢
⎥=⎢
⎥
⎣ 9 −9 ⎦ ⎣ 6 −9 ⎦ ⎣ 15 −18⎦
Thus 3(A + B) = 3A + 3B.
5⎤
⎡10
26. (2 + 3) A = 5A = ⎢
⎥
⎣15 −15⎦
3⎤ ⎡10
5⎤
⎡ 4 2⎤ ⎡6
2A + 3A = ⎢
⎥ + ⎢9 −9 ⎥ = ⎢15 −15⎥
−
6
6
⎣
⎦ ⎣
⎦ ⎣
⎦
Thus (2 + 3)A = 2A + 3A.
⎡ 2k
27. k1 ( k2 A ) = k1 ⎢ 2
⎣ 3k2
⎡ 2k k
( k1k2 ) A = ⎢ 3k1k2
k2 ⎤ ⎡ 2k1k2
=
−3k2 ⎥⎦ ⎢⎣ 3k1k2
k1k2 ⎤
−3k1k2 ⎥⎦
k1k2 ⎤
−3k1k2 ⎥⎦
⎣ 1 2
Thus k1 ( k2 A ) = ( k1k2 ) A.
182
ISM: Introductory Mathematical Analysis
Section 6.2
⎛ ⎡2
1⎤ ⎡ −12 −10 ⎤ ⎡ −2 −1⎤ ⎞
⎡ 12 10 ⎤ ⎡ 12k 10k ⎤
−⎢
+⎢
28. k ( A − 2B + C) = k ⎜⎜ ⎢
⎟⎟ = k ⎢
⎥
⎥
⎥
⎥=⎢
⎥
⎣ −4 6 ⎦ ⎣ − 4 k 6 k ⎦
⎝ ⎣ 3 −3⎦ ⎣ 4 −6 ⎦ ⎣ −3 3⎦ ⎠
k ⎤ ⎡ −12k −10k ⎤ ⎡ −2k − k ⎤ ⎡ 12k 10k ⎤
⎡ 2k
kA − 2kB + kC = ⎢
+
=
⎥−⎢
−6k ⎥⎦ ⎢⎣ −3k 3k ⎥⎦ ⎢⎣ −4k 6k ⎥⎦
⎣ 3k −3k ⎦ ⎣ 4k
Thus k(A − 2B + C) = kA − 2kB + kC.
⎡ 1 2 ⎤ ⎡ 1 1⎤ ⎡ 3 6 ⎤ ⎡ 1 1⎤ ⎡ 4 7 ⎤
29. 3A + D = 3 ⎢⎢ 0 −1⎥⎥ + ⎢⎢ 2 0 ⎥⎥ = ⎢⎢ 0 −3⎥⎥ + ⎢⎢ 2 0⎥⎥ = ⎢⎢ 2 −3⎥⎥
⎢⎣ 7 0 ⎥⎦ ⎢⎣ −1 2 ⎥⎦ ⎢⎣ 21 0 ⎥⎦ ⎢⎣ −1 2⎥⎦ ⎢⎣ 20 2 ⎥⎦
T
T
T
⎡ 0 3⎤
⎡ 0 3⎤
⎪⎧ ⎡ 1 3⎤ ⎡1 0 ⎤ ⎪⎫
30. (B − C) = ⎨ ⎢
⎥ − ⎢1 2 ⎥ ⎬ = ⎢ 3 −3⎥ = ⎢ 3 −3⎥
−
4
1
⎦ ⎣
⎦ ⎪⎭
⎣
⎦
⎣
⎦
⎩⎪ ⎣
T
⎡ 1 4⎤
⎡ 1 1 ⎤ ⎡ 2 8 ⎤ ⎡ 3 3 ⎤ ⎡ −1 5 ⎤
31. 2BT − 3CT = 2 ⎢
− 3⎢
⎥−⎢
⎥=⎢
⎥
⎥
⎥ =⎢
⎣3 −1⎦
⎣ 0 2 ⎦ ⎣ 6 −2 ⎦ ⎣ 0 6 ⎦ ⎣ 6 −8⎦
⎡ 1 3⎤ ⎡ 1 4 ⎤ ⎡ 2 6⎤ ⎡ 1 4 ⎤ ⎡ 3 10 ⎤
32. 2B + BT = 2 ⎢
⎥
⎥+⎢
⎥=⎢
⎥+⎢
⎥ =⎢
⎣ 4 −1⎦ ⎣3 −1⎦ ⎣ 8 −2⎦ ⎣3 −1⎦ ⎣11 −3⎦
T
⎡1 0 ⎤
⎡1 2 −1⎤
−⎢
33. C − D = ⎢
is impossible because CT and D are not of the same size.
⎥
⎥
⎣1 2 ⎦
⎣1 0 2 ⎦
T
34.
(
D − 2AT
)
T
T
⎧⎪ ⎡1 2 −1⎤
⎡ 1 0 7 ⎤ ⎫⎪
= ⎨⎢
−2⎢
⎥
⎥⎬
⎣ 2 −1 0 ⎦ ⎪⎭
⎩⎪ ⎣1 0 2 ⎦
T
⎧⎪ ⎡1 2 −1⎤ ⎡ 2 0 14 ⎤ ⎫⎪
⎡ −1 2 −15⎤
= ⎨⎢
⎥ − ⎢ 4 −2 0 ⎥ ⎬ = ⎢ −3 2
1
0
2
2 ⎥⎦
⎦ ⎣
⎦ ⎪⎭
⎣
⎩⎪ ⎣
⎡ −1 −3⎤
= ⎢⎢ 2 2 ⎥⎥
⎢⎣ −15 2 ⎥⎦
T
⎡ 3⎤
⎡ −4 ⎤
⎡2⎤
35. x ⎢ ⎥ − y ⎢ ⎥ = 3 ⎢ ⎥
⎣2⎦
⎣ 7⎦
⎣4⎦
⎡ 3x ⎤ ⎡ −4 y ⎤ ⎡ 6⎤ ⎡ 3 x + 4 y ⎤ ⎡ 6 ⎤
⎢ 2 x ⎥ − ⎢ 7 y ⎥ = ⎢12⎥ = ⎢ 2 x − 7 y ⎥ = ⎢12 ⎥
⎣ ⎦ ⎣
⎦ ⎣ ⎦ ⎣
⎦ ⎣ ⎦
Equating corresponding entries gives
⎧3x + 4 y = 6
⎨2 x − 7 y = 12
⎩
Multiply the first equation by 2 and the second equation by −3 to get
⎧ 6 x + 8 y = 12
⎨−6 x + 21y = −36
⎩
Now add the two equations to get
183
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
29 y = −24
24
y=−
29
Therefore
⎛ 24 ⎞ 270
3x = 6 − 4 ⎜ − ⎟ =
⎝ 29 ⎠ 29
90
x=
29
90
24
The solution is x =
, y=− .
29
29
10
⎡2⎤
⎡ −1⎤
⎡ 0⎤ ⎡
⎤
⎥
6
40. x ⎢⎢ 0 ⎥⎥ + 2 ⎢⎢ 0 ⎥⎥ + y ⎢⎢ 2 ⎥⎥ = ⎢⎢
⎥
⎢⎣ 2 ⎥⎦
⎢⎣ 6 ⎥⎦
⎢⎣ −5⎥⎦ ⎢⎣ 2 x + 12 − 5 y ⎥⎦
10
⎡ 2x − 2 ⎤ ⎡
⎤
⎢
⎥
⎢
⎥
=
2
6
y
⎢
⎥ ⎢
⎥
⎢⎣ 2 x + 12 − 5 y ⎥⎦ ⎢⎣ 2 x + 12 − 5 y ⎥⎦
2x − 2 = 10, 2x = 12, or x = 6.
2y = 6 or y = 3.
2x + 12 − 5y = 2x + 12 − 5y, which is true for all
values of x and y. Thus x = 6, y = 3.
⎡ 2 x − 4 y ⎤ ⎡16 ⎤
36. ⎢
⎥=⎢ ⎥
⎣ 5 x + 7 y ⎦ ⎣ − 3⎦
⎡ 2 x ⎤ ⎡ −4 y ⎤ ⎡16 ⎤
⎢ 5 x ⎥ + ⎢ 7 y ⎥ = ⎢ −3⎥
⎣ ⎦ ⎣
⎦ ⎣ ⎦
⎡ 30 50 ⎤ ⎡ 15 25 ⎤
41. X + Y = ⎢⎢800 720 ⎥⎥ + ⎢⎢960 800 ⎥⎥
⎢⎣ 25 30 ⎥⎦ ⎢⎣ 10
5 ⎥⎦
50 + 25 ⎤ ⎡ 45
75⎤
⎡ 30 + 15
⎢
⎥
⎢
= ⎢800 + 960 720 + 800 ⎥ = ⎢1760 1520 ⎥⎥
⎢⎣ 25 + 10
30 + 5 ⎥⎦ ⎢⎣ 35
35⎥⎦
⎡2⎤
⎡ −4 ⎤ ⎡ 16 ⎤
x⎢ ⎥+ y⎢ ⎥ = ⎢ ⎥
5
⎣ ⎦
⎣ 7 ⎦ ⎣ −3⎦
⎡ x⎤
⎡ −2 ⎤
⎡ 6⎤
37. 3 ⎢ ⎥ − 3 ⎢ ⎥ = 4 ⎢ ⎥
⎣ y⎦
⎣ 4⎦
⎣ −2 ⎦
⎡ 3x + 6 ⎤ ⎡ 24 ⎤
⎢3 y − 12 ⎥ = ⎢ −8⎥
⎣
⎦ ⎣ ⎦
3x + 6 = 24, 3x = 18, or x = 6.
4
3y – 12 = –8, 3y = 4, or y = .
3
4
Thus x = 6, y = .
3
⎡ 380 330 220 ⎤ ⎡ 400 350 150 ⎤
42. 2B − A = 2 ⎢
⎥−⎢
⎥
⎣ 460 320 750 ⎦ ⎣ 450 280 850 ⎦
⎡ 2 ⋅ 380 2 ⋅ 330 2 ⋅ 220 ⎤ ⎡ 400 350 150 ⎤
=⎢
⎥−⎢
⎥
⎣ 2 ⋅ 460 2 ⋅ 320 2 ⋅ 750 ⎦ ⎣ 450 280 850 ⎦
⎡760
=⎢
⎣920
⎡360
=⎢
⎣ 470
⎡ x⎤
⎡ 7⎤ ⎡− x ⎤
38. 3 ⎢ ⎥ − 4 ⎢ ⎥ = ⎢ ⎥
⎣2⎦
⎣− y ⎦ ⎣2 y ⎦
⎡3x − 28⎤ ⎡ − x ⎤
⎢ 6 + 4 y ⎥ = ⎢2 y ⎥
⎣
⎦ ⎣ ⎦
3x – 28 = –x, 4x = 28, or x = 7.
6 + 4y = 2y, 2y = –6, or y = –3.
Thus x = 7, y = –3.
660 440 ⎤ ⎡ 400 350 150 ⎤
−
640 1500 ⎥⎦ ⎢⎣ 450 280 850 ⎥⎦
310 290 ⎤
360 650 ⎥⎦
43. P + 0.1P = [ p1 p2 p3 ] + [0.1 p1 0.1 p2 0.1 p3 ]
= [1.1 p1 1.1 p2 1.1 p3 ] = 1.1P
Thus P must be multiplied by 1.1.
44. ( A − B)T = [ A + (−1)B]T [definition of subtraction]
= A T + [(−1)B]T [transpose of a sum]
= A T + (−1)BT [transpose of a scalar multiple]
⎡ 2⎤
⎡ x ⎤ ⎡ −10⎤
⎢
⎥
39. ⎢ 4 ⎥ + 2 ⎢⎢ y ⎥⎥ = ⎢⎢ −24⎥⎥
⎣⎢ 6 ⎦⎥
⎣⎢ 4 z ⎦⎥ ⎣⎢ 14 ⎦⎥
= A T − BT [definition of subtraction]
⎡15 −4 26 ⎤
45. ⎢
⎥
⎣ 4 7 30 ⎦
⎡ 2 + 2 x ⎤ ⎡ −10 ⎤
⎢ 4 + 2 y ⎥ = ⎢ −24 ⎥
⎢
⎥ ⎢
⎥
⎢⎣ 6 + 8 z ⎥⎦ ⎢⎣ 14 ⎥⎦
2 + 2x = –10, 2x = –12, or x = –6.
4 + 2y = –24, 2y = –28, or y = –14.
6 + 8z = 14, 8z = 8, or z = 1.
Thus x = –6, y = –14, z = 1.
⎡ −16 −11 −24 ⎤
46. ⎢
⎥
⎣ −16 −3 −36 ⎦
184
ISM: Introductory Mathematical Analysis
Section 6.3
5. c31 = 0(0) + 4(−2) + 3(3) = 1
⎡ −10 22 12 ⎤
47. ⎢
⎥
⎣ 24 36 −44 ⎦
6. c12 = 1(−2) + 3(4) + (−2)(1) = 8
Principles in Practice 6.3
7. A is 2 × 3 and E is 3 × 2, so AE is 2 × 2;
2 · 2 = 4 entries.
1. Represent the value of each book by
[ 28 22 16] and the number of each book by
8. D is 4 × 3 and E is 3 × 2, so DE is 4 × 2;
4 · 2 = 8 entries.
⎡100 ⎤
⎢ 70 ⎥ .
⎢ ⎥
⎢⎣ 90 ⎥⎦
9. E is 3 × 2 and C is 2 × 5, so EC is 3 × 5;
3 · 5 = 15 entries.
The total value is given by the following matrix
product.
⎡100 ⎤
[ 28 22 16] ⎢⎢ 70 ⎥⎥ = [2800 + 1540 + 1440]
⎢⎣ 90 ⎥⎦
= [5780]
The total value is $5780.
10. D is 4 × 3 and B is 3 × 1, so DB is 4 × 1;
4 · 1 = 4 entries.
11. F is 2 × 3 and B is 3 × 1, so FB is 2 × 1;
2 · 1 = 2 entries.
12. B is 3 × 1 and C is 2 × 5. Because the number of
columns of B does not equal the number of rows
of C, BC is not defined.
2. The total cost is given by the matrix product PQ.
⎡ 250 ⎤
PQ = [ 26.25 34.75 28.50] ⎢⎢ 325 ⎥⎥
⎢⎣175 ⎥⎦
13. E is 3 × 2, ET is 2 × 3, and B is 3 × 1, so
EET B is 3 × 1; 3 ⋅ 1 = 3 entries.
= [ 6562.5 + 11, 293.75 + 4987.5] = [22,843.75]
The total cost is $22,843.75.
14. A is 2 × 3 and E is 3 × 2, so AE is 2 × 2. Thus
E(AE) is 3 × 2; 3 · 2 = 6 entries.
15. E is 3 × 2. F is 2 × 3 and B is 3 × 1, so FB is
2 × 1. Thus E(FB) is 3 × 1; 3 · 1 = 3 entries.
3. First, write the equations with the variable terms
on the left-hand side.
8
8
⎧
⎪⎪ y + 5 x = 5
⎨
⎪y + 1 x = 5
⎪⎩
3
3
⎡8⎤
⎡1 8 ⎤
⎡ y⎤
5
5⎥
Let A = ⎢
, X = ⎢ ⎥ , and B = ⎢ ⎥ .
5⎥
⎢
⎢1 1 ⎥
x
⎣ ⎦
⎣ 3⎦
⎣3⎦
Then the pair of lines is equivalent to the matrix
⎡1 8 ⎤ ⎡ y ⎤ ⎡ 8 ⎤
5
5⎥
equation AX = B or ⎢
= ⎢ ⎥.
⎢1 1 ⎥ ⎢⎣ x ⎥⎦ ⎢ 5 ⎥
⎣ 3⎦
⎣3⎦
16. Both F and A are 2 × 3, so F + A is 2 × 3.
Because B is 3 × 1, (F + A)B is 2 × 1; 2 · 1 = 2
entries.
17. An identity matrix is a square matrix (in this
case 4 × 4) with 1's on the main diagonal and all
other entries 0's.
⎡1 0 0 0 ⎤
⎢0 1 0 0⎥
⎥
I4 = ⎢
⎢0 0 1 0⎥
⎢
⎥
⎢⎣ 0 0 0 1 ⎥⎦
Problems 6.3
⎡1
⎢0
⎢
⎢0
18. I 6 = ⎢
⎢0
⎢0
⎢
⎢⎣ 0
1. c11 = 1(0) + 3(−2) + (−2)(3) = −12
2. c23 = −2(3) + 1(−2) + (−1)(−1) = −7
3. c32 = 0(−2) + 4(4) + 3(1) = 19
4. c33 = 0(3) + 4(−2) + 3(−1) = −11
185
0 0 0 0 0⎤
1 0 0 0 0 ⎥⎥
0 1 0 0 0⎥
⎥
0 0 1 0 0⎥
0 0 0 1 0⎥
⎥
0 0 0 0 1⎥⎦
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡ 2 −4 ⎤ ⎡ 4 0 ⎤ ⎡ 2(4) + (−4)(−1) 2(0) + (−4)(3) ⎤ ⎡12 −12 ⎤
19. ⎢
=
⎥⎢
⎥=⎢
3(0) + 2(3) ⎥⎦ ⎢⎣10
6 ⎥⎦
⎣ 3 2 ⎦ ⎣ −1 3⎦ ⎣3(4) + 2(−1)
⎡ −1 1⎤
⎡ −1(1) + 1(3) −1(−2) + 1(4) ⎤ ⎡ 2 6 ⎤
⎡ 1 −2 ⎤ ⎢
⎢
⎥
20. ⎢ 0 4 ⎥ ⎢
= 0(1) + 4(3) 0(−2) + 4(4) ⎥⎥ = ⎢⎢12 16 ⎥⎥
3 4 ⎥⎦ ⎢
⎣
⎢⎣ 2(1) + 1(3) 2(−2) + 1(4) ⎥⎦ ⎢⎣ 5 0 ⎥⎦
⎣⎢ 2 1⎦⎥
⎡1 ⎤
⎡ 2 0 3⎤ ⎢ ⎥ ⎡ 2(1) + 0(4) + 3(7) ⎤ ⎡ 23⎤
21. ⎢
⎥ ⎢4⎥ = ⎢
⎥=⎢ ⎥
⎣ −1 4 5⎦ ⎢7 ⎥ ⎣ −1(1) + 4(4) + 5(7) ⎦ ⎣50 ⎦
⎣ ⎦
⎡0⎤
⎢1 ⎥
22. [1 0 6 2] ⎢ ⎥ = [1(0) + 0(1) + 6(2) + 2(3)] = [18]
⎢2⎥
⎢ ⎥
⎣⎢ 3 ⎦⎥
⎡ 1 4 −1⎤ ⎡ 2 1
23. ⎢⎢ 0 0 2 ⎥⎥ ⎢⎢ 0 −1
⎣⎢ −2 1 1⎥⎦ ⎢⎣ 1 1
⎡1(2) + 4(0) + (−1)1
= ⎢⎢ 0(2) + 0(0) + 2(1)
⎢⎣ −2(2) + 1(0) + 1(1)
0⎤
1⎥⎥
2 ⎦⎥
1(1) + 4(−1) + (−1)(1) 1(0) + 4(1) + (−1)(2) ⎤ ⎡ 1 −4 2 ⎤
0(1) + 0(−1) + 2(1)
0(0) + 0(1) + 2(2) ⎥⎥ = ⎢⎢ 2 2 4 ⎥⎥
−2(1) + 1(−1) + 1(1)
−2(0) + 1(1) + 1(2) ⎥⎦ ⎢⎣ −3 −2 3⎥⎦
⎡ 4 2 −2 ⎤ ⎡ 3 1 1
24. ⎢⎢ 3 10 0 ⎥⎥ ⎢⎢ 0 0 0
⎣⎢ 1 0 2 ⎦⎥ ⎣⎢ 0 1 0
⎡ 4(3) + 2(0) + (−2)(0)
⎢
= ⎢ 3(3) + 10(0) + 0(0)
⎢⎣ 1(3) + 0(0) + 2(0)
⎡12 2 4 −2 ⎤
= ⎢⎢ 9 3 3 0 ⎥⎥
⎢⎣ 3 3 1 2 ⎥⎦
0⎤
0 ⎥⎥
1⎦⎥
4(1) + 2(0) + (−2)(1) 4(1) + 2(0) + (−2)(0) 4(0) + 2(0) + (−2)(1) ⎤
3(1) + 10(0) + 0(1)
3(1) + 10(0) + 0(0)
3(0) + 10(0) + 0(1) ⎥⎥
1(1) + 0(0) + 2(0)
1(0) + 0(0) + 2(1) ⎥⎦
1(1) + 0(0) + 2(1)
⎡ 1 5 −2 −1⎤
1⎥
25. [1 − 2 5] ⎢ 0 0 2
⎢
⎥
1 −3⎦
⎣ −1 0
= [1 + 0 − 5 5 + 0 + 0 −2 − 4 + 5 −1 − 2 − 15]
= [−4 5 −1 −18]
26. The first matrix is 1 × 2 and the second is 3 × 2, so the product is not defined.
186
ISM: Introductory Mathematical Analysis
Section 6.3
6 −4
6⎤
⎡ 2⎤
⎡ 2(2) 2(3) 2(−2) 2(3) ⎤ ⎡ 4
⎢ 3⎥
⎢ 3(2)
⎥ ⎢ 6
9
6
9 ⎥⎥
3(3)
3(
−
2)
3(3)
−
⎥ =⎢
27. ⎢ ⎥ [ 2 3 −2 3] = ⎢
⎢ −4 ⎥
⎢ −4(2) −4(3) −4(−2) −4(3) ⎥ ⎢ −8 −12 8 −12 ⎥
⎢ ⎥
⎢
⎥ ⎢
⎥
3 −2
3⎥⎦
1(3)
1(−2)
1(3) ⎥⎦ ⎢⎣ 2
⎢⎣ 1⎥⎦
⎢⎣ 1(2)
⎡ 0 1⎤ ⎪⎧ ⎡1 0
28. ⎢
⎥ ⎨⎢
⎣ 2 3⎦ ⎩⎪ ⎣1 1
⎡ 0(1) + 1(1)
=⎢
⎣ 2(1) + 3(1)
1 ⎤ ⎡ 0 1 0 ⎤ ⎪⎫ ⎡ 0 1⎤ ⎡1 1 1⎤
+
⎬=
0 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎭⎪ ⎢⎣ 2 3⎥⎦ ⎢⎣1 1 1⎥⎦
0(1) + 1(1) 0(1) + 1(1) ⎤ ⎡1 1 1 ⎤
=
2(1) + 3(1) 2(1) + 3(1) ⎥⎦ ⎢⎣5 5 5⎥⎦
⎡1
⎡ −1 0 2 ⎤ ⎪⎫ ⎢
⎪⎧ ⎡ −2 0 2 ⎤
29. 3 ⎨ ⎢
⎥ + 2 ⎢ 1 1 −2 ⎥ ⎬ ⎢3
⎪⎩ ⎣ 3 −1 1⎦
⎣
⎦ ⎪⎭ ⎢5
⎣
⎡1
⎪⎧ ⎡ −2 0 2 ⎤ ⎡ −2 0 4 ⎤ ⎪⎫ ⎢
= 3 ⎨⎢
+
⎥ ⎢
⎥ ⎬ ⎢3
⎩⎪ ⎣ 3 −1 1⎦ ⎣ 2 2 −4 ⎦ ⎭⎪ ⎢5
⎣
1
2
⎡
⎤
⎪⎧ ⎡ −4 0 6 ⎤ ⎪⎫ ⎢
⎥ = ⎡ −12 0
= 3 ⎨⎢
3
4
⎥⎬ ⎢
⎥ ⎢
⎩⎪ ⎣ 5 1 −3⎦ ⎭⎪ ⎢5 6 ⎥ ⎣ 15 3
⎣
⎦
2⎤
4 ⎥⎥
6 ⎥⎦
2⎤
4 ⎥⎥
6 ⎥⎦
⎡1 2 ⎤
18⎤ ⎢
3 4 ⎥⎥
−9 ⎥⎦ ⎢
⎢⎣5 6 ⎥⎦
⎡ −12(1) + 0(3) + 18(5) −12(2) + 0(4) + 18(6) ⎤ ⎡ 78 84 ⎤
=⎢
⎥=⎢
⎥
⎣15(1) + 3(3) + (−9)(5) 15(2) + 3(4) + (−9)(6) ⎦ ⎣ −21 −12 ⎦
⎡ 1 −1⎤ ⎡ −1 0 −1 0 0 ⎤
30. ⎢
⎣ 0 3⎥⎦ ⎢⎣ 2 1 2 1 1⎥⎦
⎡1(−1) + (−1)(2) 1(0) + (−1)(1) 1(−1) + (−1)(2) 1(0) + (−1)(1) 1(0) + (−1)(1) ⎤
=⎢
0(0) + 3(1)
0(−1) + 3(2)
0(0) + 3(1)
0(0) + 3(1) ⎥⎦
⎣ 0(−1) + 3(2)
= ⎡ −3 1 −3 −1 −1⎤
⎢⎣ 6 3 6 3 3 ⎥⎦
⎧
⎡ 1 −2 ⎤ ⎫
1⎤ ⎢
⎡1 2 ⎤ ⎪ ⎡ 2 0
⎪ ⎡1 2⎤ ⎪⎧ ⎡ 2 + 0 + 3 −4 + 0 + 0 ⎤ ⎪⎫
1⎥⎥ ⎬ = ⎢
31. ⎢
⎥ ⎨ ⎢ 1 0 −2 ⎥ ⎢ 2
⎥ ⎨⎢
⎥⎬
3
4
⎣
⎦ ⎪⎣
⎦ ⎢ 3 0 ⎥ ⎪ ⎣3 4⎦ ⎩⎪ ⎣1 + 0 − 6 −2 + 0 + 0⎦ ⎭⎪
⎣
⎦⎭
⎩
⎡1 2 ⎤ ⎡ 5 −4 ⎤ ⎡ 5 − 10 −4 − 4 ⎤ ⎡ −5 −8⎤
=⎢
⎥⎢
⎥=⎢
⎥=⎢
⎥
⎣3 4 ⎦ ⎣ −5 −2 ⎦ ⎣15 − 20 −12 − 8⎦ ⎣ −5 −20⎦
⎧⎪ ⎡ 1 0 ⎤ ⎡ −2 4⎤ ⎫⎪ ⎡ 3 6 ⎤
⎧⎪ ⎡ −2 4 ⎤ ⎫⎪
⎡ 1 2⎤
32. 3 ⎢
− 4 ⎨⎢
− 4 ⎨I ⎢
⎬=⎢
⎥
⎥
⎢
⎥
⎥
⎥⎬
⎪⎩ ⎣0 1⎦ ⎣ 6 1⎦ ⎪⎭ ⎣ −3 12 ⎦
⎪⎩ ⎣ 6 1⎦ ⎪⎭
⎣ −1 4 ⎦
⎧⎪ ⎡ −2 4 ⎤ ⎫⎪ ⎡ 3 6 ⎤ ⎡ −8 16 ⎤ ⎡ 11 −10 ⎤
⎡ 3 6⎤
=⎢
− 4 ⎨⎢
⎥
⎥⎬ = ⎢
⎥−⎢
⎥=⎢
8⎥⎦
⎣ −3 12 ⎦
⎩⎪ ⎣ 6 1⎦ ⎭⎪ ⎣ −3 12 ⎦ ⎣ 24 4 ⎦ ⎣ −27
187
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡0 0 1 ⎤ ⎡ x ⎤ ⎡0 ⋅ x + 0 ⋅ y + 1⋅ z ⎤ ⎡ z ⎤
33. ⎢⎢ 0 1 0 ⎥⎥ ⎢⎢ y ⎥⎥ = ⎢⎢ 0 ⋅ x + 1 ⋅ y + 0 ⋅ z ⎥⎥ = ⎢⎢ y ⎥⎥
⎢⎣1 0 0 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣1 ⋅ x + 0 ⋅ y + 0 ⋅ z ⎥⎦ ⎢⎣ x ⎥⎦
a ⎤ ⎡ x ⎤ ⎡a x + a x ⎤
⎡a
34. ⎢ 11 12 ⎥ ⎢ 1 ⎥ = ⎢ 11 1 12 2 ⎥
⎣ a21 a22 ⎦ ⎣ x2 ⎦ ⎣ a21 x1 + a22 x2 ⎦
⎡ x1 ⎤
⎡ 2 1 3 ⎤ ⎢ ⎥ ⎡ 2 x1 + x2 + 3x3 ⎤
35. ⎢
⎥
⎥ ⎢ x2 ⎥ = ⎢
⎣ 4 9 7 ⎦ ⎢ x ⎥ ⎣ 4 x1 + 9 x2 + 7 x3 ⎦
⎣ 3⎦
⎡ 2 −3⎤
⎡ 2 x1 − 3 x2 ⎤
⎡ x1 ⎤ ⎢
⎢
⎥
⎥
1⎥ ⎢ ⎥ = ⎢ x2
36. ⎢ 0
⎥
x2 ⎦
⎣
⎢⎣ 2
⎢⎣ 2 x1 + x2 ⎥⎦
1⎥⎦
⎡1 0 0 ⎤
⎡3 0 0⎤
1
1
1⎢
⎢
⎥
37. D − EI = D − E = ⎢0 1 1 ⎥ − ⎢ 0 6 0 ⎥⎥
3
3
3
⎢⎣ 0 0 3 ⎦⎥
⎣⎢1 2 1 ⎦⎥
⎡1 0 0 ⎤ ⎡1 0 0 ⎤
= ⎢⎢0 1 1 ⎥⎥ − ⎢⎢0 2 0 ⎥⎥
⎢⎣1 2 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦
⎡0 0 0 ⎤
= ⎢⎢0 −1 1⎥⎥
⎢⎣ 1 2 0 ⎥⎦
⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 + 0 + 0 0 + 0 + 0 0 + 0 + 0⎤ ⎡ 1 0 0 ⎤
38. DD = ⎢⎢ 0 1 1 ⎥⎥ ⎢⎢ 0 1 1 ⎥⎥ = ⎢⎢ 0 + 0 + 1 0 + 1 + 2 0 + 1 + 1 ⎥⎥ = ⎢⎢ 1 3 2⎥⎥
⎢⎣1 2 1 ⎥⎦ ⎢⎣1 2 1 ⎥⎦ ⎢⎣1 + 0 + 1 0 + 2 + 2 0 + 2 + 1⎥⎦ ⎢⎣ 2 4 3⎥⎦
⎡ −1 1⎤
3 0⎤ ⎢
⎡ 1 −2 ⎤
⎡ −2
⎥
39. 3A − 2BC = 3 ⎢
⎥ − 2 ⎢ 1 −4 1⎥ ⎢ 0 3⎥
0
3
⎣
⎦
⎣
⎦ ⎢ 2 4⎥
⎣
⎦
⎡ 3 −6 ⎤
⎡ 2 + 0 + 0 −2 + 9 + 0 ⎤
=⎢
−2⎢
⎥
⎥
⎣0 9 ⎦
⎣ −1 + 0 + 2 1 − 12 + 4 ⎦
⎡ 3 −6 ⎤ ⎡ 4 14 ⎤ ⎡ −1 −20 ⎤
=⎢
⎥−⎢
⎥=⎢
23⎥⎦
⎣0 9 ⎦ ⎣ 2 −14 ⎦ ⎣ −2
⎡4 0 0⎤
3 0⎤ ⎢
⎡ −2
⎡ −8 + 0 + 0 0 + 21 + 0 0 + 3 + 0 ⎤
0 7 1 ⎥⎥ = ⎢
40. B(D + E) = ⎢
⎥
⎥
⎢
⎣ 1 −4 1⎦ ⎢ 1 2 4 ⎥ ⎣ 4 + 0 + 1 0 − 28 + 2 0 − 4 + 4 ⎦
⎣
⎦
21 3⎤
⎡ −8
=⎢
5
−
26
0 ⎥⎦
⎣
188
ISM: Introductory Mathematical Analysis
Section 6.3
⎡ 1 0 0⎤
⎢3
⎥ ⎡3 0 0⎤
2
2
41. 3I − FE = 3I − ⎢ 0 16 0 ⎥ ⎢⎢ 0 6 0 ⎥⎥
⎥
3
3⎢
⎢ 0 0 1 ⎥ ⎢⎣ 0 0 3 ⎥⎦
3⎦
⎣
1
⎡ ⋅3+ 0 + 0 0 + 0 + 0 0 + 0 + 0 ⎤
⎢3
⎥
2⎢
1 ⋅6 + 0
⎥
0+0+0
0
0
0
= 3I −
+
+
6
⎥
3⎢
⎢ 0 + 0 + 0 0 + 0 + 0 0 + 0 + 1 ⋅ 3⎥
3 ⎦
⎣
⎡ 2 0 0⎤ ⎡ 7
⎡1 0 0 ⎤ ⎡ 3 0 0 ⎤ ⎢ 3
⎥ ⎢3
2⎢
⎥
⎢
⎥
2
⎢
= 3I − ⎢ 0 1 0 ⎥ = ⎢ 0 3 0 ⎥ − 0 3 0 ⎥ = ⎢ 0
⎢
⎥ ⎢
3
⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 3 ⎥⎦ ⎢
0 0 23 ⎥ ⎢⎢ 0
⎣
⎦ ⎣
⎡1
⎢3
42. FE(D − I ) = ⎢ 0
⎢
⎢⎣ 0
⎡1
= ⎢0
⎢
⎣0
= ⎡0
⎢0
⎢
⎣1
0 0⎤ 3
⎥⎡
1 0 ⎢0
⎥
6
⎥⎢
0 13 ⎥⎦ ⎣ 0
0 0⎤ ⎡0
1 0⎥ ⎢0
⎥⎢
0 1⎦ ⎣ 1
0 0⎤
0 1⎥
⎥
2 0⎦
0
7
3
0
0⎤
⎥
0⎥
⎥
7⎥
3 ⎥⎦
0 0⎤ ⎛ ⎡ 1 0 0⎤ ⎡ 1 0 0⎤ ⎞
6 0 ⎥ ⎜ ⎢0 1 1⎥ − ⎢ 0 1 0 ⎥ ⎟
⎥⎜⎢
⎥ ⎢
⎥⎟
0 3⎦ ⎜⎝ ⎣ 1 2 1⎦ ⎣ 0 0 1⎦ ⎟⎠
0 0⎤
0 1⎥
⎥
2 0⎦
⎧ ⎡1 0 0 ⎤ ⎡ −1 1⎤ ⎫
⎡ −1 + 0 + 0
⎪⎢
⎪
⎥
⎢
⎥
43. (DC) A = ⎨ ⎢0 1 1 ⎥ ⎢ 0 3⎥ ⎬ A = ⎢⎢ 0 + 0 + 2
⎪ ⎢1 2 1 ⎥ ⎢ 2 4 ⎥ ⎪
⎢⎣ −1 + 0 + 2
⎦⎣
⎦⎭
⎩⎣
2 + 3 ⎤ ⎡ −1
⎡ −1 1⎤
⎡ −1 + 0
⎡ 1 −2 ⎤ ⎢
⎢
⎥
= ⎢ 2 7⎥ ⎢
= 2 + 0 −4 + 21⎥⎥ = ⎢⎢ 2
0
3⎥⎦ ⎢
⎣
⎢⎣ 1 11⎥⎦
⎢⎣ 1 + 0 −2 + 33⎥⎦ ⎢⎣ 1
1 + 0 + 0⎤
0 + 3 + 4 ⎥⎥ A
1 + 6 + 4 ⎥⎦
5⎤
17 ⎥⎥
31⎥⎦
⎧
⎡ –1 1⎤ ⎫
3 0⎤ ⎢
⎡ 2 + 0 + 0 −2 + 9 + 0 ⎤ ⎡ 1 − 2 ⎤ ⎡ 2 7 ⎤
⎪ ⎡ −2
⎪
44. A(BC) = A ⎨ ⎢
0 3⎥⎥ ⎬ = A ⎢
⎥
⎥=⎢
⎢
3⎥⎦ ⎢⎣ 1 −7 ⎥⎦
⎣ −1 + 0 + 2 1 − 12 + 4 ⎦ ⎣ 0
⎪ ⎣ 1 −4 1⎦ ⎢ 2 4 ⎥ ⎪
⎣
⎦⎭
⎩
⎡ 2 − 2 7 + 14 ⎤ ⎡ 0 21⎤
=⎢
⎥=⎢
⎥
⎣ 0 + 3 0 − 21⎦ ⎣ 3 −21⎦
45. Impossible: A is not a square matrix, so A 2 is not defined.
⎡ 1 0⎤
⎡ 1 −1 0 ⎤
⎡ 1 −1 0 ⎤ ⎢
⎢
⎥
⎥
46. A A = ⎢ −1 1⎥ ⎢
⎥ = ⎢ −1 2 1⎥
0
1
1
⎦ ⎢ 0 1 1⎥
⎢⎣ 0 1⎥⎦ ⎣
⎣
⎦
T
189
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
2
⎡ 0 0 −1⎤
⎡ 0 0 −1⎤ ⎡ 0 0 −1⎤
⎢
⎥
3
2
47. B = B B = ⎢ 2 −1 0 ⎥ B = ⎢⎢ 2 −1 0 ⎥⎥ ⎢⎢ 2 −1 0 ⎥⎥ B
⎢⎣ 0 0 2 ⎥⎦
⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦
⎡ 0 0 −2 ⎤ ⎡ 0 0 −1⎤ ⎡ 0 0 −4 ⎤
= ⎢⎢ −2 1 −2 ⎥⎥ ⎢⎢ 2 −1 0 ⎥⎥ = ⎢⎢ 2 −1 −2 ⎥⎥
⎢⎣ 0 0 4 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0
8⎥⎦
( )
⎡ 0 2 0⎤ ⎡ 0 2 0⎤
⎢
⎥⎢
⎥
48. A(B ) C = A ⎢ 0 −1 0 ⎥ ⎢ 0 −1 0 ⎥ C
⎢⎣ −1 0 2 ⎥⎦ ⎢⎣ −1 0 2 ⎥⎦
⎡ 0 −2 0 ⎤
= A ⎢⎢ 0
1 0 ⎥⎥ C
⎢⎣ −2 −2 4 ⎥⎦
⎡ 0 −2 0 ⎤
⎡ 1 −1 0 ⎤ ⎢
0
1 0 ⎥⎥ C
=⎢
⎥
⎢
⎣ 0 1 1⎦ ⎢ −2 −2 4 ⎥
⎣
⎦
T 2
⎡ 1 0⎤
⎡ 0 −3 0 ⎤ ⎢
⎥
=⎢
⎥ ⎢ 2 −1⎥
⎣ −2 −1 4 ⎦ ⎢ 0 1⎥
⎣
⎦
⎡ −6 3⎤
=⎢
⎥
⎣ −4 5 ⎦
⎛
⎡ 1 0 0⎤ ⎡ 1 0⎤ ⎞
⎡ 1 −1 0 ⎤ ⎢
0 1 0 ⎥ ⎢ 2 −1⎥ ⎟
49. ( AIC)T = ⎜ ⎢
⎜⎜ ⎣0 1 1⎥⎦ ⎢
⎥⎢
⎥ ⎟⎟
⎣ 0 0 1⎦ ⎣ 0 1⎦ ⎠
⎝
⎛
⎡1
= ⎜⎢
⎜⎜ ⎣0
⎝
⎡ −1
=⎢
⎣ 2
⎡ −1
=⎢
⎣ 1
50. A
T
( 2C )
T
⎡ 1 0⎤ ⎞
−1 0 ⎤ ⎢
2 −1⎥ ⎟
1 1⎥⎦ ⎢
⎥ ⎟⎟
⎣ 0 1⎦ ⎠
T
1⎤
0 ⎥⎦
2⎤
0 ⎥⎦
T
⎡ 1 0⎤
⎡ 2 4 0⎤
⎡2 4 0⎤ ⎢
⎢
⎥
= ⎢ −1 1⎥ ⎢
= −2 −6 2 ⎥⎥
0 −2 2 ⎥⎦ ⎢
⎣
⎢⎣ 0 1⎥⎦
⎢⎣ 0 −2 2 ⎥⎦
T
51.
( BA )
T
T
T
T
⎧ ⎡ 0 0 −1⎤ ⎡ 1 0 ⎤ ⎫
⎡ 0 −1⎤
⎡ 0 3 0⎤
⎪⎢
⎪
⎥
⎢
⎥
⎢
= ⎨ ⎢ 2 −1 0 ⎥ ⎢ −1 1⎥ ⎬ = ⎢ 3 −1⎥⎥ = ⎢
⎥
⎣ −1 −1 2 ⎦
⎪ ⎢ 0 0 2 ⎥ ⎢ 0 1⎥ ⎪
⎢
⎥
0
2
⎣
⎦
⎣
⎦
⎣
⎦
⎩
⎭
190
ISM: Introductory Mathematical Analysis
T
Section 6.3
T
⎧ ⎡ 0 0 −1⎤ ⎫
⎡ 0 0 −2 ⎤
⎡ 0 4 0⎤
⎪
⎥
⎢
⎥
⎢
⎥
T ⎪ ⎢
(2
)
2
2
1
0
4
2
0
52.
B =⎨ ⎢ −
⎥⎬ = ⎢ −
⎥ = ⎢ 0 −2 0 ⎥
⎪ ⎢0 0 2⎥ ⎪
⎢⎣ 0 0 4 ⎥⎦
⎢⎣ −2 0 4 ⎥⎦
⎦⎭
⎩ ⎣
2
⎡2 0 0⎤ ⎡2
2
2
2
53. (2I ) − 2I = (2I ) − 2I = ⎢⎢ 0 2 0 ⎥⎥ − ⎢⎢ 0
⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0
⎡2 0 0⎤ ⎡2 0 0⎤ ⎡2 0 0⎤ ⎡4
= ⎢⎢ 0 2 0 ⎥⎥ ⎢⎢ 0 2 0 ⎥⎥ − ⎢⎢ 0 2 0 ⎥⎥ = ⎢⎢ 0
⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0
0 0⎤
2 0 ⎥⎥
0 2 ⎥⎦
0 0⎤ ⎡2 0 0⎤ ⎡2 0 0⎤
4 0 ⎥⎥ − ⎢⎢ 0 2 0 ⎥⎥ = ⎢⎢ 0 2 0 ⎥⎥
0 4 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦
⎡ 1 0 0⎤
54. A T is 3 × 2, CT is 2 × 3, and B is 3 × 3, so A T CT B is 3 × 3 and ( A T CT B)0 = I = ⎢ 0 1 0 ⎥ .
⎢
⎥
⎣ 0 0 1⎦
⎡ 1 −1 0 ⎤
55. A(I – O) = A(I) = AI. Since I is 3 × 3 and A has three columns, AI = A. Thus A(I − O) = A = ⎢
⎥.
⎣ 0 1 1⎦
⎡0 0 0 ⎤
56. I O = IO = O = ⎢⎢0 0 0 ⎥⎥
⎢⎣0 0 0 ⎥⎦
T
⎡ 0 0 −1⎤
⎡ 1 −1 0 ⎤ ⎢
⎡ −2 1 −1⎤
T
2 −1 0 ⎥⎥ ( AB)T = ⎢
57. ( AB)( AB) = ⎢
⎥
⎥ ( AB)
⎢
−
0
1
1
2
1
2
⎣
⎦ ⎢ 0 0 2⎥
⎣
⎦
⎣
⎦
⎡ −2 2 ⎤
⎡ −2 1 −1⎤ ⎢
⎡ 6 −7 ⎤
=⎢
1 −1⎥⎥ = ⎢
⎥
⎢
9 ⎥⎦
⎣ 2 −1 2 ⎦ ⎢ −1 2 ⎥ ⎣ −7
⎣
⎦
T
58. B 2 − 3B + 2I
⎡ 0 0 −1⎤ ⎡ 0 0 −1⎤
⎡0
= ⎢⎢ 2 −1 0 ⎥⎥ ⎢⎢ 2 −1 0 ⎥⎥ − 3 ⎢⎢ 2
⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦
⎢⎣ 0
⎡ 0 0 −2 ⎤ ⎡ 0 0 −3⎤ ⎡ 2
= ⎢⎢ −2 1 −2 ⎥⎥ − ⎢⎢ 6 −3 0 ⎥⎥ + ⎢⎢ 0
⎢⎣ 0 0 4 ⎥⎦ ⎢⎣ 0 0 6 ⎥⎦ ⎢⎣ 0
0 −1⎤
⎡1 0 0⎤
−1 0 ⎥⎥ + 2 ⎢⎢ 0 1 0 ⎥⎥
⎢⎣ 0 0 1 ⎥⎦
0 2 ⎥⎦
0 0⎤
2 0 ⎥⎥
0 2 ⎥⎦
1⎤
1⎤ ⎡ 2 0 0 ⎤ ⎡ 2 0
⎡ 0 0
⎢
⎢
⎥
⎢
⎥
= ⎢ −8 4 −2 ⎥ + ⎢ 0 2 0 ⎥ = ⎢ −8 6 −2 ⎥⎥
⎢⎣ 0 0 −2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦
59. AX = B
1⎤
⎡3
A=⎢
⎥
−
2
9
⎣
⎦
191
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡ x⎤
X=⎢ ⎥
⎣ y⎦
⎡6⎤
B=⎢ ⎥
⎣5⎦
1⎤ ⎡ x ⎤ ⎡6 ⎤
⎡3
The system is represented by ⎢
⎥⎢ ⎥ = ⎢ ⎥ .
⎣ 2 −9 ⎦ ⎣ y ⎦ ⎣5 ⎦
60. AX = B
⎡ 3 1 1⎤
A = ⎢⎢ 1 −1 1⎥⎥
⎢⎣5 −1 2 ⎥⎦
⎡ x⎤
X = ⎢⎢ y ⎥⎥
⎢⎣ z ⎥⎦
⎡ 2⎤
B = ⎢⎢ 4 ⎥⎥
⎢⎣12 ⎥⎦
⎡ 3 1 1⎤ ⎡ x ⎤ ⎡ 2 ⎤
The system is represented by ⎢ 1 −1 1⎥ ⎢ y ⎥ = ⎢ 4 ⎥ .
⎢
⎥⎢ ⎥ ⎢ ⎥
⎣5 −1 2 ⎦ ⎣ z ⎦ ⎣12 ⎦
61. AX = B
⎡ 2 −1 3⎤
A = ⎢⎢ 5 −1 2 ⎥⎥
⎢⎣ 3 −2 2 ⎥⎦
⎡r ⎤
X = ⎢⎢ s ⎥⎥
⎢⎣ t ⎥⎦
⎡ 9⎤
B = ⎢⎢ 5⎥⎥
⎢⎣11⎥⎦
⎡ 2 −1 3⎤ ⎡ r ⎤ ⎡ 9 ⎤
The system is represented by ⎢⎢ 5 −1 2 ⎥⎥ ⎢⎢ s ⎥⎥ = ⎢⎢ 5⎥⎥ .
⎢⎣ 3 −2 2 ⎥⎦ ⎢⎣ t ⎥⎦ ⎢⎣11⎥⎦
62. “the/falcon/has/landed” converted to corresponding numbers and slashes is “20, 8, 5/ 6, 1, 12, 3, 15, 14/ 8, 1, 19/
12, 1, 14, 4, 5, 4.” Taking the numbers two at a time as 2 × 1 matrices and multiplying them by E gives:
⎡ 1 3 ⎤ ⎡ 20 ⎤ ⎡ 1 ⋅ 20 + 3 ⋅ 8 ⎤ ⎡ 20 + 24⎤ ⎡ 44⎤
⎢ 2 4 ⎥ ⎢ 8⎥ = ⎢ 2 ⋅ 20 + 4 ⋅ 8⎥ = ⎢ 40 + 32 ⎥ = ⎢ 72 ⎥
⎣
⎦⎣ ⎦ ⎣
⎦ ⎣
⎦ ⎣ ⎦
⎡ 1 3 ⎤ ⎡5 ⎤ ⎡ 1 ⋅ 5 + 3 ⋅ 6 ⎤ ⎡ 5 + 18 ⎤ ⎡ 23⎤
⎢ 2 4 ⎥ ⎢6 ⎥ = ⎢ 2 ⋅ 5 + 4 ⋅ 6 ⎥ = ⎢10 + 24⎥ = ⎢34 ⎥
⎣
⎦⎣ ⎦ ⎣
⎦ ⎣
⎦ ⎣ ⎦
192
ISM: Introductory Mathematical Analysis
⎡1
⎢2
⎣
⎡1
⎢2
⎣
3 ⎤ ⎡ 1⎤ ⎡ 1 ⋅1 + 3 ⋅12 ⎤ ⎡ 1 + 36 ⎤ ⎡37 ⎤
=
=
=
4 ⎥⎦ ⎢⎣12 ⎥⎦ ⎢⎣ 2 ⋅1 + 4 ⋅12 ⎥⎦ ⎢⎣ 2 + 48⎥⎦ ⎢⎣50 ⎥⎦
⎡1
⎢2
⎣
⎡1
⎢2
⎣
3 ⎤ ⎡14 ⎤ ⎡ 1 ⋅14 + 3 ⋅ 8 ⎤ ⎡14 + 24 ⎤ ⎡ 38 ⎤
=
=
=
4 ⎥⎦ ⎢⎣ 8⎥⎦ ⎢⎣ 2 ⋅14 + 4 ⋅ 8⎥⎦ ⎢⎣ 28 + 32 ⎥⎦ ⎢⎣ 60 ⎥⎦
⎡1
⎢2
⎣
⎡1
⎢2
⎣
3 ⎤ ⎡12 ⎤ ⎡ 1 ⋅12 + 3 ⋅1 ⎤ ⎡12 + 3 ⎤ ⎡15 ⎤
=
=
=
4 ⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣ 2 ⋅12 + 4 ⋅1⎥⎦ ⎢⎣ 24 + 4⎥⎦ ⎢⎣ 28⎥⎦
Section 6.3
3 ⎤ ⎡ 3⎤ ⎡ 1 ⋅ 3 + 3 ⋅15 ⎤ ⎡ 3 + 45 ⎤ ⎡ 48⎤
=
=
=
4 ⎥⎦ ⎢⎣15⎥⎦ ⎢⎣ 2 ⋅ 3 + 4 ⋅15⎥⎦ ⎢⎣6 + 60 ⎥⎦ ⎢⎣ 66 ⎥⎦
3 ⎤ ⎡ 1⎤ ⎡ 1 ⋅1 + 3 ⋅19 ⎤ ⎡ 1 + 57 ⎤ ⎡ 58⎤
=
=
=
4 ⎥⎦ ⎢⎣19 ⎥⎦ ⎢⎣ 2 ⋅1 + 4 ⋅19 ⎥⎦ ⎢⎣ 2 + 76 ⎥⎦ ⎢⎣ 78⎥⎦
3 ⎤ ⎡14 ⎤ ⎡ 1 ⋅14 + 3 ⋅ 4 ⎤ ⎡14 + 12 ⎤ ⎡ 26 ⎤
=
=
=
4 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ 2 ⋅14 + 4 ⋅ 4 ⎥⎦ ⎢⎣ 28 + 16⎥⎦ ⎢⎣ 44 ⎥⎦
⎡ 1 3 ⎤ ⎡ 5⎤ ⎡ 1 ⋅ 5 + 3 ⋅ 4 ⎤ ⎡ 5 + 12⎤ ⎡17 ⎤
⎢ 2 4 ⎥ ⎢ 4 ⎥ = ⎢ 2 ⋅ 5 + 4 ⋅ 4 ⎥ = ⎢10 + 16 ⎥ = ⎢ 26 ⎥
⎣
⎦⎣ ⎦ ⎣
⎦ ⎣
⎦ ⎣ ⎦
The encoded message is
44, 72, 23/ 34, 37, 50, 48, 66, 38/ 60, 58, 78/ 15, 28, 26, 44, 17, 26.
⎡ 55 ⎤
63. [ 6 10 7 ] ⎢⎢150 ⎥⎥ = [6 ⋅ 55 + 10 ⋅150 + 7 ⋅ 35]
⎢⎣ 35 ⎥⎦
= [330 + 1500 + 245]
= [2075]
The value of the inventory is $2075.
⎡100 ⎤
⎢150 ⎥
⎥ = [240, 000]
64. [ 200 300 500 250] ⎢
⎢ 200 ⎥
⎢
⎥
⎢⎣300 ⎥⎦
The total cost of the stocks is $240,000.
65. Q = [5
⎡5
R = ⎢⎢ 7
⎢⎣ 6
2 4]
20 16 7 17 ⎤
18 12 9 21⎥⎥
25 8 5 13⎥⎦
⎡ 2500 ⎤
⎢ 1200 ⎥
⎢
⎥
C = ⎢ 800 ⎥
⎢
⎥
⎢ 150 ⎥
⎢⎣ 1500 ⎥⎦
193
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡5 ⋅ 2500 + 20 ⋅1200 + 16 ⋅ 800 + 7 ⋅150 + 17 ⋅1500 ⎤
QRC = Q(RC) = Q ⎢⎢ 7 ⋅ 2500 + 18 ⋅1200 + 12 ⋅ 800 + 9 ⋅150 + 21 ⋅1500 ⎥⎥
⎢⎣ 6 ⋅ 2500 + 25 ⋅1200 + 8 ⋅ 800 + 5 ⋅150 + 13 ⋅1500 ⎥⎦
⎡75,850 ⎤
= [5 2 4] ⎢⎢ 81,550 ⎥⎥
⎢⎣ 71, 650 ⎥⎦
= [5(75,850) + 2(81,550) + 4(71, 650)]
= [828,950]
The total cost of raw materials is $828,950.
66. a.
b.
⎡3500 50 ⎤
⎢
⎥
⎡ 5 20 16 7 17 ⎤ ⎢1500 50 ⎥
RC = ⎢⎢ 7 18 12 9 21⎥⎥ ⎢1000 100 ⎥
⎢
⎥
⎢⎣ 6 25 8 5 13 ⎥⎦ ⎢ 250 10 ⎥
⎢3500
0 ⎥⎦
⎣
⎡ 17,500 + 30, 000 + 16, 000 + 1750 + 59,500 250 + 1000 + 1600 + 70 + 0 ⎤
= ⎢⎢ 24,500 + 27, 000 + 12, 000 + 2250 + 73,500 350 + 900 + 1200 + 90 + 0 ⎥⎥
⎢⎣ 21, 000 + 37,500 + 8000 + 1250 + 45,500
300 + 1250 + 800 + 50 + 0 ⎥⎦
⎡124, 750 2920 ⎤
= ⎢⎢139, 250 2540 ⎥⎥
⎢⎣113, 250 2400 ⎥⎦
⎡124, 750 2920 ⎤
QRC = Q(RC) = [5 7 12] ⎢⎢139, 250 2540 ⎥⎥
⎢⎣113, 250 2400 ⎥⎦
= [ 623, 750 + 974, 750 + 1,359, 000 14, 600 + 17, 780 + 28,800]
= [ 2,957,500 61,180]
c.
67. a.
⎡1⎤
QRCZ = (QRC)Z = [ 2,957,500 61,180] ⎢ ⎥
⎣1⎦
= [2,957,500 + 61,180] = [3,018,680]
Amount spent on goods:
⎡10, 000 ⎤
coal industry: DC P = [ 0 1 4] ⎢⎢ 20, 000 ⎥⎥ = [180, 000]
⎢⎣ 40, 000 ⎥⎦
⎡10, 000 ⎤
elec. industry: DE P = [ 20 0 8] ⎢⎢ 20, 000 ⎥⎥ = [520, 000]
⎢⎣ 40, 000 ⎥⎦
⎡10, 000 ⎤
steel industry: DS P = [30 5 0] ⎢⎢ 20, 000 ⎥⎥ = [400, 000]
⎢⎣ 40, 000 ⎥⎦
The coal industry spends $180,000, the electric industry spends $520,000, and the steel industry spends
194
ISM: Introductory Mathematical Analysis
Section 6.3
$400,000.
⎡10, 000 ⎤
consumer 1: D1P = [3 2 5] ⎢⎢ 20, 000 ⎥⎥ = [270, 000]
⎢⎣ 40, 000 ⎥⎦
⎡10, 000 ⎤
consumer 2: D2 P = [ 0 17 1] ⎢⎢ 20, 000 ⎥⎥ = [380, 000]
⎢⎣ 40, 000 ⎥⎦
⎡10, 000 ⎤
consumer 3: D3 P = [ 4 6 12] ⎢⎢ 20, 000 ⎥⎥ = [640, 000]
⎢⎣ 40, 000 ⎥⎦
Consumer 1 pays $270,000, consumer 2 pays $380,000, and consumer 3 pays $640,000.
b. From Example 3 of Sec. 6.2, the number of units sold of coal, electricity, and steel are 57, 31, and 30,
respectively. Thus the profit for coal is 10,000(57) – 180,000 = $390,000, the profit for elec. is
20,000(31) – 520,000 = $100,000, and the profit for steel is 40,000(30) – 400,000 = $800,000.
c.
From (a), the total amount of money that is paid out by all the industries and consumers is
180,000 + 520,000 + 400,000 + 270,000 + 380,000 + 640,000 = $2,390,000.
d. The proportion of the total amount in (c) paid out by the industries is
180, 000 + 520, 000 + 400, 000 110
.
=
2,390, 000
239
The proportion of the total amount in (c) paid by consumers is
270, 000 + 380, 000 + 640, 000 129
.
=
2,390, 000
239
68. (A + B)(A – B) = A(A – B) + B(A – B) [dist. prop.]
= A 2 − AB + BA − B 2 [dist prop.]
= A 2 − BA + BA − B 2 [AB = BA, given]
= A2 − B2
⎡
⎡1 2 ⎤ ⎡ 2 −3⎤ ⎢1(2) + (2)(−1) 1(−3) + 2
=
69. ⎢
⎢
⎥
⎥
3
⎣1 2 ⎦ ⎣⎢ −1 2 ⎥⎦ ⎢⎢ 1(2) + 2(−1) 1(−3) + 2
⎣
⎡a 0 0⎤
⎡d 0
⎢
⎥
70. Let D1 = ⎢ 0 b 0 ⎥ and D2 = ⎢⎢ 0 e
⎢⎣ 0 0 c ⎥⎦
⎢⎣ 0 0
a.
⎡a
D1D2 = ⎢⎢ 0
⎢⎣ 0
⎡d
D2 D1 = ⎢⎢ 0
⎢⎣ 0
Both D1D2
0 0⎤ ⎡d
b 0 ⎥⎥ ⎢⎢ 0
0 c ⎥⎦ ⎢⎣ 0
0 0 ⎤ ⎡a
e 0 ⎥⎥ ⎢⎢ 0
0 f ⎥⎦ ⎢⎣ 0
and D2 D1
( 32 )⎤⎥ = ⎡0
⎢
( 32 )⎦⎥⎥ ⎣0
0⎤
0 ⎥⎥ .
f ⎥⎦
0 ⎤ ⎡ ad 0 0 ⎤
0 ⎥⎥ = ⎢⎢ 0 be 0 ⎥⎥
0 f ⎥⎦ ⎢⎣ 0 0 cf ⎥⎦
0 0 ⎤ ⎡ ad 0 0 ⎤
b 0 ⎥⎥ = ⎢⎢ 0 be 0 ⎥⎥
0 c ⎥⎦ ⎢⎣ 0 0 cf ⎥⎦
are diagonal matrices.
0
e
195
0⎤
0 ⎦⎥
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
b. From part (a), D1D2 = D2 D1 . Thus D1 and
⎡1 0 0 5 ⎤
9 R +R
− 13
3
1
> ⎢⎢ 0 1 0 2 ⎥⎥
6 R +R
− 13
3
2
⎢⎣ 0 0 1 1 ⎥⎦
Thus there should be 5 blocks of A, 2 blocks of
B, and 1 block of C suggested.
D2 commute. [In fact, all n × n diagonal
matrices commute.]
−9.8⎤
⎡ 72.82
71. ⎢
⎥
⎣ 51.32 −36.32 ⎦
2. Let x be the number of tablets of X, y be the
number of tablets of Y, and z be the number of
tablets of Z. The system is
40x + 10y + 10z = 180
20x + 10y + 50z = 200
10x + 30y + 20z = 190
Reduce the augmented coefficient matrix of the
system.
⎡ 40 10 10 180 ⎤
⎢ 20 10 50 200 ⎥
⎢
⎥
⎢⎣10 30 20 190 ⎥⎦
⎡ 23.994 −20.832 −12.648⎤
72. ⎢
7.44 −168.64 ⎥⎦
⎣ 26.164
64.08⎤
⎡ 15.606
73. ⎢
⎥
⎣ −739.428 373.056 ⎦
54.06 ⎤
⎡11.952
74. ⎢
⎥
⎣86.496 278.648⎦
Principles in Practice 6.4
⎡10 30 20 190 ⎤
R1 ↔ R3 ⎢
> ⎢ 20 10 50 200 ⎥⎥
⎢⎣ 40 10 10 180 ⎥⎦
⎡ 1 3 2 19 ⎤
1 R
10 1
> ⎢⎢ 2 1 5 20 ⎥⎥
1 R
10 2
⎢⎣ 4 1 1 18 ⎥⎦
1 R
3
10
1. The corresponding system is
⎧ 6 A + B + 3C = 35
⎪
⎨3 A + 2 B + 3C = 22
⎪⎩ A + 5 B + 3C = 18
Reduce the augmented coefficient matrix of the
system.
⎡ 6 1 3 35 ⎤
⎢ 3 2 3 22 ⎥
⎢
⎥
⎢⎣1 5 3 18 ⎥⎦
3 2 19 ⎤
⎡1
−2R1 + R 2 ⎢
> 0 −5
1 −18⎥⎥
−4R1 + R3 ⎢
⎢⎣ 0 −11 −7 −58⎥⎦
3
2 19 ⎤
⎡1
− 15 R 2 ⎢
18 ⎥
1
1 −5
> ⎢0
5⎥
⎢
⎥
0
11
7
58
−
−
−
⎣
⎦
13
41 ⎤
⎡1 0
5
5 ⎥
⎢
−3R 2 + R1 ⎢
18
1
⎥
> 0 1 −5
5⎥
11R 2 + R3 ⎢
⎢ 0 0 − 46 − 92 ⎥
5
5 ⎦⎥
⎣⎢
13
41
⎡1 0
⎤
5
5 ⎥
5 R
⎢
− 46
3
⎥
> ⎢ 0 1 − 15 18
5⎥
⎢
⎢0 0
1 2⎥
⎣⎢
⎦⎥
⎡1 5 3 18 ⎤
R1 ↔ R3 ⎢
> ⎢ 3 2 3 22 ⎥⎥
⎢⎣ 6 1 3 35 ⎥⎦
5
3 18⎤
⎡1
−3R1 + R 2 ⎢
> ⎢0 −13 −6 −32 ⎥⎥
−6R1 + R3
⎢⎣0 −29 −15 −73⎥⎦
5
3 18⎤
⎡1
1 R
− 13
⎢
2
6
32 ⎥
> ⎢0
1
13
13 ⎥
⎢
⎥
⎣ 0 −29 −15 −73⎦
9
74 ⎤
⎡1 0
13
13 ⎥
⎢
−5R 2 + R1 ⎢
6
32 ⎥
> 0 1 13
13 ⎥
29R 2 + R 3 ⎢
⎢0 0 – 21 – 21 ⎥
⎢⎣
13
13 ⎥⎦
⎡1 0 9 74 ⎤
13 13 ⎥
13
− 21 R3 ⎢
6
32 ⎥
⎢
> 0 1 13
13 ⎥
⎢
⎢0 0 1 1 ⎥
⎣⎢
⎦⎥
⎡1 0 0 3 ⎤
− 13
R + R1 ⎢
5 3
> ⎢0 1 0 4 ⎥⎥
1R +R
2
5 3
⎢⎣0 0 1 2 ⎥⎦
She should take 3 tablets of X, 4 tablets of Y,
and 2 tablets of Z.
196
ISM: Introductory Mathematical Analysis
Section 6.4
3. Let a, b, c, and d be the number of bags of foods
A, B, C, and D, respectively. The corresponding
system is
⎧ 5a + 5b + 10c + 5d = 10, 000
⎪
⎨10a + 5b + 30c + 10d = 20, 000
⎪⎩5a + 15b + 10c + 25d = 20, 000
Problems 6.4
1. The first nonzero entry in row 2 is not to the
right of the first nonzero entry in row 1, hence
not reduced.
2. Reduced.
Reduce the augmented coefficient matrix of the
system.
⎡ 5 5 10 5 10, 000 ⎤
⎢10 5 30 10 20, 000 ⎥
⎢
⎥
⎢⎣ 5 15 10 25 20, 000 ⎥⎦
3. Reduced.
4. In row 2, the first nonzero entry is in column 2,
but not all other entries in column 2 are zeros,
hence not reduced.
2000 ⎤
⎡ 1 1 2 1
⎢
> ⎢10 5 30 10 20, 000 ⎥⎥
⎣⎢ 5 15 10 25 20, 000 ⎦⎥
1R
5 1
5. The first row consists entirely of zeros and is not
below each row containing a nonzero entry,
hence not reduced.
1
2000 ⎤
⎡1 1 2
−10R1 + R 2 ⎢
> ⎢ 0 −5 10 0
0 ⎥⎥
–5R1 + R 3
⎢⎣ 0 10 0 20 10, 000 ⎥⎦
6. The first nonzero entry of row 2 is to the left of
the first nonzero entry of row 1, hence not
reduced.
⎡1 1
> ⎢⎢ 0 1
⎢⎣ 0 10
⎡1
−R 2 + R1
> ⎢⎢ 0
−10R 2 + R 3
⎢⎣ 0
3⎤
⎡ 1 3⎤ −4R1 + R 2 ⎡ 1
> ⎢
7. ⎢
⎥
⎥
−
4
0
0
12
⎣
⎦
⎣
⎦
− 15 R 2
2
1
2000 ⎤
−2 0
0 ⎥⎥
0 20 10, 000 ⎥⎦
1 R
− 12
2
⎡1
> ⎢
⎣0
−3R 2 + R1
>
0 4
1
2000 ⎤
1 −2 0
0 ⎥⎥
0 20 20 10, 000 ⎥⎦
⎡ 0 0 4 0 2000 ⎤
R3 ⎢
> ⎢ 0 1 −2 0
0 ⎥⎥
1 1 500 ⎥⎦
⎣⎢ 0 0
0⎤
⎡ 1 0 0 −3
−4R 3 + R1 ⎢
> 0 1 0 2 1000 ⎥⎥
2R 3 + R 2 ⎢
⎢⎣ 0 0 1 1 500 ⎥⎦
1
20
3⎤
1⎥⎦
⎡1 0 ⎤
⎢0 1 ⎥
⎣
⎦
⎡ 0 −3 0 2 ⎤ R1 ↔ R 2 ⎡ 1 5 0 2 ⎤
8. ⎢
⎯⎯⎯⎯⎯→ ⎢
⎣ 1 5 0 2 ⎥⎦
⎣ 0 −3 0 2 ⎥⎦
1
− R2 ⎡ 1 5 0
2⎤
3 →
⎯⎯⎯⎯
⎢0 1 0 − 2 ⎥
3⎦
⎣
16 ⎤
−5R 2 + R1 ⎡ 1 0 0
3⎥
⎢
⎯⎯⎯⎯⎯⎯
→
⎢⎣ 0 1 0 − 23 ⎥⎦
This reduced matrix corresponds to the system
⎧ a − 3d = 0
⎪
⎨b + 2d = 1000
⎪⎩ c + d = 500
⎡2 4 6⎤
⎡1 2 3⎤
R1 ↔ R3 ⎢
⎢
⎥
9. ⎢ 1 2 3 ⎥
> ⎢ 1 2 3 ⎥⎥
⎢⎣ 1 2 3 ⎥⎦
⎢⎣ 2 4 6 ⎥⎦
⎡1 2 3 ⎤
−R1 + R 2 ⎢
> 0 0 0 ⎥⎥
–2R1 + R 3 ⎢
⎢⎣0 0 0 ⎥⎦
Letting d = r, we get the general solution of the
system:
a = 3r
b = –2r + 1000
c = –r + 500
d=r
Note that a, b, c, and d cannot be negative, given
the context, hence 0 ≤ r ≤ 500. One specific
solution is when r = 250, then a = 750, b = 500,
c = 250, and d = 250.
197
Chapter 6: Matrix Algebra
3⎤
⎡2
⎢ 1 −6 ⎥
⎥ R1 ↔ R 2>
10. ⎢
⎢ 4 8⎥
⎢
⎥
⎢⎣ 1 7 ⎥⎦
⎡ 2
⎢ 1
11. ⎢
⎢ −1
⎢
⎣⎢ 0
ISM: Introductory Mathematical Analysis
⎡ 1 −6 ⎤ –2R1 + R 2
⎢
3⎥⎥ –4R1 + R 3
⎢2
>
⎢ 4 8⎥ −R1 + R 4
⎢
⎥
⎢⎣ 1 7 ⎥⎦
2
3
1
1
2⎤
1⎥⎥
4⎥
⎥
0 ⎦⎥
⎡ 1 4 2 2⎤
⎢
⎥ 1
−2R1 + R 2 ⎢ 0 −8 −1 −3⎥ − 8 R 2
>
>
⎢0 7 3 6⎥
R1 + R3
⎢
⎥
⎣⎢ 0 2 1 0 ⎦⎥
⎡1
⎢
⎢0
⎢
⎢0
⎢⎣ 0
0
4
3
2
3
2
1
1
1⎤
2 ⎥⎥ R1 ↔ R 2
>
4⎥
⎥
0 ⎦⎥
⎡1
−4R 2 + R1 ⎢
⎢
−7 R 2 + R3 ⎢ 0
>
−2 R 2 + R 4 ⎢ 0
⎢
⎢0
⎣⎢
⎡1
⎢
1
− 8 R3 + R 2 ⎢0
> ⎢
− 34 R3 + R 4 ⎢ 0
⎢
⎢0
⎢⎣
− 23 R3 + R1
⎡1
⎢
⎢0
− 17
R
33 4
> ⎢
⎢0
⎢
⎢⎣ 0
⎡0 0
⎢2 0
12. ⎢
⎢ 0 −1
⎢
⎣⎢ 0 4
0
1
0
0
3
2
1
8
17
8
3
4
1⎤
2⎥
3⎥ 8
8 ⎥ 17
27 ⎥
8 ⎥
− 34 ⎥⎦⎥
4
0
3
2
⎡1
⎢
R3 ⎢0
> ⎢
⎢0
⎢
⎢0
⎣⎢
32 ⎤ 32
0 0 − 17
R + R1
⎥ 17 4
3
3⎥−
R + R2
1 0
17 ⎥ 17 4
27 ⎥ − 27 R 4 + R 3
0 1
17 ⎥ 17
0 0
1⎥⎦
3⎤
2⎥
⎡2 0
⎢0 0
⎢
⎢ 0 −1
⎢
⎣⎢ 0 4
⎡ 1 −6⎤ 6R 2 + R1
⎢
1⎥⎥ −32R 2 + R 3
⎢0
>
⎢ 0 32⎥ −13R 2 + R 4
⎢
⎥
⎢⎣ 0 13⎥⎦
⎡1
⎢
⎢0
⎢0
⎢
⎢⎣ 0
4 2 2⎤
⎥
1 18 83 ⎥
⎥
7 3 6⎥
2 1 0 ⎥⎦
0
1
3
2
1
8
0
1
0
3
4
1⎤
2⎥
3⎥
8⎥
27 ⎥
17 ⎥
− 34 ⎥⎦⎥
⎡1
⎢0
⎢
⎢0
⎢
⎢⎣ 0
0
1
0
0
0
0
1
0
0⎤
0 ⎥⎥
0⎥
⎥
1 ⎥⎦
⎡1 0 3 ⎤
3⎤
2⎥
⎢
2 ⎥⎥ 12 R1 ⎢ 0 0 2 ⎥ R 2 ↔ R3
> ⎢
>
⎥
0⎥
0 −1 0 ⎥
⎢
⎥
1⎦⎥
⎢⎣ 0 4 1⎥⎦
⎡1
⎢
1 0 ⎥ −4R 2 + R 4 ⎢ 0
> ⎢
⎥
0 2⎥
⎢0
⎢⎣ 0
4 1⎥⎦
0
−6 ⎤
1 R
15⎥⎥ 15
2
>
32 ⎥
⎥
13⎥⎦
32 ⎤
0 0 − 17
⎥
3⎥
1 0
17 ⎥
27 ⎥
0 1
17 ⎥
33 ⎥
0 0 − 17
⎥⎦
2⎤
3⎥⎥ R1 ↔ R 2
>
0⎥
⎥
1⎦⎥
⎡1
⎢
−R 2 ⎢0
> ⎢
⎢0
⎢⎣ 0
⎡ 1
⎢ 2
⎢
⎢ −1
⎢
⎣⎢ 0
⎡1
⎢
⎢0
⎢0
⎢
⎢⎣ 0
0
3⎤
2⎥
1 0⎥
⎥
0 2⎥
0 1⎥⎦
⎡1
⎢
1R
2 3> ⎢ 0
⎢
⎢0
⎢⎣ 0
198
3⎤
2⎥
⎡1 0 3 ⎤
2⎥
⎢
⎢0 −1 0 ⎥
⎢
⎥
⎢0 0 2⎥
⎢⎣0 4 1⎥⎦
⎡1
+ R1 ⎢ 0
1 0⎥ −
> ⎢
⎥
⎢0
0 1⎥ −R 3 + R 4
⎢
0 1⎥⎦
⎣⎢ 0
0
3R
2 3
0
1
0
0
0⎤
0 ⎥⎥
1⎥
⎥
0 ⎦⎥
0⎤
1 ⎥⎥
0⎥
⎥
0 ⎥⎦
ISM: Introductory Mathematical Analysis
Section 6.4
⎡ 2 −7 50 ⎤ ⎡ 1 3 10 ⎤
→
13. ⎢
⎣ 1 3 10 ⎥⎦ ⎢⎣ 2 −7 50 ⎥⎦
10 ⎤
⎡1 3
3 10 ⎤
⎡1
→⎢
→
⎢
30 ⎥
⎣ 0 −13 30 ⎦⎥
⎣ 0 1 − 13 ⎦
⎡ 1 0 220 ⎤
13 ⎥
→⎢
30
⎢⎣ 0 1 − 13
⎥⎦
220
30
Thus x =
and y = − .
13
13
⎡ 1 −3 −11⎤ ⎡ 1 0 − 52 ⎤
⎡ 1 −3 −11⎤ ⎡ 1 −3 −11⎤
⎢
⎥
→
→
14. ⎢
⎢
53 ⎥ → ⎢
53 ⎥
0
1 15
9 ⎦⎥ ⎣⎢ 0 15 53⎦⎥
⎣4 3
⎣⎢
⎦⎥ ⎣ 0 1 15 ⎦
2
53
Thus x = − , y =
.
5
15
4⎤
⎡1 1 4 ⎤ ⎡1 1
⎡1 1
4⎤
⎡ 3 1 4⎤ ⎡ 3 1
3
3⎥ → ⎢
3 3⎥ → ⎢
3
15. ⎢
→⎢
→⎢
⎥
⎥
⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0
⎢⎣ 0 0 −14 ⎥⎦
⎣12 4 2 ⎦ ⎣ 0 0 −14 ⎦
The last row indicates 0 = 1, which is never true, so there is no solution.
0⎤
⎥
1 ⎥⎦
⎡ 1 2 −3 0 ⎤ ⎡ 1 2 −3 0 ⎤
16. ⎢
⎥→⎢
⎥
⎣ −2 −4 6 1⎦ ⎣ 0 0 0 1⎦
The last row indicates that 0 = 1, which is never true. There is no solution.
⎡1 2 1
⎡1 2 1 4 ⎤ ⎡ 1 2 1 4 ⎤
→⎢
→⎢
17. ⎢
⎥
⎥
1
⎢⎣0 1 6
⎣3 0 2 5 ⎦ ⎣ 0 −6 −1 −7 ⎦
⎡1 0
4⎤
⎢
→
⎥
7
⎢0 1
⎥
6⎦
⎣
2
3
1
6
5⎤
3⎥
7⎥
6⎦
,
⎧x + 2 z = 5
⎪
3
3
which gives ⎨
.
1z= 7
+
y
⎪⎩
6
6
2
5
1
7
Thus, x = − r + , y = − r + , z = r, where r is any real number.
3
3
6
6
⎡ 1 0 13
2
1⎤
⎡1 3 2 1⎤ ⎡ 1 3 2 1⎤ ⎡ 1 3
2
→⎢
→⎢
18. ⎢
⎥→⎢
⎥
⎥
3
9
−
−
0
1
⎢0 1 − 3
⎣1 1 5 10 ⎦ ⎣ 0 −2 3 9 ⎦ ⎢⎣
2
2 ⎥⎦
2
⎣
Thus x = −
29 ⎤
2 ⎥
− 92 ⎥⎦
13
29
3
9
r + , y = r − , z = r, where r is any real number.
2
2
2
2
9⎤
⎡1 0
⎡ 1 −3 0 ⎤
8⎥
⎡ 1 −3 0 ⎤ ⎡ 1 −3 0 ⎤
⎢
⎢
⎥
⎢
⎥
⎢
⎥
3
3⎥
⎢
19. ⎢ 2 2 3⎥ → ⎢ 0 8 3⎥ → ⎢0
1 8⎥ → 0 1
8⎥
⎢
⎢
⎥
⎢ 0 0 − 17 ⎥
⎣⎢ 5 −1 1⎦⎥ ⎣⎢ 0 14 1⎦⎥
⎣0 14 1⎦
⎢⎣
4 ⎥⎦
From the third row, 0 = −
17
, which is never true, so there is no solution.
4
199
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
4
9⎤
⎡ 1 4 9⎤ ⎡ 1
20. ⎢3 −1 6 ⎥ → ⎢0 −13 −21⎥
⎢
⎥ ⎢
⎥
⎣ 1 −1 2 ⎦ ⎣0 −5 −7 ⎦
⎡ 1 0 33 ⎤
⎡ 1 4 9⎤
13 ⎥
⎢
21 ⎥ → 0 1 21
→ ⎢0
1 13
⎢
⎥
13
⎢
⎥
⎢
14 ⎥
⎢⎣ 0 −5 −7 ⎥⎦
⎢⎣0 0 13 ⎥⎦
⎡ 1 0 33 ⎤
13 ⎥
⎡ 1 0 0⎤
⎢
21
→ ⎢ 0 1 13 ⎥ → ⎢ 0 1 0 ⎥
⎢
⎥ ⎢ 0 0 1⎥
⎦
⎢⎣ 0 0 1⎥⎦ ⎣
The last row indicates that 0 = 1, which is never true. There is no solution.
⎡ 1 −1 −3 −5⎤ ⎡ 1 −1
21. ⎢⎢ 2 −1 −4 −8⎥⎥ → ⎢⎢ 0 1
⎢⎣ 1 1 −1 −1⎥⎦ ⎢⎣ 0 2
⎡ 1 0 −1 −3⎤ ⎡ 1 0
→ ⎢⎢ 0 1 2 2 ⎥⎥ → ⎢⎢ 0 1
⎢⎣ 0 0 1 0 ⎥⎦ ⎢⎣ 0 0
Thus, x = –3, y = 2, z = 0.
−3 − 5 ⎤ ⎡ 1 0 − 1 − 3 ⎤
2 2 ⎥⎥ → ⎢⎢ 0 1 2 2 ⎥⎥
2 4 ⎥⎦ ⎢⎣ 0 0 −2 0 ⎥⎦
0 −3⎤
0 2 ⎥⎥
1 0 ⎥⎦
7⎤ ⎡ 1
⎡ 1 1 −1 7 ⎤ ⎡ 1 1 −1
⎢
⎥
⎢
22. ⎢ 2 −3 −2 4 ⎥ → ⎢ 0 −5 0 −10 ⎥⎥ → ⎢⎢ 0
⎢⎣ 1 −1 −5 23⎥⎦ ⎢⎣ 0 −2 −4 16 ⎥⎦ ⎢⎣ 0
⎡ 1 0 −1 5⎤ ⎡ 1 0 −1 5⎤ ⎡ 1
→ ⎢⎢ 0 1 0 2 ⎥⎥ → ⎢⎢0 1 0 2 ⎥⎥ → ⎢⎢ 0
⎢⎣ 0 0 −4 20 ⎥⎦ ⎢⎣0 0 1 −5⎥⎦ ⎢⎣ 0
Thus x = 0, y = 2, z = –5.
1 −1 7 ⎤
1 0 2 ⎥⎥
−2 −4 16 ⎥⎦
0 0 0⎤
1 0 2 ⎥⎥
0 1 −5⎥⎦
⎡ 2 0 −4 8⎤ ⎡ 1 0 −2 4 ⎤ ⎡ 1 0
⎢ 1 −2 −2 14 ⎥ ⎢ 1 −2 −2 14 ⎥ ⎢ 0 −2
⎥→⎢
⎥→⎢
23. ⎢
⎢ 1 1 −2 −1⎥ ⎢ 1 1 −2 −1⎥ ⎢ 0
1
⎢
⎥ ⎢
⎥ ⎢
1 1 0 ⎦⎥ ⎣⎢3
1 1 0 ⎥⎦ ⎣⎢ 0
1
⎣⎢ 3
4 ⎤ ⎡ 1 0 −2
4⎤
⎡1
⎡ 1 0 −2
⎢0
⎢0
⎥
⎢
⎥
1 0 −5⎥ ⎢ 0 1 0 −5⎥
→⎢
→
→⎢
⎢ 0 −2 0 10 ⎥ ⎢ 0 0 0
⎢0
0⎥
⎢
⎥ ⎢
⎥
⎢
1 7 −12 ⎥⎦ ⎢⎣ 0 0 7 −7 ⎥⎦
⎢⎣ 0
⎢⎣ 0
Thus x = 2, y = –5, z = –1.
−2
4⎤
0 10 ⎥⎥
0 −5 ⎥
⎥
7 −12 ⎦⎥
0 −2 4 ⎤ ⎡ 1
1 0 −5⎥⎥ ⎢⎢ 0
→
0
1 −1⎥ ⎢ 0
⎥ ⎢
0 0 0 ⎥⎦ ⎢⎣ 0
⎡ 1 0 3 −1⎤ ⎡ 1 0 3 −1⎤ ⎡ 1 0 3 −1⎤ ⎡ 1
⎢ 3 2 11 1⎥ ⎢ 0 2 2 4 ⎥ ⎢ 0
1 1 2 ⎥⎥ ⎢⎢0
⎥→⎢
⎥→⎢
24. ⎢
→
⎢1 1 4
1⎥ ⎢ 0
1 1 2⎥ ⎢0
1 1 2⎥ ⎢0
⎢
⎥ ⎢
⎥ ⎢
⎥ ⎢
⎣⎢ 2 −3 3 −8⎦⎥ ⎣⎢ 0 −3 −3 −6 ⎦⎥ ⎣⎢ 0 −3 −3 −6 ⎦⎥ ⎣⎢0
Thus x = –3r – 1, y = –r + 2, z = r, where r is any real number.
200
0
1
0
0
0
1
0
0
0 2⎤
0 −5⎥⎥
1 −1⎥
⎥
0 0 ⎥⎦
3 −1⎤
1 2 ⎥⎥
0 0⎥
⎥
0 0 ⎦⎥
ISM: Introductory Mathematical Analysis
Section 6.4
⎡1 −1
⎢1 1
25. ⎢
⎢1 1
⎢⎣1 1
⎡1
⎢0
→⎢
⎢0
⎢⎣ 0
⎡1
⎢0
→⎢
⎢0
⎣⎢ 0
⎡1
⎢0
→⎢
⎢0
⎣⎢ 0
−1 −1 −1 0 ⎤ ⎡ 1 −1 −1 −1 −1 0 ⎤
−1 −1 −1 0 ⎥ ⎢ 0 2 0 0 0 0 ⎥
⎥→⎢
⎥
1 −1 −1 0 ⎥ ⎢ 0 2 2 0 0 0 ⎥
1 1 −1 0 ⎥⎦ ⎢⎣ 0 2 2 2 0 0 ⎥⎦
−1 −1 −1 −1 0 ⎤ ⎡ 1 0 −1 −1 −1 0 ⎤
1 0 0 0 0⎥ ⎢0 1 0 0 0 0⎥
⎥→⎢
⎥
2 2 0 0 0⎥ ⎢0 0 2 0 0 0⎥
2 2 2 0 0 ⎥⎦ ⎢⎣ 0 0 2 2 0 0 ⎥⎦
0 −1 −1 −1 0 ⎤ ⎡ 1 0 0 −1 −1 0 ⎤
1 0 0 0 0⎥ ⎢0 1 0 0 0 0⎥
⎥→⎢
⎥
0 1 0 0 0⎥ ⎢0 0 1 0 0 0⎥
0 2 2 0 0 ⎥⎦ ⎢⎣ 0 0 0 2 0 0 ⎥⎦
0 0 −1 −1 0 ⎤
⎡ 1 0 0 0 −1 0 ⎤
⎢0 1 0 0 0 0 ⎥
1 0 0 0 0⎥
⎥→⎢
⎥
0 1 0 0 0⎥
⎢0 0 1 0 0 0 ⎥
0 0 1 0 0 ⎦⎥
⎣⎢0 0 0 1 0 0 ⎦⎥
Thus, x1 = r , x2 = 0, x3 = 0, x4 = 0, and x5 = r , where r is any number.
⎡1 1
⎢1 −1
26. ⎢
⎢1 1
⎢
⎣⎢1 1
⎡1
⎢0
→⎢
⎢0
⎢
⎣⎢ 0
1
−1
−1
−1
1 1
1 0
0 1
0 0
−1
1
−1
1
−1
0⎤ ⎡ 1
0 ⎥⎥ ⎢⎢ 0
→
0⎥ ⎢0
⎥ ⎢
0 ⎦⎥ ⎣⎢ 0
0⎤
⎡1
⎢0
⎥
0 0⎥
→⎢
⎢0
0 0⎥
⎢
⎥
1 0 ⎦⎥
⎣⎢0
1 −1 0 ⎤ ⎡ 1
−2 2 0 ⎥⎥ ⎢⎢ 0
→
−2 0 0 ⎥ ⎢ 0
⎥ ⎢
−2 2 0 ⎦⎥ ⎣⎢ 0
0 0 0⎤
1 0 0 0 ⎥⎥
0 1 0 0⎥
⎥
0 0 1 0 ⎦⎥
1
−2
0
0
0
1
1
0
0
1 −1
1 −1
1 0
1 −1
0⎤ ⎡ 1
0 ⎥⎥ ⎢⎢ 0
→
0⎥ ⎢0
⎥ ⎢
0 ⎦⎥ ⎣⎢ 0
1
1
0
0
1 −1 0 ⎤
1 −1 0 ⎥⎥
1 0 0⎥
⎥
0 1 0 ⎦⎥
Thus x1 = 0, x2 = 0, x3 = 0, x4 = 0.
27. Let x = federal tax and y = state tax. Then x = 0.25(312,000 – y) and y = 0.10(312,000 – x). Equivalently,
⎧ x + 0.25 y = 78, 000
⎨
⎩0.10 x + y = 31, 200.
0.25 78, 000 ⎤ ⎡1 0.25 78, 000 ⎤
⎡ 1
⎡1 0.25 78, 000 ⎤ ⎡1 0 72, 000 ⎤
→⎢
.
→⎢
→
⎢ 0.10
⎥
⎥
1
31, 200 ⎦ ⎣ 0 0.975 23, 400 ⎦
1
24, 000 ⎥⎦ ⎢⎣ 0 1 24, 000 ⎥⎦
⎣0
⎣
Thus x = 72,000 and y = 24,000, so the federal tax is $72,000 and the state tax is $24,000.
28. x = no. of units of A to be sold and y = no. of units of B to be sold. Then x = 1.25y and 8x + 11y = 42,000.
Equivalently,
⎧ x − 1.25 y = 0,
⎨
⎩8 x + 11y = 42, 000.
0 ⎤ ⎡1 −1.25
0 ⎤
0 ⎤ ⎡1 0 2500 ⎤
⎡1 −1.25
⎡1 −1.25
.
→⎢
→
→⎢
⎢8
⎥
⎥
11
42, 000 ⎦ ⎣ 0
21
42, 000 ⎦
1
2000 ⎥⎦ ⎢⎣0 1 2000 ⎥⎦
⎣0
⎣
Thus x = 2500 and y = 2000, so 2500 units of A and 2000 units of B must be sold.
29. Let x = number of units of A produced, y = number of units of B produced, and z = number of units of C
produced. Then
no. of units: x + y + z = 11,000
total cost: 4x + 5y + 7z + 17,000 = 80,000
201
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
total profit: x + 2y + 3z = 25,000
Equivalently,
⎧ x + y + z = 11, 000
⎪
⎨4 x + 5 y + 7 z = 63, 000
⎪ x + 2 y + 3z = 25, 000
⎩
⎡ 1 1 1 11, 000 ⎤ ⎡1 1 1 11, 000 ⎤
⎢ 4 5 7 63, 000 ⎥ → ⎢ 0 1 3 19, 000 ⎥
⎢
⎥ ⎢
⎥
⎢⎣ 1 2 3 25, 000 ⎥⎦ ⎢⎣ 0 1 2 14, 000 ⎥⎦
⎡1 0 0 2000 ⎤
⎡ 1 0 −2 −8, 000 ⎤ ⎡ 1 0 −2 −8, 000 ⎤
⎢
⎥
⎢
⎥
→ ⎢ 0 1 3 19, 000 ⎥ → ⎢0 1 3 19, 000 ⎥ → ⎢⎢ 0 1 0 4000 ⎥⎥
⎢⎣ 0 0 1 5000 ⎥⎦
⎢⎣ 0 0 −1 −5, 000 ⎥⎦ ⎢⎣0 0
1 5, 000 ⎥⎦
Thus x = 2000, y = 4000, and z = 5000, so 2000 units of A, 4000 units of B and 5000 units of C should be
produced.
30. Let x = number of desks to be produced at the East Coast plant and y = number of desks to be produced at the
West Coast plant. Then x + y = 800 and 90x +20,000 = 95y + 18,000.
Equivalently,
⎧ x + y = 800
⎨
⎩90 x − 95 y = −2000.
1
800 ⎤ ⎡ 1
1
800 ⎤ ⎡ 1 1 800 ⎤ ⎡ 1 0 400⎤
⎡ 1
⎢90 −95 −2000 ⎥ → ⎢ 0 −185 −74, 000 ⎥ → ⎢ 0 1 400 ⎥ → ⎢0 1 400⎥
⎣
⎦ ⎣
⎦ ⎣
⎦ ⎣
⎦
x = 400 and y = 400
Thus the production order is 400 units at the East Coast plant and 400 units at the West Coast plant.
31. Let x = number of brand X pills, y = number of brand Y pills, and z = number of brand Z pills. Considering the
unit requirements gives the system
⎧2 x + 1y + 1z = 10 (vitamin A)
⎪
⎨3x + 3 y + 0 z = 9 (vitamin D)
⎪5 x + 4 y + 1z = 19 (vitamin E)
⎩
1
⎡ 2 1 1 10 ⎤ ⎡⎢ 1 2
⎢
⎥
⎢ 3 3 0 9 ⎥ → ⎢⎢ 3 3
⎢⎣ 5 4 1 19 ⎥⎦ ⎢5 4
⎣
5⎤ ⎡ 1
⎥ ⎢
0 9⎥ → ⎢0
⎥ ⎢
1 19 ⎥ ⎢ 0
⎦ ⎣
1
2
1
2
3
2
3
2
1
2
− 32
− 32
5⎤
⎥
−6 ⎥
⎥
−6 ⎥⎦
1
⎡1 1
5⎤ ⎡ 1 0 1 7 ⎤
2
2
⎢
⎥
→ ⎢ 0 1 −1 −4 ⎥ → ⎢⎢ 0 1 −1 −4 ⎥⎥
⎢
⎥
⎢⎣ 0 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 0 ⎥⎦
⎧x = 7 − r
⎪
Thus ⎨ y = r − 4 where r = 4, 5, 6, 7.
⎪z = r
⎩
The only solutions for the problem are z = 4, x = 3, and y = 0; z = 5, x = 2, and y = 1; z = 6, x = 1, and y = 2; z = 7,
x = 0, and y = 3. Their respective costs (in cents) are 15, 23, 31, and 39.
a.
The possible combinations are 3 of X, 4 of Z; 2 of X, 1 of Y, 5 of Z; 1 of X, 2 of Y, 6 of Z; 3 of Y, 7 of Z.
b. The combination 3 of X, 4 of Z costs 15 cents a day.
202
ISM: Introductory Mathematical Analysis
c.
Section 6.4
The least expensive combination is 3 of X, 4 of Z; the most expensive is 3 of Y, 7 of Z.
32. Let x, y, and z be the numbers of units of A, B, and C, respectively.
⎧3x + 1 y + 2 z = 490 (machine I)
⎪
⎨1x + 2 y + 1z = 310 (machine (II)
⎪2 x + 4 y + 1z = 560 (machine III)
⎩
⎡ 3 1 2 490 ⎤ ⎡ 1 2 1
⎢ 1 2 1 310 ⎥ → ⎢ 3 1 2
⎢
⎥ ⎢
⎣ 2 4 1 560 ⎦ ⎣ 2 4 1
⎡ 1 2 1 310 ⎤ ⎡ 1
→ ⎢ 0 −5 −1 −440 ⎥ → ⎢ 0
⎢
⎥ ⎢
⎣ 0 0 −1 −60 ⎦ ⎢⎣ 0
⎡1
⎢
→ ⎢0
⎢
0
⎣⎢
⎡1
→ ⎢0
⎢
⎣0
0
1
3
5
1
5
0 −1
0 0
1 0
0 1
310 ⎤
490 ⎥
⎥
560 ⎦
2 1 310 ⎤
1 15
88⎥
⎥
0 −1 −60 ⎥⎦
134 ⎤ ⎡ 1 0 53 134 ⎤
⎥ ⎢
⎥
88⎥ → ⎢ 0 1 15 88⎥
⎥ ⎢
⎥
−60 ⎥ ⎢ 0 0 1 60 ⎥
⎦ ⎣
⎦
98⎤
76 ⎥
⎥
60 ⎦
x = 98, y = 76, z = 60
Thus, 98 units of A, 76 units of B, and 60 units of C should be produced.
33. a.
Let s, d, and g represent the number of units of S, D, and G, respectively. Then
⎧12s + 20d + 32 g = 220 (stock A)
⎪
⎨16s + 12d + 28 g = 176 (stock B)
⎪8s + 28d + 36 g = 264 (stock C)
⎩
⎡12 20 32 220 ⎤
⎢16 12 28 176 ⎥
⎢
⎥
⎢⎣ 8 28 36 264 ⎥⎦
( 14 ) R1>
( 14 ) R2
( 18 ) R3
⎡3 5
⎢
⎢4 3
⎢
7
⎢⎣ 1 2
8 55⎤
⎥
7 44 ⎥
9 33⎥
⎥⎦
2
⎡ 1 7 9 33⎤
2 2
⎥
R1 ↔ R3 ⎢
> ⎢ 4 3 7 44 ⎥
⎢
⎥
⎢⎣ 3 5 8 55⎥⎦
7
9
⎡1
33⎤
2
2
⎢
⎥
−4R1 + R 2
> ⎢ 0 −11 −11 −88⎥
−3R1 + R3 ⎢
⎥
11
11
⎢⎣ 0 − 2 − 2 −44 ⎥⎦
1 R
− 11
2
7
9
⎡1
33⎤
2
2
⎢
⎥
> ⎢0
1
1
8⎥
⎢
⎥
11
11
⎢⎣ 0 − 2 − 2 −44 ⎥⎦
203
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡1 0 1 5 ⎤
− 27 R 2 + R1 ⎢
> ⎢ 0 1 1 8 ⎥⎥
11 R + R
3
2 2
⎢⎣ 0 0 0 0 ⎥⎦
Thus s = 5 – r, d = 8 – r, and g = r, where r = 0, 1, 2, 3, 4, 5.
The six possible combinations are given by
COMBINATION
r
0
1
2
3
4
5
S
5
4
3
2
1
0
D
8
7
6
5
4
3
G
0
1
2
3
4
5
b. Computing the cost of each combination, we find that they are 4700, 4600, 4500, 4400, 4300, and 4200
dollars, respectively. Buying 3 units of Deluxe and 5 units of Gold Star (s = 0, d = 3, g = 5) minimizes the
cost.
Principles in Practice 6.5
1. Write the coefficients matrix and reduce.
⎡1 3 4 ⎤
⎡5 3 4⎤ 1
5 5
⎢ 6 8 7 ⎥ 5 R1> ⎢⎢ 6 8 7 ⎥⎥ −6R1 + R 2>
⎢
⎥
⎢
⎥ −3R1 + R 3
⎣⎢ 3 1 2 ⎦⎥
⎢⎣ 3 1 2 ⎥⎦
3
⎡1
5
⎢
⎢ 0 22
5
⎢
⎢0 − 4
5
⎣
4⎤
5⎥
11 ⎥
5⎥
− 52 ⎥
⎦
3
4⎤
⎡1
⎡1 0 1 ⎤
5
5⎥ 3
2⎥
⎢
− 5 R 2 + R1 ⎢
1
1
⎢
⎥
⎢
1
> 0
> 0 1 2⎥
2⎥ 4 R + R
⎢
⎢
⎥
2
3
⎢0 − 4 − 2 ⎥ 5
⎢0 0 0 ⎥
5
5⎦
⎣
⎦
⎣
The system has infinitely many solutions since there are two nonzero rows in the reduced coefficient matrix.
1
x+ z =0
2
1
y+ z =0
2
1
1
Let z = r, so x = − r and y = − r , where r is any real number.
2
2
5 R
22 2
Problems 6.5
−1 −9 −3⎤
⎡ 1 1 −1 − 9 3 ⎤
⎡ 1 1 − 1 − 9 − 3⎤
⎡ 1 1 −1 −9 −3⎤
2 15 12 ⎥⎥ → ⎢⎢ 0 1 4 33 18⎥⎥ → ⎢⎢ 0 1 4 33 18⎥⎥ → ⎢⎢0 1 4 33 18⎥⎥
⎢⎣ 0 −1 4 23 14 ⎥⎦
⎢⎣ 0 0 8 56 32 ⎥⎦
⎢⎣0 0 1 7 4 ⎥⎦
2 5 8⎥⎦
1 0 −2 1⎤
⎡ 1 0 0 −7 −1⎤
⎥
1 0 5 2 ⎥ → ⎢⎢0 1 0
5 2 ⎥⎥
⎢⎣0 0 1 7 4 ⎥⎦
0 1 7 4 ⎥⎦
Thus w = −1 + 7r, x = 2 − 5r, y = 4 − 7r, z = r (where r is any real number).
⎡1 1
1. ⎢⎢ 2 3
⎢⎣ 2 1
⎡1
→ ⎢⎢ 0
⎢⎣ 0
204
ISM: Introductory Mathematical Analysis
Section 6.5
1 10 15 −5⎤
⎡2
⎡ 1 −5 2 15 −10 ⎤
2. ⎢ 1 −5 2 15 −10 ⎥ → ⎢ 2
1 10 15 −5⎥
⎢
⎥
⎢
⎥
9⎦
9⎦
⎣ 1 1 6 12
⎣ 1 1 6 12
15 −10 ⎤
⎡ 1 −5 2
⎡ 1 −5 2 15 −10 ⎤
⎢
6
15 ⎥
1 11 − 15
→ ⎢ 0 11 6 −15 15⎥ → 0
11
11 ⎥
⎢
⎢
⎥
⎣ 0 6 4 −3 19 ⎦
⎣⎢0 6 4 −3 19 ⎦⎥
90 − 35 ⎤
90 − 35 ⎤
⎡ 1 0 52
⎡ 1 0 52
11
11 ⎥
11
11
11 ⎥
11
⎢
⎢
15
6 − 15
15 → 0 1 6 − 15
→ ⎢ 0 1 11
⎥
⎢
⎥
11
11
11
11
11
⎢
⎥
⎢
57
119 ⎥
8
57
119
⎢⎣ 0 0 11
⎢⎣0 0 1
8
8 ⎥⎦
11
11 ⎥⎦
51
147
⎡1 0 0 −
− 2 ⎤
2
⎢
⎥
27
→ ⎢ 0 1 0 − 21
−
⎥
4
4
⎢
⎥
57
119
⎢⎣ 0 0 1
8
8 ⎥⎦
57
119
51 147
21
27
, z = r (where r is any real number).
, x= r− , y=− r+
Thus, w = r −
8
8
2
2
4
4
⎡ 1 − 1 −1 − 1 − 2 ⎤
⎡ 3 −1 −3 −1 −2 ⎤
3
3
3⎥
⎢
⎢ 2 −2 −6 −6 −4 ⎥
⎢
2
2
6
6
4⎥
−
−
−
−
⎥→
3. ⎢
⎢
⎥
⎢ 2 −1 −3 −2 −2 ⎥
⎢ 2 −1 −3 −2 −2 ⎥
⎢
⎥
1 3 7 2 ⎥⎦
⎢⎣ 3
⎢⎣ 3
1 3
7
2 ⎥⎦
⎡ 1 − 1 −1 − 1 − 2 ⎤
⎡ 1 − 1 −1 − 1 − 2 ⎤
3
3
3⎥
⎡ 1 0 0 1 0⎤
3
3
3⎥
⎢
⎢
⎢0 1 3 4 2⎥
16
8
4
⎢0 −
−4 − 3 − 3 ⎥
⎢0
1 3
4
2⎥
3
⎢
⎥
⎥→⎢
→⎢
→
⎥
1 −1 − 4 − 2
⎢
⎥
0
0
0
0
0
⎢ 0 − 1 −1 − 4 − 2 ⎥
−
0
⎢
3
3
3⎥
⎢
⎥
3
3
3⎥
⎢
⎢
⎥
⎣⎢ 0 0 0 0 0 ⎦⎥
2 6
8
4 ⎦⎥
⎢⎣ 0
2 6
8
4 ⎥⎦
⎣⎢ 0
Thus, w = –s, x = –3r – 4s + 2, y = r, z = s (where r and s are any real numbers).
5 1⎤
5 1⎤
5 1⎤
⎡1 1 0
⎡1 1 0
⎡1 1 0
⎡1
⎢ 1 0 1 2 1⎥
⎢0 −1 1 −3 0 ⎥
⎢0
⎥
⎢
1
1
3
0
−
⎥→⎢
⎥ →⎢
⎥ → ⎢0
4. ⎢
⎢ 1 −3 4 −7 1⎥
⎢0 −4 4 −12 0 ⎥
⎢ 0 −4 4 −12 0 ⎥
⎢0
⎢
⎥
⎢
⎥
⎢
⎥
⎢
1 −1 3 0 ⎥⎦
1 −1
3 0 ⎥⎦
1 −1
3 0 ⎥⎦
⎢⎣ 0
⎢⎣0
⎢⎣ 0
⎢⎣ 0
Thus, w = –r – 2s + 1, x = r – 3s, y = r, z = s (where r and s are any real numbers).
0 1 2 1⎤
1 −1 3 0 ⎥⎥
0 0 0 0⎥
⎥
0 0 0 0 ⎥⎦
2⎤
2⎤
⎡ 1 1 3 −1 2 ⎤
⎡ 1 1 3 −1
⎡ 1 1 3 −1
⎡1
⎢ 2 1 5 −2 0 ⎥
⎢ 0 −1 −1 0 −4 ⎥
⎢0
⎥
⎢0
1 1 0
4⎥
⎢
⎥
⎢
⎥
⎢
⎢
5. ⎢ 2 −1 3 −2 −8⎥ → ⎢ 0 −3 −3 0 −12 ⎥ → ⎢ 0 −3 −3 0 −12 ⎥ → ⎢ 0
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎢ 3 2 8 −3 2 ⎥
⎢ 0 −1 −1 0 −4 ⎥
⎢ 0 −1 −1 0 −4 ⎥
⎢0
⎢⎣ 1 0 2 −1 −2 ⎥⎦
⎢⎣ 0 −1 −1 0 −4 ⎥⎦
⎢⎣ 0 −1 −1 0 −4 ⎥⎦
⎢⎣ 0
Thus, w = –2r + s – 2, x = –r + 4, y = r, z = s (where r and s are any real numbers).
205
0 2 −1 −2 ⎤
1 1 0 4 ⎥⎥
0 0 0 0⎥
⎥
0 0 0 0⎥
0 0 0 0 ⎥⎦
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
2
4⎤
2
4⎤
⎡ 1 1 1 2 4⎤ ⎡ 1 1 1
⎡1 1 1
⎡1
⎢2
⎥
⎢
⎥
⎢
⎥
⎢0
1 2 2 7 ⎥ ⎢ 0 −1 0 −2 −1⎥
1 0
2
1⎥
⎢
⎢0
⎢
6. ⎢ 1 2 1 4 5⎥ → ⎢ 0
1 0
2
1⎥ → ⎢ 0
1 0
2
1⎥ → ⎢0
⎢
⎥ ⎢
⎥
⎢
⎥
⎢
⎢ 3 −2 3 −4 7 ⎥ ⎢ 0 −5 0 −10 −5⎥
⎢ 0 −5 0 −10 −5⎥
⎢0
⎢⎣ 4 −3 4 −6 9 ⎥⎦ ⎢⎣ 0 −7 0 −14 −7 ⎥⎦
⎢⎣ 0 −7 0 −14 −7 ⎥⎦
⎢⎣0
Thus, w = –r + 3, x = –2s + 1, y = r, z = s (where r and s are any real numbers).
0
1
0
0
0
1 0 3⎤
0 2 1⎥⎥
0 0 0⎥
⎥
0 0 0⎥
0 0 0 ⎥⎦
⎡ 4 −3 5 −10 11 −8⎤
⎡ 0 −5 −5 −10 5 −20 ⎤
7. ⎢
→⎢
⎥
1 5
0 3 6⎦
1 5
0 3
6 ⎥⎦
⎣2
⎣2
1 5
0 3
6⎤
⎡2
⎡2 1 5 0 3 6⎤
→⎢
→⎢
⎥
⎥
⎣ 0 −5 −5 −10 5 −20 ⎦
⎣ 0 1 1 2 −1 4 ⎦
⎡ 2 0 4 −2 4 2 ⎤ ⎡ 1 0 2 −1 2 1⎤
→⎢
⎥→⎢
⎥
⎣ 0 1 1 2 −1 4 ⎦ ⎣ 0 1 1 2 −1 4 ⎦
Thus, x1 = −2r + s − 2t + 1, x2 = − r − 2 s + t + 4, x3 = r , x4 = s, x5 = t (where r, s, and t are any real numbers).
⎡1 0
⎢0
1
8. ⎢
⎢ 2 −2
⎢
⎢⎣ 1 2
1⎤ ⎡ 1 0
0 ⎥⎥ ⎢⎢ 0
1
→
3 10 15 10 ⎥ ⎢ 0 −2
⎥ ⎢
3 −2 2 −2 ⎥⎦ ⎢⎣ 0 2
⎡1
1
4
1⎤
⎡1 0 3
⎢
⎢ 0 1 1 −2
⎥
⎢0
0
0⎥
⎢
→
→⎢
⎢ 0 0 1 −4 −7 −8⎥
⎢0
⎢
⎥
⎢
⎢⎣ 0 0 0 −7 −16 −19 ⎥⎦
⎢0
⎢⎣
72 33
18 17
Thus x1 = − + r , x2 = − r ,
7
7
7 7
3
1
4
3
1
4
1 −2
0
1 −2
0
−3
8
7
0 −3 −2
0 3 0
1 1 0
0
1 0
0 0
x3 =
1
12
7
32
7
15
7
16
7
20 15
− r,
7 7
1⎤
3
⎡1 0
⎢0 1 1
0 ⎥⎥
→⎢
⎢ 0 0 −1
8⎥
⎥
⎢
−3⎥⎦
⎢⎣0 0 −2
12
− 7 ⎤ ⎡1 0 0
⎥ ⎢
38 ⎥
⎢0 1 0
7 ⎥
→⎢
20 ⎥
⎢0 0 1
7 ⎥
⎢
⎢0 0 0
19 ⎥
7 ⎦⎥
⎣⎢
19 16
x4 = − r , and
7 7
1⎤
0 ⎥⎥
4 7 8⎥
⎥
1 −2 −3⎥⎦
⎤
− 72
0 − 33
7
7 ⎥
17
18 ⎥
0
7
7⎥
15
20
⎥
0
7
7 ⎥
16
19 ⎥
1
7
7 ⎦⎥
1
4
−2
0
x5 = r , where r is any real number.
9. The system is homogeneous with fewer equations than unknowns (2 < 3), so there are infinitely many solutions.
10. The system is homogeneous with fewer equations than unknowns (2 < 4), so there are infinitely many solutions.
5⎤
5⎤
⎡ 3 −4 ⎤
⎡ 1 5⎤
⎡1
⎡1
⎡1 0⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎢
⎥
1⎥ → ⎢⎢ 0 1 ⎥⎥ = A
11. ⎢ 1 5⎥ → ⎢ 3 −4 ⎥ → ⎢ 0 −19 ⎥ → ⎢0
⎢⎣ 4 −1⎥⎦
⎢⎣ 4 −1⎥⎦
⎢⎣ 0 −21⎥⎦
⎢⎣0 −21⎥⎦
⎢⎣ 0 0 ⎥⎦
A has k = 2 nonzero rows. Number of unknowns is n = 2. Thus k = n, so the system has the trivial solution only.
3
⎡
3
6⎤
⎡1 3
6 ⎤ ⎡ 1 0 3⎤
3 12 ⎤ ⎡ 1 2 6 ⎤ ⎢ 1
2
⎡2
2
⎥
⎢
⎥ ⎢
⎢
⎥
⎢
⎥
⎥ → ⎢0
⎥ → ⎢ 0 1 2 ⎥⎥ = A
12. ⎢ 3 −2 5⎥ → ⎢ 3 −2 5⎥ → ⎢ 0 − 13
1
2
13
−
2
⎥
⎢
⎥
⎢
⎥ ⎢
1 14 ⎦⎥ ⎢ 4
1 14 ⎥ ⎢ 0 −5 −10 ⎥
⎣⎢ 4
⎢⎣ 0 −5 −10 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦
⎣
⎦ ⎣
⎦
A has k = 2 nonzero rows. Number of unknowns is n = 3. Thus k < n, so the system has infinitely many solutions.
206
ISM: Introductory Mathematical Analysis
Section 6.5
⎡1 1 1⎤
⎡ 1 1 1⎤
⎡ 1 1 1⎤
⎡ 1 0 −1⎤
1 2 ⎥⎥ → ⎢⎢ 0 1 2 ⎥⎥ = A
13. ⎢⎢1 0 −1⎥⎥ → ⎢⎢0 −1 −2 ⎥⎥ → ⎢⎢0
⎢⎣1 −2 −5⎥⎦
⎢⎣0 −3 −6 ⎥⎦
⎢⎣0 −3 −6 ⎥⎦
⎢⎣ 0 0 0 ⎥⎦
A has k = 2 nonzero rows. Number of unknowns is n = 3. Thus k < n, so the system has infinitely many solutions.
⎡1 2 − 2 ⎤
⎡1 2 − 2⎤
3
3⎥
⎡ 3 2 −2 ⎤
⎡ 1 0 0⎤
⎡ 1 0 0⎤
3
3⎥
⎢
⎢
14. ⎢⎢ 2 2 −2 ⎥⎥ → ⎢ 2 2 −2 ⎥ → ⎢ 0 23 − 23 ⎥ → ⎢⎢0 1 −1⎥⎥ → ⎢⎢ 0 1 0 ⎥⎥ = A
⎢
⎥
⎢
⎥
5⎦⎥
5⎥
⎢ 0 −4
⎥ ⎣⎢ 0 0 1⎦⎥
5
⎣⎢ 0 −4
⎣⎢ 0 0 1⎦⎥
⎢⎣ 0 −4
⎦
⎣
⎦
A has k = 3 nonzero rows. Number of unknowns is n = 3. Thus k = n, so the system has the trivial solution only.
⎡ 1 1⎤
⎡ 1 1⎤
⎡1
15. ⎢
→⎢
→⎢
⎥
⎥
⎣3 −4 ⎦
⎣ 0 −7 ⎦
⎣0
The solution is x = 0, y = 0.
1⎤ ⎡1 0⎤
→
1⎥⎦ ⎢⎣ 0 1 ⎥⎦
⎡1 − 5 ⎤
⎡ 2 −5⎤
⎡ 2 −5⎤
2⎥
16. ⎢
→
→
⎢
⎥
⎢ 0 0⎥
8
20
−
0 ⎥⎦
⎣
⎦
⎣
⎦
⎣⎢ 0
5
The solution is x = r , y = r.
2
6⎤
⎡1 0
5⎥
⎢
⎢0 1 − 8 ⎥
15 ⎦
⎣
6 −2 ⎤ ⎡ 1 6 −2 ⎤
⎡ 1 6 −2 ⎤ ⎡ 1
→⎢
17. ⎢
⎥
⎥→⎢
8⎥→
⎣ 2 −3 4 ⎦ ⎣ 0 −15 8⎦ ⎣⎢ 0 1 − 15 ⎦⎥
6
8
The solution is x = − r , y = r , z = r.
5
15
7⎤
⎡ 1 7 ⎤ ⎡1 0 ⎤
⎡ 4 7 ⎤ ⎡ 1 74 ⎤ ⎡⎢ 1
4⎥
4⎥ →
→⎢
→⎢
18. ⎢
⎥→
⎥
⎢
⎥
⎢⎣ 0 1⎥⎦ ⎣ 0 1 ⎦
⎣ 2 3 ⎦ ⎢⎣ 2 3 ⎥⎦ ⎢⎣ 0 − 12 ⎥⎦
The solution is x = 0, y = 0.
1⎤ ⎡ 1
1⎤
⎡ 1 1⎤
⎡1
⎡1 0⎤
⎢
⎥
⎢
⎥
⎢
⎥
1⎥ → ⎢⎢ 0 1 ⎥⎥
19. ⎢ 3 −4 ⎥ → ⎢0 −7 ⎥ → ⎢ 0
⎢⎣5 −8⎥⎦
⎢⎣0 −13⎥⎦ ⎢⎣ 0 −13⎥⎦
⎢⎣ 0 0⎥⎦
The solution is x = 0, y = 0.
3⎤ ⎡ 1 0
3⎤
⎡ 2 −3 1⎤ ⎡ 0 −5 −1⎤
⎡ 1 1 1⎤
⎡1 0
⎡ 1 0 1⎤
1 −2 ⎥ → ⎢ 0
1 −2 ⎥ → ⎢ 0 1 − 2 ⎥ → ⎢ 0 1 − 2 ⎥ → ⎢ 0 1 0 ⎥
20. ⎢ 1 2 −1⎥ → ⎢ 0
⎢
⎥ ⎢
⎥
⎢
⎥
⎢
⎥ ⎢
⎥
⎢
⎥
1⎦
⎣ 1 1 1⎦ ⎣ 1 1 1⎦
⎣ 0 −5 −1⎦
⎣0 0 −11⎦ ⎣ 0 0
⎣ 0 0 1⎦
The solution is x = 0, y = 0, z = 0.
1⎤ ⎡ 1 1
1⎤ ⎡ 1
⎡1 1
⎢ 0 −7 −14 ⎥ ⎢ 0
1
2 ⎥⎥ ⎢⎢ 0
⎥→⎢
21. ⎢
→
⎢ 0 −2 −4 ⎥ ⎢ 0 −2 −4 ⎥ ⎢ 0
⎢
⎥ ⎢
⎥ ⎢
⎢⎣ 0 −5 −10 ⎥⎦ ⎢⎣ 0 −5 −10 ⎥⎦ ⎢⎣ 0
The solution is x = r, y = –2r, z = r.
0 −1⎤
1 2 ⎥⎥
0 0⎥
⎥
0 0 ⎥⎦
207
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
7⎤
7 ⎤ ⎡1
⎡ 1 1 7⎤
⎡1 1
⎡1 1
⎢ 1 −1 −1⎥
⎢ 0 −2 −8⎥
⎢0
1
4 ⎥⎥ ⎢⎢ 0
⎥→⎢
⎥→⎢
22. ⎢
→
⎢ 2 −3 −6 ⎥
⎢ 0 −5 −20 ⎥
⎢0 −5 −20 ⎥ ⎢ 0
⎢
⎥
⎢
⎥
⎢
⎥ ⎢
⎢⎣ 3 1 13⎥⎦
⎢⎣ 0 −2 −8⎥⎦
⎢⎣0 −2 −8⎥⎦ ⎢⎣ 0
The solution is x = –3r, y = –4r, z = r.
⎡1 1
⎢1 1
23. ⎢
⎢2
1
⎢
−
1
3
⎣⎢
0 3⎤
1 4 ⎥⎥
0 0⎥
⎥
0 0 ⎥⎦
4⎤
4⎤
4⎤
⎡1 1 1
⎡1 1 1
⎥
⎢
⎥
⎢
5⎥
0 0 −1
1⎥
0 −1 1 −4 ⎥⎥
→⎢
→⎢
⎢0 −1 1 −4 ⎥
⎢ 0 0 −1
3 4⎥
1⎥
⎥
⎢
⎥
⎢
⎥
2 −9 ⎦⎥
⎣⎢0 −4 1 −13⎦⎥
⎣⎢ 0 −4 1 −13⎦⎥
1
0
4⎤ ⎡ 1 0 2 0⎤ ⎡ 1
⎡1 1 1
⎢0
1 −1
4 ⎥⎥ ⎢⎢ 0 1 −1 4 ⎥⎥ ⎢⎢ 0
→⎢
→
→
⎢ 0 0 −1
1⎥ ⎢ 0 0 −1 1⎥ ⎢ 0
⎢
⎥ ⎢
⎥ ⎢
⎣⎢ 0 −4 1 −13⎦⎥ ⎣⎢ 0 0 −3 3⎦⎥ ⎣⎢ 0
The solution is w = –2r, x = –3r, y = r, z = r.
0⎤
⎡1
⎥
⎢0
4⎥
→⎢
⎢0
0
1 −1⎥
⎥
⎢
0 −3 3⎦⎥
⎣⎢ 0
0 2
1 −1
7⎤ ⎡ 1 1 2
7⎤
⎡ 1 1 2 7⎤
⎡1 1 2
⎡1
⎢ 1 −2 −1 1⎥
⎢ 0 −3 −3 −6 ⎥ ⎢ 0
⎥
⎢0
1 1
2⎥
⎢
⎥
⎢
⎥
⎢
24.
→
→
→⎢
⎢ 1 2 3 9⎥
⎢0
⎢0
1 1
2⎥ ⎢0
1 1
2⎥
⎢
⎥
⎢
⎥ ⎢
⎥
⎢
⎢⎣ 2 −3 −1 4 ⎥⎦
⎢⎣ 0 −5 −5 −10 ⎥⎦ ⎢⎣ 0 −5 −5 −10 ⎥⎦
⎢⎣ 0
The solution is w = –r – 5s, x = –r – 2s, y = r, z = s.
Principles in Practice 6.6
⎡ 1 3 ⎤ ⎡ −2 1.5⎤ ⎡1 0 ⎤
1. ⎢
⎥⎢
⎥=⎢
⎥
⎣ 2 4 ⎦ ⎣ 1 −0.5⎦ ⎣ 0 1 ⎦
Yes, they are inverses.
⎡ −2 1.5⎤ ⎡ 28⎤ ⎡13⎤ ⎡ M ⎤
2. ⎢
⎥⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥
⎣ 1 −0.5⎦ ⎣ 46 ⎦ ⎣ 5 ⎦ ⎣ E ⎦
⎡ −2 1.5⎤ ⎡ 65⎤ ⎡ 5 ⎤ ⎡ E ⎤
⎢ 1 −0.5⎥ ⎢90 ⎥ = ⎢ 20 ⎥ = ⎢ T ⎥
⎣
⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
⎡ −2
⎢ 1
⎣
⎡ −2
⎢ 1
⎣
⎡ −2
⎢ 1
⎣
⎡ −2
⎢ 1
⎣
⎡ −2
⎢ 1
⎣
1.5⎤ ⎡ 61⎤ ⎡ 1 ⎤ ⎡ A ⎤
=
=
−0.5⎥⎦ ⎢⎣82 ⎥⎦ ⎢⎣ 20 ⎥⎦ ⎢⎣ T ⎥⎦
1.5⎤ ⎡59 ⎤ ⎡14 ⎤ ⎡ N ⎤
=
=
−0.5⎥⎦ ⎢⎣88 ⎥⎦ ⎢⎣15⎥⎦ ⎢⎣ O ⎥⎦
1.5⎤ ⎡57 ⎤ ⎡15 ⎤ ⎡ O ⎤
=
=
−0.5⎥⎦ ⎢⎣86 ⎥⎦ ⎢⎣14 ⎥⎦ ⎢⎣ N ⎥⎦
1.5⎤ ⎡ 60 ⎤ ⎡ 6 ⎤ ⎡ F ⎤
=
=
−0.5⎥⎦ ⎢⎣84 ⎥⎦ ⎢⎣18⎥⎦ ⎢⎣ R ⎥⎦
1.5⎤ ⎡ 21⎤ ⎡ 9 ⎤ ⎡ I ⎤
=
=
−0.5⎥⎦ ⎢⎣34 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ D ⎥⎦
⎡ −2 1.5⎤ ⎡ 76 ⎤ ⎡ 1 ⎤ ⎡ A ⎤
⎢ 1 −0.5⎥ ⎢102 ⎥ = ⎢ 25⎥ = ⎢ Y ⎥
⎣
⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦
The message is “MEET AT NOON FRIDAY.”
208
2⎤
3⎥⎥
0 1 −1⎥
⎥
0 0 0 ⎦⎥
0 0
1 0
0 1 5⎤
1 1 2 ⎥⎥
0 0 0⎥
⎥
0 0 0 ⎥⎦
ISM: Introductory Mathematical Analysis
⎡3 1 2 1
3. ⎡⎣ E I ⎤⎦ = ⎢⎢ 2 2 2 0
⎢⎣ 2 1 3 0
⎡2 2 2
R1 ↔ R 2 ⎢
> ⎢3 1 2
⎢⎣ 2 1 3
Section 6.6
0 0⎤
1 0 ⎥⎥
0 1 ⎥⎦
0 1 0⎤
1 0 0 ⎥⎥
0 0 1 ⎥⎦
⎡ 1 1 1 1 0 0⎤
2
2
⎢
⎥
> ⎢ 3 2 3 0 1 0⎥
⎢
⎥
4 3 4 0 0 1⎥
⎣⎢
⎦
1R
2 1
⎡1
⎢
−3R1 + R 2 ⎢
> 0
−4R1 + R3 ⎢
⎢0
⎣
⎡1 1 1 0
0⎤
⎢
⎥
> ⎢3 1 2 1 0 0⎥
⎢
⎥
⎢⎣ 2 1 3 0 0 1 ⎥⎦
1
2
1R
2 1
1 0⎤
⎡1 1 1 0
2
⎢
⎥
R 2 ↔ R3
> ⎢ 0 −1 1 0 −1 1⎥
⎢
⎥
3
⎢⎣ 0 −2 −1 1 − 2 0 ⎥⎦
⎡1 0 2 0 − 1
1⎤
2
⎢
⎥
−R 2 + R1
> ⎢ 0 1 −1 0
1 −1⎥
2R 2 + R 3 ⎢
1 −2 ⎥
⎢⎣ 0 0 −3 1
⎥⎦
2
⎡1 0 2
0 − 12
1⎤
⎥
− 13 R3 ⎢
> ⎢0 1 −1
0
1 −1⎥
⎢
1
1
2⎥
⎢⎣0 0 1 − 3 − 6
3 ⎥⎦
⎡ 2
⎢ 3
E−1 = ⎢ − 13
⎢
⎢− 1
⎣ 3
⎡2
⎡⎣ F I ⎤⎦ = ⎢⎢ 3
⎢⎣ 4
− 16
5
6
− 16
1
2
3
− 16
5
6
1
−6
0
1 0
0 0⎤
⎥
1 0⎥
⎥
−2 0 1⎥
⎦
1
2
− 32
4. Let x be the number of shares of A, y be the
number of shares of B, and z be the number of
shares of C. We get the following equations
from the given conditions.
50x + 20y + 80z = 500,000
x = 2z
0.13(50x) + 0.15(20y) + 0.10(80z)
= 0.12(50x + 20y + 80z)
Simplify the first equation.
5x + 2y + 8z = 50,000
Simplify the second equation.
x – 2z = 0
Simplify the third equation.
6.5x + 3y + 8z = 6x + 2.4y + 9.6z
0.5x + 0.6y – 1.6z = 0
5x + 6y – 16z = 0
Thus, we solve the following system of
equations.
x – 2z = 0
5x + 6y – 16z = 0
5x + 2y + 8z = 50,000
⎡ 1 0 −2 ⎤
The coefficient matrix is A = ⎢⎢5 6 −16 ⎥⎥ .
⎢⎣5 2
8⎥⎦
1
⎡1 1 1 0
0⎤
2
⎢
⎥
> ⎢0
1 −1 0
1 −1⎥
⎢
⎥
3
0⎥
⎢⎣ 0 −2 −1 1 − 2
⎦
2
⎡1 0 0
3
⎢
−2R 3 + R1 ⎢
> 0 1 0 − 13
⎢
R3 + R 2
⎢0 0 1 − 1
3
⎣
1
⎡ 1 1 1 1 0 0⎤
2
2
⎥
2R 2 ⎢
> ⎢ 0 1 0 −3 2 0 ⎥
⎢
⎥
⎢⎣ 0 1 0 −2 0 1⎥⎦
⎡ 1 0 1 2 −1 0 ⎤
− 12 R 2 + R1 ⎢
> ⎢ 0 1 0 −3 2 0 ⎥⎥
−R 2 + R3
⎢⎣ 0 0 0
1 −2 1⎥⎦
F does not reduce to I so F is not invertible.
1 0⎤
⎡1 1 1 0
2
⎢
⎥
−3R1 + R 2 ⎢
> 0 −2 −1 1 − 32 0 ⎥
⎥
−2R1 + R3 ⎢
⎢0 −1 1 0 −1 1⎥
⎣
⎦
−R 2
1
2
1
2
− 13 ⎤
⎥
− 13 ⎥
⎥
2⎥
3⎦
− 13 ⎤
⎥
− 13 ⎥
⎥
2⎥
3⎦
2 1 0 0⎤
3 0 1 0 ⎥⎥
4 0 0 1 ⎥⎦
⎡ 1 0 −2 1 0 0 ⎤
⎡⎣ A I ⎤⎦ = ⎢⎢5 6 −16 0 1 0 ⎥⎥
⎢⎣5 2
8 0 0 1⎥⎦
1 0 0⎤
⎡ 1 0 −2
−5R1 + R 2 ⎢
> ⎢ 0 6 −6 −5 1 0 ⎥⎥
5
− R1 + R3
⎢⎣ 0 2 18 −5 0 1⎥⎦
209
Chapter 6: Matrix Algebra
⎡ 1 0 −2
⎢
> ⎢ 0 1 −1
⎢
⎣ 0 2 18
⎡1 0
⎢
−2 R 2 + R 3 ⎢
> 0 1
⎢
⎢0 0
⎢⎣
⎡ 1 0 −2
1 R
⎢
20 3
> ⎢ 0 1 −1
⎢
⎢0 0
1
⎣
1R
6 2
ISM: Introductory Mathematical Analysis
1 0 0⎤
1 0⎥
⎥
6
⎥
−5 0 1⎦
− 56
−2
1
−1
− 56
20 − 10
3
1
0
− 56
− 16
1
6
1
− 60
2 − 1
⎡1 0 0
3
30
⎢
2R3 + R1 ⎢
3
> 0 1 0 −1
20
R3 + R 2 ⎢
⎢0 0 1 − 1 − 1
6
60
⎣
⎡ 2 − 1
30
⎢ 3
−1 ⎢
3
A = −1
20
⎢
⎢− 1 − 1
60
⎣ 6
0 0⎤
⎥
1 0⎥
6
⎥
− 13 1⎥⎥
⎦
0⎤
⎥
0⎥
⎥
1 ⎥
20 ⎦
1⎤
10 ⎥
1 ⎥
20 ⎥
1 ⎥
20 ⎦
1⎤
10 ⎥
1 ⎥
20 ⎥
1 ⎥
20 ⎦
1⎤
⎡ 2 − 1
30 10 ⎥ ⎡ 0 ⎤ ⎡ 5000 ⎤
⎡ x⎤ ⎢ 3
⎢ ⎥ ⎢
3
1 ⎥ ⎢ 0 ⎥ = ⎢ 2500 ⎥
⎢ y ⎥ = ⎢ −1
⎥ ⎢
⎥
20 20 ⎥ ⎢
⎢⎣ z ⎥⎦ ⎢ 1
⎢⎣50, 000 ⎥⎦ ⎢⎣ 2500 ⎥⎦
1
1
⎥
−
− 60 20
⎣ 6
⎦
They should buy 5000 shares of Company A, 2500 shares of Company B, and 2500 shares of Company C.
Problems 6.6
⎡ 6 1 1 0 ⎤ ⎡ 1 16
1. ⎢
⎥→⎢
⎣ 7 1 0 1 ⎦ ⎢⎣ 7 1
1
0⎤ ⎡ 1
6
⎥→⎢
1
⎢
0 1 ⎥⎦ ⎣ 0 − 6
1
6
1
6
7
−6
0⎤
⎥
1⎥
⎦
0 −1 1⎤ ⎡ 1 0 −1 1⎤
⎡1
→⎢
⎥→
1
0 − 6 − 76 1⎥ ⎢⎣ 0 1 7 −6 ⎥⎦
⎣⎢
⎦
⎡ −1 1⎤
The inverse is ⎢
⎥.
⎣ 7 −6 ⎦
1
1
⎡1 2
⎡ 2 4 1 0 ⎤ ⎡⎢1 2 2 0 ⎤⎥
2
⎢
2. ⎢
→
→
⎣ 3 6 0 1⎥⎦ ⎢⎣1 2 0 13 ⎥⎦
⎢⎣ 0 0 − 12
The given matrix is not invertible.
0⎤
⎥
1
⎥
3⎦
⎡ 2 2 1 0 ⎤ ⎡ 2 2 1 0 ⎤ ⎡ 1 1 12 0 ⎤
→
→⎢
3. ⎢
⎥
⎣ 2 2 0 1⎥⎦ ⎢⎣ 0 0 −1 1⎥⎦ ⎣ 0 0 −1 1⎦
The given matrix is not invertible.
210
ISM: Introductory Mathematical Analysis
3
⎡1
4
8
⎢
4.
⎢0 − 1
6
⎣
Section 6.6
1 0⎤ ⎡ 1 3 4 0⎤
⎡ 1 0 4 9⎤
2
⎥→⎢
⎥→⎢
⎥
0 1⎥ ⎢⎣ 0 1 0 −6 ⎥⎦
⎣ 0 1 0 −6 ⎦
⎦
⎡4 9⎤
The inverse is ⎢
⎥.
⎣ 0 −6 ⎦
⎡
0
⎡ 1 0 0 1 0 0⎤ ⎢ 1 0 0 1
5. ⎢⎢ 0 −3 0 0 1 0 ⎥⎥ → ⎢ 0 1 0 0 − 13
⎢
⎢⎣ 0 0 4 0 0 1⎥⎦ ⎢
0
⎣0 0 1 0
⎡1
0
⎢
1
⎢
The inverse is 0 − 3
⎢
⎢0
0
⎣
0⎤
⎥
0⎥
⎥
1⎥
4⎦
0⎤
⎥
0⎥ .
⎥
1⎥
4⎦
⎡ 2 0 8 1 0 0 ⎤ ⎡⎢ 1 0 4
6. ⎢⎢ −1 4 0 0 1 0 ⎥⎥ → ⎢ −1 4 0
⎢⎣ 2 1 0 0 0 1⎥⎦ ⎢⎢ 2 1 0
⎣
0 0⎤
⎥
0 1 0⎥
⎥
0 0 1⎥
⎦
1
2
⎡ 1 0 4 1 0 0⎤
⎡1 0 4 1
2
2
⎢
⎥
⎢
1
⎢
⎥
⎢
→ 0 4 4 2 1 0 → 0 1 1 18
⎢
⎥
⎢
⎢ 0 1 −8 −1 0 1⎥
⎢0 1 −8 −1
⎣
⎦
⎣
⎡1
⎢
→ ⎢0
⎢
⎢0
⎢⎣
⎡1
⎢
→ ⎢0
⎢
⎢0
⎣
0
4
1
1
0 −9
0 0
1 0
0
0 0⎤ ⎡ 1 0 4
⎥ ⎢
1 0⎥ → ⎢0 1 1
4
⎥ ⎢
1
− 4 1⎥⎥ ⎢⎣ 0 0 1
⎦
1
2
1
8
9
−8
0 − 19
2
0
9
1 18
1
36
⎡0 − 1
9
⎢
2
⎢
The inverse is 0
9
⎢
1
⎢1
⎣ 8 36
1
2
1
8
1
8
0 0⎤
⎥
1 0⎥
4
⎥
0 1⎥
⎦
0
1
4
1
36
0⎤
⎥
0⎥
⎥
− 19 ⎥
⎦
4⎤
9⎥
1⎥
9⎥
− 19 ⎥
⎦
4⎤
9⎥
1⎥
9⎥
1
− 9⎥
⎦
.
211
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡1 2 3 1 0 0 ⎤
⎡1 2 3 1 0 0 ⎤
⎢
⎥
⎢
⎥
7. ⎢ 0 0 4 0 1 0 ⎥ → ⎢0 0 1 0 14 0 ⎥
⎢
⎥
⎣⎢ 0 0 5 0 0 1 ⎥⎦
⎣0 0 5 0 0 1 ⎦
⎡ 1 2 0 1 − 3 0⎤
4
⎢
⎥
1
⎢
⎥
→ 0 0 1 0
0
4
⎢
⎥
⎢ 0 0 0 0 − 5 1⎥
4
⎣
⎦
The given matrix is not invertible.
⎡1 0 0
⎡2 0 0 1 0 0⎤ ⎢
8. ⎢⎢ 0 0 0 0 1 0 ⎥⎥ → ⎢ 0 0 1
⎢
⎣⎢ 0 0 −4 0 0 1 ⎦⎥ ⎢ 0 0 0
⎣
The given matrix is not invertible.
0⎤
⎥
0 0 − 14 ⎥
⎥
0 1
0⎥
⎦
1
2
0
9. The matrix is not square, so it is not invertible.
⎡0 0 0⎤ ⎡0 0 0⎤
10. For any 3 × 3 matrix B, B ⎢⎢ 0 0 0 ⎥⎥ = ⎢⎢ 0 0 0 ⎥⎥ ≠ I.
⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦
Thus the matrix is not invertible.
⎡1 2 3 1
11. ⎢ 0 1 2 0
⎢
⎣0 0 1 0
⎡1 0 0
→ ⎢0 1 0
⎢
⎣0 0 1
0
1
0
1
0
0
0 ⎤ ⎡ 1 0 −1 1 −2 0 ⎤
0⎥ → ⎢0 1 2 0
1 0⎥
⎥ ⎢
⎥
1⎦ ⎣ 0 0 1 0 0 1⎦
−2
1⎤
1 −2 ⎥
⎥
0
1⎦
1⎤
⎡ 1 −2
The inverse is ⎢ 0
1 −2 ⎥ .
⎢
⎥
1⎦
⎣0 0
⎡ 1 2 −1 1 0 0 ⎤
⎡ 1 2 −1 1 0 0 ⎤
⎢
⎥
1 4 0 1 0 ⎥⎥
12. ⎢ 0 1 4 0 1 0 ⎥ → ⎢⎢0
⎢⎣ 1 −1 2 0 0 1⎥⎦
⎢⎣0 −3 3 −1 0 1⎥⎦
2
⎡1 0 0
1 −2 0 ⎤
5
⎡ 1 0 −9 1 −2 0 ⎤ ⎡ 1 0 −9
⎢
⎢
⎥
4
→ ⎢⎢ 0 1 4 0
1 0 ⎥⎥ → ⎢ 0 1 4
0
1 0⎥ → ⎢0 1 0
15
⎢
1
1
1⎥
⎢⎣ 0 0 15 −1 3 1⎥⎦ ⎢ 0 0
⎢0 0 1 − 1
1 − 15
⎢⎣
5 15 ⎥⎦
15
⎣
⎡ 2
⎢ 5
4
The inverse is ⎢ 15
⎢
⎢− 1
⎣ 15
− 15
1
5
1
5
3⎤
5⎥
4 ⎥.
− 15
⎥
1⎥
15 ⎦
212
− 15
1
5
1
5
3⎤
5⎥
4⎥
− 15
⎥
1⎥
15 ⎦
ISM: Introductory Mathematical Analysis
Section 6.6
⎡ 1 0 −2
⎡ 7 0 −2 1 0 0 ⎤
7
⎢
0
13. ⎢⎢ 0 1 0 0 1 0 ⎥⎥ → ⎢ 0 1
⎢
⎢⎣ −3 0
⎥
1 0 0 1⎦
1
⎢⎣ −3 0
⎡1 0 − 2
7
⎢
→ ⎢0 1
0
⎢
1
⎢⎣ 0 0
7
⎡1
The inverse is ⎢⎢ 0
⎣⎢ 3
1
7
0
3
7
0 0⎤ ⎡ 1 0
⎥ ⎢
1 0⎥ → ⎢0 1
⎥ ⎢
0 1⎥ ⎢ 0 0
⎦ ⎣
0 2⎤
1 0 ⎥⎥ .
0 7 ⎦⎥
⎡ 2 3 −1 1 0 0 ⎤ ⎡ 1 2
14. ⎢⎢ 1 2 1 0 1 0 ⎥⎥ → ⎢⎢ 1 1
⎢⎣ −1 −1 3 0 0 1⎥⎦ ⎢⎣ 2 3
1 0
1 0⎤ ⎡ 1
⎡1 2
→ ⎢⎢ 0 1 4 0
1 1⎥⎥ → ⎢⎢ 0
⎢⎣ 0 −1 −3 1 −2 0 ⎥⎦ ⎢⎣ 0
0 0⎤
⎥
0 1 0⎥
⎥
0 0 1⎥
⎦
1
7
1 0 2⎤
⎡1 0 0 1 0 2 ⎤
⎥
0 1 0 ⎥ → ⎢⎢ 0 1 0 0 1 0 ⎥⎥
3 0 1⎥
⎣⎢ 0 0 1 3 0 7 ⎦⎥
⎥⎦
7
0
0
1
7
1 0
1
−3 0 0
−1 1 0
2 1 0
1 4 0
0
1 1
0⎤ ⎡ 1 2
−1⎥⎥ → ⎢⎢ 0 −1
0 ⎥⎦ ⎢⎣ 0 −1
1 0⎤
⎡1 2
1 1⎥⎥ → ⎢⎢0 1
⎢⎣0 0
−1 1⎥⎦
1 0
1
−4 0
−1
−3 1 − 2
0 −1 2
0 −4
1
5
1 −1
0⎤
−1⎥⎥
0 ⎥⎦
−1⎤ ⎡ 1 0 0 7 −8 5⎤
−3⎥⎥ → ⎢⎢ 0 1 0 −4 5 −3⎥⎥
1⎥⎦ ⎢⎣ 0 0 1 1 −1 1⎥⎦
⎡ 7 −8 5⎤
The inverse is ⎢⎢ −4 5 −3⎥⎥ .
⎢⎣ 1 −1 1⎥⎦
⎡ 2 1 0 1 0 0 ⎤ ⎡ 1 −1
15. ⎢⎢ 4 −1 5 0 1 0 ⎥⎥ → ⎢⎢ 4 −1
⎢⎣ 1 −1 2 0 0 1⎥⎦ ⎢⎣ 2 1
1⎤ ⎡ 1
⎡ 1 −1 2 0 0
⎢
⎢
→ ⎢ 0 3 −3 0 1 −4 ⎥⎥ → ⎢ 0
⎢⎣ 0 3 −4 1 0 −2 ⎥⎦ ⎢ 0
⎣
1⎤
1 0 ⎥⎥
0 1 0 0 ⎥⎦
−1 2 0 0
2 0 0
5 0
1 −1 0
3 −4
1
3
1 0
⎡1 0 1 0 1 − 1⎤
⎡1 0 1 0
3
3⎥
⎢
⎢
→ ⎢ 0 1 −1 0 13 − 43 ⎥ → ⎢0 1 −1 0
⎢
⎥
⎢
⎢ 0 0 −1 1 −1
⎢0 0 1 −1
2⎥
⎣
⎦
⎣
1
3
1
3
1
1⎤
⎥
− 43 ⎥
⎥
−2 ⎦
5⎤
⎡1 0 0 1 − 2
− 13 ⎤
3
3⎥
⎢
⎥
10
4
4
⎢
⎥
− 3 → 0 1 0 −1
− 3 ⎥.
3
⎢
⎥
⎥
⎢ 0 0 1 −1
−2 ⎥
1 −2 ⎥
⎢⎣
⎦
⎦⎥
5⎤
⎡ 1 −2
3
3⎥
⎢
10
4
⎢
The inverse is −1
− 3 ⎥.
3
⎢
⎥
⎢ −1
1 −2 ⎥
⎥⎦
⎣⎢
213
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡ −1 2 −3 1 0 0 ⎤ ⎡ 1 −2 3
1 0 0 1 0⎥ → ⎢2
1 0
16. ⎢ 2
⎢
⎥ ⎢
−
−
4
2
5
0
0
1
4
2
5
⎣
⎦ ⎣
1
−
2
⎡
3 −1 0 0 ⎤
⎡ 1 −2
1
→ ⎢ 0 5 −6 2 1 0 ⎥ → ⎢ 0
⎢
⎢
⎥
0
6
7
4
0
1
−
⎣
⎦
⎣⎢ 0 6
3
1
2
⎡1 0
⎡1 0
−5
0⎤
5
5
⎢
⎥
⎢
6
2
1
→ ⎢0 1 − 5
0 ⎥ → ⎢0 1
5
5
⎢
⎢
8 − 6 1⎥
1
0 0
⎢⎣ 0 0
⎥⎦
5
5
5
⎣⎢
⎡ 1 0 0 −5 4 −3⎤
→ ⎢ 0 1 0 10 −7 6 ⎥
⎢
⎥
⎣ 0 0 1 8 −6 5 ⎦
−1
0
0
3
6
−5
0 0⎤
1 0⎥
⎥
0 1⎦
−1 0 0 ⎤
2 1 0⎥
5 5
⎥
−7 4 0 1⎦⎥
3
5
− 65
0⎤
⎥
0⎥
⎥
8 −6 5⎥
⎦
− 15
1
2
5
2
5
1
5
⎡ −5 4 −3⎤
The inverse is ⎢ 10 −7 6 ⎥ .
⎢
⎥
⎣ 8 −6 5⎦
⎡1 2 3 1 0 0 ⎤
⎡ 1 2 3 1 0 0⎤
⎢
⎥
17. ⎢1 3 5 0 1 0 ⎥ → ⎢⎢ 0 1 2 −1 1 0 ⎥⎥
⎢⎣1 5 12 0 0 1 ⎥⎦
⎢⎣ 0 3 9 −1 0 1⎥⎦
1⎤
⎡ 1 0 0 11 −3
⎡ 1 0 −1 3 −2 0 ⎤
3
3⎥
⎡ 1 0 −1 3 −2 0 ⎤
⎢
⎢
⎥
⎢
⎥
7
2
→ ⎢ 0 1 2 −1 1 0 ⎥ → ⎢0 1 2 −1 1 0 ⎥ → ⎢ 0 1 0 − 3
3 − 3⎥
⎢
⎥
⎢
⎥
2
1
2 −1
1⎥
⎢0 0 1
⎣⎢ 0 0 3 2 −3 1⎥⎦
⎢⎣0 0 1 3 −1 3 ⎥⎦
3
3⎦
⎣
1⎤
⎡ 11 −3
3⎥
⎢ 3
The inverse is ⎢ − 73
3 − 23 ⎥ .
⎢
⎥
1⎥
⎢ 2 −1
3⎦
⎣ 3
⎡ 2 −1 3 1 0 0 ⎤ ⎡ 2 −1
18. ⎢⎢ 0 2 0 0 1 0 ⎥⎥ → ⎢⎢ 0 2
⎢⎣ 2 1 1 0 0 1⎥⎦ ⎢⎣ 0 2
3
1 0 0⎤
⎡1 − 1
⎡1
2
2
2
⎢
⎥
⎢
⎢0
1 0 0 12 0 ⎥ → ⎢0
⎢
⎥
⎢
⎢0
⎢0
2 −2 −1 0 1⎥
⎣
⎦
⎣
⎡1 0 3
2
⎢
→ ⎢0 1 0
⎢
⎢0 0 1
⎣
1
2
0
1
2
1
4
1
2
1
2
1 0 0⎤
1 0 ⎥⎥
−2 −1 0 1⎥⎦
3
0
0
0⎤
⎥
1 0 0
0⎥
⎥
0 −2 −1 −1 1⎥
⎦
0
3
2
1
2
1
4
1
2
0 ⎤ ⎡ 1 0 0 − 14 − 12
⎥ ⎢
1
0⎥ → ⎢0 1 0
0
2
⎥ ⎢
1
1
1
⎥
⎢
0 0 1
−2
2
2
⎦ ⎣
⎡− 1 − 1
2
⎢ 4
1
The inverse is ⎢ 0
2
⎢
1
⎢ 1
2
⎣ 2
3⎤
4⎥
0⎥
⎥
− 12 ⎥
⎦
3⎤
4⎥
0⎥ .
⎥
− 12 ⎥
⎦
214
ISM: Introductory Mathematical Analysis
Section 6.6
⎡x ⎤
⎡1 2 ⎤ ⎡ 2 ⎤ ⎡10 ⎤
19. X = ⎢ 1 ⎥ = A −1B = ⎢
⎥ ⎢ ⎥ = ⎢ ⎥ ⇒ x1 = 10, x2 = 20
⎣8 1 ⎦ ⎣ 4 ⎦ ⎣ 20⎦
⎣ x2 ⎦
⎡ 1 0 1 ⎤ ⎡10 ⎤ ⎡ 9 ⎤
⎡ x1 ⎤
−1
20. X = ⎢ ⎥ = A B = ⎢⎢ 0 3 0 ⎥⎥ ⎢⎢ 2 ⎥⎥ = ⎢⎢ 6 ⎥⎥ ⇒ x1 = 9, x2 = 6, x3 = 16
⎣ x2 ⎦
⎢⎣ 2 0 4 ⎥⎦ ⎢⎣ −1⎥⎦ ⎢⎣16 ⎥⎦
1⎤
⎡6 5 1 0⎤
⎡1 1 0 1 ⎤ ⎡ 1 1 0
⎡ 1 1 0 1⎤ ⎡ 1 0 1 −5⎤
21. ⎢
⎥ → ⎢ 6 5 1 0 ⎥ → ⎢ 0 −1 1 −6 ⎥ → ⎢ 0 1 −1 6 ⎥ → ⎢ 0 1 −1 6 ⎥
1
1
0
1
⎣
⎦ ⎣
⎦
⎣
⎦
⎣
⎦ ⎣
⎦
x
1
−
5
2
17
⎡ ⎤
⎡
⎤⎡ ⎤ ⎡
⎤
−1
⎢ y ⎥ = A B = ⎢ −1 6 ⎥ ⎢ −3⎥ = ⎢ −20⎥ ⇒ x = 17, y = −20
⎣ ⎦
⎣
⎦⎣ ⎦ ⎣
⎦
⎡1 2
⎡ 2 4 1 0 ⎤ ⎡ 1 2 12 0 ⎤
→⎢
22. ⎢
⎥→⎢
⎥
⎢0 5
⎣ −1 3 0 1⎦ ⎢⎣ −1 3 0 1⎥⎦
⎣
⎡3
⎡ x⎤
−1
⎢ 10
=
=
A
B
⎢ y⎥
⎢1
⎣ ⎦
⎣ 10
1
2
1
2
⎡1 0 3
0⎤
10
⎥→⎢
⎢0 1 1
1⎥⎦
10
⎣
23 ⎤
− 52 ⎤ ⎡ 5⎤ ⎡ 10
⎥ ⎢ ⎥ = ⎢ ⎥ ⇒ x = 23 , y = 1
1 ⎥ −2
1 ⎥
10
10
⎣ ⎦ ⎣⎢ 10
5⎦
⎦
1 0⎤
⎡1 1
⎡ 1 1 1 0⎤ ⎡ 1 1
⎡3 1 1 0 ⎤
3
3
3
3
3
23. ⎢
→⎢
⎥→⎢
⎥→⎢
⎣3 −1 0 1⎦⎥
⎢⎣0 1
⎣3 −1 0 1⎦ ⎣ 0 −2 −1 1⎦
1⎤
⎡1 0 1
6
6⎥
→⎢
1
1
⎢⎣ 0 1 2 − 2 ⎥⎦
⎡1
⎡x⎤
−1
⎢6
=
=
A
B
⎣⎢ y ⎦⎥
⎢⎣ 12
− 52 ⎤
⎥
1⎥
5⎦
1⎤
6 ⎥ ⎡5 ⎤
1
− 2 ⎥⎦ ⎣⎢7 ⎦⎥
⎡ 3 2 1 0 ⎤ ⎡ 1 23
24. ⎢
⎥→⎢
⎣ 4 3 0 1 ⎦ ⎢⎣ 4 3
0⎤
⎥
− 12 ⎥⎦
⎡ 2⎤
= ⎢ ⎥ ⇒ x = 2, y = −1
⎣ −1⎦
0⎤ ⎡ 1 3
⎥→⎢
0 1 ⎥⎦ ⎢⎣ 0 13
1
3
1
3
1
2
2
−
1
3
4
3
0⎤
3 −2 ⎤
⎡1 0
3 −2 ⎤
⎡1 0
⎥ →⎢
⎥→⎢
1
4
−
0
1
⎥
−
0
1
4
3⎥⎦
1
⎢⎣
⎥⎦
⎣
3
3
⎦
⎡ x⎤
⎡ 3 −2 ⎤ ⎡ 26 ⎤ ⎡ 4 ⎤
−1
=
⇒ x = 4, y = 7
⎢ y ⎥ = A B = ⎢ −4
3⎥⎦ ⎢⎣37 ⎥⎦ ⎢⎣7 ⎥⎦
⎣ ⎦
⎣
25. The coefficient matrix is not invertible. The method of reduction yields
⎡ 2 6 2 ⎤ ⎡1 3 1 ⎤
⎡1 3 1 ⎤
⎢ 3 9 3 ⎥ → ⎢ 3 9 3⎥ → ⎢ 0 0 0 ⎥ .
⎣
⎦ ⎣
⎦
⎣
⎦
Thus x = –3r + 1, y = r.
26. The coefficient matrix is not invertible. The method of reduction yields
⎡ 2 6 8⎤ ⎡ 1 3 4 ⎤ ⎡ 1 3 4 ⎤
⎢
⎥→⎢
⎥→⎢
⎥.
⎣ 3 9 7 ⎦ ⎣3 9 7 ⎦ ⎣ 0 0 −5⎦
Second row indicates 0 = −5, which is never true, so there is no solution.
215
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
1 1 0 0⎤
⎡1 2 1 1 0 0 ⎤ ⎡ 1 2
27. ⎢⎢3 0 1 0 1 0 ⎥⎥ → ⎢⎢ 0 −6 −2 −3 1 0 ⎥⎥
⎢⎣1 −1 1 0 0 1 ⎥⎦ ⎢⎣ 0 −3 0 −1 0 1⎥⎦
1
⎡1 0 1 0
0 0⎤
⎡1 2 1 1
3
3
⎢
⎢
⎥
1
1
1
1
1
1
⎢
→ ⎢0
1 3 2 − 6 0⎥ → 0 1 3 2 − 6
⎢
⎢
⎥
⎢0 0 1 1 − 1
0 1⎦
⎣ 0 −3 0 −1
2
2
⎣
⎡1 0 0 − 1
6
⎢
1
→ ⎢0 1 0
3
⎢
1
⎢0 0 1
2
⎣
0⎤
⎥
0⎥
⎥
1⎥⎦
− 13 ⎤
⎥
0 − 13 ⎥
⎥
− 12
1⎥⎦
1
2
1 − 1⎤
⎡− 1
2
3 ⎥ ⎡ 4⎤ ⎡0 ⎤
⎡ x⎤
⎢ 6
⎢ y ⎥ = A −1B = ⎢ 1
1 ⎥ ⎢ 2⎥ = ⎢1 ⎥
0
−
⎢ ⎥
3⎥ ⎢ ⎥ ⎢ ⎥
⎢ 3
⎢⎣ z ⎥⎦
⎢ ⎥ ⎢ ⎥
⎢ 1 −1
1⎥⎦ ⎣ 1 ⎦ ⎣ 2 ⎦
2
⎣ 2
Thus, x = 0, y = 1, z = 2.
⎡1 1 1 1 0 0 ⎤
⎡ 1 1 1 1 0 0⎤
28. ⎢⎢1 −1 1 0 1 0 ⎥⎥ → ⎢⎢ 0 −2 0 −1 1 0 ⎥⎥
⎢⎣1 −1 −1 0 0 1⎥⎦
⎢⎣ 0 −2 −2 −1 0 1⎥⎦
1
0 0⎤ ⎡ 1 0
⎡1 1 1 1
⎢
⎢
⎥
→ ⎢0
1 0 12 − 12 0 ⎥ → ⎢ 0 1 0
⎢
⎥ ⎢
0 1⎦ ⎢ 0 0 −2
⎣ 0 −2 −2 −1
⎣
⎡1 0 1
⎢
→ ⎢0 1 0
⎢
⎢0 0 1
⎣
1
2
1
2
0
1
2
1
−2
1
2
⎡1 0 0
0⎤
⎥
⎢
0 ⎥ → ⎢0 1 0
⎥
⎢
⎢0 0 1
− 12 ⎥⎦
⎣
1
2
1
2
0⎤
⎥
0⎥
⎥
−1 1⎥
⎦
1
2
1
2
1
2
1
−2
0
1⎤
2⎥
0
0⎥
⎥
− 12 ⎥⎦
− 12
1
2
0
1⎤
⎡ ⎤
⎡1
0
2 ⎥ ⎡ 6 ⎤ ⎢ 5⎥
⎡ x⎤
⎢2
⎢ y ⎥ = A −1B = ⎢ 1 − 1
0 ⎥ ⎢⎢ −1⎥⎥ = ⎢ 72 ⎥
⎢ ⎥
2
⎢ ⎥
⎢2
⎥
⎢⎣ z ⎥⎦
1 − 1 ⎥ ⎢⎣ 4 ⎥⎦ ⎢ − 5 ⎥
⎢0
2
2⎦
⎣
⎣ 2⎦
7
5
Thus, x = 5, y = , z = − .
2
2
⎡1 1 1 1 0 0 ⎤
⎡ 1 1 1 1 0 0⎤
⎢
⎥
29. ⎢1 −1 1 0 1 0 ⎥ → ⎢⎢ 0 −2 0 −1 1 0 ⎥⎥
⎢⎣1 −1 −1 0 0 1⎥⎦
⎢⎣ 0 −2 −2 −1 0 1⎥⎦
1
0 0⎤ ⎡ 1 0
⎡1 1 1 1
⎢
⎢
⎥
→ ⎢0
1 0 12 − 12 0 ⎥ → ⎢ 0 1 0
⎢
⎥ ⎢
0 1⎦ ⎢ 0 0 −2
⎣ 0 −2 −2 −1
⎣
1
2
1
2
0
0⎤
⎥
0⎥
⎥
−1 1⎥
⎦
1
2
− 12
216
ISM: Introductory Mathematical Analysis
⎡1 0 1
⎢
→ ⎢0 1 0
⎢
⎢0 0 1
⎣
1
2
1
2
0
1
2
− 12
1
2
Section 6.6
⎡1 0 0
0⎤
⎥
⎢
0 ⎥ → ⎢0 1 0
⎥
⎢
⎢0 0 1
− 12 ⎥⎦
⎣
1
2
1
2
1⎤
2⎥
0
− 12
0
1
2
0⎥
⎥
− 12 ⎥⎦
1⎤
⎡1
⎡ ⎤
0
2 ⎥ ⎡ 2⎤ ⎢ 1 ⎥
⎡ x⎤
⎢2
⎢ y ⎥ = A −1B = ⎢ 1 − 1
0 ⎥ ⎢⎢ 1 ⎥⎥ = ⎢ 12 ⎥
⎢ ⎥
2
⎢2
⎥
⎢ ⎥
⎢⎣ z ⎥⎦
1 − 1 ⎥ ⎢⎣ 0 ⎥⎦ ⎢ 1 ⎥
⎢0
2
2⎦
⎣
⎣2⎦
1
1
Thus, x = 1, y = , z = .
2
2
⎡ 2 0 8 1 0 0 ⎤ ⎡ 1 −4 0
30. ⎢⎢ −1 4 0 0 1 0 ⎥⎥ → ⎢⎢ 2 0 8
⎢⎣ 2 1 0 0 0 1⎥⎦ ⎢⎣ 2
1 0
⎡ 1 −4
⎡ 1 −4 0 0 −1 0 ⎤
⎢
⎢
⎥
→ ⎢0 8 8 1 2 0 ⎥ → ⎢0
1
⎢
⎢⎣ 0 9 0 0 2 1⎥⎦
⎣0 9
0 −1 0 ⎤
1 0 0 ⎥⎥
0 0 1⎥⎦
0 −1 0 ⎤
⎥
1 18 14 0 ⎥
⎥
0 0 2 1⎦
0
1
⎡1 0 4
0 0⎤ ⎡ 1 0 4
2
⎢
⎥ ⎢
1
1 0⎥ → ⎢0 1 1
→ ⎢0 1 1
8
4
⎢
⎥ ⎢
⎢ 0 0 −9 − 9 − 1 1⎥ ⎢ 0 0 1
⎢⎣
⎥⎦ ⎣
8
4
4⎤
⎡0 − 1
9
9 ⎥ ⎡ 8 ⎤ ⎡0⎤
⎡ x⎤
⎢
⎢ y ⎥ A −1B = ⎢ 0
2
1 ⎥ ⎢36 ⎥ = ⎢9 ⎥
⎢ ⎥
9
9
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢⎣ z ⎥⎦
⎢ ⎥ ⎢ ⎥
1
1
⎢1
− 9 ⎥ ⎣ 9 ⎦ ⎣1 ⎦
⎣ 8 36
⎦
Thus, x = 0, y = 9, z = 1.
1
2
1
8
1
8
0
1
4
1
36
⎡1 0 0
0⎤
⎥
⎢
0⎥ → ⎢0 1 0
⎥
⎢
⎢0 0 1
− 19 ⎥
⎦
⎣
0 − 19
2
0
9
1
8
1
36
31. The coefficient matrix is not invertible. The method of reduction yields
7⎤ ⎡ 1 3 3 7⎤
⎡1 3 3 7 ⎤
⎡1 3 3
⎡1 0 0 1 ⎤
⎢ 2 1 1 4 ⎥ → ⎢ 0 −5 −5 −10 ⎥ → ⎢ 0
⎥ → ⎢0 1 1 2⎥ .
1
1
2
⎢
⎥
⎢
⎥ ⎢
⎥
⎢
⎥
⎢⎣ 1 1 1 4 ⎥⎦
⎢⎣0 −2 −2 −3⎥⎦ ⎢⎣ 0 −2 −2 −3⎥⎦
⎢⎣ 0 0 0 1 ⎥⎦
The third row indicates that 0 = 1, which is never true, so there is no solution.
32. The coefficient matrix is not invertible. The method of reduction yields
7⎤ ⎡ 1 3 3 7⎤
⎡1 3 3 7 ⎤
⎡1 3 3
⎡1 0 0 1 ⎤
⎢ 2 1 1 4 ⎥ → ⎢ 0 −5 −5 −10 ⎥ → ⎢ 0
⎥ → ⎢0 1 1 2⎥ .
1
1
2
⎢
⎥
⎢
⎥ ⎢
⎥
⎢
⎥
⎢⎣ 1 1 1 3 ⎥⎦
⎢⎣0 −2 −2 −4 ⎥⎦ ⎢⎣ 0 −2 −2 −4 ⎥⎦
⎢⎣ 0 0 0 0 ⎥⎦
Thus, x = 1, y = –r + 2, z = r.
217
4⎤
9⎥
1⎥
9⎥
− 19 ⎥
⎦
Chapter 6: Matrix Algebra
⎡1 0
⎢ 1 −1
33. ⎢
⎢2 1
⎢
⎢⎣ 1 2
⎡1
⎢0
→⎢
⎢0
⎢
⎢⎣ 0
ISM: Introductory Mathematical Analysis
2 1 1 0 0 0⎤
⎡1 0 2
⎢ 0 −1 −2
0 2 0 1 0 0 ⎥⎥
→⎢
⎢ 0 1 −4
0 1 0 0 1 0⎥
⎥
⎢
1 1 0 0 0 1⎥⎦
⎢⎣ 0 2 −1
⎡1 0
0 2 1 1 0 0 0⎤
⎢0 1
⎥
1 2 −1 1 −1 0 0 ⎥
⎢
→⎢
0 0
0 −6 0 −3 1 1 0 ⎥
⎢
⎥
0 −5 2 −3 2 0 1⎥⎦
⎢⎣ 0 0
⎡1
⎢
⎢0
→⎢
⎢0
⎢
⎢0
⎢⎣
0 0
⎡1
⎢
⎢0
→⎢
⎢0
⎢
⎢0
⎢⎣
0 0 0
1
0
1 0 −1
0
0
1
0
0 0
2
1 0 0
0
1 0
0 0
1
1
2
1
−2
1
4
1
−4
1
2
− 14
⎡ 1
⎡ w⎤
⎢ 4
⎢x⎥
⎢− 1
⎢ ⎥ = A −1B = ⎢ 4
⎢ y⎥
⎢ 1
⎢ ⎥
⎢ 2
⎢⎣ z ⎥⎦
⎢− 1
⎢⎣ 4
1 1 0 0 0⎤
1 −1 1 0 0 ⎥⎥
−1 −2 0 1 0 ⎥
⎥
0 −1 0 0 1⎥⎦
2 1 1
0
0 0⎤
2 −1 1 −1
0 0 ⎥⎥
1 0 12 − 16 − 16 0 ⎥
⎥
−5 2 −3
2
0 1⎥⎦
1
3
− 23
− 16
7
6
1
3
1
3
1
−6
− 56
⎡1 0
0⎤
⎥
⎢
⎢0 1
0⎥
⎥ →⎢
⎢0 0
0⎥
⎥
⎢
⎢0 0
1⎥⎥
⎣⎢
⎦
− 14
3
4
1
− 12
− 16
5
− 12
− 12 ⎤
⎥
1⎥
2⎥
0⎥
⎥
1⎥
2 ⎥⎦
1
− 12
− 16
7
12
− 14
1
− 12
− 16
7
12
3
4
1
− 12
− 16
5
− 12
0
1
0
0 −1
0
1
0
0
1
− 12 ⎤
⎥ ⎡ 4 ⎤ ⎡ 1⎤
1⎥⎢ ⎥ ⎢ ⎥
2 ⎥ ⎢12 ⎥ ⎢ 3⎥
=
0 ⎥ ⎢12 ⎥ ⎢ −2 ⎥
⎥⎢ ⎥ ⎢ ⎥
1 ⎥ ⎢⎣12 ⎥⎦ ⎢⎣ 7 ⎥⎦
2 ⎥⎦
Thus, w = 1, x = 3, y = –2, z = 7.
218
1
2
1
−4
1
3
− 23
− 16
7
12
1
3
1
3
1
−6
5
− 12
0⎤
⎥
0⎥
⎥
0⎥
⎥
1⎥
2 ⎦⎥
ISM: Introductory Mathematical Analysis
⎡ 1 1 −1
⎢ 0 1 1
34. ⎢
⎢ −1 1 1
⎢
⎢⎣ 1 −1 −1
⎡1
⎢0
⎢
→⎢
0
⎢
⎢0
⎢⎣
⎡1
⎢0
⎢
→⎢
0
⎢
⎢0
⎢⎣
⎡1
⎢
⎢0
→⎢
⎢0
⎢
⎢0
⎢⎣
1 0 0 0⎤
⎡ 1 1 −1
⎥
⎢0
1 0 1 0 0⎥
1 1
→⎢
⎢0 2 0
0 0 0 1 0⎥
⎥
⎢
2 0 0 0 1⎥⎦
⎢⎣ 0 −2 0
0
1 −1
1 1
1
0
1
0
1 0
0 −1 −1
1
2
1
−2
0
0
1
2
−1
1 0
1
1 1
0
1
0
1 1 − 12
1
0
0
1
1 0 0
1 0 0
1 0
0 0
1
0 0
1
2
1
2
1
−2
0
1
2
1 0
1 −1 0
0
Section 6.6
1
0
1
0 0
0⎤ ⎡ 1
0 ⎥⎥ ⎢⎢ 0
→
0⎥ ⎢0
⎥ ⎢
1⎥
⎢0
⎢⎣
2 ⎥⎦
0⎤ ⎡ 1
⎢
0 0 ⎥⎥ ⎢ 0
→⎢
− 12 0 ⎥ ⎢ 0
⎥
⎢
1 1⎥
⎢⎣ 0
2 2 ⎥⎦
⎡1
−1 − 12 ⎤
⎢
⎥
1
⎢0
⎥
0
2
⎥→⎢
⎢0
−1 − 12 ⎥
⎢
⎥
1
1⎥
⎢0
⎢⎣
2
2 ⎥⎦
0
⎡ 1 1 −1
1 0 0 0⎤
⎢0 1 1
⎥
1 0 1 0 0⎥
⎢
→⎢
0 1 0
0 1 0 1 0⎥
⎢
⎥
⎢0 −1 0
2 −1 0 0 1⎥⎦
⎣⎢
0
1 −1 0
1
0
0
1
1
0
0
0
1
0
0
0⎤
1 1
0 1
0 0 ⎥⎥
1 1 − 12 1 − 12 0 ⎥
⎥
1 2 − 12 1
0 12 ⎥⎥⎦
−1 0
1 0
0
0⎤
⎥
1
1
1 0
0 1 − 2 − 2⎥
⎥
1 0 − 12 1 −1 − 12 ⎥
⎥
1
1
0 1
0 0
2
2 ⎥⎦
0 0
0 1 − 32 − 12 ⎤
⎥
1 0
1
⎥
0 0
0
2
2
⎥
1 0 − 12 1 −1 − 12 ⎥
⎥
1
1⎥
0 1
0 0
2
2 ⎥⎦
1 0
0
⎡ 0 1 − 3 − 1⎤
2
2 ⎥ ⎡1 ⎤ ⎡ −2 ⎤
⎡ w⎤
⎢
1
⎢x⎥
⎢ 1 0
0 ⎥ ⎢ 0 ⎥ ⎢ 1⎥
2
⎢ ⎥ = A −1B = ⎢ 2
⎥⎢ ⎥ = ⎢ ⎥
⎢ y⎥
⎢ − 1 1 −1 − 1 ⎥ ⎢1 ⎥ ⎢ −2 ⎥
2⎥⎢ ⎥ ⎢ ⎥
⎢ ⎥
⎢ 2
⎢⎣ z ⎥⎦
1
1 ⎥ ⎢⎣1 ⎥⎦ ⎢⎣ 1⎥⎦
⎢ 0 0
⎢⎣
2
2 ⎥⎦
Thus, w = −2, x = 1, y = −2, z = 1.
⎡1 0 ⎤ ⎡5 −2 ⎤ ⎡ −4 2⎤
35. I − A = ⎢
⎥−⎢
⎥=⎢
⎥
⎣0 1 ⎦ ⎣ 1 2 ⎦ ⎣ −1 −1⎦
⎡ −4 2 1 0 ⎤
⎡ −1 −1 0 1⎤ ⎡ 1 1 0 −1⎤ ⎡ 1 1 0 −1⎤
⎢⎣ −1 −1 0 1⎥⎦ → ⎢⎣ −4 2 1 0 ⎥⎦ → ⎢⎣ −4 2 1 0 ⎥⎦ → ⎢⎣ 0 6 1 −4⎥⎦
⎡ 1 1 0 −1⎤ ⎡ 1 0 − 16 − 13 ⎤
⎥
→⎢
1
2⎥ → ⎢
1 −2
⎥
⎣ 0 1 6 − 3 ⎦ ⎢⎣ 0 1
6
3⎦
⎡− 1
Thus, (I − A)−1 = ⎢ 16
⎢⎣ 6
− 13 ⎤
⎥.
− 23 ⎥⎦
219
0
1 0
1
0
0
1
2
1
−2
1
0
1 0
0
1
2
0
0
0⎤
0 ⎥⎥
0⎥
⎥
1⎥
2 ⎦⎥
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡1 0 ⎤ ⎡ −3 2 ⎤ ⎡ 4 −2 ⎤
36. I − A = ⎢
⎥−⎢
⎥=⎢
⎥
⎣ 0 1 ⎦ ⎣ 4 3⎦ ⎣ −4 −2 ⎦
⎡ 4 −2 1 0 ⎤ ⎡ 1 − 12 14 0 ⎤
⎥
⎢ −4 −2 0 1⎥ → ⎢
⎣
⎦ ⎢⎣ −4 −2 0 1⎥⎦
⎡1 − 1
2
→⎢
⎢⎣ 0 −4
1
⎡1 − 1
0⎤
2
4
⎢
⎥→
1
⎢0
−
1
1 1⎥⎦
4
⎣
1
4
⎡ 1
8
Thus (I − A) −1 = ⎢
⎢− 1
⎣ 4
1
⎡1 0
0⎤
8
⎥ →⎢
1
1
⎢0 1 −
− 4 ⎦⎥
4
⎣
− 18 ⎤
⎥
− 14 ⎥⎦
− 18 ⎤
⎥
− 14 ⎥⎦
37. Let x = number of model A and y = number of model B.
a.
The system is
⎧⎪ x + y = 100 (painting)
⎨1
⎪⎩ 2 x + y = 80 (polishing)
⎡ 1 1⎤
Let A = ⎢ 1 ⎥ .
⎢⎣ 2 1⎥⎦
1 0⎤
⎡ 1 1 1 0⎤
⎡1 1
⎡ 1 1 1 0⎤
⎡ 1 0 2 −2 ⎤
→⎢
⎢1
⎥→⎢
⎥ →⎢
⎥
⎥
1
1
⎣ 0 1 −1 2 ⎦
⎣ 0 1 −1 2 ⎦
⎣⎢ 2 1 0 1 ⎦⎥
⎣⎢0 2 − 2 1⎥⎦
⎡ x⎤
−1 ⎡100 ⎤ ⎡ 2 −2 ⎤ ⎡100 ⎤ ⎡ 40 ⎤
⎢ y ⎥ = A ⎢ 80 ⎥ = ⎢ −1 2 ⎥ ⎢ 80 ⎥ = ⎢ 60 ⎥
⎣ ⎦
⎣ ⎦ ⎣
⎦⎣ ⎦ ⎣ ⎦
Thus 40 of model A and 60 of model B can be produced.
b. The system is
(widgets)
⎧10 x + 7 y = 800
⎨
x
y
(shims)
14
+
10
=
1130
⎩
⎡10 7 ⎤
Let A = ⎢
⎥.
⎣14 10 ⎦
7
⎡10 7 1 0 ⎤ ⎡ 1 10
⎢14 10 0 1⎥ → ⎢
⎣
⎦ ⎢⎣14 10
⎡1
→⎢
⎢0
⎣
7
10
1
5
1
10
− 75
0⎤
⎡1
⎥→⎢
⎢⎣ 0
1⎥
⎦
0⎤
⎥
0 1⎥⎦
1
10
⎡1 0
0⎤
5 − 72 ⎤
⎥ →⎢
⎥
⎢⎣ 0 1 −7
5⎥⎦
1 −7 5⎥⎦
7
10
1
10
⎡ 5 − 7 ⎤ ⎡ 800 ⎤ ⎡ 45⎤
⎡ x⎤
−1 ⎡ 800 ⎤
2⎥
A
=
=
⎢ ⎥
⎢
⎥ ⎢
⎢
⎥=⎢ ⎥
5⎦⎥ ⎣1130 ⎦ ⎣50 ⎦
⎣ y⎦
⎣1130 ⎦ ⎣⎢ −7
Thus 45 of model A and 50 of model B can be produced.
⎡1
⎢a
38. ⎢ 0
⎢
⎢0
⎣
0
1
b
0
0⎤ a 0 0
⎤ ⎡1 0 0 ⎤
⎥⎡
⎢
⎥
0 ⎢ 0 b 0 ⎥⎥ = ⎢⎢ 0 1 0 ⎥⎥ = I
⎥
1 ⎥ ⎢⎣ 0 0 c ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦
c⎦
220
ISM: Introductory Mathematical Analysis
39. a.
(B
−1 −1
A
) (AB) = B ( A A ) B = B
−1
−1
Section 6.6
−1
IB = B −1B = I
Since an invertible matrix has exactly one inverse, B −1A −1 is the inverse of AB.
b. From Part (a),
⎡1 1 ⎤ ⎡1 2 ⎤ ⎡ 4 6 ⎤
( AB)−1 = B −1A −1 = ⎢
⎥⎢
⎥=⎢
⎥
⎣1 2 ⎦ ⎣3 4 ⎦ ⎣ 7 10 ⎦
⎡1 − 1 ⎤
⎡1 1 ⎤
2⎥
T −1 ⎢
.
40. Left side: A T = ⎢
A
We
find
that
(
)
.
=
⎥
⎢0 1 ⎥
⎣0 2 ⎦
2 ⎦
⎣
1
⎡
⎤
1 −2
⎡ 1 0⎤
⎥.
Right side: A −1 = ⎢ 1 1 ⎥ , so ( A −1 )T = ⎢
1⎥
⎢0
⎣⎢ − 2 2 ⎦⎥
2⎦
⎣
Thus ( A T )−1 = ( A −1 )T .
⎡ 3
5
41. P T P = ⎢
⎢− 4
⎣ 5
42. a.
A
−1
4⎤⎡3
5⎥⎢5
3⎥ ⎢4
5⎦ ⎣5
− 54 ⎤ ⎡1 0 ⎤
⎥=
= I, so P T = P −1 . Yes, P is orthogonal.
3 ⎥ ⎢0 1 ⎥
⎣
⎦
5⎦
⎡ 14 −2 9 ⎤
= ⎢⎢ −6
1 −4 ⎥⎥
⎢⎣ 1 0
1⎥⎦
⎡ 14 −2 9 ⎤
R1A = [33 87 70] ⎢⎢ −6
1 −4 ⎥⎥ = [10 21 19]
⎢⎣ 1 0
1⎥⎦
⎡ 14 −2 9 ⎤
R 2 A −1 = [57 133 20] ⎢⎢ −6
1 −4 ⎥⎥ = [ 20 19 1]
⎢⎣ 1 0
1⎥⎦
⎡ 14 −2 9 ⎤
−1
R 3 A = [38 90 33] ⎢⎢ −6
1 −4 ⎥⎥ = [ 25 14 15]
⎢⎣ 1 0
1⎥⎦
−1
b. Just say no.
43. Let x be the number of shares of D, y be the number of shares of E, and z be the number of shares of F. We get the
following equations.
60x + 80y + 30z = 500,000
0.16(60x) + 0.12(80y) + 0.09(30z) = 0.1368(60x + 80y + 30z)
z = 4y
Simplify the first equation.
6x + 8y + 3z = 50,000
Simplify the second equation.
9.6x + 9.6y + 2.7z = 8.208x + 10.944y + 4.104z
1.392x – 1.344y – 1.404z = 0
1392x – 1344y – 1404z = 0
116x – 112y – 117z = 0
221
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
Simplify the third equation.
4y – z = 0
Thus we solve the following system of equations.
6x + 8y + 3z = 50,000
116x – 112y – 117z = 0
4y – z = 0
8
3⎤
⎡ 6
⎢
The coefficient matrix is A = ⎢116 −112 −117 ⎥⎥ .
⎢⎣ 0
4
−1⎥⎦
⎡ 6
8
3 1 0 0⎤
⎢
⎥
⎡⎣ A I ⎤⎦ = ⎢116 −112 −117 0 1 0 ⎥
⎢ 0
−1 0 0 1 ⎥⎦
4
⎣
4
1 1 0 0⎤
⎡ 1
3
2 6
⎢
⎥
> ⎢116 −112 −117 0 1 0 ⎥
⎢
⎥
−1 0 0 1⎥
4
⎢⎣ 0
⎦
1R
6 1
4
1
1 0 0⎤
⎡1
3
2
6
⎢
⎥
−116R1 + R 2 ⎢
800
58
> 0 − 3 −175 − 3 1 0 ⎥
⎢
⎥
⎢0
4
−1
0 0 1⎥
⎣
⎦
⎡1
3 R
⎢
− 800
2
> ⎢0
⎢
− 14 R3
⎢0
⎣
⎡1
⎢
R 2 + R3 ⎢
> 0
⎢
⎢0
⎢⎣
⎡1 4
3
32 R
⎢
29 3
⎢
> 0 1
⎢
⎢0 0
⎢⎣
0⎤
⎥
3
1
− 800
0⎥
⎥
−1
0
0 − 14 ⎥
⎦
4
1
1
⎤
0
0
3
2
6
⎥
29
3
21
⎥
1 32
−
0
400
800
⎥
29
29
3
1
0 32 400 − 800 − 4 ⎥⎥
⎦
1
1
0
0⎤
2
6
⎥
29
3
21
⎥
−
0
32 400
800
⎥
3
8 ⎥
2
− 725
− 29
1 25
⎦⎥
− 12 R 3 + R1
>
21 R + R
− 32
3
2
⎡1 4 0
3
⎢
⎢0 1 0
⎢
⎢0 0 1
⎢⎣
4
3
1
2
21
32
1
4
⎡1 0 0
− 43 R 2 + R1 ⎢
> ⎢0 1 0
⎢
⎢0 0 1
⎢⎣
1
6
29
400
19
150
1
50
2
25
1
10
1
50
2
25
0
3
1450
3
− 2900
3
− 725
1
290
3
− 2900
3
− 725
3 ⎤
1
⎡1
− 29
290
⎡ x ⎤ ⎢ 10
⎥ ⎡50, 000 ⎤ ⎡ 5000 ⎤
⎢ y⎥ = ⎢ 1 − 3
⎥ = ⎢1000 ⎥
21 ⎥ ⎢ 0
⎢ ⎥ ⎢ 50
⎥ ⎢
⎥
2900
116 ⎥ ⎢
⎢⎣ z ⎥⎦ ⎢ 2
⎢⎣ 0 ⎥⎦ ⎢⎣ 4000 ⎥⎦
3
8
⎥
− 725 − 29 ⎥
⎣⎢ 25
⎦
They should buy 5000 shares of company D, 1000
shares of company E, and 4000 shares of company F.
44. Let x be the number of shares of D, y be the number
of shares of E, and z be the number of shares of F.
We get the following conditions.
60x + 80y + 30z = 500,000
0.16(60x) + 0.12(80y) + 0.09(30z) = 0.1452(60x +
80y + 30z)
z = 2y
Simplify the first equation.
6x + 8x + 3z = 50,000
Simplify the second equation.
9.6x + 9.6y + 2.7z = 8.712x + 11.616y + 4.356z
0.888x – 2.016y – 1.656z = 0
888x – 2016y – 1656z = 0
111x – 252y – 207z = 0
Simplify the third equation.
2y – z = 0
Thus we solve the following system of equations.
6x + 8y + 3z = 50,000
111x – 252y – 207z = 0
2y – z = 0
8
3⎤
⎡ 6
⎢
The coefficient matrix is A = ⎢111 −252 −207 ⎥⎥ .
⎢⎣ 0
2
−1⎥⎦
8
3 1 0 0⎤
⎡ 6
⎡⎣ A I ⎤⎦ = ⎢⎢111 −252 −207 0 1 0 ⎥⎥
⎢⎣ 0
2
−1 0 0 1⎥⎦
4
1 1 0 0⎤
⎡ 1
3
2 6
⎢
⎥
> ⎢111 −252 −207 0 1 0 ⎥
⎢
⎥
2
−1 0 0 1⎥
⎢⎣ 0
⎦
1R
6 1
4 ⎤
29 ⎥
21 ⎥
116 ⎥
8 ⎥
− 29
⎦⎥
4
1
1 0 0⎤
⎡1
3
2
6
⎢
⎥
−111R1 + R 2 ⎢
525
37
> 0 −400 − 2 − 2 1 0 ⎥
⎢
⎥
⎢0
2
0 0 1⎥
−1
⎣
⎦
3 ⎤
− 29
⎥
21 ⎥
116 ⎥
8 ⎥
− 29
⎥⎦
⎡1 4
3
1 R
⎢
− 400
2
> ⎢0 1
⎢
− 12 R3
⎢0 −1
⎣
222
1
2
21
32
1
2
0⎤
⎥
1
0⎥
− 400
⎥
0
0 − 12 ⎥
⎦
1
6
37
800
0
ISM: Introductory Mathematical Analysis
Section 6.7
−1
1
1
⎡1 4
0
0⎤
3
2
6
⎢
⎥
R 2 + R3 ⎢
37
21
1
⎥
> 0 1 32
−
0
800
400
⎢
⎥
⎢ 0 0 37 37 − 1 − 1 ⎥
⎢⎣
32 800
400
2 ⎦⎥
1
1
⎡1 4
0
0⎤
3
2
6
32 R
⎢
⎥
37 3
37
21
1
⎥
> ⎢ 0 1 32
−
0
800
400
⎢
⎥
16 ⎥
1
2
⎢0 0
−
−
1
⎢⎣
25
925
37 ⎥⎦
8 ⎤
1
⎡ 1 4 0 11
3
75
925
37 ⎥
1
⎢
− 2 R 3 + R1
1
1
21 ⎥
> ⎢0 1 0 50
− 925
74 ⎥
21 R + R
⎢
− 32
3
2
⎢0 0 1 1 − 2 − 16 ⎥
25
925
37 ⎦⎥
⎣⎢
3
−4.7 ⎤ ⎡ 13 ⎤ ⎡ 4.78 ⎤
⎡ x ⎤ ⎡ 0.9
⎢
⎥
⎢
48. ⎢ y ⎥ = ⎢ 2 −0.4
2 ⎥⎥ ⎢⎢ 4.7 ⎥⎥ = ⎢⎢ −1.33 ⎥⎥
⎢⎣ z ⎥⎦ ⎢⎣ 1 −0.8 −0.5 ⎥⎦ ⎢⎣ 7.2 ⎥⎦ ⎢⎣ −2.70 ⎥⎦
x = 4.78, y = –1.33, z = –2.70
−1
1 − 3⎤
⎡2
14
4
2
7 ⎥ ⎡ 13 ⎤ ⎡ 14.44 ⎤
⎡ w⎤ ⎢ 5
⎢ ⎥
⎢ x ⎥ ⎢ 5 − 2 −4 −1⎥ ⎢ 7 ⎥ ⎢ 0.03⎥
9
3
⎥
⎥ ⎢8⎥=⎢
49. ⎢ ⎥ = ⎢
5 ⎥ ⎢ 9 ⎥ ⎢ −0.80 ⎥
⎢ y⎥ ⎢0
1 − 94
6⎥
⎢ ⎥ ⎢
⎢ 4 ⎥ ⎢⎢ 10.33⎥⎥
⎢⎣ z ⎥⎦ ⎢ 1
⎦
⎥
1
0
4 − 3 ⎥ ⎢⎣ 7 ⎥⎦ ⎣
⎣⎢ 2
⎦
w = 14.44, x = 0.03, y = –0.80, z = 10.33
Problems 6.7
7
6 ⎤
⎡1 0 0 3
− 37
25
2775
4
⎢
⎥
− 3 R 2 + R1
1
1
21 ⎥
> ⎢0 1 0 50
− 925
74 ⎥
⎢
⎢0 0 1 1 − 2 − 16 ⎥
⎢⎣
25
925
37 ⎦⎥
3
7
6
⎡
− 37 ⎤ 50, 000
⎡ x ⎤ ⎢ 25 2775
⎤ ⎡ 6000 ⎤
⎥⎡
⎢ y⎥ = ⎢ 1 − 1
⎥ = ⎢1000 ⎥
21 ⎥ ⎢ 0
⎢ ⎥ ⎢ 50
⎥ ⎢
⎥
925
74 ⎥ ⎢
⎢⎣ z ⎥⎦ ⎢ 1
⎢⎣ 0 ⎥⎦ ⎢⎣ 2000 ⎥⎦
16
2
⎥
− 925 − 37 ⎥
⎣⎢ 25
⎦
They should buy 6000 shares of company D, 1000
shares of company E, and 2000 shares of company
F.
45. a.
b.
46. a.
b.
⎡ 200
1200
1. A = ⎢
⎢ 400
⎣ 1200
⎡ 600 ⎤
D=⎢
⎥
⎣805 ⎦
⎡1290 ⎤
X = (I − A)−1 D = ⎢
⎥
⎣1425⎦
The total value of other production costs is
600
800
PA + PB =
(1290) +
(1425) = 1405
1200
1500
⎡ 2.05 1.28⎤
⎢ 0.73 1.71⎥
⎣
⎦
⎡ 84
⎢ 41
⎢ 30
⎣ 41
⎡ 40
200
2. A = ⎢
⎢ 120
⎣ 200
105 ⎤
82 ⎥
70 ⎥
41 ⎦
a.
18
323
11
646
23
323
120 ⎤
300 ⎥
90 ⎥
300 ⎦
⎡ 200 ⎤
D=⎢
⎥
⎣ 300 ⎦
⎡812.5⎤
X = (I − A)−1 D = ⎢
⎥
⎣ 1125 ⎦
⎡ −0.03 0.06 −0.12 ⎤
⎢ 0.13 0.02 0.05⎥
⎢
⎥
⎢⎣ −0.10 0.07
0.01⎥⎦
⎡ − 11
⎢ 323
⎢ 83
⎢ 646
⎢ − 32
⎢⎣ 323
500 ⎤
1500 ⎥
200 ⎥
1500 ⎦
b.
39 ⎤
− 323
⎥
15 ⎥
323 ⎥
4 ⎥
323 ⎥⎦
⎡ 64 ⎤
D=⎢ ⎥
⎣ 64 ⎦
⎡ 220 ⎤
X = (I − A) −1 D = ⎢
⎥
⎣ 280 ⎦
⎡ 2.75 −1.59 −1.11 ⎤
47. ⎢ −0.48 1.43 0.00 ⎥
⎢
⎥
⎣ −1.22 0.32 2.22 ⎦
223
Chapter 6: Matrix Algebra
⎡ 15
⎢ 100
25
3. A = ⎢ 100
⎢
⎢ 50
⎣⎢ 100
a.
ISM: Introductory Mathematical Analysis
45 ⎤
180 ⎥
60 ⎥
180 ⎥
60 ⎥
180 ⎦⎥
30
120
30
120
40
120
⎡ 400
⎢ 1000
200
6. A = ⎢ 1000
⎢
⎢ 200
⎣⎢ 1000
⎡1073⎤
X = (I − A)−1 D = ⎢⎢1016 ⎥⎥
⎢⎣ 952 ⎥⎦
⎡ 400 200
⎢ 1000 1000
200
400
7. A = ⎢ 1000
1000
⎢
⎢ 200 100
⎢⎣ 1000 1000
⎡ 300 ⎤
D = ⎢⎢ 400 ⎥⎥
⎢⎣ 500 ⎥⎦
⎡10 ⎤
D = ⎢⎢10 ⎥⎥
⎢⎣10 ⎥⎦
⎡ 68.59 ⎤
X = (I − A) D = ⎢⎢ 84.50 ⎥⎥
⎢⎣108.69 ⎥⎦
−1
⎡ 100
⎢ 1000
100
4. A = ⎢ 1000
⎢
⎢ 300
⎣⎢ 1000
400
800
80
800
160
800
−1
3⎤
⎡1
4⎥
8. A = ⎢ 31
⎢⎣ 4 0 ⎥⎦
⎡300 ⎤
D=⎢
⎣500 ⎥⎦
(I − A)X = D
⎡ 2 − 3 300 ⎤
4
⎥ with a calculator
Reducing ⎢ 13
1 500 ⎥⎦
⎢⎣ − 4
⎡ 1 0 1408.70 ⎤
.
results in ⎢
⎣ 0 1 852.17 ⎥⎦
Thus 1408.70 units of agriculture and 852.17
units of milling need to be produced.
⎡1559.81 ⎤
X = (I − A) D = ⎢⎢1112.44 ⎥⎥
⎢⎣1738.04 ⎥⎦
−1
200
1000
400
1000
100
1000
200 ⎤
1000 ⎥
100 ⎥
1000 ⎥
300 ⎥
1000 ⎥⎦
⎡1382 ⎤
X = (I − A) D = ⎢⎢1344 ⎥⎥
⎢⎣1301⎥⎦
240 ⎤
1200 ⎥
480 ⎥
1200 ⎥
240 ⎥
1200 ⎦⎥
⎡500 ⎤
D = ⎢⎢150 ⎥⎥
⎢⎣ 700 ⎥⎦
⎡ 400
⎢ 1000
200
5. A = ⎢ 1000
⎢
⎢ 200
⎣⎢ 1000
200 ⎤
1000 ⎥
100 ⎥
1000 ⎥
300 ⎥
1000 ⎦⎥
⎡ 250 ⎤
D = ⎢⎢ 300 ⎥⎥
⎢⎣ 350 ⎥⎦
⎡15 ⎤
D = ⎢⎢10 ⎥⎥
⎢⎣35⎥⎦
⎡134.29 ⎤
X = (I − A) −1 D = ⎢⎢162.25 ⎥⎥
⎢⎣ 234.35⎥⎦
b.
200
1000
400
1000
100
1000
200 ⎤
1000 ⎥
100 ⎥
1000 ⎥
300 ⎥
1000 ⎦⎥
⎡ 300 ⎤
D = ⎢⎢ 350 ⎥⎥
⎢⎣ 450 ⎥⎦
1
1⎤
⎡1
⎢ 10 3 4 ⎥
1
1
1
9. A = ⎢ 10
⎥
10
3
⎢1 1 1⎥
⎢⎣ 10 10 10 ⎥⎦
⎡ 300 ⎤
D = ⎢ 200 ⎥
⎢
⎥
⎣ 500 ⎦
⎡1301⎤
X = (I − A)−1 D = ⎢⎢1215⎥⎥
⎢⎣1188⎥⎦
(I − A)X = D
224
ISM: Introductory Mathematical Analysis
Chapter 6 Review
⎡ 9
− 13 − 14 300 ⎤
⎡ 1 0 0 736.39 ⎤
⎢ 10
⎥
9
1
1
−
−
200
Reducing ⎢ 10
⎥ with a calculator results in ⎢ 0 1 0 563.29 ⎥ .
10
3
⎢
⎥
⎢ 1
⎥
9
1
⎣ 0 0 1 699.96 ⎦
⎢⎣ − 10 − 10 10 500 ⎥⎦
Thus 736.39 units of coal, 563.29 units of steel, and 699.96 units of railroad services need to be produced.
Chapter 6 Review Problems
8⎤
⎡ 3 4⎤
⎡ 1 0 ⎤ ⎡ 6 8⎤ ⎡ 3 0 ⎤ ⎡ 3
− 3⎢
=⎢
−⎢
=⎢
1. 2 ⎢
⎥
⎥
⎥
⎥
⎥
⎣ −5 1⎦
⎣ 2 4 ⎦ ⎣ −10 2 ⎦ ⎣ 6 12⎦ ⎣ −16 −10 ⎦
⎡1 2⎤
⎡1 0 ⎤ ⎡ 8 16 ⎤ ⎡ 2 0 ⎤ ⎡ 6 16⎤
−2⎢
2. 8 ⎢
⎥
⎥=⎢
⎥−⎢
⎥=⎢
⎥
⎣7 0 ⎦
⎣0 1 ⎦ ⎣56 0 ⎦ ⎣ 0 2 ⎦ ⎣56 −2⎦
5⎤
⎡ 1 7⎤
⎡ 1 + 0 0 + 42 −2 + 7 ⎤ ⎡ 1 42
⎡ 1 0 −2 ⎤ ⎢
⎥ = ⎢ 2 −18 −7 ⎥
=
2
+
0
0
−
18
−
4
−
3
3. ⎢⎢ 2 −3⎥⎥ ⎢
⎥ ⎢
⎥
0 6
1⎥⎦ ⎢
⎢⎣ 1 0 ⎥⎦ ⎣
⎢⎣ 1 + 0 0 + 0 −2 + 0 ⎥⎦ ⎢⎣ 1
0 −2 ⎥⎦
⎡ 2 3⎤
4. [2 3 7] ⎢⎢ 0 −1⎥⎥ = [2(2) + 3(0) + 7(5)
⎢⎣ 5 2 ⎥⎦
2(3) + 3(−1) + 7(2)] = [39 17]
⎡ 2 3⎤ ⎛ ⎡ 2 3⎤ ⎡ 1 8⎤ ⎞ ⎡ 2 3⎤ ⎡ 1 −5⎤ ⎡11 −4⎤
5. ⎢
−
=
⎜
⎟=
⎣ −1 3⎥⎦ ⎝ ⎢⎣ 7 6 ⎥⎦ ⎢⎣ 4 4 ⎥⎦ ⎠ ⎢⎣ −1 3⎥⎦ ⎢⎣3 2 ⎥⎦ ⎢⎣ 8 11⎥⎦
⎡ 0 −5⎤ ⎪⎫
⎪⎧ ⎡ 2 0 ⎤
⎪⎧ ⎡ 2 0 ⎤ ⎡ 0 −10 ⎤ ⎪⎫
6. − ⎨ ⎢
⎥ + 2 ⎢ 6 −4 ⎥ ⎬ = − ⎨ ⎢7 8 ⎥ + ⎢12 −8⎥ ⎬
7
8
⎦
⎣
⎦ ⎭⎪
⎦ ⎣
⎦ ⎭⎪
⎩⎪ ⎣
⎩⎪ ⎣
⎡ 2 −10 ⎤ ⎡ −2 10 ⎤
= −⎢
=
0 ⎥⎦ ⎢⎣ −19 0 ⎥⎦
⎣19
2
⎡ 1 −2 ⎤
⎡ −5 −4 ⎤ ⎡ 1⎤
⎡ 3⎤ ⎡ 6 ⎤
T
1 −2] = 2 ⎢
7. 2 ⎢
= 2⎢ ⎥ = ⎢ ⎥
[
⎥
⎥
⎢
⎥
1⎦
⎣3
⎣ 6 −5⎦ ⎣ −2 ⎦
⎣16 ⎦ ⎣32 ⎦
2
T
2
⎡1 0 ⎤ ⎡1 1⎤
⎡1 0 ⎤ ⎡1 4 ⎤ ⎡1 4 ⎤
1 ⎡3 0 ⎤ ⎧⎪ ⎡1 0 ⎤ ⎫⎪
=
8.
⎨
⎬
⎢3 6 ⎥ ⎢1 3⎥
⎢1 2 ⎥ ⎢ 0 3⎥ = ⎢1 2 ⎥ ⎢ 0 9 ⎥ = ⎢1 22⎥
3⎣
⎦ ⎪⎩ ⎣
⎦ ⎪⎭
⎣
⎦⎣
⎦
⎣
⎦⎣
⎦ ⎣
⎦
⎡1 −1⎤ ⎡ 3 0 ⎤
9. (2 A)T − 3I 2 = 2A T − 3I = 2 ⎢
⎥−⎢
⎥
⎣1 2 ⎦ ⎣ 0 3⎦
⎡ 2 −2 ⎤ ⎡ 3 0 ⎤ ⎡ −1 −2 ⎤
=⎢
⎥−⎢
⎥=⎢
1⎥⎦
⎣ 2 4⎦ ⎣0 3⎦ ⎣ 2
⎡ 2 2⎤
10. A(2I ) − AOT = 2( AI ) − AO = 2 A − O = 2 A = ⎢
⎥
⎣ −2 4 ⎦
225
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
3
5
⎡1 0 ⎤ ⎡1 0 ⎤
⎡1 0 ⎤ ⎡1 0⎤ ⎡ 2 0 ⎤
11. B3 + I 5 = ⎢
⎥ + ⎢0 1 ⎥ = ⎢0 8 ⎥ + ⎢0 1⎥ = ⎢ 0 9⎥
0
2
⎣
⎦ ⎣
⎦
⎣
⎦ ⎣
⎦ ⎣
⎦
⎡0 0⎤
12. ( ABA)T − A T BT A T = A T B T A T − A T B T A T = O = ⎢
⎥
⎣0 0⎦
⎡ 5 x ⎤ ⎡15⎤
13. ⎢ ⎥ = ⎢ ⎥
⎣7 x ⎦ ⎣ y ⎦
5x = 15, or x = 3
7x = y, 7 · 3 = y, or y = 21
⎡ 2 + x 2 1 + 3 x ⎤ ⎡3 4 ⎤
14. ⎢
⎥=⎢
⎥
⎣⎢ 4 + xy 2 + 3 y ⎦⎥ ⎣3 y ⎦
2 + 3y = y, 2y = –2, or y = –1
1 + 3x = 4, 3x = 3, or x = 1
For these values of x and y, 2 + x 2 = 3 is true, and 4 + xy = 3 is true. Thus x = 1, y = –1.
4 ⎤ ⎡1 4⎤
⎡1 4 ⎤
⎡1
⎡1 0 ⎤
→⎢
→⎢
→⎢
15. ⎢
⎥
⎥
⎥
⎥
⎣5 8 ⎦
⎣0 −12 ⎦ ⎣ 0 1 ⎦
⎣0 1 ⎦
⎡0 0 7 ⎤
⎡0 5 9 ⎤ ⎡0 1
→⎢
16. ⎢
⎥
⎥→⎢
⎣0 5 9⎦
⎣0 0 7 ⎦ ⎢⎣ 0 0
⎡2 4 7⎤
⎡1 2
17. ⎢ 1 2 4 ⎥ → ⎢ 2 4
⎢
⎥
⎢
⎣ 5 8 2⎦
⎣5 8
⎡ 1 2 4⎤
⎡1
→ ⎢0 1 9 ⎥ → ⎢0
⎢
⎥
⎢
⎣ 0 0 −1⎦
⎣0
9⎤
5⎥
⎡0 1 0⎤
→⎢
⎥
1⎥⎦ ⎣ 0 0 1⎦
4⎤
4⎤
⎡1 2
⎡1 2
7 ⎥ → ⎢0 0 −1⎥ → ⎢0 −2
⎥
⎢
⎥
⎢
2⎦
⎣0 −2 −18⎦
⎣0 0
0 −14 ⎤ ⎡ 1 0 −14 ⎤ ⎡ 1
1
9⎥ → ⎢0 1
9⎥ → ⎢0
⎥ ⎢
⎥ ⎢
−1⎦ ⎣ 0 0
0
1⎦ ⎣ 0
4⎤
−18⎥
⎥
−1⎦
0 0⎤
1 0⎥
⎥
0 1⎦
⎡0 0 0 1 ⎤
⎡1 0 0 0 ⎤
18. ⎢⎢ 0 0 0 0 ⎥⎥ → ⎢⎢ 0 0 0 1 ⎥⎥
⎢⎣1 0 0 0 ⎥⎦
⎢⎣ 0 0 0 0 ⎥⎦
⎡ 2 −5 0 ⎤
⎡ 2 −5 0 ⎤ ⎡ 1 − 52 0 ⎤ ⎡1 0 0 ⎤
→⎢
19. ⎢
⎥→⎢
⎥
⎥→⎢
⎥
1 0 ⎥⎦ ⎣ 0 1 0 ⎦
⎣ 4 3 0⎦
⎣ 0 13 0 ⎦ ⎢⎣ 0
Thus x = 0, y = 0.
3
2 3⎤ ⎡ 1 0
⎡ 1 −1
3⎤
⎡ 1 −1 2 3⎤ ⎡ 1 −1 2
4
⎢
20. ⎢
⎥ → ⎢ 0 4 −5 −4 ⎥ → ⎢ 0 1 − 5 −1⎥ → ⎢
5
3
1
1
5
⎢⎣
⎥⎦ ⎣ 0 1 − 4
⎣
⎦ ⎣
⎦
4
3
5
Thus x = − r + 2 , y = r − 1 , z = r.
4
4
226
2⎤
⎥
−1⎥⎦
ISM: Introductory Mathematical Analysis
Chapter 6 Review
1⎤
2
1⎤ ⎡ 1 1 2 1⎤
⎡1 1 2
⎡1 1
⎡ 1 0 0 −1⎤
21. ⎢⎢ 3 −2 −4 −7 ⎥⎥ → ⎢⎢0 −5 −10 −10 ⎥⎥ → ⎢⎢ 0
1 2 2 ⎥⎥ → ⎢⎢ 0 1 2 2 ⎥⎥
⎢⎣ 2 −1 −2 2 ⎥⎦
⎢⎣0 −3 −6
⎢⎣ 0 0 0 6 ⎥⎦
0 ⎥⎦ ⎢⎣ 0 −3 −6 0 ⎥⎦
Row three indicates that 0 = 6, which is never true, so there is no solution.
⎡1 0
⎡ 1 −1 −1 1⎤
⎡ 1 −1 −1 1⎤
⎡ 1 −1 −1 1⎤
⎢
⎢
⎥
22. ⎢ 1 1 2 −3⎥ → ⎢0 2 3 −4 ⎥ → 0 1 32 −2 → ⎢0 1
⎢
⎥
⎢
⎥
⎢
⎥
⎢0 0
⎢⎣0 2 4 −9 ⎥⎦
⎣ 2 0 2 −7 ⎦
⎣0 2 4 −9 ⎦
⎢⎣
−1⎤ ⎡ 1 0 0 32 ⎤
⎥ ⎢
⎥
−2 ⎥ → ⎢ 0 1 0 11
⎥
2
1 −5⎥⎥ ⎢⎢ 0 0 1 −5⎥⎥
⎦ ⎣
⎦
1
2
3
2
3
11
Thus x = , y = , z = −5.
2
2
⎡1 5
1 0⎤
⎡1 5 1 0 ⎤
⎡1 5
→⎢
→⎢
23. ⎢
⎥
⎥
⎢⎣ 0 1
⎣3 9 0 1 ⎦
⎣ 0 −6 −3 1⎦
5⎤
5⎤
⎡1 0 − 3
⎡− 3
2
6⎥
2
6⎥
→⎢
⇒ A −1 = ⎢
1 − 1⎥
⎢0 1
⎢ 1 − 1⎥
2
6⎦
6⎦
⎣
⎣ 2
1
1
2
0⎤
⎥
− 16 ⎥
⎦
⎡ 0 1 1 0 ⎤ ⎡1 0 0 1 ⎤
−1 ⎡ 0 1 ⎤
24. ⎢
⎥ → ⎢ 0 1 1 0 ⎥ ⇒ A = ⎢1 0 ⎥
1
0
0
1
⎣
⎦ ⎣
⎦
⎣
⎦
3 −2
1 0 0⎤
⎡ 1 3 −2 1 0 0 ⎤ ⎡ 1
⎢
⎥
⎢
25. ⎢ 4
1 0 0 1 0 ⎥ → ⎢ 0 −11 8 −4 1 0 ⎥⎥
⎢⎣ 3 −2 2 0 0 1⎥⎦ ⎢⎣ 0 −11 8 −3 0 1⎥⎦
0 0⎤
3 −2
1 0 0 ⎤ ⎡ 1 3 −2 1
⎡1
⎢ 0 −11 8 −4 1 0 ⎥ → ⎢ 0 1 − 8 4 − 1 0 ⎥
⎥
⎢
⎥ ⎢
11 11
11
⎥
⎢⎣ 0
0 0
1 −1 1⎥⎦ ⎢ 0 0
−
0
1
1
1
⎣
⎦
3
2
1
⎡1 0
− 11
0⎤
11
11
⎢
⎥
8
4 − 1 0 ⎥ ⇒ no inverse exists
→ ⎢ 0 1 − 11
11
11
⎢
⎥
⎢0 0
0
1 −1 1⎥
⎣
⎦
⎡ 1 0 0 1 0 0⎤
5
⎡ 5 0 0 1 0 0⎤
⎡5 0 0 1 0 0⎤ ⎢
⎥
⎢
⎥
⎢
⎥
1
1
⎢
26. ⎢ −5 2 1 0 1 0 ⎥ → ⎢0 2 1 1 1 0 ⎥ → 0 1 2 2 12 0 ⎥
⎢
⎥
⎢⎣ −5 1 3 0 0 1⎥⎦
⎢⎣0 1 3 1 0 1⎥⎦ ⎢ 0 1 3 1 0 1⎥
⎣
⎦
⎡1 0 0 1
⎤
⎡1
⎡1 0 0 1
⎤
0
0
0
0⎤
0
0
5
5
⎢
⎥
⎢5
⎥
⎢
⎥
3 − 1 ⎥ ⇒ A −1 = ⎢ 2
3 − 1 ⎥.
1 0 ⎥ → ⎢0 1 0 2
→ ⎢ 0 1 12 12
2
5
5
5⎥
5
5⎥
⎢
⎢5
⎢
⎥
2⎥
2⎥
⎢0 0 1 1 − 1
⎢1 − 1
⎢ 0 0 5 1 − 1 1⎥
2 2
2
5
5
5⎦
5
5⎦
⎣
⎦
⎣
⎣5
227
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
⎡ 3 1 4 1 0 0 ⎤ ⎡1 0
27. ⎢⎢1 0 1 0 1 0 ⎥⎥ → ⎢⎢ 3 1
⎢⎣ 0 2 1 0 0 1 ⎥⎦ ⎢⎣ 0 2
1 0⎤
⎡1 0 1 0
⎡1
⎢
⎥
→ ⎢ 0 1 1 1 −3 0 ⎥ → ⎢⎢ 0
⎢⎣ 0 2 1 0 0 1⎥⎦
⎢⎣ 0
1 0 1 0⎤
4 1 0 0 ⎥⎥
1 0 0 1 ⎥⎦
0 1 0
1 0⎤
1 1 1 −3 0 ⎥⎥
0 −1 −2 6 1⎥⎦
1 0⎤
⎡1 0 1 0
⎡ 1 0 0 −2 7 1⎤
⎢
⎥
→ ⎢ 0 1 1 1 −3 0 ⎥ → ⎢⎢0 1 0 −1 3 1⎥⎥
⎢⎣ 0 0 1 2 −6 −1⎥⎦
⎢⎣0 0 1 2 −6 −1⎥⎦
⎡ x⎤
⎡ −2 7 1⎤ ⎡ 1 ⎤ ⎡ 0 ⎤
⎢ y ⎥ = A −1B = ⎢ −1 3 1⎥ ⎢ 0 ⎥ = ⎢1 ⎥
⎢ ⎥
⎢
⎥⎢ ⎥ ⎢ ⎥
⎢⎣ z ⎥⎦
⎢⎣ 2 −6 −1⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 0 ⎥⎦
Thus x = 0, y = 1, z = 0.
28. We found A −1 in Exercise 26, so
⎡3⎤
⎡1
0
0⎤ 3
⎡ x⎤
⎢5
⎥⎡ ⎤ ⎢5⎥
⎢ y ⎥ = A −1B = ⎢ 2
3 − 1 ⎥ ⎢0⎥ = ⎢ 4 ⎥
⎢ ⎥
5
5⎥ ⎢ ⎥ ⎢5⎥
⎢5
⎢⎣ z ⎥⎦
2 ⎥ ⎢⎣ 2 ⎥⎦ ⎢ 7 ⎥
⎢1 − 1
⎢⎣ 5 ⎥⎦
5
5⎦
⎣5
⎡ 0 1 1⎤ ⎡ 0
29. A = AA = ⎢ 0 0 1⎥ ⎢ 0
⎢
⎥⎢
⎣0 0 0⎦ ⎣0
⎡0 0 1⎤ ⎡0
A3 = A 2 A = ⎢0 0 0 ⎥ ⎢0
⎢
⎥⎢
⎣0 0 0 ⎦ ⎣0
2
1 1⎤ ⎡ 0
0 1⎥ = ⎢ 0
⎥ ⎢
0 0⎦ ⎣0
1 1⎤ ⎡ 0
0 1⎥ = ⎢ 0
⎥ ⎢
0 0⎦ ⎣0
0 1⎤
0 0⎥
⎥
0 0⎦
0 0⎤
0 0⎥ = O
⎥
0 0⎦
Since A3 = O, every higher power of A is also O, so A1000 = O.
⎡0 1 1 1 0 0⎤
Looking at ⎢ 0 0 1 0 1 0 ⎥ , it is clear that there is no way of transforming the left side into I3 , since there
⎢
⎥
⎣ 0 0 0 0 0 1⎦
is no way to get a nonzero entry in the first column. Thus A does not have an inverse.
⎡2 0⎤
30. A T = ⎢
⎥
⎣0 4⎦
1
−1 ⎡ 2 0 ⎤
T
⎢
⎥
=
A
⎢0 1⎥
4⎦
⎣
⎡ 1 0⎤
⎥
A −1 = ⎢ 2
⎢0 1⎥
4⎦
⎣
1 0⎤
⎡
T
⎥
A −1 = ⎢ 2
⎢0 1⎥
4⎦
⎣
( )
( )
Thus ( A T )−1 = ( A −1 )T .
228
ISM: Introductory Mathematical Analysis
31. a.
Chapter 6 Review
Let x, y, and z represent the weekly doses of capsules of brand I, II, and III, respectively. Then
(vitamin A)
⎧ x + y + 4 z = 13
⎪
(vitamin B)
+
2
+
7
=
22
x
y
z
⎨
⎪ x + 3 y + 10 z = 31 (vitamin C)
⎩
⎡1 1 4 13⎤
⎢1 2 7 22 ⎥ − R1 + R 2>
⎢
⎥ −R + R
1
3
⎢⎣1 3 10 31⎥⎦
⎡ 1 0 1 4⎤
−R 2 + R1
> ⎢ 0 1 3 9 ⎥⎥
−2R 2 + R 3 ⎢
⎢⎣ 0 0 0 0 ⎥⎦
⎡ 1 1 4 13⎤
⎢0 1 3 9⎥
⎢
⎥
⎢⎣ 0 2 6 18⎥⎦
Thus x = 4 – r, y = 9 – 3r, and z = r, where r = 0, 1, 2, 3.
The four possible combinations are
Combination
x
y
z
1
4
9
0
2
3
6
1
3
2
3
2
4
1
0
3
b. Computing the cost of each combination, we find that they are 83, 77, 71, and 65 cents, respectively. Thus
combination 4, namely x = 1, y = 0, z = 3, minimizes weekly cost.
32. a.
(A )
−1
= A −1
( ) (A A) A = (A )
( A A ) A = A IA = A A = I
3
A 3 = A −1
−1
2
−1
−1
2
−1
2
( )
IA 2 = A −1
2
A2
−1
Thus A3 is invertible.
(
) (
)
b. AB = AC. Thus A −1 ( AB) = A −1 ( AC) , A −1A B = A −1A C , IB = IC, B = C.
c.
AA = A ⇒ A −1AA = A −1A , IA = I, A = I. Thus A = I n .
⎡ 215 87 ⎤
33. ⎢
⎥
⎣ 89 141⎦
−1
⎡ x ⎤ ⎡ 7.9 −4.3 2.7 ⎤ ⎡11.1⎤ ⎡ 1.57 ⎤
34. ⎢⎢ y ⎥⎥ = ⎢⎢ 3.4 5.8 −7.6 ⎥⎥ ⎢⎢10.8⎥⎥ = ⎢⎢ −0.30 ⎥⎥
⎢⎣ z ⎥⎦ ⎢⎣ 4.5 −6.2 −7.4 ⎥⎦ ⎢⎣15.9 ⎥⎦ ⎢⎣ −0.95⎥⎦
Thus x = 1.57, y = −0.30, z = −0.95.
⎡ 10
34
35. A = ⎢
⎢ 15
⎣ 34
20 ⎤
39 ⎥
;
14 ⎥
39 ⎦
⎡10 ⎤
⎡39.7 ⎤
D = ⎢ ⎥ ; X = (I − A) −1 D = ⎢
⎥
⎣5⎦
⎣ 35.1⎦
229
Chapter 6: Matrix Algebra
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 6
⎡ 20 40 30 10 ⎤
1. A = ⎢⎢ 30 0 10 10 ⎥⎥
⎢⎣ 10 0 30 50 ⎥⎦
⎡ 7⎤
⎢10 ⎥
T=⎢ ⎥
⎢ 7⎥
⎢ ⎥
⎢⎣ 5⎥⎦
⎡ 9⎤
C = ⎢⎢ 8⎥⎥
⎢⎣10 ⎥⎦
⎧
⎡ 7⎤ ⎫
⎡800 ⎤
⎪ ⎡ 20 40 30 10 ⎤ ⎢ ⎥ ⎪
10 ⎥ ⎪
⎥
⎥
T
T ⎪⎢
T⎢
⎢
C ( AT) = C ⎨ ⎢ 30 0 10 10 ⎥
⎬ = C ⎢330 ⎥
⎢
⎥
⎪ ⎢ 10 0 30 50 ⎥ ⎢ 7 ⎥ ⎪
⎢⎣530 ⎥⎦
⎦ 5 ⎪
⎪⎣
⎣⎢ ⎦⎥ ⎭
⎩
⎡800 ⎤
= [9 8 10] ⎢⎢330 ⎥⎥ = [15,140]
⎢⎣530 ⎥⎦
The cost is $151.40.
2. To the linear system, add x1 + x2 + x3 + x4 = 52.
⎡ 20 40 30 10 ⎤
⎢ 30 0 10 10 ⎥
⎥
A=⎢
⎢ 10 0 30 50 ⎥
⎢
⎥
⎢⎣ 1 1 1 1⎥⎦
⎡1180 ⎤
⎢ 580 ⎥
⎥
B=⎢
⎢1500 ⎥
⎢
⎥
⎣⎢ 52 ⎦⎥
⎡ 8⎤
⎢ 10 ⎥
T = A −1B = ⎢ ⎥
⎢ 14 ⎥
⎢ ⎥
⎣⎢ 20 ⎦⎥
Guest 1: 8 days; guest 2: 10 days;
guest 3: 14 days; guest 4: 20 days
3. It is not possible. Different combinations of lengths of stays can cost the same. For example, guest 1 staying for
20 days and guest 3 staying for 17 days costs the same as guest 1 staying for 15 days and guest 3 staying for 21
days (each costs $214.50).
230
Chapter 7
Principles in Practice 7.1
2.
1. Let x = the number of type A magnets and
y = the number of type B magnets.
The cost for producing x type A magnets and y
type B magnets is 50 + 0.90x + 0.70y. The
revenue for selling x type A magnets and y type
B magnets is 2.00x + 1.50y.
Revenue is greater than cost when
2x + 1.5y > 50 + 0.9x + 0.7y.
0.8y > –1.1x + 50
y > –1.375x + 62.5
Sketch the dashed line y = –1.375x + 62.5 and
shade the half plane above the line. In order to
make a profit, the number of magnets of types A
and B must correspond to an ordered pair in the
shaded region. Also, to take reality into account,
both x and y must be positive (negative numbers
of magnets are not feasible).
10
y
x
10
3.
10
y
x
10
4.
2. Since negative numbers of cameras cannot be
sold, x ≥ 0 and y ≥ 0. Selling at least 50
cameras per week corresponds to x + y ≥ 50.
Selling twice as many of type I as of type II
corresponds to
x ≥ 2y. The system of inequalities is
⎧ x + y ≥ 50,
⎪⎪
x ≥ 2 y,
⎨
x ≥ 0,
⎪
y ≥ 0.
⎪⎩
y
10
x
10
5.
5
y
The region consists of points on or above the
x-axis and on or to the right of the y-axis. In
addition, the points must be on or above the line
x + y = 50 and on or below the line x = 2y.
x
4
Problems 7.1
1.
5
y
6.
10
y
x
5
x
10
231
Chapter 7: Linear Programming
7.
5
ISM: Introductory Mathematical Analysis
12.
y
y
5
x
x
5
8.
5
5
y
y
13.
6
x
5
x
5
9.
5
y
14.
5
y
x
x
5
10.
10
5
y
15.
5
y
x
10
x
5
11.
5
y
16.
5
y
x
x
5
5
232
ISM: Introductory Mathematical Analysis
17.
5
Section 7.1
y
22.
10
y
x
x
5
10
y
5
18.
23.
10
y
x
x
10
5
19.
5
24.
y
20
y
x
20
x
5
20.
25. 6x + 4y ≤ 20
y
8
7
y
x
x
5
5
21.
10
y
26. 7x + 3y ≤ 25
y
x
10
2
233
x
10
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
27. Let x be the amount purchased from supplier A,
and y the amount purchased from B. The system
of inequalities is
x + y ≤ 100,
x ≥ 0,
y ≥ 0.
100
25
y
P = 112
1
2
⎛ 47 41 ⎞
⎜ , ⎟
⎝ 3 9 ⎠
x – 3y = 2
y
x
25
(2, 0)
x
2. The feasible region appears below. The corner
1⎞ ⎛ 1 ⎞
⎛
points are (0, 0), ⎜ 0, 83 ⎟ , ⎜ 62 , 0 ⎟ .
3⎠ ⎝ 2 ⎠
⎝
Evaluating P at each corner point, we find that P
2
has a maximum value of 416 when x = 0 and
3
1
y = 83 .
3
100
28. Since negative numbers of computers cannot be
produced, x ≥ 0 and y ≥ 0. Producing at most
650 computers per week corresponds to x + y ≤
650. The system of inequalities is
⎧ x + y ≤ 650,
⎪
x ≥ 0,
⎨
⎪⎩
y ≥ 0.
y
x + 2y = 225
100
x + y = 90
1⎞
⎛
⎜ 0, 83 ⎟
3⎠
⎝
29. Since negative numbers of chairs cannot be
produced, x ≥ 0 and y ≥ 0. The inequality for
assembly time is 3x + 2y ≤ 240. The inequality
1
for painting time is x + y ≤ 80 . The system of
2
inequalities is
⎧3x + 2 y ≤ 240,
⎪1
⎪ x + y ≤ 80,
⎨2
⎪
x ≥ 0,
⎪
y ≥ 0.
⎩
4x + 3y = 250
P = 416
2
3
x
⎛ 1 ⎞ 100
⎜ 62 , 0 ⎟
⎝ 2 ⎠
3. The feasible region appears below. The corner
⎛ 10 ⎞
points are (2, 3), (0, 5), (0, 7) and ⎜ , 7 ⎟ .
⎝ 3 ⎠
Evaluating Z at each point, we find that Z has a
maximum value of –10 when x = 2 and y = 3.
The region consists of points on or above the
x-axis and on or to the right of the y-axis. In
addition, the points must be on or below the line
3x + 2y = 240 and on or below the line
1
x + y = 80 (or, equivalently x + 2y = 160).
2
10
y
y=7
3x − y = 3
Z = −10
x+y=5
(2, 3)
Problems 7.2
x
10
1. The feasible region appears below. The corner
⎛ 47 41 ⎞ ⎛ 45 ⎞
points are (2, 0), ⎜ , ⎟ , ⎜ , 0 ⎟ .
⎝ 3 9 ⎠ ⎝ 2
⎠
Evaluating P at each corner point, we find that P
1
45
has a maximum value of 112 when x =
2
2
and y = 0.
4. The feasible region appears below. The corner
⎛ 12 12 ⎞ ⎛ 99 99 ⎞
points are (8, 0), (3, 0), ⎜ , ⎟ , ⎜ , ⎟
⎝ 7 7 ⎠ ⎝ 20 20 ⎠
⎛ 27 ⎞
and ⎜ 8, ⎟ . Evaluating Z at each point, we find
⎝ 11 ⎠
that Z has a minimum value of 3 when x = 3 and
y = 0.
234
ISM: Introductory Mathematical Analysis
10
Section 7.2
y
9x + 11y = 99
x−y=0
Z=3
Z = 0.8
x=8
4x + 3y = 12
x
x
(3, 0)
9. The feasible region is unbounded with 3 corner
points. The member (see dashed line) of the
family of lines C = 3x + 2y which gives a
minimum value of C, subject to the constraints,
intersects the feasible region at corner point
23
⎛7 1⎞
Thus C has a minimum
⎜ , ⎟ where C =
3
⎝ 3 3⎠
23
7
1
value of
when x = and y = . [Note:
3
3
3
Here we chose the member of the family
1
y = (−3x + C ) whose y-intercept was closest
2
to the origin and which had at least one point in
common with the feasible region.]
y
2x − y = 2
x − 4y = 4
x
10
6. The feasible region is empty, so there is no
optimum solution.
10
y
2x + y = 10
5
y
(0, 5)
2x + y = 5
3x + y = 4
23
C=
3
⎛7 1⎞
,
⎜
⎟
⎝ 3 3⎠
8x + 7y = 56
3x + 4y = 24
x
10
x + 2y = 3
7. The feasible region is a line segment. The corner
points are (0, 1) and (4, 5). Z has a minimum
value of 3 when x = 0 and y = 1.
5
x
(3, 0)
5
10. The feasible region is unbounded with 4 corner
points. The member (see dashed line) of the
C
which gives a
family of lines y = − x +
2
minimum value of C, subject to the constraints,
intersects the feasible region at corner point
(40, 20) where C = 120. Thus C has a minimum
value of 120 when x = 40 and y = 20.
y
(4, 5)
(0, 1)
Z=3
10
(2, 0)
10
5. The feasible region is empty, so there is no
optimum solution.
10
y
10
100
x
y
5x + 2y = 200
5
8. The feasible region is a line segment. The corner
⎛ 27 21 ⎞
points are (2, 0) and ⎜ ,
⎟.
⎝ 17 17 ⎠
Z has a maximum value of 0.8 for x = 2 and
y = 0.
3x + 2y = 160
x + 2y = 80
C = 120
x
100
235
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
11. The feasible region is unbounded with 2 corner
points. The family of lines given by Z = 10x + 2y
has members (see dashed lines for two sample
members) that have arbitrarily large values of Z
and that also intersect the feasible region. Thus
no optimum solution exists.
10
50
y
3x + y = 50
Maximum Profit Line
(10, 20)
y
5x + y = 70
Z = 70
x − 2y = 0
14. Let x and y be the numbers of Vista and Xtreme
models made each day. Then we are to
maximize
P = 50x + 80y, where
x≥0
⎧
y≥0
⎪⎪
⎨ x + 3 y ≤ 24 (for machine A)
⎪
⎪⎩2 x + 2 y ≤ 24 (for machine B)
The feasible region is bounded. The corner
points are (0, 0), (0, 8), (6, 6), and (12, 0).
Evaluating P at each corner point, we find that P
is maximized at corner point (6, 6) where its
value is 780. Thus 6 of each model should be
made each day in order to give a maximum
profit of $780.
x
10
x + 2y = 4
12. The feasible region is unbounded with 3 corner
points. The family of lines given by Z = y – x has
members (see dashed lines for sample members)
that have arbitrarily small values for Z and also
intersect the feasible region. Thus no optimum
solution exists.
10
y
Z = −1
x − 3y = −6
x=3
25
y
Z = −7
x
x + 3y = 6
x
50
Z = 50
2x – 2y = 24
10
(6, 6)
13. Let x and y be the number of trucks and spinning
tops made per week, respectively. Then we are
to maximize P = 7x + 2y where
x≥0
⎧
⎪
y≥0
⎪
⎨2 x + y ≤ 80 (for machine A)
⎪ 3 x + y ≤ 50 (for machine B)
⎪ 5 x + y ≤ 70 (for finishing)
⎩
The feasible region is bounded. The corner
points are (0, 50), (14, 0) and (10, 20).
Evaluating P at each corner point, we find that P
is maximized at corner point (10, 20), where its
value is 110. Thus10 trucks and 20 spinning tops
should be made each week to give a maximum
profit of $110.
x + 3y = 24
x
25
15. Let x and y be the numbers of units of Food A
and Food B, respectively, that are purchased.
Then we are to minimize C = 1.20x + 0.80y,
where
x ≥ 0,
⎧
⎪⎪
y ≥ 0,
⎨2 x + 2 y ≥ 16 (for carbohydrates),
⎪
⎪⎩ 4 x + y ≥ 20 (for protein).
The feasible region is unbounded. The corner
points are (8, 0), (4, 4) and (0, 20). C is
minimized at corner point (4, 4) where C = 8
(see the minimum cost line). Thus 4 units of
Food A and 4 units of Food B gives a minimum
cost of $8.
236
ISM: Introductory Mathematical Analysis
30
Section 7.2
y
50
y
200x + 50y = 2500
4x + y = 20
Minimum
Cost Line
2x + 2y = 16
100x + 200y = 3000
x
x
30
Minimum
Cost Line
16. Let x and y be the numbers of units of Blend I
and Blend II, respectively, that are bought each
week. Then we are to minimize C = 8x + 10y
where
x ≥ 0,
⎧
⎪
y ≥ 0,
⎪
⎨ 2 x + 2 y ≥ 80 (for Nutrient A),
⎪ 6 x + 2 y ≥ 120 (for Nutrient B),
⎪4 x + 12 y ≥ 240 (for Nutrient C).
⎩
The feasible region is unbounded with 4 corner
points. C is minimized at the corner point
(30, 10) where C = 340 (see the minimum cost
line). thus each week the grower should buy 30
bags of Blend I and 10 bags of Blend II.
100
50
18. Let x and y be the number of days Refinery I and
Refinery II are operated, respectively. Then we
are to minimize C = 25,000x + 20,000y where
x ≥ 0,
⎧
⎪
y ≥ 0,
⎪
⎨ 2000 x + 1000 y ≥ 8000 (for low grade),
⎪3000 x + 2000 y ≥ 14, 000 (for medium grade),
⎪ 1000 x + 1000 y ≥ 5000 (for high grade).
⎩
The feasible region is unbounded with 4 corner
points. Evaluating C at each corner point, we
find that C is minimized at corner point (4, 1)
where C = 120,000. Thus, operate Refinery I for
4 days and Refinery II for 1 day for a minimum
cost of $120,000.
y
10
y
6x + 2y = 120
2000x + 1000y = 8000
2x + 2y = 80
Minimum
Cost Line
3000x + 2000y = 14,000
1000x + 1000y = 5000
x
4x + 12y = 240
x
100
10
17. Let x and y be the numbers of tons of ores I and
II, respectively, that are processed. Then we are
to minimize C = 50x + 60y, where
x ≥ 0,
⎧
y ≥ 0,
⎪⎪
⎨100 x + 200 y ≥ 3000 (for mineral A),
⎪
⎪⎩ 200 x + 50 y ≥ 2500 (for mineral B).
The feasible region is unbounded with 3 corner
points. C is minimized at the corner point
(10, 10) where C = 1100 (see the minimum cost
line). Thus 10 tons of ore I and 10 tons of ore II
give a minimum cost of $1100.
19. Let x and y be the number of chambers of type A
and B, respectively. Then we are to minimize
C = 600,000x + 300,000y, where
x ≥ 4,
⎧
⎪⎪
y ≥ 4,
⎨ 10 x + 4 y ≥ 100 (for polymer P ),
1
⎪
⎪⎩20 x + 30 y ≥ 420 (for polymer P2 ).
The feasible region is unbounded with 3 corner
points. Evaluating C at each corner point, we
find C is minimized at corner point
(6, 10) where C = 6,600,000. Thus the solution
is 6 chambers of type A and 10 chambers of
type B.
237
Chapter 7: Linear Programming
20
ISM: Introductory Mathematical Analysis
2(300 – x) + 8(200 – y ) ≥ 300,
2200 − 2 x − 8 y ≥ 300,
−2 x − 8 y ≥ –1900,
2 x + 8 y ≤ 1900.
The fifth constraint reflects the fact that
company A will not build more than 300 km
of highway, since 300 km is the total being
built; the sixth constraint is the
corresponding constraint for the amount of
expressway.
y
x=4
10x + 4y = 100
20x + 30y = 420
y=4
x
20
20. Let x and y be the number of liters produced by
the old and new processes, respectively. We
want to maximize P = 0.4x + 0.15y, where
x≥0
⎧
⎪⎪
y≥0
⎨ 25 x + 15 y ≤ 12,525 (for carbon dioxide)
⎪
⎩⎪50 x + 40 y ≤ 20, 000 (for particulate matter)
c.
The feasible region is bounded with three corner
points. Evaluating P at each corner point, we
find that P is maximized at the corner point
(400, 0), where P = 160. Thus daily production
of 400 liters by only the old process maximizes
daily profit at $160.
1000
The feasible region (see below) is bounded.
The corner points are (0, 200), (150, 200),
⎛ 650 550 ⎞
,
⎜
⎟ , (300, 100), (300, 0), and
3 ⎠
⎝ 3
(200, 0).
Evaluating D at each corner point, we find
that D is maximized at point (0, 200), where
D = 2100. That is, D is maximized when
x = 0, y = 200.
500
y
y
y = 200
2x + 8y = 1900
x + y = 400
x = 300
25x + 15y = 12,525
x + y = 200
50x + 40y = 20,000
x
22. Z = 2.71 when x = 1.14, y = 1.43
1000
21. a.
x
500
23. Z = 15.54 when x = 2.56, y = 6.74
A builds x km of highway and y km of
expressway, so B builds (300 – x) km of
highway and (200 – y) km of expressway.
Thus
D = 2x + 6y + 3(300 – x) + 5(200 – y)
= 1900 – x + y.
24. The feasible region is empty, so there is no
optimum solution.
25. Z = –75.98 when x = 9.48, y = 16.67
Principles in Practice 7.3
b. The first constraint is company A’s
construction limit.
The second constraint is company B’s
construction limit, which arises as follows:
(300 – x) + (200 – y ) ≤ 300,
500 – x – y ≤ 300,
– x – y ≤ –200,
x + y ≥ 200.
The third constraint is the minimum contract
for A.
The fourth constraint is the minimum
contract for B, which arises as follows:
1. Using the hint, the cost of shipping the TV sets
is Z = 18x + 24(25 – x) + 9y + 15(30 – y)
= 1050 – 6x – 6y.
Since negative numbers of TV sets cannot be
shipped, x ≥ 0, y ≥ 0, 25 – x ≥ 0, and
30 – y ≥ 0. Since warehouse C has only 45 TV
sets,
x + y ≤ 45. Similarly, since warehouse D has
only 40 TV sets, 25 – x + 30 – y ≤ 45 or
x + y ≥ 10.
We need to minimize Z = 1050 – 6x – 6y subject
238
ISM: Introductory Mathematical Analysis
Section 7.3
to the constraints
x + y ≤ 45,
x + y ≥ 10,
x ≤ 25,
y ≤ 30,
x ≥ 0, y ≥ 0.
2. The feasible region is a line segment. The corner
⎛ 2 16 ⎞
⎛ 16 2 ⎞
points are ⎜ , ⎟ and ⎜ , ⎟ . At each of
3
3
⎝
⎠
⎝ 3 3⎠
these points Z = 12. Thus Z is maximized at both
corner points, as well as at all points on the line
segment. Thus the solution is Z = 12 when
2 14
⎛ 2 ⎞ 16
x = (1 − t ) ⎜ ⎟ + t = + t ,
3
3
3 3
⎝ ⎠
y
50
16 14
⎛ 16 ⎞ 2
− t , and 0 ≤ t ≤ 1.
x = (1 − t ) ⎜ ⎟ + t =
3
3
3 3
⎝ ⎠
y = 30 C
B
x + y = 45
D
x = 25
A
x
x + y = 10
F
E
50
The feasible region shown has corners
A = (0, 10), B = (0, 30), C = (15, 30),
D = (25, 20), E = (25, 0), and F = (10, 0).
Evaluating the cost function at the corners gives
Z(A) = 1050 – 6(0) – 6(10) = 990
Z(B) = 1050 – 6(0) – 6(30) = 870
Z(C) = 1050 – 6(15) – 6(30) = 780
Z(D) = 1050 – 6(25) – 6(20) = 780
Z(E) = 1050 – 6(25) – 6(0) = 900
Z(F) = 1050 – 6(10) – 6(0) = 990
The minimum value of Z is 780 which occurs at
all points on the line segment joining C and D.
This is x = (1 – t)(15) + t(25) = 15 + 10t and
y = (1 – t)(30) + t(20) = 30 – 10t for 0 ≤ t ≤ 1.
Thus, ship 10t + 15 TV sets from C to A,
– 10t + 30 TV sets from C to B,
25 – (10t + 15) = –10t + 10 TV sets from D to A,
and 30 – (–10t + 30) = 10t TV sets from D to B,
for 0 ≤ t ≤ 1. The minimum cost is $780.
( 163 , 23 )
5 x
3. The feasible region appears below. The corner
⎛ 8 ⎞ ⎛ 36 4 ⎞
points are (0, 0), ⎜ 0, ⎟ , ⎜ , ⎟ and (6, 0). Z
⎝ 5⎠ ⎝ 7 7⎠
⎛ 36 4 ⎞
is maximized at ⎜ , ⎟ and (6, 0), where its
⎝ 7 7⎠
value is 84. Thus Z is also maximized at all
⎛ 36 4 ⎞
points on the line segment joining ⎜ , ⎟ and
⎝ 7 7⎠
(6, 0). The solution is Z = 84 when
6 36
⎛ 36 ⎞
x = (1 − t ) ⎜ ⎟ + 6t = t + ,
7
7
⎝ 7 ⎠
4 4
⎛4⎞
y = (1 − t ) ⎜ ⎟ + 0t = − t and 0 ≤ t ≤ 1.
7
7
7
⎝ ⎠
Problems 7.3
1. The feasible region is unbounded. Z is
minimized at corner points (2, 3) and (5, 2),
where its value is 33. Z is also minimized at all
points on the line segment joining (2, 3) and
(5, 2), so the solution is Z = 33 when
x = (1 – t)(2) + 5t = 2 + 3t
y = (1 – t)(3) + 2t = 3 – t and 0 ≤ t ≤ 1.
10
( 23 , 163 )
y
5
10
y
2x + 3y = 12
⎛ 36 4 ⎞
⎜ , ⎟
⎝ 7 7⎠
y
y = – 3x + 6
x + 5y = 8
2
y=x–3
(2, 3)
4. Using the hint, the cost of delivering the cars is
Z = 60x + 45y + 50(7 – x) + 35(4 – y)
= 490 + 10x + 10y.
Since negative numbers of cars is not possible,
x ≥ 0, y ≥ 0, 7 – x ≥ 0, and 4 – y ≥ 0. Since
(5, 2)
x
y = – 1 x + 11
3
3
x
10
10
239
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
the warehouse in Concord has only 6 cars,
x + y ≤ 6.
Similarly, since the Dublin warehouse has only 8
cars, 7 – x + 4 – y ≤ 8 or 3 ≤ x + y.
We need to minimize Z = 490 + 10x + 10y
subject to the constraints
x + y ≤ 6,
x + y ≥ 3,
x ≤ 7,
y ≤ 4,
x ≥ 0, y ≥ 0.
10
dividing by the greatest common factor of the
numbers involved. Thus, we will use
3x1 + 3 x2 + 4 x3 ≤ 300
3x1 + 3 x2 + 2 x3 ≤ 240
2 x1 + 2 x2 + 3x3 ≤ 180
x1 , x2 , x3 ≥ 0
x1
x2
x3 s1 s2 s3 P b
3
3
4 1 0 0 0 300 ⎤
s1 ⎡
⎢
3
3
2 0 1 0 0 240 ⎥⎥
s2 ⎢
2
2
3 0 0 1 0 180 ⎥
s3 ⎢
⎢
⎥
0⎥
P ⎣⎢ –150 –250 –200 0 0 0 1
⎦
x1 x2
x3 s1 s2 s3 P
b
s1 ⎡ 0 0
2 1 –1 0 0
60 ⎤
⎢
⎥
2
1
x2 ⎢ 1 1
0
0 0
80 ⎥
3
3
⎢
⎥
5
s3 ⎢ 0 0
0 – 32 1 0
20 ⎥
3
⎢
⎥
P ⎢100 0 – 100 0 250 0 1 20, 000 ⎥
3
3
⎣⎢
⎦⎥
x1 x2 x3 s1 s2 s3 P
b
y
x=7
B
A
y=4
C
x+y=6
x+y=3
x
E
D
10
The feasible region shown has corners A = (0, 3),
B = (0, 4), C = (2, 4), D = (6, 0), and E = (3, 0).
Evaluating the cost function at the corners gives
Z(A) = 490 + 10(0) + 10(3) = 520
Z(B) = 490 + 10(0) + 10(4) = 530
Z(C) = 490 + 10(2) + 10(4) = 550
Z(D) = 490 + 10(6) + 10(0) = 550
Z(E) = 490 + 10(3) + 10(0) = 520
The minimum value of Z is 520 which occurs at
all points on the line segment joining A and E.
This is x = (1 – t)(0) + t(3) = 3t and
y = (1 – t)(3) + t(0) = –3t + 3 for 0 ≤ t ≤ 1.
Thus have 3t cars delivered from Concord to
Atherton, –3t + 3 delivered from Concord to
Berkeley, 7 – 3t delivered from Dublin to
Atherton, and 4 – (–3t + 3) = 3t + 1 delivered
from Dublin to Berkeley, for 0 ≤ t ≤ 1. The
minimum cost is $520.
36 ⎤
s1 ⎡ 0 0 0 1 – 15 – 65 0
⎢
⎥
3 –2 0
x2 ⎢ 1 1 0 0
⎥
72
5
5
⎢
⎥
3 0
2
⎥
12
x3 ⎢⎢ 0 0 1 0 – 5
5
⎥
P ⎢100 0 0 0 70 20 1 20, 400 ⎥
⎣
⎦
The maximum value of P is 20,400 when
x1 = 0, x2 = 72, and x3 = 12 . The maximum
profit is $20,400 when 72 Type 2 players and
12 Type 3 players are produced and sold.
Problems 7.4
In these problems, the pivot entry is underlined.
Principles in Practice 7.4
1.
In these problems, the pivot entry is underlined.
1. Let x1 , x2 , and x3 be the numbers of Type 1,
Type 2, and Type 3 players, respectively, that
the company produces. The situation is to
maximize the profit
P = 150 x1 + 250 x2 + 200 x3 , subject to the
constraints
300 x1 + 300 x2 + 400 x3 ≤ 30, 000
15 x1 + 15 x2 + 10 x3 ≤ 1200
2 x1 + 2 x2 + 3x3 ≤ 180
x1 , x2 , x3 ≥ 0
The constraint inequalities can be simplified by
x1
⎡
s1 2
⎢
s2 ⎢ 2
Z ⎢ −1
⎣
x1
⎡4
s1 ⎢ 3
x2 ⎢⎢ 23
⎢1
Z ⎢⎣ 3
x2 s1 s2 Z
8 ⎤8
⎥
3 0 1 0 12 ⎥ 4
−2 0 0 1 0 ⎥
⎦
x2 s1 s2 Z
0 1 – 13 0 4 ⎤
⎥
1 0 4⎥
1 0
⎥
3
2 1 8⎥
0 0
⎥⎦
3
1
1 0 0
The solution is Z = 8 when x1 = 0, x2 = 4 .
240
ISM: Introductory Mathematical Analysis
2.
3.
Section 7.4
x1
x1
x2 s1 s2 Z
⎡
s1 –1 1 1 0 0 4 ⎤
⎢
⎥
1 0 1 0 6⎥ 6
s2 ⎢ 1
Z ⎢ –2 –1 0 0 1 0 ⎥
⎣
⎦
x1 x2 s1 s2 Z
⎡
s1 0 2 1 1 0 10 ⎤
⎢
⎥
x1 ⎢ 1 1 0 1 0 6 ⎥
Z ⎢ 0 1 0 2 1 12 ⎥
⎣
⎦
The solution is Z = 12 when x1 = 6, x2 = 0 .
x1
x2 s1 s2 Z
s1 ⎡ 3 2
⎢
s2 ⎢ –1 3
Z ⎢ 1 –2
⎣
x1 x2
5
1 0 0 5⎤ 2
⎥
0 1 0 3⎥ 1
0 0 1 0⎥
⎦
s1 s2 Z
s1 ⎡ 11 0 1 – 2 0 3⎤
3
⎢ 3
⎥
⎢
1
1
0 1⎥⎥
x2 ⎢ – 3 1 0
3
⎢
2 1 2⎥
Z ⎢ 13 0 0
⎥⎦
3
⎣
The solution is Z = 2 when x1 = 0, x2 = 1.
4.
x1 x2 s1
⎡
3 1
s1 2
⎢
s2 ⎢ 1 5 0
Z ⎢ −4 –7 0
⎣
x1 x2 s1
⎡ 7 0 1
s1 ⎢ 5
x2 ⎢⎢ 15 1 0
⎢
0 0
Z ⎢ – 13
⎣ 5
x1 x2
s1
⎡
5
x1 1 0
7
⎢
x2 ⎢⎢ 0 1 – 17
⎢
Z ⎢ 0 0 13
7
⎣
x2 s1 s2 s3 Z
s1 ⎡ 1 –1 1 0 0 0 1 ⎤ 1
⎢
⎥
5. s2 ⎢ 1 2 0 1 0 0 8 ⎥ 8
s3 ⎢ 1 1 0 0 1 0 5 ⎥ 5
⎢
⎥
Z ⎢⎣ –8 –2 0 0 0 1 0 ⎥⎦
x1 x2 s1 s2 s3 Z
x1 ⎡1 –1 1 0 0 0 1 ⎤
⎢
⎥
s2 ⎢ 0 3 –1 1 0 0 7 ⎥ 73
s3 ⎢ 0 2 –1 0 1 0 4 ⎥ 2
⎢
⎥
Z ⎢⎣ 0 –10 8 0 0 1 8 ⎦⎥
x1 x2 s1 s2 s3 Z
1 0
1 0
⎡
3⎤
x1 1 0
2
2
⎢
⎥
s2 ⎢ 0 0
1 1 –3 0
⎥
1
2
2
⎢
⎥
1 0
1 0
⎢
⎥
0
1
–
2
x2 ⎢
2
2
⎥
Z ⎢0 0
3 0
5 1 28⎥
⎣
⎦
The solution is Z = 28 when x1 = 3, x2 = 2 .
6.
s2 Z
9⎤ 3
⎥
1 0 10 ⎥ 2
0 1 0⎥
⎦
s2 Z
– 53 0 3⎤ 15
⎥ 7
1 0
2 ⎥⎥ 10
5
7 1 14 ⎥
⎥⎦
5
s2 Z
⎤
– 73 0 15
7⎥
2 0
11 ⎥
7
7⎥
2 1 137 ⎥
7
7 ⎥⎦
137
15
11
The solution is Z =
when x1 = , x2 = .
7
7
7
0 0
x1 x2
s1 ⎡ 1 –1
⎢
s2 ⎢ –1 1
s3 ⎢ 1 1
⎢
Z ⎣⎢ –2 6
s1
1
0
0
0
s2
0
1
0
0
s3
0
0
1
0
Z
0
0
0
1
x1 x2 s1 s2 s3 Z
⎡
x1 1 –1 1 0 0 0
⎢
s2 ⎢ 0 0 1 1 0 0
s3 ⎢ 0 2 –1 0 1 0
⎢
Z ⎢⎣ 0 4 2 0 0 1
The solution is Z = 8 when
241
4⎤ 4
4 ⎥⎥
6⎥ 6
⎥
0⎥
⎦
4⎤
8⎥⎥
2⎥
⎥
8⎥
⎦
x1 = 4, x2 = 0
Chapter 7: Linear Programming
x1
7.
x2
ISM: Introductory Mathematical Analysis
Thus the maximum value of Z is 4, when
x1 = 2, x2 = 0, x3 = 0 .
x3 s1 s2 Z
⎡
⎤
s1 ⎢ 1 2
0 1 0 0 10 ⎥ 5
s2 ⎢ 2 2
1 0 1 0 10 ⎥ 5
⎢
⎥
Z ⎢ –3 –4 – 3 0 0 1 0 ⎥
2
⎣
⎦
choosing s2 as departing variable
9. To obtain a standard linear programming
problem, we write the second constraint as
– x1 + 2 x2 + x3 ≤ 2 .
x1
s1 ⎡ 1 1 0 1 0 0 1 ⎤ 1
⎢
⎥
s2 ⎢ –1 2 1 0 1 0 2 ⎥
Z ⎢ –2 –1 1 0 0 1 0 ⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
x1 ⎡ 1 1 0 1 0 0 1⎤
⎢
⎥
s2 ⎢ 0 3 1 1 1 0 3⎥
Z ⎢0 1 1 2 0 1 2⎥
⎣
⎦
The solution is Z = 2 when x1 = 1, x2 = 0, x3 = 0 .
x1 x2 x3 s1 s2 Z
⎡
⎤
s1 ⎢ –1 0 –1 1 −1 0 0 ⎥
x2 ⎢⎢ 1 1 12 0 12 0 5⎥⎥
Z ⎢ 1 0 1 0 2 1 20 ⎥
⎢⎣
⎥⎦
2
The solution is Z = 20 when
x1 = 0, x2 = 5, x3 = 0
8. If s1 is the departing variable, then
x1 x2 x3 s1 s2 Z
10. To obtain a standard linear programming
problem, we write the third constraint as
− x1 + x2 ≤ 3.
s1 ⎡ 2 1 –1 1 0 0 4 ⎤ 2
⎢
⎥
s2 ⎢ 1 1 1 0 1 0 2 ⎥ 2
Z ⎢ –2 1 –1 0 0 1 0 ⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
1
1
1
⎡
0 0 2⎤
x1 ⎢1 2 – 2
2
⎥
3
1 1 0 0⎥ 0
s2 ⎢ 0 12
–
2
2
⎢
⎥
Z ⎢ 0 2 –2 1 0 1 4 ⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
⎡
1 1 0 2⎤
x1 1 2 0
3
3 3
⎢
⎥
x3 ⎢
1
1
2 0 0⎥
0
1
–
⎢
⎥
3
3 3
⎢
⎥
8
1
4
1 4⎥
Z ⎢0 3 0
3 3
⎣
⎦
The solution is Z = 4 when
x1 = 2, x2 = 0, x3 = 0 .
Choosing s2 as the departing variable
x1
⎡
s1 2
⎢
s2 ⎢ 1
Z ⎢ –2
⎣
x1
s1 ⎡ 0
⎢
x1 ⎢ 1
Z ⎢0
⎣
x1 x2 s1 s2 s3 s4
s1 ⎡ 1 1 1 0 0 0
s2 ⎢ 1 −1 0 1 0 0
⎢
s3 ⎢ −1
1 0 0 1 0
⎢
s4 ⎢ 1 0 0 0 0 1
Z ⎢⎢ 2 −3 0 0 0 0
⎣
x1 x2 s1 s2 s3 s4
x2 ⎡ 1 1 1 0 0 0
⎢
s2 ⎢ 2 0 1 1 0 0
s3 ⎢ −2 0 −1 0 1 0
⎢
s4 ⎢ 1 0 0 0 0 1
Z ⎢⎢⎣ 5 0 3 0 0 0
The solution is Z = 3 when
x2 x3 s1 s2 Z
1 –1
1 1
1 –1
x2 x3
–1 –3
1 1
3 1
x2 x3 s1 s2 Z
11.
1 0 0 4⎤ 2
⎥
0 1 0 2⎥ 2
0 0 1 0⎥
⎦
s1 s2 Z
1 –2 0 0 ⎤
⎥
0
1 0 2⎥
0 2 1 4⎥
⎦
242
Z
0
0
0
0
1
1⎤ 1
2 ⎥⎥
3⎥ 3
⎥
5⎥
0 ⎥⎥
⎦
Z
0
0
0
0
1
1⎤
3⎥⎥
2⎥
⎥
5⎥
3⎥⎥
⎦
x1 = 0, x2 = 1 .
x1
x2 s1 s2 s3 s4 Z
s1 ⎡ 2
⎢
s2 ⎢ –1
s3 ⎢ 5
⎢
s4 ⎢ 2
Z ⎢⎢⎣ –1
–1 1 0 0 0 0
2
0 1 0 0 0
3
1
0 0 1 0 0
0 0 0 1 0
–1 0 0 0 0 1
4 ⎤2
⎥
6⎥
20 ⎥ 4
⎥
10 ⎥ 5
0 ⎥⎥
⎦
ISM: Introductory Mathematical Analysis
Section 7.4
x1 x2 s1
choosing x1 as entering variable
s1 ⎡ 3 0
⎢ 2
x2 ⎢ 1
−
1
⎢ 2
s3 ⎢ 13 0
⎢ 2
⎢
s4 ⎢ 52 0
Z ⎢ −3 0
⎢⎣ 2
x1 x2 s1
s1 ⎡ 0 0 1
⎢
x2 ⎢ 0 1 0
⎢
x1 ⎢ 1 0 0
⎢
s4 ⎢ 0 0 0
⎢
Z ⎢0 0 0
⎢⎣
x1 x2 s1 s2 s3 s4 Z
1
1
x1 ⎡⎢1 – 2 2 0 0 0 0 2 ⎤⎥
16
1
s2 ⎢ 0 3
1 0 0 0 8⎥ 3
2
2
⎢
⎥
20
s3 ⎢ 0 11 − 5 0 1 0 0 10 ⎥ 11
2
2
⎢
⎥
s4 ⎢ 0 2
−1 0 0 1 0 6 ⎥ 3
⎢
⎥
Z ⎢0 – 3 1 0 0 0 1 2 ⎥
2
2
⎣⎢
⎦⎥
x1 x2 s1 s2 s3 s4 Z
3 0
1 0 0 32 ⎤ 32
x1 ⎡ 1 0
11
11
11 ⎥ 3
⎢
13 1 − 3 0 0 58 ⎥ 58
s2 ⎢ 0 0
11
11
11 ⎥ 13
⎢
x2 ⎢ 0 1 − 5 0
2 0 0 20 ⎥
⎢
11
11
11 ⎥
s4 ⎢ 0 0 − 1 0 − 4 1 0 26 ⎥
⎢
11
11
11 ⎥
3 0 1 52 ⎥
Z ⎢0 0 − 2 0
⎢⎣
11
11
11 ⎥⎦
x1 x2 s1 s2
s3 s4 Z
2
22 ⎤
x1 ⎡1 0 0 − 3
0 0 13
13
13
⎢
⎥
s1 ⎢ 0 0 1 11 − 3 0 0 58 ⎥
13
13
13 ⎥
⎢
5
50 ⎥
1
x2 ⎢ 0 1 0
0 0 13
13
13
⎢
⎥
5 1 0 36 ⎥
1
s4 ⎢ 0 0 0
−
13
13
13 ⎥
⎢
3
72 ⎥
2
Z ⎢0 0 0
0
1
13
13
13 ⎦⎥
⎣⎢
72
Thus the maximum value of Z is
, when
13
22
50
, x2 = . If we choose x2 as the
x1 =
13
13
entering variable, then we have:
x1
x2 s1 s2 s3 s4 Z
s1 ⎡ 2
⎢
s2 ⎢ –1
s3 ⎢ 5
⎢
s4 ⎢ 2
Z ⎢⎣⎢ –1
–1 1 0 0 0 0
2
0 1 0 0 0
3
1
0 0 1 0 0
0 0 0 1 0
–1 0 0 0 0 1
s2
s3 s4 Z
1
2
1
2
3
2
0 0 0
1
0
0 –
0 – 12
0
s2
11
13
5
13
3
– 13
1
13
2
13
The solution is Z =
x2 =
1
2
0 0 0
1 0 0
0
1 0
0 0
1
s3
s4 Z
3
− 13
1
13
2
13
5
− 13
3
13
0 0
0 0
0 0
1 0
0
1
7⎤ 3
⎥
3⎥
⎥
22
11⎥ 13
⎥
⎥
7 ⎥ 14
5
⎥
3⎥
⎦
14
58 ⎤
13 ⎥
50 ⎥
13 ⎥
22 ⎥
13 ⎥
36 ⎥
13 ⎥
72 ⎥
13 ⎦⎥
72
22
when x1 = ,
13
13
50
.
13
12. To obtain a standard linear programming
problem, we write the first constraint as
2 x1 – x2 – x3 ≤ 2 .
x1 x2 x3 s1 s2 s3 W
s1 ⎡ 2 –1 –1 1 0 0 0 2 ⎤ 1
⎢
⎥
s2 ⎢ 1 –1 1 0 1 0 0 4 ⎥ 4
s3 ⎢ 1 1 2 0 0 1 0 6 ⎥ 6
⎢
⎥
W ⎢⎣ –2 –1 2 0 0 0 1 0 ⎥⎦
x1 x2 x3 s1 s2 s3 W
x1 ⎡1 – 12 – 12 12 0 0 0 1 ⎤
⎢
⎥
3
s2 ⎢ 0 – 1
1 1 0 0 3⎥
–
2
2
2
⎢
⎥
5
1
⎢0 3
⎥ 10
–
0
1
0
5
s3 ⎢
2
2
2
⎥ 3
W ⎢⎢ 0 –2 1
1 0 0 1 2 ⎥⎥
⎣
⎦
x1 x2 x3 s1 s2 s3 W
8⎤
1
1 0 1 0
x1 ⎡ 1 0 3
3
3
3⎥
⎢
7
2
1
14
s2 ⎢ 0 0
–3 1 3 0 3⎥
3
⎢
⎥
⎢ 0 1 5 – 1 0 2 0 10 ⎥
3
3
3
3⎥
x2 ⎢
⎢
13
26
1
4
0 3 1 3⎥
W ⎢0 0 3
3
⎣
⎦⎥
4⎤
6 ⎥⎥ 3
20 ⎥ 20
⎥ 3
10 ⎥ 10
0 ⎥⎥
⎦
243
Chapter 7: Linear Programming
The solution is W =
ISM: Introductory Mathematical Analysis
26
8
10
when x1 = , x2 = , x3 = 0 .
3
3
3
13. To obtain a standard linear programming problem, we write the second constraint as − x1 – x2 + x3 ≤ 2 and the
third constraint as x1 – x2 – x3 ≤ 1 .
x1 x2 x3 s1 s2 s3 W
s1 ⎡ 4 3 –1 1 0 0 0 1 ⎤
⎢
⎥
s2 ⎢ –1 –1 1 0 1 0 0 2 ⎥ 2
s3 ⎢ 1 –1 –1 0 0 1 0 1 ⎥
⎢
⎥
W ⎣⎢ –1 12 –4 0 0 0 1 0 ⎦⎥
x1 x2 x3 s1
3
2 0 1
⎡
s1
⎢ –1 –1 1 0
x3 ⎢
s3 ⎢ 0 –2 0 0
⎢
W ⎢⎣ –5 8 0 0
s2 s3 W
1 0 0 3⎤ 1
1 0 0 2 ⎥⎥
1 1 0 3⎥
⎥
4 0 1 8⎥
⎦
x1 x2
2
x1 ⎡ 1
3
⎢
x3 ⎢0 – 1
3
s3 ⎢
⎢0 –2
⎢
34
W ⎢⎣0
3
s2 s3 W
x3 s1
1⎤
⎥
1
0 0 3⎥
⎥
0 0 1 1 0 3⎥
⎥
0 53 17
0 1 13⎥
3
⎦
The solution is W = 13 when x1 = 1, x2 = 0, x3 = 3 .
x1
14. s1 ⎡ 1
⎢
s2 ⎢ 1
s3 ⎢ 1
⎢
W ⎣⎢ –4
x1
⎡
s1 0
⎢
s2 ⎢ 0
x1 ⎢1
⎢
W ⎢⎣ 0
x1
⎡
x2 ⎢ 0
s2 ⎢ 0
⎢
x1 ⎢ 1
W ⎢0
⎣⎢
1
3
1
3
0
x2 x3
1 1
–1 1
–1 –1
0 1
1
3
4
3
s1
1
0
0
0
x2 x3 s1
2 2 1
0 2 0
–1 –1 0
–4 –3 0
x2 x3 s1 s2
1 1
1
2
0 2
0 1
0 0
1
2
0
0
0
1 2 0
0 0
s2
0
1
0
0
s3
0
0
1
0
s2
0
1
0
0
s3
–1
–1
1
4
W
0 6⎤ 6
0 10 ⎥⎥ 10
0 4⎥ 4
⎥
1 0⎥
⎦
W
0 2 ⎤1
0 6 ⎥⎥
0 4⎥
⎥
1 16 ⎥
⎦
s3 W
– 12 0
1⎤
⎥
–1 0 6 ⎥
⎥
1 0
5⎥
2
⎥
2 1 20 ⎥
⎦
The solution is W = 20 when x1 = 5, x2 = 1, x3 = 0 .
244
ISM: Introductory Mathematical Analysis
x1 x2 x3 x4
–2 0
0
s1 ⎡ 1
⎢ 1
1
0
0
s2 ⎢
15.
⎢
0
1
1
s3 0
⎢
0
0
1
–2
s4 ⎢
Z ⎢⎣⎢ –60 0 –90 0
x1 x2
–2
⎡
s1 1
⎢ 1
1
s2 ⎢
0
x3 ⎢ 0
⎢
0
s4 ⎢ 0
⎢
–60
0
Z ⎢⎣
x3 x4
0 0
0 0
1 1
0 –3
0 90
s1
1
0
0
0
0
s1
1
0
0
0
0
s2
0
1
0
0
0
s3
0
0
1
0
0
s2 s3
0 0
1 0
0 1
0 –1
0 90
s4
0
0
0
1
0
Section 7.4
Z
0
0
0
0
1
2⎤
5 ⎥⎥
4⎥ 4
⎥
7⎥ 7
0 ⎥⎥
⎦
s4
0
0
0
1
0
Z
0 2 ⎤2
0 5 ⎥⎥ 5
0 4 ⎥
⎥
0 3 ⎥
1 360 ⎥⎥
⎦
x1 x2 x3 x4 s1 s2 s3 s4 Z
x1 ⎡1 –2 0 0 1 0 0 0 0 2 ⎤
⎢
3
0 0 –1 1 0 0 0 3 ⎥⎥ 1
s2 ⎢0
0
1 1 0 0 1 0 0 4 ⎥
x3 ⎢0
⎢
⎥
0
0 –3 0 0 –1 1 0 3 ⎥
s4 ⎢0
Z ⎢⎢⎣0 –120 0 90 60 0 90 0 1 480 ⎥⎥⎦
x1 x2 x3 x4 s1 s2 s3 s4 Z
1
2
x1 ⎡ 1 0 0 0
0 0 0
4⎤
3
3
⎢
⎥
1
⎥
x2 ⎢ 0 1 0 0 – 1
0
0
0
1
3
3
⎢
⎥
x3 ⎢ 0 0 1 1
0 0 1 0 0
4⎥
⎥
s4 ⎢⎢ 0 0 0 –3
0 0 –1 1 0
3⎥
Z ⎢⎢ 0 0 0 90 20 40 90 0 1 600 ⎥⎥
⎣
⎦
The solution is Z = 600 for x1 = 4, x2 = 1, x3 = 4, x4 = 0 .
x1 x2 x3 x4 s1 s2 s3
16. s ⎡ 1
0 1 −1 1 0 0
1
⎢ 1 −1 0 1 0 1 0
s2 ⎢
s3 ⎢ 1 1 −1 1 0 0 1
⎢
Z ⎢⎣ −3 −2 2 1 0 0 0
Z
0 3⎤ 3
0 6 ⎥⎥ 6
0 5⎥ 5
⎥
1 0⎥
⎦
x1 x2 x3 x4 s1 s2 s3 Z
1 −1 1 0 0 0 3⎤
x1 ⎡ 1 0
⎢ 0 −1 −1 2 −1 1 0 0 3⎥
s2 ⎢
⎥
s3 ⎢ 0 1 −2 2 −1 0 1 0 2 ⎥ 2
⎢
⎥
Z ⎢⎣ 0 −2 5 −2 3 0 0 1 9 ⎥⎦
choosing x2 as the entering variable.
245
Chapter 7: Linear Programming
x1
x1 ⎡ 1
⎢
s2 ⎢ 0
x2 ⎢ 0
⎢
Z ⎣⎢ 0
x2 x3 x4
0
1 −1
0 −3
1 −2
0
1
ISM: Introductory Mathematical Analysis
x1 x2 s1 s2 R
⎡
⎤
x2 2 1 1 0 0
2400 ⎥
⎢
s2 ⎢ −7 0 −5 1 0 24,800 ⎥
⎢
⎥
1200 ⎥
R ⎢ 14 0 12 0 1
⎣
⎦
Thus 0 boxes from A and 2400 from B give a
maximum revenue of $1200.
s1 s2 s3 Z
1 0 0 0
3⎤
4 −2 1 1 0 5⎥⎥
2 −1 0 1 0 2 ⎥
⎥
2
1 0 2 1 13⎥
⎦
Choosing x4 as the entering variable in the
second table, we have:
x1 x2 x3 x4 s1 s2 s3 Z
1 −1 1 0 0 0 3⎤
x1 ⎡ 1 0
3
⎢ 0 −1 −1
2 −1 1 0 0 3⎥⎥ 2
s2 ⎢
1 −2 2 −1 0 1 0 2 ⎥ 1
s3 ⎢ 0
⎢
⎥
5 −2 3 0 0 1 9 ⎥
Z ⎢⎣ 0 −2
⎦
x1 x2 x3 x4 s1 s2 s3 Z
1
1 0
1 0
x1 ⎡ 1
0 0
4⎤ 8
2
2
2
⎢
⎥
s2 ⎢ 0 −2 1 0
0 1 −1 0 1⎥
⎢
⎥
x4 ⎢ 0 12 −1 1 − 12 0 12 0 1⎥ 2
⎢
⎥
2 0 1 1 11⎥
Z ⎢⎣ 0 −1 3 0
⎦
x1 x2 x3 x4 s1 s2 s3 Z
1 −1 1 0 0 0 3⎤
x1 ⎡ 1 0
⎢ 0 0 −3 4 −2 1 1 0 5⎥
s2 ⎢
⎥
x2 ⎢ 0 1 −2 2 −1 0 1 0 2 ⎥
⎢
⎥
1 2
1 0 2 1 13⎥
Z ⎣⎢ 0 0
⎦
The solution is Z = 13 when
x1 = 3, x2 = 2, x3 = 0, x4 = 0.
18. Let x, y, and z denote the numbers of units of X,
Y, and Z produced, respectively. We want to
maximize P = 6x + 8y + 12z subject to
x + 2 y + 3z ≤ 900,
4 x + 4 y + 8 z ≤ 5000,
x, y, z ≥ 0.
x y
z s1 s2 P
s1 ⎡ 1 2
3 1 0 0 900 ⎤ 300
⎢
⎥
s2 ⎢ 4 4
8 0 1 0 5000 ⎥ 625
P ⎢ –6 –8 –12 0 0 1
0 ⎥
⎣
⎦
x
y z s1 s2 P
2
z ⎡1
1 13 0 0 300 ⎤ 900
3
⎢3
⎥
⎢
8
4
4
s2 ⎢ 3 – 3 0 – 3 1 0 2600 ⎥⎥ 1950
P ⎢ –2 0 0 4 0 1 3600 ⎥
⎢⎣
⎥⎦
x y
z s1 s2 P
x ⎡1 2 3
1 0 0 900 ⎤
⎢
⎥
s2 ⎢ 0 –4 –4 –4 1 0 1400 ⎥
P ⎢ 0 4 6 6 0 1 5400 ⎥
⎣
⎦
P is maximum when x = 900, y = 0, z = 0.
This maximum profit is $5400.
17. Let x1 and x2 denote the numbers of boxes
transported from A and B, respectively. The
revenue received is R = 0.75 x1 + 0.50 x2 . We
want to maximize R subject to
2 x1 + x2 ≤ 2400 (volume),
3x1 + 5 x2 ≤ 36,800 (weight),
x1 , x2 ≥ 0.
x1 x2 s1 s2 R
⎡
⎤
s1 ⎢ 2
1 1 0 0 2400 ⎥ 1200
s2 ⎢ 3
5 0 1 0 36,800 ⎥ 12, 266 23
⎢
⎥
R ⎢– 3 – 1 0 0 1
⎥
0
2
⎣ 4
⎦
x1 x2 s1 s2 R
1
1
0 0
1200 ⎤ 2400
x1 ⎡1 2
2
⎢
⎥
⎢ 0 7 – 3 1 0 33, 200 ⎥ 9485 5
s2 ⎢
⎥
2
2
7
⎢
⎥
900 ⎥
R ⎢ 0 – 18 83 0 1
⎣
⎦
19. Let x1 , x2 , and x3 denote the numbers of chairs,
rockers, and chaise lounges produced,
respectively. We want to maximize
R = 21x1 + 24 x2 + 36 x3 subject to
x1 + x2 + x3 ≤ 400,
x1 + x2 + 2 x3 ≤ 500,
2 x1 + 3x2 + 5 x3 ≤ 1450,
x1 , x2 , x3 ≥ 0.
x1
x2
x3 s1 s2 s3 R
1
1 1 0 0 0 400 ⎤ 400
s1 ⎡ 1
⎢ 1
1
2 0 1 0 0 500 ⎥⎥ 250
s2 ⎢
3
5 0 0 1 0 1450 ⎥ 290
s3 ⎢ 2
⎢
⎥
0 ⎥
R ⎢⎣ –21 –24 –36 0 0 0 1
⎦
246
ISM: Introductory Mathematical Analysis
Section 7.5
so we check for multiple solutions. Treating x2
as an entering variable, the following table is
obtained:
x1 x2 x3
s1
s2
s3 P
b
1
0
0.39
0.73
0
0.67
0
28.18
−
⎡
⎤
x1
⎢ 0 0 1.00 1.00 1 −2.00 0 40.00 ⎥
s2 ⎢
⎥
6.36 ⎥
x2 ⎢ 0 1 0.79 −0.55 0 0.67 0
⎢
⎥
0
1 1727.27 ⎥
P ⎢⎣ 0 0 9.09 9.09 0
⎦
Another optimum solution is
x1 = 28, x2 = 6, x3 = 0, and P = 1727.
Thus, the optimum solution is for the company
to produce (1 – t)35 + 28t = 35 – 7t of device 1,
(1 – t)0 + 6t = 6t of device 2, and none of device
3, for 0 ≤ t ≤ 1.
x1 x2 x3 s1 s2 s3 R
1
⎡ 1
0 1 – 12 0 0 150 ⎤ 300
s1 ⎢ 2
2
⎥
⎢ 1
1
1
1 0 2 0 0 250 ⎥ 500
x3 ⎢ 2
2
⎥
⎢
⎥
5
1
1
s3 –
⎢ 2 2 0 0 – 2 1 0 200 ⎥ 400
R ⎢⎢ –3 –6 0 0 18 0 1 9000 ⎥⎥
⎣
⎦
x1 x2 x3 s1 s2 s3 R
300 ⎤
⎡
x2 1 1 0 2 –1 0 0
⎢ 0 0 1 –1 1 0 0
100 ⎥⎥
x3 ⎢
50 ⎥
s3 ⎢ –1 0 0 –1 –2 1 0
⎢
⎥
R ⎢⎣ 3 0 0 12 12 0 1 10,800 ⎥⎦
The production of 0 chairs, 300 rockers, and
100 chaise lounges gives the maximum revenue
of $10,800.
Problems 7.5
Principles in Practice 7.5
1. Yes; for the table, x2 is the entering variable
1. Let x1 , x2 , x3 be the numbers of device 1,
device 2, and device 3, respectively, that the
company produces. The situation is to maximize
the profit P = 50 x1 + 50 x2 + 50 x3 subject to the
constraints
5.5 x1 + 5.5 x2 + 6.5 x3 ≤ 190,
3.5 x1 + 6.5 x2 + 7.5 x3 ≤ 180,
4.5 x1 + 6.0 x2 + 6.5 x3 ≤ 165,
and the quotients
smallest.
6
3
and
tie for being the
2
1
2. Yes; the B.F.S. corresponding to the given table
has the basic variable x2 equal to 0.
x1 x2 s1
3. s ⎡ 4 –3 1
1
⎢
s2 ⎢ 3 –1 0
s3 ⎢ 5 0 0
⎢
Z ⎢⎣ –2 –7 0
s2 s3 Z
0 0 0 4⎤
1 0 0 6 ⎥⎥
0 1 0 8⎥
⎥
0 0 1 0⎥
⎦
The entering variable is x2 . Since no quotients
exist, the problem has an unbounded solution.
Thus, no optimum solution (unbounded).
and x1 , x2 , x3 ≥ 0 .
The matrices are shown rounded to 2 decimal
places, although the exact values are used in the
row operations.
Since the indicators are equal, we choose the
first column as the pivot column.
x1
x2 x3 s1 s2 s3 P b
s1 ⎡ 5.5 5.5 6.5 1 0 0 0 190 ⎤
⎢
⎥
s2 ⎢ 3.5 6.5 7.5 0 1 0 0 180 ⎥
s3 ⎢ 4.5 6.0 6.5 0 0 1 0 165⎥
⎢
⎥
0⎥
P ⎢⎣ –50 –50 –50 0 0 0 1
⎦
x1
x2 x2
s1 s2 s3 P
b
1
1
1.18
0.18
0
0
0
34.55
⎡
⎤
x1
⎢0
3 3.36 –0.64 1 0 0
59.09 ⎥⎥
s2 ⎢
9.55⎥
s3 ⎢ 0 1.50 1.18 –0.82 0 1 0
⎢
⎥
0 9.09 9.09 0 0 1 1727.27 ⎥
P ⎢⎣ 0
⎦
An optimum solution is
x1 = 35, x2 = 0, x3 = 0, and P = 1727. However,
x2 is a nonbasic variable and its indicator is 0,
4.
247
x1 x2 s1 s2 s3 s4
⎡
s1 1 −1 1 0 0 0
⎢
s2 ⎢ −1 1 0 1 0 0
s3 ⎢ 8 5 0 0 1 0
⎢
1 0 0 0 1
s4 ⎢ 2
Z ⎢⎢⎣ −2 −1 0 0 0 0
Z
0
0
0
0
1
7⎤ 7
5⎥⎥
40 ⎥ 5
⎥
6⎥ 3
0 ⎥⎥
⎦
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
Since x1 is nonbasic for the last table and its
indicator is 0, there may be multiple optimum
solutions. Treating x1 as an entering variable,
we have
x1 x2 s1
s2 s3 Z
8 − 2 0 48 ⎤
s1 ⎡ 0 0 1
7
7
7 ⎥
⎢
3
1 0 18 ⎥
x2 ⎢ 0 1 0
7
7
7⎥
⎢
8⎥
2 0
x1 ⎢ 1 0 0 − 1
7
7
7⎥
⎢
Z ⎢0 0 0
4
0 1 16 ⎥
⎣
⎦
8
18
Here Z = 16 when x1 = , x2 = . Thus
7
7
multiple optimum solutions exist. Hence Z is
8
8
maximum when x1 = (1 − t )(0) + t = t ,
7
7
18
4
x2 = (1 − t )(2) + t = 2 + t , and 0 ≤ t ≤ 1. For
7
7
the last table s3 is nonbasic and its indicator is 0.
If we continue the process for determining other
optimum solutions, we return to the second
table.
x1 x2 s1 s2 s3 s4 Z
s1 ⎡ 0 − 3 1 0 0 − 1 0 4 ⎤
2
2
⎢
⎥
3 0 1 0
1 0
⎥ 16
s2 ⎢ 0
8
2
2
⎢
⎥ 3
s3 ⎢ 0
1 0 0 1 −4 0 16 ⎥ 16
⎢
⎥
1 0 0 0
1 0
⎥6
x1 ⎢ 1
3
2
2
⎢
⎥
Z ⎢0
0 0 0 0
1 1 6⎥
⎣
⎦
The maximum value of Z is 6 when x1 = 3 and
x2 = 0. Since x2 is nonbasic for the last table
and its indicator is 0, there may be multiple
optimum solutions. Treating x2 as an entering
variable and continuing, we have
x1 x2 s1 s2 s3 s4 Z
s1 ⎡ 0 0 1
1 0
0 0 12 ⎤
⎢
2 0
1 0 16 ⎥
x2 ⎢ 0 1 0
3
3
3⎥
⎢
⎥
13
32
2
s3 ⎢ 0 0 0 −
1
−
0
3
3
3 ⎥
⎢
⎥
1
1
x1 ⎢ 1 0 0 −
0
0 13 ⎥
3
3
⎢
⎥
Z ⎢0 0 0
0 0
1 1 6⎥
⎣
⎦
1
16
Here Z = 6 when x1 = and x2 = . Thus
3
3
multiple optimum solutions exist. Hence Z is a
1
8
maximum when x1 = (1 − t )(3) + t = 3 − t ,
3
3
16
16
x2 = (1 − t )(0) + t = t , and 0 ≤ t ≤ 1. For the
3
3
last table, s2 is nonbasic and its indicator is 0. If
we continue the process for determining other
optimum solutions, we return to the second
table.
6. To obtain a standard linear programming
problem, we write the second constraint as
– x1 + x2 + x3 ≤ 4.
x1 x2 x3 s1 s2 s3 Z
s1 ⎡ 1 –1 4 1 0 0 0 6 ⎤ 6
⎢
⎥
s2 ⎢ –1 1 1 0 1 0 0 4 ⎥
s3 ⎢ 1 –6 1 0 0 1 0 8 ⎥ 8
⎢
⎥
Z ⎢⎣ –8 –2 –4 0 0 0 1 0 ⎥⎦
x1 x2 x3 s1 s2 s3 Z
x1 ⎡ 1 –1 4 1 0 0 0 6 ⎤
⎢
0 5 1 1 0 0 10 ⎥⎥
s2 ⎢ 0
s3 ⎢ 0 –5 –3 –1 0 1 0 2 ⎥
⎢
⎥
Z ⎢⎣ 0 –10 28 8 0 0 1 48⎥⎦
For the last table, x2 is the entering variable.
Since no quotients exist, the problem has an
unbounded solution.
Thus, no optimum solution (unbounded).
x1 x2 s1 s2 s3 Z
5. s1 ⎡ 2 –2 1 0 0 0 4 ⎤
s2 ⎢ –1 2 0 1 0 0 4 ⎥ 2
⎢
⎥
1 0 0 1 0 6⎥ 6
s3 ⎢ 3
⎢
⎥
Z ⎢⎣ 4 −8 0 0 0 1 0 ⎥⎦
x1 x2 s1 s2 s3 Z
s1 ⎡ 1 0 1 1 0 0 8 ⎤ 8
⎢
⎥
x2 ⎢ − 12 1 0 12 0 0 2 ⎥
⎢
⎥
s3 ⎢ 72 0 0 − 12 1 0 4 ⎥ 87
⎢
⎥
Z ⎢ 0 0 0 4 0 1 16 ⎥
⎣
⎦
Z has a maximum of 16 when x1 = 0, x2 = 2.
248
ISM: Introductory Mathematical Analysis
x1
7. s1 ⎡ 9
⎢
s2 ⎢ 4
s3 ⎢ 1
⎢
Z ⎢⎣ –5
x1
⎡3
s1 ⎢
x2 ⎢ 2
⎢
s3 ⎢ 9
Z ⎢7
⎣⎢
x2 x3 s1 s2 s3
3 –2 1 0 0
2 –1 0 1 0
–4 1 0 0 1
–6 –1 0 0 0
x2 x3 s1
0
0
x1
⎡
x2 2
⎢
s2 ⎢ 3
⎢
⎢4
s3 ⎢
Z ⎢⎢ 0
⎣
5⎤ 5
3
2 ⎥⎥ 1
3⎥
⎥
0⎥
⎦
s2 s3 Z
0 0 2⎤
⎥
1 0 0 1⎥
2
⎥
2 1 0 7⎥
⎥
3 0 1 6⎥
⎦
1 – 32
0 – 12
1 –
Z
0
0
0
1
Section 7.5
1
2
0
–1 0
–4 0
x2 x3
3 –3
–1 1
–1 2
–1 4
s1
1
0
0
0
s2
0
1
0
0
s3
0
0
1
0
x2 x3 s1 s2 s3
9 –9 1 –6 0
–1 1 0 1 0
1 0 0 –2 1
–3 6 0 2 0
x2 x3 s1
1 –1
s3 Z
– 23
1
3
0 0
0
0
0
1
– 19
– 43
1 0
0
3
1
3
0
0 1
Z has a maximum value of
0 0
0
0
1
0
3
0 0 0
1 0 0
0
1 0
0 0
1
10 ⎤
3⎥
13 ⎥
3⎥
46 ⎥
3 ⎥
10 ⎥
3 ⎦⎥
4 26
⎛ 4 ⎞ ⎛ 10 ⎞
x2 = (1 – t ) ⎜ ⎟ + ⎜ ⎟ t = + t ,
9 9
⎝9⎠ ⎝ 3 ⎠
x3 = (1 – t )(0) + 0t = 0,
and 0 ≤ t ≤ 1. For the last table, x1 is nonbasic
and its indicator is 0. if we continue the process
for determining other optimum solutions, we
return to the third table.
9. To obtain a standard linear programming
problem, we write the second constraint as
4 x1 + x 2 ≤ 6 .
Z
0 4⎤ 4
9
0 1 ⎥⎥
0 10 ⎥ 10
⎥
1 2⎥
⎦
s2
0
Z
10
10
when x1 = 0, x2 = , x3 = 0.
3
3
Thus multiple optimum solutions exist. Hence Z
is maximum when
13 13
⎛ 13 ⎞
– t,
x1 = (1 – t ) ⎜ ⎟ + 0t =
9
9 9
⎝ ⎠
Z
0 10 ⎤ 53
0 1 ⎥⎥ 1
0 12 ⎥ 6
⎥
1 0⎥
⎦
1
9
1
9
1 –1
1
3
1
3
1
3
1
3
Here Z =
For the last table, x3 is the entering variable.
Since no quotients exist, the problem has an
unbounded solution.
Thus, no optimum solution (unbounded).
x1
8. s ⎡ 6
1
⎢
s2 ⎢ 1
s3 ⎢ 2
⎢
Z ⎢⎣ –2
x1
s1 ⎡ 0
⎢
x1 ⎢1
s3 ⎢ 0
⎢
Z ⎢⎣ 0
x1
⎡
x2 ⎢ 0
x1 ⎢1
⎢
⎢
s3 ⎢ 0
⎢
Z ⎢0
⎣
x2 x3 s1 s2 s3
x1
x2
x3 s1 s2 Z
s1 ⎡ 2 1 1 1 0 0 7 ⎤ 72
⎢
⎥
s2 ⎢ 4 1 0 0 1 0 6 ⎥ 3
2
Z ⎢ –6 –2 –1 0 0 1 0 ⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
s1 ⎡ 0 1
1 1 – 12 0 4 ⎤ 4
2
⎢
⎥
⎢
3
1
1
x1 1
0 0 4 0 2 ⎥⎥
4
⎢
⎢
⎥
3
1
Z ⎢ 0 – 2 –1 0 2 1 9 ⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
x3 ⎡
0 12 1 1 – 12 0 4 ⎤ 8
⎢
⎥
x1 ⎢1 14 0 0 14 0 32 ⎥ 6
⎢
⎥
Z ⎢0 0 0 1 1 1 13⎥
⎣
⎦
Z has a maximum value of 13 when
3
x1 = , x2 = 0, x3 = 4. Since x2 is nonbasic for
2
the last table and its indicator is 0, there may be
multiple optimum solutions. Treating x2 as an
entering variable, we have
4 ⎤
9 ⎥
13 ⎥
9 ⎥ 13
3
86 ⎥
9 ⎥
10 ⎥
3 ⎥⎦
10
when
3
13
4
, x2 = , x3 = 0. Since s2 is nonbasic
9
9
for the last table and its indicator is 0, there may
be multiple optimum solutions. Treating s2 as
an entering variable, we have
x1 =
249
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
choosing s3 as departing variable
x1 x2 x3 s1 s2 Z
⎡
x3 –2 0 1 1 –1 0 1⎤
⎢
⎥
x2 ⎢ 4 1 0 0 1 0 6 ⎥
Z ⎢ 0 0 0 1 1 1 13⎥
⎣
⎦
Here Z = 13 when x1 = 0, x2 = 6, x3 = 1. Thus
multiple optimum solutions exist. Hence Z is
maximum when
3 3
⎛3⎞
x1 = (1 – t ) ⎜ ⎟ + 0t = – t ,
2 2
⎝2⎠
x1
x2
x3 s1 s2
s3 R
⎡
0 1 0 – 15 0
100 ⎤ 500
s1 ⎢
⎥ 3
2 0
⎢
⎥0
0
0
1
–
0
s2
5
⎢
⎥
⎢ 2
⎥
3
1
1 0 0 5 0
300 ⎥
x3 ⎢ 5
5
750
⎢ 24
⎥
16
48
R ⎢–
– 5 0 0 0 5 1 14, 400 ⎥
⎣ 5
⎦
x1 x2 x3 s1 s2 s3 R
100 ⎤ 100
s1 ⎡0 1 0 1 –3 1 0
⎢1 –1 0 0 5 –2 0
0 ⎥⎥ 0
x1 ⎢
300 ⎥ 300
x3 ⎢0 1 1 0 –2 1 0
⎢
⎥
R ⎢⎣0 –8 0 0 24 0 1 14, 400 ⎥⎦
x1 x2 x3 s1 s2 s3 R
100 ⎤
x2 ⎡ 0 1 0 1 –3 1 0
⎢1 0 0 1 2 –1 0
100 ⎥⎥ 50
x1 ⎢
200 ⎥ 200
x3 ⎢ 0 0 1 –1 1 0 0
⎢
⎥
R ⎢⎣ 0 0 0 8 0 8 1 15, 200 ⎦⎥
The maximum value of R is 15,200 when
x1 = 100, x2 = 100, x3 = 200. Since s2 is
nonbasic for the last table and its indicator is 0,
there may be multiple optimum solutions.
Treating s2 as an entering variable, we have
3
5
1
5
x2 = (1 – t )(0) + 6t = 6t ,
x3 = (1 – t )(4) + (1)t = 4 – 3t , and 0 ≤ t ≤ 1. For
the last table, x1 is nonbasic and its indicator is
0. If we continue the process for determining
other optimum solutions, we return to the third
table.
x1 x2 x3 x4 s1 s2 s3 P
10. s ⎡ 1 −1 0 0 1 0 0 0 2 ⎤
1
⎢
⎥
s2 ⎢ 0 1 −1 0 0 1 0 0 3⎥ 3
1 −3
1 0 0 1 0 4⎥ 4
s3 ⎢ 0
⎢
⎥
P ⎢⎣ −1 −2 −1 −2 0 0 0 1 0 ⎥⎦
x1 x2 x3 x4 s1 s2 s3 P
s1 ⎡ 1 0 −1 0 1 1 0 0 5⎤
⎢
⎥
x2 ⎢ 0 1 −1 0 0 1 0 0 3⎥
1 0 −1 1 0 1⎥
s3 ⎢ 0 0 −2
⎢
⎥
P ⎢⎣ −1 0 −3 −2 0 2 0 1 6 ⎦⎥
Now x3 is the entering variable but no quotients
exist. Thus, the feasible region is unbounded
and, hence, there is no optimum solution.
2
5
– 15
x1 x2 x3
s1 s2
5
1 0
x2 ⎡
2
⎢
1
⎢
0
0
s2
2
⎢
⎢–
0 1 – 32
x3 ⎢
8
R ⎢⎣ 0 0 0
3
2
1
2
1
2
s3 R
250 ⎤
⎥
0
50 ⎥
⎥
1 0
⎥
150
2
⎥
8 1 15, 200 ⎥
⎦
0 – 12
1 – 12
0
0
0
Here R = 15,200 when
x1 = 0, x2 = 250, x3 = 150.
Thus multiple optimum solutions exist.
Hence R is maximum when
x1 = (1 – t )(100) + 0t = 100 – 100t ,
x2 = (1 – t )(100) + 250t = 100 + 150t
x3 = (1 – t )(200) + 150t = 200 – 50t , and
0 ≤ t ≤ 1. For the last table, x1 is nonbasic and
its indicator is 0. If we continue the process for
determining other optimum solutions, we return
to the fourth table. If we were to initially choose
s2 as the departing variable, then
11. Let x1 , x2 , and x3 denote the numbers of chairs,
rockers, and chaise lounges produced,
respectively. We want to maximize
R = 24 x1 + 32 x2 + 48 x3 subject to
x1 + x2 + x3 ≤ 400,
x1 + x2 + 2 x3 ≤ 600,
2 x1 + 3x2 + 5 x3 ≤ 1500,
x1 , x2 , x3 ≥ 0.
x1 x2
x3 s1 s2 s3 R
1
1 1 0 0 0 400 ⎤ 400
s1 ⎡ 1
⎢ 1
1
2 0 1 0 0 600 ⎥⎥ 300
s2 ⎢
3
5 0 0 1 0 1500 ⎥ 300
s3 ⎢ 2
⎢
⎥
0 ⎥
R ⎣⎢ –24 –32 –48 0 0 0 1
⎦
250
ISM: Introductory Mathematical Analysis
x1
x2 x3
1
1
s1 ⎡ 1
⎢ 1
1
2
s2 ⎢
⎢
2
3
5
s3
⎢
R ⎣⎢ –24 –32 –48
Section 7.6
s1 s2 s3 R
1 0 0 0
400 ⎤ 400
0 1 0 0 600 ⎥⎥ 300
0 0 1 0 1500 ⎥ 300
⎥
0 0 0 1
0 ⎥
⎦
x1 x2 x3 s1 s2 s3 R
1
⎡
s1 12
0 1 – 12 0 0
100 ⎤ 200
2
⎢
⎥
1
1
⎥ 600
x3 ⎢ 1
1
0
0
0
300
2
2
⎢ 2
⎥
s3 ⎢ – 1 1 0 0 – 5 1 0
0 ⎥ 0
2
⎢ 2 2
⎥
R ⎢⎢ 0 –8 0 0 24 0 1 14, 400 ⎥⎥
⎣
⎦
x1 x2 x3 s1 s2
s3 R
100 ⎤ 50
s1 ⎡ 1 0 0 1 2 –1 0
⎢ 1 0 1 0 3 –1 0
300 ⎥⎥ 100
x3 ⎢
0 ⎥
x2 ⎢ –1 1 0 0 –5 2 0
⎢
⎥
R ⎢⎣ –8 0 0 0 –16 16 1 14, 400 ⎥⎦
x1 x2 x3 s1 s2 s3 R
1
1
1
⎡
50 ⎤ 100
s2 ⎢ 2 0 0 2 1 – 2 0
⎥
⎢– 1 0 1 – 3 0 1 0
150 ⎥
x3 ⎢ 2
2
2
⎥
⎢
3
5
1
x2 ⎢ 2 1 0 2 0 – 2 0
250 ⎥⎥ 500
3
R ⎢ 0 0 0 8 0 8 1 15, 200 ⎥
⎥⎦
⎣⎢
the maximum value of R is 15,200 when x1 = 0, x2 = 250, x3 = 150. For the last table, x1 is nonbasic and its
indicator is 0. Treating x1 as an entering variable, we have
x1 x2 x3 s1 s2 s3 R
100 ⎤
x1 ⎡ 1 0 0 1 2 –1 0
⎢ 0 0 1 –1 1 0 0
200 ⎥⎥
x3 ⎢
100 ⎥
x2 ⎢ 0 1 0 1 –3 1 0
⎢
⎥
R ⎢⎣ 0 0 0 8 0 8 1 15, 200 ⎥⎦
Here R = 15,200 when x1 = 100, x2 = 100, x3 = 200. For the last table, s2 is nonbasic and its indicator is 0. If we
continue the process of determining other optimum solutions, we return to the table corresponding to the solution
x1 = 0, x2 = 250, x3 = 150.
Thus, the maximum revenue is $15,200 when x1 = 100 – 100t , x2 = 100 + 150t ,
x3 = 200 – 50t , and 0 ≤ t ≤ 1
Principles in Practice 7.6
1. Using the hint, 1000 – x1 standard and 800 – x2 deluxe snowboards must be manufactured at plant II. The
constraints for plant I are x1 + x2 ≤ 1200 and x2 – x1 ≤ 200. The constraints for plant II are
(1000 – x1 ) + (800 – x2 ) ≤ 1000 or x1 + x2 ≥ 800. The quantity to be maximized is the profit
P = 40 x1 + 60 x2 + 45(1000 – x1 ) + 50(800 – x2 )
= –5 x1 + 10 x2 + 85, 000 subject to the constraints
251
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
x1 + x2 ≤ 1200,
– x1 + x2 ≤ 200,
x1 + x2 ≥ 800,
and x1 , x2 ≥ 0 .
Note that maximizing Z = –5 x1 + 10 x2 also maximizes the profit. The corresponding equations are:
x1 + x2 + s1 = 1200,
– x1 + x2 + s2 = 200,
x1 + x2 – s3 + t = 800.
The artificial objective equation is
W = –5 x1 + 10 x2 – Mt.
The augmented coefficient matrix is:
x1
x2 s1 s2 s3 t W
1 1 0 0 0 0 1200 ⎤
⎡ 1
⎢ –1
1 0 1 0 0 0 200 ⎥⎥
⎢
⎢ 1
1 0 0 –1 1 0 800 ⎥
⎢
⎥
0⎥
⎢⎣ 5 –10 0 0 0 M 1
⎦
The simplex tables follow.
x1
x2
s1 s2 s3 t W
1
1
1 0 0 0 0 1200 ⎤
⎡
s1
⎢ –1
1
0 1 0 0 0
200 ⎥⎥
s2 ⎢
1
0 0 −1 1 0
800 ⎥
t ⎢ 1
⎢
⎥
W ⎢⎣5 – M –10 – M 0 0 M 0 1 −800M ⎥⎦
x1
x2 s1
s2
s3 t W
2
0 1
0 0 0
1000
−1
⎤
s1 ⎡
⎢ –1
⎥
1
0
1
0
0
0
200
x2 ⎢
⎥
⎥
−1 1 0
2
0 0
–1
600
t ⎢
⎢
⎥
W ⎣⎢ –5 – 2M 0 0 10 + M M 0 1 2000 − 600 M ⎦⎥
x1 x2 s1 s2 s3
t W
−1 0 400 ⎤
0
1
s1 ⎡ 0 0 1
⎢
⎥
1
1
1 0
x2 ⎢ 0 1 0
−2
500 ⎥
2
2
⎢
⎥
1 0
x1 ⎢ 1 0 0 – 1 − 1
300 ⎥
2
2
2
⎢
⎥
– 52 52 + M 1 3500 ⎥
W ⎢ 0 0 0 15
2
⎣
⎦
Delete the t-column since t = 0 and return to Z.
x1 x2 s1 s2 s3 Z
s ⎡0 0 1
0 1 0 400 ⎤
3
⎢
⎥
1 0 0
700 ⎥
x2 ⎢ 0 1 12
2
⎢
⎥
1
1
x1 ⎢ 1 0 2 − 2 0 0 500 ⎥
⎢
⎥
5
15 0 0 4500
⎥
Z ⎢0 0 2
2
⎣
⎦
Thus, x1 = 500, x2 = 700, and Z = 4500. Plant I should manufacture 500 standard and 700 deluxe snowboards.
Plant II should manufacture 1000 – 500 = 500 standard and 800 – 700 = 100 deluxe snowboards. The maximum
profit is P = –5(500) + 10(700) + 85,000 = $89,500.
252
ISM: Introductory Mathematical Analysis
Section 7.6
Problems 7.6
x1 x2 s1 s2 t2 W
1. ⎡
1 1 1 0 0 0 6⎤
⎢
⎥
⎢ –1 1 0 –1 1 0 4 ⎥
⎢ –2 –1 0 0 M 1 0 ⎥
⎣
⎦
x1
x2
s1 s2 t2 W
1
1 0 0 0
6 ⎤6
s1 ⎡ 1
⎢
⎥
1
0 –1 1 0
4 ⎥4
t2 ⎢ –1
W ⎢ –2 + M –1 – M 0 M 0 1 –4M ⎥
⎣
⎦
x1 x2 s1 s2
t2
W
–1 0 2 ⎤ 1
s1 ⎡ 2 0 1 1
⎢
⎥
1
0 4⎥
x2 ⎢ –1 1 0 –1
W ⎢ –3 0 0 –1 M + 1 1 4 ⎥
⎣
⎦
x1 x2 s1 s2 Z
⎡
1 0 1⎤
x1 1 0 1
2
2
⎢
⎥
x2 ⎢ 0 1 1 – 1 0 5⎥
2
2
⎢
⎥
⎢
⎥
3
1
1 7⎥
Z ⎢0 0 2
2
⎣
⎦
The maximum is Z = 7 when x1 = 1, x2 = 5 .
x1 x2 s1 s2 t2 W
2. ⎡
1 2 1 0 0 0 8⎤
⎢
⎥
⎢ 1 6 0 –1 1 0 12 ⎥
⎢ –3 –4 0 0 M 1 0 ⎥
⎣
⎦
x1
x2
s1 s2 t2 W
2
1 0 0 0
8 ⎤4
s1 ⎡ 1
⎢
⎥
6
0 –1 1 0
12 ⎥ 2
t2 ⎢ 1
W ⎢ –3 – M –4 – 6 M 0 M 0 1 –12 M ⎥
⎣
⎦
x1 x2 s1 s2
t2 W
− 13
0 4⎤ 6
s1 ⎡ 23 0 1 13
⎢
⎥
⎢ 1 1 0 −1
1
0 2 ⎥⎥ 12
x2 ⎢ 6
6
6
⎢ 7
⎥
2 2
W ⎢− 3 0 0 − 3 3 + M 1 8⎥
⎣
⎦
x1 x2 s1 s2 Z
3
1 0
x1 ⎡ 1 0
6⎤
2
2
⎢
⎥
⎥
x2 ⎢ 0 1 – 1 – 1 0
1
4
4
⎢
⎥
⎢
7
1 1 22 ⎥
0
0
Z ⎢
⎥⎦
2
2
⎣
The maximum is Z = 22 when x1 = 6, x2 = 1 .
253
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
x1 x2 x3 s1 s2 t2 W
3. ⎡
1 2 1 1 0 0 0 5⎤
⎢
⎥
⎢ –1 1 1 0 –1 1 0 1⎥
⎢ –2 –1 1 0 0 M 1 0 ⎥
⎣
⎦
x1
x2
x3
s1 s2 t2 W
⎡
s1
1
2
1
1 0 0 0 5 ⎤ 52
⎢
⎥
t2 ⎢ –1
1
1
0 –1 1 0 1 ⎥ 1
W ⎢ –2 + M –1 – M 1 – M 0 M 0 1 – M ⎥
⎣
⎦
x1 x2 x3 s1 s2
t2 W
s1 ⎡ 3 0 –1 1 2
–2
0 3⎤ 1
⎢
⎥
x2 ⎢ –1 1 1 0 −1
1
0 1⎥
W ⎢ –3 0 2 0 –1 1 + M 1 1⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
2 0 1⎤
x1 ⎡ 1 0 – 1 1
3 3
3
⎢
⎥
2 1 – 1 0 2⎥
x2 ⎢ 0 1
3 3
3
⎢
⎥
Z ⎢0 0
1 1
1 1 4⎥
⎣
⎦
The maximum is Z = 4 when x1 = 1, x2 = 2, x3 = 0 .
x1 x2 x3 s1
s2 t2 W
⎡ 1 1 1 1 0 0 0 9⎤
⎥
4. ⎢
1 0 –1 1 0 6 ⎥
⎢ 1 –2
⎢ –1 1 –4 0 0 M 1 0 ⎥
⎣
⎦
x1
x2
x3
s1
s1 ⎡ 1
1
1
1
⎢
t2 ⎢ 1
–2
1
0
⎢
W –1 – M 1 + 2M –4 – M 0
⎣
x1 x2 x3 s1 s2
t2 W
s1 ⎡ 0 3 0 1 1
–1
0
⎢
x3 ⎢1 –2 1 0 –1
1
0
⎢
W 3 –7 0 0 –4 4 + M 1
⎣
x1 x2 x3 s1 s2 Z
x2 ⎡ 0 1 0
⎢
x3 ⎢
⎢1 0 1
⎢
Z ⎢3 0 0
⎣
1
3
1
3
2
3
7
3
– 13
– 53
s2 t2 W
0 0 0
9 ⎤9
⎥
–1 1 0
6 ⎥6
M 0 1 –6 M ⎥
⎦
3 ⎤1
⎥
6⎥
24 ⎥
⎦
1 ⎤3
⎥
0 8 ⎥⎥
⎥
1 31⎥
⎦
0
254
ISM: Introductory Mathematical Analysis
Section 7.6
x1 x2 x3 s1 s2 Z
⎡
s2 0 3 0 1 1 0 3⎤
⎢
⎥
x3 ⎢ 1 1 1 1 0 0 9 ⎥
Z ⎢ 3 5 0 4 0 1 36 ⎥
⎣
⎦
The maximum is Z = 36 when x1 = 0, x2 = 0, x3 = 9 .
x1 x2 x3 s1 t2 W
5. ⎡
1 1 1 1 0 0 10 ⎤
⎢
⎥
1 0 6⎥
⎢ 1 −1 −1 0
⎢ −3 −2 −1 0 M 1 0 ⎥
⎣
⎦
x1
x2
x3 s1 t2 W
s1 ⎡
1
1
1 1 0 0
10 ⎤ 10
⎢
⎥
−1
−1 0 1 0
t2 ⎢
1
6⎥ 6
W ⎢ −3 − M −2 + M −1 + M 0 0 1 −6M ⎥
⎣
⎦
x1 x2 x3 s1
t2 W
⎡
s1 0 2
2 1
−1 0 4 ⎤ 2
⎢
⎥
x1 ⎢ 1 −1 −1 0
1 0 6⎥
W ⎢0 −5 −4 0 3 + M 1 18⎥
⎣
⎦
x1 x2 x3 s1 W
⎡
x2 0 1 1 1 0 2 ⎤
2
⎢
⎥
x1 ⎢⎢ 1 0 0 12 0 8⎥⎥
Z ⎢⎢ 0 0 1 5 1 28⎥⎥
2
⎣
⎦
The maximum is Z = 28 when x1 = 8, x2 = 2, and x3 = 0.
x1
x2 x3
s1
t1
t2 W
⎡ 0 1 –2 −1 1 0 0 5⎤
6. ⎢
⎥
1 0 7⎥
⎢ 1 1 1 0 0
⎢ –2 –1 –3 0 M M 1 0 ⎥
⎣
⎦
x1
x2
x3
s1 t1 t2 W
t1 ⎡ 0
⎢
t2 ⎢ 1
W ⎢ –2 – M
⎣
x1
x2 ⎡ 0
⎢
t2 ⎢ 1
W ⎢ –2 – M
⎣
1
1
–1 – 2 M
x2
–2
1
–3 + M
x3
1
–2
0
3
0 –5 – 3M
s1
–1 1 0 0
5 ⎤5
⎥
0 0 1 0
7 ⎥7
M 0 0 1 –12 M ⎥
⎦
t1
t2 W
–1
1
1
–1
–1 – M 1 + 2 M
0 0
5 ⎤
⎥
1 0
2 ⎥ 23
0 1 5 – 2M ⎥
⎦
255
Chapter 7: Linear Programming
x1 x2 x3
⎡
x2 ⎢ 23
⎢
x3 ⎢ 13
⎢
1
W ⎢⎣⎢ − 3
s1
t1
1 0 – 13
0 1
1
3
1
3
1
–3
0 0
2
3
– 23 + M
x1 x2 x3
s1
ISM: Introductory Mathematical Analysis
t2 W
2
3
1
3
5
3
+M
0
19 ⎤ 19
3⎥ 2
⎥
2⎥ 6
⎥
⎥
1 25
3 ⎦⎥
0
Z
x2 ⎡ 0 1 −2 −1 0 5 ⎤
⎢
⎥
x1 ⎢1 0 3 1 0 2 ⎥
W ⎢0 0 1 1 1 9 ⎥
⎣
⎦
The maximum is Z = 9 when x1 = 2, x2 = 5, x3 = 0.
x1 x2 s1 s2 s3 t3 W
7. ⎡ 1 –1 1 0 0 0 0 1⎤
⎢ 1 2 0 1 0 0 0 8⎥
⎢
⎥
⎢ 1 1 0 0 –1 1 0 5⎥
⎢
⎥
⎢⎣ –1 10 0 0 0 M 1 0 ⎥⎦
x1
x2 s1 s2 s3 t3 W
1
–1
1 0 0 0 0
1 ⎤1
s1 ⎡
⎢ 1
2
0 1 0 0 0
8 ⎥⎥ 8
s2 ⎢
1
0 0 –1 1 0
5 ⎥5
t3 ⎢ 1
⎢
⎥
W ⎢⎣ –1 – M 10 – M 0 0 M 0 1 –5M ⎥⎦
x1
x2
s1
s2 s3 t3 W
–1
1
0 0 0 0
1 ⎤
x1 ⎡1
⎢0
⎥7
3
–1
1
0
0
0
7
s2 ⎢
⎥3
2
–1 0 –1 1 0
4 ⎥2
t3 ⎢ 0
⎢
⎥
W ⎢⎣ 0 9 – 2M 1 + M 0 M 0 1 1 – 4M ⎦⎥
x1 x2 s1 s2 s3
t3 W
1 0 –1
1 0
3⎤
x1 ⎡ 1 0
2
2
2
⎢
⎥
3
3 0
1 1
s2 ⎢ 0 0
⎥
–
1
2
2
2
⎢
⎥
1 0
⎥
x2 ⎢ 0 1 – 1 0 – 1
2
2
2
2
⎢
⎥
9 – 9 + M 1 –17 ⎥
0
W ⎢⎢ 0 0 11
2
2
2
⎣
⎦⎥
For the above table, t3 = 0. Thus W = Z.
The maximum is Z = –17 when x1 = 3, x2 = 2 .
256
ISM: Introductory Mathematical Analysis
x1 x2 x3 s1
8. ⎡ 1 1 –1 –1
⎢ 1 1 1 0
⎢
⎢ 1 –1 1 0
⎢
⎢⎣ –1 –4 1 0
x1
x2
t1 ⎡ 1
⎢
s2 ⎢ 1
t3 ⎢ 1
⎢
W ⎢⎣ –1 – 2M
x1
s2
t1
s1
1 –1 –1
1 1 0
–1 1 0
–4 1 M
x2
t3 W
0
1 0
1 0 0
0 0
1
0 M M
x3
x3
Section 7.6
0
0
0
1
s2 t1
0
1
0
0
s1
1
0
0
0
5⎤
3⎥⎥
7⎥
⎥
0⎥
⎦
t3 W
0
0
1
0
0
5 ⎤5
0
3 ⎥⎥ 3
0
7 ⎥7
⎥
1 –12M ⎥
⎦
s2 t1 t3 W
0
–2 –1
–1 1
t1 ⎡0
⎢1
1
1 0
1 0
x1 ⎢
⎢
0
–2
0
0
–1
0
t3
⎢
W ⎢⎣0 –3 + 2 M 2 + 2M M 1 + 2 M 0
There is no solution (empty feasible region).
0
0
1
0
0
2⎤
0
3⎥⎥
0
4⎥
⎥
1 3 – 6M ⎥
⎦
9. We write the third constraint as – x1 + x2 + x3 ≥ 6.
x1 x2 x3 s1 s2 s3 t2 t3 W
⎡ 1 1 1 1 0 0 0 0 0 1⎤
⎢ 1 –1 1 0 –1 0
1 0 0 2 ⎥⎥
⎢
⎢ –1 1 1 0 0 –1 0
1 0 6⎥
⎢
⎥
⎢⎣ –3 2 –1 0 0 0 M M 1 0 ⎥⎦
x1 x2
x3
s1 s2 s3 t2 t3 W
1
1
1
1 0 0 0 0 0
1 ⎤1
s1 ⎡
⎢ 1 –1
1
0 –1 0 1 0 0
2 ⎥⎥ 2
t2 ⎢
1
0 0 –1 0 1 0
6 ⎥6
t3 ⎢ –1 1
⎢
⎥
W ⎢⎣ –3 2 –1 – 2 M 0 M M 0 0 1 –8M ⎥⎦
x1
x2 x3
s1 s2 s3 t2 t3 W
1
1 1
1 0 0 0 0 0
1⎤
x3 ⎡
⎢
0
–2
0
–1
–1
0
1
0
0
1⎥⎥
t2 ⎢
–2
0 0
–1 0 –1 0 1 0
5⎥
t3 ⎢
⎢
⎥
W ⎣⎢ –2 + 2 M 3 + 2 M 0 1 + 2 M M M 0 0 1 1 – 6 M ⎦⎥
There is no solution (empty feasible region).
257
Chapter 7: Linear Programming
x1 x2
10. ⎡ 1 2
⎢ 1 6
⎢
⎢ 0
1
⎢
–1 –4
⎣⎢
x1
1
⎡
s1
⎢ 1
t2 ⎢
t3 ⎢ 0
⎢
W ⎢⎣ –1 – M
s1 s2 s3 t2 t3
1 0 0 0 0
0 –1 0
1 0
0 0 –1 0
1
0 0 0 M M
ISM: Introductory Mathematical Analysis
W
0
0
0
1
x2
s1 s2 s3
2
1 0 0
6
0 –1 0
1
0 0 –1
–4 – 7 M 0 M M
8⎤
12 ⎥⎥
2⎥
⎥
0⎥
⎦
t2
0
1
0
0
t3
0
0
1
0
W
0
8 ⎤4
0
12 ⎥⎥ 2
0
2 ⎥2
⎥
1 –14 M ⎥
⎦
Here we choose t3 as the departing variable.
x1
x2 s1 s2
s3
t2
t3
W
1
0
1
0
2
0
–2
0 4⎤ 2
⎡
s1
⎢ 1
0 0 –1
6
1
–6
0 0 ⎥⎥ 0
t2 ⎢
1 0 0
–1
0
1
0 2⎥
x2 ⎢ 0
⎢
⎥
W ⎢⎣ –1 – M 0 0 M –4 – 6M 0 4 + 7 M 1 8 ⎥⎦
x1 x2 s1 s2 s3
t2
t3 W
2
1
1
⎡
–3
0 0 4 ⎤ 12
s1 ⎢ 3 0 1 3 0
⎥
⎥
s3 ⎢ 1 0 0 – 1 1
1
–1
0
0
6
6
⎢ 6
⎥
⎢ 1
⎥
1 0
1
1
0
–
0
0
2
⎥
x2 ⎢ 6
6
6
⎢ 1
⎥
2
2
W ⎢ – 3 0 0 – 3 0 3 + M M 1 8⎥
⎣
⎦
x1 x2 s1 s2 s3 Z
⎡ 2 0 3 1 0 0 12 ⎤
s2 ⎢
⎥
1
1
s3 ⎢ 2 0 2 0 1 0 2 ⎥
⎢
⎥
x2 ⎢ 1 1 1 0 0 0 4 ⎥
2
2
Z ⎢ 1 0 2 0 0 1 16 ⎥
⎢⎣
⎥⎦
Thus the maximum value of Z is 16, when x1 = 0, x2 = 4.
If we choose t2 as the original departing variable, then
x1
x2
s1 s2 s3 t2 t3 W
1
2
1 0 0 0 0 0
8 ⎤4
⎡
s1
⎢ 1
6
0 –1 0 1 0 0
12 ⎥⎥ 2
t2 ⎢
1
0 0 –1 0 1 0
2 ⎥2
t3 ⎢ 0
⎢
⎥
W ⎢⎣ –1 – M –4 – 7 M 0 M M 0 0 1 –14 M ⎥⎦
258
ISM: Introductory Mathematical Analysis
Section 7.6
x1
x2 s1
s2
s3
t2
t3
2
1
1
⎡
0 1
0
–3
0
s1 ⎢
3
3
1
1
x2 ⎢
1 0
– 16
0
0
6
6
⎢
1
⎢ –1
0 0
–1
– 16
1
6
6
t3 ⎢
⎢
W ⎢– 1 + 1 M 0 0 – 2 – 1 M M 2 + 7 M 0
3 6
3 6
⎣ 3 6
x1 x2 s1 s2 s3 t2
t3
W
–2
0 4⎤ 2
s1 ⎡ 1 0 1 0 2 0
⎢ 0 1 0 0 –1 0
1
0
2 ⎥⎥
x2 ⎢
6
0 0⎥
s2 ⎢ –1 0 0 1 –6 –1
⎢
⎥
W ⎢⎣ –1 0 0 0 –4 M 4 + M 1 8 ⎦⎥
x1 x2 s1 s2 s3 Z
s3 ⎡ 1 0 1 0 1 0 2 ⎤
2
⎢2
⎥
x2 ⎢ 1 1 1 0 0 0 4 ⎥
2
⎢2
⎥
s2 ⎢ 2 0 3 1 0 0 12 ⎥
⎢
⎥
Z ⎢⎣ 1 0 2 0 0 1 16 ⎥⎦
The maximum is Z = 16 when x1 = 0, x2 = 4 .
W
0 4 ⎤ 12
⎥
0 2⎥
⎥
0 0⎥
⎥ 0
⎥
1 8⎥
⎦
x1 x2 s1 s3 t2 t3 W
11. ⎡ 1 –1 1 0 0 0 0 4 ⎤
⎢ –1 1 0 0
1 0 0 4 ⎥⎥
⎢
⎢ 1 0 0 –1 0
1 0 6⎥
⎢
⎥
⎢⎣ 3 –2 0 0 M M 1 0 ⎦⎥
x1
x2 s1 s3 t2 t3 W
1
–1
1 0 0 0 0
4 ⎤
⎡
s1
⎢ –1
1
0 0 1 0 0
4 ⎥⎥ 4
t2 ⎢
0
0 –1 0 1 0
6 ⎥
t3 ⎢ 1
⎢
⎥
W ⎢⎣ 3 –2 – M 0 M 0 0 1 –10 M ⎥⎦
x1
x2 s1 s3
t2
t3 W
0
0
1
0
1
0
0
8 ⎤
s1 ⎡
⎢ –1 1 0 0
1
0 0
4 ⎥⎥
x2 ⎢
0 0 –1
0
1 0
6 ⎥6
t3 ⎢ 1
⎢
⎥
W ⎢⎣1 – M 0 0 M 2 + M 0 1 8 – 6M ⎥⎦
x1 x2 s1 s3
t2
t3 W
1
0 0 8⎤
s1 ⎡ 0 0 1 0
⎢ 0 1 0 –1
1
1 0 10 ⎥⎥
x2 ⎢
0
1 0 6⎥
x1 ⎢ 1 0 0 –1
⎢
⎥
W ⎣⎢ 0 0 0 1 2 + M –1 + M 1 2 ⎦⎥
For the above table, t2 = t3 = 0. Thus W = Z.
The maximum is Z = 2 when x1 = 6, x2 = 10 .
259
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
12. We write the first constraint as − x1 + 2 x2 ≤ 12.
x1
⎡ −1
⎢ −1
⎢
⎢ 1
⎢
⎢⎣ −2
x2 s1
2
1
1
8
s2
s3
t2
t3 W
1 0 0 0 0
0 −1 0
1 0
0 0 −1 0
1
0 0 0 M M
x1
x2 s1
2
s1 ⎡ −1
⎢ −1
1
t2 ⎢
⎢
1
1
t3
⎢
W ⎢⎣ −2 8 − 2M
x1 x2
1
s1 ⎡
⎢
1
−
x2 ⎢
2
t3 ⎢
⎢
W ⎣⎢ 6 − 2M
x1 x2 s1
0
1
0
0
s2 s3
0 12 ⎤
0 2 ⎥⎥
0 10 ⎥
⎥
1 0⎥
⎦
t2 t3 W
1 0 0
0 −1 0
0 0 −1
0 M M
0
1
0
0
s1
s3
s2
0
0
1
0
0
12 ⎤ 6
0
2 ⎥⎥ 2
0
10 ⎥ 10
⎥
1 −12 M ⎥
⎦
t2 t3 W
1
2 0
−2
0
1
−1 0
0
1 −1
−1
0 8 − M M −8 + 2M
s2
s3
t2
0
0
1
0
0
8⎤ 8
0
2 ⎥⎥
0
8⎥ 4
⎥
1 −16 − 8M ⎥
⎦
t3 W
s1 ⎡ 0 0 1
− 12 0
4⎤
⎢
⎥
1 0
⎥
x2 ⎢ 0 1 0
6
2
⎢
⎥
1 0
⎥
x1 ⎢ 1 0 0
4
2
⎢
⎥
W ⎢0 0 0
5
3 −5 + M −3 + M 1 −40 ⎥
⎣
⎦
For the above table, t2 = t3 = 0. Thus W = Z. The maximum is Z = −40 when x1 = 4 and x2 = 6.
3
2
1
−2
1
2
1
2
1
−2
− 12
− 32
1
2
1
−2
13. Let x1 and x2 denote the numbers of Standard and Executive bookcases produced, respectively, each week. We
want to maximize the profit function P = 35 x1 + 40 x2 subject to
2 x1 + 3x2 ≤ 400,
3x1 + 4 x2 ≤ 500,
3x1 + 4 x2 ≥ 250,
x1 , x2 ≥ 0.
The artificial objective function is W = P – Mt3 .
x1
x2 s1 s2 s3 t3 W
3 1 0 0 0 0 400 ⎤
⎡ 2
⎢ 3
4 0 1 0 0 0 500 ⎥⎥
⎢
⎢ 3
4 0 0 –1 1 0 250 ⎥
⎢
⎥
–35 –40 0 0 0 M 1
0⎥
⎣⎢
⎦
x1
2
s1 ⎡
⎢
3
s2 ⎢
⎢
3
t3
⎢
W ⎣⎢ –35 − 3M
x2
s1 s2 s3
3
1 0
4
4
–40 – 4 M
t3 W
400 ⎤ 400
3
0 1 0 0 0
500 ⎥⎥ 125
0 0 –1 1 0
250 ⎥ 125
⎥ 2
0 0 M 0 1 –250M ⎥
⎦
0
0 0
260
ISM: Introductory Mathematical Analysis
x1 x2 s1
s1 ⎡ – 1 0 1
⎢ 4
s2 ⎢ 0 0 0
⎢
x2 ⎢ 3 1 0
4
⎢
W ⎢ –5 0 0
⎣
x1 x2 s1
s2 s3
t3
0
3
4
–
3
4
1
1
–1
0
–
1
4
1
4
0 –10 10 + M
Section 7.6
W
⎤ 850
⎥ 3
0 250 ⎥ 250
⎥
0 125
2 ⎥
⎥
1 2500 ⎥
⎦
0
425
2
s2 s3 P
s1 ⎡ – 1 0 1 – 3 0 0 25 ⎤
4
⎢ 4
⎥
s3 ⎢ 0 0 0 1 1 0 250 ⎥
⎢
⎥
x2 ⎢ 34 1 0 14 0 0 125 ⎥ 500
3
⎢
⎥
P ⎢ –5 0 0 10 0 1 5000 ⎥
⎣
⎦
x1 x2 s1 s2 s3 P
200 ⎤
s1 ⎡0 1 1 – 2 0 0
3
3
3 ⎥
⎢
s3 ⎢0 0 0
1 1 0
250 ⎥
⎢
⎥
500
1 0 0
x1 ⎢ 1 4 0
3
3
3 ⎥
⎢
⎥
⎥
P ⎢0 20
0 35
0 1 17,500
3
3
3 ⎥⎦
⎢⎣
500
2
= 166 Standard and 0 Executive
3
3
bookcases. Since an integer answer is preferable, note that x1 = 167, x2 = 0 does not satisfy the constraint
3x1 + 4 x2 ≤ 500, while x1 = 166, x2 = 0 satisfies all of the constraints. Thus the company should produce
166 Standard and 0 Executive bookcases each week.
This table indicates that, to maximize profit, the company should produce
14. Let x, y and z denote the numbers of units of products X, Y, and Z produced each week, respectively. We want to
maximize the profit function P = 50x + 60y + 75z subject to
x + 2 y + 2 z ≤ 40,
x + y + 2 z ≤ 30,
z ≥ 5,
x, y, z ≥ 0.
The artificial objective function is W = P – Mt3 .
x
y
z s1 s2 s3 t3 W
2
2 1 0 0 0 0 40 ⎤
⎡ 1
⎢ 1
1
2 0 1 0 0 0 30 ⎥⎥
⎢
⎢ 0
0
1 0 0 –1 1 0 5⎥
⎢
⎥
–50 –60 –75 0 0 0 M 1 0 ⎥
⎣⎢
⎦
x
y
z
s1
2
2
1
s1 ⎡ 1
⎢ 1
1
2
0
s2 ⎢
⎢
0
0
1
0
t3
⎢
W ⎣⎢ –50 –60 –75 – M 0
s2 s3 t3 W
0 0 0 0
40 ⎤ 20
1 0 0 0 30 ⎥⎥ 15
0 –1 1 0
5 ⎥ 5
⎥
0 M 0 1 –5M ⎥
⎦
261
Chapter 7: Linear Programming
x
y
2
s1 ⎡ 1
⎢ 1
1
s2 ⎢
⎢
0
0
z
⎢
W ⎣⎢ –50 –60
z s1 s2
0 1 0
ISM: Introductory Mathematical Analysis
s3
2
t3
–2
0 0 1
2
–2
1 0 0
–1
1
0 0 0 –75 75 + M
W
0
30 ⎤ 15
0 20 ⎥⎥ 10
0 5 ⎥
⎥
1 375⎥
⎦
x
y
z s1 s2 s3 P
1
0 1 –1 0 0 10 ⎤ 10
⎡ 0
s1 ⎢
⎥
1
1
0 0 12 1 0 10 ⎥ 20
2
s3 ⎢ 2
⎢ 1
⎥
1
1 0 12 0 0 15 ⎥ 30
z⎢ 2
2
⎢
⎥
P ⎢ – 25 – 45 0 0 75 0 1 1125⎥
2
2
2
⎣
⎦
x
y z s1 s2 s3 P
0
1 0 1 –1 0 0 10 ⎤
⎡
y
⎢ 1
⎥
0 0 – 12 1 1 0
5 ⎥ 10
s3 ⎢ 2
⎢
⎥
1
z⎢ 1
⎥ 20
0
1
–
1
0
0
10
2
⎢ 2
⎥
25
45
P ⎢⎢ – 2 0 0 2 15 0 1 1350 ⎥⎥
⎣
⎦
x y z s1 s2 s3 P
0
1 0 1 –1 0 0
10 ⎤
⎡
y
⎢ 1 0 0 –1 2 2 0
10 ⎥⎥
x⎢
5⎥
z ⎢ 0 0 1 0 0 –1 0
⎢
⎥
P ⎢⎣ 0 0 0 10 40 25 1 1475⎥⎦
The production order should be 10 units of X, 10 units of Y, and 5 units of Z for a maximum profit of $1475
15. Suppose I is the total investment. Let x1 , x2 , and x3 be the proportions invested in A, AA, and AAA bonds,
respectively. If Z is the total annual yield expressed as a proportion of I, then ZI = 0.08 x1I + 0.07 x2 I + 0.06 x3 I ,
or equivalently, Z = 0.08 x1 + 0.07 x2 + 0.06 x3 . We want to maximize Z subject to
x1 + x2 + x3 = 1,
x2 + x3 ≥ 0.50,
x1 + x2 ≤ 0.30,
x1 , x2 , x3 ≥ 0.
The artificial objective function is W = Z – Mt1 – Mt2 .
x1
x2
x3 s2 s3 t1 t2 W
1
1
0 0 1 0 0 1 ⎤
⎡ 1
⎢ 0
1
1
–1 0 0 1 0 0.5⎥⎥
⎢
⎢ 1
1
0
0 1 0 0 0 0.3⎥
⎢
⎥
⎢⎣ –0.08 –0.07 –0.06 0 0 M M 1 0 ⎥⎦
x1
x2
x3
s2 s3 t1 t2 W
1
1
1
0 0 1 0 0
1 ⎤ 1
⎡
t1
⎢
0
1
1
–1 0 0 1 0
0.5 ⎥⎥ 0.5
t2 ⎢
1
1
0
0 1 0 0 0
0.3 ⎥ 0.3
s3 ⎢
⎢
⎥
W ⎢⎣ –0.08 – M –0.07 – 2M –0.06 – 2 M M 0 0 0 1 –1.5M ⎥⎦
262
ISM: Introductory Mathematical Analysis
Section 7.7
x1
x2
x3
s2
s3
t1
0
0
1
0
–1
1
t1 ⎡
⎢
–1
0
1
–1
–1
0
t2 ⎢
⎢
1
1
0
0
1
0
x2
⎢
W ⎣⎢ –0.01 + M 0 –0.06 – 2 M M 0.07 + 2 M 0
x1
1
⎡
t1
⎢
–1
x3 ⎢
⎢
1
x2
⎢
–0.07
–M
W ⎢⎣
x2 x3
0 0
0 1
1 0
0 0
x1
x2
x3
0
–1
0
t1 ⎡
⎢
1
1
x3 ⎢0
1
0
x1 ⎢1
⎢
W ⎢⎣0 0.07 + M 0
x1
x2 x3 s2
s2
1
s3 t1
0 1
t2 W
0 0
0.7
W
0
0.5
⎤ 0.7
⎥
1 0
0.2
⎥ 0.2
⎥
0 0
0.3
⎥
0 1 0.021 – 0.9 M ⎥
⎦
t2
–1
⎤ 0.5
⎥
–1
–1 0
1
0
0.2
⎥
⎥ 0.3
0
1
0
0
0
0.3
⎥
–0.06 – M 0.01 0 0.06 + 2 M 1 0.033 – 0.5M ⎥
⎦
s2
s3
t1
t2
W
1
–1
1
–1
0
0.2
⎤ 0.2
⎥
–1
0
0
1
0
0.5
⎥
⎥
0
1
0
0
0
0.3
⎥
–0.06 – M 0.08 + M 0 0.06 + 2M 1 0.054 – 0.2 M ⎥
⎦
s3
t1 t2 W
–1 0 1
–1
1 –1 0
0.2 ⎤
s2 ⎡0
⎢0
0 1 0
–1
1 0 0
0.7 ⎥⎥
x3 ⎢
1 0 0
1
0 0 0
0.3⎥
x1 ⎢ 1
⎢
⎥
W ⎢⎣0 0.01 0 0 0.02 0.06 + M M 1 0.066 ⎥⎦
For the above table, t1 = t2 = 0. Thus W = Z.
The fund should put 30% in A bonds, 0% in AA, and 70% in AAA for a yield of 6.6%.
Problems 7.7
x1 x2 s1 s2 t1
t2 W
1. ⎡
1 −1 –1 0
1 0 0
⎢
1 0
⎢ 2 1 0 –1 0
⎢2 5 0 0 M M 1
⎣
x1
x2 s1 s2 t1
⎡
t1
1
−1 –1 0 1
⎢
t2 ⎢ 2
1 0 –1 0
⎢
W ⎢⎣ 2 − 3M 5 M M 0
x1
x2
s1
s2
3
1
t1 ⎡0
−2
–1
2
⎢
⎢
1
x1 ⎢1
0
– 12
2
W ⎢⎢0 4 + 3 M M 1 − 1 M
2
2
⎣
7⎤
⎥
9⎥
0⎥
⎦
t2 W
⎤7
⎥
1 0
9 ⎥ 92
⎥
0 1 –16M ⎥
⎦
t1
t2
W
0 0
7
1
− 12
0
1
2
−1 + 32
0
⎤5
⎥
⎥
9
0
2
⎥
⎥
1 –9 − 52 M ⎥
⎦
0
M
5
2
263
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
x1 x2 s1 s2
t1
⎡
s2 0 −3 –2 1
2
⎢
x1 ⎢ 1 −1 −1 0
1
⎢
W 0 7 2 0 –2 + M
⎣
The minimum is Z = 14 when
t2 W
5⎤
⎥
0 0
7⎥
M 1 –14 ⎥
⎦
x1 = 7, x2 = 0.
−1 0
x1 x2 s1 s2 t1
t2 W
2. ⎡
2 2 –1 0
1 0 0 1⎤
⎢
⎥
1 0 2⎥
⎢ 1 3 0 –1 0
⎢ 8 12 0 0 M M 1 0 ⎥
⎣
⎦
x1
x2
x3 s2 s3 t1 t2
W
1
t1 ⎡ 2
2
–1 0 1 0 0
1 ⎤2
⎢
⎥
t2 ⎢ 1
3
0 –1 0 1 0
2 ⎥ 23
W ⎢8 – 3M 12 – 5M M M 0 0 1 –3M ⎥
⎣
⎦
x1
x2
s1
s2
t1
t2 W
⎡
1
x2 ⎢
⎢
t2 ⎢ –2
W⎢
⎢ –4 + 2M
⎣⎢
x1 x2 s1
⎡
1 0
x2 ⎢
s1 ⎢ –
0 1
⎢
W⎢ 4 0 0
⎣
1
3
4
3
1
– 12
0
0
3
2
–1
0 6 – 32 M
s2
t1
– 13
– 23
0
–1
M
1
2
– 23
–6 + 52 M
t2 W
1
3
2
3
0
0
⎤
⎥
⎥1
1 0
⎥3
⎥
0 1 –6 – 12 M ⎥
⎦⎥
1
2
1
2
0 0
2⎤
3⎥
1⎥
3⎥
1 –8⎥
⎦
2
The minimum is Z = 8 when x1 = 0, x2 = .
3
3.
4 M
–4 + M
x1 x2
x3 s
t W
1
–1
–1
–1
1
0 18⎤
⎡
⎢12 6 3 0 M 1 0 ⎥
⎢⎣
⎦⎥
x1
x2
x3
s t W
1
–1
–1
–1
1 0
18 ⎤ 18
⎡
t
⎢12 – M 6 + M 3 + M M 0 1 –18M ⎥
W ⎣⎢
⎦⎥
x1 x2 x3 s
t W
1 0
18⎤
x1 ⎡ 1 –1 –1 –1
⎢0 18 15 12 –12 + M 1 –216 ⎥
W ⎣⎢
⎦⎥
The minimum is Z = 216 when x1 = 18, x2 = 0, x3 = 0 .
264
ISM: Introductory Mathematical Analysis
Section 7.7
x1 x2 x3 s t W
4. ⎡1 2 –1 –1 1 0 4 ⎤
⎢1 1 2 0 M 1 0 ⎥
⎢⎣
⎦⎥
x1
x2
x3
s t W
2
–1
–1 1 0
4 ⎤2
t ⎡ 1
⎢1 – M 1 – 2M 2 + M M 0 1 –4 M ⎥
W ⎢⎣
⎥⎦
1 0
x2 ⎡ 1 1 – 1 – 1
2⎤
2
2
2
⎢2
⎥
5
1 – 1 + M 1 –2 ⎥
⎢1 0
2
2
2
W ⎣⎢ 2
⎦⎥
The minimum is Z = 2 when x1 = 0, x2 = 2, x3 = 0 .
5. We write the second constraint as – x1 + x3 ≥ 4.
x1 x2 x3 s1 s2 s3 t2 W
⎡ 1 1 1 1 0 0 0 0 6⎤
⎢ –1 0 1 0 –1 0
1 0 4 ⎥⎥
⎢
⎢ 0 1 1 0 0 1 0 0 5⎥
⎢
⎥
⎢⎣ 2 3 1 0 0 0 M 1 0 ⎥⎦
x1 x2
x3 s1 s2 s3 t2 W
1
1
1
1 0 0 0 0
6 ⎤6
s1 ⎡
⎢ –1
0
1
0 –1 0 1 0
4 ⎥⎥ 4
t2 ⎢
1
1
0 0 1 0 0
5 ⎥5
s3 ⎢ 0
⎢
⎥
W ⎢⎣ 2 + M 3 1 – M 0 M 0 0 1 –4 M ⎥⎦
x1 x2 x3 s1 s2 s3
t2 W
–1 0 2 ⎤
s1 ⎡ 2 1 0 1 1 0
⎢ –1 0 1 0 –1 0
1 0 4 ⎥⎥
x3 ⎢
–1 0
1⎥
s3 ⎢ 1 1 0 0 1 1
⎢
⎥
W ⎣⎢ 3 3 0 0 1 0 –1 + M 1 –4 ⎦⎥
The minimum is Z = 4 when x1 = 0, x2 = 0, x3 = 4 .
6.
x1
⎡3
⎢0
⎢
⎢1
⎢
5
⎣⎢
x2 x3 s1 s2
1 −1 1 0
2
1
1
t3 W
0 0 4⎤
2 0 1 0 0 0 5⎥⎥
1 0 0 −1 1 0 2 ⎥
⎥
3 0 0 0 M 1 0⎥
⎦
x1
3
x2
1
s3
0
x3 s1 s2 s3 t3 W
−1 1 0 0 0 0
4⎤ 4
s1 ⎡
⎢
0
2
2 0 1 0 0 0
5⎥⎥ 52
s2 ⎢
1
1
1 0 0 −1 1 0
2⎥ 2
t3 ⎢
⎢
⎥
W ⎣⎢5 − M 1 − M 3 − M 0 0 M 0 1 −2M ⎦⎥
265
Chapter 7: Linear Programming
x1
s1 ⎡ 2
⎢
s2 ⎢ −2
x2 ⎢ 1
⎢
W ⎣⎢ 4
x2 x3 s1 s2
0 −2 1 0
ISM: Introductory Mathematical Analysis
s3
1
t3 W
−1 0
2⎤
−2 0
0 0 1 2
1⎥⎥
1 0 0 −1
1 0 2⎥
⎥
2 0 0 1 −1 + M 1 −2 ⎥
⎦
0
1
0
The minimum is Z = 2 when x1 = 0, x2 = 2, and x3 = 0.
7.
x1 x2 x3
1
⎡1 2
⎢0 1 1
⎢
⎢1 1 0
⎢
1 –1 –3
⎣⎢
s3
0
x1
1
⎡
t1
⎢ 0
t2 ⎢
s3 ⎢ 1
⎢
W ⎢⎣1 – M
x1
t1 ⎡ 1
⎢
x2 ⎢ 0
s3 ⎢ 1
⎢
W ⎢⎣1 – M
x1 x2
x1 ⎡1 0
⎢
x2 ⎢ 0 1
s3 ⎢ 0 0
⎢
W ⎣⎢ 0 0
x2
2
x1
x1 ⎡ 1
⎢
x3 ⎢ 0
s3 ⎢ 0
⎢
– Z ⎣⎢ 0
t1 t2 W
1 0 0 4⎤
0 0
1 0 1⎥⎥
1 0 0 0 6⎥
⎥
0 M M 1 0⎥
⎦
x3
1
s3 t1 t2 W
0 1 0 0
4 ⎤2
1
1
0 0 1 0
1 ⎥⎥ 1
1
0
1 0 0 0
6 ⎥6
⎥
–1 – 3M –3 – 2M 0 0 0 1 –5M ⎥
⎦
x2
x3
s3 t1
t2
W
0
–1
0 1
–2
0
2 ⎤2
1
1
0 0
1
0
1 ⎥⎥
0
–1
1 0
–1
0
5 ⎥5
⎥
0 –2 + M 0 0 1 + 3M 1 1 – 2 M ⎥
⎦
x3 s3
t1
t2
W
–1 0
1
–2
0 2⎤
1 0
0
1
0 1 ⎥⎥ 1
0 1
–1
1
0 3⎥
⎥
–1 0 –1 + M 3 + M 1 –1⎥
⎦
x2 x3 s3 – Z
1 0 0 0 3⎤
1 1 0 0 1⎥⎥
0 0 1 0 3⎥
⎥
1 0 0 1 0⎥
⎦
The minimum is Z = 0 when x1 = 3, x2 = 0, x3 = 1 .
266
ISM: Introductory Mathematical Analysis
8.
x1
x2
s1
t1
t2
Section 7.7
W
⎡ −1 1 −1 1 0 0
⎢
⎢1 1 0 0 1 0
⎢ 1 −1 0 M M 1
⎣
x1
x2
s1 t1
4⎤
⎥
1⎥
0⎥
⎦
t2 W
−1 1 0 0
t1 ⎡ −1
1
4 ⎤4
⎢
⎥
t2 ⎢ 1
1
0 0 1 0
1 ⎥1
⎢
⎥
W ⎢ 1 −1 − 2 M M 0 0 1 −5M ⎥
⎣
⎦
x1
x2 s1 t1
t2
W
t1 ⎡ −2
⎢
x2 ⎢ 1
⎢
W ⎣2 + 2M
⎤
⎥
1 0 0
1
0
1 ⎥
0 M 0 1 + 2M 1 1 − 3M ⎥
⎦
Since all of the indicators in the last table are positive, but the artificial variable t1 is 3, the feasible region is
empty. (This can also be seen graphically.)
9.
0 −1 1
x1 x2 x3 s1 s2 t1
⎡ 1 1 1 –1 0
1
⎢
⎢ –1 2 1 0 –1 0
⎢ 1 8 5 0 0 M
⎣
x1
x2
x3
1
1
t1 ⎡ 1
⎢
t2 ⎢ –1
2
1
⎢
W 1 8 – 3M 5 – 2 M
⎣
x1
x2
x3
−1
0
3
t2 W
0 0 8⎤
⎥
1 0 2⎥
M 1 0⎥
⎦
s1 s2 t1 t2 W
–1 0 1 0 0
0
M
s1
⎤8
⎥
–1 0 1 0
2 ⎥1
M 0 0 1 –10M ⎥
⎦
s2
t1
t2
W
8
1
1
⎤ 14
t1 ⎡ 3
0
–1
1
– 12
0
7
2
2
⎢ 2
⎥ 3
⎢
⎥
1
1
x2 ⎢ – 12
1
0
– 12
0
0
1
2
2
⎥
W ⎢5 – 3 M 0 1 – 1 M M 4 – 1 M 0 –4 + 3 M 1 –8 – 7 M ⎥
⎢⎣
⎥⎦
2
2
2
2
x1 x2 x3
s1
s2
t1
t2
W
⎤
x1 ⎡
1
2
1
2
– 13
0 14
⎢1 0 3 – 3 3
3
3 ⎥ 14
⎢
⎥
1
1
x2 ⎢ 0 1 23 – 13 – 13
0 10
5
3
3
3 ⎥
⎢
⎥
10
7
2
⎥
– 10
+ M – 73 + M 1 – 94
W ⎢⎢⎣ 0 0 – 3
3
3
3
3 ⎥⎦
x1 x2 x3 s1
s2
t1
t2 W
x1 ⎡
1
1
1 – 12 0 – 12
– 12 0
3⎤
2
2
⎢
⎥
3 1 –1 –1
1
1 0
⎥
x3 ⎢0
5
2
2
2
2
2
⎢
⎥
W ⎢0
1 0
3
2 –3 + M –2 + M 1 –28⎥
⎣
⎦
The minimum is Z = 28 when x1 = 3, x2 = 0, x3 = 5 .
267
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
x1 x2 x3 s1 s2 t2 W
⎡ 1 –1 –1 1 0 0 0 3⎤
⎥
10. ⎢
⎢ 1 –1 1 0 –1 1 0 3⎥
⎢4 4 6 0 0 M 1 0⎥
⎣
⎦
x1
x2
x3
s1 s2 t2 W
⎡
s1
1
–1
–1
1 0 0 0
3 ⎤3
⎢
⎥
–1
1
0 –1 1 0
3 ⎥3
t2 ⎢ 1
W ⎢ 4 – M 4 + M 6 – M 0 M 0 1 –3M ⎥
⎣
⎦
Here we choose t2 as the departing variable.
x1 x2 x3 s1 s2
t2 W
⎡
s1 0 0 –2 1 1
–1 0
0⎤
⎢
⎥
x1 ⎢ 1 –1 1 0 –1
1 0
3⎥
W ⎢0 8 2 0 4 –4 + M 1 –12 ⎥
⎣
⎦
Thus Z has a minimum value of 12 when x1 = 3, x2 = 0, x3 = 0.
If we choose s1 as the departing variable, then
x1
⎡
s1
1
⎢
t2 ⎢ 1
W ⎢4 – M
⎣
x1 x2
⎡
x1 1 –1
⎢
t2 ⎢ 0 0
W ⎢0 8
⎣
x1 x2
x1 ⎡ 1 –1
⎢
x3 ⎢0 0
⎢
W ⎢⎣0 8
x2
x3
s1 s2
t2 W
3 ⎤3
⎥
–1
1
0 –1 1 0
3 ⎥3
4 + M 6 – M 0 M 0 1 –3M ⎥
⎦
x3
s1
s2 t2 W
–1
1
0 0 0 3 ⎤
⎥
2
–1
–1 1 0 0 ⎥ 0
10 – 2 M –4 + M M 0 1 –12 ⎥
⎦
x3 s1 s2
t2 W
1 –1
1 0
0
3⎤
2
2
2
⎥
1 0
⎥
1 – 12 – 12
0
2
⎥
0
1
5 –5 + M 1 –12 ⎥
⎦
The minimum is Z = 12 when x1 = 3, x2 = 0, x3 = 0 .
–1
–1
1
0
0 0
268
ISM: Introductory Mathematical Analysis
Section 7.7
11. Let x1 , x2 , and x3 denote the annual numbers of barrels of cement produced in kilns that use device A, device B,
and no device, respectively. We want to minimize the annual emission control cost C (C in dollars) where
1
2
C = x1 + x2 + 0 x3 subject to
4
5
x1 + x2 + x3 = 3,300, 000,
1
1
x1 + x2 + 2 x3 ≤ 1, 000, 000,
2
4
x1 , x2 , x3 ≥ 0.
x1 x2 x3 s2
⎡
⎢1
⎢1
⎢2
⎢1
⎣⎢ 4
t1 W
⎤
1 0 3,300, 000 ⎥
2 1 0 0 1, 000, 000 ⎥⎥
⎥
0 0 M 1
0⎥
⎦
x2
x3 s2 t1 W
1 1 0
1
4
2
5
x1
⎡
⎤
1
1 0 1 0
3,300, 000 ⎥ 3,300, 000
t1 ⎢ 1
1
s2 ⎢⎢ 12
2 1 0 0
1, 000, 000 ⎥⎥ 500, 000
4
W ⎢ 1 – M 2 – M – M 0 0 1 –3,300, 000M ⎥
⎢⎣ 4
⎥⎦
5
x1
x2
x3 s2 t1 W
7
⎡ 3
0 – 12 1 0
2,800, 000 ⎤ 3, 200, 000
8
t1 ⎢ 4
⎥
⎥
1
1
x3 ⎢⎢ 14
1
0
0
500,
000
⎥ 4, 000, 000
8
2
W ⎢ 1 – 3 M 2 – 7 M 0 1 M 0 1 –2,800, 000 M ⎥
⎢⎣ 4 4
⎥⎦
5 8
2
x1
x2 x3 s2
t1
W
⎡ 6
⎤
8
1 0 – 74
0 3, 200, 000 ⎥ 11,200,000
x2 ⎢ 7
7
3
⎢
⎥
x3 ⎢ 17
0 1 74
– 17
0
100, 000 ⎥ 700, 000
⎥
W ⎢ 13
8
16 + M 1 –1, 280, 000 ⎥
⎢–
0
0
–
35
35
⎢⎣ 140
⎥⎦
x1 x2 x3 s2 – C
⎡
⎤
x2 ⎢ 0 1 –6 –4 0 2, 600, 000 ⎥
700, 000 ⎥
x1 ⎢ 1 0 7 4 0
⎢
⎥
3 1 –1, 215, 000 ⎥
– C ⎢ 0 0 13
20
5
⎣⎢
⎦⎥
Thus the minimum value of C is 1,215,000 when x1 = 700, 000, x2 = 2, 600, 000, x3 = 0.
The plant should install device A on kilns producing 700,000 barrels annually, and device B on kilns producing
2,600,000 barrels annually.
269
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
12. Let x1 = number of type A trucks rented,
x2 = number of type B trucks rented.
We want to minimize C = 0.40 x1 + 0.60 x2 subject to
2 x1 + 2 x2 ≥ 12,
x1 + 3 x2 ≥ 12,
x1 , x2 ≥ 0.
x1
x2
s1
s2
t1
t2 W
⎡ 2
2 –1 0
1 0 0 12 ⎤
⎢
⎥
3 0 –1 0
1 0 12 ⎥
⎢ 1
⎢ 0.40 0.60 0 0 M M 1 0 ⎥
⎣
⎦
x1
x2
s1 s2 t1 t2 W
⎡
⎤
t1 ⎢ 2
2
–1 0 1 0 0
12 ⎥ 6
t2 ⎢ 1
3
0 –1 0 1 0
12 ⎥ 4
⎢
⎥
W ⎢ 2 – 3M 3 – 5M M M 0 0 1 –24 M ⎥
5
⎣⎢ 5
⎦⎥
x1
x2 s1
s2
t1
t2
W
2
2
⎡ 4
⎤3
0 −1
1
0
4
−3
3
t1 ⎢ 3
⎥
⎢
⎥
1
1
1
1 0
0
0
4
x2 ⎢ 3
−3
⎥ 12
3
W ⎢ 1 − 4 M 0 M 1 − 2 M 0 − 1 + 5 M 1 − 12 − 4 M ⎥
⎢⎣ 5 3
⎥⎦
5 3
5 3
5
x1 x2 s1
s2
t1
t2 W
⎡
3
3
1
– 12 0 3⎤
x1 ⎢ 1 0 – 4
2
4
⎥
⎢
1
1
1
1
–2
–4
0 3⎥⎥
x2 ⎢ 0 1
4
2
W ⎢0 0
3
3 + M – 1 + M 1 –3⎥
1
– 20
⎢⎣
⎥⎦
20
10
10
The minimum value of C is 3 when x1 = 3 and x2 = 3. They should rent 3 of type A and 3 of type B. The cost per
mile is $3.00.
13. Let x1 = number of DVD players shipped from Akron to Columbus,
x2 = number of DVD players shipped from Springfield to Columbus,
x3 = number of DVD players shipped from Akron to Dayton,
x4 = number of DVD players shipped from Springfield to Dayton.
We want to minimize C = 5 x1 + 3x2 + 7 x3 + 2 x4 subject to
x1 + x2
x3 + x4
x1 + x3
x2 + x4
x1 , x2 , x3 , x4
= 150,
= 150,
≤ 200,
≤ 150,
≥ 0.
270
ISM: Introductory Mathematical Analysis
x1
⎡1
⎢0
⎢
⎢1
⎢
⎢0
⎢5
⎢⎣
x2 x3 x4 s3 s4
1 0 0 0 0
0
1 1 0 0
0
1 0
1 0
t1
1
t2 W
0 0 150 ⎤
0
1 0 150 ⎥⎥
0 0 0 200 ⎥
⎥
0 0 0 150 ⎥
M M 1
0 ⎥⎥
⎦
x3
x4 s3 s4 t1 t2 W
0
0 0 0 1 0 0
1 0
1 0
1
3 7 2 0 0
x1
x2
t1 ⎡
1
1
t2 ⎢
0
0
1
1
⎢
s3 ⎢
1
0
1
0
⎢
0
1
0
1
s4 ⎢
⎢
W ⎣⎢5 − M 3 − M 7 − M 2 − M
x1
x2
t1 ⎡
1
1
⎢
0
0
x4 ⎢
⎢
1
0
s3
⎢
0
1
s4 ⎢
W ⎢⎢⎣5 − M 3 − M
x1
t1 ⎡
1
⎢
0
x4 ⎢
⎢
s3
1
⎢
0
x2 ⎢
W ⎢⎢⎣5 − M
Section 7.7
0 0 0
1 0
1 0 0 0 0
0
1 0 0 0
0 0 0 0
1
x3 x4 s3 s4 t1
0 0 0 0 1
t2 W
0 0
1 1 0 0 0
1 0
1 0
1 0 0
−1 0 0
0 0
−1 0
1 0
5 0 0 0 0 −2 + M
1
x2
0
x3 x4 s3
1 0 0
s4 t1
−1 1
t2 W
1 0
0
0
1 1 0
1 0 1
0 0
0 0
1 0
0 0
1
−1 0 0
0 8− M
1 0 −1 0
0 0 −3 + M
0
1 1
150 ⎤
150 ⎥⎥ 150
200 ⎥
⎥
150 ⎥ 150
−300M ⎥⎥
⎦
150 ⎤ 150
150 ⎥⎥
200 ⎥
⎥
0⎥ 0
−300 − 150 M ⎥⎥
⎦
150 ⎤ 150
⎥
150 ⎥
200 ⎥ 200
⎥
0⎥
−300 − 150 M ⎥⎥
⎦
x1 x2 x3 x4 s3 s4
t1
t2 W
x1 ⎡ 1 0 1 0 0 −1
1
1 0
150 ⎤
x4 ⎢⎢ 0 0 1 1 0 0
0
1 0
150 ⎥⎥
s3 ⎢ 0 0 0 0 1 1
−1
−1 0
50 ⎥
⎢
⎥
x2 ⎢ 0 1 −1 0 0 1
0
−1 0
0⎥
W ⎢⎢⎣ 0 0 3 0 0 2 −5 + M −4 + M 1 −1050 ⎥⎥⎦
The retailer should ship as follows: to Columbus, 150 from Akron and 0 from Springfield; to Dayton, 0 from
Akron and 150 from Springfield. The transportation cost is $1050.
If s4 is chosen as the departing variable in the second table, the result is the same, although the final table is
different:
x1 x2 x3 x4 s3 s4
t1
t2 W
x1 ⎡ 1 1 0 0 0 0
1
0 0
150 ⎤
⎢
x3 ⎢ 0 −1 1 0 0 −1
0
1 0
0 ⎥⎥
s3 ⎢ 0 0 0 0 1 1
−1
−1 0
50 ⎥
⎢
⎥
x4 ⎢ 0 1 0 1 0 1
0
0 0
150 ⎥
W ⎢⎢ 0 3 0 0 0 5 −5 + M −7 + M 1 −1050 ⎥⎥
⎣
⎦
271
Chapter 7: Linear Programming
14. Let x A
xB
yA
yB
ISM: Introductory Mathematical Analysis
= number of alternators from supplier X to plant A
= number of alternators from supplier X to plant B
= number of alternators from supplier Y to plant A
= number of alternators from supplier Y to plant B
We want to minimize C = 300 x A + 320 xB + 340 y A + 280 yB
x A + y A = 7000
xB + y B = 5000
x A + xB ≥ 3000
x A + xB ≤ 5000
y A + yB ≥ 7000
x A , xB , y A , y B ≥ 0
x A xB y A y B s3 s4 s5 t1 t2 t3 t5 W
0
1
0 0 0 0
1 0 0 0 0
⎡ 1
⎢ 0
1
0
1 0 0 0 0
1 0 0 0
⎢
−
1
1
0
0
1
0
0
0
0
1 0 0
⎢
1
0
0 0 1 0 0 0 0 0 0
⎢ 1
⎢ 0
0
1
1 0 0 −1 0 0 0
1 0
⎢300 320 340 280 0 0 0 M M M M 1
⎢⎣
xA
xB
yA
yB s3
1
0
1
0 0
t1 ⎡
0
1
0
1
0
t2 ⎢
⎢
1
1
0
0 −1
t3 ⎢
1
1
0
0 0
s4 ⎢
0
0
1
1 0
t5 ⎢
⎢
W ⎣⎢300 − 2M 320 − 2 M 340 − 2 M 280 − 2M M
xA
1
t1 ⎡
0
yB ⎢
⎢
1
t3 ⎢
1
s4 ⎢
⎢
0
t5
⎢
W ⎣⎢300 − 2 M
xB
yA
0
1
1
0
1
0
1
0
−1
1
40 340 − 2 M
yB s3
0 0
1 1
0 −1
0 0
0 0
0 M
s4 s5
0 0
0 0
0 0
1 0
0 −1
0 M
xA
xB
y A yB
s3 s4
−1
1 0
1 0
t1 ⎡ 0
1
0 1
0 0
yB ⎢0
⎢
−1 0
1
0 0
xA ⎢ 1
0
0 0
1 1
s4 ⎢ 0
−1
1 0
0 0
t5 ⎢ 0
⎢
W ⎢⎣ 0 −260 + 2M 340 − 2 M 0 300 − M 0
x A xB
t1 ⎡ 0 0
yB ⎢0 1
⎢
xA ⎢ 1 1
s4 ⎢ 0 0
y A ⎢ 0 −1
⎢
W ⎢⎣ 0 80
yA
0
0
0
0
1
0
yB
s3
0
1
1
0
0
−1
0
1
0
0
0 300 − M
s4
s5
0
1
0
0
0
0
1
0
−1
0
0 340 − M
s5
0
0
0
0
−1
M
subject to
7000 ⎤
5000 ⎥
⎥
3000 ⎥
5000 ⎥
7000 ⎥
0 ⎥⎥⎦
s4 s5
0 0
0 0
0 0
1 0
0 −1
0 M
t1
t2
1
0
0
1
0
0
0
0
−1
0
0 −280 + 2 M
t1
1
0
0
0
0
0
t3
0
0
1
0
0
0
t2
0
1
0
0
0
0
t5
0
0
0
0
1
0
t3
0
0
1
0
0
0
t5
0
0
0
0
1
0
W
0
7000 ⎤
0
5000 ⎥ 5000
⎥
0
3000 ⎥
0
5000 ⎥
0
7000 ⎥ 7000
1 −22, 000 M ⎥⎦⎥
W
0
7000 ⎤ 7000
0
5000 ⎥
⎥
0
3000 ⎥ 3000
0
5000 ⎥ 5000
0
2000 ⎥
1 −1, 400, 000 − 12, 000M ⎦⎥⎥
t1
t2
t3 t5 W
−1 0 0
1
0
4000 ⎤ 4000
0
1
0 0 0
5000 ⎥
⎥
0
0
1 0 0
3000 ⎥
−1 0 0
0
0
2000 ⎥
−1
0
0 1 0
2000 ⎥ 2000
0 −280 + 2M −300 + 2 M 0 1 −2,300, 000 − 6000M ⎥⎦⎥
t1 t2
t3
t5 W
1 1
2000 ⎤ 2000
−1
−1 0
0 1
0
0 0
5000 ⎥
⎥
0 0
1
0 0
3000 ⎥
0 0
0 0
2000 ⎥ 2000
−1
0 −1
0
1 0
2000 ⎥
0 60 −300 + 2 M −340 + 2M 1 2,980, 000 − 2000 M ⎥⎥⎦
We choose t1 as the departing variable.
272
ISM: Introductory Mathematical Analysis
Section 7.7
x A xB y A yB s3 s4 s5
t1
t2 t3
t5 W
1
1
1 −1
2000 ⎤
−1 0
s3 ⎡ 0 0 0 0 1 0
0
1 0
0 0
5000 ⎥
yB ⎢0 1 0 1 0 0 0
⎢
⎥
1
1
1 0
5000 ⎥
−1 0
xA ⎢ 1 1 0 0 0 0
1 0
0⎥
−1
−1 0
s4 ⎢ 0 0 0 0 0 1 −1
−1 0
0
1 0
2000 ⎥
y A ⎢ 0 −1 1 0 0 0 −1
⎢
⎥
W ⎢⎣ 0 80 0 0 0 0 40 −300 + M −240 + M M −40 + M 1 −3,580, 000 ⎦⎥
The manufacturer should order 5000 alternators from X to be shipped to A, 2000 from Y to A, and 5000 from Y to
B. The minimum cost is $3,580,000. (Note that the same result is obtained if s4 is chosen as the departing
variable in the fifth table.)
15. a.
Roll width
⎧15" 3 2 1 0
⎨10"
0 1 3 4
⎩
Trim loss 3 8 3 8
b. We want to minimize L = 3x1 + 8 x2 + 3x3 + 8 x4 subject to
3x1 + 2 x2 + x3 ≥ 50,
x2 + 3x3 + 4 x4 ≥ 60,
x1 , x2 , x3 , x4 ≥ 0.
x1
⎡3
⎢
⎢0
⎢3
⎣
x2 x3 x4 s1
s2
t1
t2 W
0 0 50 ⎤
⎥
1 3 4 0 –1 0
1 0 60 ⎥
8 3 8 0 0 M M 1 0⎥
⎦
x1
x2
x3
x4
s1
2 1 0 –1
0
1
t1 ⎡ 3
2
1
0
⎢
t2 ⎢ 0
1
3
4
⎢
W 3 – 3M 8 – 3M 3 – 4 M 8 – 4 M
⎣
x1
x2
x3
x4
s1
⎡ 3
5
0
– 43
–1
t1 ⎢
3
1
4
1
0
x3 ⎢⎢ 0
3
3
W ⎢3 – 3M 7 – 5 M 0 4 + 4 M M
⎢⎣
3
3
x1 x2 x3 x4 s1
s2
t1
⎡1 5 0 – 4 – 1
1
1
x1 ⎢
9
9
3
9
3
⎢
1
4
1
0 –3
0
x3 ⎢0 3 1
3
W ⎢0 16 0 16
2 –1 + M
1
⎢⎣
3
3
3
x1 = 10, x2 = 0, x3 = 20, x4 = 0.
c.
s2
t1 t2 W
⎤ 50
⎥
0 –1 0 1 0
60 ⎥ 20
M M 0 0 1 –110 M ⎥
⎦
s2
t1
t2
W
–1
0
1 0 0
1
3
– 13
1 – 13 M
1
– 13
0
1
3
–1 + 43
0
t2 W
–
1
9
1
3
– 23 + M
90 in.
273
10 ⎤
⎥
0 20 ⎥⎥
⎥
1 –90 ⎥
⎦
0
50
⎤
⎥ 10
⎥
0
20
⎥
⎥
1 –60 – 30M ⎥
⎦
0
M
30
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
Principles in Practice 7.8
1. Let x1 , x2 , and x3 be the numbers respectively, of Type 1, Type 2, and Type 3 gadgets produced. The original
problem is to maximize
P = 300 x1 + 200 x2 + 200 x3 , subject to
300 x1 + 220 x2 + 180 x3 ≤ 60, 000,
20 x1 + 40 x2 + 20 x3 ≤ 2000,
3 x1 + x2 + 2 x3 ≤ 120,
and x1 , x2 , x3 ≥ 0.
The dual problem is to minimize
W = 60, 000 y1 + 2000 y2 + 120 y3 ,
subject to
300 y1 + 20 y2 + 3 y3 ≥ 300,
220 y1 + 40 y2 + y3 ≥ 200,
180 y1 + 20 y2 + 2 y3 ≥ 200,
and y1 , y2 , y3 ≥ 0.
2. Let x1 and x2 be the amounts, respectively of supplement 1 and supplement 2. The original problem is to
minimize C = 6 x1 + 2 x2 , subject to
20 x1 + 6 x2 ≥ 98,
8 x1 + 16 x2 ≥ 80,
and x1 , x2 ≥ 0.
The dual problem is to maximize
W = 98 y1 + 80 y2 , subject to
20 y1 + 8 y2 ≤ 6,
6 y1 + 16 y2 ≤ 2,
and y1 , y2 ≥ 0.
3. Let x1 , x2 , and x3 be the numbers, respectively, of devices 1, 2, and 3 produced.
The original problem is to maximize P = 30 x1 + 20 x2 + 20 x3 , subject to
30 x1 + 15 x2 + 10 x3 ≤ 300, 20 x1 + 30 x2 + 20 x3 ≤ 400, 40 x1 + 30 x2 + 25 x3 ≤ 600, and x1 , x2 , x3 ≥ 0.
The dual problem is to minimize W = 300 y1 + 400 y2 + 600 y3 , subject to
30 y1 + 20 y2 + 40 y3 ≥ 30, 15 y1 + 30 y2 + 30 y3 ≥ 20,
10 y1 + 20 y2 + 25 y3 ≥ 20,
and y1 , y2 , y3 ≥ 0.
The tablex to maximize Z = −W = −300 y1 − 400 y2 − 600 y3 follow.
y1 y2
y3 s1 s2 s3
t1 t2 t3 Z
1 0 0 0 30 ⎤
⎡ 30 20 40 −1 0 0
⎢ 15 30 30 0 −1 0 0
1 0 0 20 ⎥⎥
⎢
⎢ 10 20 25 0 0 −1 0 0
1 0 20 ⎥
⎢
⎥
300 400 600 0 0 0 M M M 1 0 ⎥
⎣⎢
⎦
274
ISM: Introductory Mathematical Analysis
y1
30
t1 ⎡
⎢
15
t2 ⎢
⎢
10
t3
⎢
Z ⎣⎢300 – 55M
Section 7.8
y2
20
y3
40
s1 s2 s3 t1 t2 t3 Z
–1 0 0 1 0 0 0
30
30
0
20
25
0
400 – 70 M
600 – 95M
M
30 ⎤
−1 0 0 1 0 0
20 ⎥⎥
0 −1 0 0 1 0
20 ⎥
⎥
M M 0 0 0 1 –70M ⎥
⎦
y1
y2
y3 s1
s2
s3 t1
t2
t3 Z
10
4
4
⎡ 10
⎤
−20
−3
0 −1
0 1
0 0
3
3
⎥
t1 ⎢
1
1
2
⎢ 1
⎥
−
1
1
0
0
0
0
0
y3 ⎢ 2
30
30
3
⎥
5
5
10
⎥
t3 ⎢ − 5
−
−
−
5
0
0
1
0
1
0
2
6
6
3
⎢
⎥
Z ⎢ 15
13
19
20
− M –200 + 25M 0 M 20 − 6 M M 0 −20 + 6 M 0 1 −400 − 3 M ⎥
⎢⎣ 2
⎥⎦
y1
y2
y3 s1
s2
s3
t1
t2
t3 Z
1
2
1
2
1
y1 ⎡1
⎤
−2
− 15
0 − 10
0
0 0
15
10
3
⎢
⎥
⎢0
⎥
1
1
1
1
1
− 10
2
1 20
0 − 20
0 0
y3 ⎢
10
2
⎥
⎥
7
7
25
1
1
t3 ⎢ 0
−
−
−
−
10
0
1
1
0
⎢
⎥
4
6
4
6
6
⎢
7M M
3 M −20 + 13 M 0 1 −400 − 25 M ⎥
1
−
+
M
M
−
0
200
10
0
20
Z ⎢
⎥⎦
4
6
4
6
6
⎣
y1
y2
y3
s1
s2 s3
t1
t2 t3 Z
15
3
3
5
s2 ⎡
⎤
−15
−4
−1 0 0
0
1 0
2
4
2
⎢
⎥
3
3
1
1
1
⎢
⎥
−
1
0
0
0
0
0
4
2
40
40
4
y3 ⎢
⎥
15
5
5
5
⎥
− 35
−
−
0
0
1
0
1
0
t3 ⎢
4
2
8
8
4
⎢
⎥
⎢
⎥
35
15
5
13
5
Z ⎢ −150 + 4 M 100 − 2 M 0 15 − 8 M 0 M −15 + 8 M M 0 1 −450 − 4 M ⎥
⎣
⎦
y1 y2 y3 s1 s2
s3
t1
t2
t3
Z
1
1
s2 ⎡ −10 0 0
−2
−1
1 −2
2
0
5 ⎤
2
⎢
⎥
1 0
1
1
1
2 ⎥
⎢ 4
−
−
0
1
0
0
15
15
15
15
3 ⎥
y3 ⎢ 3
⎥
1
2
1
2
1
y2 ⎢ − 7 1 0
− 12
0 − 15
0
0
⎢ 6
12
15
6 ⎥
⎢
⎥
− 20
+ M M − 40
+ M 1 − 1400
0 0 20
0 40
Z ⎢ − 100
⎥⎦
3
3
3
3
3
3
⎣
The t1 , t2 , and t3 columns are no longer needed.
y1 y2 y3
s1 s2
s3 Z
15
0 1 − 32 0
10 ⎤
s2 ⎡ 0 0 2
⎢
⎥
1
1⎥
y1 ⎢ 1 0 3 − 1 0
0
4
20
20
2⎥
⎢
7
3
3
1
⎢
⎥
y2 0 1
0 − 40 0
8
40
4⎥
⎢
Z ⎢ 0 0 25
5 0
15 1 −450 ⎥
⎣
⎦
From this table, the maximum profit of $450 corresponds to x1 = 5, x2 = 0, and x3 = 15.
The company should produce 5 of device 1 and 15 of device 3.
275
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
1. Minimize W = 5 y1 + 3 y2
subject to y1 − y2 ≥ 1
y1 + y2 ≥ 2
y1 , y2 ≥ 0
8. The second constraint can be written as
−8 x1 + 10 x2 ≥ −80.
Maximize W = −10 y1 − 80 y2
subject to −4 y1 − 8 y2 ≤ 5
3 y1 + 10 y2 ≤ 4
y1 , y2 ≥ 0
2. Minimize W = 3 y1 + 5 y2 subject to
9. The dual is: Maximize W = 2 y1 + 3 y2 subject to
Problems 7.8
2 y1 – y2
2 y1 + 4 y2
2 y2
y1 , y2
≥ 2,
≥ 1,
≥ –1,
≥ 0.
y1 – y2 ≤ 2,
– y1 + 2 y2 ≤ 2,
2 y1 + y2 ≤ 5,
y1 , y2 ≥ 0.
y1 y2 s1
s1 ⎡ 1 –1 1
⎢
s2 ⎢ –1 2 0
s3 ⎢ 2 1 0
⎢
W ⎢⎣ –2 –3 0
y1 y2 s1
s1 ⎡ 1 0 1
⎢ 2
y2 ⎢ – 1 1 0
⎢ 2
s3 ⎢ 52 0 0
⎢
⎢
W ⎢ – 72 0 0
⎣
y1 y2 s1
3. Maximize W = 8 y1 + 2 y2 subject to
y1 – y2
y1 + 2 y2
y1 + y2
y1 , y2
≤ 1,
≤ 8,
≤ 5,
≥ 0.
4. Maximize W = y1 + 2 y2 subject to
2 y1 + y2 ≤ 8,
2 y1 + 3 y2 ≤ 12,
y1 , y2 ≥ 0.
5. The second and third constraints can be written
as x1 − x2 ≤ −3 and − x1 − x2 ≤ −11. Minimize
W = 13 y1 – 3 y2 –11 y3 subject to
s1 ⎡ 0
⎢
y2 ⎢ 0
⎢
y1 ⎢ 1
⎢
W ⎢⎢ 0
⎣
– y1 + y2 – y3 ≥ 1,
2 y1 – y2 – y3 ≥ –1,
y1 , y2 , y3 ≥ 0.
6. The second constraint can be written as
− x1 + 2 x2 − x3 ≤ −6. Minimize W = 9 y1 – 6 y2
subject to
y1 – y2 ≥ 1,
y1 + 2 y2 ≥ –1,
y1 – y2 ≥ 4,
y1 , y2 ≥ 0.
0 1
1 0
0 0
0 0
s2
0
1
0
0
s3
0
0
1
0
W
0
0
0
1
2⎤
2 ⎥⎥ 1
5⎥ 5
⎥
0⎥
⎦
s2 s3 W
0 0 3⎤ 6
⎥
0 0 1⎥
⎥
1 0 4 ⎥ 85
⎥
⎥
0 1 3⎥
⎦
s3 W
1
2
1
2
– 12
3
2
s2
3
5
2
5
– 15
4
5
− 15
0
1
5
2
5
7
5
0
The minimum is Z =
0
1
11 ⎤
5⎥
9⎥
5⎥
8⎥
5⎥
43 ⎥
5 ⎥⎦
43
4
when x1 = 0, x2 = ,
5
5
7
x3 = .
5
10. The dual is: Maximize W = 28 y1 + 2 y2 + 16 y3
subject to
y1 + 2 y2 – 3 y3 ≤ 2,
4 y1 – y2 + 8 y3 ≤ 2,
y1 , y2 , y3 ≥ 0.
7. The first constraint can be written as
− x1 + x2 + x3 ≥ −3. Maximize W = –3 y1 + 3 y2
subject to
– y1 + y2 ≤ 4,
y1 – y2 ≤ 4,
y1 + y2 ≤ 6,
y1 , y2 ≥ 0.
y1 y2 y3
⎡
2 −3
s1 1
⎢
–1 8
s2 ⎢ 4
W ⎢ –28 –2 –16
⎣
276
s1 s2 W
1 0 0 2⎤ 2
⎥
0 1 0 2 ⎥ 12
0 0 1 0⎥
⎦
ISM: Introductory Mathematical Analysis
Section 7.8
y1 y2 y3 s1 s2 W
s1 ⎡
0 94 –5 1 – 14 0 23 ⎤ 2
⎢
⎥3
⎢
1
1
1
y1 1 – 4 2 0 4 0 2 ⎥
⎢
⎥
W ⎢ 0 –9 40 0 7 1 14 ⎥
⎢⎣
⎥⎦
y1 y2
y3 s1 s2 W
y2 ⎡
20
0 1 − 9 94 – 91 0 32 ⎤
⎢
⎥
13 1
2 0
2⎥
y1 ⎢ 1 0
9
9
9
3⎥
⎢
W ⎢0 0
20 4
6 1 20 ⎥
⎣⎢
⎦⎥
The minimum is Z = 20 when x1 = 4, x2 = 6 .
11. The dual is: Minimize W = 8 y1 + 12 y2 subject to
y1 + y2 ≥ 3,
2 y1 + 6 y2 ≥ 8,
y1 , y2 ≥ 0.
y1 y2 s1 s2 t1 t2 U
⎡ 1 1 –1 0
1 0 0 3⎤
⎢
⎥
1 0 8⎥
⎢ 2 6 0 –1 0
⎢ 8 12 0 0 M M 1 0 ⎥
⎣
⎦
y1
y2
s1 s2 t1 t2 U
1
–1 0 1 0 0
3 ⎤3
t1 ⎡ 1
⎢
⎥
6
0 –1 0 1 0
8 ⎥ 43
t2 ⎢ 2
U ⎢8 – 3M 12 – 7 M M M 0 0 1 –11M ⎥
⎣
⎦
y1 y2 s1
s2
t1
t2
U
5
1
⎡ 2
⎤5
0 –1
1
– 16
0
6
3
t1 ⎢ 3
⎥2
⎥4
1
4
y2 ⎢⎢ 13
1 0
– 16
0
0
⎥
6
3
U ⎢ 4 – 2 M 0 M 2 – 1 M 0 –2 + 7 M 1 –16 – 5 M ⎥
⎢⎣
⎥⎦
3
6
6
3
y1 y2 s1
s2
t1
t2 U
y1 ⎡
3
5⎤
1
1 0 – 32
– 14 0
4
2
2⎥
⎢
1 –1
1
1 0
1⎥
–
y2 ⎢ 0 1
2
4
2
4
2⎥
⎢
U ⎢0 0
6
1 –6 + M –1 + M 1 –26 ⎥
⎣
⎦
The maximum is Z = 26 when x1 = 6, x2 = 1 .
277
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
12. The dual is: Minimize W = 12 y1 + 8 y2 subject to
3 y1 + y2 ≥ 2,
y1 + y2 ≥ 6,
y1 , y2 ≥ 0.
y1
⎡ 3
⎢
⎢ 1
⎢12
⎣
y2 s1
s2
t1
t2 U
1 –1 0
1 0 0 2⎤
⎥
1 0 –1 0
1 0 6⎥
8 0 0 M M 1 0⎥
⎦
y1
y2
s1 s2 t1 t2 U
2
t1 ⎡ 3
1
–1 0 1 0 0
2 ⎤3
⎢
⎥
t2 ⎢ 1
1
0 –1 0 1 0
6 ⎥6
U ⎢12 – 4M 8 – 2M M M 0 0 1 –8M ⎥
⎣
⎦
y1
y2
s1
s2
t1
t2 U
1
1
2
y1 ⎡1
⎤2
– 13
0
0 0
3
3
3
⎢
⎥
⎢
⎥
16
2
1
1
t2 ⎢ 0
–1
–3
1 0
⎥8
3
3
3
U ⎢ 0 4 – 2 M 4 – 1 M M –4 + 4 M 0 1 –8 – 16 M ⎥
⎢⎣
⎥⎦
3
3
3
3
y1
y2
s1
s2
t1
t2 U
⎡
⎤
3
1
–1
0
1
0 0
2
y2
⎢
⎥
t2 ⎢
–2
0
1
–1
–1
1 0
4
⎥4
U ⎢ –12 + 2 M 0 8 – M M –8 + 2M 0 1 –16 – 4M ⎥
⎣
⎦
y1 y2 s1 s2 t1
t2 U
1 0
6⎤
y2 ⎡ 1 1 0 –1 0
⎢
⎥
s1 ⎢ –2 0 1 –1 –1
1 0
4⎥
U ⎢ 4 0 0 8 M –8 + M 1 –48⎥
⎣
⎦
The maximum is Z = 48 when x1 = 0, x2 = 8 .
13. The first constraint can be written as x1 − x2 ≥ −1. The dual is: Maximize W = – y1 + 3 y2 subject to
y1 + y2 ≤ 6,
– y1 + y2 ≤ 4,
y1 , y2 ≥ 0.
y1
⎡
s1 1
⎢
s2 ⎢ –1
W⎢1
⎣
y1
⎡
s1 2
⎢
y2 ⎢ –1
W ⎢ –2
⎣
y2 s1 s2 W
1 1 0 0 6⎤ 6
⎥
1 0 1 0 4⎥ 4
–3 0 0 1 0 ⎥
⎦
y2 s1 s2 W
0 1 –1 0 2 ⎤ 1
⎥
1 0 1 0 4⎥
0 0 3 1 12 ⎥
⎦
278
ISM: Introductory Mathematical Analysis
y1 y2 s1
Section 7.8
s2 W
y1
y2
s1
s2 W
1 ⎤ 1
⎡
80
1 – 54 0
s1 ⎢ 0
5 ⎥ 400
1
1
1 ⎥ 1
y1 ⎢1
0
0
2
50
50 ⎥ 25
⎢
W ⎢ 0 –20, 000 0 1600 1 1600 ⎥
⎣
⎦
y1 y2
s1
s2 W
1
1
1 ⎤
⎡
– 100
0
y2 0 1
80
400 ⎥
⎢
3 ⎥
1
1
y1 ⎢ 1 0 – 160
0
40
160 ⎥
⎢
W ⎢ 0 0 250 1400 1 1650 ⎥
⎣
⎦
The firm should spend $250 on newspaper
advertising and $1400 on radio advertising for a
cost of $1650.
y1 ⎡
1 0 12 – 12 0 1⎤
⎢
⎥
1 0
⎥
y2 ⎢ 0 1 12
5
2
⎢
⎥
W ⎢0 0 1
2 1 14 ⎥
⎣
⎦
The minimum is Z = 14 when x1 = 1, x2 = 2 .
14. The first constraint can be written as
−2 x1 + x2 + x3 ≥ −2. The dual is:
Maximize W = −2 y1 + 4 y2
subject to −2 y1 − y2 ≤ 2
y1 − y2 ≤ 1
y1 + 2 y2 ≤ 1
y1 , y2 ≥ 0
y1 y2 s1 s2 s3 W
s1 ⎡ −2 −1 1 0 0 0 2 ⎤
s2 ⎢ 1 −1 0 1 0 0 1⎥
⎢
⎥
s3 ⎢ 1 2 0 0 1 0 1⎥ 12
W ⎢⎣ 2 −4 0 0 0 1 0 ⎥⎦
16. Let x1 = number of type A trucks rented,
x2 = number of type B trucks rented.
We want to minimize C = 0.40 x1 + 0.60 x2
subject to
2 x1 + 2 x2 ≥ 12,
x1 + 3 x2 ≥ 12,
x1 , x2 ≥ 0.
y1 y2 s1 s2 s3 W
s1 ⎡ − 3
⎢ 2
s2 ⎢ 3
2
⎢
y2 ⎢ 1
2
W ⎢⎣ 4
0 1 0
0 0 1
1 0 0
0 0 0
1
2
1
2
1
2
0
0
5⎤
2⎥
3
⎥
2
1⎥
2⎥
The dual is: Maximize W = 12 y1 + 12 y2 subject
to
2 y1 + y2 ≤ 0.40,
2 y1 + 3 y2 ≤ 0.60,
y1 , y2 ≥ 0.
0
2 1 2 ⎥⎦
The minimum is Z = 2 when x1 = 0, x2 = 0,
x3 = 2.
y1
y2 s1 s2 W
s1 ⎡
2
1 1 0 0 52 ⎤ 15
⎢
⎥
3
s2 ⎢ 2
3 0 1 0 53 ⎥ 10
⎢
⎥
W ⎢ –12 –12 0 0 1 0 ⎥
⎣
⎦
Here we choose y1 as the entering variable.
15. Let x1 = amount spent on newspaper
advertising,
x2 = amount spent on radio advertising.
We want to minimize C = x1 + x2 subject to
40 x1 + 50 x2 ≥ 80, 000,
100 x1 + 25 x2 ≥ 60, 000,
x1 , x2 ≥ 0.
The dual is: Maximize
W = 80, 000 y1 + 60, 000 y2 subject to
y1 y2 s1 s2 W
⎡1 1
1
0 0 15 ⎤ 52
y1 ⎢
2
2
⎥
⎢
1
1
s2 ⎢ 0 2 –1 1 0 5 ⎥⎥ 10
W ⎢ 0 –6 6 0 1 12 ⎥
⎢⎣
5 ⎥⎦
y1 y2 s1 s2 W
y1 ⎡
3 –1 0
3 ⎤
1 0
4
4
20 ⎥
⎢
1 0
1⎥
y2 ⎢ 0 1 – 12
2
10 ⎥
⎢
W ⎢0 0
3
3 1 3⎥
⎣
⎦
Thus the maximum value of W, and hence the
minimum value of C, is 3.
40 y1 + 100 y2 ≤ 1,
50 y1 + 25 y2 ≤ 1,
y1 , y2 ≥ 0.
y1
y2
s1 s2 W
1
100
1 0 0 1 ⎤ 40
s1 ⎡ 40
⎢
⎥ 1
25
0 1 0 1 ⎥ 50
s2 ⎢ 50
W ⎢ –80, 000 –60, 000 0 0 1 0 ⎥
⎣
⎦
279
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
If we choose y2 as the entering variable, then
y1 y2 s1 s2 W
⎡
1 1 0 0 52 ⎤ 52
s1 ⎢ 2
⎥
3 0 1 0 53 ⎥ 15
s2 ⎢ 2
⎢
⎥
W ⎢ –12 –12 0 0 1 0 ⎥
⎣
⎦
y1 y2 s1 s2 W
⎡ 4 0 1 –1 0 1 ⎤ 3
3
5 ⎥ 20
s1 ⎢ 3
⎢
3
2
1
1
y2 ⎢ 3 1 0 3 0 5 ⎥⎥ 10
W ⎢ –4 0 0 4 1 12 ⎥
⎢⎣
5 ⎥⎦
y1 y2 s1 s2 W
3 –1 0
3 ⎤
⎡
y1 ⎢ 1 0
4
4
20 ⎥
1 0
1⎥
y2 ⎢ 0 1 – 12
2
10 ⎥
⎢
W ⎢0 0
3
3 1 3⎥
⎣
⎦
The minimum total cost per mile is $3.
17. Let y1 = number of shipping clerk apprentices,
y2 = number of shipping clerks,
y3 = number of semiskilled workers,
y4 = number of skilled workers.
We want to minimize W = 6 y1 + 9 y2 + 8 y3 + 14 y4 subject to
y1 + y2 ≥ 60,
–2 y1 + y2 ≥ 0,
y3 + y4 ≥ 90,
y3 – 2 y4 ≥ 0,
y1 , y2 , y3 , y4 ≥ 0.
The dual is: Maximize Z = 60 x1 + 0 x2 + 90 x3 + 0 x4 subject to
x1 – 2 x2 ≤ 6,
x1 + x2 ≤ 9,
x3 + x4 ≤ 8,
x3 – 2 x4 ≤ 14,
x1 , x2 , x3 , x4 ≥ 0.
x1
x2 x3 x4
0
s1 ⎡ 1 –2 0
⎢ 1
1
0
0
s2 ⎢
⎢
0
0
1
1
s3
⎢
0
1
–2
s4 ⎢ 0
⎢
Z ⎢⎣ –60 0 –90 0
s1
1
0
0
0
0
s2
0
1
0
0
0
s3
0
0
1
0
0
s4
0
0
0
1
0
Z
0 6⎤
0 9 ⎥⎥
0 8⎥ 8
⎥
0 14 ⎥ 14
1 0 ⎥⎥
⎦
280
ISM: Introductory Mathematical Analysis
x1
x2 x3 x4 s1
–2 0 0 1
s1 ⎡ 1
⎢
1 0 0 0
s2 ⎢ 1
⎢
0 1 1 0
x3 0
⎢
0
0 0 –3 0
s4 ⎢
⎢
Z ⎣⎢ –60 0 0 90 0
s2 s3
0 0
1 0
0 1
0 –1
0 90
Chapter 7 Review
s4
0
0
0
1
0
Z
0 6 ⎤6
0 9 ⎥⎥ 9
0 8 ⎥
⎥
0 6 ⎥
1 720 ⎥⎥
⎦
x1
x2 x3 x4 s1 s2 s3 s4 Z
1
–2
0 0 1 0 0 0 0
6 ⎤
⎡
x1
⎢0
3
0 0 –1 1 0 0 0
3 ⎥⎥ 1
s2 ⎢
0
1 1 0 0 1 0 0
8 ⎥
x3 ⎢0
⎢
⎥
0
0 –3 0 0 –1 1 0
6 ⎥
s4 ⎢0
Z ⎢⎢⎣0 –120 0 90 60 0 90 0 1 1080 ⎥⎥⎦
x1 x2 x3 x4 s1 s2
s3 s4 Z
1
2
⎡
0 0 0
8⎤
x1 ⎢ 1 0 0 0
3
3
⎥
1
⎥
x2 ⎢ 0 1 0 0 – 1
0
0
0
1
3
3
⎢
⎥
x3 ⎢ 0 0 1 1
0 0 1 0 0
8⎥
⎥
s4 ⎢ 0 0 0 –3
0 0 –1 1 0
6⎥
⎢
Z ⎢ 0 0 0 90 20 40 90 0 1 1200 ⎥
⎥⎦
⎣⎢
The company should employ 20 shipping clerk apprentices, 40 shipping clerks, 90 semiskilled workers, and
0 skilled workers for a total hourly wage of $1200.
Chapter 7 Review Problems
1.
5
y
x
5
2.
10
y
x
10
281
Chapter 7: Linear Programming
3.
5
ISM: Introductory Mathematical Analysis
y
8.
–5
3
5
y
x
x
5
4.
5
5
y
9.
10
y
x
x
10
5
5.
10
y
10.
10
y
x
10
6.
5
x
10
y
11. Feasible region follows. Corner points are (0, 0),
(0, 2), (1, 3), (3, 1), (3, 0). Z is maximized at
(3, 0) where its value is 3.
Thus Z = 3 when x = 3 and y = 0.
x
5
5
y
y−x=2
x+y=4
7.
5
y
x=3
(3, 0)
x
5
282
x
5
ISM: Introductory Mathematical Analysis
Chapter 7 Review
12. Feasible region follows. Corner points are (0, 1),
(0, 5), (4, 3), and (4, 1). Z is maximized at (4, 3)
where its value is 22. Thus Z = 22 when
x = 4 and y = 3.
10
y
x+y=5
y
10
⎛ 20 10 ⎞
⎜ , ⎟
⎝ 9 9⎠
x + 2y = 10
2x + 5y = 10
x
(4, 3)
(4, 0)
5x + 8y = 20
x=4
x
(5, 0)
16. Feasible region follows. Corner points are (0, 4),
(0, 6), (6, 8), (6, 0), and (4, 0). Z is minimized at
(0, 4) and (4, 0) where its value is 8. Thus Z is
minimized at all points on the line segment
joining (0, 4) and (4, 0). The solution is
Z = 8 when x = (1 – t)(0) + 4t = 4t,
y = (1 – t)(4) + 0t = 4 – 4t, and 0 ≤ t ≤ 1.
10
y=1
13. Feasible region is unbounded. Z is minimized at
the corner point (0, 2) where its value is –2.
Thus Z = –2 when x = 0 and y = 2.
5
10
y
y
10
x − y = −2
−x + 3y = 18
(6, 8)
(0, 6)
(0, 2)
(0, 4)
x+y=1
x − 2y = 2
x
17. Feasible region follows. Corner points are (0, 0),
(0, 4), (2, 3), and (4, 0). Z is maximized at (2, 3)
and (4, 0) where its value is 36. Thus Z is
maximized at all points on the line segment
joining (2, 3) and (4, 0). The solution is
Z = 36 when x = (1 – t)(2) + 4t = 2 + 2t,
y = (1 – t)(3) + 0t = 3 – 3t, and 0 ≤ t ≤ 1.
y
5
4x + 3y = 15
⎛ 10 5 ⎞
⎜ , ⎟
⎝ 3 9⎠
(0, 0)
x – 6y = 0
x
10
(4, 0)
14. Feasible region follows. Corner points are (0, 0),
⎛ 10 5 ⎞
⎜ , ⎟ , and (0, 5). Z is minimized at (0, 0)
⎝ 3 9⎠
where its value is 0. Thus Z = 0 when x = 0 and
y = 0.
(0, 5)
(6, 0)
x+y=4
5
10
x=6
(0, 4)
x
10
(0, 0)
y
x + 2y = 8
(2, 3)
3x + 2y = 12
x
(4, 0) 5
15. Feasible region follows. Corner points are
⎛ 20 10 ⎞
⎜ , ⎟ , (5, 0), and (4, 0). Z is minimized at
⎝ 9 9⎠
70
70
⎛ 20 10 ⎞
. Thus Z =
⎜ , ⎟ where its value is
9
9
⎝ 9 9⎠
20
10
when x =
and y = .
9
9
18. Feasible region is unbounded. The family of
lines given by Z = 4x + y has members having
arbitrarily large values of Z and that also
intersect the feasible region. Thus no optimum
solution exists.
283
Chapter 7: Linear Programming
20
y
ISM: Introductory Mathematical Analysis
x1
x2
3
s1 ⎡ 2
⎢ 4
3
20. s2 ⎢
⎢
0
1
s3
⎢
Z ⎣⎢ –18 –20
s1 s2 s3 Z
1 0 0 0 18 ⎤ 6
0 1 0 0 24 ⎥⎥ 8
0 0 1 0 5 ⎥5
⎥
0 0 0 1 0⎥
⎦
x1 x2 s1 s2 s3 Z
3
s1 ⎡ 2 0 1 0 –3 0 3 ⎤ 2
⎢
⎥
s2 ⎢ 4 0 0 1 –3 0 9 ⎥ 9
4
x2 ⎢ 0 1 0 0 1 0 5 ⎥
⎢
⎥
Z ⎣⎢ –18 0 0 0 20 1 100 ⎦⎥
Z = 40
(0, 12)
(4, 6)
3x + 2y = 24
x + 2y = 16
Z = 16
(16, 0) x
20
x1
⎡
s1 1
⎢
19.
s2 ⎢ 1
Z ⎢ –4
⎣
x1
x2 s1
1 0 0 12 ⎤ 2
⎥
2 0 1 0 8 ⎥4
–5 0 0 1 0 ⎥
⎦
x2 s1 s2 Z
6
x2 ⎡ 1
⎢ 6
⎢
s2 ⎢ 23
⎢
19
Z ⎢⎣⎢ – 6
x1
x2 ⎡ 0
⎢
x1 ⎢⎢1
Z ⎢0
⎣⎢
s2 Z
0
0 0
5
6
0
x2 s1
s2
Z
1
4
1
4
0
1
0 – 12
0 – 34
–
3
2
19
4
x1 x2 s1 s2
⎡
s1 0 4 1 –1
⎢
x1 ⎢ 1 2 0 1
Z ⎢0 3 0 4
⎣
Thus Z = 32 when
s3
Z
0 –
0 32 ⎤
x1 ⎡1 0
⎢
⎥
s2 ⎢ 0 0 –2 1 3 0 3 ⎥ 1
x2 ⎢⎢ 0 1 0 0 1 0 5 ⎥⎥ 5
Z ⎢ 0 0 9 0 –7 1 127 ⎥
⎣
⎦
x1 x2 s1
s2 s3 Z
1
1 0 0
3⎤
x1 ⎡⎢ 1 0 – 2
2
⎥
1 1 0
⎥
s3 ⎢ 0 0 – 2
1
3
3
⎢
⎥
2 –1 0 0
⎢0 1
⎥
4
x2 ⎢
3
3
⎥
7 0 1 134 ⎥
Z ⎢ 0 0 13
3
3
⎣⎢
⎦⎥
Thus Z = 134 when x1 = 3 and x2 = 4 .
1
2
⎤
2 ⎥ 12
⎥
1 0 4⎥ 6
⎥
0 1 10 ⎥
⎦⎥
1
6
– 13
1
x1 x2 s1 s2
1 ⎤4
⎥
0 6 ⎥⎥
⎥
1 29 ⎥
⎦
Z
4⎤
⎥
0 8⎥
1 32 ⎥
⎦
x1 = 8 and x2 = 0.
0
284
3
2
ISM: Introductory Mathematical Analysis
x1 x2 x3 s1 t1 W
21. ⎡ 1 2 3 –1 1 0 5⎤
⎢3 2 1 0 M 1 0 ⎥
⎢⎣
⎦⎥
x1
x2
x3
2
3
t1 ⎡ 1
⎢3 − M 2 − 2M 1 − 3M
W ⎢⎣
x1 x2 x3 s1
t1
1
2
1
1
⎡
x3 ⎢ 3 3 1 − 3
3
8
4
1
1
⎢
W
0
−3+M
3
⎢⎣ 3 3
5
Thus Z = when x1 = 0, x2
3
x1
⎡3
22. ⎢
⎢1
⎢1
⎣
x2
s1
s2 t1
t1 ⎡ 3
⎢
t2 ⎢ 1
W ⎢1 – 4M
⎣
x1
5
0
2 – 5M
Chapter 7 Review
s1 t1 W
5
−1 1 0
5 ⎤3
M 0 1 −5M ⎥⎥
⎦
W
5⎤
0
3⎥
5
1 − 3⎥
⎥⎦
5
= 0, and x3 = .
3
t2 W
5 –1 0
1 0 0 20 ⎤
⎥
0 0 –1 0
1 0 5⎥
2 0 0 M M 1 0⎥
⎦
x1
x2
s1 s2 t1 t2 W
–1 0 1 0 0
20 ⎤ 4
⎥
0 –1 0 1 0
5 ⎥
M M 0 0 1 –25M ⎥
⎦
s1 s2
t1
t2 W
x2
x2 ⎡ 3
1 – 15
⎢ 5
t2 ⎢ 1
0 0
⎢
1
W ⎢ − 5 − M 0 52
⎣
x1 x2 s1 s2
x2 ⎡ 0
⎢
x1 ⎢1
⎢
W ⎢0
⎢⎣
x1
⎡
s2 0
⎢
x1 ⎢⎢1
⎢
Z ⎢0
⎣
⎤ 20
⎥ 3
–1
0
1 0
5 ⎥ 5
⎥
M − 52 + M 0 1 –8 − 5M ⎥
⎦
t1
t2 W
1
5
0
0 0
1 – 15
3
5
1
5
− 53
0
0
–1
0
1
0
2
5
− 15
− 52 + M
x2
s1 s2 Z
5
3
5
3
1
3
Thus Z =
− 13 1 0
− 13
0 0
1
3
0 1
1
5
+M
4
1 ⎤5
⎥3
0 5⎥
⎥
1 –7 ⎥
⎥⎦
0
5 ⎤
3 ⎥
20 ⎥
3 ⎥
⎥
− 20
3 ⎥⎦
20
20
, x2 = 0.
when x1 =
3
3
285
Chapter 7: Linear Programming
x1 x2 s1 s2 s3 t2
⎡ 1 1 1 0 0 0
⎢
1
23. ⎢ 1 1 0 –1 0
⎢ 1 0 0 0 1 0
⎢
–1 –2 0 0 0 M
⎣⎢
x1
1
⎡
s1
⎢ 1
t2 ⎢
s3 ⎢ 1
⎢
W ⎢⎣ –1 – M
ISM: Introductory Mathematical Analysis
W
0 12 ⎤
0 5⎥⎥
0 10 ⎥
⎥
1 0⎥
⎦
x2
s1 s2
1
1 0
1
0 –1
0
0 0
–2 – M 0 M
s3
0
0
1
0
t2
0
1
0
0
W
0 12 ⎤ 12
0
5 ⎥⎥ 5
0 10 ⎥
⎥
1 –5M ⎥
⎦
x1 x2 s1 s2 s3
t2
W
0
0
1
1
0
–1
0
7 ⎤7
s1 ⎡
⎢1 1 0 –1 0
1
0 5 ⎥⎥
x2 ⎢
0
0 10 ⎥
s3 ⎢1 0 0 0 1
⎢
⎥
W ⎢⎣1 0 0 –2 0 2 + M 1 10 ⎥⎦
x1 x2 s1 s2 s3 Z
s2 ⎡ 0 0 1 1 0 0 7 ⎤
⎢
⎥
x2 ⎢ 1 1 1 0 0 0 12 ⎥
s3 ⎢ 1 0 0 0 1 0 10 ⎥
⎢
⎥
Z ⎢⎣ 1 0 2 0 0 1 24 ⎦⎥
Thus Z = 24 when x1 = 0 and x2 = 12.
x1
⎡1
24. ⎢
⎢1
⎢2
⎣
x2 s1
s2
t2 W
2 1 0 0 0 6⎤
⎥
1 0 –1 1 0 1⎥
1 0 0 M 1 0⎥
⎦
x1
x2 s1 s2 t2 W
s1 ⎡ 1
2
1 0 0 0 6 ⎤3
⎢
⎥
t2 ⎢ 1
1
0 –1 1 0 1 ⎥ 1
W ⎢2 – M 1 – M 0 M 0 1 – M ⎥
⎣
⎦
s1 ⎡ –1 0 1 2
–2 0 4 ⎤
⎢
⎥
x2 ⎢ 1 1 0 –1
1 0 1⎥
W ⎢ 1 0 0 1 –1 + M 1 –1⎥
⎣
⎦
Thus Z = 1 when x1 = 0 and x2 = 1.
25. We write the first constraint as – x1 + x2 + x3 ≥ 1.
x1
x2 x3
s1
t1
t2 W
⎡ –1 1 1 –1 1 0 0 1⎤
⎢
⎥
1 0 12 ⎥
⎢ 6 3 2 0 0
⎢ 1 2 1 0 M M 1 0⎥
⎣
⎦
286
ISM: Introductory Mathematical Analysis
Chapter 7 Review
x1
x2
x3
s1 t1 t2 W
⎡
t1
–1
1
1
–1 1 0 0
1 ⎤
⎢
⎥
t2 ⎢ 6
3
2
0 0 1 0
12 ⎥ 2
W ⎢1 – 5M 2 – 4 M 1 – 3M M 0 0 1 –13M ⎥
⎣
⎦
x1
x2
x3
s1 t1
t2
W
3
4
1
t1 ⎡0
–1 1
0
3 ⎤2
2
3
6
⎢
⎥
⎥4
1
1
1
x1 ⎢⎢1
0
0
0
2
⎥
2
3
6
W ⎢0 3 – 3 M 2 – 4 M M 0 – 1 + 5 M 1 –2 – 3M ⎥
⎢⎣
⎥⎦
2 2
3 3
6 6
x1 x2 x3
s1
t1
t2
W
2
1
x2 ⎡ 0 1 8 – 2
0 2 ⎤ 94
9
3
3
9
⎢
⎥
⎥
1
1
1
x1 ⎢⎢1 0 – 19
–
0
1
⎥
3
3
9
W ⎢ 0 0 – 2 1 –1 + M – 1 + M 1 –5⎥
⎢⎣
⎥⎦
3
3
x1 x2 x3 s1 – Z
x3 ⎡
9⎤
0 98 1 – 34 0
4⎥
⎢
⎢
⎥
5
1
1
x1 ⎢ 1 8 0
0
4
4⎥
– Z ⎢0 3 0
1 1 – 7⎥
⎢⎣
4
2
2 ⎥⎦
7
5
9
when x1 = , x2 = 0, and x3 = .
Thus Z =
2
4
4
x1
x2 x3
s1
s2 t1 W
⎡ 1 1 3 –1 0
1 0 5⎤
⎥
26. ⎢
1 4 0 1 0 0 5⎥
⎢ 2
⎢ –2 –3 –5 0 0 M 1 0 ⎥
⎣
⎦
x1
x2
x3 s1
s2 t1 W
t1 ⎡
1
1
3 −1 0 1
⎢
s2 ⎢
2
1
4
0 1 0
⎢
W −2 − M −3 − M −5 − 3M M 0 0
⎣
x1
x2 x3 s1
s2
1
1
⎡
–2
0 –1
− 34
4
t1 ⎢
1
1 1
1
x3 ⎢⎢
0
2
4
4
⎢1 1
7−1M 0 M
5+3M
M
–
+
4 4
4 4
W ⎢⎣ 2 2
We choose t1 as the departing variable.
5
0
5⎤ 3
⎥
0
5⎥ 54
1 −5 M ⎥
⎦
t1 W
5⎤5
4⎥
1 0
5⎥
4⎥5
0 0
0 1
25
4
287
⎥
– 54 M ⎥
⎦
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
Since s2 is nonbasic for the last table and its
indicator is 0, there may be multiple optimum
solutions. Treating s2 as an entering variable,
deleting the t2 -column, changing W to –Z, and
continuing, we have
x1 x2 x3 s1 s2 – Z
s2 ⎡ 1 1
0 12 1 0 1⎤
⎢2 2
⎥
x3 ⎢ 12 12 1 12 0 0 2 ⎥
⎢
⎥
– Z ⎢ 1 1 0 0 0 1 0⎥
⎣
⎦
Here Z = 0 when x1 = 0, x2 = 0, and x3 = 2. Thus
multiple optimum solutions exist. Hence Z is
minimum when
x1 = (1 – t )(0) + 0t = 0,
x2 = (1 – t )(0) + 0t = 0,
x3 = (1 – t )(1) + 2t = 1 + t ,
x1 x2 x3 s1 s2
t1 W
⎡
x2 −2 1 0 −4 −3
4 0 5⎤
⎢
⎥
x3 ⎢ 1 0 1 1
−1 0 0 ⎥ 0
1
W ⎢ −3 0 0 −7 −4 7 + M 1 15⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
⎡
x2 2 1 4 0 1 0 5⎤
⎢
⎥
s1 ⎢ 1 0 1 1 1 0 0 ⎥
Z ⎢ 4 0 7 0 3 1 15⎥
⎣
⎦
Thus Z = 15 when x1 = 0, x2 = 5, and x3 = 0.
Note that choosing x3 as the departing variable
results in the same solution.
x1 x2 x3 s1 s2 Z
⎡
s1 4 –1 0 1 0 0 2 ⎤
⎢
⎥
27.
s2 ⎢ –8 2
5 0 1 0 2⎥ 1
⎢
⎥
Z ⎢ –1 –4 –2 0 0 1 0 ⎥
⎣
⎦
x1 x2 x3 s1 s2 Z
s1 ⎡ 0 0 5 1 1 0 3⎤
2
2
⎢
⎥
5
1
⎢
x2 –4 1 2 0 2 0 1⎥
⎢
⎥
Z ⎢ –17 0 8 0 2 1 4 ⎥
⎣
⎦
For the last table, x1 is the entering variable.
Since no quotients exist, the problem has an
unbounded solution. That is, no optimum
solution (unbounded).
x1
⎡1
28. ⎢
⎢0
⎢1
⎣
x2 x3 s1
s2
and 0 ≤ t ≤ 1. For the last table, s1 is nonbasic
and its indicator is 0. If we continue the process
for determining other optimum solutions, we
return to a table corresponding to the third table.
29. The dual is: Maximize W = 35 y1 + 25 y2 subject
to
y1 + y2 ≤ 2,
2 y1 + y2 ≤ 7,
3 y1 + y2 ≤ 8,
y1 , y2 ≥ 0.
y1
y2
1
s1 ⎡ 1
⎢ 2
1
s2 ⎢
1
s3 ⎢ 3
⎢
W ⎢⎣ –35 –25
t2 W
1 2 1 0 0 0 4⎤
⎥
0 1 0 –1 1 0 1⎥
1 0 0 0 M 1 0⎥
⎦
x1 x2 x3 s1 s2 t2 W
s1 ⎡1 1 2 1 0 0 0 4 ⎤ 2
⎢
⎥
t2 ⎢0 0 1 0 –1 1 0 1 ⎥ 1
W ⎢1 1 – M 0 M 0 1 – M ⎥
⎣
⎦
x1 x2 x3 s1 s2
t2 W
⎡
s1 1 1 0 1 2 –2 0 2 ⎤ 1
⎢
⎥
x3 ⎢0 0 1 0 –1 1 0 1 ⎥
W ⎢1 1 0 0 0 M 1 0 ⎥
⎣
⎦
The minimum value of Z is 0 for
x1 = 0, x2 = 0, and x3 = 1.
s1 s2 s3 W
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
2⎤ 2
7 ⎥⎥ 72
8⎥ 8
⎥3
0⎥
⎦
y1 y2 s1 s2 s3 W
⎡
y1 1 1 1 0 0 0 2 ⎤
⎢
⎥
s2 ⎢ 0 –1 –2 1 0 0 3⎥
s3 ⎢ 0 –2 –3 0 1 0 2 ⎥
⎢
⎥
W ⎢⎣ 0 10 35 0 0 1 70 ⎥⎦
Thus Z = 70 when x1 = 35, x2 = 0, and x3 = 0.
288
ISM: Introductory Mathematical Analysis
Chapter 7 Review
30. We write the third constraint as −4 x1 − x2 ≤ −2. The dual is: Minimize W = 3 y1 + 4 y2 – 2 y3 subject to
y1 + y2 – 4 y3 ≥ 1,
– y1 + 2 y2 – y3 ≥ –2,
y1 , y2 , y3 ≥ 0.
We write the second constraint of the dual as y1 – 2 y2 + y3 ≤ 2.
y1 y2 y3 s1 s2 t1 U
⎡ 1 1 –4 –1 0
1 0 1⎤
⎢
⎥
1 0 1 0 0 2⎥
⎢ 1 –2
⎢3 4 –2 0 0 M 1 0 ⎥
⎣
⎦
y1
y2
y3
s1 s2 t1
⎡
t1
1
1
–4
–1 0 1
⎢
s2 ⎢ 1
–2
1
0 1 0
⎢
U 3 – M 4 – M –2 + 4M M 0 0
⎣
y1 y2 y3 s1 s2
t1 U
⎡
y1 1 1 –4 –1 0
1 0
1⎤
⎢
⎥
s2 ⎢ 0 –3 5 1 1
–1 0
1⎥
U ⎢0
1 10 3 0 –3 + M 1 –3⎥
⎣
⎦
Thus Z = 3 when x1 = 3 and x2 = 0.
U
0 1 ⎤1
⎥
0 2 ⎥2
1 –M ⎥
⎦
31. Let x, y, and z denote the numbers of units of X, Y, and Z produced weekly, respectively. If P is the total profit
obtained, we want to maximize
P = 10x + 15y + 22z subject to
x + 2 y + 2 z ≤ 40,
x + y + 2 z ≤ 34,
x, y, z ≥ 0.
x
y
z s1 s2 P
⎡
s1 1
2
2 1 0 0 40 ⎤ 20
⎢
⎥
s2 ⎢ 1
1
2 0 1 0 34 ⎥ 17
P ⎢ –10 –15 –22 0 0 1 0 ⎥
⎣
⎦
x y z s1 s2 P
⎡
s1 0 1 0 1 –1 0 6 ⎤ 6
⎢
⎥
z ⎢ 12 12 1 0 12 0 17 ⎥ 34
⎢
⎥
P ⎢ 1 –4 0 0 11 1 374 ⎥
⎣
⎦
1 –1 0
6⎤
y ⎡0 1 0
⎢
⎥
z ⎢ 12 0 1 – 12
1 0 14 ⎥
⎢
⎥
P⎢ 1 0 0
4 7 1 398⎥
⎣
⎦
Thus 0 units of X, 6 units of Y, and 14 units of Z give a maximum profit of $398.
289
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
32. We want to maximize P = 10x + 15y + 22z subject to
x + 2 y + 2 z ≤ 40,
x + y + 2 z ≤ 34,
x + y + z ≥ 24,
x, y, z ≥ 0.
x
y
z s1 s2 s3 t3 W
2
2 1 0 0 0 0 40 ⎤
⎡ 1
⎢ 1
1
2 0 1 0 0 0 34 ⎥⎥
⎢
⎢ 1
1
1 0 0 –1 1 0 24 ⎥
⎢
⎥
–10 –15 –22 0 0 0 M 1 0 ⎥
⎣⎢
⎦
x
y
z
s1 s2 s3 t3 W
1
2
2
1 0 0 0 0
40 ⎤ 20
s1 ⎡
⎢
1
1
2
0 1 0 0 0
34 ⎥⎥ 17
s2 ⎢
1
1
1
0 0 –1 1 0
24 ⎥ 24
t3 ⎢
⎢
⎥
W ⎢⎣ –10 – M –15 – M –22 – M 0 0 M 0 1 −24M ⎦⎥
x
y
z s1
s2
s3 t3 W
0
1
0
1
–1
0 0 0
6
⎡
⎤ 6
s1
⎢ 1
⎥
1
1
z⎢
1 0
0 0 0
17
⎥ 34
2
2
2
⎢ 1
⎥ 14
1
1
0 0
–2
–1 1 0
7
⎢ 2
⎥
2
t3 ⎢
⎥
M
M
M
W ⎢⎣1 – 2 –4 – 2 0 0 11 + 2 M 0 1 374 – 7 M ⎥⎦
x
y z
s1
s2 s3 t3 W
1 0
1
–1 0 0 0
6
⎤
y⎡ 0
⎢ 1
⎥
1
0 1
–2
1 0 0 0
14
z⎢ 2
⎥ 28
⎢ 1
⎥
0 0
– 12
0 –1 1 0
4
⎢ 2
⎥8
t3 ⎢
⎥
M
M
W ⎢⎢1 – 2 0 0 4 + 2 7 M 0 1 398 – 4M ⎥⎥
⎣
⎦
x y z
s1 s2
s3
t3 W
0 0
6⎤
y ⎡0 1 0 1 –1 0
⎢0 0 1 0 1 1
–1 0 10 ⎥⎥
z⎢
2 0
8⎥
x ⎢ 1 0 0 –1 0 –2
⎢
⎥
W ⎢⎣0 0 0 5 7 2 –2 + M 1 390 ⎥⎦
The company should produce 8 units of X, 6 units of Y, 10 units of Z, for a profit of $390.
33. Let x AC , x AD , xBC , and xBD denote the amounts (in hundreds of thousands of gallons) transported from A to C,
A to D, B to C, and B to D, respectively. If C is the total transportation cost in thousands of dollars, we want to
minimize C = x AC + 2 x AD + 2 xBC + 4 xBD subject to
x AC + x AD
xBC + xBD
x AC + xBC
x AD + xBD
x AC , x AD , xBC , xBD
≤ 6,
≤ 6,
= 5,
= 5,
≥ 0.
290
ISM: Introductory Mathematical Analysis
x AC
⎡1
⎢0
⎢
⎢1
⎢
⎢0
⎢1
⎢⎣
x AD
1
0
0
1
2
x AC
1
⎡
s1
⎢ 0
s2 ⎢
t3 ⎢ 1
⎢
t4 ⎢ 0
W ⎢⎢⎣1 – M
xBC
0
1
1
0
2
xBD
0
1
0
1
4
x AD
1
0
0
1
2–M
s1 s2
1 0
0
1
0 0
0 0
0 0
xBC
0
1
1
0
2–M
t3
0
0
1
0
M
Chapter 7 Review
t4
0
0
0
1
M
xBD
s1
0
1
1
0
0
0
1
0
4–M 0
W
0
0
0
0
1
s2
0
1
0
0
0
t3
0
0
1
0
0
6⎤
6 ⎥⎥
5⎥
⎥
5⎥
0 ⎥⎥
⎦
t4
0
0
0
1
0
W
0
6 ⎤6
0
6 ⎥⎥
0
5 ⎥5
⎥
0
5 ⎥
1 –10M ⎥⎥
⎦
W
0
1⎤ 1
0
6 ⎥⎥
0
5⎥
⎥
0
5⎥ 5
1 –5 – 5M ⎥⎥
⎦
W
0
1
⎤
⎥
0
6
⎥6
⎥5
0
5
⎥
0
4
⎥4
1 –7 – 4M ⎥⎥
⎦
x AC x AD xBC
xBD s1 s2
t3 t4
1 –1
0 1 0
–1 0
s1 ⎡ 0
⎢0
0
1
1
0
1
0 0
s2 ⎢
0 1
0 0 0
1 0
x AC ⎢ 1
⎢
1 0
1 0 0
0 1
t4 ⎢ 0
⎢
1 4 – M 0 0 –1 + M 0
W ⎢⎣ 0 2 – M
x AC x AD xBC xBD
s1
s2 t3 t4
–1
0
1
0 –1 0
x AD ⎡ 0 1
⎢0 0
1
1
0
1 0 0
s2 ⎢
⎢
1
0
1
0
0
0 1 0
x AC
⎢
1
1
–1
0 1 1
t4 ⎢ 0 0
⎢
W ⎢⎣ 0 0 3 – M 4 – M –2 + M 0 1 0
x AC x AD xBC xBD s1 s2
t3
t4 W
0
1 0
5⎤
x AD ⎡ 0 1 0 1 0 0
⎢0 0 0 0 1 1
–1
–1 0
2 ⎥⎥
s2 ⎢
0
–1 0
1⎥
x AC ⎢ 1 0 0 –1 1 0
⎢
⎥
1
1 0
4⎥
xBC ⎢ 0 0 1 1 –1 0
W ⎢⎢⎣ 0 0 0 1 1 0 –2 + M –3 + M 1 –19 ⎥⎦⎥
The minimum value of C is 19, when x AC = 1, x AD = 5, xBC = 4, and xBD = 0. Thus 100,000 gal from A to C,
500,000 gal from A to D, and 400,000 gal from B to C give a minimum cost of $19,000.
34. Let x and y be the weekly sales of Space Traders and Green Dwarf, respectively. We want to maximize
P = 5x + 9y subject to the constraints
30 x + 10 y ≤ 300
20 x + 10 y ≤ 200
10 x + 50 y ≤ 500
x, y ≥ 0
The constraints can be written as
291
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
3 x + y ≤ 30
2 x + y ≤ 20
x + 5 y ≤ 50
x, y ≥ 0
970
⎛ 50 80 ⎞
The feasible region has corner points (0, 0), (0, 10), ⎜ ,
at
⎟ , and (10, 0). P has a maximum of
9
⎝ 9 9 ⎠
⎛ 50 80 ⎞ ⎛ 5 8 ⎞
⎜ ,
⎟ = ⎜ 5 , 8 ⎟ . The possible integer values are (5, 8), (5, 9), (6, 8), and (6, 9). However, the point (6, 9)
⎝ 9 9 ⎠ ⎝ 9 9⎠
does not satisfy the second or third constraints. Evaluating P at the other three points gives that Jason should sell
5 copies of Space Trader and 9 copies of Green Dwarf, for a weekly profit of $106.
50
y
3x + y = 30
(0, 10)
2x + y = 20
⎛ 50 80 ⎞
⎜ ,
⎟
⎝ 9 9 ⎠
x + 5y = 50
x
50
(10, 0)
35. Let x and y represent daily consumption of foods A and B in 100-gram units. We want to minimize C = 8x + 22y
subject to the constraints
8 x + 4 y ≥ 176,
16 x + 32 y ≥ 1024,
2 x + 5 y ≥ 200,
x ≥ 0,
y ≥ 0.
⎛5
⎞
The feasible region is unbounded with corner points (100, 0), ⎜ , 39⎟ and (0, 44). C has a minimum value at
⎝2
⎠
(100, 0). Thus the animals should be fed 100 × 100 = 10, 000 grams = 10 kilograms of food A each day.
100
y
8x + 4y = 176
2x + 5y = 200
x
100
20
36.
0
0
50
Z = 0.89 when x = 4.78, y = 9.14
292
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 7
10
37.
0
−2
10
Z = 129.83 when x = 9.38, y = 1.63
Mathematical Snapshot Chapter 7
CURATIVE
UNITS
TOXIC
UNITS
RELATIVE
DISCOMFORT
Drug (per ounce)
500
400
1
Radiation (per min)
1000
600
1
≥ 2000
≤1400
1.
Requirement
Let x1 = number of ounces of drug and let x2 = number of minutes of radiation. We want to minimize the
discomfort D, where D = x1 + x2 , subject to
500 x1 + 1000 x2 ≥ 2000,
400 x1 + 600 x2 ≤ 1400,
where x1 , x2 ≥ 0.
5
x2
400x1 + 600x2 = 1400
(2, 1)
500x1 + 1000x2 = 2000
x1
5
⎛ 7⎞
The corner points are (0, 2), ⎜ 0, ⎟ , and (2, 1).
⎝ 3⎠
At (0, 2), D = 0 + 2 = 2;
7 7
⎛ 7⎞
at ⎜ 0, ⎟ , D = 0 + = ;
3 3
⎝ 3⎠
at (2, 1), D = 2 + 1 = 3.
Thus D is minimum at (0, 2).
The patient should get 0 ounces of drug and 2 minutes of radiation.
293
Chapter 7: Linear Programming
ISM: Introductory Mathematical Analysis
CURATIVE
UNITS
TOXIC
UNITS
RELATIVE
DISCOMFORT
Drug A (per ounce)
600
500
1
Drug B (per ounce)
500
100
2
Radiation (per min)
1000
1000
1
≥ 3000
≤ 2000
2.
Requirement
Let x1 = number of ounces of drug A,
x2 = number of ounces of drug B, and
x3 = number of minutes of radiation.
We want to minimize the discomfort D, where D = x1 + 2 x2 + x3 , subject to
600 x1 + 500 x2 + 1000 x3 ≥ 3000,
500 x1 + 100 x2 + 1000 x3 ≤ 2000,
x1 , x2 , x3 ≥ 0
To minimize D, we maximize –D by considering the artificial objective function
W = –D – Mt1 .
x1 x2
x3
s1 s2
t1 W
⎡ 600 500 1000 –1 0
1 0 3000 ⎤
⎢
⎥
⎢ 500 100 1000 0 1 0 0 2000 ⎥
⎢ 1
2
1 0 0 M 1
0⎥
⎣
⎦
x1
x2
x3
s1 s2 t1 W
t1 ⎡ 600
500
1000
–1 0 1 0
3000 ⎤ 3
⎢
⎥
s2 ⎢ 500
100
1000
0 1 0 0
2000 ⎥ 2
W ⎢1 – 600M 2 – 500M 1 –1000 M M 0 0 1 –3000M ⎥
⎣
⎦
x1
x2
x3 s1
s2
t1 W
⎤ 2.5
t1 ⎡
100
400
0 −1
1
1 0
1000
⎢
⎥
x3 ⎢
0.5
0.1
1 0
0.001
0 0
2
⎥ 20
⎢
W 0.5 –100M 1.9 – 400 M 0 M –0.001 + M 0 1 −2 − 1000 M ⎥
⎣
⎦
x1 x2 x3
s1
s2
t1
W
x2 ⎡ 0.25 1 0 –0.0025 –0.0025
0.0025 0
2.5⎤
⎢
⎥
x3 ⎢ 0.475 0 1 0.00025 0.00125
–0.00025 0 1.75⎥
W ⎢ 0.025 0 0 0.00475 0.00375 –0.00475 + M 1 –6.75⎥
⎣
⎦
The minimum value of D is 6.75 when x1 = 0, x2 = 2.5, and x3 = 1.75.
The patient should get 0 ounces of drug A, 2.5 ounces of drug B, and 1.75 minutes of radiation.
3. Answers may vary.
294
Chapter 8
Problems 8.1
3.
Assembly
Line
1.
A
Start
Finishing
Line
D
Production
Route
AD
E
AE
D
BD
E
BE
D
CD
E
CE
B
C
Red
Die
1
2
6 possible production routes
2.
BTU's
6000
Start
8000
10,000
Fan
Model
Speeds
Type
1
6000 – 1
Green
Die
1
Result
1, 1
2
1, 2
3
1, 3
4
1, 4
5
1, 5
6
1, 6
1
2, 1
2
2, 2
3
2, 3
4
2, 4
5
2, 5
6
2, 6
2
6000 – 2
1
8000 – 1
1
3, 1
2
8000 – 2
2
3, 2
1
10,000 – 1
3
3, 3
2
10,000 – 2
4
3, 4
5
3, 5
6
3, 6
3
6 model types
Start
4
5
6
1
4, 1
2
4, 2
3
4, 3
4
4, 4
5
4, 5
6
4, 6
1
5, 1
2
5, 2
3
5, 3
4
5, 4
5
5, 5
6
5, 6
1
6, 1
2
6, 2
3
6, 3
4
6, 4
5
6, 5
6
6, 6
36 possible results
295
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
4.
Toss 1
Toss 2
Toss 3
H
H
T
H
H
T
T
Start
H
H
T
T
H
T
T
Toss 4
H
Result
H, H, H, H
T
H, H, H, T
H
H, H, T, H
T
H, H, T, T
H
H, T, H, H
T
H, T, H, T
H
H, T, T, H
T
H, T, T, T
H
T, H, H, H
T
T, H, H, T
H
T, H, T, H
T
T, H, T, T
H
T, T, H, H
T
T, T, H, T
H
T, T, T, H
T
T, T, T, T
16 possible results
95!
95!
=
= 95
(95 − 1)! 94!
13.
6 P6
=
6!
6! 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
= =
= 720
(6 – 6)! 0!
1
14.
9 P4
=
9!
9!
= = 9 ⋅ 8 ⋅ 7 ⋅ 6 = 3024
(9 − 4)! 5!
15.
4 P2 ⋅ 5 P3
There are 5 roads from A to B, and 5 roads
from B to A. By the basic counting
principle, the number of possible routes for
a round trip is 5 · 5 = 25.
18.
99 P4
=
99 ⋅ 98 ⋅ 97 ⋅ 96 ⋅ 95
= 95
99 ⋅ 98 ⋅ 97 ⋅ 96
1000! 1000 ⋅ 999!
=
= 1000
999!
999!
For most calculators, attempting to evaluate
1000!
results in an error message (because of
999!
the magnitude of the numbers involved).
n Pr
n!
=
n!
( n – r )!
n!
=
1
(n – r )!
20. The number of ways to arrange 6 teams in an
order is 6 P6 = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 720 .
21. The number of ways of selecting 3 of 8
contestants in an order is 8 P3 = 8 ⋅ 7 ⋅ 6 = 336 .
8. For each of the 6 questions, there are 4 choices.
By the basic counting principle, the number of
ways to answer the questions is
22. Six out of eight items in column 2 must be
selected in an order. Thus the number of ways
the matching can be done is
8 P6 = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 = 20,160 .
4 · 4 · 4 · 4 · 4 · 4 = 46 = 4096 .
9. For each of the 10 questions, there are 2 choices.
By the basic counting principle, the number of
ways to answer the examination is
2 · 2 · ... · 2 = 210 = 1024 .
23. On each roll of a die, there are 6 possible
outcomes. By the basic counting principle, on 4
rolls the number of possible results is
6 ⋅ 6 ⋅ 6 ⋅ 6 = 64 = 1296.
10. Since there are 26 letters, there are 26 choices
for the first, third and fifth symbols. There are 10
possible digits (0 through 9) for the second,
fourth, and sixth symbols. By the basic counting
principle, the number of codes is
26 · 10 · 26 · 10 · 26 · 10 = 17,576,000.
=
99 P5
= (4 ⋅ 3)(5 ⋅ 4 ⋅ 3) = (12)(60) = 720
19. A name for the firm is an ordered arrangement
of the three last names. Thus the number of
possible firm names is 3 P3 = 3! = 3 ⋅ 2 ⋅1 = 6 .
7. There are 2 appetizers, 4 entrees, 4 desserts, and
3 beverages. By the basic counting principle, the
number of possible complete dinners is
2 · 4 · 4 · 3 = 96.
6 P3
=
17.
b. There are 5 possible roads from A to B.
Since a different road is to be used for the
return trip, there are only 4 possible roads
from B to A. By the basic counting
principle, the number of possible round-trip
routes is 5 · 4 = 20.
11.
95 P1
16.
5. There are 5 science courses and 4 humanities
courses. By the basic counting principle, the
number of selections is 5 · 4 = 20.
6. a.
12.
24. On each toss there are 2 possible outcomes. By
the basic counting principle, the number of
possible results on 8 tosses is
2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 28 = 256 .
25. The number of ways of selecting 3 of the 12
students in an order is 12 P3 = 12 ⋅11 ⋅10 = 1320 .
6!
6!
= = 6 ⋅ 5 ⋅ 4 = 120
(6 − 3)! 3!
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ISM: Introductory Mathematical Analysis
Section 8.1
26. Three of the 26 letters must be selected (without
repetition) in an order. Thus the number of
possible lock combinations is 26 P3 = 15, 600 .
35. The number of ways the waitress can place five
of the five different sandwiches (and order is
important) is 5 P5 = 5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 120.
27. The number of ways a student can choose 4 of
the 6 items in an order is 6 P4 = 6 ⋅ 5 ⋅ 4 ⋅ 3 = 360.
36. Because order is important, the number of ways
that the 5 people can line up is
5 P5 = 5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 120 .
If a woman is to be at each end, then the number
of ways to place one of the two women on the
left side is 2 P1 . Once a woman is chosen for the
left side, the other woman must be on the right
side. The number of ways to line the three men
in the middle is 3 P3 . By the basic counting
principle, the number of line ups is
2 P1 ⋅ 3 P3 = (2)(3 ⋅ 2 ⋅1) = 12 .
28. On the second roll, there are 2 possible outcomes
(a 1 or a 2). For each of the other two rolls, there
are 6 possible outcomes. By the basic counting
principle, the number of possible results for the
three rolls is 6 · 2 · 6 = 72.
29. The number of ways to select six of the six
different letters in the word MEADOW in an
order is 6 P6 = 6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 720 .
37. a.
30. The number of ways to select four of the four
different letters in the word DISC in an order is
4 P4 = 4! = 4 ⋅ 3 ⋅ 2 ⋅1 = 24 .
31. For an arrangement of books, order is important.
The number of ways to arrange 5 of 7 books is
7 P5 = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 = 2520 .
All 7 books can be arranged in
7 P7 = 7! = 5040 ways.
32. a.
To fill the four offices by different people, 4
of 12 members must be selected, and order
is important. This can be done in
12 P4 = 12 ⋅11 ⋅10 ⋅ 9 = 11,880 ways.
b. If the president and vice president must be
different members, then there are 12 choices
for president, 11 for vice president, 12 for
secretary, and 12 for treasurer. By the basic
counting principle, the offices can be filled
in 12 · 11 · 12 · 12 = 19,008 ways.
A student can enter by any of 5 doors. After
a door is chosen, the student can exit by any
of the 4 remaining doors. By the basic
counting principle, the number of ways to
enter by one door and exit by a different
door is 5 · 4 = 20.
38. a.
There are 24 possibilities for each of the
three letters in a name. By the basic
counting principle, the number of names is
24 · 24 · 24 = 243 = 13,824 .
b. There are 5 doors by which to enter and 5
doors by which to exit. By the basic
counting principle, the total number of ways
to enter and exit is 5 · 5 = 25.
b. Since the order of letters is important and no
letter is used more than one time, the
number of names is
24 P3 = 24 ⋅ 23 ⋅ 22 = 12,144 .
33. After a “four of a kind” hand is dealt, the cards
can be arranged so that the first four have the
same face value, and order is not important,
There are 13 possibilities for the first four cards
(all 2’s, all 3’s, ..., all aces). The fifth card can be
any one of the 48 cards that remain. By the basic
counting principle, the number of “four of a
kind” hands is 13 · 48 = 624.
39. There are 2 choices for the center position. After
that choice is made, to fill the remaining four
positions (and order is important), there are
4 P4 ways. By the basic counting principle, to
assign positions to the five-member team there
are 2 ⋅ 4 P4 = 2(4!) = 2(24) = 48 ways.
40. For the first letter there are two possibilities. For
the second and third letters there are 26
possibilities, and for the last letter there are 25
possibilities. By the basic counting principle, the
number of possible identifications is
2 · 26 · 26 · 25 = 33,800.
34. Five colors are available, and two are selected so
that order is important. Thus the number of ways
of placing an order is 5 P2 = 5 ⋅ 4 = 20.
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41. There are 3 P3 ways to select the first three
batters (order is important) and there are 6 P6
ways to select the remaining batters. By the basic
counting principle, the number of possible
batting orders is
3 P3 ⋅ 6 P6 = 3! ⋅ 6! = 6 ⋅ 720 = 4320.
42. a.
6.
= (4 ⋅ 3)
7.
Four of four flags can be arranged (order is
important) in 4 P4 = 4! = 24 ways. Thus 24
different signals are possible.
=
6!
6!
6 ⋅ 5 ⋅ 4! 6 ⋅ 5
=
=
=
= 15
4!(6 – 4)! 4! ⋅ 2! 4!(2 ⋅1) 2 ⋅1
2.
6 C2
=
6!
6!
6 ⋅ 5 ⋅ 4! 6 ⋅ 5
=
=
=
= 15
2!(6 – 2)! 2! ⋅ 4! (2 ⋅1)4! 2 ⋅1
3.
100 C100
4.
1001 C1
5.
=
8.
5!
3!(5 – 3)!
5 ⋅ 4 ⋅ 3!
= (12) ⋅10 = 120
3!⋅ 2!
n!
r !(n – r )!
=
n!
n!
=
⋅
(n – r )![n – (n – r )]! (n – r )!r !
n Cn
=
n!
1 1
= = =1.
n !(n – n)! 0! 1
9. The number of ways of selecting 4 of 17 people
so that order is not important is
17!
17!
=
17 C4 =
4!(17 – 4)! 4! ⋅13!
=
17 ⋅16 ⋅15 ⋅14 ⋅13!
= 2380
4 ⋅ 3 ⋅ 2 ⋅1(13!)
10. If horses A, B, and C finish in the money, then it
does not matter if A finishes in first, second, or
third place. Similarly for B and C. Thus order is
not important. The number of ways in which 3
of 8 horses finish in the money is the number of
ways of selecting 3 of the 8 without regard to
order, namely
8 ⋅ 7 ⋅ 6 ⋅ 5!
8!
8!
=
= 56 .
=
8 C3 =
3!(8 – 3)! 3! ⋅ 5! 3 ⋅ 2 ⋅1 ⋅ 5!
100!
1 1
= = =1
100!(100 – 100)! 0! 1
11. The number of ways of selecting 9 out of 13
questions (without regard to order) is
13!
13!
13 ⋅12 ⋅11 ⋅10 ⋅ 9!
=
=
13 C9 =
9!(13 − 9)! 9! ⋅ 4!
9!⋅ 4 ⋅ 3 ⋅ 2 ⋅1
= 715.
1001!
1001!
=
1!(1001 − 1)! 1!⋅1000!
1001 ⋅1000!
=
= 1001
1000!
=
5 P3 ⋅ 4 C2
=
= (4 ⋅ 3)
Thus n Cr = n Cn – r .
Problems 8.2
6 C4
n Cr
n Cn – r
b. If only one of the four flags is used, there
are 4 P1 possible signals. If exactly two
flags are used, there are 4 P2 possible
signals. Similarly, for exactly three and
exactly four flags, there are 3 P4 and 4 P4
possible signals, respectively. Thus if at
least one flag is used, the number of
possible signals is
4 P1 + 4 P2 + 4 P3 + 4 P4
=4+4·3+4·3·2+4·3·2·1
= 4 + 12 + 24 + 24 = 64.
1.
4 P2 ⋅ 5C3
12. In a deck of 52 cards, 26 of the cards are red. In
a four-card hand, the order is not important.
Thus, the number of four-card hands from the 26
red cards is
26!
26 C4 =
4!(26 − 4)!
26 ⋅ 25 ⋅ 24 ⋅ 23 ⋅ 22!
=
4!22!
= 14,950
4!
2!(4 − 2)!
4 ⋅ 3 ⋅ 2!
= 5⋅ 4⋅3
2!2!
= 60 ⋅ 6
= 360
= 5⋅ 4⋅3
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ISM: Introductory Mathematical Analysis
Section 8.2
13. The order of selecting 10 of the 74 dresses is of
no concern. Thus the number of possible
74!
74!
.
=
samples 74 C10 =
10! ⋅ (74 – 10)! 10! ⋅ 64!
18. The word STREETSBORO has 11 letters with
repetition: two S’s, two T’s, two R’s, two E’s,
one B, and two O’s. Thus the number of
distinguishable arrangements is
11!
11!
=
= 1, 247, 400.
2! ⋅ 2! ⋅ 2! ⋅ 2! ⋅1! ⋅ 2! 32
14. This situation can be considered as a two-stage
process. In the first stage, one of the 3 boxes is
selected. In the second stage, 4 of the 7 types of
jelly are selected (and order is not important),
which can be done in 7 C4 ways. By the basic
counting principle, the number of different gift
boxes that are possible is
7!
7!
3 ⋅ 7 C4 = 3 ⋅
= 3⋅
4!(7 – 4)!
4! ⋅ 3!
= 3⋅
19. The number of ways 4 heads and 3 tails can
occur in 7 tosses of a coin is the same as the
number of distinguishable permutations in the
“word” HHHHTTT, which is
7!
7 ⋅ 6 ⋅ 5 ⋅ 4!
=
= 35 .
4! ⋅ 3!
4!(6)
20. The number of ways for the given outcome to
occur is the number of distinguishable
permutations of six numbers such that two are
2’s, three are 3’s, and one is 4, which is
6!
6 ⋅ 5 ⋅ 4 ⋅ 3!
=
= 60 .
2! ⋅ 3! ⋅1!
(2)3!
7 ⋅ 6 ⋅ 5 ⋅ 4!
= 3 ⋅ 35 = 105 .
4!(3 ⋅ 2 ⋅1)
15. To score 80, 90, or 100, exactly 8, 9, or 10
questions must be correct, respectively. The
number of ways in which 8 of 10 questions can
be correct is
10!
10!
10 ⋅ 9 ⋅ 8!
=
=
= 45 .
10 C8 =
8!(10 – 8)! 8! ⋅ 2! 8! ⋅ 2 ⋅1
For 9 of 10 questions, the number of ways is
10!
10! 10 ⋅ 9!
=
=
= 10 ,
10 C9 =
9!(10 – 9)! 9! ⋅1! 9! ⋅1
and for 10 of 10 questions, it is
10!
10!
=
=1.
10 C10 =
10!(10 – 10)! 10! ⋅ 0!
Thus the number of ways to score 80 or better is
45 + 10 + 1 = 56.
21. Since the order in which the calls are made is
important, the number of possible schedules for
the 6 calls is 6 P6 = 6! = 720 .
22. The number of ways to place the 12 members in
three specific cars (cells), with 4 members in
12!
= 34, 650 .
each car, is
4! ⋅ 4! ⋅ 4!
23. The number of ways to assign 9 scientists so 3
work on project A, 3 work on B, and 3 work on
9!
= 1680 .
C is
3!3!3!
16. Each of the 11 games can be assigned to one of
three cells: a win cell, a loss cell, or a tie cell.
The number of ways to have 4 wins, 5 losses,
and 3 ties is
11!
11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5!
=
= 6930.
4! ⋅ 5! ⋅ 2!
4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 5!⋅ 2 ⋅1
24. There are 9 holly bushes, 5 of which are female,
and 4 of which are male. Then the number of
possible distinguishable arrangements is
9!
9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5!
=
= 126.
5! ⋅ 4! 5! ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
17. The word MISSISSAUGA has 11 letters with
repetition: one M, two I’s, four S’s, two A’s, one
U, and one G. Thus the number of
distinguishable arrangements is
11!
11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4!
=
1! ⋅ 2! ⋅ 4! ⋅ 2! ⋅1! ⋅1!
(2)4!(2)
= 415,800.
25. A response to the true-false questions can be
considered an ordered arrangement of 10 letters,
5 of which are T’s and 5 of which are F’s. The
number of different responses is
10! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5!
=
= 252 .
5!⋅ 5! 5!(5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1)
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26. The order in which the 7 food items are placed is
important. However, there are 3 hamburgers
(type 1), 2 cheeseburgers (type 2), and 2 steak
sandwiches (type 3). Then the number of
possible distinguishable ways of placing the
7!
= 210 .
items is
3! ⋅ 2! ⋅ 2!
12 C8 ⋅ 7 C4
12!
7!
⋅
8!(12 – 8)! 4!(7 – 4)!
12!
7!
12 ⋅11 ⋅10 ⋅ 9 ⋅ 8! 7 ⋅ 6 ⋅ 5 ⋅ 4!
⋅
=
⋅
8! ⋅ 4! 4! ⋅ 3!
8! ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 4! ⋅ 3 ⋅ 2 ⋅1
= 495 ⋅ 35 = 17,325 .
=
32. Suppose the possible games are numbered
1, 2, 3, ..., 7. The order in which four games are
won is not important. The number of ways that 4
of the possible 7 games can be won is
7!
7!
=
= 35 .
7 C4 =
4!(7 – 4)! 4! ⋅ 3!
27. The number of ways to assign 15 clients to 3
caseworkers (cells) with 5 clients to each
15!
= 756, 756.
caseworker is
5! ⋅ 5! ⋅ 5!
28. The number of ways of selecting 5 of the 10
remaining members so that order is important is
10!
10! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5!
=
=
10 P5 =
(10 − 5)! 5!
5!
= 30, 240.
29. a.
=
33. a.
Selecting 3 of the 3 males can be done in
only 1 way.
b. Selecting 4 of the 4 females can be done in
only 1 way.
Seven flags must be arranged: two are red
(type 1), three are green (type 2), and two
are yellow (type 3). Thus the number of
distinguishable arrangements (messages) is
7!
= 210.
2! ⋅ 3! ⋅ 2!
c.
b. If exactly two yellow flags are used, then
seven flags are involved and the number of
7!
= 210. If all
different messages is
2! ⋅ 3! ⋅ 2!
three yellow flags are used, then eight flags
are involved and the number of different
8!
= 560. Thus if at least
messages is
2! ⋅ 3! ⋅ 3!
two yellow flags are used, the number of
different messages is 210 + 560 = 770.
Selecting 2 males and 2 females can be
considered as a two-stage process. In the
first stage, 2 of the 3 males are selected (and
order is not important), which can be done
in 3 C2 ways. In the second stage, 2 of the
4 females are selected, which can be done in
4 C2 ways. By the basic counting principle,
the ways of selecting the subcommittee is
3!
4!
⋅
3 C2 ⋅ 4 C2 =
2!(3 – 2)! 2!(4 – 2)!
=
3!
4!
⋅
= 3 ⋅ 6 = 18
2! ⋅1! 2! ⋅ 2!
34. Exactly 2, 3, or 4 females can serve on the
subcommittee. Following the procedure in
Problem 33(c), the number of ways exactly 2
females can serve is 4 C2 ⋅ 4C2 . The number of
ways exactly 3 females can serve is 4 C3 ⋅ 4 C1 .
The number of ways exactly four females can
serve is 1. Thus the number of ways that at least
2 females can serve on the subcommittee is
4 C2 ⋅ 4 C2 + 4 C3 ⋅ 4 C1 + 1
30. Of the 10 applicants, 4 will be hired for the
assembly department (cell 1), 2 for the shipping
department (cell 2), and 4 will not be hired
(cell 3). Thus the number of ways to fill the
10!
= 3150 .
positions is
4! ⋅ 2! ⋅ 4!
4!
4!
4!
4!
⋅
+
⋅
+1
2! ⋅ 2! 2! ⋅ 2! 3! ⋅1! 1! ⋅ 3!
= 6 · 6 + 4 · 4 + 1 = 36 + 16 + 1 = 53.
=
31. The order in which the securities go into the
portfolio is not important. The number of ways
to select 8 of 12 stocks is 12 C8 . The number of
ways to select 4 of 7 bonds is 7 C4 . By the basic
counting principle, the number of ways to create
the portfolio is
35. There are 4 cards of a given denomination and
the number of ways of selecting 3 cards of that
denomination is 4 C3 .
Since there are 13 denominations, the number of
ways of selecting 3 cards of one denomination is
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ISM: Introductory Mathematical Analysis
Section 8.3
13 ⋅ 4C3 . After that selection is made, the 2 other
cards must be of the same denomination (of
which 12 denominations remain). Thus for the
remaining 2 cards there are 12 ⋅ 4 C2 selections.
By the basic counting principle, the number of
possible full-house hands is
4!
4!
13 ⋅ 4C3 ⋅12 ⋅ 4C2 = 13 ⋅
⋅12 ⋅
3! ⋅1!
2! ⋅ 2!
= 13 · 4 · 12 · 6 = 3744.
Principles in Practice 8.3
1. This is a combination problem because the order
in which the videos are selected is not important.
The number of possible choices is the number of
ways 3 videos can be selected from 400 without
regard to order.
400!
400!
=
400 C3 =
3!(400 – 3)! 3!397!
400 ⋅ 399 ⋅ 398 ⋅ 397!
3!397!
400 ⋅ 399 ⋅ 398
=
3⋅ 2
= 10,586,800
=
36. There are 13 denominations and four cards of
each denomination. The number of ways to get a
pair of 8’s is 4 C2 . For the other pair, there are
12 denominations left to choose from, so 12 C1
possibilities, with 4 C2 ways to get such a pair.
For the last card there are 11 denominations left,
with 4 cards in each denomination. By the basic
counting principle the number of two-pair hands
where one pair is 8’s is
4 C2 ⋅ 12 C1 ⋅ 4 C2 ⋅11 ⋅ 4
4!
12!
4!
=
⋅
⋅
⋅11 ⋅ 4
2! ⋅ 2! 1! ⋅11! 2! ⋅ 2!
= 19, 008.
Problems 8.3
1. {9D, 9H, 9C, 9S]
2. {HHHH, HHHT, HHTH, HHTT, HTHH,
HTHT, HTTH, HTTT, THHH, THHT, THTH,
THTT, TTHH, TTHT, TTTH, TTTT}
3. {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H,
6T}
37. This situation can be considered as placing 18
tourists into 3 cells: 7 tourist go to the
7-passenger tram, 8 go to the 8-passenger tram,
and 3 tourists remain at the bottom of the
mountain. This can be done in
18!
= 5,250,960 ways.
7! ⋅ 8! ⋅ 3!
4. {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
38. a.
7. a.
5. [64, 69, 60, 61, 46, 49, 40, 41, 96, 94, 90, 91, 06,
04, 09, 01, 16, 14, 19, 10]
6. {BBBB, BBBG, BBGB, BBGG, BGBB, BGBG,
BGGB, BGGG, GBBB, GBBG, GBGB, GBGG,
GGBB, GGBG, GGGB, GGGG}
The 10 students are to be placed in 3 groups,
with 4 in group A, 3 in group B, and 3 in
group C. This can be done in
10!
= 4200 ways.
4! ⋅ 3! ⋅ 3!
{RR, RW, RB, WR, WW, WB, BR, BW,
BB};
b. {RW, RB, WR, WB, BR, BW}
8. {ADF, ADG, AEF, AEG, BDF, BDG, BEF,
BEG, CDF, CDG, CEF, CEG}
b. For a given assignment of students to the
three groups, the number of ways of
selecting a group leader and a secretary for
group A (order is important) is 4 P2 ; for
group B, it is 3 P2 ; and for group C it is
3 P2 . Thus the number of ways that the
instructor can split the class into 3 groups
and designate a group leader and secretary
in each group is
4200 ⋅ 4 P2 ⋅ 3 P2 ⋅ 3 P2
= 4200(4 ⋅ 3)(3 ⋅ 2)(3 ⋅ 2) = 1,814,400.
9. Sample space consists of ordered sets of six
elements and each element is H or T. Since there
are two possibilities for each toss (H or T), and
there are six tosses, by he basic counting
principle, the number of sample points is
2 · 2 · 2 · 2 · 2 · 2 = 26 = 64 .
10. Sample space consists of ordered sets of five
elements where each element is an integer
between 1 and 6 inclusive. Since there are six
possibilities for each die, and there are 5 dice, by
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ISM: Introductory Mathematical Analysis
22. ( E ∪ G ) ∩ F ′
= {1, 2,3, 4,5, 6,8} ∩ {1, 2, 4, 6,8,10}
= {1, 2, 4, 6, 8}
the basic counting principle, the number of
sample points is 6 · 6 · 6 · 6 · 6 = 6 = 7776 .
5
11. Sample space consists of ordered pairs where the
first element indicates the card drawn (52
possibilities) and the second element indicates
the number on the die (6 possibilities). By the
basic counting principle, the number of sample
points is 52 · 6 = 312.
23. E1 ∩ E2 ≠ ∅ ; E1 ∩ E3 ≠ ∅ ; E1 ∩ E4 = ∅ ;
E2 ∩ E3 = ∅ ; E2 ∩ E4 ≠ ∅ ; E3 ∩ E4 = ∅ .
Thus E1 and E4 , E2 and E3 , and E3 and E4
are mutually exclusive.
12. Sample space consists of ordered sets of four
elements where the elements and their position
indicate the rabbit selected on the respective
draw. Since the rabbits are not replaced, for the
first draw there are 9 possibilities, for the second
draw there are 8 possibilities, and for the third
and fourth there are 7 and 6 possibilities,
respectively. By the basic counting principle, the
number of sample points is 9 · 8 · 7 · 6 = 3024.
24. If both cards are jacks, then both cards can
neither be clubs nor 3’s. Thus E J ∩ EC = ∅ and
E J ∩ E3 = ∅. If both cards are clubs, then both
cards cannot be 3’s. Thus EC ∩ E3 = ∅.
E J and EC , E J and E3 , EC and E3 are
mutually exclusive.
25. E ∩ F ≠ ∅ , E ∩ G = ∅ , E ∩ H ≠ ∅,
E ∩ I ≠ ∅ , F ∩G ≠ ∅ , F ∩ H ≠ ∅
F ∩ I = ∅, G ∩ H = ∅ , G ∩ I = ∅,
H ∩ I ≠ ∅. Thus E and G, F and I, G and H, and
G and I are mutually exclusive.
13. Sample space consists of combinations of
52 cards taken 10 at a time. Thus the number of
sample points is 52 C10 .
14. Sample space consists of all four letter “words.”
For each of the four letters there are 26
possibilities. By the basic counting principle, the
number of sample points is
26. E ∩ F = ∅ , E ∩ G = ∅ , E ∩ H ≠ ∅ ,
E ∩ I ≠ ∅ , F ∩G ≠ ∅ , F ∩ H ≠ ∅ ,
F ∩I =∅, G∩H =∅, G∩I =∅,
H ∩I =∅.
Thus E and F, E and G, F and I, G and H,
G and I, H and I are mutually exclusive.
26 · 26 · 26 · 26 = 264 = 456,976 .
15. The sample points that are either in E, or in F, or
in both E and F are 1, 3, 5, 7, and 9. Thus
E ∪ F = {1, 3, 5, 7, 9}.
27. a.
16. The sample points in S that are not in G are
1, 3, 5, 7, 9, and 10. Thus
G ′= {1, 3, 5, 7, 9, 10}.
S = {HHH, HHT, HTH, HTT, THH, THT,
TTH, TTT}
b.
E1 = {HHH, HHT, HTH, HTT, THH, THT,
TTH}
c.
E2 = {HHT, HTH, HTT, THH, THT, TTH,
TTT}
d.
18. F ′ = {1, 2, 4, 6, 8, 10} and
G ′ = {1, 3, 5, 7, 9, 10}, so F ′ ∩ G ′ = {1, 10}.
E1 ∪ E2 = {HHH, HHT, HTH, HTT, THH,
THT, TTH, TTT} = S
e.
E1 ∩ E2 = {HHT, HTH, HTT, THH, THT,
TTH}
19. The sample points in S that are not in F are
1, 2, 4, 6, 8, and 10. Thus F′ = {1, 2, 4, 6, 8, 10}.
f.
( E1 ∪ E2 )′ = S ′ = ∅
20. ( E ∪ F )′ = {1,3,5, 7,9}′ = {2, 4, 6,8,10}
g.
( E1 ∩ E2 )′ = {HHT, HTH, HTT, THH,
17. The sample points in S that are not in E are 2, 4,
6, 7, 8, 9, and 10. Thus E ′ = {2, 4, 6, 7,8,9,10}.
The sample points common to both E ′ and F are
7 and 9. Thus E ′ ∩ F = {7, 9}.
THT, TTH}′ = {HHH, TTT}
21. ( F ∩ G )′ = ∅′ = S
28. a.
302
{BB, BG, GB, GG}
ISM: Introductory Mathematical Analysis
Section 8.4
b. {BG, GB, GG}
c.
Problems 8.4
{BB, BG, GB}
1. 3000P(E) = 3000(0.25) = 750
d. No; {BG, GB, GG}′ = {BB} ≠ event in (c)
29. a.
2. 3000P(E) = 3000[1 – P(E′)] = 3000(1 – 0.45)
= 3000(0.55) = 1650
{ABC, ACB, BAC, BCA, CAB, CBA}
3. a.
b. {ABC, ACB}
c.
30. a.
b.
{BAC, BCA, CAB, CBA}
{UUV, UUW, UUX, UUZ, UVV, UVW,
UVX, UVZ, UXV, UXW, UXX, UXZ,
UYV, UYW, UYX, UYZ, VUV, VUW,
VUX, VUZ, VVV, VVW, VVX, VVZ,
VXV, VXW, VXX, VXZ, VYV, VYW,
VYX, VYZ, WUV, WUW, WUX, WUZ,
WVV, WVW, WVX, WVZ, WXV, WXW,
WXX, WXZ, WYV, WYW, WYX, WYZ}
4. a.
b.
P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F )
= 0.2 + 0.3 – 0.1 = 0.4
1 3
=
4 4
P(E´) = 1 – P(E) = 1 –
P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F )
=
1 1 1 5
+ – =
4 2 8 8
5. If E and F are mutually exclusive, then
E∩F =∅ .
Thus P ( E ∩ F ) = P (∅) = 0 . Since it is given
that P ( E ∩ F ) = 0.831 ≠ 0 , E and F are not
mutually exclusive.
b. {VVV}
c.
P(E′) = 1 – P(E) = 1 – 0.2 = 0.8
{UUV, UUW, UUX, UUZ, UVV, UVW,
UVX, UVZ, UXV, UXW, UXX, UXZ,
UYV, UYW, UYX, UYZ, VUV, VUW,
VUX, VUZ, VVW, VVX, VVZ, VXV,
VXW, VXX, VXZ, VYV, VYW, VYX,
VYZ, WUV, WUW, WUX, WUZ, WVV,
WVW, WVX, WVZ, WXV, WXW, WXX,
WXZ, WYV, WYW, WYX, WYZ}
More than one supplier is used.
6. P ( E ∪ F ) = P ( E ) + P( F ) − P( E ∩ F )
Thus P ( F ) = P ( E ∪ F ) + P( E ∩ F ) − P( E )
13 1 1 1
=
+ − = .
20 10 2 4
31. Using the properties in Table 8.1, we have
( E ∩ F ) ∩ ( E ∩ F ′)
= ( E ∩ F ∩ E ) ∩ F ′ [property 15]
= ( E ∩ E ∩ F ) ∩ F ′ [property 11]
= ( E ∩ E ) ∩ ( F ∩ F ′) [porperty 15]
[property 5]
= E ∩∅
[property 9]
=∅
Thus
( E ∩ F ) ∩ ( E ∩ F ′) = ∅, so E ∩ F and E ∩ F ′
are mutually exclusive.
7. a.
E8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
P ( E8 ) =
b.
n( E8 ) 5
=
n( S ) 36
E2 or 3 = {(1, 1), (1, 2), (2, 1)}
P ( E2 or 3 ) =
c.
32. Using the properties in Table 8.1, we have
( E ∩ F ) ∪ ( E ∩ F ′)
[property 16]
= E ∩ ( F ∪ F ′)
[property 4]
= E∩S
[property 7]
=E
n( E2 or 3 ) 3
1
=
=
36 12
n( S )
E3, 4, or 5 = {(1, 2), (2,1), (1,3), (2, 2), (3,1),
(1, 4), (2,3), (3, 2), (4,1)}
P ( E3, 4, or 5 ) =
d.
n( E3, 4,
or 5 )
n( S )
=
9 1
=
36 4
E12 or 13 = E12 , since E13 is an impossible
event.
E12 = {(6, 6)}
P ( E12 or 13 ) =
303
n( E12 or 13 ) 1
=
n( S )
36
Chapter 8: Introduction to Probability and Statistics
e.
ISM: Introductory Mathematical Analysis
E2 = {(1,1)}
e.
E4 = {(1,3), (2, 2), (3,1)}
E6 = {(1,5), (2, 4), (3,3, ), (4, 2), (5,1)}
E8 = {(2, 6), (3,5), (4, 4), (5,3), (6, 2)}
E10 = {(4, 6), (5,5), (6, 4)}
E12 = {(6, 6)}
Because a heart is not a club,
Eheart ∩ Eclub = ∅ .
Thus
P ( Eheart or club ) = P( Eheart ∪ Eclub )
= P ( Eheart ) + P ( Eclub )
=
n( Eheart ) n( Eclub ) 13 13
+
=
+
n( S )
n( S )
52 52
=
26 1
=
52 2
P ( Eeven ) = P( E2 ) + P ( E4 )
+ P ( E6 ) + P ( E8 ) + P ( E10 ) + P ( E12 )
=
1
3
5
5
3
1 18 1
+
+
+
+
+
=
=
36 36 36 36 36 36 36 2
f.
P ( Eclub and 4 ) =
f.
1 1
P ( Eodd ) = 1 – P ( Eeven ) = 1 – =
2 2
g.
′ than 10 = E10 ∪ E11 ∪ E12
Eless
= {(4, 6), (5,5), (6, 4)} ∪ {(5, 6), (6,5)} ∪ {(6, 6)}
= {(4, 6), (5,5), (6, 4), (5, 6), (6,5), (6, 6)} .
′ than 10 )
P ( Eless than 10 ) = 1 – P ( Eless
=1–
6 30 5
=
= .
36 36 6
g.
P(club or 4)
= P(club) + P(4) – P(club and 4)
13 4
1 16 4
=
+
–
=
=
52 52 52 52 13
h.
Ered and king = {KH, KD}
=
i.
b. P(diamond) =
n( S )
=
1
52
n( E jack )
=
b.
d
n( Ered ) 26 1
P(red) =
=
=
n( S )
52 2
n( S )
Espade and heart = ∅
EH,5 = {H5}
c.
304
n( S )
=
1
12
n( Ehead ) 6 1
=
=
n( S )
12 2
n( E3 ) = 2 ⋅1 = 2
P(3) =
d.
n( EH,5 )
n( Ehead ) = 1 ⋅ 6 = 6 .
P(head) =
4
1
=
52 13
P(jack) =
2
1
=
52 26
P(head and 5) =
n( E diamond ) 13 1
=
=
n( S )
52 4
c.
n( S )
10. n(S) = 2 · 6 = 12
a.
n( Eking of hearts )
n(Ered and king )
Thus P(spade and heart) = 0
9. n(S) = 52.
P(king of hearts) =
n( Eclub and 4 ) 1
=
n( S )
52
P (red and king)=
8. E2 or 3 shows = {(2, 1), (2, 2), (2, 3), (2, 4),
(2, 5), (2, 6), (3, 1), (3, 2), (3, 3),
(3, 4), (3, 5), (3, 6), (1, 2), (4, 2),
(5, 2), (6, 2), (1, 3), (4, 3), (5, 3),
(6, 3)}
n( E2 or 3 shows ) 20 5
=
=
P ( E2 or 3 shows ) =
n( S )
36 9
a.
Eclub and 4 = {4C}
n( E3 ) 2
1
=
=−
n( S ) 12
6
n( Ehead and even ) = 1 ⋅ 3 = 3
P(head and even)
n( Ehead and even ) 3 1
=
=
=
n( S )
12 4
ISM: Introductory Mathematical Analysis
Section 8.4
11. n(S) = 2 · 6 · 52 = 624
a.
14. n(S) = 52 · 52 = 2704
P(tail, 3, queen of hearts)
n( ET,3,QH ) 1 ⋅1 ⋅1
1
=
=
=
n( S )
624 624
a.
=
b. P(tail, 3, queen)
n( ET,3,Q ) 1 ⋅1 ⋅ 4
1
=
=
=
n( S )
624 156
c.
P(head, 2 or 3, queen)
n( EH,2 or 3,Q ) 1 ⋅ 2 ⋅ 4 1
=
=
=
n( S )
624
78
b.
a.
E3 heads = {HHH}
n( E3 heads ) 1
=
n(S )
8
d.
E1 tail = {HHT,HTH,THH} .
b.
P(no more than 2 heads) = 1 – P(3 heads)
1 7
= 1– =
8 8
c.
b.
P (all hearts) =
n( E3 girls )
n( S )
=
1
8
E1 boy = {BGG,GBG,GGB}
n( E1 boy )
=
3
8
n( Eno girl )
=
n( S )
Eno girl = {BBB}
P(no girl) =
E no more than 1 tail = E0 tails ∪ E1 tail
= {HHH} ∪ {HHT,HTH,THH}
= {HHH, HHT, HTH, THH}.
P(no more than 1 tail)
n( Eno more than 1 tail ) 4 1
= =
=
8 2
n( S )
P (all kings) =
E3 girls = {GGG}
P(1 boy) =
n( S )
1
8
d. P(at least 1 girl) = 1 – P(no girl)
1 7
=1– =
8 8
16. The sample space consists of 18 jelly beans.
Thus n(S) = 18.
13. n(S) = 52 · 51 · 50 = 132,600
a.
1
169
P(3 girls) =
n( E1 tail ) 3
P(1 tail) =
=
n( S )
8
c.
4⋅4
2704
15. n(S) = 2 · 2 · 2 = 8
12. n(S) = 8
P(3 heads) =
n( S )
=
b. Number of ways both cards are king of
hearts: 1. Number of ways either first card is
king of hearts and second card is a different
heart, or vice versa: 2(1 · 12) = 24. Number
of ways either first card is king of diamonds,
clubs, or spades, and second card is a heart,
or vice versa: 2(3 · 13) = 78. Thus, number
ways one card is a king and the other is a
heart is 1 + 24 + 78 = 103, so probability of
103
.
given event is
2704
d. P(head, even, diamond)
n( EH,E,D ) 1 ⋅ 3 ⋅13 1
=
=
=
n( S )
624
16
a.
n( Eboth kings )
P(both kings) =
a.
4⋅3⋅ 2
1
=
132, 600 5525
P(blue) =
n( Eblue ) 8 4
=
=
n( S )
18 9
b. P(not red) = 1 – P(red)
n( Ered )
7 11
=1–
=1–
=
n( S )
18 18
13 ⋅12 ⋅11 11
=
132, 600 850
305
Chapter 8: Introduction to Probability and Statistics
c.
ISM: Introductory Mathematical Analysis
c.
The events of drawing a red jelly bean and
drawing a white jelly bean are mutually
exclusive. Thus
P(red or white) = P(red) + P(white)
7 3 10 5
+
=
=
=
18 18 18 9
d. P(neither red nor blue) = P(white) =
e.
Eyellow = ∅ . Thus P(yellow) = 0
f.
Ered ∩ Eyellow = ∅
d. P(no F) = 1 – P(F) = 1 –
3 1
=
18 6
=1–
e.
Thus P(red or yellow) = P(red) + P(yellow)
7
7
.
+0 =
=
18
18
17. The sample space consists of 60 stocks. Thus
n(S) = 60.
a.
P(6% or more) =
=
n( E6% or more )
n( S )
48 4
=
60 5
b. P(less than 6%) = 1 – P(6% or more)
4 1
=1– =
5 5
18. Let N = number of ties. Then the number of pure
silk ties is 0.4N.
a.
b.
P (100% pure silk) =
0.4 N
= 0.4
N
a.
P (not 100% silk) = 1 − P(100% pure silk)
= 1 − 0.4 = 0.6
P(A) =
=
2 38
=
= 0.95
40 40
Let N = number of students. Then n(S) = N.
Of the N students, 0.10N received an A,
0.25N a B, 0.35N a C, 0.25N a D, 0.05N an
F.
0.10 N
= 0.1
P(A) =
N
0.10 N + 0.25 N
P(A or B) =
N
0.35 N
=
= 0.35
N
P(neither D nor F) = P(A, B, or C)
0.10 N + 0.25 N + 0.35 N
=
N
0.70 N
=
= 0.7
N
P(no F) = 1 – P(F)
0.05 N
= 1 – 0.05 = 0.95
=1–
N
P(both red) =
n( ER,R )
n( S )
=
3⋅ 4 4
=
45 15
b. P(one red and other green)
n( ER,G ) + n( EG , R ) 3 ⋅ 5 + 2 ⋅ 4
=
=
n( S )
45
=
n( EA ) 4
1
=
=
= 0.1
n( S )
40 10
b. P(A or B) =
n ( EF )
n( S )
20. Bag 1 contains 5 jelly beans, and Bag 2 contains
9.
n(S) = 5 · 9 = 45.
19. n(S) = 40
Of the 40 students, 4 received an A, 10 a B, 14 a
C, 10 a D, and 2 an F.
a.
P(neither D nor F) = P(A, B, or C)
n( EA,B, or C ) 4 + 10 + 14 28
=
=
= 0.7
=
n( S )
40
40
n( EA or B ) 4 + 10
=
n( S )
40
14
= 0.35
40
306
15 + 8 23
=
45
45
ISM: Introductory Mathematical Analysis
Section 8.4
21. The sample space consists of combinations of 2
people selected from 5. Thus
5!
5⋅4
=
= 10 . Because there
n(S) = 5 C2 =
2! ⋅ 3!
2
are only 2 women in the group, the number of
possible 2-woman committees is 1. Thus
n( E2 women ) 1
=
.
P(2 women) =
n( S )
10
4
22. Because there are 3 men and 2 women, the
number of possible committees consisting of a
man and a woman is 3 · 2 = 6.
Thus
n( Eman and woman )
.
P(man and woman) =
n( S )
=
34 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4!
34 2 ⋅ 7 ⋅ 5
⋅
⋅
=
1
48 4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 4!
48
2835
.
=
32, 768
=
6 3
= .
10 5
25. A poker hand is a 5-card deal from 52 cards.
Thus n( S ) = 52 C5 . In 52 cards, there are 4 cards
of a particular denomination. Thus, for a four of
a kind, the number of ways of selecting 4 of 4
cards of a particular denomination is 4 C4 . Since
there are 13 denominations, 4 cards of the same
denomination can be dealt in 13 ⋅ 4C4 ways. For
the remaining card, there are 12 denominations
that are possible, and for each denomination
there are 4 C1 ways of dealing a card. Thus
23. Number of ways to answer exam is
210 = 1024 = n( S ) .
a.
There is only one way to achieve 100 points,
namely to answer each question correctly.
Thus
n( E100 points )
1
=
.
P(100 points) =
n( S )
1024
n( Efour of a kind )
n( S )
13 ⋅ 4C4 ⋅12 ⋅ 4C1
=
52 C5
13 ⋅12 ⋅ 4
=
52 C5
P(four of a kind) =
b. Number of ways to score
90 points = number of ways that exactly one
question is answered incorrectly = 10.
Thus
P(90 or more points)
= P(90 points) + P(100 points)
10
1
11
+
=
.
=
1024 1024 1024
26. a.
24. Number of ways to answer exam is
48 = 65,536 = n( S ) .
a.
4
34
⎛1⎞ ⎛3⎞
. Since there
incorrectly is ⎜ ⎟ ⎜ ⎟ =
⎝4⎠ ⎝4⎠
48
are 8 C4 distinguishable orders in which one
can arrange 4 correct and 4 incorrect
answers, and since each arrangement has the
same overall probability of occurring, the
probability of 4 correct and 4 incorrect
34
34 8!
⋅ 8 C4 =
⋅
answers is
48
48 4!4!
b.
n( Eall correct )
1
=
P(all correct) =
n( S )
65,536
b. The probability of answering one question
correctly when answering in a random
1
and the probability of
fashion is
4
3
answering incorrectly is . Thus, the
4
probability of answering the first four
questions correctly and the last four
P( E ∪ F ) = P( E ) + P( F ) − P( E ∩ F )
Thus P ( F ) = P ( E ∪ F ) + P ( E ∩ F ) − P ( E )
5 1 1 1
=
+ − = .
14 7 4 4
P ( E ′ ∪ F ) = P ( E ′) + P( F ) – P( E ′ ∩ F )
⎛ 1⎞ 1
= ⎜1 − ⎟ + − P( E ′ ∩ F )
⎝ 4⎠ 4
= 1 − P( E ′ ∩ F )
Since F = ( E ∩ F ) ∪ ( E ′ ∩ F )
and E ∩ F and E ′ ∩ F are mutually
exclusive P ( F ) = P( E ∩ F ) + P( E ′ ∩ F ) ,
1 1
= + P( E ′ ∩ F )
4 7
307
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
30. Here Shiloh needs to win 5 more rounds to win
the game and Caitlin needs to win 8 more
rounds. Shiloh’s probability of winning is
7
C
3302 1651
∑ 12212k = 4096 = 2048 . Thus Shiloh’s share
k =0
1 1 3
− = . Hence,
4 7 28
3 25
P( E ′ ∪ F ) = 1 −
=
.
28 28
Thus P ( E ′ ∩ F ) =
27. n(S) =
a.
100 C3
=
100!
= 161, 700
3! ⋅ 97!
of the pot is
31. Let p = P(1) = P(3) = P(5). Then
2p = P(2) = P(4) = P(6). Since P(S) = 1, then
1
3(p) + 3(2p) = 1, 9p = 1, p = p (1) = .
9
35!
= 6545
3!⋅ 32!
n( E3 females )
P ( E3 females ) =
n( S )
n( E3 females ) = 35C3 =
=
32. Let p1 = P (a) = P(b) = P(c) = P(d ) = P(e), and
p2 = P ( f ) = P( g ). Then
6545
≈ 0.040
161, 700
1 5
− p1.
2 2
Since p1 is not known, it is not possible to
determine P ( f ) = p2 . If it is also known that
P ( S ) = 5( p1 ) + 2( p2 ) = 1, p2 =
b. The number of ways of selecting one
professor is 15; the number of ways of
selecting two associate professors is 24 C2 .
Thus n( E1 professor & 2 associate professors )
1
P ({a, f }) = , then we have
3
24!
= 15 ⋅ 276 = 4140 .
2! ⋅ 22!
Therefore,
P ( E1 professor & 2 associate professers )
= 15 ⋅
=
1
P ({a, f }) = P(a ) + P( f ) = p1 + p2 = .
3
1
1 5⎛1
⎞
Thus p1 = − p2 and p2 = − ⎜ − p2 ⎟ .
3
2 2⎝3
⎠
4140
≈ 0.026 .
161, 700
−
28. P(even number) = P(2) + P(4) + P(6)
2 1 1
4 2
=
+ +
=
=
10 10 10 10 5
33. a.
29. Shiloh needs to win 3 more rounds to win the
game and Caitlin needs to win 5 more rounds.
Shiloh’s probability of winning is
4
∑
7 Ck
7
3
1
2
2
p2 = − or p2 = and so P ( f ) = .
2
3
9
9
Of the 100 voters, 51 favor the tax increase.
51
Thus P(favors tax increase) =
= 0.51 .
100
b. Of the 100 voters, 44 oppose the tax
increase. Thus
44
P(opposes tax increase) =
= 0.44 .
100
1 4
∑ 7 Ck
27 k = 0
1
=
(7 C0 + 7 C1 + 7 C2 + 7 C3 + 7 C4 )
27
1
=
(1 + 7 + 21 + 35 + 35)
27
99
=
128
Shiloh’s share of the pot is then
99
($25) ≈ $19.34.
128
k =0
=
1651
($50) ≈ $40.31.
2048
2
308
c.
Of the 100 voters, 3 are Republican with no
opinion. Thus
3
P(is a Republican with no opinion) =
100
= 0.03 .
34. a.
For the chain, the total average number of
sales is 170 units. For brand B, 65 units per
month are sold. Thus
65 13
=
≈ 0.38 .
P(sale is for brand B) =
170 34
ISM: Introductory Mathematical Analysis
Section 8.5
b. Since 95 units per month are sold at the
Exton store, and 30 are of brand C,
P(sale is for brand C given that it is at Exton
30 6
=
≈ 0.32 .
store) =
95 19
Problems 8.5
1. a.
P(E F ) =
n( E ∩ F ) 1
=
n( F )
5
b. Using the result of part (a),
35.
P( E )
P( E )
=
=
P( E ′) 1 – P( E ) 1 –
4
5
( 54 )
=
4
5
1
5
=
P( E ′ | F ) = 1 – P( E | F ) = 1 –
4
1
The odds are 4:1.
c.
1
P( E )
P( E )
=
= 6 =
36.
P( E ′) 1 – P( E ) 1 – 1
6
( )
37.
38.
1
6
5
6
1
=
5
F ′ = {3, 7,8, 9} so
P ( E | F ′) =
n( E ∩ F ′) 1
= .
n( F ′)
4
n( F ∩ E ) 1
=
n( E )
2
The odds are 1:5.
d.
P( F | E ) =
P( E )
P( E )
0.7
0.7 7
=
=
=
=
P( E ′) 1 – P( E ) 1 – 0.7 0.3 3
The odds are 7:3.
e.
F ∩ G = {5, 6} so
P( E | F ∩ G ) =
P( E )
P( E )
0.001
0.001
1
=
=
=
=
P( E ′) 1 – P( E ) 1 – 0.001 0.999 999
The odds are 1:999.
2. a.
1 4
= .
5 5
P( E ) =
n( E ∩ ( F ∩ G )) 0
= = 0.
n( F ∩ G )
2
n( E ) 2
=
n( S ) 5
39. P ( E ) =
7
7
=
7 + 5 12
b.
P( E | F ) =
n( E ∩ F ) 0
= =0
n( F )
2
40. P ( E ) =
100
100
=
100 + 1 101
c.
P( E | G ) =
n( E ∩ G ) 2
=
n(G )
3
41. P( E ) =
4
4 2
=
=
4 + 10 14 7
d.
P (G | E ) =
n(G ∩ E ) 2
= =1
n( E )
2
42. P( E ) =
a
a
1
=
=
a + a 2a 2
e.
F ′ = {1, 2,5}
P (G | F ′) =
43. Odds that it will rain tomorrow
P (rain)
0.75
0.75
=
=
=
=3.
P(no rain) 1 − 0.75 0.25
The odds are 3:1.
f.
P( E )
=
P ( E ′)
1
P ( E ′)
P( E )
=
1
3
5
E ′ = {3, 4,5}
P ( E ′ | F ′) =
44. The odds of E not occurring are the odds of
P ( E ′) P( E ′) 3
=
= . Then
event E ′ which is
P ( E ′′) P( E ) 5
n( E ′ ∩ F ′) 1
=
n( F ′)
3
3. P( E | E ) =
P( E ∩ E ) P( E )
=
=1
P( E )
P( E )
4. P(∅ | E ) =
P(∅ ∩ E ) P(∅)
0
=
=
=0
P( E )
P( E ) P( E )
5
= , so the odds that E does
3
occur are 5:3.
In general, if the odds of E not occurring are a:b,
then the odds that E does occur are b:a.
n(G ∩ F ′) 2
=
n( F ′)
3
5. P ( E ′ | F ) = 1 – P( E | F ) = 1 – 0.57 = 0.43
309
Chapter 8: Introduction to Probability and Statistics
6. P ( F | G ) =
7. a.
b.
ISM: Introductory Mathematical Analysis
10. P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F ), so
P ( E ∩ F ) = P( E ) + P( F ) – P( E ∪ F )
P ( F ∩ G ) P(∅)
0
=
=
=0
P (G )
P (G ) P (G )
3 3 7 1
+ –
=
5 10 10 5
P( E ∩ F ) 1/ 5
2
Then P(E|F) =
=
= .
P( F )
3 /10 3
=
P ( E ∩ F ) 1/ 6 1
P( E | F ) =
=
=
P( F )
1/ 3 2
P( F | E ) =
P ( F ∩ E ) 1/ 6 2
=
=
P( E )
1/ 4 3
11. a.
P( F ) =
8. First we find P ( E ∩ F ) :
P( E | F ) =
P( E ∩ F )
,
P( F )
P( E ∩ F ) = P ( E F ) P( F ) =
3 1 1
⋅ = .
4 3 4
Then
P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F )
1 1 1
= + −
4 3 4
1
= .
3
9. a.
b.
P( F | E ) =
P ( F ∩ E ) 1/ 6 2
=
=
P( E )
1/ 4 3
c.
d.
P ( F | II) =
c.
P(O | I) =
d.
P (III) =
e.
P (III | O) =
n(III ∩ O) 10
=
n(O)
47
f.
P (II | N ′) =
n(II ∩ N ′)
n( N ′)
=
12. a.
From part (b) P( F ) =
b.
P ( E ) = P( E ∩ F ) + P( E ∩ F ′)
=
c.
d.
n(Public ∩ Middle)
n(Middle)
55 11
=
80 16
n(High ∩ Private)
n(Private)
14 2
=
49 7
n(Private ∩ High)
n(High)
14
25
P (Public ∪ Low)
=P (Public) + P (Low) – P (Public ∩ Low)
=
310
35 + 15
50 25
=
=
125 + 47 172 86
P(Private|High)=
=
1/12
1/12 1
=
= .
1 − 1/ 2 1/ 2 6
64
8
=
200 25
P(High|Private)=
=
1 1
= + P ( E ∩ F ′)
4 6
1 1 1
so P ( E ∩ F ′) = – = .
4 6 12
P ( E ∩ F ′)
Then P( E | F ′) =
P( F ′)
n(O ∩ I) 22 11
=
=
n(I)
78 39
P (Public|Middle)=
=
1
.
2
P ( E ∩ F ) 1/ 6 1
Then P( E | F ) =
=
= .
P( F )
1/ 2 3
n( F ∩ II) 35
=
n(II)
58
b.
P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F )
7 1
1
= + P( F ) –
12 4
6
7 1 1 1
Thus P ( F ) =
− + = ..
12 4 6 2
125 5
=
200 8
126 70 60 136
+
–
=
175 175 175 175
ISM: Introductory Mathematical Analysis
Section 8.5
18. S = {HHHH, HHHT, HHTH, HHTT, HTHH,
HTHT, HTTH, HTTT, THHH, THHT, THTH,
THTT, TTHH, TTHT, TTTH, TTTT}.
Let E = {four tails } = {TTTT}, F = {first toss is
a tail} = {THHH, THHT, THTH, THTT, TTHH,
TTHT, TTTH, TTTT}.
n( E ∩ F ) 1
= , the
Since P ( E | F ) =
n( F )
8
corresponding odds are
P( E | F )
1/ 8
1
=
= ; that is, 1 to 7.
P( E ′ | F ) 1 – (1/ 8) 7
13. a.
P(A | B) =
P(A ∩ B) 0.20 1
=
=
P(B)
0.40 2
b.
P(B | A) =
P(B ∩ A) 0.20 4
=
=
P (A)
0.45 9
14. P (scratched screen|def. ear pieces)
P(scratched screen ∩ def. ear pieces)
=
P(def. ear pieces)
0.13 13
=
=
0.19 19
19. P (< 4 | odd) =
15. S = {BB, BG, GG, GB}
Let E = {at least one girl} = {BG, GG, GB},
F = {at least one boy} = {BB, BG, GB}.
n( E ∩ F ) 2
= .
Thus P ( E | F ) =
n( F )
3
20. Let F denote face card. There are 3 face cards
for each suit. Let R denote red card. Half the
cards are red, so there are 26.
n( F ∩ R ) 6
3
P ( F R) =
=
= .
n( R )
26 13
16. S = {BBB, BBG, BGB, BGG, GBB, GBG,
GGB, GGG}
Let
E = {at least two girls}
= {BGG, GBG, GGB, GGG},
F = {at least one boy}
= {BBB, BBG, BGB, BGG, GBB, GBG, GGB},
G = {oldest is a girl}
= {GBB, GBG, GGB, GGG}.
a.
P( E | F ) =
n( E ∩ F ) 3
=
n( F )
7
b.
P( E | G ) =
n( E ∩ G ) 3
=
n(G )
4
21. Method 1. The usual sample space has 36
outcomes, where the event
“two 1’s” is {(1, 1)}. Note that
{at least one 1}′ ={no 1's} , and the event
“no 1’s” occurs in 5 · 5 = 25 ways. Thus
P(two 1’s | at least one 1)
n(two 1's ∩ at least one 1) n({(1,1)}) 1
=
=
=
36 – 25
11
n(at least one 1)
Method 2. From the usual sample space, we find
that the reduced sample space for “at least one
1” (which has 11 outcomes) is {(1, 1), (1, 2),
(1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1),
(5, 1)}.
1
.
Thus P(two 1’s | at least one 1) =
11
17. S = {HHH, HHT, HTH, THH, THT, TTH,
TTT}.
Let E = {exactly two tails}
= {HTT, THT, TTH},
F = {second toss is a tail}
= {HTH, HTT, TTH, TTT},
G = {second toss is a head}
= {HHH, HHT, THH, THT}.
a.
P( E | F ) =
n( E ∩ F ) 2 1
= =
n( F )
4 2
b.
P( E | G ) =
n( E ∩ G ) 1
=
n(G )
4
n(< 4 ∩ odd)
n({1,3})
2
=
=
n(odd)
n({1,3,5}) 3
22. Method 1. The reduced sample space, having 6
outcomes, is {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),
(5, 6)}, where, in each pair, the outcome 5 on the
red die is given first. Two pairs have a sum
greater than 9, namely (5, 5) and
2 1
(5, 6). Thus P (sum>9|5 on red) = = .
6 3
Method 2. The usual sample space has 36
outcomes. Let E = {5 on red}. Then n(E) = 6.
Let F = {sum > 9}. Then n( E ∩ F ) = 2 , namely
(red 5, green 5) and (red 5, green 6). Thus
n( E ∩ F ) 2 1
P( F | E ) =
= = .
n( E )
6 3
311
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
29. Let E = {second card is not a face card} and
F = {first card is a face card}.
51−11
40
n( E ∩ F ) 12 ⋅ 51
P( E | F ) =
=
=
12
51
n( F )
23. The usual sample space consists of ordered pairs
(R, G), where R = no. on red die and G = no. on
green die. Now, n(green is even) = 6 · 3 = 18,
because the red die can show any of six numbers
and the green any of three: 2, 4, or 6. Also,
n(total of 7 ∩ green even)
= n({(5, 2), (3, 4), (1, 6)}) = 3.
Thus
P (total of 7|green even)
=
n(total of 7 ∩ green even)
n(green even)
=
3 1
= .
18 6
30. a.
=
b.
P( E | F ) =
b.
P( E ∩ F ) =
12 12 3 3
9
⋅
= ⋅ =
52 52 13 13 169
31. P ( K1 ∩ Q2 ∩ J 3 )
= P ( K1 ) P(Q2 | K1 ) P ( J 3 | ( K1 ∩ Q2 ))
=
4 4 4
8
⋅ ⋅
=
52 51 50 16,575
32. P ( AS1 ∩ AH 2 ∩ AD2 )
= P ( AS1 ) P ( AH 2 AS1 ) P ( AD2 ( AS1 ∩ AH 2 ) )
1 1 1
1
=
⋅ ⋅
=
.
52 51 50 132, 600
n( E ∩ F )
3 1
=
=
n( F )
24 8
a.
12 11 11
⋅ =
52 51 221
P ( F1 ∩ F2 ) = P ( F1 ) P( F2 | F1 )
=
24. The usual sample space S consists of 36 ordered
pairs. Let E = {sum is 6} and
F = {second toss is neither 2 nor 4}.
Then n(F) = 6 · 4 = 24 and
n( E ∩ F ) = n{(5, 1), (3, 3), (1, 5)} = 3.
n( E ∩ F ) 3
1
=
=
n( S )
36 12
33. P ( J1 ∩ J 2 ∩ J 3 )
= P ( J1 ) P ( J 2 | J1 ) P( J 3 | ( J1 ∩ J 2 ))
25. The usual sample space consists of 36 ordered
pairs. Let E = {total > 7} and
F = {first toss > 3}. Then n( F ) = 3 ⋅ 6 = 18 and
n( E ∩ F )
= n({(4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5),
(5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)})
= 12
n( E ∩ F ) 12 2
=
= .
Thus P ( E F ) =
n( F )
18 3
=
4 3 2
1
⋅ ⋅ =
52 51 50 5525
34. Using a probability tree, we find that there are
two possible paths such that the second card is a
heart, namely, a heart followed by a heart, or a
nonheart followed by a heart. Thus
P ( H 2 ) = P( H1 ∩ H 2 ) + P ( H1′ ∩ H 2 )
= P ( H1 ) P( H 2 | H1 ) + P ( H1′ ) P ( H 2 | H1′ )
26. Let the sample space consist of ordered pairs
(c, d), where c is T or H, and d is the number
showing on the die. Let E = {tails shows} and
F = {die shows odd number). Then
N(F) = 2 · 3 = 6 and n( E ∩ F ) = 1 ⋅ 3 = 3 . Thus
=
13 12 39 13 1
⋅ + ⋅ = .
52 51 52 51 4
35. Let D = {two diamonds} and
R = {first card red}. We have
D ∩ R = {two diamonds} = D and
n( E ∩ F ) 3 1
P( E | F ) =
= = .
n( F )
6 2
P(D) =
n( K ∩ H ) 1
27. P ( K | H ) =
=
n( H )
13
28. P ( H | F ) =
P ( F1 ∩ F2 ) = P ( F1 ) P( F2 | F1 )
13 12
⋅ .
52 51
Thus P ( D | R ) =
n( H ∩ F ) 3 1
=
=
n( F )
12 4
312
P( D ∩ R)
=
P( R)
13 ⋅ 12
52 51
26
52
=
2
.
17
ISM: Introductory Mathematical Analysis
Section 8.5
36. Using a probability tree, we find that there are two possible paths such that she will be on time, namely, she gets
the call and she is on time, or she doesn’t get the call and she is on time.
P (T ) = P (C ∩ T ) + P(C ′ ∩ T )
= P (C ) P (T | C ) + P (C ′) P(T | C ′)
= (0.9)(0.9) + (0.1)(0.4) = 0.85
37. a.
b.
38. a.
P (U ) = P( F ∩ U ) + P (O ∩ U ) + P ( N ∩ U )
= P(F)P(U|F) + P(O)P(U|O) + P(N)P(U|N)
= (0.60)(0.45) + (0.30)(0.55) + (0.10)(0.35)
47
= 0.47 =
100
P( F | U ) =
P( F ∩ U ) (0.60)(0.45) 27
=
=
P (U )
0.47
47
P (contact ∩ purchase) =P (contact)P (purchase|contact)
= (0.02)(0.014) = 0.00028
b. 100,000(0.00028) = 28
39. a.
After the first draw, if the rabbit drawn is red, then 4 rabbits remain, 3 of which are yellow.
3
P(second is yellow | first is red) =
4
b. After red rabbit is replaced, 5 rabbits remain, 3 of which are yellow.
3
P(second is yellow | first is red) =
5
40. P (G2 ) = P (G1 ∩ G2 ) + P ( R1 ∩ G2 ) = P (G1 ) P (G2 | G1 ) + P( R1 ) P (G2 | R1 ) =
4 4 3 3 25
⋅ + ⋅ =
7 7 7 7 49
41. P (W ) = P (Box 1 ∩ W) + P (Box 2 ∩ W) = P(Box 1)P(W | Box 1) + P(Box 2)P(W | Box 2) =
42. a.
P (W ) = P ( B1 ∩ W ) + P ( B 2 ∩ W ) + P( B3 ∩ W )
= P(B1)P(W | B1) + P(B2)P(W | B2) + P(B3)P(W | B3)
1 3 1 4 1 2 158
= ⋅ + ⋅ + ⋅ =
3 5 3 7 3 6 315
b.
P ( R ) = P ( B1 ∩ R) + P( B 2 ∩ R) + P( B3 ∩ R)
= P(B1)P(R | B1) + P(B2)P(R | B2) + P(B3)P(R | B3)
1 2 1 3 1 2 122
= ⋅ + ⋅ + ⋅ =
3 5 3 7 3 6 315
c.
1 1 1
P (G ) = P ( B3 ∩ G ) = P ( B3) P(G | B3) = ⋅ =
3 3 9
313
1 2 1 2 9
⋅ + ⋅ =
2 5 2 4 20
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
43. P (W2 ) = P( B1 ∩ G1 ∩ W2 ) + P ( B1 ∩ R1 ∩ W2 ) + P( B 2 ∩ W1 ∩ W2 )
= P ( B1) P ( G1 B1) P (W2 (G1 ∩ B1) ) + P( B1) P ( R1 B1) P (W2 ( R1 ∩ B1) ) + P( B 2) P (W1 B 2 ) P (W2 (W1 ∩ B 2) )
=
1 1 1 1 1 1 1 1 1 1
⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ =
2 2 3 2 2 3 2 2 3 4
44. P ( D1 ∩ D2 ∩ D3 ∩ D4 ) = P ( D1 ) P ( D2 D1 ) P ( D3 ( D1 ∩ D2 ) ) P ( D4 ( D1 ∩ D2 ∩ D3 ) )
5 4 3 2 1
= ⋅ ⋅ ⋅ =
10 9 8 7 42
45. P (Und.) = P (MS ∩ Und.) + P(DS ∩ Und.)
= P (MS)P (Und.|MS) + P(DS)P(Und|DS)
20, 000 1
40, 000 3
=
⋅
+
⋅
60, 000 100 60, 000 100
7
=
300
46. P (5000) = P ( B1 ∩ 5000) + P ( B 2 ∩ 5000) + P( B3 ∩ 5000)
= P(B1)P(5000|B1) + P(B2)P(5000|B2) + P(B3)P(5000|B3)
1 1 1 2 1 1 11
= ⋅ + ⋅ + ⋅ =
3 2 3 8 3 6 36
47. P (Def) = P(A ∩ Def)+P (B ∩ Def)+P(C ∩ Def)
= P(A)P(Def | A) + P(B)P(Def | B) + P(C)P(Def | C)
= (0.10)(0.06) + (0.20)(0.04) + (0.70)(0.05) = 0.049
48. P (Def) = P(A ∩ Def) +P(B ∩ Def)+P(C ∩ Def)+P(D ∩ Def)
= P(A)P(Def | A) + P(B)P(Def | B) + P(C)P(Def | C) + P(D)P(Def | D)
= (0.30)(0.06) + (0.20)(0.03) + (0.35)(0.02) + (0.15)(0.05)
= 0.0385
49. a.
b.
P ( D ∩ V ) = P( D ) P(V | D) = (0.40)(0.15) = 0.06
P (V ) = P ( D ∩ V ) + P ( R ∩ V ) + P ( I ∩ V )
= P(D)P(V | D) + P(R)P(V | R) + P(I)P(V | I)
= (0.40)(0.15) + (0.35)(0.20) + (0.25)(0.10)
= 0.155
50. Because Richard was not hired, the number of sample points in the reduced sample space is 7 C4 = 35, of which
Allison, Lesley, Tom, and Bronwyn form one sample point. Thus
1
P (Allison, Lesley, Tom, and Bronwyn were hired) = .
35
314
ISM: Introductory Mathematical Analysis
Section 8.6
3. P ( E ∩ F ) = P ( E ) P ( F ) ,
51. P(3 Fem|at least one Fem)
P (3 Fem ∩ at least one Fem)
=
P (at least one Fem)
6 C3
1 2
1 7 7
= ⋅ P ( F ) so P ( F ) = ⋅ =
9 7
9 2 18
4
P (3 Fem)
4
C
=
= 11 3 = 33 =
2
C
5
3
1–P (no Fem) 1 –
1 – 33 31
C
4. P ( E ) = P ( E | F ) =
1
,
3
11 3
so P ( E ′) = 1 – P ( E ) = 1 –
Problems 8.6
1. a.
b.
P( E ∩ F ) = P( E ) P( F ) =
3 8 2
⋅ = = P( E ∩ F )
4 9 3
Since P ( E ) P ( F ) = P ( E ∩ F ) , events E and F
are independent.
1 3 1
⋅ =
3 4 4
5. P ( E ) P ( F ) =
P( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F )
=
1 3 1 5
+ – =
3 4 4 6
6. P(E)P(F) = (0.28)(0.15) = 0.042 ≠ P ( E ∩ F ) ,
so E and F are dependent events.
1
3
c.
P( E | F ) = P( E ) =
d.
P( E ′ | F ) = 1 – P( E | F ) = 1 –
e.
P ( E ∩ F ′) = P ( E ) P ( F ′) =
f.
P ( E ∪ F ′) = P ( E ) + P ( F ′) – P( E ∩ F ′)
7. Let F = {full service} and
I = {increase in value}.
400 2
=
P( F ) =
600 3
n( F ∩ I ) 320 2
and P ( F | I ) =
=
=
n( I )
480 3
Since P(F | I) = P(F), events F and I are
independent.
1 2
=
3 3
1 1 1
⋅ =
3 4 12
8. Let M = {male} and C = {cruncher}.
130 26
and
P(M ) =
=
175 35
n( M ∩ C ) 55 11
P(M | C ) =
=
=
n(C )
80 16
Since P(M | C) ≠ P(M), events M and C are
dependent.
1 1 1 1
= + –
=
3 4 12 2
g.
P ( E | F ′) =
P ( E ∩ F ′) 1/12 1
=
=
P ( F ′)
1/ 4 3
2. a.
P ( E ∩ F ) = P ( E ) P ( F ) = (0.1)(0.3) = 0.03
b.
P ( F ∩ G ) = P ( F ) P(G ) = (0.3)(0.6) = 0.18
c.
P ( E ∩ F ∩ G ) = P ( E ) P ( F ) P (G )
= (0.1)(0.3)(0.6) = 0.018
d.
e.
1 2
= .
3 3
9. Let S be the usual sample space consisting of
ordered pairs of the form (R, G), where the first
component of each pair represents the number
showing on the red die, and the second
component represents the number on the green
die. Then n( S ) = 6 ⋅ 6 = 36. For E, any number
of four can occur on the red die, and any number
on the green die. Thus n( E ) = 4 ⋅ 6 = 24. For F
we have F = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)},
so n(F) = 5.
Also, E ∩ F = {(4, 4), (5, 3), (6, 2)}, so
P( E ∩ F ∩ G )
P( F ∩ G)
0.018
=
= 0.1
0.18
P ( E (F ∩ G) ) =
P ( E ′ ∩ F ∩ G ′) = P ( E ′) P ( F ) P(G ′)
= (0.9)(0.3)(0.4) = 0.108
n( E ∩ F ) = 3. Thus P ( E ) P ( F ) =
and P ( E ∩ F ) =
315
3
1
= . Since
36 12
24 5
5
⋅
=
36 36 54
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
P ( E ) P ( F ) ≠ P ( E ∩ F ), events E and F are
dependent.
a.
E ∩ F = {(3,3)}, so P(E ∩ F )=
1
. Since
49
7 7
1
⋅
=
= P( E ∩ F ) ,
49 49 49
events E and F are independent.
P( E ) P( F ) =
26 1
=
52 2
12 3
6
3
P( F ) =
= , and P ( E ∩ F ) =
=
.
52 13
52 26
1 3
3
Because P ( E ) P ( F ) = ⋅ =
= P( E ∩ F ) ,
2 13 26
events E and F are independent.
10. P ( E ) =
b.
E ∩ G = {(3, 2), (3, 4), (3, 6)},
3
. Since
49
7 24 24
P ( E ) P (G ) =
⋅
=
≠ P( E ∩ G ) ,
49 49 343
events E and G are dependent.
so P( E ∩ G ) =
11. S = {HH, HT, TH, TT},
E = {HT, TH, TT},
F = {HT, TH}, and E ∩ F = {HT, TH} .
c.
3
Thus P ( E ) =
4
2 1
P ( F ) = = , and
4 2
2 1
P ( E ∩ F ) = = . We have
4 2
3 1 3
P ( E ) P ( F ) = ⋅ = ≠ P ( E ∩ F ) , so events E
4 2 8
and F are dependent.
F ∩ G = {(2,3), (4,3), (6,3)}
so P( F ∩ G ) =
3
.
49
Since
7 24 24
⋅
=
≠ P( F ∩ G ) .
49 49 343
Events F and G are dependent.
P ( F ) P(G ) =
d.
12. S = {HHH, HHT, HTH, THH, HTT, THT, TTH,
TTT}
and n(S) = 8.
E = {HTT, THT, TTH, TTT} and n(E) = 4.
F = {HHT, HTH, THH, HTT, THT, TTH} and
n(F) = 6.
E ∩ F = {HTT, THT, TTH} and n( E ∩ F ) = 3 .
14. a.
4 6 3
⋅ = = P( E ∩ F ) , so E
8 8 8
and F are independent.
Thus P ( E ) P ( F ) =
b.
E ∩ F ∩ G = ∅, so P ( E ∩ F ∩ G ) = 0 .
However,
P ( E ) P ( F ) P(G ) ≠ 0 = P ( E ∩ F ∩ G ) ,
so events E, F and G are not independent.
E = {3}
F = {5}
E ∩ F = ∅ , so E and F are mutually
exclusive.
P( E ) = P( F ) =
1
6
P( E ∩ F ) = 0
1 1 1
⋅ =
≠ P( E ∩ F )
6 6 36
Thus E and F are not independent.
P( E ) P( F ) =
13. Let S be the set of ordered pairs whose first
(second) component represents the number on
the first (second) chip. Then n(S) = 7 · 7 = 49,
n(E) = 1 · 7 = 7, and n(F) = 7 · 1 = 7. For G, if
the first chip is 1, 3, 5 or 7, then the second chip
must be 2, 4 or 6; if the first chip is 2, 4 or 6,
the second must be 1, 3, 5 or 7. Thus
n(G) = 4 · 3 + 3 · 4 = 24.
15. P ( E ∩ F ) = P ( E ) P ( F | E ), thus
P ( E ∩ F ) 0.3
=
= 0.75
P( F | E ) 0.4
Since P(E) = 0.75 ≠ 0.5 = P(E | F), E and F are
dependent.
P(E) =
316
ISM: Introductory Mathematical Analysis
Section 8.6
16. P ( E ∩ F ) = P( F ) P ( E | F ) ,
thus P ( F ) =
P( E ∩ F )
=
P( E | F )
5
9
2
3
=
5
6
P ( E ∪ F ) = P ( E ) + P ( F ) – P ( E ∩ F ), so
P( E ) = P( E ∪ F ) – P( F ) + P( E ∩ F ) =
Since P ( E ) =
17 5 5 2
– + =
18 6 9 3
2
= P ( E | F ) , events E and F are independent.
3
17. Let E = {red 4} and F = {green > 4}. Assume E and F are independent.
1 1 1
P( E ∩ F ) = P( E ) P( F ) = ⋅ =
6 3 18
18. Ei = {2 or 3 shows on ith roll}, where i = 1, 2, 3. Assume the Ei 's are independent.
⎛ 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ 1
P ( E1 ∩ E2 ∩ E3 ) = P ( E1 ) P ( E2 ) P ( E3 ) = ⎜ ⎟ ⎜ ⎟⎜ ⎟ =
⎝ 3 ⎠ ⎝ 3 ⎠⎝ 3 ⎠ 27
19. Let F = {first person attends regularly} and S = {second person attends regularly}.
1 1 1
Then P ( F ∩ S ) = P ( F ) P( S ) = ⋅ =
.
5 5 25
6 1
=
36 6
Assume that the throws are independent.
P(double on all three throws) = P(double on 1st) · P(double on 2nd) · P(double on 3rd)
1 1 1
1
.
= ⋅ ⋅ =
6 6 6 216
20. P(double on any throw) =
21. Because of replacement, assume the cards selected on the draws are independent events.
P(ace, then face card, then spade) = P(ace) · P(face card) · P(spade)
4 12 13
3
⋅ ⋅
=
=
52 52 52 676
22. Assume the outcomes on the rolls are independent events.
a.
P (> 4, > 4, > 4, > 4, > 4, > 4, > 4) =
2 2 2 2 2 2 2
1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
6 6 6 6 6 6 6 2187
b.
P (< 4, < 4, < 4, < 4, < 4, < 4, < 4) =
3 3 3 3 3 3 3
1
⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =
6 6 6 6 6 6 6 128
23. a.
P (Bill gets A ∩ Jim gets A ∩ Linda gets A)
= P(Bill gets A) · P(Jim gets A) · P(Linda gets A)
3 1 4 3
.
= ⋅ ⋅ =
4 2 5 10
317
Chapter 8: Introduction to Probability and Statistics
b.
P (Bill no A ∩ Jim no A ∩ Linda no A)
= P(Bill no A) · P(Jim no A) · P(Linda no A)
1 1 1 1
= ⋅ ⋅ =
4 2 5 40
c.
P (Bill no A ∩ Jim no A ∩ Linda gets A)
= P(Bill no A) · P(Jim no A) · P(Linda gets A)
1 1 4 1
= ⋅ ⋅ =
4 2 5 10
ISM: Introductory Mathematical Analysis
24. Assume independence of rolls.
⎛ 5 ⎞ ⎛ 5 ⎞ ⎛ 5 ⎞ 91
P(at least one 6) = 1 – P(no 6’s) = 1 – ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ =
⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ 216
25. Let A = {A survives 15 more years},
B = {B survives 15 more years}.
2 3 2
⋅ =
3 5 5
a.
P ( A ∩ B ) = P ( A) P ( B) =
b.
P ( A′ ∩ B) = P( A′) P ( B ) =
c.
A ∩ B ′ and A′ ∩ B are mutually exclusive.
1 3 1
⋅ =
3 5 5
P[( A ∩ B ′) ∪ ( A′ ∩ B)] = P ( A) P ( B ′) + P ( A′) P( B ) =
2 2 1 3 7
⋅ + ⋅ =
3 5 3 5 15
d. P(at least one survives) = P(exactly one survives) + P(both survive) =
e.
P(neither survives) = 1 – P(at least one survives) = 1 –
7 2 13
.
+ =
15 5 15
13 2
.
=
15 15
26. Assume that drawing a particular size of paper and a particular size of envelope are independent events.
P (paper A ∩ envelope A) + P (paper B ∩ envelope B) = (0.63)(0.57) + (0.37)(0.43) ≈ 0.52
27. Assume the colors selected on the draws are independent events.
7 6
7
⋅ =
18 18 54
a.
P (W1 ∩ G2 ) = P (W1 ) P (G2 ) =
b.
P[( R1 ∩ W2 ) ∪ (W1 ∩ R2 )] = P ( R1 ) P (W2 ) + P (W1 ) P ( R2 ) =
318
5 7 7 5
35
⋅ + ⋅ =
18 18 18 18 162
ISM: Introductory Mathematical Analysis
Section 8.6
28. Assume the rolls are independent.
P(7 on a roll) = P{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} =
P(12 on a roll) = P{(6, 6)} =
6 1
=
36 6
1
36
P(7 on one roll and 12 on the other) =
1 1
1 1
1
⋅ + ⋅ =
6 36 36 6 108
29. Assume that the selections are independent.
3 3 7 7 9 9 139
P (both red ∪ both white ∪ both green) = ⋅ + ⋅ + ⋅ =
19 19 19 19 19 19 361
30. Assume the throws are independent. For a particular number,
3
1 1 1 ⎛1⎞
.
⋅ ⋅ =
6 6 6 ⎜⎝ 6 ⎟⎠
Since the particular number can be any of 6 numbers,
P(particular number on three throws) =
3
1
⎛1⎞
.
P(same number in 3 throws) = 6 ⎜ ⎟ =
6
36
⎝ ⎠
31. Assume that the draws are independent.
P (particular 1st ticket ∩ particular 2nd ticket)
1 1
1
⋅
=
20 20 400
P(sum is 35) = P{(20, 15), (19, 16), (18, 17), (17, 18), (16, 19), (15, 20)}
3
⎛ 1 ⎞
= 6⎜
=
⎟
⎝ 400 ⎠ 200
=
32. a.
P({TT33}) = P(T on 1st coin) P(T on 2nd coin) P(3 on 1st die) P(3 on 2nd die)
1 1 1 1
1
= ⋅ ⋅ ⋅ =
2 2 6 6 144
b. P(two heads, one 4 and one 6)
= P(H on 1st coin) P(H on 2nd coin) P(4 on 1st die) P(6 on 2nd die)
+ P(H on 1st coin)P(H on 2nd coin) P(6 on 1st die) P(4 on 2nd die)
1
1
⎛ 1 1 1⎞
= ⎜ ⋅ ⋅ ⋅ ⎟⋅2 =
72
⎝2 2 6 6⎠
33. a.
1 1 1
1
⋅ ⋅ =
12 12 12 1728
b. To get exactly one even, there are 3 C1 = 3 ways.
P(one even and two odd) = 3[P(even 1st spin) · P(odd 2nd spin)·P(odd 3rd spin)]
⎛ 6 6 6⎞ 3
= 3⎜ ⋅ ⋅ ⎟ = .
⎝ 12 12 12 ⎠ 8
34. a.
4 13 2
1
⋅ ⋅
=
52 52 52 1352
319
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
b.
4 4 4
1
⋅ ⋅ =
52 52 52 2197
c.
The queen, spade, and black ace can be drawn in any order, so there are 3! = 6 orders, thus
4 13 2
3
6⋅ ⋅ ⋅
=
.
52 52 52 676
d. The ace can come first, second, or third, so 3 ⋅
35. a.
4 48 48 432
⋅ ⋅
=
.
52 52 52 2197
The number of ways of getting exactly four correct answers out of five is 5 C4 = 5 . Each of these ways has a
1 1 1 1 3
3
. Thus
⋅ ⋅ ⋅ ⋅ =
4 4 4 4 4 1024
3
15
P(exactly 4 correct) = 5 ⋅
.
=
1024 1024
probability of
b. P(at least 4 correct) = P(exactly 4) + P(exactly 5)
15
1 1 1 1 1 1
=
+ ⋅ ⋅ ⋅ ⋅ =
1024 4 4 4 4 4 64
c.
36. a.
The number of ways of getting exactly three correct answers out of five is
1 1 1 3 3
9
, so
⋅ ⋅ ⋅ ⋅ =
5 C3 = 10 . Each of these ways has a probability of
4 4 4 4 4 1024
9
45
P(exactly 3 correct) = 10 ⋅
. Thus
=
1024 512
P(3 or more correct) = P(exactly 3) + P(at least 4)
45
1
53
.
=
+
=
512 64 512
P(none hit) = (0.5)(0.6)(0.3) = 0.09
b. P(only Linda hits) = (0.5)(0.6)(0.7) = 0.21
c.
P(exactly one hits target) = P(only Bill) + P(only Jim) + P(only Linda)
= (0.5)(0.6)(0.3) + (0.5)(0.4)(0.3) + (0.5)(0.6)(0.7) = 0.36
d. P(exactly 2) = P(not Bill) + P(not Jim) + P(not Linda)
= (0.5)(0.4)(0.7) + (0.5)(0.6)(0.7) + (0.5)(0.4)(0.3) = 0.41
e.
P(all hit) = (0.5)(0.4)(0.7) = 0.14
37. A wrong majority decision can occur in one of two mutually exclusive ways: exactly two wrong
recommendations, or three wrong recommendations. Exactly two wrong recommendations can occur in
3 C2 = 3 mutually exclusive ways. Thus
P(wrong majority decision)
= [(0.04)(0.05)(0.9) + (0.04)(0.95)(0.1) + (0.96)(0.05)(0.1)] + (0.04)(0.05)(0.1)
= 0.0106.
320
ISM: Introductory Mathematical Analysis
Section 8.7
Problems 8.7
1. P ( E | D) =
P( E ) P( D | E )
=
P( E ) P( D | E ) + P( F ) P( D | F )
2⋅ 1
5 10
2⋅ 1 + 3⋅1
5 10 5 5
For the second part, P ( D′ | F ) = 1 – P ( D | F ) = 1 –
P( D′ | E ) = 1 – P( D | E ) = 1 –
P ( F | D ′) =
2. P ( E1 | S ) =
P ( E3 | S ′) =
1
4
1 4
= , and
5 5
1
9
. Then
=
10 10
P( F ) P( D′ | F )
=
P( E ) P( D′ | E ) + P( F ) P( D ′ | F )
3⋅4
5 5
2⋅ 9 + 3⋅4
5 10 5 5
=
4
.
7
P ( E1 ) P ( S | E1 )
=
P ( E1 ) P ( S | E1 ) + P( E2 ) P( S | E2 ) + P( E3 ) P( S | E3 )
1⋅2
5 5
1⋅2+ 3 ⋅ 7
5 5 10 10
P ( E3 ) P ( S ′ | E3 )
=
P ( E1 ) P ( S ′ | E1 ) + P ( E2 ) P ( S ′ | E2 ) + P( E3 ) P( S ′ | E3 )
3. D = {is Democrat},
R = {is Republican},
I = {is Independent},
V = {voted}.
P ( D) P (V | D)
P ( D) P(V | D) + P( R) P(V | R) + P( I ) P(V | I )
(0.42)(0.25)
=
(0.42)(0.25) + (0.33)(0.27) + (0.25)(0.15)
175
=
≈ 0.453
386
P( D | V ) =
4. D = {tire is domestic}
I = {tire is imported}
S = {tire is all-season}
2000 2
1000 1
= and P ( I ) =
= .
P( D) =
3000 3
3000 3
2
1
Note: 40% =
and 10% =
.
5
10
P( I ) P( S | I )
P(I|S) =
P( I ) P( S | I ) + P( D) P( S | D)
=
=
1⋅ 1
3 10
1⋅ 1 + 2⋅2
3 10 3 5
=
1
9
321
+
1⋅1
2 2
1⋅3+ 3 ⋅ 3
5 5 10 10
1⋅1
2 2
+
=
1⋅1
2 2
4
.
27
=
25
.
46
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
5. D = {has the disease}
D′ = {does not have the disease}
R = {positive reaction}
N = {negative reaction} = R ′
a.
P( D | R) =
P( D) P( R | D)
(0.03)(0.86)
258
=
=
≈ 0.275
P ( D) P( R | D) + P ( D′) P ( R | D′) (0.03)(0.86) + (0.97)(0.07) 937
b.
P( D | N ) =
P( D) P( N | D)
(0.03)(0.14)
14
=
=
≈ 0.005
P ( D) P ( N | D) + P ( D′) P ( N | D′) (0.03)(0.14) + (0.97)(0.93) 3021
6. I = {increase in earnings}
D = {declare a dividend}
3
1
and 10% =
.
Note: 60% =
5
10
P( I | D) =
P( I ) P( D | I )
=
P ( I ) P ( D | I ) + P ( I ′) P ( D | I ′)
1⋅3
3 5
1⋅3+ 2⋅ 1
3 5 3 10
=
3
= 75%
4
7. B1 = {first bag selected}
B2 = {second bag selected}
R = {red jelly bean drawn}
1
P ( B1 ) = P ( B2 ) = .
2
P( B1 ) P( R | B1 )
=
P ( B1 ) P ( R | B1 ) + P ( B2 ) P ( R | B2 )
P ( B1 | R ) =
1⋅4
2 6
1⋅4+1⋅2
2 6 2 5
=
5
.
8
8. B1 = {Bowl I selected}
B2 = {Bowl II selected}
B3 = {Bowl III selected}
W = {white ball selected}
P ( B1 ) = P ( B2 ) = P ( B3 ) =
P ( B1 | W ) =
1
3
P ( B1 ) P (W | B1 )
=
P( B1 ) P (W | B1 ) + P ( B2 ) P(W | B2 ) + P( B3 ) P(W | B3 )
9. A = {unit from line A}
B = {unit from line B}
D = {defective unit}.
300 3
=
P ( A) =
800 8
500 5
P( B) =
=
800 8
P( A | D) =
P ( A) P ( D | A)
=
P ( A) P ( D | A) + P ( B) P( D | B )
3⋅ 2
8 100
3⋅ 2 + 5⋅ 5
8 100 8 100
322
=
6
31
1⋅3
3 5
+
1⋅3
3 5
1⋅3
3 7
+
1⋅2
3 6
=
63
143
ISM: Introductory Mathematical Analysis
Section 8.7
10. A = {unit from line A}
B = {unit from line B}
C = {unit from line C}
D = {unit from line D}
F = {defective unit}
a.
P( A | F ) =
P ( A) P ( F | A)
P ( A) P ( F | A) + P ( B ) P ( F | B ) + P (C ) P ( F | C ) + P ( D) P( F | D)
(0.35)(0.02)
7
=
(0.35)(0.02) + (0.20)(0.05) + (0.30)(0.03) + (0.15)(0.04) 32
Parts (b), (c), and (d) are similarly determined.
=
b.
10 5
=
32 16
c.
9
32
d.
6
3
=
32 16
11. C = {call made}
T = {on time for meeting}
P (C ) P (T | C )
P (C | T ) =
P (C ) P (T | C ) + P (C ′) P (T | C ′)
=
(0.95)(0.9)
114
=
≈ 0.958
(0.95)(0.9) + (0.05)(0.75) 119
12. J D = {jar with dark chocolate only selected}
J M = {jar with dark and milk chocolates selected}
D = {dark chocolate selected}
1
P( J D ) = P( J M ) =
2
P( J D | D) =
P( J D ) P( D | J D )
=
P( J D ) P( D | J D ) + P( J M ) P( D | J M )
1 ⋅ 50
2 50
1 ⋅ 50 + 1 ⋅ 20
2 50 2 50
=
5
7
13. W = {walking reported}
B = {bicycling reported}
R = {running reported}
C = {completed requirement}
P (W | C ) =
P(W ) P (C | W )
=
P(W ) P (C | W ) + P ( B ) P (C | B) + P ( R ) P (C | R)
55.1% would be expected to report walking.
323
1⋅ 9
2 10
1⋅ 9 +1⋅4
2 10 4 5
+
1⋅2
4 3
=
27
≈ 0.551
49
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
14. C = {charges battery}
S = {car starts}
P (C ′ | S ′) =
P (C ′) P ( S ′ | C ′)
=
P (C ′) P( S ′ | C ′) + P(C ) P( S ′ | C )
1 ⋅4
10 5
1 ⋅4 + 9 ⋅1
10 5 10 8
=
32
≈ 0.416
77
15. J = {had Japanese-made car}
E = {had European-made car}
A = {had American-made car}
B = {buy same make again}
P( J | B) =
P( J ) P( B | J )
=
P ( J ) P ( B | J ) + P ( E ) P ( B | E ) + P( A) P( B | A)
3 ⋅ 85
5 100
3 ⋅ 85 + 1 ⋅ 50
5 100 10 100
3 ⋅ 40
+ 10
100
=
3
4
16. D = {dalhousium is present}
P = {positive test}
N = {negative test} = P ′
a.
P( D | P) =
P( D) P( P | D)
(0.005)(0.80)
400
=
=
≈ 0.0261
P ( D) P( P | D) + P ( D′) P( P | D′) (0.005)(0.80) + (0.995)(0.15) 15,325
b.
P( D | N ) =
P( D) P( N | D)
(0.005)(0.20)
100
=
=
≈ 0.0012
P ( D) P ( N | D) + P( D′) P ( N | D′) (0.005)(0.20) + (0.995)(0.85) 84, 675
17. P = {pass the exam}
A = {answer every question}
P ( A) P ( P | A)
(0.75)(0.8)
24
=
=
≈ 0.828
P( A | P) =
P ( A) P( P | A) + P( A′) P ( P | A′) (0.75)(0.8) + (0.25)(0.50) 29
18. P = {predicted smoking}
S = {smoking now}
P( P) P( S ′ | P)
(0.75)(0.7)
7
=
=
= 70%
P ( P | S ′) =
P( P ) P ( S ′ | P ) + P ( P′) P ( S ′ | P′) (0.75)(0.7) + (0.25)(0.9) 10
19. S = {signals sent}
D = {signals detected}
P( S | D) =
P( S ) P( D | S )
=
P ( S ) P ( D | S ) + P ( S ′) P ( D | S ′)
2⋅3
5 5
2⋅3+ 3⋅ 1
5 5 5 10
=
4
5
20. AM = {A average at midterm}
A = {A for course}
′ )
P ( A′M ) P( A | AM
(0.4)(0.6)
4
P ( A′M | A) =
=
=
≈ 0.364
P( A′M ) P ( A | A′M ) + P ( AM ) P( A | AM ) (0.4)(0.6) + (0.6)(0.7) 11
21. S = {movie is a success}
U = {“Two Thumbs Up”}
P(S | U ) =
P( S ) P (U | S )
=
P ( S ) P (U | S ) + P ( S ′) P(U | S ′)
8 ⋅ 70
10 100
8 ⋅ 70 + 2 ⋅ 20
10 100 10 100
324
=
14
≈ 0.933
15
ISM: Introductory Mathematical Analysis
Section 8.7
22. G1 = {green ball drawn from Bowl 1}
R1 = {red ball drawn from Bowl 1}
G2 = {green ball drawn from Bowl 2}
P (G1 ) P (G2 | G1 )
=
P (G1 | G2 ) =
P (G1 ) P(G2 | G1 ) + P( R1 ) P (G2 | R1 )
5⋅4
9 8
5⋅4 + 4⋅3
9 8 9 8
=
5
8
23. S = {is substandard request}
C = {is considered substandard request by Blackwell}
a.
P (C ) = P ( S ) P(C | S ) + P( S ′) P(C | S ′) = (0.20)(0.75) + (0.8)(0.15) = 0.27 =
b.
P(S | C ) =
c.
P(Error) = P (C ′ ∩ S ) + P (C ∩ S ′)
= P ( S ) P(C ′ | S ) + P( S ′) P(C | S ′)
27
100
(0.20)(0.75) 0.15 15
P ( S ) P (C | S )
=
=
=
≈ 0.556
0.27
0.27 27
P ( S ) P (C | S ) + P ( S ′) P (C | S ′)
= (0.20)(0.25) + (0.80)(0.15) = 0.17 =
17
100
24. I = {first chest selected}
II = {second chest selected}
III = {third chest selected}
G = {gold coin found}.
For the coin in the other drawer to be silver, we want the probability that the third chest was selected given that a
gold coin was found.
1⋅1
P ( III ) P (G | III )
1
3 2
=
=
P ( III | G ) =
P( I ) P (G | I ) + P ( II ) P(G | II ) + P( III ) P(G | III ) 1 ⋅1 + 1 ⋅ 0 + 1 ⋅ 1 3
3
25. a.
P( L | E ) =
=
b.
(0.25)(0.49)
≈ 0.18
(0.25)(0.49) + (0.25)(0.64) + (0.5)(0.81)
P(M | E ) =
=
c.
P( L) P( E | L)
P ( L) P ( E | L) + P ( M ) P ( E | M ) + P ( H ) P( E | H )
(0.25)(0.64)
≈ 0.23
(0.25)(0.49) + (0.25)(0.64) + (0.5)(0.81)
P( H | E ) =
=
P( M ) P( E | M )
P( L) P( E | L) + P( M ) P( E | M ) + P( H ) P( E | H )
P( H ) P( E | H )
P( L) P ( E | L) + P( M ) P( E | M ) + P ( H ) P ( E | H )
(0.5)(0.81)
≈ 0.59
(0.25)(0.49) + (0.25)(0.64) + (0.5)(0.81)
d. High quality
325
3
3 2
Chapter 8: Introduction to Probability and Statistics
26. a.
P( L | E ) =
(a)
=
P( L) P( E | L)
P ( L) P ( E | L) + P ( M ) P ( E | M ) + P ( H ) P( E | H )
(0.25)(0.44)
≈ 0.39
(0.25)(0.44) + (0.25)(0.32) + (0.5)(0.18)
P(M | E ) =
(b)
=
=
P( M ) P( E | M )
P( L) P( E | L) + P( M ) P( E | M ) + P( H ) P( E | H )
(0.25)(0.32)
≈ 0.29
(0.25)(0.44) + (0.25)(0.32) + (0.5)(0.18)
P( H ) P( E | H )
P( L) P ( E | L) + P( M ) P( E | M ) + P ( H ) P ( E | H )
P( H | E ) =
(c)
ISM: Introductory Mathematical Analysis
(0.5)(0.18)
≈ 0.32 .
(0.25)(0.44) + (0.25)(0.32) + (0.5)(0.18)
(d) Low quality
P( L | E ) =
b. (a)
=
P( L) P( E | L)
P ( L) P ( E | L) + P ( M ) P ( E | M ) + P ( H ) P( E | H )
(0.25)(0.07)
≈ 0.54
(0.25)(0.07) + (0.25)(0.04) + (0.5)(0.01)
P(M | E ) =
(b)
=
(0.25)(0.04)
≈ 0.31
(0.25)(0.07) + (0.25)(0.04) + (0.5)(0.01)
P( H | E ) =
(c)
P( M ) P( E | M )
P( L) P( E | L) + P( M ) P( E | M ) + P( H ) P( E | H )
P( H ) P( E | H )
P( L) P ( E | L) + P( M ) P( E | M ) + P ( H ) P ( E | H )
(0.5)(0.01)
≈ 0.15
(0.25)(0.07) + (0.25)(0.04) + (0.5)(0.01)
(d) Low quality
=
27. F = {fair weather}
I = {inclement weather}
W = {predict fair weather}.
P ( F ) P(W | F )
(0.6)(0.7)
7
=
= ≈ 0.78
P( F | W ) =
P( F ) P (W | F ) + P ( I ) P (W | I ) (0.6)(0.7) + (0.4)(0.3) 9
Chapter 8 Review Problems
= 8 ⋅ 7 ⋅ 6 = 336
1.
8 P3
2.
20 P1
= 20
3.
9 C7
=
9!
9!
9 ⋅ 8 ⋅ 7! 9 ⋅ 8
=
=
=
= 36
7!(9 – 7)! 7!⋅ 2! 7!⋅ 2 ⋅1
2
326
ISM: Introductory Mathematical Analysis
4.
12 C5
=
Chapter 8 Review
of the same face value, another two with a
different face value, and the last with yet another
face value is
4!
4!
⋅12 ⋅
⋅ 44
13 ⋅ 4C2 ⋅12 ⋅ 4C 2 ⋅ 44 = 13 ⋅
2!⋅ 2!
2!2!
= 13 ⋅ 6 ⋅12 ⋅ 6 ⋅ 44 = 247,104.
12!
12 ⋅11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7!
=
= 792
5!(12 − 5)!
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 7!
5. For each of the first 3 characters there are 26
choices, while for each of the last 3 characters
there are 10 choices. By the basic counting
principle, the number of license plates that are
possible is
26 ⋅ 26 ⋅ 26 ⋅ 10 ⋅ 10 ⋅ 10 = 17,576,000.
13. a.
6. The number of choices for appetizers is 2, for
the entrée it is 4, and for the dessert it is 3. By
the basic counting principle, the number of
complete dinners that are possible is
2 · 4 · 3 = 24.
Three bulbs are selected from 24, and the
order of selection is not important. Thus the
number of possible selections is
24!
24!
=
24 C3 =
3!(24 – 3)! 3!⋅ 21!
=
7. Each of the five switches has 2 possible
positions. By the basic counting principle, the
number of different codes is
24 ⋅ 23 ⋅ 22 ⋅ 21! 24 ⋅ 23 ⋅ 22
=
= 2024 .
3 ⋅ 2 ⋅1 ⋅ 21!
3 ⋅ 2 ⋅1
b. Only one bulb is defective and that bulb
must be included in the selection. The other
two bulbs must be selected from the 23
remaining bulbs and there are 23 C2 such
selections possible. Thus the number of
ways of selecting three bulbs such that one
is defective is
23!
23!
=
1 ⋅ 23C2 = 23C2 =
2!(23 – 2)! 2!⋅ 21!
2 · 2 · 2 · 2 · 2 = 25 = 32 .
8. A batting order consists of nine names selected
from nine names such that order is important.
The number of such selections is
9 P9 = 9! = 362,880 .
23 ⋅ 22 ⋅ 21! 23 ⋅ 22
=
= 253 .
2 ⋅1 ⋅ 21!
2 ⋅1
9. A possibility for first, second, and third place is
a selection of three of the seven teams so that
order is important. Thus the number of ways the
season can end is 7 P3 = 7 ⋅ 6 ⋅ 5 = 210 .
14. To score 90, exactly nine questions must be
correct; to score 100, all ten questions must be
correct. If exactly nine questions are answered
correctly, there are three ways of answering the
tenth question incorrectly. But the number of
ways of selecting nine of ten items is 10 C9 .
Thus the number of ways to score 90 is 3 ⋅ 10C9 .
The number of ways to answer all ten questions
correctly is 10 C10 , or more simply, 1. Thus the
number of ways to score 90 or better is
10!
+1
3 ⋅ 10C9 + 1 = 3 ⋅
9! ⋅1!
= 3 · 10 + 1 = 31.
10. Nine of the nine trophies can be arranged so that
order is important. The first two can be placed
on the top shelf, the next three on the middle
shelf, and the last four on the bottom shelf. The
number of such arrangements is
9 P9 = 9! = 362,880 .
11. The order of the group is not important. Thus the
number of groups that can board is
11!
11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6!
=
= 462.
11 C6 =
6! ⋅ 5!
5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 6!
12. There are four cards with a particular face value
and there are 4 C2 ways of selecting two of
them. Because there are 13 different face values,
the number of ways of selecting two cards with
the same face value is 13 ⋅ 4C2 . There are 12
remaining face values, so there are 12 ⋅ 4 C2 ways
of selecting two cards having a different face
value. After making these selections, there are
44 cards available with a different face value.
Thus the number of 5-card hands with two cards
15. In the word MISSISSIPPI, there are 11 letters
with repetition: 1 M, 4 I’s, 4 S’s, and 2 P’s. Thus
the number of distinguishable permutations is
11!
= 34, 650 .
1!⋅ 4!⋅ 4!⋅ 2!
327
Chapter 8: Introduction to Probability and Statistics
ISM: Introductory Mathematical Analysis
16. Nine flags must be arranged: two are red (type
1), three are green (type 2) and four are white
(type 3). Thus the number of distinguishable
9!
= 1260.
permutations is
2! ⋅ 3! ⋅ 4!
c.
22. P ( E1 ∪ E2 ) = P( E1 ) + P( E2 ) – P( E1 ∩ E2 )
0.7 = 0.6 + P ( E2 ) – 0.2
P ( E2 ) = 0.3
17. Of the nine professors, four go to Dalhousie
University (Cell A), three go to St. Mary’s (Cell
B), and two are not assigned (Cell C). The
number of possible assignments is
9!
= 1260.
4! ⋅ 3! ⋅ 2!
10! 10 ⋅ 9 ⋅ 8! 10 ⋅ 9
=
=
= 45
2!⋅ 8! 2 ⋅1 ⋅ 8!
2 ⋅1
Let E be the event that box is rejected. If box is
rejected, the one defective chip must be in the
two-chip sample and there are nine possibilities
for the other chip. Thus
n(E) = 9
n( E ) 9 1
=
= = 0.2 .
and P(E) =
n( S ) 45 5
23. n(S) =
18. Two of the three vans can be selected in 3 C2
ways. After two vans are chosen, the operator
must assign 14 people so that 7 go to one van
(cell 1) and 7 go to the other van (cell 2). This
14!
ways. By the basic
can be done in
7! ⋅ 7!
counting principle, the number of ways to assign
the people to two vans is
14!
3!
14!
=
⋅
= 10, 296 .
3 C2 ⋅
7! ⋅ 7! 2! ⋅1! 7! ⋅ 7!
19. a.
E1 ∩ E2 = {4, 5, 6}
c.
E1′ ∪ E2 = {7,8} ∪ {4,5, 6, 7} = {4,5, 6, 7,8}
d. The intersection of any event and its
complement is ∅.
e.
( E1 ∩ E2′ )′ = ({1, 2,3, 4,5, 6} ∩ {1, 2,3,8})′
= {1, 2, 3}′ = {4, 5, 6, 7, 8}
f.
From (b), E1 ∩ E2 ≠ ∅ , so E1 and E2 are
not mutually exclusive.
20. a.
{1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T,
5T, 6T}
21. a.
b.
=
25. Number of ways to answer exam is
45 = 1024 = n( S ) . Let
E = {exactly two questions are incorrect). The
number of ways of selecting two of the five
5!
= 10 .
questions that are incorrect is 5 C2 =
2! ⋅ 3!
However, there are three ways to answer a
question incorrectly. Since two questions are
incorrect n(E) = 10 · 3 · 3 = 90. Thus
n( E )
90
45
=
=
.
P(E) =
n( S ) 1024 512
b. {2H, 2T}
c.
10 C2
24. Percentage of rats given drug
D = 100 – (35 + 25 + 15) = 25%.
Number of rats given C = 100(0.15) = 15.
Number of rats given D = 100(0.25) = 25.
If E = event that rat was injected with C or D,
n( E ) 15 + 25
=
= 0.40.
then P(E) =
n( S )
100
If the experiment is repeated on a larger group of
300 rats but with the drugs given in the same
proportion, then the number of rats given drug C
is 300(0.15) = 45 and the number of rats given
drug D is 300(0.25) = 75 and
n( E ) 45 + 75
=
= 0.40. Thus there is no
P( E ) =
n( S )
300
effect on the previous probability.
E1 ∪ E2 = {1, 2, 3, 4, 5, 6, 7}
b.
{R1R 2 R 3 , G1G 2 G 3 }
{2H, 4H, 6H}
26. a.
{R1R 2 R 3 , R1R 2 G 3 , R1G 2 R 3 , R1G 2 G 3 ,
G1R 2 R 3 , G1R 2 G 3 , G1G 2 R 3 , G1G 2 G 3 }
{R1R 2 G 3 , R1G 2 R 3 , G1R 2 R 3 }
328
Of the 200 cola drinkers, 35 like both A and
B. Thus
35
7
.
=
P(likes both A and B) =
200 40
ISM: Introductory Mathematical Analysis
Chapter 8 Review
b. If a person likes A but not B, then the person likes A only, and conversely. Thus
70
7
=
.
P(likes A, but not B) =
200 20
27. a.
There are 10 jelly beans in the bag.
n(S) = 10 · 10 = 100
n( Eboth red ) = 4 ⋅ 4 = 16
Thus P ( Eboth red ) =
n( Eboth red ) 16
4
=
=
.
n( S )
100 25
b. n(S) = 10 · 9 = 90
n( Eboth red ) = 4 ⋅ 3 = 12
Thus P ( Eboth red ) =
12 2
=
.
90 15
28. n(S) = 6 · 6 = 36
a.
E2 or 7 = {(1, 1), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)}
P ( E2 or 7 ) =
n( E2 or 7 ) 7
=
n( S )
36
b.
Emultiple of 3 = E3, 6, 9 or 12
= {(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)}
n( Emultiple of 3 ) 12 1
=
=
P ( Emultiple of 3 ) =
36 3
n( S )
c.
Eno less than 7 = E7, 8, 9, 10, 11, or 12 = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (2, 6), (6, 2), (3, 5), (5, 3),
(4, 4), (3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
n( E7, 8, 9, 10, 11, or 12 ) 21 7
=
=
P ( Eno less than 7 ) =
36 12
n( S )
29. n(S) = 52 · 52 ⋅ 52.
a.
There are 26 black cards in a deck. Thus n( Eall black ) = 26 ⋅ 26 ⋅ 26 and
P ( Eall black ) =
26 ⋅ 26 ⋅ 26 1
= .
52 ⋅ 52 ⋅ 52 8
b. There are 13 diamonds in a deck, none of which are black. If E = event that two cards are black and the other
is a diamond, then E occurs if the diamond is the first, second, or third card. Thus
3 ⋅13 ⋅ 26 ⋅ 26 3
n(E) = 13 ⋅ 26 ⋅ 26 + 26 ⋅ 13 ⋅ 26 + 26 ⋅ 26 ⋅ 13 = 3 ⋅ 13 ⋅ 26 ⋅ 26 and P ( E ) =
= .
52 ⋅ 52 ⋅ 52
16
329
Chapter 8: Introduction to Probability and Statistics
30. n(S) =
a.
52 C2
=
ISM: Introductory Mathematical Analysis
39. a.
52!
= 1326
2!⋅ 50!
There are 13 hearts in a deck. Thus
13!
n( Eboth hearts ) = 13C2 =
= 78
2! ⋅11!
and P ( Eboth hearts ) =
78
1
=
.
1326 17
b. Out of 36 sample points, the event
{getting a total of 7 and having a 4 show} is
{(4, 3), (3, 4)}. Thus the probability of this
2
1
=
.
event is
36 18
b. There are four aces and two red kings, and
no red king is an ace. If E = event that one
card is an ace and the other is a red king,
then n(E) = 4 · 2 = 8 and
8
4
=
≈ 0.006.
P(E) =
1326 663
3
3
8
5
8
31.
P( E )
= 8 =
P( E ′) 1 – 3
8
32.
P( E )
0.92
0.92 92 23
=
=
=
=
or 23:2
P( E′) 1 – 0.92 0.08 8
2
()
33. P ( E ) =
6
6
=
6 +1 7
34. P ( E ) =
3
3
=
3+ 4 7
35. P ( F ′ H ) =
=
40. The reduced sample space consists of
{(3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5),
(5, 6), (6, 5), (6, 6)}. Out of these 10 points, only
one has a first toss that is less than 4. Thus the
1
.
conditional probability is
10
3
or 3:5
5
P( F ′ ∩ H )
=
P( H )
10
52
1
4
=
41. The second number must be a 1 or 2, so the
reduced sample space has 6 · 2 = 12 sample
points. Of these, the event
{first number ≤ second number} consists of
(1, 1), (1, 2), and (2, 2). Thus the conditional
3 1
= .
probability is
12 4
42. It does not matter whether the first two cards are
drawn or are left in place. Thus, imagine that
they are merely lifted high enough for the third
card to be drawn. The probability that this card is
1
a heart is .
4
10
13
36. The reduced sample space consists of {(6, 1),
(6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, 6), (2, 6),
(3, 6), (4, 6), (5, 6)}.
In none of these 11 points, is the sum less than 7.
Thus P(sum < 7 | a 6 shows) = 0.
37. P ( S ∩ M ) = P ( S ) P ( M | S ) = (0.6)(0.7) = 0.42
43. a.
b.
38. P (Q ∩ H ∩ AC ) = P (Q) P ( H ) P ( AC )
=
The reduced sample space consists of
{(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}.
In two of these 11 points, the sum of the
components is 7. Thus
2
.
P(sum = 7 | a 4 shows) =
11
4 13 1
1
⋅ ⋅
=
52 52 52 2704
P ( L′ | F ) =
n( L′ ∩ F ) 160 1
=
=
n( F )
480 3
400 2
= and
600 3
n( L ∩ M ) 80 2
=
= .
P(L|M) =
n( M )
120 3
Since P(L|M) = P(L), events L and M are
independent.
P ( L) =
44. E = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
F = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)}
a.
330
Since E ∩ F = {(4, 4)} ≠ ∅ , E and F are not
mutually exclusive.
ISM: Introductory Mathematical Analysis
b.
P( E ) =
6 1
P( E ∩ F )
=
= and P ( E | F ) =
36 6
P( F )
Chapter 8 Review
1
36
6
36
=
1
.
6
Since P(E) = P(E | F), events E and F are independent.
45. P = {attend public college}
M = {from middle-class family}
125 5
=
P( P) =
175 7
n( P ∩ M ) 55 11
=
=
P( P | M ) =
n( M )
80 16
Since P(P | M) ≠ P(P), events P and M are dependent.
46. P ( E | F ) =
P( E ∩ F )
P( F )
so P ( E ∩ F ) = P ( E F ) P ( F ) =
1 1 1
⋅ = , thus
6 3 18
P( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F )
=
47. a.
1 1 1 19
+ − = .
4 3 18 36
P(none take root) = (0.3)(0.3)(0.3)(0.3) = 0.0081
b. The probability that a particular two shrubs take root and the remaining two do not is (0.7)(0.7)(0.3)(0.3).
The number of ways the two that take root can be chosen from the four shrubs is 4 C2 . Thus
P(exactly two take root) = 4 C2 (0.7)2 (0.3)2 = 0.2646 .
c.
For at most two shrubs to take root, either none does, exactly one does, or exactly two do.
P(none) + P(exactly one) + P(exactly two)
= 0.0081 + 4C1 (0.7)(0.3)3 + 0.2646
= 0.0081 + 0.0756 + 0.2646
= 0.3483
48. Being effective for at least three of the persons means that it is effective for exactly three of them or for all four of
them. Thus
P(exactly three) + P(all four)
= 4C3 (0.75)(0.75)(0.75)(0.25) + (0.75)(0.75)(0.75)(0.75)
≈ 0.738
49. P ( RII ) = P (GI ) P ( RII | GI ) + P ( RI ) P( RII | RI )
3 4 2 5 22
.
= ⋅ + ⋅ =
5 9 5 9 45
50. a.
P (W ) = P ( BI ) P (W | BI ) + P ( BII ) P(W | BII )
=
1 2 1 3 1 3
7
⋅ + ⋅ = +
=
2 6 2 5 6 10 15
331
Chapter 8: Introduction to Probability and Statistics
b.
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 8
P ( BII | W )
=
P ( BII ) P (W | BII )
P ( BI ) P (W | BI ) + P( BII ) P (W | BII )
=
1⋅3
2 5
7
15
51. P (G | A) =
=
1. Trial and error should yield a critical value of
around 0.645.
9
14
2. Possible answers: One could use cellular
automata to model disease spread. The rules
would be similar to the fad model, since a person
who recovers from a disease is generally
immune for some time afterward. One could also
use cellular automata to model the formation of
political opinion blocks. Each cell could be in
one of three of four states, and a cell could be
influenced by its neighbors. Some cells could be
highly subject to neighbor influence while others
were relatively immune.
P (G ∩ A) 0.1 1
=
=
P ( A)
0.4 4
52. S = {live within the state} and
F = {first time attending}.
P( S ′) P( F ′ | S ′)
P ( S ′ | F ′) =
P ( S ) P ( F ′ | S ) + P ( S ′) P ( F ′ | S ′)
=
53. a.
b.
98 ⋅ 27
507 100
409 ⋅ 60 + 98 ⋅ 27
507 100 507 100
=
441
≈ 0.097
4531
F = {produced by first shift}
S = {produced by second shift}
D = {scratched}
P(D) = P(F)P(D|F) + P(S)P(D|S)
3000
5000
=
⋅ (0.01) +
⋅ (0.02)
8000
8000
= 0.00375 + 0.0125 = 0.01625
P( F ) P( D | F )
P( F ) P( D | F ) + P( S ) P( D | S )
0.00375 3
=
=
≈ 0.23
0.01625 13
P( F | D) =
54. E = {passed the exam}
S = {satisfactory performance}.
P( E ) P( S | E )
P( E | S ) =
P( E ) P( S | E ) + P ( E′) P ( S | E′)
=
(0.35)(0.8)
0.28 56
=
=
≈ 0.59
(0.35)(0.8) + (0.65)(0.3) 0.475 95
332
Chapter 9
Problems 9.1
1. µ =
∑ x f ( x) = 0(0.1) + 1(0.4) + 2(0.2) + 3(0.3) = 1.7
x
Var( X ) =
∑ x2 f ( x) − µ 2 = [02 (0.1) + 12 (0.4) + 22 (0.2) + 32 (0.3)] − (1.7)2 = 1.01
x
σ = Var( X ) = 1.01 ≈ 1.00
0.5
f(x)
0.4
0.3
0.2
0.1
x
0
2. µ =
1
2
3
∑ x f ( x) = 4(0.4) + 5(0.6) = 4.6
x
Var(X) = [42 (0.4) + 52 (0.6)] − (4.6)2 = 0.24
σ = 0.24 ≈ 0.49
1.0
f(x)
0.8
0.6
0.4
0.2
x
5
3. µ =
⎛1⎞
⎛1⎞
⎛1⎞
∑ x f ( x) = 1⎜⎝ 4 ⎟⎠ + 2 ⎜⎝ 4 ⎟⎠ + 3 ⎜⎝ 2 ⎟⎠ = 4 = 2.25
9
x
Var( X ) =
⎡ ⎛1⎞
⎛1⎞
⎛ 1 ⎞⎤ ⎛ 9 ⎞
∑ x2 f ( x) − µ 2 = ⎢⎣12 ⎜⎝ 4 ⎟⎠ + 22 ⎜⎝ 4 ⎟⎠ + 32 ⎜⎝ 2 ⎟⎠⎥⎦ − ⎜⎝ 4 ⎟⎠
x
σ=
11
11
=
≈ 0.83
16
4
333
2
=
11
= 0.6875
16
Chapter 9: Additional Topics in Probability
⎛1⎞
⎛2⎞
⎛1⎞
ISM: Introductory Mathematical Analysis
⎛2⎞
⎛1⎞
∑ x f ( x) = 0 ⎜⎝ 7 ⎟⎠ + 1⎜⎝ 7 ⎟⎠ + 2 ⎜⎝ 7 ⎟⎠ + 3 ⎜⎝ 7 ⎟⎠ + 4 ⎜⎝ 7 ⎟⎠ = 7
4. µ =
14
=2
x
⎡ ⎛1⎞
12
⎛2⎞
⎛1⎞
⎛2⎞
⎛ 1 ⎞⎤
≈ 1.71
Var( X ) = ⎢ 02 ⎜ ⎟ + 12 ⎜ ⎟ + 22 ⎜ ⎟ + 32 ⎜ ⎟ + 42 ⎜ ⎟ ⎥ − 22 =
7
7
7
7
7
7
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝ ⎠⎦
⎣ ⎝ ⎠
12
≈ 1.31
7
σ=
P(X = 3) = 1 − [P(X = 5) + P(X = 6) + P(X = 7)] = 1 − [0.3 + 0.2 + 0.4]= 0.1
5. a.
µ=
b.
∑ x f ( x) = 3(0.1) + 5(0.3) + 6(0.2) + 7(0.4) = 5.8
x
σ2 =
c.
∑ x2 f ( x) − µ 2 = [32 (0.1) + 52 (0.3) + 62 (0.2) + 72 (0.4)] − (5.8)2 = 1.56
x
6a + 2a + 0.2 = 1 ⇒ a = 0.1
Thus P(X = 2) = 6(0.1) = 0.6, and P(X = 4) = 2(0.1) = 0.2.
6. a.
µ = 2(0.6) + 4(0.2) + 6(0.2) = 3.2.
b.
7. Distribution of X:
1
3
3
1
f (0) = , f (1) = , f (2) = , f (3) =
8
8
8
8
⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ 12 3
= = 1.5
E( X ) =
x f ( x ) = 0 ⎜ ⎟ + 1⎜ ⎟ + 2 ⎜ ⎟ + 3 ⎜ ⎟ =
⎝8⎠ ⎝8⎠ ⎝8⎠ ⎝8⎠ 8 2
∑
x
σ = Var( X ) =
2
∑ x2 f ( x) − [ E ( x)]2
x
⎡ ⎛1⎞
⎛3⎞
⎛3⎞
⎛ 1 ⎞⎤ ⎛ 3 ⎞
= ⎢ 02 ⎜ ⎟ + 12 ⎜ ⎟ + 22 ⎜ ⎟ + 32 ⎜ ⎟ ⎥ − ⎜ ⎟
⎝8⎠
⎝8⎠
⎝ 8 ⎠⎦ ⎝ 2 ⎠
⎣ ⎝8⎠
24 9 6 3
=
− = = = 0.75
8 4 8 4
σ=
2
3
3
=
≈ 0.87
4
2
8. Distribution of X: f (1) =
4 2
2 1
= , f (2) = =
6 3
6 3
⎛2⎞
⎛1⎞ 4
E ( X ) = 1⎜ ⎟ + 2 ⎜ ⎟ = ≈ 1.33
⎝3⎠
⎝3⎠ 3
σ2 =
⎡ ⎛2⎞
⎛ 1 ⎞⎤ ⎛ 4 ⎞
∑ x2 f ( x) − [ E ( x)]2 = ⎢⎣12 ⎜⎝ 3 ⎟⎠ + 22 ⎜⎝ 3 ⎟⎠⎥⎦ − ⎜⎝ 3 ⎟⎠
2
= 2−
x
σ=
2
≈ 0.47
9
334
16 2
= ≈ 0.22
9 9
ISM: Introductory Mathematical Analysis
Section 9.1
9. The number of outcomes in the sample space is
5 C2 = 10 .
Distribution of X:
C ⋅ C
C
1
3
f (0) = 2 2 = , f (1) = 2 1 3 1 = ,
10
10
10
5
C
3
f (2) = 3 2 =
10
10
⎛ 1 ⎞ ⎛3⎞
⎛ 3⎞
E( X ) =
x f ( x ) = 0 ⎜ ⎟ + 1⎜ ⎟ + 2 ⎜ ⎟
⎝ 10 ⎠ ⎝ 5 ⎠
⎝ 10 ⎠
13. a.
If X is the gain (in dollars), then
X = –2 or 4998.
Distribution of X:
7999
1
, f (4998) =
f (−2) =
8000
8000
E ( x) = ∑ xf ( x)
x
∑
7999
1
+ 4998 ⋅
8000
8000
11, 000
=−
≈ −$1.38 (a loss)
8000
= −2 ⋅
x
6
= = 1.2
5
σ2 =
∑ x2 f ( x) − [ E ( x)]2
b. Here X = –4 or 4996. Distribution of X:
7998
2
f (−4) =
, f (4996) =
8000
8000
x
⎡ ⎛ 1 ⎞
⎛3⎞
⎛ 3 ⎞⎤ ⎛ 6 ⎞
= ⎢ 02 ⎜ ⎟ + 12 ⎜ ⎟ + 22 ⎜ ⎟ ⎥ − ⎜ ⎟
10
5
⎝ ⎠
⎝ 10 ⎠ ⎦ ⎝ 5 ⎠
⎣ ⎝ ⎠
9 36 9
= −
=
= 0.36
5 25 25
σ=
2
E ( X ) = ∑ xf ( x)
x
7998
2
+ 4996 ⋅
8000
8000
= −$2.75 (a loss)
= −4 ⋅
9
3
= = 0.6
25 5
14. If X is the gain (in dollars) per game, then
X = 10 or –6.
Distribution of X:
2 1
6 3
f (10) = = , f (−6) = =
8 4
8 4
1
3
E( X ) =
x f ( x) = 10 ⋅ + (−6) ⋅
4
4
10. Distribution of X:
9
12
4
f (0) =
, f (1) =
, f (2) =
25
25
25
⎛ 9 ⎞ ⎛ 12 ⎞
⎛ 4 ⎞ 20 4
E ( X ) = 0 ⎜ ⎟ + 1⎜ ⎟ + 2 ⎜ ⎟ =
=
25
25
⎝ ⎠ ⎝ ⎠
⎝ 25 ⎠ 25 5
= 0.8
⎡
⎛ 9 ⎞ 2 ⎛ 12 ⎞ 2 ⎛ 4 ⎞ ⎤ ⎛ 4 ⎞
⎟ + 1 ⎜ 25 ⎟ + 2 ⎜ 25 ⎟ ⎥ − ⎜ 5 ⎟
⎝ ⎠
⎝ ⎠⎦ ⎝ ⎠
⎣ ⎝ 25 ⎠
28 16 12
=
−
=
= 0.48
25 25 25
σ 2 = ⎢ 02 ⎜
σ=
∑
2
x
= −$2 (a loss)
15. Let X = daily earnings (in dollars).
Distribution of X:
4
3
f (200) = , f (−30) =
7
7
12 2 3
=
≈ 0.69
25
5
E( X ) =
11.
C
1
f (0) = P ( X = 0) = 2 2 =
10
5 C2
x
4
3
= 200 ⋅ + (−30) ⋅
7
7
710
=
≈ $101.43
7
C ⋅ C
6 3
f (1) = P ( X = 1) = 3 1 2 1 =
=
10 5
5 C2
C
3
f (2) = P( X = 2) = 3 2 =
10
5 C2
12. P ( X = x) =
∑ x f ( x)
16. Let X = gain (in dollars) to the chain of a
restaurant in a shopping center.
Distribution of X:
f (75, 000) = 0.65, f (−20, 000) = 0.35
E(X) = 75,000(0.65) + (–20,000)(0.35)
= $41,750.
⋅ 6C3− x
10 C3
4 Cx
335
Chapter 9: Additional Topics in Probability
ISM: Introductory Mathematical Analysis
17. The probability that a person in the group is not hospitalized is
1 – (0.001 + 0.002 + 0.003 + 0.004 + 0.008) = 0.982.
Let X = gain (in dollars) to the company from a policy.
Distribution of X:
f (10) = 0.982, f (−90) = 0.001, f (−190) = 0.002, f (−290) = 0.003, f (−390) = 0.004, f (−490) = 0.008
E ( X ) = 10(0.982) + (−90)(0.001) + (−190)(0.002) + (−290)(0.003) + (–390)(0.004) + (–490)(0.008)
= $3.00
18. E(X) = 0(0.05) + 1(0.10) + 2(0.15) + 3(0.20) + 4(0.15) + 5(0.15) + 6(0.10) + 7(0.05) + 8(0.05) = 3.70
19. Let p = the annual premium (in dollars) per policy. If X = gain (in dollars) to the company from a policy, then
either X = p or X = –(180,000 – p). We set E(X) = 50:
−(180, 000 − p )(0.002) + p (0.998) = 50
−360 + 0.002 p + 0.998 p = 50
−360 + p = 50
p = $410
20. Let X = player’s gain (in dollars) per play.
Distribution of X:
1
36
f (35) =
, f (−1) =
37
37
1
36
1
E ( X ) = 35 ⋅ + (−1) ⋅
=−
≈ −$0.03 (a loss)
37
37
37
21. Let X = gain (in dollars) on a play.
5
If 0 heads show, then X = 0 − 1.25 = − .
4
1
If exactly 1 head shows, then X = 1.00 − 1.25 = − .
4
3
If 2 heads show, then X = 2.00 − 1.25 = .
4
Distribution of X:
⎛ 5⎞ 1 ⎛ 1⎞ 1 ⎛3⎞ 1
f ⎜− ⎟ = , f ⎜− ⎟ = , f ⎜ ⎟ =
⎝ 4⎠ 4 ⎝ 4⎠ 2 ⎝4⎠ 4
1
⎛ 5 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ ⎛ 3 ⎞⎛ 1 ⎞
E ( X ) = ⎜ − ⎟ ⎜ ⎟ + ⎜ − ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ = − = −0.25
4
⎝ 4 ⎠⎝ 4 ⎠ ⎝ 4 ⎠⎝ 2 ⎠ ⎝ 4 ⎠⎝ 4 ⎠
Thus there is an expected loss of $0.25 on each play.
For a fair game, let p = amount (in dollars) paid to play.
Distribution of X:
1
1
1
f (− p ) = , f (1 − p ) = , f (2 − p ) =
4
2
4
We set E(X) = 0:
1
1
1
(− p ) + (1 − p ) + (2 − p ) = 0
4
2
4
p 1 p 1 p
− + − + − =0
4 2 2 2 4
1− p = 0
p =1
Thus you should pay $1 for a fair game.
336
ISM: Introductory Mathematical Analysis
Section 9.2
Principles in Practice 9.2
0
2.
1. Here p = 0.30, q = 1 – p = 0.70, and
n = 4.
1
P ( X = 0) = 4C0 (0.3)0 (0.7)4 = 0.2401
2401
10, 000
4116
10, 000
2646
10, 000
3.
756
10, 000
0
81
10, 000
3
0
2
2!
16
⎛1⎞ ⎛4⎞
f (0) = 2 C0 ⎜ ⎟ ⎜ ⎟ =
⋅1 ⋅
5
5
0!
⋅
2!
25
⎝ ⎠ ⎝ ⎠
16 16
= 1⋅1⋅
=
25 25
1
2
1
3
0
3! 8
⎛2⎞ ⎛1⎞
⋅ ⋅1
f (3) = 3C3 ⎜ ⎟ ⎜ ⎟ =
3!⋅ 0! 27
⎝ 3⎠ ⎝3⎠
8
8
= 1 ⋅ ⋅1 =
27
27
1
2! 1 4
⎛1⎞ ⎛4⎞
f (1) = 2C1 ⎜ ⎟ ⎜ ⎟ =
⋅ ⋅
⎝ 5 ⎠ ⎝ 5 ⎠ 1!⋅1! 5 5
1 4 8
= 2⋅ ⋅ =
5 5 25
2
2
3! 4 1
⎛2⎞ ⎛1⎞
⋅ ⋅
f (2) = 3C2 ⎜ ⎟ ⎜ ⎟ =
2!⋅1! 9 3
⎝ 3⎠ ⎝3⎠
4 1 4
= 3⋅ ⋅ =
9 3 9
Problems 9.2
µ = np = 3 ⋅
0
2! 1
⎛1⎞ ⎛4⎞
f (2) = 2C2 ⎜ ⎟ ⎜ ⎟ =
⋅ ⋅1
5
5
2!
⋅ 0! 25
⎝ ⎠ ⎝ ⎠
1
1
.
= 1 ⋅ ⋅1 =
25
25
1 2
µ = np = 2 ⋅ =
5 5
1 4
σ = npq = 2 ⋅ ⋅
5 5
=
0
1
1
⎛ 2⎞ ⎛1⎞
f (0) = 3C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅
=
27 27
⎝ 3⎠ ⎝ 3⎠
3! 2 1
⎛ 2⎞ ⎛1⎞
⋅ ⋅
f (1) = 3C1 ⎜ ⎟ ⎜ ⎟ =
1!⋅ 2! 3 9
⎝ 3⎠ ⎝3⎠
2 1 2
= 3⋅ ⋅ =
3 9 9
P ( X = 4) = 4 C4 (0.3) (0.7) = 0.0081
1.
0
1
4
=
3
1
1
⎛1⎞
⎜ 2 ⎟ = 1 ⋅ 8 ⋅1 = 8
⎝ ⎠
3
2
1 1
3
σ = npq = 3 ⋅ ⋅ =
2 2
2
P ( X = 3) = 4C3 (0.3)3 (0.7)1 = 0.0756
=
1
⎛1⎞
f (3) = 3C3 ⎜ ⎟
⎝2⎠
1
µ = np = 3 ⋅ =
2
P ( X = 2) = 4C2 (0.3) 2 (0.7) 2 = 0.2646
=
2
1 1 3
⎛1⎞ ⎛1⎞
f (2) = 3C2 ⎜ ⎟ ⎜ ⎟ = 3 ⋅ ⋅ =
2
2
4 2 8
⎝ ⎠ ⎝ ⎠
P ( X = 1) = 4C1 (0.3)1 (0.7)3 = 0.4116
=
2
1 1 3
⎛1⎞ ⎛1⎞
f (1) = 3C1 ⎜ ⎟ ⎜ ⎟ = 3 ⋅ ⋅ =
2 4 8
⎝2⎠ ⎝2⎠
P ( X = x) = n C x p x q n − x , x = 0, 1, 2, 3, 4
=
3
1 1
⎛1⎞ ⎛1⎞
f (0) = 3C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅ =
2
2
8 8
⎝ ⎠ ⎝ ⎠
=
4.
2
2 1
= 2; σ = npq = 3 ⋅ ⋅
3
3 3
2
6
=
3
3
f (0) = 4C0 (0.4)0 (0.6)4 =
4!
⋅1 ⋅ (0.6)4
0!⋅ 4!
= 1 ⋅1 ⋅ (0.6)4 = 0.1296
f (1) = 4C1 (0.4)1 (0.6)3 =
= 4(0.4)(0.6)3 = 0.3456
8
2 2
=
25
5
337
4!
(0.4)(0.6)3
1!⋅ 3!
Chapter 9: Additional Topics in Probability
f (2) = 4C2 (0.4)2 (0.6) 2 =
ISM: Introductory Mathematical Analysis
11. Let X = number of heads that occurs.
1
p = , n = 11
2
4!
(0.4) 2 (0.6) 2
2!⋅ 2!
= 6(0.4) 2 (0.6) 2 = 0.3456
8
⎛1⎞ ⎛1⎞
P ( X = 8) = 11C8 ⎜ ⎟ ⎜ ⎟
⎝2⎠ ⎝2⎠
1 1
= 165 ⋅
⋅
256 8
165
=
≈ 0.081
2048
4!
f (3) = 4C3 (0.4) (0.6) =
(0.4)3 (0.6)
3!⋅1!
3
1
= 4(0.4)3 (0.6) = 0.1536
f (4) = 4C4 (0.4)4 (0.6)0 =
4!
(0.4) 4 ⋅1
4!⋅ 0!
= 1(0.4)4 ⋅1 = 0.0256
µ = np = 4(0.4) = 1.6
12. Let X = number of correct answers. p =
σ = npq = 4(0.4)(0.6) ≈ 0.98
3
5
1
2
3
⎛1⎞ ⎛ 2⎞
6. P ( X = 2) = 5C2 ⎜ ⎟ ⎜ ⎟
⎝3⎠ ⎝ 3⎠
1 8
80
= 10 ⋅ ⋅
=
≈ 0.3292
9 27 243
2
14. Let X = number of aces selected. The probability
4
1
of selecting an ace on any draw is p =
= .
52 13
n=3
9. P ( X < 2) = P ( X = 0) + P ( X = 1)
1
⎛1⎞ ⎛1⎞
⎛1⎞ ⎛1⎞
= 5C0 ⎜ ⎟ ⎜ ⎟ + 5C1 ⎜ ⎟ ⎜ ⎟
2
2
⎝ ⎠ ⎝ ⎠
⎝2⎠ ⎝2⎠
1
1 1
6
3
= 1⋅1⋅ + 5 ⋅ ⋅ =
=
32
2 16 32 16
2
⎛ 7 ⎞ ⎛ 5 ⎞
P ( X = 2) = 4 C2 ⎜ ⎟ ⎜ ⎟
⎝ 12 ⎠ ⎝ 12 ⎠
49 25 1225
= 6⋅
⋅
=
≈ 0.3545
144 144 3456
2
8. P ( X = 4) = 7 C4 (0.2)4 (0.8)3
= 35(0.0016)(0.512) = 0.028672
5
3
13. Let X = number of green marbles drawn. The
probability of selecting a green marble on any
7
draw is
, n = 4.
12
16 1
⎛4⎞ ⎛1⎞
7. P ( X = 2) = 4 C2 ⎜ ⎟ ⎜ ⎟ = 6 ⋅ ⋅
5
5
25 25
⎝ ⎠ ⎝ ⎠
96
=
= 0.1536
625
0
1
,n =6
4
27
⎛1⎞ ⎛3⎞
P ( X = 3) = 6C3 ⎜ ⎟ ⎜ ⎟ = 20 ⋅
⎝4⎠ ⎝4⎠
46
540
=
≈ 0.132
4096
5. P(X = 5) = 6 C5 (0.2) (0.8)
= 6(0.00032)(0.8) = 0.001536
2
3
2
1
1 12
⎛ 1 ⎞ ⎛ 12 ⎞
P ( X = 2) = 3C2 ⎜ ⎟ ⎜ ⎟ = 3 ⋅
⋅
169 13
⎝ 13 ⎠ ⎝ 13 ⎠
36
=
≈ 0.016
2197
4
15. Let X = number of defective switches selected.
The probability that a switch is defective is
p = 0.02, n = 4.
10. P ( X ≥ 2) = 1 − [ P ( X = 0) + P ( X = 1)]
0
6
1
5
⎡
⎛ 2⎞ ⎛1⎞
⎛ 2⎞ ⎛1⎞ ⎤
= 1 − ⎢ 6 C0 ⎜ ⎟ ⎜ ⎟ + 6 C1 ⎜ ⎟ ⎜ ⎟ ⎥
⎢⎣
⎝ 3⎠ ⎝ 3⎠
⎝ 3 ⎠ ⎝ 3 ⎠ ⎥⎦
2 ⎤
13
⎡ 1
= 1 − ⎢1 ⋅
+ 6⋅
⎥ = 1 − 729
729
729
⎣
⎦
716
=
≈ 0.982
729
P ( X = 2) = 4C2 (0.02) 2 (0.98) 2
= 6(0.0004)(0.9604) ≈ 0.002
16. p = 0.2, n = 3
P ( X = x) = 3C x (0.2) x (0.8)3− x
338
ISM: Introductory Mathematical Analysis
Section 9.2
17. Let X = number of heads that occurs. p =
a.
b.
2
1
3
0
1
,n =3
4
9
1 3
⎛1⎞ ⎛3⎞
P ( X = 2) = 3C2 ⎜ ⎟ ⎜ ⎟ = 3 ⋅ ⋅ =
64
4
4
16
4
⎝ ⎠ ⎝ ⎠
1
1
⎛1⎞ ⎛3⎞
P ( X = 3) = 3C3 ⎜ ⎟ ⎜ ⎟ = 1 ⋅ ⋅1 =
64
4
4
64
⎝ ⎠ ⎝ ⎠
Thus
9
1
10
5
P ( X = 2) + P ( X = 3) =
+
=
=
64 64 64 32
18. Let X = number of hearts selected.
13 1
p=
= ,n =7
52 4
4
3
a.
1 27
945
⎛1⎞ ⎛3⎞
P ( X = 4) = 7 C4 ⎜ ⎟ ⎜ ⎟ = 35 ⋅
=
≈ 0.058
⋅
256 64 16,384
⎝4⎠ ⎝4⎠
b.
P ( X ≥ 4) = P ( X = 4) + P ( X = 5) + P( X = 6) + P( X = 7)
5
2
6
1
7
945
⎛1⎞ ⎛3⎞
⎛1⎞ ⎛3⎞
⎛1⎞ ⎛3⎞
+ 7 C5 ⎜ ⎟ ⎜ ⎟ + 7 C6 ⎜ ⎟ ⎜ ⎟ + 7 C7 ⎜ ⎟ ⎜ ⎟
16,384
4
4
4
4
⎝ ⎠ ⎝ ⎠
⎝ ⎠ ⎝ ⎠
⎝4⎠ ⎝4⎠
945
1
9
1 3
1
=
+ 21 ⋅
⋅ + 7⋅
⋅ + 1⋅
⋅1
16,384
1024 16
4096 4
16,384
1156
289
=
=
≈ 0.071
16,384 4096
=
19. Let X = number of defective in sample.
1
p= , n=6
5
P ( X ≤ 1) = P ( X = 0) + P ( X = 1)
0
6
1
⎛1⎞ ⎛4⎞
⎛1⎞ ⎛4⎞
= 6C0 ⎜ ⎟ ⎜ ⎟ + 6 C1 ⎜ ⎟ ⎜ ⎟
5
5
⎝ ⎠ ⎝ ⎠
⎝5⎠ ⎝5⎠
4096
1 1024
= 1 ⋅1 ⋅
+ 6⋅ ⋅
15, 625
5 3125
10, 240 2048
=
=
≈ 0.655
15, 625 3125
5
20. Let X = number of persons with computer.
p = 0.7, n = 5
P ( X ≥ 3) = P ( X = 3) + P ( X = 4) + P ( X = 5)
= 5C3 (0.7)3 (0.3) 2 + 5C4 (0.7) 4 (0.3)1 + 5C5 (0.7)5 (0.3)0
= 10(0.343)(0.09) + 5(0.2401)(0.3) + 1(0.16807)(1)
= 0.3087 + 0.36015 + 0.16807
= 0.83692
339
0
Chapter 9: Additional Topics in Probability
ISM: Introductory Mathematical Analysis
21. Let X = number of hits in four at-bats.
p = 0.300, n = 4
P ( X ≥ 1) = 1 − P( X = 0) = 1 − 4C0 (0.300)0 (0.700) 4 = 1 − 1 ⋅1 ⋅ (0.2401) = 0.7599
22. Let X = number of stocks that increase in value. The probability that a stock increases in value is p = 0.6.
Here n = 4. We must find
P(X ≥ 2) = 1 – P(X < 2) = 1 − [P(X = 0) + P(X = 1)].
P ( X = 0) = 4 C0 (0.6)0 (0.4) 4 = 1 ⋅1 ⋅ (0.0256) = 0.0256
P ( X = 1) = 4C1 (0.6)1 (0.4)3 = 4(0.6)(0.064) = 0.1536
P ( X ≥ 2) = 1 − P ( X < 2) = 1 − [0.0256 + 0.1536] = 1 − 0.1792 ≈ 0.82
23. Let X = number of girls. The probability that a child is a girl is p =
P(X ≥ 2) = 1 – P(X < 2) = 1 – [P(X = 0) + P(X = 1)].
0
1
. Here n = 5. We must find
2
5
1
1
⎛1⎞ ⎛1⎞
P ( X = 0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅
=
2
2
32
32
⎝ ⎠ ⎝ ⎠
1
4
1 1
5
⎛1⎞ ⎛1⎞
P ( X = 1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ =
2 16 32
⎝2⎠ ⎝2⎠
Thus,
5⎤
3 13
⎡1
=
P ( X ≥ 2) = 1 − [ P( X = 0) + P ( X = 1)] = 1 − ⎢ + ⎥ = 1 −
32
32
16
16
⎣
⎦
2
2 3
, n = 50, q = 1 − p = 1 − =
5
5 5
2 3
σ 2 = npq = 50 ⋅ ⋅ = 12
5 5
24. p =
25. µ = 3, σ 2 = 2
Since µ = np, then np = 3. Since σ 2 = npq, then (np )q = 2, or 3q = 2, so q =
Since np = 3, then n ⋅
2
1
= 3, or n = 9. Thus
3
7
1 128
512
⎛1⎞ ⎛2⎞
P ( X = 2) = 9C2 ⎜ ⎟ ⎜ ⎟ = 36 ⋅ ⋅
=
≈ 0.234.
9 2187 2187
⎝3⎠ ⎝ 3⎠
26. a.
E ( X ) = µ = np = 15(0.06) = 0.9
b.
Var( X ) = σ 2 = npq = 15(0.06)(0.94) = 0.846
c.
P ( X ≤ 1) = P( X = 0) + P ( X = 1)
= 15C0 (0.06)0 (0.94)15 + 15C1 (0.06)1 (0.94)14
= 1 ⋅1 ⋅ (0.94)15 + 15(0.06)(0.94)14 ≈ 0.77
340
2
2 1
. Thus, p = 1 − q = 1 − = .
3
3 3
ISM: Introductory Mathematical Analysis
Section 9.3
Problems 9.3
⎡a
8. ⎢ 5
⎣⎢ 12
⎡ 1 2⎤
2 3⎥
1. ⎢
⎢− 3 1 ⎥
⎣ 2 3⎦
No, since the entry at row 2 column 1 is
negative.
b⎤
⎥
a ⎦⎥
5
5
7
= 1, so a = 1 −
= .
12
12 12
7
5
b + a = 1, so b = 1 −
= .
12 12
a+
⎡ 0.1 1 ⎤
2. ⎢
⎥
⎣ 0.9 0 ⎦
Yes, since all entries are nonnegative and the
sum of the entries in each column is 1.
⎡ 0.4 a a ⎤
9. ⎢⎢ a 0.1 b ⎥⎥
⎢⎣ 0.3 b c ⎥⎦
0.4 + a + 0.3 = 1, so a = 0.3.
a + 0.1 + b = 1, 0.3 + 0.1 + b = 1, so b = 0.6.
a + b + c = 1, 0.3 + 0.6 + c = 1, so c = 0.1.
⎡ 1 1 1⎤
⎢ 2 8 3⎥
3. ⎢ − 14 85 13 ⎥
⎢
⎥
⎢ 3 1 1⎥
⎣⎢ 4 4 3 ⎦⎥
No, since there is a negative entry.
⎡a
⎢
10. ⎢ a
⎢
⎣⎢ a
⎡ 0.2 0.6 0 ⎤
4. ⎢⎢ 0.7 0.2 0 ⎥⎥
⎢⎣ 0.1 0.2 0 ⎥⎦
No, since the sum of the entries in column 3 is
not 1.
a
b
1
4
a⎤
⎥
b⎥
⎥
c ⎦⎥
a + a + a = 1, 3a = 1, a =
1
3
1
1
1
5
= 1, + b + = 1, b =
4
3
4
12
1 5
1
a + b + c = 1, + + c = 1, c =
3 12
4
a+b+
⎡ 0.4 0 0.5 ⎤
5. ⎢⎢ 0.2 0.1 0.3⎥⎥
⎢⎣ 0.4 0.9 0.2 ⎥⎦
Yes, since all entries are nonnegative and the
sum of the entries in each column is 1.
⎡ 0.4 ⎤
11. ⎢ ⎥
⎣ 0.6 ⎦
Yes, all entries are nonnegative and their sum is
1.
⎡1 ⎤
12. ⎢ ⎥
⎣0⎦
Yes, all entries are nonnegative and their sum is
1.
⎡ 0.5 0.1 0.3⎤
6. ⎢⎢ 0.4 0.3 0.3⎥⎥
⎢⎣ 0.6 0.6 0.4 ⎥⎦
No, since the sum of the entries in column 1 is
not 1.
⎡ 0.2 ⎤
13. ⎢ 0.7 ⎥
⎢ ⎥
⎣ 0.5 ⎦
No, the sum of the entries is not 1.
⎡ 2 b⎤
3
⎥
7. ⎢
⎢a 1 ⎥
4⎦
⎣
2
1
+ a = 1, so a = .
3
3
1
3
b + = 1, so b = .
4
4
⎡ 0.9 ⎤
14. ⎢ −0.1⎥
⎢
⎥
⎣ 0.2 ⎦
No, the entry in the second row is negative.
341
Chapter 9: Additional Topics in Probability
ISM: Introductory Mathematical Analysis
⎡ 2 1⎤ ⎡ 1 ⎤ ⎡ 11 ⎤
⎥ ⎢ 4 ⎥ = ⎢ 12 ⎥
15. X1 = TX0 = ⎢ 31
1
⎢⎣ 3 0 ⎥⎦ ⎢⎣ 34 ⎥⎦ ⎢⎣ 12
⎥⎦
⎡ 2 1⎤ ⎡ 11 ⎤ ⎡ 25 ⎤
⎥ ⎢ 12 ⎥ = ⎢ 36 ⎥
X 2 = TX1 = ⎢ 31
1
0
⎥
⎢⎣ 3
⎥⎦ ⎢⎣ 12
⎥⎦ ⎢⎣ 11
36 ⎦
⎡ 2 1⎤ ⎡ 25 ⎤ ⎡ 83 ⎤
⎥ ⎢ 36 ⎥ = ⎢ 108 ⎥
X3 = TX2 = ⎢ 31
⎥ ⎢ 25 ⎥
⎢⎣ 3 0 ⎥⎦ ⎢⎣ 11
36 ⎦ ⎣ 108 ⎦
⎡1
16. X1 = TX0 = ⎢ 12
⎢⎣ 2
1⎤ ⎡1⎤
4⎥ ⎢2⎥
3 1
⎥⎢ ⎥
4⎦ ⎣2⎦
⎡3⎤
= ⎢ 85 ⎥
⎢⎣ 8 ⎥⎦
⎡1
X 2 = TX1 = ⎢ 12
⎢⎣ 2
1 ⎤ ⎡3⎤
4 ⎥ ⎢8 ⎥
3 5
⎥⎢ ⎥
4 ⎦ ⎣8 ⎦
⎡ 11 ⎤
⎥
= ⎢ 32
21
⎢⎣ 32
⎥⎦
⎡1
X3 = TX2 = ⎢ 12
⎢⎣ 2
1 ⎤ ⎡ 11 ⎤
4 ⎥ ⎢ 32 ⎥
3
⎥ ⎢ 21 ⎥
4 ⎦ ⎣ 32 ⎦
⎡ 0.3
17. X1 = TX0 = ⎢
⎣ 0.7
⎡ 0.3
X 2 = TX1 = ⎢
⎣ 0.7
⎡ 43 ⎤
⎥
= ⎢ 128
85
⎢⎣ 128
⎥⎦
0.5⎤ ⎡ 0.4 ⎤ ⎡ 0.42 ⎤
=
0.5⎥⎦ ⎢⎣ 0.6 ⎥⎦ ⎢⎣ 0.58 ⎥⎦
0.5⎤ ⎡ 0.42 ⎤ ⎡0.416 ⎤
=
0.5⎥⎦ ⎢⎣ 0.58 ⎥⎦ ⎢⎣0.584 ⎥⎦
⎡ 0.3 0.5⎤ ⎡0.416 ⎤ ⎡ 0.4168 ⎤
=
X3 = TX2 = ⎢
⎣0.7 0.5⎥⎦ ⎢⎣0.584 ⎥⎦ ⎢⎣ 0.5832 ⎥⎦
⎡ 0.1
18. X1 = TX0 = ⎢
⎣ 0.9
⎡ 0.1
X 2 = TX1 = ⎢
⎣ 0.9
⎡ 0.1
X3 = TX2 = ⎢
⎣0.9
0.9 ⎤ ⎡ 0.2 ⎤ ⎡ 0.74 ⎤
=
0.1⎥⎦ ⎢⎣ 0.8⎥⎦ ⎢⎣ 0.26 ⎥⎦
0.9 ⎤ ⎡ 0.74 ⎤ ⎡ 0.308⎤
=
0.1⎥⎦ ⎢⎣ 0.26 ⎥⎦ ⎢⎣0.692 ⎥⎦
0.9 ⎤ ⎡ 0.308 ⎤ ⎡0.6536 ⎤
=
0.1⎥⎦ ⎢⎣0.692 ⎥⎦ ⎢⎣ 0.3464 ⎥⎦
⎡ 0.1
19. X1 = TX0 = ⎢ 0.2
⎢
⎣ 0.7
⎡ 0.1
X 2 = TX1 = ⎢ 0.2
⎢
⎣ 0.7
⎡ 0.1
X3 = TX2 = ⎢ 0.2
⎢
⎣ 0.7
0
0.4
0.6
0
0.4
0.6
0
0.4
0.6
0.3⎤ ⎡0.2 ⎤ ⎡ 0.26 ⎤
0.3⎥ ⎢ 0 ⎥ = ⎢ 0.28 ⎥
⎥⎢ ⎥ ⎢
⎥
0.4 ⎦ ⎣ 0.8⎦ ⎣ 0.46 ⎦
0.3⎤ ⎡ 0.26 ⎤ ⎡ 0.164 ⎤
0.3⎥ ⎢ 0.28⎥ = ⎢ 0.302⎥
⎥⎢
⎥ ⎢
⎥
0.4 ⎦ ⎣ 0.46⎦ ⎣ 0.534⎦
0.3⎤ ⎡ 0.164 ⎤ ⎡ 0.1766 ⎤
0.3⎥ ⎢ 0.302 ⎥ = ⎢ 0.3138 ⎥
⎥ ⎢
⎥
⎥⎢
0.4 ⎦ ⎣ 0.534 ⎦ ⎣ 0.5096 ⎦
342
ISM: Introductory Mathematical Analysis
Section 9.3
⎡ 0.4
⎢ 0
20. X1 = TX0 = ⎢
⎢ 0.4
⎢⎣ 0.2
0.1
0.1
0.7
0.1
0.2
0.3
0.4
0.1
0.1⎤ ⎡ 0.1⎤ ⎡ 0.17 ⎤
0.3⎥ ⎢ 0.3⎥ ⎢ 0.21⎥
⎥⎢ ⎥ = ⎢
⎥
0.4 ⎥ ⎢0.4 ⎥ ⎢ 0.49 ⎥
0.2 ⎥⎦ ⎢⎣0.2 ⎥⎦ ⎢⎣ 0.13⎥⎦
⎡ 0.4
⎢ 0
X 2 = TX1 = ⎢
⎢ 0.4
⎢⎣ 0.2
0.1
0.1
0.7
0.1
0.2
0.3
0.4
0.1
0.1⎤ ⎡0.17 ⎤ ⎡ 0.200 ⎤
0.3⎥ ⎢ 0.21⎥ ⎢0.207 ⎥
⎥⎢
⎥=⎢
⎥
0.4 ⎥ ⎢ 0.49 ⎥ ⎢ 0.463⎥
0.2 ⎥⎦ ⎢⎣ 0.13⎥⎦ ⎢⎣ 0.130 ⎥⎦
⎡0.4
⎢ 0
X3 = TX2 = ⎢
⎢0.4
⎣⎢0.2
0.1
0.1
0.7
0.1
0.2
0.3
0.4
0.1
0.1⎤ ⎡ 0.200 ⎤ ⎡ 0.2063⎤
0.3⎥ ⎢ 0.207 ⎥ ⎢ 0.1986 ⎥
⎥⎢
⎥=⎢
⎥
0.4 ⎥ ⎢ 0.463 ⎥ ⎢ 0.4621⎥
0.2 ⎦⎥ ⎣⎢ 0.130 ⎦⎥ ⎣⎢ 0.1330 ⎦⎥
21. a.
⎡1
T2 = ⎢ 4
⎢3
⎣4
3⎤ ⎡1
4⎥⎢4
1⎥⎢3
4⎦⎣4
⎡5
8
T3 = T2 T = ⎢
⎢3
⎣8
3⎤
4⎥
1⎥
4⎦
⎡5
8
=⎢
⎢3
⎣8
3⎤ ⎡1
8⎥ ⎢4
5⎥ ⎢3
8⎦ ⎣4
3⎤
4⎥
1⎥
4⎦
3⎤
8⎥
5⎥
8⎦
⎡7
16
=⎢
⎢9
⎣ 16
9⎤
16 ⎥
.
7⎥
16 ⎦
3
.
8
b. Entry in row 2, column 1, of T2 is
c.
22. a.
Entry in row 1, column 2 of T3 is
⎡1
3
T2 = ⎢
⎢2
⎣3
1⎤⎡1
2⎥⎢3
1⎥⎢2
2⎦⎣3
⎡4
9
T3 = T2 T = ⎢
⎢5
⎣9
⎡4 5 ⎤
9 12 ⎥
=⎢
⎢5 7 ⎥
⎣ 9 12 ⎦
5 ⎤ ⎡ 1 1 ⎤ ⎡ 23
12 ⎥ ⎢ 3 2 ⎥ ⎢ 54
=
7 ⎥ ⎢ 2 1 ⎥ ⎢ 31
12 ⎦ ⎣ 3 2 ⎦ ⎣ 54
1⎤
2⎥
1⎥
2⎦
b. Entry in row 2, column 1, of T2 is
c.
23. a.
9
.
16
Entry in row 1, column 2 of T3 is
31 ⎤
72 ⎥
41 ⎥
72 ⎦
5
.
9
31
.
72
⎡ 0 0.5 0.3⎤ ⎡0 0.5 0.3⎤ ⎡ 0.50 0.23 0.27 ⎤
T = ⎢⎢1 0.4 0.3⎥⎥ ⎢⎢1 0.4 0.3⎥⎥ = ⎢⎢ 0.40 0.69 0.54 ⎥⎥
⎢⎣ 0 0.1 0.4 ⎥⎦ ⎢⎣ 0 0.1 0.4 ⎥⎦ ⎢⎣0.10 0.08 0.19 ⎥⎦
⎡ 0.50 0.23 0.27 ⎤ ⎡0 0.5 0.3⎤ ⎡ 0.230 0.369 0.327 ⎤
3
2
T = T T = ⎢⎢ 0.40 0.69 0.54 ⎥⎥ ⎢⎢1 0.4 0.3⎥⎥ = ⎢⎢ 0.690 0.530 0.543 ⎥⎥
⎢⎣ 0.10 0.08 0.19 ⎥⎦ ⎢⎣0 0.1 0.4 ⎥⎦ ⎢⎣0.080 0.101 0.130 ⎥⎦
2
b. Entry in row 2, column 1, of T2 is 0.40.
343
Chapter 9: Additional Topics in Probability
c.
ISM: Introductory Mathematical Analysis
Entry in row 1, column 2 of T3 is 0.369.
⎡ 0.1 0.1 0.1⎤ ⎡ 0.1 0.1 0.1⎤ ⎡ 0.10 0.10 0.10 ⎤
T = ⎢⎢ 0.2 0.1 0.1⎥⎥ ⎢⎢ 0.2 0.1 0.1⎥⎥ = ⎢⎢ 0.11 0.11 0.11⎥⎥
⎢⎣ 0.7 0.8 0.8⎥⎦ ⎢⎣ 0.7 0.8 0.8⎥⎦ ⎢⎣0.79 0.79 0.79⎥⎦
⎡0.10 0.10 0.10 ⎤ ⎡ 0.1 0.1 0.1⎤ ⎡ 0.10 0.10 0.10 ⎤
3
2
T = T T = ⎢⎢ 0.11 0.11 0.11⎥⎥ ⎢⎢ 0.2 0.1 0.1⎥⎥ = ⎢⎢ 0.11 0.11 0.11⎥⎥ .
⎢⎣0.79 0.79 0.79 ⎥⎦ ⎢⎣0.7 0.8 0.8⎥⎦ ⎢⎣ 0.79 0.79 0.79 ⎥⎦
2
24. a.
b. Entry in row 2, column 1, of T2 is 0.11.
c.
Entry in row 1, column 2 of T3 is 0.10.
2⎤
⎡ 1 2 ⎤ ⎡1 0 ⎤ ⎡ − 1
2 3⎥
2
3⎥
⎢
⎢
25. T − I =
−
=
⎢ 1 1 ⎥ ⎢⎣ 0 1 ⎥⎦ ⎢ 1 − 2 ⎥
3⎦
⎣2 3⎦
⎣ 2
⎡ 1
⎡1 0 4 ⎤
1 1⎤
7⎥
⎢
⎥
⎢
2 0⎥ → " → ⎢0 1 3 ⎥
⎢− 1
3
7⎥
⎢ 2
⎥
⎢
⎢ 1 − 2 0⎥
⎢0 0 0 ⎥
3
⎣ 2
⎦
⎣
⎦
⎡4⎤
Q = ⎢ 73 ⎥
⎢⎣ 7 ⎥⎦
⎡1
2
26. T − I = ⎢
⎢1
⎣2
⎡ 1
⎢
⎢− 1
⎢ 2
⎢ 1
⎣ 2
1
1
4
− 14
1⎤
4 ⎥ ⎡1
−
3 ⎥ ⎢0
⎣
4⎦
1
0⎤ ⎡ − 2
⎢
=
1 ⎥⎦ ⎢ 1
⎣ 2
1 ⎤
4 ⎥
− 14 ⎥⎦
⎡1 0 1 ⎤
1⎤
3⎥
⎢
⎥
2
0⎥ → " → ⎢0 1 3 ⎥
⎢
⎥
⎥
⎢0 0 0 ⎥
0 ⎥⎦
⎣
⎦
⎡1⎤
3
Q=⎢ ⎥
⎢1⎥
⎣2⎦
⎡ 1 3 ⎤ ⎡1 0 ⎤ ⎡ − 4 3 ⎤
5
5 ⎥
5 5⎥
⎢
27. T − I =
−⎢
=⎢
⎥
4
2
4
⎢
⎢
⎥ 0 1⎦
− 53 ⎥
⎣5 5⎦ ⎣
⎣ 5
⎦
⎡ 1
⎡1 0 3 ⎤
1 1⎤
7⎥
⎢
⎥
⎢
⎢ − 4 3 0⎥ → " → ⎢0 1 4 ⎥
7⎥
⎢ 5 5
⎥
⎢
⎢ 4 − 3 0⎥
⎢0 0 0 ⎥
⎢⎣ 5
⎥⎦
5
⎣
⎦
⎡3⎤
7
Q=⎢ ⎥
⎢4⎥
⎣7⎦
344
ISM: Introductory Mathematical Analysis
Section 9.3
⎡ 1 1 ⎤ ⎡1 0 ⎤ ⎡ − 3 1 ⎤
4 3⎥
4
3 ⎥
⎢
28. T − I =
−⎢
=⎢
⎥
3
3
2
⎢
⎥ 0 1⎦ ⎢
− 13 ⎥
⎣4 3⎦ ⎣
⎣ 4
⎦
⎡ 1
⎡1 0 4 ⎤
1 1⎤
13 ⎥
⎢
⎥
⎢
⎢ − 3 1 0 ⎥ → " → ⎢0 1 9 ⎥
13 ⎥
⎢ 4 3
⎥
⎢
⎢ 3 − 1 0⎥
⎢0 0 0 ⎥
⎢⎣ 4
⎥⎦
3
⎣
⎦
⎡4⎤
13
Q=⎢ ⎥
⎢9⎥
⎣ 13 ⎦
⎡ 0.4
29. T − I = ⎢⎢ 0.3
⎢⎣ 0.3
1
⎡ 1
⎢ −0.6 0.6
⎢
⎢ 0.3 −0.7
⎢
0.1
⎢⎣ 0.3
0.6 0.6 ⎤ ⎡1 0 0 ⎤ ⎡ −0.6 0.6 0.6 ⎤
0.3 0.1⎥⎥ − ⎢⎢0 1 0 ⎥⎥ = ⎢⎢ 0.3 −0.7 0.1 ⎥⎥
0.1 0.3⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0.3
0.1 −0.7 ⎥⎦
1
1⎤
⎡1 0 0 0.5 ⎤
⎢0 1 0 0.25⎥
0.6 0 ⎥⎥
⎥
→" → ⎢
⎢0 0 1 0.25⎥
0.1 0 ⎥
⎥
⎢
⎥
−0.7 0 ⎥⎦
0 ⎥⎦
⎢⎣0 0 0
⎡ 0.5 ⎤
Q = ⎢⎢ 0.25⎥⎥
⎢⎣ 0.25⎥⎦
⎡ 0.1 0.4 0.3⎤ ⎡1 0 0 ⎤ ⎡ −0.9 0.4 0.3 ⎤
30. T − I = ⎢⎢ 0.2 0.2 0.3⎥⎥ − ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ 0.2 −0.8 0.3 ⎥⎥
⎢⎣ 0.7 0.4 0.4 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0.7 0.4 −0.6 ⎥⎦
1
1
1⎤
⎡ 1
⎡1 0 0 0.2707 ⎤
⎢ −0.9 0.4 0.3 0 ⎥
⎢
⎥
⎢
⎥ → " → ⎢ 0 1 0 0.2481⎥
⎢ 0.2 −0.8 0.3 0 ⎥
⎢0 0 1 0.4812 ⎥
⎢
⎥
⎢
⎥
0 ⎥⎦
⎢⎣ 0.7 0.4 −0.6 0 ⎥⎦
⎢⎣ 0 0 0
⎡ 0.2707 ⎤
Q ≈ ⎢⎢ 0.2481⎥⎥
⎢⎣ 0.4812 ⎥⎦
Flu
31. a.
b.
No flu
Flu ⎡ 0.1 0.2 ⎤
T=
No flu ⎢⎣ 0.9 0.8 ⎥⎦
⎡ 120 ⎤ ⎡ 0.6 ⎤
200
X0 = ⎢ ⎥ = ⎢ ⎥ .
⎢ 80 ⎥ ⎣ 0.4 ⎦
⎣ 200 ⎦
If a period is 4 days, then 8 days corresponds to 2 periods, and 12 days corresponds to 3 periods. The state
vector corresponding to 8 days from now is
⎡ 0.19 0.18⎤ ⎡ 0.6 ⎤ ⎡ 0.186 ⎤
X 2 = T2 X0 = ⎢
⎥⎢ ⎥ = ⎢
⎥.
⎣ 0.81 0.82 ⎦ ⎣ 0.4 ⎦ ⎣ 0.814 ⎦
345
Chapter 9: Additional Topics in Probability
ISM: Introductory Mathematical Analysis
Thus 0.186(200) ≈ 37 students can be
expected to have the flu 8 days from now.
The state vector corresponding to 12 days
from now is
⎡ 0.181 0.182 ⎤ ⎡ 0.6 ⎤
X3 = T3 X0 = ⎢
⎥⎢ ⎥
⎣0.819 0.818⎦ ⎣ 0.4 ⎦
⎡0.1814 ⎤
=⎢
⎥.
⎣0.8186 ⎦
Thus 0.1814(200) ≈ 36 students can be
expected to have the flu 12 days from now.
b.
35% to location 1, 45.8% to location 2,
19.2% to location 3
H
L
H ⎡0.55 0.25⎤
32. T = ⎢
L ⎣0.45 0.75⎥⎦
⎡ 0.65⎤
X0 = ⎢
⎥
⎣ 0.35⎦
35. a.
2
c.
A B
A ⎡0.7 0.4 ⎤
T= ⎢
B ⎣ 0.3 0.6 ⎥⎦
b. Wednesday corresponds to step 2.
⎡ 0.61 0.52 ⎤
T2 = ⎢
⎥.
⎣ 0.39 0.48⎦
The probability is 0.61.
34. a.
D R
O
D ⎡0.8 0.1 0.3 ⎤
T= ⎢
R ⎢0.1 0.8 0.2 ⎥⎥
O ⎢⎣0.1 0.1 0.5 ⎥⎦
⎡ 0.68 0.19 0.41 ⎤
b. T = ⎢⎢ 0.18 0.67 0.29 ⎥⎥
⎢⎣ 0.14 0.14 0.30 ⎥⎦
The probability is 0.19.
⎡ 0.415 0.325⎤ ⎡ 0.65⎤
X 2 = T X0 = ⎢
⎥⎢
⎥
⎣ 0.585 0.675⎦ ⎣ 0.35⎦
⎡ 0.3835⎤
=⎢
⎥
⎣ 0.6165⎦
38.35% of the members will be performing highimpact exercising.
2
33. a.
X 2 = TX1
⎡0.7 0.2 0.2 ⎤ ⎡ 0.30 ⎤
= ⎢⎢ 0.1 0.8 0.2 ⎥⎥ ⎢⎢ 0.48⎥⎥
⎢⎣ 0.2 0 0.6 ⎥⎦ ⎢⎣0.22 ⎥⎦
⎡0.350 ⎤
= ⎢⎢ 0.458 ⎥⎥
⎢⎣0.192 ⎥⎦
X1 = TX0
⎡0.8 0.1 0.3 ⎤ ⎡ 0.40 ⎤
= ⎢⎢0.1 0.8 0.2 ⎥⎥ ⎢⎢ 0.40 ⎥⎥
⎢⎣0.1 0.1 0.5 ⎥⎦ ⎢⎣ 0.20 ⎥⎦
⎡0.42 ⎤
= ⎢⎢0.40 ⎥⎥
⎢⎣ 0.18⎥⎦
40% are expected to be Republican.
U
S R
U ⎡0.7 0.1 0.1⎤
36. T = ⎢
S ⎢0.1 0.8 0.1⎥⎥
R ⎢⎣0.2 0.1 0.8⎥⎦
X1 = TX0
⎡0.7 0.2 0.2 ⎤ ⎡ 0.2 ⎤
= ⎢⎢ 0.1 0.8 0.2 ⎥⎥ ⎢⎢ 0.5⎥⎥
⎢⎣ 0.2 0 0.6 ⎥⎦ ⎢⎣ 0.3⎥⎦
⎡0.30 ⎤
= ⎢⎢ 0.48⎥⎥
⎢⎣0.22 ⎥⎦
a.
30% to location 1, 48% to location 2, 22%
to location 3
15 years corresponds to step 3.
⎡ 0.412 0.196 0.196 ⎤
T3 = ⎢⎢ 0.219 0.562 0.219 ⎥⎥
⎢⎣ 0.369 0.242 0.585 ⎥⎦
The entry in row 3, column 2 of T3 is
0.242, so the probability is 0.242.
346
ISM: Introductory Mathematical Analysis
b.
Section 9.3
X3 = T3 X0
⎡0.412 0.196 0.196 ⎤ ⎡ 0.50 ⎤ ⎡ 0.304 ⎤
= ⎢⎢0.219 0.562 0.219 ⎥⎥ ⎢⎢ 0.25⎥⎥ = ⎢⎢0.30475⎥⎥
⎣⎢0.369 0.242 0.585 ⎦⎥ ⎣⎢ 0.25⎦⎥ ⎣⎢0.39125⎦⎥
The population is expected to be 30.4% urban, 30.475% suburban, 39.125% rural.
37. a.
b.
c.
A Compet.
A ⎡ 0.8 0.3⎤
T=
Compet. ⎢⎣ 0.2 0.7 ⎥⎦
⎡ 0.8 0.3⎤ ⎡ 0.70 ⎤
X1 = TX0 = ⎢
⎥⎢
⎥
⎣ 0.2 0.7 ⎦ ⎣ 0.30 ⎦
⎡0.65⎤
=⎢
⎥
⎣0.35⎦
A is expected to control 65% of the market.
⎡ 0.8
T−I = ⎢
⎣ 0.2
1
⎡ 1
⎢ −0.2 0.3
⎢
⎢⎣ 0.2 −0.3
0.3 ⎤ ⎡1 0 ⎤ ⎡ −0.2 0.3⎤
−
=
0.7 ⎥⎦ ⎢⎣ 0 1 ⎥⎦ ⎢⎣ 0.2 −0.3⎥⎦
1⎤
⎡ 1 0 0.6 ⎤
⎥
0 ⎥ → … → ⎢⎢ 0 1 0.4⎥⎥
⎢⎣ 0 0
0 ⎥⎦
0 ⎥⎦
⎡0.6 ⎤
Q=⎢ ⎥
⎣0.4 ⎦
In the long run, A can expect to control 60% of the market.
38. a.
b.
Fords Non-Fords
Fords ⎡ 0.75 0.35⎤
T=
Non-fords ⎢⎣ 0.25 0.65⎥⎦
⎡ 0.75 0.35⎤ ⎡1 0 ⎤ ⎡ −0.25 0.35⎤
T−I = ⎢
⎥−⎢
⎥=⎢
⎥
⎣ 0.25 0.65⎦ ⎣ 0 1 ⎦ ⎣ 0.25 −0.35⎦
1
1⎤
⎡ 1
⎡1 0 0.5833⎤
⎢ −0.25 0.35 0 ⎥ → … → ⎢ 0 1 0.4167 ⎥
⎢
⎥
⎢
⎥
⎢⎣ 0.25 −0.35 0 ⎥⎦
⎢⎣ 0 0
0 ⎥⎦
⎡ 0.5833⎤
Q≈⎢
⎣ 0.4167 ⎥⎦
In the long run, 58.33% of car purchases in the region are expected to be Fords.
39. a.
T=
1 2
5 3⎤
⎡
1 7 7
⎢
⎥
2⎢2 4⎥
⎣7 7⎦
347
Chapter 9: Additional Topics in Probability
ISM: Introductory Mathematical Analysis
⎡ 31 27 ⎤ ⎡ 1 ⎤ ⎡ 29 ⎤ ⎡ 0.5918 ⎤
49 49 ⎥ ⎢ 2 ⎥ ⎢ 49 ⎥
b. X 2 = T X0 = ⎢
=
≈
18
22 ⎥ ⎢ 1 ⎥ ⎢ 20 ⎥ ⎢ 0.4082 ⎥⎦
⎢
⎣ 49 49 ⎦ ⎣ 2 ⎦ ⎣ 49 ⎦ ⎣
About 59.18% in compartment 1 and 40.82% in compartment 2.
2
c.
40. a.
b.
c.
3⎤
⎡ 5 3 ⎤ ⎡1 0 ⎤ ⎡ − 2
7
7⎥
7 7⎥
⎢
⎢
T−I =
−
=
⎢ 2 4 ⎥ ⎢⎣ 0 1 ⎥⎦ ⎢ 2 − 3 ⎥
7⎦
⎣7 7⎦
⎣ 7
⎡ 1
⎡1 0 3⎤
1 1⎤
5⎥
⎢
⎥
⎢
⎢ − 2 3 0⎥ → … → ⎢0 1 2 ⎥
5⎥
⎢ 7 7
⎥
⎢
⎢ 2 − 3 0⎥
⎢0 0 0⎥
⎢⎣ 7
⎥⎦
7
⎣
⎦
⎡ 3 ⎤ ⎡ 0.6 ⎤
5
Q=⎢ ⎥=⎢ ⎥
⎢ 2 ⎥ ⎣ 0.4 ⎦
⎣5⎦
In the long run, there will be 60% in compartment 1 and 40% in compartment 2.
Doesn't
Works Work
T=
Works ⎡ 0.8
0.1⎤
⎢
Doesn't Work ⎣ 0.2
0.9 ⎥⎦
⎡ 0.562 0.219 ⎤
T3 = ⎢
⎥
⎣ 0.438 0.781 ⎦
The probability is 0.562.
⎡ 0.8 0.1⎤ ⎡1 0 ⎤ ⎡ −0.2 0.1⎤
T−I = ⎢
⎥−⎢
⎥=⎢
⎥
⎣ 0.2 0.9 ⎦ ⎣ 0 1 ⎦ ⎣ 0.2 −0.1⎦
⎡1 0 1 ⎤
1
1⎤
3⎥
⎡ 1
⎢
⎢ −0.2 0.1 0 ⎥ → … → ⎢0 1 2 ⎥
⎢
⎥
3⎥
⎢
⎢⎣ 0.2 −0.1 0 ⎥⎦
⎢0 0 0 ⎥
⎣
⎦
⎡1⎤
3
Q=⎢ ⎥
⎢2⎥
⎣3⎦
⎛1⎞
In the long run, the number of machines working properly is ⎜ ⎟ (42) = 14 .
⎝3⎠
348
ISM: Introductory Mathematical Analysis
41. a.
⎡3
⎢
T−I = 4
⎢1
⎣4
1⎤
2 ⎥ ⎡1
−
1 ⎥ ⎢0
⎣
2⎦
⎡ 1
⎢
⎢− 1
⎢ 4
⎢ 1
⎣ 4
⎡1 0
1⎤
⎢
⎥
0⎥ → … → ⎢0 1
⎢
⎥
⎢0 0
0 ⎥⎦
⎣
1
1
2
− 12
0⎤ ⎡ − 4
=⎢
1 ⎥⎦ ⎢ 1
⎣ 4
1
Section 9.3
1⎤
2⎥
1
− 2 ⎥⎦
2⎤
3⎥
1⎥
3⎥
0⎥
⎦
⎡2⎤
3
Q=⎢ ⎥
⎢1⎥
⎣3⎦
b. Presently, A accounts for 50% of sales and in long run A will account for
percentage increase in sales above the present level is
42. a.
b.
c.
66 23 − 50
50
⋅100% =
2
2
, or 66 % , of sales. Thus the
3
3
16 23
50
1
⋅100% = 33 % .
3
A
B C
A ⎡ 0.8 0.2 0.2 ⎤
T= ⎢
B ⎢ 0.1 0.7 0.1 ⎥⎥
C ⎢⎣ 0.1 0.1 0.7 ⎥⎦
⎡ 0.68 0.32 0.32 ⎤
T2 = ⎢⎢ 0.16 0.52 0.16 ⎥⎥
⎢⎣ 0.16 0.16 0.52 ⎥⎦
The probability is 0.52.
Initially 500 customers are to be considered. The probability that a customer goes to branch A is
⎡ 0.4 ⎤
200
100
to branch B,
= 0.4 ; and to branch C,
= 0.2 . Thus X0 = ⎢⎢ 0.4 ⎥⎥ .
500
500
⎢⎣ 0.2 ⎥⎦
⎡ 0.8 0.2 0.2 ⎤ ⎡0.4 ⎤ ⎡ 0.44 ⎤
X1 = TX0 = ⎢⎢ 0.1 0.7 0.1⎥⎥ ⎢⎢0.4 ⎥⎥ = ⎢⎢ 0.34 ⎥⎥
⎢⎣ 0.1 0.1 0.7 ⎥⎦ ⎢⎣0.2 ⎥⎦ ⎢⎣ 0.22 ⎥⎦
Thus 0.44(500) = 220 customers can be expected to go to A on their next visit,
0.34(500) = 170 to B, and 0.22(500) = 110 to C.
349
200
= 0.4 ;
500
Chapter 9: Additional Topics in Probability
d.
ISM: Introductory Mathematical Analysis
⎡ 0.8 0.2 0.2 ⎤ ⎡1 0 0 ⎤ ⎡ −0.2 0.2 0.2 ⎤
T − I = ⎢⎢ 0.1 0.7 0.1 ⎥⎥ − ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ 0.1 −0.3 0.1⎥⎥
⎢⎣ 0.1 0.1 0.7 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0.1 0.1 −0.3⎥⎦
1
1
1⎤
⎡ 1
⎡1 0 0 0.50 ⎤
⎢ −0.2 0.2 0.2 0 ⎥
⎢
⎥
⎢
⎥ → … → ⎢0 1 0 0.25⎥
⎢ 0.1 −0.3 0.1 0 ⎥
⎢0 0 1 0.25⎥
⎢
⎥
⎢
⎥
0 ⎦⎥
⎣⎢ 0.1 0.1 −0.3 0 ⎦⎥
⎣⎢0 0 0
In the long run, 0.50(500) = 250 can be expected to go to A, 0.25(500) = 125 to B, and 0.25(500) to C.
⎡1
43. T2 = TT = ⎢ 2
⎢1
⎣2
1 ⎤ ⎡ 12
⎥⎢
0 ⎥⎦ ⎢⎣ 12
1 ⎤ ⎡ 34
⎥=⎢
0 ⎥⎦ ⎢⎣ 14
1⎤
2⎥
1⎥
2⎦
Since all entries of T2 are positive, T is regular.
⎡0 1 ⎤
2
n
n
44. For the matrix A = ⎢
⎥ , A = I (the 2 × 2 identity matris). Thus A = I if n is even, and A = A if n is odd.
1
0
⎣
⎦
In either case there are nonpositive entries, and thus A is not regular.
Chapter 9 Review Problems
1. µ =
∑ xf ( x) = 1⋅ f (1) + 2 ⋅ f (2) + 3 ⋅ f (3) = 1(0.7) + 2(0.1) + 3(0.2) = 1.5
x
Var( X ) =
∑ x2 f ( x) − µ 2 = ⎡⎣12 (0.7) + 22 (0.1) + 32 (0.2)⎤⎦ − (1.5)2 = 0.65
x
σ = Var( X ) = 0.65 ≈ 0.81
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
f(x)
x
1
2. µ =
2
3
∑ xf ( x) = 0 ⋅ 6 + 1⋅ 2 + 2 ⋅ 3 = 6
1
x
Var( X ) =
1
1
7
∑ x 2 f ( x) − µ 2
x
1
1
1⎤ ⎛ 7 ⎞
⎡
= ⎢ 02 ⋅ + 12 ⋅ + 22 ⋅ ⎥ − ⎜ ⎟
6
2
3⎦ ⎝ 6 ⎠
⎣
11 49 17
= −
=
6 36 36
σ=
2
17
17
=
≈ 0.69
36
6
350
ISM: Introductory Mathematical Analysis
1/2
Chapter 9 Review
f(x)
192
32
,
=
1326 221
6
1
.
f (2) = P ( E2 aces ) =
=
1326 221
f (1) = P ( E1 ace ) =
1/3
1/6
b.
E( X ) =
x
0
3. a.
1
=
n(S) = 2 · 6 = 12
E0 = {H1}, E1 = {T1, H2} , E2 = {T 2, H 3} ,
E3 = {T3, H4} , E4 = {T4, H5} ,
E5 = {T5, H6} , E6 = {T6}
f (1) = P ( E1 ) =
n ( E0 )
n( S )
n ( E1 )
n( S )
=
=
b.
E( X ) =
n ( E6 )
n( S )
2 1
=
12 6
=
1
.
6
1
12
n ( Etwo 10's ) = 4 ⋅ 4 .
∑ xf ( x)
Dist. of X:
⎛ 1 ⎞ 48 ⋅ 48 144
,
f ⎜− ⎟ =
=
⎝ 4 ⎠ 52 ⋅ 52 169
1 1+ 2 + 3 + 4 + 5
1
+
+ 6⋅
12
6
12
15 6 36
= 0+ +
=
=3
6 12 12
= 0⋅
⎛ 3 ⎞ 2 ⋅ 4 ⋅ 48 24
,
f ⎜ ⎟=
=
⎝ 4 ⎠ 52 ⋅ 52 169
4⋅4
1
⎛7⎞
.
f ⎜ ⎟=
=
⎝ 4 ⎠ 52 ⋅ 52 169
1 144 3 24 7 1
E( X ) = − ⋅
+ ⋅
+ ⋅
4 169 4 169 4 169
−144 + 72 + 7
65
5
=
=−
=−
≈ −0.10
4 ⋅169
676
52
There is a loss of $0.10 per play.
52!
52 ⋅ 51
n( S ) = 52 C2 =
=
= 1326 . In a
2!⋅ 50!
2
deck there are 4 aces and 48 non-aces. Thus
48!
48 ⋅ 47
n ( E0 aces ) = 48 C2 =
=
2!⋅ 46!
2
= 1128 .
For E1 ace to occur, one card is an ace and
the other is non-ace. Thus
n ( E1 ace ) = 4 ⋅ 48 = 192 .
n ( E2 aces ) = 4 C2 =
6. Let X = gain (in dollars) to company.
Dist. of X: f(40,000) = 0.45,
f(–10,000) = 1 – 0.45 = 0.55
E(X) = (40,000)(0.45) + (–10,000)(0.55)
= 18,000 – 5500 = $12,500 per station
4!
4⋅3
=
=6.
2!⋅ 2!
2
Therefore,
f (0) = P ( E0 aces ) =
1
34
2
=
221 13
Eone 10 occurs if the first card is a 10 and the
second is a non-10, or vice versa. Thus
n ( Eone 10 ) = 4 ⋅ 48 + 48 ⋅ 4 = 2 ⋅ 4 ⋅ 48 .
x
4. a.
32
5. Let X = gain (in dollars) on a play. If no 10
1
1
appears, then X = 0 − = − ; if exactly one 10
4
4
1 3
appears, then X = 1 − = ; if two 10’s appear,
4 4
1 7
then X = 2 − = .
4 4
n(S) = 52 · 52. In a deck, there are 4 10’s and 48
non 10’s. Thus n ( Eno 10 ) = 48 ⋅ 48 . The event
1
12
Similarly, f(2), f(3), f(4), and f(5) equal
f (6) = P ( E6 ) =
188
x
2
f (0) = P ( E0 ) =
∑ xf ( x) = 0 ⋅ 221 + 1⋅ 221 + 2 ⋅ 221
1128 188
,
=
1326 221
351
Chapter 9: Additional Topics in Probability
7. a.
ISM: Introductory Mathematical Analysis
Let X = gain (in dollars) on each unit
shipped. Then P(X = –100) = 0.08 and
P(X = 200) = 1 – 0.08 = 0.92.
E(X) = –100f(–100) + 200f(200)
= –100(0.08) + 200(0.92)
= $176 per unit
0
10.
1
2
5
0
⎛1⎞
f (5) = 5C5 ⎜ ⎟
⎝ 3⎠
1
µ = np = 5 ⋅ =
3
1
1
⎛2⎞
⋅1 =
⎜ ⎟ = 1⋅
243
243
⎝3⎠
5
3
1 2
10
10
=
=
≈ 1.05
3 3
9
3
σ = npq = 5 ⋅ ⋅
11. P(X ≤ 1) = P(X = 0) + P(X = 1)
0
5
1
4
⎛3⎞ ⎛1⎞
⎛3⎞ ⎛1⎞
= 5C0 ⎜ ⎟ ⎜ ⎟ + 5C1 ⎜ ⎟ ⎜ ⎟
4
4
⎝ ⎠ ⎝ ⎠
⎝4⎠ ⎝4⎠
1
3 1
16
1
= 1⋅1⋅
+ 5⋅ ⋅
=
=
1024
4 256 1024 64
⎛ 40,999,999 ⎞
– 1.00 ⎜
⎟ ≈ –0.63
⎝ 41, 000, 000 ⎠
There is a loss of about $0.63 per play.
= 0.522
3
1 4 40
⎛1⎞ ⎛ 2⎞
f (3) = 5C3 ⎜ ⎟ ⎜ ⎟ = 10 ⋅ ⋅ =
27 9 243
⎝3⎠ ⎝ 3⎠
4
1
1 2 10
⎛1⎞ ⎛ 2⎞
f (4) = 5C4 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ =
3
3
81
3 243
⎝ ⎠ ⎝ ⎠
8. There are 41 million combinations from which
to choose. Let x = gain (in dollars) per play. If
the player wins, then
x = 15,000,000 – 1.00 = 14,999,999 and
1
. If the player
P(X = 14,999,999) =
41, 000, 000
loses, then X = –1.00 and
1
40,999,999
.
P(X = –1.00) = 1 –
=
41, 000, 000 41, 000, 000
E(X) = 14,999,999f (14,999,999) – 1.00f (–1.00)
⎛
1
⎞
= 14,999,999 ⎜
⎟
⎝ 41, 000, 000 ⎠
f (0) = 4C0 (0.15)0 (0.85)4 ≈
4
1 16 80
⎛1⎞ ⎛ 2⎞
f (1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ =
3 81 243
⎝3⎠ ⎝ 3⎠
2
3
1 8
80
⎛1⎞ ⎛ 2⎞
=
f (2) = 5C2 ⎜ ⎟ ⎜ ⎟ = 10 ⋅ ⋅
9 27 243
⎝ 3⎠ ⎝ 3⎠
b. Since the expected gain per unit is $176 and
4000 units are shipped per year, then
expected annual profit is
4000(176) = $704,000.
9.
5
32
32
⎛1⎞ ⎛2⎞
f (0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅
=
3
3
243
243
⎝ ⎠ ⎝ ⎠
0
4!
⋅1(0.522)
0!4!
6
6!
⎛ 2⎞ ⎛1⎞
⎛ 1 ⎞
(1) ⎜
12. P ( X = 0) = 6 C0 ⎜ ⎟ ⎜ ⎟ =
⎟
3
3
0!
⋅
6!
⎝ ⎠ ⎝ ⎠
⎝ 729 ⎠
1
⎛ 1 ⎞
= 1(1) ⎜
⎟=
⎝ 729 ⎠ 729
f (1) = 4 C1 (0.15)1 (0.85)3
4!
≈
⋅ (0.15)(0.614) = 0.368
1!3!
1
5
6! ⎛ 2 ⎞ ⎛ 1 ⎞
⎛2⎞ ⎛1⎞
P ( X = 1) = 6C1 ⎜ ⎟ ⎜ ⎟ =
1!⋅ 5! ⎜⎝ 3 ⎟⎠ ⎜⎝ 243 ⎟⎠
⎝ 3⎠ ⎝3⎠
f (2) = 4 C2 (0.15)2 (0.85)2
4!
=
⋅ (0.0225)(0.7225) ≈ 0.098
2!2!
⎛ 2 ⎞ ⎛ 1 ⎞ 12
= 6⎜ ⎟⎜
⎟=
⎝ 3 ⎠ ⎝ 243 ⎠ 729
f (3) = 4C3 (0.15)3 (0.85)1
4!
=
⋅ (0.003375)(0.85) ≈ 0.011
3!1!
2
4
6! ⎛ 4 ⎞⎛ 1 ⎞
⎛2⎞ ⎛1⎞
P ( X = 2) = 6C2 ⎜ ⎟ ⎜ ⎟ =
⎟
2!⋅ 4! ⎜⎝ 9 ⎟⎜
⎝ 3⎠ ⎝3⎠
⎠⎝ 81 ⎠
6 ⋅ 5 ⋅ 4! ⎛ 4 ⎞⎛ 1 ⎞
⎛ 4 ⎞ ⎛ 1 ⎞ 60
= 15 ⎜ ⎟ ⎜ ⎟ =
⎜
⎟⎜
⎟
2 ⋅1 ⋅ 4! ⎝ 9 ⎠⎝ 81 ⎠
⎝ 9 ⎠ ⎝ 81 ⎠ 729
P(X > 2) = 1 – P(X ≤ 2)
= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]
12
60 ⎤
73 656
⎡ 1
= 1− ⎢
+
+
= 1−
=
⎥
729 729
⎣ 729 729 729 ⎦
f (4) = 4C4 (0.15) 4 (0.85)0
4!
≈
⋅ (0.000506)1 = 0.0005
4!0!
µ = np = 4(0.15) = 0.6
=
σ = npq = 4(0.15)(0.85) ≈ 0.71
352
ISM: Introductory Mathematical Analysis
Chapter 9 Review
13. The probability that a 2 or 3 results on one roll is
2 1
= . Let X = number of 2’s or 3’s that appear
6 3
1
on 4 rolls. Then X is binomial with p = and
3
n = 4.
3
2
1
1 2 8
⎛1⎞ ⎛2⎞
P ( X = 3) = 4C3 ⎜ ⎟ ⎜ ⎟ = 4 ⋅ ⋅ =
27 3 81
⎝3⎠ ⎝ 3⎠
=
P ( X = 0) = 4C0 (0.9)0 (0.1) 4 = 0.0001
15. Let X = number of heads that occur. Then X is
binomial.
5
243
243
⎛ 2⎞ ⎛3⎞
P ( X = 0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅
=
3125 3125
⎝5⎠ ⎝5⎠
1
4
2 81
810
⎛ 2⎞ ⎛3⎞
P ( X = 1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅
=
5 625 3125
⎝ 5⎠ ⎝5⎠
P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)]
810 ⎤
1053 2072
⎡ 243
= 1− ⎢
+
= 1−
=
⎥
3125 3125
⎣ 3125 3125 ⎦
18. From column 1, a + a + 0.2 = 1, so 2a = 0.8, or
a = 0.4.
From column 3, b + b + a = 1, so 2b = 1 – a, or
1 − a 1 − 0.4
b=
=
= 0.3.
2
2
From column 2, a + b + c = 1, so
c = 1 – a – b = 1 – 0.4 – 0.3 = 0.3.
⎡ 0.1 0.3 0.1⎤ ⎡ 0.5⎤ ⎡ 0.10 ⎤
19. X1 = TX0 = ⎢ 0.2 0.4 0.1⎥ ⎢ 0 ⎥ = ⎢ 0.15 ⎥
⎢
⎥⎢ ⎥ ⎢
⎥
⎣ 0.7 0.3 0.8⎦ ⎣ 0.5⎦ ⎣ 0.75⎦
⎡ 0.1 0.3 0.1⎤ ⎡0.10 ⎤ ⎡ 0.130 ⎤
X 2 = TX1 = ⎢ 0.2 0.4 0.1⎥ ⎢ 0.15⎥ = ⎢ 0.155 ⎥
⎢
⎥⎢
⎥ ⎢
⎥
⎣ 0.7 0.3 0.8⎦ ⎣ 0.75⎦ ⎣ 0.715 ⎦
16. On any draw, the probability of selecting a red
2 1
= . Let
jelly bean is
10 5
X = number of red jelly beans selected in five
1
draws. Then X is binomial with p = and
5
n = 5.
0
⎡ 0.1 0.3 0.1⎤ ⎡ 0.130 ⎤
X3 = TX2 = ⎢ 0.2 0.4 0.1⎥ ⎢ 0.155⎥
⎢
⎥⎢
⎥
⎣ 0.7 0.3 0.8⎦ ⎣ 0.715⎦
⎡0.1310 ⎤
= ⎢ 0.1595⎥
⎢
⎥
⎣ 0.7095⎦
5
1024
⎛1⎞ ⎛4⎞
P ( X = 0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅
5
5
3125
⎝ ⎠ ⎝ ⎠
1024
=
3125
1
1024 1280 640 2944
+
+
=
= 0.94208
3125 3125 3125 3125
17. From column 1, 0.1 + a + 0.6 = 1, so a = 0.3.
From column 2, 2a + b + b = 1, so 2b = 1 – 2a,
1 − 2a 1 − 2(0.3)
=
= 0.2.
or b =
2
2
From column 3, a + b + c = 1, so c = 1 – a – b,
or
c = 1 – 0.3 – 0.2 = 0.5.
14. Let X = number of bushes that live. Then X is
binomial.
0
3
1 64
⎛1⎞ ⎛4⎞
P ( X = 2) = 5C2 ⎜ ⎟ ⎜ ⎟ = 10 ⋅ ⋅
5
5
25
125
⎝ ⎠ ⎝ ⎠
640
=
3125
P ( X ≤ 2) = P ( X = 0) + P( X = 1) + P( X = 2)
4
1 256
⎛1⎞ ⎛4⎞
P ( X = 1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅
5 625
⎝5⎠ ⎝5⎠
1280
=
3125
353
Chapter 9: Additional Topics in Probability
ISM: Introductory Mathematical Analysis
⎡ 0.4 0.1 0.1⎤ ⎡ 0.1⎤ ⎡ 0.13⎤
20. X1 = TX0 = ⎢ 0.2 0.6 0.5 ⎥ ⎢ 0.3⎥ = ⎢ 0.50 ⎥
⎢
⎥⎢ ⎥ ⎢
⎥
⎣ 0.4 0.3 0.4 ⎦ ⎣ 0.6 ⎦ ⎣ 0.37 ⎦
⎡ 0.4 0.1 0.1⎤ ⎡ 0.13⎤ ⎡0.139 ⎤
X 2 = TX1 = ⎢ 0.2 0.6 0.5 ⎥ ⎢ 0.50 ⎥ = ⎢ 0.511⎥
⎢
⎥⎢
⎥ ⎢
⎥
⎣ 0.4 0.3 0.4 ⎦ ⎣0.37 ⎦ ⎣0.350 ⎦
c.
⎡ 1 2 ⎤ ⎡1 0 ⎤ ⎡ − 2 2 ⎤
3 3⎥
3
3 ⎥
−
=⎢
23. T − I = ⎢
⎢ 2 1 ⎥ ⎢⎣ 0 1 ⎥⎦ ⎢ 2 − 2 ⎥
3⎦
⎣3 3⎦
⎣ 3
1
⎡ 1
⎤
⎡
⎤
1 0 2
1 1
⎢
⎥
⎢
⎥
⎢ − 2 2 0⎥ → … → ⎢0 1 1 ⎥
2⎥
⎢ 3 3
⎥
⎢
⎢ 2 − 2 0⎥
⎢0 0 0 ⎥
3
⎣
⎦
⎣ 3
⎦
1
⎡ ⎤
Q = ⎢2⎥
⎢1⎥
⎣2⎦
⎡ 0.4 0.1 0.1⎤ ⎡ 0.139⎤
X3 = TX2 = ⎢ 0.2 0.6 0.5⎥ ⎢ 0.511⎥
⎢
⎥⎢
⎥
⎣ 0.4 0.3 0.4 ⎦ ⎣ 0.350 ⎦
⎡0.1417 ⎤
= ⎢ 0.5094 ⎥
⎢
⎥
⎣ 0.3489 ⎦
21. a.
⎡1
7
T2 = TT = ⎢
⎢6
⎣7
⎡ 19
49
T =T T=⎢
⎢ 30
⎣ 49
⎡ 109 117 ⎤
343 343 ⎥
=⎢
⎢ 234 226 ⎥
⎣ 343 343 ⎦
3
2
3⎤ ⎡1
7⎥ ⎢7
4⎥⎢6
7⎦ ⎣7
3⎤
7⎥
4⎥
7⎦
15 ⎤ ⎡ 1
49 ⎥ ⎢ 7
34 ⎥ ⎢ 6
49 ⎦ ⎣ 7
⎡ 19
49
=⎢
⎢ 30
⎣ 49
3⎤
7⎥
4⎥
7⎦
15 ⎤
49 ⎥
34 ⎥
49 ⎦
b. From T2 , entry in row 1, column 2, is
c.
22. a.
⎡ 0.4 0.4 0.3⎤ ⎡1 0 0 ⎤
24. T − I = ⎢⎢ 0.3 0.2 0.3⎥⎥ − ⎢⎢ 0 1 0 ⎥⎥
⎢⎣ 0.3 0.4 0.4 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦
0.3⎤
⎡ −0.6 0.4
⎢
= ⎢ 0.3 −0.8 0.3⎥⎥
⎢⎣ 0.3 0.4 −0.6 ⎥⎦
1
1
1⎤
⎡ 1
⎢ −0.6 0.4 0.3 0 ⎥
⎢
⎥ →…
⎢ 0.3 −0.8 0.3 0 ⎥
⎢
⎥
⎣⎢ 0.3 0.4 −0.6 0 ⎦⎥
⎡ 1 0 0 0.36 ⎤
⎢ 0 1 0 0.27 ⎥
⎥
→⎢
⎢ 0 0 1 0.36 ⎥
⎢
⎥
⎣⎢ 0 0 0 0 ⎦⎥
15
.
49
From T3 , entry in row 2, column 1, is
234
.
343
⎡ 0.36 ⎤
Q ≈ ⎢ 0.27 ⎥
⎢
⎥
⎣ 0.36 ⎦
T2 = TT
⎡ 0 0.4 0.3⎤ ⎡ 0 0.4 0.3⎤
= ⎢⎢ 0 0.3 0.5⎥⎥ ⎢⎢ 0 0.3 0.5⎥⎥
⎢⎣1 0.3 0.2 ⎥⎦ ⎢⎣1 0.3 0.2 ⎥⎦
= ⎡ 0.3 0.21 0.26 ⎤
⎢ 0.5 0.24 0.25 ⎥
⎢
⎥
⎣⎢ 0.2 0.55 0.49 ⎦⎥
T3 = T2 T
⎡0.3
= ⎢⎢0.5
⎢⎣0.2
⎡0.26
= ⎢⎢0.25
⎢⎣0.49
From T3 , entry in row 2, column 1, is 0.25.
Japanese Non-Japanese
Japanese ⎡ 0.8
0.6⎤
25. T =
Non-Japanese ⎢⎣ 0.2
0.4⎥⎦
a.
0.21 0.26 ⎤ ⎡ 0 0.4 0.3⎤
0.24 0.25 ⎥⎥ ⎢⎢0 0.3 0.5 ⎥⎥
0.55 0.49 ⎥⎦ ⎢⎣1 0.3 0.2 ⎥⎦
0.261 0.247 ⎤
0.347 0.32 ⎥⎥
0.392 0.433 ⎥⎦
b. From T2 , entry in row 1, column 2, is 0.21.
354
⎡ 0.8 0.6 ⎤ ⎡ 0.8 0.6 ⎤
T2 = ⎢
⎥⎢
⎥
⎣ 0.2 0.4 ⎦ ⎣ 0.2 0.4⎦
⎡0.76 0.72 ⎤
=⎢
⎥
⎣0.24 0.28 ⎦
From row 1, column 1, the probability that a
person who currently owns a Japanese car
will buy a Japanese car two cars later is
0.76. Thus 76% of people who currently
own Japanese cars will own Japanese cars
two cars later.
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 9
⎡ 0.76 0.72 ⎤ ⎡ 0.6 ⎤ ⎡ 0.744 ⎤
X 2 = T2 X0 = ⎢
⎥⎢ ⎥ = ⎢
⎥
⎣ 0.24 0.28⎦ ⎣ 0.4 ⎦ ⎣ 0.256 ⎦
Two cars from now, we expect 74.4% Japanese, 25.6% non-Japanese.
b.
⎡ 0.8 0.6 ⎤ ⎡1 0 ⎤ ⎡ −0.2 0.6 ⎤
T−I = ⎢
⎥−⎢
⎥ =⎢
⎥
⎣ 0.2 0.4 ⎦ ⎣ 0 1 ⎦ ⎣ 0.2 −0.6 ⎦
1
1⎤
⎡ 1
⎡1 0 0.75⎤
⎢ −0.2 0.6 0 ⎥ → … → ⎢0 1 0.25⎥
⎢
⎥
⎢
⎥
⎢⎣ 0.2 −0.6 0 ⎥⎦
⎢⎣0 0 0 ⎥⎦
⎡ 0.75⎤
Q=⎢
⎣ 0.25⎥⎦
In the long run, 75% Japanese cars, 25% non-Japanese cars.
c.
⎡0.7 0.4 0.1⎤ ⎡ 0.5 ⎤ ⎡ 0.49 ⎤
X1 = TX0 = ⎢⎢ 0.2 0.5 0.1⎥⎥ ⎢⎢ 0.3 ⎥⎥ = ⎢⎢ 0.27 ⎥⎥
⎢⎣ 0.1 0.1 0.8⎥⎦ ⎢⎣ 0.2 ⎥⎦ ⎢⎣ 0.24 ⎥⎦
49% are expected to vote for party 1, 27% for party 2, 24% for party 3.
26. a.
⎡ 0.7 0.4 0.1⎤ ⎡ 1
T − I = ⎢ 0.2 0.5 0.1⎥ − ⎢ 0
⎢
⎥ ⎢
⎣ 0.1 0.1 0.8⎦ ⎣ 0
b.
1
1
⎡ 1
⎢ −0.3 0.4 0.1
⎢
⎢ 0.2 −0.5 0.1
⎢
⎢⎣ 0.1 0.1 −0.2
0
1
0
0 ⎤ ⎡ −0.3 0.4 0.1 ⎤
0 ⎥ = ⎢ 0.2 −0.5 0.1 ⎥
⎥ ⎢
⎥
1 ⎦ ⎣ 0.1 0.1 −0.2 ⎦
⎡1
1⎤
⎢
⎢0
0 ⎥⎥
→… → ⎢
0⎥
⎢0
⎥
⎢
0 ⎥⎦
⎢⎣ 0
0 0
1 0
0 1
0 0
3⎤
7 ⎥
5⎥
21 ⎥
1 ⎥
3⎥
0 ⎥⎦
⎡0.429 ⎤
Q ≈ ⎢0.238 ⎥
⎢
⎥
⎣0.333 ⎦
In the long run, 43% will vote for party 1, 24% for party 2, and 33% for party 3.
Mathematical Snapshot Chapter 9
⎡0 ⎤
⎡0⎤
⎡0⎤
⎢1 ⎥
⎢0⎥
⎢ 1⎥
1. For X0 = ⎢ ⎥ or ⎢ ⎥ , the first entry of the state vector is greater than 0.5 for n = 7 or greater. If X0 = ⎢ ⎥ , then
⎢0⎥
⎢0 ⎥
⎢ 1⎥
⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
⎢⎣ 0 ⎥⎦
⎡ 0.5217 ⎤
⎢ 0.0000 ⎥
T X0 ≈ ⎢
⎥.
⎢ 0.4783⎥
⎢⎣ 0.0000 ⎥⎦
7
355
Chapter 9: Additional Topics in Probability
⎡ 1 0.1 0.1 0.01 ⎤ ⎡1
⎢ 0 0 0.9 0.09 ⎥ ⎢0
⎥−⎢
2. T − I = ⎢
⎢ 0 0.9 0 0.09 ⎥ ⎢0
⎢
⎥ ⎢
0 0.81 ⎥⎦ ⎢⎣0
⎢⎣ 0 0
1
1
1⎤
⎡1 1
⎢ 0 0.1 0.1 0.01 0 ⎥
⎢
⎥
⎢ 0 −1 0.9 0.09 0 ⎥ → … →
⎢
⎥
⎢ 0 0.9 −1 0.09 0 ⎥
⎢⎣ 0 0
0 −0.19 0 ⎥⎦
0
1
0
0
⎡1
⎢0
⎢
⎢0
⎢
⎢0
⎢0
⎣
ISM: Introductory Mathematical Analysis
0 ⎤ ⎡0 0.1 0.1 0.01 ⎤
0 ⎥⎥ ⎢⎢ 0 −1 0.9 0.09 ⎥⎥
=
0 ⎥ ⎢ 0 0.9 −1 0.09 ⎥
⎥ ⎢
⎥
1 ⎥⎦ ⎢⎣ 0 0
0 −0.19 ⎥⎦
0
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
0
1⎤
0 ⎥⎥
0⎥ .
⎥
0⎥
0 ⎥⎦
3. Against Always Defect,
1 2 3 4
1 ⎡0 0 0 0 ⎤
2 ⎢1 0.1 1 0.1⎥ .
T= ⎢
⎥
3 ⎢0 0 0 0 ⎥
4 ⎢⎣ 0 0.9 0 0.9 ⎥⎦
Against Always Cooperate,
1 2 3 4
1 ⎡1 0.1 1 0.1⎤
2 ⎢ 0 0 0 0 ⎥⎥ .
T= ⎢
3 ⎢ 0 0.9 0 0.9 ⎥
⎢
⎥
4 ⎣⎢ 0 0 0 0 ⎦⎥
Against regular Tit-for-tat,
1 2 3 4
1 ⎡1 0.1 0 0 ⎤
2 ⎢ 0 0 1 0.1⎥⎥ .
T= ⎢
3 ⎢ 0 0.9 0 0 ⎥
⎢
⎥
4 ⎣⎢ 0 0 0 0.9 ⎦⎥
4. With Player 2 always defecting, after one round the game is in a stable pattern of Player 1 cooperating with
⎡ 0 ⎤
⎢ 0.1⎥
probability 0.1 and defecting with probability 0.9. The steady state vector in this case is ⎢ ⎥ .
⎢ 0 ⎥
⎢⎣ 0.9 ⎥⎦
With Player 2 always cooperating, after one round the game settles into steady mutual cooperation.
⎡1 ⎤
⎢0⎥
With Player 2 playing standard Tit-for-tat, the probabilities gradually tilt toward mutual cooperation: ⎢ ⎥ is the
⎢0⎥
⎣⎢ 0 ⎦⎥
steady state vector. In this case, it takes only one “forgiving” Tit-for-tat-er to guarantee mutual cooperation in the
long run.
356
Chapter 10
Problems 10.1
Principles in Practice 10.1
1. a.
1. The graph of the greatest integer function is
shown.
b. 0
10
10
–10
lim f ( x) does not exist when a is an integer
x →a
since the limits are different depending on the
side from which you approach the integer.
lim f ( x) exists for all numbers which are not
integers.
2. a.
2
c.
2
3. a.
1
c.
4. a.
4
4
2. lim V ( r ) = lim π r 3 = π lim r 3
3 r →1
r →1
r →1 3
4
4
= π (1)3 = π
3
3
(
3. lim R ( x) = lim 500 x − 6 x 2
x →8
x →8
c.
)
x →8
= 500 lim x − 6 lim x 2 = 500(8) − 6(8) 2
x →8
x →8
= 4000 − 384 = 3616
4. lim p = lim
) = lim ⎡⎣⎢50 (t + 4t )⎤⎦⎥
+ 3t + 20 lim ( t + 3t + 20 )
(
50 t 2 + 4t
t →2 t 2
3
–1
b. does not exist
= lim 500 x − lim 6 x 2
=
1
b. does not exist
x →a
t →2
c.
b. 1
–10
x →8
1
2
t →2
2
t →2
50 ⎡ 22 + 4(2) ⎤ 600
⎣
⎦=
= 20
2
30
2 + 3(2) + 20
5. As h → 0 , both the numerator and denominator
approach 0. For h ≠ 0,
125 + 2( x + h) − (125 + 2 x)
lim
h
h →0
125 + 2 x + 2h − 125 − 2 x
2h
= lim
= lim
h
h→0
h →0 h
= lim 2 = 2 .
1
5. f(−0.9) = −3.7
f(−0.99) = −3.97
f(−0.999) = −3.997
estimate of limit: −4
f(−1.1) = −4.3
f(−1.01) = −4.03
f(−1.001) = −4.003
6. f(−3.1) = −6.1
f(−3.01) = −6.01
f(−3.001) = −6.001
estimate of limit: −6
f(−2.9) = −5.9
f(−2.99) = −5.99
f(−2.999) = −5.999
7. f(−0.1) ≈ 0.9516
f(−0.01) ≈ 0.9950
f(−0.001) ≈ 0.9995
estimate of limit: 1
f(0.1) ≈ 1.0517
f(0.01) ≈ 1.0050
f(0.001) ≈ 1.0005
8. f(−0.1) ≈ 0.5132
f(−0.01) ≈ 0.5013
f(−0.001) ≈ 0.5001
estimate of limit: 0.5
f(0.1) ≈ 0.4881
f(0.01) ≈ 0.4988
f(0.001) ≈ 0.4999
9. lim 16 = 16
x→2
10. lim 2 x = 2(3) = 6
x →3
h→0
357
Chapter 10: Limits and Continuity
11.
12.
13.
(
ISM: Introductory Mathematical Analysis
)
lim t 2 − 5 = (−5) 2 − 5 = 25 − 5 = 20
t →−5
21.
16
⎛1⎞
lim (5t − 7) = 5 ⎜ ⎟ − 7 = −
3
3
⎝ ⎠
22.
t →1/ 3
lim (3 x3 − 4 x 2 + 2 x − 3)
24. lim
(
lim x 2 + 6
lim ( x − 6)
) = (−6)
x →−6
+6
(−6) − 6
2
t2 − 4
(t + 2)(t − 2)
= lim
= lim(t + 2) = 4
t−2
t →2 t − 2
t →2
t →2
26. lim
0 − 7(0) + 1
=
(
h →0
)
x2 − 2 x
x( x − 2)
= lim
= lim ( x − 2) = −2
x
x
x →0
x →0
x →0
lim h 2 − 7h + 1
h→0
28. lim
=0
z 2 − 5z − 4
z →0
z +1
2
29. lim
= z →0
lim ( z 2 + 1)
lim
p2 + p + 5 =
= lim
2
z →0
02 − 5(0) − 4
= −4
p →4
x 2 − 9 x + 20
x − 3x − 4
x−5
1
= lim
=−
5
x→4 x + 1
lim ( z 2 − 5 z − 4)
=
19.
−9
x−3
1
1
= lim
=
x →3 ( x + 3)( x − 3) x →3 x + 3 6
= lim
lim h
x→4
18. lim
x−3
x →3 x 2
0
30.
02 + 1
x→4
x 4 − 81
lim
x →−3 x 2
+ 8 x + 15
lim
y →15
y+3 =
( x − 4)( x − 5)
( x − 4)( x + 1)
( x 2 + 9)( x 2 − 9)
x →−3 ( x + 3)( x + 5)
= lim
( x 2 + 9)( x + 3)( x − 3)
( x + 3)( x + 5)
x →−3
= lim
(
lim p 2 + p + 5
p →4
)
( x 2 + 9)( x − 3)
x+5
x →−3
= −54
= lim
= 42 + 4 + 5 = 25 = 5
20.
t −3
3
=−
4
4
t
−
t →0
= lim
=5
27. lim
=
t →0 t 2 (t − 4)
x →3
x →−6
− 7h + 1
t 2 (t + 3)
x2 − x − 6
( x − 3)( x + 2)
= lim
x −3
x−3
x →3
x →3
= lim ( x + 2)
42
7
=−
−12
2
2
− 4t
2
= lim
25. lim
t →−3
h
x →−1
t 3 + 3t 2
t →0 t 3
lim (t − 2)
t − 2 t →−3
−3 − 2 −5
5
15. lim
=
=
=
=−
lim (t + 5) −3 + 5
2
2
t →−3 t + 5
h →0 h 2
= lim 1 = 1
x →2
4r − 3 4(9) − 3 36 − 3 33
=
=
=
=3
14. lim
11
11
11
r →9 11
17. lim
x +1
x2 − x − 2
( x − 2)( x + 1)
= lim
x−2
x−2
x→2
x →2
= lim ( x + 1) = 3
= 3(−2)3 − 4(−2) 2 + 2(−2) − 3
= −24 − 16 − 4 − 3
= −47
=
lim
x →−1 x + 1
23. lim
x →−2
x2 + 6
16. lim
=
t →−6 x − 6
x2 + 2 x
x( x + 2)
= lim
= lim x = −2
x →−2 x + 2
x →−2 x + 2
x →−2
lim
lim ( y + 3) = 15 + 3 = 18
31. lim
y →15
3x 2 − x − 10
=3 2
358
(3x + 5)( x − 2)
x → 2 ( x + 7)( x − 2)
= lim
+ 5 x − 14
3 x + 5 11
= lim
=
9
x→2 x + 7
x→2 x 2
ISM: Introductory Mathematical Analysis
32.
x2 + 2 x − 8
lim
x →−4 x 2
+ 5x + 4
Section 10.1
x−2
( x + 4)( x − 2)
= lim
=2
x →−4 x + 1
x →−4 ( x + 4)( x + 1)
= lim
⎡ 4 + 4h + h 2 ⎤ − 4
4h + h 2
h(4 + h)
(2 + h) 2 − 22
⎦
33. lim
= lim (4 + h) = 4
= lim
= lim
= lim ⎣
h
h
h
h
h →0
h →0
h→0
h →0
h →0
( x + 2) 2 − 4
x2 + 4 x
= lim ( x + 4) = 4
= lim
x
x
x →0
x →0
x →0
34. lim
( x + h) 2 − x 2
2 xh + h 2
= lim (2 x + h) = 2 x
= lim
h
h
h →0
h→0
h→0
35. lim
3( x + h) 2 + 7( x + h) − 3x 2 − 7 x
3 x 2 + 6 xh + 3h 2 + 7 x + 7 h − 3 x 2 − 7 x
= lim
h
h
h →0
h →0
6 xh + 3h 2 + 7h
h(6 x + 3h + 7)
= lim
= lim
h
h
h →0
h →0
= lim (6 x + 3h + 7) = 6 x + 7
36. lim
h →0
f ( x + h) − f ( x )
[7 − 3( x + h)] − (7 − 3x)
−3h
= lim
= lim
= lim − 3 = −3
h
h
h →0
h →0
h →0 h
h →0
37. lim
38. lim
h →0
f ( x + h) − f ( x )
[2( x + h) + 3] − (2 x + 3)
2h
= lim
= lim
= lim 2 = 2
h
h
h →0
h →0 h
h→0
(
⎡ ( x + h ) 2 − 3⎤ − x 2 − 3
f ( x + h) − f ( x )
⎦
39. lim
= lim ⎣
h
h
h →0
h →0
= lim
h →0
(
x 2 + 2 xh + h 2 − 3 − x 2 − 3
) = lim 2 xh + h
h →0
h
h
2
)
= lim (2 x + h) = 2 x
h →0
(
)
⎡ ( x + h)2 + ( x + h) + 1⎤ − x 2 + x + 1
f ( x + h) − f ( x )
2 xh + h 2 + h
⎦
= lim
= lim (2 x + h + 1) = 2 x + 1
= lim ⎣
h
h
h
h →0
h →0
h →0
h →0
40. lim
f ( x + h) − f ( x )
[( x + h)3 − 4( x + h)2 ] − [ x3 − 4 x 2 ]
= lim
h
h
h →0
h →0
3
2
2
x + 3 x h + 3xh + h3 − 4 x 2 − 8 xh − 4h 2 − x3 + 4 x 2
= lim
h
h →0
2
2
3
3 x h + 3xh + h − 8 xh − 4h 2
= lim
h
h →0
h(3x 2 + 3xh + h 2 − 8 x − 4h)
= lim
h
h →0
= lim (3 x 2 + 3 xh + h 2 − 8 x − 4h) = 3x 2 − 8 x
41. lim
h →0
359
Chapter 10: Limits and Continuity
ISM: Introductory Mathematical Analysis
f ( x + h) − f ( x )
[3 − ( x + h) + 4( x + h)2 ] − (3 − x + 4 x 2 )
= lim
h
h
h →0
h →0
3 − x − h + 4 x 2 + 8 xh + 4h 2 − (3 − x + 4 x 2 )
= lim
h
h →0
− h + 8 xh + 4h 2
= lim
h
h →0
h(−1 + 8 x + 4h)
= lim
h
h →0
= lim (−1 + 8 x + 4h)
42. lim
h →0
= −1 + 8 x
x−2 −2
= lim
x−6
x →6
x →6
43. lim
x →6 ( x − 6)
x →6
x−2 −2
( x − 6)
( x − 2) − 4
= lim
= lim
(
(
x−2 +2
1
x−2 +2
44. For lim
=
x−2 +2
x−2 +2
= lim
x →6 ( x − 6)
)
)
x−6
(
x−2 +2
)
1
4
x2 + x + c
x →3 x 2
)
(
)(
− 5x + 6
x2 + x + c
to exist, x – 3 must be a factor of the numerator x 2 + x + c :
x →3 ( x − 3)( x − 2)
= lim
x + x + c = ( x − 3)( x + r ) = x 2 + (r − 3) x − 3r
Thus r – 3 = 1, or r = 4. So c = –3r = –3(4) = –12.
For c = –12,
x2 + x + c
x 2 + x − 12
( x − 3)( x + 4)
x+4 7
= lim
= lim
= lim
= =7
lim
2
2
1
x →3 x − 5 x + 6 x →3 x − 5 x + 6
x →3 ( x − 3)( x − 2) x →3 x − 2
2
Th − Tc Th − 0 Th
=
=
=1
Th
Th
Tc →0 Th
45. a.
lim
Th − Tc Th − Th
0
=
=
=0
Th
Th
Tc →Th Th
b.
46.
lim
lim
r →7.5×107
47.
E=
lim
r →7.5×107
−
7.0
7.0 × 1017
7.0 × 1017
=−
=−
× 1010 ≈ −9.33 × 109 ft-lb
7
r
7.5
7.5 × 10
15
0
0
5
11.00
360
ISM: Introductory Mathematical Analysis
Section 10.2
5
48.
lim P ( x) = 2343.056
x →53.2
lim P ( x) = 224(53.2) − 3.1(53.2)2 − 800
x →53.2
5
–5
Principles in Practice 10.2
–5
1
1. The graph of p(x) is shown.
10,000
10
49.
0
5
15
0
x →∞
quickly drops down toward zero. According to
this function, a low price corresponds to a high
demand and a high price corresponds to a low
demand.
2
0
0
10
0
From the graph, it is apparent that
lim p( x) = 0 . The graph starts out high and
4.00
50.
= 2343.056
2. The graph of y(x) is shown.
5
550
0.80
51. The graph of C(p) is shown. (Negative amounts
of impurities and money are not reasonable, so
only the first quadrant is shown.)
0
100,000
1000
0
From the graph, it is apparent that
lim y ( x) = 500 . The greatest yearly sales that
x →∞
0
0
the company can expect is $500,000, even with
unlimited spending on advertising.
5
As p gets closer and closer to 0, the values of
C(p) increase without bound, so lim C ( p) does
3. The graph of C(x) is shown.
1,000,000
p →0
not exist.
52. The graph of P(x) is shown with the value
x = 53.2 indicated.
0
3500
1000
0
From the graph it is apparent that
lim C ( x) = ∞ . This indicates that the cost
x →∞
increases without bound the more units that you
make.
0
0
75
4. lim f ( x) does not exist since
x →1
lim f ( x) = lim 100 = 100 while
x →1−
361
x →1−
Chapter 10: Limits and Continuity
ISM: Introductory Mathematical Analysis
lim f ( x) = lim 175 = 175 .
x →1+
6.
x →1+
lim 19 = 19
x →−∞
lim f ( x) = 250 since
x → 2.5
lim
x →2.5−
f ( x) = lim
x →2.5+
7.
f ( x) = 250
6x
lim
x →0
−
x
4
= lim
x →0
−
6
x3
= −∞ since x3 is negative
and close to 0 for x → 0− .
Problems 10.2
1. a.
lim 7
7
7
x→2
8. lim
=
= =7
lim ( x − 1) 1
x→2 x − 1
2
x →2
b. 3
c.
does not exist
d.
−∞
e.
∞
f.
∞
g.
∞
9.
10.
11.
1
j.
1
12.
13.
k. 1
2. a.
14.
0
b. –∞
c.
15.
does not exist
3.
2
f.
1
g.
1
h is close to 0 when h is
5−h = 0
lim
h →5−
lim
x →−2
−
−3
=∞
x+2
lim 21/ 2 = 21/ 2
x →0−
(
)
lim 4 x − 1 . As x → 1+ , then x – 1
x →1+
(
)
lim 4 x − 1 = 4 ⋅ lim
16.
lim ( x − 2)
17.
x →3+
As x → 3 , then x − 2 → 1 .
5.
h = 0 since
lim
h →0 +
x →1+
+
4.
lim (t − 1)3 = ∞
t →∞
approaches 0 through positive values. So
x − 1 → 0 . Thus
d. ∞
e.
x →−∞
positive and close to 0.
h. 0
i.
lim x 2 = ∞ since x 2 is positive for x → −∞ .
x →1+
x −1 = 4 ⋅ 0 = 0 .
lim ⎛⎜ x x 2 − 4 ⎞⎟ = 0
⎠
x → 2+ ⎝
lim
x →∞
x + 10
As x becomes very large, so does x + 10.
Because square roots of very large numbers are
very large, lim x + 10 = ∞ .
lim (1 − x 2 ) = 0
x →−1+
x →∞
18.
lim 5 x
x →−∞
As x becomes very negative, so does 5x. Thus
lim 5 x = −∞ .
lim − 1 − 10 x
x →−∞
As x becomes very negative,
1 − 10x becomes very positive. Because square
roots of very large numbers are very large,
lim − 1 − 10 x = −∞.
x →−∞
x →−∞
362
ISM: Introductory Mathematical Analysis
19.
20.
21.
22.
23.
24.
3
lim
x →∞
x
= 3 lim
x →∞
−6
lim
x →∞ 5 x
3
=−
x
1
x
Section 10.2
= 3⋅ 0 = 0
1
2
30.
6
1
6
lim
= − ⋅0 = 0
4
/
3
5 x→∞ x
5
31.
x +8
x
= lim = lim 1 = 1
x →∞ x − 3 x →∞ x
x →∞
lim
lim
x →∞
2x − 4
2x
= lim
= lim (−1) = −1
3 − 2 x x →∞ −2 x x →∞
x2 − 1
lim
x →−∞ x3
r3
lim
r →∞ r 2
25. lim
+ 4x − 3
+1
= lim
= lim
= lim
x →−∞ x3
r3
r →∞ r 2
3t 3 + 2t 2 + 9t − 1
5t − 5
2
t →∞
x2
x →−∞
1
=0
x
33.
r →∞
3t 3
t →∞ 5t
7 − 5 x3 + 2 x 2
2 2
= lim
=
5
x →−∞ 5
x →−∞
x+3
lim
−
x →3
34.
2
3t
t →∞ 5
3
= lim t
5 t →∞
=∞
27.
28.
= lim
35.
36.
5x
−x +4
5
5
1 5
= lim
= ⋅ lim
= ⋅0 = 0
6
3 x →∞ x 6 3
x →∞ 3 x
x →∞ 3 x 7
3
5w + 7 w − 1
2
w→∞
4 − 3 x3
lim
x3 − 1
x →∞
x →∞
4 + 5x − 7 x
3 x − x3
lim
x →−∞ x3
x →∞ 3 x 7
37.
lim
−3 x3
x3
x →∞
+ x +1
2
x →−3
lim
2
2
(4 x − 1)
⋅ lim
3
= lim
1
43 x →−∞ x3
lim
3 − 4x − 2x
3
5x − 8x + 1
−2
2
= lim
=−
5
x→∞ 5
x →∞
3
2
43
= lim
x →∞
38. lim
x + 3x
t 2 − 4t + 3
= lim (−1) = −1
x →−∞
= lim
−
(t − 1)(t − 3)
t →3 (t + 1)(t − 3)
− 2t − 3
t −1 2 1
= lim
= =
2
t →3 t + 1 4
t →3 t 2
−2 x
x3
x
= −∞
x →∞ −7
= lim
2
(5 x − 1)( x + 3)
x( x + 3)
x →−3
5x − 1
= lim
x
x →−3−
−16
=
−3
16
=
3
2
⋅0 = 0
−7 x
= lim
2
x →−∞ 43 x 3
=
2 2
=
5 5
= lim (−3) = −3
− x3
x →−∞
7
7
7
1 7
= lim
= ⋅ lim = ⋅ 0 = 0
2 x →∞ x 2
x →∞ 2 x + 1 x →∞ 2 x
x →−∞
w→∞
x3
x →∞
= lim
5w
= lim
2
x →∞
= lim
5 x 2 + 14 x − 3
−
2 w2
w→∞
= lim
6 − 4 x 2 + x3
lim
= lim
lim
=
29.
5x
lim
x+3
x →3 ( x + 3)( x − 3)
−
2 w2 − 3w + 4
lim
= lim
26.
−5 x3
x →−∞
= lim
x −9
1
= lim
= −∞
x →3− x − 3
2
−2 x3
= lim
32. As x → −3− , then 3x → −9 and 9 − x 2 → 0
through negative values. Thus
3x
= ∞.
lim
−
x →−3 9 − x 2
= lim r = ∞
= lim
3 − 2 x − 2 x3
lim
3
5 x3
39. lim
x →1
363
x 2 − 3x + 1
x2 + 1
=
(
) = −1 = − 1
lim x 2 − 3 x + 1
x →1
(
)
lim x 2 + 1
x →1
2
2
Chapter 10: Limits and Continuity
3 x3 − x 2
=
40. lim
x →−1 2 x + 1
(
lim 3 x3 − x 2
x →−1
lim (2 x + 1)
ISM: Introductory Mathematical Analysis
) = −4 = 4
48.
−1
x →−1
1
→ ∞ . Thus
x −1
41. As x → 1+ , then
49. As x → 1+ , then 1 − x → 0 through negative
−5
= ∞.
values. Thus, lim
+ 1− x
x →1
1 ⎤
⎡
lim ⎢1 +
=∞
+
x − 1 ⎥⎦
x →1 ⎣
42.
lim −
x5 + 2 x3 − 1
x5 − 4 x 2
= lim (−1) = −1
x →−∞
= lim −
x →−∞
x5
x5
50.
x →−∞
43.
lim
x →−7
x2 + 1
−
x − 49
2
51.
x 2 + 1 → 50 and x 2 − 49
approaches 0 through positive values. Thus
x2 + 1
→∞.
x 2 − 49
lim x = lim x = 0
x →0 +
x →0 +
lim x = lim (− x) = 0
x →0 −
x →0 −
Thus, lim x = 0.
x →0
16 − x 4 → 0
52.
lim
1
1
= lim = ∞
+
x x →0 x
lim
1
⎛ 1⎞
= lim −
=∞
x x→0− ⎜⎝ x ⎟⎠
x →0
x →0
45. As x → 0+ , x + x 2 approaches 0 through
5
positive values. Thus
→∞.
x + x2
46. As x → ∞ , then
7 ⎞
⎛
lim ⎜ −
= −∞
+
− 3 ⎟⎠
x
x →3 ⎝
7 ⎞
⎛
lim ⎜ −
⎟ = +∞
−
x →3 ⎝ x − 3 ⎠
Answer: does not exist.
. As x → −7 − , then
44. As x → −2+ , then x → −2 and
through positive values. Thus,
x
lim
= −∞.
+
x →−2
16 − x 4
1
= −∞
2x −1
1
lim
=∞
+ 2x −1
x →1/ 2
Answer: does not exist
lim
x →1/ 2−
+
−
Thus, lim
x →0
53.
1
→ 0 . Thus
x
54.
1⎞
⎛
lim ⎜ x + ⎟ = ∞ .
x⎠
x →∞ ⎝
lim
x →−∞
x +1
x
= lim
= lim 1 = 1
x
x →−∞ x x →−∞
⎡ 3 2 x2 ⎤
3 x 2 + 3 − 2 x3
lim ⎢ −
= lim
⎥
x →∞ ⎢ x x 2 + 1 ⎥ x →∞
x3 + x
⎣
⎦
= lim
x →∞
x
47. lim
= −∞
x →1− x − 1
x
lim
=∞
+ x −1
x →1
Answer: does not exist
55.
−2 x3
x3
= lim (−2) = −2
x →∞
⎧2 if x ≤ 2
f ( x) = ⎨
⎩1 if x > 2
a.
b.
364
1
= ∞.
x
lim f ( x) = lim 1 = 1
x →2+
x → 2+
lim f ( x) = lim 2 = 2
x →2−
x →2−
ISM: Introductory Mathematical Analysis
c.
Section 10.2
lim f ( x) ≠ lim f ( x) , so lim f ( x) does
x →2+
x → 2−
lim g ( x) = lim x = 0
a.
x →2
x →0 +
x →0 +
not exist.
d.
e.
56.
x →∞
b.
c.
x →−∞
d.
e.
x →0 +
x →−∞
lim g ( x) = 0
if x ≤ 2
x →∞
x →2−
59.
x →−∞
b.
c.
x →−∞
⎛ 5000
⎞
lim c = lim ⎜
+ 6⎟ = 0 + 6 = 6
q →−∞
q →∞ ⎝ q
⎠
lim f ( x) = lim f ( x) = lim f ( x) = 2
c
x →2−
lim f ( x) = lim (−2 + 4 x − x 2 ) = −∞
x →∞
x →∞
lim c = 6
q→∞
lim f ( x) = lim x = −∞
x →−∞
x →−∞
6
q
if x < 0
⎧x
57. g ( x) = ⎨
⎩− x if x > 0
a.
x →∞
lim g ( x) = lim x 2 = ∞
e.
lim f ( x) = lim x = 2
x → 2+
lim g ( x) = lim x = ∞
d.
if x > 2
x →2+
x →2−
x →0−
x →0
lim f ( x) = lim (−2 + 4 x − x 2 ) = 2
x → 2+
x →0 −
lim g ( x) = lim g ( x ) = 0 , so
c.
lim f ( x) = lim 2 = 2
x→2
x →0 −
x →∞
⎧⎪ x
f ( x) = ⎨
2
⎪⎩−2 + 4 x − x
a.
lim g ( x) = lim x 2 = 0
b.
lim f ( x) = lim 1 = 1
5000
60.
lim g ( x) = lim (− x) = 0
x →0 +
x →0 +
q →∞
7q + 12, 000
q
⎛
12, 000 ⎞
= lim ⎜ 7 +
⎟ = 7+0 = 7
q ⎠
q →∞ ⎝
lim g ( x) = lim x = 0
x →0 −
lim c = lim
q →∞
x →0 −
c
lim g ( x) = lim g ( x ) = 0 , so
x →0 +
x →0−
c=7+
lim g ( x) = 0
x →0
12,000
q
lim c = 7
q→∞
d.
lim g ( x) = lim (− x ) = −∞
x →∞
7
x →∞
q
e.
lim g ( x) = lim x = −∞
x →−∞
⎪⎧ x 2
58. g ( x) = ⎨
⎪⎩ x
12,000
x →−∞
2000 ⎞
⎛
61. lim ⎜ 50, 000 −
= 50, 000 − 0 = 50, 000
t + 1 ⎟⎠
t →∞ ⎝
if x < 0
if x > 0
365
Chapter 10: Limits and Continuity
62.
lim ⎛⎜ x 2 + x − x ⎞⎟
x →∞ ⎝
⎠
⎛ x2 + x − x ⎞ ⎛ x2 + x + x ⎞
⎜
⎟⎜
⎟
⎠⎝
⎠
= lim ⎝
x →∞
(x
= lim
x →∞
= lim
x →∞
2
)
x2 + x + x
–5
= lim
x →∞
x 1+
1
+x
x
x
⎛
⎞
1
x ⎜⎜ 1 + + 1⎟⎟
x
⎝
⎠
1
1
=
=
1+ 0 +1 2
–10
x
–∞
⎛ 1⎞
x 2 ⎜1 + ⎟ + x
x⎠
⎝
69.
= lim
20
x →∞
0
1
1+
1
+1
x
c.
1.
f ( x) = x3 − 5 x; x = 2
(i) f is defined at x = 2: f(2) = −2
⎧⎪ 2 − x
if x < 2
f ( x) = ⎨
3
⎪⎩ x + k ( x + 1) if x ≥ 2
(ii) lim f ( x) = lim ( x3 − 5 x ) = 23 − 5(2) = −2,
x→2
2− x = 0
x→2
which exists.
(iii) lim f ( x) = −2 = f (2)
lim f ( x) = lim ⎡ x3 + k ( x + 1) ⎤ = 8 + 3k
⎦
x → 2+ ⎣
x→2
x →2+
Thus f is continuous at x = 2.
8
If lim f ( x) exists, then 8 + 3k = 0. So k = − .
3
x →2
2.
f ( x) =
65. 1, 0.5, 0.525, 0.631, 0.912, 0.986, 0.998;
conclude limit is 1.
x −3
; x = −3
5x
(i) f is defined at x = −3: f (−3) =
−44
66. 0.368, 0.135, 0.00674, 0.0000454, 3.72 × 10 ,
can’t do last two. Conclude that the limit is 0.
(ii)
4
67.
does not exist
Problems 10.3
x→∞
x → 2−
11
b. 9
900 x
900 x
= lim
x →∞
x →∞ 10 + 45 x x →∞ 45 x
= lim 20 = 20
lim f ( x) = lim
5
0
a.
lim y = lim
x → 2−
0
x
x →∞
64.
10
68.
x2 + x + x
+ x − x2
= lim
63.
ISM: Introductory Mathematical Analysis
(iii) lim f ( x) =
0
x →−3
1
x−3 2
= , which exists
5
x →−3 5 x
lim f ( x ) = lim
x →−3
2
= f (−3)
5
Thus f is continuous at x = −3.
–1
0
366
−6 2
=
−15 5
ISM: Introductory Mathematical Analysis
Section 10.3
8. Continuous at 2 and –2 because f is a polynomial
function (which is continuous everywhere).
3. g ( x) = 2 − 3 x ; x = 0
(i) g is defined at x = 0; g (0) = 2 .
9. Discontinuous at 3 and −3 because at both points
the denominator of this rational function is 0.
(ii) lim g ( x) = lim 2 − 3 x = 2 , which exists
x →0
x →0
10. Continuous at 2 and –2 because f is a rational
function and at neither point is the denominator
zero.
(iii) lim g ( x) = 2 = g (0)
x →0
Thus g is continuous at x = 0.
4.
f ( x) =
11.
x
;x=2
8
x → 2+
2 1
(i) f is defined at x = 2; f (2) = = .
8 4
x →2
lim f ( x) = lim x 2 = 0 . Since
x →0
x →0
lim f ( x) = 4 = f (2) and lim f ( x) = 0 = f (0) ,
x →2
x →0
f is continuous at both 2 and 0.
Answer: Continuous at 2 and 0.
x−4
;x=4
x+4
12.
(i) h is defined at x = 4, h (4) = 0.
1
⎪⎧
f ( x) = ⎨ x
⎪⎩0
if x ≠ 0
if x = 0
1
= ∞ , lim f ( x)
x →0
x →0
x →0 x
does not exist. Thus f is discontinuous at x = 0.
At x = –1, f is defined; f(–1) = –1.
1
lim f ( x) = lim = −1 . Since
x →−1
x →−1 x
lim f ( x) = −1 = f (−1) , f is continuous at
Because lim f ( x) = lim
x−4 0
(ii) lim h( x) = lim
= = 0 , which exists.
8
x→4
x→4 x + 4
+
(iii) lim h( x ) = 0 = h(4)
x→4
Thus h is continuous at x = 4.
6.
x →2−
lim f ( x) = 4 . In addition,
x 2 1
= = , which exists.
8 4
x→2 8
1
(iii) lim f ( x) = = f (2) .
4
x→2
Thus f is continuous at x = 2.
5. h( x) =
x →2+
lim f ( x) = lim x 2 = 4 , we have
x →2−
(ii) lim f ( x ) = lim
x→2
⎧⎪ x + 2 if x ≥ 2
f ( x) = ⎨ 2
if x < 2
⎪⎩ x
f is defined at x = 2 and x = 0; f(2) = 4, f(0) = 0.
Because lim f ( x) = lim ( x + 2) = 4 and
+
x →−1
f ( x) = 3 x ; x = –1
x = –1.
Answer: Discontinuous at 0, continuous at –1.
(i) f is defined at x = –1; f(–1) = –1.
13. f is a polynomial function.
(ii)
lim f ( x) = lim
x →−1
x →−1
3
x = 3 −1 = −1 , which
14. f is a polynomial function
2 3
1 2⎤
⎡
⎢ f ( x) = 5 + 5 x − 5 x ⎥ .
⎣
⎦
exists.
(iii) lim f ( x) = −1 = f (−1)
x →−1
15. f is a rational function and the denominator is
never zero.
Thus f is continuous at x = –1.
7. Continuous at –2 and 0 because f is a rational
function and at neither point is the denominator
zero.
16. f is a polynomial function ⎡ f ( x) = x − x 2 ⎤ .
⎣
⎦
17. None, because f is a polynomial function.
367
Chapter 10: Limits and Continuity
ISM: Introductory Mathematical Analysis
18. None, because h is a polynomial function.
30.
19. The denominator of this rational function is zero
only when x = –4. Thus f is discontinuous only
at x = –4.
20. The denominator of this rational function is zero
only when x = ±2. Thus f is discontinuous only
at x = ±2.
x →−1−
31.
23. x 2 + 2 x − 15 = 0 , (x + 5)(x – 3) = 0, x = –5 or 3.
Discontinuous at –5 and 3.
24. x 2 + x = 0 , x(x + 1) = 0, x = 0 or –1.
Discontinuous at 0 and –1.
(
lim f ( x) = lim ( x − 1) = 0 , then lim f ( x) = 0 .
x →1
32.
27. x 2 + 1 = 0 has no real roots, so no discontinuity
exists.
)
28. x 4 − 1 = 0 , x 2 + 1 x 2 − 1 = 0 ,
( x + 1) ( x + 1)( x − 1) = 0 , x = ±1.
2
33.
Discontinuous at ±1.
⎧1 if x ≥ 0
f ( x) = ⎨
⎩−1 if x < 0
For x < 0, f(x) = –1, which is a polynomial and
hence continuous. For x > 0, f(x) = 1, which is a
polynomial and hence continuous. Because
lim f ( x) = lim (−1) = −1 and
x →0 −
x →0 +
⎧ x − 3 if x > 2
f ( x) = ⎨
⎩3 − 2 x if x < 2
For x < 2, f(x) = 3 – 2x, which is a polynomial
and hence continuous. For x > 2, f(x) = x – 3,
which is a polynomial and hence continuous.
Because f is not defined at x = 2, it is
discontinuous there.
⎪⎧ x 2 + 1 if x > 2
f ( x) = ⎨
if x < 2
⎪⎩8 x
For x < 2, f(x) = 8x, which is a polynomial and
hence continuous. For x > 2, f ( x) = x 2 + 1,
which is a polynomial and hence continuous.
Because f is not defined at x = 2, it is
discontinuous there.
34.
lim f ( x) = lim 1 = 1 , lim f ( x) does not
x →0 +
x →1
x = 1.
f has no discontinuities.
3
26. Discontinuous at x = , for which the
2
denominator is zero.
x →0 −
x →1+
Since lim f ( x ) = 0 = f (0) , f is continuous at
x = 0, ±1. Discontinuous at 0, ±1.
29.
x →1−
x →1+
)
)(
x →−1
if x ≤ 1
⎧0
f ( x) = ⎨
⎩ x − 1 if x > 1
For x < 1, f(x) = 0, which is a polynomial and
hence continuous. For x > 1, f(x) = x – 1, which
is a polynomial and hence continuous. For x = 1,
f is defined [f(1) = 0]. Because
lim f ( x) = lim 0 = 0 and
x →1−
25. x − x = 0 , x x − 1 = 0 , x(x + 1)(x – 1) = 0,
(
x →−1+
does not exist.
Thus f is discontinuous at x = –1.
22. None, because f is a polynomial function.
2
x →−1−
lim f ( x ) = lim (2 x + 1) = −1 , lim f ( x)
x →−1+
21. None, because g is a polynomial function.
8 6 12 4 18 2 9 ⎤
⎡
⎢ g ( x) = 15 x − 5 x + 5 x − 5 ⎥
⎣
⎦
3
⎧2 x + 1 if x ≥ −1
f ( x) = ⎨
if x < −1
⎩1
For x < –1, f(x) = 1, which is a polynomial and
hence continuous. For x > –1, f(x) = 2x + 1,
which is a polynomial and hence continuous.
Because lim f ( x) = lim 1 = 1 and
x →0
exist.
Thus f is discontinuous at x = 0.
⎧ 16
if
⎪
f ( x) = ⎨ x 2
⎪3x − 2 if
⎩
x≥2
x<2
For x < 2, f(x) = 3x − 2, which is a polynomial
16
,
and hence continuous. For x > 2, f ( x) =
x2
which is continuous because x > 2 means that the
denominator is never zero.
368
ISM: Introductory Mathematical Analysis
Section 10.4
For x = 2, f is defined [f(2) = 4]. Because
lim f ( x) = lim (3 x − 2) = 4 and
x →2−
38.
x →2−
lim f ( x) = lim
16
+
x →2
+
x→2
x = 2.
f has no discontinuities.
35.
0.34
Answer: Yes
–2
Principles in Practice 10.4
y
1. We need to solve V(x) > 0. The zeros of V(x)
occur when x = 0, 8 – 2x = 0, and 10 – 2x = 0, or
x = 0, 4, and 5. These zeros determine the
intervals (–∞, 0) (0, 4), (4, 5), and (5, ∞). Using
x = –1, 1, 4.5, and 6 for test points, we find the
sign of V(x):
V(–1) = (–)(+)(+) = –, so V(x) < 0 on (–∞, 0);
V(1) = (+)(+)(+) = +, so V(x) > 0 on (0, 4);
V(4.5) = (+)(–)(+) = –, so V(x) < 0 on (4, 5);
V(6) = (+)(–)(–) = +, so V(x) > 0 on (5, ∞).
The volume is positive when 0 < x < 4 or 5 < x.
However, x > 5 is unrealistic (as is x < 0) since
the longest side of the piece of metal has length
2(5) = 10 inches. Thus, the volume is positive
when 0 < x < 4.
0.28
0.22
0.16
0.10
x
1
2
3
4
5
Discontinuous at 1, 2, 3, 4.
36.
10
–10
= 4, then lim f ( x) = 4.
x→2
x2
Since lim f ( x) = 4 = f (2), f is continuous at
x→2
2
5
y
x
5
Problems 10.4
1. x 2 − 3 x − 4 > 0
f ( x) = x 2 − 3x − 4 = ( x + 1)( x − 4) has zeros –1
For –3.5 ≤ x ≤ 3.5, discontinuities at –3, –2, –
1, 0, 1, 2, 3.
37.
1000
and 4. By considering the intervals (–∞, –1),
(–1, 4), and (4, ∞), we find f(x) > 0 on (–∞, –1)
and (4, ∞).
Answer: (–∞, –1), (4, ∞)
y
2. x 2 − 8 x + 15 > 0
f ( x) = x 2 − 8 x + 15 = ( x − 3)( x − 5) has zeros 3
and 5. By considering the intervals (–∞, 3),
(3, 5), and (5, ∞), we find f(x) > 0 on (–∞, 3) and
(5, ∞).
Answer: (–∞, 3), (5, ∞)
x
f is continuous at 2.
f is discontinuous at 5.
f is discontinuous at 10.
20
3. x 2 − 3x − 10 ≤ 0
f(x) = (x + 2)(x − 5) has zeros −2 and 5. By
considering the intervals (−∞, −2), (−2, 5), and
(5, ∞), we find f(x) < 0 on (−2, 5).
Answer: [−2, 5]
369
Chapter 10: Limits and Continuity
ISM: Introductory Mathematical Analysis
9. (x + 2)(x – 3)(x + 6) ≤ 0
f(x) = (x + 2)(x – 3)(x + 6) has zeros –2, 3, and
–6. By considering the intervals (–∞, –6),
(–6, –2), (–2, 3), and (3, ∞), we find f(x) < 0 on
(–∞, –6) and (–2, 3).
Answer: (–∞, –6], [–2, 3]
4. 14 − 5 x − x 2 ≤ 0 , or equivalently,
x 2 + 5 x − 14 ≥ 0
f ( x) = x 2 + 5 x − 14 = ( x + 7)( x − 2) has zeros –7
and 2. By considering the intervals (–∞, –7),
(–7, 2), and (2, ∞), we find f(x) ≥ 0 on (–∞, –7)
and (2, ∞).
Answer: (–∞, –7], [2, ∞)
10. (x + 5)(x + 2)(x − 7) ≤ 0
f(x) = (x + 5)(x + 2)(x − 7) has zeros −5, −2 and
7. By considering the intervals (−∞, −5),
(−5, −2), (−2, 7) and (7, ∞), we find f(x) < 0 on
(−∞, −5) and (−2, 7).
Answer: (−∞, −5], [−2, 7]
5. 2 x 2 + 11x + 14 < 0
f ( x) = 2 x 2 + 11x + 14 = (2 x + 7)( x + 2) has zeros
7
and –2. By considering the intervals
2
7⎞ ⎛ 7
⎛
⎞
⎜ −∞, − 2 ⎟ , ⎜ − 2 , − 2 ⎟ , and (–2, ∞), we find
⎝
⎠ ⎝
⎠
−
11. –x(x – 5)(x + 4) > 0, or equivalently,
x(x – 5)(x + 4) < 0.
f(x) = x(x – 5)(x + 4) has zeros, 0, 5, and –4. By
considering the intervals (–∞, –4), (–4, 0), (0, 5),
and (5, ∞), we find f(x) < 0 on (–∞, –4) and
(0, 5).
Answer: (–∞, –4), (0, 5)
⎛ 7
⎞
f(x) < 0 on ⎜ − , − 2 ⎟ .
⎝ 2
⎠
⎛ 7
⎞
Answer: ⎜ − , − 2 ⎟
2
⎝
⎠
12. ( x + 2)2 > 0
f ( x) = ( x + 2) 2 has –2 as zero. By considering
6. x 2 − 4 < 0 . f ( x) = x 2 − 4 = ( x + 2)( x − 2) has
the intervals (–∞, –2) and (–2, ∞), we find
f(x) > 0 on both intervals.
Answer: (–∞, –2), (–2, ∞)
zeros ±2. By considering the intervals (–∞, –2),
(–2, 2), and (2, ∞), we find f(x) < 0 on (–2, 2).
Answer: (–2, 2)
13. x3 + 4 x ≥ 0
7. x 2 + 4 < 0 . Since x 2 + 4 is always positive, the
(
By considering the intervals (–∞, 0) and (0, ∞),
we find f(x) > 0 on (0, ∞).
Answer: [0, ∞)
8. 2 x 2 − x − 2 ≤ 0 . f ( x) = 2 x 2 − x − 2 has zeros
(
)
f ( x) = ( x + 2) ( x − 1) = ( x + 2) ( x + 1)( x − 1)
1 ± 17
. By considering the intervals
4
⎛
1 − 17 ⎞ ⎛ 1 − 17 1 + 17 ⎞
,
⎜⎜ −∞,
⎟, ⎜
⎟ , and
4 ⎟⎠ ⎜⎝ 4
4 ⎟⎠
⎝
14. ( x + 2)2 x 2 − 1 < 0
2
2
2
has zeros –2, –1, and 1. By considering the
intervals (–∞, –2), (–2, –1), (–1, 1), and (1, ∞),
we find f(x) < 0 on (–1, 1).
Answer: (–1, 1)
⎛ 1 + 17
⎞
, ∞ ⎟ , we find f(x) < 0 on
⎜⎜
⎟
⎝ 4
⎠
⎛ 1 − 17 1 + 17
,
⎜⎜
4
⎝ 4
)
f ( x) = x x 2 + 4 has 0 as the only (real) zero.
inequality x 2 + 4 < 0 has no solution.
Answer: no solution
⎞
⎟⎟ .
⎠
15. x3 + 8 x 2 + 15 x ≤ 0
f(x) = x(x + 3)(x + 5) has zeros 0, –3, and −5. By
considering the intervals (−∞, −5), (−5, −3),
(−3, 0), and (0, ∞), we find f(x) < 0 on (−∞, −5)
and (−3, 0).
Answer: (−∞, −5], [−3, 0]
⎡ 1 − 17 1 + 17 ⎤
Answer: ⎢
,
⎥
4 ⎦⎥
⎣⎢ 4
370
ISM: Introductory Mathematical Analysis
Section 10.4
discontinuous at x = –5 and x = 1; f has zeros 3
and –2. By considering the intervals (–∞, –5),
(–5, –2), (–2, 1), (1, 3), and (3, ∞), we find
f(x) > 0 on (–∞, –5), (–2, 1), and (3, ∞).
Answer: (–∞, –5), [–2, 1), [3, ∞)
16. x3 + 6 x 2 + 9 x < 0
f ( x) = x( x 2 + 6 x + 9) = x( x + 3)2 has zeros −3
and 0. By considering the intervals (∞, −3),
(−3, 0) and (0, ∞), we find f(x) < 0 on (−∞, −3)
and (−3, 0).
Answer: (−∞, −3), (−3, 0)
17.
x
x −9
2
22.
<0
x2 − 1
<0
x
23.
24.
4
≥0
x −1
3
x − 5x + 6
x2 + 4 x − 5
f ( x) =
≤0
2x +1
x2
3
2
=
≤0
2x +1
x2
is discontinuous at x = 0 and f has
1⎤
⎛
Answer: ⎜ −∞, − ⎥
2⎦
⎝
25. x 2 + 2 x ≥ 2 , or equivalently, x 2 + 2 x − 2 ≥ 0 .
f ( x) = x 2 + 2 x − 2 has zeros −1 ± 3 . By
(
)
( −1 − 3, − 1 + 3 ) , and ( −1 + 3, ∞ ) , we find
f(x) > 0 on ( −∞, − 1 − 3 ) and ( −1 + 3, ∞ ) .
Answer: ( −∞, − 1 − 3 ⎤ , ⎡ –1 + 3, ∞ )
⎦ ⎣
considering the intervals – ∞, – 1 – 3
≥0
x + 4x − 5
2
x + 6x + 5
1⎞
⎛
f(x) < 0 on ⎜ −∞, − ⎟ .
2⎠
⎝
>0
x2 − x − 6
( x + 5)( x − 1)
is
( x + 2)( x + 1)
1
as a zero. By considering the intervals
2
1⎞ ⎛ 1 ⎞
⎛
⎜ −∞, − 2 ⎟ , ⎜ − 2 , 0 ⎟ , and (0, ∞), we find
⎝
⎠ ⎝
⎠
3
is never zero, but is
( x − 2)( x − 3)
discontinuous when x = 2, 3. By considering the
intervals (–∞, 2), (2, 3),and (3, ∞), we find
f(x) > 0 on (–∞, 2) and (3, ∞).
Answer: (–∞, 2), (3, ∞)
21.
=
−
f ( x) =
x2 − x − 6
3
f ( x) =
4
is discontinuous when x = 1, and
x −1
f(x) = 0 has no root. By considering the intervals
(–∞, 1) and (1, ∞), we find f(x) > 0 on (1, ∞).
Note also that f(x) ≠ 0 for any x.
Answer: (1, ∞)
2
2
3
is never zero,
x + 6 x + 5 ( x + 5)( x + 1)
but is discontinuous at x = –5 and x = –1. By
considering the intervals (–∞, –5), (–5, –1), and
(–1, ∞), we find that f(x) < 0 on (–5, –1).
Answer: (–5, –1)
f ( x) =
20.
x2 + 4 x − 5
2
f ( x) =
x2 − 1
is discontinuous at x = 0; f has
x
zeros at ±1. By considering the intervals
(−∞, −1), (−1, 0), (0, 1), and (1, ∞), we find
f(x) < 0 on (–∞, –1) and (0, 1).
Answer: (–∞, –1), (0, 1)
f ( x) =
19.
≤0
x + 3x + 2
discontinuous at x = −1 and −2; f has zeros −5
and 1. By considering the intervals (−∞, −5),
(−5, −2), (−2, −1), (−1, 1), and (1, ∞), we find
f(x) < 0 on (−5, −2) and (−1, 1).
Answer: [−5, −2), (−1, 1]
is discontinuous when x = ±3;
x2 − 9
f has 0 as a zero. By considering the intervals
(–∞, –3), (–3, 0), (0, 3), and (3, ∞), we find
f(x) < 0 on (–∞, –3) and (0, 3).
Answer: (–∞, –3), (0, 3)
18.
x 2 + 3x + 2
f ( x) =
x
f ( x) =
x2 + 4 x − 5
=
( x − 3)( x + 2)
is
( x + 5)( x − 1)
371
Chapter 10: Limits and Continuity
26. x 4 − 16 ≥ 0 .
(
ISM: Introductory Mathematical Analysis
4( x − 8)2 ≥ 324
)
f ( x) = x 2 + 4 ( x + 2)( x − 2) has (real) zeros –2
( x − 8)2 ≥ 81
x 2 − 16 x − 17 ≥ 0
(x – 17)(x + 1) ≥ 0
Solving gives x ≤ –1 or x ≥ 17. Since x must
be positive, we have x ≥ 17.
Answer: 17 in. by 17 in.
and 2. By considering the intervals (–∞, –2),
(–2, 2), and (2, ∞), we find f(x) > 0 on (–∞, –2)
and (2, ∞).
Answer: (–∞, –2], [2, ∞)
27. Revenue = (no. of units)(price per unit). We
q (28 − 0.2q ) ≥ 750
want
30. Let n = no. of persons over the 50 that attend.
Then each of 50 + n persons will pay 50 – 0.50n.
We want
(50 + n)(50 – 0.50n) ≥ 50(50)
1
25n − n 2 ≥ 0
2
1 ⎞
⎛
n ⎜ 25 − n ⎟ ≥ 0
2 ⎠
⎝
Solving gives 0 ≤ n ≤ 50. Thus the size of the
group is between 50 and 100 inclusive.
Answer: 50 ≤ size of group ≤ 100
0.2q − 28q + 750 ≤ 0
2
q 2 − 140q + 3750 ≤ 0
Using the quadratic formula,
q 2 − 140q + 3750 = 0 when q ≈ 36.09, 103.91.
Thus q 2 − 140q + 3750 ≤ 0 when
36.09 ≤ q ≤ 103.91, so sales revenue will be at
least $750 when between 37 and 103 units,
inclusive, are sold.
x
28.
x
x
–10
2
(2 − 2 x)(1 − 2 x) ≥
21
16
–10
64 x − 96 x + 11 ≥ 0
(8x – 11)(8x – 1) ≥ 0
1
11
Solving gives x ≤ or x ≥ . From the
8
8
1
diagram, clearly, x cannot exceed . Thus
2
1
x≤ .
8
1
mi
Answer:
8
4
4
5
32.
–5
5
–5
(–∞, 1.51)
5
33.
x–8
4
10
(–∞, –7.72]
2
29.
10
31.
x 1
–5
5
4
–5
x–8
(–∞, –0.5), (0.667, ∞)
4
4
4
4
If x is the length of a side of the piece of
aluminum, then the box will be 4 by x – 8 by
x – 8.
372
ISM: Introductory Mathematical Analysis
Chapter 10 Review
5
34.
9. As x → ∞, x + 1 → ∞ . Thus lim
x →∞
5
–5
10.
–5
(−2, −1.62), (0.62, 1)
11.
x2 + 1
lim
x →∞
2x
2
x2
= lim
x →∞
2x
2
= lim
x →∞
2
=0.
x +1
1 1
=
2 2
2x + 5
2x
2 2
= lim
= lim =
x →∞ 7 x − 4 x →∞ 7 x x →∞ 7 7
lim
Chapter 10 Review Problems
1.
(
)
12.
lim 2 x + 6 x − 1 = 2(−1) + 6(−1) − 1 = −5
x →−1
2. lim
2
2 x − 3x + 1
2
2 x2 − 2
x →0
=
2
(
)
13.
lim 2 x 2 − 3x + 1
= x →0
lim 2 x 2 − 2
x →0
(
)
1
1
=−
−2
2
14.
x −9
( x + 3)( x − 3)
3. lim
= lim
2
x( x − 3)
x →3 x − 3 x x →3
x+3 6
= lim
= =2
3
x →3 x
2
2x + 3
15.
2t − 3
2t − 3
= −∞ and lim
= ∞ . Thus
+ t −3
3
−
t
t →3
t →3
2t − 3
lim
does not exist.
t →3 t − 3
lim
−
x6
lim
= lim x = −∞
x →−∞ x 5
x →−∞
x+3
lim
x →−∞ 1 − x
x→4
lim (2 x + 3)
17.
5. lim ( x + h) = x + 0 = x
x2 − 1
lim
x →∞
(3 x + 2)2
x
2
x →∞ 9 x
x −4
2
= lim
− 3x + 2
x+2 4
= lim
= =4
1
x→2 x − 1
x →2 x 2
x→2
x3 + 4 x 2
lim
x →−4 x 2
+ 2x − 8
( x + 2)( x − 2)
( x − 2)( x − 1)
2
= lim
x →∞ 9 x 2
x2 − 1
+ 12 x + 4
1 1
=
9
x →∞ 9
= lim
x2 + x − 2
( x + 2)( x − 1)
= lim
x −1
x −1
x →1
x →1
= lim ( x + 2) = 3
x →1
x 2 ( x + 4)
x →−4 ( x + 4)( x − 2)
19.
= lim
x2
16
8
=
=−
3
−6
x →−4 x − 2
lim
−
x+3
x+3
x →3 ( x + 3)( x − 3)
= lim
x −9
1
= lim
= −∞
− x−3
x →3
x →3
2
−
2− x
⎡ x − 2⎤
= lim ⎢ −
⎥ = lim (−1) = −1
x→2 x − 2 x→2 ⎣ x − 2 ⎦ x →2
( x − 1)( x + 2)
x →1 ( x − 1)( x + 5)
20. lim
= lim
+ 4x − 5
x+2 3 1
= lim
= =
6 2
x →1 x + 5
x →1 x 2
x
= lim (−1) = −1
− x x →−∞
18. lim
lim
x2 + x − 2
x →−∞
x→4
= lim
h →0
6. lim
= lim
x →−4
x →−4
8. lim
=0
x →−∞ x 4
16. lim 3 64 = lim 4 = 4
−5
5
4. lim
=
=
=−
2
12
12
x →−4 x 2 − 4
lim ( x − 4)
7.
1
lim
21. As x becomes large, so does 3x. Because the square
roots of large numbers are also large,
lim 3x = ∞ .
x →∞
373
Chapter 10: Limits and Continuity
ISM: Introductory Mathematical Analysis
22. As y → 5+ , y – 5 approaches 0 through positive
y → 5+
23.
lim
π− x
x →∞
= lim
x
97
x →∞
(
x3 x100 + 13
x →∞
x103 + 1
= lim
πx3 − x100
h →0
y −5 = 0.
values. Thus lim
x100 + 13
f ( x + h) − f ( x )
h
[8( x + h) − 2] − [8 x − 2]
= lim
h
h→0
8h
= lim
= lim 8 = 8
h →0 h
h →0
29. lim
)
)
f ( x + h) − f ( x )
h
h →0
x103
= lim
30. lim
− x100
x →∞
( )
(
x3 π − x
x
97
⎡ 2( x + h)2 − 3⎤ − ⎡ 2 x 2 − 3⎤
⎦ ⎣
⎦
= lim ⎣
h
h→0
= lim − x3 = −∞
x →∞
24.
ex 2 − x 4
lim
x →−∞ 31x − 2 x3
4 xh + 2h 2
h
h→0
h(4 x + 2h)
= lim
= lim (4 x + 2h) = 4 x
h
h→0
h →0
− x4
= lim
x →−∞
= lim
−2 x3
x
= −∞
x →−∞ 2
= lim
25.
1 ⎞
⎛
31. y = 23 ⎜ 1 −
⎟
⎝ 1 + 2x ⎠
1
Considering
, we have
1 + 2x
1
1
1 1
lim
= ⋅ lim = ⋅ 0 = 0 . Thus
2 x →∞ x 2
x →∞ 1 + 2 x
lim f ( x ) = lim x 2 = 1
x →1−
x →1−
lim f ( x) = lim x = 1
x →1+
x →1+
Thus lim f ( x) = 1 .
x →1
26.
lim f ( x) = lim ( x + 5) = 8
x →3−
x →3−
⎡ ⎛
1 ⎞⎤
lim y = lim ⎢ 23 ⎜ 1 −
⎥ = 23(1 − 0) = 23
1 + 2 x ⎟⎠ ⎦
x →∞
x →∞ ⎣ ⎝
Answer: 23
lim f ( x ) = lim 6 = 6
x →3+
x →3+
Because lim f ( x) ≠ lim f ( x) ,
x →3−
x →3+
lim f ( x ) does not exist.
32.
x →3
27.
x 2 − 16
= lim
4− x
x → 4+
lim
x→4
+
= lim −
x→4
+
33. f(x) = x + 5; x = 7
(i) f is defined at x = 7; f(7) = 12
x − 4 approaches 0 through
(ii) lim f ( x) = lim ( x + 5) = 7 + 5 = 12 , which
x + 4 → 8 . Thus
x →7
x+4
lim
x →5
+
x 2 − 3x − 10
x −5
10 x
x →∞ 0.1x
x →∞
x →7
exists
→ −∞ .
x−4
Answer: –∞
28.
= lim
Answer: 100
x−4
positive values and
−
10 x
x →∞ 1 + 0.1x
= lim 100 = 100
x−4 x+4
−( x − 4)
x+4
As x → 4+ ,
lim y = lim
x →∞
(iii) lim f ( x ) = 12 = f (7)
x →7
= lim
x →5
+
= lim
x →5+
Thus f is continuous at x = 7.
( x − 5)( x + 2)
x−5
x − 5( x + 2)
= 0⋅7
=0
374
ISM: Introductory Mathematical Analysis
34.
x −5
x +2
2
Chapter 10 Review
; x=5
43.
(i) f is defined at x = 5; f(5) = 0
(ii) lim f ( x) = lim
x −5
x →5 x 2
x →5
+2
=
0
= 0, which
27
⎧ x + 4 if x > −2
f ( x) = ⎨
⎩3x + 6 if x ≤ −2
For x < –2, f(x) = 3x + 6, which is a polynomial
and hence continuous. For x > –2, f(x) = x + 4,
which is a polynomial and hence continuous.
Because lim f ( x) = lim (3x + 6) = 0 and
x →−2−
x →−2−
lim f ( x) = lim ( x + 4) = 2 , lim f ( x ) does
exists
x →−2+
x →−2+
x →−2
not exist. Thus f is discontinuous at x = –2.
(iii) lim f ( x ) = 0 = f (5)
x →5
Thus f is continuous at x = 5.
44.
1
35. Since f ( x) = x 2 is polynomial function, it is
5
continuous everywhere.
38.
x2
is a rational function and the
x+3
denominator is zero at x = –3. Thus f is
discontinuous at x = –3.
x →1+
x →1
45. x 2 + 4 x − 12 > 0
f ( x) = x 2 + 4 x − 12 = ( x + 6)( x − 2) has zeros –6
and 2. By considering the intervals (–∞, –6),
(–6, 2), and (2, ∞), we find f(x) > 0 on (–∞, –6)
and (2, ∞).
Answer: (–∞, –6), (2, ∞)
is a rational function
2x + 3
whose denominator is never zero, f is continuous
everywhere.
2
46. 3x 2 − 3 x − 6 ≤ 0
40. Since f ( x) = (2 − 3 x)3 is a polynomial function,
it is continuous everywhere.
f ( x) = 3x 2 − 3 x − 6 = 3( x − 2)( x + 1) has zeros
−1 and 2. By considering the intervals (−∞, −1),
(−1, 2), and (2, ∞), we find f(x) < 0 on (−1, 2).
Answer: [−1, 2]
4 − x2
4 − x2
is a rational
41. f ( x) =
=
x 2 + 3 x − 4 ( x + 4)( x − 1)
function and the denominator is zero only when
x = –4 or x = 1, so f is discontinuous at x = –4, 1.
42.
f ( x) =
2x + 6
x +x
3
=
2x + 6
(
)
x x2 + 1
x →1
x = 1.
f is discontinuous at x = 0.
is a rational function and the
x3
denominator is zero at x = 0. Thus f is
discontinuous at x = 0.
x −1
x →1+
Since lim f ( x) = 1 = f (1) , f is continuous at
0
39. Since f ( x) =
if x ≥ 1
1
, which is a rational
x
function whose denominator is zero when x = 0.
Thus f is discontinuous at x = 0. If x > 1, then
f(x) = 1, which a polynomial function and hence
continuous. At x = 1, f is defined [f(x) = 1].
1
Because lim f ( x) = lim = 1 and
x →1−
x →1− x
lim f ( x) = lim 1 = 1 , then lim f ( x) = 1 .
f ( x) =
f ( x) =
if x < 1
If x < 1, then f ( x) =
36. Since f ( x) = x 2 − 2 is a polynomial function, it
is continuous everywhere.
37.
⎧⎪ 1
f ( x) = ⎨ x
⎪⎩1
47. x5 ≤ 7 x 4 , x5 − 7 x 4 ≤ 0
f ( x) = x5 − 7 x 4 = x 4 ( x − 7) has zeros 0 and 7.
By considering the intervals (−∞, 0), (0, 7), and
(7, ∞), we find f(x) < 0 on (−∞, 0) and (0, 7).
Answer: (−∞, 7]
is a rational function
and the denominator is zero only when x = 0, so
f is discontinuous at x = 0.
375
Chapter 10: Limits and Continuity
ISM: Introductory Mathematical Analysis
f ( x) = x3 + 8 x 2 + 15 x = x( x + 5)( x + 3) has zeros
0, –5, and –3. By considering the intervals
(–∞, –5), (–5, –3), (–3, 0), and (0, ∞), we find
f(x) > 0 and (–5, –3) and (0, ∞).
Answer: [–5, –3], [0, ∞)
49.
x+5
x2 − 1
5
53.
48. x3 + 8 x 2 + 15 x ≥ 0
0
3
–5
1.00
<0
1
54.
x+5
is discontinuous when
( x + 1)( x − 1)
x = ±1, and f has –5 as a zero. By considering the
intervals (–∞, –5), (–5, –1), (–1, 1), and (1, ∞),
we find f(x) < 0 on (–∞, –5) and (–1, 1).
Answer: (–∞, –5), (–1, 1)
f ( x) =
50.
51.
0
–1
0.25
x( x + 5)( x + 8)
<0
3
x( x + 5)( x + 8)
f ( x) =
has zeros 0, –5, and –8.
3
By considering the intervals (–∞, –8), (–8, –5),
(–5, 0), and (0, ∞), we find f(x) < 0 on (–∞, –8)
and (–5, 0).
Answer: (–∞, –8), (–5, 0)
x 2 + 3x
x2 + 2 x − 8
f ( x) =
x2 − 9
x 2 − 16
f ( x) =
5
55.
–5
0
≥0
2
=
x( x + 3)
is
( x + 4)( x − 2)
0.50
=
5
–5
10
57.
–5
≤0
[2.00, ∞)
( x + 3)( x − 3)
is discontinuous
( x + 4)( x − 4)
5
–10
10
58.
x − 16
when x = ±4 and has zeros x = ±3. By
considering the intervals (−∞, −4), (−4, −3),
(−3, 3), (3, 4), and (4, ∞) we find f(x) < 0 on
(−4, −3) and (3, 4).
Answer: (−4, −3], [3, 4)
2
–5
–5
x2 + 3x
x2 − 9
5
5
56.
x + 2x − 8
discontinuous when x = −4, 2 and has zeros
x = −3, 0. By considering the intervals (–∞, –4),
(–4, –3), (–3, 0), (0, 2), and (2, ∞) we find
f(x) > 0 on (–∞, –4), (–3, 0), and (2, ∞).
Answer: (–∞, –4), [–3, 0], (2, ∞)
52.
5
–5
5
–10
(−1, 1.32]
376
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 10
Mathematical Snapshot Chapter 10
1. D = 8432e − rt
A year from now, t = 1 and D = 8000. Thus
8000 = 8432e − r
8000
e−r =
8432
8000
−r = ln
8432
8000
r = − ln
≈ 0.053
8432
The rate is 5.3%.
2. D = 8432e −0.06t
We want to find t when D =
8432
.
2
8432
= 8432e −0.06t
2
1
= e−0.06t
2
1
−0.06t = ln
2
1
ln 2
ln 2
t=
=
≈ 12
−0.06 0.06
It would take about 12 years.
3. An exponential model assumes a fixed
repayment rate. In reality, the repayment rate is
constantly changing as a result of changing
fiscal policy and other factors.
377
Chapter 11
Principles in Practice 11.1
1.
(
)
dH d
=
6 + 40t − 16t 2
dt
dt
H (t + h) − H (t )
= lim
h
h →0
⎡ 6 + 40(t + h) − 16(t + h)2 ⎤ − 6 + 40t − 16t 2
⎦
= lim ⎣
h
h→0
(
)
6 + 40t + 40h − 16t 2 − 32th − 16h 2 − 6 − 40t + 16t 2
h
h →0
= lim
40h − 32th − 16h 2
= lim (40 − 32t − 16h)
h
h →0
h →0
= 40 – 32t
dH
= 40 − 32t
dt
= lim
Problems 11.1
1. a.
f ( x) = x3 + 3 , P = (−2, −5)
[(−3)3 + 3] − (−5)
[(−2.5)3 + 3] − (−5)
= 19. If x = −2.5, then mPQ =
= 15.25.
−3 − (−2)
−2.5 − (−2)
Continuing in this manner, we complete the table:
To begin, if x = −3, then mPQ =
b.
2. a.
x-value of Q
−3
−2.5
−2.2
−2.1
−2.01
−2.001
mPQ
19
15.25
13.24
12.61
12.0601
12.0060
We estimate that mtan at P is 12.
f ( x) = e 2 x , P = (0, 1)
e 2(1) − 1
e 2(0.5) − 1
≈ 6.3891 . If x = 0.5, then mPQ =
≈ 3.4366 .
1− 0
0.5 − 0
Continuing in this manner, we complete the table:
To begin, if x = 1, then mPQ =
x-value of Q
mPQ
1
0.5
0.2
0.1
0.01
0.001
6.3891
3.4366
2.4591
2.2140
2.0201
2.0020
b. We estimate that mtan at P is 2.
3. f(x) = x
f ( x + h) − f ( x )
( x + h) − x
h
= lim
= lim = lim 1 = 1
h
h
h →0
h →0
h → 0 h h →0
f ′( x) = lim
378
ISM: Introductory Mathematical Analysis
Section 11.1
4. f(x) = 4x – 1
10. f(x) = 7.01
f ( x + h) − f ( x )
h
7.01 − 7.01
= lim
h
h →0
0
= lim = lim 0 = 0
h → 0 h h →0
f ( x + h) − f ( x )
h
[4( x + h) − 1] − [4 x − 1]
= lim
h
h →0
4h
= lim
= lim 4 = 4
h →0 h
h →0
f ′( x) = lim
f ′( x) = lim
h →0
h →0
5. y = 3x + 5. Let y = f(x).
dy
f ( x + h) − f ( x )
= lim
dx h→0
h
[3( x + h) + 5] − [3x + 5]
= lim
h
h →0
3h
= lim
= lim 3 = 3
h →0 h
h →0
11. Let f ( x) = x 2 + 4 x − 8.
(
6. y = –5x. Let y = f(x).
dy
f ( x + h) − f ( x )
= lim
dx h→0
h
[−5( x + h)] − [−5 x]
= lim
h
h →0
−5h
= lim
= lim (−5) = −5
h →0 h
h →0
x 2 + 2 xh + h 2 + 4 x + 4h − 8 − x 2 − 4 x + 8
h
h →0
= lim
2 xh + h 2 + 4h
h
h →0
= lim (2 x + h + 4) = 2 x + 0 + 4 = 2 x + 4
= lim
h →0
12. y = f ( x) = x 2 + 5 x + 1
7. Let f(x) = 5 – 4x.
d
f ( x + h) − f ( x )
(5 − 4 x) = lim
dx
h
h →0
[5 − 4( x + h)] − [5 − 4 x]
= lim
h
h →0
−4h
= lim
= lim (−4) = −4
h →0 h
h→0
8. Let f ( x) = 1 −
f ( x + h) − f ( x )
h
[( x + h) 2 + 5( x + h) + 1] − [ x 2 + 5 x + 1]
= lim
h
h →0
x 2 + 2 xh + h 2 + 5 x + 5h + 1 − x 2 − 5 x − 1
= lim
h
h →0
2 xh + h 2 + 5h
= lim
h
h →0
= lim (2 x + h + 5) = 2 x + 0 + 5 = 2 x + 5
y ′ = lim
h →0
x
2
h →0
⎡1 − x + h ⎤ − ⎡1 − x ⎤
d ⎛ x⎞
2 ⎦ ⎣
2⎦
⎣
−
=
1
lim
⎜
⎟
dx ⎝ 2 ⎠ h→0
h
= lim
− h2
h →0
h
)
d 2
x + 4x − 8
dx
f ( x + h) − f ( x )
= lim
h
h →0
⎡ ( x + h)2 + 4( x + h) − 8⎤ − ⎡ x 2 + 4 x − 8⎤
⎦ ⎣
⎦
= lim ⎣
h
h →0
13. p = f (q) = 3q 2 + 2q + 1
dp
f ( q + h) − f ( q )
= lim
dq h→0
h
1
⎛ 1⎞
= lim ⎜ − ⎟ = −
2
h→0 ⎝ 2 ⎠
⎡3(q + h)2 + 2(q + h) + 1⎤ − ⎡3q 2 + 2q + 1⎤
⎦ ⎣
⎦
= lim ⎣
h
h →0
9. f(x) = 3
f ( x + h) − f ( x )
h
h →0
3−3
0
= lim
= lim = lim 0 = 0
h →0 h
h →0 h h → 0
f ′( x) = lim
6qh + 3h 2 + 2h
h
h →0
= lim (6q + 3h + 2) = 6q + 0 + 2 = 6q + 2
= lim
h →0
379
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
14. Let f ( x) = x 2 − x − 3.
(
x+h+2 − x+2
h
)
d 2
x − x−3
dx
f ( x + h) − f ( x )
= lim
h
⎡ ( x + h ) 2 − ( x + h ) − 3⎤ − ⎡ x 2 − x − 3⎤
⎦ ⎣
⎦
= lim ⎣
h
h →0
x+h+2 − x+2 x+h+2 + x+2
⋅
h
x+h+2 + x+2
( x + h + 2) − ( x + 2)
1
=
=
x+h+2 + x+2
h x+h+2 + x+2
=
(
Thus f ′( x) = lim
2 xh + h 2 − h
= lim
= lim (2 x + h − 1) = 2 x − 1
h
h →0
h →0
15. y = f ( x) =
)
h →0
18. H ( x) =
6
x
6
x+h
=
3
x−2
H ( x + h) − H ( x )
h
h →0
3 − 3
= lim x + h − 2 x − 2
h
h →0
Multiplying the numerator and denominator by
(x + h − 2)(x − 2) gives
3( x − 2) − 3( x + h − 2)
H ′( x ) = lim
h →0 h( x + h − 2)( x − 2)
−3h
= lim
h →0 h( x + h − 2)( x − 2)
−3
3
= lim
=−
h →0 ( x + h − 2)( x − 2)
( x − 2) 2
⎡
6 ⎤
6
6
= lim ⎢ −
=−
⎥=−
x( x + 0)
h → 0 ⎣ x ( x + h) ⎦
x2
16. C = f (q) = 7 + 2q − 3q 2
dC
f ( q + h) − f ( q )
= lim
dq h→0
h
19. y = f ( x) = x 2 + 4
⎡ 7 + 2(q + h) − 3(q + h)2 ⎤ − ⎡ 7 + 2q − 3q 2 ⎤
⎦ ⎣
⎦
= lim ⎣
h
h →0
f ( x + h) − f ( x )
h
2
⎡ ( x + h) + 4 ⎤ − ⎡ x 2 + 4 ⎤
⎦ ⎣
⎦
= lim ⎣
h
h→0
y ′ = lim
h →0
2h − 6qh − 3h 2
= lim (2 − 6q − 3h)
h
h →0
h →0
= 2 − 6q
= lim
2 xh + h 2
= lim (2 x + h) = 2 x + 0 = 2 x
h
h→0
h →0
The slope at (–2, 8) is y ′(−2) = 2(−2) = −4 .
= lim
17.
f ( x) = x + 2
f ′( x) = lim
h →0
1
2 x+2
H ′( x ) = lim
6
x
−
f ( x + h) − f ( x )
= lim
h
h
h →0
h →0
Multiplying the numerator and denominator by
x(x + h) gives
6 x − 6( x + h)
−6h
y ′ = lim
= lim
h →0 hx( x + h)
h→0 hx( x + h)
y ′ = lim
1
x+h+2 + x+2
f ( x + h) − f ( x )
h
20. y = f ( x) = 1 − x 2
f ( x + h) − f ( x )
y ′ = lim
h
h →0
⎡1 − ( x + h)2 ⎤ − ⎡1 − x 2 ⎤
⎦ ⎣
⎦
= lim ⎣
h
h →0
x+h+2 − x+2
h
h →0
Rationalizing the numerator gives
= lim
−2 xh − h 2
h
h →0
= lim (−2 x − h) = −2 x
= lim
h →0
The slope at (1, 0) is y ′(1) = −2(1) = −2.
380
ISM: Introductory Mathematical Analysis
Section 11.1
26. y = ( x − 7) 2 = x 2 − 14 x + 49
21. y = f ( x) = 4 x 2 − 5
f ( x + h) − f ( x )
y ′ = lim
h
h →0
⎡ 4( x + h) 2 − 5⎤ − ⎡ 4 x 2 − 5⎤
⎦ ⎣
⎦
= lim ⎣
h
h →0
⎡ ( x + h)2 − 14( x + h) + 49 ⎤ − ⎡ x 2 − 14 x + 49 ⎤
⎦ ⎣
⎦
′
=
y lim ⎣
h
h →0
2 xh + h 2 − 14h
= lim (2 x + h − 14) = 2 x − 14
h
h →0
h →0
If x = 6, then y ′ = 2(6) − 14 = −2 . The tangent line
at (6, 1) is y – 1 = –2(x – 6), or y = –2x + 13.
= lim
8 xh + 4h 2
= lim (8 x + 4h) = 8 x
h
h→0
h →0
The slope when x = 0 is y ′(0) = 8(0) = 0 .
= lim
d
22. As shown in Example 5,
dx
If x = 1, the slope is y ′(1) =
( x) = 2
27. y =
1
x
.
3
x −1
y ′ = lim
1
.
2
3
− 3
( x + h ) −1 x −1
h
h →0
3( x −1) −3( x + h −1)
( x + h −1)( x −1)
= lim
23. y = x + 4
h
−3h
−3
= lim
= lim
h→0 h( x + h − 1)( x − 1) h→0 ( x + h − 1)( x − 1)
h →0
[( x + h) + 4] − [ x + 4]
h
= lim = 1
h
h →0
h →0 h
If x = 3, then y ′ = 1 . The tangent line at the
point (3, 7) is y – 7 = 1(x – 3), or y = x + 4.
y ′ = lim
=−
2
24. y = 3 x − 4
3
( x − 1)2
3
If x = 2, then y ′ = − = −3 . The tangent line at
1
(2, 3) is y − 3 = −3( x − 2) , or y = −3 x + 9 .
[3( x + h) 2 − 4] − [3x 2 − 4]
y ′ = lim
h
h →0
6 xh + 3h 2
= lim
= lim (6 x + 3h) = 6 x
h
h →0
h →0
If x = 1, then y ′ = 6(1) = 6.
28. y =
The tangent line at (1, −1) is y + 1 = 6(x − 1) or
y = 6x − 7.
5
1 − 3x
y ′ = lim
5
1−3( x + h )
− 1−53 x
h
5(1 − 3 x) − 5[1 − 3( x + h)]
= lim
h→0 h[1 − 3( x + h)](1 − 3 x)
h →0
25. y = x 2 + 2 x + 3
⎡ ( x + h)2 + 2( x + h) + 3⎤ − ⎡ x 2 + 2 x + 3⎤
⎦ ⎣
⎦
y ′ = lim ⎣
h
h →0
15h
h→0 h[1 − 3( x + h)](1 − 3 x)
= lim
2 xh + h 2 + 2h
h
h →0
= lim (2 x + h + 2) = 2 x + 2
15
h→0 [1 − 3( x + h)](1 − 3 x)
= lim
= lim
h →0
=
If x = 1, then y ′ = 2(1) + 2 = 4. The tangent line
at the point (1, 6) is y − 6 = 4(x − 1), or
y = 4x + 2.
15
(1 − 3 x)2
15 3
= . The tangent line at
25 5
3
3
11
(2, –1) is y + 1 = ( x − 2) , or y = x − .
5
5
5
If x = 2, then y ′ =
381
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
⎛ η ⎞⎛
dC ⎞
29. r = ⎜
⎟ ⎜ rL −
⎟
η
dD
+
1
⎠
⎝
⎠⎝
dC ⎞
⎛
(1 + η )r = η ⎜ rL −
⎟
dD ⎠
⎝
dC ⎞
⎛
r + η r = η ⎜ rL −
⎟
dD ⎠
⎝
dC ⎞
⎛
r = η ⎜ rL −
⎟ −η r
dD ⎠
⎝
dC
⎛
⎞
−r⎟
r = η ⎜ rL −
dD
⎝
⎠
r
η=
dC
rL − r − dD
30. 1.565, 1.470
31. –3.000, 13.445
32. 0.680, 1820.369
33. –5.120, 0.038
34. y = f ( x) = x 2 + x
f ( x + h) − f ( x )
h
2
⎡ ( x + h) + ( x + h) ⎤ − ⎡ x 2 + x ⎤
⎦ ⎣
⎦
= lim ⎣
h
h →0
f ′( x) = lim
h →0
2 xh + h 2 + h
= lim (2 x + h + 1) = 2 x + 1
h
h →0
h →0
If x = –2, then f ′( x) = −3 . The tangent line at the point (–2, 2) is y – 2 = –3(x + 2), or
y = –3x – 4.
= lim
7
–5
5
–3
5
35.
–5
5
–5
382
ISM: Introductory Mathematical Analysis
Section 11.1
For the x-values of the points where the tangent to the graph of f is horizontal, the corresponding values of f ′( x)
are 0. This is expected because the slope of a horizontal line is zero and the derivative gives the slope of the
tangent line.
3
36. n = 4: ( z − x) ∑ xi z 3−i = ( z − x)( z 3 + xz 2 + x 2 z + x3 )
i =0
= z 4 − xz 3 + xz 3 − x 2 z 2 + x 2 z 2 − x3 z + x3 z − x 4
= z 4 − x4
2
n = 3: ( z − x) ∑ xi z 2−i = ( z − x)( z 2 + xz + x 2 )
i =0
= z 3 − xz 2 + xz 2 − x 2 z + x 2 z − x3
= z 3 − x3
1
n = 2: ( z − x) ∑ xi z1−i = ( z − x)( z + x) = z 2 − x 2
4
i =0
3
f ( x) = 2 x + x − 3 x 2
f ( z ) − f ( x)
z−x
z→x
2 z 4 + z 3 − 3z 2 − (2 x 4 + x3 − 3x 2 )
= lim
z−x
z→x
2( z 4 − x 4 ) + ( z 3 − x3 ) − 3( z 2 − x 2 )
= lim
z−x
z→x
2( z − x)( z 3 + xz 2 + x 2 z + x3 ) + ( z − x)( z 2 + xz + x 2 ) − 3( z − x)( z + x)
= lim
z−x
z→x
= lim [2( z 3 + xz 2 + x 2 z + x3 ) + ( z 2 + xz + x 2 ) − 3( z + x)]
f ′( x) = lim
z→x
= 2(4 x3 ) + (3 x 2 ) − 3(2 x)
= 8 x3 + 3 x 2 − 6 x
4
37. n = 5: ( z − x) ∑ xi z 4−i = ( z − x)( z 4 + xz 3 + x 2 z 2 + x3 z + x 4 )
i =0
= z 5 − xz 4 + xz 4 − x 2 z 3 + x 2 z 3 − x3 z 2 + x3 z 2 − x 4 z + x 4 z − x5
= z 5 − x5
2
n = 3: ( z − x) ∑ xi z 2−i = ( z − x)( z 2 + xz + x 2 )
i =0
= z 3 − xz 2 + xz 2 − x 2 z + x 2 z − x3
= z 3 − x3
f ( x ) = 4 x5 − 3 x3
383
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
f ( z ) − f ( x)
z−x
z→x
4 z 5 − 3z 3 − (4 x5 − 3x3 )
= lim
z−x
z→x
4( z 5 − x5 ) − 3( z 3 − x3 )
= lim
z−x
z→x
4( z − x)( z 4 + xz 3 + x 2 z 2 + x3 z + x 4 ) − 3( z − x)( z 2 + xz + x 2 )
= lim
z−x
z→x
= lim [4( z 4 + xz 3 + x 2 z 2 + x3 z + x 4 ) − 3( z 2 + xz + x 2 )]
f ′( x) = lim
z→x
= 4(5 x 4 ) − 3(3x 2 )
= 20 x 4 − 9 x 2
Principles in Practice 11.2
d
(50q − 0.3q 2 )
dq
d
d
=
(50q ) −
0.3q 2
dq
dq
1. r ′(q) =
(
)
d
d
= 50 (q ) − 0.3 ( q 2 )
dq
dq
= 50(1) – 0.3(2q) = 50 – 0.6q
The marginal revenue is r ′(q) = 50 − 0.6q .
Problems 11.2
1. f(x) = 5 is a constant function, so f ′( x) = 0
2.
⎛6⎞
f ( x) = ⎜ ⎟
⎝7⎠
2/3
is a constant function, so f ′( x) = 0
3. y = x6 , y ′ = 6 x6−1 = 6 x5
4.
f ′( x) = 21x 21−1 = 21x 20
5. y = x80 ,
dy
= 80 x80−1 = 80 x79
dx
6. y = x5.3 , y ′ = 5.3 x5.3−1 = 5.3 x 4.3
7.
(
)
f ( x) = 9 x 2 , f ′( x) = 9 2 x 2−1 = 18 x
(
)
8. y ′ = 4 3 x3−1 = 12 x 2
9. g ( w) = 8w7 , g ′( w) = 8(7 w7 −1 ) = 56w6
384
ISM: Introductory Mathematical Analysis
Section 11.2
10. v ′( x) = exe −1
11. y =
28.
(
)
2 4
2
8
x , y ′ = 4 x 4−1 = x3
3
3
3
(
f ′( p ) = 3 4 p 4−1 = 4 3 p3
13.
t7
1
7
f (t ) = , f ′(t ) =
(7t 7 −1 ) = t 6
25
25
25
14. y ′ =
(
(
f ′( x) = 3(1) − 0 = 3
17.
f ′( x) = 4(2 x) − 2(1) + 0 = 8 x − 2
( ) ( )
32. p ′( x) =
)
f ′(t ) = −13 ( 2t ) + 14(1) + 0 = −26t + 14
1
⎛ 1 1 −1 ⎞
21. y ′ = 3 x3−1 − ⎜ x 2 ⎟ = 3 x 2 −
2 x
⎝2
⎠
(
)
34.
28 − 19
⎛ 14 ⎞ ( − 14 )−1
f ′( x) = 2 ⎜ − ⎟ x 5
=− x 5
5
⎝ 5⎠
37.
2
= −39 x + 28 x − 2
(
f ′( x) = 2 0 − 4 x
)
) = −8x
38. y = x7 / 2 , y ′ =
3
26. φ ′(t ) = 5(3t 3−1 − 0) = 15t 2
(
(
39.
)
1
13 − x 4 ,
3
1
4
g ′( x) = 0 − 4 x 4−1 = − x3
3
3
27. g ( x) =
1
f ( x) = 11 x = 11x 2 ,
11
⎛ 1 ⎞ ( 1 )−1 11 − 1
f ′( x) = 11⎜ ⎟ x 2 = x 2 =
2
2 x
⎝2⎠
24. V ′(r ) = 8r 8−1 − 7 6r 6−1 + 3(2r ) + 0 = 8r 7 − 42r 5 + 6r
25.
⎛ 5 ( 5 )−1 ⎞ 3 − 1 10 2
3 ( 34 )−1
+ 2⎜ x 3 ⎟ = x 4 + x3
x
4
3
⎝3
⎠ 4
2 −7
⎛ 2 ⎞ −7
36. y ′ = 5(3x 2 ) − ⎜ − ⎟ x 5 = 15 x 2 + x 5
5
⎝ 5⎠
)
4 −1
3 35 −1 3 −2 / 5
= x
x
5
5
f ′( x) =
23. y ′ = −13 3x3−1 + 14(2 x) − 2(1) + 0
(
( )
1
2
2
7 x 6 + (1) = x 6 +
7
3
3
33.
35. y ′ =
22. y ′ = −8 4 x 4−1 + 0 = −32 x3
(
)
3 4 7 3
x + x
10
3
3
7
6
f ′( x) =
4 x3 + 3 x 2 = x3 + 7 x 2
10
3
5
19. g ′( p ) = 4 p 4−1 − 3 3 p3−1 − 0 = 4 p3 − 9 p 2
20.
9 2
x + 8x
2
9
+ 3 x3−1 − (2 x) + 8(1)
2
f ( x) =
18. F ′( x) = 5(2 x) − 9(1) = 10 x − 9
(
)
5
5
30. k ′( x) = −2(2 x) + (1) + 0 = −4 x +
3
3
31.
16.
(
= 16 x3 + 3 x 2 − 9 x + 8
)
f ( x) = x + 3, f ′( x) = 1 + 0 = 1
)
h′( x) = 4 4 x 4−1
1
7 x 7 −1 = x 6
7
15.
(
5 4
5
x − 6 , f ′( x) = 4 x 4−1 − 0 = 10 x3
2
2
29. h( x) = 4 x 4 + x3 −
)
12.
f ( x) =
7 72 −1 7 5 / 2
= x
x
2
2
1
−2
⎛ 1 −2 ⎞
f (r ) = 6r 3 , f ′(r ) = 6 ⎜ r 3 ⎟ = 2r 3
⎝3
⎠
1
−3
⎛ 1 −3 ⎞
40. y = 4 x 4 , y ′ = 4 ⎜ x 4 ⎟ = x 4
⎝4
⎠
)
41.
385
f ( x) = x −4 , f ′( x) = −4 x −4−1 = −4 x −5
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
42.
f ′( s ) = 2(−3s −4 ) = −6 s −4
43.
f ( x) = x −3 + x −5 − 2 x −6 ,
(
53.
) (
f ′( x) = −3x −3−1 + −5 x −5−1 − 2 −6 x −6−1
= −3 x
44.
−4
− 5x
−6
+ 12 x
(
(
)
54. Φ ( x) =
)
− 12
55.
56.
f ( x) = 2 x −3
f ′( x) = 2(−3 x
47. y =
8
−4
) = −6 x
(
)
1
5
=
58.
4x
1
5
y ′ = −5 x −6 = − x −6
4
4
49. g ( x) =
)
4
3
=
50. y =
51.
1
x
2
1
=
2/3
=
1 −2 / 3
x
2
3
4 3
= 3x
− 34
2
1
= 2x
− 12
x2
−3
⎛ 1 −3 ⎞
y′ = 2 ⎜ − x 2 ⎟ = − x 2
⎝ 2
⎠
4 −3
x
3
)
1 − 12
x
2
1 −3
y′ = − x 2
4
60. y =
= x −2 , y ′ = −2 x −3
5
2+( 1 )
⎛ 1⎞
61. y = x 2 x = x 2 ⎜ x 2 ⎟ = x 2 = x 2
⎝ ⎠
5 32
y′ = x
2
1 ⎛1⎞ 1
f (t ) = ⎜ ⎟ = t −1
2⎝t ⎠ 2
1
1
f ′(t ) = −1 ⋅ t −2 = − t −2
2
2
(
f ( x) =
59. y =
3x
4
g ′( x) = −3 x −4 = −4 x −4
3
(
3 2
x
9 −7
⎛ 3 −7 ⎞
f ′( x) = 3 ⎜ − x 4 ⎟ = − x 4
4
⎝ 4
⎠
1 −5
x
4
(
1
3
2x
8 x
1⎛ 2
1
⎞
q ′( x) = ⎜ − x −5 / 3 ⎟ = − x −5 / 3
2⎝ 3
3
⎠
y ′ = 8 −5 x −6 = −40 x −6
48. y =
−7
⎛ 1 −3 ⎞
⎛ 3 −7 ⎞ 3 −3
f ′( z ) = 3 ⎜ z 4 ⎟ − 0 − 8 ⎜ − z 4 ⎟ = z 4 + 6 z 4
⎝4
⎠
⎝ 4
⎠ 4
57. q( x) =
−4
= 8 x −5
x5
f ( x) = −9 x1/ 3 + 5 x −2 / 5 ,
−2
−7
⎛ 1 −2 ⎞ ⎛ 2 −7 ⎞
f ′( x) = −9 ⎜ x 3 ⎟ + 5 ⎜ − x 5 ⎟ = −3 x 3 − 2 x 5
⎝3
⎠ ⎝ 5
⎠
1
45. y = = x −1
x
dy
1
= −1 ⋅ x −1−1 = − x −2 = −
dx
x2
46.
)
1 3
x − 3 x −3 ,
3
1
Φ ′( x) = (3x 2 ) − 3(−3x −4 ) = x 2 + 9 x −4
3
−7
⎛ 1 −1 ⎞
f ′( x) = 100 −3 x −4 + 10 ⎜ x 2 ⎟
⎝2
⎠
= −300 x −4 + 5 x
1
x + 7 x −1
7
1
1
f ′( x) = (1) + 7 −1x −2 = − 7 x −2
7
7
f ( x) =
)
7 −1
x
9
7
7
g ′( x) = (−1x −2 ) = − x −2
9
9
62.
52. g ( x) =
386
f ( x) = (8 x5 ), f ′( x) = 40 x 4
ISM: Introductory Mathematical Analysis
63.
(
Section 11.2
)
f ( x) = x 3 x 2 − 10 x + 7 = 3x3 − 10 x 2 + 7 x
72.
f ( x ) = 3 x9 − 5 x5 + 4 x3
73. w( x ) =
(
65.
3
2
3
)
( ) = 9x
f ( x) = x (3 x) = x 9 x
2
74.
5
f ′( x) = 45 x 4
66. s ( x) = 3 x
( 4 x − 6 x + 3)
= x1/ 3 ( x1/ 4 − 6 x + 3)
= x 7 /12 − 6 x 4 / 3 + 3x1/ 3
7
s ′( x) = x −5 /12 − 8 x1/ 3 + x −2 / 3
12
67. v( x) = x
v′( x) =
68.
− 23
1
( x + 5) = x 3 + 5 x
(
)
13
f (q) =
8
3
f ( w) =
w−5
w5
7 x3 + x
6 x
1 ⎛ 7 x3
x ⎞
= ⎜
+
⎟
1/
2
1/
⎜
6⎝ x
x 2 ⎟⎠
1
= (7 x5 / 2 + x1/ 2 )
6
1 ⎛ 35
1
⎞
f ′( x) = ⎜ x3 / 2 + x −1/ 2 ⎟
6⎝ 2
2
⎠
1 1/ 2
−1
= x (35 x + x )
12
39
2
′
y x =−3 = −60
y ′ x =3 / 2 = −
77. y is a constant, so y ′ = 0 for all x.
−1
78. y ′ = 3 − 2 x −1/ 2 = 3 −
2
q2
y ′ x =4 = 2
7
y ′ x =9 =
3
13
y ′ x = 25 =
5
= w−4 − 5w−5
f ′( w) = −4w−5 + 25w−6 = − w−6 (4w − 25)
71.
= 1+ x
y ′ x =0 = −6
3q 2 + 4q − 2 3q 2 4q 2
=
+
−
q
q
q q2
f ′(q ) = 3(1) + 0 − 2(− q −2 ) = 3 + 2q −2 = 3 +
70.
x2
76. y ′ = −6 − 6 x 2
)
= 3q + 4 − 2q
f ( x) =
x3
+
y ′ x = 2 = 16
y ′ x =−3 = −14
13 8 56 3 33 − 2
f ′( x) = x 5 + x 5 + x 5
5
5
5
1 − 52
= x 13x 2 + 56 x + 33
5
69.
x2
y ′ x =0 = 4
f ( x) = x 5 x 2 + 7 x + 11 = x 5 + 7 x 5 + 11x 5
(
=
75. y ′ = 6 x + 4
− 32
1 − 23 10 − 53 1 − 35
x − x = x ( x − 10)
3
3
3
3
x 2 + x3
x2
x2
w′( x) = 0 + 1 = 1
f ′( x) = 27 x8 − 25 x 4 + 12 x 2
= x 2 27 x6 − 25 x 2 + 12
(
f ′( x) = 4 x3 + 6 x 2 − 16 x = 2 x 2 x 2 + 3 x − 8
f ′( x) = 9 x 2 − 20 x + 7
64.
f ( x) = x 2 ( x − 2)( x + 4) = x 4 + 2 x3 − 8 x 2
f ( x) = ( x + 1)( x + 3) = x 2 + 4 x + 3
f ′( x) = 2 x + 4 = 2( x + 2)
387
2
x
)
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
79. y = 4 x 2 + 5 x + 6
y′ = 8x + 5
84. y =
y ′ x =1 = 13
(
1
x3
)
85. y =
( 2 − x2 ) = 2 x−
1
2
3
− x2 .
3 12
x
2
5 2
x − x3
2
5 x − 3x 2 = 0. Then x(5 − 3x) = 0, x = 0 or
5
x= .
3
5
125
. This
If x = 0, then y = 0. If x = , y =
3
54
⎛ 5 125 ⎞
gives the points (0, 0) and ⎜ ,
⎟.
⎝ 3 54 ⎠
3
x4
3
16
An equation of the tangent line is
1
3
3
1
y − = − ( x − 2), or y = − x + .
8
16
16
2
y ′ x =2 = −
82. y = − 3 x = − x
−
1
2
y ′ = 5 x − 3x 2
A horizontal tangent line has slope 0, so we set
= x −3
y ′ = −3x −4 = −
− 32
) = x−
1
25
y ′ x =4 = − − 3 = −
8
8
When x = 4, then y = –7. The tangent line is
25
25
11
y + 7 = − ( x − 4) , or y = − x + .
8
8
2
1
1 − x2
5
1
y ′ = ( −2 x )
5
8
y ′ x =4 = −
5
An equation of the tangent line is
8
8
17
y + 3 = − ( x − 4) , or y = − x + .
5
5
5
81. y =
x
y′ = − x
An equation of the tangent line is
y – 15 = 13(x – 1), or y = 13x + 2.
80. y =
(
x 2 − x2
86. y =
x5
− x +1
5
y′ = x4 − 1
A horizontal tangent line has slope 0, so we set
1
3
x 4 − 1 = 0 . Then x 4 = 1 , so x = 1 or x = –1. If
1
9
x = 1, then y = ; if x = –1, then y = . This
5
5
9⎞
⎛ 1⎞
⎛
gives the points ⎜1, ⎟ and ⎜ −1, ⎟ .
5⎠
⎝ 5⎠
⎝
1 −2
1
y′ = − x 3 = − 2
3
3x 3
1
1
1
y ′ x =8 = −
=−
=−
3⋅ 4
12
⎛ 23 ⎞
3⎜ 8 ⎟
⎝ ⎠
An equation of the tangent line is
1
4
1
y + 2 = − ( x − 8) , or y = − x − .
12
3
12
87. y = x 2 − 5 x + 3
y′ = 2 x − 5
Setting 2x – 5 = 1 gives 2x = 6, x = 3. When
x = 3, then y = –3. This gives the point (3, –3).
83. y = 3 + x − 5 x 2 + x 4
y ′ = 1 − 10 x + 4 x3 .
88. y = x 4 − 31x + 11
When x = 0, then y = 3 and y ′ = 1 . Thus an
equation of the tangent line is y – 3 = 1(x – 0), or
y = x + 3.
y ′ = 4 x3 − 31
If 4 x3 − 31 = 1, then x3 = 8, x = 2. When x = 2,
then y = −35. This gives the point (2, −35).
388
ISM: Introductory Mathematical Analysis
89.
1
f ( x) = x +
x
1
= x2 + x
Section 11.3
Principles in Practice 11.3
− 12
1. Here
1 − 32
1
1
x −1
−
=
x =
2
2 x 2x x 2x x
x −1
x −1
− f ′( x) =
−
=0.
Thus
2x x
2x x 2x x
1
x
2
x −1
f ′( x) =
− 12
−
dP
= 5 and ∆p = 25.5 – 25 = 0.5.
dp
dP
∆p = 5(0.5) = 2.5
dp
The profit increases by 2.5 units when the price
is changed from 25 to 25.5 per unit.
∆P ≈
90. z = (1 + b) w p − bwc
dw p
dz
= (1 + b)
−b
dwc
dwc
Rewriting the right side and factoring out 1 + b
dw p b(1 + b)
dz
,
gives
= (1 + b)
−
1+ b
dwc
dwc
2.
(
)
dy d
=
16t − 16t 2 = 16 − 16(2t ) = 16 − 32t
dt dt
dy
= 16 − 32(0.5) = 16 − 16 = 0
dt t =0.5
The graph of y(t) is shown.
5
⎡ dw p
dz
b ⎤
= (1 + b) ⎢
−
⎥.
dwc
⎣ dwc 1 + b ⎦
91. y = x3 − 3x
y ′( x ) = 3x 2 − 3
0
( )−3 = 9
y ′ x =2 = 3 2
1
0
2
When t = 0.5, the object is at the peak of its
flight.
The tangent line at (2, 2) is given by
y – 2 = 9(x – 2), or y = 9x – 16.
3. V ′(r ) =
8
( )
4
π 3r 2 + 4π(2r ) = 4πr 2 + 8πr
3
When r = 2, V ′(r ) = 4π(2) 2 + 8π(2) = 32π and
4
32π
80
π(2)3 + 4π(2)2 =
+ 16π = π .
3
3
3
The relative rate of change of the volume when
V ′(2) 32π 6
=
= = 1.2 . Multiplying 1.2
r = 2 is
V (2) 80 π 5
V (r ) =
5
–5
–2
92. y = 3 x = x1/ 3
3
by 100 gives the percentage rate of change:
(1.2)(100) = 120%.
1 −2 / 3
1
x
=
3 2
3
3 x
1
y ′ x =−8 =
12
The tangent line at (−8, −2) is given by
1
4
1
y + 2 = ( x + 8), or y = x − .
12
3
12
y ′( x ) =
Problems 11.3
1. s = f (t ) = 2t 2 + 3t
If ∆t = 1, then over [1, 2] we have
∆s f (2) − f (1) 14 − 5
=
=
= 9.
2 −1
1
∆t
If ∆t = 0.5, then over [1, 1.5] we have
∆s f (1.5) − f (1) 9 − 5
=
=
= 8.
1.5 − 1
0.5
∆t
Continuing this way, we obtain the following
table:
1
–15
5
–3
389
Chapter 11: Differentiation
∆t
∆s
∆t
1
0.5
0.2
9
8
7.4
ISM: Introductory Mathematical Analysis
0.1
7.2
0.01
0.001
7.02
7.002
We estimate the velocity when t = 1 to be 7 m/s. With differentiation we get v =
ds
= 4(1) + 3 = 7 m/s.
dt t =1
2. y = f ( x) = 2 x + 5 .
If ∆x = 1, then over [3, 4] we have
∆y f (4) − f (3)
13 − 11
=
=
≈ 0.2889
∆x
∆x
1
If ∆x = 0.5, then over [3, 3.5] we have
∆y f (3.5) − f (3)
12 − 11
=
=
≈ 0.2950
∆x
∆x
0.5
Continuing in this way we obtain the following table:
∆x
1
0.5
0.2
0.1
0.01
0.001
∆y
∆x
0.2889
0.2950
0.2988
0.3002
0.3014
0.3015
We estimate the rate of change to be 0.3015.
1
⎛
⎞
≈ 0.3015. ⎟
⎜ Note: The actual rate of change is
11
⎝
⎠
3. s = f (t ) = 2t 2 − 4t
a.
When t = 7, then s = 2(7 2 ) − 4(7) = 70 m.
b.
∆s f (7.5) − f (7) [2(7.5)2 − 4(7.5)] − 70
=
=
= 25 m/s
∆t
0.5
0.5
c.
v=
ds
= 4t − 4. If t = 7, then v = 4(7) − 4 = 24 m/s
dt
1
4. s = f (t ) = t + 1
2
a.
When t = 2, s =
1
(2) + 1 = 2 m.
2
b.
1
∆s f (2.1) − f (1) ⎡⎣ 2 (2.1) + 1⎤⎦ − 2
=
=
= 0.5 m/s
0.1
0.1
∆t
c.
v=
ds 1
1
= . If t = 2, then v = m/s
dt 2
2
390
ds
= 4t + 3,
dt
ISM: Introductory Mathematical Analysis
Section 11.3
5. s = f (t ) = 2t 3 + 6
a.
b.
c.
b.
When t = 1, s = 2(1)3 + 6 = 8 m.
=
∆s f (1.02) − f (1)
=
∆t
0.02
⎡ 2(1.02)3 + 6 ⎤ − 8
⎦
=⎣
0.02
= 6.1208 m/s
v=
c.
ds
= 6t 2 . If t = 1, then
dt
9.
v = 6(1)2 = 6 m/s
6. s = f (t ) = −3t 2 + 2t + 1
a.
b.
c.
( )
When t = 1, s = −3 12 + 2(1) + 1 = 0 m.
∆s f (1.25) − f (1)
=
∆t
0.25
⎡ −3(1.25) 2 + 2(1.25) + 1⎤ − 0
⎦
=⎣
= −4.75 m/s
0.25
v=
b.
c.
4
7/2⎤
⎥⎦ − 0
1
4
ds
7
= 12t 3 − t 5 / 2 . If t = 0, then
dt
2
7
v = 12(0)3 − (0)5 / 2 = 0 m/s.
2
v=
dy 25 32
dy 25
=
(27) = 337.50 .
=
x . If x = 9,
dx 2
dx 2
dT
= 0 + 0.27(1 − 0) = 0.27
dTe
12.
dV
= 4πr 2
dr
13. c = 500 + 10q,
dc
= 10 . When q = 100,
dq
dc
= 10 .
dq
∆s f (2.1) − f (2)
=
∆t
0.1
⎡(2.1)4 − 2(2.1)3 + 2.1⎤ − 2
⎦
=⎣
= 10.261 m/s
0.1
14. c = 5000 + 6q,
15.
ds
= 4t 3 − 6t 2 + 1. If t = 2, then
dt
dc
dc
= 6 . When q = 36,
=6.
dq
dq
dc
= 0.1(2q ) + 3 = 0.2q + 3 . When q = 5,
dq
dc
= 0.2(5) + 3 = 4.
dq
( ) ( )
v = 4 23 − 6 22 + 1 = 9 m/s
a.
1
m/s
64
( 14 ) − ( 14 )
11.
( )
8. s = f (t ) = 3t − t
1
4
⎡
⎢3 ⋅
=⎣
dA
dA
= 2πr . If r = 3,
= 2π(3) = 6π .
dr
dr
ds
= −6t + 2. If t = 1, v = –4 m/s
dt
4
− f (0)
10.
When t = 2, s = 24 − 2 23 + 2 = 2 m.
v=
( )
1
4
When r = 6.3 × 10−4 ,
dV
= 4π[6.3 × 10−4 ]2 = 158.76π×10−8
dr
≈ 4.988 × 10−6.
7. s = f (t ) = t 4 − 2t 3 + t
a.
∆s
=
∆t
f
16.
7/2
When t = 0, s = 3 ⋅ 04 = 07 / 2 = 0.
17.
391
dc
dc
= 0.2q + 3 . When q = 3,
= 3.6 .
dq
dq
dc
= 2q + 50 . Evaluating when q = 15, 16 and
dq
17 gives 80, 82 and 84, respectively.
Chapter 11: Differentiation
18.
ISM: Introductory Mathematical Analysis
1 ⎞
1
⎛
24. r = q ⎜ 15 − q ⎟ = 15q − q 2
30
30
⎝
⎠
dr
1
= 15 − q
dq
15
dc
= 0.12q 2 − q + 4.4
dq
Evaluating when q = 5, 25, and 1000 gives 2.4,
54.4 and 119,004.4, respectively.
19. c = 0.01q + 5 +
500
q
For q = 5,
c = cq = 0.01q 2 + 5q + 500
q = 150,
dc
= 0.02q + 5
dq
dr
= 250 + 90q − 3q 2 . Evaluating when
dq
q = 5, 10 and 25 gives 625, 850 and 625,
respectively.
dc
=7
dq q =100
26. r = 60q − 0.2q 2
1000
20. c = 2 +
q
c = cq = 2q + 1000
dr
= 60 − 0.4q
dq
Evaluating when q = 10 and 20 gives 56 and 52,
respectively.
dc
= 2 for all q
dq
3
27.
2
21. c = cq = 0.00002q − 0.01q + 6q + 20, 000
dc
= 0.00006q 2 − 0.02q + 6
dq
dc
= 6.750 − 0.000328(2q) = 6.750 − 0.000656q
dq
dc
= 6.750 − 0.000656(2000) = 5.438
dq q = 2000
c −10, 484.69
c= =
+ 6.750 − 0.000328q
q
q
dc
= 4.6 . If q = 500, then
dq
dc
= 11 .
dq
−10, 484.69
+ 6.750 − 0.000328(2000)
2000
= 0.851655
c (2000) =
22. c = cq = 0.002q3 − 0.5q 2 + 60q + 7000
28.
dc
= 0.006q 2 − q + 60
dq
If q = 15, then
dr
=5.
dq
25. r = 250q + 45q 2 − q3
dc
=6
dq q =50
If q = 100, then
dr 44
dr
=
; for q =15,
= 14 ; for
dq 3
dq
dc
= −0.79 + 0.04284q − 0.0003q 2
dq
dc
= 0.7388
dq q =70
dc
= 46.35. If q = 25, then
dq
dc
= 38.75.
dq
29. PR 0.93 = 5, 000, 000
P = 5, 000, 000 R −0.93
dP
= −4, 650, 000 R −1.93
dR
23. r = 0.8q
dr
= 0.8 for all q.
dq
30.
392
dv
= −10,500 for all t.
dt
ISM: Introductory Mathematical Analysis
31. a.
Section 11.3
e.
dy
= −1.5 − x
dx
dy
= −1.5 − 6 = −7.5
dx x =6
36. a.
b. Setting –1.5 – x = –6 gives x = 4.5.
32. c = f (q ) = 0.4q 2 + 4q + 5
dc
= 0.8q + 4
dq
If q = 2, then
dc
= 5.6 . Over the interval [2, 3],
dq
∆c f (3) − f (2) 20.6 − 14.6
=
=
=6.
∆q
3− 2
1
33. a.
b.
y′
1
=
y x+4
c.
y ′(5) = 1
d.
1
1
= ≈ 0.111
5+ 4 9
e.
11.1%
34. a.
−9 x 2
y′
=
y 5 − 3x3
c.
y ′(1) = −9
d.
−9
9
= − = −4.5
5−3
2
e.
−450%
y′ = −3 x 2
b.
y′ −3 x 2
=
y 8 − x3
c.
y ′(1) = −3
d.
−3
3
= − ≈ −0.429
8 −1
7
e.
–42.9%
38. a.
y ′ = −3
y ′ = −9 x 2
b.
37. a.
y′ = 1
63.2%
y′ = 2 x + 3
b.
y′
2x + 3
=
y x2 + 3x − 4
c.
y ′(−1) = 2(−1) + 3 = 1
d.
1
1
= − ≈ −0.167
1− 3 − 4
6
e.
–16.7%
b.
y′
−3
3
=
=
y 7 − 3x 3x − 7
c.
y ′(6) = −3
d.
3
3
= ≈ 0.2727
3(6) − 7 11
e.
27.27%
39. c = 0.3q 2 + 3.5q + 9
35. a.
y′ = 6 x
dc
= 0.6q + 3.5
dq
b.
y′
6x
=
y 3x 2 + 7
dc
= 0.6(10) + 3.5 = 9.5. If
dq
q = 10, then c = 74 and
c.
y ′(2) = 6(2) = 12
d.
12
12
=
≈ 0.632
12 + 7 19
If q = 10, then
dc
dq
c
393
(100) =
9.5
(100) ≈ 12.8% .
74
Chapter 11: Differentiation
40. y =
ISM: Introductory Mathematical Analysis
100
= 100 x −1
x
dy
100
= −100 x −2 = −
dx
x2
dy
100
y′
−1
=−
= −1 and (100) = (100) = −10% .
If x = 10,
dx
100
y
10
41. a.
dr
= 30 − 0.6q
dq
b. If q = 10,
c.
42. a.
9%
dq
= 10 − 0.4q
dr
b. If q = 25,
c.
43.
r′
30 − 6
24
4
=
=
=
≈ 0.09 .
r 300 − 30 270 45
r′
10 − 0.4(25)
=
= 0.
r 10(25) − 0.2(25) 2
0%
W ′ 0.864t −0.568 0.432
=
=
W
t
2t 0.432
1.3 I 0.3
44. a.
R1′ 1855.24 1.3
= 1.3 =
I
R1
I
1855.24
1.3 I 0.3
R2′ 1101.29 1.3
= 1.3 =
I
R2
I
1101.29
b. They are equal.
c.
f ′ x nC1 x n −1 n
=
=
f ( x)
x
C1 x n
g ′( x ) nC2 x n −1 n
=
=
g ( x)
x
C2 x n
The rates are equal.
45. The cost of q = 20 bikes is qc = 20(150) = $3000 . The marginal cost, $125, is the approximate cost of one
additional bike. Thus the approximate cost of producing 21 bikes is $3000 + $125 = $3125.
394
ISM: Introductory Mathematical Analysis
Section 11.4
dc
dq
dc
1 dq 1
dc c
= = c , and the marginal cost
46. The relative rate of change of c is
, which is given to be :
= . Thus
c
dq q
q c q
⎛ dc ⎞
function ⎜ ⎟ and the average cost function (c ) are equal.
⎝ dq ⎠
47. $5.07 per unit
48. 11,275 people per year
Principles in Practice 11.4
1.
dR
d
d
= (2 − 0.15 x) (225 + 20 x) + (225 + 20 x) (2 − 0.15 x)
dx
dx
dx
= (2 – 0.15x)(20) + (225 + 20x)(–0.15)
= 40 – 3x – 33.75 – 3x = 6.25 – 6x
dR
= 6.25 − 6 x
dx
1
2. T ( x) = x 2 − x3
3
T ′( x ) = 2x − x 2
When the dosage is 1 milligram the sensitivity is T ′(1) = 2(1) − 12 = 1 .
Problems 11.4
1.
f ′( x) = (4 x + 1)(6) + (6 x + 3)(4) = 24x + 6 + 24x + 12 = 48x + 18 = 6(8x + 3)
2.
f ′( x) = (3x − 1)(7) + (7 x + 2)(3) = 42 x − 1
3. s ′(t ) = (5 − 3t )(3t 2 − 4t ) + (t 3 − 2t 2 )(−3) = 15t 2 − 20t − 9t 3 + 12t 2 − 3t 3 + 6t 2 = −12t 3 + 33t 2 − 20t
4. Q′( x) = (3 + x)(10 x) + (5 x 2 − 2)(1) = 15 x 2 + 30 x − 2
5.
(
)
(
(
)
(
)
f ′(r ) = 3r 2 − 4 (2r − 5) + r 2 − 5r + 1 (6r ) = 6r 3 − 15r 2 − 8r + 20 + 6r 3 − 30r 2 + 6r = 12r 3 − 45r 2 − 2r + 20
)
(
6. C ′( I ) = 2 I 2 − 3 (6 I − 4) + 3I 2 − 4 I + 1 (4 I ) = 12 I 3 − 8I 2 − 18I + 12 + 12 I 3 − 16 I 2 + 4 I = 2 12 I 3 − 12 I 2 − 7 I + 6
7. Without the product rule we have
(
)
f ( x) = x 2 2 x 2 − 5 = 2 x 4 − 5 x 2
f ′( x) = 8 x3 − 10 x
8. Without the product rule we have
(
)
f ( x ) = 3 x3 x 2 − 2 x + 2 = 3 x5 − 6 x 4 + 6 x3
f ′( x) = 15 x 4 − 24 x3 + 18 x 2
395
)
Chapter 11: Differentiation
(
)
ISM: Introductory Mathematical Analysis
(
)
9. y′ = x 2 + 3x − 2 (4 x − 1) + 2 x 2 − x − 3 (2 x + 3)
(
) (
= 4 x3 + 12 x 2 − 8 x − x 2 − 3x + 2 + 4 x3 − 2 x 2 − 6 x + 6 x 2 − 3x − 9
)
= 8 x3 + 15 x 2 − 20 x − 7
10. φ ′( x) = (3 − 5 x + 2 x 2 )(1 − 8 x) + (2 + x − 4 x 2 )(−5 + 4 x)
= 3 − 5 x + 2 x 2 − 24 x + 40 x 2 − 16 x3 − 10 − 5 x + 20 x 2 + 8 x + 4 x 2 − 16 x3
= −32 x3 + 66 x 2 − 26 x − 7
11.
f ′( w) = ( w2 + 3w − 7)(6 w2 ) + (2 w3 − 4)(2w + 3)
= 6 w4 + 18w3 − 42 w2 + 4 w4 + 6 w3 − 8w − 12
= 10w4 + 24w3 − 42 w2 − 8w − 12
12.
(
)
(
)
f ′( x) = 3 x − x 2 (−1 − 2 x) + 3 − x − x 2 (3 − 2 x)
= −3 x − 5 x 2 + 2 x3 + 9 − 3x − 3 x 2 − 6 x + 2 x 2 + 2 x3
= 4 x3 − 6 x 2 − 12 x + 9
(
)(
) (
)
13. y′ = x 2 − 1 9 x 2 − 6 + 3 x3 − 6 x + 5 (2 x) − 4(8 x + 2)
= 9 x 4 − 15 x 2 + 6 + 6 x 4 − 12 x 2 + 10 x − 32 x − 8
= 15 x 4 − 27 x 2 − 22 x − 2
( ) (
)
14. h′( x) = 4 5 x 4 + 3 ⎡⎢ 8 x 2 − 5 (2) + (2 x + 2)(16 x) ⎤⎥
⎣
⎦
= 20 x 4 + 3(16 x 2 − 10 + 32 x 2 + 32 x)
= 20 x 4 + 144 x 2 + 96 x − 30
3 ⎡ 1/ 2
⎛ 1
⎞⎤
(5 p − 2)(3) + (3 p − 1) ⎜ 5 ⋅ p −1/ 2 ⎟ ⎥
⎢
2⎣
⎝ 2
⎠⎦
3⎡
15
5
⎤
= ⎢15 p1/ 2 − 6 + p1/ 2 − p −1/ 2 ⎥
2⎣
2
2
⎦
3
−1/ 2
1/ 2
= [45 p − 12 − 5 p
]
4
15. F ′( p ) =
3
⎛1
⎞
⎛1
⎞
16. g ′( x) = ( x1/ 2 + 5 x − 2) ⎜ x −2 / 3 − x −1/ 2 ⎟ + ( x1/ 3 − 3 x1/ 2 ) ⎜ x −1/ 2 + 5 ⎟
2
⎝3
⎠
⎝2
⎠
1 −1/ 6 5 1/ 3 2 −2 / 3 3 15 1/ 2
1
3
= x
+ x − x
− − x + 3 x −1/ 2 + x −1/ 6 + 5 x1/ 3 − − 15 x1/ 2
3
3
3
2 2
2
2
1
−1/ 6
−1/ 2
−2 / 3
1/ 2
1/ 3
= (−135 x + 40 x + 5 x
+ 18 x
− 4x
− 18)
6
17. y = 7 ⋅
2
is a constant function, so y′ = 0 .
3
18. y = x3 − 6 x 2 + 11x − 6
y′ = 3x 2 − 12 x + 11
396
ISM: Introductory Mathematical Analysis
Section 11.4
19. y = 6 x3 + 47 x 2 + 31x − 28
y′ = 18 x 2 + 94 x + 31
20.
21.
dy (4 x + 1)(2) − (2 x − 3)(4) 8 x + 2 − 8 x + 12
=
=
dx
(4 x + 1)2
(4 x + 1)2
14
=
(4 x + 1)2
( x − 1)(5) − (5 x)(1)
f ′( x) =
=−
( x − 1)
2
=
5x − 5 − 5x
( x − 1)2
5
( x − 1)2
(5 − x)(−5) − (−5 x)(−1)
22. H ′( x) =
(5 − x)2
−25 + 5 x − 5 x
25
=
=−
2
(5 − x)
(5 − x) 2
23.
24.
−13
f ( x) =
5
=−
13 −5
x
3
3x
13
65
f ′( x ) = − (−5 x −6 ) =
3
3x6
(
)
5 2
x −2
7
5
10
f ′( x) = (2 x) = x
7
7
f ( x) =
25. y′ =
=
( x − 1)(1) − ( x + 2)(1)
( x − 1)2
x −1− x − 2
( x − 1)2
=−
3
( x − 1)2
26. h′( w) =
=
=
(
)
( w − 3)(6w + 5) − 3w2 + 5w − 1 (1)
( w − 3)2
6w2 − 13w − 15 − 3w2 − 5w + 1
( w − 3) 2
3w2 − 18w − 14
( w − 3)2
397
Chapter 11: Differentiation
27.
ISM: Introductory Mathematical Analysis
z 2 − 4 ) (−2) − (6 − 2 z )(2 z )
(
h′( z ) =
2
( z2 − 4)
=
−2 z 2 + 8 − 12 z + 4 z 2
( z2 − 4)
2 ( z2 − 6z + 4)
=
2
( z2 − 4)
2
2 z 2 − 12 z + 8
( z2 − 4)
2
(3 x 2 + 5 x + 3)(4 x + 5) − (2 x 2 + 5 x − 2)(6 x + 5)
28. z ′ =
=
=
(3x 2 + 5 x + 3) 2
12 x3 + 35 x 2 + 37 x + 15 − (12 x3 + 40 x 2 + 13x − 10)
(3x 2 + 5 x + 3)2
2
−5 x + 24 x + 25
=
(3x 2 + 5 x + 3) 2
( x2 − 5x ) (16 x − 2) − (8x2 − 2x + 1) (2 x − 5)
2
( x2 − 5x )
16 x3 − 82 x 2 + 10 x − (16 x3 − 44 x 2 + 12 x − 5 ) −38 x 2 − 2 x + 5
=
=
2
2
2
x
−
5
x
(
)
( x2 − 5x )
29. y′ =
30.
x 2 + 1)( 3x 2 − 2 x ) − ( x3 − x 2 + 1) (2 x)
(
f ′( x) =
2
( x2 + 1)
=
=
31.
3 x 4 − 2 x3 + 3 x 2 − 2 x − 2 x 4 + 2 x3 − 2 x
(
x x3 + 3 x − 4
( x2 + 1)
)
( x 2 + 1) 2
2
2 x 2 − 3 x + 2 ) (2 x − 4) − ( x 2 − 4 x + 3) (4 x − 3)
(
y′ =
2
( 2 x 2 − 3x + 2 )
4 x3 − 14 x 2 + 16 x − 8 − ( 4 x3 − 19 x 2 + 24 x − 9 )
=
2
( 2 x 2 − 3x + 2 )
=
5x2 − 8x + 1
( 2 x 2 − 3x + 2 )
2
398
ISM: Introductory Mathematical Analysis
Section 11.4
32. The quotient rule can be used, or we can write
z4 + 4 1 3
= z + 4 z −1 ,
F ( z) =
3z
3
(
so F ′( z ) =
33.
)
38.
1
3z 4 − 4
3 z 2 − 4 z −2 =
.
3
3z 2
(
)
=
x100 + 7 ) (0) − (1) (100 x99 )
(
100 x99
g ′( x) =
=−
2
2
( x100 + 7 )
( x100 + 7 )
=
−9
9
= − x −5
5
2
2x
45 −6
y′ =
x
2
34. y =
35. u (v) =
x
41. y′ =
3x2 − x − 1
1
5
3
⎞ x+5
⎟=
3
⎟
⎠ 16 x 2
2
3
= 3x − x − x
=
− 13
=
x3
2
2 − 13 1 − 43
2
1
x + x = 5x 3 − 1 + 4
3
3
3x 3 3x 3
15 x 2 − 2 x + 1
3x
2
(3 x − 5)(5) − (5 x + 1)(3)
(3 x − 5)2
28
(3 x − 5)
2
+ 6 x −4
+ 6 x −4
[( x + 2)( x − 4)](1) − ( x − 5)(2 x − 2)
[( x + 2)( x − 4)]2
)
(
[( x + 2)( x − 4)]
− x 2 − 10 x + 18
)
[( x + 2)( x − 4)]2
(9 x − 1)(3 x + 2) 27 x 2 + 15 x − 2
=
4 − 5x
4 − 5x
(
)
(4 − 5 x)(54 x + 15) − 27 x 2 + 15 x − 2 (−5)
(4 − 5 x)
2
−270 x 2 + 141x + 60 + 135 x 2 + 75 x − 10
=−
399
(3 x + 1)2
2
y′ =
=
(3x + 1)(2) − (2 x)(3)
(
42. y =
4
3
+
x 2 − 2 x − 8 − 2 x 2 − 12 x + 10
2
y′ = 5 x 3 −
=
)
)
2
( x − 8)(0) − (4)(1)
= 6 x2 −
1⎛ 1
−1 ⎞
= ⎜ x 2 − 5x 2 ⎟
⎠
8 x 8⎝
3
(
x0.7 2 x 2.1 + 1
40. q ′( x) = 6 x 2 +
x−5
=
0.3 1 + 28 x1.8 − 12 x 2.1
⎞ 2(v3 + 4)
⎟=
v2
⎠
1 ⎛ 1 −1 5 −3 ⎞ 1 ⎛ 1
5
y′ = ⎜ x 2 + x 2 ⎟ = ⎜ 1 + 3
⎜
8⎝ 2
2
⎠ 16 ⎝ x 2 x 2
37. y =
(
(2 x 2.1 + 1) 2
( x − 8)
4
2
=
+
2
( x − 8)
(3 x + 1) 2
v3 − 8 v3 8
=
− = v 2 − 8v −1
v
v v
3x 2 − x − 1
0.6 x1.4 + 0.3x −0.7 − 4.2 x1.4 + 8.4 x1.1
39. y′ = −
4
⎛
u′(v) = 2v + 8v −2 = 2 ⎜ v +
v2
⎝
36. y =
2 x 2.1 + 1)( 0.3 x −0.7 ) − ( x 0.3 − 2 )( 4.2 x1.1 )
(
y′ =
2
( 2x2.1 + 1)
(4 − 5 x) 2
135 x − 216 x − 50
(4 − 5 x)2
Chapter 11: Differentiation
(
)(
ISM: Introductory Mathematical Analysis
)
(
)(
⎡ t 2 − 1 t 3 + 7 ⎤ (2t + 3) − t 2 + 3t 5t 4 − 3t 2 + 14t
⎢
⎦⎥
43. s′(t ) = ⎣
2
⎡ t 2 −1 t3 + 7 ⎤
⎢⎣
⎥⎦
(
=
44.
)(
)
)
−3t 6 − 12t 5 + t 4 + 6t 3 − 21t 2 − 14t − 21
(
)(
)
⎡ t 2 −1 t3 + 7 ⎤
⎣⎢
⎦⎥
2
17
f ( s) =
3
5s − 10 s 2 + 4s
0 − 17 ⎡15s 2 − 20s + 4 ⎤
17 15s 2 − 20s + 4
⎣
⎦
=−
f ′( s ) =
2
2
5s3 − 10s 2 + 4 s
5s3 − 10s 2 + 4 s
(
45. y = 3x −
2
x
(
(
)
)
)
2( x −1) −3 x
x ( x −1)
− x3−1
= 3x −
x−2
x−2
x+2
x+2
= 3x +
= 3x +
3
x( x − 1)( x − 2)
x − 3x2 + 2 x
y′ = 3 +
= 3−
( x3 − 3 x 2 + 2 x)(1) − ( x + 2)(3 x 2 − 6 x + 2)
[ x( x − 1)( x − 2)]2
2 x3 + 3 x 2 − 12 x + 4
[ x( x − 1)( x − 2)]2
3
46. y = 3 − 12 x +
y ′ = −36 x 2 +
47.
f ′( x ) =
1 − 25
x +2
x2 + 5
3
= 3 − 12 x +
x 2 + 2 −5
x2 + 2
2
x +5
= 3 − 12 x3 +
( x 4 + 7 x 2 + 10)(2 x) − ( x 2 − 3)(4 x3 + 14 x)
( x 4 + 7 x 2 + 10) 2
(a − x)(1) − (a + x)(−1)
(a − x)
2
=
x2 − 3
x 4 + 7 x 2 + 10
= −36 x 2 +
2a
(a − x)2
x −1 + a −1 ax a + x
⋅ =
x −1 − a −1 ax a − x
(a − x)(1) − (a + x)(−1)
2a
f ′( x ) =
=
2
(a − x)
(a − x)2
48. Simplifying, f ( x) =
(
)(
)
y′ = ( 4 x 2 + 2 x − 5 )( 3 x 2 + 7 ) + ( x3 + 7 x + 4 ) (8 x + 2)
49. y = 4 x 2 + 2 x − 5 x3 + 7 x + 4
y′(−1) = (−3)(10) + (−4)(−6) = −6
400
−2 x5 + 12 x3 + 62 x
[( x 2 + 2)( x 2 + 5)]2
ISM: Introductory Mathematical Analysis
50. y =
y′ =
x3
x4 + 1
( x 4 + 1)(3 x 2 ) − ( x3 )(4 x3 )
( x 4 + 1)2
y′(−1) =
51. y =
y′ =
Section 11.4
(2)(3) − (−1)(−4)
(2)
6
x −1
( x − 1)(0) − (6)(1)
( x − 1)
y′(3) = −
6
2
2
2
=−
=
2
1
2
6
=−
( x − 1) 2
3
2
3
3
15
The tangent line is y − 3 = − ( x − 3) , or y = − x + .
2
2
2
52. y =
x+5
x2
= x −1 + 5 x −2
y ′ = − x −2 − 10 x −3 = −
y ′(1) = −1 − 10 = −11
1
x
2
−
10
x3
The tangent line is y − 6 = −11(x − 1) or y = −11x + 17.
(
)
y′ = (2 x + 3) ⎡⎢ 2 ( 4 x3 − 10 x ) ⎤⎥ + ⎡⎢ 2 ( x 4 − 5 x 2 + 4 ) ⎤⎥ (2)
⎣
⎦ ⎣
⎦
53. y = (2 x + 3) ⎡⎢ 2 x 4 − 5 x 2 + 4 ⎤⎥
⎣
⎦
y′(0) = (3)(0) + [2(4)](2) = 16
The tangent line is y – 24 = 16(x – 0), or y = 16x + 24.
54. y =
x +1
2
x ( x − 4)
=
x +1
3
x − 4 x2
x3 − 4 x 2 ) (1) − ( x + 1) ( 3 x 2 − 8 x )
(
y′ =
2
( x3 − 4 x 2 )
y′(2) =
(−8)(1) − (3)(–4)
(−8)
2
The tangent line is y +
=
4
1
=
64 16
3 1
1
1
= ( x − 2) , or y = x − .
8 16
16
2
401
Chapter 11: Differentiation
55. y =
y′ =
x
2x − 6
(2 x − 6)(1) − x(2)
(2 x − 6)
2
ISM: Introductory Mathematical Analysis
500
q
r = pq = 500
60. p =
=
−6
(2 x − 6) 2
dr
=0
dq
1
1
If x = 1, then y =
= − and
2−6
4
−6
−6
3
y′ =
=
=− .
8
(−4)2 16
61. p =
3
y′ − 8 3
Thus
=
= = 1.5 .
y −1 2
4
56. y =
y′ =
r = pq =
1− x
1+ x
(1 + x)(−1) − (1 − x)(1)
(1 + x) 2
v=
2
=
=−
2
t +1
58. s =
v=
=
6
= −1.5 m/s.
4
ds (t 2 + 7)(1) − (t + 3)(2t )
=
dt
(t 2 + 7) 2
2
(t + 7)
2
=
(7 + t )(1 − t )
2
(t + 7)
q 2 + 750q
q + 50
(
)
2
dr (q + 50)(2q + 750) − q + 750q (1)
=
dq
(q + 50)2
q 2 + 100q + 37,500
(q + 50)2
63.
dC
= 0.672
dI
64.
dC
= 0.712
dI
t2 + 7
7 − 6t − t 2
−3
q + 750
q + 50
r = pq =
t +3
=
(q + 2)2
62. p =
ds (t 3 + 1)(0) − 2(3t 2 )
6t 2
=
=−
dt
(t 3 + 1)2
(t 3 + 1)2
If t = 1, then v = −
216
(1 + x )2
. When t = 1, then s = 1 m.
3
108q
− 3q
q+2
dr (q + 2)(108) − (108q)(1)
=
−3
dq
(q + 2) 2
1
y′ − 18 1
When x = 5, then
.
=
=
y − 2 12
3
57. s =
108
−3
q+2
65. C = 3 + I 1/ 2 + 2 I 1/ 3
dC
1
2
1
2
= 0 + I −1/ 2 + I −2 / 3 =
+
dI
2
3
2 I 33 I 2
dC 1 2 7
= + = .
When I = 1, then
dI 2 3 6
dS
dC
1
2
= 1−
= 1−
−
dI
dI
2 I 33 I 2
2
v = 0 when t = −7 or t = 1. Since t is positive, we
choose t = 1.
59. p = 50 − 0.01q
r = pq = 50q − 0.01q 2
dr
= 50 − 0.02q
dq
When I = 1, then 1 −
402
dC
7
1
= 1− = − .
dI
6
6
ISM: Introductory Mathematical Analysis
66.
Section 11.4
dC 3
1
= −
dI 4 6 I
dC
43
dS
43 17
=
, so
= 1−
=
dI I = 25 60
dI I = 25
60 60
dC
67.
=
dI
(
I +4
( )
)
) ( 8I + 1.2
I − 0.2 − ⎛⎜16 I + 0.8 I 3 − 0.2 I ⎞⎟ 1
⎝
⎠ 2 I
(
I +4
)
2
dS
dC
≈ 0.615 , so
≈ 1 − 0.615 = 0.385 when I = 36.
dI
dI I =36
dC
68.
=
dI
(
I +5
) ( 10I + 0.75
( )
)
I − 0.4 − ⎛⎜ 20 I + 0.5 I 3 − 0.4 I ⎞⎟ 1
⎝
⎠ 2 I
(
I +5
)
2
dS
dC
≈ 0.393 , so
≈ 1 − 0.393 = 0.607 when I = 100.
dI
dI I =100
1
69. Simplifying gives C = 10 + 0.7 I − 0.2 I 2
a.
b.
dC
0.1
−1
= 0.7 − 0.1I 2 = 0.7 −
dI
I
dS
dC
0.1
= 1−
= 0.3 +
dI
dI
I
dS
0.1
= 0.3 +
= 0.32
dI I = 25
5
dC
dI
C
when I = 25 is
0.7 − 0.1
5
10 + 0.7(25) − 0.2(5)
70. Simplifying S gives
S=
I − 2 I −8
(
I +2
)(
I −4
)=
I −4
I +2
I +2
dS 1 −1/ 2
1
Thus
.
= I
=
dI 2
2 I
dS
1
dC
=
≈ 0.04082 and
≈ 1 − 0.04082 ≈ 0.9592.
dI I =150 2 ⋅ 150
dI I =150
71.
=
≈ 0.026
dc
(q + 2)(2q ) − q 2 (1)
q 2 + 4q 6q(q + 4)
= 6⋅
= 6⋅
=
dq
(q + 2) 2
(q + 2) 2
(q + 2) 2
403
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
dc
dc
d ⎛ c ⎞ q ⋅ dq − c (1)
d
=
72. We assume that
(c ) = 0 . Thus 0 =
.
⎜ ⎟=
dq dq ⎝ q ⎠
dq
q2
This implies that q ⋅
dc
dc
dc c
−c = 0 , q⋅
= = c , so the marginal cost function
=c,
dq
dq
dq q
cost function ( c ) are equal.
⎛ dc ⎞
⎜ ⎟ and the average
⎝ dq ⎠
900 x
10 + 45 x
dy (10 + 45 x)(900) − (900 x)(45)
=
dx
(10 + 45 x )2
73. y =
dy
(100)(900) − (1800)(45) 9
=
=
dx x = 2
10
(100)2
0.05V
A + xV
d
( A + xV )(0.05) − (0.05V )( x)
(RT) =
dV
( A + xV ) 2
74. RT =
=
0.05 A
( A + xV ) 2
Both numerator and denominator are always positive, so
increases by one unit, RT increases.
0.7355 x
1 + 0.02744 x
dy (1 + 0.02744 x)(0.7355) − (0.7355 x)(0.02744)
=
dx
(1 + 0.02744 x )2
75. y =
=
76.
0.7355
(1 + 0.02744 x)2
a (1 + x) − b(2 + n) x
a (2 + n)(1 + x ) − b(2 + n) x
For convenience let c = 2 + n.
a (1 + x) − bcx 1 a (1 + x) − bcx
= ⋅
.
Then f ( x) =
ac(1 + x) − bcx c a(1 + x) − bx
f ( x) =
1 [a (1 + x) − bx](a − bc) − [a (1 + x) − bcx](a − b)
f ′( x) = ⋅
c
[a (1 + x) − bx]2
1
1
(−c + 1)ab
−abc + ab
= ⋅
= ⋅
c [a (1 + x) − bx]2 c [a(1 + x) − bx]2
=
1 [−1(2 + n) + 1]ab
−(1 + n)ab
⋅
=
2
2 + n [a(1 + x) − bx]
[a(1 + x) − bx]2 (2 + n)
g ( x) =
A + Bx
C + Dx
404
d
(RT) > 0 . This rate of change means that if V
dV
ISM: Introductory Mathematical Analysis
g ′( x) =
=
=
Section 11.5
(C + Dx )( B ) − ( A + Bx)( D)
(C + Dx) 2
CB + BDx + AD − BDx
(C + Dx)2
BC − AD
(C + Dx )2
C⎞
⎛
Thus, g ′( x) has the form given. When g ′( x) is defined ⎜ for x ≠ ⎟ , its sign is constant.
D⎠
⎝
dc
d ⎛c⎞
77.
=
⎜ ⎟=
dq dq ⎝ q ⎠
78.
dc − c(1)
q ⋅ dq
q
2
. When q = 20 we have
dc
dq
c
dc − c
q⋅ dq
=
q2
=
c
20(125) − 20(150)
(20)2
150
=−
1
120
dy
= (3)(2 x − 1)( x − 4) + (3 x + 1)(2)( x − 4) + (3 x + 1)(2 x − 1)(1)
dx
= 18 x 2 − 50 x + 3
Principles in Practice 11.5
1. By the chain rule,
dy dy dx d
d
=
⋅ =
4 x 2 ⋅ (6t ) = (8 x)(6) = 48 x .
dt dx dt dx
dt
dy
Since x = 6t,
= 48(6t ) = 288t .
dt
( )
Problems 11.5
(
)
(
)
1.
dy dy du
=
⋅
= (2u − 2)(2 x − 1) = ⎡⎢ 2 x 2 − x − 2 ⎤⎥ (2 x − 1) = 2 x 2 − 2 x − 2 (2 x − 1) = 4 x3 − 6 x 2 − 2 x + 2
⎣
⎦
dx du dx
2.
dy dy du
=
⋅
= 6u 2 − 8 7 − 3x 2 = 2 3x6 − 42 x 4 + 147 x 2 − 4 7 − 3 x 2
dx du dx
3.
dy dy dw ⎛ 2 ⎞
2
2
=
⋅
= ⎜−
⎟ (−1) = 3 =
3
dx dw dx ⎝ w ⎠
w
(2 − x)3
4.
5.
(
)(
) (
)(
)
dy dy dz 1 −3 / 4 4
5 x 4 − 4 x3
=
⋅ = z
(5 x − 4 x3 ) =
dx dz dx 4
4 ⎛⎜ 4 ( x5 − x 4 + 3)3 ⎞⎟
⎝
⎠
⎡ (t + 1) − (t − 1) ⎤
dw dw du
2 ⎤
1 −1
dw
⎡2⎤
2⎡
= 0, so
= 3(0) 2 ⎢ ⎥ = 0 .
=
⋅
= (3u 2 ) ⎢
⎥ = 3u ⎢
⎥ . If t = 1, then u =
2
2
dt du dt
+
1
1
dt
⎣4⎦
t =1
⎣⎢ (t + 1)
⎦⎥
⎣⎢ (t + 1) ⎦⎥
405
Chapter 11: Differentiation
6.
ISM: Introductory Mathematical Analysis
dz dz du ⎛
1 ⎞
=
⋅
= ⎜ 2u +
⎟ (4 s ) . If s = –1, then
ds du ds ⎝
2 u⎠
dz
⎛5⎞
u = 1, so
= ⎜ ⎟ (−4) = −10
ds s =−1 ⎝ 2 ⎠
14.
)
3
(
(
= −3 x 2 − 2
(
(
)
(
3
2
)
2
)
(
=
= 4(2 x + 1)( x 2 + x)3
)(
)
− 52
(
d 3
x − 8x2 + x
dx
(
1
2
)
(
)
−1
1
(10 x − 1) 5 x 2 − x 2
2
(
)
)
−1
1
5 x 2 − x 2 (10 x − 1)
2
20. y = 3x 2 − 7 = 3 x 2 − 7
y′ =
(
) (3x − 16 x + 1)
99
= 200 ( 3 x 2 − 16 x + 1)( x3 − 8 x 2 + x )
= 200 x − 8 x + x
99
−4
− 52
(
d 2
( x + x)
dx
= 4( x 2 + x)3 (2 x + 1)
2
)
) ( 7 − 4 x3 )
19. y = 5 x 2 − x = 5 x 2 − x
12. y ′ = 4( x 2 + x)3
3
)
(
= −6 7 − 4 x3 7 x − x 4
y′ =
⋅
(
d 2
x −2
dx
(2 x) = −6 x x 2 − 2
(
= 30 x 2 (3 + 2 x3 )
99
⋅
⎛ 3⎞
18. y′ = 4 ⎜ − ⎟ 7 x − x 4
⎝ 2⎠
3
d
(3 + 2 x3 )
dx
= 5(3 + 2 x3 ) 4 (6 x 2 )
)
−4
−4
d
(2 x3 − 8 x)
dx
= −12(6 x 2 − 8)(2 x3 − 8 x) −13
11. y ′ = 5(3 + 2 x3 ) 4 ⋅
(
)
)
3
d
⎛ 5⎞
17. y′ = 2 ⎜ − ⎟ ( x 2 + 5 x − 2)−12 / 7 ⋅ ( x 2 + 5 x − 2)
dx
⎝ 7⎠
10
2
−12 / 7
= − (2 x + 5)( x + 5 x − 2)
7
= 4 x − 4 (2 x) = 8 x x − 4
13. y′ = 2 ⋅100 x3 − 8 x 2 + x
)
16. y ′ = −12(2 x3 − 8 x) −13 ⋅
d
(3 x + 2)
dx
d 2
x −4
dx
)
15. y′ = −3 x 2 − 2
= 6(3x + 2)5 (3) = 18(3x + 2)5
(
)
(
)
10. y′ = 4 x 2 − 4 ⋅
(
4
1
2x2 + 1
2
2
3 d
1
2
y′ = ⋅ 4 2 x + 1
(2 x 2 + 1)
2
dx
=
= 2(2 x 2 + 1)3 (4 x) = 8 x 2 x 2 + 1
dy dy du
8.
=
⋅
= 9u 2 − 2u + 7 (5) . If x = 1, then
dx du dx
dy
u = 3, so
= (82)(5) = 410
dx x =1
9. y′ = 6(3 x + 2)5 ⋅
4
(
dy dy dw
7.
=
⋅
= (6w − 8)(4 x) . If x = 0, then
dx dw dx
dy
= 0.
dx
(
2 x 2 + 1)
(
y=
2
(
)
)
1
2
(
)
−1
−1
1
3 x 2 − 7 2 (6 x) = 3 x 3 x 2 − 7 2
2
1
21. y = 4 2 x − 1 = (2 x − 1) 4
y′ =
1
1
−3
−3
(2 x − 1) 4 (2) = (2 x − 1) 4
4
2
(
3
)
22. y = 8 x 2 − 1 = 8 x 2 − 1
y′ =
406
(
)
1
3
(
)
−2
−2
1
16
8 x 2 − 1 3 (16 x) = x 8 x 2 − 1 3
3
3
ISM: Introductory Mathematical Analysis
(
)
( )
−
−
12
⎛2⎞
y′ = 2 ⎜ ⎟ ( x3 + 1) ( 3 x 2 ) = x 2 ( x3 + 1)
5
⎝5⎠
23. y = 2 5 x3 + 1
2
= 2 x3 + 1
Section 11.5
2
5
3
5
6
2
2x − x +1
(
(
)
= 6 2x2 − x + 1
)
2
y′ = 6(−1) 2 x − x + 1
−2
(
)
= −6(4 x − 1) 2 x 2 − x + 1
26. y =
(
3
= 3 x4 + 2
4
x +2
(
y′ = 3(−1) x 4 + 2
27. y =
1
(
x 2 − 3x
)
(
2
28. y =
1
(2 + x)
4
32. y = 2 x +
)
= (2 x)
4
2
1
1
= (2 x) 2 + (2 x)
− 12
− 12
− (2 x)
− 32
33. y′ = x 2 ⎡5( x − 4) 4 (1) ⎤ + ( x − 4)5 (2 x)
⎣
⎦
−2
= x( x − 4)4 [5 x + 2( x − 4)]
= x( x − 4)4 (7 x − 8)
−2
34. y ′ = x ⎡ 4( x + 4)3 (1) ⎤ + ( x + 4)4 (1)
⎣
⎦
= ( x + 4)3 (4 x + x + 4) = ( x + 4)3 (5 x + 4)
−3
1
35. y = 4 x 2 5 x + 1 = 4 x 2 (5 x + 1) 2
−1
⎛1
⎞
y ′ = 4 x 2 ⎜ (5 x + 1) 2 (5) ⎟ + 5 x + 1(8 x)
2
⎝
⎠
= (2 + x) −4
= 10 x 2 (5 x + 1)
y ′ = −4(2 + x)−5 (1) = −4(2 + x)−5
29. y =
– 53
2x
1
−1
−3
⎛ ⎞
⎛ 1⎞
y′ = ⎜ ⎟ (2 x) 2 (2) + ⎜ − ⎟ (2 x) 2 (2)
⎝2⎠
⎝ 2⎠
(2 x − 3)
= −2(2 x − 3) x 2 − 3x
(6 x − 1)
1
7
−2
−2
y′ = (7 x) 3 (7) + 3 7(1) = (7 x) 3 + 3 7
3
3
−1
) ( 4x3 ) = −12 x3 ( x4 + 2)
−3
− 23
1
−2
)
)
5
3
)
31. y = 3 7 x + 3 7 x = (7 x) 3 + 3 7 x
−2
(
)
(
−1
2
2
3
= −2(6 x − 1) 3 x 2 − x
(4 x − 1)
= x − 3x
y′ = −2 x 2 − 3x
(
)
(
= 3 3x 2 − x
( 3x2 − x )
−
⎛ 2⎞
y′ = 3 ⎜ − ⎟ ( 3 x 2 − x )
⎝ 3⎠
3
5
24. y = 7 3 ( x5 − 3)5 = 7( x5 − 3)5 / 3
5
y ′ = 7 ⋅ ( x5 − 3) 2 / 3 (5 x 4 )
3
175 4 5
x ( x − 3) 2 / 3
=
3
25. y =
3
30. y =
= 4(9 x 2 + 1)−1/ 2
9x +1
⎛ 1⎞
y′ = 4 ⎜ − ⎟ (9 x 2 + 1) −3 / 2 (18 x )
⎝ 2⎠
= −36 x(9 x 2 + 1)−3 / 2
407
− 12
+ 8x 5x + 1
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
1
36. y = 4 x3 1 − x 2 = 4 x3 (1 − x 2 ) 2
⎡⎛ 1 ⎞
⎤
y′ = 4 x3 ⎢⎜ ⎟ (1 − x 2 ) −1/ 2 (−2 x) ⎥ + 1 − x 2 (12 x 2 )
⎣⎝ 2 ⎠
⎦
4
4x
=−
+ 12 x 2 1 − x 2
2
1− x
(
)
(
)
3
2
⎡
⎤
37. y′ = x 2 + 2 x − 1 (5) + (5 x) ⎢3 x 2 + 2 x − 1 (2 x + 2) ⎥
⎣
⎦
(
) (
)
2
= 5 ( x 2 + 2 x − 1) ( 7 x 2 + 8 x − 1)
2
= 5 x 2 + 2 x − 1 ⎡⎢ x 2 + 2 x − 1 + 3 x(2 x + 2) ⎤⎥
⎣
⎦
(
)( ) (
)
3
4
⎡
⎤
38. y′ = x 2 ⎢ 4 x3 − 1 3 x 2 ⎥ + x3 − 1 (2 x)
⎣
⎦
(
)
(
3
)⎤⎥⎦ = 2 x ( 7 x3 − 1)( x3 − 1)
3
= 2 x x3 − 1 ⎡⎢ 6 x3 + x3 − 1
⎣
39. y′ = (8 x − 1)3 ⎡ 4(2 x + 1)3 (2) ⎤ + (2 x + 1) 4 ⎡3(8 x − 1)2 (8) ⎤
⎣
⎦
⎣
⎦
= 8(8 x − 1) 2 (2 x + 1)3 [(8 x − 1) + 3(2 x + 1)]
= 8(8 x − 1) 2 (2 x + 1)3 (14 x + 2)
= 16(8 x − 1)2 (2 x + 1)3 (7 x + 1)
40. y ′ = (3x + 2)5 [2(4 x − 5)(4)] + (4 x − 5)2 [5(3 x + 2) 4 (3)]
= (3x + 2) 4 (4 x − 5)[8(3 x + 2) + 15(4 x − 5)]
= (3x + 2) 4 (4 x − 5)(84 x − 59)
11
⎛ x − 3 ⎞ ⎡ ( x + 2)(1) − ( x − 3)(1) ⎤
41. y′ = 12 ⎜
⎥
⎟ ⎢
⎝ x + 2 ⎠ ⎣⎢
( x + 2)2
⎦⎥
11 ⎡
5 ⎤
⎛ x −3 ⎞
= 12 ⎜
⎥
⎟ ⎢
⎝ x + 2 ⎠ ⎢⎣ ( x + 2) 2 ⎥⎦
60( x − 3)11
=
( x + 2)13
⎛ 2x ⎞
42. y′ = 4 ⎜
⎟
⎝ x+2⎠
3
1⎛ x−2⎞
43. y′ = ⎜
⎟
2⎝ x+3⎠
⎡ ( x + 2)(2) − 2 x(1) ⎤ 128 x3
⎢
⎥=
( x + 2)2
⎢⎣
⎥⎦ ( x + 2)5
− 12
⎡ ( x + 3)(1) − ( x − 2)(1) ⎤
⎢
⎥
( x + 3)2
⎣⎢
⎦⎥
⎛ x−2⎞
=
2 ⎜⎝ x + 3 ⎟⎠
2( x + 3)
5
− 12
=
5
2( x + 3)
2
x+3
x−2
408
ISM: Introductory Mathematical Analysis
1 ⎛ 8x2 − 3 ⎞
44. y′ = ⎜
⎟
3 ⎜⎝ x 2 + 2 ⎟⎠
1 ⎛ 8x2 − 3 ⎞
= ⎜
⎟
3 ⎜⎝ x 2 + 2 ⎟⎠
=
(
)
(
Section 11.5
)
⎡ 2
⎤
2
⎢ x + 2 (16 x) − 8 x − 3 (2 x) ⎥
⎢
⎥
2
⎢
⎥
x2 + 2
⎣
⎦
− 23
− 23
(
)
38 x
( x2 + 2)
2
38 x
(
3 8x2 − 3
) ( x2 + 2)
2
3
( x2 + 4)
3
45. y′ =
4
3
(
⎡
(2) − (2 x − 5) ⎢3 x 2 + 4
⎣
)
2
⎤
(2 x ) ⎥
⎦
( x2 + 4)
2
x 2 + 4 ) {( x 2 + 4 ) (2) − (2 x − 5)[3(2 x)]}
(
=
6
( x2 + 4)
=
=
2 x 2 + 8 − 12 x 2 + 30 x
(
=
=
47. y′ =
=
( x2 + 4)
4
−2 5 x 2 − 15 x − 4
46. y ′ =
=
6
( x2 + 4)
=
−10 x 2 + 30 x + 8
)
( x2 + 4)
4
4
(3x 2 + 7)[4(4 x − 2)3 (4)] − (4 x − 2)4 (6 x)
(3x 2 + 7) 2
3
(4 x − 2) [16(3x 2 + 7) − 6 x (4 x − 2)]
(3 x 2 + 7)2
(4 x − 2)3 (24 x 2 + 12 x + 112)
(3 x 2 + 7)2
(3 x − 1)3 ⎡5(8 x − 1)4 (8) ⎤ − (8 x − 1)5 ⎡3(3x − 1)2 (3) ⎤
⎣
⎦
⎣
⎦
6
(3 x − 1)
(3 x − 1)2 (8 x − 1) 4 [(3x − 1)(40) − (8 x − 1)(9)]
(3 x − 1)6
(8 x − 1)4 (48 x − 31)
(3x − 1) 4
409
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
48. y = 3 ( x − 2) 2 ( x + 2) = [( x − 2) 2 ( x + 2)]1/ 3
1
y′ = [( x − 2) 2 ( x + 2)]−2 / 3 [(1)( x − 2)2 + 2( x − 2)( x + 2)]
3
1
= [( x − 2) 2 ( x + 2)]−2 / 3 ( x − 2)[ x − 2 + 2( x + 2)]
3
1
= [( x − 2) 2 ( x + 2)]−2 / 3 ( x − 2)(3 x + 2)
3
1
= ( x − 2)−1/ 3 ( x + 2)−2 / 3 (3 x + 2)
3
(
49. y = 6 5 x 2 + 2
1
⎡
⎤
x4 + 5 = 6 ⎢ 5x2 + 2 x4 + 5 2 ⎥
⎣
⎦
)
(
)(
)
1
−1
⎡
⎤
1
y′ = 6 ⎢ 5 x 2 + 2 ⋅ x 4 + 5 2 4 x3 + x 4 + 5 2 (10 x) ⎥
2
⎣
⎦
1
1
−
⎡
⎤
= 6 ⎢ 5 x 2 + 2 x 4 + 5 2 2 x3 + x 4 + 5 2 (10 x) ⎥
⎣
⎦
1
1
−
⎡ 2
⎤
= 12 x ⎢ 5 x + 2 x 4 + 5 2 x 2 + x 4 + 5 2 (5) ⎥
⎣
⎦
(
) (
(
)(
(
) ( ) (
) ( ) (
)(
)
) ( ) (
(
Factoring out x 4 + 5
(
)
)
− 12
)
gives
) ⎡⎢⎣(5x2 + 2)( x2 ) + ( x4 + 5) (5)⎤⎥⎦
−
= 12 x ( x 4 + 5 ) (10 x 4 + 2 x 2 + 25 )
y′ = 12 x x 4 + 5
− 12
1
2
50. y′ = 3 − 4 ⎡ x(2)(7 x + 1)(7) + (7 x + 1)2 (1) ⎤
⎣
⎦
= 3 − 4 ⎡147 x 2 + 28 x + 1⎤ = −588 x 2 − 112 x − 1
⎣
⎦
51. y′ = 8 +
= 8+
(t + 4)(1) − (t − 1)(1)
(t + 4)
5
(t + 4)
2
2
⎛ 8t − 7 ⎞⎛ 1 ⎞
− 2⎜
⎟⎜ ⋅ 8 ⎟
⎝ 4 ⎠⎝ 4 ⎠
− (8t − 7) = 15 − 8t +
5
(t + 4)2
410
ISM: Introductory Mathematical Analysis
52. y =
y′ =
=
=
=
53. y′ =
(2 x3 + 6)(7 x − 5)
(2 x + 4) 2
=
Section 11.5
14 x 4 − 10 x3 + 42 x − 30
(2 x + 4)2
(2 x + 4)2 (56 x3 − 30 x 2 + 42) − (14 x 4 − 10 x3 + 42 x − 30)[2(2 x + 4)(2)]
(2 x + 4) 4
(2 x + 4)[(2 x + 4)(56 x3 − 30 x 2 + 42) − 4(14 x 4 − 10 x3 + 42 x − 30)]
(2 x + 4) 4
112 x 4 − 60 x3 + 84 x + 224 x3 − 120 x 2 + 168 − 56 x 4 − 40 x3 − 168 x + 120
(2 x + 4)3
4(14 x 4 + 51x3 − 30 x 2 − 21x + 72)
(2 x + 4)3
( x3 − 5)5 [(2 x + 1)3 (2)( x + 3)(1) + ( x + 3)2 (3)(2 x + 1)2 (2)] − (2 x + 1)3 ( x + 3)2 [5( x3 − 5)4 (3 x 2 )]
( x3 − 5)10
(
)
(
) ( 12 ) ( x + 2)−
2
⎡
(9 x − 3) ⎢ x + 2(2) 4 x 2 − 1 (8 x) + 4 x 2 − 1
⎣
54. y′ =
(9 x − 3)2
55.
(
)
dy
= 0.
dx
dz dz dy dx
=
⋅ ⋅ = (4 y − 4)(6)(2)
dt dy dx dt
When t = 1, then x = 2 and y = 7. Thus
(
57. y′ = 3 x 2 − 7 x − 8
)
2
dz
= (24)(6)(2) = 288 .
dt t =1
(2 x − 7)
If x = 8, then slope = y′ = 3(64 − 56 − 8) 2 (16 − 7) = 0 .
1
58. y = ( x + 1) 2
1
−1
( x + 1) 2
2
1
If x = 8, y′ = .
6
y′ =
411
(
)
2
⎤
2
⎥ − x + 2 4 x − 1 (9)
⎦
3
dy dy du ⎡
⎡
⎤
=
⋅
= 3(5u + 6)2 (5) ⎤ ⎢ 4 x 2 + 1 (2 x) ⎥
⎣
⎦
dx du dx
⎣
⎦
When x = 0, then
56.
1
2
Chapter 11: Differentiation
(
)
59. y = x 2 − 8
y′ =
ISM: Introductory Mathematical Analysis
2
3
(
2 2
x −8
3
64. y =
)
− 13
4x
(2 x) =
(
3 x2 − 8
)
If x = −1, y ′ = 3(2)2 = 12.
=
( x + 1)
( ) (7 x + 2)
m = 6, then q = 30, so
(7) − 7 x + 2(1)
Also,
( x + 1) 2
( 72 )
1
−
7 x+2
2
( x + 1)
If x = 1, then y′ =
7x + 2
( )( ) − 3 = − 1 . The
2
1
3
4
6
3
1
tangent line is y − = − ( x − 1) , or
2
6
1
5
y = − x+ .
6
3
(
)
62. y = −3 3 x 2 + 1
(
)
−4
(
)
3
(
)
2
(
)
dr
= −0.2q + 70 . If
dq
dq
dq
1
=6.
=
(200 − 2m) . When m = 40,
dm
dm 20
dr
Thus
= (6)(6) = 36 .
dm m = 40
(6 x)
and y′ = 6 x x 2 + 9
dq
dr
= 5. Thus
= (26)(5) = 130.
dm
dm m =6
m = 40, then q = 320, so
dr
= −64 + 70 = 6 .
dq m = 40
If x = 0, then y′ = 0 . The tangent line is
y + 3 = 0(x – 0), or y = –3.
63. y = x 2 + 9
= −24 + 50 = 26.
m =6
r = pq = −0.1q 2 + 70q , so
−3
y′ = −3(−3) 3x 2 + 1
dr
dq
dr
= −0.8q + 50, . For
dq
1
200m − m 2
20
p = –0.1q + 70; m = 40
dr dr dq
=
⋅
dm dq dm
66. q =
7
2
( x − 1)4
1
12
4
and y ′ = −
= − , so
4
27
27
3
r = pq = −0.4q 2 + 50q,
The tangent line is y − 8 = 12(x + 1) or
y = 12x + 20.
( x + 1)
6x
2
65. q = 5m, p = –0.4q + 50; m = 6
dr dr dq
=
⋅
dm dq dm
60. y ′ = 3( x + 3)2 (1) = 3( x + 3) 2
61. y′ =
and y ′ = −
⎛ y′ ⎞
4
⎜ ⎟ (100) = − ⋅ 27(100) = −400%
y
27
⎝ ⎠
12
= 4 . Thus the tangent line
3(1)
is y – 1 = 4(x – 3), or y = 4x – 11.
1
2
( x − 1)
3
When x = 2, y =
1
3
If x = 3, then y′ =
− 12
1
2
. When
67. q =
10m 2
m2 + 9
525
;m=4
p=
q+3
x = 4, then y = (25)3 and y′ = 6(4)(25) 2 , so
y′
6(4)(25)2
24
(100) =
(100) =
(100) = 96%
3
y
25
(25)
dr dr dq
=
⋅
dm dq dm
r = pq =
525q
, so
q+3
dr
(q + 3)(1) − q(1)
1575
.
= 525 ⋅
=
2
dq
(q + 3)
(q + 3)2
412
ISM: Introductory Mathematical Analysis
dr
dq
If m = 4, then q = 32, so
(
2
m +9
dq
=
dm
m2 + 9 )
(
=
=
)
1
2
=
m=4
(20m) − 10m2 ⋅ 12
2
Section 11.5
1575 9
= .
1225 7
(m
2
+9
)
− 12
b.
(
=−
)
⎡ 20m m 2 + 9 − 10m3 ⎤
⎢⎣
⎥⎦
2
m +9
3
10m + 180m
(
m2 + 9
)
=−
c.
3
2
70. p =
q
q 2 + 20 ⎛⎜ 100 − q 2 + 20 ⎞⎟
⎝
⎠
q
100 q 2 + 20 − q 2 − 20
r = pq = 100q − q q 2 + 20
dr
45, 000
4500q
, so
.
=
dq (q + 10) 2
q + 10
dr
dq
If m = 9, then q = 90, so
( m2 + 19)
3
2
=
m =9
71.
9
.
2
. When m = 9, then
dq 19
.
=
dm 10
72.
dr
9 19
= ⋅ = 8.55 .
dm m =9 2 10
(
dp
1
= 0 − q 2 + 20
dq
2
2
q + 20
− 12
⎤
(2q) + q 2 + 20(1) ⎥
⎦
− q 2 + 20
k
; q = f(m)
q
r = pq = k, so
1900
q2
)
dr dr dq
=
⋅
dm dq dm
dr dr dq
=
⋅
dm dq dm
69. a.
100 − q 2 + 20
= 100 −
m 2 + 19
4500
;m=9
p=
q + 10
Thus
q 2 + 20
(
100m
dq
=
dm
=
⎡ 1
= 100 − ⎢ q ⋅ q 2 + 20
⎣ 2
dr
9 272
= ⋅
≈ 13.99 .
dm m = 4 7 25
r = pq =
−q
dr
dq
When m = 4, then
dq 10(64) + 180(4) 1360 272
. Thus
=
=
=
3
125
25
dm
(25) 2
68. q =
p
(2m)
m +9
− 12
dp
dq
)
− 12
(2q ) =
dr
dq
dr
= 0 . Thus
= 0⋅
=0.
dm
dm
dq
dc dc dq
=
⋅
= (12 + 0.4q )(−1.5)
dp dq dp
When p = 85, then q = 772.5, so
dc
= −481.5.
dp p =85
⎛ 250 ⎞
f (t ) = 1 − ⎜
⎟
⎝ 250 + t ⎠
3
2
250 ⎤
⎛ 250 ⎞ ⎡
f ′(t ) = −3 ⎜
⎥
⎟ ⎢−
⎝ 250 + t ⎠ ⎣⎢ (250 + t )2 ⎦⎥
−q
2
⎛ 250 ⎞ ⎡ 250 ⎤
f ′(100) = −3 ⎜
⎟ ⎢−
⎥
⎝ 350 ⎠ ⎣ 3502 ⎦
⎛ 25 ⎞ ⎛ 1 ⎞
= −3 ⎜ ⎟ ⎜ −
⎟
⎝ 49 ⎠ ⎝ 490 ⎠
15
.
=
4802
Thus when t increases from 100 to 101, the
proportion discharged increases by
15
.
approximately
4802
2
q + 20
413
Chapter 11: Differentiation
73.
dc
=
dq
( q 2 + 3)
1
2
ISM: Introductory Mathematical Analysis
⎡
(10q ) − 5q 2 ⎢ 12 q 2 + 3
⎣
q2 + 3
( ) (
)
− 12
⎤
(2q ) ⎥
⎦
(
Multiplying numerator and denominator by q 2 + 3
(
)
(
(
)
1
2
gives
)
q 2 + 3 (10q) − 5q 2 (q ) 5q3 + 30q 5q q 2 + 6
dc
.
=
=
=
3
3
3
dq
2
2
2
2
2
2
q +3
q +3
q +3
74. a.
(
)
)
(
)
dS
dS
= 680 E − 4360 . If E = 16,
= 6520 .
dE
dE
b. Solving 680 E − 4360 = 5000 gives 680 E = 9360, E ≈ 13.8.
75.
(
)
( )
dV dV dr
=
⋅ = 4πr 2 ⎡10−8 (2t ) + 10−7 ⎤ . When t = 10, then r = 10−8 102 + 10−7 (10) = 10−6 + 10−6 = 2(10)−6 .
⎣
⎦
dt
dr dt
Thus
2
dV
= 4π ⎡ 2(10) −6 ⎤ ⎡10−8 (2)(10) + 10−7 ⎤ = 4π ⎡ 4(10)−12 ⎤ ⎡⎢3 10−7 ⎤⎥ = 48π(10)−19
⎣
⎦
⎣
⎦
⎣
⎦⎣
⎦
dt t =10
(
76. a.
b.
77. a.
)
dp 1
−1
−1
= (2 ρVI ) 2 (2 ρV ) = ρV (2 ρVI ) 2
dI 2
dp
dI
p
=
ρV (2 ρVI )
(2 ρVI )
− 12
1
2
=
1
2I
d
( I x ) = −0.001416 x3 + 0.01356 x3 + 1.696 x − 34.9
dx
d
( I x ) = −256.238.
If x = 65,
dx
b. If x = 65,
d (I )
dx x
Ix
≈
−256.238
≈ −0.01578
16, 236.484
If x = 65, the percentage rate of change is
d (I )
dx x
Ix
⋅=
78. (P + a)(v + b) = k
k
v+b =
P+a
k
v=
−b
P+a
v = k ( P + a ) −1 − b
dv
k
= k (−1)( P + a )−2 = −
dP
( P + a)2
414
−25, 623.8
= −1.578%.
16, 236.484
ISM: Introductory Mathematical Analysis
79. By the chain rule,
Section 11.5
dq
−100
dc dc dq
100
. We are given that q =
= 100 p −1 , so
. Thus
=
⋅
= −100 p −2 =
dp
dp dq dp
p
p2
100 1
dc
dc dc ⎡ −100 ⎤
= and we are given that
= 0.01 . Therefore
=
⎢ 2 ⎥ . When q = 200, then p =
dp dq ⎢⎣ p ⎥⎦
200 2
dq
⎡
⎤
dc
−100 ⎥
⎢
= 0.01 ⎢
= −4 .
2 ⎥
dp
⎢⎣ 12 ⎥⎦
( )
80. a.
b.
When m = 12, then q = 3000, so r = 1500.
r 1500 1
Thus p = =
= = $0.50 .
q 3000 2
( )
1000 + 3q (50) − 50q 12 (1000 + 3q)
dr
=
1000 + 3q
dq
dr
dq
c.
=
q =3000
− 12
(3)
2750
11
=
10, 000 40
dr dr dq
dr
. From part (b) we know
. Now,
=
⋅
dm dq dm
dq
3
1
dq
dq
⎛3⎞
= 610 .
= (2m) ⎜ ⎟ (2m + 1) 2 (2) + (2m + 1) 2 (2) , so
dm
dm m =12
⎝2⎠
Thus
81.
dr
11
671
=
⋅ 610 =
.
dm m =12 40
4
dy dy dx
=
⋅ = f ′( x) g ′(t ) . We are given that g(2) = 3, so x = 3 when t = 2. Thus
dt dx dt
dy
dy
dx
=
⋅
= f ′(3) g ′(2) = 10(4) = 40 .
dt t = 2 dx x = g (2) dt t = 2
82. a.
b.
⎛ 324
5 19 ⎞
19 19
+ + ⎟ = 0+0+ =
lim c = lim ⎜
⎜
⎟
2
q 18
18 18
q →∞
q →∞
⎝ q + 35
⎠
c = cq =
dc
=
dq
dc
dq
c.
324q
q 2 + 35
+5+
19
q
18
q 2 + 35(324) − 324q
( 12 ) ( q2 + 35)
q 2 + 35
− 12
(2q ) 19
+
18
=3
q =17
From part (b) the increase in cost of the additional unit is approximately $300. Since the corresponding
revenue increases by $275, the move should not be made.
415
Chapter 11: Differentiation
ISM: Introductory Mathematical Analysis
83. 86,111.37
84. 5.25
Chapter 11 Review Problems
1.
f ( x) = 2 − x 2
(
⎡ 2 − ( x + h) 2 ⎤ − 2 − x 2
f ( x + h) − f ( x )
⎦
f ′( x) = lim
= lim ⎣
h
h
h →0
h →0
⎡ 2 − x 2 − 2hx − h 2 ⎤ − 2 − x 2
−2hx − h 2
⎦
= lim ⎣
= lim
h
h
h→0
h →0
−h(2 x + h)
= lim
= lim − (2 x + h) = −2 x
h
h→0
h →0
(
2.
)
f ( x) = 2 x 2 − 3 x + 1
f ′( x) = lim
h →0
f ( x + h) − f ( x )
h
(
)
⎡ 2( x + h) 2 − 3( x + h) + 1⎤ − 2 x 2 − 3x + 1
⎦
= lim ⎣
h
h→0
⎡ 2 x 2 + 4hx + 2h 2 − 3 x − 3h + 1⎤ − 2 x 2 − 3x + 1
⎦
= lim ⎣
h
h→0
(
)
4hx + 2h 2 − 3h
h(4 x + 2h − 3)
= lim
h
h
h→0
h→0
= lim (4 x + 2h − 3) = 4 x − 3
= lim
h→0
3.
f ( x) = 3x
f ′( x) = lim
h →0
3( x + h) − 3x 3( x + h) + 3 x
⋅
h
3( x + h) + 3 x
= lim
h→0
= lim
h→0 h
= lim
h→0
=
3( x + h) − 3x
f ( x + h) − f ( x )
= lim
h
h
h →0
(
3( x + h) − 3 x
3( x + h) + 3x
)
= lim
h→0 h
(
3h
3( x + h) + 3x
)
3
3( x + h) + 3x
3
3x + 3x
=
3
2 3x
=
3
2 x
416
)
ISM: Introductory Mathematical Analysis
4.
f ( x) =
Chapter 11 Review
2
1 + 4x
2
2
−
f ( x + h) − f ( x )
1+ 4( x + h ) 1+ 4 x
= lim
f ′( x) = lim
h
h
h →0
h →0
2(1 + 4 x) − 2[1 + 4( x + h)]
= lim
h→0 h[1 + 4( x + h)](1 + 4 x)
−8h
h→0 h[1 + 4( x + h)](1 + 4 x)
= lim
−8
−8
=
x
h
x
x
[1
+
4(
+
)](1
+
4
)
[1
+
4(
)](1 + 4 x)
h→0
= lim
=−
8
(1 + 4 x)2
5. y is a constant function, so y′ = 0 .
6. y′ = e(1) x1−1 = ex0 = e
( ) ( )
7. y′ = 7 4 x3 − 6 3 x 2 + 5(2 x) + 0
(
= 28 x3 − 18 x 2 + 10 x = 2 x 14 x 2 − 9 x + 5
)
8. y′ = 4(2 x + 0) − 7(1) = 8 x − 7
9.
(
)
f ( s ) = s 2 s 2 + 2 = s 4 + 2s 2
(
)
f ′( s ) = 4s 3 + 2(2 s ) = 4 s3 + 4 s = 4 s s 2 + 1
1
10. y = ( x + 3) 2
y′ =
1
1
−1
−1
( x + 3) 2 (1) = ( x + 3) 2
2
2
(
)
1 2
x +1
5
1
2x
y′ = (2 x) =
5
5
11. y =
12. y = −
2
2x
2
= − x −2 , so y ′ = −1(−2) x −3 = 2 x −3 .
13. y ′ = ( x3 + 7 x 2 )(3x 2 − 2 x) + ( x3 − x 2 + 5)(3 x 2 + 14 x)
= 3 x5 + 19 x 4 − 14 x3 + 3x5 + 11x 4 − 14 x3 + 15 x 2 + 70 x
= 6 x5 + 30 x 4 − 28 x3 + 15 x 2 + 70 x
(
)
14. y′ = x 2 + 1
100
(
)
(1) + ( x − 6)(100) x 2 + 1
99
(
) ( 201x2 − 1200 x + 1)
(2 x) = ( x 2 + 1)99 [ x 2 + 1 + 200 x( x − 6)] = x 2 + 1
417
99
Chapter 11: Differentiation
(
ISM: Introductory Mathematical Analysis
)
99
(4 x + 4) = 400( x + 1)[(2 x)( x + 2)]99
15.
f ′( x) = 100 2 x 2 + 4 x
16.
f ( w) = w w + w2 = w 2 + w2 f ′( w) =
3
3 12
w + 2w
2
17. y = 3(2 x + 1)−1
y′ = 3(−1)(2 x + 1) −2 (2) = −
6
(2 x + 1)2
5x2 − 8x 5
= x−4
2x
2
5
y′ =
2
18. y =
(
)
(
)
3
4
⎡
⎤
19. y′ = (8 + 2 x) ⎢ (4) x 2 + 1 (2 x) ⎥ + x 2 + 1 (2)
⎣
⎦
( )
( )
3
= 2 ( x 2 + 1) ( 32 x + 8 x 2 + x 2 + 1)
3
= 2 ( x 2 + 1) ( 9 x 2 + 32 x + 1)
3
= 2 x 2 + 1 ⎡⎢ 4 x(8 + 2 x) + x 2 + 1 ⎤⎥
⎣
⎦
6
−2
−2
⎛3⎞
20. g ′( z ) = ⎜ ⎟ (2 z ) 5 (2) + 0 = (2 z ) 5
5
⎝5⎠
21.
z 2 + 4 ) (2 z ) − ( z 2 − 1) (2 z )
(
10 z
f ′( z ) =
=
2
2
( z2 + 4)
( z2 + 4)
22. y′ =
( x + 2)2 (1) − ( x − 5)(2)( x + 2)
( x + 2)
4
=
12 − x
( x + 2)3
1
23. y = (4 x − 1) 3
1
4
−2
−2
y′ = (4 x − 1) 3 (4) = (4 x − 1) 3
3
3
24. f is a constant function, so f ′( x) = 0.
25. y = (1 − x 2 )
− 12
−3
−3
⎛ 1⎞
y′ = ⎜ − ⎟ (1 − x 2 ) 2 (−2 x) = x(1 − x 2 ) 2
⎝ 2⎠
418
ISM: Introductory Mathematical Analysis
26. y =
Chapter 11 Review
x2 + x
2 x2 + 3
2 x 2 + 3) (2 x + 1) − ( x 2 + x ) (4 x) −2 x 2 + 6 x + 3
(
y′ =
=
2
2
2
2
x
+
3
(
)
( 2 x 2 + 3)
2
27. h′( x) = ( x − 6) 4 ⎢⎡3 ( x + 5 ) ⎥⎤ + ( x + 5)3 ⎡ 4( x − 6)3 ⎤
⎣
⎦
⎣
⎦
= ( x − 6)3 ( x + 5)2 [3( x − 6) + 4( x + 5)]
= ( x − 6)3 ( x + 5)2 (7 x + 2)
28. y′ =
29. y′ =
30.
x(5)( x + 3) 4 − ( x + 3)5 (1)
=
x2
( x + 6)(5) − (5 x − 4)(1)
( x + 6)
2
=
( x + 3)4 (4 x − 3)
x2
34
( x + 6)2
f ( x) = 5 x3 3 + 2 x 4 = 5 x3 (3 + 2 x 4 )1/ 2
⎡1
⎤
f ′( x) = (3 + 2 x 4 )1/ 2 (15 x 2 ) + 5 x3 ⎢ (3 + 2 x 4 ) −1/ 2 (8 x3 ) ⎥
2
⎣
⎦
= 15 x 2 (3 + 2 x 4 )1/ 2 + 20 x6 (3 + 2 x 4 )−1/ 2
3 − 11 3 ⎛ − 11 ⎞ − 11
− 11
⎛ 3 ⎞ − 11 ⎛ 3 ⎞
31. y′ = 2 ⎜ − ⎟ x 8 + ⎜ − ⎟ (2 x) 8 (2) = − x 8 − ⎜ 2 8 ⎟ x 8
4
4⎝
⎠
⎝ 8⎠
⎝ 8⎠
11
11
11
11
3 − ⎛
3⎛
− ⎞
− ⎞ −
= − x 8 ⎜1 + 2 8 ⎟ = − ⎜1 + 2 8 ⎟ x 8
4
4⎝
⎝
⎠
⎠
1⎛ x⎞
32. y′ = ⎜ ⎟
2⎝2⎠
− 12
⎛1⎞ 1⎛2⎞
⎜ ⎟+ ⎜ ⎟
⎝2⎠ 2⎝ x⎠
− 12
(
1
)
1 ⎛ 2 ⎞2 1 ⎛ 2 ⎞
−2 x −2 = ⎜ ⎟ − ⎜ ⎟
4⎝ x⎠
x2 ⎝ x ⎠
− 12
⎛2⎞
=⎜ ⎟
⎝ x⎠
1
1 ⎤ x−2 x
⎛ x ⎞2 ⎡ 1
=⎜ ⎟ ⎢ − ⎥=
⎝ 2 ⎠ ⎣ 2 x x2 ⎦ 2 x2 2
33.
x2 + 5)
(
y′ =
1
2
(
(2 x) − x 2 + 6
) ( 12 ) ( x2 + 5)
− 12
(2 x)
x2 + 5
(
Multiplying the numerator and denominator by x 2 + 5
x 2 + 5 ) (2 x) − x ( x 2 + 6 )
x ( x2 + 4 )
(
x3 + 4 x
=
=
y′ =
2
2
( x + 5)
( x + 5) ( x 2 + 5)
3
2
3
2
)
3
2
419
1
2
gives
− 12
⎡1 ⎛ 2 ⎞ 1 ⎤
⎢4 ⎜ x ⎟ − 2 ⎥
⎣ ⎝ ⎠ x ⎦
Chapter 11: Differentiation
(
34. y = 7 − 3 x 2
)
2
3
(
2
y′ = 7 − 3 x 2
3
35. y′ =
ISM: Introductory Mathematical Analysis
1
41. y = x 3
)
− 13
(
(
( − 6 x ) = −4 x 7 − 3 x
3 3
x + 6 x2 + 9
5
(
)
)
y′ =
1
2 −3
1
. An
12
1
equation of the tangent line is y − 2 = ( x − 8) ,
12
1
4
or y = x + .
12
3
When x = 8, then y = 2 and y′ =
) (3x2 + 12 x )
− 52
2
−
3 3
x + 6 x 2 + 9 5 (3x)( x + 4)
5
−2
9
= x( x + 4) x3 + 6 x 2 + 9 5
5
=
(
)
x2
x − 12
42. y =
36. z ′ = 0.4[ x 2 (−3)( x + 1) −4 (1) + ( x + 1) −3 (2 x)] + 0
= 0.4( x + 1)−4 [−3x 2 + ( x + 1)(2 x)]
y′ =
37. g ( z ) = − z ( z − 1)2 = − z 3 + 2 z 2 − z
g ′( z ) = −3z 2 + 4 z − 1
(
=
(
)
5
( z5 + 2 z − 5)
43.
)
−4
3 5
z + 2z − 5
4
3
g ′( z ) = − (−4) z 5 + 2 z − 5
4
(
( x − 12)(2 x) − x 2 (1)
=
x 2 − 24 x
( x − 12) 2
( x − 12)2
When x = 13, then y = 169 and y ′ = −143. An
equation of the tangent line is
y − 169 = −143(x − 13) or y = −143x + 2028.
= 0.4( x + 1)−4 (− x 2 + 2 x)
38. g ( z ) = −
1 − 23
x
3
) (5z 4 + 2)
−5
f ( x) = 4 x 2 + 2 x + 8
f ′( x) = 8 x + 2
f(1) = 14 and f ′(1) = 10 . The relative rate of
f ′(1) 10 5
=
= ≈ 0.714 , so the
f (1) 14 7
percentage rate of change is 71.4%.
change is
3 5z 4 + 2
44.
39. y = x 2 − 6 x + 4
y′ = 2 x − 6
When x = 1, then y = –1 and y′ = −4 . An
equation of the tangent line is
y – (–1) = –4(x – 1), or y = –4x + 3.
f ( x) =
f ′( x) =
x
x+4
( x + 4)(1) − x(1)
( x + 4)
2
=
4
( x + 4)2
1
4
and f ′(1) =
. The relative rate of
5
25
f ′(1) 4
change is
= = 0.8 , so the percentage rate
f (1) 5
of change is 80%.
f (1) =
40. y = −2 x3 + 6 x + 1
y′ = −6 x 2 + 6
When x = 2, then y = –3 and y′ = −18 . An
equation of the tangent line is
y – (–3) = –18(x – 2), or y = –18x + 33.
45. r = q (20 − 0.1q) = 20q − 0.1q 2
dr
= 20 − 0.2q
dq
46.
dc
= 0.0003q 2 − 0.04q + 3
dq
dc
dq
420
=2
q =100
ISM: Introductory Mathematical Analysis
47.
Chapter 11 Review
1
dC
⎛ 1 ⎞ −1
= 0.6 − 0.25 ⎜ ⎟ I 2 = 0.6 −
2
dI
I
8
⎝ ⎠
dC
≈ 0.569
dI I =16
Thus the marginal propensity to consume is
0.569, so the marginal propensity to save is
1 – 0.569 = 0.431.
54. y = 12 −
dy
36
= −12(−1)(1 + 3 x)−2 (3) =
dx
(1 + 3x)2
36
1
Setting
= gives (1 + 3x) 2 = 108,
2
3
(1 + 3 x)
1 + 3 x = ±6 3, x =
dp (q + 5)(1) − (q + 12)(1)
7
=
=−
48.
2
dq
(q + 5)
(q + 5)2
dr
= 500 − 0.2q.
dq
dt
dT
d
dT
55. a.
3
50. Since c = 0.03q + 1.2 + , then
q
c = qc = 0.03q 2 + 1.2q + 3 . Thus
dc
dc
= 0.06q + 1.2 , so
dq
dq
51.
= 7.2 .
q =100
when T = 38 is
175 ⎤
4
4
⎡4
=
= .
⎢3T − 4 ⎥
⎣
⎦ T =38 3 T =38 3
when T = 35 is
11 ⎤
1
1
⎡1
=
=
.
⎢ 24 T + 4 ⎥
⎣
⎦ T =35 24 T =35 24
(
56. s = 9 2t 2 + 3
= 0.7396
v=
q = 70
52. q = 50m − m2
p = –0.01q + 9; m = 10
dr dr dq
=
⋅
dm dq dm
)
−1
(
ds
= −9 2t 2 + 3
dt
If t = 1, then v = −
57. V ′ =
)
−2
(4t ) =
−36t
( 2t 2 + 3)
2
36
m/s.
25
ft 3
1 2
πd . If d = 4 ft, then V ′ = 8π
.
ft
2
r = pq = −0.01q 2 + 9q , so
dr
= −0.02q + 9 .
dq
58. v = 128 – 32t. Set 128 – 32t = 64 to get t = 2.
If m = 10, then q = 400, so
dr
dq
59. c = cq = 2q 2 +
= −8 + 9 = 1 .
m =10
dq
dq
= 50 − 2m . When m = 10,
= 30 .
dm
dm
dr
Thus
= (1)(30) = 30 .
dm m =10
53.
dt
dT
d
dT
b.
dc
= 0.125 + 0.00878q
dq
dc
dq
−1 ± 6 3
, x ≈ 3.13 or
3
x ≈ −3.80.
Because we must have x ≥ 0, then x ≈ 3.13.
49. Since p = −0.1q + 500, then
r = pq = −0.1q 2 + 500q. Thus
12
1 + 3x
10, 000
= 2q 2 + 10, 000q −1
q
dc
10, 000
= 4q − 10, 000q −2 = 4q −
dq
q2
dy
= 42 x 2 − 34 x − 16
dx
dy
= 84 eggs/mm
dx x = 2
421
Chapter 11: Differentiation
60. y =
( x3 + 2) x + 1
x4 + 2 x
=
ISM: Introductory Mathematical Analysis
( x3 + 2) x + 1
x( x3 + 2)
=
65. –0.32
x +1
x
Mathematical Snapshot Chapter 11
−1
⎛
⎞
x ⎜ 12 ( x + 1) 2 (1) ⎟ − x + 1(1)
dy
⎠
= ⎝
dx
x2
dy
3
=−
2 and y = 2 when x = 1. An
dx x =1
4
equation of the tangent line is
3
3
7
y− 2 =−
2( x − 1) or y = −
2x +
2.
4
4
4
61. a.
1. In Problems 63 and 64 of Sec. 11.4, the slope is
≈ 0.7 . In Fig. 11.15 the slope is above 0.9. More
is spent; less is saved.
2. In the lowest quintile, the average family spends
more than it earns, thus accumulating debt.
3. The slope of the family consumption curve is
112, 040
, which for
1.9667 × 1010 + 224, 080x
x = 25,000 equals about 0.705. You would
expect the family to spend $705 and save $295.
q = 10 m 2 + 4900 − 700
p = 19,300 − 8q ; m = 240
dr dr dq
.
=
⋅
dm dq dm
4. For x = 90,000, the slope of the consumption
curve is 0.561. You would expect the family to
spend $561 and save $439.
r = pq = q 19,300 − 8q , so
dr
−1
⎛1⎞
= q ⎜ ⎟ (19,300 − 8q ) 2 (−8) + 19,300 − 8q (1).
dq
⎝2⎠
If m = 240, then q = 1800, so
dr
230
=−
≈ −32.86.
dq m = 240
7
(
dq
1
= 10 ⋅ m2 + 4900
dm
2
dq
= 9.6. Thus
dm m = 240
)
− 12
(2m) .
dr
≈ (−32.86)(9.6) = −315.456
dm m = 240
b.
dr
dm
r
=
m = 240
=
−315.456
r
q =1800
−315.456
1800 4900
= −0.0025
c.
dr
< 0, there would be no
dm
additional revenue generated to offset the
cost of $400.
No. Since
62. 21.094
63. 0.305
64. $5.05
422
5. Answers may vary.
Chapter 12
Principles in Practice 12.1
1.
(
)
dq d ⎡
=
25 + 2 ln 3 p 2 + 4 ⎤
⎦⎥
dp dp ⎣⎢
= 0+2
(
)
d ⎡
ln 3 p 2 + 4 ⎤
⎦⎥
dp ⎣⎢
⎛
1 ⎞ d
2
3 p2 + 4 =
(6 p)
= 2⎜
⎟
2
⎜ 3 p 2 + 4 ⎟ dp
3p + 4
⎝
⎠
12 p
=
3 p2 + 4
(
)
2. With I 0 = 1 , R(I) = log I.
dR d
d ⎡ ln I ⎤
= [log I ] =
dI dI
dI ⎢⎣ ln10 ⎥⎦
1 1
1
=
⋅ =
ln10 I I ln10
Problems 12.1
1.
dy
d
1 4
= 4 ⋅ (ln x) = 4 ⋅ =
dx
dx
x x
2.
dy 5 ⎛ 1 ⎞ 5
=
=
dx 9 ⎜⎝ x ⎟⎠ 9 x
3.
dy
1
3
=
(3) =
dx 3x − 7
3x − 7
4.
dy
1
5
=
(5) =
dx 5 x − 6
5x − 6
5. y = ln x 2 = 2 ln x
dy
1 2
= 2⋅ =
dx
x x
6.
dy
1
6x + 2
(6 x + 2) =
=
2
dx 3x 2 + 2 x + 1
3x + 2 x + 1
7.
dy
1
2x
( −2 x ) = −
=
dx 1 − x 2
1 − x2
423
Chapter 12: Additional Differentiation Topics
8.
9.
dy
1
−2 x + 6
(−2 x + 6) =
=
dx − x 2 + 6 x
− x2 + 6 x
−2( x − 3) 2( x − 3)
=
=
− x( x − 6) x( x − 6)
f ′( X ) =
=
=
=
10.
ISM: Introductory Mathematical Analysis
f ′(r ) =
=
=
1
6
3
(24 X 5 + 6 X 2 )
4X + 2X
24 X 5 + 6 X 2
4X 6 + 2X 3
6 X 2 (4 X 3 + 1)
2 X 3 (2 X 3 + 1)
3(4 X 3 + 1)
X (2 X 3 + 1)
( 8r
+ 2r + 1
1
4
2r − 3r
2
3
− 6r + 2
)
8r 3 − 6r + 2
2r 4 − 3r 2 + 2r + 1
(
)
2 4r 3 − 3r + 1
2r 4 − 3r 2 + 2r + 1
11.
⎛1⎞
f ′(t ) = t ⎜ ⎟ + (ln t )(1) = 1 + ln t
⎝t ⎠
12.
dy
⎛1⎞
= x 2 ⎜ ⎟ + (ln x)(2 x) = x + 2 x ln x
dx
⎝x⎠
= x(1 + 2 ln x)
13.
dy
⎡ 1
⎤
= x3 ⎢
(2) ⎥ + ln(2 x + 5) ⋅ 3 x 2
dx
⎣ 2x + 5 ⎦
=
14.
2 x3
+ 3x 2 ln(2 x + 5)
2x + 5
⎡ 1
⎤
dy
= (ax + b)3 ⎢
(a ) ⎥ + [ln(ax + b)]3(ax + b) 2 (a )
dx
⎣ (ax + b) ⎦
= a(ax + b)2 + 3a (ax + b)2 ln(ax + b)
= a(ax + b)2 [1 + 3ln(ax + b)]
424
ISM: Introductory Mathematical Analysis
15. y = log3 (8 x − 1) =
Section 12.1
ln(8 x − 1)
ln 3
21.
dy
1 d
=
⋅ [ln(8 x − 1)]
dx ln 3 dx
1
1
8
=
⋅
(8) =
ln 3 8 x − 1
(8 x − 1)(ln 3)
16.
(
)
=
(
f ( w) = log w2 + w = log10 w2 + w
=
(
ln w2 + w
)
)
(
23. y = ln x 2 + 4 x + 5
(
(
)
2
2
(
ln x 2 + 4
)
(
)
ln 2
25. y = 9 ln 1 + x 2 =
ln x
1
=
( x 2 ln x)
ln 2 ln 2
⎤
dy
1 ⎡ 2⎛1⎞
=
+ ln x(2 x) ⎥
⎢x
dx ln 2 ⎣ ⎜⎝ x ⎟⎠
⎦
x
=
(1 + 2 ln x)
ln 2
20.
26.
( 1z ) − (ln z)(1) = 1 − ln z
z
2
z
( )
2 x ln x − x
2
ln x
=
)
(
27.
x[2 ln x − 1]
ln 2 x
425
)
⎛
⎞
t5
f (t ) = ln ⎜
⎟ = 5ln t − ln(1 + 3t 2 + t 4 )
⎜ 1 + 3t 2 + t 4 ⎟
⎝
⎠
1
1
⎛ ⎞
f ′(t ) = 5 ⎜ ⎟ −
(6t + 4t 3 )
⎝ t ⎠ 1 + 3t 2 + t 4
5(1 + 3t 2 + t 4 ) − t (6t + 4t 3 )
=
t (1 + 3t 2 + t 4 )
=
2
2 1
dy (ln x)(2 x) − x x
=
dx
(ln x)2
=
(
= 3ln x 2 + 4 x + 5
9
ln 1 + x 2
2
dy 9
1
9x
= ⋅
(2 x) =
dx 2 1 + x 2
1 + x2
⎤
⎥
⎥
⎦⎥
18. y = x 2 log 2 x = x 2 ⋅
z
3
1
24. y = 6 ln 3 x = 6 ⋅ ln x = 2 ln x
3
dy
1 2
= 2⋅ =
dx
x x
⎤
1 ⎡ 1
dy
(2 x) ⎥
= 2x +
dx
ln 2 ⎢⎣ x 2 + 4
⎦
⎡
1
= 2 x ⎢1 +
⎢
(ln 2) x 2 + 4
⎣⎢
)
dy
1
(2 x + 4)
= 3⋅
2
dx
x + 4x + 5
3(2 x + 4)
6( x + 2)
=
=
2
2
x + 4x + 5 x + 4x + 5
)
17. y = x + log 2 x + 4 = x +
f ′( z ) =
x(ln x )3
dy
1 100
= 100 ⋅ =
dx
x
x
1
1
⋅
f ′( w) =
(2 w + 1)
2
ln10 w + w
2w + 1
=
(ln10) w2 + w
19.
2 x 2 ln x − 2( x 2 + 3)
22. y = ln x100 = 100 ln x
ln10
2
2
2
1
dy (ln x) (2 x) − ( x + 3)2(ln x) x
=
dx
(ln x) 4
t 4 + 9t 2 + 5
t (1 + 3t 2 + t 4 )
⎛ 1+ l ⎞
f (l ) = ln ⎜
⎟ = ln(1 + l ) − ln(1 − l )
⎝ 1− l ⎠
1
1
f ′(l ) =
(−1)
−
1+ l 1− l
(1 − l ) + (1 + l )
2
=
=
(1 + l )(1 − l )
1− l2
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
(
⎛ 2x + 3 ⎞
28. y = ln ⎜
⎟ = ln(2 x + 3) − ln(3 x − 4)
⎝ 3x − 4 ⎠
dy
2
3
=
−
dx 2 x + 3 3x − 4
2(3x − 4) − 3(2 x + 3)
17
=
=−
(2 x + 3)(3 x − 4)
(2 x + 3)(3 x − 4)
29. y = ln 4
1 + x2
=
(
)
(
2
2
⎡
1 ⎢ 2x 1 − x + 2x 1 + x
=
4⎢
1 + x2 1 − x2
⎢⎣
(
)(
)
) ⎤⎥ =
⎥
⎥⎦
= 13ln x + 13ln(5 x + 2)1/ 3
13
= 26 ln x + ln(5 x + 2)
3
dy
1
1
26
65
⎛ ⎞ 13
= 26 ⎜ ⎟ + ⋅
(5) =
+
dx
x 3(5 x + 2)
⎝ x ⎠ 3 5x + 2
)
1
= 6 ln x − 6 ln(2 x + 1) 2
2x +1
= 6 ln x − 3ln(2 x + 1)
dy 6
1
6
6
= − 3⋅
(2) = −
dx x
2x + 1
x 2x +1
x
1 − x4
35.
1
= [ln( x3 − 1) − ln( x3 + 1)]
30. y = ln 3
3
x +1 3
(
36.
(
) + 2 x ln(2 x + 1)
2 x2 + 1
2x + 1
dy
⎡1
⎤
= (ax + b) ⎢ (a ) ⎥ + ln(ax ) ⋅ (a )
dx
⎣ ax
⎦
=
2 x2
)
dy
⎡ 1
⎤
(2) ⎥ + ln(2 x + 1) ⋅ (2 x)
= x2 + 1 ⎢
dx
⎣ 2x + 1 ⎦
=
dy 1 ⎡ 3x 2
3x 2 ⎤
= ⎢
−
⎥
dx 3 ⎢⎣ x3 − 1 x3 + 1 ⎦⎥
1 ⎡ 3x 2 ( x3 + 1) − 3 x 2 ( x3 − 1) ⎤
= ⎢
⎥
3 ⎢⎣
( x3 − 1)( x3 + 1)
⎥⎦
ax + b
+ a ln(ax)
x
37. y = ln x3 + ln 3 x = 3ln x + (ln x)3
x6 − 1
(
dy
1
1 3 3(ln x) 2
= 3 ⋅ + 3(ln x )2 ⋅ = +
dx
x
x x
x
) ( x + x − 1)⎤⎥⎦
= 2 ln ( x + 2 ) + ln ( x + x − 1)
⎡
31. y = ln ⎢ x 2 + 2
⎣
2
3
2
=
3
(
)
dy
1
1
= 2⋅
(2 x) +
3x 2 + 1
2
3
dx
x +2
x + x −1
=
x
34. y = 6 ln
x3 − 1
=
)
2
1⎡
ln 1 + x 2 − ln 1 − x 2 ⎤
⎦⎥
4 ⎣⎢
1 − x2
dy 1 ⎡ 2 x
−2 x ⎤
= ⎢
−
⎥
2
dx 4 ⎣ 1 + x
1 − x2 ⎦
(
) (
33. y = 13ln x 2 3 5 x + 2
4x
x2 + 2
+
38.
3x2 + 1
(
3 1 + ln 2 x
)
x
dy
= (ln 2) x (ln 2) −1
dx
39. y = ln 4 (ax) = [ln(ax)]4
x3 + x − 1
3
dy
⎛ 1 ⎞ 4 ln (ax)
= 4[ln(ax )]3 ⎜ ⋅ a ⎟ =
dx
x
⎝ ax ⎠
32. y = ln ⎡ (5 x + 2)4 (8 x − 3)6 ⎤
⎣
⎦
= 4 ln(5 x + 2) + 6 ln(8 x − 3)
40. y = ln 2 (2 x + 11) = [ln(2 x + 11)]2
dy
1
1
= 4⋅
(5) + 6 ⋅
(8)
dx
5x + 2
8x − 3
20
48
=
+
5x + 2 8x − 3
dy
1
4 ln(2 x + 11)
= 2[ln(2 x + 11)] ⋅
(2) =
dx
2 x + 11
2 x + 11
426
ISM: Introductory Mathematical Analysis
41. y = x ln x − 1 =
Section 12.1
1
x ln( x − 1)
2
47. y =
⎤
dy 1 ⎡ ⎛ 1 ⎞
= ⎢x
+ ln( x − 1) ⋅ (1) ⎥
dx 2 ⎣ ⎜⎝ x − 1 ⎟⎠
⎦
x
=
+ ln x − 1
2( x − 1)
(
y′ =
)
1
ln(2 x + 1)
4
dy
1 1
1
3
1
= 3⋅ + ⋅
(2) = +
dx
x 4 2x + 1
x 2(2 x + 1)
x
1+ x 2
=
=
x + 1 + x2
1
=
(ln 3) − 1
ln 2 3
.
( )
= 25 ⋅
(q + 2) ln(q + 2) − q
(q + 2) ln 2 (q + 2)
1
q+2
.
49. c = 25 ln(q + 1) + 12
dc
25
dc
25
, so
.
=
=
dq q + 1
dq q =6 7
−1
⎡ 1
⎤
2 2
x
1
1
(2 x) ⎥
+
+
⎢
⎥⎦
⎣⎢ 2
)
1 + x2 + x
1 + x 2 ⎜⎛ x + 1 + x 2 ⎟⎞
⎝
⎠
50. c =
500
ln(q + 20)
c = cq =
1 + x2
500q
ln(q + 20)
( )
1
[ln(q + 20)](1) − q q + 20
dc
= 500 ⋅
dq
[ln(q + 20)]2
45. y = ln( x 2 − 3 x − 3)
y′ =
ln 2 x
ln(q + 2)(1) − q
dr
= 25 ⋅
dq
ln 2 (q + 2)
3
dy 1
−1 3
= (4 + 3ln x) 2 ⋅ =
dx 2
x 2 x 4 + 3ln x
1+
ln x
25
25q
, so r = pq =
. Thus the
ln(q + 2)
ln(q + 2)
marginal revenue is
1
(
2
48. p =
43. y = 4 + 3ln x = (4 + 3ln x ) 2
dy
1
=
dx x + 1 + x 2
( ) = ln x − 1
(ln x)(1) − x 1x
When x = 3 the slope is y′(3) =
42. y = ln x3 4 2 x + 1 = 3ln x +
44.
x
ln x
2x − 3
2
x − 3x − 3
The slope of the tangent line at x = 4 is
8−3
y′(4) =
= 5. Also, if x = 4, then
16 − 12 − 3
y = ln(16 − 12 − 3) = ln 1 = 0. Thus an equation
of the tangent line is y – 0 = 5(x – 4), or
y = 5x − 20.
50
ln 70 − 70
dc
= 500 ⋅
≈ $97.90
dq q =50
(ln 70)2
51.
dq d
=
[25 + 10 ln(2 p + 1)]
dp dp
= 0 + 10
46. y = x[ln(x) – 1]
⎛1⎞
y′ = x ⎜ ⎟ + [ln( x) − 1](1) = ln x
⎝x⎠
When x = e, y = 0 and y′ = 1 . The equation of
the tangent line is y – 0 = 1(x – e), or y = x – e.
=
427
⎛ 1 ⎞ d
d
[ln(2 p + 1)] = 10 ⎜
⎟ [2 p + 1]
dp
⎝ 2 p + 1 ⎠ dp
10
20
(2) =
2 p +1
2 p +1
Chapter 12: Additional Differentiation Topics
52. With I 0 = 17, L( I ) = 10 log
ISM: Introductory Mathematical Analysis
Principles in Practice 12.2
I
.
17
1. The rate of change of temperature with respect to
dT
. T(t) has the form Ceu where C is a
time is
dt
constant and u = kt.
dT d ⎡ kt ⎤
d
Ce = C ⎡ ekt ⎤
=
⎦
dt dt ⎣
dt ⎣ ⎦
dL d ⎡
I ⎤
d
=
10 log ⎥ = 10 [log I − log17]
dI dI ⎢⎣
17 ⎦
dI
= 10
=
d
dI
⎡ ln I
⎤
⎡ 1 1
⎤
⎢ ln10 − log17 ⎥ = 10 ⎢ ln10 ⋅ I − 0 ⎥
⎣
⎦
⎣
⎦
10
I ln10
( ) dtd [kt ] = Ce
= C ekt
⎛ T
⎞
53. A = 6 ln ⎜
− a ⎟ . Rate of change of A with
⎝ a −T
⎠
respect to T:
⎡ (a − T )(1) − T (−1) ⎤
dA
1
= 6⋅
⎢
⎥
T −a
dT
(a − T )2
⎢⎣
⎥⎦
a −T
= 6⋅
= 6⋅
=
1
T − a ( a −T )
a −T
⋅
1. y ′ = 5 ⋅
2. y′ =
a
T − a 2 + aT (a − T ) 2
6a
(T − a
2
)
d x
(e ) = 5e x
dx
2e x
5
3. y′ = e 2 x
2
+3
(4 x) = 4 xe2 x
2
+3
4. y′ = e 2 x
2
+5
(4 x) = 4 xe2 x
2
+5
+ aT (a − T )
dy
1
f ′( x)
=
f ′( x) =
,
dx f ( x)
f ( x)
which is the relative rate of change of y = f(x)
with respect to x.
6.
7.
1 d
1 ⎛ 1 du ⎞
⋅ (ln u ) =
⋅
ln b dx
ln b ⎜⎝ u dx ⎟⎠
x4
f ′( x) = 0 for x ≈ −1.65, 1.65
2
2
+ 4r + 4
+ 6 x3 +1
( −3q
2
+6
)
+ 6 q −1
(6r + 4) = 2(3r + 2)e3r
2
+ 4r + 4
(2 x + 18 x 2 )
2
+ 6 x3 +1
( )
9. y′ = x e x + e x (1) = e x ( x + 1)
10. y′ = 3x 4 ⎡ e− x (−1) ⎤ + e− x (12 x3 ) = 3 x3e− x (4 − x)
⎣
⎦
57. Note that f(x) is defined for all x ≠ 0.
x
f ′(r ) = e3r
3
= 2 x(1 + 9 x)e x
f ′( x) = x (1 + 3ln x)
f ′( x) = 0 for x ≈ 0.72
f ′( x) =
+ 6 q −1
)
8. y ′ = e x
2
x 2 ⋅ 12 (2 x) − ln( x 2 ) ⋅ 2 x
3
(
du
⎛ 1 du ⎞ 1
= ( logb e ) ⎜ ⋅ ⎟ = ( logb e )
dx
⎝ u dx ⎠ u
56.
f ′(q) = e− q
= −3 q 2 − 2 e − q
d
d ln u
( logb u ) = ⎛⎜ ⎞⎟
dx
dx ⎝ ln b ⎠
=
d
( 9 − 5 x ) = e9−5 x (−5) = −5e9−5 x
dx
5. y′ = e9−5 x ⋅
54. If y = ln f(x), then
55.
(k ) = Ckekt
Problems 12.2
⎡
⎤
a
⎢
⎥
2
⎢⎣ (a − T ) ⎥⎦
a −T
kt
=
2
2
11. y′ = x 2 ⎡ e − x (−2 x) ⎤ + e− x (2 x)
⎣⎢
⎦⎥
2 − 2 ln( x 2 )
x3
2
(
= 2 xe− x 1 − x 2
)
12. y′ = x ⎡ e3 x (3) ⎤ + e3 x (1) = e3 x (3 x + 1)
⎣
⎦
428
ISM: Introductory Mathematical Analysis
(
1 x
e + e− x
3
13. y =
y′ =
14.
)
1⎡ x
e x − e− x
e + e − x (−1) ⎤ =
⎦
3⎣
3
dy (e x + e − x )[e x − e− x (−1)] − (e x − e− x )[e x + e− x (−1)]
=
dx
(e x + e − x ) 2
=
15.
Section 12.2
(e x + e − x ) 2 − (e x − e − x ) 2
x
(e + e
−x 2
)
=
4
(e + e − x ) 2
x
( )
d 2 x3
d ⎡ (ln 5)2 x3 ⎤
=
e
5
⎥⎦
dx
dx ⎢⎣
3
= e(ln 5)2 x [(ln 5)6 x 2 ]
3
= (6 x 2 )52 x ln 5
16. y = 2 x x 2 = e(ln 2) x x 2
y ′ = e(ln 2) x (2 x) + x 2 ⎡ e(ln 2) x (ln 2) ⎤
⎣
⎦
( )
( )
( )
= 2 x 2 x + x 2 2 x (ln 2) = x 2 x (2 + x ln 2)
17.
w2 ⎡ e2 w (2) ⎤ − e2 w [2 w]
⎣
⎦
f ′( w) =
w4
=
2e 2 w ( w − 1)
w3
18. y′ = e x −
x
19. y′ = e1+
x
1 ⎞
⎛ 1 − 12 ⎞
x− x ⎛
⎜1 −
⎟
⎜1 − 2 x ⎟ = e
⎝
⎠
⎝ 2 x⎠
1+ x
⎛ 1 − 12 ⎞ e
⎜2x ⎟=
⎝
⎠ 2 x
20. y ′ = 3(e2 x + 1) 2 (e2 x (2) + 0) = 6e2 x (e2 x + 1)2
21. y = x5 − 5 x = x5 − e(ln 5) x
y′ = 5 x 4 − e(ln 5) x (ln 5) = 5 x 4 − 5 x ln 5
22.
2
f ( z ) = e−1/ z = e− z
2
−2
f ′( z ) = e−1/ z [−(−2 z −3 )] =
2 −1/ z 2
e
z3
429
Chapter 12: Additional Differentiation Topics
(
)
(
ISM: Introductory Mathematical Analysis
)
e x + 1 ⎡e x ⎤ − e x − 1 ⎡e x ⎤
dy
⎣ ⎦
⎣ ⎦
23.
=
2
dx
ex +1
=
2e
(
)
33.
dp
= −0.015e−0.5
dq q =500
x
( e + 1)
x
dp
= 15e−0.001q (−0.001) = −0.015e−0.001q
dq
2
34.
24. y′ = e2 x [1] + ( x + 6) ⎡ e2 x (2) ⎤ = e2 x (2 x + 13)
⎣
⎦
dp
⎛ 5 ⎞
−5q / 750
= 9e−5q / 750 ⎜ −
⎟ = −0.06e
dq
⎝ 750 ⎠
dp
= −0.06e−2
dq q =300
25. y = ln e x = x so y ′ = 1.
q
(
q
7000e 700
35. c =
, so c = cq = 7000e 700 . The
q
)
1
⎛1
⎞
26. y′ = e− x ⋅ + (ln x) −e− x = e− x ⎜ − ln x ⎟
x
⎝x
⎠
27. y ′ = e x
2
ln x 2
= 2 xe
⎡ 2 1
⎤
2
⎢ x ⋅ 2 (2 x) + (ln x )(2 x) ⎥
x
⎣
⎦
x 2 ln x 2
q
= 10e 700 . Thus
2
(1 + ln x )
28. y = ln e4 x +1 = 4 x + 1 , so
29.
marginal cost function is
2
f ( x) = ee x e x = e1+ x + x
dc
= 10e0.5 and
dq q =350
dc
= 10e .
dq q =700
dy
= 4.
dx
2
2q+6
2
f ′( x) = e1+ x + x (1 + 2 x) = (1 + 2 x)e1+ x + x
850
e 800
36. c =
+ 4000
q
q
2
2
f ′(−1) = [1 + 2(−1)]e1+ ( −1) + ( −1) = −e
30.
f ( x) = 5
x 2 ln x
( )
= e
ln 5
x 2 ln x
=e
c = cq = 850 + 4000e
37. w = e x
f ′(1) = e0 (ln 5)[1 + 0] = ln 5
y′ = e−2 . Thus an equation of the tangent line is
y−e
3
−4 x
q +3
dc
= 10e 400 .
dq
+ x ln( x − 1) and x =
By the chain rule,
31. y = e x , y′ = e x . When x = –2, then y = e−2 and
−2
q +3
= 850 + 4000e 400
dc
dc
= 10e0.25 and
= 10e0.5 .
dq q =97
dq q =197
2
= e(ln 5) x ln x (ln 5)[ x + 2 x ln x]
−2
2q+6
800
The marginal cost function is
(ln 5) x 2 ln x
2
⎧
1
⎡
⎤⎫
f ′( x) = e(ln 5) x ln x ⎨(ln 5) ⎢ x 2 ⋅ + (ln x)(2 x) ⎥ ⎬
x
⎣
⎦⎭
⎩
−2
q
dc
⎛ 1 ⎞
= 7000e 700 ⎜
⎟
dq
⎝ 700 ⎠
(
t +1
t −1
dw dw dx
=
⋅
dt
dx dt
)
⎡ 3
⎤
⎛ 1 ⎞
= ⎢e x − 4 x 3 x 2 − 4 + x ⎜
+ [ln( x − 1)(1)]⎥
⎟
⎝ x −1 ⎠
⎣
⎦
−2
= e ( x + 2) , or y = e x + 3e .
⎡ (t − 1)(1) − (t + 1)(1) ⎤
· ⎢
⎥
(t − 1) 2
⎢⎣
⎥⎦
32. y ′ = e x
When x = 1, y = e and y ′ = e. Thus an equation
3
x
⎡
⎤ ⎡ −2 ⎤
= ⎢ 3x2 − 4 e x −4 x +
+ ln( x − 1) ⎥ ⎢
⎥.
x −1
⎣
⎦ ⎣⎢ (t − 1) 2 ⎦⎥
(
of the tangent line is y − e = e(x − 1) or y = ex.
430
)
ISM: Introductory Mathematical Analysis
When t = 3, then x =
Section 12.2
3 +1 4
= = 2 and
3 −1 2
42.
dw
⎡ 1⎤
= [8 + 2 + 0] ⎢ − ⎥ = −5 .
dt
⎣ 2⎦
38.
(
)
( )
f ( x) = 10
t
t
t
= −(ln10)10
(
45. Since S = Pert , then
dS
dt
x−2
S
(
rPert
Pert
(
dS
= Pert r = rPert . Thus
dt
= r.
)
dy
= K ⎡ −e− ax (−a ) ⎤ = aKe− ax
⎣
⎦
dx
Solving the original equation for e− ax gives
y
e− ax = − + 1 . Thus substitution,
K
dy
⎛ y
⎞
= aK ⎜ − + 1⎟ = a(− y + K ) = a (K – y), as
dx
⎝ K
⎠
was to be shown.
1
+ 10
dq
= 500 −e−0.2t
dt
dq
Thus
= 100e−2 .
dt t =10
=
46. y = K 1 − e− ax
1
+ 0.01e x − 2
8+ x
)
) (−0.2) = 100e
d (ln β )t
[e
]
dt
t
+ 0.01
f ′(2) −(ln10)10
=
≈ 0.0374
2
−
f (2)
10 + ln(10) + 0.01
41. q = 500 1 − e−0.2t
d t
[β ]
dt
= kα β ( β t ln α ) ln β
1
+
+ 0.01e x − 2
8+ x
−2
t
= kα β (ln α )e(ln β )t (ln β )
= e(ln10)( − x ) + ln(8 + x) + 0.01e x − 2
−x
/2
Y ′ = ke(ln α ) β (ln α )
= kα β (ln α )
+ ln(8 + x) + 0.01e
2
t
−x ⎥
⎦
f ′( x) = e(ln10)( − x ) (− ln10) +
e− x
/2
44. Y = kα β = ke(ln α ) β
c⎤
d x
(c − x c )
= (ln c)c − c
dx
x =1
If this is zero, (ln c)c – c = 0, or c[ln(c) – 1] = 0.
Since c > 0, we must have ln(c) – 1 = 0, ln c = 1,
or c = e.
40.
2
43. P = 1.92e0.0176t
dP
= 1.92e0.0176t (0.0176) = P(0.0176)
dt
= 0.0176P = kP for k = 0.0176.
= (ln c)e(ln c ) x − cx c −1 = (ln c)c x − cxc −1
−x
e− x
(− x)
2π
1 −1/ 2
e
(−1) ≈ −0.242
2π
f ′(1) =
d
dy
[ f (u )] =
and by the chain rule
dx
dx
d
dy dy du
du
[ f (u )] =
=
⋅
= f ′(u )
= u3 ⋅ e x
dx
dx du dx
dx
= ( e x )3 ⋅ e x = e 3 x ⋅ e x = e 4 x
x
2π
1
f ′( x) =
f ′( x) = x3 and u = e x . Let y = f(u). Then
d x
d ⎡ ln c
39.
c − xc =
e
dx
dx ⎢⎣
d ⎡ (ln c ) x
=
e
− xc ⎤
⎦
dx ⎣
1
f ( x) =
−0.2t
47. N = 10 A10−bM = 10 A−bM = e(ln10)( A−bM )
dN
= e(ln10)( A−bM ) (ln10)(−b) , so
dM
dN
= 10 A−bM (ln10)(−b) = −b 10 A−bM ln10
dM
(
431
)
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
48. p = 0.89 ⎡ 0.01 + 0.99(0.85)t ⎤
⎣
⎦
a.
52. S = ln
dP
= 0.89 ⎡0.99(0.85)t ln(0.85) ⎤
⎣
⎦
dt
a.
= 0.8811(0.85)t ln(0.85)
This represents the rate of change of
proportion of correct recalls with respect to
length of recall interval.
−
b. If
50. C (t ) =
a.
b.
⎞
⎟ C (t )
⎠
=
7
≈ $0.847 billion
3
= $(0.847)(1000) million
= $847 million
R⎡
R
1 − e0 ⎤ = [1 − 1] = 0
⎣
⎦
r
r
53.
3
2
f ′( x) = (6 x 2 + 2 x − 3)e2 x + x −3 x
f ′( x) = 0 for x ≈ −0.89, 0.56
54.
f ′( x) = 1 − e− x
f ′( x) = 0 gives e x = 1 or x = 0.
− ( r )t ⎞ ⎤
R⎡ r R⎛
⎢1 − ⋅ ⎜ 1 − e V ⎟ ⎥
V ⎢⎣ R r ⎝
⎠ ⎥⎦
Problems 12.3
R⎡ r
⎤ R r
= ⎢1 − C (t ) ⎥ = − C (t )
V⎣ R
⎦ V V
1. η =
p
q
dp
dq
=
p
q
−2
.
When q = 5 then p = 40 – 2(5) = 30, so
f (t ) = 1 − e−0.008t
f ′(t ) = 0.008e
−I
I = ln
− ( r )t ⎞ ⎤
R⎡ ⎛
= ⎢1 − ⎜ 1 − e V ⎟ ⎥
V ⎣⎢ ⎝
⎠ ⎦⎥
51.
dS 1
e− I
1
= , then
= .
−I
8
dI 8
3+ e
1
(e )(3 + e ) 8
1
1
=
I
3e + 1 8
3e I + 1 = 8
7
eI =
3
dC R ⎡ r −( Vr )t ⎤ R −( Vr )t
= ⎢ e
⎥= e
dt
r ⎣V
⎦ V
=
dC
dS
e− I
3
.
= 1−
= 1−
=
−I
dI
dI
3+ e
3 + e− I
I
−( Vr )t ⎤
R⎡
⎢1 − e
⎥
r ⎣
⎦
C (0) =
dC
dS
= 1−
.
dI
dI
(e I ) e − I
− ( r )t ⎛ r ⎞
dC
= C0 e V ⎜ − ⎟
dt
⎝ V⎠
⎞
⎛r
⎟ = −⎜ V
⎠
⎝
Recall that
Thus
( Vr )t
⎛ r
= [C (t ) ] ⎜ −
⎝ V
3+e
= ln 5 − ln(3 + e− I )
−I
dS
1
e− I
(e− I )(−1) =
=−
dI
3 + e− I
3 + e− I
b. If t = 2, then
dp
= 0.8811(0.85)2 ln(0.85) ≈ −0.10
dt
49. C (t ) = C0 e
5
η=
30
5
= −3
−2
Because η > 1 , demand is elastic.
−0.008t
f ′(100) = 0.008e−0.8 ≈ 0.0036
p
q
6
= 100 = −1.5
−0.04 −0.04
Because η > 1, demand is elastic.
2. η =
432
ISM: Introductory Mathematical Analysis
3. p =
Section 12.3
9. q = 1200 − 150p
3500
= 3500q −1
q
η=
dp
3500
= −3500q −2 = −
dq
q2
η=
p
q
dp
dq
=
p
q
− 3500
q2
=
(3500 / q )
q
− 3500
q2
4. η =
5. η =
p
q
dp
dq
(500 / q )
= −1
=
=
p
q
− 500 2
( q + 2)
q
1000
− 3
q
=
1
= − , inelastic
2
[500 /( q + 2)]
q
− 500 2
( q + 2)
=
=−
η=
6. η =
=
800 /(2 q +1)
q
1600
−
(2 q +1) 2
When q = 24, η = −
7. η =
p
q
dp
dq
=
=−
q+2
q
η=
2q + 1
2q
η=
49
, elastic
48
8. η =
−e
−
q
200
p
q
dp
dq
=
p dq
⋅
q dp
p⎛ 1 ⎞
p
⎜− ⎟ = − 2
q ⎝ 2q ⎠
2q
If p = 400, then q = 500 − 400 = 10, so
η=−
p
q
400
= −2. η > 1 , so demand is elastic.
200
q
12. q = 2500 − p 2
⎛ 150 ⎞
= −⎜
− 1⎟ . Because η > 1 ,
⎝ e
⎠
η=
−
=
100e
q
2
−e
−
=−
200
q
=
2
When q = 200, η = −
=
p dq
⋅
q dp
)
− 12
dq 1
= 2500 − p 2
(−2 p)
dp 2
q
200
q
200
p
q
dp
dq
(
demand is elastic.
p
q
p dq
⋅
q dp
−1
1
dq 1
−1
= (500 − p ) 2 (−1) =
=−
2q
dp 2
2 500 − p
When q = 100, then p = 150 – e and
η=
=
11. q = 500 − p
−e100
100
150 −e
100
e
− 100
p
q
dp
dq
50
dq
= −1 , so η =
(−1) = −1 , unit elasticity.
50
dp
106
53
When q = 104, then η = −
= − . Because
104
52
η > 1 , demand is elastic.
p
q
− 1600 2
(2 q +1)
p dq p
⋅
= (−150)
q dp q
10. q = 100 – p
When p = 50, then q = 50.
2
p
q
−1000
q3
=
If p = 4, then q = 1200 − 150(4) = 600, so
4
η=
(−150) = −1. Since η = 1, demand has
600
unit elasticity.
Because η = 1 , demand has unit elasticity.
p
q
dp
dq
p
q
dp
dq
η=
200
= −1, so demand has
200
−p
2500 − p 2
=−
p⎛ p⎞
⎜− ⎟ = −
q⎝ q⎠
p
q
p2
q2
If p = 20, then q = 2100 , so we have
unit elasticity.
η=−
433
400
4
= − , inelastic.
2100
21
Chapter 12: Additional Differentiation Topics
13. q =
ISM: Introductory Mathematical Analysis
( p − 100) 2
2
p
q
dp
dq
η=
=
η=
p dq
⋅
q dp
η=
(20 − 100) 2
= 3200 . Thus
2
20
1
η=
(20 − 100) = − . Demand is inelastic.
3200
2
q=
=
p dq
⋅
q dp
=−
18. η =
η
demand
10
60
− 10
3
elastic
3
− 10
inelastic
−1
unit elasticity
130
η=
η=
Setting
=
=
p
q
dp
dq
=
p dq
⋅
q dp
−p
2500 − p
p ⎛ −p ⎞
⎜
⎟=−
q⎝ q ⎠
2
=
p2
q2
−p
, so
q
.
Now, if p = 30, then q = 2500 − 302 = 40, so
(30) 2
9
=
−
= − . If the price of 30
p =30
2
16
(40)
decreases to 28.5, that is, it changes by
−1.5
= −5%, then demand would change by
30
⎛ 9 ⎞
approximately −5 ⎜ − ⎟ %, or 2.8%. (That is,
⎝ 16 ⎠
demand increases by 2.8%.)
p = 36 − 0.25q
p
q
dp
dq
p =15
dq
=
dp
η
16. a.
p dq
⋅
q dp
q = 2500 − p 2
q
6.50
=
1
%,
2
then the change in demand is approximately
⎛1 ⎞
⎜ 2 % ⎟ (−1.2) = −0.6%. Thus demand decreases
⎝
⎠
approximately 0.6%.
p
0.05q
200
p
q
dp
dq
demand. Thus if the price of 15 increases
p
3
= −1 yields q = ±10.
15
6
(30 − 40) = − = −1.2. Now,
125
5
(% change in price) · (η ) = % change in
so η
15. p = 13 – 0.05q
η=
−2q 2
When p = 15, then q = 500 − 40(15) + 152 = 125,
dq
p
= 2 p − 50, so η = (2 p − 50).
dp
q
If p = 20, then q = 250, and
20
200
4
η=
= − , inelastic.
(40 − 50) = −
250
250
5
p
q
dp
dq
300 − q 2
dq
p
= −40 + 2 p , so η = (2 p − 40).
dp
q
14. q = p 2 − 50 p + 850
η=
=
17. q = 500 − 40 p + p 2
p
( p − 100) . If p = 20, then
q
p
q
dp
dq
p
q
dp
dq
Since q > 0, we must have q = 10.
dq 1
= (2)( p − 100)(1) = p − 100 , so
dp 2
η=
p = 300 − q 2
b.
36 − 0.25q
−0.25q
36 − 0.25q
= −1 yields q = 72.
−0.25q
434
ISM: Introductory Mathematical Analysis
Section 12.3
22. p = mq + b
19. p = 500 – 2q
η=
p
q
dp
dq
=
500 − 2 q
q
−2
=
Note: q =
q − 250
q
q − 250
< −1. For
q
q > 0, we have q – 250 < –q, 2q < 250, so
q < 125. Thus, if 0 < q < 125, demand is elastic.
q − 250
> −1.
If demand is inelastic, then η =
q
For q > 0, the inequality implies q > 125. Thus if
125 < q < 250, then demand is inelastic.
If demand is elastic, then η =
a.
b.
23. a.
3q − 50
=
=
−3
3q
b.
c.
1000
q2
b + cq
2
= a (b + cq 2 ) −1/ 2
p
q
dp
dq
=
a(b + cq 2 ) −1/ 2
−acq 2 (b + cq 2 ) −3 / 2
=−
b + cq 2
cq 2
⎛ b
⎞
= −⎜
+ 1⎟
2
⎜
⎟
cq
⎝ cq
⎠
b
If b, c > 0, then
+ 1 > 1 so |η| > 1 and
cq 2
demand is elastic.
η=−
b + cq 2
2
If |η| = 1, then
b
cq 2
+ 1 = 1, which can only
occur if b = 0.
1000
r = pq =
q
24. η =
dr
1000
= −1000q −2 = −
dq
q2
=
a
p=
Thus η does not depend on a.
⎛ 1⎞
⎛
3q ⎞
p ⎜ 1 + ⎟ = (50 − 3q) ⎜ 1 +
⎟
η
3
− 50 ⎠
q
⎝
⎠
⎝
⎛ 3q − 50 + 3q ⎞
= (50 − 3q) ⎜
⎟
⎝ 3q − 50 ⎠
dr
= 50 − 6q =
dq
η=
p
p −b
Thus if p = 0, then η = 0.
η=
η=
50 −3q
q
1000
q3
− 2000
q3
−
= −acq (b + cq 2 )−3 / 2
dr
= 50 − 6q
dq
p
q
dp
dq
m
p →b −
dp
1
= − a (b + cq 2 )−3 / 2 (2cq)
dq
2
r = pq = 50q − 3q 2
21. p =
p →b −
p
( p −b ) / m
= lim
p
= −∞
−b
p
p →b
20. p = 50 − 3q
η=
p
q
p →b − dp
dq
lim η = lim
= lim
Since Total Revenue = r = pq = 500q − 2q 2 ,
then r ′ = 500 − 4q = 4(125 − q ). If
0 < q < 125, then r ′ > 0, so r is increasing. If
125 < q < 250, then r ′ < 0, so r is decreasing.
p
q
dp
dq
p−b
m
p
q
dp
dq
=
p dq
⋅
q dp
We differentiate implicitly for
1
=−
2
dq
.
dp
q 2 (1 + p) 2 = p
(
)
⎛ dq ⎞
q 2 ⋅ 2(1 + p)(1) + 1 + p 2 ⎜ 2q ⎟ = 1
⎝ dp ⎠
dq
2q 2 (1 + p) + 2q (1 + p) 2
=1
dp
⎛ 1 ⎞ 1000
1000 dr
p ⎜1 + ⎟ =
(1 − 2) = −
=
2
η
dq
q
q2
⎝
⎠
435
Chapter 12: Additional Differentiation Topics
Thus
ISM: Introductory Mathematical Analysis
dq 1 − 2q 2 (1 + p)
=
dp
2q(1 + p) 2
q 2 (1 + p) 2 1 − 2q 2 (1 + p) 1 − 2q 2 (1 + p)
⋅
=
q
2
2q(1 + p) 2
If p = 9, we find q from the given equation:
Hence η =
q 2 (1 + 9)2 = 9
q2 =
9
100
( )
2
3
1 − 2 10
(1 + 9)
3
since q > 0. Thus η p =9 =
= −0.4
q=
10
2
25. a.
(
q=
60
+ ln 65 − p3
p
η=
p
q
dp
dq
=
)
p dq p ⎡ 60
3 p2 ⎤
= ⎢−
−
⎥
q dp q ⎢⎣ p 2 65 − p3 ⎦⎥
If p = 4, then q =
60
4 ⎡ 60 3(16) ⎤
207
=−
≈ −13.8, and demand is elastic.
+ ln1 = 15 , so η = ⎢ − −
4
15 ⎣ 16 65 − 64 ⎥⎦
15
b. The percentage change in q is (–2)(–13.8) = 27.6%, so q increases by approximately 27.6%.
c.
26. a.
Lowering the price increases revenue because demand is elastic.
0.02 q +19 ⎤
p = 50 ⎡⎢ (151 − q )
⎥⎦
⎣
ln p = ln 50 + 0.02 q + 19 ln(151 − q)
⎡ q + 19
⎤
1 dp
1
(−1) + ln(151 − q) ⋅
= 0 + 0.02 ⎢
⎥
p dq
2 q + 19 ⎦⎥
⎣⎢ 151 − q
When q = 150, then p = 50, so
b.
η q =150 =
p
q
dp
dq q =150
dp
⎡ 13 0 ⎤
= 0.02(50) ⎢ − + ⎥ = −13
dq q =150
⎣ 1 26 ⎦
50
= 150 ≈ −0.0256
−13
Thus demand is inelastic.
c.
(elasticity)(% change in price) = % change in demand
−10
(−0.0256)(% change in price) =
⋅100
150
100 ⎛
1
⎞
% change in price = −
= 260%
15 ⎜⎝ −0.0256 ⎟⎠
Thus price per unit of $50 changes by 2.6(50) = $130, so it is approximately 50 + 130 = $180.
d. The manufacturer should increase the price because demand is inelastic.
436
ISM: Introductory Mathematical Analysis
Section 12.4
−5
25
50
⋅100 = − % and the percentage change in quantity is
⋅100 = 10%.
80
4
500
Thus, since (elasticity)(% change in price) ≈ % change in demand,
⎛ 25 ⎞
(elasticity) ⎜ − ⎟ ≈ 10.
⎝ 4 ⎠
27. The percentage change in price is
40
8
= − = −1.6
25
5
dr
To estimate
when p = 80, we have
dq
elasticity ≈ −
⎛
⎛ 1⎞
dr
1 ⎞
⎟ = 30.
= p ⎜ 1 + ⎟ = 80 ⎜1 +
⎜ −8 ⎟
dq
⎝ η⎠
5⎠
⎝
28. η =
p
q
dp
dq
=
2000 − q 2
−2q
2
=
For 5 ≤ q ≤ 40, η =
1 1000
−
2 q2
1000
2
−
1
2000
and η ′ = −
. Since η ′ < 0 , η is decreasing on [5, 40] and thus η is
2
q3
q
maximum at q = 5 and a minimum at q = 40.
29.
dp
−200
= 200(−1)(q + 5) −2 =
dq
(q + 5)2
Thus η =
p
q
dp
dq
=
200
q ( q + 5)
− 200 2
( q + 5)
For 5 ≤ q ≤ 95, η =
=−
q+5
.
q
5
q+5
5
= 1 + and η ′ = −
.
q
q
q2
Since η ′ < 0, η is decreasing on [5, 95], and thus η is maximum at q = 5 and minimum at q = 95.
Principles in Practice 12.4
⎛ P ⎞
1. Assume that P is a function of t and differentiate both sides of ln ⎜
⎟ = 0.5t with respect to t.
⎝ 1− P ⎠
d
dt
⎡ ⎛ P ⎞⎤ d
⎢ ln ⎜
⎟ ⎥ = [0.5t ]
⎣ ⎝ 1 − P ⎠ ⎦ dt
⎛ 1 ⎞d ⎡ P ⎤
⎜
⎟
= 0.5
⎜ P ⎟ dt ⎢⎣ 1 − P ⎥⎦
⎝ 1− P ⎠
1 − P (1)(1 − P) − P (−1) dP
⋅
⋅
= 0.5
P
dt
(1 − P )2
1 − P + P dP
⋅
= 0.5
P (1 − P ) dt
dP
= 0.5P (1 − P )
dt
437
Chapter 12: Additional Differentiation Topics
2.
ISM: Introductory Mathematical Analysis
( )
3. 6 y 2 y ′ − 14 x = 0
14 x
7x
y′ =
=
2
6y
3y2
dV
d ⎡4
dr
dr
⎤ 4
= ⎢ πr 3 ⎥ = π 3r 2
= 4πr 2
dt
dt ⎣ 3
3
dt
dt
⎦
dr
= 5 and r = 12,
When
dt
dV
= 4π(12) 2 (5) = 2880π . The balloon is
dt
increasing at the rate of 2880π cubic
inches/minute.
4. 4 x − 6 yy′ = 0
y′ =
5.
3. The hypotenuse is the length of the ladder, so
x 2 + y 2 = 100 . Differentiate both sides of the
equation with respect to t.
d ⎡ 2
d
x + y 2 ⎤ = [100]
⎦ dt
dt ⎣
dx
dy
2x + 2 y
=0
dt
dt
When y = 8, we can find x by using the
Pythagorean theorem.
2x
3y
x1/ 3 + y1/ 3 = 3
1 −2 / 3 1 −2 / 3
x
+ y
y′ = 0
3
3
y −2 / 3 y ′ = − x − 2 / 3
y′ = −
=−
x −2 / 3
y −2 / 3
y2 / 3
x2 / 3
3
x + 8 = 100
=−
x 2 = 100 − 64 = 36
x=6
= −3
2
2
When x = 6, y = 8, and
2(6)(3) + 2(8)
dx
= 3 , we have
dt
y2
x2
⎛ 1 ⎞ −4 ⎛ 1 ⎞ −4
6. ⎜ ⎟ x 5 + ⎜ ⎟ y 5 y′ = 0
⎝5⎠
⎝5⎠
dy
=0
dt
dy
=0
dt
dy
36
9
=−
=−
dt
16
4
dy
9
= − , thus the top of the ladder is sliding
dt
4
9
down the wall at the rate of
feet/sec.
4
4
y′ = −
y5
4
x5
4
⎛ y ⎞5
= −⎜ ⎟
⎝x⎠
⎛ 3 ⎞ −1 ⎛ 3 ⎞ −1
7. ⎜ ⎟ x 4 + ⎜ ⎟ y 4 y′ = 0
⎝4⎠
⎝4⎠
1
y′ = −
y4
1
x4
Problems 12.4
8. 3 y 2 y′ = 4
1. 2 x + 8 yy′ = 0
x + 4 yy′ = 0
4 yy′ = − x
y′ =
4
3y2
9. By the product rule xy′ + y (1) = 0 , xy′ = − y ,
x
4y
y′ = −
2. 6 x + 12 yy′ = 0
y′ = −
3 2
x
36 + 16
y′ = −
y2
x
2y
438
y
x
ISM: Introductory Mathematical Analysis
Section 12.4
10. 2 x + xy ′ + y (1) − 4 yy ′ = 0
xy ′ − 4 yy ′ = −2 x − y
−2 x − y
2x + y
=
y′ =
−x + 4 y
x − 4y
11. xy′ + y (1) − y′ − 11 = 0
y′( x − 1) = 11 − y
y′ =
11 − y
x −1
3x 2 − 3 y 2 y ′ = 3x 2 y ′ + 6 xy − 3x(2 yy ′) − 3 y 2
12.
y ′(−3 y 2 − 3x 2 + 6 xy ) = 6 xy − 3 y 2 − 3 x 2
y′ = 1
13. 6 x 2 + 3 y 2 y′ − 12 ( xy′ + y ) = 0
3 y 2 y′ − 12 xy′ = 12 y − 6 x 2
(
y′ ( y
)
y′ 3 y 2 − 12 x = 12 y − 6 x 2
y′ =
)
2
− 4 x = 4 y − 2 x2
4 y − 2x2
y2 − 4x
14. 6 x 2 + (3x) y′ + y (3) + 3 y 2 y′ = 0
(
)
y′ ( x + y ) = −2 x
y ′ 3 x + 3 y 2 = −6 x 2 − 3 y
2
y′ = −
15. x =
2
−y
2x 2 + y
x + y2
y + 4 y = y1/ 2 + y1/ 4
1 −1/ 2
1
y
y ′ + y −3 / 4 y ′
2
4
⎛ 2 y1/ 4 + 1 ⎞
⎛ 1
1 ⎞
= y′ ⎜
+
= y′ ⎜
⎟
⎟
⎜ 2 y1/ 2 4 y 3 / 4 ⎟
⎜ 4 y3 / 4 ⎟
⎝
⎠
⎝
⎠
1=
y′ =
4 y3 / 4
2 y1/ 4 + 1
(
)
( )
16. x3 3 y 2 y′ + y 3 3 x 2 + 1 = 0
y′ = −
1 + 3x2 y3
3 x3 y 2
439
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
(
)( ) x +1 y (1 + y′)
6e (1 + e ) ( x + y ) = 1 + y′
y′ = 6e (1 + e ) ( x + y ) − 1
17. 5 x3 (4 y 3 y ′) + 15 x 2 y 4 − 1 + 2 yy ′ = 0
y ′(20 x3 y 3 + 2 y ) = 1 − 15 x 2 y 4
y′ =
23. 2 1 + e3 x 3e3 x =
1 − 15 x 2 y 4
3x
20 x3 y 3 + 2 y
3x
1
x
1
(2 y + 1) y′ =
x
1
y′ =
x(2 y + 1)
18. 2 yy′ + y′ =
24.
( )
⎛1⎞
19. y ⎜ ⎟ + (ln x ) y′ = x e y y′ + e y (1)
⎝x⎠
⎡ ln( x ) − xe y ⎤ y′ = e y − y
⎣
⎦
x
20.
1
(1 + y ′)
x+ y
1
y′
e x + y + y ′e x + y =
+
x+ y x+ y
⎛
1 ⎞
1
y′ ⎜ e x+ y −
− e x+ y
⎟=
x
y
x
y
+
+
⎝
⎠
y ′ = −1
e x + y (1 + y ′) =
y′ = −
xe y − y
xy′ = − y ( x + 1)
y ( x + 1)
x
( )
21. ⎡ x e y y′ + e y (1) ⎤ + y′ = 0
⎣⎢
⎦⎥
⎛
⎞
x
y
− x + 1 ⎟ y′ =
− y +1
⎜
⎜ 2 y +1
⎟
2 x +1
⎝
⎠
xe y y′ + e y + y′ = 0
y
y′ = −
1+ 2
3
=− .
1+ 4
5
⎛
⎞
1
⋅ y′ ⎟ + y + 1(1)
26. x ⎜
⎜ 2 y +1
⎟
⎝
⎠
⎛ 1 ⎞
= y⎜
⎟ + x + 1( y′)
⎝ 2 x +1 ⎠
x
y
⋅ y′ − x + 1 ⋅ y′ =
− y +1
2 y +1
2 x +1
xy′ + y (1)
+1 = 0
xy
xy′ + y + xy = 0
( xe + 1) y′ = −e
1+ y
x + 2y
At the point (1, 2), y′ = −
x ⎡ ln( x) − xe y ⎤
⎣
⎦
y′ = −
3x
25. 1 + [ xy′ + y (1)] + 2 yy′ = 0
xy′ + 2 yy′ = −1 − y
( x + 2 y ) y′ = −(1 + y )
y
⎡ ln( x ) − xe y ⎤ y′ = xe − y
⎣
⎦
x
y′ =
3x
y
y′ =
ey
xe y + 1
y
−
2 x +1
x
−
2 y +1
y +1
x +1
3
At (3, 3),
22. 8 x + 18 yy ′ = 0
8 x = −18 yy ′
8x
4x
y′ = −
=−
18 y
9y
dy 4 − 2
=
= 1.
dx 3 − 2
4
440
ISM: Introductory Mathematical Analysis
Section 12.4
27. 8 x + 18 yy′ = 0
32. p = 400 − q
8x
4x
y′ = −
=−
18 y
9y
(
d
d
( p) =
400 − q
dp
dp
⎛ 1⎞
Thus at ⎜ 0, ⎟ , y′ = 0; at ( x0 , y0 ) ,
⎝ 3⎠
4 x0
y′ = −
.
9 y0
28.
1= −
2
2
33. p =
3
x + x yy ′ + xy + y y ′ = 2 yy ′
( x 2 y + y 3 − 2 y ) y ′ = − x3 − xy 2
y′ =
− x( x 2 + y 2 )
y ( x 2 + y 2 − 2)
40
(q + 5)
3
⋅
dq
dp
dq
(q + 5)3
=−
dp
40
3x2 + y
y′ = −
x + 2y
At (−1, 1), y ′ = −4 and the tangent line is given
34.
by y − 1 = −4[x − (−1)], or y = −4x − 3.
30. 2 yy′ + [ xy′ + y (1)] − 2 x = 0
2x − y
2y + x
1
and the tangent line is given by
2
1
1
y − 3 = ( x − 4), or y = x + 1.
2
2
At (4, 3), y′ =
p=
10
2
q +3
d
d ⎡ 10 ⎤
( p) =
⎢
⎥
dp
dp ⎢⎣ q 2 + 3 ⎥⎦
20q
dq
1= −
⋅
2
2 dp
(q + 3)
dq
(q 2 + 3)2
=−
dp
20q
From the original equation, we have
10
dq
q 2 + 3 = . Thus we can write
as
p
dp
31. p = 100 − q 2
(
(q + 5)2
1= −
29. 3x 2 + xy ′ + y + 2 y ′ = 0
d
d
( p) =
100 − q 2
dp
dp
20
d
d ⎡ 20 ⎤
( p) =
⎢
⎥
dp
dp ⎢⎣ (q + 5)2 ⎥⎦
d
d ⎡
( p) =
20(q + 5)−2 ⎤
⎦
dp
dp ⎣
At (0, 2), y ′ = 0.
y′ =
1 dq
⋅
2 q dp
dq
= −2 q
dp
2( x 2 + y 2 )(2 x + 2 yy ′) = 8 yy ′
( x 2 + y 2 )( x + yy ′) = 2 yy ′
3
)
)
( )
dq
=−
10
p
dq
1 = −2q ⋅
dp
dp
dq
1
=−
dp
2q
35. ln
2
20q
=−
I
= −λ t
I0
ln I − ln I 0 = −λ t
1 dI
= −λ
I dt
dI
= −λ I
dt
441
5
qp 2
.
Chapter 12: Additional Differentiation Topics
⎛
⎞
E
36. 1.5M = log ⎜
11 ⎟
⎝ 2.5 × 10 ⎠
(
1.5M = log E − log 2.5 × 1011
d
(1.5M ) =
dM
d
(1.5M ) =
dM
1.5 =
d
dM
d
dM
ISM: Introductory Mathematical Analysis
1 2
I = SI + I . Differentiating implicitly
4
with respect to I:
dS 1
dS ⎤
⎡
2S
+ I = ⎢ S (1) + I
+ 1,
dI 2
dI ⎥⎦
⎣
39. S 2 +
)
(
)
⎡ ln E
⎤
⎢ ln10 − log ( 2.5 × 10 ) ⎥
⎣
⎦
⎡log E − log 2.5 × 1011 ⎤
⎣⎢
⎦⎥
dS
dS
I
−I
= S +1− ,
dI
dI
2
dS 2S + 2 − I dS 2S + 2 − I
(2 S − I )
=
,
=
.
dI
2
dI
2(2S − I )
2S
11
1 ⎛ 1 dE ⎞
⋅
ln10 ⎜⎝ E dM ⎟⎠
Marginal propensity to consume =
dE
= 1.5E ln10
dM
d
d ⎡ ln E
⎤
(1.5M ) =
− log(2.5 × 1011 ) ⎥
dE
dE ⎢⎣ ln10
⎦
dM
1 1
1.5
=
⋅
dE ln10 E
dM
1
=
dE 1.5E ln10
Thus
df
df
f
,
=− .
dλ dλ
λ
Solving v = f λ for f and differentiating: f =
λ
f (t )
1
+σ
= C1 + C2t. Thus
1 − f (t )
1 − f (t )
f ′(t )
f ′(t )
σ f ′(t )
+
+
= C2
f (t ) 1 − f (t ) [1 − f (t )]2
to λ:
v
dC
24 + 2 − 16
10 6 3
= 1−
= 1−
=
= .
dI
2(24 − 16)
16 16 8
ln f (t ) − ln[1 − f (t )] + σ [1 − f (t )]−1 = C1 + C2t ,
37. v = f λ. Differentiating implicitly with respect
0 = f (1) + λ
dC
2S + 2 − I
= 1−
. When I = 16 and
dI
2(2 S − I )
S = 12,
40. ln
dC
dS
.
= 1−
dI
dI
⎡ 1
⎤
1
σ
f ′(t ) ⎢
+
+
⎥ = C2
2
⎣⎢ f (t ) 1 − f (t ) [1 − f (t )] ⎦⎥
,
⎡ [1 − f (t )]2 + f (t )[1 − f (t )] + σ f (t ) ⎤
f ′(t ) ⎢
⎥ = C2
f (t )[1 − f (t )]2
⎢⎣
⎥⎦
⎡ [1 − f (t )][1 − f (t ) + f (t )] + σ f (t ) ⎤
f ′(t ) ⎢
⎥ = C2
f (t )[1 − f (t )]2
⎢⎣
⎥⎦
⎡ [1 − f (t )] + σ f (t ) ⎤
f ′(t ) ⎢
⎥ = C2
2
⎣⎢ f (t )[1 − f (t )] ⎦⎥
df
v
fλ
f
=−
=−
= − , which is the same
2
2
λ
dλ
λ
λ
as before.
so
38. (P + a)(v + b) = k
d
d
[( P + a )(v + b)] =
(k )
dP
dP
dv
( P + a)
+ (v + b)(1) = 0
dP
dv
v+b
=−
. From the original equation,
dP
P+a
dv
k
v+b =
. Thus we can write
as
dP
( P + a)
Thus f ′(t ) =
dv
k
=−
.
dP
( P + a)2
442
C2 f (t )[1 − f (t )]2
σ f (t ) + [1 − f (t )]
ISM: Introductory Mathematical Analysis
Section 12.5
Problems 12.5
(
)
1. y = ( x + 1)2 ( x − 2) x 2 + 3 . Take natural logarithms of both sides,
(
)
ln y = ln ⎡ ( x + 1) 2 ( x − 2) x 2 + 3 ⎤ .
⎢⎣
⎥⎦
Using properties of logarithms on the right side gives
(
)
ln y = 2 ln( x + 1) + ln( x − 2) + ln x 2 + 3 .
Differentiating both sides with respect to x,
y′
2
1
2x
=
+
+
.
2
y x +1 x − 2 x + 3
Solving for y′ ,
⎡ 2
1
2x ⎤
+
+
y′ = y ⎢
⎥.
2
x
+
x
−
1
2
x + 3⎦
⎣
Expressing y′ in terms of x,
(
)
⎡ 2
1
2x ⎤
+
+
y′ = ( x + 1)2 ( x − 2) x 2 + 3 ⎢
⎥
2
1
2
x
+
x
−
x + 3⎦
⎣
(
)
4⎤
⎡
2. ln y = ln ⎢ (3 x + 4)(8 x − 1) 2 3x 2 + 1 ⎥
⎣
⎦
(
)
= ln(3x + 4) + 2 ln(8 x − 1) + 4 ln 3 x 2 + 1
y′
3
8
6x
=
+ 2⋅
+ 4⋅
y 3x + 4
8x −1
3x2 + 1
⎡ 3
16
24 x ⎤
+
+
y′ = y ⎢
⎥
⎣ 3x + 4 8 x − 1 3x2 + 1 ⎦
4 ⎡ 3
16
24 x ⎤
+
+
= (3 x + 4)(8 x − 1)2 3 x 2 + 1 ⎢
⎥
x
+
x
−
3
4
8
1
3x2 + 1 ⎦
⎣
(
(
)
)
2
⎡
⎤
3. ln y = ln ⎢ 3x3 − 1 (2 x + 5)3 ⎥
⎣
⎦
(
)
= 2 ln 3 x3 − 1 + 3ln(2 x + 5)
y′
9 x2
2
= 2⋅
+ 3⋅
3
+5
y
2
x
3x − 1
⎡ 18 x 2
6 ⎤
y′ = y ⎢
+
⎥
3
⎢⎣ 3 x − 1 2 x + 5 ⎥⎦
⎡ 18 x 2
2
6 ⎤
y′ = 3 x3 − 1 (2 x + 5)3 ⎢
+
⎥
3
⎢⎣ 3x − 1 2 x + 5 ⎥⎦
(
)
443
Chapter 12: Additional Differentiation Topics
4.
ISM: Introductory Mathematical Analysis
y = (2 x 2 + 1) 8 x 2 − 1
ln y = ln ⎡⎢ (2 x 2 + 1) 8 x 2 − 1 ⎤⎥
⎣
⎦
1
= ln(2 x 2 + 1) + ln(8 x 2 − 1)
2
y′
4x
1 16 x
=
+ ⋅
y 2 x2 + 1 2 8x2 − 1
⎡ 4x
8x ⎤
y′ = y ⎢
+
⎥
2
⎣ 2 x + 1 8x2 − 1 ⎦
⎡ 4x
8x ⎤
= (2 x 2 + 1) 8 x 2 − 1 ⎢
+
⎥
2
⎣ 2 x + 1 8x2 − 1 ⎦
5. y = x + 1 x 2 − 2 x + 4
ln y = ln ⎛⎜ x + 1 x 2 − 2 x + 4 ⎞⎟
⎝
⎠
1
1
1
2
ln y = ln( x + 1) + ln x − 2 + ln( x + 4)
2
2
2
y′ 1 ⎡ 1
2x
1 ⎤
=
+
+
y 2 ⎢⎣ x + 1 x 2 − 2 x + 4 ⎥⎦
(
y′ =
=
)
y⎡ 1
2x
1 ⎤
+
+
2 ⎢⎣ x + 1 x 2 − 2 x + 4 ⎥⎦
x + 1 x2 − 2 x + 4 ⎡ 1
2x
1 ⎤
+
+
⎢
⎥
2
2
⎣ x +1 x − 2 x + 4 ⎦
6. ln y = ln ⎡ (2 x + 1) x3 + 2 3 2 x + 5 ⎤
⎢⎣
⎥⎦
1
1
= ln(2 x + 1) + ln( x3 + 2) + ln(2 x + 5)
2
3
2
1 3x 2
1
2
y′
=
+ ⋅
+ ⋅
3
y 2x + 1 2 x + 2 3 2x + 5
⎡ 2
⎤
3x2
2
+
+
y′ = y ⎢
⎥
3
⎣⎢ 2 x + 1 2( x + 2) 3(2 x + 5) ⎦⎥
⎡ 2
⎤
3x 2
2
= (2 x + 1) x3 + 2 3 2 x + 5 ⎢
+
+
⎥
3
⎢⎣ 2 x + 1 2( x + 2) 3(2 x + 5) ⎥⎦
444
ISM: Introductory Mathematical Analysis
Section 12.5
1 − x2 1
7. ln y = ln
= ln 1 − x 2 − ln(1 − 2 x )
1− 2x
2
y′ 1 −2 x
−2
= ⋅
−
y 2 1 − x2 1 − 2 x
⎡
x
2 ⎤
+
y′ = y ⎢ −
2 1 − 2x ⎥
⎣ 1− x
⎦
(
y′ =
)
10. ln y = ln
(
)
y′ =
9. y =
11.
x2 + 5 ⎡ 2 x
1 ⎤
−
x + 9 ⎢⎣ x 2 + 5 x + 9 ⎥⎦
1
2
( 2x
2
+2
)
2
( x + 1)2 (3x + 2)
(
)
2 ⎤
⎡
2
⎢ 2x + 2
⎥
ln y = ln ⎢
⎥
2
⎢ ( x + 1) (3x + 2) ⎥
⎢⎣
⎥⎦
(
)
2
= 2 ln 2 x + 2 − 2 ln( x + 1) − ln(3 x + 2)
+2
)
(
)
)
2
⎡1
4x
x ⎤
+
−
⎢
⎥
2
2 + x2 ⎦
2 + x2 ⎣ x 1 + x
( x + 3)( x − 2)
2x −1
( x + 3)( x − 2)
ln y = ln
2x −1
1
1
1
= ln( x + 3) + ln( x − 2) − ln(2 x − 1)
2
2
2
1 1
1
2
y′ 1 1
= ⋅
+ ⋅
− ⋅
y 2 x + 3 2 x − 2 2 2x −1
1
2 ⎤
y⎡ 1
+
−
y′ = ⎢
2 ⎣ x + 3 x − 2 2 x − 1 ⎥⎦
1 ( x + 3)( x − 2) ⎡ 1
1
2 ⎤
=
+
−
⎢
⎥
2
2x −1
⎣ x + 3 x − 2 2x −1⎦
y=
3
(
)
6 x3 + 1
2
x 6 e−4 x
(
(
)
)
1⎡
ln(6) + 2 ln x3 + 1 − 6 ln( x) − (−4 x ) ln e ⎤
⎦⎥
3 ⎣⎢
1
= ⎡ ln(6) + 2 ln x3 + 1 − 6 ln( x) + 4 x ⎤
⎦⎥
3 ⎣⎢
=
⎡ 8x
2
3 ⎤
y′ = y ⎢
−
−
2
x
1
3
x
+
+ 2 ⎦⎥
⎣ 2x + 2
2
(
x 1 + x2
12. ln y = ln
y′
4x
1
3
= 2⋅
− 2⋅
−
2
y
x + 1 3x + 2
2x + 2
( 2x
)
⎡1
4x
x ⎤
y′ = y ⎢ +
−
⎥
2
x
1+ x
2 + x2 ⎦
⎣
y′ 1 ⎡ 2 x
1 ⎤
= ⎢
−
2
y 2 ⎣ x + 5 x + 9 ⎥⎦
y ⎡ 2x
1 ⎤
y′ = ⎢
−
2 ⎣ x 2 + 5 x + 9 ⎥⎦
y′ =
2
2 + x2
(
x +5 1 ⎡
=
ln x 2 + 5 − ln( x + 9) ⎤
⎦⎥
x+9
2 ⎣⎢
8. ln y = ln
)
1
= ln x + 2 ln 1 + x 2 − ln 2 + x 2
2
y′ 1
2x
1 2x
= + 2⋅
− ⋅
y x
1 + x2 2 2 + x2
1 − x2 ⎡ x
2 ⎤
+
⎢
2
1 − 2 x ⎣ x − 1 1 − 2 x ⎥⎦
2
(
x 1 + x2
2
⎤
y′ 1 ⎡ 3 x 2
6
= ⎢2 ⋅
− + 4⎥
y 3 ⎢⎣ x3 + 1 x
⎦⎥
2
⎤
y ⎡ 6x
6
− + 4⎥
y′ = ⎢
3
3 ⎣⎢ x + 1 x
⎦⎥
⎡ 4x
2
3 ⎤
=
−
−
⎢ 2
⎥
2
( x + 1) (3x + 2) ⎣ x + 1 x + 1 3 x + 2 ⎦
(
)
3
1 3 6 x +1
y=
3
x 6 e−4 x
445
2
⎡ 6 x2
⎤
6
− + 4⎥
⎢ 3
⎣⎢ x + 1 x
⎦⎥
Chapter 12: Additional Differentiation Topics
2
ISM: Introductory Mathematical Analysis
17. y = (3 x + 1) 2 x . Thus
2
13. y = x x +1 , thus ln y = ln x x +1 = ( x 2 + 1) ln x.
y′
1
= ( x 2 + 1) ⋅ + (ln x)(2 x)
y
x
2
⎛ x +1
⎞
+ 2 x ln x ⎟
y′ = y ⎜
⎜ x
⎟
⎝
⎠
2
⎛
⎞
2
+
x
1
= x x +1 ⎜
+ 2 x ln x ⎟
⎜ x
⎟
⎝
⎠
14. y = (2 x)
x
ln y = ln ⎡( 3x + 1)
⎣⎢
⎡ 3x
⎤
= 2(3 x + 1) 2 x ⎢
+ ln(3x + 1) ⎥
+
3
x
1
⎣
⎦
18. y = ( x 2 + 1) x +1 , thus
ln y = ln( x 2 + 1) x +1 = ( x + 1) ln( x 2 + 1).
y′
2x
= x + 1⋅
+ ln( x 2 + 1) ⋅1
2
y
x +1
⎡ 2 x( x + 1)
⎤
+ ln( x 2 + 1) ⎥
y′ = y ⎢
2
⎣ x +1
⎦
+
2
x
(
x
1)
⎡
⎤
= ( x 2 + 1) x +1 ⎢
+ ln( x 2 + 1) ⎥
2
⎣ x +1
⎦
y′
1
⎡1⎤
= x ⎢ ⎥ + [ln 2 + ln x] ⋅
y
2 x
⎣x⎦
⎡ 1 ln(2 x) ⎤
y′ = y ⎢
+
⎥
2 x ⎦
⎣ x
⎡ 2 + ln(2 x) ⎤
y′ = (2 x) x ⎢
⎥
⎣ 2 x ⎦
( )
1
ln x
ln x =
.
x
x
19. y = 4e x x3 x . Thus
(
= ln 4 + x + 3 x ln x.
⎡ ⎛1⎞
⎤
y′
= 1 + 3 ⎢ x ⎜ ⎟ + (ln x)(1) ⎥
y
x
⎣ ⎝ ⎠
⎦
y′ = y (4 + 3ln x)
1
x x (1 − ln x)
y′ = 4e x x3 x (4 + 3ln x)
x2
x
20. y = (ln x)e . Thus ln y = e x ln(ln x).
x
⎛ 3 ⎞
16. y = ⎜ ⎟ . Thus
⎝ x2 ⎠
⎛ 3 ⎞
ln y = x ln ⎜ ⎟ = x[ln 3 − 2 ln x].
⎝ x2 ⎠
y′
⎛ 2⎞
= x ⎜ − ⎟ + (ln 3 − 2 ln x )(1)
y
⎝ x⎠
⎛ 3 ⎞
= −2 + ln ⎜ ⎟
⎝ x2 ⎠
⎡
⎛ 3
y ′ = y ⎢ −2 + ln ⎜
⎝ x2
⎣
)
ln y = ln 4 + ln e x x3 x = ln 4 + ln e x + ln x3 x
1
y′ x x − (ln x)(1)
=
y
x2
⎡ 1 − ln x ⎤
y′ = y ⎢
⎥
⎣ x2 ⎦
y′ =
= 2 x ln(3 x + 1)
⎡ 3x
⎤
+ ln(3 x + 1) ⎥
y′ = 2 y ⎢
+
3
x
1
⎣
⎦
ln y = ln(2 x) x = x [ln 2 + ln x ].
1
⎦⎥
⎧ ⎛ 3 ⎞
⎫
y′
= 2 ⎨x ⎜
⎟ + [ln(3x + 1)](1) ⎬
+
y
3
x
1
⎠
⎩ ⎝
⎭
. Thus
15. y = x x . Thus ln y =
2x ⎤
x
⎞⎤ ⎛ 3 ⎞ ⎡
⎛ 3
⎟ ⎥ = ⎜ 2 ⎟ ⎢ −2 + ln ⎜ 2
⎠⎦ ⎝ x ⎠ ⎣
⎝x
y′
⎡ 1 ⎤
x
= ex ⎢
⎥ + [ln(ln x)]e
y
⎣ x ln x ⎦
⎡ 1
⎤
+ ln(ln x) ⎥ e x
y′ = y ⎢
x
x
ln
⎣
⎦
x ⎡
1
⎤
= (ln x)e ⎢
+ ln(ln x ) ⎥ e x
x
x
ln
⎣
⎦
⎞⎤
⎟⎥
⎠⎦
446
ISM: Introductory Mathematical Analysis
Section 12.5
21. y = (4 x − 3) 2 x +1
25. y = e x ( x 2 + 1) x
ln y = ln(4 x − 3) 2 x +1 = (2 x + 1) ln(4 x − 3)
ln y = ln e x + ln( x 2 + 1) x
(
y′
⎡ 4 ⎤
= (2 x + 1) ⎢
⎥ + [ln(4 x − 3)](2)
y
⎣ 4x − 3 ⎦
(
)
⎡ ⎛ 2 x ⎞⎤ ⎡
y′
2
⎤
= 1+ ⎢x ⎜
⎟ ⎥ + ⎣⎢ln x + 1 (1) ⎥⎦
2
y
⎣ ⎝ x + 1 ⎠⎦
⎡ 4(2 x + 1)
⎤
+ 2 ln(4 x − 3) ⎥
y′ = y ⎢
−
4
x
3
⎣
⎦
⎡
⎤
2 x2
+ ln x 2 + 1 ⎥
y′ = y ⎢1 +
2
⎣⎢ x + 1
⎦⎥
When x = 1, then y = 2e and
y′ = 2e[1 + 1 + ln(2)] = 2e(2 + ln 2). Thus an
equation of the tangent line is
y − 2e = 2e(2 + ln 2)(x − 1), or
y = (4e + 2e ln 2)x − 2e − 2e ln 2.
(
dy
⎡ 12
⎤
When x = 1, then
= 1 ⎢ + 2 ln(1) ⎥ = 12.
dx
⎣1
⎦
22. y = (ln x)ln x
ln y = ln(ln x)ln x = (ln x) ln(ln x)
y′
⎡ 1 1⎤
⎛1⎞
= (ln x ) ⎢
⋅ ⎥ + [ln(ln x)] ⎜ ⎟
y
⎣ ln x x ⎦
⎝x⎠
26.
⎡ 1 ln(ln x) ⎤
y′ = y ⎢ +
x ⎥⎦
⎣x
⎡ 1 + ln(ln x) ⎤
y′ = (ln x)ln x ⎢
⎥
x
⎣
⎦
)
y = xx
ln y = x ln x
y′
1
= x ⋅ + (ln x )(1) = 1 + ln x
y
x
When x = 1,
dy 1 ⎡ 1 + ln(1) ⎤
−1
=1 ⎢
When x = e,
⎥=e .
dx
⎣ e ⎦
y′
= 1 + ln1 = 1 + 0 = 1.
y
27. y = (3x )−2 x
ln y = –2x ln(3x)
⎧ ⎡1
⎫
y′
⎤
= −2 ⎨ x ⎢ (3) ⎥ + [ln(3 x)](1) ⎬
y
⎩ ⎣ 3x ⎦
⎭
23. y = ( x + 1)( x + 2) 2 ( x + 3) 2
ln y = ln(x + 1) + 2 ln(x + 2) + 2 ln(x + 3)
y′
1
2
2
=
+
+
y x +1 x + 2 x + 3
= –2[1 + ln(3x)]
y′
⋅100 gives the percentage rate of change.
y
Thus –2[1 + ln(3x)](100) = 60
1 + ln(3x) = –0.3
ln(3x) = –1.3
2
2 ⎤
⎡ 1
y′ = y ⎢
+
+
⎥
⎣ x +1 x + 2 x + 3⎦
When x = 0, then y = 36 and y′ = 96. Thus an
equation of the tangent line is
y – 36 = 96(x – 0), or y = 96x + 36.
24.
)
= x + x ln x 2 + 1
3x = e −1.3
1
x=
3e1.3
y = xx
ln y = x ln x
y′
1
= x ⋅ + (ln x )(1) = 1 + ln x
y
x
y ′ = y (1 + ln x) = x x (1 + ln x)
When x = 1, then y = 1 and
y ′ = 11 (1 + ln1) = 1(1 + 0) = 1. An equation of the
28. y = [ f ( x)]g ( x )
ln y = g(x) ln[f(x)]
⎛ 1
⎞
y′
= g ( x) ⎜
⋅ f ′( x) ⎟ + ln[ f ( x)]g ′( x)
y
⎝ f ( x)
⎠
tangent line is y − 1 = 1(x − 1) or y = x.
⎛
⎞
g ( x)
y′ = y ⎜ f ′( x)
+ g ′( x) ln[ f ( x)] ⎟
f ( x)
⎝
⎠
⎛
⎞
g ( x)
y′ = [ f ( x)]g ( x ) ⎜ f ′( x)
+ g ′( x) ln[ f ( x)] ⎟
f ( x)
⎝
⎠
447
Chapter 12: Additional Differentiation Topics
29.
ISM: Introductory Mathematical Analysis
r′
p′
q′
⋅100% = ⋅100% + ⋅100%
r
p
q
p′
= (1 + η ) 100%
p
where η =
η=
p
q
dp
dq
=
p dq
⋅ .
q dp
p
500 − 40 p + p 2
⋅ (−40 + 2 p)
1
⎛1 ⎞
% increase in price will result in a (1 − 1.2) ⎜ % ⎟ = −0.1% change in
2
⎝2 ⎠
revenue, which is a 0.1% decrease in revenue.
When p = 15, then η = −1.2 and a
30.
r′
p′
q′
⋅100% = ⋅100% + ⋅100%
r
p
q
p′
= (1 + η ) 100%
p
where η =
p
q
dp
dq
=
p dq
⋅ .
q dp
p
η=
⋅ (−40 + 2 p)
500 − 40 p + p 2
When p = 15, then η = −1.2 and a 10% decrease in price will result in a (1 − 1.2)(−10%) = 2% change in revenue,
which is a 2% increase in revenue.
Principles in Practice 12.6
3
1. Let f ( x) = 20 x − 0.01x 2 − 850 + 3ln x, then f ′( x) = 20 − 0.02 x + . f(10) ≈ –644 and f(50) ≈ 137,
x
so we use 50 to be the first approximation, x1 , to find the break-even quantity between 10 and 50.
xn +1 = xn −
= xn −
=
=
f ( xn )
f ′ ( xn )
= xn −
20 xn − 0.01xn2 − 850 + 3ln xn
20 − 0.02 xn + 3 xn−1
20 xn2 − 0.01xn3 − 850 xn + 3xn ln xn
20 xn − 0.02 xn2 + 3
(
20 xn2 − 0.02 xn3 + 3xn − 20 xn2 − 0.01xn3 − 850 xn + 3 xn ln xn
20 xn − 0.02 xn2
+3
−0.01xn3 + 853 xn − 3xn ln xn
20 xn − 0.02 xn2 + 3
x2 = 50 −
f (50)
≈ 42.82602
f ′(50)
x3 = 42.82602 −
f (42.82602)
≈ 42.85459
f ′(42.82602)
448
)
ISM: Introductory Mathematical Analysis
x4 = 42.85459 −
Section 12.6
f (42.85459)
≈ 42.85459
f ′(42.85459)
Since the values of x3 and x4 differ by less than 0.0001, we take the first break-even quantity
to be x ≈ 42.85459 or 43 televisions.
f(1900) ≈ 1073 and f(2000) ≈ –827, so we use 2000 to be
the first approximation, x1 , for the break-even quantity between 1900 and 2000.
x2 = 2000 −
f (2000)
≈ 1958.63703
f ′(2000)
x3 = 1958.63703 −
f (1958.63703)
≈ 1957.74457
f ′(1958.63703)
x4 = 1957.74457 −
f (1957.74457)
≈ 1957.74415
f ′(1957.74457)
x5 = 1957.74415 −
f (1957.74415)
≈ 1957.74415
f ′(1957.74415)
Since the values of x4 and x5 differ by less than 0.0001, we take the second break-even quantity to be
x ≈ 1957.74415 or 1958 televisions.
Problems 12.6
1. We want a root of f ( x) = x3 − 4 x + 1 = 0. We see that f(0) = 1 and f(1) = –2 have opposite signs, so there must be
a root between 0 and 1. Moreover, f(0) is closer to 0 than is f(1), so we select x1 = 0 as our initial estimate. Since
f ′( x) = 3x 2 − 4, the recursion formula is
xn +1 = xn −
f ( xn )
x3 − 4 xn + 1
= xn − n
.
f ′ ( xn )
3 xn2 − 4
Simplifying gives xn +1 =
2 xn3 − 1
3 xn2 − 4
. Thus we obtain:
n
xn
xn +1
1
0.00000
0.25000
2
0.25000
0.25410
3
0.25410
0.25410
Because x4 − x3 < 0.0001, the root is approximately x4 = 0.25410.
3
⎛1⎞
2. Let f ( x) = x3 + 2 x 2 − 1. f ⎜ ⎟ = − and
2
8
⎝ ⎠
1
⎛1⎞
f(1) = 2 (note the sign change). Since f ⎜ ⎟ is closer to 0 than is f(1), we select x1 = . We have
2
⎝2⎠
x3 + 2 xn2 − 1
f ′( x) = 3x 2 + 4 x, so the recursion formula is xn +1 = xn − n
3 xn2 + 4 xn
449
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
n
xn
xn +1
xn +1
1
2.50000
2.58974
0.50000
0.63636
2
2.58974
2.58425
2
0.63636
0.61838
3
2.58425
2.58423
3
0.61838
0.61803
4
2.58423
2.58423
4
0.61803
0.61803
n
xn
1
Since x5 − x4 < 0.0001, the root is
Because x5 − x4 < 0.0001, the root is
approximately x5 = 2.58423.
approximately x5 = 0.61803.
(Note that f ′(0) = 0, so we cannot use 0 for
x1.)
5. Let f ( x) = x3 + x + 1. We have f(−1) = −1 and
f(0) = 1 (note the sign change). Choose x1 = −1.
Since f ′( x) = 3 x 2 + 1, the recursion formula is
3
3. Let f ( x) = x − x − 1. We have f(1) = –1 and
f(2) = 5 (note the sign change). Since f(1) is
closer to 0 than is f(2), we choose x1 = 1. We
xn +1 = xn −
have f ′( x) = 3 x 2 − 1, so the recursion formula is
xn +1 = xn −
=
f ( xn )
f ′ ( xn )
2 xn3 + 1
3xn2
= xn −
=
xn3 − xn − 1
3xn2 − 1
−1
n
xn
xn +1
1
1.00000
1.50000
2
1.50000
1.34783
3
1.34783
1.32520
4
1.32520
1.32472
5
1.32472
1.32472
3xn2 + 1
n
xn
xn +1
1
−1
−0.75000
2
−0.75000
−0.68605
3
−0.68605
−0.68234
4
−0.68234
−0.68233
approximately x5 = −0.68233.
6. x3 = 2 x + 5, so use f ( x) = x3 − 2 x − 5 = 0. We
have f(2) = –1 and f(3) = 16, so f(2) is closer to 0
than is f(3). We choose x1 = 2. Since
f ′( x) = 3x 2 − 2, the recursion formula is
approximately x6 = 1.32472.
xn +1 = xn −
4. Let f ( x) = x3 − 9 x + 6. We have f(2.5) = –0.875
and f(3) = 6. Since f(2.5) is closer to 0 than is
f(3), we choose x1 = 2.5. We have
f ′( x) = 3x − 9 , so xn +1 = xn −
2 xn3 − 1
Because x5 − x4 < 0.0001, the root is
Since x6 − x5 < 0.0001, the root is
2
xn3 − 9 xn + 6
3xn2 − 9
f ( xn )
x3 + xn + 1
= xn − n
f ′( xn )
3 xn2 + 1
.
xn3 − 2 xn − 5
3 xn2 − 2
=
2 xn3 + 5
3 xn2 − 2
n
xn
xn +1
1
2.00000
2.10000
2
2.10000
2.09457
3
2.09457
2.09455
Because x4 − x3 < 0.0001, the root is
approximately x4 = 2.09455.
450
ISM: Introductory Mathematical Analysis
Section 12.6
7. x 4 = 3x − 1 , so use f ( x) = x 4 − 3x + 1 = 0 . Since
f(0) = 1 and f(1) = –1 (note the sign change), f(0)
and f(1) are equally close to 0. We shall choose
9. Let f ( x) = x 4 − 2 x3 + x 2 − 3. f(1) = –3 and
f(2) = 1 (note the sign change), so f(2) is closer
to 0 than is f(1). We choose x1 = 2. Since
x1 = 0. Since f ′( x) = 4 x3 − 3, the recursion
formula is
f ( xn )
x 4 − 3 xn + 1
= xn − n
xn +1 = xn −
f ′ ( xn )
4 xn3 − 3
=
f ′( x) = 4 x3 − 6 x 2 + 2 x, the recursion formula is
xn +1 = xn −
x 4 − 2 xn3 + xn2 − 3
f ( x)
= xn − n
f ′ ( xn )
4 xn3 − 6 xn2 + 2 xn
3xn4 − 1
n
xn
xn +1
4 xn3 − 3
1
2.00000
1.91667
2
1.91667
1.90794
3
1.90794
1.90785
n
xn
xn +1
1
0.00000
0.33333
2
0.33333
0.33766
3
0.33766
0.33767
Because x4 − x3 < 0.0001, the root is
approximately x4 = 1.90785.
10. Let f ( x) = x 4 − x3 + x − 2. f(1) = –1 and
f(2) = 8, so f(1) is closer to 0 than is f(2). We
Because x4 − x3 < 0.0001, the root is
approximately x4 = 0.33767.
choose x1 = 1. Since f ′( x) = 4 x3 − 3x 2 + 1, the
recursion formula is
8. Let f ( x) = x 4 + 4 x − 1. Since f(–2) = 7 and
f(–1) = –4, f(–1) is closer to 0 than is f(–2).
However, f ′(−1) = 0, so we shall choose
xn +1 = xn −
x1 = −2. Since f ′( x) = 4 x3 + 4, the recursion
formula is
xn +1 = xn −
xn4 + 4 xn − 1
4 xn3 + 4
=
3xn4 + 1
4 xn3 + 4
xn4 − xn3 + xn − 2
4 xn3 − 3 xn2 + 1
n
xn
xn +1
1
1.00000
1.50000
2
1.50000
1.34677
n
3
1.34677
1.31040
xn
xn +1
4
1.31040
1.30858
1
−2.00000
−1.75000
5
1.30858
1.30857
2
−1.75000
−1.67092
3
−1.67092
−1.66332
4
−1.66332
−1.66325
Because x6 − x5 < 0.0001, the root is
approximately x6 = 1.30857.
11. The desired number is x, where x3 = 71, or
Because x5 − x4 < 0.0001, the root is
x3 − 71 = 0. Thus we want to find a root of
approximately x5 = −1.66325.
f ( x) = x3 − 71 = 0. Since 43 = 64 , the solution
should be close to 4, so we choose x1 = 4 as our
initial estimate. We have f ′( x) = 3x 2 , so the
recursion formula is
f ( xn )
x3 − 71 2 xn3 + 71
= xn − n
=
xn +1 = xn −
f ′ ( xn )
3xn2
3xn2
451
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
n
xn
xn +1
n
xn
xn +1
1
4
4.146
1
3
2.37
2
4.146
4.141
2
2.37
2.03
4
4.141
4.141
3
2.03
1.94
4
1.94
1.94
Thus to three decimal places,
3
71 = 4.141.
Thus the solutions are –4.99 and 1.94.
12. The desired number is x, where x 4 = 19, or
14. We must solve ln x = 5 – x. That is, we must
determine all roots of f(x) = ln(x) + x – 5 = 0. A
rough sketch shows that the graph of the
logarithmic function y = ln x intersects the line
y = 5 – x at one point, where x is between 3 and
1
4. We choose x1 = 3. Since f ′( x) = + 1, the
x
recursion formula is
f ( xn )
ln( xn ) + xn − 5
= xn −
xn +1 = xn −
1 +1
f ′ ( xn )
x 4 − 19 = 0. Thus we want to find a root of
f ( x) = x 4 − 19. Since 24 = 16, the solution
should be close to 2, so we choose x1 = 2 as our
initial estimate. We have f ′( x) = 4 x3 , so the
recursion formula is
xn +1 = xn −
=
f ( xn )
x 4 − 19
= xn − n
f ′( xn )
4 xn3
3 xn4 + 19
xn
4 xn3
n
xn
xn +1
n
xn
xn +1
1
3
3.676
1
2
2.09
2
2.676
3.693
2
2.09
2.09
3
3.693
3.693
Thus to two decimal places,
4 19
Thus the solution is approximately 3.693.
= 2.09.
15. The break-even quantity is the value of q when
total revenue and total cost are equal: r = c, or
r – c = 0. Thus we must find a root of
13. We want real solutions to e x = x + 5. Thus we
want to find roots of f ( x) = e x − x − 5 = 0. A
(
f (q) = q − 250 + 0.1q3 = 0, so f ′(q) = 1 + 0.3q 2 .
The recursion formula is
f ( qn )
q − 250 + 0.1qn3
= qn − n
qn +1 = qn −
f ′ ( qn )
1 + 0.3qn 2
two roots. Since f ′( x) = e x − 1, the recursion
formula is
f ( xn )
e xn − xn − 5
= xn −
xn +1 = xn −
f ′ ( xn )
e xn − 1
We choose q1 = 13, as suggested.
If x1 = −5, we obtain
n
xn
xn +1
1
−5
−4.99
2
−4.99
−4.99
)
3q − 250 + 2q − 0.1q3 = 0, or
rough sketch of the exponential function y = e x
and the line y = x + 5 shows that there are two
intersection points: one when x is near –5, and
the other when x is near 3. Thus we must find
n
qn
qn +1
1
13
13.33
2
13.33
13.33
Thus q ≈ 13.33.
If x1 = 3, we obtain:
452
ISM: Introductory Mathematical Analysis
16. a.
Section 12.6
The break-even quantity is the value of q
when total cost = total revenue: c = r,
c – r = 0. Thus we solve
q2
1
40 + 3q +
+ = 7 q. Multiplying both
1000 q
sides by q and simplifying, we see that the
problem is equivalent to solving
q3
f (q) =
− 4q 2 + 40q + 1 = 0.
1000
= qn −
qn
qn +1
1
3
2.875
2
2.875
2.880
3
2.880
2.880
Thus q ≈ 2.880.
18. In the same manner as problem 17, we must find
a root of f (q) = 0.2q3 + 1.5q − 8 = 0, so
3q 2
− 8q + 40, the recursion
b. Since f ′(q) =
1000
formula is
f ( qn )
qn +1 = qn −
f ′ ( qn )
qn3
− 4qn 2 + 40qn
1000
3qn2
− 8qn + 40
1000
n
f ′(q ) = 0.6q 2 + 1.5. The recursion formula is
qn +1 = qn −
f (qn )
0.2qn3 + 1.5qn − 8
= qn −
f ′(qn )
0.6qn2 + 1.5
We select q1 = 5 as suggested.
+1
We select q1 = 10 as suggested.
n
qn
qn +1
1
5
3.54
2
3.54
2.85
n
qn
qn +1
3
2.85
2.71
1
10
10.05
4
2.71
2.70
2
10.05
10.05
5
2.70
2.70
Thus q = 2.70, so p = 10 − 2.70 = 7.30 (from the
demand equation).
Thus q ≈ 10.05.
17. The equilibrium quantity is the value of q for
which supply and demand are equal, that is, it is
100
, or of
a root of 2q + 5 =
q2 + 1
100
f ( q ) = 2q + 5 −
= 0. Since
q2 + 1
200q
f ′(q) = 2 +
, the recursion formula is
2
2
q +1
(
qn +1 = qn −
19. For a critical value of f ( x) =
we want a root of f ′( x) = x 2 − 2 x − 5 = 0. Since
d
[ f ′( x)] = 2 x − 2, the recursion formula is
dx
xn 2 − 2 xn − 5
.
2 xn − 2
For the given interval [3, 4], note that
f ′(3) = −2 and f ′(4) = 3 have opposite signs.
Thus there is a root x between 3 and 4. Since 3 is
closer to 0, we shall select x1 = 3.
xn +1 = xn −
)
f ( qn )
f ′ ( qn )
= qn −
100
qn 2 +1
200 qn
2qn + 5 −
2+
( q +1)
n
2
x3
− x 2 − 5 x + 1,
3
2
A rough sketch shows that the graph of the
supply equation intersects the graph of the
demand equation when q is near 3. Thus we
select q1 = 3.
n
xn
xn +1
1
3.0
3.5
2
3.5
3.45
3
3.45
3.45
Thus x ≈ 3.45.
453
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
Principles in Practice 12.7
1.
6.
dh
= 0 − 16(2t ) = −32t ft/sec
dt
d 2F
dq
d 2h
d
= [−32t ] = −32 feet/sec2
2
dt
dt
The acceleration of the rock at time t is
–32 feet/sec2 or 32 feet/sec2 downward.
dq
7.
2
(q + 1)3
1
= x −1
x
y′ = − x −2
y ′′ = 2 x −3
y ′′′ = −6 x −4 = −
2. y′ = 5 x 4 − 8 x3 + 14 x
9.
y ′′ = 20 x3 − 24 x 2 + 14
2
y ′′′ = 60 x − 48 x
f (q) =
1
2q
4
=
6
x4
1 −4
q
2
f ′(q) = −2q −5
f ′′(q ) = 10q −6
dy
3.
= −1
dx
f ′′′(q ) = −60q −7 = −
=0
10.
dy
= −1 − 2 x
dx
dx 2
(q + 1) 2
f ( x) = x 2 ln x
8. y =
1. y′ = 12 x 2 − 24 x + 6
y ′′ = 24 x − 24
y ′′′ = 24
d2y
=
1
⎛2⎞
f ′′( x) = x ⎜ ⎟ + (1 + 2 ln x)(1) = 3 + 2 ln x
⎝x⎠
Problems 12.7
4.
3
=−
⎛1⎞
f ′( x) = x 2 ⎜ ⎟ + (ln x)(2 x) = x(1 + 2 ln x)
⎝ x⎠
c′(q) = 14q + 11
c′′ = 14
When x = 3, the rate of change of the marginal
cost function is 14 dollars/unit2.
dx 2
2
d 3F
2. The rate of change of the marginal cost function
with respect to x is c′′(q) .
d2y
dF
1
=
dq q + 1
60
q7
1
f ( x) = x = x 2
1 − 12
x
2
1 −3
1
f ′′( x) = − x 2 = − 3
4
4x 2
f ′( x) =
= −2
5. y′ = 3x 2 + e x
11.
y ′′ = 6 x + e x
1
f (r ) = 9 − r = (9 − r ) 2
1
−1
f ′(r ) = − (9 − r ) 2
2
1
1
−3
f ′′(r ) = − (9 − r ) 2 = −
3
4
4(9 − r ) 2
y ′′′ = 6 + e x
y (4) = e x
454
ISM: Introductory Mathematical Analysis
12. y = e−4 x
(2 x + 5)(5 x − 2)
x +1
= ln(2 x + 5) + ln(5 x − 2) − ln( x + 1)
2
18. y = ln
y′ = −8 xe−4 x
2
2
⎡
y ′′ = −8 ⎢ x ⎜⎛ −8 xe−4 x
⎣ ⎝
= 8e−4 x
13. y =
2
Section 12.7
(8x − 1)
⎞ −4 x 2 (1) ⎤
⎟+e
⎥⎦
⎠
2
5
1
+
−
2 x + 5 5x − 2 x + 1
4
25
1
y ′′ = −
−
+
2
2
(2 x + 5)
(5 x − 2)
( x + 1) 2
y′ =
2
1
= (2 x + 3)−1
2x + 3
19.
dx 2
= 8(2 x + 3)
−3
z
=
8
20. y =
4
y′ =
=−
( x − 1)
dx
= −2( x − 1)−2
y ′′ = 4( x − 1) −3 =
1
2
16. y = 2 x + (2 x)
( )
2
=
e x (−1) − (1 − x)e x
(e )
x
2
=
x−2
ex
21. y = e2 x + e3 x
dy
= 2e2 x + 3e3 x
dx
4
( x − 1)3
d2y
dx 2
1
2
1
−1
−1
−1
y′ = x + (2 x) 2 (2) = x 2 + (2 x) 2
2
⎡
1 −3 1
1
1
−3
y ′′ = − x 2 − (2 x) 2 (2) = − ⎢ 3 +
3
⎢ 2
2
2
(2 x) 2
⎣ 2x
ex
d2y
( x − 1) 2
2
x
( )
x +1
x −1
( x − 1)(1) − ( x + 1)(1)
2
z
x
x
dy e (1) − x e
1− x
=
=
2
dx
ex
ex
y ′′ = 180(3x + 7)3
15. y =
z
2
z
(2 x + 3)3
14. y = (3 x + 7)5
y′ = 15(3 x + 7)
( )
( )
f ′′( z ) = ( ze ) (1) + ( z + 2) ⎡ ze + e (1) ⎤
⎣
⎦
= e ( z + 4z + 2)
f ′( z ) = z 2 e z + e z (2 z ) = ze z ( z + 2)
dy
= −2(2 x + 3)−2
dx
d2y
f ( z) = z 2e z
d3y
− 12
dx3
d4y
⎤
⎥
⎥
⎦
dx
4
d5 y
dx
17. y = ln[ x( x + 6)] = ln( x) + ln( x + 6)
5
= 4e2 x + 9e3 x
= 8e2 x + 27e3 x
= 16e2 x + 81e3 x
= 32e2 x + 243e3 x
d5 y
1
1
+
= x −1 + ( x + 6)−1
x x+6
⎡ 1
1 ⎤
y ′′ = − x −2 + (−1)( x + 6) −2 = − ⎢ +
⎥
2
( x + 6)2 ⎥⎦
⎢⎣ x
y′ =
dx
455
5
= 32e0 + 243e0 = 32 + 243 = 275
x =0
Chapter 12: Additional Differentiation Topics
22. y = e
(
) = eln( x +1)
2 ln x3 +1
3
(
2
(
)
= x3 + 1
)
ISM: Introductory Mathematical Analysis
26. 9 x 2 + 16 y 2 = 25
18 x + 32 yy ′ = 0
2
y′ = 6 x 2 x 3 + 1 = 6 x 5 + 6 x 2
y′ = −
y ′′ = 30 x 4 + 12 x
When x = 1, then y ′′ = 30 + 12 = 42.
27.
( ) = − 4y
16 y 2
=−
2
+ x2
16 y3
−1
y3
y ′′ =
=
x
y
y (1) − x( y′)
y
y 2 − x2
y
3
2
=
−16
y
3
=
y−x
=−
y
1
⎡ 12 ⎛ 1 − 12 ⎞
⎛ −1 ⎞ ⎤
x ⎜ 2 y y′ ⎟ − y 2 ⎜ 12 x 2 ⎟ ⎥
⎢
1
⎠
⎝
⎠⎥
y ′′ = − ⎢ ⎝
4⎢
x
⎥
⎢
⎥
⎣
⎦
1
1 ⎤
⎡ 1⎛
1
2 ⎞
⎡
⎢ x2 ⎜ − y ⎟ − y2 ⎥
1 − y2
1
1
⎢
1
−
⎢ ⎜
⎥
⎟
1
4x 2 ⎠ x 2 ⎥
1 y2
1 ⎢ 4 x2
=− ⎢ ⎝
=− ⎢
⎥
8⎢
8⎢
x
x
⎢
⎥
⎢
⎢
⎥
⎣⎢
⎢⎣
⎥⎦
1
⎡
2 ⎤
⎢ 1 + y1 ⎥
1
⎡ 12
4
2 ⎤
1⎢
x2 ⎥ 1 ⎢ x + 4y ⎥
= ⎢
=
⎥
8 ⎢ x ⎥ 8 ⎢ 4 x 32 ⎥
⎣
⎦
⎢
⎥
⎢⎣
⎥⎦
1⎡ 4 ⎤
1
= ⎢ 3⎥= 3
8 ⎢ 4 x 2 ⎥ 8x 2
⎣
⎦
()
x
y
2
16
y3
25. y 2 = 4 x
2 yy′ = 4
y′ =
2
= 2 y −1
y
(
)
y ′′ = −2 y −2 y′ = −2 y −2 2 y −1 = −
1
1 x 2
1 y2
y′ = − ⋅
=− ⋅ 1
1
2 2 y− 2
4 x2
1
24. x 2 − y 2 = 16
2 x − 2 yy′ = 0
y′ =
1
1 − 12
−1
x + 2 y 2 y′ = 0
2
1 −1
−1
2 y 2 y′ = − x 2
2
16 y 2
16 y 3
9 16 y 2 + 9 x 2
225
⋅
=−
3
16
16 y
256 y 3
x + 4y 2 = 4
4 y (1) − x(4 y′)
16
)
x +4 y = 4
1
2
4 y − 4 x − 4xy
=−
=−
=−
x
4y
y ′′ = −
(
9x
9 y (1) − xy ′
9 y − x − 16 y
y ′′ = − ⋅
=− ⋅
16
16
y2
y2
23. x 2 + 4 y 2 − 16 = 0
2 x + 8 yy′ = 0
8 yy′ = −2 x
y′ = −
9x
16 y
4
y3
456
⎤
⎥
⎥
⎥
⎥
⎥
⎦⎥
ISM: Introductory Mathematical Analysis
Section 12.7
31. y = e x + y
28. y 2 − 6 xy = 4
2 yy′ − 6[ xy′ + y (1)] = 0
2 yy′ − 6 xy′ = 6 y
(2 y − 6 x) y′ = 6 y
y′ = e x + y (1 + y′)
y ′ − e x + y y′ = e x + y
(
y ′′ = 3 ⋅
( y − 3 x) y′ − y ( y′ − 3)
( y − 3x) 2
= 9⋅
= 9⋅
y − 6 xy
( y − 3 x)
3
= 9⋅
4
( y − 3x)
3
=
y − xy′
y ′′ =
36
3
=
29. xy + y – x = 4
xy′ + y (1) + y′ − 1 = 0
xy′ + y′ = 1 − y
( x + 1) y′ = 1 − y
=
(1 + x)2
=
(1 − y )
y
1− y
(1 − y )
2
=
2
=
y′
(1 − y ) 2
y
(1 − y )3
e x − e y y′ = 2 x + 2 yy′
y′ =
−2(1 − y )
(1 + x) 2
ex − 2x
ey + 2y
(e
y ′′ =
y
) (
)(
x
x
y
2( y − 1)
(1 + x) 2
y
30. x 2 + 2 xy + y 2 = 1
2 x + 2 y + 2 xy ′ + 2 yy ′ = 0
( x + y ) y ′ = −( x + y )
y ′ = −1
y ′′ = 0
2
y
y
=
)(
+ 2 y e x − 2 − e x − 2 x e y y ′ + 2 y′
(e + 2 y )
( e + 2 y )( e − 2) − ( e − 2 x )( e + 2) y′
=
(e + 2 y )
( e + 2 y ) ( e − 2) − ( e − 2x ) ( e + 2)
=
(e + 2 y )
(1− y )
(1 + x) ⎡⎢ − (1+ x ) ⎤⎥ − (1 − y )
⎣
⎦
(1 + x)2
−(1 − y ) − (1 − y )
(1 − y ) y′ − y (− y′)
32. e x − e y = x 2 + y 2
1− y
y′ =
1+ x
(1 + x)(− y′) − (1 − y )(1)
y ′′ =
(1 + x) 2
=
x+ y
1 − e x+ y
y
y′ =
1− y
( y − 3x) 2
( y − 3x)
e
y′ =
3y
y − x ⎡⎢ y −3 x ⎤⎥
⎣
⎦ = 9 ⋅ y ( y − 3 x) − 3 xy
= 9⋅
2
( y − 3 x)
( y − 3 x )3
2
)
y′ 1 − e x + y = e x + y
6y
3y
y′ =
=
2 y − 6 x y − 3x
2
x
y
2
x
y
2
33. x 2 + 3x + y 2 = 4 y
2 x + 3 + 2 yy ′ = 4 y ′
2 yy ′ − 4 y ′ = −2 x − 3
2x + 3 2x + 3
y′ = −
=
2y − 4 4 − 2y
457
2
y
)
Chapter 12: Additional Differentiation Topics
y ′′ =
=
=
(4 − 2 y )(2) − (2 x + 3)(−2 y ′)
(4 − 2 y ) 2
2(4 − 2 y ) + 2(2 x + 3)
( )
2 x +3
4− 2 y
(4 − 2 y )2
2(4 − 2 y ) 2 + 2(2 x + 3)2
(4 − 2 y )3
When x = 0 and y = 0, then
34.
ISM: Introductory Mathematical Analysis
d2y
dx
2
=
2(4)2 + 2(3) 2
3
4
=
25
.
32
f ( x) = (3 x − 5)e−2 x
f ′( x) = (3x − 5) ⎡ −2e−2 x ⎤ + e−2 x [3] . Thus,
⎣
⎦
f ′( x) = e−2 x [−2(3x − 5) + 3] = (13 − 6 x)e−2 x
f ′′( x) = (13 − 6 x) ⎡ −2e−2 x ⎤ + e−2 x [−6]
⎣
⎦
= 2e−2 x [−(13 − 6 x) − 3]
= 4(3 x − 8)e−2 x
f ′′( x) + 4 f ′( x) + 4 f ( x)
= 4(3 x − 8)e−2 x + 4 ⎡ (13 − 6 x)e−2 x ⎤ + 4 ⎡ (3x − 5)e−2 x ⎤ = [4(3 x − 8) + 4(13 − 6 x) + 4(3 x − 5)]e−2 x
⎣
⎦
⎣
⎦
= [0]e−2 x = 0, as was to be shown.
35.
f ( x) = (5 x − 3)4
f ′( x) = 20(5 x − 3)3
f ′′( x) = 300(5 x − 3)2
36.
1
2
f ( x) = 6 x +
f ′( x) = 3x
− 12
x
− 12
6
−3
x 2
−
12
−5
3 −3 x 2
f ′′( x) = − x 2 +
2
8
−7
9 − 5 5x 2
f ′′′( x) = x 2 −
4
16
458
ISM: Introductory Mathematical Analysis
37.
Chapter 12 Review
dc
= 0.6q + 2
dq
d 2c
dq 2
= 0.6
d 2c
dq 2
42.
= 0.6
q =100
2
39.
2
2
= 3e x + 2 xe x + e2 xe
d r
dq 2
= −104.
f ( x) = x 4 − 6 x 2 + 5 x − 6
f ′( x) = 4 x − 12 x + 5
f ′′( x) = 12 x 2 − 12 = 12( x + 1)( x − 1)
Clearly f ′′( x) = 0 when x = ±1.
a.
y
=
b.
y ′′ =
=
41.
e y − 2ey
−2
y
( y − 2)2
( ) =−
y
y −2
2
( y − 2)
x
2y
( y − 2)3
=
=
2
3r + 7r + 1
+ 4 x +5
2
+ 4 x +5
(2 x + 4) = 2( x + 2)e x
f (t ) = log 6 t 2 + 1 =
)
(
3
8. y = 35 x = e(ln 3)5 x
(
2
+ 4 x +5
)
1
log 6 t 2 + 1
2
)
(
)
)
3
3
3
y ′ = e(ln 3)5 x (ln 3)(15 x 2 ) = 15 x 2 ln 3 ⎛⎜ 35 x ⎞⎟
⎝
⎠
−2 y′
( y − 2)2
9. y = ( x − 6)( x + 5)(9 − x)
2y
ln y = ln ( x − 6)( x + 5)(9 − x)
(2 − y )3
1
= [ln( x − 6) + ln( x + 5) + ln(9 − x)]
2
y′ 1 ⎡ 1
1
−1 ⎤
=
+
+
y 2 ⎢⎣ x − 6 x + 5 9 − x ⎥⎦
f ′( x) = 6e − 3x − 30 x
(
6r + 7
2
7. y′ = e x (2 x) + x 2 + 2 e x = e x x 2 + 2 x + 2
2
f ′′( x) = 6 e x − x − 5
3r + 7 r + 1
(6r + 7) =
(
1
y
=
=
2
−
y
2
1− y
( y − 2)( y′) − y ( y′)
1
2
(
2 x
⎛ y⎞
y2 ⎜ e 2 ⎟
y e
⎝y ⎠
y′ =
=
y
x
⎛ y⎞
e − 2 ye
ey − 2y ⎜ e2 ⎟
⎝y ⎠
e
f ′(r ) =
2
1 ln t + 1
= ⋅
. Thus
2
ln 6
⎞
1⎛ 1
1
t
.
f ′(t ) = ⎜
⋅
⋅ [2t ] ⎟ =
2 ⎝ ln 6 t 2 + 1
⎠ (ln 6) t 2 + 1
2 x
y
)
3.
6.
x
x
−1
f ′( w) = we w + e w + 2 w = we w + e w + 2 w
y′ = e x
( e ) + e (2 yy′)
( e − 2 ye ) y′ = y e
e y′ = y
x
−1
2.
5. y = e x
40. e y = y 2 e x
2
(
2
2
4. y = eln x = x. Thus y′ = 1.
3
y
x 4 x3 5 x 2
+
+
+x
4
3
2
1. y ′ = 3e x + 0 + e x (2 x) + (e2 ) xe
= −80 − 6q
When q = 4,
f ′( x) =
Chapter 12 Review Problems
dr
= 400 − 80q − 3q 2
dq
dq 2
x5 x 4 5 x3 x 2
+
+
+
20 12
6
2
f ′′( x) = x3 + x 2 + 5 x + 1
f ′′( x) = 0 when x ≈ −0.21.
38. r = pq = 400q − 40q 2 − q3
d 2r
f ( x) =
)
f ′′( x) = 0 when x ≈ –4.99 or 1.94.
459
Chapter 12: Additional Differentiation Topics
y′ =
=
10.
=
ex
e1/ t
t
2
18. y′ =
=
e2 x
xe
2x
(
=
1 − x ln x
xe
)
x 2 e x − e− x − e x + e− x (2 x)
x4
x 2 e x − x 2 e− x − 2 xe x − 2 xe− x
x3
f (q) = ln ⎡(q + 1) 2 (q + 2)3 ⎤
⎣
⎦
= 2 ln(q + 1) + 3ln(q + 2)
21.
14. y = ( x + 2)3 ( x + 1)4 ( x − 2)2
ln y = 3 ln(x + 2) + 4 ln(x + 1) + 2 ln(x − 2)
y′
3
4
2
=
+
+
y x + 2 x +1 x − 2
4
2 ⎤
⎡ 3
+
+
y′ = y ⎢
⎥
⎣ x + 2 x +1 x − 2 ⎦
4
2 ⎤
⎡ 3
= ( x + 2)3 ( x + 1)4 ( x − 2)2 ⎢
+
+
x
+
x
+
x
−
2
1
2 ⎥⎦
⎣
ln 2 x
e x ( x ln x − 1)
x ln 2 x
(
)
1
⎡1 + 2l + 3l 2 ⎤
⎦
1 + l + l2 + l3 ⎣
1 + 2l + 3l 2
1 + l + l 2 + l3
22. y = ( x 2 ) x
2
ln y = x 2 ln x 2 = 2 x 2 ln x
y′
⎛1⎞
= 2 x 2 ⎜ ⎟ + (ln x) ( 4 x )
y
⎝x⎠
y′ = 2 xy (1 + 2 ln x)
2
y′ = 2 x( x 2 ) x (1 + 2 ln x)
+ 2 x −5)(ln 2)
= (4 x + 2)(ln 2)2
( 1x ) = e x ( ln x − 1x )
f (l ) = ln 1 + l + l 2 + l 3
=
y′ = e
(ln x) 2
f ′(l ) =
2
3
f ′(q) =
+
q +1 q + 2
(2 x 2 + 2 x −5)(ln 2)
(ln x)e x − e x
⎞
⎟ = ln 5 − 2 ln x
⎠
1
2
y′ = 0 − 2 ⋅ = −
x
x
e x ( x − 2) − e− x ( x + 2)
2
4e2 x +1
x
⎛ 5
20. y = ln ⎜
⎝ x2
x4
15. y = e(2 x
=
19. y = log 2 (8 x + 5) 2 = 2 log 2 (8 x + 5)
ln(8 x + 5)
= 2⋅
ln 2
1
8
16
y′ = 2 ⋅
⋅
=
ln 2 8 x + 5 (8 x + 5) ln 2
x
) (
4e3 x
xe x −1
x ⎡ e2 x +1 (2) ⎤ − e2 x +1[1] 4e2 x +1 (2 x − 1)
⎦
=
y′ = 4 ⋅ ⎣
x2
x2
( 1x ) − (ln x) ( e x )
e x − xe x ln x
12. y′ =
=
17. y =
( x − 6)( x + 5)(9 − x) ⎡ 1
1
1 ⎤
+
+
⎢
⎥
2
⎣ x −6 x +5 x −9⎦
11. y′ =
13.
y⎡ 1
1
1 ⎤
+
−
2 ⎣⎢ x − 6 x + 5 9 − x ⎥⎦
f ′(t ) = e1/ t (−1 ⋅ t −2 ) = −
=
ISM: Introductory Mathematical Analysis
23. y = ( x + 1) x +1
ln y = ( x + 1) ln( x + 1)
(4 x + 2)(ln 2)
2 x 2 + 2 x −5
y′
1
= ( x + 1)
+ ln( x + 1)[1] = 1 + ln( x + 1)
y
x +1
16. y is a constant, so y′ = 0.
y′ = y[1 + ln( x + 1)] = ( x + 1) x +1[1 + ln( x + 1)]
460
ISM: Introductory Mathematical Analysis
(1 − e ) e − (1 + e )( −e ) = 2e
y′ =
(1 − e )
(1 − e )
x
24.
Chapter 12 Review
x
x
x
x
x
2
x
2
1
25. φ (t ) = ln ⎛⎜ t 4 − t 2 ⎞⎟ = ln t + ln(4 − t 2 )
2
⎝
⎠
1 1
1
1
t
φ ′(t ) = + ⋅
⋅ (−2t ) = −
2
t 2 4−t
t 4 − t2
26. y = ( x + 3)ln x
ln y = [ln x] ln(x + 3)
1
1
y′
= (ln x)
+ ln( x + 3) ⋅
y
x+3
x
⎡ ln x ln( x + 3) ⎤
y′ = y ⎢
+
⎥
x
⎣x+3
⎦
ln( x + 3) ⎤
ln x ⎡ ln x
= ( x + 3) ⎢
+
⎥
x
⎣x+3
⎦
27.
y=
ln y =
y′
=
y
y′ =
=
28. y′ =
( x 2 + 1)1/ 2 ( x 2 + 2)1/ 3
(2 x3 + 6 x) 2 / 5
1
1
2
ln( x 2 + 1) + ln( x 2 + 2) − ln(2 x3 + 6 x)
2
3
5
1⎛ 1 ⎞
1⎛ 1 ⎞
2⎛
1
⎞
2
⎜
⎟ (2 x) + ⎜ 2
⎟ (2 x) − ⎜ 3
⎟ (6 x + 6)
2 ⎝ x2 + 1 ⎠
3⎝ x + 2 ⎠
5 ⎝ 2x + 6x ⎠
⎡ x
2x
6( x 2 + 1) ⎤
y⎢
+
−
⎥
2
2
3
⎢⎣ x + 1 3( x + 2) 5( x + 3 x) ⎥⎦
2x
6( x 2 + 1) ⎤
( x 2 + 1)1/ 2 ( x 2 + 2)1/ 3 ⎡ x
+
−
⎢
⎥
2
2
3
(2 x3 + 6 x) 2 / 5
⎣⎢ x + 1 3( x + 2) 5( x + 3x) ⎦⎥
x
( 1x ) − ln x ( 12 ) x−
1
2
=
x
( )
29. y = x x
x
= xx
2 − ln x
3
2x 2
2
2
ln y = ln x x = x 2 ln x
y′
⎛1⎞
= x 2 ⎜ ⎟ + (ln x)(2 x) = x + 2 x ln x
y
⎝x⎠
( )
y′ = y ( x + 2 x ln x) = x x
x
( x + 2 x ln x)
461
Chapter 12: Additional Differentiation Topics
30. y = x
ISM: Introductory Mathematical Analysis
(x )
x
ln y = ln x
( x ) = x x ln x
x
( )
y′
d x
⎛1⎞
= x x ⎜ ⎟ + (ln x )
x
y
x
dx
⎝ ⎠
Note: If v = x x , then ln v = ln x x = x ln x;
v′
⎛1⎞
= x ⎜ ⎟ + (ln x)(1) = 1 + ln x
v
⎝x⎠
d x
v′ =
x = v(1 + ln x) = x x (1 + ln x)
dx
y′
⎛1⎞
Thus
= x x ⎜ ⎟ + (ln x) ⎡ x x (1 + ln x) ⎤
⎣
⎦
y
⎝x⎠
( )
⎡1
⎤
= x x ⎢ + (1 + ln x) ln x ⎥
⎣x
⎦
⎡1
⎤
y′ = yx x ⎢ + (1 + ln x) ln x ⎥
⎣x
⎦
=x
( x ) x x ⎡ 1 + (1 + ln x) ln x ⎤
x
⎢x
⎣
⎥
⎦
31. y = ( x + 1) ln x 2 = 2( x + 1) ln x
⎡
⎤
⎛1⎞
⎡ x +1
⎤
y′ = 2 ⎢( x + 1) ⎜ ⎟ + (ln x)(1) ⎥ = 2 ⎢
+ ln x ⎥
⎝x⎠
⎣ x
⎦
⎣
⎦
⎡2
⎤
When x = 1, then y′ = 2 ⎢ + ln1⎥ = 4.
⎣1
⎦
32. y =
ex
2
+1
x2 + 1
2
1
1
ln y = ln ⎛⎜ e x +1 ⎞⎟ − ln( x 2 + 1) = x 2 + 1 − ln( x 2 + 1)
2
⎝
⎠ 2
⎡
y′
1
1
1 ⎤
= 2x − ⋅
(2 x) = x ⎢ 2 −
⎥
2
y
2 x2 + 1
x + 1⎦
⎣
⎡
1 ⎤
y ′ = yx ⎢ 2 −
⎥
2
x + 1⎦
⎣
y′ =
ex
2
+1
⎡
1 ⎤
x ⎢2 −
⎥
2
x + 1⎦
x +1 ⎣
2
When x = 1, then y ′ =
e1+1
1 ⎤ 3e2 2
⎡
(1) ⎢ 2 −
⎥= 4 .
1 + 1 ⎣ 1 + 1⎦
462
ISM: Introductory Mathematical Analysis
33. y = e
e + x ln
Chapter 12 Review
( 1x ) = ee− x ln x
⎛ ⎡ ⎛1⎞
⎤⎞
y′ = ee− x ln x ⎜⎜ − ⎢ x ⎜ ⎟ + (ln x)(1) ⎥ ⎟⎟
⎦⎠
⎝ ⎣ ⎝x⎠
= −(1 + ln x)ee− x ln x
When x = e, then y′ = −(1 + ln e)ee −e ln e = −(2)e0 = −2.
⎡ 25 x ( x 2 − 3x + 5)1/ 3 ⎤
34. y = ⎢
⎥
2
3
⎣⎢ ( x − 3 x + 7)
⎦⎥
−1
1
⎡
⎤
ln y = −1 ⎢5 x ln 2 + ln( x 2 − 3x + 5) − 3ln( x 2 − 3 x + 7) ⎥
3
⎣
⎦
⎡
y′
1
2x − 3
2x − 3 ⎤
= − ⎢5ln 2 + ⋅
− 3⋅
⎥
2
2
y
3 x − 3x + 5
x − 3x + 7 ⎦
⎣
⎡
2x − 3
3(2 x − 3) ⎤
y′ = − y ⎢5ln 2 +
−
⎥
3( x 2 − 3 x + 5) x 2 − 3 x + 7 ⎦⎥
⎣⎢
⎡ 25 x ( x 2 − 3x + 5)1/ 3 ⎤
y′ = (−1) ⎢
⎥
2
3
⎣⎢ ( x − 3 x + 7)
⎦⎥
When x = 0, then y′ = −
−1
⎡
2x − 3
3(2 x − 3) ⎤
⎢5ln 2 +
⎥
2
3( x − 3x + 5) x 2 − 3x + 7 ⎦⎥
⎣⎢
343 ⎡
1 9⎤
1862
5ln 2 − + ⎥ = −343(ln 2)52 / 3 −
.
1/ 3 ⎢
5
7
⎦
5 ⎣
54 / 3
35. y = 3e x
y′ = 3e x
If x = ln 2, then y = 3eln 2 = 6 and y′ = 3eln 2 = 6.
An equation of the tangent line is y – 6 = 6(x – ln 2), y = 6x + 6 – 6 ln 2, y = 6x + 6(1 – ln 2). Alternatively, since
6 ln 2 = ln 26 = ln 64, the tangent line can be written as y = 6x + 6 – ln 64.
36. y = x + x 2 ln x
⎡ ⎛1⎞
⎤
y′ = 1 + ⎢ x 2 ⎜ ⎟ + (ln x)(2 x) ⎥ = 1 + x + 2 x ln x
x
⎣ ⎝ ⎠
⎦
When x = 1, then y = 1 + 1(0) = 1 and y′ = 1 + 1 + 2(0) = 2 . Thus an equation of the tangent line is y – 1 = 2(x – 1),
or y = 2x – 1.
2
37. y = x ⎛⎜ 22− x ⎞⎟ . To find y′ we shall use logarithmic differentiation.
⎝
⎠
⎡ ⎛ 2− x 2 ⎞ ⎤
2
ln y = ln ⎢ x ⎜ 2
⎟ ⎥ = ln x + 2 − x ln 2
⎠⎦
⎣ ⎝
y′ 1
= + (−2 x) ln 2
y x
(
)
⎡1
⎤
y′ = y ⎢ − 2(ln 2) x ⎥
⎣x
⎦
463
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
When x = 1, then y = 2 and y′ = 2(1 − 2 ln 2). The equation of the tangent line is
y – 2 = 2(1 – 2 ln 2)(x – 1). The y-intercept of the tangent line corresponds to the point where x = 0:
y – 2 = 2(1 – 2 ln 2)(–1) = –2 + 4 ln 2
Thus y = 4 ln 2 and the y-intercept is (0, 4 ln 2).
38. w = 2 x +1 + ln(1 + x 2 ) = e(ln 2)( x +1) + ln(1 + x 2 )
2
x = log 2 (t 2 + 1) − e(t −1) =
dw dw dx
=
⋅
dt
dx dt
⎛
2x
= ⎜ 2 x +1 (ln 2) +
1 + x2
⎝
ln(t 2 + 1) (t −1)2
−e
ln 2
⎞
2
⎞⎛
2t
− e(t −1) [2(t − 1)] ⎟
⎟ ⎜⎜
2
⎟
⎠ ⎝ (ln 2)(t + 1)
⎠
2
When t = 1, x = log 2 (1 + 1) − e(1−1) = 1 − 1 = 0, w = 21 + ln1 = 2 + 0 = 2, and
39. y = e x
2
y′ = e x
− 2 x +1
2
− 2 x +1
[2 x − 2] = (2 x − 2)e x
y ′′ = 2( x − 1)e x
= 2e x
2
2
− 2 x +1
− 2 x +1
2
− 2 x +1
(2 x − 2) + 2e x
2
− 2 x +1
(2( x − 1)2 + 1)
At (1, 1), y ′′ = 2e0 (2(0) + 1) = 2.
40. y = x 2 e x
(
y′ = x 2 e x + e x (2 x) = e x x 2 + 2 x
(
y ′′′ = e (2 x + 4) + ( x
)
)
(
y ′′ = e x (2 x + 2) + x 2 + 2 x e x = e x x 2 + 4 x + 2
x
2
)
(
)
+ 4 x + 2 e x = e x x2 + 6 x + 6
)
At (1, e), y ′′′ = e(1 + 6 + 6) = 13e
41. y = ln(2x)
1
y′ =
(2) = x −1
2x
y ′′ = −1 ⋅ x −2 = − x −2
y ′′′ = −(−2) x −3 =
At (1, ln 2), y ′′′ =
2
x3
2
13
=2
464
dw
⎛ 2
⎞
= (21 ln 2 + 0) ⎜
− 1(0) ⎟ = 2.
dt
⎝ 2 ln 2
⎠
ISM: Introductory Mathematical Analysis
Chapter 12 Review
42. y = x ln x
1
y′ = x ⋅ + (ln x )(1) = 1 + ln x
x
1 1
y ′′ = 0 + =
x x
1
At (1, 0), y ′′ = = 1
1
47. x + xy + y = 5
1 + xy′ + y (1) + y′ = 0
( x + 1) y′ = −1 − y
1+ y
x +1
( x + 1) y′ − (1 + y )
y ′′ = −
( x + 1) 2
1+1
2
At (2, 1), y′ = −
= − and
2 +1
3
y′ = −
43. 2 xy + y 2 = 10
2 ( xy′ + y ) + 2 yy′ = 0
2 xy′ + 2 yy′ = −2 y
( x + y ) y′ = − y
y′ = −
y ′′ = −
y
x+ y
y′ =
−3 x 2 y 3
3 x3 y 2
y′ = −
=−
y ′′ = −
y
x
( )
ln x + 2 ln y = xy
1 2
+ y′ = xy′ + y
x y
y ′′ = −
2
)
(e
2
xy − y
2 x − x2 y
y ′[ y (2 + y ln x)] = −
y2
x(2 + y ln x)
4
27
9
)
− e x y′ = ( y + 1)e x
( y + 1)e x
2
⎡
⎛ 1
⎞⎤
y 2 ⎢e y ln x ⎜ y ⋅ + (ln x) y′ ⎟ ⎥ + e y ln x [ 2 yy′] = 0
⎝ x
⎠⎦
⎣
y′ = −
y
y′ =
y 2 (ln x) y′ + 2 yy′ = −
( ) ( )=
3 − 23 − 2 − 13
e y y′ − e x y′ = ( y + 1)e x
2x − x 2 y y′ = xy 2 − y
=e
( x + 2 y ) y′ − y (1 + 2 y′)
e y y′ = ( y + 1)e x + e x ( y′)
2
2xy′ − x yy′ = xy − y
46. y e
y
x + 2y
49. e y = ( y + 1)e x
y + 2 xy′ = x 2 yy′ + xy 2
2 y ln x
4
9
( x + 2 y )2
2
At (–1, 2), y′ = − and
3
45. ln xy 2 = xy
y′ =
9
=
48. xy + y 2 = 2
x( y′) + y (1) + 2 yy′ = 0
44. 3x 2 y3 + 3 x3 y 2 y ′ = 0
y ′(3x3 y 2 ) = −3x 2 y3
(
( )
3 − 23 − 2
=
y3
x
1
1−
y ′′ =
3
y
x
465
e y − ex
1
y +1
=
=
( )=
−(
) e
( y + 1)
ey
ey
y +1
ey
y +1
ey
y
−
ey
y +1
y +1
y
y ( y′) − ( y + 1)( y′)
y
2
=
− y′
y
2
=−
y +1
y
2
y
=−
y +1
y3
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
50. x1/ 2 + y1/ 2 = 1
54.
1 −1/ 2 1 −1/ 2 dy
x
+ y
⋅
=0
2
2
dx
y
dy
=−
dx
x
( x ) ⎛⎜⎝ 2 1 y ⋅ dydx ⎞⎟⎠ − ( y ) ( 2 1 x )
=−
2
dx 2
( x)
=
51.
+
y
2 x
x
=
x+ y
2x x
=
x5 x 4 2 x3
+
+
+ x2 + 1
10 6
3
f ′( x) =
x 4 2 x3
+
+ 2 x2 + 2 x
2
3
f ′′( x) = 2 x3 + 2 x 2 + 4 x + 2
f ′′( x) = 0 when x ≈ −0.57.
d2y
1
2
f ( x) =
55. p =
1
η=
2x x
f ′(t )
= ⎡ −0.8e−0.01t (−0.01) − 0.2e−0.0002t (−0.0002) ⎤
⎣
⎦
=
500 / q
q
− 500
q2
= −1
56. p = 900 − q 2
52. log N = A – bM
d
d
(log N ) =
( A − bM )
dM
dM
d ⎛ ln N ⎞
d
( A − bM )
=
⎜
⎟
dM ⎝ ln10 ⎠ dM
η=
p
q
dp
dq
=
900 − q 2
q
−2q
=−
900 − q 2
2q 2
When q = 10, then η = −4. Since η > 1,
demand is elastic.
1
1 dN
⋅
= −b
ln10 N dM
1 dN
(log e)
= −b
N dM
dN
bN
−
=
dM log e
57. p = 18 – 0.02q
η=
=
18−0.02 q
q
−0.02
=−
18 − 0.02q
0.02q
58. p = 20 − 2 q
η=
p
q
dp
dq
a.
When p = 8, then η =
b.
η=
⎛b⎞
⎛b⎞
= log ⎜ ⎟ + ( A − bM ) = A + log ⎜ ⎟ − bM
q
⎝ ⎠
⎝q⎠
4
p
q
dp
dq
When q = 600, then η = −0.5. Because η < 1,
demand is inelastic.
⎛ b
⎞
⎛ dN ⎞
log ⎜ −
= log ⎜
⋅N⎟
⎟
⎝ dM ⎠
⎝ log e
⎠
⎛ b ⎞
= log ⎜
⎟ + log N
⎝ log e ⎠
f ′( x) = (12 x3 + 6 x 2 − 25)e3 x
f ′( x) = 0 when x ≈ 1.13.
p
q
dp
dq
Since η = 1, demand has unit elasticity when
q = 200.
= 0.008e−0.01t + 0.00004e−0.0002t
53.
500
q
3
+ 2 x − 25 x
=
p
q
− 1
q
=
−p
q
=
−p
10 −
p
2
=
2p
p − 20
2(8)
4
=− .
8 − 20
3
2p
p − 20
20
, then η < −1, so η > 1 and
3
demand is elastic.
If p >
466
ISM: Introductory Mathematical Analysis
59. η =
p
q
dp
dq
=
Chapter 12 Review
40
1
=−
80 − 200
3
% change in q ≈ (% change in price) (η )
η
b.
p dq
⋅
q dp
η=
−p
2500 − p
2
=
−p
, so
q
61. We want a root of f ( x) = x3 − 2 x − 2 = 0. We
have f(1) = –3 and f(2) = 2 (note the sign
change). Since f(2) is closer to 0 than is f(1), we
p ⎛ −p ⎞
p2
⎜
⎟ = − 2 . Now, if p = 30, then
q⎝ q ⎠
q
q = 2500 − 302 = 40, so
η
p =30
=−
(30) 2
(40)
2
=−
choose x1 = 2. We have f ′( x) = 3x 2 − 2, so the
recursion formula is
f ( xn )
x3 − 2 xn − 2
= xn − n
xn +1 = xn −
f ′ ( xn )
3xn2 − 2
9
16
2
%, then demand
3
would change by approximately
3
⎛ 2 ⎞⎛ 9 ⎞
⎜ − 3 ⎟ ⎜ − 16 ⎟ %, or 8 %. (That is, demand
⎝
⎠⎝
⎠
If the price of 30 decreases
=
3
increases by approximately %.)
8
60. a.
η=
p
q
dp
dq
=
5
⎛ 1⎞
= 5 ⎜ − ⎟ % = − % = −1.67%. Thus
3
3
⎝
⎠
demand decreases by approximately 1.67%.
q = 2500 − p 2
dq
=
dp
p = 40
p dq
= ⋅
q dp
2 xn3 + 2
3xn2 − 2
n
xn
xn +1
1
2.00000
1.80000
2
1.80000
1.76995
3
1.76995
1.76929
4
1.76929
1.76929
q = 100 − p , where 0 < p < 100.
Because x5 − x4 < 0.0001, the root is
dq
−1
=
. Thus
dp 2 100 − p
approximately x5 = 1.7693.
η=
p
62. We want real solutions of e x = 3 x. Thus we
−1
⋅
100 − p 2 100 − p
want to find roots of f ( x) = e x − 3 x = 0. A
rough sketch of the exponential function y = e x
and the line y = 3x shows that there are two
intersection points: one when x is near 0.5, and
the other when x is near 1.5. Thus we must find
−p
p
=
=
2(100 − p) 2 p − 200
p
< −1.
2 p − 200
Noting that the denominator is negative for
0 < p < 100, we multiply both sides of the
inequality by 2p – 200 and reverse the
direction of the inequality
200
p > −2 p + 200, 3 p > 200, p >
3
200
< p < 100 for elastic demand.
Thus
3
For elastic demand we want
two roots. Since f ′( x) = e x − 3, the recursion
formula is xn +1 = xn −
f ( xn )
f ′ ( xn )
= xn −
If x1 = 0.5, we obtain
467
n
xn
xn +1
1
0.5
0.610
2
0.610
0.619
3
0.619
0.619
e xn − 3 xn
e xn − 3
Chapter 12: Additional Differentiation Topics
ISM: Introductory Mathematical Analysis
If x1 = 1.5, we obtain
n
xn
xn +1
1
1.5
1.512
2
1.512
1.512
Thus the solutions are 0.619 and 1.512.
Mathematical Snapshot Chapter 12
1. F = 25, D = 3400, V = 36.5, R = 0.05.
2 FD
2(25)(3400)
q=
=
≈ 305.2
RV
(0.05)(36.5)
The economic order quantity is 305 units.
2. If the number of units maintained as a safety
margin is denoted by m, then the amount in
stock at any time is increased by m units. The
average inventory level is thus increased by m
q
units, to m + units. The carrying cost is then
2
FD
q⎞
⎛
C (q) =
+ RV ⎜ m + ⎟
q
2⎠
⎝
FD RVq
=
+
+ RVm
q
2
d
( RVm) = 0, the maintenance of a
dq
safetly margin does not affect the calculation of
the economic order quantity.
Since
3. Answers may vary.
468
Chapter 13
maximum, since C ′(t ) changes from + to –. The
drug is at its greatest concentration 2 hours after
the injection.
Principles in Practice 13.1
1. The graph of c(q ) = 2q3 − 21q 2 + 60q + 500 is
shown.
Problems 13.1
C(q)
1. Decreasing on (–∞, –1) and (3, ∞); increasing on
(–1, 3); relative minimum (–1, –1); relative
maximum (3, 4).
2. Decreasing on (–∞, –1) and (0, 1); increasing on
(–1, 0) and (1, ∞); relative minima (–1, –1) and
(1, –1); relative maximum (0, 0).
q
2 5
There looks to be a relative maximum at q = 2
and a relative minimum at q = 5.
c ′(q) = 6q 2 − 42q + 60 = 6(q 2 − 7q + 10)
= 6(q – 5)(q – 2)
c ′(q) = 0 when q = 2 or q = 5. If q < 2, then
c ′(q) = 6(−)(−) = + , so c(q) is increasing. If
2 < q < 5, then c′(q) = 6(−)(+ ) = − , so c(q) is
decreasing. If 5 < q, then c′(q) = 6(+)(+) = + , so
c(q) is increasing. When q = 2, there is a relative
maximum, since c′(q) changes from + to –. The
relative maximum value is
3. Decreasing on (–∞, –2) and (0, 2); increasing on
(–2, 0) and (2, ∞); relative minima (–2, 1) and
(2, 1); no relative maximum.
4. Decreasing on (–∞, 0) and (0, ∞); never
increasing; no relative maximum; no relative
minimum.
In the following problems, we denote the critical value
by CV.
5.
2(2)3 − 21(2) 2 + 60(2) + 500 = 552 . When q = 5,
there is a relative minimum, since c′(q) changes
from – to +. The relative minimum value is
CV: x = −3, 1, 2
−
C ′(t ) =
0.14t
(t + 2) 2
−
+
−3
2(5)3 − 21(5)2 + 60(5) + 500 = 525.
2. First, find C ′(t ) , with C (t ) =
f ′( x) = ( x + 3)( x − 1)( x − 2)
f ′( x) = 0 when x = −3, 1, 2
1
+
2
Increasing on (−3, 1) and (2, ∞); decreasing on
(−∞, −3) and (1, 2); relative maximum when
x = 1; relative minima when x = −3, 2.
.
6.
0.14(t + 2)2 − 0.14t (2)(t + 2)
(t + 2)4
0.14(t + 2) − 0.28t 0.28 − 0.14t
=
=
(t + 2)3
(t + 2)3
0.14(2 − t )
=
(t + 2)3
C ′(t ) = 0 when t = 2 and is undefined when
t = –2. However, since t denotes the number of
hours after an injection, negative values of t are
+
not reasonable. If 0 ≤ t < 2, C ′(t ) = = + , so
+
−
C(t) is increasing. If 2 < t, C ′(t ) = = − , so C(t)
+
is decreasing. When t = 2, there is a relative
f ′( x) = 2 x( x − 1)3
CV: x = 0, 1
+
–
0
+
1
Increasing on (–∞, 0) and (1, ∞); decreasing on
(0, 1); relative maximum when x = 0; relative
minimum when x = 1.
7.
f ′( x) = ( x + 1)( x − 3)2
CV: x = –1, 3
–
–1
+
+
3
Decreasing on (–∞, –1); increasing on (–1, 3)
and (3, ∞); relative minimum when x = –1.
469
Chapter 13: Curve Sketching
8.
f ′( x ) =
ISM: Introductory Mathematical Analysis
x( x + 2)
1
⎛ 1 ⎞
on ⎜ − , 2 ⎟ ; relative maximum when x = − ;
3
3
⎝
⎠
relative minimum when x = 2.
2
x +1
CV: x = 0, –2
+
–
–2
+
0
13. y = −
Increasing on (–∞, –2) and (0, ∞); decreasing on
(–2, 0); relative maximum when x = –2; relative
minimum when x = 0.
(
y′ = − x 2 − 4 x + 5 = − x 2 + 4 x − 5
–
y′ = 6 x2
CV: x = 0
–5
+
0
14. y =
10. y = x 2 + 4 x + 3
y′ = 2 x + 4 = 2( x + 2)
CV: x = –2
–
+
(
)
y′ = 4 x3 − 4 x = 4 x x 2 − 1 = 4 x( x + 1)( x − 1)
CV: x = 0, ±1
–
–
+
–1
1
2
–
0
+
1
Decreasing on (–∞, –1) and (0, 1); increasing on
(–1, 0) and (1, ∞); relative maximum when x =
0; relative minima when x = ±1.
1⎞
⎛
⎛1
⎞
Increasing on ⎜ −∞, ⎟ ; decreasing on ⎜ , ∞ ⎟ ;
2⎠
⎝
⎝2
⎠
1
relative maximum when x = .
2
16. y = −3 + 12 x − x3
(
)
y′ = 12 − 3 x 2 = 3 4 − x 2 = 3(2 + x)(2 − x)
5 2
x − 2x + 6
2
CV: x = ±2
–
y′ = 3x 2 − 5 x − 2 = (3x + 1)( x − 2)
–2
+
–
2
Decreasing on (–∞, –2) and (2, ∞); increasing on
(–2, 2); relative minimum when x = –2; relative
maximum when x = 2.
1
CV: x = − , 2
3
−
0
15. y = x 4 − 2 x 2
1
CV: x =
2
12. y = x3 −
+
Decreasing on (−∞, −3); increasing on (−3, 0)
and (0, ∞); relative minimum at x = −3.
11. y = x − x 2 + 2
y′ = 1 − 2 x
3
x4
+ x3
4
–3
–2
−1
1
CV: x = −3, 0
Decreasing on (–∞, –2); increasing on (–2, ∞);
relative minimum when x = –2.
+
–
y ′ = x3 + 3x 2 = x 2 ( x + 3)
+
+
+
Decreasing on (–∞, –5) and (1, ∞); increasing on
(–5, 1); relative minimum when x = –5; relative
maximum when x = 1.
Increasing on (−∞, 0); increasing on (0, ∞); no
relative maximum or minimum
–
)
= −( x + 5)( x − 1)
CV: x = –5, 1
9. y = 2 x3 + 1
+
x3
− 2 x2 + 5x − 2
3
+
17. y = x3 −
2
1⎞
⎛
Increasing on ⎜ −∞, − ⎟ and (2, ∞); decreasing
3⎠
⎝
7 2
x + 2x − 5
2
y′ = 3x 2 − 7 x + 2 = (3 x − 1)( x − 2)
1
CV: x = , 2
3
470
ISM: Introductory Mathematical Analysis
−
+
1
3
Section 13.1
21. y =
2
y ′ = x 2 − 10 x + 22
1⎞
⎛
Increasing on ⎜ −∞, ⎟ and (2, ∞); decreasing
3⎠
⎝
1
⎛1 ⎞
on ⎜ , 2 ⎟ ; relative maximum when x = ,
3
3
⎝
⎠
relative minimum when x = 2.
18. y = x3 − 6 x 2 + 12 x − 6
(
By the quadratic formula, y ′ = 0 when
x=
)
(5 −
Increasing on (–∞, 2) and (2, ∞); no relative
maximum or relative minimum.
11 2
x − 10 x + 2
2
22. y =
+
)(
–
– 5
(
)
(
)(
–
2
,± 5
3
+
2
3
)( x − 5 )( x + 5 )
)
–
+
5
2
3
⎛
2
2⎞
Increasing on −∞, − 5 , ⎜⎜ −
,
⎟ , and
3 ⎟⎠
⎝ 3
⎛
2⎞
5, ∞ ; decreasing on ⎜⎜ − 5, −
⎟ and
3 ⎟⎠
⎝
⎛ 2
⎞
, 5 ⎟⎟ ; relative maxima when x = − 5 ,
⎜⎜
3
⎝
⎠
(
(
2
y′ = −15 x 2 + 2 x + 1 = −(5 x + 1)(3 x − 1)
1 1
CV: − ,
5 3
5
9 5 47 3
x −
x + 10 x
5
3
CV: x = ±
20. y = −5 x + x + x − 7
–1
)
3, 5 + 3 ; increasing on 5 + 3, ∞ ;
(
5
2
+
)
= 3x − 2 3x + 2
+
3
(
y′ = 9 x 4 − 47 x 2 + 10 = 9 x 2 − 2 x 2 − 5
2⎞
⎛
⎛5
⎞
Increasing on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ;
3⎠
⎝
⎝2
⎠
2
5
⎛
⎞
decreasing on ⎜ − , ⎟ ; relative maximum
⎝ 3 2⎠
2
5
when x = − ; relative minimum when x = .
3
2
–
5+ 3
minimum at x = 5 + 3.
2 5
CV: x = − ,
3 2
3
+
relative maximum at x = 5 − 3; relative
y′ = 6 x 2 − 11x − 10 = (2 x − 5)(3 x + 2)
–2
or x = 5 ± 3.
Increasing on −∞, 5 − 3 ; decreasing on
2
–
–
5– 3
+
19. y = 2 x3 −
2(1)
+
CV: x = 2
+
10 ± (−10) 2 − 4(1)(22)
CV: x = 5 ± 3
y′ = 3x 2 − 12 x + 12 = 3 x 2 − 4 x + 4 = 3( x − 2) 2
+
x3
− 5 x 2 + 22 x + 1
3
+
)
)
2
2
; relative minima when x = −
, 5.
3
3
–
1
3
23. y = 3x5 − 5 x3
1⎞
⎛
⎛1
⎞
Decreasing on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ;
5
3
⎝
⎠
⎝
⎠
⎛ 1 1⎞
increasing on ⎜ − , ⎟ ; relative minimum when
⎝ 5 3⎠
1
1
x = − ; relative maximum when x = .
5
3
y′ = 15 x 4 − 15 x 2 = 15 x 2 ( x + 1)( x − 1)
CV: x = 0, ±1
+
–
–1
–
0
+
1
Increasing on (–∞, –1) and (1, ∞); decreasing on
(–1, 0) and (0, 1); relative maximum when
x = –1; relative minimum when x = 1.
471
Chapter 13: Curve Sketching
24. y = 3x −
ISM: Introductory Mathematical Analysis
x6
2
28. y =
y′ = 3 − 3x5 = 3(1 − x5 )
4
3
(
2
1
+
0
−
(
)
+
0
1
30. y = 3 x ( x − 2)
27. y = 8 x 4 − x8
(
= 8 x3 ( 2 + x 2 )( 2 − x 2 )
= 8 x3 ( 2 + x 2 ) ( 2 − x )(
y′ = 32 x3 − 8 x7 = 8 x3 4 − x 4
)
2+x
y′ =
(
decreasing on ( −
)
–
–
0
)
(
)
(
)
+
1
2
1
2
⎛ 1⎞
Decreasing on (–∞, 0) and ⎜ 0, ⎟ ; increasing
⎝ 2⎠
1
⎛1
⎞
on ⎜ , ∞ ⎟ ; relative minimum when x = ; no
2
⎝2
⎠
relative maximum.
)
Increasing on −∞, – 2 and 0, 2 ;
2, 0 and
2
CV: x = 0,
–
2
2(2 x − 1)
3x 3
CV: x = 0, ± 2
0
)
Increasing on (−1, 0) and (1, ∞); decreasing on
(−∞, −1) and (0, 1); relative maximum when
x = 0; relative minima when x = ±1.
2
Decreasing on (−∞, 0) and (0, 2); increasing on
(2, ∞); relative minimum at x = 2.
– 2
−
+
−1
+
3
y′ = 8 x( x 2 − 1)3 = 8 x( x + 1)3 ( x − 1)3
CV: 0, –1, 1
+
–
1
2
+
29. y = ( x 2 − 1) 4
3x4
− 4 x3 + 17
26. y =
2
y ′ = 6 x3 − 12 x 2 = 6 x 2 ( x − 2)
CV: x = 0, 2
+
–
(
Decreasing on (–∞, –4) and (0, ∞); increasing on
(–4, 0); relative minimum when x = –4; relative
maximum when x = 0.
0
)
⎛ 1 1⎞
Increasing on −∞, − 3 , ⎜ − , ⎟ , 3, ∞ ;
⎝ 2 2⎠
1
⎛
⎞
⎛1
⎞
decreasing on ⎜ − 3, − ⎟ and ⎜ , 3 ⎟ ;
2⎠
⎝
⎝2
⎠
1
relative maxima when x = − 3, ; relative
2
1
minima when x = − , 3 .
2
–
–
+
2
y′ = −5 x 4 − 20 x3 = −5 x3 ( x + 4)
CV: x = 0, –4
–
–
– 3 –1
25. y = − x5 − 5 x 4 + 200
–4
)(
)
1
CV: x = ± , ± 3
2
–
+
)(
= (2 x − 1)(2 x + 1) x + 3 x − 3
Increasing on (–∞, 1); decreasing on (1, ∞);
relative maximum when x = 1.
–
(
y′ = 4 x 4 − 13x 2 + 3 = 4 x 2 − 1 x 2 − 3
= 3(1 − x)( x + x + x + x + 1)
CV: x = 1
+
4 5 13 3
x − x + 3x + 4
5
3
2, ∞ ; relative
maxima when x = ± 2 , relative minimum when
x = 0.
472
ISM: Introductory Mathematical Analysis
Section 13.1
sign chart because it is a point of discontinuity of
y.
5
= 5( x − 1)−1
x −1
31. y =
y′ = −5( x − 1)
−2
=−
–
5
1
y′ = 8 x −
3
x2
CV: None, but x = 0 must be included in the sign
chart because it is a point of discontinuity of y.
−
x
= 10 x
y′ = −5 x
− 32
− 12
. [Note: x > 0]
5
y′ =
3x
2x + 5
3(2 x + 5) − (3 x)(2)
(2 x + 5)2
(2 x + 5) 2
5
must be included in the
2
sign chart because it is a point of discontinuity of
y.
5⎞
⎛
⎛ 5
⎞
Increasing on ⎜ −∞, − ⎟ and ⎜ − , ∞ ⎟ ; no
2⎠
⎝
⎝ 2
⎠
relative extremum
x + 4x + 3
2
–
38. y =
y′ =
=
=
( x + 1)( x + 3)
–
+
–1
2x2
4 x 2 − 25
(4 x 2 − 25)(4 x) − (2 x 2 )(8 x)
(4 x 2 − 25)2
100 x
100 x
=−
=−
2
2
(4 x − 25)
(2 x − 5)2 (2 x + 5)2
x2
2− x
2
( x + 2)
–3 –2
2
(2 − x)(2 x) − x 2 (−1)
)
2
Increasing on (–∞, –3) and (–1, ∞); decreasing
on (–3, –2) and (–2, –1); relative maximum
when x = –3; relative minimum when x = –1.
–5
y′ =
(
( x + 2)(2 x ) − x 2 − 3 (1)
2
+
+
35. y =
+
( x + 2)
( x + 2)2
CV: x = –3, –1, but x = –2 must be included in
the sign chart because it is a point of
discontinuity of y.
15
CV: None but x = −
+
(2 x − 1)(4 x 2 + 2 x + 1)
1
2
y′ =
=
=
=
x2 − 3
x+2
37. y =
< 0 for x > 0.
x3
Decreasing on (0, ∞); no relative extremum.
34. y =
=−
1
⎛1
⎞
Increasing on ⎜ , ∞ ⎟ ; decreasing on (−∞, 0)
2
⎝
⎠
1
⎛ 1⎞
and ⎜ 0, ⎟ ; relative minimum when x = .
2
⎝ 2⎠
0
Decreasing on (–∞, 0) and (0, ∞); no relative
extremum.
10
−
0
–
33. y =
1
x
x2
x2
1
CV: x = , but x = 0 must be included in the
2
sign chart because it is a point of discontinuity of
y.
3
= 3 x −1
x
–
4
36. y = 4 x 2 +
Decreasing on (–∞, 1) and (1, ∞); no relative
extremum.
y′ = −3x −2 = −
–
Decreasing on (–∞, 0) and (4, ∞); increasing on
(0, 2) and (2, 4); relative minimum when x = 0;
relative maximum when x = 4.
–
32. y =
+
2
0
( x − 1)2
CV: None, but x = 1 must be included in the sign
chart because it is a point of discontinuity of y.
–
+
x(4 − x)
(2 − x)
(2 − x)2
CV: x = 0, 4, but x = 2 must be included in the
CV: x = 0, but x = ±
473
5
must be included in the
2
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
(
sign chart because they are points of
discontinuity of y.
+
–
+
–5
2
–
5⎞
⎛
⎛ 5
Increasing on ⎜ −∞, − ⎟ and ⎜ − ,
2⎠
⎝
⎝ 2
⎞
0⎟;
⎠
y′ =
2
2
( x − 1)−1/ 3 =
3
3
3 x −1
CV: x = 1
y′ =
−
5x + 2
1
( x2 + 1) (5) − (5x + 2)(2 x) = −5x2 − 4 x + 5
2
2
( x2 + 1)
( x2 + 1)
42. y = x 2 ( x + 3) 4
y′ = x 2 (4)( x + 3)3 + ( x + 3) 4 (2 x)
y′ = 0 when −5 x 2 − 4 x + 5 = 0 ; by the quadratic
= 2 x( x + 3)3 [2 x + ( x + 3)]
−2 ± 29
5
formula, x =
–
= 2 x( x + 3)3 (3x + 3) = 6 x( x + 3)3 ( x + 1)
CV: x = 0, –3, –1
–
−2 ± 29
5
+
5
⎛
−2 − 29 ⎞
Decreasing on ⎜⎜ −∞,
⎟⎟ and
5
⎝
⎠
⎛ −2 + 29
⎞
, ∞ ⎟⎟ ; increasing on
⎜⎜
5
⎝
⎠
⎛ −2 − 29 −2 + 29 ⎞
,
⎜⎜
⎟⎟ ; relative minimum
5
5
⎝
⎠
) ( 3x 2 − 9 ) =
− 23
= x 2 ( x − 6)3 (7 x − 18)
CV: x = 0, 6,
+
+
–
–3 – 3 0
–
+
3
( x + 3 )( x − 3 )
2
[ x( x + 3)( x − 3)] 3
(
)(
Increasing on (–∞, –3), −3, − 3 ,
–
18
7
18
7
+
6
⎛ 18 ⎞
Increasing on (–∞, 0), ⎜ 0, ⎟ , and (6, ∞);
⎝ 7⎠
⎛ 18 ⎞
decreasing on ⎜ , 6 ⎟ ; relative maximum when
⎝7
⎠
18
; relative minimum when x = 6.
x=
7
+
3
+
0
CV: x = ± 3, 0, ± 3
+
( )
= x 2 ( x − 6)3 [4 x + 3( x − 6)]
3
40. y = x3 − 9 x
(
0
y′ = x3 ⎡ 4( x − 6)3 ⎤ + ( x − 6) 4 3 x 2
⎣
⎦
−2 − 29
; relative maximum when
5
1 3
x − 9x
3
+
43. y = x3 ( x − 6) 4
−2 + 29
x=
.
5
y′ =
–
–3 –1
–2 – 29 –2 + 29
when x =
+
Increasing on (–3, –1) and (0, ∞); decreasing on
(–∞, –3) and (–1, 0); relative maximum when
x = –1; relative minima when x = –3 and x = 0.
–
5
+
Increasing on (1, ∞); decreasing on (−∞, 1);
relative minimum when x = 1.
x2 + 1
CV: x =
)
41. y = ( x − 1) 2 / 3
⎛ 5⎞
⎛5
⎞
decreasing on ⎜ 0, ⎟ and ⎜ , ∞ ⎟ ; relative
⎝ 2⎠
⎝2
⎠
maximum at x = 0.
39. y =
(
relative maximum when x = − 3 ; relative
minimum when x = 3 .
5
2
0
)
(3, ∞); decreasing on − 3, 0 and 0, 3 ;
)
3, 3 , and
474
ISM: Introductory Mathematical Analysis
Section 13.1
⎛ 3 2⎞
Decreasing on ⎜⎜ 0,
⎟ ; increasing on
2 ⎟⎠
⎝
⎛3 2 ⎞
3 2
.
, ∞⎟⎟ ; relative minimum when x =
⎜⎜
2
⎝ 2
⎠
2
44. y = x(1 − x) 5
2
−3 ⎤
⎡ 2
y′ = x ⎢ − (1 − x) 5 ⎥ + (1 − x) 5 (1)
⎣ 5
⎦
− 35 ⎡ 2
−3 ⎛ 7
⎤
⎞
= (1 − x) ⎢ − x + (1 − x) ⎥ = −(1 − x) 5 ⎜ x − 1⎟
⎣ 5
⎦
⎝5
⎠
5
CV: x = , 1
7
+
–
48. y = x −1e x
⎛1 1 ⎞
⎛ x −1 ⎞
y ′ = x −1e x − x −2 e x = e x ⎜ − ⎟ = e x ⎜
⎟
2
x
x ⎠
⎝
⎝ x2 ⎠
CV: x = 1, but x = 0 must also be included in the
sign chart because it is a point of discontinuity of
y.
+
1
5
7
5⎞
⎛
Increasing on ⎜ −∞, ⎟ and (1, ∞); decreasing
7⎠
⎝
5
5
⎛
⎞
on ⎜ , 1⎟ ; relative maximum when x = ;
7
⎝7 ⎠
relative minimum when x = 1.
−πx
+π
y ′ = −πe
−πx
45. y = e
−
49. y = e x + e− x
y′ = e x − e − x
< 0 for all x. Thus decreasing on
Setting y′ = 0 gives e x − e− x = 0 , e x = e− x ,
x = –x, x = 0
CV: x = 0
–
Decreasing on (–∞, 0); increasing on (0, ∞);
relative minimum when x = 0.
1
x=e =
e
1
CV: x =
e
50. y = e− x
2
1
e
+
⎛ 1⎞
Decreasing on ⎜ 0, ⎟ ; increasing on
⎝ e⎠
1
relative minimum when x = .
e
⎛1
⎞
⎜ , ∞⎟ ;
⎝e
⎠
2
/2
–
0
Increasing on (−∞, 0); decreasing on (0, ∞);
relative maximum at x = 0
51. y = x ln x – x. [Note: x > 0.]
⎡ 1
⎤
y′ = ⎢ x ⋅ + (ln x)(1) ⎥ − 1 = ln x
x
⎣
⎦
CV: x = 1
47. y = x 2 − 9 ln x . [Note: x > 0.]
y′ = 2 x −
/2
y ′ = − xe− x
CV: x = 0
+
–
+
0
−1
CV: x =
1
0
46. y = x ln x. (Note: x > 0.)
y′ = 1 + ln x
y′ = 0 when 1 + ln x = 0, ln x = –1, or
0
+
Increasing on (1, ∞); decreasing on (−∞, 0) and
(0, 1); relative minimum when x = 1.
(−∞, ∞); no relative extremum.
–
−
9 2 x2 − 9
=
x
x
–
0
3 2
2
+
1
Decreasing on (0, 1); increasing on (1, ∞);
relative minimum when x = 1; no relative
maximum.
+
0 3 2
5
475
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
( )
y′ = ( x 2 + 1)( −e− x ) + e− x (2 x)
= −e− x ⎡⎢( x 2 + 1) − 2 x ⎤⎥ = −e− x ( x − 1) 2
⎣
⎦
52. y = x 2 + 1 e− x
4
x
5
CV: x = 1
–
y
–12
–
1
Decreasing on (–∞, 1) and (1, ∞); never
increasing; no relative extremum.
55. y = 3x − x3 = x
(
)(
3−x
)
)
Symmetric about origin.
Intercepts (−2, 0), (5, 0), (0, −10)
y′ = 2 x − 3
y′ = 3 − 3 x 2 = 3(1 + x )(1 − x)
CV: x = ±1
Decreasing on (–∞, –1) and (1, ∞); increasing on
(–1, 1); relative minimum when x = –1; relative
maximum when x = 1.
3
2
3⎞
⎛
Decreasing on ⎜ −∞, ⎟ ; increasing on
2⎠
⎝
3
⎛3
⎞
⎜ , ∞ ⎟ ; relative minimum when x = .
2
⎝2
⎠
y
3+x
Intercepts: (0, 0), ± 3, 0
53. y = x 2 − 3x − 10 = ( x + 2)( x − 5)
CV: x =
(
5
y
x
5
5
x
6
(
)
56. y = x 4 − 16 = x 2 + 4 ( x + 2)( x − 2)
Intercepts (±2, 0), (0, –16)
Symmetric about y-axis.
54. y = 2 x 2 − 5 x − 12 = (2 x + 3)( x − 4)
y′ = 4 x 3
CV: x = 0
Decreasing on (–∞, 0); increasing on (0, ∞);
relative minimum when x = 0.
⎛ 3 ⎞
Intercepts ⎜ − , 0 ⎟ , (4, 0), (0, –12)
⎝ 2 ⎠
5⎞
⎛
y′ = 4 x − 5 = 4 ⎜ x − ⎟
4⎠
⎝
5
CV: x =
4
5⎞
⎛
⎛5
⎞
Decreasing on ⎜ −∞, ⎟ ; increasing on ⎜ , ∞ ⎟ ;
4
4
⎝
⎠
⎝
⎠
5
relative minimum when x = .
4
4
y
x
–2
2
–16
476
5
ISM: Introductory Mathematical Analysis
(
57. y = 2 x3 − 9 x 2 + 12 x = x 2 x 2 − 9 x + 12
Section 13.1
)
y
8
Note that 2 x 2 − 9 x + 12 = 0 has no real roots.
The only intercept is (0, 0).
(
y′ = 6 x 2 − 18 x + 12 = 6 x 2 − 3 x + 2
)
x
= 6(x – 2)(x – 1)
CV: x = 1, 2
Increasing on (–∞, 1) and (2, ∞); decreasing on
(1, 2); relative maximum when x = 1; relative
minimum when x = 2.
8
–2 –1
6
6⎞
⎛
60. y = x 6 − x5 = x5 ⎜ x − ⎟
5
5⎠
⎝
⎛6 ⎞
Intercepts (0, 0), ⎜ , 0 ⎟
⎝5 ⎠
y
5
4
y′ = 6 x5 − 6 x 4 = 6 x 4 ( x − 1)
CV: x = 0, 1
Increasing on (1, ∞); decreasing on (−∞, 0) and
(0, 1); relative minimum when x = 1.
x
1 2
2
5
y
2
58. y = 2 x3 − x 2 − 4 x + 4
The x-intercept is not convenient to find.
y-intercept is (0, 4).
y ′ = 6 x 2 − 2 x − 4 = 2(3 x + 2)( x − 1)
x
3
2
CV: x = − , 1
3
2⎞
⎛
Increasing on ⎜ −∞, − ⎟ and (1, ∞); decreasing
3⎠
⎝
61. y = ( x − 1) 2 ( x + 2)2
Intercepts: (1, 0), (–2, 0), (0, 4)
y′ = ( x − 1) 2 ⋅ 2( x + 2) + ( x + 2) 2 ⋅ 2( x − 1)
= 2(x – 1)(x + 2)[(x – 1) + (x + 2)]
= 2(x – 1)(x + 2)(2x + 1)
1
CV: x = 1, –2, −
2
⎛ 1 ⎞
Decreasing on (–∞, –2) and ⎜ − , 1⎟ ; increasing
⎝ 2 ⎠
1
⎛
⎞
on ⎜ −2, − ⎟ and (1, ∞); relative minima when
2⎠
⎝
x = –2 or x = 1; relative maximum when
1
x=− .
2
2
⎛ 2 ⎞
on ⎜ − , 1⎟ ; relative maximum when x = − ;
3
3
⎝
⎠
relative minimum when x = 1.
8
y
x
5
59. y = x 4 + 4 x3 + 4 x 2 = x 2 ( x + 2)2
Intercepts (0, 0), (–2, 0)
8
y′ = 4 x3 + 12 x 2 + 8 x = 4 x( x + 1)( x + 2)
CV: x = 0, –1, –2
Increasing on (–2, –1) and (0, ∞); decreasing on
(–∞, –2) and (–1, 0); relative maximum when
x = –1; relative minima when x = –2 or x = 0.
y
4
x
–2
477
1
5
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
62. y = x ( x 2 − x − 2) = x ( x − 2)( x + 1)
[Note x ≥ 0.]
Intercepts (0, 0), (2, 0)
10
y = x5 / 2 − x3 / 2 − 2 x1/ 2
y
x
10
–2
5 3 / 2 3 1/ 2
1
− x − 2 ⋅ x −1/ 2
x
2
2
2
1
2
=
(5 x − 3 x − 2)
2 x
1
=
(5 x + 2)( x − 1)
2 x
CV: x = 0, 1 (x ≥ 0)
Decreasing on (0, 1); increasing on (1, ∞);
relative minimum when x = 1.
y′ =
5
65.
y
10
y
x
10
66.
x
5
y
5
2
(
)
x
63. y = 2 x − x = x 2 − x . [Note: x ≥ 0.]
1
Intercepts (0, 0), (4, 0)
1
1− x
y′ =
−1 =
x
x
CV: x = 0, 1
Increasing on (0, 1); decreasing on (1, ∞);
relative maximum when x = 1.
5
4
67. c f = 25, 000
cf =
cf
q
=
25, 000
q
d
25, 000
< 0 for q > 0, so c f is a
cf = −
dq
q2
( )
y
decreasing function for q > 0.
1
68. c = 3q − 3q 2 + q3
x
1
4
8
Marginal cost is given by
5
2
dc
= 3 − 6q + 3q 2 .
dq
dc ⎤
d ⎡⎢ dq
dc
⎣
⎦⎥ < 0, that is,
Thus
is increasing when
dq
dq
2
64. y = x 3 + 5 x 3 = x 3 ( x + 5)
Intercepts (0, 0), (–5, 0)
5 2 10 5( x + 2)
y′ = x 3 + 1 =
1
3
3x 3
3x 3
when −6 + 6q > 0. Hence q > 1.
CV: x = 0, –2
Increasing on (–∞, –2) and (0, ∞); decreasing on
(–2, 0); relative maximum when x = –2; relative
minimum when x = 0.
478
ISM: Introductory Mathematical Analysis
Section 13.1
69. p = 400 – 2q
Revenue is given by
r = pq = (400 − 2q )q
⎛ T ⎞
73. E = 0.71⎜1 − c ⎟
⎝ Th ⎠
⎛T
dE
= 0.71⎜ c
⎜T2
dTh
⎝ h
= 400q − 2q 2
Marginal revenue is r ′ = 400 − 4q . Marginal
revenue is increasing when its derivative is
positive. But (r′)′ = −4 < 0 . Thus marginal
revenue is never increasing.
increases.
a2
⎛ a⎞
74. r = 2 F + ⎜ 1 − ⎟ p − p 2 +
b
⎝ b⎠
dr ⎛ a ⎞
b−a
⎛b−a
= ⎜1 − ⎟ − 2 p =
− 2 p = 2⎜
−
dp ⎝ b ⎠
b
⎝ 2b
dr
Setting
= 0 gives the critical value
dp
70. c = q
dc
1
Marginal cost =
. Since
=
dq 2 q
dc ⎤
d ⎡⎢ dq
⎣ ⎥⎦ = − 1 < 0 for q > 0, marginal cost is
dq
4 q3
decreasing for q > 0.
Average cost = c =
p=
c
1
. Since
=
q
q
144 ⎤
⎡
75. C (k ) = 100 ⎢100 + 9k +
, 1 ≤ k ≤ 100
k ⎥⎦
⎣
r ′ = 240 + 114q – 3q 2 = 3(40 − q )(2 + q )
Since q ≥ 0, we have q = 40 as the only CV.
Since r is increasing on (0, 40) and decreasing
on (40, ∞), r is a maximum when output is 40.
a.
C(1) = 25,300
b.
⎡ 9k 2 − 144 ⎤
⎡ 144 ⎤
C ′(k ) = 100 ⎢9 −
= 100 ⎢
⎥
⎥
k2 ⎦
k2
⎣
⎢⎣
⎥⎦
72. z = (1 + b) w p − bwc , w p is function of wc , and
⎡ 9(k + 4)(k − 4) ⎤
= 100 ⎢
⎥
k2
⎣
⎦
Since k ≥ 1, the only critical value is k = 4.
If 1 ≤ k < 4, then C ′(k ) < 0 and C is
decreasing. If 4 < k ≤ 100, then C ′(k ) > 0
and C is increasing. Thus C has an absolute
minimum for k = 4.
b > 0.
dw p
dz
= (1 + b)
− b(1)
dwc
dwc
⎡ dw p
b ⎤
= (1 + b) ⎢
−
⎥ (factoring)
⎣ dwc 1 + b ⎦
dw p
b−a
b−a
dr
. If p <
, then
> 0 and r is
2b
2b
dp
b−a
dr
, then
< 0 and r is
2b
dp
decreasing. Thus revenue is maximum for
b−a
p=
.
2b
71. r = 240q + 57 q 2 − q3
c.
dw p
b
b
, then
−
<0.
dwc b + 1
dwc b + 1
Because b > 0, then 1 + b > 0. Thus from
dz
part (a),
< 0 so z is a decreasing
dwc
b. If
⎞
p⎟
⎠
increasing. If p >
dc
1
=−
< 0 for q > 0, average cost is
dq
2 q3
decreasing for q > 0.
a.
⎞
⎟ > 0 , so as Th increases, E
⎟
⎠
<
function of wc .
479
C(4) = 17,200
Chapter 13: Curve Sketching
76. P =
ISM: Introductory Mathematical Analysis
100
1 + 100, 000e−0.36h
(
dP d ⎡
100 1 + 100, 000e−0.36 h
=
dh dh ⎢⎣
3, 600, 000
=
2
e0.36h 1 + 100, 000e−0.36 h
(
Since
)
−1 ⎤
–5
⎥
⎦
)
dP
> 0 , P is an increasing function of h.
dh
84.
Problems 13.2
79. Relative maximum: (2.74, 3.74); relative
minimum: (–2.74, –3.74)
1.
f ( x) = x 2 − 2 x + 3 and f is continuous over
[0, 3].
f ′( x) = 2 x − 2 = 2( x − 1)
The only critical value on (0, 3) is x = 1. We
evaluate f at this point and at the endpoints:
f(0) = 3, f(1) = 2, and f(3) = 6.
Absolute maximum: f(3) = 6;
absolute minimum: f(1) = 2
2.
f ( x) = −2 x 2 − 6 x + 5 and f is continuous over
[–3, 2].
3⎞
⎛
f ′( x) = −4 x − 6 = −4 ⎜ x + ⎟
2⎠
⎝
3
The only critical value on (–3, 2) is x = − . We
2
3
19
⎛
⎞
have f(–3) = 5, f ⎜ − ⎟ = , and f(2) = –15.
⎝ 2⎠ 2
⎛ 3 ⎞ 19
Absolute maximum: f ⎜ − ⎟ = ;
⎝ 2⎠ 2
absolute minimum: f(2) = –15
80. Relative maximum: (0.05, 3.05)
81. Relative minima: 0, 1.50, 2.00; relative maxima:
0.57, 1.77
82. f has relative extrema when x ≈ 0.38, 1.18;
f ′( x) = 0 when x ≈ 0.38, 1.18.
4
2.5
–2.5
f ′( x) = 4 − 6 x − 3 x 2
10
b.
3.
5
–5
–5
c.
f ′( x) = 4 x3 − 2 x − 2( x + 2)
= 4 x3 − 4 x − 4
CV: x ≈ 1.32
78. Relative minimum: (1.26, −5.74)
–1
5
–10
77. Relative minimum: (−3.83, 0.69)
83. a.
10
d.
1 3 1 2
x + x − 2 x + 1 and f is continuous
3
2
over [−1, 0].
f ′( x) = x 2 + x − 2 = ( x + 2)( x − 1)
f ( x) =
There are no critical values on (−1, 0), so we
only have to evaluate f at the endpoints:
19
and f(0) = 1.
f (−1) =
6
19
Absolute maximum: f (−1) =
6
Absolute minimum: f(0) = 1
f ′( x) > 0 on (–2.53, 0.53); f ′( x) < 0 on
(–∞, –2.53), (0.53, ∞), f is inc. on
(–2.53, 0.53); f is dec. on (–∞, –2.53),
(0.53, ∞).
480
ISM: Introductory Mathematical Analysis
4.
f ( x) =
[0, 1].
Section 13.2
1 4 3 2
x − x and f is continuous over
4
2
(
)(
f ′( x) = x3 − 3 x = x x + 3 x − 3
8.
)
f ′( x) = 7 x 2 + 4 x − 3 = (7 x − 3)( x + 1)
There are no critical values on (0, 1), so we only
have to evaluate f at the end points: f(0) = 0 and
5
f (1) = −
4
Absolute maximum: f(0) = 0;
5
absolute minimum: f (1) = −
4
5.
2
(
2
9.
)
f ( x) = 3 x 4 − x6 and f is continuous over [–1, 2].
= 6x
3
(
(
2−x
)(
2+x
)
)
= 6(2 x + 3)( x − 1)
The only critical values on (–1, 2) are x = 0,
⎛1 ⎞
The only critical value on ⎜ , 3 ⎟ is x = 1. We
⎝2 ⎠
evaluate f at this point and the endpoints:
19
⎛1⎞
f ⎜ ⎟ = − ; f(1) = –8, f(3) = 84.
4
⎝2⎠
Absolute maximum: f(3) = 84;
absolute minimum: f(1) = –8
We have f(–1) = 2, f(0) = 0, f
f(2) = –16.
Absolute maximum: f
10.
2 − 13
x .
3
The only critical value on (–8, 8) is x = 0. We
have f(–8) = 4, f(0) = 0, and f(8) = 4. Thus there
is an absolute maximum when x = –8 or
x = 8, and an absolute minimum when x = 0.
Absolute maximum: f(–8) = f(8) = 4;
absolute minimum: f(0) = 0
f ( x) =
1 4 1 2
x − x + 3 and f is continuous over
4
2
The critical values of f on (−2, 3) are x = −1, 0,
11
1. We have f(−2) = 5, f (−1) = , f(0) = 3,
4
11
75
and f (3) = .
f (1) =
4
4
75
Absolute maximum: f (3) =
4
11
Absolute minimum: f (−1) = f (1) =
4
f ( x) = −3x5 + 5 x3 and f is continuous over
[–2, 0].
(
( 2) = 4 ;
[−2, 3].
f ′( x) = x3 − x = x( x 2 − 1) = x( x − 1)( x + 1)
2
f ( x) = x 3 and f is continuous over [–8, 8].
f ′( x) = −15 x 4 + 15 x 2 = 15 x 2 1 − x 2
( 2 ) = 4 , and
2.
absolute minimum: f(2) = –16
f ′( x) =
7.
3
. We
7
⎛ 3 ⎞ 13
have f(0) = 1, f ⎜ ⎟ =
, and f(3) = 73.
⎝ 7 ⎠ 49
Absolute maximum: f(3) = 73;
⎛ 3 ⎞ 13
absolute minimum: f ⎜ ⎟ =
⎝ 7 ⎠ 49
f ′( x) = 12 x3 − 6 x5 = 6 x3 2 − x 2
f ′( x) = 12 x + 6 x − 18 = 6 2 x + x − 3
6.
The only critical value on (0, 3) is x =
f ( x) = 4 x3 + 3 x 2 − 18 x + 3 and f is continuous
⎡1 ⎤
over ⎢ , 3⎥ .
⎣2 ⎦
7 3
x + 2 x 2 − 3x + 1 and f is continuous
3
over [0, 3].
f ( x) =
)
11.
= 15 x 2 (1 + x)(1 − x)
The only critical value on (–2, 0) is x = –1. We
have f(–2) = 56, f(–1) = –2, and f(0) = 0.
Absolute maximum: f(–2) = 56;
absolute minimum: f(–1) = –2.
f ( x) = x 4 − 9 x 2 + 2 and f is continuous over
[–1, 3].
(
f ′( x) = 4 x3 − 18 x = 2 x 2 x 2 − 9
= 2x
(
2x − 3
)(
2x + 3
)
)
The only critical values on (–1, 3) are x = 0 and
3
3 2
. We have f(–1) = –6, f(0) = 2,
x=
=
2
2
481
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
⎛3 2 ⎞
73
f ⎜⎜
⎟⎟ = − , and f(3) = 2.
4
⎝ 2 ⎠
Absolute maximum: f(0) = f(3) = 2;
⎛3 2 ⎞
73
absolute minimum: f ⎜⎜
⎟⎟ = −
4
⎝ 2 ⎠
12.
f ( x) =
–4
x
and f is continuous over [0, 2].
2
x +1
a.
c.
d. 14,283
Problems 13.3
1.
3
−4
+
0
3⎞
⎛
Concave up on ⎜ −∞, − ⎟ and (0, ∞); concave
4⎠
⎝
⎛ 3 ⎞
down on ⎜ − , 0 ⎟ . Inflection points when
⎝ 4 ⎠
3
x = − , 0. .
4
10
28
–1
2.
Absolute maximum: f(−26) = f(28) = 9;
absolute minimum: f(1) = 0
f ( x) =
x5 x 4
+
− 2 x2
20 4
f ′′( x) = ( x − 1)( x + 2) 2
f ′′( x) is 0 when x = 1, –2. Sign chart for f ′′ :
f ( x) = 0.2 x3 − 3.6 x 2 + 2 x + 1 and f is continuous
–
over [−1, 2].
5
–1
−
+
2
–26
f ( x ) = 2 x 4 + 3 x3 + 2 x − 3
f ′′( x) = 6 x(4 x + 3)
3
f ′′( x) is 0 when x = 0, − . Sign chart for f ′′ :
4
f ( x) = ( x − 1) 3 and f is continuous over
[−26, 28].
14.
9
2
The only critical value on (0, 2) is x = 1. We
2
1
have f(0) = 0, f (1) = ,and f (2) = .
5
2
1
Absolute maximum: f (1) = ;
2
absolute minimum: f(0) = 0
13.
–3.22, –0.78
b. 2.75
(1 + x)(1 − x)
( x2 + 1)
9
0
x 2 + 1) − x(2 x)
(
1 − x2
f ′( x) =
=
2
2
( x2 + 1)
( x2 + 1)
=
10
15.
–
–2
+
1
Concave down on (–∞, –2) and (–2, 1); concave
up on (1, ∞). Inflection point when x = 1.
2
3.
–10
f ( x) =
f ′′( x) =
Absolute maximum f(0.28) ≈ 1.28; absolute
minimum f(2) = −7.8
2 + x − x2
x2 − 2 x + 1
2(7 − x)
( x − 1) 4
f ′′( x) is 0 when x = 7. Although f ′′ is not
defined when x = 1, f is not continuous at x = 1.
So there is no inflection point when x = 1, but
x = 1 must be considered in concavity analysis.
482
ISM: Introductory Mathematical Analysis
Section 13.3
Sign chart for f ′′ :
+
+
6.
–
1
7
f ′′( x) =
Concave up on (–∞, 1) and (1, 7); concave down
on (7, ∞). Inflection point when x = 7.
4.
( x − 1)2
2(2 x + 1)
f ′′( x) =
( x − 1) 4
+
1
f ′′( x) = 0 when x = − . Although f ′′ is not
2
defined when x = 1, f is not continuous at x = 1.
So there is no inflection point when x = 1, but
x = 1 must be considered in concavity analysis.
Sign chart of f ′′ :
–
+
–1
–2
+
8. y = −74 x 2 + 19 x − 37
y′ = −148 x + 19
y ′′ = −148 < 0 for all x. Thus the graph is
concave down on (–∞, ∞).
1
2
9. y = 4 x3 + 12 x 2 − 12 x
2
x +1
f ( x) =
y′ = 12 x 2 + 24 x − 12
y ′′ = 24 x + 24 = 24( x + 1)
Possible inflection point when x = –1. Concave
down on (–∞, –1): concave up on (–1, ∞);
inflection point when x = –1.
2
x −2
f ′′( x) =
(
) = 6 ( 3x 2 + 2 )
3
3
( x2 − 2) ⎡⎣( x − 2 )( x + 2 )⎤⎦
6 3x2 + 2
f ′′( x) is never 0. Although f ′′ is not defined
10. y = x3 − 6 x 2 + 9 x + 1
when x = ± 2 , f is not continuous at x = ± 2 .
So there is no inflection point when x = ± 2 ,
but x = ± 2 must be considered in concavity
analysis. Sign chart of f ′′ :
+
–
– 2
y′ = 3x 2 − 12 x + 9
y ′′ = 6 x − 12 = 6( x − 2)
Possible inflection point when x = 2. Concave
down on (–∞, 2); concave up on (2, ∞);
inflection point when x = 2.
+
2
(
)
Concave up on −∞, − 2 and
(
2
down for all x, that is, on (–∞, ∞).
1⎞
⎛
down on ⎜ −∞, − ⎟ .
2⎠
⎝
Inflection point when x =
–
0
7. y = −2 x 2 + 4 x
y′ = −4 x + 4
y ′′ = −4 < 0 for all x, so the graph is concave
⎛ 1 ⎞
Concave up on ⎜ − , 1⎟ and (1, ∞); concave
⎝ 2 ⎠
5.
( 4 − x2 )
)
3
2
Concave up on (–2, 0); concave down on (0, 2).
Inflection point when x = 0.
1
2
(
2 x x2 − 6
Note that the domain of f is [–2, 2]. f ′′( x) is 0
only when x = 0; f ′′ is not defined when x = ±2,
which are the endpoints of the domain of f. The
only possible point of inflection occurs when
x = 0. Sign chart for f ′′ :
x2
f ( x) =
f ( x) = x 4 − x 2
)
(
11. y = 2 x3 − 5 x 2 + 5 x − 2
)
2, ∞ ;
y ′ = 6 x 2 − 10 x + 5
concave down on − 2, 2 . No inflection
5⎞
⎛
y ′′ = 12 x − 10 = 12 ⎜ x − ⎟
6⎠
⎝
point.
5
Possible inflection point when x = . Concave
6
483
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
1
5⎞
⎛
⎛5
⎞
down on ⎜ −∞, ⎟ ; concave up on ⎜ , ∞ ⎟ ;
6
6
⎝
⎠
⎝
⎠
5
inflection point when x =
6
15. y = 2 x 5
2 − 54
x
5
8 −9
8
y ′′ = − x 5 = −
9
25
25 x 5
y′ =
12. y = x 4 − 8 x 2 − 6
y ′′ is not defined when x = 0 and y is continuous
there. Thus there is a possible inflection point
when x = 0. Concave up on (–∞, 0); concave
down on (0, ∞); inflection point when x = 0.
y′ = 4 x3 − 16 x
4⎞
⎛
y ′′ = 12 x 2 − 16 = 12 ⎜ x 2 − ⎟
3⎠
⎝
⎛
2 3 ⎞⎛
2 3⎞
= 12 ⎜⎜ x −
⎟⎜ x +
⎟
⎟⎜
3 ⎠⎝
3 ⎟⎠
⎝
16. y =
x 4 19 x3 7 x 2
+
−
+ x+5
2
6
2
19
y′ = 2 x 3 + x 2 − 7 x + 1
2
2
′′
y = 6 x + 19 x − 7 = (3 x − 1)(2 x + 7)
17. y =
y ′ = 8 x3 − 96 x + 7
y ′′ = 24 x 2 − 96 = 24( x 2 − 4) = 24( x + 2)( x − 2)
Possible inflection points when x = ±2. Concave
up on (−∞, −2) and (2, ∞); concave down on
(−2, 2); inflection points when x = ±2.
7 1
Possible inflection points when x = − , .
2 3
7⎞
⎛
⎛1
⎞
Concave up on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ;
2⎠
⎝
⎝3
⎠
⎛ 7 1⎞
concave down on ⎜ − , ⎟ ; inflection points
⎝ 2 3⎠
7 1
when x = − , .
2 3
x4 9 x2
+
+ 2x
4
2
y′ = − x 3 + 9 x + 2
y ′′ = −3x 2 + 9 = −3( x 2 − 3)
(
)(
)
Possible inflection points when x = ± 3 .
(
)
Concave down on – ∞, − 3 and
(
84
x5
Although y ′′ is not defined when x = 0, y is not
continuous there. Thus there is no possible
inflection point. However, x = 0 must be
considered in concavity analysis. Concave down
on (−∞, 0); concave up on (0, ∞); no inflection
point
2 3
.
3
= −3 x + 3 x − 3
= 7 x −3
y ′′ = 84 x −5 =
13. y = 2 x 4 − 48 x 2 + 7 x + 3
14. y = −
x
3
y ′ = −21x −4
2 3
Possible inflection points x = ±
. Concave
3
⎛
⎛2 3
⎞
2 3⎞
up on ⎜⎜ −∞, –
, ∞ ⎟⎟ ; concave
⎟⎟ and ⎜⎜
3 ⎠
⎝
⎝ 3
⎠
⎛ 2 3 2 3⎞
down on ⎜⎜ −
,
⎟ ; inflection points when
3
3 ⎟⎠
⎝
x=±
7
)
(
)
5
1
1
1
2
18. y = − x 4 − x3 + x 2 + x −
2
6
2
3
5
1
1
y′ = −10 x3 − x 2 + x +
2
3
3, ∞ ;
concave up on − 3, 3 ; inflection points
when x = ± 3 .
y ′′ = −30 x 2 − x + 1 = −(5 x + 1)(6 x − 1)
1 1
Possible inflection points when x = − , .
5 6
1⎞
⎛
⎛1
⎞
Concave down on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ;
5⎠
⎝
⎝6
⎠
484
ISM: Introductory Mathematical Analysis
Section 13.3
(
(
⎛ 1 1⎞
concave up on ⎜ − , ⎟ ; inflection points when
⎝ 5 6⎠
1 1
x=− , .
5 6
1 5 1 4 1 3 1
2
x − x + x − x−
20
4
6
2
3
1 4
1
1
y′ = x − x 3 + x 2 −
4
2
2
6⎞
⎛
y ′′ = 30 x 4 − 36 x 2 = 30 x 2 ⎜ x 2 − ⎟
5⎠
⎝
⎛
6 ⎞⎛
6⎞
= 30 x 2 ⎜⎜ x −
⎟⎟ ⎜⎜ x +
⎟
5 ⎠⎝
5 ⎟⎠
⎝
)
y ′′ is 0 when x = 0 or x − 3x + 1 = 0 . Using the
quadratic formula to solve x 2 − 3x + 1 = 0 gives
6
.
5
⎛
⎛ 6
⎞
6⎞
Concave up on ⎜⎜ −∞, −
, ∞ ⎟⎟ ;
⎟⎟ and ⎜⎜
5
5
⎝
⎠
⎝
⎠
⎛ 6 ⎞
⎛
6⎞
concave down on ⎜⎜ − , 0 ⎟⎟ and ⎜⎜ 0,
⎟.
5 ⎟⎠
⎝ 5 ⎠
⎝
Possible inflection points when x = 0, ±
3± 5
. Thus possible inflection points
2
3± 5
occur when x = 0,
. Concave down on
2
⎛ 3− 5 3+ 5 ⎞
(–∞, 0) and ⎜⎜
,
⎟ ; concave up on
2 ⎟⎠
⎝ 2
⎛ 3− 5 ⎞
⎛ 3+ 5
⎞
⎜⎜ 0, 2 ⎟⎟ and ⎜⎜ 2 , ∞ ⎟⎟ ; inflection points
⎝
⎠
⎝
⎠
x=
Inflection points when x = ±
3± 5
.
2
23. y =
y′ =
1 5
x − 3x3 + 17 x + 43
10
1
y′ = x 4 − 9 x 2 + 17
2
y ′′ = 2 x3 − 18 x = 2 x( x 2 − 9)
20. y =
1 6 7 4
x − x + 5x2 + 2 x − 1
21. y =
30
12
1
7
y′ = x5 − x3 + 10 x + 2
5
3
)(
y ′′ = x 4 − 7 x 2 + 10 = x 2 − 2 x 2 − 5
4
24. y = 1 −
y′ =
x2
x3
6
x4
No possible inflection point, but we must
consider x = 0 in the concavity analysis.
Concave down on (−∞, 0) and (0, ∞).
)
Possible inflection points when x = ± 2, ± 5 .
)(
1
2
y ′′ = −
)( x − 2 )( x + 5 )( x − 5 )
(
( x − 1)2
( x − 1)3
No possible inflection point, but we consider
x = 1 in the concavity analysis. Concave down
on (–∞, 1); concave up on (1, ∞).
Possible inflection points when x = 0, ±3.
Concave down on (−∞, −3) and (0, 3); concave
up on (−3, 0) and (3, ∞); inflection points when
x = 0, ±3.
(
6
.
5
x +1
x −1
−2
y ′′ =
= 2 x( x + 3)( x − 3)
= x+ 2
)
2, 5 ; inflection points when
y′ = 6 x5 − 12 x3
2
(
)
22. y = x 6 − 3 x 4
y ′′ = x3 − 3x 2 + x = x x 2 − 3 x + 1
when x = 0,
(
x = ± 5, ± 2 .
19. y =
(
)
5, ∞ ; concave down on − 5, − 2 and
)
Concave up on −∞, − 5 , − 2, 2 , and
485
Chapter 13: Curve Sketching
25. y =
ISM: Introductory Mathematical Analysis
x2
27. y =
2
x +1
x 2 + 1) (2 x) − x 2 (2 x)
(
2x
y′ =
=
2
2
( x2 + 1)
( x2 + 1)
2
x 2 + 1) (2) − 2 x(2) ( x 2 + 1) (2 x)
(
y ′′ =
4
( x2 + 1)
x 2 + 1) (2) − 8 x 2
(
=
3
( x2 + 1)
2 (1 − 3x 2 ) 2 (1 + 3x )(1 − 3 x )
=
=
3
3
2
x
1
+
( )
( x2 + 1)
Possible inflection points when x = ±
y′ =
4 x2
x+3
y′ =
y ′′ =
=
( x + 3)(8 x) − 4 x 2 (1)
( x + 3) 2
=
(
4 x2 + 6 x
6( x + 3)2
1 ( x + 3) 2 (21) − (21x + 40)[2( x + 3)]
⋅
6
( x + 3)4
=
1 ( x + 3)(21) − (21x + 40)(2)
⋅
6
( x + 3)3
=
1 −21x − 17
1 21x + 17
⋅
=− ⋅
3
6 ( x + 3)
6 ( x + 3)3
3
2⎤
⎡
1 ( x + 3) (21) − (21x + 17) ⎣3( x + 3) ⎦
′′
y =− ⋅
6
( x + 3)6
1 ( x + 3)(21) − (21x + 17)(3)
=− ⋅
6
( x + 3) 4
1 −42 x + 12
7x − 2
=− ⋅
=
4
6 ( x + 3)
( x + 3)4
1
.
3
⎛
⎛ 1
⎞
1 ⎞
, ∞⎟ ;
Concave down on ⎜ −∞, −
⎟ and ⎜
3
3
⎝
⎠
⎝
⎠
⎛ 1
1 ⎞
,
concave up on ⎜ −
⎟ ; inflection points
3
3⎠
⎝
1
when x = ±
.
3
26. y =
21x + 40
2
(x = –3
7
must be considered in concavity analysis).
2⎞
⎛
Concave down on (–∞, –3) and ⎜ −3, ⎟ ;
7⎠
⎝
⎛2
⎞
concave up on ⎜ , ∞ ⎟ ; inflection point when
⎝7
⎠
2
x= .
7
Possible inflection point when x =
28. y = 3( x 2 − 2) 2
)
y ′ = 12 x( x 2 − 2) = 12( x3 − 2 x)
2⎞
⎛
y ′′ = 12(3x 2 − 2) = 36 ⎜ x 2 − ⎟
3⎠
⎝
⎛
6 ⎞⎛
6⎞
= 36 ⎜ x −
⎟⎜ x +
⎟
⎜
3 ⎟⎠ ⎜⎝
3 ⎟⎠
⎝
( x + 3) 2
( x + 3)2 (4)(2 x + 6) − 4( x 2 + 6 x)(2)( x + 3)
( x + 3)4
72
( x + 3)3
No possible inflection point, but we must include
x = –3 in the concavity analysis. Concave down
on (–∞, –3); concave up on (–3, ∞).
Possible inflection points when x = ±
6
.
3
⎛
⎛ 6 ⎞
6⎞
Concave up on ⎜ −∞, −
, 0⎟;
⎟⎟ and ⎜⎜
⎜
⎟
3 ⎠
⎝
⎝ 3
⎠
⎛
6
6⎞
,
concave down on ⎜ −
⎟ ; inflection
⎜ 3
3 ⎟⎠
⎝
points when x = ±
486
6
.
3
ISM: Introductory Mathematical Analysis
Section 13.3
29. y = 5e x
3
⎛
⎞
Concave down on ⎜ 0, e 2 ⎟ ; concave up on
⎝
⎠
3
3
⎛ 2
⎞
2
⎜ e , ∞ ⎟ ; inflection point when x = e .
⎝
⎠
y′ = 5e x
y ′′ = 5e x
Thus y ′′ > 0 for all x. Concave up on (–∞, ∞).
34. y =
x
−x
x
−x
x
−x
30. y = e − e
y′ = e + e
y ′′ = e − e
y′ =
Setting y ′′ = 0 gives e x = e − x or, equivalently,
=
x = 0. Concave down on (–∞, 0); concave up on
(0, ∞); inflection point when
x = 0.
y′ = 3xe x + 3e x = 3e x ( x + 1)
=
y ′′ = 3e x (1) + 3( x + 1)e x = 3e x ( x + 2)
y ′′ = 0 if x = –2. Concave down on (–∞, –2);
=
2
2
2
2
y′ = 2 x 2 e x + e x = e x (2 x 2 + 1)
x2
2
y ′′ = e (4 x) + 2 x(2 x + 1)e
x2
x2
(
)
3e x (2 x) − x 2 + 1 3e x
9e
2x
=
(
3e
3e x
(
)
3e x (2 − 2 x) − 2 x − x 2 − 1 3e x
9e 2 x
(
)
(2 − 2 x) − 2 x − x 2 − 1
3e x
x2 − 4 x + 3
x
=
( x − 1)( x − 3)
35. y = x 2 − x − 6 = ( x − 3)( x + 2)
3
= e (4 x + 6 x)
Intercepts: (0, −6), (3, 0) and (−2, 0)
1⎞
⎛
y′ = 2x − 1 = 2 ⎜ x − ⎟
2⎠
⎝
concave up on (0, ∞); inflection point when
x = 0.
CV: x =
33. y =
1
2
1⎞
⎛
Decreasing on ⎜ −∞, ⎟ ; increasing on
2⎠
⎝
2 x ⋅ 1x − (ln x)(2)
25 ⎞
⎛1
⎞
⎛1
⎜ 2 , ∞ ⎟ ; relative minimum at ⎜ 2 , − 4 ⎟ .
⎝
⎠
⎝
⎠
y ′′ = 2
No possible inflection point. Concave up on
(−∞, ∞).
y ′′ =
=
2
ln x
. (Note: x > 0.)
2x
y′ =
=
x
2x − x −1
= 2 xe (2 x + 3)
y ′′ = 0 when x = 0. Concave down on (–∞, 0);
x2
)
2 x − x2 + 1
3e
3e x
Possible inflection points when x = 1, 3.
Concave up on (–∞, 1) and (3, ∞); concave
down on
(1, 3); inflection point when x = 1, 3.
concave up on (–2, ∞); inflection point when
x = –2.
32. y = xe x
3e x
2
y ′′ =
31. y = 3xe x
x2 + 1
4x2
( )
=
1 − ln x
2x2
2 x 2 − 1x − (1 − ln x)(4 x)
4 x4
−2 x − (1 − ln x)(4 x)
4
4
4x
−1 − (1 − ln x)(2)
2 x3
=
x
10
2 ln( x) − 3
2 x3
y ′′ is 0 if 2ln(x) – 3 = 0, ln x =
y
3
3
, x = e2 .
2
487
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
⎛1 9⎞
relative maximum at ⎜ , ⎟
⎝2 4⎠
y ′′ = −2
No possible inflection point. Concave down on
(–∞, ∞).
36. y = x 2 + 2
Intercept (0, 2)
y′ = 2 x
CV: x = 0
Decreasing on (–∞, 0); increasing on (0, ∞);
relative minimum at (0, 2).
y ′′ = 2
No possible inflection point. Concave up on
(–∞, ∞). Symmetric about the y-axis.
8
5
x
y
5
39. y = x3 − 9 x 2 + 24 x − 19
The x-intercepts are not convenient to find; the
y-intercept is (0, –19).
x
5
y′ = 3x 2 − 18 x + 24 = 3( x − 2)( x − 4)
CV: x = 2, x = 4
Increasing on (–∞, 2) and (4, ∞); decreasing on
(2, 4); relative maximum at (2, 1); relative
minimum at (4, –3).
y ′′ = 6 x − 18 = 6( x − 3)
Possible inflection point when x = 3. Concave
down on (–∞, 3); concave up on (3, ∞);
inflection point at (3, –1).
37. y = 5 x − 2 x 2 = x(5 − 2 x)
⎛5 ⎞
Intercepts (0, 0) and ⎜ , 0 ⎟
⎝2 ⎠
y′ = 5 − 4 x
CV: x =
5
4
5⎞
⎛
⎛5
⎞
Increasing on ⎜ −∞, ⎟ ; decreasing on ⎜ , ∞ ⎟ ;
4
4
⎝
⎠
⎝
⎠
⎛ 5 25 ⎞
relative maximum at ⎜ ,
⎟.
⎝4 8 ⎠
y ′′ = −4
No possible inflection point. Concave down on
(–∞, ∞).
y
8
y
x
2
8
5
40. y = x3 − 25 x 2 = x 2 ( x − 25)
Intercepts: (0, 0) and (25, 0)
50 ⎞
⎛
y ′ = 3 x 2 − 50 x = 3 x ⎜ x − ⎟
3 ⎠
⎝
x
5
CV: x = 0,
38. y = x − x 2 + 2 = −( x − 2)( x + 1)
Intercepts (2, 0), (–1, 0), and (0, 2)
y′ = 1 − 2 x
CV: x =
y
50
3
⎛ 50
⎞
Increasing on (−∞, 0) and ⎜ , ∞ ⎟ ; decreasing
⎝ 3
⎠
⎛ 50 ⎞
on ⎜ 0,
⎟ ; relative maximum at (0, 0); relative
⎝ 3 ⎠
1
2
62,500 ⎞
⎛ 50
.
minimum at ⎜ , −
3
27 ⎟⎠
⎝
1⎞
⎛
⎛1
⎞
Increasing on ⎜ −∞, ⎟ ; decreasing on ⎜ , ∞ ⎟ ;
2⎠
⎝
⎝2
⎠
488
ISM: Introductory Mathematical Analysis
Section 13.3
42. y = x3 − 6 x 2 + 9 x = x( x − 3) 2
Intercepts (0, 0) and (3, 0)
25 ⎞
⎛
y ′′ = 6 x − 50 = 6 ⎜ x − ⎟
3 ⎠
⎝
y′ = 3x 2 − 12 x + 9 = 3( x − 1)( x − 3)
CV: x = 1 and x = 3
Increasing on (–∞, 1) and (3, ∞); decreasing on
(1, 3); relative maximum at (1, 4); relative
minimum at (3, 0).
y ′′ = 6 x − 12 = 6( x − 2)
Possible inflection point when x = 2. Concave
down on (–∞, 2); concave up on (2, ∞);
inflection point at (2, 2).
25
. Concave
3
⎛ 25
⎞
concave up on ⎜ , ∞ ⎟ ;
3
⎝
⎠
Possible inflection point when x =
25 ⎞
⎛
;
down on ⎜ −∞,
3 ⎟⎠
⎝
31, 250 ⎞
⎛ 25
.
inflection point at ⎜ , −
27 ⎟⎠
⎝ 3
1200
y
x
48
5
y
x
8
41. y =
x3
x3 − 12 x
− 4x =
3
3
(
)(
1
= x x+2 3 x−2 3
3
(
43. y = x3 − 3 x 2 + 3 x − 3
Intercept (0, –3)
)
Intercepts (0, 0) and ±2 3, 0
y′ = 3x 2 − 6 x + 3 = 3( x − 1) 2
CV: x = 1
Increasing on (–∞, 1) and (1, ∞); no relative
maximum or minimum
y ′′ = 6( x − 1)
Possible inflection point when x = 1. Concave
down on (–∞, 1); concave up on (1, ∞);
inflection point at (1, –2).
)
y′ = x 2 − 4 = ( x + 2)( x − 2)
CV: x = ±2
Increasing on (–∞, –2) and (2, ∞); decreasing on
16 ⎞
⎛
(–2, 2); relative maximum at ⎜ −2, ⎟ ; relative
3⎠
⎝
16 ⎞
⎛
minimum at ⎜ 2, − ⎟ .
3⎠
⎝
y ′′ = 2 x
Possible inflection point when x = 0. Concave
down on (–∞, 0); concave up on (0, ∞); inflection
point at (0, 0). Symmetric about the origin.
10
5
y
x
5
y
5 2
5
⎛
⎞
x + 2 x = x ⎜ 2x2 + x + 2 ⎟
2
2
⎝
⎠
Intercept (0, 0)
44. y = 2 x3 +
x
10
y′ = 6 x 2 + 5 x + 2
CV: none
Increasing on (–∞, ∞).
5⎤
⎡
y ′′ = 12 x + 5 = 12 ⎢ x + ⎥
⎣ 12 ⎦
Possible inflection point at x = −
489
5
. Concave
12
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
5⎞
⎛
down on ⎜ −∞, − ⎟ ; concave up on
12
⎝
⎠
235 ⎞
⎛ 5
⎞
⎛ 5
⎜ − , ∞ ⎟ ; inflection point at ⎜ − , −
⎟.
12
12
432
⎝
⎠
⎝
⎠
y
⎛1 ⎞
on ⎜ , 1⎟ ; relative minimum at
⎝3 ⎠
relative maximum at (1, 4)
2⎞
⎛
y ′′ = −6 x + 4 = −6 ⎜ x − ⎟
3⎠
⎝
3
⎛ 1 104 ⎞
⎜ 3 , 27 ⎟ ;
⎝
⎠
2
. Concave
3
⎛2
⎞
concave down on ⎜ , ∞ ⎟ ;
⎝3
⎠
Possible inflection point when x =
x
2⎞
⎛
up on ⎜ −∞, ⎟ ;
3⎠
⎝
3
⎛ 2 106 ⎞
inflection point at ⎜ ,
⎟
⎝ 3 27 ⎠
45. y = 4 x3 − 3 x 4 = x3 (4 − 3 x)
8
y
⎛4 ⎞
Intercepts (0, 0), ⎜ , 0 ⎟
⎝3 ⎠
y′ = 12 x 2 − 12 x3 = 12 x 2 (1 − x)
CV: x = 0 and x = 1
Increasing on (–∞, 0) and (0, 1); decreasing on
(1, ∞); relative maximum at (1, 1).
y ′′ = 24 x − 36 x 2 = 12 x(2 − 3x)
x
5
47. y = −2 + 12 x − x3
Intercept (0, –2)
2
Possible inflection points at x = 0 and x = .
3
⎛2
⎞
Concave down on (–∞, 0) and ⎜ , ∞ ⎟ ; concave
⎝3
⎠
⎛ 2⎞
up on ⎜ 0, ⎟ ; inflection points at (0, 0) and
⎝ 3⎠
⎛ 2 16 ⎞
⎜ ,
⎟
⎝ 3 27 ⎠
3
y′ = 12 − 3x 2 = 3(2 + x)(2 − x)
CV: x = ±2
Decreasing on (–∞, –2) and (2, ∞); increasing on
(–2, 2); relative minimum at (–2, –18); relative
maximum at (2, 14).
y ′′ = −6 x
Possible inflection point when x = 0. Concave up
on (–∞, 0); concave down on (0, ∞); inflection
point at (0, –2).
y
y
20
x
3
x
5
46. y = − x3 + 2 x 2 − x + 4
Intercept (0, 4)
y ′ = −3x 2 + 4 x − 1 = −(3 x − 1)( x − 1)
48. y = (3 + 2 x)3
1
CV: x = , 1
3
⎛ 3 ⎞
Intercepts (0, 27), ⎜ − , 0 ⎟
⎝ 2 ⎠
1⎞
⎛
Decreasing on ⎜ −∞, ⎟ and (1, ∞); increasing
3⎠
⎝
y′ = 6(3 + 2 x)2
CV: x = −
490
3
2
ISM: Introductory Mathematical Analysis
Section 13.3
x3 3 x 2 x 2
( x − 3)
−
=
5
5
5
Possible inflection points when x = 0 and x = 3.
Concave down on (–∞, 0) and (0, 3); concave up
on (3, ∞); inflection point at (3, –1.62).
3⎞
⎛
⎛ 3
⎞
Increasing on ⎜ −∞, − ⎟ and ⎜ − , ∞ ⎟ ; no
2
2
⎝
⎠
⎝
⎠
relative maximum or minimum.
y ′′ = 24(3 + 2 x)
y ′′ =
3
.Concave
2
3⎞
⎛
⎛ 3
⎞
down on ⎜ −∞, − ⎟ ; concave up on ⎜ − , ∞ ⎟ ;
2⎠
⎝
⎝ 2
⎠
3
⎛
⎞
inflection point at ⎜ − , 0 ⎟ .
⎝ 2 ⎠
Possible inflection point at x = −
40
10
y
x
10
y
(
)
= x ( 5 + x 2 )( 5 − x 2 )
= x ( 5 + x 2 ) ( 4 5 + x )( 4 5 − x )
Intercepts (0, 0) and ( ± 4 5, 0 )
51. y = 5 x − x5 = x 5 − x 4
x
5
49. y = 2 x3 − 6 x 2 + 6 x − 2 = 2( x − 1)3
Intercepts (0, –2), (1, 0)
Symmetric about the origin.
( ) (
= 5(1 − x)(1 + x) (1 + x 2 )
)(
y′ = 5 − 5 x 4 = 5 1 − x 4 = 5 1 − x 2 1 + x 2
2
y′ = 6( x − 1)
CV: x = 1
Increasing on (–∞, 1) and (1, ∞); no relative
maximum or minimum.
y ′′ = 12( x − 1)
Possible inflection point when x = 1. Concave
down on (–∞, 1); concave up on (1, ∞);
inflection point at (1, 0).
)
CV: x = ±1
Decreasing on (–∞, –1) and (1, ∞); increasing on
(–1, 1); relative minimum at (–1, –4); relative
maximum at (1, 4).
y ′′ = −20 x3
Possible inflection point when x = 0. Concave
up on (–∞, 0); concave down on (0, ∞);
inflection point at (0, 0).
y
3
5
x
y
3
x
5
x5 x 4
x4
−
=
( x − 5)
100 20 100
Intercepts (0, 0), (5, 0)
x 4 x3 x3
y′ =
−
=
( x − 4)
20 5 20
CV: x = 0 and x = 4
Increasing on (–∞, 0) and (4, ∞); decreasing on
(0, 4); relative maximum at (0, 0); relative
minimum at (4, –2.56).
50. y =
52. y = x 2 ( x − 1)2
Intercepts: (0, 0), (1, 0)
y ′ = x 2 [2( x − 1)(1)] + 2 x( x − 1) 2
= 2 x( x − 1)(2 x − 1)
= 4 x3 − 6 x 2 + 2 x
CV: x = 0, 1 and x =
491
1
2
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
⎛1 ⎞
Decreasing on (−∞, 0) and ⎜ , 1⎟ ; increasing
⎝2 ⎠
3
y
⎛ 1⎞
on ⎜ 0, ⎟ and (1, ∞); relative minima at (0, 0)
⎝ 2⎠
x
3
⎛1 1 ⎞
and (1, 0); relative maximum at ⎜ , ⎟
⎝ 2 16 ⎠
y ′′ = 12 x 2 − 12 x + 2 = 2(6 x 2 − 6 x + 1)
From the quadratic formula, there are possible
3± 3
. Concave up
inflection points when x =
6
⎛
⎛ 3+ 3
⎞
3− 3 ⎞
on ⎜⎜ −∞,
, ∞ ⎟⎟ ; concave
⎟⎟ and ⎜⎜
6 ⎠
⎝
⎝ 6
⎠
⎛ 3− 3 3+ 3 ⎞
down on ⎜⎜
,
⎟ ; inflection points at
6 ⎟⎠
⎝ 6
⎛ 3− 3 1 ⎞
⎛ 3+ 3 1 ⎞
,
,
⎜⎜
⎟⎟ and ⎜⎜
⎟.
36 ⎠
36 ⎟⎠
⎝ 6
⎝ 6
5
5⎤
⎡
54. y = 3x5 − 5 x3 = 3x3 ⎢ x 2 − ⎥
3⎦
⎣
⎛
5 ⎞⎛
5⎞
= 3x3 ⎜⎜ x +
⎟⎟ ⎜⎜ x −
⎟
3 ⎠⎝
3 ⎟⎠
⎝
⎛ 5 ⎞
Intercepts (0, 0) and ⎜⎜ ± , 0 ⎟⎟
⎝ 3 ⎠
Symmetric about the origin.
y′ = 15 x 4 − 15 x 2 = 15 x 2 ( x + 1)( x − 1)
CV: x = 0 and x = ±1
Increasing on (–∞, –1) and (1, ∞); decreasing on
(–1, 0) and (0, 1); relative maximum at (–1, 2);
relative minimum at (1, –2).
⎡
2⎤⎡
2⎤
y ′′ = 60 x3 − 30 x = 60 x ⎢ x +
⎥ ⎢x −
⎥
2 ⎦⎣
2 ⎦
⎣
y
x
5
2
.
2
⎛
⎛
2⎞
2⎞
Concave down on ⎜⎜ −∞, −
⎟;
⎟⎟ and ⎜⎜ 0,
2 ⎠
2 ⎟⎠
⎝
⎝
⎛
⎛ 2
⎞
2 ⎞
concave up on ⎜⎜ −
, 0 ⎟⎟ and ⎜⎜
, ∞ ⎟⎟ ;
⎝ 2
⎠
⎝ 2
⎠
⎛ 2
⎞
7 2
inflection points at ⎜⎜
,−
⎟ , (0, 0), and
2
8 ⎟⎠
⎝
⎛
2 7 2⎞
,
⎜⎜ −
⎟.
2
8 ⎟⎠
⎝
Possible inflection points at x = 0 and x = ±
53. y = 3x 4 − 4 x3 + 1
Intercepts (0, 1) and (1, 0) [the latter is found by
inspection of the equation]. No symmetry.
y′ = 12 x3 − 12 x 2 = 12 x 2 ( x − 1)
CV: x = 0 and x = 1
Decreasing on (–∞, 0) and (0, 1); increasing on
(1, ∞); relative minimum at (1, 0).
y ′′ = 36 x 2 − 24 x = 12 x(3 x − 2)
Possible inflection points at x = 0 and x =
2
.
3
3
⎛2
⎞
Concave up on (–∞, 0) and ⎜ , ∞ ⎟ ; concave
3
⎝
⎠
⎛ 2⎞
down on ⎜ 0, ⎟ ; inflection points at (0, 1) and
⎝ 3⎠
⎛ 2 11 ⎞
⎜ ,
⎟.
⎝ 3 27 ⎠
y
x
3
492
ISM: Introductory Mathematical Analysis
Section 13.3
55. y = 4 x 2 − x 4 = x 2 (2 + x)(2 − x)
Intercepts (0, 0) and (±2, 0)
Symmetric about the y-axis.
(
3
y′ = 8 x − 4 x = 4 x 2 − x
= 4x
(
2+x
)(
2−x
)
2
1⎞
⎛ 1
,− ⎟
⎜±
4⎠
2
⎝
1⎞
⎛
y ′′ = 12 x 2 − 2 = 12 ⎜ x 2 − ⎟
6⎠
⎝
⎡
1 ⎤⎡
1 ⎤
= 12 ⎢ x +
⎥ ⎢x −
⎥
6⎦⎣
6⎦
⎣
)
CV: x = 0, ± 2
(
) ( )
decreasing on ( − 2, 0 ) and ( 2, ∞ ) ; relative
maxima at ( ± 2, 4 ) ; relative minimum at (0, 0).
Possible inflection points when x = ±
Increasing on −∞, − 2 and 0, 2 ;
y
2
.
x=±
3
⎛ 2
⎞
, ∞ ⎟⎟ ;
and ⎜⎜
⎝ 3
⎠
⎛
2⎞
Concave down on ⎜⎜ −∞, −
⎟
3 ⎟⎠
⎝
⎛ 2 2⎞
concave up on ⎜⎜ − ,
⎟⎟ ; inflection points at
⎝ 3 3⎠
⎛ 2 20 ⎞
⎜⎜ ± ,
⎟⎟ .
⎝ 3 9 ⎠
5
57. y = x1/ 3 ( x − 8) = x 4 / 3 − 8 x1/ 3
Intercepts (0, 0) and (8, 0)
4
8
y′ = x1/ 3 − x −2 / 3
3
3
4⎡
2 ⎤ 4( x − 2)
= ⎢ x1/ 3 −
⎥=
2
3⎣
x /3 ⎦
3x2 / 3
CV: x = 0, 2
Decreasing on (–∞, 0) and (0, 2); increasing on
(2, ∞); relative minimum at
x
( 2, − 63 2 ) ≈ ( 2, − 7.56) .
2
56. y = x − x = x ( x + 1)( x − 1)
Intercepts (0, 0), (−1, 0), and (1, 0)
Symmetric about the y-axis.
y′ = 4 x 3 − 2 x = 2 x
CV: x = 0, ±
(
)(
2x + 1
3
x
y
2
.
2
5
4
6
⎛
⎛ 1
⎞
1 ⎞
Concave up on ⎜ −∞, −
, ∞⎟ ;
⎟ and ⎜
6⎠
⎝ 6
⎠
⎝
⎛ 1
1 ⎞
concave down on ⎜ −
,
⎟ ; inflection
6
6⎠
⎝
⎛ 1
5 ⎞
points at ⎜ ±
, − ⎟.
36 ⎠
6
⎝
⎡2
⎤
y ′′ = 8 − 12 x 2 = 12 ⎢ − x 2 ⎥
⎣3
⎦
⎛ 2
⎞⎛ 2
⎞
= 12 ⎜⎜
− x ⎟⎟ ⎜⎜
+ x ⎟⎟
⎝ 3
⎠⎝ 3
⎠
Possible inflection points when
1
4 −2 / 3 16 −5 / 3
+ x
x
9
9
4⎡ 1
4 ⎤ 4( x + 4)
= ⎢
+
=
9 ⎣ x 2 / 3 x5 / 3 ⎥⎦
9 x5 / 3
Possible inflection points when x = –4, 0.
Concave up on (–∞, –4) and (0, ∞); concave
y ′′ =
)
2x −1
1
2
(
1 ⎞
1 ⎞
⎛
⎛
Decreasing on ⎜ −∞, −
⎟ and ⎜ 0,
⎟;
2⎠
2⎠
⎝
⎝
⎛ 1
⎞
⎛ 1
⎞
increasing on ⎜ −
, 0 ⎟ and ⎜
, ∞ ⎟;
2 ⎠
⎝
⎝ 2
⎠
relative maximum at (0, 0); relative minima at
down on (–4, 0); inflection points at −4, 12 3 4
)
and (0, 0). Observe that at the origin the tangent
line exists but it is vertical.
493
Chapter 13: Curve Sketching
21
ISM: Introductory Mathematical Analysis
y
2
59. y = 4 x1/ 3 + x 4 / 3 = x1/ 3 (4 + x)
Intercepts (0, 0) and (–4, 0)
⎤
4
4
4⎡ 1
y′ = x −2 / 3 + x1/ 3 = ⎢
+ x1/ 3 ⎥
2
/
3
3
3
3 ⎣x
⎦
4(1 + x)
=
3x 2 / 3
CV: x = 0, –1
Decreasing on (–∞, –1); increasing on (–1, 0)
and (0, ∞); rel. min at (–1, –3)
8
4
4⎡ 1
2 ⎤
−
y ′′ = − x −5 / 3 + x −2 / 3 = ⎢
⎥
2
/
3
5
9
9
9 ⎣x
x /3 ⎦
x
16
8
3
(2, –6 2 )
58. y = ( x − 1) 2 ( x + 2)2
Intercepts (0, 4), (1, 0), (–2, 0)
y′ = ( x − 1)2 [2( x + 2)] + ( x + 2)2 [2( x − 1)]
= 2(x – 1)(x + 2)(2x + 1)
1
CV: x = −2, − , 1
2
⎛ 1 ⎞
Decreasing on (–∞, –2) and ⎜ − , 1⎟ ; increasing
⎝ 2 ⎠
=
4( x − 2)
9 x5 / 3
Possible inflection points when x = 0, 2.
Concave up on (–∞, 0) and (2, ∞); concave
down on
(
)
(0, 2); inflection point at (0, 0) and 2, 6 3 2 .
1⎞
⎛
on ⎜ −2, − ⎟ and (1, ∞); relative maximum at
2⎠
⎝
Observe that at the origin the tangent line exists
but it is vertical.
⎛ 1 81 ⎞
⎜ − 2 , 16 ⎟ ; relative minima at
⎝
⎠
y
7
3
(2, 6 2 )
(–2, 0) and (1, 0); y′ = 2(2 x3 + 3 x 2 − 3x − 2), so
(
)
y ′′ = 6 2 x 2 + 2 x − 1 . Setting y ′′ = 0 and using
x
the quadratic formula gives possible inflection
−1 ± 3
. Concave up on
points at x =
2
⎛
⎛ −1 + 3
⎞
−1 − 3 ⎞
, ∞ ⎟ ; concave
⎜⎜ −∞,
⎟⎟ and ⎜⎜
⎟
2 ⎠
2
⎝
⎝
⎠
–4
Intercepts (0, 2), (−1, 0) and (−4, 0)
1
y ′ = ( x + 1) ⋅
+ x + 4(1)
2 x+4
1
[( x + 1) + 2( x + 4)]
=
2 x+4
3( x + 3)
=
2 x+4
CV: x = −3, −4
Decreasing on (−4, −3); increasing on (−3, ∞);
relative minimum at (−3, −2)
1
3 x + 4(1) − ( x + 3) ⋅ 2 x + 4
y ′′ = ⋅
2
2
x+4
−1 ± 3
2
8
5
60. y = ( x + 1) x + 4 [Note: x ≥ −4]
⎛ −1 − 3 −1 + 3 ⎞
down on ⎜
,
⎟ ; inflection points
⎜
2
2 ⎟⎠
⎝
when x =
–1
y
x
(
5
)
3 2( x + 4) − ( x + 3)
3( x + 5)
⋅
=
3/ 2
4
( x + 4)
4( x + 4)3 / 2
No possible inflection point. Concave up on
(−4, ∞).
=
494
ISM: Introductory Mathematical Analysis
5
Section 13.3
Observe that at the origin the tangent line exists
but it is vertical.
y
8
y
x
5
x
61. y = 6 x 2 / 3 −
8
⎛ x1/ 3 ⎞
x
= 6 x 2 / 3 ⎜1 −
⎟
⎜
2
12 ⎟⎠
⎝
63.
Intercepts (0, 0) and (1728, 0)
⎞ 1⎛8− 3 x ⎞
1 1⎛ 8
= ⎜
− 1⎟ = ⎜
⎟
2 2 ⎝ 3 x ⎠ 2 ⎜⎝ 3 x ⎟⎠
CV: x = 0, 512
Increasing on (0, 512); decreasing on (−∞, 0)
and (512, ∞); relative maximum at (512, 128);
relative minimum at (0, 0).
4
4
y ′′ = − x −4 / 3 = −
3
3x 4 / 3
Possible inflection point at x = 0. Concave down
on (−∞, 0) and (0, ∞). Observe that at the origin
the tangent line exists but it is vertical.
y ′ = 4 x −1/ 3 −
y
8
y
4
x
2
64.
8
8
y
4
x
300
4
x
2500
65.
5
8
y
1
x
1
62. y = 5 x 2 / 3 − x5 / 3 = x 2 / 3 (5 − x)
Intercepts (0, 0) and (5, 0)
⎤
10 −1/ 3 5 2 / 3 5 ⎡ 2
− x
= ⎢
− x2 / 3 ⎥
y′ =
x
1/
3
3
3
3⎣x
⎦
=
66.
8
5
y
5(2 − x)
3 x1/ 3
CV: x = 0, 2
Increasing on (0, 2); decreasing on (–∞, 0) and
(2, ∞); relative minimum at (0, 0); relative
(
3
4
x
)
3
maximum at 2, 3 4 ≈ (2, 4.76)
10 −4 / 3 10 −1/ 3
10(1 + x)
− x
=−
x
9
9
9 x4 / 3
Possible inflection point when x = 0, –1.
Concave up on (–∞, –1); concave down on
(–1, 0), and (0, ∞); inflection point at (–1, 6).
y ′′ = −
495
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
100
q+2
67. p =
g ′′( x) = −
dp
100
=−
< 0 for q > 0, so p is
dq
(q + 2)2
d2 p
=
200
=
> 0 for q > 0,
dq 2 (q + 2)3
the demand curve is concave up.
decreasing. Since
=
c ′′ =
1
e
U0
A
2
2
⎡
−x ⎛ x ⎞
−x ⎤
⎢x ⋅ e 2A ⎜ − ⎟ + e 2A ⎥
A ⎣⎢
⎝ A⎠
⎦⎥
2
⋅e
− 2x A
⋅e
− 2x A
2
(x
2
−A
)
( x + A )( x − A )
A , then g ′′( x) < 0 , so the graph is
71. y = 12.5 + 5.8(0.42) x
q2
y′ = 5.8(0.42) x ln(0.42)
Since ln(0.42) < 0, we have y′ < 0 , so the
function is decreasing.
y ′′ = 5.8(0.42) x ln 2 (0.42) > 0 , so the function is
concave up.
2
q3
Since c′′ > 0 for q > 0, the graph of the average
cost function is concave up for q > 0.
69. S = f ( A) = 12 4 A , 0 ≤ A ≤ 625 . For the given
72. H = 1.00 ⎡1 − e −(0.0464t + 0.0670 ⎤
⎣
⎦
dH
− (0.0464t + 0.0670)
= 0.0464e
> 0 , so H is
dt
increasing.
d 2H
= −(0.0464)2 e −(0.0464t + 0.0670) < 0 , so H is
dt 2
concave down.
− 34
values of A we have S ′ = 3 A > 0 and
⎛ 9 ⎞ −7
S ′′ = − ⎜ ⎟ A 4 < 0 . Thus y is increasing and
⎝4⎠
concave down.
60
A2
U0
A
concave down. If x > A , then g ′′( x) > 0 , so
the graph is concave up.
c
1
= q+2+
q
q
c′ = 1−
U0
A
A2
If 0 ≤ x <
68. c = q 2 + 2q + 1
c=
e
e
S
73. n = f (r ) = 0.1ln(r ) +
a.
A
625
U0
−x
2
70. g ( x) = e A e 2 A , A > 0, x ≥ 0 (since x
represents quantity).
g ′( x) = −
e
U0
A
⎡ − x2 ⎤
⎢ xe 2 A ⎥
A ⎢⎣
⎥⎦
496
7
− 0.8 , 1 ≤ r ≤ 10
r
dn 0.1 7 0.1r − 7 0.1(r − 70)
=
−
=
=
<0
dr
r
r2
r2
r2
for 1 ≤ r ≤ 10. Thus the graph of f is
always falling. Also,
d 2n
0.1 14 14 − 0.1r
=−
+
=
2
dr
r 2 r3
r3
0.1(140 − r )
=
>0
r3
for 1 ≤ r ≤ 10. Thus the graph is concave
up.
ISM: Introductory Mathematical Analysis
b.
10
Section 13.3
f(r)
⎛ 2a ⎞
concave downon ⎜ , 0 ⎟ . In either case, y has
⎝ 3
⎠
2a
two points of inflection, when x = 0 and x =
.
3
6
2
76.
r
1
10
–3
c.
dn
dr
3
= −0.26 , so the rate of decrease is
r =5
0.26.
–2
74.
Two inflection points
150
77. y = x3 − 2 x 2 + x + 3
y′ = 3 x 2 − 4 x + 1
When x = 2, then y = 5 and y′ = 5 . Thus an
equation of the tangent line at x = 2 is
y – 5 = 5(x – 2), or y = 5x – 5. Graphing the
curve and the tangent line indicates that the
curve lies above the tangent line around x = 2.
Thus the curve is concave up at x = 2.
20
–20
–50
a.
One relative maximum point
10
b. One relative minimum point
c.
One inflection point
10
75.
–2
–2
3
0
4
78.
f ( x ) = 2 x3 + 3 x 2 − 6 x + 1
f ′( x) = 6 x 2 + 6 x − 6
f ′′( x) = 12 x + 6
–12
Two inflection points
The relative minimum of f ′ occurs at a value of
x for which ( f ′( x))′ = f ′′( x) = 0. Around this
value of x, ( f ′( x))′ goes from − to +. Since
( f ′( x))′ = f ′′( x), the concavity of f must change
from concave down to concave up.
y = x5 ( x − a) = x 6 − ax5
y ′ = 6 x5 − 5ax 4
y ′′ = 30 x 4 − 20ax3 = 10 x3 (3 x − 2a)
Possible inflection points when x = 0 and
2a
x=
. If a > 0, y is concave up on (−∞, 0) and
3
⎛ 2a
⎞
⎛ 2a ⎞
⎜ , ∞ ⎟ ; concave down on ⎜ 0,
⎟ . If a < 0,
3 ⎠
⎝ 3
⎠
⎝
2a ⎞
⎛
y is concave up on ⎜ −∞,
⎟ and (0, ∞);
3 ⎠
⎝
79.
f ( x ) = x 6 + 3 x5 − 4 x 4 + 2 x 2 + 1
f ′( x) = 6 x5 + 15 x 4 − 16 x3 + 4 x
f ′′( x) = 30 x 4 + 60 x3 − 48 x 2 + 4
Inflection points of f when x ≈ –2.61, –0.26.
497
Chapter 13: Curve Sketching
80.
f ( x) =
ISM: Introductory Mathematical Analysis
x +1
1
.
4
Because there is only one relative extremum and
f is continuous, the relative maximum is an
absolute maximum.
Thus there is a relative maximum when x =
2
x +1
f ′( x ) = −
x2 + 2 x − 1
( x + 1)
2 ( x + 3 x − 3x − 1)
f ′′( x) =
( x + 1)
2
2
3
4. y = 3x 2 − 5 x + 6
y′ = 6 x − 5
2
2
3
CV: x =
Inflection points of f when
x ≈ −3.73, −0.27, 1.00.
y ′′ = 6
⎛5⎞
y ′′ ⎜ ⎟ = 6 > 0
⎝6⎠
Problems 13.4
1. y = x 2 − 5 x + 6
y′ = 2 x − 5
CV: x =
y ′′ = 2
5
6
5
.
6
Because there is only one relative extremum and
f is continuous, the relative minimum is an
absolute minimum.
Thus there is a relative minimum when x =
5
2
⎛5⎞
y ′′ ⎜ ⎟ = 2 > 0
⎝2⎠
5. y =
1 3
x + 2 x2 − 5x + 1
3
y ′ = x 2 + 4 x − 5 = ( x + 5)( x − 1)
5
.
2
Because there is only one relative extremum and
f is continuous, the relative minimum is an
absolute minimum.
Thus there is a relative minimum when x =
CV: x = −5, 1
y ′′ = 2 x + 4
y ′′(−5) = −6 < 0 ⇒ relative maximum when
x = −5
y ′′(1) = 6 > 0 ⇒ relative minimum when x = 1
2
2. y = 5 x + 20 x + 2
y′ = 10 x + 20
CV: x = −2
y ′′ = 10
y ′′(−2) = 10 > 0
6. y = x3 − 12 x + 1
y′ = 3x 2 − 12 = 3( x + 2)( x − 2)
CV: x = ±2
y ′′ = 6 x
y ′′(−2) = −12 < 0 ⇒ relative maximum when
x = –2
y ′′(2) = 12 > 0 ⇒ relative minimum when
x=2
Thus there is a relative minimum when x = −2.
Because there is only one relative extremum and
f is continuous, the relative minimum is an
absolute minimum.
3. y = −4 x 2 + 2 x − 8
y′ = −8 x + 2
7. y = − x3 + 3 x 2 + 1
1
CV: x =
4
y ′′ = −8
y′ = −3 x 2 + 6 x = −3x( x − 2)
CV: x = 0, 2
y ′′ = −6 x + 6
y ′′(0) = 6 > 0 ⇒ relative minimum when x = 0
⎛1⎞
y ′′ ⎜ ⎟ = −8 < 0
⎝4⎠
y ′′(2) = −6 < 0 ⇒ relative maximum when x = 2
498
ISM: Introductory Mathematical Analysis
Section 13.4
8. y = x 4 − 2 x 2 + 4
⎛1⎞
y ′′ ⎜ ⎟ = 60 > 0 ⇒ relative minimum when
⎝3⎠
1
x=
3
y′ = 4 x3 − 4 x = 4 x( x + 1)( x − 1)
CV: = 0, ±1
y ′′ = 12 x 2 − 4
y ′′(0) = −4 < 0 ⇒ relative maximum when x = 0
y ′′(1) = 8 > 0 ⇒ relative minimum when x = 1
12. y =
y ′′(−1) = 8 > 0 ⇒ relative minimum when
x = –1
y ′ = 55 x 2 − 2 x − 21 = (5 x + 3)(11x − 7)
3 7
CV: x = − ,
5 11
y ′′ = 110 x − 2
9. y = 7 − 2 x 4
y′ = −8 x3
CV: x = 0
y ′′ = −24 x 2
⎛ 3⎞
y ′′ ⎜ − ⎟ = −68 < 0 ⇒ relative maximum when
⎝ 5⎠
3
x=−
5
⎛7⎞
y ′′ ⎜ ⎟ = 68 > 0 ⇒ relative minimum when
⎝ 11 ⎠
Since y ′′(0) = 0 , the second-derivative test fails.
Using the first-derivative test, we see that f
increases for x < 0 and f decreases for x > 0, so
there is a relative maximum when x = 0.
x=
10. y = −2 x 7
y′ = −14 x 6
CV: x = 0
y ′′ = −84 x5
(
11. y = 81x5 − 5 x
(
)(
2
)
CV: x = –2, –5, −
)
(
7
2
)
y ′′ = 2 ⎡ x 2 + 7 x + 10 (2) + (2 x + 7)(2 x + 7) ⎤
⎣⎢
⎦⎥
′′
y (−5) = 18 > 0 ⇒ relative minimum when
x = –5
⎛ 7⎞
y ′′ ⎜ − ⎟ = −9 < 0 ⇒ relative maximum when
⎝ 2⎠
)
)
= 5(3 x + 1)(3x − 1) 9 x 2 + 1
CV: x = ±
2
= 2( x + 2)( x + 5)(2 x + 7)
= 5 9x −1 9x + 1
(
(
)
y′ = 2 x 2 + 7 x + 10 (2 x + 7)
y′ = 81 ⋅ 5 x 4 − 5 = 5 81x 4 − 1
(
11
7
13. y = x 2 + 7 x + 10
Since y ′′(0) = 0 , the second-derivative test fails.
However, using the first-derivative test, we see
that f decreases for x < 0 and for x > 0, so there is
neither a relative maximum nor a relative
minimum when x = 0.
2
55 3
x − x 2 − 21x − 3
3
7
2
y ′′(−2) = 18 > 0 ⇒ relative minimum when
x = –2
x=−
1
3
y ′′ = 81 ⋅ 5 ⋅ 4 x3
14. y = − x3 + 3 x 2 + 9 x − 2
⎛ 1⎞
y ′′ ⎜ − ⎟ = −60 < 0 ⇒ relative maximum when
⎝ 3⎠
1
x=−
3
y ′ = −3x 2 + 6 x + 9 = −3( x 2 − 2 x − 3)
= −3( x + 1)( x − 3)
CV: x = −1, 3
y ′′ = −6 x + 6
y ′′(−1) = 12 > 0 ⇒ relative minimum when
x = −1
499
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
y ′′(3) = −12 < 0 ⇒ relative maximum when
x=3
4
x
When x = 0 the denominator is zero but the
numerator is not zero, so x = 0 is a vertical
asymptote.
⎛4⎞
⎛4⎞
lim ⎜ ⎟ = 0. Similarly, lim ⎜ ⎟ = 0, so
x →∞ ⎝ x ⎠
x →−∞ ⎝ x ⎠
y = 0 is a horizontal asymptote.
5. y = f ( x) =
Problems 13.5
x
x −1
When x = 1 the denominator is zero but the
numerator is not zero. Thus x = 1 is a vertical
asymptote.
x
x
lim
= lim = lim 1 = 1 .
x →∞ x − 1 x →∞ x
x →∞
Similarly lim f ( x) = 1 . Thus the line y = 1 is a
1. y = f ( x) =
=
x2 − 2
x2
x2
When x = 0 the denominator is zero but the
numerator is not. Thus x = 0 is a vertical
⎛
2 ⎞
asymptote. lim ⎜ 1 − ⎟ = 1 − 0 = 1. Similarly
x →∞ ⎝
x2 ⎠
lim f ( x) = 1, so y = 1 is a horizontal
x →−∞
horizontal asymptote.
x +1
x
When x = 0 the denominator is zero but the
numerator is not. Thus x = 0 is a vertical
x +1
x
= lim = lim 1 = 1 .
asymptote. lim
x →∞ x
x →∞ x
x →∞
Similarly lim f ( x) = 1 . Thus y = 1 is a
2. y = f ( x) =
x →−∞
asymptote.
1
(
x
−
1)(
x + 1)
x −1
Vertical asymptotes are x = 1 and x = –1.
1
1
lim
= lim
= 0 . Similarly,
2
x →∞ x − 1 x →∞ x 2
lim f ( x ) = 0 . Thus y = 0 is a horizontal
7. y = f ( x) =
x →−∞
horizontal asymptote.
3.
2
6. y = f ( x) = 1 −
x+2
3x − 5
5
When x = the denominator is zero but the
3
5
numerator is not. Thus x = is a vertical
3
x
1 1
asymptote. lim f ( x) = lim
= lim = .
x →∞
x →∞ 3 x
x →∞ 3 3
1
1
Similarly lim f ( x) = . Thus y = is a
3
3
x →−∞
horizontal asymptote.
f ( x) =
1
=
2
x →−∞
asymptote.
x
(
x
−
2)(
x + 2)
x −4
Vertical asymptotes: x = 2, x = –2.
x
x
1
lim
= lim
= lim = 0 . Similarly,
2
2
x →∞ x − 4 x →∞ x
x →∞ x
lim f ( x ) = 0 . Thus y = 0 is a horizontal
8. y = f ( x) =
x
=
2
x →−∞
asymptote.
9. y = f ( x) = x 2 − 5 x + 5 is a polynomial function,
so there are no horizontal or vertical asymptotes.
2x + 1
4. y = f ( x) =
2x + 1
Observe that both the numerator and
1
1
denominator are zero for x = − . For x ≠ − ,
2
2
we have f(x) = 1. Thus f is a constant function
1
for x ≠ − . Hence there are no vertical or
2
horizontal asymptotes.
10. y = f ( x) =
=
500
x4
x3 − 4
=
x4
x3 −
(3 4)
3
x4
( x − 22 / 3 )( x 2 + 22 / 3 x + 24 / 3 )
=
x4
x3 − (22 / 3 )3
ISM: Introductory Mathematical Analysis
Section 13.5
Vertical asymptote: x = 22 / 3.
x4
4x
= x+
so the line y = x is an
x3 − 4
x3 − 4
oblique asymptote.
11.
2 x2
16.
=
17.
asymptote.
x3
is a polynomial function, so there are
5
no horizontal or vertical asymptotes.
f ( x) =
13. y =
2 x 2 + 3x + 1
2
x −5
=
2 x2 + 3x + 1
( x − 5 )( x + 5 )
2 x2
x →∞
x →−∞
asymptote.
14. y = f ( x) =
x2 − 1
(2 x − 1)( x − 4)
3 − x4
=
3 − x4
x3 + x 2 x 2 ( x + 1)
Vertical asymptotes are x = 0 and x = –1.
3 − x4
3 − x2
so the line y = −x + 1
= −x +1+
x3 + x 2
x3 + x 2
is an oblique asyptote.
x 2 + 4 x3 + 6 x 4
3x2
Observe that both the numerator and the
denominator are zero when x = 0. For x ≠ 0, we
have
x2
1
f ( x) =
(1 + 4 x + 6 x 2 ) = (1 + 4 x + 6 x 2 ).
2
3
3x
Thus f is a polynomial function for x ≠ 0. Hence
there are neither horizontal nor vertical
asymptotes.
= lim 2 = 2
x →∞
x2
Similarly, lim = 2. Thus y = 2 is a horizontal
x →∞
f ( x) =
18. y = f ( x) =
Vertical asymptotes are x = − 5 and x = 5.
lim f ( x) = lim
2x2 − 9 x + 4
=
1
and x = 4.
2
x2
1 1
= lim = ,and
lim f ( x) = lim
2
x →∞
x →∞ 2 x 2
x →∞ 2
1
1
lim f ( x) = . Thus y = is a horizontal
2
2
x →−∞
asymptote.
x →−∞
12.
x2 − 1
Vertical asymptotes are x =
2 x2
x 2 + x − 6 ( x + 3)( x − 2)
Vertical asymptotes are x = –3 and x = 2.
2 x2
lim f ( x) = lim
= lim 2 = 2 , and
x →∞
x →∞ x 2
x →∞
lim f ( x) = 2 . Thus y = 2 is a horizontal
f ( x) =
f ( x) =
2 x3 + 1
3x(2 x − 1)(4 x − 3)
19. y = f ( x) =
1
, and
2
3
2 x3
1
1
= lim
=
.
x = . lim f ( x) = lim
4 x→∞
x →∞ 24 x3
x →∞ 12 12
1
1
. Thus y =
is a
Similarly, lim f ( x) =
12
12
x →−∞
horizontal asymptote.
x 2 − 3x − 4
x2 − 3x − 4
(1 + 2 x)2
1
From the denominator, x = − is a vertical
2
asymptote.
x2
1 1
= lim = , and
lim f ( x) = lim
2
4
x →∞
x →∞ 4 x
x →∞ 4
1
1
lim f ( x) = , so y = is a horizontal
4
4
x →−∞
asymptote.
Vertical asymptotes are x = 0, x =
2
5 x − 13
+5 =
x−3
x−3
From the denominator, x = 3 is a vertical
asymptote.
5x
= lim 5 = 5, and
lim f ( x) = lim
x →∞
x →∞ x
x →∞
lim f ( x) = 5. Thus, y = 5 is a horizontal
15. y = f ( x) =
20. y = f ( x) =
1 + 4 x + 4 x2
=
x4 + 1
1 − x4
=
x4 + 1
(1 + x ) (1 − x)(1 + x)
2
From the denominator, vertical asymptotes are
x = 1 and x = –1.
x →−∞
asymptote.
501
Chapter 13: Curve Sketching
lim f ( x) = lim
x4
ISM: Introductory Mathematical Analysis
24.
= lim − 1 = −1 , and
− x 4 x →∞
lim f ( x) = −1 . Thus y = –1 is a horizontal
x →∞
x →∞
x →∞
asymptote.
9 x 2 − 16
=
(3 x + 4)(3 x − 4)
2(3 x + 4)
2(3 x + 4)2
4
When x = − , both the numerator and
3
denominator are zero. Since
3x − 4
lim f ( x) = lim
= −∞ , the
+
+ 2(3 x + 4)
x →−4 / 3
x →−4 / 3
line x = −
lim
x →∞
2
3
x
Symmetric about the origin. Vertical asymptote
3
3
, so y = 0 is a
is x = 0. lim = 0 = lim
x →∞ x
x →−∞ x
horizontal asymptote.
3
y′ = −
x2
CV: None, however x = 0 must be included in
the inc.-dec. analysis. Decreasing on (–∞, 0) and
(0, ∞).
6
y ′′ =
x3
No possible inflection point, but we include x =
0 in the concavity analysis. Concave down on
(–∞, 0); concave up on (0, ∞).
25. y =
4
is a vertical asymptote.
3
9 x 2 − 16
2(3x + 4)
= lim
2
9 x2
x →∞ 18 x
Similarly, lim f ( x) =
x →−∞
2
1 1
= .
2
x →∞ 2
= lim
1
1
. Thus y =
is a
2
2
horizontal asymptote.
2
2x
24 x 2 + 20 x − 4
+
=
5 12 x 2 + 5 x − 2 5(12 x 2 + 5 x − 2)
4( x + 1)(6 x − 1)
=
5(3 x + 2)(4 x − 1)
22. y = f ( x) =
5
5
1
24 x 2
2 2
= lim = .
x = . lim f ( x) = lim
2
4 x→∞
x →∞ 60 x
x →∞ 5 5
2
2
Similarly, lim f ( x) = . Thus, y = is a
5
5
x →−∞
horizontal asymptote.
26. y =
2
2x − 3
2⎞
⎛
Intercept: ⎜ 0, − ⎟
3⎠
⎝
3
.
2
lim y = 0 = lim y, so y = 0 is a horizontal
Vertical asymptote is x =
23. y = f ( x) = 2e x + 2 + 4
We have lim f ( x) = +∞ and
x →∞
x →−∞
asymptote.
4
y′ = −
(2 x − 3) 2
x →∞
lim f ( x) = 2 ⋅ lim e x + lim 4
x →−∞
y
x
2
1
When x = − or x = , the denominator is 0,
3
4
but the numerator is not. Thus, vertical
2
asymptotes are x = − and
3
x →−∞
x →−∞
is a horizontal asymptote. There is no vertical
asymptote because f(x) neither increases nor
decreases without bound around any fixed value
of x.
x →−∞
21. y = f ( x) =
f ( x) = 12e− x
lim f ( x) = 0 and lim f ( x) = +∞ . Thus y = 0
x →−∞
= 2(0) + 4 = 4
Thus y = 4 is a horizontal asymptote. There is no
vertical asymptote because f(x) neither increases
nor decreases without bound around any fixed
value of x.
502
ISM: Introductory Mathematical Analysis
Section 13.5
3
must be considered in the
2
3⎞
⎛
inc. dec. analysis. Decreasing on ⎜ −∞, ⎟ and
2⎠
⎝
(Note: x > 0)
x
lim y = 0 , so y = 0 is a horizontal asymptote.
x →∞
lim y = +∞ , so the line x = 0 is a vertical
x →0+
⎛3
⎞
⎜ 2 , ∞ ⎟.
⎝
⎠
16
y ′′ =
(2 x − 3)3
asymptote.
5
y′ = −
< 0 for x > 0. Decreasing on (0, ∞).
x3
15
y ′′ =
> 0 for x > 0. Concave up on (0, ∞).
2 x5
3
must be
2
considered in the concavity analysis. Concave
3⎞
⎛
⎛3
⎞
down on ⎜ −∞, ⎟ ; concave up on ⎜ , ∞ ⎟ .
2⎠
⎝
⎝2
⎠
No possible inflection point, but x =
5
10
28. y =
CV: None, but x =
16
y
y
x
16
x
5
29. y = x 2 +
x
x −1
Intercept (0, 0)
Vertical asymptote is x = 1
lim y = 1 = lim y , so y = 1 is a horizontal
x4 + 1
x 2 = 0 gives x = 0 as the only vertical
asymptote. Because the degree of the numerator
is greater than the degree of the denominator, no
horizontal asymptote exists.
x →−∞
asymptote.
( x − 1)(1) − x(1)
1
y′ =
=−
2
( x − 1)
( x − 1)2
CV: None, but x = 1 must be included in the
inc.-dec. analysis. Decreasing on (–∞, 1) and
(1, ∞).
2
y ′′ =
( x − 1)3
No possible inflection point, but x = 1 must be
included in concavity analysis. Concave up on
(1, ∞), concave down on (−∞, 1).
y
=
x2
x2
x ≠ 0, so there is no y-intercept. Setting
y = 0 ⇒ no x-intercept. Replacing x by –x
yields symmetry about the y-axis. Setting
27. y =
x →∞
1
y = x 2 + x −2
y′ = 2 x − 2 x
=
(
2
)
−3
= 2x −
2
x
3
2 x + 1 ( x + 1)( x − 1)
=
2 x4 − 2
x
3
=
(
)
2 x4 − 1
x
3
.
x3
CV: x = ±1, but x = 0 must be included in the
inc.-dec. analysis. Decreasing on (–∞, –1) and
(0, 1); increasing on (–1, 0) and (1, ∞); relative
minima at (–1, 2) and (1, 2),
6
y ′′ = 2 +
>0
x4
for all x ≠ 0. Concave up on (–∞, 0) and
(0, ∞).
5
x
5
503
Chapter 13: Curve Sketching
8
ISM: Introductory Mathematical Analysis
y
y
50
x
5
x
5
3x2 − 5 x − 1
x−2
⎛ 1⎞
Intercept: ⎜ 0, ⎟
⎝ 2⎠
Vertical asymptote is x = 2.
3x 2 − 5 x − 1
1
= 3x + 1 +
so y = 3x + 1 is an
x−2
x−2
oblique asymptote.
1
(
x
1)(
x − 1)
+
x −1
Intercept (0, –1)
Symmetric about the y-axis.
Vertical asymptotes are x = –1 and x = 1.
1
1
= 0 = lim
, so y = 0 is a
lim
x →∞ x 2 − 1
x →−∞ x 2 − 1
horizontal asymptote.
2x
y′ = −
2
2
x −1
31. y =
30. y =
y′ =
=
( x − 2)(6 x − 5) − (3 x 2 − 5 x − 1)(1)
( x − 2)2
6± 3
,
3
but x = 2 must be included in the inc.-dec.
⎛
6− 3 ⎞
analysis. Increasing on ⎜⎜ −∞,
⎟ and
3 ⎟⎠
⎝
( x − 1)
y ′′ = −2 ⋅
2
=
2
(
)
(1) − x ⎡ 4 x x 2 − 1 ⎤
⎢⎣
⎥⎦
( x − 1)
( x − 1) ⎡⎣⎢( x − 1) − 4x ⎤⎦⎥
= −2 ⋅
( x − 1)
2 ( 3 x + 1)
2 ( 3 x + 1)
=
=
( x − 1) [( x + 1)( x − 1)]
2
⎛6+ 3
⎞
⎛6− 3 ⎞
, ∞ ⎟⎟ ; decreasing on ⎜⎜
, 2 ⎟⎟ and
⎜⎜
⎝ 3
⎠
⎝ 3
⎠
⎛ 6+ 3 ⎞
⎜⎜ 2,
⎟ ; relative maximum at
3 ⎟⎠
⎝
⎛6− 3
⎞
, 7 − 2 3 ⎟⎟ ; relative minimum at
⎜⎜
⎝ 3
⎠
⎛6+ 3
⎞
, 7 + 2 3 ⎟⎟ .
⎜⎜
⎝ 3
⎠
=
)
CV: x = 0, but x = ±1 must be included in the
inc.-dec. analysis. Increasing on (–∞, –1) and
(–1, 0); decreasing on (0, 1) and (1, ∞); relative
maximum at (0, –1).
3 x 2 − 12 x + 11
From the quadratic formula, CV: x =
y ′′ =
=
(
( x − 2)2
2
1
2
2
2
2
2
4
2
2
4
2
3
3
No possible inflection point, but x = ±1 must be
considered in the concavity analysis. Concave up
on (–∞, –1) and (1, ∞); concave down on (–1, 1).
5
2
( x − 2) (6 x − 12) − (3 x − 12 x + 11)2( x − 2)
y
( x − 2)4
( x − 2)(6 x − 12) − 2(3 x 2 − 12 x + 11)
x
5
( x − 2)3
2
( x − 2)3
No possible inflection point, but x = 2 must be
included in the concavity analysis. Concave
down on (−∞, 2); concave up on (2, ∞)
504
ISM: Introductory Mathematical Analysis
Section 13.5
1
32. y =
y ′′ =
2
(1 − x )3
No possible inflection point, but x = 1 must be
included in the concavity analysis. Concave up
on (–∞, 1); concave down on (1, ∞).
x +1
Intercept (0, 1)
Symmetric about the y-axis.
1
1
= 0 = lim
, so y = 0 is a
lim
2
2
x →∞ x + 1
x →−∞ x + 1
horizontal asymptote.
−2 x
y′ =
2
2
x +1
(
4
y
3
5x
)
CV: x = 0
Increasing on (–∞, 0); decreasing on (0, ∞);
relative maximum at (0, 1)
y ′′ =
(
)
2 3x 2 − 1
(
)
x2 + 1
34. y =
3
Possible inflection points at x = ±
1
3
x2
Intercept is (−1, 0)
Vertical asymptote is x = 0.
1+ x
x
1
= lim
= lim = 0
lim
x →∞ x 2
x →∞ x 2
x →∞ x
1+ x
= lim
, so y = 0 is the only horizontal
x →−∞ x 2
asymptote.
x+2
y′ = −
x3
CV: x = −2, but x = 0 must be included in the
inc-dec. analysis. Increasing on (−2, 0);
decreasing on (−∞, −2) and (0, ∞); relative
1⎞
⎛
minimum at ⎜ −2, − ⎟ .
4⎠
⎝
. Concave
⎛
⎛ 1
⎞
1 ⎞
up on ⎜ −∞, −
, ∞ ⎟ ; concave
⎟ and ⎜
3
3
⎝
⎠
⎝
⎠
⎛ 1
1 ⎞
down on ⎜ −
,
⎟ ; inflection points at
3
3⎠
⎝
⎛ 1 3⎞
, ⎟
⎜±
3 4⎠
⎝
5
1+ x
y
x
5
y ′′ =
2(3 + x)
x4
Possible inflection point when x = 3, but x = 0
must be included in the concavity analysis.
Concave up on (−3, 0) and (0, ∞); concave down
2⎞
⎛
on (−∞, −3); inflection point at ⎜ −3, − ⎟ .
9⎠
⎝
1+ x
1− x
Intercepts: (0, 1) and (–1, 0).
x = 1 is the only vertical asymptote. Since
1+ x
x
lim
= lim
= lim − 1 = −1
x →∞ 1 − x
x →∞ − x x →∞
1+ x
= lim
x →−∞ 1 − x
the only horizontal asymptote is y = –1.
(1 − x)(1) − (1 + x )(−1)
2
=
y′ =
2
(1 − x)
(1 − x) 2
No critical values, but x = 1 must be considered
in the ind.-dec. analysis. Increasing on (–∞, 1)
and (1, ∞).
33. y =
y
5
x
5
505
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
x2
7x + 4
Intercept: (0, 0)
x3 + 1
x
Intercept: (–1, 0)
Vertical asymptote is x = 0. Because the degree
of the numerator is greater than the degree of the
denominator, no horizontal asymptote exists.
35. y =
36. y =
Vertical asymptote is x = −
4
.
7
1
4
x2
1
4
16
so y = x −
= x− +
7
49
7x + 4 7
49 49(7 x + 4)
is an oblique asymptote.
(7 x + 4)(2 x) − x 2 (7)
y′ =
(7 x + 4)2
=
7 x2 + 8x
(7 x + 4)
2
=
Since y = x 2 + x −1 ,
y′ = 2 x − x −2 = 2 x −
x(7 x + 8)
(7 x + 4) 2
⎛ 1
1⎞
minimum at ⎜ 3 , 33 ⎟ .
⎜ 2
4 ⎟⎠
⎝
4⎞
⎛ 8
and (0, ∞); decreasing on ⎜ − , − ⎟ and
7⎠
⎝ 7
y ′′ = 2 + 2 x
)
2
(
)
(14 x + 8) − 7 x 2 + 8 x [14(7 x + 4)]
(7 x + 4)4
(
−3
2
= 2+
3
10
=
(
)
2 x3 + 1
3
y
)
(7 x + 4) ⎡ (7 x + 4)(14 x + 8) − 14 7 x 2 + 8 x ⎤
⎢⎣
⎥⎦
=
4
(7 x + 4)
=
.
x2
x
x
Possible inflection point when x = –1, but x = 0
must be included in concavity analysis. Concave
up on (–∞, –1) and (0, ∞); concave down on
(–1, 0); inflection point at (–1, 0).
16 ⎞
⎛ 4 ⎞
⎛ 8
⎜ − 7 , 0 ⎟ ; relative maximum at ⎜ − 7 , − 49 ⎟ ;
⎝
⎠
⎝
⎠
relative minimum at (0, 0).
+4
x2
2 x3 − 1
1
, but x = 0 must be included in inc.2
dec. analysis. Decreasing on (–∞, 0) and
⎛
⎛ 1
⎞
1⎞
⎜⎜ 0, 3 ⎟⎟ ; increasing on ⎜⎜ 3 , ∞ ⎟⎟ ; relative
2⎠
⎝
⎝ 2
⎠
8
4
, but x = − must be included in
7
7
8⎞
⎛
the inc.-dec. analysis. Increasing on ⎜ −∞, − ⎟
7⎠
⎝
2
=
CV: x = 3
CV: x = 0, −
(7x
y ′′ =
1
x
10
32
(7 x + 4)3
4
must be
7
included in concavity analysis. Concave down
4⎞
⎛
⎛ 4
⎞
on ⎜ −∞, − ⎟ ; concave up on ⎜ − , ∞ ⎟ .
7
7
⎝
⎠
⎝
⎠
No possible inflection point but x = −
37. y =
9
2
9x − 6x − 8
9
(3x + 2)(3 x − 4)
=
9⎞
⎛
Intercept: ⎜ 0, − ⎟
8⎠
⎝
2
4
Vertical asymptotes: x = − , x =
3
3
9
1
lim y = lim
= lim
= 0 = lim y
x →∞
x →∞ 9 x 2
x →∞ x 2
x →−∞
Thus y = 0 is a horizontal asymptote. Since
y
3
x
3
(
y = 9 9 x2 − 6 x − 8
506
)
−1
,
ISM: Introductory Mathematical Analysis
(
y′ = 9(−1) 9 x 2 − 6 x − 8
=−
)
−2
Section 13.5
(18 x − 6)
54(3x − 1)
[(3x + 2)(3x − 4)]2
1
2
4
, but x = − and x = must be included in inc.-dec. analysis.
3
3
3
2⎞
⎛
⎛ 2 1⎞
⎛1 4⎞
⎛4
⎞
Increasing on ⎜ −∞, − ⎟ and ⎜ − , ⎟ ; decreasing on ⎜ , ⎟ and ⎜ , ∞ ⎟ ;
3⎠
⎝3
⎠
⎝
⎝ 3 3⎠
⎝3 3⎠
CV: x =
⎛1
⎞
relative maximum at ⎜ , – 1⎟ . Finding y ′′ gives:
3
⎝
⎠
(9x
y ′′ = −54 ⋅
2
− 6x − 8
)
2
(
)
(3) − (3 x − 1) ⎡ 2 9 x 2 − 6 x − 8 (18 x − 6) ⎤
⎢⎣
⎥⎦
(9x
2
− 6x − 8
)
4
)(
)
( 9 x − 6 x − 8)
−162 ( −27 x + 18 x − 12 ) 486 ( 9 x − 6 x + 4 )
=
=
[(3x + 2)(3 x − 4)]
9
x
−
6
x
−
8
(
)
= −54 ⋅
(
3 9 x 2 − 6 x − 8 ⎡ 9 x 2 − 6 x − 8 − 4(3x − 1)(3x − 1) ⎤
⎣⎢
⎦⎥
4
2
2
2
3
2
3
Since 9 x 2 − 6 x + 4 = 0 has no real roots, y ′′ is never zero. No possible inflection points,
but x = −
2
4
and x = must be included in concavity analysis. Concave up on
3
3
⎛4
⎞
and ⎜ , ∞ ⎟ ; concave down on
⎝3
⎠
3
2⎞
⎛
⎜ −∞, − 3 ⎟
⎝
⎠
⎛ 2 4⎞
⎜− 3 , 3 ⎟ .
⎝
⎠
y
x
3
38. y =
8 x2 + 3x + 1
2x2
8 x 2 + 3 x + 1 is never 0 and x cannot be zero. Thus no intercepts. Vertical asymptote is x = 0.
lim y = lim
x →∞
x →∞
8x2
2x2
= lim 4 = 4 = lim y
x →∞
x →−∞
Thus y = 4 is a horizontal asymptote. Since y = 4 +
y′ = −
3 −1 1 −2
x + x , we have
2
2
3 −2
1
3x + 2
x − x −3 = − x −3 (3 x + 2) = −
2
2
2 x3
507
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
1⎞
⎛ 4
⎜ − , − ⎟.
3
12
⎝
⎠
2
CV: x = − , but x = 0 must be included in the
3
2⎞
⎛
inc. dec. analysis. Decreasing on ⎜ −∞, − ⎟ and
3⎠
⎝
⎛ 2
(0, ∞); increasing on ⎜ − ,
⎝ 3
y ′′ = −3 ⋅
⎞
0 ⎟ ; relative
⎠
= −3 ⋅
y
y
2
3x + 1
(3 x − 2)2
x
3
⎛ 1 ⎞ ⎛ 1⎞
Intercepts: ⎜ − , 0 ⎟ , ⎜ 0, ⎟
⎝ 3 ⎠ ⎝ 4⎠
2
Vertical asymptote is x = .
3
3x
1
lim y = lim
= lim
= 0 = lim y
x →∞
x →∞ 9 x 2
x →∞ 3 x
x →−∞
Thus y = 0 is a horizontal asymptote.
y′ =
=
3(3 x − 2)2 [(3 x − 2) − 3(3x + 4)]
7
Possible inflection point when x = − , but
3
2
x = must be included in concavity analysis.
3
7⎞
⎛
Concave down on ⎜ −∞, − ⎟ ; concave up on
3⎠
⎝
⎛ 7 2⎞
⎛2
⎞
⎜ − , ⎟ and ⎜ , ∞ ⎟ ; inflection point at
3
3
3
⎝
⎠
⎝
⎠
2 ⎞
⎛ 7
⎜ − , − ⎟.
27 ⎠
⎝ 3
x
10
39. y =
(3x − 2)6
(3x − 2)6
3(−6 x − 14) 18(3 x + 7)
= −3 ⋅
=
(3x − 2) 4
(3x − 2) 4
⎛ 2 23 ⎞
minimum at ⎜ − ,
⎟.
⎝ 3 8 ⎠
3
y ′′ = 3 x −3 + 3x −4 =
( x + 1).
x4
Possible inflection point when x = −1, but x = 0
must be included in the concavity analysis.
Concave down on (−∞, −1); concave up on
(−1, 0) and (0, ∞); inflection point at (−1, 3).
16
(3x − 2)3 (3) − (3x + 4)(3)(3x − 2)2 (3)
40. y =
(6 x + 5)2
⎛ 1 ⎞ ⎛ 1 ⎞
Intercepts: ⎜ − , 0 ⎟ , ⎜ 0,
⎟
⎝ 3 ⎠ ⎝ 25 ⎠
5
Vertical asymptote is x = − .
6
3x
1
= lim
= 0 = lim y
lim y = lim
2
x →∞
x →∞ 36 x
x →∞ 12 x
x →−∞
Thus y = 0 is horizontal asymptote.
(6 x + 5)2 (3) − (3 x + 1)[12(6 x + 5)]
y′ =
(6 x + 5)4
3(6 x + 5)[(6 x + 5) − 4(3 x + 1)]
=
(6 x + 5)4
3(−6 x + 1) −3(6 x − 1)
=
=
(6 x + 5)3
(6 x + 5)3
(3x − 2)2 (3) − (3 x + 1)(2)(3x − 2)(3)
(3 x − 2) 4
3(3 x − 2)[(3 x − 2) − 2(3 x + 1)]
=−
3x + 1
(3 x − 2) 4
3(3 x + 4)
(3 x − 2)3
4
2
CV: x = − , but x = must be included in
3
3
inc.-dec. analysis.
4⎞
⎛
⎛2
⎞
Decreasing on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ;
3⎠
⎝
⎝3
⎠
4
2
⎛
⎞
increasing on ⎜ − , ⎟ ; relative minimum at
⎝ 3 3⎠
508
ISM: Introductory Mathematical Analysis
Section 13.5
1
5
, but x = − must be included in
6
6
5⎞
⎛
inc.-dec. analysis. Decreasing on ⎜ −∞, − ⎟ and
6⎠
⎝
CV: x =
⎛1
⎞
⎜ 6 , ∞ ⎟ ; increasing on
⎝
⎠
41. y =
⎛ 5 1⎞
⎜ − 6 , 6 ⎟ ; relative
⎝
⎠
= −3 ⋅
(
(
)
inc.-dec. analysis. Increasing on − 3, 0 and
( 0, 3 ) ; decreasing on ( −∞, − 3 ) and
4
(
(6 x + 5)4
2
Possible inflection point when x = , but
3
5
x = − must be included in concavity analysis.
6
5⎞
⎛
⎛ 5 2⎞
Concave down on ⎜ −∞, − ⎟ and ⎜ − , ⎟ ;
6
⎝
⎠
⎝ 6 3⎠
)
(
)
y ′′ = 2 x −3 − 12 x −5 = 2 x −5 x 2 − 6 =
(
2 x2 − 6
5
)
x
Possible inflection points when x = ± 6 , but
x = 0 must be included in the concavity analysis.
(
)
concave up on ( − 6, 0 ) and (
(
)
Concave down on −∞, − 6 and 0, 6 ;
⎛2 1 ⎞
⎜ 3 , 27 ⎟ .
⎝
⎠
y
⎛
−5 6 ⎞
⎜⎜ − 6,
⎟.
36 ⎟⎠
⎝
x
3
3
y
3
– 3
509
x
3
)
6, ∞ ;
⎛
5 6⎞
inflection points at ⎜ 6,
⎟ and
⎜
36 ⎟⎠
⎝
( 16 , 241 ) ( 23 , 271 )
6
⎛
2 3⎞
3, ∞ ; relative maximum at ⎜ 3,
⎟;
⎜
9 ⎟⎠
⎝
⎛
2 3⎞
relative minimum at ⎜ − 3, −
⎟.
⎜
9 ⎟⎠
⎝
⎛2
⎞
concave up on ⎜ , ∞ ⎟ ; inflection point at
⎝3
⎠
–5
3 − x2
x4
CV: x = ± 3 , but x = 0 must be included in the
−12 x + 8
3
)
y′ = − x −2 + 3x −4 = x −4 − x 2 + 3 =
(6 x + 5)6
(6 x + 5)
3x − 2
( x + 1)( x − 1)
asymptote. Since y = x −1 − x −3 , then
6(6 x + 5)2 [(6 x + 5) − 3(6 x − 1)]
= −18 ⋅
= 72 ⋅
(6 x + 5)3 (6) − (6 x − 1) ⎡18(6 x + 5)2 ⎤
⎣
⎦
6
(6 x + 5)
3
=
x
x3
Intercepts are (–1, 0) and (1, 0).
Symmetric about the origin.
Vertical asymptote x = 0.
x2 − 1
x2
1
lim
= lim
= lim
3
3
x →∞ x
x →∞ x
x →∞ x
1− x
= 0 = lim
, so y = 0 is the only horizontal
x →−∞ x 2
⎛1 1 ⎞
maximum at ⎜ ,
⎟ . Finding y ′′ gives:
⎝ 6 24 ⎠
y ′′ = −3 ⋅
x2 − 1
Chapter 13: Curve Sketching
42. y =
ISM: Introductory Mathematical Analysis
at (0, 1).
3x
( x − 2)2
Intercept (0, 0)
Vertical asymptote at x = 2
3x
3x
3
= lim
= lim = 0 and
lim
2
2
x →∞ x − 4 x + 4 x →∞ x
x →∞ x
3x
= 0, so y = 0 is the only
lim
x →−∞ x 2 − 4 x + 4
horizontal asymptote.
−3( x + 2)
y′ =
( x − 2)3
CV: x = −2, but x = 2 must be included in the
inc.-dec. analysis. Decreasing on (−∞, −2) and
(2, ∞); increasing on (−2, 2); relative maximum
3⎞
⎛
at ⎜ −2, − ⎟
8⎠
⎝
y ′′ =
y ′′ =
=
(
(
)
( x + 1)(2 x + 2) − x 2 + 2 x [2]
3
=
2
y
5
x
5
6( x + 4)
16
44. y =
3x 4 + 1
x3
No intercepts
Symmetric about the origin.
Vertical asymptote is x = 0.
y
Since y = 3 x + x −3 ,
1
x2 + x + 1
=
x +1
x +1
Intercept: (0, 1). x = –1 is the only vertical
asymptote. y = x is an oblique asymptote.
43. y = x +
)
( x + 1)(2 x + 1) − x 2 + x + 1
( x + 1)
x2 + 2 x
2
=
−4
= 3−
3
2
10
4
=
(
3
= 3x +
1
x3
so
)
3 x 2 + 1 ( x + 1)( x − 1)
x
x4
CV: ±1, but x = 0 must be considered in the inc.dec. analysis. Increasing on (–∞, –1) and (1, ∞);
decreasing on (–1, 0) and (0, 1); relative
maximum at (–1, –4); relative minimum at
(1, 4).
12
y ′′ =
x5
No possible inflection point, but x = 0 must be
included in the concavity analysis. Concave
down on (–∞, 0); concave up on (0, ∞).
x
10
(
3x 4 + 1
x
y = 3x is an oblique asymptote.
y′ = 3 − 3 x
=
( x + 1)
4
( x + 1)
( x + 1)3
No possible inflection point, but x = –1 must be
included in the concavity analysis. Concave
down on (–∞, –1); concave up on (–1, ∞).
( x − 2)4
Possible inflection point when x = −4, but x = 2
must be included in the concavity analysis.
Concave down on (−∞, −4); concave up on
1⎞
⎛
(−4, 2) and (2, ∞); inflection point at ⎜ −4, − ⎟ .
3⎠
⎝
y′ =
)
( x + 1)2 (2 x + 2) − x 2 + 2 x [2( x + 1)]
y
x( x + 2)
( x + 1)
( x + 1)2
CV: 0 and –2, but x = –1 must be included in the
inc.-dec. analysis. Increasing on (–∞, –2) and
(0, ∞); decreasing on (–2, –1) and (–1, 0);
relative maximum at (–2, –3); relative minimum
x
10
510
ISM: Introductory Mathematical Analysis
45. y =
−3x 2 + 2 x − 5
−3 x 2 + 2 x − 5
(3 x + 1)( x − 1)
=
3x2 − 2 x − 1
Section 13.5
Note that −3x 2 + 2 x − 5 is never zero.
Intercept: (0, 5)
1
Vertical asymptotes are x = − and x = 1.
3
lim y = lim
−3 x 2
= lim − 1 = −1 = lim y
x →∞
x →−∞
3x 2
Thus y = –1 is horizontal asymptote.
x →∞
x →∞
( 3x
y′ =
2
)
(
(3x − 2 x − 1)
2(3x − 1) ⎡( 3x − 2 x − 1) (−1) − ( −3 x
⎢⎣
=
(3x − 2 x − 1)
2
2
2
12(3x − 1)
( 3x
2
)
− 2x −1
2
2
)
+ 2x − 5 ⎤
⎥⎦
2
2
=
)
− 2 x − 1 (−6 x + 2) − −3x 2 + 2 x − 5 (6 x − 2)
12(3 x − 1)
=
[(3x + 1)( x − 1)]2
1
1
, but x = − and x = 1 must be included in inc.-dec. analysis.
3
3
1⎞
⎛
⎛ 1 1⎞
⎛1 ⎞
Decreasing on ⎜ −∞, − ⎟ and ⎜ − , ⎟ ; increasing on ⎜ , 1⎟ and (1, ∞); relative minimum at
3
3
3
⎝
⎠
⎝
⎠
⎝3 ⎠
CV: x =
( 3x
y ′′ = 12 ⋅
( 3x
= 12 ⋅
= 12 ⋅
2
2
)
( 3x
) (
2
)
− 2x −1
4
)
− 2 x − 1 ⎡3 3 x 2 − 2 x − 1 − 2(3 x − 1)(6 x − 2) ⎤
⎣⎢
⎦⎥
( 3x
2
)
− 2 x − 1 (3) − (3 x − 1) ⎡ 2 3 x 2 − 2 x − 1 (6 x − 2) ⎤
⎢⎣
⎥⎦
−27 x 2 + 18 x − 7
( 3x
(
2
⎛1 7⎞
⎜3, 2⎟.
⎝
⎠
)
− 2x −1
3
=
2
)
− 2x −1
(
4
−12 27 x 2 − 18 x + 7
)
[(3x + 1)( x − 1)]3
1
and x = 1 must be included
3
1⎞
⎛
⎛ 1 ⎞
in concavity analysis. Concave down on ⎜ −∞, − ⎟ and (1, ∞); concave up on ⎜ − , 1⎟ .
3⎠
⎝
⎝ 3 ⎠
Since 27 x 2 − 18 x + 7 is never zero, there is no possible inflection point, but x = −
511
Chapter 13: Curve Sketching
10
ISM: Introductory Mathematical Analysis
y
–1
3
5x
1
(3 x + 2)2 + 1
=
3x + 2
3x + 2
9 x 2 + 12 x + 5
=
3x + 2
46. y = 3x + 2 +
Note that 9 x 2 + 12 x + 5 is never zero.
⎛ 5⎞
Intercept: ⎜ 0, ⎟
⎝ 2⎠
2
Vertical asymptote is x = − ; oblique asymptote is y = 3x + 2.
3
y′ = 3 −
= 3⋅
3
= 3⋅
(3x + 2) 2
2
9 x + 12 x + 3
(3 x + 2)
2
(3 x + 2)2 − 1
(3 x + 2)2
= 9⋅
(3 x + 1)( x + 1)
(3x + 2)2
1
2
and x = −1, but x = − must be included in inc.-dec. analysis. Increasing on (−∞, −1) and
3
3
1
2
⎛
⎞
⎛
⎞
⎛ 2 1⎞
⎜ − , ∞ ⎟ ; decreasing on ⎜ −1, − ⎟ and ⎜ − , − ⎟ ; relative maximum at (−1, −2); relative minimum at
3⎠
3⎠
⎝ 3
⎠
⎝
⎝ 3
⎛ 1 ⎞
⎜ − , 2 ⎟.
⎝ 3 ⎠
18
y ′′ = −3(−2)(3 x + 2)−3 (3) =
(3 x + 2)3
2
2⎞
⎛
No possible inflection point, but x = − must be included in concavity analysis. Concave down on ⎜ −∞, − ⎟ ;
3⎠
3
⎝
⎛ 2
⎞
concave up on ⎜ − , ∞ ⎟ .
3
⎝
⎠
CV: x = −
y
10
x
5
512
ISM: Introductory Mathematical Analysis
47.
5
Section 13.5
y
52. For y = 6 − 3e− x we have
(
)
⎛
3 ⎞
lim 6 − 3e− x = lim ⎜ 6 − ⎟ = 6 − 3(0) = 6
x →∞
x →∞ ⎝
ex ⎠
Thus the line y = 6 is a horizontal asymptote for
x
5
the graph of y = 6 − 3e− x . For y = 6 + 3e− x , we
(
)
obtain lim 6 + 3e− x = 6 + 3(0) = 6 , so the line
x →∞
48.
5
y = 6 is also a horizontal asymptote for the graph
y
of y = 6 + 3e− x .
y
16
x
5
y = 6 + 3e–x
8
4
49.
5
y = 6 – 3e–x
y
(
)
⎛
76 ⎞
53. lim 150 − 76e−t = lim ⎜150 − ⎟
t →∞
t →∞ ⎝
et ⎠
= 150 – 0 = 150
Thus y = 150 is a horizontal asymptote.
x
5
54.
50.
5
x
16
1
y
–15
15
x
5
–1
x ≈ –0.08, y = 0
55.
a
a
51. When x = − , then a + bx = 0 so x = − is a
b
b
vertical asymptote.
x
x
1 1
= lim
= lim =
lim
b
x →∞ a + bx x →∞ bx
x →∞ b
1
Thus y = is a horizontal asymptote.
b
8
–5
5
–8
x ≈ ±2.45, x ≈ 0.67, y = 2
56.
10
–10
10
–10
513
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
0.15625
relative and absolute maximum when x =
The other number is 20 − x =
–1.54255
–6.54255
In the standard window, two vertical asymptotes
of the form x = k, where k > 0, are apparent
(x ≈ 0.68 and x ≈ 7.32). By zooming around
x = –4, another vertical asymptote is apparent
(x = –4). Thus three vertical asymptotes exist.
(
A′ =
0
25
0
From the graph, it appears that lim y ≈ 0.48 .
x →∞
Stream
Thus a horizontal asymptote is y ≈ 0.48.
Algebraically, we have
lim
)
1
(9000 − 30 x)
18
Setting A′ = 0 ⇒ x = 300 . Since
1
A′′(300) = (−30) < 0 , we have a maximum at
18
9000 − 15(300)
x = 300. Thus y =
= 250 . The
18
dimensions are 300 ft by 250 ft.
1
x →∞
40
.
3
3. We are given that 15x + 9(2y) = 9000, or
9000 − 15 x
y=
. We want to maximize area A,
18
where A = xy.
⎛ 9000 − 15 x ⎞ 1
2
A = xy = x ⎜
⎟ = 18 9000 x − 15 x
18
⎝
⎠
–0.15625
57.
20
.
3
0.34e
0.7 x
4.2 + 0.71e0.7 x
0.34
= lim
x →∞ 4.2
e0.7 x
+ 0.71
=
=
y
0.34e0.7 x
e0.7 x
lim
0.7 x
x →∞ 4.2 + 0.71e
0.7 x
e
y
x
1200
, and
x
want to minimize N = 2x + 6y. We have
⎛ 1200 ⎞
N = 2x + 6 y = 2x + 6 ⎜
⎟, x> 0
⎝ x ⎠
4. We are given that xy = 1200, or y =
0.34
≈ 0.48
0 + 0.71
Problems 13.6
1. Let the numbers be x and 82 − x. Then if
P = x(82 − x) = 82 x − x 2 , we have P ′ = 82 − 2 x.
Setting P ′ = 0 ⇒ x = 41. Since P ′′ = −2 < 0,
there is a maximum when x = 41. Because
82 − x = 41, the required numbers are 41 and 41.
N′ = 2 −
7200
x2
Setting N′ = 0 yields x 2 = 3600 , so x = 60. We
14, 400
have N ′′ =
, so N ′′(60) > 0 and we have
x3
a minimum. If x = 60, then y = 20. Thus
N = 2(60) + 6(20) = 240 ft.
2. Let the numbers be x and 20 – x, where
0 ≤ x ≤ 20. Let
P = (2 x)(20 − x) 2 = 2 x3 − 80 x 2 + 800 x .
y
dP
= 0 gives
Setting
dx
y
y
y
x
2
P′ = 6 x − 160 x + 800 = 2(3 x − 20)( x − 20) = 0 ,
20
or x = 20. P′ > 0 on
3
⎛ 20 ⎞
⎛ 20
⎞
⎜ 0, 3 ⎟ and P′ < 0 on ⎜ 3 , 20 ⎟ . Thus P has a
⎝
⎠
⎝
⎠
from which x =
514
y
y
ISM: Introductory Mathematical Analysis
Section 13.6
5. c = 0.05q 2 + 5q + 500
Avg. cost per unit = c =
c ′ = 0.05 −
500
q2
500
1
. The answer
B
does not depend on A because A represents the
initial value of q, so it doesn’t change q over
time.
revenue is maximum when p =
c
500
= 0.05q + 5 +
q
q
. Setting c ′ = 0 yields
9.
2
, q = 10, 000, q = ±100 . We
q2
exclude q = –100 because q represents the
1000
number of units. Since c′′ =
> 0 for q > 0,
q3
c is an absolute minimum when q = 100 units.
0.05 =
a.
Setting f ′( p) = 0 gives −1 +
900
( p + 10)2
=0,
= 1 , ( p + 10)2 = 900,
( p + 10)2
p + 10 = ±30 , from which p = 20.
Since f ′′( p ) =
−1800
< 0 for p = 20, we
( p + 10)3
have an absolute maximum of
f(20) = 110 grams.
d 2C
= −0.0024 < 0 , a maximum
ds 2
occurs when s = 50. Thus a minimum can occur
only at an endpoint of the domain. If s = 0, then
C = 0.08; if s = 60, then C = 2.96. Thus the
minimum cost of $0.08 per hour occurs for
s = 0 mi/h and might be due to depreciation,
insurance, and so on.
9
, so we have an
11
9
absolute minimum of f (100) = 51 grams.
11
b. f(0) = 70 and f (100) = 51
2
D3
⎛ C D ⎞ CD
10. R = D 2 ⎜ − ⎟ =
−
2
3
⎝2 3⎠
dR
= CD − D 2 . This
The rate of change of R is
dD
is the function to be maximized. Setting
C
d ⎛ dR ⎞
= C − 2 D = 0 gives D = . Since
⎜
⎟
dD ⎝ dD ⎠
2
7. p = –5q + 30
Since total revenue = (price)(quantity),
r = pq = (−5q + 30)q = −5q 2 + 30q
Setting r ′ = −10q + 30 = 0 ⇒ q = 3 . Since
r ′′ = −10 < 0 , r is maximum at q = 3 units, for
which the corresponding price is
p = –5(3) + 30 = $15.
8. q = Ae
900
, where 0 ≤ p ≤ 100.
p + 10
900
6. C = 0.12s − 0.0012 s 2 + 0.08 , where 0 ≤ s ≤
dC
60. Setting
= 0 gives 0.12 – 0.0024s = 0 ⇒
ds
s = 50. Since
f ( p) = 160 − p −
d 2 ⎛ dR ⎞
⎜
⎟ = −2 < 0 , the maximum rate of
dD 2 ⎝ dD ⎠
C
change occurs when D = .
2
− Bp
Revenue = r = pq = pAe− Bp
r ′ = A[e− Bp (1) + pe− Bp (− B)]
= A(1 − Bp )e− Bp
⎛1
⎞
= AB ⎜ − p ⎟ e− Bp
⎝B
⎠
1
Critical value: p =
B
1
If p < , then r ′ > 0 and r is increasing. If
B
1
p > , then r ′ < 0 and r is decreasing. Thus
B
11. p = 85 − 0.05q
c = 600 + 35q
Profit = Total Revenue – Total Cost
P = pq − c = (85 − 0.05q )q − (600 + 35q )
= −(0.05q 2 − 50q + 600)
Setting P ′ = −(0.1q − 50) = 0 yields q = 500.
Since P ′′(500) = −0.1 < 0, P is a maximum
when q = 500 units. This corresponds to a price
of p = 85 − 0.05(500) = $60 and a profit of
P = $11,900.
515
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
12. Cost per unit = $3
10
p=
q
MR =
20 1
= = MC.
60 3
2 2
10, 000
q − 40q +
3
q
Profit = Total Revenue – Total Cost
Since total revenue r = pq and
total cost = c = cq ,
P = pq − cq
c=
⎛2
⎞
= q3 − 100q 2 + 3200q − ⎜ q3 − 40q 2 + 10, 000 ⎟
⎝3
⎠
1
= q3 − 60q 2 + 3200q − 10, 000
3
13. p = 42 – 4q
80
c = 2+
q
Total Cost = c = cq = 2q + 80
Profit = Total Revenue – Total Cost
P = pq – c = (42 – 4q)q – (2q + 80)
P′ = q 2 − 120q + 3200 = (q − 40)(q − 80)
Setting P′ = 0 gives q = 40 or 80. Evaluating
profit at q = 0, 40, 80, and 120 gives
P(0) = –10,000
130, 000
1
P (40) =
= 43,333
3
3
98, 000
2
P (80) =
= 32, 666
3
3
P(120) = 86,000
Thus the profit maximizing output is q = 120
units, and the corresponding maximum profit is
$86,000.
)
P′ = −(8q − 40)
Setting P′ = −(8q − 40) = 0 gives q = 5. We find
that P ′′ = −8 < 0 , so P has a maximum value
when q = 5. The corresponding price p is
42 – 4(5) = $22.
14. p =
1
and MC = , then for q = 3600 we
3
15. p = q 2 − 100q + 3200 on [0, 120]
5 −3
Moreover, we have P ′′ = − q 2 < 0 for q > 0,
2
25
. The
so P is maximum when q =
9
corresponding price is p = $6.
(
q
have MR =
Profit = Total Revenue – Total Cost
P = pq – c
⎛ 10 ⎞
P=⎜
⎟ q − (3q) = 10 q − 3q
⎜ q⎟
⎝
⎠
25
5
− 3 = 0 yields q =
Setting P′ =
.
9
q
= − 4q 2 − 40q + 80
20
40
16. a.
q
c = cq = 2q3 − 42q 2 + 228q + 210
dc
= 6q 2 − 84q + 228 = 6(q 2 − 14q + 38)
dq
Using the quadratic formula to solve
dc
= 0 gives q = 7 − 11 ≈ 3.68 or
dq
1 2000
c= +
3
q
q
+ 2000
3
Profit = Total Revenue − Total Cost
q
P = pq − c = 40 q − − 2000
3
20 1
− = 0 yields q = 3600.
Setting P ′ =
q 3
Total cost = c = cq =
q = 7 + 11 ≈ 10.32. Evaluating c at q = 3,
7 − 11, 7 + 11, and 12 gives
570, 434 + 44 11 ≈ 579.93,
434 − 44 11 ≈ 288.07, and 354,
respectively. Thus the minimum cost is
when q = 7 + 11 ≈ 10.32.
c(10) = 290 and c(11) = 298, so production
should be fixed at q = 10 for a minimum
cost of $290.
Since P ′′ = −10q −3 / 2 < 0 for q > 0, it follows
that P is a maximum when q = 3600. The
40
≈ $0.67. Since
corresponding price is p =
60
516
ISM: Introductory Mathematical Analysis
Section 13.6
b. c(7) = 434, so the minimum cost still occurs
when q = 7 + 11 ≈ 10.32 and production
should again be fixed at 10 units.
20. Note that as the number of units produced and
sold increases from 0 to 600, the profit increases
from 0 to (600)(400) = $24,000. Let
q = number of units produced and sold beyond
600. Then the total profit P is given by
P = (600)(40) + (40 − 0.05q )q
17. Total fixed costs = $1200,
material-labor costs/unit = $2, and the demand
100
.
equation is p =
q
= 24, 000 + 40q − 0.05q 2
P ′ = 40 − 0.10q
Setting P ′ = 0 yields q = 400. Since
P ′′ = −0.10 < 0, P is a maximum when q = 400,
that is, the total number of units = 600 + 400
= 1000.
Profit = Total Revenue – Total Cost
P = pq – c
100
P=
⋅ q − (2q + 1200)
q
= 100 q − 2q − 1200
(
= 2 50 q − q − 600
21. See the figure in the text. Given that x 2 y = 32 ,
)
we want to minimize S = 4( xy ) + x 2 . Since
⎛ 25
⎞
− 1⎟ = 0 yields q = 625. We
Setting P′ = 2 ⎜
⎜ q
⎟
⎝
⎠
y=
32
x2
, where x > 0, we have
⎛ 32 ⎞
128
S = 4 x ⎜ ⎟ + x2 =
+ x 2 , from which
2
x
⎝x ⎠
128
S′ = −
+ 2 x . Setting S′ = 0 gives
x2
256
2 x3 = 128 , x3 = 64 , x = 4. Since S ′′ =
+2,
x3
we get S ′′(4) > 0 , so x = 4 gives a minimum. If
− 32
see that P ′′ = −25q < 0 for q > 0, so P is
maximum when q = 625. When q = 625,
50
MR =
= 2 = MC. When q = 625, then
625
p = $4.
18. Let x = number of $10 per month increases
so the monthly rate is 400 + 10x and the number
of rented apartments is 100 – 2x. Monthly
revenue r is given by
r = (rent/apt.) (no. of apt. rented)
r = (400 + 10x)(100 – 2x)
r ′ = (400 + 10 x)(−2) + (100 − 2 x)(10)
= 200 – 40x = 40(5 – x)
Setting r′ = 0 yields x = 5. Since r ′′ = −40 < 0 ,
then r is maximum when x = 5. This results in a
monthly rate for an apartment of
400 + 10(5) = $450.
32
= 2 . The dimensions are
16
4 ft × 4 ft × 2 ft.
x = 4, then y =
22. See the figure in the text. We want to maximize
V = x 2 y given that 4 xy + x 2 = 192 , or
192 − x 2
4x
⎛
192 − x 2 ⎞ 1
V = x2 ⎜
⎟ = 192 x − x3 , x > 0
⎜ 4x ⎟ 4
⎝
⎠
1
3
2
V ′ = 192 − 3 x = 64 − x 2
4
4
Setting V ′ = 0 gives x = 8. Since
⎛3⎞
V ′′ = ⎜ ⎟ (−2 x) , then V ′′(8) < 0 , so x = 8 gives
⎝4⎠
a maximum. If x = 8, then y = 4.
The dimensions are 8 ft × 8 ft × 4 ft.
y=
(
19. If x = number of $0.50 decreases, where
0 ≤ x ≤ 36, then the monthly fee for each
subscriber is 18 – 0.50x, and the total number of
subscribers is 4800 + 150x. Let r be the total
(monthly) revenue.
revenue = (monthly rate)(number of subscribers)
r = (18 – 0.50x)(4800 +150x)
r ′ = (18 − 0.50 x)(150) + (4800 + 150 x)(−0.50)
= 300 –150x = 150(2 – x)
Setting r ′ = 0 yields x = 2.
Evaluating r when x = 0, 2, and 36, we find that
r is a maximum when x = 2. This corresponds to
a monthly fee of 18 – 0.50(2) = $17 and a
monthly revenue r of $86,700.
(
) (
The volume is 82 (4) = 256 ft 3 .
517
)
)
Chapter 13: Curve Sketching
23. V = x ( L − 2 x )
ISM: Introductory Mathematical Analysis
2
If S′ = 0 , then πr 3 − K = 0 , πr 3 = K ,
= L2 x − 4 Lx 2 + 4 x3
L
where 0 < x < .
2
r=3
π
= 12 x 2 − 8 Lx + L2
= (2x − L)(6x − L)
L
L
For 0 < x < , setting V ′ = 0 gives x = .
6
2
L
⎛
⎞
Since V ′ > 0 on ⎜ 0, ⎟ and V ′ < 0 on
⎝ 6⎠
L
⎛L L⎞
⎜ , ⎟ , V is maximum when x = . Thus the
6
2
6
⎝
⎠
L
length of the side of the square must be
in.,
6
which results in a volume of
2
V = πr 2 h
240
, x > 0. We want
x
(
K − πr 2
. Thus Equation
2πr
Setting V ′ = 0 gives r =
h=
=
K − π 3Kπ
2π 3Kπ
2
3
2π
K
K
3π
⋅
=
K
3π
K
3π
2
3
K
. Thus
3π
K
2π 3Kπ
=
K
3π
Note that since V ′′ = −3πr < 0 for r > 0, we have
a maximum.
27. p = 600 − 2q
c = 0.2q 2 + 28q + 200
Profit = Total Revenue – Total Cost
P = pq – c
(
P = (600 − 2q )q − 0.2q 2 + 28q + 200
(2)
K
πr
> 0 for r > 0, we
(2) becomes
Kr − πr 3
V=
2
dV K − 3πr 2
.
=
2
dr
25. See the figure in the text.
V = K = πr 2 h
(1)
becomes
2K
S=
+ πr 2
r
r3
(2)
From Equation (1), h =
⎛ 240
⎞
A = ( x + 10)( y + 6) = ( x + 10) ⎜
+ 6⎟
⎝ x
⎠
2400
= 300 + 6 x +
x
2400
A′ = 6 −
x2
Setting A′ = 0 gives x = 20. Since
4800
A′′ =
> 0 for x = 20, we have a minimum.
x3
Thus y = 12, so the dimensions are 20 + 10 by
12 + 6, that is, 30 in. × 18 in.
From Equation (1) h =
4K
26. See the figure in the text.
S = K = 2πrh + πr 2
(1)
to minimize A where
S = 2πrh + πr
2
3
have a minimum.
L⎛
L⎞
2L
in 3 .
⎜L− ⎟ =
6⎝
3⎠
27
2
( Kπ )
K
⎛ K ⎞3
.
=⎜ ⎟ =3
π
π
⎝ ⎠
Note that since S ′′ = 2π +
3
24. Since xy = 240, then y =
1
K
h=
V ′ = L2 − 8Lx + 12 x 2
K
. Thus
π
2
(
= − 2.2q 2 − 572q + 200
. Thus Equation (2)
)
)
P′ = −(4.4q − 572)
Setting P′ = 0 yields q = 130. Since
P ′′ = −4.4 < 0 , P is maximum when q = 130
units. The corresponding price is
p = 600 – 2(130) = $340, and the profit is
P = $36,980. If a tax of $22/unit is imposed on
the manufacturer, then the cost equation is
)
2 πr 3 − K
dS
2K
=−
+ 2πr =
.
dr
r2
r2
518
ISM: Introductory Mathematical Analysis
Section 13.6
minimize the sum C of carrying costs and set-up
costs.
⎡ ⎛ q ⎞⎤
⎛ 1000 ⎞
C = 0.128 ⎢10 ⎜ ⎟ ⎥ + 40 ⎜
⎟
2
⎣ ⎝ ⎠⎦
⎝ q ⎠
c1 = 0.2q 2 + 28q + 200 + 22q
= 0.2q 2 + 50q + 200 .
The demand equation remains the same. Thus
P1 = pq − c1
(
= (600 − 2q)q − 0.2q 2 + 50q + 200
(
= − 2.2q 2 − 550q + 200
)
)
= 0.64q +
C ′ = 0.64 −
P′1 = −(4.4q − 550)
c = 0.2q 2 + 28q + 200 . Revenue, both before
and after the license fee, is given by
30. c = 0.004q3 + 20q + 5000
p = 450 – 4q
Profit = Total Revenue – Total Cost
P = pq − c
r = pq = 600q − 2q 2 . After the license fee, the
cost equation is
(
c1 = c + 1000 = 0.2q 2 + 28q + 1200 and the profit
is
P1 = r − c1
(
= 600q − 2q
) − ( 0.2q
+ 28q + 1200
q2
40, 000
= 62,500 ,
0.64
80, 000
>0,
q = 250 (since q > 0). Since C ′′ =
q3
C is minimum when q = 250. Thus the economic
lot size is 250/lot (4 lots).
28. Original data: p = 600 –2q,
2
40, 000
Setting C′ = 0 yields q 2 =
Setting P′1 = 0 yields q = 125. Since
P1′′ = −4.4 < 0 , P1 is maximum when q = 125
units. The corresponding price is p = $350 and
the profit is P1 = $34,175 .
2
40, 000
q
= (450 − 4q )q − 0.004q3 + 20q + 5000
(
P′ = − ( 0.012q + 8q − 430 )
= −2 ( 0.006q + 4q − 215 )
P = − 0.004q3 + 4q 2 − 430q + 5000
)
)
)
2
As in Problem 27, we find that P1 has a
maximum when q = 130 units, which gives
p = $340. Thus the profit-maximizing price and
output remain the same. Since
Profit
= r − c1 = r − (c + 1000) = (r − c) − 1000, when
q = 130 we have
Profit = 36,980 – 1000 (from Problem 27)
= $35,980
2
Setting P′ = 0 yields
0.006q 2 + 4q − 215 = 0
−4 ± 21.16 −4 ± 4.6
=
0.012
0.012
−4 + 4.6
= 50 . Since P
Since q ≥ 0, choose q =
0.012
is increasing on [0, 50) and decreasing on
(50, ∞), P is maximum when q = 50 units.
q=
29. Let q = number of units in a production run.
Since inventory is depleted at a uniform rate,
q
assume that the average inventory is . The
2
q
⎛ ⎞
value of average inventory is 10 ⎜ ⎟ , and
⎝2⎠
31. Let x = number of people over the 30.
Note: 0 ≤ x ≤ 10.
Revenue = r
= (number attending)(charge/person)
= (30 + x)(50 – 1.25x)
⎡ ⎛ q ⎞⎤
carrying costs are 0.128 ⎢10 ⎜ ⎟ ⎥ . The number
⎣ ⎝ 2 ⎠⎦
1000
of production runs per year is
, and total
q
= 1500 + 12.5 x − 1.25 x 2
r ′ = 12.5 − 2.5 x
Setting r′ = 0 yields x = 5. Since r ′′ = −2.5 < 0 ,
r is maximum when x = 5, that is, when 35
attend.
⎛ 1000 ⎞
set-up costs are 40 ⎜
⎟ . We want to
⎝ q ⎠
519
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
34. Let q = level of production.
Total Cost
Average Cost = c =
q
For 0 ≤ q ≤ 5000, we have
30q + 10q + 20, 000
20, 000
.
c=
= 40 +
q
q
Note that total cost for 5000 units is 220,000.
For
q > 5000,
⎛ cost for those
⎞
(cost for first 5000) + ⎜
⎟
units
beyond
5000
⎝
⎠
c=
q
32. Let N = horsepower of motor.
(Total annual cost) = C = (Annual cost to lease)
+ (Annual operating cost)
⎛ 0.008 ⎞
C = (200 + 0.40 N ) + 80, 000 ⎜
⎟
⎝ N ⎠
640
= 200 + 0.04 N +
N
640
C ′ = 0.4 −
N2
Setting C ′ = 0 yields N 2 = 1600, so N = 40
(since N > 0). Since C ′′ =
1280
> 0 for N > 0, C
N3
is a minimum when N = 40 horsepower.
=
33. The cost per mile of operating the truck is
s
0.165 +
. Driver’s salary is $18/hr. The
200
700
number of hours for 700 mi trip is
. Driver’s
s
12, 600
⎛ 700 ⎞
salary for trip is 18 ⎜
, or
. The cost
⎟
s
⎝ s ⎠
of operating the truck for the trip is
s ⎤
⎡
700 ⎢0.165 +
.
200 ⎥⎦
⎣
Total cost of trip is
12, 600
s ⎞
⎛
C=
+ 700 ⎜ 0.165 +
200 ⎟⎠
s
⎝
220, 000 + [45(q − 5000) + 10(q − 5000)]
q
c = 55 −
55, 000
q
If 0 < q ≤ 5000, then c ′ = −
20, 000
< 0 and
q2
thus c is decreasing. If q > 5000, then
55, 000
> 0 and thus c is increasing.
c′ =
q2
Hence c is minimum when q = 5000 units.
35. Profit P is given by
P = Total revenue – Total cost
= Total revenue – (salaries + fixed cost)
= 50q – (1000m + 3000)
(
)
= 50 ( m − 15m + 72m − 60 ) , where 0 ≤ m ≤ 8
P′ = 50 ( 3m − 30m + 72 )
= 150 ( m − 10m + 24 ) = 150(m − 4)(m − 6)
= 50 m3 − 15m2 + 92m − 1000m − 3000
12, 600
7
Setting C ′ = −
+ = 0 yields s 2 = 3600 ,
2
2
s
25, 200
>0
or s = 60 (since s > 0). Since C ′′ =
s3
for s > 0, C is a minimum when s = 60 mi/h.
3
2
2
2
Setting P′ = 0 gives the critical values 4 and 6.
We now evaluate P at these critical values and
also at the endpoints 0 and 8.
P(0) = –3000
P(4) = 2600
P(6) = 2400
P(8) = 3400
Thus Ms. Jones should hire 8 salespeople to
obtain a maximum weekly profit of $3400.
520
ISM: Introductory Mathematical Analysis
Section 13.6
36. Profit P is given by
P = Total revenue – Total cost = pq – Total cost
P
/ton, then
2
P ⎛ 24 − 6 x ⎞
⎡ 12 − 3 x ⎤
= P ⎢x +
PT = Px + ⎜
2 ⎝ 5 − x ⎟⎠
5 − x ⎦⎥
⎣
= 400q − 50q 2 − Total cost. (q in hundreds)
dP
d
= 400 − 100q −
(Total cost)
dq
dq
= 400 – 100q – Marginal cost
800
= 400 − 100q −
q+5
=
=
⎡ x 2 − 10 x + 22 ⎤
P′T = P ⎢
⎥
2
⎢⎣ (5 − x)
⎥⎦
Setting P′T = 0 and using an argument similar
to that above, we find that PT is a maximum
400(q + 5) − 100q(q + 5) − 800
q+5
when x = 5 − 3 tons.
2
−100q − 100q + 1200
q+5
38. x = number of floors. Let R = rate of return.
Total Revenue
R=
Total Cost
60, 000 x
=
(10 x)[120, 000 + 3000( x − 1)] + 1, 440, 000
2x
=
2
x + 39 x + 48
−100(q + 4)(q − 3)
q+5
Setting P′ = 0 gives the critical value 3 (since
q > 0). We find that P′ > 0 for 0 < q < 3, and
P′ < 0 for q > 3. Thus there is a maximum profit
when q = 3000 jackets.
=
R′ = 2 ⋅
37. x = tons of chemical A (x ≤ 4),
24 − 6 x
y=
= tons of chemical B, profit on
5− x
A = $2000/ton, and profit on B = $1000/ton.
⎛ 24 − 6 x ⎞
Total Profit = PT = 2000 x + 1000 ⎜
⎟
⎝ 5− x ⎠
( x 2 + 39 x + 48) 2
R ′ = 0 when x = 48 = 4 3 (x ≥ 0). Since R is
(
)
increasing on 0, 4 3 and decreasing on
(4
)
3, ∞ , R is a maximum when
x = 4 3 ≈ 6.93. The number of floors in the
building must be an integer, so we evaluate R
when x = 6 and x = 7: R(6) ≈ 0.0377;
R(7) ≈ 0.0378. Thus 7 floors should be built to
maximize the rate of return.
⎡ 12 − 3 x ⎤
= 2000 ⎢ x +
5 − x ⎥⎦
⎣
⎡ (5 − x)(−3) − (12 − 3x)(−1) ⎤
P′T = 2000 ⎢1 +
⎥
(5 − x) 2
⎣⎢
⎦⎥
⎡
3 ⎤
= 2000 ⎢1 −
⎥
2
⎣⎢ (5 − x ) ⎦⎥
39. P ( j ) = Aj
L4
V 3 L2
+B
V
1+ j
dP AL4 BV 3 L2
=
−
=0
dj
V
(1 + j )2
⎡ x 2 − 10 x + 22 ⎤
= 2000 ⎢
⎥
2
⎢⎣ (5 − x)
⎥⎦
Setting P′T = 0 yields (by the quadratic
formula)
10 ± 2 3
x=
= 5± 3
2
Because x ≤ 4, choose x = 5 − 3 . Since PT is
Solving for (1 + j ) 2 gives (1 + j ) 2 =
40. a.
BV 4
AL2
d ⎛
2al ⎞
2al
−2atr + v −
= 1+
= 0 when
⎜
⎟
dv ⎝
v ⎠
v2
v = −2al . Note that
)
increasing on ⎡ 0, 5 − 3 and decreasing on
⎣
5 − 3, 4 ⎤ , PT is a maximum for x = 5 − 3
⎦
tons. If profit on A is P/ton and profit on B is
(
48 − x 2
d2 ⎛
2al ⎞ −4al
−2atr + v −
=
> 0 for
2⎜
v ⎟⎠
dv ⎝
v3
521
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
a < 0, l > 0, and v > 0. Thus −2atr + v −
42. The profit function is given by
2al
v
P = TR − TC = q3 − 20q 2 + 160q − (30q + 50)
= q3 − 20q 2 + 130q − 50
where P is in thousands of dollars, q is in tons,
and 0 ≤ q ≤ 12. From the graph, the maximum
profit occurs when q = 12 tons. The
corresponding maximum profit is $358,000 and
the selling price per ton is $64,000.
is a minimum for v = −2al .
b.
v = −2(−19.6)(20) = 784 = 28 ft/s.
c.
N=
−2(−19.6)
(−2)(−19.6)(0.5) + 28 −
2( −19.6)(20)
28
500
≈ 0.5 cars/s = 0.5(3600) cars/h = 1800
cars/h
d. When v = −2al , then
−2a
N = N (l ) =
−2atr + −2al +
−2a
=
–100
a
=
−2atr + 2 −2al atr − −2al
The relative change in N when l is reduced
N (15) − N (20)
.
from 20 ft to 15 ft is
N (20)
Chapter 13 Review Problems
3x 2
x 2 − 16 ( x + 4)( x − 4)
When x = ±4 the denominator is zero and the
numerator is not zero. Thus x = 4 and x = –4 are
vertical asymptotes.
3x 2
3x 2
lim
= lim
= lim 3 = 3
x →∞ x 2 − 16 x →∞ x 2
x →∞
Similarly, lim y = 3 . Thus y = 3 is the only
1. y =
With a = −19.6 ft/s 2 and tr = 0.5 s, then
−19.6
N (20) =
(−19.6)(0.5) − −2(−19.6)(20)
≈ 0.5185
−19.6
N (15) =
(−19.6)(0.5) − −2(−19.6)(15)
≈ 0.5756
The relative change is
N (15) − N (20) 0.5756 − 0.5158
≈
≈ 0.1101
N (20)
0.5158
41. c =
=
(
=
x →−∞
x+3
x
3
(3 − x)
9 x − 3x
When x = 0 or x = 3, the denominator is zero and
the numerator is not zero. Thus x = 0 and x = 3
are vertical asymptotes.
x
1
1
lim y = lim
= − lim = 0
3 x →∞ x
x →∞
x →∞ −3 x 2
Similarly, lim y = 0. Thus y = 0 is the only
2. y =
dc
18 120 3q 2 − 18q − 120
= 3− −
=
dq
q q2
q2
=
3x 2
horizontal asymptote.
c
120
= 3q + 50 − 18ln(q ) +
,q>0
q
q
3 q 2 − 6q − 40
12
0
−2 al
−2 al
x+3
2
=
x →−∞
)
horizontal asymptote.
2
q
3(q − 10)(q + 4)
3. y =
5x2 − 3
=
5x2 − 3
(3x + 2) 2 9 x 2 + 12 x + 4
2
When x = − , the denominator is zero and the
3
2
numerator is not zero. Thus x = − is a vertical
3
asymptote.
2
q
Critical value is q = 10 since q ≥ 0.
dc
dc
Since
< 0 for 0 < q < 10, and
> 0 for
dq
dq
q > 10, we have a minimum when q = 10 cases.
This minimum average cost is
3(10) + 50 – 18 ln 10 + 12 ≈ $50.55.
522
ISM: Introductory Mathematical Analysis
5x2
Chapter 13 Review
5 5
=
9
x →∞
x →∞ 9 x
x →∞ 9
5
5
Similarly, lim y = . Thus y = is the only horizontal asymptote.
9
9
x →−∞
lim y = lim
4. y =
2
= lim
− x 2 − 30 x − 6
4 x + 1 3x + 1
−
=
3x − 5 2 x − 11 (3x − 5)(2 x − 11)
5
5
11
11
or x =
, the denominator is zero and the numerator is not zero. Thus x = and x =
are
3
3
2
2
vertical asymptotes.
− x2
1
⎛ 1⎞
lim y = lim
= lim ⎜ − ⎟ = −
2
6
x →∞
x →∞ 6 x
x →∞ ⎝ 6 ⎠
When x =
Similarly, lim y = −
x →−∞
5.
f ( x) =
f ′( x) =
1
1
. Thus y = − is the only horizontal asymptote.
6
6
5x2
3 − x2
(3 − x 2 )(10 x) − 5 x 2 (−2 x)
2 2
=
(3 − x )
Thus x = 0 is the only critical value.
(
10 x(3 − x 2 + x 2 )
2 2
(3 − x )
=
30 x
(3 − x 2 ) 2
)
Note: Although f ′ ± 3 is not defined, ± 3 are not critical values because ± 3 are not in the domain of f.
6.
f ( x) = 8( x − 1)2 ( x + 6)4
f ′( x) = 8(2)( x − 1)( x + 6) 4 + 8( x − 1)2 (4)( x + 6)3
= 16( x − 1)( x + 6)3 [ x + 6 + 2( x − 1)]
= 16( x − 1)( x + 6)3 (3x + 4)
Thus x = 1, x = –6, and x = −
7.
f ( x) =
4
are the critical values.
3
3
x +1
3 − 4x
1
−2 ⎤
⎡
(3 − 4 x) ⎢ 13 ( x + 1) 3 ⎥ − ( x + 1) 3 (−4)
⎣
⎦
f ′( x) =
=
(3 − 4 x) 2
2
1 ( x + 1) − 3 [(3 − 4 x) + 12( x + 1)]
3
2
(3 − 4 x)
=
8 x + 15
2
3( x + 1) 3 (3 − 4 x)2
15
3
3
f ′( x) is zero when x = − ; f ′( x) is not defined when x = –1 or x = . However
is not in the domain of f.
8
4
4
15
Thus x = −
and x = –1 are critical values.
8
523
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
− 5x
8.
13xe 6
f ( x) =
6x + 5
⎡ ⎛
⎤
− 5x ⎞
− 5x
− 5x
(6 x + 5) ⎢ x ⎜ − 56 e 6 ⎟ + e 6 (1) ⎥ − xe 6 (6)
⎠
⎣ ⎝
⎦
f ′( x ) = 13 ⋅
(6 x + 5) 2
13 −e
= ⋅
6
− 56x
{(6 x + 5)[5 x − 6] + 36 x}
(6 x + 5)2
(
=
)
{
}
2
13 − 30 x + 25 x − 30
⋅
5x
6
e 6 (6 x + 5) 2
2
13 −5 6 x + 5 x − 6
−65(2 x + 3)(3x − 2)
= ⋅
=
5x
5x
6
2
e 6 (6 x + 5)
6e 6 (6 x + 5) 2
2
5
5
3
or x = . Although f ′( x) is not defined when x = − , − is not in the
3
6
6
2
2
3
domain of f. Thus x = − and x = are the only critical values.
3
2
f ′( x) is zero when x = −
9.
5
f ( x) = − x3 + 15 x 2 + 35 x + 10
3
f ′( x) = −5 x 2 + 30 x + 35
= −5( x 2 − 6 x − 7) = −5( x − 7)( x + 1)
CV: x = −1 and x = 7. Decreasing on (−∞, −1) and (7, ∞); increasing on (−1, 7)
10.
2 x2
f ( x) =
( x + 1) 2
4 x( x + 1)2 − 2 x 2 (2)( x + 1)
f ′( x) =
4x
=
( x + 1)
( x + 1)3
CV: x = 0, but x = –1 is also considered in the inc.-dec. analysis. Increasing on (–∞, –1) and
(0, ∞); decreasing on (–1, 0).
11.
f ( x) =
4
6x4
x2 − 3
( x − 3)( 4 x ) − x (2 x)
f ′( x) = 6 ⋅
( x − 3)
12 x ⎡ 2 ( x − 3) − x ⎤ 12 x ( x − 6 )
⎣⎢
⎦⎥ =
=
( x − 3)
( x − 3)
2
3
2
2
3
2
2
=
(
4
2
2
3
2
2
2
)( x − 6 )
2
3 )( x − 3 ) ⎤
⎦
12 x3 x + 6
(
⎡ x+
⎣
CV: x = 0, ± 6 , but x = ± 3 must also be considered in the inc.-dec. analysis. Decreasing on
( −∞, − 6 ) , ( 0, 3 ) , and (
)
(
)(
)
3, 6 ; increasing on − 6, − 3 , − 3, 0 and
524
(
)
6, ∞ .
ISM: Introductory Mathematical Analysis
12.
Chapter 13 Review
1
. Concave
2
1⎞
⎛
⎛1
⎞
down on ⎜ −∞, ⎟ ; concave up on ⎜ , ∞ ⎟ .
2
2
⎝
⎠
⎝
⎠
3
f ( x ) = 4 5 x3 − 7 x
f ′′( x) is not defined when x =
1
f ′( x) = 4 ⋅ (5 x3 − 7 x) −2 / 3 (15 x 2 − 7)
3
4(15 x 2 − 7)
=
3(5 x3 − 7 x)2 / 3
=
=
4
(
15 x + 7
)(
15 x − 7
2
(
3⎡x (
⎣
4
16.
)
f ′( x) = 3x 2 + 4 x − 5
f ′′( x) = 6 x + 4 = 2(3 x + 2)
2/3
3[ x (5 x − 7)]
)( 15x − 7 )
2/3
7 )( 5 x − 7 ) ⎤
⎦
15 x + 7
5x +
2
. Concave down on
3
2⎞
⎛
⎛ 2
⎞
⎜ −∞, − 3 ⎟ ; concave up on ⎜ − 3 , ∞ ⎟ .
⎝
⎠
⎝
⎠
f ′′( x) = 0 when x = −
7
7
, 0, ±
15
5
⎛
7⎞ ⎛ 7
7 ⎞
Increasing on ⎜⎜ −∞, −
⎟⎟ , ⎜⎜ − , −
⎟⎟ ,
5
5
15
⎝
⎠ ⎝
⎠
⎛ 7
⎛ 7
⎞
7⎞
,
, ∞ ⎟⎟ ; decreasing on
⎜⎜
⎟⎟ , and ⎜⎜
⎝ 15 5 ⎠
⎝ 5
⎠
⎛
⎛
7 ⎞
7 ⎞
, 0 ⎟⎟ and ⎜⎜ 0,
⎜⎜ −
⎟.
15 ⎟⎠
⎝ 15 ⎠
⎝
CV: x = ±
13.
17.
= 3(2 x + 1)2 (2 x + 1 + 6 x + 4)
= 3(2 x + 1)2 (8 x + 5)
f ′′( x) = 3{(2 x + 1) 2 (8) + (8 x + 5)[2(2 x + 1)(2)]}
= 12(2 x + 1)[2(2 x + 1) + 8 x + 5]
= 12(2 x + 1)(12 x + 7)
f ( x) = x 4 − x3 − 14
1
7
or x = − . Concave
2
12
7⎞
⎛
⎛ 1
⎞
up on ⎜ −∞, − ⎟ and ⎜ − , ∞ ⎟ ; concave
12 ⎠
⎝
⎝ 2
⎠
1⎞
⎛ 7
down on ⎜ − , − ⎟ .
2⎠
⎝ 12
f ′′( x) = 0 when x = −
f ′′( x) = 12 x 2 − 6 x = 6 x(2 x − 1)
1
. Concave up on
2
⎛1
⎞
⎛ 1⎞
(–∞, 0) and ⎜ , ∞ ⎟ ; concave down on ⎜ 0, ⎟ .
⎝2
⎠
⎝ 2⎠
f ′′( x) = 0 when x = 0 or x =
f ( x) =
f ′( x) =
x−2
x+2
( x + 2)(1) − ( x − 2)(1)
f ′′( x) = −
( x + 2)2
18.
=
)
(
2
)
(
)
= 2 2 x3 − 3 x 2 − x + 1
4
( x + 2) 2
(
)
f ′′( x) = 2 6 x 2 − 6 x − 1
8
f ′′( x) = 0 when 6 x 2 − 6 x − 1 = 0 ; by the
quadratic formula x =
on (−∞, −2); concave down on (−2, ∞)
1
15
. Concave up on
±
2
6
⎛
⎛1
⎞
1
15 ⎞
15
, ∞ ⎟ ; concave
⎜⎜ −∞, −
⎟⎟ and ⎜⎜ +
⎟
2
6 ⎠
6
⎝
⎝2
⎠
1
= (2 x − 1) −1
f ( x) =
2x −1
⎛1
15 1
15 ⎞
down on ⎜ −
, +
⎟.
⎜2
6 2
6 ⎟⎠
⎝
f ′( x) = −2(2 x − 1) −2
f ′′ = 8(2 x − 1)3 =
(
f ( x) = x 2 − x − 1
f ′( x) = 2 x 2 − x − 1 (2 x − 1)
( x + 2)3
f ′′( x) is not defined when x = −2. Concave up
15.
f ( x) = (2 x + 1)3 (3 x + 2)
f ′( x) = (2 x + 1)3 (3) + (3 x + 2)[3(2 x + 1) 2 (2)]
f ′( x) = 4 x3 − 3x 2
14.
f ( x ) = x3 + 2 x 2 − 5 x + 2
8
(2 x − 1)3
525
Chapter 13: Curve Sketching
19.
ISM: Introductory Mathematical Analysis
f ( x) = 2 x3 − 9 x 2 + 12 x + 7
(
f ′( x) = 6 x 2 − 18 x + 12 = 6 x 2 − 3 x + 2
23.
)
f ( x) =
f ′( x) =
5 23 2 − 13 1 − 13
5x + 2
x + x = x (5 x + 2) =
1
3
3
3
3x 3
2
CV: x = 0 and x = −
5
2⎞
⎛
Increasing on ⎜ −∞, − ⎟ and (0, ∞); decreasing
5⎠
⎝
2x +1
2
⎛ 2 ⎞
on ⎜ − , 0 ⎟ . Relative maximum when x = − ;
5
5
⎝
⎠
relative minimum when x = 0.
x2
x 2 (2) − (2 x + 1)(2 x)
x4
2 x[ x − (2 x + 1)] 2(− x − 1) −2( x + 1)
=
=
=
x4
x3
x3
CV: x = –1, but x = 0 must be considered in inc.dec. analysis. Decreasing on (–∞, –1) and
(0, ∞); increasing on (–1, 0). Relative minimum
when x = –1.
21.
f ( x) =
24.
= x 2 ( x − 2)3 [4 x + 3( x − 2)]
= x 2 ( x − 2)3 (7 x − 6)
CV: x = 0, 2,
x10 x5
+
10
5
f ′( x) =
=
x2
25. y = x5 − 5 x 4 + 3x
x2 − 4
(
)
( x − 4)
2
( )
( x − 4)
2 x ⎡ x2 − 4 − x2 ⎤
⎣⎢
⎦⎥
=−
y′ = 5 x 4 − 20 x3 + 3
x 2 − 4 (2 x) − x 2 (2 x)
2
6
7
⎛ 6⎞
Increasing on (−∞, 0), ⎜ 0, ⎟ , and (2, ∞);
⎝ 7⎠
⎛6 ⎞
decreasing on ⎜ , 2 ⎟ . Relative maximum when
⎝7 ⎠
6
x = ; relative minimum when x = 2.
7
CV: x = 0 and x = −1
Decreasing on (−∞, −1); increasing on (−1, 0)
and (0, ∞); relative minimum when x = −1
f ( x) =
f ( x) = x3 ( x − 2) 4
f ′( x) = x3 [4( x − 2)3 (1)] + ( x − 2)4 (3x 2 )
f ′( x) = x9 + x 4 = x 4 ( x5 + 1)
22.
2
f ′( x) =
= 6(x – 1)(x – 2)
CV: x = 1 and x = 2
Increasing on (–∞, 1) and (2, ∞); decreasing on
(1, 2). Relative maximum when x = 1; relative
minimum when x = 2.
20.
5
2
f ( x) = x 3 ( x + 1) = x 3 + x 3
2
y ′′ = 20 x3 − 60 x 2 = 20 x 2 ( x − 3)
Possible inflection points occur when x = 0 or
x = 3. Concave down on (–∞, 0) and (0, 3);
concave up on (3, ∞). Concavity changes at
x = 3, so there is an inflection point when x = 3.
2
=
−8 x
( x − 4)
2
2
x2 + 2 1
2
= x + x −1
5x
5
5
1
y ′ = (1 − 2 x −2 )
5
4
4
y ′′ = x −3 =
5
5 x3
y ′′ is never zero. Although y ′′ is not defined
when x = 0, y is not continuous there. Thus there
is no inflection point.
26. y =
8x
[( x + 2)( x − 2)]2
CV: x = 0, but x ±2 must be considered in inc.dec. analysis. Increasing on (–∞, –2) and
(–2, 0); decreasing on (0, 2) and (2, ∞). Relative
maximum when x = 0.
526
ISM: Introductory Mathematical Analysis
(
Chapter 13 Review
)
Possible inflections points occur when x = ±2 or
2
2 5
x=±
=±
. Concave up on (–∞, –2),
5
5
27. y = 4(3 x − 5) x 4 + 2 = 12 x5 − 20 x 4 + 24 x − 40
y′ = 60 x 4 − 80 x3 + 24
y ′′ = 240 x3 − 240 x 2 = 240 x 2 ( x − 1)
Possible inflection points occur when x = 0 or
x = 1. Concave down on (–∞, 0) and (0, 1);
concave up on (1, ∞). Inflection point when
x = 1.
28. y = x 2 + 2 ln(− x)
y′ = 2 x +
⎛
2 5⎞
⎜⎜ −2, −
⎟ , and
5 ⎟⎠
⎝
(Note: x < 0)
when x = ±2, ±
2
x
2
y ′′ = 2 −
⎛ 2 5 2 5⎞
,
⎜⎜ −
⎟ , and (2, ∞); concave down on
5
5 ⎟⎠
⎝
2x − 2
=
x3
e
x
2
=
31.
2( x + 1)( x − 1)
= x3e − x
32.
y ′ = x3 (−e− x ) + e− x (3x 2 ) = −e− x ( x3 − 3 x 2 )
= e − x ( x3 − 6 x 2 + 6 x )
= xe− x ( x 2 − 6 x + 6)
y ′′ is defined for all x and y ′′ is zero only when
x = 0 or x 2 − 6 x + 6 = 0. Using the quadratic
formula on the second equation, the possible
points of inflection occur when x = 0, 3 ± 3.
)
(
)
33.
Concave up on 0, 3 − 3 and 3 + 3, ∞ ;
(
)
(
)
(
)
2
(
)
= 36( x + 2)( x − 2)
x
and f is continuous on [–2, 0].
(5 x − 6) 2
(5 x − 6) 2 (1) − x[10(5 x − 6)]
(5 x − 6) 4
(5 x − 6)[(5 x − 6) − 10 x]
(5 x − 6)
4
=
−5 x − 6
(5 x − 6)3
5x + 6
(5 x − 6)3
6
.
5
Evaluating f at this value and at the endpoints
1
1
⎛ 6⎞
and
, f ⎜− ⎟ = −
gives f (−2) = −
128
120
⎝ 5⎠
f(0) = 0. Absolute maximum: f(0) = 0; absolute
1
⎛ 6⎞
.
minimum: f ⎜ − ⎟ = −
5
120
⎝
⎠
The only critical value on (–2, 0) is x = −
) (
)
2
⎧
⎫
y ′′ = 36 ⎨ x ⎡ 4 x x 2 − 4 ⎤ + x 2 − 4 (1) ⎬
⎢
⎥
⎣
⎦
⎩
⎭
2
2
2
2
= 36 x − 4 ⎡ 4 x + x − 4 ⎤ = 36 x − 4 5 x 2 − 4
⎣⎢
⎦⎥
(
f ( x) =
=−
3
y′ = 36 x x 2 − 4
f ( x) = 2 x3 − 15 x 2 + 36 x and f is continuous on
[0, 3].
=
Inflection points when x = 0, 3 ± 3.
30. y = 6 x − 4
f ( x) = 3x 4 − 4 x3 and f is continuous on [0, 2].
f ′( x) =
concave down on (−∞, 0) and 3 − 3, 3 + 3 .
2
2 5
.
5
f ′( x) = 6 x 2 − 30 x + 36 = 6( x − 2)( x − 3)
The only critical value on (0, 3) is x = 2.
Evaluating f at this value and at the endpoints
gives f(0) = 0, f(2) = 28, f(3) = 27. Absolute
maximum: f(2) = 28; absolute minimum: f(0) =
0.
y ′′ = −e− x (3 x 2 − 6 x) − ( x3 − 3 x 2 )(−e− x )
(
⎞
2 ⎟ . Inflection points
⎟
⎠
f ′( x) = 12 x3 − 12 x 2 = 12 x 2 ( x − 1)
The only critical value on (0, 2) is x = 1.
Evaluating f at this value and at the endpoints
gives f(0) = 0, f(1) = –1, and f(2) = 16. Absolute
maximum: f(2) = 16; absolute minimum: f(1) = –1.
x
x
x2
Possible inflection point occurs when x = –1.
Concave up on (–∞, –1); concave down on
(–1, 0). Inflection point when x = – 1.
29. y =
2
2
⎛2 5
,
⎜⎜
⎝ 5
(
(
5x + 2
)
)(
(
5x − 2
)
)(
)
527
Chapter 13: Curve Sketching
34.
ISM: Introductory Mathematical Analysis
f ( x) = ( x + 1)2 ( x − 1) 2 / 3 and f is continuous on [2, 3].
⎡2
⎤
f ′( x) = ( x + 1) 2 ⎢ ( x − 1) −1/ 3 ⎥ + ( x − 1)2 / 3 [2( x + 1)]
⎣3
⎦
2
−1/ 3
[( x + 1) + 3( x − 1)]
= ( x + 1)( x − 1)
3
4
4( x + 1)(2 x − 1)
= ( x + 1)( x − 1) −1/ 3 (2 x − 1) =
3
3( x − 1)1/ 3
There are no critical values on [2, 3]. Evaluating f at the endpoints gives f(2) = 9 and f (3) = 16(22 / 3 ) ≈ 25.4.
Absolute maximum f (3) = 16(22 / 3 ) ≈ 25.4; absolute minimum: f(2) = 9
35.
(
)
f ( x) = x 2 + 1 e− x
a.
( )( )
⎡ x + 1 − 2 x ⎤ = −e
) ⎦⎥
(x
⎣⎢(
f ′( x) = x 2 + 1 −e− x + e− x (2 x)
= −e − x
−x
2
2
)
− 2x + 1
= −e− x ( x − 1)2
CV: x = 1
Decreasing on (–∞, 1) and (1, ∞). No relative extrema.
b.
{
(
f ′′( x) = − e− x [2( x − 1)] + ( x − 1)2 −e− x
)}
= e − x ( x − 1)[−2 + ( x − 1)]
= e − x ( x − 1)( x − 3)
Possible inflection points when x = 1, 3. Concave up on (–∞, 1) and (3, ∞); concave down on (1, 3).
(
)
(
)
Inflection points at (1, f (1)) = 1, 2e−1 and (3, f (3)) = 3, 10e−3 .
36. Let y = f ( x) =
a.
x
2
x −1
.
Replacing x by –x and y by –y yields − y =
x2 + 1
=−
x2 + 1
( x 2 − 1) 2
[( x + 1)( x − 1)]2
decreasing on (−∞, −1), (−1, 1), and (1, ∞).
d.
2
, or y =
x
2
x −1
(− x) − 1
graph is symmetric about the origin. No other symmetry exists.
b. Since f ′( x) = −
c.
−x
, which is the original equation. Thus the
, there are no critical values. f ′( x) < 0 for all x, so f(x) is
From (b), There are no relative extrema.
1
= 0 . Similarly, lim f ( x) = 0 . Thus the line
x →−∞
x →−∞ x
x →∞
y = 0 is a horizontal asymptote to the graph of f.
lim f ( x) = lim
x
x →−∞ x 2
= lim
528
ISM: Introductory Mathematical Analysis
y
e.
Chapter 13 Review
5
y
25
x
x
5
f.
5
From the graph it is clear that no absolute
extrema exist.
39. y = x3 − 12 x + 20
Intercept: (0, 20)
No symmetry; no asymptotes
37. y = x 2 − 2 x − 24 = ( x + 4)( x − 6)
Intercepts: (–4, 0), (6, 0), (0, –24)
No symmetry. No asymptotes.
y′ = 2 x − 2 = 2( x − 1)
CV: x = 1
Increasing on (1, ∞); decreasing on (–∞, 1);
relative minimum at
(1, –25).
y ′′ = 2
No possible inflection point. Concave up on
(–∞, ∞).
25
y′ = 3x 2 − 12
(
)
= 3 x 2 − 4 = 3( x + 2)( x − 2)
CV: x = ±2
Increasing on (–∞, –2) and (2, ∞); decreasing on
(–2, 2); relative maximum at (–2, 36); relative
minimum at (2, 4).
y ′′ = 6 x
Possible inflection point when x = 0. Concave up
on (0, ∞); concave down on (–∞, 0); inflection
point at (0, 20).
y
40
x
25
20
–2
(1, –25)
38. y = 2 x3 + 15 x 2 + 36 x + 9
Intercept: (0, 9)
No symmetry; no asymptotes
2
x
10
40. y = x 4 − 4 x3 − 20 x 2 + 150
Intercept: (0, 150)
No symmetry. No asymptotes.
y ′ = 6 x 2 + 30 x + 36 = 6( x 2 + 5 x + 6)
= 6( x + 3)( x + 2)
y′ = 4 x3 − 12 x 2 − 40 x = 4 x( x 2 − 3 x − 10)
= 4 x( x + 2)( x − 5)
CV: x = 0, –2, 5. Increasing on (–2, 0) and
(5, ∞); decreasing on (–∞, –2) and (0, 5);
relative maximum at (0, 150); relative minima at
(–2, 118) and (5, –225).
CV: x = −3, −2
Increasing on (−∞, −3) and (−2, ∞); decreasing
on (−3, −2); relative maximum at (−3, −18);
relative minimum at (−2, −19)
y ′′ = 12 x + 30 = 6(2 x + 5)
(
y ′′ = 12 x 2 − 24 x − 40 = 4 3x 2 − 6 x − 10
5
Possible inflection point when x = − .
2
5⎞
⎛
Concave down on ⎜ −∞, − ⎟ ; concave up on
2⎠
⎝
⎛ 5
⎞
⎜ − 2 , ∞ ⎟ ; inflection point at
⎝
⎠
y
Possible inflection points when x = 1 ±
⎛
39 ⎞
Concave up on ⎜⎜ −∞, 1 −
⎟ and
3 ⎟⎠
⎝
⎛
⎞
39
, ∞ ⎟⎟ ; concave down on
⎜⎜1 +
3
⎝
⎠
37 ⎞
⎛ 5
⎜− 2 , − 2 ⎟
⎝
⎠
529
)
39
.
3
Chapter 13: Curve Sketching
⎛
⎜⎜1 −
⎝
⎛
⎜⎜1 −
⎝
⎛
⎜⎜1 +
⎝
ISM: Introductory Mathematical Analysis
horizontal asymptote.
5
y′ = −
( x − 3)2
CV: None, but x = 3 must be considered in the
inc.-dec. analysis. Decreasing on (–∞, 3) and
(3, ∞).
10
y ′′ =
( x − 3)3
No possible inflection point, but x = 3 must be
considered in concavity analysis. Concave up on
(3, ∞); concave down on (–∞, 3).
39
39 ⎞
, 1+
⎟ ; inflection points at
3
3 ⎟⎠
⎞
39 298
,
+ 16 39 ⎟⎟ ≈ (−1.08, 133.03) and
3
9
⎠
⎞
39 298
,
− 16 39 ⎟⎟ ≈ (3.08, − 66.81) .
3
9
⎠
300
y
(0, 150)
x
10
5
y
(5, –225)
(
x
8
)
41. y = x3 − x = x x 2 − 1 = x( x + 1)( x − 1)
Intercepts (0, 0), (−1, 0), and (1, 0)
Symmetric about the origin. No asymptotes.
y′ = 3 x 2 − 1 =
CV: ±
(
)(
3x + 1
)
3x − 1
43. y = f ( x) =
x2
Intercept: (–5, 0)
No symmetry.
x = 0 is the only vertical asymptote.
x
1
lim y = 100 lim
= 100 lim = 0 , and
2
x →∞
x →∞ x
x →∞ x
lim y = 0 , so y = 0 is the only horizontal
3
3
⎛
⎛ 3
⎞
3⎞
Increasing on ⎜⎜ −∞, −
, ∞ ⎟⎟ ;
⎟⎟ and ⎜⎜
3 ⎠
⎝
⎝ 3
⎠
⎛
3
3⎞
decreasing on ⎜⎜ −
,
⎟.
3 ⎟⎠
⎝ 3
y ′′ = 6 x
Possible inflection point when x = 0. Concave
down on (–∞, 0); concave up on (0, ∞);
inflection point at (0, 0).
y
x →−∞
asymptote.
y = 100 ⎡ x −1 + 5 x −2 ⎤
⎣
⎦
⎡ 1 10 ⎤
y′ = 100 ⎡ − x −2 − 10 x −3 ⎤ = −100 ⎢ + ⎥
⎣
⎦
⎣ x 2 x3 ⎦
−100( x + 10)
=
x3
CV: x = –10 but x = 0 must be included in inc.dec. analysis. Increasing on (–10, 0); decreasing
on (–∞, –10) and (0, ∞); relative minimum at
(–10, –5).
⎡ 1 15 ⎤
y ′′ = 100 ⎡ 2 x −3 + 30 x −4 ⎤ = 200 ⎢ + ⎥
⎣
⎦
⎣ x3 x 4 ⎦
5
x
5
42. y =
100( x + 5)
x+2
x −3
=
200( x + 15)
x4
Possible inflection point when x = –15, but x = 0
must also be considered in concavity analysis.
Concave up on (–15, 0) and (0, ∞); concave
down on (–∞, –15); inflection point at
2⎞
⎛
Intercepts: ⎜ 0, − ⎟ , (–2, 0)
3⎠
⎝
Vertical asymptote is x = 3.
x+2
x+2
lim
= 1 = lim
, so y = 1 is a
x →∞ x − 3
x →−∞ x − 3
530
ISM: Introductory Mathematical Analysis
Chapter 13 Review
40 ⎞
⎛
⎜ −15, − ⎟
9 ⎠
⎝
20
lim y = lim
x →∞
f(x)
(
=
)
(
y ′′ = −2 ⋅
= −2 ⋅
)
3
(3x − 1)8
6(3 x − 1)3 [(3 x − 1) − 2(6 x + 1)]
1
Possible inflection point when x = − , but
3
1
x = must be considered in concavity analysis.
3
1⎞
⎛
⎛1
⎞
Concave up on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ;
3
3
⎝
⎠
⎝
⎠
⎛ 1 1⎞
concave down on ⎜ − , ⎟ ; inflection point at
⎝ 3 3⎠
⎛ 1 1⎞
⎜ − , ⎟.
⎝ 3 12 ⎠
No possible inflection point, but x = ±1 must be
considered in concavity analysis. Concave up on
(–1, 1); concave down on (–∞, –1) and (1, ∞).
10
(3x − 1)4 (6) − (6 x + 1)[4(3x − 1)3 (3)]
(3 x − 1)8
−12(−9 x − 3) 36(3x + 1)
=
=
(3 x − 1)5
(3 x − 1)5
−6 3 x 2 + 1
( x2 − 1)
2(3x − 1) 2 [(3x − 1) − 9 x]
1
1
CV: x = − , but x = must be considered in
3
6
1⎞
⎛
inc.-dec. analysis. Increasing on ⎜ −∞, − ⎟ ;
6⎠
⎝
⎛ 1 1⎞
⎛1
⎞
decreasing on ⎜ − , ⎟ and ⎜ , ∞ ⎟ ; relative
6
3
3
⎝
⎠
⎝
⎠
⎛ 1 8⎞
maximum at ⎜ − , ⎟ .
⎝ 6 81 ⎠
CV: x = 0 but x = ±1 must also be considered in
inc.-dec. analysis. Increasing on (0, 1) and (1,
∞); decreasing on (–∞, –1) and (–1, 0); relative
minimum at (0, 4).
y ′′ =
2
1
=0
lim
27 x→∞ x 2
=
(3 x − 1)6
2(−6 x − 1) −2(6 x + 1)
=
=
(3x − 1) 4
(3 x − 1)4
( x + 2)( x − 2)
x − 1 ( x + 1)( x − 1)
Intercepts: (0, 4), (2, 0), (–2, 0)
Symmetric about the y-axis. Vertical asymptotes
are x = 1 and x = –1.
x2
lim y = lim
= 1 = lim y , so y = 1 is the
x →−∞
x →−∞ x 2
x →∞
only horizontal asymptote.
6x
y′ =
2
x2 − 1
2
27 x
= lim y,
3
x →−∞
=
x2 − 4
2x
so y = 0 is a horizontal asymptote.
(3x − 1)3 (2) − 2 x[3(3x − 1)2 (3)]
y′ =
(3 x − 1)6
x
20
44. y =
x →∞
y
x
5
y
45. y =
1
2x
(3x − 1)3
Intercept: (0, 0)
No symmetry
x
2
1
Vertical asymptote is x = .
3
531
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
1
46. y = 6 x 3 (2 x − 1)
47.
⎛1 ⎞
Intercepts: (0, 0), ⎜ , 0 ⎟
⎝2 ⎠
No symmetry. No vertical asymptote.
As x → ∞ , both 6x1/ 3 and 2 x − 1 → ∞ . As
1
Setting f ′( x) = 0 ⇒ e x = e− x ⇒ x = − x ⇒ x = 0
CV: x = 0
Increasing on (0, ∞); decreasing on (–∞, 0);
relative minimum at (0, 1). Finding f ′′( x) gives:
x → −∞ , both 6x 3 and 2 x − 1 → −∞ . Thus
lim y = ∞ = lim y . So no horizontal
x →∞
x →−∞
(
)
asymptote exists. Since y = 6 2 x 4 / 3 − x1/ 3 ,
e x + e− x
. f ′′( x) > 0 for all x. No
2
possible inflection point. Concave up on
(–∞, ∞).
1
⎛8
⎞
y′ = 6 ⎜ x1/ 3 − x −2 / 3 ⎟ = 2 x −2 / 3 (8 x − 1)
3
3
⎝
⎠
2(8 x − 1)
=
x2 / 3
1
CV: x = 0,
8
⎛ 1⎞
Decreasing on (–∞, 0) and ⎜ 0, ⎟ ; increasing on
⎝ 8⎠
1
9⎞
⎛
⎞
⎛1
⎜ , ∞ ⎟ ; relative minimum at ⎜ , − ⎟ .
4⎠
⎝8
⎠
⎝8
2
⎛8
⎞ 4
y ′′ = 2 ⎜ x −2 / 3 + x −5 / 3 ⎟ = x −5 / 3 (4 x + 1)
3
3
⎝
⎠ 3
4(4 x + 1)
=
3 x5 / 3
1
Possible inflection points when x = − , 0 .
4
1⎞
⎛
Concave up on ⎜ −∞, − ⎟ and (0, ∞); concave
4⎠
⎝
⎛ 1 ⎞
down on ⎜ − , 0 ⎟ ; inflection points at
⎝ 4 ⎠
3
⎛ 1 9 2⎞
⎜⎜ − ,
⎟⎟ and (0, 0).
⎝ 4 2 ⎠
10
e x + e− x
2
Intercept: (0, 1)
Symmetric about the y-axis. No asymptotes.
e x − e− x
f ′( x) =
2
f ( x) =
f ′′( x) =
3
f(x)
x
3
48. y = f ( x) = 1 − ln( x3 ) = 1 − 3ln x
( )
x-intercept is ( e1/ 3 , 0 ) . Since x ≠ 0, there is no
y = 0 ⇒ ln x3 = 1 ⇒ x3 = e ⇒ x = e1/ 3 , so the
y-intercept. No symmetry. Since lim y = ∞,
x →0+
x = 0 is a vertical asymptote. No horizontal
asymptote.
3
f ′( x) = −
x
CV: None. Decreasing on (0, ∞).
3
f ′′( x) =
x2
No possible inflection points.
Concave up on (0, ∞).
y
y
x
5
1
x
5
532
ISM: Introductory Mathematical Analysis
49. a.
Chapter 13 Review
False. f ′ ( x0 ) = 0 only indicates the
2
b.
possibility of a relative extremum at x0 , For
example, if f ( x) = x3 , then f ′( x) = 3x 2
and f ′(0) = 0 . However there is no relative
extremum at x = 0.
Then x1 < x2 and f ( x1 ) = −1 < f ( x2 ) = 1 .
c.
True. The absolute minimum is f(0) = 0 and
the absolute maximum is f(1) = 1.
3
f(x)
1
d.
x
–1
1
f ′( x) =
2π
f ′ is defined for all x; f ′( x) = 0 only when
x = 0. Thus x = 0 is a critical value. If x < 0,
then f ′( x) > 0 ; if x > 0 then f ′( x) < 0 .
3
From (b), f has a relative maximum when
x = 0. The coordinates of this relative
⎛
1 ⎞
maximum are ⎜ 0,
⎟.
2π ⎠
⎝
x →−∞
e.
=
x0 = 0 , then f ′′ ( x0 ) = 0 , but ( x0 , f ( x0 ) )
− x2
− x2
− x2
1
=
=
2π
1
(0) = 0
(0) = 0
2π
2
2
⎤
1 ⎡ − x2
−x
⎢ xe (− x) + e 2 (1) ⎥
2π ⎣⎢
⎦⎥
( x2 − 1) = e−
x2
2
( x + 1)( x − 1)
2π
2π
′′
f is defined for all x; f ′′( x) = 0 when
is not an inflection point. See graph in part
(c).
x = ±1. f is concave up on (–∞, –1) and
(1, ∞); f is concave down on (–1, 1).
False. Consider the function f whose graph
is shown. On (–2, 2) it has exactly one
relative maximum [at the point (0, 1)] but no
absolute maximum.
3
e
f ′′( x) = −
e
e
2
2π
2
point. For example, consider f ( x) = x 4 . If
2π
1
x →∞
d. False. If concavity does not change around
x0 , then ( x0 , f ( x0 ) ) is not an inflection
2
1
lim
lim
e.
− x2
Thus f is increasing on (–∞, 0) and is
decreasing on (0, ∞).
b. False. For example, let x1 = −1 and x2 = 1 .
c.
− xe
f.
f(x)
x
3
From (e), f changes concavity at x = ±1.
Also f is continuous there. Thus f has
inflection points at x = ±1; the coordinates
−1 ⎞
−1 ⎞
⎛
⎛
e 2 ⎟
e 2 ⎟
⎜
⎜
and 1,
are −1,
⎜
⎜
2π ⎟
2π ⎟
⎝
⎠
⎝
⎠
g.
f(x)
1
x
50. Let y = f ( x) =
a.
1
2π
2
e
− x2
3
.
h. Absolute maximum: f (0) =
Replacing x by –x yields the original
equation. Thus the graph is symmetric about
the y-axis. No other symmetry exists.
No absolute minimum.
533
1
2π
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
54. a.
51. c = q3 − 6q 2 + 12q + 18
dc
= 3q 2 − 12q + 12 . Marginal
dq
cost is increasing when its derivative, which is
d 2c
, is positive.
dq 2
b. R(0.5) ≈ 18.5%
Marginal cost =
d 2c
dq
2
c.
d 2R
= 6q − 12 = 6(q − 2)
R
1
x
52. r = 320q3 / 2 − 2q 2
1
dr
= 480q1/ 2 − 4q.
dq
Marginal revenue is increasing when its
d 2r
is positive.
derivative, which is
dq 2
Marginal revenue =
dq 2
d 2r
2
= 240q
=0⇒
240
−4 =
240
q
55.
f (t ) = At 3 + Bt 2 + Ct + D
f ′(t ) = 3 At 2 + 2 Bt + C
f ′′(t ) = 6 At + 2 B , which gives an inflection
B
.
3A
This value of a must be such that f ′(a) = 0 .
point when 6At + 2B = 0, that is for a = −
−4
2
⎛ B ⎞
⎛ B ⎞
3A ⎜ −
⎟ + 2B ⎜ −
⎟+C = 0
⎝ 3A ⎠
⎝ 3A ⎠
1 ⎛ B2 ⎞ 2 ⎛ B2 ⎞
⎜
⎟− ⎜
⎟+C = 0
3 ⎜⎝ A ⎟⎠ 3 ⎜⎝ A ⎟⎠
1 ⎛ B2 ⎞
C= ⎜
⎟
3 ⎜⎝ A ⎟⎠
− 4 = 0 ⇒ 240 = 4 q
q
dq
⇒ q = 60 ⇒ q = 3600
d 2r
> 0 for 0 < q < 3600. Thus marginal
dq 2
revenue is increasing on (0, 3600).
53. p = 200 −
=
29.92
> 0 for 0 ≤ x ≤ 1.
dx
(4.4 − 3.4 x)3
We obtain the following graph:
2
> 0 for q > 2. Thus marginal cost is
dq 2
increasing for q > 2.
−1/ 2
R(1) = 100%
dR
4.4
=
> 0 for 0 ≤ x ≤ 1
dx (4.4 − 3.4 x)2
d 2c
d 2r
R(0) = 0%
3AC = B 2 ,
which was to be shown.
q
, q > 0. The revenue function r is
5
given by
56. a.
3
⎛
q⎞
q2
r = pq = ⎜ 200 −
q = 200q −
.
⎟
⎜
5 ⎟⎠
5
⎝
3 1
r ′ = 200 − q 2
10
3 −1
3
r ′′ = − q 2 = −
20
20 q
Let a = p + q. Then S becomes
⎡
⎤
⎥
ma 2 ⎢
e − at
S=
⎢
⎥
2
p ⎢ q − at
⎥
e
+
1
⎣⎢ p
⎦⎥
(
Since r ′′ < 0 for q > 0, the graph of the revenue
function is concave down for q > 0.
534
)
ISM: Introductory Mathematical Analysis
⎡
dS ma ⎢
=
⎢
dt
p ⎢
⎢⎣
2
(
(
b.
+1
2
− at
( e
+ 1)
+ 1) + 2
e + 1)
− at
⎡2
⎣⎢
q − at
p
ma 2
=
ae − at
p
=
) ( −ae ) − e
( e
⎛q
⎞ −( e
e
+
1
) ⎜⎝ p
⎟
⎠
(
q − at
e
p
ma3 − at
=
e
p
Chapter 13 Review
(
q − at
p
)(
+1
− aq − at
e
p
4
q − at
e
p
4
)
⎡ q e−( p + q )t − 1⎤
⎢⎣ p
⎥⎦
q − ( p + q )t
e
p
)
+1
3
dS
= 0 when
dt
⎡q
⎤
m
( p + q )3 e−( p + q )t ⎢ e− ( q + p )t − 1⎥ = 0
p
⎣p
⎦
Since m, p + q, and e −( p + q )t are nonzero, we must have
q − ( p + q )t
−1 = 0
e
p
e − ( p + q )t =
p
q
⎛ p⎞
−( p + q )t = ln ⎜ ⎟
⎝q⎠
() ()
p
q
ln q
ln p
t=−
=
p+q
p+q
15
57.
–5
)⎤⎦⎥ ⎤⎥⎥
⎥
⎥⎦
q − at
e a −1
p
3
q − at
e +1
p
m ( p + q )3 e − ( p + q ) t
p
(
q − at
p
− at
q − at
p
5
–5
Relative maximum (–1.32, 12.28); relative minimum (0.44, 1.29)
535
Chapter 13: Curve Sketching
58.
ISM: Introductory Mathematical Analysis
5
63. p = 500 − q , where 100 ≤ q ≤ 200.
Total revenue = r = pq = q 500 − q
–1
⎛1⎞
r ′ = q ⎜ ⎟ (500 − q )−1/ 2 (−1) + 500 − q (1)
⎝2⎠
1
= (500 − q )−1/ 2 [− q + 2(500 − q)]
2
1000 − 3q
=
2 500 − q
1
–5
Maximum: (1, 1); minimum: (−0.60, −2.24)
5
59.
=
–5
(
3 1000
−q
3
)
2 500 − q
No critical values on (100, 200). r(100) = 2000;
r(200) ≈ 3464, so 200 units should be produced
for maximum revenue.
5
–5
The x-value of the inflection point of f
corresponds to the x-intercept of f ′′ . Thus the
64. c = 0.01q 2 + 5q + 100
Avg. cost c =
x-value of the inflection point is x ≈ –0.60.
60.
dc
100
1 100 q 2 − 1002
= 0.01 −
=
−
=
dq
q 2 100 q 2
100q 2
5
=
10
–10
c
100
= 0.01q + 5 +
q
q
(q − 100)(q + 100)
100q 2
We find that c is decreasing on (0, 100) and
increasing on (100, ∞), so average cost is
minimum when q = 100.
–5
Horizontal asymptote y = 0;
vertical asymptote x ≈ –0.25
65. p = 500 − 3q
61. q = 80m2 − 0.1m 4
c = q + 200 +
(
)
dq
= 160m − 0.4m3 = 0.4m 400 − m2
dm
dq
= 0.4m(20 + m)(20 − m) . Setting
= 0 yields
dm
m = 0 or m = 20 (for m ≥ 0). We find that q is
increasing on (0, 20) and decreasing on (20, ∞),
so q is maximum at m = 20.
1000
q
Total Cost = c = cq = q 2 + 200q + 1000
Profit = Total Revenue – Total Cost
P = pq − c = (500 − 3q )q − (q 2 + 200q + 1000)
= −4(q 2 − 75q + 250)
P ′ = −4(2q − 75)
Setting P ′ = 0 yields q = 37.5. Since
P ′′ = −8 < 0, P is maximum when q = 37.5. In
reality, whole units are likely. Since
P(37) = P(38) = 4624, the maximum profit is
$4624.
62. p = 100e −0.1q
Total revenue = r = pq = 100qe−0.1q
r ′ = 100[e −0.1q (1) + q(−0.1)e−0.1q ]
= 10e −0.1q (10 − q )
r ′ = 0 when q = 10. Since r is increasing when
q < 10 and decreasing when q > 10, revenue is
maximized when q = 10.
66. V = (10 – 2x)(16 – 2x)x
(
= 4 x3 − 13x 2 + 40 x
Note: 0 < x < 5.
536
)
ISM: Introductory Mathematical Analysis
(
V ′ = 4 3x 2 − 26 x + 40
Chapter 13 Review
)
= 4(x – 2)(3x – 20)
20
. On (0, 5), x = 2 is the only critical value. At x = 2 in.,
3
V ′′ = 4(6 x − 26) = 4(12 − 26) = −56 < 0 , so V is maximum at x = 2 in.
Setting V ′ = 0 gives x = 2 or x =
x
x
x
x
10 – 2x
x
x
x
x
16 – 2x
67. 2x + 4y = 800; thus x = 400 – 2y
Area = A = xy = (400 − 2 y ) y
= 400 y − 2 y 2
dA
= 400 − 4 y = 4(100 − y )
dy
d2A
dA
= −4 < 0 , A is maximum when y = 100. When y = 100, then x = 200.
= 0 gives y = 100. Since
dy
dy 2
The dimensions are 200 ft by 100 ft.
Setting
y
y
y
y
x
500
x
Printed area = A = (x – 8)(y – 10)
⎛ 500
⎞
= ( x − 8) ⎜
− 10 ⎟
x
⎝
⎠
68. xy = 500, so y =
4000
,x>0
x
4000
= 580 − 10x −
A′ = −10 +
x2
Setting A′ = 0 gives x = 20. When x = 20, A′′ = −
8000
x
3
Thus the dimensions are 20 in. by 25 in.
6
y
4
4
4
x
537
< 0 , so A is maximum. When x = 20, then y =
500
= 25 .
20
Chapter 13: Curve Sketching
ISM: Introductory Mathematical Analysis
c = 2q3 − 9q 2 + 12q + 20 , where
69. a.
(
3
≤ q ≤ 6.
4
)
dc
= 6q 2 − 18q + 12 = 6 q 2 − 3q + 2
dq
= 6(q – 1)(q – 2)
dc
Setting
= 0 gives q = 1 or 2. Evaluating c at these critical values and the endpoints:
dq
⎛ 3 ⎞ 793
c⎜ ⎟ =
≈ 24.78 , c(1) = 25, c(2) = 24, c(6) = 200. Thus a minimum occurs at q = 2, which corresponds
⎝ 4 ⎠ 32
24, 000
= $120 .
to 200 stands and a total cost of $24,000. This gives an average cost per stand of
200
b. There are no critical values of c in 3 ≤ q ≤ 6, so we only evaluate c at the endpoints:
c(3) = 29, c(6) = 200. Thus a minimum occurs at q = 3, which gives 300 stands.
70. N =
N′ =
=
12,100 + 110t + 100t 2
121 + t 2
, where t ≥ 0.
(121 + t 2 )(110 + 200t ) − (12,100 + 110t + 100t 2 )(2t )
(121 + t 2 ) 2
110(121 − t 2 )
121 + t 2
Setting N ′ = 0 gives t = 11, from which N = 105. Since N ′ > 0 for 0 ≤ t < 11 and N ′ < 0 for t > 11, there is an
absolute maximum when t = 11.
Mathematical Snapshot Chapter 13
1. Figure 13.74 does not readily show how long it takes for the population to reach its final size. Figure 13.75 shows
that this takes about 45 days.
2. The population declines until it stabilizes between 311 and 312 (as can be verified by inspecting the final state of
dP
< 0 for all P ≥ 312 .
the variable P). This is consistent with the fact that
dt
3. Even if the graph starts out exactly coinciding with the ideal curve, a line segment tangent to the curve at one end
must (in general) lie slightly off the curve at the other end. This introduces errors that accumulate over successive
iterations. The amount of cumulative error could be reduced by taking smaller time steps, such as 1 month instead
of 1 year, and correspondingly drawing shorter line segments.
538
Chapter 14
Problems 14.1
11. ∆y = [4 – 7(3.02)] – [4 – 7(3)] = –0.14
dy = –7 dx = –7(0.02) = –0.14
1. y = 5x – 7
d
dy =
(5 x − 7)dx = 5 dx
dx
12. ∆y = ⎡5(−1.02) 2 − 5(−1)2 ⎤ = 0.202
⎣
⎦
dy = 10x dx = 10(–1)(–0.02) = 0.2
2. dy = y′dx = 0 dx = 0
13. ∆y
= [2(−1.9)2 + 5(−1.9) − 7] − [2(−2) 2 + 5(−2) − 7]
= −0.28
dy = (4x + 5)dx = [4(−2) + 5](0.1) = −0.3
1
−1
3. d [ f ( x)] = f ′( x)dx = ( x 4 − 9) 2 (4 x3 )dx
2
2 x3
dx
=
x4 − 9
4. d [ f ( x)] = f ′( x)dx
(
= 3(8 x − 5) 4 x 2 − 5 x + 2
5. u = x
du =
)
2
14. ∆y = [3(−1.03) + 2] − [3(−1) + 2]2 = 0.1881
dy = 6(3x + 2) dx = 6[3(–1) + 2](–0.03) = 0.18
2
15. ∆y = 32 − (3.95) 2 − 32 − (42 ) ≈ 0.049
dx
( )
d −2
2
x
dx = −2 x −3 dx = − dx
dx
x3
du = u′dx = −
1 −3 / 2
x
dx
2
(
17. a.
)
d ⎡
1
ln x 2 + 7 ⎤ dx =
(2 x)dx
⎢
⎥
2
⎣
⎦
dx
x +7
2x
=
dx
2
x +7
7. dp =
f ′(1) =
+3
= 3e 2 x
2
+3
=
18. a.
[(3x + 1)(4 x) + 3]dx
(12x
2
dx =
−4
(−0.05) = 0.050
16
x+5
x +1
( x + 1)(1) − ( x + 5)(1)
( x + 1)
2
=
−4
( x + 1) 2
−4
= −1
4
b. We use f(x + dx) ≈ f(x) + dy with x = 1,
dx = 0.1.
f (1.1) = f (1 + 0.1) ≈ f (1) + f ′(1)dx
9. dy = y′dx
2
2
= ⎡(9 x + 3)e2 x +3 (4 x) + e2 x +3 (9) ⎤ dx
⎢⎣
⎥⎦
2
f ( x) =
f ′( x) =
d ⎛ x 3 + 2 x −5 ⎞
2
x3 + 2 x −5
dx
⎜e
⎟ dx = (3x + 2)e
dx ⎝
⎠
= 3e 2 x
32 − x
2
16. ∆y = ln 4.9 – ln 5 ≈ –0.0202
1
1
1
dy =
(−1)dx = dx =
(0.1) = −0.02
−x
x
−5
6. u = x −1/ 2
8. dp =
−x
dy =
−2
)
+ 4 x + 3 dx
1
ln( x 2 + 12)
2
1
1
x
dy = ⋅
(2 x )dx =
dx
2
2 x 2 + 12
x + 12
10. y = ln x 2 + 12 =
6
+ (−1)(0.1) = 2.9
2
y = f ( x ) = x3 x
Using logarithmic differentiation,
ln y = 3x ln x,
1 dy
⎛1⎞
⋅
= 3 x ⎜ ⎟ + (ln x)(3) = 3(1 + ln x)
y dx
⎝x⎠
dy
= y[3(1 + ln x)] = 3 x3 x (1 + ln x)
dx
f ′(1) = 3(1)(1 + 0) = 3
539
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
23. Let y = f(x) = ln x
b. We use f(x + dx) ≈ f(x) + dy with x = 1,
dx = –0.02
f (0.98) = f (1 − 0.02) ≈ f (1) + f ′(1)dx
f ( x + dx) ≈ f ( x) + dy = ln( x) +
= 13 + (3)(−0.02) = 0.94
If x = 1 and dx = –0.03, then
ln(0.97) = f (1 + (−0.03))
1
≈ ln(1) + (−0.03) = −0.03
1
19. Let y = f ( x) = x
f ( x + dx) ≈ f ( x) + dy = x +
1
dx
2 x
24. Let y = f(x) = ln x
If x = 289 and dx = ⫺1, then
288 = f (289 − 1)
1
≈ 289 +
(−1)
2 289
577
=
34
≈ 16.97
f ( x + dx) ≈ f ( x) + dy = ln( x) +
1
ln1.01 = f (1 + 0.01) ≈ ln(1) + (0.01) = 0.01
1
25. Let y = f ( x) = e x
f ( x + dx) ≈ f ( x) + dy = x +
f ( x + dx) ≈ f ( x) + dy = e x + e x dx
If x = 0 and dx = 0.001, then
1
dx
2 x
e0.001 = f (0 + 0.001) ≈ e0 + e0 (0.001) = 1.001
If x = 121 and dx = 1, then
= 11
1
2 121
26. Let y = f ( x) = e x
(1)
f ( x + dx) ≈ f ( x) + dy = e x + e x dx
If x = 0 and dx = –0.01, then
1
.
22
e −0.01 = f (0 + (–0.01)) ≈ e0 + e0 (−0.01) = 0.99
21. Let y = f ( x) = 3 x
1
65.5 = f (64 + 1.5) ≈ 3 64 +
1.5
3 ⋅ 42
=4
dx
2
( )
2
3 3 64
22. Let y = f ( x) = 4 x .
1
4x
If x = 16 and dx = 0.3, then
4 16.3
4
0.3
3
2
29.
dq
= 6 p p2 + 5
dp
30.
dq
1
dp
=
, so
= 2 p+5
dp 2 p + 5
dq
(
31. q = p −1 ,
( )
4 16
=2
dy
dx
1
= 10 x + 3, so
=
dx
dy 10 x + 3
dx
1
= f (16 + 0.3) ≈ 4 16 +
= 2+
3
4
28.
(1.5)
1
32
f ( x + dx) = f ( x) + dy = 4 x +
dx
1 1
dy
= 2 , so
=
=
dy
dy
2
dx
dx
3x 3
If x = 64 and dx = 1.5, then
= 4+
27.
1
f ( x + dx ) ≈ f ( x) + dy = 3 x +
3
1
dx
x
If x = 1 and dx = 0.01, then
20. Let y = f ( x) = x
122 = f (121 + 1) ≈ 121 +
1
dx
x
3
(0.3)
32.
3
320
540
)
2
, so
dp
=
dq
1
(
6 p p2 + 5
)
2
dq
−1
dp
= −1 p −2 =
= − p2
, so
2
dp
dq
p
dq
dp
1
1
= −2e4− 2 p , so
=−
= − e2 p −4
−
p
4
2
dp
dq
2
2e
ISM: Introductory Mathematical Analysis
33.
Section 14.1
dx
1
1
=
=
dy dy 14 x − 6
39. p =
10
q
dx
dx 1
=
dy 36
If x = 3,
34.
p(q + dq ) ≈ p + dp =
x
dx 3
= .
If x = 3,
dy 2
35. p =
= 2+
500
q+2
q =18
(q + 2) 2
500
=−
q =18
4
5
36. p = 50 − q
dp
1
=−
dq
2 q
5
q3
dq
1
51
=
= 2.04
25 25
q +8
q4
+ 3q + 400
2
If q = 10 and dq = 2,
41. c =
dq
= −2 q
dp
dq
dp
−
We approximate p when q = 40.
200
100
p(q + dq ) ≈ p + dp =
−
dq
3
q + 8 (q + 8) 2
If q = 41 and dq = 1, then
200
100
p(40) = p(41 − 1) ≈
−
(1)
3
49 (49) 2
200 100 9700
=
−
=
≈ 28.28
7
343 343
dq
(q + 2) 2
=−
dp
500
=−
q
200
40. p =
dp
−500
=
dq (q + 2) 2
dq
dp
10
If q = 25 and dq = –1, then
10
5
−
p(24) = p(25 + (−1)) ≈
(−1)
25
(25)3
dx
1
1 x
=
= =
dy dy 2 2
dx
. We approximate p when q = 24.
= −2 q
q =100
q =100
= −20
3
4
q
2
=
+ 3q + 400
(2003)(2)
≈ 0.7
5430
42. S = 20 I , I decreases from 45 to 44
37. P = 397 q − 2.3q 2 − 400, q changes from 90 to
91.
∆P ≈ dP = P ′dq = (397 − 4.6q )dq
Choosing q = 90 and dq = 1,
∆P ≈ [397 − 4.6(90)](1) = −17.
True change is
P(91) − P(90) = 16,680.7 − 16,700 = −19.3.
∆S ≈ dS = S ′dI =
10
I
dI
Choosing I = 45 and dI = −
∆S ≈
38. r = 250q + 45q 2 − q3 , q increases from 40 to 41.
(
( 2q + 3) dq
dc
=
c
)
43. V =
∆r ≈ dr = r ′dq = 250 + 90q − 3q 2 dq
Choosing q = 40 and dq = 1,
∆r ≈ (–950)(1) = –950
True change is
r(41) – r(40) = 16,974 – 18,000 = –1026
1
, then
2
10 ⎛ 1 ⎞
⎜ − ⎟ ≈ −0.745.
45 ⎝ 2 ⎠
4 3
πr
3
∆V ≈ dV = V ′dr = 4πr 2 dr
(
) (
dr = 6.6 × 10−4 − 6.5 × 10−4
= 0.1× 10−4 = 10−5
541
)
1
.
2
Chapter 14: Integration
(
∆V ≈ 4π 6.5 × 10−4
ISM: Introductory Mathematical Analysis
) (10 ) = (1.69 ×10 ) π cm
2
−5
−11
3
.
44. (P + a)(v + b) = k
k
P=
−a
v+b
dP = −k (v + b)−2 dv
45. a.
We substitute q = 40 and p = 20
402 4000
=
2+
200 202
2 + 8 = 10
10 = 10
b. We differentiate implicitly with respect to p.
1 ⎛ dq ⎞
8000
0+
⎜ 2q ⎟ = − 3
200 ⎝ dp ⎠
p
From part (a) q = 40 when p = 20. Substituting gives
dq ⎞
1 ⎛
8000
⎜ 2 ⋅ 40 ⎟ = − 3
dp ⎠
200 ⎝
20
dq
= −2.5
dp
c.
46. a.
q( p + dp) ≈ q ( p) + dq = q( p ) + q′( p)dp
q(19.20) = q (20 + (−0.8))
≈ q (20) + q′(20)dp
= 40 + (−2.5)(−0.8)
= 42 units
Profit = TR − TC = pq − cq
P=
1 3
80, 000 ⎞ 1 3
⎛
q − 66q 2 + 7000q − ⎜ 500q − q 2 +
= q − 65q 2 + 6500q − 40, 000
2
2 ⎟⎠ 2
⎝
If q = 100, then P =
1
(100)3 − 65(100) 2 + 6500(100) − 40, 000 = 460, 000
2
b. We use P(q + dq) ≈ P(q) + dP with q = 100 and dq = −2.
P (98) = P (100 + (−2))
⎛3
⎞
≈ P (100) + ⎜ q 2 − 130q + 6500 ⎟ dq
⎝2
⎠
⎡3
⎤
2
= 460, 000 + ⎢ (100) − 130(100) + 6500 ⎥ (−2)
2
⎣
⎦
= $443, 000
542
ISM: Introductory Mathematical Analysis
Section 14.2
Principles in Practice 14.2
1.
2.
4.
24
24
∫ 5 x dx = 5∫ x dx = 5 ⋅
5.
∫ 5x
t3
+ C = 0.04t 3 + C
3
The form of the revenue function is
x 24+1
+C
24 + 1
x 25
x 25
= 5⋅
+C =
+C
25
5
2
∫ 0.12t dt = 0.12
R (t ) = 0.04t 3 + C .
3. Let S(t) = the number of subscribers t months
after the competition entered the market, then
480
S ′(t ) = −
.
t3
480
S (t ) = ∫ −
dt = −480 ∫ t −3 dt
t3
⎛ t −2 ⎞
240
= −480 ⎜
⎟ + C = 240t −2 + C = 2 + C
⎜ −2 ⎟
t
⎝
⎠
240
+C .
The number of subscribers is S (t ) =
t2
4.
8
∫ x dx =
∫ 28.3 dq = 28.3q + C
The form of the cost function is 28.3q + C.
x8+1
x9
+C =
+C
8 +1
9
3.
∫(
= 5⋅
6.
7.
)
3
= 500t +
3
2
8.
2 3
The population is N (t ) = 500t + t 2 + C
3
∫ ( 2.1t
2
)
dS
∫ dt dt .
9.
dx = 2 ∫ x −10 dx = 2 ⋅
7
∫ x4
dx = 7 ∫ x −4 dx = 7 ⋅
7
3 x3
1
∫ t 7 / 4 dt = ∫ t
4
3t
3/ 4
S (t ) = 0.7t 3 − 32.7t 2 + 491.6t + C
1 1
t −7 / 4+1
− 74 + 1
−9
−5
4
=−
∫ 2 x dx = 2 ∫ x dx = 2 ln x + C
dt =
+C =
+C
7 x 4
= ⋅
+C
2 −5
Problems 14.2
2.
−7 / 4
+1
7 −9
7 x4
10. ∫ 9 dx = ∫ x 4 dx = ⋅
+C
2
2 −9 + 1
2x 4
4
= 0.7t − 32.7t + 491.6t + C
The amount of money saved is
∫ 7 dx = 7 x + C
x −4+1
7 x −3
+C =
+C
−4 + 1
−3
+C
7
2
1.
x −10+1
+C
−10 + 1
2 x −9
2
+C = −
+C
−9
9 x9
=−
⎛ t3 ⎞
⎛ t2 ⎞
= 2.1⎜ ⎟ − 65.4 ⎜ ⎟ + 491.6t + C
⎜3⎟
⎜2⎟
⎝ ⎠
⎝ ⎠
1
x −6
5
+C = −
+C
−6
6 x6
∫ x10
=−
− 65.4t + 491.6 dt
3
x −7 +1
+C
−7 + 1
z −3
1
1 z −3+1
dz = ∫ z −3 dz = ⋅
+C
3
3
3 −3 + 1
2
=
2 3
+ C = 500t + t 2 + C
3
5. The amount of money saved is
∫
dx = 5∫ x −7 dx = 5 ⋅
1 z −2
1
= ⋅
+C = −
+C
3 −2
6z2
1
⎛
⎞
500 + 300 t dt = ∫ ⎜ 500 + 300t 2 ⎟ dt
⎝
⎠
t2
−7
14
5
5x 4
1
543
+C
t −3 / 4
− 43
+C
Chapter 14: Integration
11.
∫ (4 + t )dt = ∫ 4 dt + ∫ t dt = 4t +
= 4t +
12.
13.
ISM: Introductory Mathematical Analysis
∫(
t2
+C
2
19.
)
r 3 + 2r dr = ∫ r 3 + 2 ∫ r dr
=
r 3+1
r1+1
+ 2⋅
+C
3 +1
1+1
=
r4
+ r2 + C
4
∫( y
=
t t +1
+C
1+1
20.
)
5
− 5 y dy = ∫ y 5 dy − ∫ 5 y dy
y5+1
y1+1
− 5⋅
+C
5 +1
1+1
y6
y2
y6 5 y2
=
− 5⋅
+C =
−
+C
6
2
6
2
14.
∫ ( 5 − 2w − 6w ) dw
= ∫ 5 dw − 2∫ w dw − 6 ∫ w2 dw
2
= 5w − 2 ⋅
15.
∫ ( 3t
= 3⋅
16.
23.
2
6
24.
∫ (5 − 2
−1
=
2 x3 8 x5
⋅ − ⋅ +C
7 3 3 5
=
2 x3 8 x5
−
+C
21 15
x
dx = π ∫ e x dx = πe x + C
⎛ ex
⎞
1
+ 2 x ⎟ dx = ∫ e x dx + 2∫ x dx
⎟
3
3
⎝
⎠
ex
+ x2 + C
3
∫(x
8.3
)
− 9 x6 + 3 x −4 + x −3 dx
=
x9.3
x7
x −3 x −2
− 9⋅
+ 3⋅
+
+C
9.3
7
−3 −2
=
x9.3 9 x 7 1
1
−
−
−
+C
3
9.3
7
x
2 x2
∫ (0.7 y
3
+ 10 + 2 y −3 )dy
y4
y −2
+ 10 y + 2 ⋅
+C
4
−2
1
= 0.175 y 4 + 10 y −
+C
y2
t3 t5 t7
+ + +C
3 5 7
= 0.7 ⋅
17. Since 7 + e is a constant,
∫ (7 + e)dx = (7 + e) x + C .
18.
⎛ 2 x2 8 4 ⎞
2 2
8 4
∫ ⎜⎜ 7 − 3 x ⎟⎟ dx = 7 ∫ x dx − 3 ∫ x dx
⎝
⎠
=
∫ (1 + t + t + t )dt
= ∫ 1 dt + ∫ t 2 dt + ∫ t 4 dt + ∫ t 6 dt
=t+
x 2 3 x5
−
+C
14 20
∫ ⎜⎜
t
t
− 4 ⋅ + 5t + C = t 3 − 2t 2 + 5t + C
3
2
4
=
22.
2
2
1
3 4
⎟ dx = ∫ x dx − ∫ x dx
7
4
⎠
1
x2
= ex + 2 ⋅
+C
3
2
− 4t + 5 dt = 3∫ t dt − 4 ∫ t dt + ∫ 5 dt
3
4⎞
1 x 2 3 x5
⋅
− ⋅ +C
7 2 4 5
∫ πe
w2
w3
− 6⋅
+C
2
3
)
3
=
21.
= 5w − w2 − 2w3 + C
2
⎛x
∫ ⎜⎝ 7 − 4 x
1 +1
−2 x
2 1
2 x2
25. ∫
+C
dx = − ∫ x 2 dx = − ⋅
3
3
3 1 +1
)
1⎞
9
9
⎛
dx = ∫ ⎜ 5 − ⎟ dx = ∫ dx = x + C
2⎠
2
2
⎝
2
3
2
3
2
2 x
4x
=− ⋅
+C = −
+C
3 3
9
2
26.
544
∫ dz = ∫ 1 dz = 1⋅ z + C = z + C
ISM: Introductory Mathematical Analysis
Section 14.2
− 1 +1
1
1 −1
1 x 4
dx = ∫ x 4 dx = ⋅
27. ∫
+C
8
4
4 − 1 +1
4 x2
4
3
=
33.
3
1 x4
x4
⋅
+C =
+C
4 3
3
(
4
28.
−4
−4
4
∫ (3x)3 dx = ∫ 27 x3 dx = − 27 ∫ x
−3
4 x
⋅
+C
27 −3 + 1
4 x −2
2
=− ⋅
+C =
+C
27 −2
27 x 2
29.
⎛ x3
∫ ⎜⎜
⎝ 3
−
)
(
dx
−3+1
=−
(
3u − 4
1
1
du = ∫ (3u − 4)du = 3∫ u du − 4∫ du
5
5
5
2
⎛
⎞
1 u
3
4
= ⎜ 3 − 4u ⎟ + C = u 2 − u + C
⎟
5 ⎜⎝ 2
10
5
⎠
1
= 2∫ z dz − ∫ 5 dz
7
⎞
1 ⎛ z2
1
= ⎜ 2 ⋅ − 5z ⎟ + C = z 2 − 5z + C
⎟
7 ⎜⎝ 2
7
⎠
∫
34.
3 ⎞
1
⎟ dx = ∫ x3 dx − 3∫ x −3 dx
3⎟
3
x ⎠
35.
1 ⎛1
1 x
⎞
⎟ dx = ∫ e dx
36
⎠
1
1
=
e x dx = e x + C
36 ∫
36
∫ 12 ⎜⎝ 3 e
∫ (u
e
x
+ eu )du = ∫ u e du + ∫ eu du
1 x3+1
x −3+1
= ⋅
− 3⋅
+C
−3 + 1
3 3 +1
=
1 x4
3
x −2
x4
= ⋅
− 3⋅
+C =
+
+C
−2
3 4
12 2 x 2
30.
⎛ 1
1 ⎞
1
∫ ⎜⎝ 2 x3 − x 4 ⎟⎠ dx = 2 ∫ x
−3
36.
dx − ∫ x −4 dx
1 x −2 x −3
⋅
−
+C
2 −2 −3
1
1
=−
+
+C
2
4x
3 x3
31.
⎛ 3w2
2
∫ ⎜⎜ 2 − 3w2
⎝
32.
∫ e− s
y
⎛ 3
2 e
y
−
y
+
3
2
⎜
∫⎜
6
⎝
= 3⋅
=
⎞
3
2
⎟ dw = ∫ w2 dw − ∫ w−2 dw
⎟
2
3
⎠
37.
1 y
e dy
6∫
y4
y3 1 y
− 2⋅
+ ⋅e + C
4
3 6
3 y 4 2 y3 e y
−
+
+C
4
3
6
∫ (2
)
1
⎛ 1
⎞
x − 3 4 x dx = ∫ ⎜ 2 x 2 − 3x 4 ⎟ dx
⎝
⎠
1
1
3
= 2⋅
s
⎞
⎟ dy
⎟
⎠
= 2 ∫ x 2 dx − 3∫ x 4 dx
3 w3 2 w−1
w3 2
= ⋅
− ⋅
+C =
+
+C
2 3 3 −1
2 3w
4
u e+1 u
+e +C
e +1
= 3∫ y 3 dy − 2∫ y 2 dy +
=
)
s
ds = 4∫ e ds = 4e + C
38.
545
x2
3
2
5
− 3⋅
x4
5
4
3
5
4 x 2 12 x 4
+C =
−
+C
3
5
∫ 0 dt = 0 ⋅ t + C = C
)
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
⎛ 3 x2
⎞
7
39. ∫ ⎜ −
−
+ 6 x ⎟dx
⎜
⎟
5
2 x
⎝
⎠
2
1
−
⎛
⎞
x 3 7x 2
= ∫⎜−
−
+ 6 x ⎟ dx
⎜ 5
⎟
2
⎝
⎠
2
1
1
7 −
= − ∫ x 3 dx − ∫ x 2 dx + 6 ∫ x dx
5
2
5
44.
45.
1
1 x3 7 x2
x2
=− ⋅
− ⋅
+ 6⋅
+C
5 5 2 1
2
3x
− 7 x + 3x2 + C
25
=−
40.
⎛3
∫ ⎜⎝
u+
1
= ∫ u 3 du + ∫ u
=
u
+
4
3
=
z3
+ 2z2 + 4z + C
3
∫ (3u + 2)
3
du = ∫ (27u 3 + 54u 2 + 36u + 8)du
u4
u3
u2
+ 54 ⋅ + 36 ⋅
+ 8u + C
4
3
2
27 4
u + 18u 3 + 18u 2 + 8u + C
4
u
1
2
1
2
46.
⎞
⎟ du
⎠
− 12
du
2
⎛ 2
⎞
⎛ − 15 ⎞
1
dx
2
x
1
−
=
−
⎟ dx
∫ ⎜⎝ 5 x ⎟⎠
∫ ⎜⎝
⎠
−1
⎛ − 25
⎞
= ∫ ⎜ 4 x − 4 x 5 + 1⎟ dx
⎝
⎠
3
= 4⋅
+C
∫(x
=
=
42.
2
)
(
)
47.
x4
x3
x2
− 3⋅ + 5⋅
− 15 x + C
4
3
2
4
=
( x3 + 8x2 + 7 ) dx = ∫ ( x7 + 8x6 + 7 x4 ) dx
x8
x7
x5
=
+ 8⋅
+ 7⋅ +C
8
7
5
x8 8 x 7 7 x 5
=
+
+
+C
8
7
5
43.
∫
48.
x2
5
2
x2
3
2
+ x+C
v3
v −4
+ 3v − 2 ⋅
+C
3
−4
2v 3
1
+ 3v +
+C
3
2v 4
∫ ⎡⎣6e
u
(
− u3
)
7
⎡
⎤
u + 1 ⎤ du = ∫ ⎢ 6eu − u 2 − u 3 ⎥du
⎦
⎣
⎦
9
= 6⋅e −
u2
9
2
9
−
u4
+C
4
2u 2 u 4
= 6e −
−
+C
9
4
u
3
+ 3⋅
4
5
−2
u
1
⎛ 3
⎞
x ( x + 3)dx = ∫ ⎜ x 2 + 3x 2 ⎟ dx
⎝
⎠
5
=
− 4⋅
( 2v4 + 3v2 − 2v−3 ) dv
= ∫ ( 2v 2 + 3 − 2v −5 ) dv
∫v
= 2⋅
2
x
5x
− x3 +
− 15 x + C
4
2
∫x
3
5
4
x5
4
20 x 5
=
− 5x 5 + x + C
3
+ 5 ( x − 3)dx = ∫ x3 − 3x 2 + 5 x − 15 dx
4
x5
3
4
1
3u 3
=
+ 2u 2 + C
4
41.
)
2
1 ⎞
−1
⎛ 1
⎟ du = ∫ ⎜ u 3 + u 2
u⎠
⎝
4
3
(
dz = ∫ z 2 + 4 z + 4 dz
z3
z2
+ 4 ⋅ + 4z + C
3
2
=
1
2
2
=
= 27 ⋅
2
3
5
3
∫ ( z + 2)
+C
5
3
2x 2
=
+ 2x 2 + C
5
546
ISM: Introductory Mathematical Analysis
49.
∫
z 4 + 10 z 3
2z2
dz =
Section 14.3
Principles in Practice 14.3
1 ⎛ z 4 10 z 3 ⎞
⎜ + 2 ⎟ dz
2 ∫ ⎜⎝ z 2
z ⎟⎠
(
1. N (t ) = ∫
)
1
= ∫ z 2 + 10 z dz
2
1 ⎛ z3
z2 ⎞
= ⎜ + 10 ⋅ ⎟ + C
2 ⎜⎝ 3
2 ⎟⎠
z3 5z 2
=
+
+C
6
2
50.
51.
x4 − 5x2 + 2 x
∫
e x + e2 x
ex
40, 000 = 800(5) + 200e5 + C , so
(
C = 40, 000 − 4000 + 200e5
N (t ) = 800t + 200et + 6317.37
d
( y′ ) = 84t + 24
dt
⎛ t2 ⎞
y′ = ∫ (84t + 24)dt = 84 ⎜ ⎟ + 24t + C1
⎜2⎟
⎝ ⎠
2. Since y ′′ =
⎛ e x e2 x ⎞
dx = ∫ ⎜ +
⎟ dx
⎜ ex ex ⎟
⎝
⎠
x
= ∫ 1 + e dx
= 42t 2 + 24t + C1
Since y′(8) = 2891 , we have
)
2891 = 42(8) 2 + 24(8) + C1 = 2880 + C1 , so
= x + ex + C
52.
∫
3
( x + 1)
x
2
2
dx = ∫
6
C1 = 2891 − 2880 = 11 , and y′ = 42t 2 + 24t + 11 .
3
x + 2x +1
(
)
y (t ) = ∫ y′dt = ∫ 42t 2 + 24t + 11 dt
dx
x2
= ∫ ( x 4 + 2 x + x −2 )dx
⎛ t3 ⎞
⎛ t2 ⎞
= 42 ⎜ ⎟ + 24 ⎜ ⎟ + 11t + C2
⎜3⎟
⎜2⎟
⎝ ⎠
⎝ ⎠
x5
x 2 x −1
+ 2⋅
+
+C
5
2
−1
x5
1
=
+ x2 − + C
x
5
=
= 14t 3 + 12t 2 + 11t + C2
Since y(2) = 185, we have
185 = 14(2)3 + 12(2)2 + 11(2) + C2
= 182 + C2 , so C2 = 185 − 182 = 3 .
53. No, F(x) – G(x) might be a nonzero constant.
54. a.
)
= 36, 000 − 200e5 ≈ 6317.37
dx =
(
)
= 800t + 200et + C
Since N(5) = 40,000, we have
1 ⎛ 2
2⎞
⎜ x − 5 + ⎟ dx
∫
2
5
x⎠
⎝
5x
3
⎞
1⎛ x
= ⎜ − 5 x + 2 ln x ⎟ + C
⎜
⎟
5⎝ 3
⎠
∫
(
dN
dt = ∫ 800 + 200et dt
dt
y (t ) = 14t 3 + 12t 2 + 11t + 3
( )
d
F ( x) =
xe x = xe x + e x (1) = e x ( x + 1)
dx
b. There is only one.
55. Because an antiderivative of the derivative of a
function is the function itself, we have
d ⎡ 1 ⎤
1
∫ dx ⎢⎢ 2 ⎥⎥ dx = 2 + C .
x +1
⎣ x +1 ⎦
547
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
Problems 14.3
1.
5. y ′′ = −3x 2 + 4 x
y ′ = ∫ (−3x 2 + 4 x)dx = − x3 + 2 x 2 + C1
dy
= 3x − 4
dx
y = ∫ (3 x − 4)dx =
Using y (−1) =
y ′(1) = 2 implies 2 = −1 + 2 + C1 , so C1 = 1.
2
3x
− 4x + C
2
y = ∫ (− x3 + 2 x 2 + 1)dx = −
13
gives
2
1 2
y(1) = 3 implies 3 = − + + 1 + C2 , so
4 3
13 3(−1) 2
=
− 4(−1) + C
2
2
13 11
= +C
2
2
3x 2
Thus C = 1, so y =
− 4x + 1 .
2
2.
C2 =
(
)
Using y(3) =
x2
+ x + C1
2
y′(0) = 0 implies 0 = 0 + 0 + C1 , so C1 = 0 .
y′ = ∫ ( x + 1)dx =
⎡ x2
⎤
x3 x 2
y = ∫ ⎢ + x ⎥ dx =
+
+ C2 .
6
2
⎣⎢ 2
⎦⎥
y(0) = 5 implies 5 = 0 + 0 + C2 , so C2 = 5 . Thus
x3 x 2
−
+C
3
2
19 33 32
19
gives
= − +C
2
3 2
2
y=
19 9
= +C
2 2
y ′′ = ∫ 2 x dx = x 2 + C1
x3 x 2
−
+5.
3
2
y ′′(−1) = 3 implies that 3 = 1 + C1 , so C1 = 2 .
x3
+ 2 x + C2
3
y′(3) = 10 implies 10 = 9 + 6 + C2 , so C2 = −5 .
(
5
y=∫
x
)
y′ = ∫ x 2 + 2 dx =
x
5
x3 x 2
+
+5.
6
2
7. y ′′′ = 2 x
Thus, C = 5, so y =
3. y′ =
x 4 2 x3
19
19
. Thus y = −
+
+ x+ .
4
3
12
12
6. y ′′ = x + 1
dy
= x2 − x
dx
y = ∫ x 2 − x dx =
x 4 2 x3
+
+ x + C2
4
3
dx = ∫ 5 x
− 12
1
dx = 5 ⋅
x2
1
2
⎛ x3
⎞
x4
y = ∫ ⎜ + 2 x − 5 ⎟ dx =
+ x 2 − 5 x + C3 .
⎜ 3
⎟
12
⎝
⎠
y(0) = 13 implies that 13 = 0 + 0 − 0 + C3 , so
+ C = 10 x + C
y(9) = 50 implies 50 = 10 9 + C , 50 = 30 + C,
C = 20.
Thus y = 10 x + 20.
C3 = 13 . Therefore y =
y(16) = 10 ⋅ 4 + 20 = 60
8. y ′′′ = e x + 1
4. y′ = − x 2 + 2 x
(
x4
+ x 2 − 5 x + 13 .
12
)
y ′′ = ∫ e x + 1 dx = e x + x + C1
x3
y = ∫ − x + 2 x dx = − + x 2 + C
3
8
1
y(2) = 1 implies 1 = − + 4 + C , so C = − .
3
3
y′ = ∫ e x + x dx = e x +
x3
1
+ x2 − .
3
3
1
1 1
y (1) = − + 1 − =
3
3 3
⎡
x2 ⎤
x3
y = ∫ ⎢e x +
+ 1⎥ dx = e x +
+ x + C3
2
6
⎥⎦
⎣⎢
(
2
)
y ′′(0) = 1 implies 1 = 1 + 0 + C1 , so C1 = 0 .
x2
+ C2
2
y′(0) = 2 implies 2 = 1 + 0 + C2 , so C2 = 1 .
(
Thus y = −
548
)
ISM: Introductory Mathematical Analysis
Section 14.3
y(0) = 3 implies that 3 = 1 + 0 + 0 + C3 , so
C3 = 2 . Thus y = e x +
9.
p=
x3
+ x+2.
6
function is p = 5000 − 3q −
dr
= 0.7
dq
13.
r = ∫ 0.7 dq = 0.7 q + C
14.
dr
1
= 10 − q
dq
16
(
When q = 0, then c = 2000, so C = 2000. Thus
the cost function is c = q 2 + 75q + 2000 .
15.
(
)
0.08 3
q − 0.8q 2 + 6.5q + C. If q = 0, then
3
c = 8000, from which C = 8000. Hence
0.08 3
c=
q − 0.8q 2 + 6.5q + 8000. If q = 25,
3
1
substituting gives c(25) = 8079 or $8079.17.
6
)
= 275q − 0.5q 2 − 0.1q3 + C . When q = 0, r must
2
dc
= 0.08q 2 − 1.6q + 6.5
dq
c = ∫ 0.08q 2 − 1.6q + 6.5 dq
Thus r = ∫ 275 − q − 0.3q 2 dq
16.
3
be 0, so C = 0 and r = 275q − 0.5q − 0.1q .
dc
= 0.000204q 2 − 0.046q + 6
dq
c = ∫ (0.000204q 2 − 0.046q + 6)dq
r
= 275 − 0.5q − 0.1q 2 .
q
Thus the demand function is
Since r = pq, then p =
= 0.000068q3 − 0.023q 2 + 6q + C
When q = 0, then c = 15,000, from which
C = 15,000. The cost function is
p = 275 − 0.5q − 0.1q 2 .
12.
dc
= 2q + 75
dq
c = ∫ (2q + 75)dq = q 2 + 75q + C
1
p = 10 − q .
32
dr
= 275 − q − 0.3q 2
dq
dc
= 1.35
dq
When q = 0, then c = 200, so 200 = 0 + C, or
C = 200. Thus c = 1.35q + 200.
1 ⎤
1
⎡
r = ∫ ⎢10 − q ⎥ dq = 10q − q 2 + C
16 ⎦
32
⎣
When q = 0, then r = 0, so C = 0 and
1
r = 10q − q 2 . Since r = pq, then
32
r
1
p = = 10 − q . The demand function is
q
32
11.
3q3
.
2
c = ∫ 1.35dq = 1.35q + C
If q = 0, r must be 0, so 0 = 0 + C, C = 0. Thus
r = 0.7q. Since r = pq, we have
r 0.7q
p= =
= 0.7 . The demand function is
q
q
p = 0.7.
10.
r
3q3
= 5000 − 3q −
. Therefore the demand
q
2
c = 0.000068q3 − 0.023q 2 + 6q + 15, 000. When
q = 200, substitution gives c(200) = 15,824.
dr
= 5000 − 3(2q + 2q3 ), so
dq
P2
⎡ P
⎤
+ 2P + C
17. G = ∫ ⎢ − + 2 ⎥ dP = −
50
⎣ 25
⎦
When P = 10, then G = 38, so 38 = –2 + 20 + C,
from which C = 20. Thus
1
G = − P 2 + 2 P + 20 .
50
r = ∫ (5000 − 6q − 6q3 )dq
3q 4
+C
2
When q = 0, then r = 0, so C = 0 and
3q 4
r = 5000q − 3q 2 −
. Since r = pq, then
2
= 5000q − 3q 2 −
549
Chapter 14: Integration
18.
ISM: Introductory Mathematical Analysis
dy
= −1.5 − x
dx
x2
+C
2
When x = 1, then y = 57.3, so
y = ∫ (−1.5 − x )dx = −1.5 x −
57.3 = –1.5 – 0.5 + C, or C = 59.3. Thus y = −1.5 x − 0.5 x 2 + 59.3 .
19. v = ∫ −
( P1 − P2 ) r
dr = −
2lη
( P1 − P2 ) r 2
4lη
Since v = 0 when r = R, then 0 = −
v=−
20.
( P1 − P2 ) r 2 ( P1 − P2 ) R 2
+
4lη
dr
= 100 − 3q 2
dq
(
4lη
=
+C
( P1 − P2 ) R 2
4lη
+ C , so C =
( P1 − P2 ) ( R 2 − r 2 )
4lη
( P1 − P2 ) R 2
4lη
. Thus
.
)
r = ∫ 100 − 3q 2 dq = 100q − q3 + C
When q = 0, then r = 0, so C = 0 and r = 100q − q3 . Since r = pq, then p =
η=
p
q
dp
dq
=
p
q
−2q
=−
p
2q 2
When q = 5, then p = 75, so η =
21.
r
= 100 − q 2 .
q
dc
= 0.003q 2 − 0.4q + 40
dq
(
−75
3
=− .
2(25)
2
)
c = ∫ 0.003q 2 − 0.4q + 40 dq = 0.001q3 − 0.2q 2 + 40q + C
When q = 0, then c = 5000, so
5000 = 0 – 0 + 0 + C, or C = 5000. Thus c = 0.001q3 − 0.2q 2 + 40q + 5000 . When q = 100, then c = 8000. Since
8000
Total Cost c
dc
= $80 . (Observe that knowing
= , when q = 100, we have c =
= 27.50
100
Quantity
q
dq
when q = 50 is not relevant to the problem.)
Avg. Cost = c =
22.
f ′′( x) = 30 x 4 + 12 x
f ′( x) = ∫ (30 x 4 + 12 x)dx = 6 x5 + 6 x 2 + C1
f ′(1) = 10, so 10 = 6 + 6 + C1 and C1 = −2.
f ′( x) = 6 x5 + 6 x 2 − 2
f ( x) = ∫ (6 x5 + 6 x 2 − 2)dx = x6 + 2 x3 − 2 x + C2
550
ISM: Introductory Mathematical Analysis
Section 14.4
Thus
f (965.335245) − f (−965.335245)
= [(965.335245)6 + 2(965.335245)3 − 2(965.335245) + C2 ]
− [(−965.335245)6 + 2(−965.335245)3 − 2(−965.335245) + C2 ]
= 3,598, 280, 000
Principles in Practice 14.4
1. Using the values given,
dT
= −0.5(70 − 60)e−0.5t = −5e−0.5t
dt
dT
dt = ∫ −5e −0.5t dt = 10e−0.5t + C
dt
T (t ) = ∫
2. The number of words memorized is v(t).
35
v(t ) = ∫
dt = 35ln t + 1 + C .
t +1
Problems 14.4
1. Let u = x + 5 ⇒ du = 1dx = dx
∫ ( x + 5)
2.
7
[dx] = ∫ u 7 du =
∫ 15( x + 2)
4
u8
( x + 5)8
+C =
+C
8
8
dx = 15∫ ( x + 2)4 [dx] = 15 ⋅
( x + 2)5
+ C = 3( x + 2)5 + C
5
3. Let u = x 2 + 3 ⇒ du = 2 x dx
(
(
)
5
)
5
2
2
5
∫ 2 x x + 3 dx = ∫ x + 3 [2 x dx] = ∫ u du =
( x 2 + 3)
=
u6
+C
6
6
6
+C
4. Let u = x3 + 5 x 2 + 6 ⇒ du = (3x 2 + 10 x)dx.
∫ ( 3x
2
)(
)
+ 10 x x3 + 5 x 2 + 6 dx
(
) (
1
)
= ∫ x3 + 5 x 2 + 6 ⎡⎢ 3x 2 + 10 x dx ⎤⎥
⎣
⎦
= ∫ u du =
u2
+C
2
( x3 + 5 x 2 + 6 )
=
2
2
+C
551
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
(
)
10. Let u = x − 5 ⇒ du = dx.
1
−1
∫ x − 5 dx = ∫ ( x − 5) 2 [dx]
5. Let u = y 3 + 3 y 2 + 1 ⇒ du = 3 y 2 + 6 y dy
∫(
3 y2 + 6 y
)(
)
y3 + 3 y 2 + 1
(
= ∫ u du =
5
3
(
=
+C
)
3 3
y + 3 y2 + 1
5
2
5
3
3
2
(5t 3 − 3t 2 + t )18
+C
18
12.
5
5
1
∫ (3x − 1)3 dx = 3 ∫ (3x − 1)3 [3 dx ]
5 1
5
du = ∫ u −3 du
∫
3
3 u
3
∫
4x
(
dx = ∫ 2 x 2 − 7
( 2x2 − 7 )
−9
2 x2 − 7 )
(
=−
+C
10
)
−10
dx =
1
(7 x − 6)4 [7 dx]
7∫
=
1 4
1 u5
u
du
=
⋅ +C
7∫
7 5
=
(7 x − 6)5
+C
35
(3x3 + 7 ) dx = 19 ∫ (3x3 + 7 )
4
3
1 ( 3x + 7 )
= ⋅
+C
∫x
3
2
3
⎡9 x 2 dx ⎤
⎣
⎦
4
4
+C
36
13. Let v = 5u 2 − 9 ⇒ dv = 10u du
[4 x dx]
∫ u (5u
2
− 9)14 du =
1
(5u 2 − 9)14 [10u du ]
10 ∫
1 14
1 v15
(5u 2 − 9)15
v dv = ⋅
+C =
+C
∫
10
10 15
150
9. Let u = 2 x − 1 ⇒ du = 2 dx .
14.
1
2
2 x − 1dx = ∫ (2 x − 1) dx
2
∫ 9 x 1 + 2 x dx =
(
1
1
= ∫ (2 x − 1) 2 [2 dx]
2
=
4
3 x3 + 7 )
(
=
9
∫
+C = 2 x−5 +C
1
2
9
5 u −2
5(3 x − 1)−2
= ⋅
+C = −
+C
3 −2
6
8.
( x − 5) 2
∫ (7 x − 6)
17
7. Let u = 3 x − 1 ⇒ du = 3 dx
=
+C
11. Let u = 7 x − 6 ⇒ du = 7 dx
+C
∫ (15t − 6t + 1)(5t − 3t + t ) dt
= ∫ (5t 3 − 3t 2 + t )17 [(15t 2 − 6t + 1)dt ]
=
1
2
1
5
u3
u1/ 2
−1/ 2
∫ u du =
) ⎡⎢⎣(3 y 2 + 6 y ) dy ⎤⎥⎦
2
3
6.
dy
2
3
= ∫ y3 + 3 y 2 + 1
=
2
3
2
9 1 + 2x
= ⋅
3
4
3
2
)
3
2
+C
2
3
1 12
1 u
1
u du = ⋅
+ C = (2 x − 1) 2 + C
∫
2
2 3
3
2
=
552
(
3 1 + 2 x2
2
)
(
9
1 + 2 x2
4∫
3
2
+C
)
1
2
[4 x dx]
ISM: Introductory Mathematical Analysis
15. Let u = 27 + x5 ⇒ du = 5 x 4 dx
1
4
4
5 3
5
∫ 4 x 27 + x dx = 5 ∫ 27 + x
(
)
(
)
1
3
Section 14.4
24.
⎡5 x 4 dx ⎤
⎣
⎦
4
(
)
4
3
+C
16. Let u = 4 − 5x ⇒ du = −5dx.
1
9
9
∫ (4 − 5 x) dx = − 5 ∫ (4 − 5 x) [−5 dx]
1
1 u10
1
= − ∫ u 9 du = − ⋅
+ C = − (4 − 5 x)10 + C
5
5 10
50
26.
dt =
∫
t 2 +t
20.
2 − w3
∫ −3w e
2
[(2t + 1) dt ]
+t
+C
3
3 4 x4
3
4
dx = ∫
1
(
)
⎡ 3 x 2 + 4 x3 dx ⎤
⎥⎦
x + x ⎣⎢
3
4
6 x2 − 6 x
∫ 1 − 3x2 + 2 x3 dx
3
dw = ∫ e− w ⎡ −3w2 dw⎤ = e− w + C
⎣
⎦
dx =
3 x 2 + 4 x3
28. Let u = 1 − 3 x 2 + 2 x3 ⇒ du = (−6 x + 6 x 2 )dx.
1
=∫
2
3
[(−6 x + 6 x 2 )dx
1 − 3x + 2 x
1
= ∫ du = ln u + C = ln 1 − 3x 2 + 2 x3 + C
u
21. Let u = 7 x ⇒ du = 14 x dx
1 7 x2
1 u
7 x2
∫ xe dx = 14 ∫ e [14 x dx] = 14 ∫ e du
2
1
1
= eu + C = e7 x + C
14
14
∫x e
[(1 + 2 x + 6 x 2 )dx]
3
= ln x3 + x 4 + C
2
22.
2
dx
x +x
1
= ∫ du = ln u + C
u
19. Let u = t 2 + t ⇒ du = (2t + 1)dt
t 2 +t
x + x 2 + 2 x3
2
27. Let u = x3 + x 4 ⇒ du = (3x 2 + 4 x3 )dx
5 3t + 7
5
e
[3 dt ] = e3t + 7 + C
∫
3
3
∫ (2t + 1)e dt = ∫ e
= ∫ eu du = eu + C = et
∫
12 x 2 + 4 x + 2
= ln[( x + x 2 + 2 x3 )2 ] + C
3x
3t + 7
∫ 5e
1 −6 x 5 ⎡
−30 x 4 dx ⎤
e
⎣
⎦
30 ∫
x + x + 2x
= 2 ln x + x 2 + 2 x3 + C
∫ 3e dx = ∫ e [3 dx]
= ∫ eu du = eu + C = e3 x + C
18.
dx = −
1 −6 x5
e
+C
30
=∫
17. Let u = 3 x ⇒ du = 3 dx
3x
e
25. Let u = x + 5 ⇒ du = dx
1
1
∫ x + 5 [dx] = ∫ u du = ln u + C = ln x + 5 + C
3
3
27 + x5
5
4 −6 x5
=−
4 1
4 u3
+C
= ∫ u 3 du = ⋅
5
5 4
=
∫x
29. Let u = z 2 − 6 ⇒ du = 2 z dz
6z
∫ ( z 2 − 6)5 = 3∫ ( z
1 4 x4 ⎡ 3 ⎤
16 x dx
e
⎣
⎦
16 ∫
= 3∫ u −5 du = 3
2
− 6)−5 [2 z dz ]
u −4
3
+ C = − ( z 2 − 6)−4 + C
−4
4
4
4
1
e4 x
= ⋅ e4 x + C =
+C
16
16
30.
3
3
∫ (5v − 1)4 dv = 5 ∫ (5v − 1)
−4
[5 dv]
3 (5v − 1)−3
⋅
+C
−3
5
1
= − (5v − 1)−3 + C
5
23. Let u = −3x ⇒ du = −3dx.
4 −3 x
−3 x
∫ 4e dx = − 3 ∫ e [−3 dx]
4
4
4
= − ∫ eu du = − eu + C = − e−3 x + C
3
3
3
=
553
Chapter 14: Integration
31.
4
ISM: Introductory Mathematical Analysis
39. Let u = x 2 − 4 ⇒ du = 2 x dx
1
∫ x dx = 4∫ x dx = 4 ln x + C
x
∫
3
1
1
32. ∫
[2 dy ]
dy = 3 ⋅ ∫
1+ 2y
2 1+ 2 y
3
= ln 1 + 2 y + C
2
x2 − 4
dx =
(
)
− 12
1
2
−
4
[2 x dx]
x
2∫
1
=
1 − 12
1 u2
+C
u du = ⋅
∫
2
2 1
2
2
= x −4 +C
33. Let u = s 3 + 5 ⇒ du = 3s 2 ds
s2
1
1
∫ s3 + 5 ds = 3 ∫ s3 + 5 ⎡⎣3s
=
34.
2
40. Let u = 1 − 3x ⇒ du = −3 dx.
9
1
∫ 1 − 3x dx = −3∫ 1 − 3x [−3 dx]
1
= −3∫ du = −3ln u + C = −3ln 1 − 3x + C
u
ds ⎤
⎦
1 1
1
1
du = ln u + C = ln s3 + 5 + C
∫
3 u
3
3
2 x2
⎛
1⎞
1
∫ 3 − 4 x3 dx = 2 ⎜⎝ − 12 ⎟⎠ ∫ 3 − 4 x3 ⎡⎣−12 x
2
dx ⎤
⎦
41. Let u = y 4 + 1 ⇒ du = 4 y3 dy
∫ 2y
1
= − ln 3 − 4 x3 + C
6
1
1
36. ∫
dt = 7 ⋅ ∫
[10t dt ]
2
2
10
5t − 6
5t − 6
7
= ln 5t 2 − 6 + C
10
=
5 x dx = 5 ∫ x1/ 2 dx = 5
4
dy = 2 ∫ y 3e y +1dy
1 y 4 +1 ⎡ 3 ⎤
e
4 y dy
⎣
⎦
4∫
1
1
= ∫ eu du = eu + C
2
2
1 y 4 +1
= e
+C
2
7t
∫
e
= 2⋅
35. Let u = 4 − 2x ⇒ du = −2 dx
5
5
1
∫ 4 − 2 x dx = − 2 ∫ 4 − 2 x [−2 dx]
5 1
5
5
= − ∫ du = − ln u + C = − ln 4 − 2 x + C
2 u
2
2
37.
3 y 4 +1
x
3
2
3
2
42.
∫2
1
2 x − 1dx = ∫ (2 x − 1) 2 [2 dx]
3
=
=
+C
(2 x − 1) 2
3
2
+C
3
2
(2 x − 1) 2 + C
3
43. Let u = −2v3 + 1 ⇒ du = −6v 2 dv
3
1
2 −2v3 +1
dv = − ∫ e−2v +1 ⎡ −6v 2 dv ⎤
∫v e
⎣
⎦
6
2 5 32
x +C
3
1 u
1
e du = − eu + C
6∫
6
3
1
= − e−2v +1 + C
6
=−
1
1
38. ∫
dx = ∫ (3 x)−6 [3 dx]
6
3
(3x )
1 (3 x)−5
⋅
+C
3 −5
1
= − (3 x)−5 + C
15
=
554
ISM: Introductory Mathematical Analysis
44.
∫3
x2
2 x3 + 9
(
1
= ⋅
6
=
45.
dx =
2 x3 + 9
)
(
1
2 x3 + 9
4
)
)
− 13
1
3
⎡6 x 2 dx ⎤
2
9
x
+
⎣
⎦
6∫
52.
2
3
=
+C
53.
−5 x
x
−5 x
x
∫ ( e + 2e ) dx = ∫ e dx + 2∫ e dx
∫ 4 3 y + 1dy = 4∫ ( y + 1)
= 4⋅
47.
( y + 1)
4
3
4
3
1
3
54.
[dy ]
3
55.
4
(7 − 2 x − 5 x)
+C
4
3 y2
∫ 2 ye
2
49.
x +2
dy = 2 ⋅
∫ x3 + 6 x
dx =
2
1 3 y2
1
e [6 y dy ] = e3 y + C
∫
6
3
(
56.
)
1
1 ⎡ 2
3 x + 6 dx ⎤⎥
∫
3
⎦
3 x + 6 x ⎣⎢
x
−3 x
5x
∫ (e + 2e − e )dx
2
1
= ∫ e x dx − ∫ e−3 x [(−3)dx] − ∫ e5 x [5 dx]
3
5
57.
16 s − 4
5
∫ (8w
∫ −(x
)
+1
−1
dx = ∫
x
2
2x + 1
dx
)
+ w2 − 2)(6w − w3 − 4 w6 ) −4 dw
)( x3 − x6 ) dx
−10
1
⎡ 3 x 2 − 6 x5 dx ⎤
= − ∫ ( x3 − x 6 )
) ⎦⎥
⎣⎢(
3
3
6 −9
−9
1 (x − x )
1 3
x − x6 ) + C
=− ⋅
+C =
(
−9
3
27
2
3
−10
− 2 x5
∫ 5 (v − 2)e
∫ ( 2x
2 − 4v + v 2
dv
3
)(
)
+ x x 4 + x 2 dx
(
) ⎡⎣⎢( 4 x3 + 2 x ) dx ⎤⎦⎥
4
2 2
2
1 (x + x )
1
= ⋅
+ C = ( x4 + x2 ) + C
2
2
4
=
2
1
= e x − e−3 x − e5 x + C
3
5
51.
2
2
3 1
= ⋅ ∫ e2− 4v + v [(2v − 4) dv]
5 2
2
3
= e2− 4v + v + C
10
1
= ln x3 + 6 x + C
3
50.
∫ x (2x
1
(6 w − w3 − 4 w6 ) −4 [(6 − 3w2 − 24w5 )dw]
3∫
1 (6 w − w3 − 4w6 )−3
=− ⋅
+C
3
−3
1
= (6 w − w3 − 4w6 )−3 + C
9
1
= − (7 − 2 x 2 − 5 x) 4 + C
2
48.
2 3 2
(t + t + 1)7 + C
7
=−
∫ (8 x + 10)(7 − 2 x − 5 x) dx
= −2 ∫ (7 − 2 x 2 − 5 x)3 [(−4 x − 5)dx]
= −2 ⋅
6
(t 3 + t 2 + 1)7
+C
7
(
4
2
2
1
1
[4 x dx]
∫
2
4 2x +1
1
= ln 2 x 2 + 1 + C
4
+ C = 3( y + 1) 3 + C
2
3
=
1
= − ∫ e −5 x [−5 dx] + 2∫ e x dx
5
1
= − e−5 x + 2e x + C
5
46.
2
∫ (6t + 4t )(t + t + 1) dt
= 2 ∫ (t 3 + t 2 + 1)6 [(3t 2 + 2t )dt ]
= 2⋅
+C
2
3
2
3
(
Section 14.4
1
∫ 3 − 2s + 4s 2 dx = 2∫ 3 − 2s + 4s 2 [(8s − 2)ds]
= 2 ln 3 − 2 s + 4s 2 + C
555
1
x4 + x2
2∫
1
Chapter 14: Integration
58.
∫ (e
)
3.1 2
ISM: Introductory Mathematical Analysis
dx = ∫ e6.2 dx = e6.2 x + C , because e6.2
65.
is a constant.
7 + 14 x
59.
dx
2 5
x
−x
∫ (e − e )
2
1 ⎞
1
dx
⎟ dx = ∫ 2 xdx − ∫
2x ⎠
2x
1
1
1
−1
= ∫ (2 x) 2 [2 dx] − ∫ (2 x) 2 [2 dx]
2
2
2x −
3
∫ (4 − x − x ) 2 −5
= −7 ∫ (4 − x − x ) [(−1 − 2 x)dx]
2
2
1
2 2 32
=
x − 2x 2 + C
3
(
66.
)
dx = ∫ e2 x − 2 + e−2 x dx
1 2x
⎛ 1⎞
e [2 dx] − ∫ 2 dx + ⎜ − ⎟ ∫ e−2 x [−2 dx]
2∫
⎝ 2⎠
1
1
= e2 x − 2 x − e−2 x + C
2
2
1 2 x −2 x
= e −e
− 2x + C
2
∫3
x4
ex
5
dx = 3∫ x 4 e− x dx = −
5
67.
)
=
61. u = 4 x3 + 3x 2 − 4
(
68.
)
du = 12 x 2 + 6 x dx = 6 x(2 x + 1)dx
4x
∫ x(2 x + 1)e
3
2
+3 x − 4
63.
3
(
2
1 8 − 5x
=− ⋅
5
10
)
5
2
2
64.
∫e
− 7x
dx = −7 ∫ e
− 7x
+C = −
(
3
2
69.
⎡
)
5
2
2
− 16
)
2
−
1 ⎤
dx
2 x + 5 ⎥⎦
2
1
1
1
x 2 − 16 [2 x dx] − ∫
[2 dx]
∫
2
2 2x + 5
⎡
2
)
3
1
− ln 2 x + 5 + C
2
⎤
x5
x
∫ ⎢⎢ x2 + 1 + ( x6 + 1)2 ⎥⎥dx
⎣
=∫
[−10 x dx]
1
8 − 5x2
25
)
x5 2 x3
+
+ x+C
5
3
(
u 4 1 6−3u 2
+
e
[−6u du ]
4 6∫
2
1
1
= u 4 + e6−3u + C
4
6
(8 − 5x2 ) dx = − 101 ∫ (8 − 5x2 )
(
2
2
3
1
= x 2 − 16
6
3
6 −3u
∫ (u − ue )du =
∫x
)
+ 1 dx = ∫ x 4 + 2 x 2 + 1 dx
(
)
3
2
1 ( x − 16 )
1
= ⋅
− ln 2 x + 5 + C
dx
1 4 x3 + 3 x 2 − 4
e
[6 x(2 x + 1)dx]
6∫
3
2
1
1
1
= ∫ eu du = eu + C = e4 x +3 x − 4 + C
6
6
6
2
2
∫ ⎢⎣ x ( x
=
=
62.
∫(x
3 − x5
e [−5 x 4 dx]
5∫
5
3
= − e− x + C
5
=
(
3
1
1 (2 x) 2 1 (2 x) 2
(2 x) 2
= ⋅
− ⋅
+C =
− 2x + C
3
1
2
2
3
(4 − x − x 2 ) −4
= −7
+C
−4
7
= (4 − x − x 2 )−4 + C
4
60.
⎛
∫ ⎜⎝
=
⎦
x
x2 + 1
dx + ∫
x5
(
)
x6 + 1
2
dx
(
)
1
1
1
[2 x dx] + ∫ x6 + 1
∫
2
2 x +1
6
(
)
−1
−2
6
1
1 x +1
2
= ln x + 1 + ⋅
+C
−1
2
6
1
1
= ln x 2 + 1 −
+C
6
2
6 x +1
+C
(
(
− 7x
⎡ 1 ⎤
−
=
−
+C
dx
e
7
⎢ 7 ⎥
⎣
⎦
556
)
)
(
)
⎡6 x5 dx ⎤
⎣
⎦
ISM: Introductory Mathematical Analysis
70.
⎡ 3
3
∫ ⎢⎢ x − 1 + ( x − 1)2 ⎥⎥ dx = ∫ x − 1 [dx] + ∫ ( x − 1)
⎣
⎦
= 3ln x − 1 +
71.
⎤
1
Section 14.4
⎡ 2
−2
[dx]
( x − 1) −1
1
+ C = 3ln x − 1 −
+C
x −1
−1
⎤
− 8 x5 )( x3 − x6 )−8 ⎥ dx
⎦
1
1
4
3
6 −8
= ∫
[4 dx] − ∫ ( x − x ) [(3 x 2 − 6 x5 )dx]
2 4x + 1
3
1
4 ( x3 − x 6 )−7
= ln 4 x + 1 − ⋅
+C
−7
2
3
1
4
= ln 4 x + 1 + ( x3 − x 6 ) −7 + C
2
21
∫ ⎢⎣ 4 x + 1 − (4 x
72.
∫ (r
73.
∫ ⎢⎣
⎡
3
+5
)
2
2
(
)
dr = ∫ r 6 + 10r 3 + 25 dr =
3x + 1 −
1 7 5 4
r + r + 25r + C
7
2
1
1
⎤
x
1
1
1
2
dx = ∫ (3 x + 1) 2 [3 dx] − ∫
[2 x dx]
⎥ dx = ∫ (3x + 1) dx − ∫ 2
2
3
2 x +3
x +3
x + 3⎦
x
2
3
(
)
3
2
1 (3 x + 1) 2 1
= ⋅
− ln x 2 + 3 + C = (3 x + 1) 2 − ln x 2 + 3 + C
3
9
3
2
2
74.
⎡
x2
x
⎤
1
1
1
∫ ⎢⎢ 3x2 + 5 − ( x3 + 1)3 ⎥⎥ dx = 6 ∫ 3x2 + 5 [6 x dx] − 3 ∫ ( x
3
+ 1) −3 [3x 2 dx]
⎣
⎦
1
1 ( x3 + 1)−2
1
1
= ln 3x 2 + 5 − ⋅
+ C = ln 3x 2 + 5 + ( x3 + 1)−2 + C
6
3
6
6
−2
75. Let u = x ⇒ du =
∫
1 − 12
1
x dx =
dx .
2
2 x
e x
⎡ 1
⎤
dx = 2∫ e x ⎢
dx ⎥
x
⎣2 x ⎦
= 2 ∫ eu du = 2eu + C = 2e x + C
76.
∫ (e
77.
∫
5
)
(
)
− 3e dx = e5 − 3e x + C , because e5 − 3e is a constant.
1 + e2 x
4e x
(
dx =
1 ⎛ 1 e2 x ⎞
⎜ +
⎟ dx
4 ∫ ⎜⎝ e x e x ⎟⎠
)
1
e− x + e x dx
∫
4
1
1
= − ∫ e− x [−1 dx] + ∫ e x dx
4
4
1 −x 1 x
= − e + e +C
4
4
=
557
Chapter 14: Integration
78.
ISM: Introductory Mathematical Analysis
1
1
⎛1 ⎞2 ⎡ 1 ⎤
+ 9 dt = −2∫ ⎜ + 9 ⎟ ⎢ − dt ⎥
t
⎝t
⎠ ⎣ t2 ⎦
2
∫ t2
= −2
(
1 +9
t
3
2
)
83. y ′′ =
+C
y′(−2) = 3 implies 3 =
3
)
1
∫ x2 + 2 x ln ( x
x +1
2
(
2
x + 2x
)
(2 x + 2)dx
+ 2 x dx
)
=
⎡ 2x + 2 ⎤
1
dx ⎥
ln x 2 + 2 x ⎢
2∫
⎣ x2 + 2 x ⎦
=
1
1 u2
1
u du = ⋅
+ C = ln 2 x 2 + 2 x + C
∫
2
2 2
4
(
4
3
80. Let u = 8 x 4 = 2 x 3 ⇒ du =
)
8 13
x dx
3
84. y ′′ = ( x + 1)3 / 2
3
y ′ = ∫ ( x + 1) 2 dx =
4
3 2x3 ⎡8 1 ⎤ 3 u
∫ xe dx = 8 ∫ e ⎢⎣ 3 x 3 dx ⎥⎦ = 8 ∫ e du
3
3 3 4
= eu + C = e 8 x + C
8
8
3
3
8 x4
2
64
⋅ 32 + C1 ⇒ C1 = − , so
5
5
5
2
64
y ′ = ( x + 1) 2 −
5
5
5
64 ⎤
⎡2
y = ∫ ⎢ ( x + 1) 2 − ⎥ dx
5
5⎦
⎣
1
(3 − 2 x)2 [−2 dx]
2∫
1 (3 − 2 x)3
1
=− ⋅
+ C = − (3 − 2 x)3 + C
2
3
6
1
11
y(0) = 1 implies 1 = − (27) + C , so C = .
6
2
1
3 11
Thus y = − (3 − 2 x) + .
6
2
(
7
=
y = ln
2 ( x + 1) 2 64
⋅
−
x + C2
7
5
5
2
7
4
64
=
x + C2
( x + 1) 2 −
35
5
4
64
y(3) = 0 implies 0 =
⋅128 − (3) + C2 , so
35
5
7
832
4
64
832
C2 =
. Thus y =
( x + 1) 2 −
x+
.
35
35
5
35
)
1
1
1
[2 x dx] = ln x 2 + 6 + C
∫
2
2 x +6
2
1
1
y(1) = 0 implies 0 = ln(7) + C , so C = − ln 7 .
2
2
1⎡
Thus y = ⎢ln x 2 + 6 − ln 7 ⎤⎥ , or
⎦
2⎣
(
5
2
( x + 1) 2 + C1
5
y ′(3) = 0 ⇒ 0 =
81. y = ∫ (3 − 2 x) 2 dx = −
82. y =
1
5
+ C1 , so C1 = . Thus
2
2
5
y′ = − x −1 + .
2
5⎞
1
5
⎛
y = ∫ ⎜ − x −1 + ⎟ dx = − ∫ dx + ∫ dx
2⎠
2
x
⎝
5
= − ln x + x + C2
2
5
y(1) = 2 implies that 2 = 0 + + C2 , so
2
1
C2 = − . Thus
2
5
1
1 5
1
y = − ln x + x − = ln + x − .
2
2
x 2
2
4 ⎛1 ⎞2
= − ⎜ + 9⎟ + C
3⎝t
⎠
(
x2
y′ = ∫ x −2 dx = − x −1 + C1
3
2
79. Let u = ln x 2 + 2 x ⇒ du =
1
)
x2 + 6
7
558
ISM: Introductory Mathematical Analysis
85. V (t ) = ∫
=
Section 14.5
Problems 14.5
dV
dt = ∫ 8e0.05t dt
dt
8
e0.05t [0.05 dt ]
0.05 ∫
1.
dx
2x2
⎛ 2 x6 8 x 4
4x
= ∫⎜
+
−
⎜ 2 x2 2 x2 2 x2
⎝
⎞
⎟ dx
⎟
⎠
1
4
2
= ∫ x dx + 4 ∫ x dx − 2∫ dx
x
x5 4 3
=
+ x − 2 ln x + C
5 3
= 160e0.05t + C
The house cost $350,000 to build, so V(0) = 350.
350 = 160e0 + C = 160 + C
190 = C
V (t ) = 160e0.05t + 190
dl
12
dt = ∫
dt
dt
2t + 50
= 6 ln 2t + 50 + C
Since the expected life span was 63 years in
1940, l(0) = 63.
63 = 6 ln 50 + C
C = 63 – 6 ln 50 ≈ 39.53
l (t ) = 6 ln 2t + 50 + 39.53
86. l (t ) = ∫
2.
3.
9 x2 + 5
5 ⎞
⎛
∫ 3x dx = ∫ ⎜⎝ 3x + 3x ⎟⎠ dx
3
5
= x 2 + ln x + C
2
3
∫ ( 3x
2
+2
)
2 x3 + 4 x + 1dx
(
) ⎡⎢⎣( 6 x2 + 4) dx ⎤⎥⎦
3
1 ( 2 x + 4 x + 1)
= ⋅
+C
=
l (58) = 6 ln 166 + 39.53 ≈ 70.20
The expected life span for people born in 1998
(58 years after 1940) is about 70 years.
1
2
1
2 x3 + 4 x + 1
2∫
3
2
3
2
2
87. Note that r > 0.
B
Rr
⎡ Rr B1 ⎤
C = ∫⎢
+ ⎥ dr = ∫
dr + ∫ 1 dr
2K
r
⎣ 2K r ⎦
1
R
r dr + B1 ∫ dr
=
2K ∫
r
2
R r
=
⋅ + B1 ln r + B2
2K 2
Thus we obtain C =
∫
2 x6 + 8 x 4 − 4 x
=
4.
(
x
∫4
x2 + 1
dx =
2
1 ( x + 1)
= ⋅
2
Rr
+ B1 ln r + B2 .
4K
3
4
2
1
3
− 3x)dx = e3 x + 2 − x 2 + C
88. f ( x) = ∫ (e
3
2
1
1
1
⎛ ⎞
f ⎜ ⎟ = 2 implies 2 = e3 − + C , so
3
6
⎝3⎠
13 1
C = − e3 . Thus,
6 3
1
3
13 1
f ( x ) = e3 x + 2 − x 2 + − e3 ,
3
2
6 3
1
13 1
f (2) = e8 − 6 + − e3
3
6 3
1 8
3
= (2e − 2e − 23) ≈ 983.12
6
=
3x+2
5.
)
1
2 x3 + 4 x + 1
3
(
)
2 2
x +1
3
3
4
3
2
+C
(
)
−1
1
x 2 + 1 4 [2 x dx]
∫
2
3
4
+C
+C
9
dx = 9∫ (2 − 3 x)−1/ 2 dx
2 − 3x
⎛ 1⎞
= 9 ⎜ − ⎟ ∫ (2 − 3x) −1/ 2 [−3 dx]
⎝ 3⎠
(2 − 3 x)1/ 2
= −3
+ C = −6 2 − 3 x + C
∫
1
2
6.
∫
2 xe x
e
x2
−2
= ln e
559
2
x2
dx = ∫
1
e
−2 +C
x2
⎡ 2 xe x 2 dx ⎤
⎢
⎥⎦
−2⎣
Chapter 14: Integration
7.
∫4
7x
( )
dx = ∫ eln 4
ISM: Introductory Mathematical Analysis
7x
dx = ∫ e(ln 4)(7 x ) dx
13.
1
e(ln 4)(7 x ) [7 ln 4 dx]
7 ln 4 ∫
1
=
⋅ e(ln 4)(7 x ) + C
7 ln 4
5e 2 x
5
=
=
8.
9.
( )
1
e
7 ln 4
t
∫5
ln 4 7 x
+C =
=
14.
7x
4
+C
7 ln 4
7
1
1 (ln 5)t
=
e(ln 5)t [ln 5 dt ] =
⋅e
+C
ln 5 ∫
ln 5
5t
=
+C
ln 5
x2
⎞
⎛
4
dx
=
14
x
−
2
xe
⎟⎟
∫ ⎜⎜
⎠
⎝
15.
x
x
x2
− 4 ⋅ e 4 + C = 7 x 2 − 4e 4 + C
2
2
e⎞
⎛ x
+ x e + ex + ⎟ dx
x⎠
1
= ∫ e x dx + ∫ x e dx + e ∫ x dx + e ∫ dx
x
x e+1 ex 2
+
+ e ln x + C
e +1
2
x
2
dx = −
+C
⎡ 7
⎤
⎢ − 2 dx ⎥
⎣ x
⎦
1 7x
e
7∫
Note that since x 2 + 9 > 0 for all values of x, the
absolute value bars are not needed.
6 x 2 − 11x + 5
2
= 2x − 3 +
.
3x − 1
3x − 1
18. By using long division on the integrand,
5 − 4x2
4 ⎞
⎛
∫ 3 + 2 x dx = ∫ ⎜⎝ −2 x + 3 − 3 + 2 x ⎟⎠ dx
1
[2 dx]
= ∫ (−2 x + 3)dx − 2 ∫
3 + 2x
6 x 2 − 11x + 5
2 ⎞
⎛
dx = ∫ ⎜ 2 x − 3 +
⎟ dx
3x − 1
3
x
−1 ⎠
⎝
1
1
[3 dx]
= 2 ∫ x dx − ∫ 3 dx + 2 ⋅ ∫
3 3x − 1
2
= x 2 − 3x + ln 3x − 1 + C
3
Thus
∫ x2
1
2
17. By using long division on the integrand,
5 x3
45 x ⎞
⎛
∫ x2 + 9 dx = ∫ ⎜⎝ 5 x − x2 + 9 ⎟⎠ dx
45
1
[2 x dx]
= ∫ 5 x dx − ∫
2
2 x +9
5
45
= x 2 − ln( x 2 + 9) + C
2
2
∫ ⎜⎝ e
11. By long division,
7
dx = ∫ e x ⋅
)
16. By using long division on the integrand,
2 x 4 − 6 x3 + x − 2
dx
∫
x−2
16 ⎞
⎛
= ∫ ⎜ 2 x3 − 2 x 2 − 4 x − 7 −
⎟ dx
x−2⎠
⎝
1
2
= x 4 − x3 − 2 x 2 − 7 x − 16 ln x − 2 + C .
2
3
⎞
⎟⎟ dx
⎠
2
= ex +
dx = − ∫ e8−6 x [−6 dx]
1
= − e +C
7
x
⎡1
⎤
= 14 ∫ x dx − 2 ⋅ 2∫ e 4 ⎢ x dx ⎥
2
⎣
⎦
10.
ex
(2)dx]
7
x
x2
2
)
4 −3 x 2
(
= 14 ∫ x dx − 2∫ xe 4 dx
= 14 ⋅
∫ 6 (e
2x
5
ln(7e2 x + 4) + C
14
= −e8−6 x + C = − e4−3 x
dt = ∫ (eln 5 )t dt = ∫ e(ln 5)t dt
x2
⎛
2
x
7
−
e
∫ ⎜⎜ 4
⎝
1
∫ 7e2 x + 4 dx = 14 ∫ 7e2 x + 4[7e
∫
= − x 2 + 3x − 2 ln 3 + 2 x + C
19.
(3 x + 2)( x − 4)
3 x 2 − 10 x − 8
dx = ∫
dx
12. ∫
x −3
x−3
11 ⎞
3 2
⎛
= ∫ ⎜ 3x − 1 −
⎟ dx = x − x − 11ln x − 3 + C
2
x −3⎠
⎝
∫
(
x +2
2
3 x
2
= ⋅
3
560
)
(
dx =
x +2
3
)
2
3∫
(
x +2
3
+C =
2
9
(
)
2
⎡ 1
⎤
dx ⎥
⎢
⎣2 x ⎦
x +2
)
3
+C
ISM: Introductory Mathematical Analysis
20.
5e s
∫ 1 + 3es
Section 14.5
( )
5
1
[3e s ds ]
∫
3 1 + 3e s
5
= ln(1 + 3e s ) + C
3
4
⎛ 1
⎞
5⎜ x3 + 2 ⎟
4
⎠ dx = 3 5 ⎛ x 13 + 2 ⎞ ⎡ 1 x − 23 dx ⎤
21. ∫ ⎝
⎟ ⎢
∫ ⎜⎝
⎥
3 2
⎠ ⎣3
⎦
x
=
5
⎛ 1
⎞
= 3⎜ x 3 + 2 ⎟ + C
⎝
⎠
22.
∫
1
1+ x
1 2
⎛
⎞
dx = 2∫ ⎜ 1 + x 2 ⎟
⎝
⎠
x
)
3
2
+C
2
ln x
⎡ 1 ⎤ (ln x)
dx = ∫ (ln x) ⎢ dx ⎥ =
+C
x
2
⎣x ⎦
1
= ln 2 x + C
2
24.
0.6
dt = −
2
⎡ 3
⎤
(3 − t 3 / 2 )0.6 ⎢ − t1/ 2 dt ⎥
∫
3
⎣ 2
⎦
31.
1.6
2 (3 − t 3 / 2 )1.6
5
3−t t
=− ⋅
+C = −
+C
3
1.6
12
(
25.
26.
)
∫
32.
2
9 x5 − 6 x 4 − ex3
7x
2
dx = 2 ∫
1
2
⎡2 ⎤
⎢ x dx ⎥
⎣
⎦
3
3
e x +1 dx = ∫ x 2 (e x +1 )1/ 2 dx
3
3
2 x 2+1 ⎡ 3 2 ⎤ 2 x 2+1
e
⎢ 2 x dx ⎥ = 3 e dx
3∫
⎣
⎦
8
⎡ 1
1
⎤
∫ ( x + 3) ln( x + 3) dx = 8∫ ln( x + 3) ⎢⎣ x + 3 dx ⎥⎦
= 8ln ln( x + 3) + C
ln (r + 1)
⎛ 1
⎞
dr = ∫ [ln(r + 1)]2 ⎜
dr ⎟
r +1
r
+
1
⎝
⎠
1 3
= ln (r + 1) + C
3
∫
2
2
30. By using long division on the integrand,
x+3
3 ⎞
⎛
∫ x + 6 dx = ∫ ⎜⎝1 − x + 6 ⎟⎠ dx = x − 3ln x + 6 + C
)
∫ t (3 − t t )
4
∫x
=
∫
(
∫ x ln
29.
2
23.
( )
ln ( 2 x )
( 2x )
= 2 ln ln ( 2x 2 ) + C
3
(
1 ln 3 ln x
3ln x
e
+C =
+C
ln 3
ln 3
28.
⎡ 1 − 12 ⎤
⎢ 2 x dx ⎥
⎣
⎦
1 2
⎛
2 ⎞
⎜1 + x ⎟
⎝
⎠ + C = 4 1+ x
= 2⋅
3
3
ln x
eln 3
3ln x
27. ∫
dx = ∫
dx
x
x
1
⎡ ln 3 ⎤
=
e(ln 3) ln x ⎢
dx ⎥
∫
ln 3
⎣ x
⎦
1 (ln 3) ln x
=
⋅e
+C
ln 3
ds =
⎛ e2 + x e − 2 x ⎞ dx = ee2 x + 1 xe +1 − x 2 + C
⎟
⎠
e +1
∫ ⎜⎝ e
33. By using long division on the integrand,
2x ⎞
x3 + x 2 − x − 3
⎛
∫ x2 − 3 dx = ∫ ⎜⎝ x + 1 + x 2 − 3 ⎟⎠ dx
1
[2 x dx]
= ∫ ( x + 1)dx + ∫
2
x −3
x2
=
+ x + ln x 2 − 3 + C
2
e ⎞
6
⎛9
dx = ∫ ⎜ x3 − x 2 − x ⎟ dx
7
7 ⎠
⎝7
9 4 2 3 e 2
=
x − x − x +C
28
7
14
34.
∫
4 x ln 1 + x 2
1 + x2
(
= ∫ ln 1 + x
561
2
)
dx = ∫
(
4 x ⋅ 12 ln 1 + x 2
1 + x2
(
) dx
)
2
2
⎡ 2x
⎤ ln 1 + x
dx ⎥ =
+C
⎢
2
⎣1 + x 2 ⎦
Chapter 14: Integration
35.
∫
6 x 2 ln( x3 + 1) 2
3
x +1
ISM: Introductory Mathematical Analysis
42.
dx
⎡ 6 x2
⎤
= ∫ [2 ln( x3 + 1)]1/ 2 ⎢
dx ⎥
3
⎣⎢ x + 1 ⎦⎥
=
36.
[2 ln( x3 + 1)]3 / 2
3
2
(
2
∫3 x + 2
= 3∫ e
37.
38.
⎛
)
− 12
( x +2)
2
1
2
2
xe x + 2 dx
43.
− 12
2
⎡
⎤
2
x
x
2
dx ⎥ = 3e x + 2 + C
+
⎢
⎣
⎦
∫ x2 + 2 x −1 dx = 2 ∫ x2 + 2 x−1 ⎡⎢⎣( 2 x − 2 x
=
∫
)
=
⎞
− ln 7 ⎟ dx
⎟
4
⎝ x − 4x
⎠
1
1
4
2
= ∫ ( x − 4 x) [(4 x3 − 4)dx] − ln 7 ∫ dx
4
3
1 4
= ( x − 4 x) 2 − (ln 7) x + C
6
x − x −2
1
( x + 1) ln ( x2 + 1)
1
dx
⎡ 2x
⎤
dx ⎥
⎢ 2
ln x + 1 ⎣ x + 1 ⎦
(
1
)
2
(
)
= ln ln x 2 + 1 + C
x3 − 1
∫ ⎜⎜
2x
2
=∫
2
+ C = ln 3 / 2 ( x3 + 1) 2 + C
3
(
∫
44.
xe x
2
2
ex + 2
1
⋅
2
(
(
)
− 12
1
x2
⎡ 2 xe x 2 dx ⎤
dx = ∫ e + 2
⎣⎢
⎦⎥
2
2
ex + 2
)
1
2
2
+ C = ex + 2 + C
1
2
5
∫ (3x + 1)[1 + ln(3x + 1)]2 dx
5
⎡ 1
⎤
⋅ 3 dx ⎥
[1 + ln(3 x + 1)]−2 ⎢
∫
3
⎣ 3x + 1
⎦
5
=−
+C
3[1 + ln(3 x + 1)]
=
−2
)⎤⎥⎦ dx
1
ln x 2 + 2 x −1 + C
2
45.
∫
( e− x + 6 )
e
x
2
(
dx = − ∫ e − x + 6
e− x + 6 )
(
=−
)
2
⎡ −e− x dx ⎤
⎣
⎦
3
39.
∫
2 x 4 − 8 x3 − 6 x 2 + 4
x3
dx
6 4 ⎞
⎛
= ∫ ⎜ 2 x − 8 − + ⎟ dx
x x3 ⎠
⎝
1
= 2 ∫ x dx − ∫ 8 dx − 6 ∫ dx + 4 ∫ x −3 dx
x
⎡
⎤
⎢ 1
⎥
1
46. ∫ ⎢
−
⎥ dx
2
⎢ 8 x + 1 e x 8 + e− x ⎥
⎣
⎦
−2
1
1
⎡ −e− x dx ⎤
= ∫
[8 dx ] − (−1) ∫ 8 + e − x
⎣
⎦
8 8x + 1
(
−2
2
40.
e x + e− x
∫ e x − e− x dx = ∫ e x − e− x ⎡⎢⎣( e
1
x
)
(
x
x
− 8 x − 6 ln x + 4 ⋅
+C
2
−2
2
= x 2 − 8 x − 6 ln x −
+C
x2
= 2⋅
+C
3
(
)
−1
8 + e− x
1
= ln 8 x + 1 +
+C
8
−1
1
1
= ln 8 x + 1 −
+C
8
8 + e− x
)
+ e− x dx ⎤⎥
⎦
= ln e x − e − x + C
41. By using long division on the integrand,
x
1 ⎞
⎛
∫ x + 1 dx = ∫ ⎜⎝1 − x + 1 ⎟⎠ dx = x − ln x + 1 + C
562
)
ISM: Introductory Mathematical Analysis
47.
∫(x
3
+ ex
)
(
x 2 + e dx = ∫ x x 2 + e x 2 + e
(
1
= ∫ x2 + e
2
=
48.
(
1 2
x +e
5
∫3
x ln x
)(
2
1 ( x + e)
[2 x dx] = ⋅
Section 14.5
)
)
5
2
3
2
1
2
53. eln( x
+C
54.
+C
55.
( )
x ln x
(1 + ln x )dx = ∫ eln 3
(1 + ln x )dx
1
e(ln 3) x ln x [(ln 3)(1 + ln x)dx]
ln 3 ∫
1 (ln 3) x ln x
1 ln 3 x ln x
=
⋅e
+C =
e
+C
ln 3
ln 3
=
49.
∫
=
56.
3x ln x
+C
ln 3
1
2
⎛
2
3⋅8
3
2
3
2
3
3
2
1
2
57.
1
2
⎞ ⎡ 32 3 12 ⎤
⎢8 ⋅ 2 ⋅ x dx ⎥
⎣
⎦
3
∫ ⎜⎝ 8 2 x 2 + 3 ⎟⎠
=
3 ⋅16 2
⎛ 32 32
⎞2
⎜ 8 x + 3⎟
⎠ +C
⋅⎝
50.
3
2
− 23
∫
∫e
f ( x ) + ln( f ′( x ))
dx = ∫ e f ( x ) ⋅ eln( f ′( x )) dx
= ∫ e f ( x ) [ f ′( x) dx]
dr
200
=
dq (q + 2)2
r = pq, then p =
⎡1 ⎤
⎢ x dx ⎥
⎣
⎦
r 100
=
.
q q+2
The demand function is p =
1
3
= 6(ln x) + C
s
3
2
2
ds = − ∫ e− s
51. ∫
3
3
e s
3
2
= − e− s + C
3
(q + 2)−1
+C
−1
200
+C
q+2
When q = 0, then r = 0, so 0 = –100 + C, or
200
100q
+ 100 =
. Since
C = 100. Hence r = −
q+2
q+2
3
2
∫ x(ln x)2 / 3 dx = 2∫ (ln x)
ln x + ln e x
dx
x
x
ln x + x
⎛ ln x ⎞
=∫
+ 1⎟ dx
dx = ∫ ⎜
x
⎝ x
⎠
∫
=−
3
2
⎡
2 + 3⎤ + C
(8
)
=
x
⎢
⎥
⎦
36 2 ⎣
1
( ) dx =
ln xe x
r = ∫ 200(q + 2)−2 dq = 200 ⋅
3
2
∫ dx = ∫ 1 dx = x + C
= e f ( x) + C
⎛
⎞
x (8 x ) + 3dx = ∫ ⎜ 8 x + 3 ⎟ ⋅ x dx
⎝
⎠
3
2
+1)
ln 2 x
⎡1 ⎤
= ∫ (ln x) ⎢ dx ⎥ + ∫ 1 dx =
+ x+C
2
⎣x ⎦
( )
=
2
is simply x 2 + 1. Thus
1
ln( x 2 +1)
dx = ∫ ( x 2 + 1)dx = x3 + x + C
∫e
3
dx
5
2
5
2
2
)
58.
⎡ 3 12 ⎤
⎢ − 2 s ds ⎥
⎣
⎦
100
.
q+2
dr
900
=
dq (2q + 3)3
r = ∫ 900(2q + 3)−3 dq
ln 3 x
1
⎡1 ⎤
dx = ∫ (ln x)3 ⎢ dx ⎥
52. ∫
3x
3
⎣x ⎦
= 900 ⋅
1
(2q + 3) −3 [2 dq]
2∫
= 450 ⋅
(2q + 3)−2
225
+C = −
+C
−2
(2q + 3)2
When q = 0, then r = 0, so 0 = –25 + C or
225
C = 25. Hence r = −
+ 25 . Since
(2q + 3) 2
1 (ln x)4
1
= ⋅
+ C = ln 4 x + C
3
4
12
563
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
r 25
225
=
−
q q q(2q + 3)2
r = pq, then p =
62.
−1 ⎞
⎛
1 (2 I ) 2 ⎟
⎜
−
C=∫
dI
⎜2
2 ⎟
⎝
⎠
1
1
−1
= ∫ dI − ∫ (2 I ) 2 [2 dI ]
2
4
⎤
25 ⎡
9
The demand function is p =
⎢1 −
⎥.
2
q ⎢⎣ (2q + 3) ⎥⎦
59.
dc
20
=
dq q + 5
1
1
1 (2 I ) 2
= ⋅I − ⋅
+ C1
1
2
4
20
1
dq = 20 ∫
dq = 20 ln q + 5 + C
q+5
q+5
When q = 0, then c = 2000, so
2000 = 20 ln(5) + C, or C = 2000 – 20 ln 5.
Hence c = 20 ln q + 5 + 2000 − 20 ln 5
c=∫
2
2I
I
= −
+ C1
2
2
3
3
4
+ C1 , so
implies = 1 −
4
4
2
I
The consumption function is C = −
2
C (2) =
q+5
= 20 ( ln q + 5 − ln 5 ) + 2000 = 20 ln
+ 2000
5
q+5
+ 2000.
The cost function is c = 20 ln
5
60.
63.
dc
= 3e0.002q
dq
c = ∫ 3e0.002 q dq = 3 ⋅
1
e0.002q [0.002 dq]
∫
0.002
The cost function is c = 1500e
0.002 q
C = ∫I
⎤
1
⎥ dI = 3 dI − 1 I − 2 dI
∫
∫
⎥
4
6
⎦
1
3
1 I2
3
I
= I− ⋅
+ C1 = I −
+ C1
1
4
6
4
3
2
+ 500.
Thus C =
dC
1
=
dI
I
− 12
3
I
+ C1 .
I−
4
3
C(25) = 23 implies that 23 =
1
dI =
I2
1
2
71
.
12
The consumption function is
3
1
71
C= I−
I+ .
4
3
12
C(9) = 8 implies that 8 = 2 ⋅ 3 + C1 , or C1 = 2 .
(
)
I +1 .
The consumption function is C = 2
3
5
⋅ 25 − + C1 , so
4
3
C1 =
+ C1 = 2 I + C1
Thus C = 2 I + 2 = 2
3
.
4
2I 3
+ .
2
4
C1 =
dC 3
1
= −
dI 4 6 I
−1
⎡
3 I 2
C = ∫⎢ −
⎢4
6
⎣
= 1500e0.002q + C
When q = 0, then c = 2000, so 2000 = 1500 + C,
or C = 500.
61.
dC 1
1
= −
dI 2 2 2 I
(
)
I +1 .
64.
dc
100
= 10 −
dq
q + 10
⎛
100 ⎞
c = ∫ ⎜ 10 −
⎟ dq = 10q − 100 ln q + 10 + C
q
+ 10 ⎠
⎝
ln q + 10 C
c
Avg. cost = = 10 − 100
+
q
q
q
When q = 100, then avg. cost = 50, so
ln(110) C
+
50 = 10 − 100
, or
100
100
C = 100(40 + ln (110)). Thus
c = 10q − 100 ln q + 10 + 100(40 + ln(110))
564
ISM: Introductory Mathematical Analysis
Section 14.5
Evaluating c when q = 0 gives fixed cost:
c(0) = –100 ln(10) + 100(40 + ln (110)) ≈ 4240.
The fixed cost is $4240.
b.
1
dc 100q − 3998q + 60
=
dq
q 2 − 40q + 1
a.
3
⎛
⎞2
0.04q 2 + 4 ⎟
⎜
0.9 ⎝
⎠ +C
=
⋅
3
0.06
dc
100(40)2 − 3998(40) + 60
=
dq q = 40
(40) 2 − 40(40) + 1
= $140 per unit
2
3
3
⎛
⎞2
Thus c = 10 ⎜ 0.04q 2 + 4 ⎟ + C . When
⎝
⎠
dc
by using long
dq
3
q = 0, then c = 360, so 360 = 10(4) 2 + C , or
3
division:
100q 2 − 3998q + 60
c=∫
dq
q 2 − 40q + 1
3
⎛
⎞2
C = 280. Hence c = 10 ⎜ 0.04q 2 + 4 ⎟ + 280 .
⎝
⎠
3
When q = 25, then c = 10(9) 2 + 280 = $550 .
⎛
2q − 40 ⎞
= ∫ ⎜ 100 +
⎟ dq
2
⎜
q − 40q + 1 ⎟⎠
⎝
1
= ∫ 100 dq + ∫
[(2q − 40)dq]
2
q − 40q + 1
c.
Thus c = 100q + ln q 2 − 40q + 1 + C. When
q = 0, then c = 10,000, so
10,000 = 0 + ln(1) + C, so C = 10,000.
If c = f(q), then f(q + dq) ≈ f(q) + dc
dc
= f (q) +
dq . Letting q = 25 and
dq
dq = –2, we have
dc
f (23) = f (25 − 2) ≈ f (25) +
⋅ (−2)
dq q = 25
= 550 + 13.50(–2) = $523
2
Hence c = 100q + ln q − 40q + 1 + 10, 000 .
When q = 40, then
c = 4000 + ln(1) + 10,000 = $14,000.
67.
dV
8t 3
=
dt
0.2t 4 + 8000
8t 3
V =∫
If c = f(q), then
dc
f (q + dq ) ≈ f (q ) + dc = f (q) +
dq
dq
Letting q = 40 and dq = 2, we have
dc
f (42) = f (40 + 2) ≈ f (40) +
⋅ (2)
dq q = 40
0.2t 4 + 8000
(
= 10 ∫ 0.2t 4 + 8000
( 0.2t 4 + 8000)
= 10
1
2
= 14,000 + 140(2) = $14,280
)
dt
− 12
⎡ 0.8t 3 ⎤ dt
⎣
⎦
1
2
+C
Thus V = 20 0.2t 4 + 8000 + C . If t = 0, then
V = 500, so 500 = 20 8000 + C ,
500 = 20 1600 ⋅ 5 + C , 500 = 800 5 + C , or
C = 500 − 800 5 . Hence
3
dc 9
=
q 0.04q 2 + 4
66.
dq 10
a.
1
⎡
2 dq ⎤
0.06
q
⎢⎣
⎥⎦
3
b. To find c, we integrate
c.
3
9
q 0.04q 2 + 4dq
10
3
2
0.9 ⎛
2 + 4⎞
=
0.04
q
⎜
⎟
∫
0.06 ⎝
⎠
2
65.
c=∫
dc
9
9
27
25 9 = ⋅ 5 ⋅ 3 =
=
10
2
dq q = 25 10
= $13.50 per unit
V = 20 0.2t 4 + 8000 + 500 − 800 5 .
When t = 10, then
V = 20 10, 000 + 500 − 800 5
= 20(100) + 500 − 800 5 ≈ $711 per acre.
565
Chapter 14: Integration
68.
ISM: Introductory Mathematical Analysis
dr
a
ae− q
ae− q
=
=
=
dq e q + b (eq + b)e− q 1 + be− q
r=∫
ae− q
−q
b.
1
⎛ 1⎞
dq = ⎜ − ⎟ a ∫
[−be − q dq ]
⎝ b ⎠ 1 + be− q
1
Thus S =
I
I
− 5.4 3 + C1 . When I = 24,
2
3
then S = 3, so 3 = 12 − 5.4 3 8 + C1 , or
I
I
− 5.4 3 + 1.8 . If C is
2
3
the total national consumption (in billions of
dollars), then C + S = I, or
⎛I
⎞
I
C = I − S = I − ⎜⎜ − 5.4 3 + 1.8 ⎟⎟ .
3
⎝2
⎠
C1 = 1.8 . Thus S =
dS
5
dI = ∫
dI
dI
( I + 2) 2
Therefore, C =
( I + 2) −1
+ C1
−1
c.
5
+ C1 . If C is the total national
I +2
consumption (in billions of dollars), then
5
− C1 .
C + S = I, or C = I – S. Hence C = I +
I +2
1
When I = 8, then C = 7.5, so 7.5 = 8 + − C1 , or
2
5
C1 = 1 . Thus S = 1 −
. If S = 0, then
I +2
5
5
0 = 1−
⇒
=1⇒ 5 = I + 2 ⇒ I = 3
I +2
I +2
Thus S = −
70. a.
⎞
⎟ dI
⎟
⎠
I 1.8 I 3
I 5.4 3
= −
⋅ + C1 = −
⋅ I + C1
2 33 1
2 33
3
a
Now r = 0 when q = 0, so 0 = − ln(1 + b) + C ,
b
a
or C = ln(1 + b). Hence
b
a
a
r = − ln(1 + be − q ) + ln(1 + b)
b
b
a
1+ b
= ln
b 1 + be − q
r
a
1+ b
p= =
ln
q bq 1 + be − q
= 5∫ ( I + 2) −2 dI = 5 ⋅
⎛1
1.8
dS
dI = ∫ ⎜ −
⎜2 3 2
dI
3I
⎝
⎛ 1 1.8 − 23 ⎞
I ⎟ dI
= ∫⎜ −
3
3
⎝2
⎠
1 + be
a
= − ln(1 + be − q ) + C
b
69. S = ∫
S=∫
I
I
+ 5.4 3 − 1.8 .
2
3
From (b), when I = 81, then
81
81
C = + 5.4 3
− 1.8 = 40.5 + 16.2 − 1.8
2
3
= 54.9
Thus consumption is $54.9 billion when
income is $81 billion.
d. If C = f(I), then
f(I + dI) ≈ f(I) + dC = f ( I ) +
dC
dI . Let
dI
I = 81 and dI = –3. Then
f (78) = f (81 − 3) ≈ f (81) +
dC
(−3)
dI I =81
17
(−3) = 53.2 . Thus when income
30
is $78 billion, then consumption is
approximately $53.2 billion.
= 54.9 +
If C is total national consumption (in
billions of dollars), then
⎛1
dC
dS
1.8 ⎞
⎟ . Thus
= 1−
= 1− ⎜ −
⎜2 3 2 ⎟
dI
dI
3I ⎠
⎝
⎛1
dC
1.8 ⎞⎟
= 1− ⎜ −
⎜ 2 3 3(81) 2 ⎟
dI I =81
⎝
⎠
⎛ 1 1.8 ⎞ 17
= 1− ⎜ −
.
⎟=
⎝ 2 27 ⎠ 30
566
ISM: Introductory Mathematical Analysis
Section 14.6
Principles in Practice 14.6
1. Divide the interval [0, 10] into n subintervals of equal length ∆x, so ∆x =
10
. The endpoints of the subintervals
n
10 ⎛ 10 ⎞ ⎛ 10 ⎞
⎛ 10 ⎞
⎛ 10 ⎞
, 2 ⎜ ⎟ , 3 ⎜ ⎟ , ... , (n − 1) ⎜ ⎟ , and n ⎜ ⎟ = 10 . Letting Sn denote the sum of the areas of the
n
⎝ n⎠ ⎝ n⎠
⎝ n⎠
⎝ n⎠
rectangles corresponding to right-hand endpoints, we have
10 ⎛ 10 ⎞ 10 ⎡ ⎛ 10 ⎞ ⎤
10 ⎡ ⎛ 10 ⎞ ⎤
Sn = R ⎜ ⎟ + R ⎢ 2 ⎜ ⎟ ⎥ +…+ R ⎢ n ⎜ ⎟ ⎥
n ⎝ n ⎠ n ⎣ ⎝ n ⎠⎦
n ⎣ ⎝ n ⎠⎦
are 0,
=
10 ⎡ ⎧
⎧
⎛ 10 ⎞ ⎫ ⎧
⎛ 10 ⎞ ⎫
⎛ 10 ⎞ ⎫⎤
⎢ ⎨600 − 0.5 ⎜ ⎟ ⎬ + ⎨600 − 0.5(2) ⎜ ⎟ ⎬ +… + ⎨600 − 0.5(n) ⎜ ⎟ ⎬⎥
n ⎣⎩
⎝ n ⎠⎭ ⎩
⎝ n ⎠⎭
⎝ n ⎠ ⎭⎦
⎩
=
⎤
10 ⎡
⎛ 10 ⎞
600n − 0.5 ⎜ ⎟ {1 + 2 +…+ n}⎥
⎢
n ⎣
⎝ n⎠
⎦
=
10 ⎡
⎛ 10 ⎞ n(n + 1) ⎤
600n − 0.5 ⎜ ⎟
⎢
⎥
n ⎣
⎝ n⎠ 2 ⎦
10
[600n − 2.5(n + 1)]
n
⎛ n +1⎞
= 6000 − 25 ⎜
⎟
⎝ n ⎠
Now take the limit of Sn as n → ∞
=
⎡
⎡
⎛ n + 1 ⎞⎤
⎛ 1 ⎞⎤
lim Sn = lim ⎢6000 − 25 ⎜
⎟ ⎥ = lim ⎢6000 − 25 ⎜ 1 + ⎟ ⎥ = 6000 − 25 = 5975
n
n →∞ ⎣
n
→∞
⎝
⎠⎦
⎝ n ⎠⎦
⎣
The total revenue for selling 10 units is $5975.
n →∞
Problems 14.6
1. f(x) = x, y = 0, x = 1
1
S3 , ∆x =
3
1 ⎛1⎞ 1 ⎛ 2⎞ 1
S3 = f ⎜ ⎟ + f ⎜ ⎟ +
3 ⎝3⎠ 3 ⎝ 3⎠ 3
⎛ 3 ⎞ 1 ⎡1 2 3⎤ 1 6 2
f⎜ ⎟ = ⎢ + + ⎥= ⋅ =
⎝ 3 ⎠ 3 ⎣3 3 3⎦ 3 3 3
2
The area is approximately
sq unit.
3
2. f(x) = 3x, y = 0, x = 1
1
S5 , ∆x =
5
1 ⎛1⎞ 1 ⎛2⎞ 1
S5 = f ⎜ ⎟ + f ⎜ ⎟ + f
5 ⎝5⎠ 5 ⎝ 5⎠ 5
9
The area is approximately
5
⎛ 3 ⎞ 1 ⎛ 4 ⎞ 1 ⎛ 5 ⎞ 1 ⎡ 3 6 9 12 15 ⎤ 1 45 9
=
⎜ ⎟ + f ⎜ ⎟+ f ⎜ ⎟ = ⎢ + + + + ⎥ = ⋅
⎝ 5 ⎠ 5 ⎝ 5 ⎠ 5 ⎝ 5 ⎠ 5 ⎣5 5 5 5 5 ⎦ 5 5 5
sq units.
567
Chapter 14: Integration
3.
ISM: Introductory Mathematical Analysis
6. f(x) = 3x + 2; [0, 3]
3
∆x =
n
3 ⎛3⎞
3 ⎛ 3⎞
Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟
n ⎝n⎠
n ⎝ n⎠
3⎡ ⎛3⎞
⎛ 3 ⎞⎤
= ⎢ f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ ⎥
n⎣ ⎝n⎠
⎝ n ⎠⎦
f ( x) = x 2 , y = 0, x = 1
1
4
1 ⎛1⎞ 1 ⎛2⎞ 1
S4 = f ⎜ ⎟ + f ⎜ ⎟ + f
4 ⎝4⎠ 4 ⎝4⎠ 4
1⎡1
4 9 16 ⎤
= ⎢ + + + ⎥
4 ⎣ 16 16 16 16 ⎦
1 30 15
= ⋅
=
4 16 32
15
The area is approximately
32
S 4 , ∆x =
4.
⎛3⎞ 1 ⎛4⎞
⎜ 4⎟+ 4 f ⎜ 4⎟
⎝ ⎠
⎝ ⎠
=
3 ⎧9
⎫
⎨ (1 +…+ n) + 2n ⎬
n ⎩n
⎭
3 ⎧ 9 n(n + 1)
⎫ 27(n + 1)
= ⎨ ⋅
+ 2n ⎬ =
+6
n ⎩n
2
2n
⎭
sq units.
=
f ( x) = x 2 + 1 , y = 0, x = 0, x = 1
1
2
1 ⎛1⎞ 1
S2 = f ⎜ ⎟ +
2 ⎝2⎠ 2
3 ⎧⎡ ⎛ 3 ⎞ ⎤
⎡ ⎛ 3 ⎞ ⎤⎫
⎨ ⎢3 ⎜ ⎟ + 2 ⎥ +…+ ⎢3 ⎜ n ⋅ ⎟ + 2 ⎥ ⎬
n ⎩⎣ ⎝ n ⎠ ⎦
⎣ ⎝ n ⎠ ⎦⎭
S 2 , ∆x =
7. a.
⎛ 2 ⎞ 1 ⎡ 5 8 ⎤ 1 13 13
f ⎜ ⎟= ⎢ + ⎥= ⋅ =
8
⎝ 2 ⎠ 2 ⎣4 4⎦ 2 4
Sn =
1 ⎡⎛ 1 ⎞ ⎛ 2 ⎞
⎛ n ⎞⎤
⎜ + 1⎟ + ⎜ + 1⎟ +…+ ⎜ + 1⎟ ⎥
⎢
n ⎣⎝ n ⎠ ⎝ n ⎠
⎝ n ⎠⎦
1 ⎡1
⎤
(1 + 2 +…+ n) + n ⎥
n ⎢⎣ n
⎦
1 ⎡ 1 n(n + 1)
⎤
= ⎢ ⋅
+ n⎥
n ⎣n
2
⎦
n +1
=
+1
2n
=
The area is approximately
13
sq units.
8
5. f(x) = 4x; [0, 1]
1
∆x =
n
1 ⎛1⎞
1 ⎛ 1⎞
Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟
n ⎝n⎠
n ⎝ n⎠
1⎡ ⎛1⎞
⎛ 1 ⎞⎤
= ⎢ f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ ⎥
n⎣ ⎝n⎠
⎝ n ⎠⎦
b.
1⎡ 1
n⎤
4 ⋅ +…+ 4 ⋅ ⎥
n ⎢⎣ n
n⎦
4
4 n(n + 1)
=
⋅
[1 +…+ n] =
2
2
n
n2
2(n + 1)
=
n
=
⎡ n +1 ⎤
lim Sn = lim ⎢
+ 1⎥
n →∞ ⎣ 2n
⎦
1
1
⎡
⎤
= lim ⎢ +
+ 1⎥
n→∞ ⎣ 2 2n
⎦
1
3
= + 0 +1 =
2
2
n →∞
2 ⎡⎛ 2 ⎞ ⎛ 2 ⎞
⎛ 2⎞
⎢⎜ ⎟ + ⎜ 2 ⋅ ⎟ +…+ ⎜ n ⋅ ⎟
n ⎢⎝ n ⎠ ⎝ n ⎠
⎝ n⎠
⎣
2
8. a.
Sn =
2
2 22 ⎡ 2
1 + 22 +…+ n 2 ⎤
⋅
⎦
n n2 ⎣
8 n(n + 1)(2n + 1)
=
⋅
6
n3
=
=
568
4(n + 1)(2n + 1)
3n 2
2⎤
⎥
⎥⎦
ISM: Introductory Mathematical Analysis
lim Sn = lim
b.
4(n + 1)(2n + 1)
Section 14.6
4 ⎡ 2n 2 + 3n + 1⎤
⎦
= lim ⎣
n→∞
3n 2
3n 2
⎡4 ⎛
3 1 ⎞⎤ 4
8
= lim ⎢ ⎜ 2 + +
⎟ ⎥ = (2 + 0 + 0) =
2
n n ⎠⎦ 3
3
n→∞ ⎣ 3 ⎝
n →∞
n →∞
9. f(x) = x, y = 0, x = 1
1
∆x =
n
1 ⎛1⎞
1
Sn = f ⎜ ⎟ +…+
n ⎝n⎠
n
1 n +1 1 ⎡ 1 ⎤
= ⋅
= ⎢1 + ⎥
2 n
2 ⎣ n⎦
1
lim Sn =
2
n →∞
1
The area is
sq unit.
2
n⎤ 1
1 n(n + 1)
⎛ 1 ⎞ 1 ⎡1
[1 +…+ n] =
f ⎜ n ⋅ ⎟ = ⎢ +…+ ⎥ =
⋅
2
n⎦ n
2
⎝ n ⎠ n ⎣n
n2
10. f(x) = 3x, y = 0, x = 1
1
∆x =
n
1 ⎛1⎞
1 ⎛ 1⎞ 1⎡ 1
n⎤
Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ = ⎢3 ⋅ +…+ 3 ⋅ ⎥
n ⎝n⎠
n ⎝ n⎠ n⎣ n
n⎦
3
3 n(n + 1) 3 n + 1 3 ⎡ 1 ⎤
=
⋅
[1 +…+ n] =
= ⋅
= ⎢1 + ⎥
2
2
2 n
2 ⎣ n⎦
n
n2
3
lim Sn =
2
n →∞
3
The area is
sq units.
2
11.
f ( x) = x 2 , y = 0, x = 1
∆x =
1
n
2
2
1 ⎛1⎞
1 ⎛ 1 ⎞ 1 ⎡⎛ 1 ⎞
⎛ 1⎞ ⎤
f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ = ⎢⎜ ⎟ +…+ ⎜ n ⋅ ⎟ ⎥
n ⎝n⎠
n ⎝ n ⎠ n ⎢⎝ n ⎠
⎝ n ⎠ ⎥⎦
⎣
1 ⎡2
1 n(n + 1)(2n + 1)
=
1 +…+ n 2 ⎤ =
⋅
3⎣
⎦
6
n
n3
Sn =
=
1 2n 2 + 3n + 1 1 ⎡
3 1 ⎤
⋅
= ⎢2 + + ⎥
2
6
6⎣
n n2 ⎦
n
lim Sn =
n →∞
1
3
The area is
1
sq unit.
3
569
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
12. y = x 2 , y = 0, x = 1, x = 2
1
n
1 ⎛ 1⎞ 1 ⎛
1⎞
1 ⎛
1⎞
Sn = f ⎜1 + ⎟ + f ⎜1 + 2 ⋅ ⎟ + " f ⎜ 1 + n ⋅ ⎟
n ⎝ n⎠ n ⎝
n⎠
n ⎝
n⎠
2
2⎫
⎧
1 ⎪⎡ 1 ⎤
1⎤ ⎪
⎡
= ⎨ ⎢1 + ⎥ + " + ⎢1 + n ⋅ ⎥ ⎬
n ⎪⎣ n ⎦
n⎦ ⎪
⎣
⎩
⎭
⎧
⎡
⎤
⎡
1⎪
1 1
1
1 ⎤ ⎫⎪
= ⎨ ⎢1 + 2 ⋅ + ⎥ + " + ⎢1 + 2n ⋅ + n 2 ⋅ ⎥ ⎬
2
n ⎪⎩ ⎣
n n ⎦
n
n 2 ⎦ ⎪⎭
⎣
∆x =
1⎧
2
1 2
2 ⎫
⎨n + (1 + " + n) + 2 (1 + " + n ) ⎬
n⎩
n
n
⎭
2 n(n + 1) 1 n(n + 1)(2n + 1)
= 1+
⋅
+
⋅
2
6
n2
n3
=
n + 1 1 2n 2 + 3n + 1
+ ⋅
n
6
n2
3 1 ⎤
⎡ 1⎤ 1 ⎡
= 1 + ⎢1 + ⎥ + ⎢ 2 + + ⎥
n n2 ⎦
⎣ n⎦ 6 ⎣
= 1+
lim Sn = 1 + 1 +
n →∞
The area is
13.
1 7
=
3 3
7
sq units.
3
f ( x) = 3x 2 , y = 0, x = 1
∆x =
1
n
Sn =
2
2
1 ⎛1⎞
1 ⎛ 1⎞ 1⎡ ⎛1⎞
⎛ 1⎞ ⎤
f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ = ⎢3 ⎜ ⎟ +…+ 3 ⎜ n ⋅ ⎟ ⎥
n ⎝n⎠
n ⎝ n⎠ n⎢ ⎝n⎠
⎝ n ⎠ ⎥⎦
⎣
3 ⎡2
3 n(n + 1)(2n + 1) 1 2n 2 + 3n + 1 1 ⎡
3 1 ⎤
1 +…+ n 2 ⎤ =
⋅
= ⋅
= ⎢2 + + ⎥
3
2
3⎣
⎦
6
2
2⎣
n n2 ⎦
n
n
n
lim Sn = 1
=
n →∞
The area is 1 sq unit.
570
ISM: Introductory Mathematical Analysis
14.
Section 14.6
f ( x) = 9 − x 2 , y = 0, x = 0
20
3
n
3 ⎛3⎞
3 ⎛ 3⎞
Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟
n ⎝n⎠
n ⎝ n⎠
2
⎡ ⎛ 3 ⎞ 2 ⎤ ⎫⎪
3 ⎧⎪ ⎡ ⎛ 3 ⎞ ⎤
= ⎨ ⎢9 − ⎜ ⎟ ⎥ +…+ ⎢9 − ⎜ n ⋅ ⎟ ⎥ ⎬
n ⎪⎢ ⎝ n ⎠ ⎥
n ⎠ ⎥⎪
⎦
⎣⎢ ⎝
⎦⎭
⎩⎣
2
⎫⎪
3 ⎧⎪
⎛3⎞
= ⎨9n − ⎜ ⎟ ⎡12 +…+ n 2 ⎤ ⎬
⎦
n⎪
⎝n⎠ ⎣
⎩
⎭⎪
y
∆x =
x
16.
5
4
∫0 9 dx
Let f(x) = 9
4
∆x =
n
4 ⎛4⎞
4 ⎛ 4⎞
Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟
n ⎝n⎠
n ⎝ n⎠
4
4
= [9 +…+ 9] = ⋅ 9n = 36
n
n
⎡ 27 n(n + 1)(2n + 1) ⎤
= 27 − ⎢ ⋅
⎥
6
⎣ n3
⎦
2
⎡
⎤
9 2n + 3n + 1
9⎡
3 1 ⎤
= 27 − ⎢
⎥ = 27 − ⎢ 2 + + 2 ⎥
2
2 ⎢⎣
2
n
n
n ⎦
⎣
⎥⎦
lim Sn = 27 − 9 = 18
n →∞
4
Sn = 36
∫0 9 dx = nlim
→∞
The area is 18 sq units.
15.
3
1
3
∫1 5x dx
10
Let f(x) = 5x.
2
∆x =
n
2 ⎛ 2⎞
2 ⎛
2⎞
Sn = f ⎜1 + ⎟ + " + f ⎜1 + n ⋅ ⎟
n ⎝ n⎠
n ⎝
n⎠
2⎡ ⎛ 2⎞
2
⎛
⎞⎤
= ⎢5 ⎜ 1 + ⎟ + " + 5 ⎜ 1 + n ⋅ ⎟ ⎥
n⎣ ⎝ n⎠
n ⎠⎦
⎝
10 ⎡
2
⎤
= ⎢ (1 + " + 1) + (1 + " + n) ⎥
n ⎣
n
⎦
10 ⎡
2 n(n + 1) ⎤
= ⎢n + ⋅
n ⎣
n
2 ⎥⎦
10
= [n + n + 1]
n
10
= (2n + 1)
n
10
= 20 +
n
y
4
17.
x
10
3
∫0 −4x dx
Let f(x) = –4x.
3
∆x =
n
3 ⎛3⎞
3 ⎛ 3⎞
Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟
n ⎝n⎠
n ⎝ n⎠
3⎡ ⎛3⎞
36
⎛ 3 ⎞⎤
= ⎢ −4 ⎜ ⎟ −…− 4 ⎜ n ⋅ ⎟ ⎥ = − [1 +…+ n]
n⎣ ⎝n⎠
⎝ n ⎠⎦
n2
=−
3
Sn = 20
∫1 5 x dx = nlim
→∞
3
36 n(n + 1)
n +1
⎡ 1⎤
⋅
= −18 ⋅
= −18 ⎢1 + ⎥
2
2
n
⎣ n⎦
n
Sn = −18
∫0 −4 x dx = nlim
→∞
571
Chapter 14: Integration
15
ISM: Introductory Mathematical Analysis
y
Sn =
3
2
⎡⎛ 1 ⎞ 2
1 ⎧⎪ ⎡⎛ 1 ⎞
1⎤
1 ⎤ ⎫⎪
⎨ ⎢⎜ ⎟ + ⎥ +…+ ⎢⎜ n ⋅ ⎟ + n ⋅ ⎥ ⎬
n ⎪ ⎢⎝ n ⎠
n⎥
n ⎥⎪
⎢⎣⎝ n ⎠
⎦
⎦⎭
⎩⎣
=
x
15
1 ⎛1⎞
1 ⎛ 1⎞
f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟
n ⎝n⎠
n ⎝ n⎠
2
⎫⎪
1 ⎧⎪⎛ 1 ⎞ ⎡ 2
1
2
⎨⎜ ⎟ ⎣1 +…+ n ⎤⎦ + [1 +…+ n]⎬
n ⎪⎝ n ⎠
n
⎩
⎭⎪
1 n(n + 1)(2n + 1) 1 n(n + 1)
=
⋅
+
⋅
6
2
n3
n2
=
18.
4
∫1 (2 x + 1)dx
Let f(x) = 2x + 1.
4 −1 3
∆x =
=
n
n
3 ⎛ 3⎞
3 ⎛
3⎞
Sn = f ⎜1 + ⎟ + " + f ⎜1 + n ⋅ ⎟
n ⎝ n⎠
n ⎝
n⎠
⎡ ⎛
3 ⎧⎡ ⎛ 3 ⎞ ⎤
3 ⎞ ⎤⎫
= ⎨ ⎢ 2 ⎜1 + ⎟ + 1⎥ + " + ⎢ 2 ⎜1 + n ⋅ ⎟ + 1⎥ ⎬
n ⎩⎣ ⎝ n ⎠ ⎦
n ⎠ ⎦⎭
⎣ ⎝
3⎧
6
⎫
= ⎨2n + (1 + 2 + " + n) + n ⎬
n⎩
n
⎭
18 n(n + 1)
= 6+
⋅
+3
2
n2
n +1
= 9 + 9⋅
n
⎛ 1⎞
= 9 + 9 ⎜1 + ⎟
⎝ n⎠
4
∫1
)
+ x dx = lim Sn =
n→∞
1 1 5
+ =
3 2 6
y
x
20.
5
2
∫1 ( x + 2)dx
Let f(x) = x + 2.
1
∆x =
n
1 ⎛ 1⎞
1 ⎛
1⎞
Sn = f ⎜ 1 + ⎟ +…+ f ⎜ 1 + n ⋅ ⎟
n ⎝ n⎠
n ⎝
n⎠
1 ⎧ ⎡⎛ 1 ⎞ ⎤
⎡⎛
1 ⎞ ⎤⎫
= ⎨ ⎢⎜1 + ⎟ + 2 ⎥ +…+ ⎢⎜ 1 + n ⋅ ⎟ + 2 ⎥ ⎬
n ⎩ ⎣⎝ n ⎠ ⎦
n ⎠ ⎦⎭
⎣⎝
1⎧
1
⎫
= ⎨n + (1 + " + n) + 2n ⎬
n⎩
n
⎭
1 n(n + 1)
= 1+
⋅
+2
2
n2
1 n +1
= 3+ ⋅
2 n
1 ⎡ 1⎤
= 3 + ⎢1 + ⎥
2 ⎣ n⎦
2
1 7
Sn = 3 + =
∫1 ( x + 2)dx = nlim
2 2
→∞
x
6
)
+ x dx
Let f ( x) = x 2 + x .
∆x =
2
1
n →∞
2
1⎡
3 1 ⎤ 1 ⎡ 1⎤
2 + + ⎥ + ⎢1 + ⎥
⎢
6⎣
n n2 ⎦ 2 ⎣ n ⎦
5
10
1
=
1
(2 x + 1)dx = lim Sn = 9 + 9 = 18
∫0 ( x
1 2n 2 + 3n + 1 1 n + 1
⋅
+ ⋅
6
2 n
n2
∫0 ( x
y
19.
=
1
n
572
ISM: Introductory Mathematical Analysis
Section 14.7
y
5
area is composed of three subareas, A1 , A2 , and
A3 and A = A1 + A2 + A3 .
A1 = area of rectangle = (2)(1) = 2 sq units
x
21.
3
1
1
(1)(1) = sq unit
2
2
1 ⎛1⎞ 1
A3 = area of triangle = (1) ⎜ ⎟ = sq unit
2 ⎝2⎠ 4
Since
1 1 11
sq units, we
A = A1 + A2 + A3 = 2 + + =
2 4 4
3
11
have ∫ f ( x)dx = .
−1
4
A2 = area of triangle =
5
1 2
x 2 + 1 dx is simply a real number. Thus
∫2
3
Dx ⎡⎢ ∫ x 2 + 1dx ⎤⎥ = Dx (real number) = 0.
⎣ 2
⎦
22.
0 ≤ x <1
2 if
⎧
⎪
f ( x) = ⎨ 4 − 2 x if
⎪5 x − 10 if
⎩
y
3
1≤ x < 2
y = f(x) A3
2≤ x≤3
1
A 1 2 A2
f is continuous and f(x) ≥ 0 on [0, 3]. Thus
x
3
3
∫0 f ( x) dx gies the area A bounded by y = f(x),
y = 0, x = 0 and x = 3. From the diagram, this
area is composed of three subareas, A1 , A2 and
A3 , and A = A1 + A2 + A3 .
A1 = area of rectangle = (1)(2) = 2 sq unit
24. 44.6 sq units
25. 14.77 sq units
1
(1)(2) = 1 sq unit
2
1
A3 = area of triangle = (1)(10) = 5 sq unit
2
Since A = A1 + A2 + A3 = 2 + 1 + 5 = 8 sq units,
A2 = area of triangle =
we have
10
26. 1.7 sq units
27. 2.4
28. 0.7
3
∫0 f ( x) dx = 8.
29. –25.5
30. 0.39
y
Principles in Practice 14.7
5
y = f(x)
1.
A3
A1
A2
1
x
2
3
∫3 10, 000e
(
0.02t
⎛
e0.02t
dt = ⎜ 10, 000
⎜
0.02
⎝
= 500, 000e0.02t
3
6
⎞
⎟
⎟
⎠3
)3
6
(
)
= 500, 000 ( e0.12 − e0.06 ) ≈ 32,830
= 500, 000 e0.02(6) − e0.02(3)
⎧1
if x ≤ 1
⎪⎪
23. f ( x) = ⎨2 − x if 1 ≤ x ≤ 2
⎪
x
⎪⎩−1 + 2 if x > 2
f is continuous and f(x) ≥ 0 on [–1, 3]. Thus
∫−1 f ( x)dx
6
The total income for the chain between the third
and sixth years was about $32,830.
gives the area A bounded by y = f(x),
y = 0, x = –1, and x = 3. From the diagram, this
573
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
2. The total cost for the first 5 years is M(5) or
5
9.
M (5) − M (0) = ∫ M ′( x )dx
0
5
⎛ x3
⎞
90
x
+
5000
dx
=
⎜ 90 + 5000 x ⎟
∫0
⎜
⎟
3
⎝
⎠0
5
(
)
2
(
= 30 x3 + 5000 x
) 0 = 30(5)3 + 5000(5) − 0
5
Problems 14.7
2.
3
∫0
∫2
(1 − e)dx = (1 − e) x
2
∫1
x2
5 x dx = 5 ⋅
2
2
11.
∫1 3t
1
∫−3 (2 x − 3)dx = ( x
1
13.
3
3
1 ⎛ 3⎞ 4
dt = − ⋅ t −2 = − − ⎜ − ⎟ =
1
2
6 ⎝ 2⎠ 3
2 x −2
x −1
dx = −
2
2
∫1
2
2
=−
1
∫2 (
3
1
2x 1
∫4 (2t − 3t
= −160 − (−10) = −150
2
2
)
)dt = (t
−8
8
− 3x
1
) −3 = −2 − 18 = −20
14.
3
3
∫1/ 2 (
3/ 2
=
1
⎞
1 ⎛ 17 ⎞
⎟ = − −⎜− ⎟
⎟
2 ⎝ 2⎠
⎠ −1
1
y 2 − 2 y + 1 dy = ∫ ( y − 1)2 dy = ( y − 1)3
2
3
2
2
8
x dx = ∫ x 4 / 3 dx
8
−8
3 ⋅128 3(−128)
=
−
7
7
768
=
7
8 1 7
− =
3 3 3
1
8 3 4
∫−8
3x7 / 3
=
7
16
=
=8
2
8.
−3
1 ⎛ 1⎞ 1
= − −⎜− ⎟ =
4 ⎝ 2⎠ 4
4
⎛
1
9 y2
6. ∫ (4 − 9 y ) = ⎜ 4 y −
⎜
−1
2
⎝
=
3
5 15
= 10 − =
2 2
x2
4. ∫ −5 x dx = −5 ⋅
2
2
7.
9
9
2
8
5.
9
∫8 dt = ∫8 1 dt = t 8 = 9 − 8 = 1
12.
= 4(1 − e) − 2(1 − e) = 2(1 − e)
3.
)
10.
3
5 dx = 5 x = 5(3) − 5(0) = 15 − 0 = 15
0
4
−1
⎛
⎞
w2
3w − w − 1 dw = ⎜ w3 −
− w⎟
⎜
⎟
2
⎝
⎠ −2
2
1
15
= − − (−8) =
2
2
= 3750 + 25,000 = 28,750
The total cost for the first 5 years is $28,750.
1.
∫−2 (
−1
3/ 2
⎛ x3 x 2
⎞
x + x + 1 dx = ⎜ +
+ x⎟
⎜ 3
⎟
2
⎝
⎠ 1/ 2
)
2
15 2 37
− =
4 3 12
3
1
15.
∫1/ 2 x2
16.
∫9 (
36
3
dx = −
1
1
5
= − − (−2) =
x 1/ 2
3
3
)
( z + 1)6
17. ∫ ( z + 1) dz =
−1
6
1
2
1
− t3 )
4
= 0 − (−48) = 48
574
36
⎛2 3
⎞
x − 2 dx = ⎜ x 2 − 2 x ⎟ = 72 − 0 = 72
3
⎝
⎠9
1
5
=
−1
32
32
−0 =
3
3
ISM: Introductory Mathematical Analysis
Section 14.7
8
18.
8⎛
∫1 ⎜⎝ x
1
3
−x
− 13
2
⎛ 43
⎞
3x
3x 3 ⎟
⎞
⎜
−
⎟ dx = ⎜
4
2 ⎟
⎠
⎝
⎠1
27.
⎛ 3 ⎞ 27
= 6−⎜− ⎟ =
⎝ 4⎠ 4
19.
1
∫0 2 x
=
20.
21.
2
( x3 − 1)
3
(
)
1 3
x −1
6
3
∫2
(
)
3
2 1 3
x − 1 ⎡3x 2 dx ⎤
∫
⎣
⎦
3 0
= 0−
1
1
=−
6
6
0
( x + 2)4
4
3
28.
2
dy = 4 ln y
29.
−1
24.
25.
26.
∫
e +1
∫2
30.
1
1
e dx =
2
)(
+ 4 x x3 + 2 x 2
(
= ∫ x3 + 2 x 2
0
) (
4
)
4
31.
∫1/ 3
1
∫−1
10 − 3 pdp = −
(
q q 2 + 3dq =
q2 + 3
)
3
2
1
1 2
2 [ −3 dp ]
(10
−
3
p
)
3 ∫1/ 3
(
)
1
1 1 2
2
+
q
3
[2q dq]
2 ∫−1
1
=
3
8 8
− =0
3 3
∫
=
0
(
(
)
)
3
1 7x +1
⋅
4
21
4
3
1
( 7 x + 1)
3
=
4
3
0
dx
)
32.
243
243
−0 =
5
5
0
16 1 15
−
=
28 28 28
⎞
x
⎜⎜ 3x − 2
⎟ dx
( x + 2)4 / 3 ⎟⎠
⎝
7
1 7
= ∫ 3 x dx − ∫ ( x 2 + 2)−4 / 3 [2 x dx]
0
2 0
∫0
7⎛
3
= x2
2
0
7
0
1 ( x 2 + 2)−1/ 3
− ⋅
2
−1
3
7
0
3
3
= (7 − 0) + [9−1/ 3 − 2−1/ 3 ]
2
2
3⎛
1
1 ⎞
= ⎜7 +
−
⎟
3
3
2⎝
9
2⎠
575
1
28
3
=
1
3
1 1
+ 1dx = ∫ 7 x3 + 1 ⎡ 21x 2 dx ⎤
⎣
⎦
0
21
1 23
x 7 x3
0
1
⎡ 3 x 2 + 4 x dx ⎤
⎢⎣
⎥⎦
=
5
1
1 20 / 3
2 [3 dx ]
x
+
(3
5)
3 ∫−1/ 3
−1
51
x3 + 2 x 2 )
(
=
4
3 x + 5 dx =
0
2 x3
1
2
= e5 x = e5 − 0 = e5
1
e +1
dx = ln x − 1
= ln e − ln1 = 1 − 0 = 1
2
x −1
∫0 ( 3x
∫−1/ 3
=
5 1 x3 2
5 3
e [3 x dx] = e x
∫
3 0
3
5 1 0
5
= (e − e ) = (e − 1)
3
3
1
∫0 5 x
20 / 3
1
3
= − − (−1) =
4
4
3
2
2
38
= − (10 − 3 p) 2
= − (8 − 27) =
9
9
9
1/ 3
1
1 5
e dx
0
( x − 3)
2
2
= 4(ln 8 − ln1)
6
22. ∫ e dx = 6 ln x
= 6 ln1 − 6 ln ee
−e x
e
−e
= 0 − 6e = −6e
23.
1
20
=4(ln 8 – 0) = 4 ln 8
−1
5
5
−3
3 3
2
(3x + 5) 2
9
− 13
2
= (125 − 8) = 26
9
625
369
− 64 =
4
4
=
5
=
8
y
∫4
( x − 3)−2
dx = 2 ∫ ( x − 3) dx = 2 ⋅
4
−2
( x − 3)3
2
=−
dx =
( x + 2)3 dx =
84
∫1
41
5
4
Chapter 14: Integration
33.
2 x3 + x
1
∫0 x 2 + x4 + 1
ISM: Introductory Mathematical Analysis
39.
dx
(
)
=
1 1
1
⎡ 4 x3 + 2 x dx ⎤
⎥⎦
2 ∫0 x 4 + x 2 + 1 ⎣⎢
=
1
ln x 4 + x 2 + 1
2
(
1
(
n
= m(b − a ) + b 2 − a 2
2
36.
37.
1 ex
∫0
⎞
n
⎟ = my + y 2
⎟
2
⎠a
a
0
a
)
x
95
∫1
ln e
42.
2
1
∫−11 + e x
= −2 ∫
−2
⎞
⎟ = 8 ⎧[ 0 − (−2) ] + ⎛ 1 − 0 ⎞ ⎫
⎨
⎜
⎟⎬
⎟⎟
⎝2
⎠⎭
⎩
0⎠
= 2 ln
43.
2
∫0
+1
1
1
⋅
e− x
e− x
dx
dx
(
)
1/ 2
1/ 2
2
+x
= 9 2 −6
4 x 2 dx + ∫
0
1/ 2
3
2
2 x dx
1/ 2
2
1 ⎞ 47
⎛1
⎞ ⎛
= ⎜ − 0⎟ + ⎜ 4 − ⎟ =
6
4 ⎠ 12
⎝
⎠ ⎝
3
44. ⎛⎜ ∫ x dx ⎞⎟ − ∫
⎝ 1
⎠
)
3 3
x dx
1
3
⎛ 2 3⎞
x ⎟
x4
⎜
=
−
⎜⎜ 2 ⎟⎟
4
1
1
⎝
⎠
3
⎛ 9 1 ⎞ ⎛ 81 1 ⎞
= ⎜ − ⎟ −⎜ − ⎟
⎝2 2⎠ ⎝ 4 4⎠
= 43 − 20
= 44
576
e +1
1 +1
e
e2 + e
e(e + 1)
= 2 ln
= 2 ln e = 2
1+ e
1+ e
0
3
1
) −1
⎡ −e− x dx ⎤ = −2 ln e− x + 1
⎦
+1 ⎣
f ( x)dx = ∫
4 x3
=
3
1 ⎞
38. ∫ ⎜ 6 x −
⎟ dx
1 ⎝
2x ⎠
2 1
1 2
−1
= 6 ∫ x 2 dx − ∫ (2 x) 2 [2 dx]
1
2 1
) (
−1 1 + e x
(
2⎛
(
1
2
1
= −2 ln e−1 + 1 + 2 ln(e + 1) = 2 ln
⎛ x −1 x − 2 x − 3 ⎞
−2
−3
−4
+
−
=
+
−
x
x
x
dx
3(
)
3
⎜
⎟
∫π
⎜ −1 −2 −3 ⎟
⎝
⎠π
⎛ 1
1
1 ⎞ ⎛ 1
1
1 ⎞
= 3⎜ − −
+
+
⎟ − 3⎜ − −
⎟
⎝ e 2e 2 3e3 ⎠ ⎝ π 2π2 3π3 ⎠
1 3
3
3
3
1
= − −
+ +
−
3 e
2 π
2
π3
2e
2π
e
e
dx = ∫
−1 e− x
e
2
95
1
−1 e− x
1
1
⎡ 3
⎤
= ⎢ 4 x 2 − (2 x) 2 ⎥ = 8 2 − 2 − 4 − 2
⎣
⎦1
)
95
x
95
dx = ∫ 1 dx = x 1
1
x
= 95 − 1 = 94
dx = ∫
e− x
⎛ 2
x
= 8⎜ −
⎜⎜ 2
⎝
x2
2
(
= (6 + ln 19) – 0 = 6 + ln 19
= 2∫
+
)
2
∫−2 8 x dx = 8 ⎜⎝ ∫−2 − x dx + ∫0 x dx ⎟⎠
0
x
(
⎡ x4 x2
⎤
=⎢ +
+ ln x3 + 5 x + 1 ⎥
2
⎣⎢ 4
⎦⎥ 0
⎞
1
1 3 x2 + 2 x
[(2 x + 2)dx]
e
2 ∫1
41. Using long division on the integrand
2 x 6 + 6 x 4 + x3 + 8 x 2 + x + 5
dx
∫0
x3 + 5 x + 1
2⎡
3x 2 + 5 ⎤
= ∫ ⎢ x3 + x +
⎥dx
0
x3 + 5 x + 1 ⎥⎦
⎢⎣
1
⎛
dx =
1 x2 + 2 x
1
e3 12
e
e −1
= e15 − e3 =
2
2
2
1
b
b
b
40.
− e− x
1
dx = (e x + e− x )
2
2
0
1
= [(e + e −1 ) + (1 + 1)]
2
1⎛
1
⎞
= ⎜e + + 2⎟
2⎝
e
⎠
1
x2 + 2 x
3
=
) 0 = 12 [ln 3 − ln1] = 12 ln 3
⎛
b
ny 2
34. ∫ (m + ny )dy = ⎜ my +
⎜
a
2
⎝
35.
3
∫1 ( x + 1)e
ISM: Introductory Mathematical Analysis
45.
x
1
1
t2
f ( x) = ∫ 3
x
dt = −
1⎛
1
Section 14.7
3
3
3
= − + 3 = 3−
t1
x
x
3⎞
∫e f ( x)dx = ∫e ⎜⎝ 3 − x ⎟⎠ dx = ( 3x − 3ln x )
51.
46.
∫
∫
1
3 2
dx = 0 +
=
=
1
1
3 2
1
3 2
1
=
3
47.
3
∫1
2
∫1
2
3
1
3
2
3
1
2
∫0
3 2
2
e
2
3
1 dx
2
48.
∫1
4
∫1
0
2 −0
)
49.
3 x3
3⎛
d
3 x
∫2 ⎜⎝ dx ∫2 e
50.
f ( x) = ∫
3
x et
54.
2
3
4
1
2
2
et + e − t
x
1
e
=∫
e
t
e +e
−t
= ln et + e−t
=
1
e + e− x
e x − e− x
x
10
∫0
−4
x
− 12
(
3
∫2 f ( x) dx = 7 − 6 = 1 .
55.
d 3 x3
e dx = 0. Thus
dx ∫2
31
1⎛
1⎞
⎝
⎠
2
0
1 ⎡ 1 ⎛ 1 ⎞⎤ 1
= ⎢ − ⎜ − ⎟ ⎥ = . Thus
3 ⎣ 8 ⎝ 8 ⎠ ⎦ 12
dx =
)
x
10−4
1
2
1
2 0
5
∫0 2000e
−0.06t
=2 x
dt = 2000 ⋅
2000 −0.06t
e
0.06
≈ $8639
5
=−
dt
[(et − e−t ) dt ]
x
e
= ln(e x + e− x ) − ln(ee − e−e )
f ′( x) =
0
1
1
−0 =
2
2
10−4
0
= 2 10−4 − 0
= 2 10−2 = 0.02
3
⎞
3
dx ⎟ dx = ∫ 0 dx = C 2 = C − C = 0
2
⎠
− e −t
=
a
1
3
dx is a constant, so
0
53. The total number receiving between a and b
dollars equals the number N(a) receiving a or
more dollars minus the number N(b) receiving b
or more dollars. Thus
f ( x)dx = ∫ f ( x)dx − ∫ f ( x)dx + ∫ f ( x)dx
∫2 e
0
1
x2
2
1
1
µ = ; σ2 =
2
12
f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx
3
1
b
4
2
1
N (a) − N (b) = ∫ − Ax − B dx .
2
6 = 2 − ∫ f ( x)dx + 5 , so
5
0
1⎛
1⎞
= ⎜x− ⎟
3⎝
2⎠
= ∫ f ( x)dx + ∫ f ( x)dx = 4 + 3 = 7
4
5
= α 2T − 0 = α 2T
1
f ( x)dx = ∫ f ( x)dx − ∫ f ( x)dx
3
T
σ 2 = ∫ ( x − µ ) 2 f ( x)dx = ∫ ⎜ x − ⎟ dx
0
0
2
f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx , so
1
5
52. µ = ∫ ( x ⋅1)dx = ∫ x dx =
x
(
∫0
5
α 2 dt = α 2 t
1
= (3 – 0) – (3e – 3) = 6 – 3e
7 x2
2
e dx +
7
0
T
[e x + e− x (−1)] − 0
e x + e− x
577
=−
0
5 −0.06t
1
e
[−0.06 dt ]
−0.06 ∫0
(
)
2000 −0.03
e
−1
0.06
Chapter 14: Integration
56.
∫0 ( e
t
− aτ
ISM: Introductory Mathematical Analysis
)
− e−bτ dτ
61.
1 t − aτ
1 t −bτ
e [− a dτ ] −
e [−b dτ ]
−a ∫0
−b ∫0
=
⎛ e− aτ e−bτ ⎞
= ⎜−
+
⎟
⎜
a
b ⎟⎠
⎝
0
⎛ e− at e−bt ⎞ ⎛ 1 1 ⎞
= ⎜−
+
⎟−⎜− + ⎟
⎜
a
b ⎟⎠ ⎝ a b ⎠
⎝
=
57.
− at
1− e
a
64
−
∫36 10, 000
62.
1− e
b
64
36
63.
1
2
(100 − t ) [(−1) dt ]
3
2
= − (10, 000)(100 − t ) 2
3
t
∫0
64
36
3000e0.05τ dτ = 3000 ⋅
= 60, 000e0.05τ
59.
t
0
64.
180
∫90
3
800
=
q
3
∫10 ( 250 + 90q − 3q
20
dq
200 q1/ 2
=
⋅
1
3
2
400
500
(
2
800
500
)
800 − 500 ≈ $1367.99
20
) dq = ( 250q + 45q2 − q3 ) 10
(
2
+ 8q
(
)6
12
(8t + 10)dt = 4t 2 + 10t
81× 106
∫0
(300 + t )
)
4
12
(
dt = 81× 106
(300 + t ) −3
−3
= 696 − 0 = 696
= 696 − 204 = 492
) ∫0700 (300 + t )−4 dt
700
0
700
(
) (3001+ t )3
(
)
(
)
= − 27 × 106
)
)0
(8t + 10)dt = 4t 2 + 10t
700
12
(
12
∫0
(
1 t 0.05τ
e
[0.05 dτ ]
0.05 ∫0
0
1 ⎞
⎛ 1
= − 27 × 106 ⎜
−
⎟
3
3003 ⎠
⎝ 1000
1 ⎞
⎛ 1
= − 27 × 106 ⎜
−
⎟
9
27 ⋅106 ⎠
⎝ 10
27
27
973
=−
+1 = −
+1 =
= 0.973
3
1000
1000
10
) 65
75
= 1162.5 – 942.5 = $220
60.
∫
= 81× 106
= 60, 000 e0.05t − 1
∫65 (0.2q + 8)dq = ( 0.1q
75
2000
10 3q
800 −1/ 2
q
dq
500
400
∫6
2
= − (10, 000)[216 − 512]
3
≈ 1,973,333
58.
800
500
dq = ∫
= 15,000 − 6000 = $9000
100 − tdt
= (−1)(10, 000) ∫
3
=
−bt
300q
200
=
t
2000
800
∫500
(0.004q 2 − 0.5q + 50) dq
180
0.004 3
q − 0.25q 2 + 50q
3
90
= 8676 − 3447
= $5229
=
65. G = ∫
R
−R
578
i dx = ix
R
−R
= iR − (−iR) = 2 Ri
ISM: Introductory Mathematical Analysis
66. E = ∫
Section 14.7
i ⎡ −k ( R − x) −k ( R+ x) ⎤
e
dx
+e
⎦
R
−R 2 ⎣
i ⎡ 1 R −k ( R− x)
1 R −k ( R+ x)
⎤
e
[k dx] +
e
[− k dx]⎥
∫
⎢
−
R
2 ⎣k
−k ∫− R
⎦
R
i ⎡ R −k ( R− x)
=
[k dx] − ∫ e− k ( R + x ) [− k dx]⎤⎥
e
−R
2k ⎢⎣ ∫− R
⎦
=
R
=
i ⎡ −k ( R − x) −k ( R + x) ⎤
e
−e
⎦
2k ⎣
−R
i
2k
i
=
2k
=
(
) (
)
⎡ 2 − 2e −2 kR ⎤ = i (1 − e−2kR )
⎣
⎦ k
⎡ 1 − e − k (2 R ) − e− k (2 R ) − 1 ⎤
⎣⎢
⎦⎥
∫ (m + x)[1 − (m + x)]dx = ∫0 ( m + x − m − 2mx − x
A= 0
R
R
∫0 [1 − (m + x)]dx
∫0 (1 − m − x)dx
R
R
67.
=
⎡ mx +
⎢⎣
x2
2
− m2 x − mx 2 −
x3 ⎤
3 ⎥⎦
2
R
0
R
⎡ x − mx − x 2 ⎤
2 ⎦⎥
⎣⎢
0
⎡ mR + R 2 − m2 R − mR 2 − R3 ⎤ − 0
2
3 ⎥⎦
⎢
=⎣
2
⎡ R − mR − R ⎤ − 0
2 ⎦⎥
⎣⎢
2
R ⎡⎢ m + R2 − m2 − mR − R3 ⎤⎥ m + R − m 2 − mR − R 2
⎣
⎦=
2
3
=
R
R
1− m − 2
R ⎡⎣1 − m − 2 ⎤⎦
68.
69.
3.5
∫2.5 (1 + 2 x + 3x
4
∫0
2
)dx = ( x + x 2 + x3 )
3.5
2.5
= 58.625 − 24.375
= 34.25
1 4
1 (4 x + 4) −1
dx = ∫ (4 x + 4)−2 [4 dx] = ⋅
4 0
4
−1
(4 x + 4) 2
1
4
4
0
4
1
1
1 1
1 ⎛1 ⎞ 1
=− ⋅
=− ⋅
= − ⎜ − 1⎟ =
= 0.05
4 4x + 4 0
16 x + 1 0
16 ⎝ 5 ⎠ 20
70.
1 3t
e dt
0
∫
=
1 1 3t
e3t
e [3 dt ] =
∫
3 0
3
1
=
0
(
)
1 3
e − 1 ≈ 6.36
3
71. 3.52
72. 0.23
579
2
) dx
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
estimate of the amount the culture grew over the
first four hours is
4
0.5
0.2t 2
∫0 0.3e dt ≈ 3 (34.5956) ≈ 5.77 grams.
73. 14.34
74. 3.64
Principles in Practice 14.8
Problems 14.8
60
1. In this case, f (t ) =
, n = 5, a = 0, and
t2 + 9
b−a 5−0
b = 5. Thus h =
=
= 1 . The terms to
n
5
be added are
60
60
f (0) =
=
= 20
2
3
0 +9
2 f (1) =
2 f (2) =
2 f (3) =
2 f (4) =
2(60)
=
2
1 +9
2(60)
2
3 +9
2(60)
10
=
2
2 +9
2(60)
120
13
120
=
18
=
2
120
1.
≈ 37.9473
≈ 33.2820
≈ 28.2843
2.
2
2. In this case, f (t ) = 0.3e0.2t , n = 8, a = 0, and
b−a 4
= = 0.5 . The terms to be
8
n
added are
f (0) = 0.3e0 = 0.3
4 f (0.5) = 4(0.3)e
2 f (1) = 2(0.3)e
0.2
4 f (1.5) = 4(0.3)e
2 f (2) = 2(0.3)e
1.8
2 f (3) = 2(0.3)e
4 f (3.5) = 4(0.3)e
≈ 1.2615
≈ 1.8820
≈ 1.3353
4 f (2.5) = 4(0.3)e
≈ 4.1884
≈ 3.6298
2.45
1
170
, n = 6, a = –2, b = 4
1 + x2
Simpson’s
b − a 4 − (−2) 6
=
= =1
h=
n
6
6
f (−2) = 34 = 34
4 f (−1) = 4(85) = 340
2 f (0) = 2(170) = 340
4 f (1) = 4(85) = 340
2 f (2) = 2(34) = 68
4 f (3) = 4(17) = 68
f (4) = 10 = 10
1200
170
1
∫−2 1 + x2 dx ≈ 3 (1200) = 400
≈ 0.7328
1.25
170
f ( x) =
4
0.45
0.8
, n = 6, a = –2, b = 4. Trapezoidal
1 + x2
b − a 4 − (−2) 6
h=
=
= =1
n
6
6
f (−2) = 34 = 34
2 f (−1) = 2(85) = 170
2 f (0) = 2(170) = 340
2 f (1) = 2(85) = 170
2 f (2) = 2(34) = 68
2 f (3) = 2(17) = 34
f (4) = 10 = 10
826
4
120
= 24
5
0.05
170
∫−2 1 + x2 dx ≈ 2 (826) = 413
4 +9
60
60
f (5) =
=
≈ 10.2899
2
34
5 +9
The sum of the above terms is 153.8035. The
estimate of the radius after 5 seconds is
5 60
1
∫0 2 dt ≈ 2 (153.8035) ≈ 76.90 feet.
t +9
b = 4. Thus, h =
f ( x) =
≈ 13.9060
3.2
f (4) = 0.3e ≈ 7.3598
The sum of the above terms is 34.5956. The
580
ISM: Introductory Mathematical Analysis
3.
1 2
dx ≈
4
∫1
6.
0.2
(3.4000) = 0.340
2
Actual value:
1 2
x dx
0
∫
x3
=
3
1
=
0
1
≈ 0.333
3
f ( x) = x 2 , n = 4, a = 0, b = 1
Simpson’s
b − a 1− 0
h=
=
= 0.25
n
4
f (0) = 0.0000
4 f (0.25) = 0.2500
2 f (0.50) = 0.5000
4 f (0.75) = 2.2500
f (1) = 1.0000
4.0000
1 2
∫0 x
dx ≈
f ( x) =
1 2
x dx
0
∫
x3
=
3
1
=
0
4
1
x
2
1
1
= − − (−1) = 0.750
x1
4
1
, n = 6, a = 1, b = 4
x
Trapezoidal
b − a 4 −1
h=
=
= 0.5
n
6
f (1) = 1.0000
2 f (1.5) = 1.3333
2 f (2) = 1.0000
2 f (2.5) = 0.8000
2 f (3) = 0.6667
2 f (3.5) = 0.5714
f (4) = 0.2500
5.6214
f ( x) =
0.5
dx ≈
(5.6214) ≈ 1.405
x
2
Actual value:
41
∫1
4
dx = ln x
x
≈ 1.386
7.
1
≈ 0.333
3
1
dx ≈
x2
4
dx = −
41
, n = 4, a = 1, b = 4
x2
Simpson’s
b − a 4 −1
h=
=
= 0.75
n
4
f (1) = 1.0000
4 f (1.75) = 1.3061
2 f (2.50) = 0.3200
4 f (3.25) = 0.3787
f (4) = 0.0625
3.0673
∫1
1
∫1
0.25
1
(4.0000) = ≈ 0.333
3
3
Actual value:
5.
Actual value:
f ( x) = x 2 , n = 5, a = 0, b = 1
Trapezoidal
b − a 1− 0 1
h=
=
= = 0.2
n
5
5
f (0) = 0.0000
2 f (0.2) = 0.0800
2 f (0.4) = 0.3200
2 f (0.6) = 0.7200
2 f (0.8) = 1.2800
f (1) = 1.0000
3.4000
∫0 x
4.
Section 14.8
= ln 4 − ln1 = ln 4 – 0 = ln 4
1
x
, n = 4, a = 0, b = 2
x +1
Trapezoidal
b−a 2−0
h=
=
= 0.5
n
4
f (0) = 0.0000
2 f (0.5) = 0.6667
2 f (1) = 1.0000
2 f (1.5) = 1.2000
f (2) = 0.6667
3.5334
f ( x) =
Thus
2 x
0.5
∫0 x + 1 dx ≈ 2 (3.5334) ≈ 0.883
8.
0.75
(3.0673) ≈ 0.767
3
581
f ( x) =
1
, n = 4, a = 2, b = 4
x + x2
Simpson’s
b−a 4−2
h=
=
= 0.5
n
4
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
12. a = 2, b = 5, h = 0.5
0
f (2) =
4 f (2.5) = 24
2 f (3) = 20
4 f (3.5) = 44
2 f (4) = 28
4 f (4.5) = 60
f (5) = 16
192
f (2) = 0.1667
4 f (2.5) = 0.4571
2 f (3) = 0.1667
4 f (3.5) = 0.2540
f (4) = 0.0500
1.0945
4
dx
∫2 x + x2 ≈
9.
70
∫45 l (t )dt ,
0.5
(1.0945) ≈ 0.182
3
0.5
(192) = 32
3
The area is about 32 square units.
5
∫2 f ( x) dx ≈
males, n = 5, a = 45, b = 70
70 − 45
=5
5
l (45) = 93, 717
2l (50) = 183, 232
2l (55) = 177, 292
2l (60) = 168,376
2l (65) = 155, 094
l (70) = 68,375
846, 086
h=
70
13.
55
∫35
3 −1
= 0.5
4
=
f (1) = 1
4 f (1.5) = 4(2) =
2 f (2) = 2(2) =
4 f (2.5) = 4(0.5) =
=
f (3) = 1
5
3
∫1
l (t )dt , females, n = 4, a = 35, b = 55
55 − 35
=5
4
l (35) = 97,964
2l (40) = 194, 796
2l (45) = 193,164
2l (50) = 190, 784
l (55) = 93,562
770, 270
h=
55
f ( x)dx , n = 4, a = 1, b = 3
h=
∫45 l (t )dt ≈ 2 (846, 086) = 2,115, 215
10.
3
∫1
14.
f ( x)dx ≈
f ( x) =
1
8
4
2
1
16
0.5
8
(16) =
3
3
2
1+ x
, a = 1, b = 3, n = 4
3 −1
= 0.5
4
Simpson’s
f (1) ≈ 1.4142
4 f (1.5) ≈ 5.0596
2 f (2) ≈ 2.3094
4 f (2.5) ≈ 4.2762
f (3) = 1.0000
14.0594
h=
5
∫35 l (t )dt ≈ 2 (770, 270) = 1,925, 675
11. a = 1, b = 5, h = 1
f (1) = 0.4 = 0.4
4 f (2) = 4(0.6) = 2.4
2 f (3) = 2(1.2) = 2.4
4 f (4) = 4(0.8) = 3.2
f (5) = 0.5 = 0.5
8.9
2
0.5
(14.0594) ≈ 2.343
dx ≈
3
1+ x
For the actual value, we have
3 2
3
−1/ 2
∫1 1 + x dx = 2∫1 (1 + x) dx
3
∫1
3
(
)
= 2[2(1 + x)1/ 2 ] = 4 2 − 2 ≈ 2.343
1
∫1 f ( x)dx ≈ 3 (8.9) ≈ 3.0
The area is about 3.0 square units.
5
1
582
ISM: Introductory Mathematical Analysis
15.
Section 14.8
=0
=
f (0) = 0.5 − 0.5
4 f (0.5) = 4(2.3 − 0.3) = 4(2) =
2 f (1) = 2(2.2 − 0.7) = 2(1.5) =
= 4(2) =
4 f (1.5) = 4(3 − 1)
2 f (2) = 2(2.5 − 0.5) = 2(2) =
4 f (2.5) = 4(2.2 − 0.2) = 4(2) =
2 f (3) = 2(1.5 − 0.5) = 2(1) =
4 f (3.5) = 4(1.3) − 0.8) = 4(0.5) =
=0
=
f (4) = 1 − 1
f ( x) = 1 − x 2 , a = 0, b = 1, n = 4
1− 0
= 0.25
4
Simpson’s
f (0) = 1.0000
4 f (0.25) = 3.8730
2 f (0.50) = 1.7321
4 f (0.75) = 2.6458
f (1) = 0.0000
9.2509
h=
1
∫0
4
Area ≈ ∫ f ( x)dx ≈
0.25
1 − x dx ≈
(9.2509) ≈ 0.771
3
2
0
18. a.
dr
16. ∫
dq = r (80) − r (0) = r (80)
0 dq
[since r(0) = 0]
Using Simpson’s rule with h = 10 and
dr
f (q) =
:
dq
80
f (0) = 10
=
4 f (10) = 4(9) =
2 f (20) = 2(8.5) =
4 f (30) = 4(8) =
2 f (40) = 2(8.5) =
4 f (50) = 4(7.5) =
2 f (60) = 2(7) =
4 f (70) = 4(6.5) =
f (80) = 7
=
MC =
0
8
3
8
4
8
2
2
0
35
0.5
35
km2
(35) =
3
6
dc
dq
dc
dq
dq
= c (100) − c(0)
= (total cost of 100 units) − (fixed costs)
= total variable costs of 100 units
Using the trapezoidal rule with h = 20 and
dc
f (q) =
to estimate the integral:
dq
f (0) = 260
2 f (20) = 500
2 f (40) = 480
2 f (60) = 400
2 f (80) = 480
f (100) = 250
2370
100 dc
20
∫0 dq dq ≈ 2 (2370) = $23, 700
100
∫0
10
36
17
32
17
30
14
26
7
189
dr
10
dq ≈ (189) = 630
dq
3
The total revenue is about $630.
80
∫0
17. Let f(x) = distance from near to far shore at point
b.
4
x on highway. Then area ≈ ∫ f ( x)dx . Using
0
MR =
dr
dq
dr
dq = r (100) − r (0) = r (100)
dq
[since r(0) = 0]
= total revenue from sale of 100 units
Using the trapezoidal rule with h = 20 and
dr
g (q) =
to estimate the integral:
dq
100
∫0
Simpson’s rule with h = 0.5:
583
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
=
=
=
=
=
=
g (0)
2 g (20)
2 g (40)
2 g (60)
2 g (80)
g (100)
3. y = 5x + 2, x = 1, x = 4
410
700
600
500
540
250
3000
100 dr
20
∫0 dq dq ≈ 2 (3000) = $30, 000
c.
4
Area = ∫ (5 x + 2)dx
1
4
⎛ 5x2
⎞
9 87
=⎜
+ 2 x ⎟ = 48 − =
⎜ 2
⎟
2 2
⎝
⎠1
25
y
x
5
At q = 100: total revenue = 30,000
total cost = (total var. costs) + (fixed costs)
= 23, 700 + 2000 = 25, 700
Thus maximum profit
= (total revenue) − (total costs)
= 30,000 − 25,700 = $4300.
4. y = x + 5, x = 2, x = 4
4
⎛ x2
⎞
Area = ∫ ( x + 5)dx = ⎜
+ 5x ⎟
⎜
⎟
2
⎝ 2
⎠2
= 28 – 12 = 16
4
Problems 14.9
10
In Problems 1–34, answers are assumed to be
expressed in square units.
y
1. y = 4x, x = 2
x
2
Area = ∫ 4 x dx =
0
10
2
2 x2
0
5
=8–0=8
y
5. y = x – 1, x = 5
5
⎛ x2
⎞
5
Area = ∫ ( x − 1)dx = ⎜
− x⎟
⎜ 2
⎟
1
⎝
⎠1
x
2
2. y =
=
5
5
3
x + 1, x = 0, x = 16
4
y
16
⎛ 3x 2
⎞
⎞
Area = ∫ ⎜ x + 1⎟ dx = ⎜
+ x⎟
⎜
⎟
0 ⎝4
⎠
⎝ 8
⎠0
= 112 – 0 = 112
16 ⎛ 3
20
15 ⎛ 1 ⎞ 16
−⎜− ⎟ =
=8
2 ⎝ 2⎠ 2
x
1
y
x
20
584
5
8
ISM: Introductory Mathematical Analysis
Section 14.9
6. y = 3 x 2 , x = 1, x = 3
9. y = x 2 + 2 , x = –1, x = 2
3
3
Area = ∫ 3 x 2 dx = x3 = 27 − 1 = 26
1
40
Area = ∫
1
2
−1
y
=
(
2
⎛ x3
⎞
x + 2 dx = ⎜ + 2 x ⎟
⎜ 3
⎟
⎝
⎠ −1
)
2
20 ⎛ 7 ⎞ 27
−⎜− ⎟ =
=9
3 ⎝ 3⎠ 3
y
8
x
5
x
7. y = x 2 , x = 2, x = 3
3 2
x dx
2
Area = ∫
16
x3
=
3
–1
3
= 9−
2
8 19
=
3 3
2
5
10. y = x + x 2 + x3 , x = 1
y
1
⎛ x 2 x3 x 4 ⎞
1
Area = ∫ ( x + x 2 + x3 )dx = ⎜
+
+
⎟
⎜ 2
0
3
4 ⎟⎠
⎝
0
=
x
2 3
13
13
−0 =
12
12
3
5
y
8. y = 2 x 2 − x , x = –2, x = –1
Area = ∫
−1
−2
(
x
3
−1
⎛ 2 x3 x 2 ⎞
2 x − x dx = ⎜
− ⎟
⎜ 3
2 ⎟⎠
⎝
−2
2
)
7 ⎛ 44 ⎞ 37
= − −⎜− ⎟ =
6 ⎝ 6 ⎠ 6
16
11. y = x 2 − 2 x , x = –3, x = –1
y
Area = ∫
−1
−3
1
)
4
50
= − − (−18) =
3
3
x
–2 –1
(
−1
⎛ x3
⎞
x − 2 x dx = ⎜ − x 2 ⎟
⎜ 3
⎟
⎝
⎠ −3
2
18
5
y
x
–3
585
–1
2
5
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
12. y = 3x 2 − 4 x , x = –2, x = –1
Area = ∫
−1
−2
15. y = 2 − x − x3 , x = –3, x = 0
(3x2 − 4 x ) dx = ( x3 − 2 x2 ) −2
−1
Area = ∫
y
(2 − x − x )
3
−3
= –3 – (–16) = 13
32
0
0
⎛
x2 x4 ⎞
dx = ⎜ 2 x −
− ⎟
⎜
2
4 ⎟⎠
⎝
−3
⎛ 123 ⎞ 123
= 0−⎜−
⎟=
4
⎝ 4 ⎠
y
40
x
–2 –1
5
x
13. y = 2 − x − x
2
5
1
⎛
x 2 x3 ⎞
Area = ∫ (2 − x − x )dx = ⎜ 2 x −
− ⎟
⎜
−2
2
3 ⎟⎠
⎝
−2
7 ⎛ 10 ⎞
= −⎜− ⎟
6 ⎝ 3⎠
9
=
2
1
5
2
16. y = e x , x = 1, x = 3
3
3
1
1
Area = ∫ e x dx = e x
32
= e3 − e
y
y
x
1
x
3
5
17. A = 3 + 2 x − x 2
4
, x = 1, x = 2
x
24
Area = ∫ dx = 4 ln x
1 x
= ln 16
Area = ∫
−1
(3 + 2x − x2 ) dx
3
14. y =
5
3
2
1
⎛
x3 ⎞
= ⎜ 3x + x 2 − ⎟
⎜
3 ⎟⎠
⎝
−1
= 4 ln(2) – 0 = 4 ln 2
⎛ 5 ⎞ 32
= 9−⎜− ⎟ =
⎝ 3⎠ 3
y
5
y
x
1 2
x
5
5
586
ISM: Introductory Mathematical Analysis
18. y =
1
( x − 1)
Area = ∫
2
21. y = x + 9 , x = –9, x = 0
, x = 2, x = 3
1
3
Area = ∫
( x − 1)
( x − 1)−1
=
−1
3
dx = ∫ ( x − 1) −2 dx
2
=
3
⎛ 1 ⎞
= ⎜−
⎟
⎝ x −1 ⎠ 2
2
1
1
= − − (−1) =
2
2
5
0
( x + 9)
−9
0
3
2
3
2
1
0
x + 9dx = ∫ ( x + 9) 2 dx
−9
3
2
2
Section 14.9
3
−9
2( x + 9) 2
=
3
= 18 – 0 = 18
10
0
−9
y
y
x
–9
6
x
5
22. y = x 2 − 4 x , x = 2, x = 6
5
4
6
2
4
Area = ∫ −( x 2 − 4 x)dx + ∫ ( x 2 − 4 x)dx
1
19. y = , x = 1, x = e
x
e1
e
Area = ∫ dx = ln x = ln e – ln 1 = 1 – 0 = 1
1
1 x
4
y
20
y
x
1
e
5
x
7
1
, x = 1, x = e2
x
e2 1
Area = ∫
dx = ln x
1 x
20. y =
23. y = 2 x − 1 , x = 1, x = 5
e2
1
Area = ∫
2
5
1
= ln e − ln1 = 2 − 0 = 2
y
=
1
3
e2
587
2 x − 1 dx
1
1 5
(2 x − 1) 2 [2 dx ]
∫
1
2
(2 x − 1) 2
=
3
x
1
6
⎛ x3
⎞
⎛ x3
⎞
= ⎜ − + 2x2 ⎟ + ⎜ − 2x2 ⎟
⎜ 3
⎟
⎜ 3
⎟
⎝
⎠2 ⎝
⎠4
⎡ 32 16 ⎤ ⎡ ⎛ 32 ⎞ ⎤
= ⎢ − ⎥ + ⎢0 − ⎜ − ⎟ ⎥ = 16
⎣ 3 3 ⎦ ⎣ ⎝ 3 ⎠⎦
5
= 9−
1
1 26
=
3 3
Chapter 14: Integration
5
ISM: Introductory Mathematical Analysis
y
26. y = x 2 + 4 x − 5, x = −5, x = 1
1
Area = ∫ −( x 2 + 4 x − 5) dx
−5
x
1
5
24. y = x3 + 3x 2 , x = –2, x = 2
10
2
⎛ x4
⎞
2
+ x3 ⎟
Area = ∫ x3 + 3 x 2 dx = ⎜
⎜ 4
⎟
−2
⎝
⎠ −2
= 12 – (–4) = 16
(
25
1
⎛ x3
⎞
= − ⎜ + 2 x2 − 5x ⎟
⎜ 3
⎟
⎝
⎠ −5
8
100
⎛
⎞
= −⎜− −
⎟
⎝ 3 3 ⎠
= 36
)
y
x
10
y
x
27. y = e x + 1, x = 0, x = 1
5
1
1
Area = ∫ (e x + 1)dx = (e x + x) = (e1 + 1) − 1 = e
0
0
8
y
25. y = 3 x , x = 2
Area = ∫
23
0
4
3x 3
xdx = ∫ x dx =
0
4
2
1
3
( ) = 332
2
0
4
3(2) 3
=
−0
4
x
5
3
=
3 2 2
4
2
3
28. y = x , x = –2, x = 2
y
Area = ∫
2
−2
x2
=−
2
x
2
0
−2
0
2
−2
0
x dx = ∫ (− x)dx + ∫ x dx
x2
+
2
2
0
= [0 – (–2)] + [2 – 0] = 4
3
y
x
–2
588
2
ISM: Introductory Mathematical Analysis
29. y = x +
Section 14.9
2
, x = 1, x = 2
x
32. y = x − 2 , x = 2, x = 6
3
2( x − 2) 2
Area = ∫ x − 2 dx =
2
3
2
= (2 + 2 ln 2) −
5
y
1 3
3
= + 2 ln 2 = + ln 4
2 2
2
y
x
x
2
2
5
6
8
33. y = 2 x − x 2 , x = 1, x = 3
30. y = 4 + 3 x − x 2
Area = ∫
4
Area = ∫ (4 + 3 x − x 2 )dx
2
1
−1
4
⎛
3 x 2 x3 ⎞
= ⎜ 4x +
− ⎟
⎜
2
3 ⎟⎠
⎝
−1
56 ⎛ 13 ⎞
=
−⎜− ⎟
3 ⎝ 6⎠
125
=
6
10
2
16
16
= −0 =
3
3
8
1
6
6
⎛x
⎞
2⎛
2⎞
+ 2 ln x ⎟
Area = ∫ ⎜ x + ⎟ dx = ⎜
⎜
⎟
1 ⎝
x⎠
⎝ 2
⎠1
2
( 2 x − x2 ) dx + ∫23 − ( 2 x − x2 ) dx
2
⎛
⎛
x3 ⎞
x3 ⎞
= ⎜ x2 − ⎟ − ⎜ x2 − ⎟
⎜
⎜
3 ⎟⎠
3 ⎟⎠
⎝
1 ⎝
3
2
⎡4 2⎤ ⎡ 4⎤ 6
= ⎢ − ⎥ − ⎢0 − ⎥ = = 2
⎣3 3⎦ ⎣ 3⎦ 3
5
y
y
3
x
1 2
x
10
5
34. y = x 2 + 1 , x = 0, x = 4
31. y = x3 , x = –2, x = 4
0
4 3
x dx
0
3
Area = ∫ − x dx + ∫
−2
x4
=−
4
0
−2
x4
+
4
4
Area = ∫
0
=
0
= [0 – (–4)] + [64 – 0] = 68
64
4
y
(
)
76
76
−0 =
3
3
25
–2
4
⎛ x3
⎞
x + 1 dx = ⎜ + x ⎟
⎜ 3
⎟
⎝
⎠0
2
y
x
x
4
4
589
Chapter 14: Integration
35.
ISM: Introductory Mathematical Analysis
⎧ 2
if
f ( x) = ⎨3x
⎩16 − 2 x if
0≤ x<2
x≥2
3
2
38. a.
Area = ∫ f ( x)dx = ∫ 3x dx + ∫ (16 − 2 x)dx
0
=
2
x3
0
0
(
+ 16 x − x 2
2
)2
3
3
(1 − x)2 dx
1
1
= − (−1 − 0) =
9
9
= [8 – 0] + [39 – 28] = 19 sq units
16
1
2
1
1 (1 − x)3
= (−1) ∫ (1 − x)2 [− dx] = − ⋅
1
3
3
3
3
2
21
P (1 ≤ x ≤ 2) = ∫
y
b.
36. y =
5
c.
1
b−a
1
x
dx =
ab−a
b−a
Area = ∫
=
t
t
d.
a
39. a.
x
a
37. a.
c.
x2
P (0 ≤ x ≤ 1) = ∫ x dx =
08
16
1
16
1
=
0
x2
P (2 ≤ x ≤ 4) = ∫ x dx =
28
16
41
P ( x ≥ 3) = ∫
41
3
8
x dx =
3
1
3
∫0 f ( x)dx = ∫0 f ( x)dx + ∫1
2 4
x
16
3
f ( x)dx
1
+ P( x ≥ 1)
9
P (3 ≤ x ≤ 7) = ∫
4
2
1
−0
16
b.
c.
1 3
= 1− =
4 4
= 1−
71
3
= ln 7 − ln 3 = ln
b
11
=
b.
t
11
8
Thus, P ( x ≥ 1) = .
9
y
1
b–a
1
1
(1 − x)2 dx = − (1 − x)3
03
9
0
1
1
= − (0 − 1) =
9
9
P ( x ≤ 1) = ∫
1=
t
a
t −a
−
=
sq units
b−a b−a b−a
51
P ( x ≤ 5) = ∫
x
dx = ln x
7
3
dx = ln x
e x
= ln(5) − ln e = ln(5) − 1
P ( x ≥ 4) = ∫
e2
4
7
3
5
e
1
dx = ln x
x
e2
4
2
= ln e − ln 4 = 2 − ln 4
d.
9
7
=
16 16
(
)
P e ≤ x ≤ e2 = ∫
= ln x
e2
e
e2
e
1
dx
x
= ln e 2 − ln e
=2–1=1
40. a.
590
1
5/ 2 1
5⎞
⎛
P ⎜1 ≤ x ≤ ⎟ = ∫
(1 − x )2 dx
1
2⎠
3
⎝
5/ 2
1
1 ⎛ 27
⎞ 3
= − (1 − x)3
= − ⎜ − − 0⎟ =
9
9⎝ 8
⎠ 8
1
x
2 3
2
r
∫1
1
x2
r
dx = −
1
1
1
= − +1 = 1−
x1
r
r
ISM: Introductory Mathematical Analysis
Section 14.9
y
b.
y = 12
x
x
1
c.
r
⎛ 1⎞
dx = lim ⎜1 − ⎟ [from part (a)]
→∞
r
⎝ r⎠
x
=1– 0=1
r
1
r →∞ 1
2
∫
lim
d.
y
y = 12
x
x
1
41. 1.89 sq units
42. 7.18 sq units
43. The x-intercept on [1, 3] is A ≈ 2.190327947
(
A
)
Area = ∫ − x 4 − 2 x3 − 2 dx + ∫
1
3
A
( x4 − 2 x3 − 2) dx
≈ 11.41 sq units
y
50
x
5
44. The x-intercepts are A ≈ –0.3294085282 and B ≈ 1.539613346
Area = ∫
B
A
(1 + 3x − x4 ) dx ≈ 3.53 sq units
10
y
x
5
–20
591
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
Problems 14.10
1. Area = ∫
b
a
3
( yUPPER − yLOWER ) dx = ∫−2 ⎡⎣( x + 6) − x 2 ⎤⎦ dx
( yUPPER − yLOWER ) dx = ∫0 ( 2x − x 2 ) dx
a
2. Area = ∫
2
b
3. Intersection points:
x 2 − x = 2 x, x 2 − 3 x = 0, x( x − 3) = 0 ⇒ x = 0 or x = 3
Area = ∫
3
0
4
( yUPPER − yLOWER ) dx + ∫3 ( yUPPER − yLOWER )dx
(
)
(
)
3
4
= ∫ ⎡⎢ 2 x − x 2 − x ⎤⎥ dx + ∫ ⎡⎢ x 2 − x − 2 x ⎤⎥ dx
0⎣
3 ⎣
⎦
⎦
4. Intersection points: x( x − 3) 2 = 2 x, x( x − 3) 2 − 2 x = 0 , x ⎡ ( x − 3) 2 − 2 ⎤ = 0 , x( x 2 − 6 x + 7) = 0 ⇒ x = 0 , 3 ± 2
⎣
⎦
(from the quadratic formula)
Area = ∫
3− 2
0
=∫
3− 2 ⎡
0
3+ 2
( yUPPER − yLOWER ) dx + ∫3− 2 ( yUPPER − yLOWER ) dx
3+
2
x ( x − 3) − 2 x ⎥⎤dx + ∫
3−
⎣⎢
⎦
2
2
⎡ 2 x − x( x − 3)2 ⎤ dx
⎣
⎦
5. The graphs of y = 1 − x 2 and y = x – 1 intersect when 1 − x 2 = x − 1 , 0 = x 2 + x − 2 , 0 = ( x − 1)( x + 2) ⇒ x = 1 or
x = –2. When x = 1, then y = 0. We use horizontal elements, where y ranges from 0 to 1. Solving y = x – 1 for x
gives x = y + 1, and solving y = 1 − x 2 for x gives x 2 = 1 − y , x = ± 1 − y . We must choose x = 1 − y because
x is not negative over the given region.
1
1
Area = ∫ ( xRIGHT − xLEFT )dy = ∫ ⎡⎣ ( y + 1) − 1 − y ⎤⎦ dy
0
0
6. The graphs of y = 2x and y = –2x – 8 intersect when 2x = –2x – 8, 4x = –8, x = –2. When x = –2, then y = –4. We
y
use horizontal elements, where y ranges from –4 to 4. Solving y = 2x for x gives x = ; solving y = –2x – 8 for x
2
−y −8
gives 2x = –y – 8, x =
.
2
4
4 ⎡ y ⎛ − y − 8 ⎞⎤
Area = ∫ ( xRIGHT − xLEFT )dy = ∫ ⎢ − ⎜
⎟ dy
−4
−4 ⎣ 2 ⎝
2 ⎠ ⎥⎦
592
ISM: Introductory Mathematical Analysis
Section 14.10
7. The graphs of y = x 2 − 5 and y = 7 − 2 x 2
8
y
intersect when x 2 − 5 = 7 − 2 x 2 , 3x 2 = 12,
x 2 = 4, so x = ± 4 = ±2. We use vertical
elements.
2
Area = ∫ ( yUPPER − yLOWER ) dx
x
1
2
5
= ∫ [(7 − 2 x 2 ) − ( x 2 − 5)] dx
1
10
10. y = x, y = –x + 3, y = 0. Region appears below.
3
Intersection: x = –x + 3, 2x = 3, x =
2
y
Area = ∫
x
10
3/ 2
0
x2
=
2
3/ 2
0
x dx + ∫
3
(− x + 3)dx
3/ 2
3
⎛ x2
⎞
+⎜−
+ 3x ⎟
⎜ 2
⎟
⎝
⎠ 3/ 2
⎡ 9 ⎤ ⎡⎛ 9
⎞ ⎛ 9 9 ⎞⎤ 9
= ⎢ − 0 ⎥ + ⎢⎜ − + 9 ⎟ − ⎜ − + ⎟ ⎥ =
⎣ 8 ⎦ ⎣⎝ 2
⎠ ⎝ 8 2 ⎠⎦ 4
8. The curves y 2 = x and 2y = 3 − x (or x = 3 − 2y)
intersect when y 2 = 3 − 2 y, y 2 + 2 y − 3 = 0,
5
(y + 3)(y − 1) = 0 ⇒ y = −3 or 1. We use
horizontal elements.
y
1
Area = ∫ ( xRIGHT − xLEFT ) dy
0
1
= ∫ [(3 − 2 y ) − y 2 ] dy
0
5
x
3
2
y
5
11. y = x 2 + 1, x ≥ 0, x = 0, y = 3. Region appears
below.
x
5
Intersection: x 2 + 1 = 3 , so x = ± 2
2
Area = ∫
0
[3 − ( x 2 + 1)]dx = ∫
0
2
9. y = x 2 , y = 2 x
Region appears below.
5
y
Intersection: x 2 = 2 x, x 2 − 2 x = 0 , x(x – 2) = 0,
so x = 0 or 2.
Area = ∫
0
(2x − x )
2
(2 − x 2 )dx
⎛
x3 ⎞
4 2
4 2
= ⎜ 2x − ⎟
=
−0 =
⎜
⎟
3
3
3
⎝
⎠0
In Problems 9–34, the answers are assumed to be
expressed in square units.
2
2
2
⎛
x3 ⎞
dx = ⎜ x 2 − ⎟
⎜
3 ⎟⎠
⎝
0
x
3
8⎞
4
⎛
= ⎜4− ⎟−0 =
3
3
⎝
⎠
593
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
14. y 2 = x + 1 , x = 1. Region appears below.
12. y = x 2 + 1 , y = x + 3. Region appears below.
Intersection: y 2 = 2, y = ± 2
Intersection: x 2 + 1 = x + 3, x 2 − x − 2 = 0 ,
(x + 1)(x – 2) = 0, so x = –1, 2
(
2
)
Area = ∫ ⎡⎢ ( x + 3) − x 2 + 1 ⎤⎥ dx
−1 ⎣
⎦
x + 2 − x 2 ) dx
−1 (
=∫
−
2
(
)
⎛ x2
x3 ⎞
=⎜
+ 2x − ⎟
⎜ 2
3 ⎟⎠
⎝
−1
5
8⎞ ⎛1
1⎞ 9
⎛
= ⎜2+ 4− ⎟−⎜ − 2+ ⎟ =
3
2
3⎠ 2
⎝
⎠ ⎝
y
x
5
– 2
15. x = 8 + 2y, x = 0, y = –1, y = 3. Region appears
below.
5
(
3
Area = ∫ (8 + 2 y )dy = 8 y + y 2
−1
2
10
Intersection: 10 − x 2 = 4, x 2 = 6, so x = ± 6
6
3
) −1
= (24 + 9) – (–8 + 1) = 40
13. y = 10 − x , y = 4. Region appears below.
− 6
6
y
[(10 − x 2 ) − 4] dx
=∫
− 6
2
y
2
x
Area = ∫
2
⎛
2 2⎞ ⎛
2 2⎞ 8 2
= ⎜⎜ 2 2 −
⎟⎟ − ⎜⎜ −2 2 +
⎟=
3 ⎠ ⎝
3 ⎟⎠
3
⎝
2
8
3⎞
⎛
⎡1 − y 2 − 1 ⎤ dy = ⎜ 2 y − y ⎟
⎥⎦
⎜
2 ⎣⎢
3 ⎟⎠
⎝
−
2
Area = ∫
x
16
2
(6 − x ) dx
6
⎛
x3 ⎞
= ⎜ 6x − ⎟
⎜
3 ⎟⎠
⎝
− 6
⎛
6 6⎞ ⎛
6 6⎞
= ⎜⎜ 6 6 −
⎟⎟ − ⎜⎜ −6 6 +
⎟
3
3 ⎟⎠
⎝
⎠ ⎝
=8 6
16. y = x − 6, y 2 = x . Region appears below.
Intersection: y 2 = y + 6, y 2 − y − 6 = 0,
(y + 2)(y − 3) = 0, so y = −2, 3.
3
Area = ∫ ⎡ ( y + 6) − ( y 2 ) ⎤dy
⎦
−2 ⎣
y
3
10
⎛ y2
y3 ⎞
=⎜
+ 6y −
⎟
⎜ 2
3 ⎟⎠
⎝
−2
x
10
8 ⎞ 125
⎛9
⎞ ⎛
= ⎜ + 18 − 9 ⎟ − ⎜ 2 − 12 + ⎟ =
2
3⎠
6
⎝
⎠ ⎝
594
ISM: Introductory Mathematical Analysis
10
Section 14.10
y
19. y 2 = 4 x, y = 2x − 4. Region appears below.
⎛y
⎞
Intersection: y 2 = 4 ⎜ + 2 ⎟ , y 2 − 2 y − 8 = 0,
⎝2
⎠
(y + 2)(y − 4) = 0, so y = −2 or 4.
2
4 ⎡⎛ y
⎞ y ⎤
Area = ∫ ⎢⎜ + 2 ⎟ − ⎥ dy
−2 ⎝ 2
⎠ 4 ⎥⎦
⎢⎣
(9, 3)
x
16
(4, –2)
4
⎛ y2
y3 ⎞
=⎜
+ 2y − ⎟
⎜ 4
12 ⎟⎠
⎝
−2
16 ⎞ ⎛
2⎞
⎛
= ⎜ 4 + 8 − ⎟ − ⎜1 − 4 + ⎟
3⎠ ⎝
3⎠
⎝
=9
17. y = 4 − x 2 , y = –3x. Region appears below.
Intersection: −3x = 4 − x 2 , x 2 − 3x − 4 = 0 ,
(x + 1)(x – 4) = 0, so x = –1 or 4.
Area
∫−1 ⎡⎣⎢( 4 − x
4
2
) − (−3x)⎤⎦⎥ dx
5
4
y
⎛
x3 3 x 2 ⎞
= ⎜ 4x −
+
⎟
⎜
3
2 ⎟⎠
⎝
−1
(4, 4)
x
64
1 3 ⎞ 125
⎛
⎞ ⎛
= ⎜ 16 − + 24 ⎟ − ⎜ −4 + + ⎟ =
3
3 2⎠
6
⎝
⎠ ⎝
6
(1, –2)
5
y
20. y = x3 , y = x + 6, x = 0
Region appears below.
x
10
Intersection: x3 = x + 6, x3 − x − 6 = 0,
( x − 2)( x 2 + 2 x + 3) = 0 ⇒ x = 2
x3 = 0 ⇒ x = 0
2
Area = ∫ [( x + 6) − x3 ] dx
2
18. x = y + 2, x = 6 . Region appears below.
2
0
Intersection: y + 2 = 6, y = 4 , y = ±2
(
)
(
)
2
2
Area = ∫ ⎡⎢ 6 − y 2 + 2 ⎤⎥ dy = ∫ 4 − y 2 dy
−2 ⎣
−2
⎦
2
⎛
y3 ⎞
8 ⎞ 32
⎛ 8⎞ ⎛
= ⎜ 4y −
⎟ = ⎜ 8 − ⎟ − ⎜ −8 + ⎟ =
⎜
⎟
3 ⎠
3⎠ ⎝
3⎠ 3
⎝
⎝
−2
5
2
⎛ x2
x4 ⎞
=⎜
+ 6x − ⎟
⎜ 2
4 ⎟⎠
⎝
0
= (2 + 12 − 4) − (0) = 10
2
10
y
x
10
y
x
8
595
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
2 y + 15 ⎞
⎛
23. y 2 = 3 x, 3x − 2y = 15 ⎜ or x =
⎟ . Region
3 ⎠
⎝
appears below.
⎛ 2 y + 15 ⎞
Intersection: y 2 = 3 ⎜
⎟,
⎝ 3 ⎠
21. 2 y = 4 x − x 2 , 2y = x – 4. Region appears below.
Intersection: x − 4 = 4 x − x 2 , x 2 − 3 x − 4 = 0 ,
(x + 1)(x – 4) = 0, so x = –1 or 4. Note that the
4 x − x2
y-values of the curves are given by y =
2
x−4
and y =
.
2
4 ⎡⎛ 4 x − x 2 ⎞ ⎛ x − 4 ⎞ ⎤
Area = ∫ ⎢⎜
⎟−⎜
⎟ ⎥ dx
⎟
−1 ⎢⎜
⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦
⎞
4 ⎛3
x2
= ∫ ⎜ x−
+ 2 ⎟ dx
⎟
−1 ⎜ 2
2
⎝
⎠
y 2 − 2 y − 15 = 0, (y + 3)(y − 5) = 0, so y = −3 or
5.
2
5 ⎡⎛ 2
⎞ y ⎤
Area = ∫ ⎢⎜ y + 5 ⎟ − ⎥ dy
−3 ⎝ 3
⎠ 3 ⎥⎦
⎢⎣
5
⎛1
y3 ⎞
= ⎜ y2 + 5 y − ⎟
⎜3
9 ⎟⎠
⎝
−3
125 ⎞
⎛ 25
= ⎜ + 25 −
⎟ − (3 − 15 + 3)
9 ⎠
⎝ 3
256
=
9
4
⎛ 3x 2 x3
⎞
=⎜
− + 2x ⎟
⎜ 4
⎟
6
⎝
⎠ −1
64
⎛
⎞ ⎛3 1
⎞
= ⎜ 12 − + 8 ⎟ − ⎜ + − 2 ⎟
6
⎝
⎠ ⎝4 6
⎠
125
=
12
5
8
y
( 253 , 5)
y
x
9
(3, –3)
x
8
24. y = 2 − x 2 , y = x. Region appears below.
Intersection: x = 2 − x 2 , x 2 + x − 2 = 0 ,
( x + 2)( x − 1) = 0 ⇒ x = −2 or 1.
22. y = x , y = x 2 . Region appears below.
2
4
Intersection: x = x , x = x , x − x = 0 ,
(
)
x x3 − 1 = 0 , so x = 0, 1.
Area = ∫
1
0
(
x−x
2
)
(
)
1 1⎞ ⎛
8
⎛
⎞ 9
= ⎜ 2 − − ⎟ − ⎜ −4 + − 2 ⎟ =
3 2⎠ ⎝
3
⎝
⎠ 2
1
⎛ 32
⎞
2x
x3
dx = ⎜
− ⎟
⎜ 3
3 ⎟
⎝
⎠0
5
1
⎛ 2 1⎞
= ⎜ − ⎟−0 =
3
3
3
⎝
⎠
5
1
⎛
x3 x 2 ⎞
Area ∫ ⎡⎢ 2 − x 2 − x ⎤⎥ dx = ⎜ 2 x −
− ⎟
⎜
−2 ⎣
⎦
3
2 ⎟⎠
⎝
−2
1
4
y
(1, 1)
(–2, –2)
y
(1, 1)
x
5
596
x
5
ISM: Introductory Mathematical Analysis
Section 14.10
25. y = 8 − x 2 , y = x 2 , x = –1, x = 1. Region appears
below.
27. y = x 2 , y = 2, y = 5. Region appears below.
Area
Intersection: x 2 = 8 − x 2 , 2 x 2 = 8, x 2 = 4 , so
x = ±2.
(
1
Area = ∫ ⎡⎢ 8 − x
−1 ⎣
⎛
2x
= ⎜ 8x −
⎜
3
⎝
3
8
2
)− x
1
2⎤
(
dx = ∫ 8 − 2 x
−1
⎦⎥
2
(
3
)
) dx
2 2
3
1
4y2
=
3
⎞
2⎞ ⎛
2 ⎞ 44
⎛
⎟ = ⎜ 8 − ⎟ − ⎜ −8 + ⎟ =
⎟
3⎠ ⎝
3⎠ 3
⎠ −1 ⎝
5
4⋅5 5 4⋅ 2 2 4
−
= 5 5−2 2
3
3
3
(
=
2
y
y
8
x=– y
x= y
x
x
–1
1
5
5
28. y = x3 + x, y = 0 (x-axis), x = −1, x = 2
Region appears below.
26. y 2 = 6 − x, 3y = x + 12. Region appears below.
y 2 = 6 − (3 y − 12), y 2 + 3 y − 18 = 0,
3
Area = ∫ [(6 − y 2 ) − (3 y − 12)] dy
2
0
0
2
⎛ x4 x2 ⎞
⎛ x4 x2 ⎞
= ⎜−
− ⎟ +⎜
+
⎟
⎜ 4
⎜ 4
2 ⎟⎠
2 ⎟⎠
⎝
⎝
−1
0
⎡ ⎛ 1 1 ⎞⎤
= ⎢ 0 − ⎜ − − ⎟ ⎥ + [(4 + 2) − 0]
⎣ ⎝ 4 2 ⎠⎦
27
=
4
−6
3
= ∫ (18 − y 2 − 3 y ) dy
3
⎛
y3 3 y 2 ⎞
= ⎜18 y −
−
⎟
⎜
3
2 ⎟⎠
⎝
−6
27 ⎞
⎛
= ⎜ 54 − 9 − ⎟ − (−108 + 72 − 54)
2 ⎠
⎝
243
=
2
10
0
−1
Area = ∫ −( x3 + x)dx + ∫ ( x3 + x)dx
(y + 6)(y − 3) = 0, so y = −6, 3
−6
5
5
y2
= ∫ ⎡ y − − y ⎤ dy = ∫ 2 ydy = 2 ⋅
3
⎣
⎦
2
2
5
15
y
x
y
3
x
50
597
)
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
(
)
29. y = x3 − 1 , y = x ⫺ 1. Region appears below. Intersection: x3 − 1 = x − 1, x3 − x = 0 , x x 2 − 1 = 0 ,
x(x + 1)(x – 1) = 0, so x = 0 or x = ±1.
0
1
−1
0
Area = ∫ [ x3 − 1 − ( x − 1)] dx + ∫ [ x − 1 − ( x3 − 1)] dx
=∫
0
−1
( x3 − x ) dx + ∫01( x − x3 ) dx
0
1
⎛ x4 x2 ⎞
⎛ x2 x4 ⎞
=⎜
− ⎟ +⎜
− ⎟
⎜ 4
⎜ 2
2 ⎟⎠
4 ⎟⎠
⎝
⎝
−1
0
⎡ ⎛ 1 1 ⎞ ⎤ ⎡⎛ 1 1 ⎞ ⎤ 1
= ⎢ 0 − ⎜ − ⎟ ⎥ + ⎢⎜ − ⎟ − 0 ⎥ =
⎣ ⎝ 4 2 ⎠ ⎦ ⎣⎝ 2 4 ⎠ ⎦ 2
2
y
x
3
(
)
30. y = x3 , y = x . Region appears below. Intersection: x3 = x , x6 = x, x6 − x = 0 , x x5 − 1 = 0 , x = 0, 1
Area = ∫
1
0
(
x−x
3
)
1
⎛ 32
⎞
2x
x4
dx = ⎜
− ⎟
⎜ 3
4 ⎟
⎝
⎠0
5
⎛2 1⎞
= ⎜ − ⎟−0 =
12
⎝3 4⎠
3
y
(1, 1)
x
3
1
−17 − 4 x 1
. Region appears below. Intersection:
= , −17 x − 4 x 2 = 4 , 4 x 2 + 17 x + 4 = 0 ,
x
x
4
1
(4x + 1)(x + 4) = 0, so x = − or –4.
4
31. 4 x + 4 y + 17 = 0, y =
598
ISM: Introductory Mathematical Analysis
Area = ∫
Section 14.10
−1/ 4
⎛
17
x2 ⎞
⎛ −17 − 4 x ⎞ ⎤
−
dx
ln
x
x
=
+
+
⎜
⎟
⎟⎥
⎢x ⎜
⎜
4
2 ⎟⎠
4
⎝
⎠⎦
⎣
⎝
−4
−1/ 4 ⎡ 1
−4
255
⎛ 1 17 1 ⎞
= ⎜ ln − + ⎟ − ( ln 4 − 17 + 8 ) =
− 4 ln 2
4
16
32
32
⎝
⎠
1
y
x
–1
1
4
32. y 2 = − x − 2, x − y = 5, y = −1, y = 1.
Region appears below.
Intersection: y 2 = − x − 2 intersects y = ±1 when x = −3; x − y = 5 intersects y = 1 when x = 6;
x − y = 5 intersects y = −1 when x = 4
1
⎛ y2
1
1
y3 ⎞
Area = ∫ [( y + 5) − (− y 2 − 2)] dy = ∫ ( y + 7 + y 2 ) dy = ⎜
+ 7y +
⎟
⎜ 2
−1
−1
3 ⎟⎠
⎝
−1
1⎞ ⎛1
1 ⎞ 44
⎛1
= ⎜ +7+ ⎟−⎜ −7− ⎟ =
3⎠ ⎝ 2
3⎠ 3
⎝2
y
5
x
5
33. y = x – 1, y = 5 – 2x. Region appears below.
Intersection: x – 1 = 5 – 2x, 3x = 6, so x = 2.
2
4
2
4
0
2
0
2
Area = ∫ [(5 − 2 x) − ( x − 1)]dx + ∫ [( x − 1) − (5 − 2 x)]dx = ∫ (6 − 3x)dx + ∫ (3x − 6)dx
=−
1 2
1 4
(6 − 3 x)2
(6 − 3x)[−3 dx] + ∫ (3x − 6)[3 dx] = −
∫
3 0
3 2
6
= –[0 – 6] + [6 – 0] = 6 + 6 = 12
5
y
y = 5 – 2x
y=x–1
x
8
599
2
+
0
(3x − 6) 2
6
4
2
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
34. y = x 2 − 4 x + 4, y = 10 − x 2 . Region appears below.
Intersection: x 2 − 4 x + 4 = 10 − x 2 , 2 x 2 − 4 x − 6 = 0 , x 2 − 2 x − 3 = 0 , (x – 3)(x + 1) = 0, so x = 3, –1.
(
) (
)
(
) (
)
3
4
Area = ∫ ⎡⎢ 10 − x 2 − x 2 − 4 x + 4 ⎤⎥ dx + ∫ ⎡⎢ x 2 − 4 x + 4 − 10 − x 2 ⎤⎥dx
2⎣
3 ⎣
⎦
⎦
2(
)
3
3
4
2 x 2 − 4 x − 6 ) dx = 2 {∫ ( 3 + 2 x − x 2 ) dx + ∫ ( x 2 − 2 x − 3) dx}
(
3
2
3
= ∫ 6 + 4 x − 2 x 2 dx + ∫
4
3
4
⎧⎛
3⎞
⎛ x3
⎞ ⎫⎪
⎧ ⎡ 22 ⎤ ⎡ 20
⎪
⎤⎫
2 x
2
= 2 ⎨⎜ 3x + x − ⎟ + ⎜ − x − 3 x ⎟ ⎬ = 2 ⎨ ⎢9 − ⎥ + ⎢ − − (−9) ⎥ ⎬ = 2{4} = 8
⎜
⎟
⎜
⎟
3 ⎠
3
3⎦ ⎣ 3
⎦⎭
⎩⎣
⎪⎝
⎠ 3 ⎪⎭
2 ⎝
⎩
y
10
x
5
x
∫0 ⎡⎢⎣ x − ( 14
15
1
Area between curve and diag.
35.
=
Area under diagonal
2
)
1 x ⎤ dx
+ 15
⎥⎦
1
∫0 x dx
1
14 ⎤
14 1
14 ⎛ x 2 x3 ⎞
14 ⎡⎛ 1 1 ⎞ ⎤ 14 1 7
Numerator = ∫ ⎢ x − x 2 ⎥ dx = ∫ x − x 2 dx = ⎜
− ⎟ = ⎢⎜ − ⎟ − 0 ⎥ = ⋅ =
⎜
⎟
0 ⎣ 15
15 ⎦
15 0
15 ⎝ 2
3 ⎠
15 ⎣⎝ 2 3 ⎠ ⎦ 15 6 45
0
(
1 ⎡14
x2
Denominator = ∫ x dx =
0
2
1
Coefficient of inequality =
1
0
7
45
1
2
=
1
2
=
14
45
)
11 x
∫0 ⎡⎣⎢ x − ( 12
1
Area between curve and diag.
36.
=
Area under diagonal
2
)
1 x ⎤ dx
+ 12
⎦⎥
1
∫0 x dx
1
11 1
11 ⎛ x 2 x3 ⎞
11 ⎡⎛ 1 1 ⎞ ⎤ 11 1 11
Numerator = ∫ x − x 2 dx = ⎜
− ⎟ = ⎢⎜ − ⎟ − 0 ⎥ = ⋅ =
⎜
⎟
0
12
12 ⎝ 2
3 ⎠
12 ⎣⎝ 2 3 ⎠ ⎦ 12 6 72
0
(
Denominator =
)
1
(see Problem 35).
2
Coefficient of inequality =
11
72
1
2
=
11
36
600
ISM: Introductory Mathematical Analysis
Section 14.10
37. y 2 = 3 x, y = mx
Intersection: (mx)2 = 3 x, m2 x 2 = 3x
3
m2
3
If x = 0, then y = 0; if x =
m2
, then y =
28
3
32
3
3
.
m
With horizontal elements,
3 / m2
0
5
(
2
3
⋅100 = 87.5%
x 2 = k , x = ± k . Equating areas gives
∫−
k
k
( k − x2 ) dx = 12 ∫−22 ( 4 − x2 ) dx
⎛
x3 ⎞
⎜ kx − ⎟
⎜
3 ⎟⎠
⎝
−
)
3 x − mx dx.
2
k
k
1⎛
x3 ⎞
= ⎜ 4x − ⎟
2 ⎜⎝
3 ⎟⎠
−2
4
16
k =
3
3
3
2
y
3
2
( )
k 2 = 4 ⇒ k = 4 3 = 22
x
8
5
2
3
y
y = x 2 − 1 , y = 2x + 2
x
2
– k
Intersection: x − 1 = 2 x + 2 ,
x 2 − 2 x − 3 = 0 , (x – 3)(x + 1), so x = 3 and
–1. The area is
∫−1 ⎡⎣⎢2 x + 2 − ( x
2
− 1 ⎤⎥dx
⎦
41. 4.76 sq units
42. Two integrals are involved.
Answer: 36.65 sq units
3
⎛ x3
⎞
32
= ⎜ − + x2 + 3x ⎟ =
⎜ 3
⎟
⎝
⎠ −1 3
8
5
40. 0.23 sq units
)
3
= ∫ ( − x 2 + 2 x + 3) dx
−1
3
k
43. Two integrals are involved.
Answer: 7.26 sq units
y
44. Three integrals are involved.
Answer: 358.18 sq units
x
5
601
4
= 2 3 ≈ 2.52
k
38. a.
) dx = 43 . Thus
39. y = x 2 and y = k intersect when
3
m
⎛y
y y ⎞
y ⎞
−
⎜ −
⎟ dy = ⎜
⎟
⎜m 3 ⎟
⎜ 2m 9 ⎟
0
⎝
⎠
⎝
⎠0
9
3
3
=
−
=
square units
2 m3 m3 2 m3
Note: With vertical elements,
Area = ∫
3/ m ⎛
2
2
32 4 28
. Hence the
− =
3 3 3
percentage above the x-axis is
.
Area = ∫
1
the area above is
m2 x 2 − 3x = 0, x(m2 x − 3) = 0, x = 0 or
x=
∫−1(1 − x
b. The area below is
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
Problems 14.11
PS = ∫
1. D : p = 22 − 0.8q ⎫
S : p = 6 + 1.2q ⎬⎭
CS = ∫
5
⎛
q2 ⎞
= ⎜ 0.5q −
⎟ = (2.5 − 1.25) − 0 = 1.25
⎜
20 ⎟⎠
⎝
0
[ f (q) − p0 ] dq
0
8
8
= ∫ [(22 − 0.8q ) − 15.6] = ∫ (6.4 − 0.8q) dq
0
0
(
= 6.4q − 0.4q 2
q0
PS = ∫
0
4. D : p = 400 − q 2 ⎫
⎬
S : p = 20q + 100 ⎭
) 0 = (51.2 − 25.6) − 0 = 25.6
8
Equilibrium pt. ( q0 , p0 ) = (10, 300)
[ p0 − g (q)] dq
8
8
0
0
(
(
=∫
8
10
PS = ∫
=∫
(900 − q 2 ) dq
30
0
5
5
3.5
3.5
= 100[(50 − 25) − (35 − 12.25)]
= 225
PS = ∫
3.5
0.5
5
Equilibrium pt. = ( q0 , p0 ) = (5, 5)
CS = ∫
= (2000 − 1000) − 0 = 1000
CS = ∫ 100(10 − 2 p )dp = 100(10 p − p 2 )
50
⎫
⎪
q +5
⎬
q
+ 4.5⎪⎭
10
q0
10
Equilibrium pt. = ( q0 , p0 ) = (300, 3.5)
We use horizontal strips and integrate with
respect to p.
⎛
q3 ⎞
= ⎜ 900q − ⎟
⎜
3 ⎟⎠
⎝
0
= (27, 000 − 9000) − 0
= 18, 000
S: p=
)0
5. D : q = 100(10 − 2 p ) ⎫
S : q = 50(2 p − 1) ⎬⎭
30
3. D : p =
(200 − 20q )dq
(
PS = ∫ [1300 − (400 + q 2 )] dq
0
[300 − (20q + 100)]dq
= 200q − 10q 2
30
⎛
q3 ⎞
= ∫ (900 − q 2 ) dq = ⎜ 900q − ⎟
⎜
0
3 ⎟⎠
⎝
0
= (27, 000 − 9000) − 0 = 18, 000
=∫
10
0
CS = ∫ [(2200 − q ) − 1300]
0
30
10
0
2
30
(100 − q2 ) dq
⎛
q3 ⎞
1000 ⎞
2000
⎛
= ⎜ 100q − ⎟ = ⎜ 1000 −
⎟−0 =
⎜
⎟
3 ⎠
3 ⎠
3
⎝
⎝
0
2. D: p = 2200 − q 2 ⎪⎫
⎬
S : p = 400 + q 2 ⎪⎭
Equilibrium point = (q0 , p0 ) = (30, 1300)
0
10
0
) 0 = (76.8 − 38.4) − 0 = 38.4
30
)
10
CS = ∫ ⎡⎢ 400 − q 2 − 300 ⎤⎥ dq
0 ⎣
⎦
= ∫ [15.6 − (6 + 1.2q )]dq = ∫ (9.6 − 1.2q )dq
= 9.6q − 0.6q 2
[ p0 − g (q)] dq
5⎡
5⎛
q⎞
⎛q
⎞⎤
= ∫ ⎢5 − ⎜ + 4.5 ⎟ ⎥ dq = ∫ ⎜ 0.5 − ⎟ dq
0⎣
0
10 ⎠
⎝ 10
⎠⎦
⎝
Equilibrium pt. = ( q0 , p0 ) = ( 8, 15.6 )
q0
q0
0
p
50(2 p − 1)dp = 50( p 2 − p )
= 50[(12.25 − 3.5) − (0.25 − 0.5)]
= 450
CS
S
(300, 3.5)
PS
[ f (q) − p0 ] dq
D
5⎡
5
⎤
50
=∫ ⎢
− 5⎥ dq = ( 50 ln q + 5 − 5q )
0 q+5
0
⎣
⎦
= [50 ln(10) – 25] – [50 ln(5)]
= 50[ln(10) – ln(5)] – 25 = 50 ln(2) – 25
q
1000
602
3.5
0.5
ISM: Introductory Mathematical Analysis
Section 14.11
6. D : q = 100 − p ⎫⎪
⎬
p
S : q = 2 − 10 ⎪⎭
Equilibrium pt. = ( q0 , p0 ) = (8, 36)
Integrating with respect to p,
CS = ∫
100
3
2
= − (100 − p) 2
3
300
⎛
q2 ⎞
= ⎜ 5q −
= (1500 − 750) − 0 = 750
⎟
⎜
120 ⎟⎠
⎝
0
For CS we integrate with respect to p. From the
demand equation, q = 0 ⇒ p = 20 .
100
20
⎛
p3 ⎞
CS = ∫ 400 − p dp = ⎜ 400 p −
⎟
⎜
10
3 ⎟⎠
⎝
10
2
8000 ⎞ ⎛
1000 ⎞
⎛
= ⎜ 8000 −
⎟ − ⎜ 4000 −
⎟ = 1666
3
3 ⎠ ⎝
3 ⎠
⎝
20
36
⎛ 2
⎞ 1024
= 0 − ⎜ − ⋅ 512 ⎟ =
3
3
⎝
⎠
36 ⎡ p
⎤
PS = ∫ ⎢ − 10 ⎥ dp
20 2
⎣
⎦
p = 211−5 = 64
CS = ∫
20
=−
S
(8, 36)
10. a.
10
7. We integrate with respect to p. From the demand
equation, when q = 0, then p = 100.
100
84
=∫
100
84
=−
1
100
(10 + 10)(30 + 20) = 1000, (20)(50) = 1000,
1000 = 1000
30 − 4(10) + 10 = 0, 30 − 40 + 10 = 0, 0 = 0
1000
,
q + 20
1000
− 10
q + 20
⎤
⎞
1000
− 10 ⎟ − 10 ⎥ dq
⎢⎜
⎠
⎣⎝ q + 20
⎦
30
= [1000 ln(q + 20) − 20q ] 0
= 1000 ln(50) − 600 − [1000 ln(20)]
⎛ 50 ⎞
= 1000 ln ⎜ ⎟ − 600
⎝ 20 ⎠
⎛5⎞
= 1000 ln ⎜ ⎟ − 600
⎝2⎠
CS = ∫
30 ⎡⎛
0
84
3
20 ⎡
20
⎤
= − ⎢0 − (16) 2 ⎥ = − (−64)
3 ⎣
3
⎦
2
= 426 ≈ $426.67
3
400 − p 2
+5,
60
60 p = 400 − p 2 + 300 , p 2 + 60 p − 700 = 0 ,
11. CS ≈ 1197; PS ≈ 477
( p + 70)( p − 10) = 0 ⇒ p = 10 and
q = 400 − 102 = 300 , so equilibrium pt. is
( q0 ,
)
⎛ 211
⎞
64
− 320 − ⎜ −
− 0⎟
⎜ ln 2
⎟
ln 2
⎝
⎠
p=
−10(100 − p ) 2 [− dp ]
8. At equilibrium, p =
5
⎛ 211− q
⎞
− 64 dq = ⎜ −
− 64q ⎟
⎜ ln 2
⎟
⎝
⎠0
b. (p + 10)(q + 20) = 1000, p + 10 =
10 100 − pdp
3
20
(100 − p ) 2
3
(2
11− q
≈ 2542.307 hundred ≈ $254,000
q
CS = ∫
5
0
D
PS
)
211− q = 2q +1 ⇒ 11 − q = q + 1 ⇒ q = 5 , so
p
CS
(
2
9. At equilibrium,
36
⎛ p2
⎞
=⎜
− 10 p ⎟ = (324 − 360) − (100 − 200)
⎜ 4
⎟
⎝
⎠ 20
= 64
100
⎛ q
⎞⎤
⎢10 − ⎜ 60 + 5 ⎟ ⎥ dq
⎝
⎠⎦
⎣
0
100 − p dp
36
300 ⎡
PS = ∫
p0 ) = (300, 10) .
603
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
12. Let p = f(q).
5.
40
PS = ∫ [80 − f (q )] dq
0
40
=∫
0
80 dq − ∫
= 3200 − ∫
40
0
40
0
f (q ) dq
6
∫ ( x + 5)3 dx = 6∫ ( x + 5)
−3
dx
6( x + 5)−2
+C
−2
=
= −3( x + 5)−2 + C
f (q ) dq
Use the trapezoid rule with h = 10 to estimate
40
∫0
f (q ) dq :
f (0)
2 f (10)
2 f (20)
2 f (30)
f (40)
6.
=
25 =
= 2(49) =
= 2(59) =
= 2(71) =
=
80 =
25
98
118
142
80
463
7.
8.
Chapter 14 Review Problems
)
∫ dx = ∫ 1 dx = 1⋅ x + C = x + C
3.
∫0 (
)
8⎛
9.
⎛ 1⎞
2
∫0 xe
4− x 2
dx = −
=0
2
)
− 6 dx ⎤⎥
⎦
1 2 4− x 2
[−2 x dx]
e
2 ∫0
2
0
13
) (
∫0
3t + 8dt =
4
3
8
4
3
(−3)
−
302
302
1
1 (3t + 8) 3
= ⋅
4
3
3
⎛
⎞
2x 2
⎜
= 2⋅
+ x2 ⎟
⎜
⎟
3
⎝
⎠0
3
2
⎛
⎞
= ⎜ 2 ⋅ 2 2 + 64 ⎟ − 0
3
⎝
⎠
64
256
=
+ 64 =
3
3
4.
3
302
∫ x3 − 6 x + 1 dx = 2∫ x3 − 6 x + 1 ⎡⎢⎣( 3x
(
1
2
(
9
)
1
1
= − e0 − e 4 = e 4 − 1
2
2
⎞
2 x + 2 x dx = ∫ ⎜ 2 x + 2 x ⎟ dx
0⎝
⎠
8
302
6 x 2 − 12
2
1
= − e4− x
2
x4
x2
+ 2⋅
− 7x + C
1. ∫ x3 + 2 x − 7 dx =
4
2
x4
=
+ x2 − 7 x + C
4
2.
( y − 6)302
dy =
302
= 2 ln x3 − 6 x + 1 + C
f (q ) dq ≈
(
∫3 ( y − 6)
301
=
10
(463) = 2315
2
Thus PS = 3200 − 2315 = $885.
40
∫0
9
(3t + 8)
4
=
)
1
∫ 5 − 3x dx = 4 ⎜⎝ − 3 ⎟⎠ ∫ 5 − 3x [−3 dx]
∫
11.
∫ y( y + 1)
604
0
1
=
0
113 11
−4
4
1
∫0
2
(
)
dy = ∫ y3 + 2 y 2 + y dy
y 4 2 y3 y 2
+
+
+C
4
3
2
=
12.
1
4 − 2x
4
1
⎛4 2 ⎞
dx = ∫ ⎜ − x ⎟ dx = x − x 2 + C
7
7
7
⎝7 7 ⎠
10.
4
= − ln 5 − 3x + C
3
4
3
1
1 1
3 [3 dt ]
(3
t
+
8)
3 ∫0
1
10−8 dx = 10−8 x = 10−8 − 0 = 10−8
0
ISM: Introductory Mathematical Analysis
13.
⎛ t1/ 5 t1/ 3 ⎞
dt
=
∫ t
∫ ⎜⎜ t1/ 2 − t1/ 2 ⎟⎟ dt
⎝
⎠
7 /10
t
t5 / 6
= ∫ (t −3 /10 − t −1/ 6 )dt =
−
+C
5
t −3t
10
6
= t 7 /10 − t 5 / 6 + C
7
5
14.
7
10
20.
5
6
21.
3
∫1
2t
22.
2
1 3 1
dt = ∫
[6t 2 dt ]
3
3 1 3 + 2t 3
3 + 2t
3
1
= ln(3 + 2t 3 )
3
1
1
1 ⎛ 57 ⎞
= [ln(57) − ln(5)] = ln ⎜ ⎟
3
3 ⎝ 5 ⎠
2
3
∫ x 3x + 2dx =
(
3
1 3x + 2
= ⋅
3
9
)
3
2
2
18.
3
)
1
2
(
4
)
3
2
∫ (e
4
2 7/2
(x + x )
2y
7
2
24.
+C
25.
= ln x + 2 ⋅
x −1
+C
−1
2
−2
+C
dx
2
+C
x
3e3 x
1
2
∫0 1 + e3x dx = ∫0 1 + e3x [3e
(
3x
dx]
2
0
70
∫7
2
dx = x
∫1 5 x
)
4
70
= 70 − 7 = 63
7
5 − x 2 dx = −
5 (5 − x 2 )3 / 2
=− ⋅
3
2
4 4
( x + x 2 )7 / 2 + C
7
2
2
1
5 2
(5 − x 2 )1/ 2 [−2 x dx]
2 ∫1
)
1 2y
⎛ 1⎞
e [2 dy ] − ⎜ − ⎟ ∫ e−2 y [−2 dy ]
2∫
⎝ 2⎠
1
1
1
= e2 y + e−2 y + C = e2 y + e−2 y + C
2
2
2
=
)
605
2
5
= − (5 − x 2 )3 / 2
3
1
5
5
35
= − (13 / 2 − 43 / 2 ) = − (1 − 8) =
3
3
3
− e−2 y dy
(
1
2
3
1
⎛ y5 y 2
⎞
23. ∫ 10 y − y + 1 dy = 10 ⎜
−
+ y⎟
⎜ 5
⎟
−2
2
⎝
⎠ −2
⎛1 1 ⎞
⎛ 32
⎞
= 10 ⎜ − + 1⎟ − 10 ⎜ − − 2 − 2 ⎟ = 111
5
2
5
⎝
⎠
⎝
⎠
2 5/ 2
+C =
)
+ C = − 7 − 2 x2
2 ⎞
1
⎡9 x 2 dx ⎤
⎣
⎦
2
+C =
3 x3 + 2
27
(
2
3
= ln(1 + e6 ) − ln(1 + 1)
⎛ 1 + e6 ⎞
= ln ⎜
⎟
⎜ 2 ⎟
⎝
⎠
∫ (8 x + 4 x)( x + x ) dx
= 2 ∫ ( x 4 + x 2 )5 / 2 [(4 x3 + 2 x)dx]
= 2⋅
19.
(
1
3 x3 + 2
9∫
⎛1
)
= ln(1 + e3 x )
4 x2 − x
16. ∫
dx = ∫ (4 x − 1)dx = 2 x 2 − x + C
x
17.
)
∫ ⎜⎝ x + x 2 ⎟⎠ dx = ∫ x dx + 2∫ x
= ln x −
1 (0.5 x − 0.1)5
⋅
+ C = (0.5 x − 0.1)5 + C
0.2
5
(
−1
8⎛ 1⎞
dx = ⎜ − ⎟ ∫ 7 − 2 x 2 3 [−4 x dx]
3
3⎝ 4⎠
3 7 − 2 x2
8x
(
(0.5 x − 0.1)
dx
0.4
1 1
=
⋅
( 0.5 x − 0.1)4 [0.5 dx]
0.4 0.5 ∫
∫
∫
2
2 3 7 − 2x
=− ⋅
3
2
4
=
15.
Chapter 14 Review
Chapter 14: Integration
26.
1
∫0
(
(2 x + 1) x 2 + x
)
4
ISM: Introductory Mathematical Analysis
0(
)
1
4
dx = ∫ x 2 + x [(2 x + 1) dx] =
(
x2 + x
51
)
=
5
25
32
−0 =
5
5
0
⎡
1
27. ∫ ⎢ 2 x −
2
0⎢
( x + 1) 3
⎣
1
1
1
1
⎡
⎤
2
1
3 ⎤
1
1
x
(
x
1)
+
− 23
⎢
⎥ = ⎡ x 2 − 3( x + 1) 3 ⎤
⎥ dx = 2 x dx − ( x + 1) [dx] = 2 ⋅
−
∫0
∫0
⎢⎣
⎥⎦
1
⎢ 2
⎥
⎥
0
3
⎦
⎣
⎦0
= ⎡1 − 33 2 ⎤ − [ 0 − 3] = 4 − 33 2
⎣
⎦
28.
∫3 3 (
27
)
3 x − 2 x + 1 dx = 3∫
27 ⎛
3
⎜ 3x
⎝
1
2
3
⎡⎛
2
⎞ ⎛
= 3 ⎢⎜ 3 ⋅ 3 3 − 729 + 27 ⎟ − ⎜
3
⎠ ⎝
⎣⎝
( )
29.
30.
t −3
∫
t2
27
2 3
⎛
⎞
⎞
− 2 x + 1⎟ dx = 3 ⎜ 3 ⋅ x 2 − x 2 + x ⎟
3
⎠
⎝
⎠3
3
2
⎤
⎞
3⋅
3 − 9 + 3 ⎟ ⎥ = 3(−540) = −1620
3
⎠⎦
( )
−1
⎡ 12
⎤
t 2
t −1
t
3⎥
3 2
−1
⎛ − 32
−2 ⎞
⎢
− 3⋅
+ C = −2t 2 + 3t −1 + C = −
+C
dt = ∫
dt = ∫ ⎜ t − 3t ⎟ dt =
−
1
⎢ t2 t2 ⎥
t
−
1
−2
⎝
⎠
t
⎣
⎦
3z3
1 ⎞
⎛ 2
+ z +1+
⎟ dz
z −1 ⎠
⎛ z3 z 2
⎞
= 3⎜ +
+ z + ln z − 1 ⎟ + C
⎜ 3
⎟
2
⎝
⎠
∫ z − 1 dz = 3∫ ⎜⎝ z
0
⎛ x2
⎞
0 ⎛
x2 + 4 x − 1
5 ⎞
31. ∫
dx = ∫ ⎜ x + 2 −
dx
=
+ 2 x − 5ln x + 2 ⎟
⎜
⎟
⎜
⎟
−1
−1 ⎝
x+2
x+2⎠
⎝ 2
⎠ −1
0
⎛1
⎞ 3
= (−5ln 2) − ⎜ − 2 − 0 ⎟ = − 5ln 2
2
⎝
⎠ 2
32.
∫
( x2 + 4)
(
2
x2
dx = ∫
x 4 + 8 x 2 + 16
)
x2
dx
= ∫ x 2 + 8 + 16 x −2 dx
=
x3
x −1
x3
16
+ 8 x + 16
+C =
+ 8x − + C
3
3
−1
x
1
2 ⎛ 3
⎞2 ⎡ 3 1 ⎤
33. ∫ 9 x x + 1 dx = 9 ⋅ ∫ ⎜ x 2 + 1⎟ ⎢ x 2 dx ⎥
3 ⎝
⎠ ⎣2
⎦
3
2
3
⎛ 32
⎞2
3
⎜ x + 1⎟
⎛ 32
⎞2
⎝
⎠
= 6⋅
+ C = 4 ⎜ x + 1⎟ + C
3
⎝
⎠
2
606
ISM: Introductory Mathematical Analysis
Chapter 14 Review
1
34.
e
∫
5x
∫
3
dx =
3x
5x 2
e
1
x
40.
dx
1
2
1
⎤
2 ⎡ 5 −1
1∫ e 5 x ⎢
x 2 dx ⎥
3⋅ 5
⎣ 2
⎦
1
⎛
⎞
2
2
⎜ e 5x ⎟ + C
=
⎟
15 ⎜⎝
⎠
2
e 5x + C
=
15
2
=
35.
36.
e eln x
∫1
x
= ln e – ln 1
=1–0=1
∫
6x2 + 4
e
3
x +2x
= −2e
e
x
1
2
dx = ∫
2
x
dx = ∫
dx = −2∫ e
(
− x3 + 2 x
(
∫
(1 + e2 x )3
e
−2 x
) +C =
dx =
=
38.
∫
3
−2
e
6 + e−3 x
2
∫3
=
1
(3x2 + 2) dx⎤⎦⎥
+C
x+5
⎛ 5⎞
dx = ∫ ⎜ 1 + ⎟ dx = x + 5ln x + C
x
⎝ x⎠
y(1) = 3 implies 3 = 1 + 0 + C, so C = 2. Thus
y = x + 5ln x + 2
42. y = ∫
(1 + e2 x )3
+C
8
In Problems 43–58, answers are assumed to be
expressed in square units.
dx
(
)
−2
= − ∫ ( 6 + e−3 x ) ⎡ −3e−3 x dx ⎤
⎣
⎦
−
1
6 + e−3 x )
(
1
=−
+C =
3x
6 + e−3 x
3 x ln10
103 x dx = 3∫ e 2
)
1 2x
e [2 dx] + ∫ 3 dx
2∫
1
= e2 x + 3 x + C
2
1
1 1
y (0) = − implies that − = + 0 + C , so
2
2 2
1
C = –1. Thus y = e2 x + 3 x − 1
2
dx = ln x
⎣⎢
x3 + 2 x
(
e
) ⎡−
dx
x 2 + 3x + 7
⎛ 5 x3 + 15 x 2 + 35 x
2x + 3 ⎞
= ∫⎜
+
⎟ dx
⎜ x 2 + 3x + 7
x 2 + 3x + 7 ⎟⎠
⎝
1
[(2 x + 3) dx]
= ∫ 5 x dx + ∫
2
x + 3x + 7
5x2
=
+ ln x 2 + 3x + 7 + C
2
41. y = ∫ e2 x + 3 dx = ∫ e2 x dx + ∫ 3 dx
1
(1 + e2 x )3 [2e2 x dx]
2∫
−1
39.
x
− x3 + 2 x
e
37.
e1
1
∫
5 x3 + 15 x 2 + 37 x + 3
43. y = x 2 − 1 , x = 2, y ≥ 0. Region appears below.
Area = ∫
2
1
( x2 − 1) dx
2
⎛ x3
⎞
= ⎜ − x⎟
⎜ 3
⎟
⎝
⎠1
⎛8
⎞ ⎛1 ⎞ 4
= ⎜ − 2 ⎟ − ⎜ − 1⎟ =
⎝3
⎠ ⎝3 ⎠ 3
+C
dx
8
3x
2
ln10 ⎡ 3ln10
⎤
e2
= 3⋅
∫
⎢ 2 dx ⎥
3ln10
⎣
⎦
=
3x
2 32x ln10
2
e
+C =
10 2 + C
ln10
ln10
=
2 103 x
+C
ln10
y
x
2
607
5
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
44. y = 4e x , x = 0, x = 3. Region appears below.
3
3
0
0
Area = ∫ 4e x dx = 4e x
100
= 4(e3 − 1)
y
x
3
5
45. y = x + 4 , x = 0. Region appears below.
Area = ∫
0
x + 4dx = ∫ ( x + 4) [dx] =
−4
5
3
1
2
0
−4
y
( x + 4) 2
3
2
0
3
−4
2( x + 4) 2
=
3
0
=
−4
16
16
−0 =
3
3
x
5
46. y = x 2 − x − 6, x = −4, x = 3. Region appears below.
−2
3
⎛ x3 x 2
⎞
⎛ x3 x 2
⎞
Area = ∫ ( x − x − 6) dx + ∫ −( x − x − 6) dx = ⎜ −
− 6x ⎟ − ⎜ −
− 6x ⎟
⎜ 3
⎟
⎜
⎟
−4
−2
2
2
⎝
⎠ −4 ⎝ 3
⎠ −2
9
⎡⎛ 8
⎞ ⎛ 64
⎞ ⎤ ⎡⎛
⎞ ⎛ 8
⎞ ⎤ 67
= ⎢⎜ − − 2 + 12 ⎟ − ⎜ − − 8 + 24 ⎟ ⎥ − ⎢⎜ 9 − − 18 ⎟ − ⎜ − − 2 + 12 ⎟ ⎥ =
2
⎠ ⎝ 3
⎠ ⎦ ⎣⎝
⎠ ⎝ 3
⎠⎦ 2
⎣⎝ 3
−2
10
3
2
2
y
x
10
608
ISM: Introductory Mathematical Analysis
Chapter 14 Review
47. y = 5 x − x 2 . Region appears below.
50. y = x3 − 1 , x = –1. Region appears below.
5
1
⎛ 5 x 2 x3 ⎞
5
− ⎟
Area = ∫ 5 x − x 2 dx = ⎜
⎜ 2
0
3 ⎟⎠
⎝
0
⎛ x4
⎞
1
Area = ∫ − x3 − 1 dx = − ⎜
− x⎟
⎜ 4
⎟
−1
⎝
⎠ −1
125
⎛ 125 125 ⎞
=⎜
−
⎟−0 =
3 ⎠
6
⎝ 2
⎛ 3⎞ ⎛5⎞
= −⎜ − ⎟ + ⎜ ⎟ = 2
⎝ 4⎠ ⎝4⎠
(
)
(
y
10
)
y
3
x
3
x
10
–5
48. y = 4 x , x = 1, x = 16. Region appears below.
Area = ∫
16 4
1
=
xdx = ∫
16
1
5
4x 4
x dx =
5
1
4
51. y 2 = 4 x , x = 0, y = 2. Region appears below.
16
Area = ∫
0
1
5
128 4 124
− =
5 5
5
4
y2
y3
dy =
4
12
2
2
=
0
8
2
−0 =
12
3
y
2
y
x
5
2
x
16
8
52. y = 3x 2 − 5, x = 0, y = 4. Region appears below.
3x 2 − 5 = 4, 3x 2 = 9, x 2 = 3, so x = ± 3.
1
+ 2, x = 1, x = 4. Region appears below.
x
4⎛ 1
4
⎞
Area = ∫ ⎜ + 2 ⎟ dx = ( ln x + 2 x )
1 ⎝x
1
⎠
= [ln(4) + 8] − [0 + 2] = 6 + ln 4
49. y =
Area = ∫
3
[4 − (3x 2 − 5)] dx
=∫
3
[9 − 3 x 2 ] dx = (9 x − x3 )
0
(
0
)
= 9 3 −3 3 −0 = 6 3
y
5
5
y
x
5
x
0
1
4
5
609
3
0
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
53. y = x 2 + 4 x − 5 , y = 0. Region appears below.
Area = ∫
−2
5/ 2
x 2 + 4 x − 5 = 0 , (x + 5)(x – 1) = 0, so x = –5, 1.
1
(
=∫
)
2
−2
Area = ∫ − x + 4 x − 5 dx
−5
[(10 − x 2 ) − ( x 2 − x)]dx
(10 + x − 2 x 2 )dx
5/ 2
⎛
x 2 2 x3 ⎞
= ⎜10 x +
−
⎟
⎜
2
3 ⎟⎠
⎝
−2
25 125 ⎞ ⎛
16 ⎞
⎛
= ⎜ 25 + −
⎟ − ⎜ −20 + 2 + ⎟
8 12 ⎠ ⎝
3⎠
⎝
243
=
8
1
⎛ x3
⎞
= − ⎜ + 2 x2 − 5x ⎟
⎜ 3
⎟
⎝
⎠ −5
⎛1
⎞ ⎛ 125
⎞
= − ⎜ + 2 − 5⎟ + ⎜ −
+ 50 + 25 ⎟ = 36
⎝3
⎠ ⎝ 3
⎠
y
4
5/ 2
15
x
y
4
x
5
54. y = 2 x 2 , y = x 2 + 9 . Region appears below.
56. y = x , x = 0, y = 3. Region appears below.
2 x 2 = x 2 + 9, x 2 = 9 , so x = ±3
3
(
x = 3 , so x = 9.
) ( )
Area = ∫ ⎡⎢ x 2 + 9 − 2 x 2 ⎤⎥ dx
−3 ⎣
⎦
3
⎛
2x 2
⎜
Area = ∫ 3 − x dx = 3 x −
0
⎜
3
⎝
= (27 − 18) − 0 = 9
9
3
⎛
x3 ⎞
= ∫ 9 − x 2 dx = ⎜ 9 x − ⎟
⎜
−3
3 ⎟⎠
⎝
−3
= (27 – 9) – (–27 + 9) = 36
3
(
)
18
5
y
(
)
9
⎞
⎟
⎟
⎠0
y
3
x
16
x
–3
2
3
5
57. y = ln x, x = 0, y = 0, y = 1. Region appears
below.
2
55. y = x − x, y = 10 − x . Region appears below.
y = ln x ⇒ x = e y
x 2 − x = 10 − x 2 , 2 x 2 − x − 10 = 0,
1
1
0
0
Area = ∫ e y dy = e y
5
(x + 2)(2x − 5) = 0, so x = −2 or .
2
5
= e −1
y
1
x
5
610
ISM: Introductory Mathematical Analysis
Chapter 14 Review
58. y = 2 − x, y = x − 3, y = 0, y = 2. Region appears
below.
62.
1000
33
∫10
3q + 70
2
Area = ∫ [( y + 3) − (2 − y )] dy
2
0
0
= (4 + 2) − 0 = 6
3
1000 (3q + 70) 2
=
⋅
1
3
2
33
x
63.
100
∫0
0.008e−0.008t dt = − ∫
1
3
3
⎛
⎞
2q ⎟ dq = ∫ 100dq −
2 ∫ q 2 dq
59. r = ∫ ⎜ 100 −
2
2
⎝
⎠
64.
3
3
3
q2
= 100q −
2⋅
+ C = 100q − 2q 2 + C
3
2
5
∫0 4000e
=
2
100
0
0.05t
4000 0.05t
e
0.05
When q = 0, then r = 0. Thus 0 = 0 – 0 + C, so
1
r
= 100 − 2q 2 = 100 − 2q . Thus
q
5
=
3
4
3
4
= ∫ (9 − 3x)dx + ∫ (3x − 9)dx
0
q3 7 2
+ q + 6q + C
60. c = ∫ q + 7q + 6 dq =
3 2
When q = 0, then c = 2500. Thus
2500 = 0 + 0 + 0 + C, so C = 2500. Hence
q3 7 2
c=
+ q + 6q + 2500 . When q = 6, then
3 2
c = $2734.
)
3
⎛
3x2
= ⎜ 9x −
⎜
2
⎝
3
4
⎞
⎛ 3x 2
⎞
− 9x ⎟
⎟ +⎜
⎟
⎜
⎟
⎠0 ⎝ 2
⎠3
⎡⎛
27 ⎞ ⎤ ⎡
⎛ 27
⎞⎤
= ⎢⎜ 27 − ⎟ − 0 ⎥ + ⎢(24 − 36) − ⎜ − 27 ⎟ ⎥
2
2
⎝
⎠
⎝
⎠⎦
⎣
⎦ ⎣
= 15 square units
16
2
4000 ⎡ 0.25 ⎤
e
− 1 ≈ $22, 722
⎦
0.05 ⎣
0
3
∫15 (250 − q − 0.2q
1 5 0.05t
[0.05 dt ]
e
0.05 ∫0
Area = ∫ [(9 − 2 x) − x]dx + ∫ [ x − (9 − 2 x)]dx
p = 100 − 2q .
2
[−0.008 dt ]
= −e−0.8 + 1 ≈ 0.5507
dt = 4000 ⋅
0
e
65. y = 9 – 2x, y = x; from x = 0 to x = 4. Region
appears below. Intersection: x = 9 – 2x, 3x = 9, so
x = 3.
3
C = 0. Hence r = 100q − 2q 2 . Since r = pq,
then p =
100 0.008t
0
= −e−0.008t
25
10
2000
3q + 70
3
10
2000
[13 − 10] = $2000
=
3
5
61.
33
=
y
(
1 33
−1
(3q + 70) 2 [3 dq ]
∫
3 10
1
0
2
= ∫ (2 y + 1) dy = ( y 2 + y )
dq = 1000 ⋅
y
)dq
y = 9 – 2x
25
⎛
q 2 0.2q3 ⎞
= ⎜ 250q −
−
⎟
⎜
2
3 ⎟⎠
⎝
15
3125 ⎞
⎛
= ⎜ 6250 − 312.5 −
⎟ − (3750 − 112.5 − 225)
3 ⎠
⎝
≈ $1483.33
y=x
x
4
611
8
Chapter 14: Integration
ISM: Introductory Mathematical Analysis
1
66. y = 2 x 2 , y = 2 − 5x; from x = −1 to x = .
3
Region appears below.
68. D : p = (q − 5) 2 ⎪⎫
⎬
S : p = q 2 + q + 3⎪⎭
Equilibrium pt. = ( q0 , p0 ) = (2, 9)
2 x 2 = 2 − 5 x, 2 x 2 + 5 x − 2 = 0,
−5 ± 41
x=
(from the quadratic formula),
4
x ≈ −2.85 or 0.35.
Area = ∫
1/ 3
−1
CS = ∫
2
0
2
3
⎡
⎤
⎡ (q − 5)2 − 9 ⎤ dq = ⎢ (q − 5) − 9q ⎥
⎣
⎦
⎣⎢ 3
⎦⎥ 0
2
⎛ −27
⎞ ⎛ 125
⎞
=⎜
− 18 ⎟ − ⎜ −
− 0 ⎟ = $14 thousand
3
3
3
⎝
⎠ ⎝
⎠
≈ $14,667
[(2 − 5 x) − 2 x 2 ]dx
1/ 3
⎛
5 x 2 2 x3 ⎞
= ⎜ 2x −
−
⎟
⎜
2
3 ⎟⎠
⎝
−1
5 2⎞
⎛2 5 2 ⎞ ⎛
= ⎜ − − ⎟ − ⎜ −2 − + ⎟
2 3⎠
⎝ 3 18 81 ⎠ ⎝
340
=
square units
81
69.
qn
∫q0
n
dq
= −(u + v) ∫ dt
0
q − qˆ
ln q − qˆ
qn
q0
n
= −(u + v)t 0
ln qn − qˆ − ln q0 − qˆ = −(u + v)n
y
ln q0 − qˆ − ln qn − qˆ = (u + v)n
15
ln
q0 − qˆ
= (u + v)n
qn − qˆ
n=
x
5
q − qˆ
1
ln 0
u + v qn − qˆ
as was to be shown.
67. D : p = 0.01q 2 − 1.1q + 30 ⎫⎪
⎬
S : p = 0.01q 2 + 8
⎪⎭
Equilibrium pt. = ( q0 , p0 ) = (20, 12)
CS = ∫
q0
0
=∫
20
=∫
20
0
0
=
=
)
⎡ 0.01q 2 − 1.1q + 30 − 12 ⎤ dq
⎣⎢
⎦⎥
( 0.01q
2
)
20
q0
0
=
=∫
20
0
20
[ p0 − g (q)] dq = ∫0
( 4 − 0.01q )
2
4η l
π ( P1 − P2 )
R
∫0
2η l
π ( P1 − P2 ) R
2η l
(
)
r R 2 − r 2 dr
∫0 ( R
2
)
r − r 3 dr
R
⎛ 0.01q3 1.1q 2
⎞
=⎜
−
+ 18q ⎟
⎜ 3
⎟
2
⎝
⎠0
PS = ∫
0
π ( P1 − P2 ) ⎛ R 2 r 2 r 4 ⎞
=
− ⎟
⎜
⎜ 2
2η l
4 ⎟⎠
⎝
0
− 1.1q + 18 dq
2
⎛ 80
⎞
= ⎜ − 220 + 360 ⎟ − 0 = 166
3
⎝ 3
⎠
( P1 − P2 ) ( R 2 − r 2 )
70. Q = ∫ 2π rv dr = 2π∫ r ⋅
[ f (q) − p0 ] dq
(
R
π ( P1 − P2 ) ⎛ R 4 ⎞ πR 4 ( P1 − P2 )
⎜
⎟=
⎜ 4 ⎟
2η l
8η l
⎝
⎠
As was to be shown.
=
(
π ( P1 − P2 ) ⎡⎛ R 4 R 4 ⎞ ⎤
−
⎢⎜
⎟ − 0⎥
2η l
4 ⎟⎠ ⎦⎥
⎢⎣⎜⎝ 2
)
⎡12 − 0.01q 2 + 8 ⎤ dq
⎣⎢
⎦⎥
20
⎛
0.01q3 ⎞
dq = ⎜ 4q −
⎟
⎜
3 ⎟⎠
⎝
0
80 ⎞
1
⎛
= ⎜ 80 − ⎟ − 0 = 53
3 ⎠
3
⎝
612
dr
ISM: Introductory Mathematical Analysis
Mathematical Snapshot Chapter 14
71. Case 1. r ≠ –1
b. Total number of units sold
r +1 1/ x
1/ x
1 1/ x
u
g ( x) = ∫ ku r du = ∫ u r du =
1
1
k
r +1
(
)
80
0
(
1
= 40t − 0.25t 2
1
x − r −1 − 1
r +1
1 ⎡
1
−(r + 1) x − r − 2 ⎤ = −
g ′( x) =
r
⎣
⎦
r +1
x +2
Case 2. r = –1
1/ x 1
1 1/ x
g ( x) = ∫ ku −1du = ∫
du
1
1
k
u
1/ x
⎛1⎞
= ln u
= ln ⎜ ⎟ − 0 = − ln x
1
⎝x⎠
1
1
g ′( x) = − = −
r
x
x +2
=
R
0
= ∫ f (t )dt = ∫
c.
3. a.
Total revenue
R
30
0
30
0
=∫
(
5
0
5
)0
c.
= (500 – 25) – 0 = 475
25
20
(100 − 2t )dt = (100t − t 2 )
= (2500 − 625) − (2000 − 400) = 275
2. a.
25
20
R
0
80
0
80
=∫
0
(50 + 0.2t ) ⋅ (40 − 0.5t )dt
( 2000 − 17t − 0.1t 2 ) dt
80
17
1 ⎞
⎛
= ⎜ 2000t − t 2 − t 3 ⎟
2
30
⎝
⎠0
= 160, 000 − 54, 400 −
(900 − t 2 ) dt
Average delivered price
total revenue
=
total number of units sold
2, 002,500
=
= $111.25
18, 000
4. Answers may vary.
Total revenue = ∫ (m + st ) f (t )dt
=∫
30
0
30
1 ⎞
⎛
= ⎜ 900t − t 3 ⎟ = 27, 000 − 9000 = 18, 000
3 ⎠0
⎝
f (t )dt = ∫ (100 − 2t )dt = 100t − t 2
f (t )dt = ∫
R
0
= ∫ f (t )dt = ∫
Mathematical Snapshot Chapter 14
25
30
b. Total number of units sold
75. CS ≈ 1148; PS ≈ 251
∫20
(90, 000 + 900t − 100t 2 − t3 ) dt
100 3 1 4
t − t
3
4 0
= 2,700,000 + 405,000 – 900,000 – 202,500
= $2,002,500
74. Two integrals are needed.
Answer: 32.75
b.
(100 + t )(900 − t 2 )dt
= 90, 000t + 450t 2 −
73. Two integrals are involved.
Answer: 15.08 sq units
∫0
= 3200 − 1600 = 1600
= ∫ (m + st ) f (t )dt = ∫
72. Two integrals are needed.
Answer: 101.75 sq units
5
80
Average delivered price
total revenue
=
total number of units sold
88,533.33
≈
≈ $55.33
1600
0
1. a.
)0
(40 − 0.5t )dt
51, 200
≈ $88,533.33
3
613
Chapter 15
Thus,
Principles in Practice 15.1
2
0.1t
(
) (
)
1.
e0.1t
+C
0.1
= −40te0.1t + 400e0.1t + C
S (t ) = −40te0.1t + 400e0.1t + C and S(0) = 5000
5000 = 0 + 400e0 + C
C = 4600
2.
2. P (t ) = ∫ 0.1t (ln t )2 dt
Let u = (ln t ) 2 and dv = 0.1t dt, so
2 ln t
⎛1⎞
du = 2(ln t ) ⎜ ⎟ dt =
dt and
t
t
⎝ ⎠
2
t2
= 0.05t 2
2
dt
(
)
⎛ 2 ln t ⎞
= 0.05t 2 (ln t ) 2 − ∫ 0.05t 2 ⎜
⎟ dt
⎝ t ⎠
3.
(
−x
dx
−x
dx = − xe− x − ∫ −e− x dx
= − xe − x − ∫ e− x [− dx] = − xe− x − e− x + C
)
⎛1⎞
ln t − ∫ 0.05t 2 ⎜ ⎟ dt
⎝t ⎠
= −e − x ( x + 1) + C
= 0.05t 2 ln t − ∫ 0.05t dt
= 0.05t 2 ln t − 0.05
∫ xe
∫ xe
1
du = dt and v = 0.05t 2 .
t
2
dx
v = −e − x .
∫ 0.1t ln t dt , let u = ln t and dv = 0.1t dt, so
∫ 0.1t ln t dt = 0.05t
3 x +1
Letting u = x, dv = e− x dx , then du = dx,
= 0.05(t ln t )2 − ∫ 0.1t ln t dt
For
∫ xe
If u = x and dv = e3 x +1dx, then du = dx and
1
v = e3 x +1.
3
x 3 x +1
1 3 x +1
3 x +1
∫ xe dx = 3 e − ∫ 3 e dx
x
1 1
= e3 x +1 − ⋅ ∫ e3 x +1[3 dx]
3
3 3
x 3 x +1 1 3 x +1
= e
− e
+C
3
9
1
= e3 x +1 (3x − 1) + C
9
S (t ) = −40te0.1t + 400e0.1t + 4600
∫ 0.1t (ln t )
∫ f ( x)dx = uv − ∫ v du
3
3
2
2
= x ⋅ ( x + 5) 2 − ∫ ( x + 5) 2 dx
3
3
3
5
2
2
2
= x( x + 5) 2 − ⋅ ( x + 5) 2 + C
3
3 5
3
5
2
4
= x ( x + 5) 2 − ( x + 5) 2 + C
3
15
= −40te0.1t + ∫ 40e0.1t dt
v = ∫ 0.1t dt = 0.1
)
2
Problems 15.1
dt = (−4t ) 10e0.1t − ∫ 10e0.1t (−4)dt
= −40te0.1t + 40
2
= 0.05(t ln t ) − 0.05t ln t + 0.025t + C
Let u = –4t and dv = e0.1t dt , so du = –4 dt, and
1 0.1t
v = ∫ e0.1t dt =
e
= 10e0.1t .
0.1
∫ −4te
(
P (t ) = 0.05(t ln t )2 − 0.05t 2 ln t − 0.025t 2 + C
1. S (t ) = ∫ −4te0.1t dt
t2
+C
2
= 0.05t 2 ln t − 0.025t 2 + C
614
ISM: Introductory Mathematical Analysis
4.
∫ xe
−5x
Section 15.1
dx
8.
Letting u = x, dv = e−5 x dx, then du = dx,
Letting u = t, dv = e−t dt , then du = dt, v = −e −t
⎛ t ⎞
−t
−t
∫ ⎜⎝ et ⎟⎠ dt = −te − ∫ −e dt
1
v = − e−5 x .
5
∫ xe
−5 x
xe−5 x
1
− ∫ − e−5 x dx
5
5
−5 x
−5 x
xe
e
=−
+
+C
5
5(−5)
dx = −
=−
5.
∫y
3
e
= −te −t − e−t + C = −e−t (t + 1) + C
9.
−5 x
1⎞
⎛
x+ ⎟+C
5 ⎜⎝
5⎠
∫y
6.
3
∫ 3x
y 4 ln y
y4 ⎛ 1 ⎞
−∫
⎜ dy ⎟
4
4 ⎝y ⎠
y 4 ln y
y3
y 4 ln y y 4
−∫
dy =
−
+C
4
4
4
16
=
y4
4
∫x
2
1⎤
⎡
⎢ ln( y ) − 4 ⎥ + C
⎣
⎦
10.
ln x dx
∫x
=
7.
2
1
dx ,
x
12 x
1+ 4x
dx
then du = 12dx, v =
x3
3
ln x dx =
∫
Letting u = 12x, dv = (1 + 4 x)
Letting u = ln x, dv = x 2 dx , then du =
v=
2 x + 3 dx
1
1
= 3x ⋅ (2 x + 3)3 / 2 − ∫ (2 x + 3)3 / 2 ⋅ 3 dx
3
3
3/ 2 1
= x(2 x + 3)
− (2 x + 3)5 / 2 + C
5
1
3/ 2
= (2 x + 3) [5 x − (2 x + 3)] + C
5
1
= (2 x + 3)3 / 2 (3x − 3) + C
5
3
= (2 x + 3)3 / 2 ( x − 1) + C
5
y4
4
=
2 x + 3 dx
1
v = (2 x + 3)3 / 2 .
3
ln y dy
ln y dy =
∫ 3x
Letting u = 3x, dv = 2 x + 3 dx, then du = 3dx,
⎛1⎞
Letting u = ln y, dv = y3 dy , then du = ⎜ ⎟ dy ,
⎝ y⎠
v=
⎛ t ⎞
∫ ⎜⎝ et ⎟⎠ dt
∫
3
3
x ln x
x ⎛1 ⎞
− ∫ ⎜ dx ⎟
3
3 ⎝x ⎠
1
(1 + 4 x)
2
1
3
= 4 x − 1[6 x − (1 + 4 x)] + C
= (2 x − 1) 4 x + 1 + C
∫ ln(4 x) dx
⎛1⎞
Letting u = ln(4x), dv = dx, then du = ⎜ ⎟ dx ,
⎝x⎠
v = x.
⎛1 ⎞
∫ ln(4 x)dx = x ln(4 x) − ∫ x ⎜⎝ x dx ⎟⎠
= x ln(4 x) − ∫ dx = x ln(4 x) − x + C
= x[ln(4x) – 1] + C
615
dx ,
1
2
1 + 4x
(1 + 4 x) 2
dx = 12 x ⋅
−∫
⋅12dx
2
2
1+ 4x
12 x
= 6 x 1 + 4 x − (1 + 4 x ) 2 + C
x3 ln x x3
x3 ⎡
1⎤
−
+C =
ln( x) − ⎥ + C
3
9
3 ⎢⎣
3⎦
− 12
Chapter 15: Methods and Applications of Integration
11.
ISM: Introductory Mathematical Analysis
x
∫ (5 x + 2)3 dx
14.
1
(5 x + 3)−2 .
10
x
∫ (5 x + 2)3 dx
=−
x
10(5 x + 3)
−∫−
2
1
(5 x + 3)−2 dx
10
1 (5 x + 3)−1
⋅
+C
5(−1)
10(5 x + 3) 2 10
1
x
=−
−
+C
2 50(5 x + 3)
10(5 x + 3)
x
=−
12.
+
ln( x + 1)
⎡ 1
1
⎤
∫ 2( x + 1) dx = 2 ∫ ln( x + 1) ⎢⎣ x + 1 dx ⎥⎦
15.
n
(Form: ∫ u du )
ln( x + 1)
∫ 2( x + 1) dx =
3x + 5
e2 x
2
∫1 4 xe
∫
ln x
x
2
ln( x + 1)
+C
4
∫1
dx
Letting u = ln x, dv = x −2 dx , then du =
2x
dx
Letting u = 4x, dv = e2 x dx , then du = 4dx,
1
v = e2 x
2
2
2
13.
dx
Letting u = 3x + 5, dv = e−2 x dx, then du = 3dx
1
and v = − e−2 x .
2
3x + 5
3x + 5
1 −2 x
∫ e2 x dx = − 2e2 x − ∫ − 2 e ⋅ 3 dx
3 x + 5 3 −2 x
=−
+ ∫ e dx
2e 2 x 2
3 x + 5 3 ⎛ 1 −2 x ⎞
=−
+ ⎜− e
⎟+C
⎠
2e 2 x 2 ⎝ 2
1
[2(3x + 5) + 3] + C
=−
4e 2 x
1
(6 x + 13) + C
=−
4e 2 x
Letting u = x, dv = (5 x + 3)−3 dx, then du = dx
and v = −
∫
2
4 xe 2 x dx = ⎡ 2 xe2 x − ∫ 2e2 x dx ⎤
⎣
⎦1
2
2
= ⎡ 2 xe2 x − e2 x ⎤ = e2 x (2 x − 1)
⎣
⎦1
1
1
dx ,
x
(
)
= e 4 (3) − e2 (1) = e2 3e2 − 1
−1
v = −x .
ln x
ln x
⎞
−1 ⎛ 1
∫ x 2 dx = − x − ∫ − x ⎜⎝ x dx ⎟⎠
ln x
ln x 1
=−
+ ∫ x −2 dx = −
− +C
x
x
x
1
= − (1 + ln x) + C
x
16.
2
∫1 2 xe
−3 x
dx
Letting u = 2x, dv = e−3 x dx, then du = 2 dx and
1
v = − e−3 x .
3
616
ISM: Introductory Mathematical Analysis
2
∫1 2 xe
−3 x
Section 15.1
dx
19.
2
⎡ 2 xe−3 x
⎤
2
= ⎢−
− ∫ − e−3 x dx ⎥
3
3
⎢⎣
⎥⎦ 1
1
1
⎡
⎤
= ⎢ −6 x(4 − x) 2 − ∫ −2(4 − x) 2 (3 dx) ⎥
⎣
⎦1
2
3
1
⎡
⎤
= ⎢ −6 x(4 − x) 2 − 4(4 − x ) 2 ⎥
⎣
⎦1
2
1
1 ⎞ ⎤ ⎡ 2e −3
⎛
⎜ 2 + 3 ⎟⎥ − ⎢− 3
⎝
⎠ ⎦⎥ ⎣⎢
{
17.
∫0 xe
− x2
⎛ 1 ⎞⎤
⎜1 + 3 ⎟ ⎥
⎝
⎠ ⎦⎥
{
18.
∫
(
= 2 9 3 − 10 2
20.
1
3 x3
4 − x2
=−
0
(
) (
1 −1
1
e − 1 = 1 − e −1
2
2
∫e
u
du )
)
∫ (ln x)
)
− 12
4 − x2
)
1
2
2
1
2
1
2
= − 4 − x 2 ⎡3 x 2
⎣⎢
(
)
⎡ 2 ln x ⎤
dx = x(ln x) 2 − ∫ x ⎢
dx ⎥
⎣ x
⎦
∫ ln( x)dx , let u = ln x, dv = dx. Then
⎛1
Thus
( )
(4 − x )
= − x2 + 8
2
⎞
∫ ln( x)dx = x ln x − ∫ x ⎜⎝ x dx ⎟⎠ = x[ln( x) − 1] + C1 .
.
dx
= −3 x 2 4 − x 2
= −3 x 2
dx
⎛1⎞
du ⎜ ⎟ dx , v = x, so
⎝x⎠
dx , then
du = 6x dx,
∫
)
= x(ln x)2 − 2 ∫ ln( x)dx
(
(
2 −9 3
⎡ 2 ln x ⎤
du = ⎢
⎥ dx , v = x.
⎣ x ⎦
For
v = − 4 − x2
2
2
Letting u = (ln x)2 , dv = dx, then
dx
Letting u = 3 x 2 , dv = x 4 − x 2
3 x3
∫ (ln x)
)
2
} 1 = −2 (10
= −2 4 − x ( x + 8)
1 1 2
dx = − ∫ e− x (−2 x dx) (Form:
2 0
2
1
= − e− x
2
}1
= −2 4 − x [3 x + 2(4 − x)]
2e ⎡ 7 3 ⎛ 4 ⎞ ⎤
=−
⎢ − e ⎜ ⎟⎥
3 ⎣3
⎝ 3 ⎠⎦
2
=−
[7 − 4e3 ]
6
9e
1
dx , then du = 3dx,
2
2
−6
− 12
v = −2(4 − x) .
2 3x
∫1 4 − x dx
⎡ 2 xe−3 x 2e−3 x ⎤
= ⎢−
−
⎥
3
9 ⎥⎦
⎢⎣
1
⎡ 2e−6
= ⎢−
3
⎣⎢
dx
1
2
2
1 ⎞⎤
⎛
⎜ x + 3 ⎟⎥
⎝
⎠ ⎦⎥
4− x
Letting u = 3x, dv = (4 − x)
⎡ 2 xe−3 x 2 e−3 x ⎤
= ⎢−
+ ⋅
⎥
3
3 −3 ⎥⎦
⎢⎣
1
⎡ 2e−3 x
= ⎢−
3
⎣⎢
3x
2
∫1
(
− ∫ − 4 − x2
)
1
2
(6 x dx)
( ) +C
+ 2 ( 4 − x )⎤ + C
⎦⎥
− 2 4 − x2
3
2
2
4 − x2 + C
617
∫ (ln x)
2
dx = x ⎡ (ln x)2 − 2 ln( x) + 2 ⎤ + C .
⎣
⎦
Chapter 15: Methods and Applications of Integration
21.
ISM: Introductory Mathematical Analysis
∫ 3(2 x − 2) ln( x − 2)dx
24.
1
e
Letting u = ln x and dv = x 3 dx, then du =
∫ 3(2 x − 2) ln( x − 2)dx
and v =
= 3x ( x − 2) ln( x − 2) − ∫ 3 x dx
1
3
dx
x−2
3
Letting u = xe x , dv = ( x + 1)−2 dx , then
du = ( x + 1)e x dx , v = −( x + 1) −1 .
∫ ( x + 1)2
=−
xe x
+ e x dx
x +1 ∫
xe x
+ ex + C
x +1
x ⎞
ex
⎛
⎛ x +1− x ⎞
= e x ⎜1 −
= ex ⎜
+C =
+C
⎟
⎟
x +1
⎝ x +1 ⎠
⎝ x +1 ⎠
∫x
25.
2 x
Letting u = x 2 , dv = e x dx , then du = 2x dx and
v=e .
∫(
2 x
2 x
x
∫ x e dx = x e − ∫ e (2 x dx)
26.
x
x
∫ xe dx , let u = x, dv = e dx . Then du = dx,
dx = xe x − ∫ e x dx = xe x − e x + C1
= e x ( x − 1) + C1 .
Thus
(
∫x
2
(
)
dx = ∫ x 2 − 2 xe− x + e−2 x dx
x3 e −2 x
−
− 2∫ xe− x dx
3
2
x − e− x
)
∫x
2
dx =
∫ xe
−x
dx ,
x3 e −2 x
−
+ 2e− x ( x + 1) + C
3
2
2 3x
e dx
Letting u = x 2 , dv = e3 x dx, then du = 2x dx and
1
v = e3 x .
3
1 2 3x
1 3x
2 3x
∫ x e dx = 3 x e − ∫ 3 e ⋅ 2 x dx
1
2
= x 2 e3 x − ∫ xe3 x dx
3
3
v = e x and
x
)
x
= x e − 2∫ xe dx
∫ xe
−x
Using Problem 3 for
x
For
∫(x −e
=
e dx
2 x
1
3
⎛
3 43
3 43 1 ⎤ ⎞⎟
⎡
⎜
= 5 ⎢ ln x ⋅ x − ∫ x ⋅ dx ⎥
⎜⎣
x ⎦e⎟
4
4
⎝
⎠
3⎞
⎛ 3ln x 4 3 1
⎡
⎤
= 5⎜ ⎢
x 3 − ∫ x 3 dx ⎥ ⎟
⎜⎣ 4
4
⎦e⎟
⎝
⎠
⎛ 3ln x 4 9 4 3 ⎞
⎡
⎤
= 5⎜ ⎢
x3 − x3 ⎥ ⎟
⎜⎣ 4
16 ⎦ e ⎟
⎝
⎠
⎛ ⎡ 3ln 3 43 9 43 ⎤ ⎡ 3ln e 43 9 43 ⎤ ⎞
= 5⎜ ⎢
e − e ⎥⎟
3 − 3 ⎥−⎢
16 ⎦ ⎣ 4
16 ⎦ ⎠
⎝⎣ 4
4
4
⎛3 ⎡
3⎤ 3 ⎡1⎤⎞
= 5 ⎜ 3 3 ⎢ln 3 − ⎥ − e 3 ⎢ ⎥ ⎟
4
4⎦ 4 ⎣4⎦⎠
⎣
⎝
∫ ( x + 1)2 dx
dx = −
3 43
x .
4
e
xe x
xe x
1
dx
x
5∫ x 3 ln x dx
3
= 3x ( x − 2) ln( x − 2) − x 2 + C
2
23.
3
x ln( x5 ) dx = 5∫ x 3 ln x dx
Letting u = 3 ln(x − 2), dv = (2x − 2)dx, then
3
du =
dx and v = x 2 − 2 x = x( x − 2).
x−2
= 3x ( x − 2) ln( x − 2) − ∫ x( x − 2) ⋅
22.
33
∫e
e dx = x 2 e x − 2 ⎡ e x ( x − 1) ⎤ + C
⎣
⎦
2 x
)
= e x x2 − 2x + 2 + C .
For
∫ xe
3x
dx, let u = x, dv = e3 x dx, then
1
du = dx, v = e3 x , and
3
1 3x
1 3x
3x
∫ xe dx = 3 xe − ∫ 3 e dx
1
1
= xe3 x − e3 x + C1.
3
9
618
ISM: Introductory Mathematical Analysis
Section 15.1
∫ (2
Thus,
1 2 3x 2 ⎛ 1 3x 1 3x ⎞
∫ x e dx = 3 x e − 3 ⎜⎝ 3 xe − 9 e ⎟⎠ + C
1 3x
e (9 x 2 − 6 x − 2) + C
=
27
2 3x
27.
3 x2
∫x e
=
Letting u = x 2 , dv = xe x dx , then du = 2x dx,
⎛1⎞ 2
v = ⎜ ⎟ ex .
⎝2⎠
2
2
=
28.
30.
2
x2e x
ex
−∫
(2 x dx)
2
2
2
2
x2e x
ex
ex
−
+C =
2
2
2
( x − 1) + C
=
2
5 x
∫ x e dx
2
x4
2
x4
=
2
2
e x − 2 ∫ x3e x dx
Using Problem 27 for
∫x
5 x2
e dx =
3 x2
∫x e
4
x x2
⎡1 2
⎤
e − 2 ⋅ ⎢ e x ( x 2 − 1) ⎥ + C
2
⎣2
⎦
29.
x
⎛
− 12
dx
1
−2
2
= x ln ⎛⎜ x + x 2
[2 x dx]
⎝
= x ln ⎛⎜ x + x 2 + 1 ⎞⎟ − x 2 + 1 + C
⎝
⎠
e3
31. Area = ∫ (ln x)dx . Letting u = ln x, dv = dx,
1
∫ ( 2 + x ) dx = ∫ ( 2 + 2 x2 + x ) dx
= ∫ 22 x dx + ∫ x 2 x +1 dx + ∫ x 2 dx
For ∫ x 2 x +1 dx , let u = x, dv = 2 x +1 dx . Then
2x
x +1
( )
1
+ 1 ⎞⎟ − ∫ ( x + 1)
⎠ 2
2
x4 x2
e − e x ( x 2 − 1) + C
2
1 x2 4
= e ( x − 2 x 2 + 2) + C
2
2
1
2
∫ ln ⎜⎝ x +
= x ln ⎛⎜ x + x 2 + 1 ⎞⎟ − ∫ x x 2 + 1
⎝
⎠
dx,
=
x
⎛ x2 + 1 + x ⎞
⎜
⎟=
x + x 2 + 1 ⎜⎝
x 2 + 1 ⎟⎠
1
x 2 + 1 ⎞⎟ dx , let
⎠
⎛
⎞
2
u = ln ⎜ x + x + 1 ⎟ , dv = dx. Then
⎝
⎠
1
du =
dx , v = x, and
x2 + 1
⎛
⎞
2
∫ ln ⎜⎝ x + x + 1 ⎟⎠ dx
2
1 2
e x − ∫ e x ⋅ 4 x3 dx
2
2
dx = ∫ 22 x dx + ∫ x 2 x +1 dx + ∫ x 2 dx
1
x
1
x3
⋅ 22 x −1 +
⋅ 2 x +1 −
⋅ 2 x +1 +
+C
ln 2
ln 2
3
ln 2 2
For
x2
Letting u = x and dv = xe dx, then
1 2
du = 4 x3 dx and v = e x .
2
5 x
∫ x e dx =
2
d ⎡ ⎛
⎤
ln ⎜ x + x 2 + 1 ⎞⎟ ⎥
dx ⎢⎣ ⎝
⎠⎦
⎛
1
x ⎞
⎜1 +
⎟
=
x + x 2 + 1 ⎝⎜
x 2 + 1 ⎠⎟
2
4
)
1 2x
2 [2 dx] + ∫ x 2 x +1 dx + ∫ x 2 dx
2∫
1
x
1
x3
=
⋅ 22 x +
⋅ 2 x +1 −
⋅ 2 x +1 +
+C
2 ln 2
ln 2
3
ln 2 2
2
2
+x
=
dx
3 x
∫ x e dx =
x
2
⎛1⎞
then du = ⎜ ⎟ dx , v = x.
⎝x⎠
e3
e3
1 ⎤
⎡
∫1 (ln x)dx = ⎢⎣( x ln x) − ∫ x ⋅ x dx ⎥⎦
1
1
⋅ 2 x +1 and
ln 2
x
1
x +1
x +1
x +1
∫ x2 dx = ln 2 ⋅ 2 − ln 2 ∫ 2 dx
x
1
=
⋅ 2 x +1 −
⋅ 2 x +1 + C1 . Thus
2
ln 2
ln 2
du = dx, v =
e3
e3
= ⎡( x ln x ) − ∫ dx ⎤ = [ x ln( x) − x]
⎣
⎦1
= ⎡e ⋅ 3 − e − [1 ⋅ 0 − 1] = 2e + 1
⎣
⎦
3
3⎤
3
The area is (2e3 + 1) sq units.
619
1
Chapter 15: Methods and Applications of Integration
ISM: Introductory Mathematical Analysis
du = dq, v = −10e− (0.1q +1) , and
1
32. Area = ∫ x 2 e x dx.
0
20
I1 = ⎡ −10(q + 10)e −(0.1q +1) + 10 ∫ e− (0.1q +1) dq ⎤
⎣
⎦0
Letting u = x 2 , dv = e x dx, then du = 2x dx and
v = ex .
∫x
20
= ⎡ −10(q + 10)e− (0.1q +1) − 100e− (0.1q +1) ⎤
⎣
⎦0
2 x
For
e = x 2 e x − 2 ∫ xe x dx
∫ xe
x
= −10e −(0.1q +1) [(q + 10) + 10]
dx, let u = x and dv = e x dx, then
du = dx and v = e x .
∫ xe
x
Thus
x
x
x
x
= −10e
x
= xe − ∫ e dx = xe − e = e ( x − 1).
1 2 x
x e
0
∫
2 x
x
dx = ( x e − 2[e ( x − 1)])
= (e x [ x 2 − 2 x + 2])
= e−2
The area is (e − 2) sq units.
1
0
Thus
CS = 10 I1 − 300e −3 I 2
(
0
−3
35. a.
1
dx,
x
−1
≈ 237.89
Consider
∫ p dq . Letting u = p, dv = dq,
then du =
dp
dq , v = q. Thus
dq
dp
2 2
x ln x dx
1
2
(since r = pq).
dp
dq .
dq
Combining the integrals gives
⎛
dp ⎞
r = ∫ ⎜ p + q ⎟ dq .
dq ⎠
⎝
b. From (a), r = ∫ p dq + ∫ q
2
⎛x
⎞
1
= ⎜ ln x − ∫ x 2 dx ⎟
⎜ 3
⎟
3
⎝
⎠1
2
⎛ x3
1 ⎞
= ⎜ ln x − x3 ⎟
⎜ 3
9 ⎟⎠
⎝
1
8⎞ ⎛
1⎞
⎛8
= ⎜ ln 2 − ⎟ − ⎜ 0 − ⎟
9⎠ ⎝
9⎠
⎝3
8
7
= ln(2) −
3
9
7⎞
⎛8
The area is ⎜ ln(2) − ⎟ sq units.
3
9⎠
⎝
c.
When q = 20, then p = 300e
dp ⎞
⎜ p + q ⎟ dq
dq ⎠
⎝
q dr
=∫ 0
dq = r ( q0 ) − r (0) = r ( q0 )
0 dq
[since r(0) = 0].
.
dx
x
x
x
∫ f ( x)e dx = f ( x)e − ∫ f ′( x)e dx . Thus
x
x
x
∫ f ( x)e dx + ∫ f ′( x)e dx = f ( x)e + C
20
For I1 , let u = q + 10, dv = e
x
v = e x . Using integration by parts,
= 10 ∫ (q + 10)e −(01.q +1) dq − 300e−3 ∫ dq
0
0
I2
− (0.1q +1)
∫ f ( x )e
Letting u = f(x), dv = e x dx , then du = f ′( x)dx ,
⎡10(q + 10)e −(0.1q +1) − 300e−3 ⎤ dq
⎣
⎦
I1
dr
dp
. Thus
= p+q
dq
dq
q0 ⎛
36.
20
From (b),
∫0
34. p = 10(q + 10)e−(0.1q +1)
−3
dp
∫ p dq = pq − ∫ q dq dq = r − ∫ q dq dq
⎛ x3
x3 1 ⎞
= ⎜ ln x − ∫ ⋅ dx ⎟
⎜ 3
3 x ⎟⎠
⎝
1
3
0
= −400e−3 + 200e−1
)
= −10, 000e + 2000e
CS ≈ $237.89
x3
.
3
CS = ∫
0
0
= 10 −400e −3 + 200e−1 − 300e−3 (20)
Letting u = ln x, dv = x 2 dx, then du =
20
20
20
I 2 = q 0 = 20 − 0 = 20
33. Area = ∫
∫
(q + 20)
1
2 2
x ln x dx.
1
v=
− (0.1q +1)
20
dq . Then
620
ISM: Introductory Mathematical Analysis
Section 15.2
37. f and its inverse f −1 satisfy the equation
r (q) = ∫
−1
f ( f ( x)) = x. Differentiating this equation
using the Chain Rule we get:
f ′( f −1 ( x)) ⋅ ( f −1 )′( x) = 1. Thus
( f −1 )′( x) =
∫
1
f ′( f −1 ( x))
=∫
. Now to evaluate
=
letting u = f −1 ( x) and dv = dx. Then
du =
So
f ′( f −1 ( x))
∫f
−1
( x) dx = xf
To evaluate
( x) − ∫
x
∫ f ′( f −1 ( x)) dx
x
f ′( f −1 ( x))
1
f ′( f −1 ( x))
dx.
we will use the fact
∫ f ′( f −1 ( x))
.
300t 3
t2 + 6
dx = ∫ f ( f −1 ( x)) ⋅ ( f −1 )′( x) dx
=
5(q + 4)
2
=
300t 3 + 1800t − 1800t
(
)
t2 + 6
300t t 2 + 6 − 1800t
2
t +6
= 300t −
1800t
u = t 2 + 6 , so du = 2t dt
Principles in Practice 15.2
Express
300t 3
t2 + 6
1800t
t 2 + 6 is irreducible. To integrate
, let
t2 + 6
( x) dx = xf −1 ( x) − F ( f −1 ( x)) + C.
1. r (q ) = ∫ r ′(q)dq = ∫
5 3(q + 1)3
ln
.
2
q+3
300t 3 by t 2 + 6 to reduce the fraction.
since F ′ = f . Finally,
−1
5 1
5
ln + C so C = ln 3 and
2 3
2
dt
t2 + 6
Since the degree of the numerator is greater than
the degree of the denominator, we first divide
= F ( f −1 ( x))
∫f
5
2
2. V (t ) = ∫ V ′(t )dt = ∫
Hence
x
dx
5 (q + 1)3
+C
ln
2
q+3
r (q) =
that x = f ( f −1 ( x)) and
( f −1 )′( x) =
dq − ∫
Since r(0) = 0, 0 =
dx and v = x.
−1
q + 4q + 3
dq
q +1
q+3
15
5
= ln q + 1 − ln q + 3 + C
2
2
f −1 ( x) dx we will use integration by parts,
1
15
2
5(q + 4)
2
300t 3
5(q + 4)
2
q + 4q + 3
1800t
∫ t 2 + 6 dt = ∫ 300t dt − ∫ t 2 + 6 dt
dq
= 150t 2 − 900 ln t 2 + 6 + C
(
)
V (t ) = 150t 2 − 900 ln t 2 + 6 + C
as a sum of partial
q + 4q + 3
fractions.
5(q + 4)
5(q + 4)
A
B
=
=
+
2
q + 4q + 3 (q + 1)(q + 3) q + 1 q + 3
Problems 15.2
1.
5(q + 4) = A(q + 3) + B(q + 1)
5
.
2
15
When q = –1, we get 5(3) = A(2), so A =
.
2
When q = –3, we get 5(1) = –2B, so B = −
621
10 x
2
=
10 x
A
B
=
+
( x + 6)( x + 1) x + 6 x + 1
x + 7x + 6
10 x = A( x + 1) + B ( x + 6)
If x = –1, then –10 = 5B, or B = –2. If x = –6,
then –60 = –5A, or A = 12.
12
2
−
Answer
x + 6 x +1
Chapter 15: Methods and Applications of Integration
2.
3.
4.
5.
6.
x+5
ISM: Introductory Mathematical Analysis
x+5
A
B
=
+
2
x
−
x
+
x
−
x
+1
(
1)(
1)
1
x −1
x + 5 = A(x + 1) + B(x – 1)
If x = –1, then 4 = –2B, or B = –2. If x = 1, then
6 = 2A, or A = 3.
3
2
−
Answer
x −1 x +1
=
2 x2
2
= 2+
7.
2 x 2 − 15
= 2+
−10 x − 12
8.
=
x+4
2
=
2x + 3
x ( x − 1)
=
x +x
x x +1
2
3x2 + 5
(x
2
+4
)
2
=
Ax + B
2
x +4
(
(
9.
A
B
+
x + 2 ( x + 2) 2
x + 4 x + 4 ( x + 2)
x + 4 = A(x + 2) + B
If x = −2, then 2 = B. If x = 0, then 4 = 2A + B,
2A = 4 − B = 4 − 2 = 2, or A = 1.
1
2
+
Answer:
x + 2 ( x + 2)2
2
A Bx + C
+
x x2 + 1
+
Cx + D
(x
2
+4
)
)
2
3x 2 + 5 = Ax3 + Bx 2 + (4 A + C ) x + (4 B + D)
Thus A = 0, B = 3, 4A + C = 0, 4B + D = 5. This
gives A = 0, B = 3, C = 0, D = –7.
3
7
−
Answer:
2
2
x +4
x2 + 4
−10 x − 15
x+4
2
=
2
3x 2 + 5 = ( Ax + B ) x 2 + 4 + (Cx + D)
x + 5x
−10 x − 15
f ( x) =
x2 + 3
x 2 + 3 = ( A + B ) x 2 + Cx + A
Thus A + B = 1, C = 0, A = 3. This gives A = 3,
B = –2, C = 0.
3
2x
Answer: −
2
x x +1
(by long division).
x2 + 5x
−10 x − 15 A
B
=
= +
2
x
(
x
+
5)
x
x
+5
x + 5x
–10x – 15 = A(x + 5) + Bx. If x = 0, then
–15 = 5A, or A = –3. If x = –5, then 35 = –5B,
or B = –7.
3
7
Answer: 2 − −
x x+5
2
=
( )
+ 3 = A ( x + 1) + ( Bx + C ) x
3
x2
x2 + 5x + 6
−10 x − 12
A
B
=
=
+
2
(
x
+
2)(
x
+
3)
x
+
2
x
+3
x + 5x + 6
−10x − 12 = A(x + 3) + B(x + 2)
If x = −3, then 18 = −B, or B = −18.
If x = −2, then 8 = A.
8
18
−
Answer: 2 +
x+2 x+3
x + 5x + 6
−10 x − 12
x2 + 3
5x − 2
)
5x − 2
A
B
= +
x
x
−
x
x
−1
(
1)
x −x
5x – 2 = A(x – 1) + Bx
If x = 1, then 3 = B. If x = 0, then –2 = –A, or
A = 2.
5x − 2
3 ⎞
⎛2
∫ x2 − x dx = ∫ ⎜⎝ x + x − 1 ⎟⎠ dx
2
=
= 2 ln x + 3ln x − 1 + C = ln x 2 ( x − 1)3 + C
10.
A B
C
+
+
2
x x
x −1
2 x + 3 = Ax( x − 1) + B( x − 1) + Cx 2
If x = 0, then 3 = –B, or B = –3. If x = 1, then
5 = C. If x = –1, then 1 = 2A – 2B + C,
1 = 2A + 6 + 5, or A = –5.
5 3
5
+
Answer: − −
x x2 x − 1
7x + 6
7x + 6
A
B
= +
+
+3
x
(
x
3)
x
x
x + 3x
7x + 6 = A(x + 3) + Bx
If x = −3, then −15 = −3B, or B = 5.
If x = 0, then 6 = 3A, or A = 2.
7x + 6
5 ⎞
⎛2
∫ x2 + 3x dx = ∫ ⎜⎝ x + x + 3 ⎟⎠ dx
= 2 ln x + 5ln x + 3 + C
2
=
= ln x 2 ( x + 3)5 + C
622
ISM: Introductory Mathematical Analysis
11.
x + 10
x + 10
A
B
=
+
(
x
+
1)(
x
−
2)
x
+
1
x
−2
x −x−2
x + 10 = A(x – 2) + B(x + 1)
If x = 2, then 12 = 3B, or B = 4. If x = –1, then 9 = –3A, or A = –3.
x + 10
4 ⎞
⎛ −3
∫ x2 − x − 2 dx = ∫ ⎜⎝ x + 1 + x − 2 ⎟⎠ dx
2
=
= −3ln x + 1 + 4 ln x − 2 + C = ln
12.
Section 15.2
2x −1
2
=
( x − 2) 4
( x + 1)3
+C
2x −1
A
B
=
+
( x − 4)( x + 3) x − 4 x + 3
x − x − 12
2x − 1 = A(x + 3) + B(x − 4)
If x = −3, then −7 = −7B, or B = 1. If x = 4, then 7 = 7A, or A = 1.
2x −1
1 ⎞
⎛ 1
∫ x2 − x − 12 dx = ∫ ⎜⎝ x − 4 + x + 3 ⎟⎠ dx
= ln x − 4 + ln x + 3 + C = ln ( x − 4)( x + 3) + C
13.
3 x3 − 3 x + 4
1 3 x3 − 3 x + 4
⋅
4
x2 − 1
4 x2 − 4
1⎛
4 ⎞
= ⎜ 3x +
⎟
2
4⎝
x −1 ⎠
4
4
A
B
=
=
+
2
x
−
x
+
x
−
x
+1
(
1)(
1)
1
x −1
4 = A(x + 1) + B(x – 1)
If x = –1, then 4 = –2B, or B = –2. If x = 1, then 4 = 2A, or A = 2.
3 x3 − 3 x + 4
1 ⎛
2
−2 ⎞
∫ 4 x 2 − 4 dx = 4 ∫ ⎜⎝ 3x + x − 1 + x + 1 ⎟⎠ dx
=
2
⎤
⎛ 1 ⎞ ⎡ 3x
= ⎜ ⎟⎢
+ 2 ln x − 1 − 2 ln x + 1 ⎥ + C
⎝ 4 ⎠ ⎣⎢ 2
⎦⎥
2
2
x −1 ⎤
⎛ 1 ⎞ ⎡ 3x
⎢
⎥+C
=⎜ ⎟
+ ln
x +1 ⎥
⎝ 4 ⎠ ⎢⎣ 2
⎦
14.
7(4 − x 2 )
7(2 + x)(2 − x)
=
( x − 4)( x − 2)( x + 3) ( x − 4)( x − 2)( x + 3)
−7( x + 2)
A
B
=
=
+
( x − 4)( x + 3) x − 4 x + 3
−7(x + 2) = A(x + 3) + B(x − 4)
If x = −3, then 7 = −7B, or B = −1. If x = 4, then −42 = 7A, or A = −6.
7(4 − x 2 )
−1 ⎞
⎛ −6
∫ ( x − 4)( x − 2)( x + 3) dx = ∫ ⎜⎝ x − 4 + x + 3 ⎟⎠ dx
= −6 ln x − 4 − ln x + 3 + C
= − ln ( x − 4)6 ( x + 3) + C
623
Chapter 15: Methods and Applications of Integration
15.
3x − 4
3
=
2
3x − 4
A
B
C
= +
+
x( x + 1)( x − 2) x x + 1 x − 2
x − x − 2x
3x − 4 = A(x + 1)(x − 2) + Bx(x − 2) + Cx(x + 1)
If x = 0, then −4 = −2A, or A = 2.
7
If x = −1, then −7 = 3B, or B = − .
3
1
If x = 2, then 2 = 6C, or C = .
3
7
1 ⎞
⎛
3x − 4
2 −3
3
⎜
dx
=
+
+
∫ x3 − x 2 − 2 x
∫ ⎜ x x + 1 x − 2 ⎟⎟ dx
⎝
⎠
7
1
= 2 ln x − ln x + 1 + ln x − 2 + C
3
3
= ln
16.
ISM: Introductory Mathematical Analysis
4− x
4
x −x
2
=
x2 3 x − 2
3
( x + 1)7
4− x
2
x ( x + 1)( x − 1)
=
+C
A B
C
D
+
+
+
x x2 x + 1 x − 1
4 − x = Ax( x + 1)( x − 1) + B( x + 1)( x − 1) + Cx 2 ( x − 1) + Dx 2 ( x + 1)
5
3
If x = 0, then 4 = −B, or B = −4. If x = −1, then 5 = −2C, or C = − . If x = 1, then 3 = 2D, or D = . If x = 2,
2
2
then 2 = 6A + 3B + 4C + 12D, 2 = 6A − 12 − 10 + 18, or 2 = 6A − 4, so A = 1.
3 ⎞
⎛1 4
− 52
4− x
2
⎜
dx
=
−
+
+
∫ x4 − x2
∫ ⎜ x x2 x + 1 x − 1 ⎟⎟ dx
⎝
⎠
4 5
3
= ln x + − ln x + 1 + ln x − 1 + C
x 2
2
=
17.
4 1 x 2 ( x − 1)3
+ ln
+C
x 2
( x + 1)5
2(3x5 + 4 x3 − x)
∫ x6 + 2 x4 − x2 − 2dx = ∫ x6 + 2 x 4 − x 2 − 2 ⎡⎣⎢( 6 x
1
5
)
+ 8 x3 − 2 x dx ⎤
⎦⎥
⎛
⎛1⎞ ⎞
⎜ Form: ∫ ⎜ ⎟ du ⎟ (Partial fractions not required.)
⎝u⎠ ⎠
⎝
Answer: ln x6 + 2 x 4 − x 2 − 2 + C
18.
x 4 − 2 x3 + 6 x 2 − 11x + 2
= x +1+
7 x 2 − 13x + 2
x3 − 3 x 2 + 2 x
x3 − 3 x 2 + 2 x
7 x 2 − 13 x + 2 7 x 2 − 13 x + 2 A
B
C
=
= +
+
3
2
x − 3 x + 2 x x( x − 1)( x − 2) x x − 1 x − 2
7 x 2 − 13x + 2 = A( x − 1)( x − 2) + Bx( x − 2) + Cx( x − 1)
If x = 0, then 2 = 2A, or A = 1. If x = 1, then −4 = −B, or B = 4. If x = 2, then 4 = 2C, or C = 2.
624
ISM: Introductory Mathematical Analysis
∫
x 4 − 2 x3 + 6 x 2 − 11x + 2
3
2
x − 3x + 2 x
Section 15.2
1
4
2 ⎞
⎛
= ∫ ⎜ x +1+ +
+
⎟ dx
x x −1 x − 2 ⎠
⎝
x2
+ x + ln x + 4 ln x − 1 + 2 ln x − 2 + C
2
x2
=
+ x + ln x( x − 1)4 ( x − 2)2 + C
2
=
19.
2 x2 − 5x − 2
A
B
C
+
+
x − 1 x − 2 ( x − 2) 2
=
2
( x − 2) ( x − 1)
2 x 2 − 5 x − 2 = A( x − 2) 2 + B ( x − 1)( x − 2) + C ( x − 1)
If x = 1, then –5 = A. If x = 2, then –4 = C.
If x = 0, then –2 = 4A + 2B – C, –2 = –20 + 2B + 4, or B = 7.
⎡ −5
−4 ⎤
2x2 − 5x − 2
7
∫ ( x − 2)2 ( x − 1) dx = ∫ ⎢⎢ x − 1 + x − 2 + ( x − 2)2 ⎥⎥ dx
⎣
⎦
= −5ln x − 1 + 7 ln x − 2 +
20.
−3x3 + 2 x − 3
2
2
x ( x − 1)
=
4
4
( x − 2)7
+C =
+ ln
+C
x−2
x−2
( x − 1)5
−3 x 3 + 2 x − 3
2
x ( x + 1)( x − 1)
=
A B
C
D
+
+
+
2
x x
x +1 x −1
3
−3x + 2 x − 3 = Ax( x + 1)( x − 1) + B( x + 1)( x − 1) + Cx 2 ( x − 1) + Dx 2 ( x + 1)
If x = 0, then −3 = −B, or B = 3. If x = −1, then −2 = −2C, or C = 1. If x = 1, then −4 = 2D, or D = −2. If x = 2, then
−23 = 6A + 3B + 4C + 12D, −23 = 6A + 9 + 4 − 24, or A = −2.
⎛ −2 3
−3x3 + 2 x − 3
−2 ⎞
1
∫ x 2 ( x 2 − 1) dx = ∫ ⎝⎜ x + x 2 + x + 1 + x − 1 ⎠⎟ dx
= −2 ln x −
21.
2( x 2 + 8)
3
x + 4x
=
3
3
x +1
+ ln x + 1 − 2 ln x − 1 + C = − + ln
+C
2
x
x
x ( x − 1)2
2 x 2 + 16
(
2
x x +4
(
)
)
=
A Bx + C
+
x x2 + 4
2 x 2 + 16 = A x 2 + 4 + ( Bx + C ) x
2 x 2 + 16 = ( A + B) x 2 + Cx + 4 A
Thus A + B = 2, C = 0, 4A = 16. This gives A = 4, B = –2, C = 0.
∫
22.
⎡ x4 ⎤
⎛4
−2 x ⎞
1
1
2
=
−
=
4
ln
x
−
ln
x
+
4
+
C
=
ln
4
dx
[2
x
dx
]
dx = ∫ ⎜ +
dx
⎢ 2
⎥+C
⎟
∫ x ∫ x2 + 4
x3 + 4 x
⎝ x x2 + 4 ⎠
⎣⎢ x + 4 ⎦⎥
2( x 2 + 8)
4 x3 − 3 x 2 + 2 x − 3
2
( x + 3)( x + 1)( x − 2)
(
=
Ax + B
2
x +3
+
C
D
+
x +1 x − 2
4 x3 − 3 x 2 + 2 x − 3 = ( Ax + B)( x + 1)( x − 2) + C ( x 2 + 3)( x − 2) + D( x 2 + 3)( x + 1)
If x = −1, then −12 = −12C, or C = 1.
If x = 2, then 21 = 21D, or D = 1.
If x = 0, then −3 = −2B − 6C + 3D, −3 = −2B − 6 + 3, 0 = −2B, or B = 0.
625
)
Chapter 15: Methods and Applications of Integration
ISM: Introductory Mathematical Analysis
If x = 1, then 0 = −2(A + B) − 4C + 8D, 0 = −2A − 4 + 8, −4 = −2A, or A = 2.
⎛ 2x
4 x3 − 3 x 2 + 2 x − 3
1
1 ⎞
∫ ( x 2 + 3)( x + 1)( x − 2) dx = ∫ ⎝⎜ x2 + 3 + x + 1 + x − 2 ⎠⎟ dx
= ln( x 2 + 3) + ln x + 1 + ln x − 2 + C
= ln ( x 2 + 3)( x + 1)( x − 2) + C
23.
− x3 + 8 x 2 − 9 x + 2
( x + 1) ( x − 3)
2
2
Ax + B
=
2
x +1
+
C
D
+
x − 3 ( x − 3) 2
(
= ( Ax + B) ( x − 6 x + 9 ) +C ( x
) (
)
− x3 + 8 x 2 − 9 x + 2 = ( Ax + B )( x − 3)2 + C ( x − 3) x 2 + 1 + D x 2 + 1
2
3
3
) (
)
− 3x 2 + x − 3 + D x 2 + 1
2
= ( A + C ) x + ( B − 6 A − 3C + D) x +(9 A − 6 B + C ) x + (9 B − 3C + D)
Thus A + C = –1, B – 6A – 3C + D = 8, 9A – 6B + C = –9, 9B – 3C + D = 2. This gives A = –1, B = 0, C = 0,
D = 2.
⎛ −x
− x3 + 8 x 2 − 9 x + 2
1
2
0
2 ⎞
2
∫ x2 + 1 ( x − 3)2 dx = ∫ ⎜⎜ x2 + 1 + x − 3 + ( x − 3)2 ⎟⎟ dx = − 2 ln x + 1 − x − 3 + C
⎝
⎠
(
24.
(
)
5x4 + 9 x2 + 3
2
x( x + 1)
=
2
)
A Bx + C Dx + E
+
+
x x 2 + 1 ( x 2 + 1)2
5 x 4 + 9 x 2 + 3 = A( x 2 + 1) 2 + ( Bx + C ) x( x 2 + 1) + ( Dx + E ) x
= A( x 4 + 2 x 2 + 1) + ( Bx + C )( x3 + x) + Dx 2 + Ex
= ( A + B) x 4 + Cx3 + (2 A + B + D) x 2 + (C + E ) x + A
Thus, A + B = 5, C = 0, 2A + B + D = 9, C + E = 0, and A = 3. This gives A = 3, B = 2, C = 0, D = 1, and E = 0.
⎛3
⎞
x
5x4 + 9 x2 + 3
2x
∫ x( x2 + 1)2 dx = ∫ ⎜⎜ x + x2 + 1 + ( x2 + 1)2 ⎟⎟ dx
⎝
⎠
1
2
= 3ln x + ln x + 1 −
+C
2( x 2 + 1)
1
= ln x3 ( x 2 + 1) −
+C
2
2( x + 1)
25.
14 x3 + 24 x
( x + 1)( x
2
2
+2
(
)
=
Ax + B
2
x +1
)
+
Cx + D
x2 + 2
(
)
14 x3 + 24 x = x 2 + 2 ( Ax + B) + x 2 + 1 (Cx + D) = ( A + C ) x3 + ( B + D) x 2 + (2 A + C ) x + (2 B + D)
Thus A + C = 14, B + D = 0, 2A + C = 24, 2B + D = 0.
This gives A = 10, B = 0, C = 4, D = 0.
626
ISM: Introductory Mathematical Analysis
∫
(
14 x3 + 24 x
Section 15.2
⎛ 10 x
4x ⎞
dx = ∫ ⎜
+
⎟ dx
2
2
⎝ x +1 x + 2 ⎠
x +1 x + 2
2
)(
2
)
= 5∫
1
2
x +1
[2 dx] + 2 ∫
1
[2 dx]
2
x +2
( ) ( )
⎡
⎤
= ln ⎢( x + 1) ( x + 2 ) ⎥ + C
⎣
⎦
= 5ln x 2 + 1 + 2 ln x 2 + 2 + C
5
2
26.
2
2
12 x3 + 20 x 2 + 28 x + 4
1 ⎛ Ax + B
Cx + D ⎞
= ⎜
+
⎟
2
3( x + 2 x + 3)( x + 1) 3 ⎝ x + 2 x + 3 x 2 + 1 ⎠
2
2
12 x3 + 20 x 2 + 28 x + 4 = ( Ax + B )( x 2 + 1) + ( x 2 + 2 x + 3)(Cx + D)
= ( A + C ) x3 + ( B + D + 2C ) x 2 + ( A + 2 D + 3C ) x + ( B + 3D )
Thus, A + C = 12, B + D + 2C = 20, A + 2D + 3C = 28, B + 3D = 4. This gives A = 4, B = 4, C = 8, D = 0.
12 x3 + 20 x 2 + 28 x + 4
1 ⎛ 4x + 4
8x ⎞
∫ 3 x2 + 2 x + 3 x2 + 1 dx = 3 ∫ ⎜⎝ x2 + 2 x + 3 + x 2 + 1 ⎟⎠ dx
(
)(
)
=
(
⎡
= ln ⎢ x 2 + 2 x + 3
⎢⎣
(
27.
3 x3 + 8 x
2
( x + 2)
2
=
Ax + B
2
x +2
+
)
(
)
1⎡
2 ln x 2 + 2 x + 3 + 4 ln x 2 + 1 ⎤ + C
⎦⎥
3 ⎣⎢
)(
2
3
4
3 ⎤
x2 + 1 ⎥ + C
⎥⎦
)
Cx + D
( x 2 + 2) 2
3x3 + 8 x = ( Ax + B)( x 2 + 2) + Cx + D
= Ax3 + Bx 2 + (2 A + C ) x + (2 B + D)
Thus, A = 3, B = 0, 2A + C = 8, 2B + D = 0.
This gives A = 3, B = 0, C = 2, D = 0.
⎛ 3x
⎞
3 x3 + 8 x
2x
3
1
2
∫ ( x 2 + 2)2 dx = ∫ ⎜⎜ x2 + 2 + ( x2 + 2)2 ⎟⎟ dx = 2 ln( x + 2) − x2 + 2 + C
⎝
⎠
28.
3x2 − 8 x + 4
∫ x3 − 4 x2 + 4 x − 6 dx = ∫ x3 − 4 x 2 + 4 x − 6 ⎡⎣⎢( 3x
1
2
)
− 8 x + 4 dx ⎤
⎦⎥
⎛
⎛1⎞ ⎞
⎜ Form: ∫ ⎜ ⎟ du ⎟ (Partial fractions not required.)
⎝u⎠ ⎠
⎝
Answer: ln x3 − 4 x 2 + 4 x − 6 + C
627
Chapter 15: Methods and Applications of Integration
29.
2 − 2x
ISM: Introductory Mathematical Analysis
If x = –2, then 30 = B. If x = 0, then –18 = 2A + B,
–18 = 2A + 30, or A = –24.
2 − 2x
A
B
=
+
2
(
x
+
3)(
x
+
4)
x
+
3
x
+4
x + 7 x + 12
2 – 2x = A(x + 4) + B(x + 3)
If x = –4, then 10 = –B, or B = –10. If x = –3,
then 8 = A.
1
1⎛ 8
2 − 2x
−10 ⎞
∫0 x2 + 7 x + 12 dx = ∫0 ⎜⎝ x + 3 + x + 4 ⎟⎠ dx
=
= ⎡⎣8ln x + 3 − 10 ln x + 4 ⎤⎦
16
∫0
1
30.
x2 + 4 x + 3
= 3+
= 6 − 24 ln 3 − 10 − (−24 ln 2 − 15)
= 11 + 24 ln
3x + 4
x2 + 4 x + 3
3x + 4
= 3+
( x + 1)( x + 3)
200(q + 3)
2
10
3250 ⎤
⎡
= ⎢120 ln(16) + 80 ln(11) −
− [120 ln(6)]
22 ⎥⎦
⎣
8
1625
= 120 ln + 80 ln(11) −
≈ $161.80
3
11
) ≥ 0 on [0, 1].
+ 1)
( x + 2)2
0
6( x 2 + 1)
Problems 15.3
6 x2 + 1
1 6( x 2
= 6+
1. Let u = x, a 2 = 9 . Then du = dx.
dx
x
+C
∫ (9 − x2 )3 / 2 =
9 9 − x2
dx
−24 x − 18
(by long division)
( x + 2) 2
A
B
=
+
2
x + 2 ( x + 2)2
( x + 2)
–24x – 18 = A(x + 2) + B
2
200(q + 3)
A
B
=
+
(q + 6)(q + 1) q + 6 q + 1
325 ⎤
⎡
= ⎢120 ln q + 6 − 80 ln q + 1 −
q
22 ⎥⎦ 0
⎣
1
5
1
5
⎛
⎞
= 6 + ln 3 + ln 5 − ⎜ 3 + ln 2 + ln 4 ⎟
2
2
2
2
⎝
⎠
1
5
1
5
= 3 + ln 3 + ln 5 − ln 2 − ln 4
2
2
2
2
Area = ∫
=
q + 7q + 6
200(q + 3) = A(q + 1) + B(q + 6)
If q = –1, then 400 = 5B, or B = 80. If q = –6,
then –600 = –5A, or A = 120.
10 ⎡ 120
80 325 ⎤
CS = ∫ ⎢
+
−
⎥ dq
0 q + 6 q +1
22 ⎦
⎣
2
( x + 2)
2
sq units.
3
10 ⎡ 200( q + 3)
325 ⎤
−
32. CS = ∫ ⎢
⎥ dq
0 q 2 + 7q + 6
22 ⎦⎥
⎣⎢
1
5
⎛
⎞
= ⎜ 3x + ln x + 1 + ln x + 3 ⎟
2
2
⎝
⎠1
2
2
3
The area is 11 + 24 ln
1
If x = −1, then 1 = 2A, or A = . If x = −3, then
2
5
−5 = −2B, or B = .
2
2
2 3 x + 15 x + 13
∫1 x2 + 4 x + 3 dx
2⎛
1 1
5 1 ⎞
= ∫ ⎜3+ ⋅
+ ⋅
dx
1 ⎝
2 x + 1 2 x + 3 ⎟⎠
(
( x + 2)2
1⎡
−24
30 ⎤
= ∫ ⎢6 +
+
⎥ dx
0
x + 2 ( x + 2) 2 ⎥⎦
⎢⎣
1
3x + 4
A
B
=
+
( x + 1)( x + 3) x + 1 x + 3
3x + 4 = A(x + 3) + B(x + 1)
31. Note that
2
30 ⎤
⎡
= ⎢6 x − 24 ln x + 2 −
x
+ 2 ⎥⎦ 0
⎣
0
= 8 ln 4 − 10 ln 5 − (8 ln 3 − 10 ln 4)
= 18 ln(4) – 10 ln(5) – 8 ln(3)
3x 2 + 15 x + 13
( x + 1) dx
( x + 2)
−24 x − 18
628
ISM: Introductory Mathematical Analysis
Section 15.3
2. Let u = 2x, a 2 = 25 . Then du = 2dx.
dx
1
(2dx)
= ∫
∫
3
3
2
2
⎡ 25 − (2 x) 2 ⎤ 2
25 − 4 x 2
⎣
⎦
⎡
⎤
1
(2 x)
⎥+C
= ⎢
2 ⎢ 25 25 − (2 x) 2 ⎥
⎣
⎦
(
=
8. Formula 32 with u = x, a 2 = 7 . Then du = dx.
dx
x
∫ ( x2 + 7)3 / 2 = 2 + C
7 x +7
)
x
25 25 − 4 x 2
9. Formula 12 with u = x, a = 2, b = 3, c = 4, k = 5.
Then du = dx.
x dx
∫ (2 + 3x)(4 + 5 x)
+C
=
3. Let u = 4x, a 2 = 3 . Then du = 4 dx.
dx
(4 dx)
= 4∫
∫ 2
2
2
x 16 x + 3
(4 x) (4 x)2 + 3
10. Formula 37 with u = 5x, a = 2. Then du = 5 dx.
1 5x
1 25 x
5x
∫ 2 dx = 5 ∫ 2 (5 dx) = 5 ⋅ ln 2 + C
⎡ (4 x)2 + 3 ⎤
⎥+C
= 4 ⎢−
⎢
3(4 x) ⎥
⎣
⎦
11. Formula 45 with u = x, a = 5, b = 2, c = 3. Then
dx
1
3 x − ln 5 x + 2e3 x + C
du = dx. ∫
=
3x
15
5 + 2e
12. Formula 14 with u = x, a = 1, b = 1. Then
du = dx.
4. Let u = x 2 , a 2 = 9. Then du = 2x dx.
3 dx
3
(2 x dx)
∫ 3 4 = 2∫ 2 2 2 2
x x −9
(x ) (x ) − 9
=
∫x
2 2
⎤
3 ⎡⎢ − ( x ) − 9
−
+ C⎥
⎥
2⎢
9 x2
⎣
⎦
x4 − 9
6 x2
+C
3
105
+C
du = 11 dx .
∫
dx
x 5 − 11x 2
=−
7. Formula 28 with u = x, a = 3. Then du = dx.
∫
1 + x dx =
)
14. Formula 20 with u = 11x , a = 5 . Then
6. Formula 8 with u = x, a = 2, b = 5. Then du = dx.
⎡
x 2 dx ⎤
3 x 2 dx
∫ (2 + 5 x)2 = 3 ⎢⎢ ∫ (2 + 5 x)2 ⎥⎥
⎣
⎦
⎡ x
⎤
4
4
ln 2 + 5 x ⎥ + C
= 3⎢ −
−
25
125(2
5
)
125
x
+
⎣
⎦
1
= ln
2
x x +9 3
2
(
2 8 − 12 x + 15 x 2 (1 + x) 2
13. Formula 9 with u = x, a = 5, b = 2. Then du = dx.
⎡
⎤
dx
7 dx
∫ x(5 + 2 x)2 = 7 ⎢⎢ ∫ x(5 + 2 x)2 ⎥⎥
⎣
⎦
⎡
1
1
x ⎤
= 7⎢
+ ln
⎥+C
⎣ 5(5 + 2 x) 25 5 + 2 x ⎦
5. Formula 5 with u = x, a = 6, b = 7. Then du = dx.
dx
1
x
∫ x(6 + 7 x) = 6 ln 6 + 7 x + C
dx
)
(
16 x 2 + 3
=−
+C
3x
=
1 ⎡4
2
⎤
ln 4 + 5 x − ln 2 + 3x ⎥ + C
2 ⎢⎣ 5
3
⎦
x2 + 9 − 3
+C
x
629
1
5
ln
=∫
11dx
(
11x
) ( 5) −(
5 + 5 − 11x 2
11x
2
+C
11x
)
2
Chapter 15: Methods and Applications of Integration
ISM: Introductory Mathematical Analysis
15. Formula 3 with u = x, a = 2, b = 1. Then du = dx.
1
1 x dx
⎛4⎞
∫0 2 + x = ( x − 2 ln 2 + x ) 0 = 1 − 2 ln 3 + 2 ln 2 = 1 – ln 9 + ln 4 = 1 + ln ⎜⎝ 9 ⎟⎠
16. Formula 4 with u = x, a = 3, b = 7. Then du = dx.
⎛ x2 3x
⎞
2 x 2 dx
x 2 dx
9
=
=
2
2
∫ 3 + 7 x ∫ 3 + 7 x ⎜⎜ 14 − 49 + 343 ln 3 + 7 x ⎟⎟ + C
⎝
⎠
17. Formula 23 with u = x, a 2 = 3 . Then du = dx.
1⎛
⎞
2
2
2
∫ x − 3 dx = 2 ⎜⎝ x x − 3 − 3ln x + x − 3 ⎟⎠ + C
18. Formula 11 with u = x, a = 1, b = 5, c = 3, k = 2. Then du = dx.
dx
1
1 + 5x
∫ (1 + 5 x)(2 x + 3) = 13 ln 2 x + 3 + C
19. Formula 38 with u = x, a = 12. Then du = dx.
1/12
∫0
12 x
xe
e12 x
dx =
(12 x − 1)
144
1/12
=
0
1
1
[e(0) − 1(−1)] =
144
144
20. Formula 46 with u = 3x, a = 2, b = 5.
Then du = 3 dx.
1
2 + 3x
1
2 + 3x
∫ 5 + 3x dx = 3 ∫ 5 + 3x (3 dx) = 3 ⎡⎣ (2 + 3x)(5 + 3x) − 3ln
(
)
2 + 3x + 5 + 3x ⎤ + C
⎦
21. Formula 39 with u = x, n = 2, a = 1. Then du = dx.
∫x
2 x
e dx = x 2 e x − 2 ∫ xe x dx
Applying Formula 38 on
∫x
∫ xe
x
dx with u = x, a = 1 (so du = dx) gives
(
∫ xe
x
dx = e x ( x − 1) + C1 . Thus
)
e dx = x 2 e x − 2 ⎡ e x ( x − 1) ⎤ + C = e x ⎡ x 2 − 2( x − 1) ⎤ + C = e x x 2 − 2 x + 2 + C
⎣
⎦
⎣
⎦
2 x
22. Formula 6 with u = x, a = 1, b = 1. Then du = dx.
2
∫1
4 dx
x 2 (1 + x )
= 4∫
2
1
2
⎛ 1
1+ x ⎞
3⎞
3
⎛ 1
= 4 ⎜ − + ln
⎟ = 4 ⎜ − + ln ⎟ − 4(−1 + ln 2) = 2 + 4 ln
2
2
2
4
x
x
⎝
⎠
x (1 + x)
⎝
⎠1
dx
23. Formula 26 with u = 5 x, a 2 = 1. Then du = 5 dx.
∫
5x2 + 1
2x2
5
5x2 + 1
(
)
5 dx
∫ 5x2
2 5
⎞
5 ⎛⎜
5x2 + 1
=
−
+ ln 5 x + 5 x 2 + 1 ⎟ + C
⎟
2 ⎜
5x
⎝
⎠
dx =
630
ISM: Introductory Mathematical Analysis
Section 15.3
24. Formula 17 with u = x, a = 2, b = –1. Then du = dx.
∫x
dx
2− x
=
1
2
2− x − 2
ln
2− x + 2
+C
25. Formula 7 with u = x, a = 1, b = 3. Then du = dx.
x dx
1⎛
1 ⎞
∫ (1 + 3x)2 = 9 ⎜⎝ ln 1 + 3x + 1 + 3x ⎟⎠ + C
26. Formula 47 with u = 3x, a = 5, b = 6. Then du = 3 dx.
3 dx
11
∫ (5 + 3x)(6 + 3x) = ln 2 + 3x + (5 + 3x)(6 + 3x) + C
27. Formula 34 with u = 5 x, a = 7 . Then du = 5dx
dx
∫ 7 − 5x2
=
1
5
∫
1
( 7 ) − ( 5x)
2
2(
)
5dx =
1 ⎛ 1
⎜
ln
5 ⎜⎝ 2 7
7 + 5x ⎞
⎟+C
7 − 5 x ⎟⎠
28. Formula 24 with u = 3 x, a 2 = 6 . Then du = 3dx .
∫ 7x
2
3x 2 − 6 dx =
=
7
( 3)
3
∫(
3x
) ( 3x )
2
2
−6
(
3 dx
)
⎤
7 ⎡ 3x
36
(6 x 2 − 6) 3 x 2 − 6 − ln 3 x + 3 x 2 − 6 ⎥ + C
⎢
8
3 3⎣ 8
⎦
29. Formula 42 with u = 3x, n = 5. Then du = 3 dx.
36
5
5
5
∫ 36 x ln(3x)dx = 36∫ x ln(3x)dx = 36 ∫ (3x) ln(3x)(3 dx)
4 ⎡ (3x)6 ln(3 x) (3 x)6 ⎤
6
= ⎢
−
⎥ + C = x [6 ln(3x) − 1] + C
81 ⎣⎢
6
36 ⎦⎥
30. Formula 10 with u = x, a = 3, b = 2. Then du = dx.
⎡
⎤
⎡
5 dx
dx
3 + 4x
4
3 + 2x
∫ x2 (3 + 2 x)2 = 5 ⎢⎢ ∫ x2 (3 + 2 x)2 ⎥⎥ = 5 ⎢⎣− 9 x(3 + 2 x) + 27 ln x
⎣
⎦
⎤
⎥+C
⎦
31. Formula 13 with u = x, a = 1, b = 3. Then
du = dx.
3
⎡
2
2(9
x
−
2)(1
+
3
x
)
∫ 270 x 1 + 3xdx = 270∫ x 1 + 3x dx = 270 ⎢⎢
15 ⋅ 9
⎣
⎤
⎥+C
⎥
⎦
3
= 4(9 x − 2)(1 + 3 x) 2 + C
32. Formula 42 with u = x, n = 2. Then du = dx.
⎛ x3 ln x x3 ⎞
2
2
9
x
ln
x
dx
=
9
x
ln
x
dx
=
9
− ⎟ + C = 3x3 (ln x) − x3 + C
⎜
∫
∫
⎜ 3
9 ⎟⎠
⎝
631
Chapter 15: Methods and Applications of Integration
ISM: Introductory Mathematical Analysis
33. Formula 27 with u = 2x, a 2 = 13 . Then
du = 2 dx.
dx
1
1
= ∫
(2 dx)
∫ 2
4 x − 13 2
(2 x)2 − 13
=
38. Formula 2 with u = x3 , a = 1, b = 2. Then
du = 3 x 2 dx.
1 3x2
1
3
∫0 1 + 2 x3 2 ln 1 + 2 x 0
1
1
= ln 3 − ln 1 = ln 3
2
2
1
ln 2 x + 4 x 2 − 13 + C
2
34. Formula 44 with u = 2x. Then du = 2 dx.
dx
(2 dx)
∫ x ln(2 x) = ∫ (2 x) ln(2 x)
x dx
=
∫
(
1
2 x
dx
16 − 9 x 2
+C
8x
( 2 ) − ( 3x )
∫ 6x
x π + 7e
4 x
2 + 2 − 3x2
3x
)
π + 7e
dx = 3 ⋅
u
du.
2 x5 / 2
e
5∫
6 x5 / 2
e
+C
5
2 x 2 + 1dx =
2
(
3dx
)
2
3
2
(
)
= 2x2 + 1
+C
⎡ 5 3/ 2 ⎤
⎢ 2 x dx ⎥
⎣
⎦
⎛ 1
⎞
dx ⎟
⎜
⎝2 x ⎠
3
2
(
)
3
2x2 + 1
2∫
1
2
(4 x dx)
3
2
+C
+C
5 x3 − x
1 −1 ⎞
⎛5
dx = ∫ ⎜ x 2 − x 2 ⎟ dx
2x
2
⎝2
⎠
5 3
= x − x +C
6
42.
∫
43.
∫ x2 − 5 x + 6 dx = ∫ ( x − 3)( x − 2) dx
dx
4 x
5/ 2
3 ( 2 x + 1)
= ⋅
2
3x
1
xe x
∫e
41. Can be put in the form ∫ u n du .
2
= 2∫
)
=
37. Formula 45 with u = x , a = π, b = 7, c = 4 .
Then du =
(
1
ln x 2 + 1 + C
2
∫ 3x
du = 3dx .
= 2 − 3 x 2 − 2 ln
1
40. Can be put in the form
36. Formula 22 with u = 3 x, a = 2 . Then
2 − 3x 2
dx = ∫
x
1
1
∫ u du .
∫ x2 + 1 = 2 ∫ x2 + 1 (2 x dx)
35. Formula 21 with u = 3x, a 2 = 16. Then
du = 3 dx.
2 dx
(3 dx)
= 2(3) ∫
∫ 2
x 16 − 9 x 2
(3 x)2 16 − (3 x)2
⎛ 16 − 9 x 2 ⎞
⎟+C
= 6⎜ −
⎜
16(3x) ⎟
⎝
⎠
∫
=
39. Can be put in the form
= ln ln(2 x) + C
=−
1
dx
1
1
Formula 11 with u = x, a = −3, b = 1, c = −2, and
k = 1. Then du = dx.
1
1
∫ x2 − 5 x + 6 dx = ∫ ( x − 3)( x − 2) dx
x−3
= ln
+C
x−2
⎡ 1 ⎛
⎞⎤
= 2 ⎢ ⎜ 4 x − ln π + 7e4 x ⎟ ⎥ + C
π
4
⎝
⎠
⎣
⎦
1 ⎛
⎞
4 x
=
⎜ 4 x − ln π + 7e
⎟+C
2π ⎝
⎠
632
ISM: Introductory Mathematical Analysis
Section 15.3
50. Formula 41 with u = x 2 . Then du = 2x dx.
44. Can be put in the form ∫ u n du .
e2 x
(
3
∫1 3x ln x
)
− 12
1
dx = ∫ e2 x + 3
(2e2 x dx)
2x
2
e +3
∫
2
dx
e
45. Formula 42 with u = x and n = 3. Then du = dx.
x4 ⎡
1⎤
3
∫ x ln x dx = 4 ⎢⎣ln( x) − 4 ⎥⎦ + C
3 e
3
ln( x 2 )[2 x dx] = [ x 2 ln( x 2 ) − x 2 ]
∫
1
2
2
1
3 2
2
2
= [(e ln(e ) − e ) − (1 ⋅ ln1 − 1)]
2
3
= (e2 + 1)
2
46. Formula 38 with u = x and a = −1. Then du = dx.
51. Formula 15 with u = x, a = 4 and b = –1. Then
du = dx.
=
= e2 x + 3 + C
3
∫0
3
xe− x dx = e− x (− x − 1) = e−3 (−4) − 1(−1)
∫1
0
= 1 − 4e
−3
=
2
47. Formula 38 with u = x and a = 3. Then
du = 2x dx.
2
x dx
2
=
4− x
2(− x − 8) 4 − x
3
(
2
9 3 − 10 2
3
⎡ e3 x 2
⎤
= 2⎢
(3 x 2 − 1) ⎥ + C
⎢ 9
⎥
⎣
⎦
2
2 3x
2
= e (3x − 1) + C
9
3
= 35 ⋅
2
3 + 2 xdx = 35
(
2 72 − 72 x + 60 x
2
2 2
∫1 x
∫2
) (3 + 2 x)
(
1
2
∫0
840
2 x dx
1
8 − x2
=−
49. Formula 43 and then Formula 41. For Formula
43, let u = x, n = 0, and m = 2. Then du = dx.
(8 − x )
(
Now we apply Formula 41 to the last integral
with u = x (so du = dx).
0
= −2
(
633
− 12
(−2 x dx)
1
1
2
(
)
0
1 1
2
= −2
0
7 −2 2
=2 2 2− 7
x dx = x(ln x)2 − 2 x(ln x) + 2 x + C
)
1
2
= −2 8 − x 2
2
2
∫ ln x dx = x ln x − 2∫ ln x dx
(
= −∫ 8 − x2
2
= 98 7 − 25 5
2
2
)
53. Can be put in the form ∫ u n du .
1
∫ ln
3
2
=
[23(11)3 / 2 − 14(8)3 / 2 ]
135
2
=
253 11 − 224 2
135
3 + 2 xdx
3
2
)
2(9 x − 4)(2 + 3 x)3 / 2
135
x 2 + 3 x dx =
48. Formula 14 with u = x, a = 3 and b = 2. Then
du = dx.
2
1
52. Formula 13 with u = x, a = 2, and b = 3. Then
du = dx.
2
3 3x
2 3x
∫ 4 x e dx = 2∫ x e [2 x dx]
∫1 35 x
2
)
)
(
7− 8
)
Chapter 15: Methods and Applications of Integration
ISM: Introductory Mathematical Analysis
58. Formula 6 with u = q, a = 1 and b = –1. Then
du = dq.
1 0.1 dq
n=−
0.4 ∫0.3 q 2 (1 − q )
54. Formula 39 with u = x, n = 2, a = 3. Then
du = dx.
x 2 e3 x 2
2 3x
3x
∫ x e dx = 3 − 3 ∫ xe dx
For
3x
∫ xe
0.1
dx, use Formula 38 with u = x and
a = 3. Then du = dx.
⎤
x 2 e3 x 2 ⎡ e3 x
− ⎢
(3x − 1) ⎥
3
3 ⎢⎣ 9
⎦⎥
3x
e
[9 x 2 − 6 x + 2]
=
27
2 3x
∫ x e dx =
=−
1 ⎡ 1
1− q ⎤
⎢ − − ln
⎥
0.4 ⎣⎢ q
q ⎥⎦
0.3
=−
1 ⎧
7 ⎤⎫
⎡ 10
⎨[−10 − ln 9] − ⎢ − − ln ⎥ ⎬
0.4 ⎩
3
3 ⎦⎭
⎣
=−
1 ⎛ 20
7⎞
−
− ln 9 + ln ⎟ ≈ 20
0.4 ⎜⎝ 3
3⎠
ln 2
ln 2 2 3 x
x e
0
∫
⎛ e3 x
⎞
[9 x 2 − 6 x + 2] ⎟
dx = ⎜
⎜ 27
⎟
⎝
⎠0
8
1
[9(ln 2) 2 − 6 ln 2 + 2] − [2]
=
27
27
2
2
[36(ln 2) − 24 ln 2 + 7]
=
27
59. a.
9
∫0 1000e
∫1
2
2
1
b. For
10
∫0
qn
0
= ln
du can
dt
10
∫0
500te−0.06t dt use Formula 38 with
500te−0.06t dt
10
1
qn
= ln
10
⎡ e−0.06t
⎤
(−0.06t − 1) ⎥
= 500 ⎢
⎣⎢ 0.0036
⎦⎥ 0
500
[e−0.6 (−1.6) − (−1)]
=
0.0036
≈ $16,930.75
∫ k dx .
q0
−0.04t
0
2
dq
q
= ln
1− q
q (1 − q)
u
= 500 ∫ te−0.06t dt
60.
57. Formula 5 with u = q, a = 1, and b = –1. Then
du = dq.
∫q
∫e
t = u and a = −0.06, so du = dt.
dx = ∫ 1 dx = x = 2 − 1 = 1
1
dt , the form
1000 9 −0.04t
e
(−0.04 dt )
−0.04 ∫0
9
1000 −0.04t
e
=−
0.04
0
1000 −0.36
(e
=−
− 1)
0.04
≈ $7558.09
( )
2
−0.04t
=
1
1
= 2 ln(4) − 1 − ln(2) +
2
4
1
3
= 2 ln 22 − ln(2) −
2
4
1
3
= 4 ln(2) − ln(2) −
2
4
7
3
= (ln 2) −
2
4
56. Can be put in the form
9
∫0 1000e
be applied.
55. Integration by parts or Formula 42. For Formula
42, let u = 2x, n = 1. Then du = 2 dx.
2
1 2
∫1 x ln(2 x)dx = 4 ∫1 (2 x) ln(2 x)[2 dx]
1 ⎡ (2 x) 2 ln(2 x) (2 x)2 ⎤
= ⎢
−
⎥
4 ⎢⎣
2
4 ⎥⎦
For
qn
q0
− ln
1 − qn
1 − q0
T
∫0 ke
=−
qn (1 − q0 )
q0 (1 − qn )
634
− rt
ke
−ke− rt
⎛ 1⎞ T
dt = k ⎜ − ⎟ ∫ e− rt (− r dt ) =
r
⎝ r⎠ 0
− rT
r
⎛ 1 − e− rT
k
+ = k⎜
⎜
r
r
⎝
⎞
⎟
⎟
⎠
T
0
ISM: Introductory Mathematical Analysis
10
∫0
61. a.
Section 15.4
10
400e0.06(10−t ) dt = 400 ∫ e0.6−0.06t dt
3.
0
10 0.6 −0.06t
e e
dt
0
0.6 10 −0.06t
= 400∫
∫0
= 400e
e
=
dt
⎛ 1 ⎞ 10 −0.06t
= 400e0.6 ⎜
(−0.06 dt )
⎟∫ e
⎝ −0.06 ⎠ 0
=
400e0.6 −0.06t
e
−0.06
≈ $5481
10
=
0
4.
5
=
0.04(5−t )
5.
f =
5
dt = 40∫ te0.2 e−0.04t dt
=
0
⎡ e−0.04t
⎤
(−0.04t − 1) ⎥
⎢
⎢⎣ 0.0016
⎥⎦ 0
6.
0.2
40e ⎡ −0.2
e
(−0.2 − 1) − 1(−1) ⎤ ≈ $535
⎦
0.0016 ⎣
62. Use Formula 38 with u = t and a = −0.07, so
du = dt.
5
∫0 50, 000te
−0.07t
3 2
1
1 x3
x dx = ⋅
∫
3 − (−1) −1
4 3
7
=
3
0
5
f =
)
3
1
2t 5 dt
∫
3 − (−3) −3
1 t6
⋅
6 3
3
−3
1 4 2
t t + 9 dt
4 − 0 ∫0
⎛ 1 ⎞⎛ 1 ⎞ 4
= ⎜ ⎟ ⎜ ⎟ ∫ t 2 + 9[2t dt ]
⎝ 4 ⎠⎝ 2 ⎠ 0
f =
)
3
2
4
⎤
⎥
49
⎥ =
6
⎥
⎥⎦
0
7.
1 9
f =
6 xdx =
9 − 1 ∫1
8.
f =
9
1 ⎛ 32 ⎞
⎜ 4 x ⎟ = 13
8⎝
⎠1
3
3
=
−1
1 3 5
1 5
1⎛ 5
⎞
= ⎜ − + 5⎟
dx = ⋅ −
∫
2
1
3 −1 x
2 x1 2⎝ 3
⎠
5
=
3
1⎛
−1 ⎞
9− ⎟
4 ⎜⎝
3 ⎠
9. P =
=
2
2.
= −1
−1
(
(
dt = 50, 000∫ te−0.07t dt
Problems 15.4
f =
2
1 3 2
x + x + 1 dx
3 − 1 ∫1
⎡ 2
1 ⎢2 t +9
= ⎢
8⎢
3
⎢⎣
5
⎡ e−0.07t
⎤
= 50, 000 ⎢
(−0.07t − 1) ⎥
⎢⎣ 0.0049
⎥⎦ 0
50, 000 −0.35
[e
(−1.35) − 1(−1)]
=
0.0049
= $496, 640
1.
)
1
= [36 − (−3)6 ]
18
=0
5
=
)
⎞
1 ⎛ x3 x 2
22
= ⎜
+
+ x⎟ =
⎜
⎟
2⎝ 3
2
3
⎠1
∫
= 40e
(
1
2 x − x3
3
f =
5
40e0.2 te−0.04t dt
0
0.2
(
2
1
2 − 3 x 2 dx
2 − (−1) ∫−1
3
400e0.6 ⎡ −0.6 ⎤
−1
e
⎦
−0.06 ⎣
b. Use Formula 38 with u = t and a = –0.04, so
du = dt.
∫0 40te
f =
⎛ 3x 2
⎞
1 2
7
(3x − 1)dx = ⎜
− x⎟ =
∫
⎜ 2
⎟
2 −1 1
⎝
⎠1 2
(
)
100
1
369q − 2.1q 2 − 400 dq
∫
100 − 0 0
(
1
184.5q 2 − 0.7 q3 − 400q
100
)
100
0
1
=
(1,845, 000 − 700, 000 − 40, 000) − 0
100
= 11,050
Answer: $11,050
635
Chapter 15: Methods and Applications of Integration
10. c =
(
ISM: Introductory Mathematical Analysis
)
500
1
4000 + 10q + 0.1q 2 dq
500 − 100 ∫100
1 ⎛
0.1q3 ⎞
=
⎜ 4000q + 5q 2 +
⎟
400 ⎜⎝
3 ⎟⎠
14.
Principles in Practice 15.5
≈ 17,333.33
1. Separating variables, we have
dI
= −0.0085I
dx
dI
= −0.0085dx
I
1
∫ I dI = − ∫ 0.0085 dx
ln I = −0.0085 x + C1
To solve for I, we convert to exponential
Formula
100
1 2
3000e0.05t dt
2 − 0 ∫0
3000 1 2 0.05t
=
⋅
e
[0.05 dt ]
2 0.05 ∫0
= 30, 000e0.05t
2
0
(
)
T
R (1 + α t ) 2
dt
F1
= 30, 000 e0.1 − 1 ≈ 3155.13
Answer: $3155.13
1 T R
1
12. C =
dt =
∫
0
T −0
F (t )
T
R 1
⋅
TF1 α
=
R
α TF1
T
∫0
∫0
R
(1 + α t ) [α dt ] =
α TF1
2
I = e−0.0085 x +C1 = Ce−0.0085 x . Since I = I 0
when x = 0, I 0 = Ce0 = C , so
I ( x) = I 0 e−0.0085 x .
T
⎡ (1 + α t )3 ⎤
⎢
⎥
3
⎣⎢
⎦⎥ 0
Problems 15.5
⎡ (1 + α T )
1⎤
− ⎥
⎢
3
3 ⎥⎦
⎢⎣
3
1. y′ = 2 xy 2
dy
= 2 xy 2
dx
dy
= 2 x dx
y2
R ⎡
=
1 + 3α T + 3α 2T 2 + α 3T 3 − 1⎤
⎦
3α TF1 ⎣
=
=
R
1
⎛
⎞
(3α T ) ⎜1 + α T + α 2T 2 ⎟
3α TF1
3
⎝
⎠
(
2 2
R 1 + α T + 13 α T
∫y
)
−
F1
r ( q0 )
r ( q0 )
q0
−2
dy = ∫ 2 x dx
1
= x2 + C
y
y=−
1
dr
13. Average value =
dq .
∫
q0 − 0 dq
1
⎡ r ( q0 ) − r (0) ⎤⎦
=
q0 ⎣
But r(0) = 0, so avg. value =
1 1
1
dx ≈ 0.32
1 − 0 ∫0 x 2 − 4 x + 5
500
Answer: $17,333.33
11.
f =
1
2
x +C
2. y ′ = x 2 y 2
dy
= x2 y 2
dx
dy
= x 2 dx
y2
. Since
= [price per unit when q0 units are sold] ⋅q0 ,
we have
⎡ price per unit ⎤
⎢ when q0 units ⎥ ⋅ q0
⎢
⎥
are sold
⎦
avg. value = ⎣
q0
= price per unit when ⋅q0 units are sold.
636
ISM: Introductory Mathematical Analysis
dy
∫ y2 = ∫ x
2
Section 15.5
6. y ′ = e x y 3
dy
= e x y3
dx
dy
= e x dx
3
y
dy
x
∫ y3 = ∫ e dx
1
−
= ex + C
2
2y
1
2
y =−
x
2(e + C )
dx
1 x3
=
+ C1
y
3
1 1
− = ( x3 + 3C1 )
y 3
1
1
= − ( x3 + C )
3
y
3
y=−
3
x +C
−
3.
dy
− 3x x 2 + 1 = 0
dx
(
)
dy = 3x x 2 + 1
∫ dy = 3∫ x ( x
1
2
dx
dy
dx
dy
y
) dx
3
∫ dy = 2 ∫ ( x + 1) [2 x dx]
3 ( x + 1)
y= ⋅
+C
2
+1
1
2
1
2
2
∫
3
2
2
2
3
2
(
)
y = x2 + 1
3
2
dy
dx
=
y ∫ x
ln y = ln x + C1
ln y = ln x + ln C, where C > 0.
ln y = ln(Cx) ⇒ y = Cx , where C > 0.
+C
8.
dy x
=
4.
dx y
y dy = x dx
∫ y dy = ∫ x dx
dy
+ xe x = 0
dx
dy = − xe x dx
x
∫ dy = ∫ − xe dx
y = ∫ − xe x dx
y 2 x2
=
+ C1
2
2
Using integration by parts or formula 38 gives
y = (1 − x)e x + C
y 2 = x 2 + 2C1
y 2 = x2 + C
5.
y
, where x, y > 0.
x
y
=
x
dx
=
x
7. y′ =
9. y ′ =
dy
= y , where y > 0.
dx
dy
= dx
y
1
where y(1) = 1.
y2
dy
1
=
dx y 2
y 2 dy = dx
∫y
dy
∫ y = ∫ dx
ln y = x + C1
y = e x +C1 = eC1 e x = Ce x , where C = eC1 . Thus
2
dy = ∫ dx
y3
= x+C
3
Given y(1) = 1, we obtain
y = Ce x , where C > 0.
637
13
= 1 + C , so
3
Chapter 15: Methods and Applications of Integration
ISM: Introductory Mathematical Analysis
3
13. (3x 2 + 2)3 y ′ − xy 2 = 0, where y (0) = .
2
dy
(3x 2 + 2)3
= xy 2
dx
dy
x
=
2
2
y
(3x + 2)3
dy
x
∫ y 2 = ∫ (3x2 + 2)3 dx
1
2
−2
−3
∫ y dy = 6 ∫ (3x + 2) [6 x dx]
1
1
− =−
+C
2
y
12(3 x + 2)2
2
2⎞
⎛
C = − . Thus y 3 = 3 ⎜ x − ⎟ = 3 x − 2,
3
3⎠
⎝
y = 3 3x − 2.
10. y′ = e x − y , where y(0) = 0
dy e x
=
dx e y
e y dy = e x dx
∫e
y
dy = ∫ e x dx
e y = ex + C
Since y(0) = 0, we have e0 = e0 + C , 1 = 1 + C,
3
we have
2
2
1
1
1
− =−
+ C , − = − + C , so
3
2
3
48
2(2)
2
Given that y (0) =
C = 0. Thus e y = e x , so y = x.
11. e y y′ − x 2 = 0 , where y = 0 when x = 0.
ey
dy
= x2
dx
31
. Thus,
48
1
1
31
− =−
−
2
2
y
48
12(3 x + 2)
C=−
e y dy = x 2 dx
∫e
y
dy = ∫ x 2 dx
x3
ey =
+C
3
=−
Given that y(0) = 0, we have e0 = 0 + C , so
1 = C ⇒ ey =
3
1
y2
3
x
x +3
+1 , ey =
, so
3
3
.
48(3 x 2 + 2)2
4 + 31(3x 2 + 2) 2
.
14. y ′ + x3 y = 0 and y = e when x = 0.
dy
= − x3 y
dx
dy
= − x3 dx
y
dy
3
∫ y = − ∫ x dx
= 0, where y(1) = 2
dy
1
=−
dx
y2
dx
y 2 dy = −
x2
dx
2
∫ y dy = − ∫ x2
y3 1
= +C
3
x
x2
Now, y(1) = 2 implies C =
48(3 x 2 + 2)2
Hence, y =
x3 + 3
y = ln
.
3
12. x 2 y′ +
4 + 31(3 x 2 + 2)2
x4
+C
4
Given y(0) = e, ln e = 0 + C, so C = 1.
4
x4
− x +1
Thus ln y = −
+ 1, so y = e 4 .
4
ln y = −
5
. Thus
3
y3 1 5
3
3
= + , y 3 = + 5, y = 3 + 5.
3
x 3
x
x
638
ISM: Introductory Mathematical Analysis
15.
Section 15.5
2
dy 3x 1 + y
=
, where y > 0 and y (1) = 8 .
dx
y
y dy
1 + y2
17. 2
)
(
∫ xe
∫(
2
+2
2
+9
) dydx = 3xy
y2 + 9
)
(
)
2 2
y +9
3
1
2
3x2 + 2
3
2
x3 + 2 x + 1
[2 y dy ] = ∫
, where y(0) = 0.
2
e− x = 2( y 2 + 1) −1/ 2 − 1.
dx
1
(
19. (q + 1) 2
)
⎡ 3x 2 + 2 dx ⎤
3
⎥⎦
x + 2 x + 1 ⎣⎢
1
dc
= cq
dq
q
∫ c dc = ∫ (q + 1)2 dq
3
= ln x + 2 x + 1 + C
Using partial fractions or Formula 7 for
q
∫ (q + 1)2 dq , we obtain
1
ln c = ln(q + 1) +
+ C . Now, fixed cost is
q +1
given to be e, which means that c = e when q =
0. This implies 1 = 0 + 1 + C, so C = 0. Thus
ln( q +1) + q1+1
1
ln c = ln(q + 1) +
,
⇒c=e
q +1
2
Now y(0) = 0 implies that (27) = ln(1) + C , so
3
C = 18. Thus
3
2 2
2
y + 9 = ln x3 + 2 x + 1 + 18 .
3
(
dx = ∫ y ( y 2 + 1)−3 / 2 dy
1 − x2
1
e [−2 x dx] = ∫ ( y 2 + 1) −3 / 2 [2 y dy ]
∫
2
2
1 − x 2 1 ( y 2 + 1)−1/ 2
− e
=
+C
2
2
−1/ 2
1 − x2
e
= ( y 2 + 1)−1/ 2 + C
2
1 1
1
Now y(0) = 0 gives =
+ C , so C = − .
2
2
1
1 − x2
1
Thus e
= ( y 2 + 1) −1/ 2 − or
2
2
2
16. 2 y x3 + 2 x + 1
− x2
−
⎡ 3x2 3 ⎤
Since y > 0, y = ⎢
+ ⎥ −1 .
2 ⎥⎦
⎣⎢ 2
y + 9 dy = ∫
)
2
⎡ 3x 2 3 ⎤
y2 = ⎢
+ ⎥ −1
2 ⎥⎦
⎣⎢ 2
∫ 2y
(
xe− x dx = y ( y 2 + 1)−3 / 2 dy
2
2
)
18. x( y 2 + 1)3 / 2 dx = e x y dy, where y(0) = 0.
2
(
(
)
2
)
⎡ 3x2 3 ⎤
1 + y2 = ⎢
+ ⎥
2 ⎥⎦
⎣⎢ 2
)
(
)
−1
1
2 2
1
y
[2 y dy ] = 3∫ x dx
+
2∫
1
3x 2
2
1 + y2 =
+C
2
1
3
y (1) = 8 ⇒ (1 + 8) 2 = + C
2
3
C=
2
Thus
1
3x 2 3
2
1 + y2 =
+
2
2
(
(
1
−2
1
x x2