C H A P T E R 1
Preparation for Calculus
Section 1.1
Graphs and Models.................................................................................2
Section 1.2
Linear Models and Rates of Change....................................................11
Section 1.3
Functions and Their Graphs.................................................................22
Section 1.4
Fitting Models to Data..........................................................................34
Section 1.5
Inverse Functions..................................................................................37
Section 1.6
Exponential and Logarithmic Functions .............................................54
Review Exercises ..........................................................................................................63
Problem Solving ...........................................................................................................73
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 1
Preparation for Calculus
Section 1.1 Graphs and Models
1. y = − 32 x + 3
7. y = 4 − x 2
x-intercept: (2, 0)
x
−3
−2
0
2
3
y
−5
0
4
0
−5
y-intercept: (0, 3)
Matches graph (b).
y
2. y =
9 − x2
6
x-intercepts: ( −3, 0), (3, 0)
(0, 4)
y-intercept: (0, 3)
2
(−2, 0)
−6
Matches graph (d).
(2, 0)
x
−4
4
(−3, −5)
3. y = 3 − x 2
6
−2
(3, −5)
−4
−6
(
x-intercepts:
)(
3, 0 , − 3, 0
)
8. y = ( x − 3)
y-intercept: (0, 3)
Matches graph (a).
4. y = x − x
3
2
x
0
1
2
3
4
5
6
y
9
4
1
0
1
4
9
x-intercepts: (0, 0), ( −1, 0), (1, 0)
y
y-intercept: (0, 0)
10
(0, 9)
(6, 9)
8
Matches graph (c).
6
5. y =
+ 2
1x
2
4
2
(1, 4)
(2, 1)
(5, 4)
(4, 1)
x
−4
−2
0
2
4
y
0
1
2
3
4
−6 −4 −2
x
−2
4
2
6
(3, 0)
9. y = x + 2
y
6
x
−5
−4
−3
−2
−1
0
1
y
3
2
1
0
1
2
3
(4, 4)
4
(2, 3)
(0, 2)
(−2, 1)
−4
y
x
−2
2
(−4, 0)
4
6
−2
4
(− 5, 3)
6. y = 5 − 2 x
(− 4, 2) 2
(− 1, 1)
(− 3, 1)
x
−1
0
1
2
5
2
y
7
5
3
1
0
3
4
−1
−3
−6
−4
(1, 3)
(0, 2)
x
(− 2, 0)
2
−2
y
8
(−1, 7)
(0, 5)
4
2
−6 −4 −2
−2
−4
2
(1, 3)
(2, 1)
( (
5,0
2
(3, −1)
x
(4, −3)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.1
10. y = x − 1
14. y =
x
−3
−2
−1
0
1
2
3
y
2
1
0
−1
0
1
2
1
x+ 2
x
−6
−4
−3
−2
−1
y
− 14
− 12
−1
Undef.
1
y
5
4
3
2
3
(3, 2)
2
(− 2, 1)
(−1, 1)
(2, 1)
1
−1
(− 1, 0)
2
3
−1
(− 6, − 14 )
(− 4, − 12 )
(1, 0)
(0, − 1)
−2
11. y =
2
1
2
1
4
(0, 12 )
(2, 14 )
x
x
−3 −2
0
y
4
(− 3, 2)
3
Graphs and Models
1 2 3
−2
−3
−4
−5
(−3, − 1)
x −6
x
0
1
4
9
16
y
−6
−5
−4
−3
−2
15. y =
5− x
5
(−4.00, 3)
(2, 1.73)
y
−6
6
2
−3
x
−4
4
−2
8
12
(9, −3)
16
(16, −2)
−4
(4, −4)
(1, −5)
−6
(0, −6)
−8
12. y =
(a)
(2, y)
= ( 2, 1.73)
(b)
( x, 3)
= ( −4, 3)
(y =
(3 =
5−2 =
5 − ( − 4)
)
16. y = x 5 − 5 x
x + 2
6
x
−2
−1
y
0
1
0
2
2
7
14
2
3
4
(− 0.5, 2.47)
−9
9
(1, − 4)
y
−6
5
4
(14, 4)
3
(7, 3)
(− 1, 1)
2
(2, 2)
(0, 2 )
13. y =
5
(−0.5, y )
(b)
( x , − 4)
= ( −0.5, 2.47)
= ( −1.65, − 4) and ( x, − 4) = (1, − 4)
17. y = 2 x − 5
y-intercept: y = 2(0) − 5 = −5; (0, − 5)
x
(− 2, 0)
(a)
10
15
20
x-intercept: 0 = 2 x − 5
3
x
5 = 2x
x
−3
−2
−1
0
1
2
3
y
−1
− 32
−3
Undef.
3
3
2
1
x =
5;
2
( 52 , 0)
18. y = 4 x 2 + 3
y-intercept: y = 4(0) + 3 = 3; (0, 3)
2
y
x-intercept: 0 = 4 x 2 + 3
(1, 3)
3
(2, 32 (
2
(3, 1)
(−3, −1)
1
−3 −2 −1
−1
−2
x
1
2
3
)
3 ≈ 1.73
−3 = 4 x 2
None. y cannot equal 0.
(−2, − 32 (
(−1, −3)
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4
Chapter 1
Preparation for Calculus
19. y = x 2 + x − 2
24. y =
y-intercept: y = 02 + 0 − 2
y = −2; (0, − 2)
x 2 + 3x
(3 x
+ 1)
y-intercept: y =
x-intercepts: 0 = x 2 + x − 2
x = −2, 1; ( −2, 0), (1, 0)
x-intercepts: 0 =
20. y = x − 4 x
3
0 =
y-intercept: y 2 = 03 − 4(0)
y = 0; (0, 0)
x =
x-intercepts: 0 = x3 − 4 x
⎡⎣3(0) + 1⎤⎦
2
x 2 + 3x
(3x + 1)
x( x + 3)
2
(3x + 1)
0, − 3; (0, 0), ( −3, 0)
2
25. x 2 y − x 2 + 4 y = 0
0 = x( x − 2)( x + 2)
x = 0, ± 2; (0, 0), ( ± 2, 0)
21. y = x 16 − x
02 + 3(0)
y = 0; (0, 0)
0 = ( x + 2)( x − 1)
2
2
y-intercept: 02 ( y ) − 02 + 4 y = 0
y = 0; (0, 0)
x-intercept: x 2 (0) − x 2 + 4(0) = 0
2
x = 0; (0, 0)
y-intercept: y = 0 16 − 02 = 0; (0, 0)
x-intercepts: 0 = x 16 − x 2
0 = x
(4 − x)(4 + x)
x = 0, 4, − 4; (0, 0), ( 4, 0), ( − 4, 0)
22. y = ( x − 1)
x2 + 1
26. y = 2 x −
x2 + 1
y-intercept: y = 2(0) −
y = −1; (0, −1)
0 = 2x −
x-intercept:
2x =
y-intercept: y = (0 − 1) 02 + 1
x2 + 1
3x 2 = 1
x2 + 1
x2 =
x = 1; (1, 0)
1
3
x = ±
2− x
23. y =
5x + 1
2− 0
= 2;
y -intercept: y =
5(0) + 1
2− x
x-intercept: 0 =
5x + 1
0 = 2−
x
x = 4;
( 4, 0)
x2 + 1
4x2 = x2 + 1
y = −1; (0, −1)
x-intercept: 0 = ( x − 1)
02 + 1
x =
(0, 2)
Note: x = −
3
3
3 ⎛ 3
;⎜
,
3 ⎜⎝ 3
⎞
0 ⎟⎟
⎠
3 3 is an extraneous solution.
27. Symmetric with respect to the y-axis because
y = ( − x) − 6 = x 2 − 6.
2
28. y = x 2 − x
No symmetry with respect to either axis or the origin.
29. Symmetric with respect to the x-axis because
(− y )2
= y 2 = x 3 − 8 x.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.1
40. y =
30. Symmetric with respect to the origin because
( − y ) = (− x) + (− x)
3
− y = −x − x
3
y = x + x.
3
2x
3
+1
( 0)
y =
2
3
0 =
2x
3
+ 1 = 1, y -intercept
+ 1 ⇒ − 23 x = 1 ⇒ x = − 23 , x-intercept
(
)
Intercepts: (0, 1), − 32 , 0
31. Symmetric with respect to the origin because
(− x)(− y ) = xy = 4.
5
Graphs and Models
y
2
Symmetry: none
(0, 1)
(− 32 , 0)
x
32. Symmetric with respect to the x-axis because
−1
x( − y ) = xy 2 = −10.
2
1
2
−1
−2
33. y = 4 −
x +3
No symmetry with respect to either axis or the origin.
y = 9 − (0) = 9, y -intercept
2
34. Symmetric with respect to the origin because
(− x)( − y ) −
4 − ( − x)
xy −
2
41. y = 9 − x 2
0 = 9 − x 2 ⇒ x 2 = 9 ⇒ x = ± 3, x-intercepts
= 0
Intercepts: (0, 9), (3, 0), ( −3, 0)
4 − x 2 = 0.
y
10
y = 9 − ( − x) = 9 − x 2
(0, 9)
2
35. Symmetric with respect to the origin because
−y =
6
Symmetry: y-axis
−x
( − x)
2
2
+1
(−3, 0)
x
2
−2
4
6
42. y = 2 x 2 + x = x( 2 x + 1)
2
x
is symmetric with respect to the y-axis
x2 + 1
( − x)
2
( − x) + 1
2
because y =
(3, 0)
−6 −4 −2
x
.
y = 2
x +1
36. y =
4
=
y = 0( 2(0) + 1) = 0, y -intercept
0 = x( 2 x + 1) ⇒ x = 0, − 12 , x-intercepts
x2
.
x +1
(
2
)
Intercepts: (0, 0), − 12 , 0
37. y = x 3 + x is symmetric with respect to the y-axis
y
5
Symmetry: none
4
3
because y = ( − x ) + (− x ) = −( x 3 + x) = x3 + x .
3
2
(− 12 , 0)
38. y − x = 3 is symmetric with respect to the x-axis
−3
−2
1
(0, 0)
x
−1
1
2
3
because
−y − x = 3
43. y = x3 + 2
y − x = 3.
y = 03 + 2 = 2, y -intercept
0 = x3 + 2 ⇒ x3 = − 2 ⇒ x = − 3 2, x-intercept
39. y = 2 − 3 x
(
0 = 2 − 3( x) ⇒ 3x = 2 ⇒ x =
Intercepts: (0, 2),
(
2,
3
0
)
Intercepts: − 3 2, 0 , (0, 2)
y = 2 − 3(0) = 2, y -intercept
)
2
,
3
x-intercept
y
Symmetry: none
5
4
y
3
(0, 2)
Symmetry: none
2
1
−1
(− 3 2, 0)
(0, 2)
−3 −2
( (
2
,0
3
1
x
−1
1
2
3
x
2
3
−1
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6
Chapter 1
Preparation for Calculus
44. y = x 3 − 4 x
47. x = y 3
y = 03 − 4(0) = 0, y -intercept
y 3 = 0 ⇒ y = 0, y -intercept
x = 0, x-intercept
x3 − 4 x = 0
x ( x 2 − 4) = 0
Intercept: (0, 0)
− x = (− y ) ⇒ − x = − y 3
3
x( x + 2)( x − 2) = 0
Symmetry: origin
x = 0, ± 2, x-intercepts
y
Intercepts: (0, 0), ( 2, 0), ( −2, 0)
4
y = ( − x) − 4( − x) = − x3 + 4 x = −( x3 − 4 x)
3
3
2
(0, 0)
Symmetry: origin
y
2
3
4
−3
−4
(−2, 0)
(0, 0)
−1
1
−2
3
−3
x
−4 −3 −2 −1
(2, 0)
1
−1
48. x = y 2 − 4
x
3
y2 − 4 = 0
−2
(y
−3
+ 2)( y − 2) = 0
y = ± 2, y -intercepts
45. y = x
x +5
x = 02 − 4 = − 4, x-intercept
Intercepts: (0, 2), (0, − 2), ( − 4, 0)
y = 0 0 + 5 = 0, y -intercept
x
x + 5 = 0 ⇒ x = 0, − 5, x -intercepts
Intercepts: (0, 0), ( −5, 0)
x = (− y) − 4 = y 2 − 4
2
y
y
Symmetry: x-axis
3
Symmetry: none
3
2
(−5, 0)
(0, 0)
−4 −3 −2 −1
(0, 2)
x
1
(− 4, 0)
2
−5
−2
x
−1
1
(0, − 2)
−3
−4
−3
46. y =
y =
25 − x 2
25 − 0
=
2
25 = 5, y -intercept
y =
25 − x 2 = 0
25 − x 2 = 0
(5 + x)(5 − x)
= 0
Intercepts: (0, 5), (5, 0), ( −5, 0)
25 − ( − x )
2
8
x
8
⇒ Undefined ⇒ no y -intercept
0
8
= 0 ⇒ No solution ⇒ no x-intercept
x
x = ± 5, x-intercept
y =
49. y =
=
25 − x 2
Intercepts: none
−y =
8
8
⇒ y =
x
−x
Symmetry: origin
y
Symmetry: y-axis
8
y
6
7
6
(−5, 0)
4
3
2
1
−4 −3 −2 −1
4
(0, 5)
2
−2
x
2
4
6
8
(5, 0)
x
1 2 3 4 5
−2
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.1
50. y =
7
53. y 2 − x = 9
10
x2 + 1
y2 = x + 9
10
y = 2
= 10, y -intercept
0 +1
y = ±
±
10
2
10
x2 + 1
=
10
9 = ± 3, y -intercepts
x +9 = 0
12
( − x) + 1
0+9 = ±
x +9 = 0
y
Intercept: (0, 10)
x+9
y = ±
10
= 0 ⇒ No solution ⇒ no x-intercepts
x2 + 1
y =
Graphs and Models
x = − 9, x-intercept
(0, 10)
Intercepts: (0, 3), (0, − 3), ( −9, 0)
Symmetry: y-axis
(− y )2
− x = 9 ⇒ y2 − x = 9
2
−6 −4 −2
x
2
4
Symmetry: x-axis
6
y
51. y = 6 − x
6
4
y = 6 − 0 = 6, y -intercept
(− 9, 0)
6− x = 0
−10
2
−6 −4 −2
(0, 3)
x
−2
2
(0, − 3)
−4
6 = x
−6
x = ± 6, x-intercepts
Intercepts: (0, 6), ( −6, 0), (6, 0)
y = 6 − −x = 6 − x
54. x 2 + 4 y 2 = 4 ⇒ y = ±
Symmetry: y-axis
y = ±
y
2
(0, 6)
x2 = 4
4
2
(− 6, 0)
−8
−4 −2
−2
4 − 02
4
= ±
= ± 1, y -intercepts
2
2
x 2 + 4(0) = 4
8
6
4 − x2
2
(6, 0)
x = ± 2, x-intercepts
x
4
2
6
8
Intercepts: ( −2, 0), ( 2, 0), (0, −1), (0, 1)
−4
−6
( − x)2
−8
52. y = 6 − x
+ 4( − y ) = 4 ⇒ x 2 + 4 y 2 = 4
2
Symmetry: origin and both axes
y
y = 6 − 0 = 6 = 6, y -intercept
3
6 − x = 0
2
(0, 1)
6 − x = 0
(2, 0)
(−2, 0)
6 = x, x -intercept
Intercepts: (0, 6), (6, 0)
−3
−1
−2
x
1
3
(0, −1)
−3
Symmetry: none
y
8
(0, 6)
4
2
(6, 0)
x
2
4
6
8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
8
Chapter 1
Preparation for Calculus
55. x + 3 y 2 = 6
57.
4x − y = 7 ⇒ y = 4x − 7
3y2 = 6 − x
8 − x = 4x − 7
6 − x
3
y = ±
15 = 5 x
6−0
= ±
3
y = ±
3 = x
2, y -intercepts
The corresponding y-value is y = 5.
x + 3(0) = 6
Point of intersection: (3, 5)
2
x = 6, x-intercept
(
)(
Intercepts: (6, 0), 0,
2 , 0, −
2
)
3x + 4
2
− 4 x − 10
4 x + 2 y = −10 ⇒ y =
2
58. 3x − 2 y = − 4 ⇒ y =
x + 3( − y ) = 6 ⇒ x + 3 y 2 = 6
2
3x + 4
− 4 x − 10
=
2
2
3 x + 4 = − 4 x − 10
Symmetry: x-axis
y
4
7 x = −14
3
( 0,
2
2)
1
x = −2
(6, 0)
x
−1
1
−2
2
3
( 0, −
6
The corresponding y-value is y = −1.
7
2)
Point of intersection: ( −2, −1)
−3
−4
59. x 2 + y = 6 ⇒ y = 6 − x 2
56. 3x − 4 y 2 = 8
x + y = 4 ⇒ y = 4− x
4 y 2 = 3x − 8
y = ±
3x
4
6 − x2 = 4 − x
−2
0 = x2 − x − 2
0 = ( x − 2)( x + 1)
3
( 0) − 2 = ± − 2
4
⇒ no solution ⇒ no y -intercepts
y = ±
x = 2, −1
The corresponding y-values are y = 2 (for x = 2) and
y = 5 (for x = −1).
3 x − 4(0) = 8
2
Points of intersection: ( 2, 2), ( −1, 5)
3x = 8
8
x = , x-intercept
3
Intercept:
( 83 , 0)
2
Symmetry: x-axis
3 − x = ( x − 1)
2
3 − x = x2 − 2x + 1
x = −1 or x = 2
6
4
−4
x = 3 − y2 ⇒ y2 = 3 − x
0 = x 2 − x − 2 = ( x + 1)( x − 2)
y
−2
−2
60.
y = x −1
3x − 4( − y ) = 8 ⇒ 3 x − 4 y 2 = 8
2
x + y = 8 ⇒ y = 8− x
The corresponding y-values are y = −2 (for x = −1)
( 83, 0)
x
2
6
8
10
and y = 1 (for x = 2).
Points of intersection: ( −1, − 2), ( 2, 1)
−6
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.1
x − y = 1 ⇒ y = x −1
y = 1 − x2
2
y = x 4 − 2x 2 + 1
2
5 − x = x − 2x + 1
2
9
64. y = x 4 − 2 x 2 + 1
61. x 2 + y 2 = 5 ⇒ y 2 = 5 − x 2
5 − x 2 = ( x − 1)
Graphs and Models
2
0 = 2 x 2 − 2 x − 4 = 2( x + 1)( x − 2)
(0, 1)
−3
x = −1 or x = 2
(−1, 0)
The corresponding y-values are y = −2 (for x = −1)
−2
y = 1 − x2
and y = 1 (for x = 2).
Points of intersection: ( −1, 0), (0, 1), (1, 0)
Points of intersection: ( −1, − 2), ( 2, 1)
Analytically, 1 − x 2 = x 4 − 2 x 2 + 1
62. x 2 + y 2 = 25 ⇒ y 2 = 25 − x 2
0 = x4 − x2
−3 x + y = 15 ⇒ y = 3 x + 15
25 − x 2 = (3x + 15)
3
(1, 0)
0 = x 2 ( x + 1)( x − 1)
2
x = −1, 0, 1.
25 − x 2 = 9 x 2 + 90 x + 225
65. y =
0 = 10 x 2 + 90 x + 200
x +6
y =
0 = x 2 + 9 x + 20
− x2 − 4 x
0 = ( x + 5)( x + 4)
4
x = − 4 or x = −5
y=
The corresponding y-values are y = 3 (for x = − 4)
x+6
(3,
(− 2, 2)
−7
and y = 0 (for x = −5).
3)
2
− x 2 − 4x
y=
−2
Points of intersection: ( − 4, 3), ( −5, 0)
(
Points of intersection: ( −2, 2), −3,
63. y = x3 − 2 x 2 + x − 1
x+6 =
Analytically,
−4
x2 + 5x + 6 = 0
y = x3 − 2x2 + x − 1
(x
(2, 1)
+ 3)( x + 2) = 0
x = −3, − 2.
6
(0, −1)
− x2 − 4x
x + 6 = − x2 − 4 x
y = − x2 + 3x − 1
4
)
3 ≈ (−3, 1.732)
(−1, −5)
66. y = − 2 x − 3 + 6
−8
y = −x2 + 3x − 1
y = 6− x
Points of intersection: ( −1, − 5), (0, −1), ( 2, 1)
7
Analytically, x − 2 x + x − 1 = − x + 3 x − 1
3
2
2
x( x − 2)( x + 1) = 0
x = −1, 0, 2.
y=6−x
(1, 5)
x3 − x 2 − 2 x = 0
(3, 3)
−4
8
−1
y = −⏐2x − 3⏐+ 6
Points of intersection: (3, 3), (1, 5)
Analytically, − 2 x − 3 + 6 = 6 − x
2x − 3 = x
2 x − 3 = x or 2 x − 3 = − x
x = 3 or
x = 1.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10
Chapter 1
Preparation for Calculus
67. (a) Using a graphing utility, you obtain
y = 0.005t 2 + 0.27t + 2.7.
(b)
71. y = kx 3
(a) (1, 4):
30
0
(c) For 2020, t = 40.
y = 0.005( 40) + 0.27( 40) + 2.7
3
(b)
(− 2, 1):
1 = k ( − 2) = − 8k ⇒ k = − 18
(c)
(0, 0):
0 = k (0) ⇒ k can be any real number.
(d)
(−1, −1):
−1 = k ( −1) = − k ⇒ k = 1
16
0
4 = k (1) ⇒ k = 4
3
3
3
72. y 2 = 4kx
2
= 21.5
(a) (1, 1):
12 = 4k (1)
1 = 4k
The GDP in 2020 will be $21.5 trillion.
68. (a) Using a graphing utility, you obtain
y = 0.24t 2 + 12.6t − 40.
(b)
(b)
(2, 4):
330
(c)
5
(0, 0):
k =
1
4
( 4)
= 4 k ( 2)
2
16
= 8k
k
= 2
0 = 4 k ( 0)
2
k can be any real number.
20
30
The model is a good fit for the data.
(d)
(3, 3):
(3)
(c) For 2020, t = 30.
y = 0.24(30) + 12.6(30) − 40
2
2
= 4k (3)
9
= 12k
k
=
9
12
=
3
4
= 554
The number of cellular phone subscribers in 2020
will be 554 million.
2.04 x + 5600 = 3.29 x
5600 = 3.29 x − 2.04 x
5600 = 1.25 x
x =
5600
= 4480
1.25
To break even, 4480 units must be sold.
74. Answers may vary. Sample answer:
(
y = x +
3
2
)( x − 4)( x − 52 ) has intercepts at
x = − 32 , x = 4, and x = 52 .
75. (a) If (x, y) is on the graph, then so is ( − x, y ) by y-axis
symmetry. Because ( − x, y ) is on the graph, then so
is ( − x, − y ) by x-axis symmetry. So, the graph is
10,770
70. y =
− 0.37
x2
symmetric with respect to the origin. The converse is
not true. For example, y = x3 has origin symmetry
but is not symmetric with respect to either the x-axis
or the y-axis.
400
0
y = ( x + 4)( x − 3)( x − 8) has intercepts at
x = − 4, x = 3, and x = 8.
C = R
69.
73. Answers may vary. Sample answer:
100
0
If the diameter is doubled, the resistance is changed by
approximately a factor of 14. For instance,
y ( 20) ≈ 26.555 and y ( 40) ≈ 6.36125.
(b) Assume that the graph has x-axis and origin
symmetry. If (x, y) is on the graph, so is ( x, − y ) by
x-axis symmetry. Because ( x, − y ) is on the graph,
then so is ( − x, − ( − y )) = (− x, y ) by origin
symmetry. Therefore, the graph is symmetric with
respect to the y-axis. The argument is similar for
y-axis and origin symmetry.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.2
11
Linear Models and Rates of Change
76. (a) Intercepts for y = x 3 − x :
y = 03 − 0 = 0 ; (0, 0)
y -intercept:
x-intercepts: 0 = x 3 − x = x( x 2 − 1) = x( x − 1)( x + 1) ;
(0, 0), (1, 0) (−1, 0)
Intercepts for y = x 2 + 2:
y = 0 + 2 = 2 ; (0, 2)
y -intercept:
x-intercepts: 0 = x 2 + 2
None. y cannot equal 0.
(b) Symmetry with respect to the origin for y = x3 − x because
− y = ( − x) − ( − x) = − x3 + x.
3
Symmetry with respect to the y-axis for y = x 2 + 2 because
y = ( − x) + 2 = x 2 + 2.
2
x3 − x = x 2 + 2
(c)
x3 − x 2 − x − 2 = 0
(x
− 2)( x 2 + x + 1) = 0
x = 2 ⇒ y = 6
Point of intersection : (2, 6)
Note: The polynomial x 2 + x + 1 has no real roots.
77. False. x-axis symmetry means that if ( − 4, − 5) is on the
graph, then ( − 4, 5) is also on the graph. For example,
(4, − 5) is not on the graph of
(− 4, − 5) is on the graph.
x = y 2 − 29, whereas
⎛ −b ±
79. True. The x-intercepts are ⎜
⎜
⎝
b 2 − 4ac
,
2a
⎞
0 ⎟.
⎟
⎠
⎛ b
⎞
80. True. The x-intercept is ⎜ − , 0 ⎟.
2
a
⎝
⎠
78. True. f ( 4) = f ( −4).
Section 1.2 Linear Models and Rates of Change
1. m = 2
6. m =
2. m = 0
7 −1
6
=
= −2
−2 − 1
−3
y
3. m = −1
(−2, 7)
7
6
5
4. m = −12
3
2 − ( −4)
6
5. m =
=
= 3
5−3
2
2
(1, 1)
1
−4 −3 −2 −1
x
1
3
4
y
3
2
(5, 2)
1
−1
x
1
2
3
5
6
7
−2
−3
−4
(3, −4)
−5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12
Chapter 1
Preparation for Calculus
1−6
−5
=
, undefined.
4−4
0
7. m =
11.
y
m = −2
m is undefined.
The line is vertical.
m=
−3
2
m=1
8
y
6
(3, 4)
4
7
2
(4, 6)
6
5
x
−6 −4
2
−2
4
4
8 10
3
2
m = −3
x
−2 −1
1
2
y
12.
(4, 1)
1
3
5
6
1
m=3
(−2, 5)
−5 − ( −5)
0
8. m =
=
= 0
5−3
2
6
m=0
4
m=3
x
−6
The line is horizontal.
−2
2
4
−2
y
1
13. Because the slope is 0, the line is horizontal and its
equation is y = 2. Therefore, three additional points are
x
−1
−1
1
2
3
4
5
6
(0, 2), (1, 2), (5, 2).
−2
−3
−4
14. Because the slope is undefined, the line is vertical and its
equation is x = − 4. Therefore, three additional points
(3, − 5) (5, −5)
−6
are ( − 4, 0), ( − 4, 1), ( − 4, 2).
9. m =
2 1
1
−
3 6
= 2 = 2
1
1 ⎛ 3⎞
− − ⎜− ⎟
4
2 ⎝ 4⎠
15. The equation of this line is
y − 7 = −3( x − 1)
y = −3x + 10.
y
Therefore, three additional points are (0, 10), (2, 4), and
(3, 1).
3
2
(− 12 , 23 )
−3
(− 34 , 16 )
16. The equation of this line is
y + 2 = 2( x + 2)
x
−2
1
−1
2
3
y = 2 x + 2.
−2
Therefore, three additional points are ( −3, − 4), ( −1, 0),
−3
and (0, 2).
⎛ 3⎞ ⎛ 1⎞
⎜ ⎟ − ⎜− ⎟
1
8
4
4
= −
10. m = ⎝ ⎠ ⎝ ⎠ =
3
⎛7⎞ ⎛5⎞
3
−
⎜ ⎟−⎜ ⎟
8
⎝8⎠ ⎝ 4⎠
17.
y =
3x
4
+3
4 y = 3x + 12
0 = 3x − 4 y + 12
y
y
5
3
4
(0, 3)
2
1
−2
2
( 78 , 34 )
1
x
−1
1
−1
( 54 , − 14 )
−4 −3 −2 −1
x
1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.2
Linear Models and Rates of Change
13
21. y + 2 = 3( x − 3)
18. The slope is undefined so the line is vertical.
x = −5
y + 2 = 3x − 9
x +5 = 0
y = 3 x − 11
0 = 3x − y − 11
y
1
y
x
−4 −3 −2 −1
(−5, −2)
1
−1
3
2
−2
1
−3
x
−2 −1
−1
−4
1
2
3
5
6
(3, −2)
−2
−5
4
−3
−4
−5
19.
y =
2x
3
3y = 2x
y − 4 = − 53 ( x + 2)
22.
0 = 2x − 3y
5 y − 20 = −3 x − 6
3x + 5 y − 14 = 0
y
4
y
3
5
2
4
(−2, 4)
(0, 0)
x
1
2
3
4
2
−1
1
−3
y = 4
20.
y −4 = 0
−2
x
−1
1
23. (a) Slope =
y
2
∆y
1
=
∆x
3
(b)
5
x
10 ft
(0, 4)
3
30 ft
2
By the Pythagorean Theorem,
1
−3
−2
x 2 = 302 + 102 = 1000
x
−1
1
2
x = 10 10 ≈ 31.623 feet.
y
Population (in millions)
24. (a)
(b) The slopes are:
310
(9, 307)
305
300
(5, 295.8)
295
290
(8, 304.4)
(7, 301.6)
(6, 298.6)
(4, 293)
t
4
5
6
7
8
9
Year (4 ↔ 2004)
295.8
5
298.6
6
301.6
7
304.4
8
307.0
9
−
−
−
−
−
−
−
−
−
−
293.0
4
295.8
5
298.6
6
301.6
7
304.4
8
= 2.8
= 2.8
= 3.0
= 2.8
= 2.6
The population increased least rapidly from 2008 to 2009.
(c) Average rate of change from 2004 to 2009:
307.0 − 293.0
14
=
9− 4
5
= 2.8 million per yr
(d)
For 2020, t = 20 and y ≈ 16( 2.8) + 293.0 = 337.8 million.
⎡⎣Equivalently, y ≈ 11( 2.8) + 307.0 = 337.8.⎤⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14
Chapter 1
Preparation for Calculus
25. y = 4 x − 3
33. y = −2 x + 1
The slope is m = 4 and the y-intercept is (0, − 3).
y
3
26. − x + y = 1
y = x +1
1
The slope is m = 1 and the y-intercept is (0, 1).
−2
x
−1
1
27. x + 5 y = 20
2
−1
y = − 15 x + 4
Therefore, the slope is m = − 15 and the y-intercept is
34. y =
−1
1x
3
y
(0, 4).
2
28. 6 x − 5 y = 15
y =
6x
5
1
−3
x
−3 −2 −1
Therefore, the slope is m =
(0, − 3).
3
(0, −1)
6
5
−2
and the y-intercept is
−3
−4
29. x = 4
35. y − 2 =
The line is vertical. Therefore, the slope is undefined and
there is no y-intercept.
3
2
y =
(x
− 1)
+
3x
2
1
2
y
30. y = −1
4
The line is horizontal. Therefore, the slope is m = 0 and
the y-intercept is (0, −1).
3
2
1
x
−4 −3 −2
31. y = −3
1
2
3
4
−2
−3
y
−4
2
1
x
−3 −2 −1
1
2
3
4
36. y − 1 = 3( x + 4)
5
−2
y = 3 x + 13
−4
y
−5
−6
16
12
32. x = 4
y
−16 −12 −8
3
x
4
−4
8
−8
2
1
x
1
2
3
37. 2 x − y − 3 = 0
5
y = 2x − 3
−1
−2
y
1
x
−2
−1
2
3
−1
−2
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.2
38. x + 2 y + 6 = 0
− 12 x
y =
8−3
5
= , undefined
6−6
0
43. m =
−3
15
Linear Models and Rates of Change
The line is horizontal.
y
x = 6
4
x−6 = 0
2
− 10
−8
−6
y
x
−2
(6, 8)
8
−4
6
−6
4
(6, 3)
2
8−0
m =
39.
= 2
4−0
y − 0 = 2( x − 0)
8
y = 2x
4
0 = 2x − y
2
x
−2
y
2
8
−2
(4, 8)
6
−2 − ( −2)
0
=
= 0
3−1
2
44. m =
(0, 0)
−4
4
y = −2
x
−2
2
4
6
y + 2 = 0
y
40. m =
7 − ( −2)
9
=
= 3
1 − ( −2)
3
1
y
8
y − ( −2) = 3( x − ( −2))
0 = 3x − y + 4
41. m =
8−0
8
= −
2−5
3
8
( x − 5)
3
8
40
y = − x+
3
3
8 x + 3 y − 40 = 0
y −0 = −
42. m =
2
3
4
6
4
−3
x
−6 −4
2
(−2, −2)
4
(1, −2)
(3, − 2)
−4
6
−4
7 3
11
−
11
2
4
45. m =
= 4 =
1
1
2
−0
2
2
y
9
8
7
6
5
4
3
2
1
−1
(2, 8)
(5, 0)
x
6 7 8 9
−2
7
6
5
y − 2 = −x + 1
(1, 2)
2
1
x
−4 −3 −2 −1
( 12 , 72 )
2
1
2
3
( 0, 34 )
1
x
−4 −3 −2 −1
1
2
3
4
y
1
−8 ⎛
5⎞
=
y +
⎜x − ⎟
4
3⎝
4⎠
3
x + y −3 = 0
3
11
y −
= ( x − 0)
4
2
11
3
y =
x +
2
4
0 = 22 x − 4 y + 3
⎛ 3⎞ ⎛ 1⎞
⎜ ⎟ − ⎜− ⎟
1
8
4
4
46. m = ⎝ ⎠ ⎝ ⎠ =
= −
3
⎛7⎞ ⎛5⎞
3
−
⎜ ⎟−⎜ ⎟
8
⎝8⎠ ⎝ 4⎠
y
(−3, 6)
y
4
3
1 2 3 4
6−2
4
=
= −1
−3 − 1
−4
y − 2 = −1( x − 1)
1
−1
y + 2 = 3( x + 2)
y = 3x + 4
x
−1
(1, 7)
3
2
12 y + 3 = −32 x + 40
32 x + 12 y − 37 = 0
1
−2
( 78 , 34 )
x
−1
1
−1
( 54 , − 14 )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16
Chapter 1
Preparation for Calculus
x = 3
47.
52.
x −3 = 0
y
2
1
(3, 0)
1
2
x
4
x
y
+
=
a
a
−3 4
+
=
a
a
1
=
a
a =
1
1
1
1⇒ x + y =1
x + y −1 = 0
−1
−2
53.
48. m = −
b
a
−b
x +b
a
y =
b
x+ y = b
a
x
y
+
=1
a
b
x
y
+
2a
a
9
−2
+
2a
a
9− 4
2a
5
x
+
=1
=1
=1
= 2a
a =
5
2
y
=1
() ()
2
y
5
2
5
2
2y
x
+
=1
5
5
x + 2y = 5
(0, b)
x + 2y − 5 = 0
(a, 0)
x
y
+
=1
a
−a
x
54.
x
y
49.
+
=1
2
3
3x + 2 y − 6 = 0
50.
x
y
+
=1
2
2
−
−
3
y
− 3x
−
=1
2
2
3 x + y = −2
3x + y + 2 = 0
51.
x
y
+
=
a
a
1
2
+
=
a
a
3
=
a
a =
1
(− 23 ) + (− 2)
−a
a
−
x
+
=1
2
+ 2 = a
3
4
a =
3
y
( 43 ) (− 43 )
=1
4
3
3x − 3 y − 4 = 0
x − y =
55. The given line is vertical.
(a) x = −7, or x + 7 = 0
1
(b) y = −2, or y + 2 = 0
1
56. The given line is horizontal.
3 ⇒ x+ y = 3
x + y −3 = 0
(a) y = 0
(b) x = −1, or x + 1 = 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.2
57. x − y = − 2
y =
m =1
y − 5 = 1( x − 2)
y −5 = x − 2
(a)
7
8
=
5
3
( x − 34 )
y −
(
= − 53 x −
7
8
3
4
)
24 x + 40 y − 53 = 0
y = −x + 7
62. 3x + 4 y = 7
m = −1
(a)
y − 2 = −1( x + 3)
4 y = −3 x + 7
y = − 34 x +
y − 2 = −x − 3
(b) y − 2 = 1( x + 3)
y −2 = x +3
0 = x − y +5
7
4
m = − 34
x + y +1 = 0
(a)
y − ( −5) = − 34 ( x − 4)
y + 5 = − 34 x + 3
4 y + 20 = −3 x + 12
3x + 4 y + 8 = 0
59. 4 x − 2 y = 3
y = 2x −
3
2
m = 2
(a) y − 1 = 2( x − 2)
y − 1 = 2x − 4
0 = 2x − y − 3
y − 1 = − 12 ( x − 2)
2 y − 2 = −x + 2
x + 2y − 4 = 0
60. 7 x + 4 y = 8
4 y = −7x + 8
−7
x + 2
4
7
m = −
4
y =
1
−7⎛
5⎞
=
y +
⎜x − ⎟
2
4 ⎝
6⎠
−7
1
35
=
x +
2
4
24
24 y + 12 = − 42 x + 35
y +
42 x + 24 y − 23 = 0
(b)
y −
40 y − 35 = −24 x + 18
58. x + y = 7
(a)
5
3
(b)
x + y −7 = 0
(b)
m =
0 = 40 x − 24 y − 9
y − 5 = −1( x − 2)
y − 5 = −x + 2
5x
3
24 y − 21 = 40 x − 30
x − y + 3 = 0
(b)
17
61. 5 x − 3 y = 0
y = x + 2
(a)
Linear Models and Rates of Change
1
4⎛
5⎞
y +
= ⎜x − ⎟
2
7⎝
6⎠
42 y + 21 = 24 x − 20
(b) y − ( −5) =
y +5 =
4
3
(x
4x
3
− 4)
−
16
3
3 y + 15 = 4 x − 16
0 = 4 x − 3 y − 31
63. The slope is 250.
V = 1850 when t = 2.
V = 250(t − 2) + 1850
= 250t + 1350
64. The slope is 4.50.
V = 156 when t = 2.
V = 4.5(t − 2) + 156
= 4.5t + 147
65. The slope is −1600.
V = 17,200 when t = 2.
V = −1600(t − 2) + 17,200
= −1600t + 20,400
66. The slope is − 5600.
V = 245,000 when t = 2.
V = − 5600(t − 2) + 245,000
= − 5600t + 256,200
24 x − 42 y − 41 = 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18
Chapter 1
Preparation for Calculus
67. m1 =
1−0
= −1
−2 − ( −1)
m2 =
−2 − 0
2
= −
2 − ( −1)
3
71. Equations of altitudes:
a −b
y =
( x + a)
c
x = b
a +b
y = −
( x − a)
c
Solving simultaneously, the point of intersection is
⎛ a 2 − b2 ⎞
⎜ b,
⎟.
c
⎝
⎠
m1 ≠ m2
The points are not collinear.
10
−6 − 4
= −
7−0
7
11 − 4
7
m2 =
= −
−5 − 0
5
m1 ≠ m2
68. m1 =
y
(b, c)
The points are not collinear.
(a, 0)
69. Equations of perpendicular bisectors:
x
(− a, 0)
c
a − b⎛
a + b⎞
y −
=
⎜x −
⎟
2
2 ⎠
c ⎝
y −
⎛b c⎞
72. The slope of the line segment from ⎜ , ⎟ to
⎝ 3 3⎠
c
a + b⎛
b − a⎞
=
⎜x −
⎟
2
−c ⎝
2 ⎠
Setting the right-hand sides of the two equations equal
and solving for x yields x = 0.
Letting x = 0 in either equation gives the point of
intersection:
⎛ −a 2 + b2 + c 2 ⎞
⎜ 0,
⎟.
2c
⎝
⎠
3
3a 2 − 3b 2 − c 2
2bc
⎡( − a 2 + b 2 + c 2 ) ( 2c)⎤ − (c 3)
⎦
m2 = ⎣
0 − (b 3)
( a +2 b , 2c )
x
(−a, 0)
(2b)
=
⎛ −a 2 + b2 + c 2 ⎞
⎜ 0,
⎟ is:
2c
⎝
⎠
(b, c)
)
(3a 2 − 3b2 − c 2 ) (3c)
⎛b c⎞
The slope of the line segment from ⎜ , ⎟ to
⎝ 3 3⎠
y
(
⎡( a 2 − b 2 ) c⎤ − (c 3)
⎦
m1 = ⎣
b − (b 3)
=
This point lies on the third perpendicular bisector,
x = 0.
b − a, c
2
2
⎛ a2 − b2 ⎞
⎜ b,
⎟ is:
c
⎝
⎠
=
(a, 0)
(−3a 2 + 3b 2 + 3c 2 − 2c 2 ) (6c) = 3a 2 − 3b2 − c 2
−b 3
2bc
m1 = m2
70. Equations of medians:
y
Therefore, the points are collinear.
c
y = x
b
c
( x + a)
3a + b
c
y =
( x − a)
−3a + b
y =
( b −2 a , 2c )
(b, c)
( a +2 b , 2c )
x
(− a, 0)
(0, 0) (a, 0)
⎛b c⎞
Solving simultaneously, the point of intersection is ⎜ , ⎟.
⎝ 3 3⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.2
73. ax + by = 4
Linear Models and Rates of Change
74. (a) Lines c, d, e and f have positive slopes.
(a) The line is parallel to the x-axis if a = 0 and
b ≠ 0.
(c) Lines c and e appear parallel.
(b) The line is parallel to the y-axis if b = 0 and
a ≠ 0.
(d) Lines b and f appear perpendicular.
(b) Lines a and b have negative slopes.
Lines d and f appear parallel.
(c) Answers will vary. Sample answer: a = −5 and
b = 8.
Lines b and d appear perpendicular.
75. Find the equation of the line through the points (0, 32)
and (100, 212).
−5 x + 8 y = 4
y =
1
8
(5 x
+ 4) =
5x
8
+
1
2
m =
F − 32 =
(d) The slope must be − 52 .
F =
Answers will vary. Sample answer: a = 5 and
b = 2.
y =
5x
2
5
2
1
2
180 = 9
100
5
9 C − 0
5
9 C + 32
5
(
)
or
5x + 2 y = 4
(e) a =
19
C =
(−5 x
+ 4) =
− 52 x
+ 2
(5 F
− 160)
5F − 9C − 160 = 0
For F = 72°, C ≈ 22.2°.
and b = 3.
76. C = 0.51x + 200
+ 3y = 4
For x = 137, C = 0.51(137) + 200 = $269.87.
5x + 6 y = 8
77. (a) Current job:
1
9
W1 = 0.07 s + 2000
New job offer: W2 = 0.05s + 2300
(b)
3500
(15,000, 3050)
0
1500
20,000
Using a graphing utility, the point of intersection is (15,000, 3050).
Analytically, W1 = W2
0.07 s + 2000 = 0.05s + 2300
0.02 s = 300
s = 15,000
So, W1 = W2 = 0.07(15,000) + 2000 = 3050.
When sales exceed $15,000, the current job pays more.
(c) No, if you can sell $20,000 worth of goods, then W1 > W2 .
(Note: W1 = 3400 and W2 = 3300 when s = 20,000.)
78. (a) Depreciation per year:
875
5
1000
= $175
y = 875 − 175 x
where 0 ≤ x ≤ 5.
0
6
0
(b) y = 875 − 175( 2) = $525
(c)
200 = 875 − 175 x
175 x = 675
x ≈ 3.86 years
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20
Chapter 1
Preparation for Calculus
79. (a) Two points are (50, 780) and (47, 825).
The slope is
825 − 780
45
m =
=
= −15.
47 − 50
−3
82. The tangent line is perpendicular to the line joining the
point ( 4, − 3) and the center of the circle, (1, 1).
y
p − 780 = −15( x − 50)
4
p = −15 x + 750 + 780 = −15 x + 1530
or
−6
x =
(b)
2
1
(1530 − p)
15
−2
(1, 1)
x
2
−2
4
(4, −3)
−6
50
Slope of the line joining (1, 1) and ( 4, − 3) is
1+3
−4
=
.
1− 4
3
0
1600
Tangent line:
0
If p = 855, then x = 45 units.
1
(c) If p = 795, then x =
(1530 − 795) = 49 units
15
3
( x − 4)
4
3
y = x−6
4
0 = 3 x − 4 y − 24
y +3 =
80. (a) y = 18.91 + 3.97 x
(x
(b)
= quiz score, y = test score)
1( −2) + (−1)(1) − 2
83. x − y − 2 = 0 ⇒ d =
100
12 + 12
5
5 2
=
2
2
=
0
20
4( 2) + 3(3) − 10
0
(c) If x = 17, y = 18.91 + 3.97(17) = 86.4.
(d) The slope shows the average increase in exam score
for each unit increase in quiz score.
(e) The points would shift vertically upward 4 units.
The new regression line would have a y-intercept
4 greater than before: y = 22.91 + 3.97 x.
84. 4 x + 3 y − 10 = 0 ⇒ d =
7
5
d =
1(0) + 1(1) − 5
=
12 + 12
1−5
2
=
4
= 2 2.
2
86. A point on the line 3x − 4 y = 1 is ( −1, −1). The
distance from the point ( −1, −1) to 3x − 4 y − 10 = 0 is
(5, 12)
8
4
−8 −4
=
85. A point on the line x + y = 1 is (0, 1). The distance
from the point (0, 1) to x + y − 5 = 0 is
81. The tangent line is perpendicular to the line joining the
point (5, 12) and the center (0, 0).
y
42 + 32
(0, 0) 8
x
d =
16
−8
3( −1) − 4( −1) − 10
32 + ( −4)
2
=
−3 + 4 − 10
9
= .
5
5
− 16
Slope of the line joining (5, 12) and (0, 0) is
12
.
5
The equation of the tangent line is
−5
y − 12 =
( x − 5)
12
−5
169
y =
x +
12
12
5 x + 12 y − 169 = 0.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.2
Linear Models and Rates of Change
21
87. If A = 0, then By + C = 0 is the horizontal line y = −C B. The distance to ( x1 , y1 ) is
By1 + C
Ax1 + By1 + C
⎛ −C ⎞
.
d = y1 − ⎜
=
⎟ =
B
⎝ B ⎠
A2 + B 2
If B = 0, then Ax + C = 0 is the vertical line x = −C A. The distance to ( x1 , y1 ) is
Ax1 + C
Ax1 + By1 + C
⎛ −C ⎞
.
d = x1 − ⎜
=
⎟ =
A
A
⎝
⎠
A2 + B 2
(Note that A and B cannot both be zero.) The slope of the line Ax + By + C = 0 is − A B.
The equation of the line through ( x1 , y1 ) perpendicular to Ax + By + C = 0 is:
B
( x − x1 )
A
Ay − Ay1 = Bx − Bx1
y − y1 =
Bx1 − Ay1 = Bx − Ay
The point of intersection of these two lines is:
(1)
( 2)
Ax + By = −C
⇒ A2 x + ABy = − AC
Bx − Ay = Bx1 − Ay1 ⇒ B 2 x − ABy = B 2 x1 − ABy1
( A2 + B 2 ) x
x =
Ax + By = −C
(By adding equations (1) and (2))
= − AC + B 2 x1 − ABy1
− AC + B x1 − ABy1
A2 + B 2
2
(3)
( 4)
⇒
ABx + B 2 y = − BC
Bx − Ay = Bx1 − Ay1 ⇒ − ABx + A2 y = − ABx1 + A2 y1
( A2 + B 2 ) y
= − BC − ABx1 + A2 y1 ( By adding equations (3) and ( 4))
y =
− BC − ABx1 + A2 y1
A2 + B 2
⎛ − AC + B 2 x1 − ABy1 − BC − ABx1 + A2 y1 ⎞
,
⎜
⎟ point of intersection
A2 + B 2
A2 + B 2
⎝
⎠
The distance between ( x1 , y1 ) and this point gives you the distance between ( x1 , y1 ) and the line Ax + By + C = 0.
2
d =
⎡ − AC + B 2 x1 − ABy1
⎤
⎡ − BC − ABx1 + A2 y1
⎤
− x1 ⎥ + ⎢
− y1 ⎥
⎢
2
2
2
2
A
B
A
B
+
+
⎣
⎦
⎣
⎦
2
=
⎡ − AC − ABy1 − A2 x1 ⎤
⎡ − BC − ABx1 − B 2 y1 ⎤
⎢
⎥ + ⎢
⎥
2
2
A + B
A2 + B 2
⎣
⎦
⎣
⎦
=
⎡− A(C + By1 + Ax1 ) ⎤
⎡− B(C + Ax1 + By1 ) ⎤
⎢
⎥ + ⎢
⎥
2
2
A + B
A2 + B 2
⎣
⎦
⎣
⎦
2
2
2
2
=
( A2 + B 2 )(C + Ax1 + By1 )2
2
( A2 + B 2 )
=
Ax1 + By1 + C
A2 + B 2
88. y = mx + 4 ⇒ mx + (−1) y + 4 = 0
d =
Ax1 + By1 + C
A + B
2
2
=
m3 + ( −1)(1) + 4
m + ( −1)
2
2
=
3m + 3
m2 + 1
The distance is 0 when m = −1. In this case, the line y = − x + 4 contains the point (3, 1).
8
−9
(−1, 0)
9
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
22
Chapter 1
Preparation for Calculus
89. For simplicity, let the vertices of the rhombus be (0, 0),
(a, 0), (b, c), and ( a + b, c), as shown in the figure.
The slopes of the diagonals are then m1 =
m2 =
c
and
a +b
c
. Because the sides of the rhombus are
b−a
91. Consider the figure below in which the four points are
collinear. Because the triangles are similar, the result
immediately follows.
y2* − y1*
y − y1
= 2
x2* − x1*
x2 − x1
y
equal, a = b + c , and you have
2
2
2
c
c
c2
c2
m1m2 =
⋅
= 2
=
= −1.
a +b b−a
b − a2
−c 2
Therefore, the diagonals are perpendicular.
(x 2 , y2 )
(x *2 , y*2 )
(x1, y1 )
(x *1, y*1 )
y
x
(b, c)
92. If m1 = −1 m2 , then m1m2 = −1. Let L3 be a line with
(a + b, c)
slope m3 that is perpendicular to L1. Then m1m3 = −1.
x
(0, 0)
So, m2 = m3 ⇒ L 2 and L 3 are parallel. Therefore,
(a , 0)
L 2 and L1 are also perpendicular.
90. For simplicity, let the vertices of the quadrilateral be
(0, 0), (a, 0), (b, c), and (d, e), as shown in the figure.
The midpoints of the sides are
⎛a
⎜ ,
⎝2
⎞ ⎛ a + b c ⎞ ⎛b + d c + e⎞
⎛d e⎞
0 ⎟, ⎜
, ⎟, ⎜
,
⎟, and ⎜ , ⎟.
2⎠ ⎝ 2
2 ⎠
⎠ ⎝ 2
⎝ 2 2⎠
The slope of the opposite sides are equal:
c
c +e
−0
−
2
2
=
a +b a
b+ d
−
−
2
2
2
e
c
c+e
0−
−
2 =
2
2
a
d
a +b b+
−
−
2
2
2
2
e
2 = c
d
b
2
93. True.
a
c
a
ax + by = c1 ⇒ y = − x + 1 ⇒ m1 = −
b
b
b
b
c
b
bx − ay = c2 ⇒ y = x − 2 ⇒ m2 =
a
a
a
1
m2 = −
m1
94. False; if m1 is positive, then m2 = −1 m1 is negative.
95. True. The slope must be positive.
d
= −
e
a −d
96. True. The general form Ax + By + C = 0 includes both
horizontal and vertical lines.
Therefore, the figure is a parallelogram.
y
(d, e)
( b +2 d ,
c+e
2
)
(b, c)
( d2 , 2e )
(a +2 b , 2c )
x
(0, 0)
( a2 , 0)
(a, 0)
Section 1.3 Functions and Their Graphs
1. (a) f (0) = 7(0) − 4 = −4
2. (a) f ( −4) =
−4 + 5 =
(b) f ( −3) = 7( −3) − 4 = −25
(b) f (11) =
11 + 5 =
(c) f (b) = 7(b) − 4 = 7b − 4
(c) f ( 4) =
4+ 5 =
(d) f ( x − 1) = 7( x − 1) − 4 = 7 x − 11
(d) f ( x + ∆ x) =
1 =1
16 = 4
9 = 3
x + ∆x + 5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.3
( 5) = 5 − ( 5)
2
⎛ ⎛ π ⎞⎞
⎛ π⎞
⎛ π⎞
(b) f ⎜ − ⎟ = cos⎜ 2⎜ − ⎟ ⎟ = cos⎜ − ⎟ = 0
⎝ 4⎠
⎝ 2⎠
⎝ ⎝ 4 ⎠⎠
= 5−5 = 0
(c) g ( −2) = 5 − ( −2) = 5 − 4 = 1
2
⎛ ⎛ π ⎞⎞
2π
1
⎛π ⎞
= −
(c) f ⎜ ⎟ = cos⎜ 2⎜ ⎟ ⎟ = cos
3
3
3
2
⎝ ⎠
⎝ ⎝ ⎠⎠
(d) g (t − 1) = 5 − (t − 1) = 5 − (t 2 − 2t + 1)
2
(d) f (π ) = cos( 2(π )) = 1
= 4 + 2t − t 2
4. (a) g ( 4) = 42 ( 4 − 4) = 0
(b) g
( )=( )(
3
2
3
2
2
3
2
)
−4 =
9
4
( )=
− 52
(c) g (c) = c (c − 4) = c − 4c
2
3
6. (a) f (π ) = sin π = 0
− 45
8
− 2
⎛ 5π ⎞
⎛ 5π ⎞
(b) f ⎜ ⎟ = sin ⎜ ⎟ =
4
4
2
⎝ ⎠
⎝ ⎠
2
⎛ 2π ⎞
⎛ 2π ⎞
(c) f ⎜ ⎟ = sin ⎜ ⎟ =
3
⎝ ⎠
⎝ 3 ⎠
(d) g (t + 4) = (t + 4) (t + 4 − 4)
2
= (t + 4) t = t 3 + 8t 2 + 16t
2
7.
8.
9.
f ( x + ∆x ) − f ( x )
∆x
f ( x ) − f (1)
x −1
f ( x ) − f ( 2)
x − 2
=
=
=
10.
f ( x ) − f (1)
x −1
=
(x
+ ∆x ) − x 3
3
∆x
3 x − 1 − (3 − 1)
x −1
(1
=
1
⎛ π⎞
⎛ π⎞
(d) f ⎜ − ⎟ = sin ⎜ − ⎟ = −
2
⎝ 6⎠
⎝ 6⎠
x 3 + 3 x 2 ∆x + 3 x 2 ( ∆ x ) + ( ∆ x ) − x 3
2
=
3
= 3 x 2 + 3x∆x + ( ∆x) , ∆x ≠ 0
2
∆x
3( x − 1)
x −1
3
2
= 3, x ≠ 1
)
x −1 −1
x − 2
1− x −1
1+
⋅
− 2) x − 1 1 +
(x
x −1
=
x −1
( x − 2)
2− x
(
x −11+
x −1
)
=
(
−1
x −11 +
x −1
)
, x ≠ 2
x( x + 1)( x − 1)
x3 − x − 0
=
= x( x + 1), x ≠ 1
x −1
x −1
11. f ( x ) = 4 x 2
Domain: ( −∞, ∞)
Range: [0, ∞)
12. g ( x) = x 2 − 5
Domain: ( −∞, ∞)
Range: [−5, ∞)
13. f ( x) = x3
Domain: ( −∞, ∞)
Range: ( −∞, ∞)
14. h( x) = 4 − x 2
Domain: ( −∞, ∞ )
Range: ( −∞, 4]
=
23
5. (a) f (0) = cos( 2(0)) = cos 0 = 1
3. (a) g (0) = 5 − 02 = 5
(b) g
Functions and Their Graphs
15. g ( x) =
6x
Domain: 6 x ≥ 0
x ≥ 0 ⇒ [0, ∞ )
Range: [0, ∞)
16. h( x) = −
x +3
Domain: x + 3 ≥ 0 ⇒ [−3, ∞ )
Range: ( −∞, 0]
17. f ( x ) =
16 − x 2
16 − x 2 ≥ 0 ⇒ x 2 ≤ 16
Domain: [− 4, 4]
Range: [0, 4]
Note: y =
16 − x 2 is a semicircle of radius 4.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
24
Chapter 1
Preparation for Calculus
18. f ( x ) = x − 3
Domain: ( −∞, ∞ )
Range: [0, ∞)
19. f (t ) = sec
πt
4
≠
( 2n
26. h( x) =
1
sin x − (1 2)
1
≠ 0
2
1
sin x ≠
2
sin x −
πt
4
+ 1)π
⇒ t ≠ 4n + 2
2
Domain: all t ≠ 4n + 2, n an integer
Range: ( −∞, − 1] ∪ [1, ∞)
Domain: all x ≠
27. f ( x) =
π
6
5π
+ 2nπ , n integer
6
+ 2nπ ,
1
x+3
x+3 ≠ 0
20. h(t ) = cot t
x +3 ≠ 0
Domain: all t = nπ , n an integer
Domain: all x ≠ −3
Range: ( −∞, ∞)
Domain: ( −∞, − 3) ∪ ( −3, ∞)
21. f ( x ) =
28. g ( x) =
3
x
Domain: all x ≠ 0 ⇒ ( −∞, 0) ∪ (0, ∞ )
Range: ( −∞, 0) ∪ (0, ∞)
x − 2
x + 4
Domain: all x ≠ − 4
22. f ( x) =
Range: all y ≠ 1
[Note: You can see that the range is all y ≠ 1 by
graphing f.]
23. f ( x ) =
x +
1− x
x ≥ 0 and 1 − x ≥ 0
x ≥ 0 and
x ≤1
Domain: 0 ≤ x ≤ 1 ⇒ [0, 1]
24. f ( x ) =
x − 3x + 2
2
x − 3x + 2 ≥ 0
1
x2 − 4
x2 − 4 ≠ 0
(x
− 2)( x + 2) ≠ 0
Domain: all x ≠ ± 2
Domain: ( − ∞, − 2) ∪ ( − 2, 2) ∪ ( 2, ∞)
⎧2 x + 1, x < 0
29. f ( x ) = ⎨
⎩2 x + 2, x ≥ 0
(a) f ( −1) = 2( −1) + 1 = −1
(b) f (0) = 2(0) + 2 = 2
(c) f ( 2) = 2( 2) + 2 = 6
(d) f (t 2 + 1) = 2(t 2 + 1) + 2 = 2t 2 + 4
(Note: t 2 + 1 ≥ 0 for all t.)
Domain: ( −∞, ∞)
Range: ( −∞, 1) ∪ [2, ∞)
2
(x
− 2)( x − 1) ≥ 0
Domain: x ≥ 2 or x ≤ 1
Domain: ( −∞, 1] ∪ [2, ∞)
2
25. g ( x) =
1 − cos x
1 − cos x ≠ 0
cos x ≠ 1
Domain: all x ≠ 2nπ , n an integer
2
⎪⎧ x + 2, x ≤ 1
30. f ( x) = ⎨ 2
⎪⎩2 x + 2, x > 1
(a) f ( −2) = ( −2) + 2 = 6
2
(b) f (0) = 02 + 2 = 2
(c) f (1) = 12 + 2 = 3
(d) f ( s 2 + 2) = 2( s 2 + 2) + 2 = 2 s 4 + 8s 2 + 10
2
(Note: s 2 + 2 > 1 for all s.)
Domain: ( −∞, ∞)
Range: [2, ∞)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.3
36. f ( x ) =
⎪⎧ x + 1, x < 1
31. f ( x ) = ⎨
⎪⎩− x + 1, x ≥ 1
y
5
(b) f (1) = −1 + 1 = 0
4
(c) f (3) = −3 + 1 = −2
2
1
(d) f (b + 1) = − (b + 1) + 1 = − b
2
2
−3
Domain: ( − ∞, ∞ )
37. f ( x ) =
−3 + 4 =
0+ 4 = 2
(c) f (5) =
5+ 4 = 3
1
2
3
Range: ( −∞, ∞)
⎧⎪ x + 4, x ≤ 5
32. f ( x ) = ⎨
2
⎪⎩( x − 5) , x > 5
(b) f (0) =
x
−1
−1
Domain: ( −∞, ∞)
Range: ( − ∞, 0] ∪ [1, ∞)
(a) f ( −3) =
+3
1 x3
4
(a) f ( −3) = − 3 + 1 = 4
2
25
Functions and Their Graphs
y
9 − x2
5
Domain: [−3, 3]
1 =1
4
Range: [0, 3]
2
1
−4 −3 −2 −1
x
1
2
3
4
−2
−3
(d) f (10) = (10 − 5) = 25
2
38. f ( x ) = x +
Domain: [−4, ∞)
4 − x2
Domain: [−2, 2]
Range: [0, ∞)
33. f ( x) = 4 − x
Range: ⎡⎣−2, 2 2 ⎤⎦ ≈ [−2, 2.83]
y
Domain: ( −∞, ∞)
y-intercept: (0, 2)
8
(
6
Range: ( −∞, ∞)
x-intercept: −
4
4
x
−2
2
3
4
(−
4
34. g ( x) =
x
(0, 2)
2, 0(
x
y
−4 −3 −2
1
−1
2
3
4
−2
6
Domain: ( −∞, 0) ∪ (0, ∞)
)
y
2
−4
2, 0
−3
4
−4
2
Range: ( −∞, 0) ∪ (0, ∞)
x
2
4
6
39. g (t ) = 3 sin π t
y
3
2
35. h( x) =
1
y
x −6
t
1
3
3
Domain:
x −6 ≥ 0
x ≥ 6 ⇒ [6, ∞ )
Range: [0, ∞)
2
1
x
3
6
9
12
Domain: ( −∞, ∞)
Range: [−3, 3]
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
26
Chapter 1
Preparation for Calculus
40. h(θ ) = −5 cos
θ
49. y 2 = x 2 − 1 ⇒ y = ±
2
x2 − 1
y is not a function of x because there are two values of y
for some x.
Domain: ( −∞, ∞ )
Range: [−5, 5]
x2
x2 + 4
50. x 2 y − x 2 + 4 y = 0 ⇒ y =
y
5
4
3
2
1
y is a function of x because there is one value of y for
each x.
−2π
51. The transformation is a horizontal shift two units to the
right.
θ
2π
Shifted function: y =
x − 2
−5
2−0
1
=
mi min during the first
4−0
2
4 minutes. The student is stationary for the next
2 minutes. Finally, the student travels
6−2
= 1 mi min during the final 4 minutes.
10 − 6
41. The student travels
42.
d
52. The transformation is a vertical shift 4 units upward.
Shifted function: y = sin x + 4
53. The transformation is a horizontal shift 2 units to the
right and a vertical shift 1 unit downward.
Shifted function: y = ( x − 2) − 1
2
54. The transformation is a horizontal shift 1 unit to the left
and a vertical shift 2 units upward.
27
Shifted function: y = ( x + 1) + 2
3
18
55. y = f ( x + 5) is a horizontal shift 5 units to the left.
9
Matches d.
t1
t2
t3
t
43. x − y = 0 ⇒ y = ±
2
56. y = f ( x) − 5 is a vertical shift 5 units downward.
Matches b.
x
y is not a function of x. Some vertical lines intersect the
graph twice.
x2 − 4 − y = 0 ⇒ y =
44.
x2 − 4
y is a function of x. Vertical lines intersect the graph at
most once.
57. y = − f ( − x) − 2 is a reflection in the y-axis, a
reflection in the x-axis, and a vertical shift downward
2 units. Matches c.
58. y = − f ( x − 4) is a horizontal shift 4 units to the right,
followed by a reflection in the x-axis. Matches a.
45. y is a function of x. Vertical lines intersect the graph at
most once.
59. y = f ( x + 6) + 2 is a horizontal shift to the left
46. x + y = 4
60. y = f ( x − 1) + 3 is a horizontal shift to the right 1 unit,
2
2
y = ±
4 − x2
6 units, and a vertical shift upward 2 units. Matches e.
and a vertical shift upward 3 units. Matches g.
y is not a function of x. Some vertical lines intersect the
graph twice.
47. x 2 + y 2 = 16 ⇒ y = ± 16 − x 2
y is not a function of x because there are two values of y
for some x.
48. x 2 + y = 16 ⇒ y = 16 − x 2
y is a function of x because there is one value of y for
each x.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.3
61. (a) The graph is shifted 3 units to the left.
Functions and Their Graphs
27
(f) The graph is stretched vertically by a factor of 14 .
y
y
4
4
2
−6
−4
x
−2
2
4
−4
−2
x
−2
2
4
6
−4
−6
−6
(b) The graph is shifted 1 unit to the right.
(g) The graph is a reflection in the x-axis.
y
y
4
2
2
x
−2
2
4
6
8
−4
−2
x
−2
2
4
6
−2
−4
−4
−6
(h) The graph is a reflection about the origin.
y
(c) The graph is shifted 2 units upward.
6
y
4
6
4
−6
2
−4
x
−4
2
4
−2
x
−2
2
4
6
−4
−2
(d) The graph is shifted 4 units downward.
y
−4
x
−2
2
4
6
−2
−4
−6
−8
(e) The graph is stretched vertically by a factor of 3.
y
−4
x
−2
4
6
−2
−4
−6
−8
−10
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
28
Chapter 1
Preparation for Calculus
62. (a) g ( x) = f ( x − 4)
(e)
g ( 6) = f ( 2) = 1
g ( x) = 2 f ( x)
g ( 2) = 2 f ( 2) = 2
g ( 0 ) = f ( −4 ) = − 3
g ( −4) = 2 f ( −4) = −6
The graph is shifted 4 units to the right.
The graph is stretched vertically by a factor of 2.
y
y
(2, 2)
2
4
1
3
2
x
−1
1
2
3
5
x
−5 −4 −3 − 2 − 1
(6, 1)
1
6
1
2
3
1
2
−3
7
−4
−2
−5
(0, −3)
−4
(−4, −6)
g ( x) =
1
2
f ( x)
g ( 2) =
f ( 2) =
g ( − 6) = f ( − 4) = − 3
1
2
g ( −4 ) =
1
2
f ( −4) = − 32
The graph is shifted 2 units to the left.
The graph is stretched vertically by a factor of 12 .
g ( x ) = f ( x + 2)
(b)
(f )
g ( 0) = f ( 2) = 1
y
y
4
3
2
2
(2, 12 )
1
(0, 1)
x
5
x
−7 −6 −5 −4 −3
−1
1
3
2
3
−2
−3
−3
(−6, −3)
1
−1
( −4, − 32 )
−2
−4
−4
−5
−6
g ( x) = f ( x) + 4
(c)
4
g ( 2) = f ( 2) + 4 = 5
(g)
g ( x) = f ( − x)
g ( − 2) = f ( 2) = 1
g ( − 4) = f ( − 4) + 4 = 1
g ( 4) = f ( − 4) = − 3
The graph is shifted 4 units upward.
The graph is a reflection in the y-axis.
y
y
6
(2, 5)
5
3
4
(−2, 1) 2
1
2
−3 −2 −1
−1
1
(−4, 1)
1
2
−3
3
g ( 2) = f ( 2) − 1 = 0
g ( − 4) = f ( − 4) − 1 = − 4
The graph is shifted 1 unit downward.
(h)
g ( 2) = f ( 2) = − 1
g ( − 4) = f ( − 4) = 3
The graph is a reflection in the x-axis.
y
y
5
(2, 0)
x
4
3
2
1
5
g ( x) = − f ( x)
2
5
4
−5
g ( x) = f ( x) − 1
1
3
(4, −3)
−4
−2
(d)
2
−2
x
−5 −4 −3 −2 −1
x
1
2
3
(−4, 3)
4
3
2
−3
(− 4, −4)
−4
1
−5
−5 −4 −3 −2 −1
−1
−6
−2
x
3
(2, −1)
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.3
Functions and Their Graphs
29
63. f ( x) = 3x − 4, g ( x) = 4
(a) f ( x) + g ( x) = (3x − 4) + 4 = 3 x
(b) f ( x) − g ( x) = (3 x − 4) − 4 = 3 x − 8
(c) f ( x) ⋅ g ( x) = (3 x − 4)( 4) = 12 x − 16
(d) f ( x ) g ( x) =
3x − 4
3
= x −1
4
4
64. f ( x ) = x 2 + 5 x + 4, g ( x) = x + 1
(a) f ( x ) + g ( x) = ( x 2 + 5 x + 4) + ( x + 1) = x 2 + 6 x + 5
(b) f ( x ) − g ( x) = ( x 2 + 5 x + 4) − ( x + 1) = x 2 + 4 x + 3
(c) f ( x ) ⋅ g ( x) = ( x 2 + 5 x + 4)( x + 1)
= x3 + 5 x 2 + 4 x + x 2 + 5 x + 4
= x3 + 6 x 2 + 9 x + 4
(d) f ( x) g ( x) =
x2 + 5x + 4
( x + 4)( x + 1) = x + 4, x ≠ −1
=
x +1
x +1
65. (a) f ( g (1)) = f (0) = 0
67. f ( x ) = x 2 , g ( x) =
(b) g ( f (1)) = g (1) = 0
(f
D g )( x ) = f ( g ( x ))
(c) g ( f (0)) = g (0) = −1
(d) f ( g ( −4)) = f (15) =
= f
(f) g ( f ( x)) = g
( x) = ( x)
( x) = ( x)
2
= x, x ≥ 0
Domain: [0, ∞)
15
(e) f ( g ( x)) = f ( x 2 − 1) =
x
D f )( x) = g ( f ( x)) = g ( x 2 ) =
x2 − 1
(g
2
Domain: ( −∞, ∞)
− 1 = x − 1, ( x ≥ 0)
66. f ( x) = sin x, g ( x) = π x
(a) f ( g ( 2)) = f ( 2π ) = sin ( 2π ) = 0
⎛ ⎛ 1 ⎞⎞
⎛π ⎞
⎛π ⎞
(b) f ⎜ g ⎜ ⎟ ⎟ = f ⎜ ⎟ = sin ⎜ ⎟ = 1
2
2
⎝
⎠
⎝
⎠
⎝2⎠
⎝
⎠
(c) g ( f (0)) = g (0) = 0
⎛ ⎛ π ⎞⎞
⎛ ⎛ π ⎞⎞
(d) g ⎜ f ⎜ ⎟ ⎟ = g ⎜ sin ⎜ ⎟ ⎟
⎝ ⎝ 4 ⎠⎠
⎝ ⎝ 4 ⎠⎠
⎛ 2⎞
⎛ 2⎞
π 2
= g ⎜⎜
⎟⎟ = π ⎜⎜
⎟⎟ =
2
2
2
⎝
⎠
⎝
⎠
x2 = x
No. Their domains are different. ( f D g ) = ( g D f )
for x ≥ 0.
68. f ( x ) = x 2 − 1, g ( x) = cos x
(f
D g )( x ) = f ( g ( x)) = f (cos x ) = cos 2 x − 1
Domain: ( −∞, ∞)
(g
D f )( x ) = g ( x 2 − 1) = cos( x 2 − 1)
Domain: ( −∞, ∞)
No, f D g ≠ g D f .
(e) f ( g ( x)) = f (π x) = sin (π x)
(f) g ( f ( x)) = g (sin x ) = π sin x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
30
Chapter 1
69. f ( x ) =
(f
Preparation for Calculus
3
, g ( x) = x 2 − 1
x
70.
D g )( x) = f ( g ( x)) = f ( x 2 − 1) =
(g
D f )( x ) = g ( f ( x))
(
)
x+ 2 =
1
x+ 2
⎛1⎞
D f )( x) = g ⎜ ⎟ =
⎝ x⎠
1
+ 2 =
x
1 + 2x
x
You can find the domain of g D f by determining the
2
9
9− x
⎛ 3⎞ ⎛ 3⎞
= g⎜ ⎟ = ⎜ ⎟ − 1 = 2 − 1 =
x
x
x
x2
⎝ ⎠ ⎝ ⎠
intervals where (1 + 2x) and x are both positive, or both
2
negative.
Domain: all x ≠ 0 ⇒ ( −∞, 0) ∪ (0, ∞)
+
−2
No, f D g ≠ g D f .
71. (a)
D g )( x ) = f
Domain: ( −2, ∞)
3
x2 − 1
Domain: all x ≠ ±1 ⇒ (−∞, −1) ∪ ( −1, 1) ∪ (1, ∞ )
(g
(f
+
+ − − +
−1 − 1
2
0
+
+
1
+
x
2
(
Domain: −∞, − 12 ⎦⎤, (0, ∞)
D g )(3) = f ( g (3)) = f ( −1) = 4
(f
(b) g ( f ( 2)) = g (1) = −2
(c) g ( f (5)) = g ( −5), which is undefined
72.
(d)
(f
D g )( −3) = f ( g ( −3)) = f (−2) = 3
(e)
(g
D f )( −1) = g ( f ( −1)) = g ( 4) = 2
(f)
f ( g ( −1)) = f ( −4), which is undefined
( A D r )(t )
= A( r (t )) = A(0.6t ) = π (0.6t ) = 0.36π t 2
2
( A D r )(t ) represents the area of the circle at time t.
73. F ( x ) =
2x − 2
Let h( x) = 2 x, g ( x) = x − 2 and f ( x) =
x.
Then, ( f D g D h)( x) = f ( g ( 2 x)) = f (( 2 x) − 2) =
( 2 x) −
2 =
2 x − 2 = F ( x).
[Other answers possible]
74. F ( x) = −4 sin (1 − x)
Let f ( x) = −4 x, g ( x) = sin x and h( x) = 1 − x. Then,
(f
D g D h)( x ) = f ( g (1 − x)) = f (sin (1 − x)) = −4 sin (1 − x) = F ( x).
[Other answers possible]
( 32 , 4) is on the graph.
f is odd, then ( 32 , − 4) is on the graph.
75. (a) If f is even, then
(b) If
76. (a) If f is even, then ( −4, 9) is on the graph.
(b) If f is odd, then ( −4, − 9) is on the graph.
77. f is even because the graph is symmetric about the y-axis. g is neither even nor odd. h is odd because the graph is symmetric
about the origin.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.3
78. (a) If f is even, then the graph is symmetric about the
y-axis.
79. f ( x ) = x 2 ( 4 − x 2 )
2
f
(
f ( − x ) = ( − x) 4 − ( − x)
y
6
f is even.
4
f ( x) = x 2 (4 − x 2 ) = 0
2
2
4
2
) = x (4 − x ) =
2
2
f ( x)
x 2 ( 2 − x)( 2 + x) = 0
x
−6 −4 −2
−2
31
Functions and Their Graphs
6
Zeros: x = 0, − 2, 2
−4
−6
80. f ( x) =
(b) If f is odd, then the graph is symmetric about the
origin.
f ( − x) =
x
( − x)
3
= − 3 x = − f ( x)
f is odd.
y
f ( x) =
6
f
3
3
x = 0 ⇒ x = 0 is the zero.
4
2
x
−6 −4 −2
−2
2
4
81. f ( x ) = x cos x
f ( − x ) = ( − x) cos ( − x) = − x cos x = − f ( x)
6
−4
f is odd.
−6
f ( x ) = x cos x = 0
π
Zeros: x = 0,
2
+ nπ , where n is an integer
82. f ( x) = sin 2 x
f ( − x ) = sin 2 ( − x ) = sin ( − x ) sin ( − x) = ( − sin x)( − sin x) = sin 2 x
f is even.
sin 2 x = 0 ⇒ sin x = 0
Zeros: x = nπ , where n is an integer
4 − ( −6)
10
=
= −5
−2 − 0
−2
83. Slope =
y − 4 = −5( x − ( −2))
y − 4 = −5 x − 10
y = −5 x − 6
For the line segment, you must restrict the domain.
f ( x) = −5 x − 6, −2 ≤ x ≤ 0
y
6
(−2, 4)
8−1
7
=
5−3
2
7
y − 1 = ( x − 3)
2
7
21
y −1 = x −
2
2
7
19
y = x−
2
2
For the line segment, you must restrict the domain.
7
19
f ( x) = x − , 3 ≤ x ≤ 5
2
2
84. Slope =
4
y
2
−6 −4 −2
4
(5, 8)
8
x
2
6
6
−4
−6
4
(0, −6)
2
(3, 1)
x
−2
2
4
6
8
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 1
Preparation for Calculus
85. x + y 2 = 0
89. Answers will vary. Sample answer: In general, as the
price decreases, the store will sell more.
y2 = −x
y = −
−x
f ( x) = −
y
Number of sneakers sold
32
−x, x ≤ 0
y
3
2
1
x
x
−5 −4 −3 −2 −1
Price (in dollars)
1
−2
90. Answers will vary. Sample answer: As time goes on, the
value of the car will decrease
−3
y
86. x 2 + y 2 = 36
y = −
Value
y 2 = 36 − x 2
36 − x 2 , − 6 ≤ x ≤ 6
y
t
8
4
2
−4 −2
−2
x
2
4
y =
91.
c − x2
y 2 = c − x2
−4
x 2 + y 2 = c, a circle.
87. Answers will vary. Sample answer: Speed begins and
ends at 0. The speed might be constant in the middle:
Speed (in miles per hour)
y
For the domain to be [−5, 5], c = 25.
92. For the domain to be the set of all real numbers, you
must require that x 2 + 3cx + 6 ≠ 0. So, the
discriminant must be less than zero:
(3c)2
− 4(6) < 0
9c 2 < 24
c2 <
x
Time (in hours)
−
88. Answers will vary. Sample answer: Height begins a few
feet above 0, and ends at 0.
y
− 23
8
3
8
3
< c <
6 < c <
8
3
2
3
6
93. (a) T ( 4) = 16°, T (15) ≈ 23°
(b) If H (t ) = T (t − 1), then the changes in temperature
Height
will occur 1 hour later.
(c) If H (t ) = T (t ) − 1, then the overall temperature
would be 1 degree lower.
x
Distance
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.3
33
97. f ( x) = x + x − 2
94. (a) For each time t, there corresponds a depth d.
(b) Domain: 0 ≤ t ≤ 5
If x < 0, then f ( x) = − x − ( x − 2) = −2 x + 2.
Range: 0 ≤ d ≤ 30
(c)
Functions and Their Graphs
If 0 ≤ x < 2, then f ( x) = x − ( x − 2) = 2.
d
If x ≥ 2, then f ( x) = x + ( x − 2) = 2 x − 2.
30
25
So,
20
⎧−2 x + 2, x ≤ 0
⎪
0 < x < 2.
f ( x ) = ⎨2,
⎪2 x − 2,
x
≥ 2
⎩
15
10
5
t
1
2
3
4
5
6
98. p1 ( x) = x3 − x + 1 has one zero. p2 ( x) = x3 − x has
(d) d ( 4) ≈ 18. At time 4 seconds, the depth is
three zeros. Every cubic polynomial has at least one
zero. Given p( x) = Ax3 + Bx 2 + Cx + D, you have
approximately 18 cm.
y
Average number of
acres per farm
95. (a)
p → −∞ as x → −∞ and p → ∞ as x → ∞ if
500
400
A > 0. Furthermore, p → ∞ as x → −∞ and
300
p → −∞ as x → ∞ if A < 0. Because the graph has
200
no breaks, the graph must cross the x-axis at least one time.
100
99. f ( − x) = a2 n +1 ( − x)
x
10 20 30 40 50 60
Year (0 ↔ 1960)
+ " + a3 (− x) + a1 ( − x)
3
= − ⎡⎣a2 n +1 x 2 n +1 + " + a3 x 3 + a1x⎤⎦
(b) A( 25) ≈ 445 ( Answers will vary.)
96. (a)
2 n +1
= − f ( x)
Odd
25
100
0
0
2
⎛ x ⎞
⎛ x ⎞
⎛ x ⎞
(b) H ⎜ ⎟ = 0.002⎜ ⎟ + 0.005⎜ ⎟ − 0.029
⎝ 1.6 ⎠
⎝ 1.6 ⎠
⎝ 1.6 ⎠
= 0.00078125 x 2 + 0.003125 x − 0.029
100. f ( − x) = a2 n ( − x)
2n
+ a2 n − 2 ( − x )
2n − 2
+ " + a2 ( − x) + a0
2
= a2 n x 2 n + a2 n − 2 x 2 n − 2 + " + a2 x 2 + a0
= f ( x)
Even
101. Let F ( x) = f ( x) g ( x) where f and g are even. Then F ( − x) = f ( − x) g ( − x) = f ( x) g ( x) = F ( x).
So, F ( x) is even. Let F ( x) = f ( x) g ( x) where f and g are odd. Then
F ( − x) = f ( − x) g ( − x) = ⎡−
⎣ f ( x)⎤⎡−
⎦⎣ g ( x)⎤⎦ = f ( x) g ( x) = F ( x).
So, F ( x) is even.
102. Let F ( x) = f ( x) g ( x) where f is even and g is odd. Then
F ( − x) = f ( − x) g ( − x) = f ( x) ⎡−
⎣ g ( x)⎤⎦ = − f ( x) g ( x) = − F ( x).
So, F ( x) is odd.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
34
Chapter 1
Preparation for Calculus
103. By equating slopes,
y −2
0−2
=
x −3
0−3
6
y −2 =
x −3
6
2x
+ 2 =
,
y =
x −3
x −3
⎛ 2x ⎞
x2 + ⎜
⎟ .
⎝ x − 3⎠
x2 + y2 =
104. (a) V = x( 24 − 2 x)
108. False. If f ( x) = x 2 then, f (3x ) = (3 x) = 9 x 2 and
2
3 f ( x) = 3x 2 . So, 3 f ( x) ≠ f (3x).
with respect to the x-axis.
110. True. If the domain is {a}, then the range is { f ( a )}.
2
Domain: 0 < x < 12
(b)
107. True. The function is even.
109. False. The constant function f ( x ) = 0 has symmetry
2
L =
106. True
111. First consider the portion of R in the first quadrant:
x ≥ 0, 0 ≤ y ≤ 1 and x − y ≤ 1; shown below.
1100
y
12
(0, 1)
−100
(2, 1)
1
Maximum volume occurs at x = 4. So, the
dimensions of the box would be 4 × 16 × 16 cm.
(c)
The area of this region is
1 + 12 = 32 .
2
−1
x
length and width
volume
1
24 − 2(1)
1⎡⎣24 − 2(1)⎤⎦ = 484
2
24 − 2( 2)
2 ⎡⎣24 − 2( 2)⎤⎦ = 800
3
24 − 2(3)
3⎡⎣24 − 2(3)⎤⎦ = 972
4
24 − 2( 4)
4 ⎡⎣24 − 2( 4)⎤⎦ = 1024
5
24 − 2(5)
5⎡⎣24 − 2(5)⎤⎦ = 980
6
24 − 2(6)
6 ⎡⎣24 − 2(6)⎤⎦ = 864
−1
x
(0, 0)
(1, 0) 2
−1
By symmetry, you obtain the entire region R:
2
y
2
2
2
2
( 32 ) = 6.
The area of R is 4
2
(−2, 1)
(2, 1)
x
−2
1
2
(2, −1)
(−2, −1)
−2
2
The dimensions of the box that yield a maximum
volume appear to be 4 × 16 × 16 cm.
105. False. If f ( x ) = x 2 , then f ( −3) = f (3) = 9, but
−3 ≠ 3.
112. Let g ( x) = c be constant polynomial.
Then f ( g ( x)) = f (c) and g ( f ( x)) = c.
So, f (c) = c. Because this is true for all real numbers c,
f is the identity function: f ( x) = x.
Section 1.4 Fitting Models to Data
1. (a) and (b)
2. (a)
y
y
15
1000
14
13
900
12
11
800
10
700
9
8
600
7
x
x
900
1050
1200
1350
Yes, the data appear to be approximately linear.
The data can be modeled by equation
y = 0.6 x + 150. (Answers will vary).
(c) When x = 1075, y = 0.6(1075) + 150 = 795.
7
8
9 10 11 12 13 14 15
The data do not appear to be linear.
(b) Quiz scores are dependent on several variables such
as study time, class attendance, and so on. These
variables may change from one quiz to the next.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.4
3. (a) d = 0.066 F
(b)
35
7. (a) Using graphing utility,
S = 180.89 x 2 − 205.79 x + 272.
10
(b)
d = 0.066F
0
Fitting Models to Data
25,000
110
0
14
0
0
The model fits the data well.
(c) If F = 55, then d ≈ 0.066(55) = 3.63 cm.
(d)
4. (a) s = 9.7t + 0.4
(b)
(c) When x = 2, S ≈ 583.98 pounds.
The breaking strength is approximately 4 times
greater.
45
(e)
−1
5
The model fits the data well.
(c) If t = 2.5, s = 24.65 meters second.
5. (a) Using a graphing utility, y = 0.122 x + 2.07
The correlation coefficient is r ≈ 0.87.
23,860
≈ 4.37
5460
When the height is doubled, the breaking strength
increases approximately by a factor of 4.
−5
(b)
2370
≈ 4.06
584
8. (a) Using a graphing utility
t = 0.0013s 2 + 0.005s + 1.48.
(b)
15
60
25
95
0
0
500
0
(c) Greater per capita energy consumption by a country
tends to correspond to greater per capita gross
national income. The three countries that most
differ from the linear model are Canada, Japan, and
Italy.
(d) Using a graphing utility, the new model
is y = 0.142 x − 1.66.
The correlation coefficient is r ≈ 0.97.
(c) According to the model, the times required to attain
speeds of less than 20 miles per hour are all about
the same. Furthermore, it takes 1.48 seconds to reach
0 miles per hour, which does not make sense.
(d) Adding (0, 0) to the data produces
t = 0.0009 s 2 + 0.053s + 0.10.
(e) Yes. Now the car starts at rest.
9. (a) y = −1.806 x3 + 14.58 x 2 + 16.4 x + 10
(b)
300
6. (a) Trigonometric function
(b) Quadratic function
(c) No relationship
0
(d) Linear function
7
0
(c) If x = 4.5, y ≈ 214 horsepower.
10. (a) T = 2.9856 × 10−4 p 3 − 0.0641 p 2 + 5.282 p + 143.1
(b)
350
0
150
110
(c) For T = 300° F , p ≈ 68.29 lb in.2 .
(d) The model is based on data up to 100 pounds per
square inch.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36
Chapter 1
Preparation for Calculus
11. (a) y1 = − 0.0172t 3 + 0.305t 2 − 0.87t + 7.3
y2 = − 0.038t 2 + 0.45t + 3.5
y3 = 0.0063t 3 + −0.072t 2 + 0.02t + 1.8
(b)
20
y1 + y2 + y3
y1
y2
y3
0
11
0
y1 + y2 + y3 = − 0.0109t 3 + 0.195t 2 − 0.40t + 12.6
For 2014, t = 14. So,
y1 + y2 + y3 = − 0.0109(14) + 0.195(14) − 0.40(14) + 12.6
3
2
≈ 15.31 cents/mile
12. (a) N1 = 1.89t + 46.8
Linear model
N 2 = 0.0485t − 2.015t + 27.00t − 42.3
3
(b)
2
Cubic model
100
N1
N2
0
20
40
(c) The cubic model is the better model.
(d) N 3 = − 0.414t 2 + 11.00t + 4.4
Quadratic model
100
0
20
40
The model does not fit the data well.
(e) For 2014, t = 24 and
N1 ≈ 92.16 million
N 2 ≈ 115.524 million
The linear model seems too high. The cubic model is better.
(f) Answers will vary.
13. (a) Yes, y is a function of t. At each time t, there is one
and only one displacement y.
(c) One model is y = 0.35 sin ( 4π t ) + 2.
(d)
4
(b) The amplitude is approximately
(2.35 − 1.65)
(0.125, 2.35)
2 = 0.35.
(0.375, 1.65)
The period is approximately
2(0.375 − 0.125) = 0.5.
0
0.9
0
The model appears to fit the data.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
14. (a) S (t ) = 56.37 + 25.47 sin (0.5080t − 2.07)
(b)
100
Inverse Functions
37
(d) The average is the constant term in each model.
83.70°F for Miami and 56.37°F for Syracuse.
(e) The period for Miami is 2π 0.4912 ≈ 12.8. The
period for Syracuse is 2π 0.5080 ≈ 12.4. In both
M(t)
cases the period is approximately 12, or one year.
0
(f ) Syracuse has greater variability because
25.47 > 7.46.
13
0
The model is a good fit.
(c)
15. Answers will vary.
100
16. Answers will vary.
S(t)
0
13
0
The model is a good fit.
17. Yes, A1 ≤ A2 . To see this, consider the two triangles of areas A1 and A2 :
T2
T1
γ1
a1
β1
γ2
a2
b1
α1
b2
β2
c1
α2
c2
For i = 1, 2, the angles satisfy α i + β i + γ i = π . At least one of α1 ≤ α 2 , β1 ≤ β 2 , γ 1 ≤ γ 2 must hold.
Assume α1 ≤ α 2 . Because α 2 ≤ π 2 (acute triangle), and the sine function increases on [0, π 2], you have
A1 = 12 b1c1 sin α1 ≤
≤
1
2
b2c2 sin α1
sin α 2 = A2
1b c
2 2 2
Section 1.5 Inverse Functions
f ( x) = 5 x + 1
1. (a)
x −1
g ( x) =
5
⎛ x − 1⎞
⎛ x − 1⎞
f ( g ( x)) = f ⎜
⎟ = 5⎜
⎟+1 = x
⎝ 5 ⎠
⎝ 5 ⎠
g ( f ( x)) = g (5 x + 1) =
(b)
f ( x) = 3 − 4 x
2. (a)
(5 x
f ( g ( x ))
+ 1) − 1
= x
5
g ( f ( x ) ) = g (3 − 4 x ) =
(b)
y
3 − (3 − 4 x )
4
= x
y
3
8
f
2
f
1
−3
3− x
4
⎛3 − x⎞
⎛3 − x⎞
= f⎜
⎟ = 3 − 4⎜
⎟ = x
⎝ 4 ⎠
⎝ 4 ⎠
g ( x) =
g
4
x
1
2
3
2
x
−2
2
−2
4
g
8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
38
Chapter 1
Preparation for Calculus
f ( x) = x3
3. (a)
g ( x) =
3
g ( x) =
x
( x) = ( x)
f ( g ( x)) = f
3
3
g ( f ( x)) = g ( x3 ) =
3
3
=
f
x
1
2
(
16 − x
)
2
16 − (16 − x 2 )
x2 = x
y
(b)
g
−3 −2
)
16 − x = 16 −
g ( f ( x)) = g (16 − x 2 ) =
2
1
(
= 16 − (16 − x) = x
x3 = x
3
x ≥ 0
16 − x
f ( g ( x)) = f
= x
y
(b)
f ( x) = 16 − x 2 ,
6. (a)
20
3
16
−2
12
−3
f
8
g
f ( x) = 1 − x3
4. (a)
g ( x) =
x
8
(
f ( g ( x )) = f
)
1− x = 1−
3
(
3
3
x3 = x
1− x
)
3
7. (a)
= 1 − (1 − x) = x
g ( f ( x)) = g (1 − x 3 )
=
1 − (1 − x 3 ) =
3
2
3
x
−1
2
2
1
−2
−1
f ( x) =
g ( x) = x + 4,
2
3
x ≥ 0
f ( g ( x)) = f ( x + 4)
2
8. (a)
( x 2 + 4) − 4
=
g ( f ( x)) = g
=
x
1
x−4
2
(
(
x−4
x −4
)
=
x2 = x
)
2
+ 4 = x−4+ 4 = x
y
12
f=g
3
−1
(b)
1
= x
1x
y
(b)
g
5. (a)
20
1
x
1
g ( x) =
x
1
f ( g ( x)) =
= x
1x
f 3
−2
16
f ( x) =
g ( f ( x)) =
y
(b)
12
1− x
3
g
10
8
6
4
1
,
x≥0
1+ x
1− x
g ( x) =
,
0 < x ≤1
x
1
1
⎛1 − x ⎞
f ( g ( x)) = f ⎜
=
= x
⎟=
x
−
1
1
x
⎝
⎠ 1+
x
x
1
1−
⎛ 1 ⎞
1+ x = x ⋅1+ x = x
g ( f ( x)) = g ⎜
⎟=
1
1+ x
1
⎝1 + x ⎠
1+ x
f ( x) =
y
(b)
f
2
3
g
x
2
4
6
8
10
12
2
1
f
x
1
2
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
20. f ( x) = 5 x
9. Matches (c)
10. Matches (b)
Inverse Functions
39
x −1
12
11. Matches (a)
12. Matches (d)
0
13. f ( x) =
3
x
4
6
0
+ 6
One-to-one; has an inverse
One-to-one; has an inverse
21. g ( x) = ( x + 5)
14. f ( x) = 5 x − 3
3
200
One-to-one; has an inverse
15. f (θ ) = sin θ
− 10
2
Not one-to-one; does not have an inverse
− 50
x2
16. f ( x) = 2
x + 4
One-to-one; has an inverse
Not one-to-one; does not have an inverse
22. h( x) = x + 4 − x − 4
9
1
17. h( s ) =
−3
s − 2
−9
9
1
−4
8
−9
Not one-to-one; does not have an inverse
−7
One-to-one; has an inverse
18. f ( x ) =
23. f ( x ) =
x4
− 2 x2
4
Not one-to-one; f does not have an inverse.
6x
x + 4
2
24. f ( x ) = sin
5
3x
2
Not one-to-one; f does not have an inverse.
−7
8
−5
One-to-one; has an inverse
Not one-to-one; does not have an inverse
19. g (t ) =
25. f ( x ) = 2 − x − x3
1
26. f ( x) =
3
x +1
One-to-one; has an inverse
t2 + 1
3
−3
3
−1
Not one-to-one; does not have an inverse
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
40
Chapter 1
Preparation for Calculus
27. (a) f ( x) = 2 x − 3 = y
29. (a) f ( x) = x5 = y
y +3
2
x +3
y =
2
x +3
−1
f ( x) =
2
x =
f
−1
x =
5
y
y =
5
x
5
x = x1 5
( x)
(b)
=
y
f
2
(b)
f −1
y
1
4
f
x
−2
2
1
2
−1
x
−2
−2
−2
4
2
f
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(d) Domain of f :
(d) Domain of f :
all real numbers
Range of f :
Domain of f
all real numbers
all real numbers
x =
y
(b)
: all real numbers
all real numbers
30. (a) f ( x) = x3 − 1 = y
x =
3
y +1
y =
3
x +1
f −1 ( x) =
3
x + 1 = ( x + 1)
28. (a) f ( x) = 7 − 4 x = y
7 − y
4
7 − x
y =
4
7 − x
−1
f ( x) =
4
all real numbers
−1
Range of f −1 :
Domain of f −1 : all real numbers
Range of f −1 :
all real numbers
Range of f :
(b)
13
y
5
4
3
2
f −1
x
−5 −4 −3
2 3 4 5
5
4
f
f
3
f −1
1
−2 −1
−1
x
1
3
4
5
−2
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(d) Domain of f :
(c) The graphs of f and f −1 are reflections of each other
across the line y = x.
(d) Domain of f :
Range of f :
Domain of f
−4
−5
all real numbers
Range of f :
Domain of f
all real numbers
all real numbers
−1
Range of f −1 :
: all real numbers
all real numbers
all real numbers
−1
Range of f −1 :
: all real numbers
all real numbers
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
f ( x) =
31. (a)
33. (a) f ( x) =
x = y
x = y
2
y = x
2
4 − x 2 = y,
4− x = y
2
x ≥ 0
y
0 ≤ x ≤ 2
2
x =
4 − y2
y =
4 − x2
f −1 ( x ) =
3
f −1
2
41
x2 = 4 − y 2
f −1 ( x) = x 2 ,
(b)
Inverse Functions
(b)
f
4 − x2 ,
0 ≤ x ≤ 2
y
1
3
x
1
2
f = f −1
2
3
1
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
x ≥ 0
(d) Domain of f :
y ≥ 0
Range of f :
Domain of f −1 : x ≥ 0
Range of f −1 :
y ≥ 0
x
1
2
3
(c) The graphs of f and f −1 are reflections of each other
in the line y = x. In fact, the graphs are identical.
0 ≤ x ≤ 2
(d) Domain of f :
0 ≤ y ≤ 2
Range of f :
32. (a) f ( x) = x = y, x ≥ 0
2
(b)
x =
y
y =
x
f −1 ( x) =
x
Domain of f
−1
: 0 ≤ x ≤ 2
Range of f −1 :
34. (a) f ( x ) =
0 ≤ y ≤ 2
x 2 − 4 = y,
x
y
f
4
3
f
2
(b)
1
x
1
2
3
x ≥ 0
y ≥ 0
Domain of f −1 : x ≥ 0
Range of f −1 :
y2 + 4
y =
x2 + 4
y ≥ 0
x 2 + 4,
x ≥ 0
5
f
4
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
Range of f :
x =
y
4
(d) Domain of f :
2
f −1 ( x) =
−1
x ≥ 2
= y + 4
2
−1
f
3
2
1
x
1
2
3
4
5
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(d) Domain of f :
y ≥ 0
Range of f :
Domain of f
x ≥ 2
−1
Range of f −1 :
: x ≥ 0
y ≥ 2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
42
Chapter 1
35. (a) f ( x ) =
Preparation for Calculus
3
37. (a) f ( x) = x 2 3 = y,
x −1 = y
x −1 = y
x = y
3
y = x3 2
x = y +1
3
f −1 ( x ) = x 3 2 , x ≥ 0
y = x3 + 1
f −1 ( x ) = x 3 + 1
(b)
x ≥ 0
32
(b)
4
f
2
−1
f
f −1
f
−3
3
0
6
0
−2
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(d) Domain of f :
Range of f :
Domain of f
Range of f
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
−1
:
Domain of f
y ≥ 0
x = y5 3
y = x5 3
5
y5
⎛ y⎞
2x − 1 = ⎜ ⎟ =
243
⎝3⎠
y 5 + 243
243
y 5 + 243
x =
486
(b)
: x ≥ 0
38. (a) f ( x) = x3 5 = y
36. (a) f ( x) = 3 5 2 x − 1 = y
2x =
−1
Range of f −1 :
: all real numbers
all real numbers
y ≥ 0
Range of f :
all real numbers
all real numbers
−1
x ≥ 0
(d) Domain of f :
y =
x5 + 243
486
f −1 ( x) =
x5 + 243
486
f −1 ( x) = x5 3
(b)
2
f −1
f
−3
3
−2
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(d) Domain of f :
6
f
Range of f :
f −1
−4
all real numbers
8
all real numbers
Domain of f −1 : all real numbers
Range of f −1 :
all real numbers
−6
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(d) Domain of f :
Range of f :
all real numbers
all real numbers
Domain of f −1 : all real numbers
Range of f −1 :
all real numbers
39. (a) f ( x ) =
x
x2 + 7
= y
x = y
x2 + 7
x 2 = y 2 ( x 2 + 7) = y 2 x 2 + 7 y 2
x 2 (1 − y 2 ) = 7 y 2
x =
y =
f −1 ( x) =
7y
1 − y2
7x
1 − x2
7x
1 − x2
,
−1 < x < 1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
(b)
f −1
f
3
y = 1.25 x + 1.60(50 − x)
−2
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
(d) Domain of f :
= −0.35 x + 80,
y = −0.35 x + 80
0.35 x = 80 − y
−1 < y < 1
x =
Domain of f −1 : −1 < x < 1
Range of f −1 :
− x) =
y =
44. C =
5
9
100
35
(F
9
C
5
(a)
x ≠ 1
(80
− 73) =
− 32),
100
5
f
6
F ≥ −459.6
= F − 32
(c) For F ≥ −459.6, C =
−4
(c) The graphs of f and f −1 are reflections of each other
in the line y = x.
all x ≠ 0
Domain of f −1 : all x ≠ 1
41.
0
f ( x)
1
2
4
f
−1
( x)
2
3
4
0
1
2
4
f
(3, 2)
(1, 0)
x
2
3
4
f ( x)
x
f
−1
( x)
0
2
6
4
2
0
0
2
4
6
2
0
= x 2 + 2, x ≥ 0
9 − x 2 is not one-to-one.
For example, f (3) = f ( − 3) = 0.
Not one-to-one; does not have an inverse.
8
6
( x)
47. f ( x) = −3
y
x
−1
46. f ( x ) =
(2, 1)
1
1
42.
x − 2, x ≥ 2, y ≥ 0
x = y2 + 2
(4, 4)
3
1
= 71.6°F.
y = x − 2
2
x
(22)
2
3
3
9
5
x − 2, x ≥ 2
y =
4
1
− 32) ≥ −273.1 1.
f is one-to-one; has an inverse.
all y ≠ 0
2
(F
(d) If C = 22°, then F = 32 +
y
x
5
9
So, the domain is C ≥ −273. 1 = −273 19 .
45. f ( x ) =
all y ≠ 1
Range of f −1 :
= 20 pounds.
(b) The inverse function gives the Fahrenheit
temperature F corresponding to the Celsius
temperature C.
−1
Range of f :
(80 − x)
F = 32 + 95 C
4
(d) Domain of f :
20
7
(d) If x = 73 in the inverse function,
2
y −1
−6
(80
The total cost will be between $62.50 and $80.00.
2
x −1
2
−1
,
f ( x) =
x −1
f
100
35
(c) Domain of inverse is 62.5 ≤ x ≤ 80.
y =
(b)
(80 − y )
x represents cost and y represents pounds.
x ≠ 0
x(1 − y ) = −2
x =
100
35
Inverse: y =
all real numbers
x + 2
= y,
x
x + 2 = yx
0 ≤ x ≤ 50.
(b) Find the inverse of the original function.
all real numbers
Range of f :
40. (a) f ( x ) =
43
43. (a) Let x be the number of pounds of the commodity
costing 1.25 per pound. Because there are 50 pounds
total, the amount of the second commodity
is 50 − x. The total cost is
2
−3
Inverse Functions
(0, 6)
4
(2, 2)
2
(4, 0)
2
4
6
x
8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
44
Chapter 1
P reparation for Calculus
48. f ( x ) = x − 2 , x ≤ 2
⎡ π⎞
56. f ( x) = sec x on ⎢0, ⎟
⎣ 2⎠
= − ( x − 2)
f passes the Horizontal Line Test on [0, π 2), so it is
= 2 − x
f is one-to-one; has an inverse.
2− x = y
one-to-one.
57. f ( x) = ( x − 3) is one-to-one for x ≥ 3.
2
2− y = x
f −1 ( x ) = 2 − x,
(x
x ≥ 0
− 3) = y
2
x −3 =
49. f ( x) = ax + b
x =
y + 3
y =
x + 3
f −1 ( x) =
x + 3,
f is one-to-one; has an inverse.
ax + b = y
y −
a
x −
y =
a
x
−
f −1 ( x) =
a
x =
b
b
, a ≠ 0
58. f ( x ) = x − 3 is one-to-one for x ≥ 3.
x −3 = y
x = y +3
y = x +3
3
f
f is one-to-one; has an inverse.
y = ( x + a) + b
3
y − b = ( x + a)
3
x + a =
3
y −b
x =
3
y −b − a
3
x −b − a
f
( x)
=
x ≥ 0
(Answer is not unique.)
.
b
50. f ( x ) = ( x + a ) + b
−1
y
−1
( x)
= x + 3,
x ≥ 0
(Answer is not unique.)
59. (a) f ( x) = ( x + 5)
2
y
8
6
4
51. f ( x ) = ( x − 4) on [4, ∞ )
2
2
f passes the Horizontal Line Test on [4, ∞), so it is
one-to-one.
52. f ( x) = x + 2 on [−2, ∞)
f passes the Horizontal Line Test on [−2, ∞), so it is
x
−12 −10 −8 −6 −4 −2
−2
2
(b) f is one-to-one on [−5, ∞). (Note that f is also
one-to-one on ( −∞, −5]. )
(c) f ( x) = ( x + 5) = y,
2
x +5 =
one-to-one.
x =
53. f ( x) =
4
on (0, ∞)
x2
f passes the Horizontal Line Test on (0, ∞), so it is
one-to-one.
f
−1
(d) Domain of f
x ≥ −5
y
y −5
y =
x −5
( x) =
x −5
−1
: x ≥ 0
54. f ( x ) = cot x on (0, π )
f passes the Horizontal Line Test on (0, π ), so it is
one-to-one.
55. f ( x ) = cos x on [0, π ]
f passes the Horizontal Line Test on [0, π ], so it is
one-to-one.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
60. (a) f ( x ) = (7 − x) = ( x − 7)
2
61. (a) f ( x ) =
2
Inverse Functions
45
x2 − 4x
y
y
12
8
10
6
8
4
6
4
x
−4 −2
2
x
2
4
6
8
10
(c) f ( x) = ( x − 7) = y,
2
x −7 =
x2 − 4 x = y2
y
y
y = 7+
x
= 7+
x 2 − 4 x = y, x ≥ 4
x2 − 4 x + 4 = y2 + 4
(x
− 2) = y 2 + 4
2
x − 2 =
x
−1
: x ≥ 0
(d) Domain of f
f
(d) Domain of f
62. (a) f ( x ) = −
8
one-to-one on ( −∞, 0]. )
(c) f ( x) =
x ≥ 7
x = 7+
( x)
6
(b) f is one-to-one on [4, ∞). (Note that f is also
one-to-one on ( −∞, 7]. )
−1
4
12
(b) f is one-to-one on [7, ∞). (Note that f is also
f
2
−2
−1
y2 + 4
x = 2+
y2 + 4
y = 2+
x2 + 4
( x)
= 2+
x2 + 4
−1
: x ≥ 0
25 − x 2
y
6
4
2
x
−6
−4 −2
−2
2
4
6
−4
−6
(b) f is one-to-one on [0, 5]. (Note that f is also one-to-one on [−5, 0]. )
(c) f ( x ) = −
25 − x 2 = y,
25 − x = y
2
0 ≤ x ≤ 5, − 5 ≤ y ≤ 0
2
x 2 = 25 − y 2
f
(d) Domain of f
−1
x =
25 − y 2
y =
25 − x 2
( x)
=
25 − x 2
−1
: −5 ≤ x ≤ 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
46
Chapter 1
Preparation for Calculus
67. f ( x) = sin x
63. (a) f ( x) = 3 cos x
1
⎛π ⎞
f⎜ ⎟ =
= a ⇒ f
2
⎝6⎠
y
4
3
6
⎜ ⎟ =
6
⎝ 2⎠
68. f ( x) = cos 2 x
x
−6
π
−1 ⎛ 1 ⎞
10
f ( 0) = 1 = a ⇒ f
−1
(1)
= 0
−3
−4
(b) f is one-to-one on [0, π ]. (other answers possible)
69. f ( x ) = x3 −
f ( 2) = 6 = a ⇒ f
(c) f ( x) = 3 cos x = y
y
3
cos x =
70. f ( x ) =
⎛ y⎞
x = arccos ⎜ ⎟
⎝ 3⎠
⎛ x⎞
y = arccos ⎜ ⎟
⎝ 3⎠
−1
f
( x)
4
x
⎛ x⎞
= arccos ⎜ ⎟
⎝ 3⎠
( 6)
= 2
( 2)
= 8
−1
x−4
f (8) = 2 = a ⇒ f
−1
In Exercises 71–74, use the following.
f ( x) =
1x
8
− 3 and g ( x ) = x 3
f −1 ( x ) = 8( x + 3) and g −1 ( x ) =
3
x
−1
71.
( f −1
g −1 )(1) = f −1 ( g −1 (1)) = f −1 (1) = 32
64. (a) f ( x) = 2 sin x
72.
( g −1
f −1 )( −3) = g −1 ( f −1 ( −3)) = g −1 (0) = 0
73.
( f −1
f −1 )(6) = f −1 ( f −1 (6)) = f −1 (72) = 600
74.
( g −1
g −1 )( −4) = g −1 ( g −1 ( −4)) = g −1
(d) Domain of f
: −3 ≤ x ≤ 3
y
2
1
π
−
2
x
π
2
=
3 3
(
3
−4
)
−4 = − 9 4
−2
In Exercises 75–78, use the following.
⎡ π π⎤
(b) f is one-to-one on ⎢− , ⎥. (other answers possible)
⎣ 2 2⎦
(c) f ( x) = 2 sin x = y
y
2
sin x =
f ( x ) = x + 4 and g ( x ) = 2 x − 5
f −1 ( x ) = x − 4 and g −1 ( x ) =
75.
( g −1
=
⎛ x⎞
y = arcsin ⎜ ⎟
⎝ 2⎠
−1
(d) Domain of f
( x)
⎛ x⎞
= arcsin ⎜ ⎟
⎝ 2⎠
−1
: −2 ≤ x ≤ 2
65. f ( x) = x + 2 x − 1
3
f (1) = 2 = a ⇒ f
−1
( 2)
=1
66. f ( x ) = 2 x5 + x3 + 1
f ( −1) = −2 = a ⇒ f
−1
( − 2)
f −1 )( x) = g −1 ( f −1 ( x))
= g −1 ( x − 4)
⎛ y⎞
x = arcsin ⎜ ⎟
⎝ 2⎠
f
x + 5
2
(x
− 4) + 5
2
x +1
=
2
76.
( f −1
g −1 )( x) = f −1 ( g −1 ( x ))
⎛ x + 5⎞
= f −1 ⎜
⎟
⎝ 2 ⎠
x +5
=
− 4
2
x −3
=
2
= −1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
77.
(f
g )( x) = f ( g ( x))
g)
Note: ( f
78.
(g
= ( 2 x − 5) + 4
(b) The domain of f
= 2x − 1
(c) f
−1
( x)
g)
−1
x +1
.
2
=
= g −1
f −1
Note: ( g
( 2)
(c) f
−1
( 2)
= 2( x + 4) − 5
= 2x + 3
2
( x)
f)
−1
−1
= f
g
is the range of f : [−3, 3].
≈ 1.73 because f (1.73) ≈ 2.
f −1
3
x −3
.
2
=
−1
y
81.
4
−1
is the range of f : [−2, 2].
= − 4 because f ( − 4) = 2.
(b) The domain of f
= g ( x + 4)
f)
−1
−1
80. (a) f is one-to-one because it passes the Horizontal Line
Test.
f )( x) = g ( f ( x))
So, ( g
47
79. (a) f is one-to-one because it passes the Horizontal Line
Test.
= f ( 2 x − 5)
So, ( f
Inverse Functions
f
x
−4 −3 −2 −1
1
2
3
4
−2
−3
−1
−4
82.
y
f −1
4
3
f
2
x
−4 −3 −2
1
2
3
4
−2
−3
−4
83. y = arcsin x
(a)
x
−1
− 0.8
− 0.6
− 0.4
− 0.2
0
0.2
0.4
0.6
0.8
1
y
−1.571
− 0.927
− 0.644
− 0.412
− 0.201
0
0.201
0.412
0.644
0.927
1.571
y
(b)
2
(c)
π
2
(d) Symmetric about origin:
arcsin ( − x) = −arcsin x
−1
Intercept: (0, 0)
1
x
−1
1
−2
−
π
2
84. y = arccos x
(a)
x
−1
− 0.8
− 0.6
− 0.4
− 0.2
0
0.2
0.4
0.6
0.8
1
y
3.142
2.498
2.214
1.982
1.772
1.571
1.369
1.159
0.927
0.644
0
y
(b)
⎛ π⎞
(d) Intercepts: ⎜ 0, ⎟ and (1, 0)
⎝ 2⎠
␲
(c)
π
−1
1
0
−1
x
1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
48
Chapter 1
Preparation for Calculus
85. y = arccos x
⎛
2
2 3π ⎞
⎛ 3π ⎞
⎜⎜ − 2 , 4 ⎟⎟ because cos⎜⎝ 4 ⎟⎠ = − 2 .
⎝
⎠
1
⎛1 π ⎞
⎛π ⎞
⎜ , ⎟ because cos⎜ ⎟ = .
2
⎝2 3⎠
⎝4⎠
⎛ 3 π⎞
⎛π ⎞
⎜⎜ 2 , 6 ⎟⎟ because cos⎜⎝ 6 ⎟⎠ =
⎝
⎠
102. sin y =
1 − x2
103. tan y =
1 − x2
x
3
.
2
105. sec y =
86. No, g is not the inverse of f. f ( x) = sin x is not one-toone. The graph of g is not the graph of a function.
1
π
=
2
6
87. arcsin
x
104. cot y =
1 − x2
1
x
1
106. csc y =
1 − x2
3⎞
3
⎛
107. (a) sin ⎜ arctan ⎟ =
4⎠
5
⎝
88. arcsin 0 = 0
1
π
=
2
3
89. arccos
5
3
90. arccos 1 = 0
θ
4
3
π
91. arctan
=
3
6
)
5π
6
)
π
(
3 =
(
2 = −
92. arccot −
93. arccsc −
(
94. arcsec −
)
2 =
4⎞
5
⎛
(b) sec⎜ arcsin ⎟ =
5⎠
3
⎝
5
4
4
θ
3π
4
95. arccos (0.8) ≈ 2.50
3
⎛
2⎞
⎛π ⎞
108. (a) tan ⎜⎜ arccos
⎟⎟ = tan ⎜⎝ 4 ⎟⎠ = 1
2
⎝
⎠
96. arcsin ( − 0.39) ≈ − 0.40
⎛ 1 ⎞
97. arcsec (1.269) = arccos⎜
⎟ ≈ 0.66
⎝ 1.269 ⎠
98. arctan ( − 5) ≈ −1.37
99. cos ⎡⎣arccos( − 0.1)⎤⎦ = − 0.1
100. arcsin (sin 3π ) = arcsin (0) = 0
In Exercises 101–106, use the triangle.
1
2
2
θ
2
5⎞
12
⎛
(b) cos⎜ arcsin ⎟ =
13 ⎠
13
⎝
13
5
θ
12
1 − x2
y
x
101.
y = arccos x
cos y = x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
⎡
⎛ 1 ⎞⎤
⎛ π⎞
109. (a) cot ⎢arcsin ⎜ − ⎟⎥ = cot ⎜ − ⎟ = − 3
⎝ 2 ⎠⎦
⎝ 6⎠
⎣
113. y = sin (arcsec x)
θ = arcsec x, 0 ≤ θ ≤ π , θ =/
3
θ
2
⎡
13
⎛ 5 ⎞⎤
(b) csc ⎢arctan ⎜ − ⎟⎥ = −
12
5
⎝
⎠
⎣
⎦
π
2
x2 − 1
x
y = sin θ =
1
49
Inverse Functions
The absolute value bars on x are necessary because of the
restriction 0 ≤ θ ≤ π , θ =/ π 2, and sin θ for this
domain must always be nonnegative.
x
12
x2 − 1
θ
−5
13
⎡
⎛ 3 ⎞⎤
110. (a) sec ⎢arctan ⎜ − ⎟⎥ =
⎝ 5 ⎠⎦
⎣
34
5
5
θ
1
114. y = sec ⎡⎣arcsin ( x − 1)⎤⎦
θ = arcsin ( x − 1)
1
y = sec θ =
θ
1
2x − x2
−3
x−1
θ
2x − x 2
34
⎡
5 11
⎛ 5 ⎞⎤
(b) tan ⎢arcsin ⎜ − ⎟⎥ = −
6
11
⎝ ⎠⎦
⎣
x⎞
⎛
115. y = tan ⎜ arcsec ⎟
3
⎝
⎠
θ = arcsec
11
x
x −9
y = tan θ =
3
2
θ
6
x
3
111. y = cos(arcsin 2x)
θ = arctan
x ⎞
⎟
2⎠
x
2
x2 + 2
x
θ = arcsin 2x
1
3
−5
⎛
116. y = csc⎜ arctan
⎝
y = cos θ =
x2 − 9
θ
1 − 4x2
2x
θ
y = csc θ =
x2 + 2
x
117. arcsin (3 x − π ) =
θ
2
1
2
( 12 )
1 ⎡sin 1 + π ⎤
( 2) ⎦
3⎣
3x − π = sin
1− 4x 2
x =
112. sec(arctan 4x)
118. arctan ( 2 x − 5) = −1
θ = arctan 4x
y = sec θ =
≈ 1.207
16 x 2 + 1
2 x − 5 = tan ( −1)
x =
1
2
(5 + tan(−1))
≈ 1.721
1 + 16x 2
4x
θ
1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
50
Chapter 1
Preparation for Calculus
2 x = arccos
119. arcsin
(
2 x = sin arccos
2x =
120. arccos x = arcsec x
x
x
x = cos(arcsec x)
)
x =
1 − x, 0 ≤ x ≤ 1
2x = 1 − x
x =
1−x
θ
1
3
x
x2 = 1
1
3x = 1
1
x
x2 − 1
θ
x = ±1
1
x
121. y = arccos x
y = arctan x
The point of intersection is given by
f ( x) = arccos x − arctan x = 0, cos(arccos x) = cos(arctan x).
y
π
1
x =
1 + x2
π
2
x 2 (1 + x 2 ) = 1
−1 +
x 4 + x 2 − 1 = 0 when x 2 =
2
−1 +
2
So, x = ±
5
5
−3
−2
.
(0.7862, 0.6662)
x
−1
−
1
2
3
π
2
≈ ± 0.7862.
Point of intersection: (0.7862, 0.6662) ⎣⎡Because f ( − 0.7862) = π =/ 0.⎤⎦
122. y = arcsin x
y = arccos x
y
The point of intersection is given by
f ( x) = arcsin x − arccos x = 0, sin (arcsin x) = sin (arccos x).
π
π
2
x = 1 − x2
x2 = 1 − x2
−3
Point of intersection:
(
2 2, π 4
)
(
⎡Because f −
⎣
of y. Interchange x and y to get y = f
domain of f
−1
3
3
y
y =
3
x
( x)
=
( x). Let the
−1
( f ( x)) = x.
)
124. The graphs of f and f −1 are mirror images with respect
to the line y = x.
125. The trigonometric functions are not one-to-one. So, their
domains must be restricted to define the inverse
trigonometric functions.
⎧arctan (1 x) + π , −∞ < x < 0
⎪
f ( x) = ⎨π 2,
x = 0
⎪arctan 1 x ,
0 < x < ∞
( )
⎩
y = x3
−1
2
126. You could graph f ( x) = arccot ( x) as follows.
f ( x) = x3
x =
−1
be the range of f . Verify that
( x)) = x and f
Example:
f
x
1
2 2 = −π =/ 0.⎤
⎦
123. Let y = f ( x) be one-to-one. Solve for x as a function
−1
−1
)
1
2
= ±
2
2
x = ±
f (f
−2
2, π
2 4
(
3
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
127. arctan
9
= arcsin
x
9
131.
x 2 + 81
x + 81
2
9
f ( x ) = arcsin ( x − 1)
x
36 − x 2
π
2
,
1
.
x
⎛1⎞
So, y = arcsin ⎜ ⎟. Therefore,
⎝ x⎠
⎛1⎞
arccsc x = arcsin ⎜ ⎟.
⎝ x⎠
(b) arctan x + arctan
1
π
= , x > 0
x
2
Let y = arctan x + arctan (1 x).
Then tan y =
tan (arctan x) + tan ⎡⎣arctan (1 x )⎤⎦
1 − tan (arctan x ) tan ⎡⎣arctan (1 x)⎤⎦
x + (1 x)
=
1 − x(1 x)
=
2
π
y
2
π
3π
4
π
2
Domain: ( −∞, ∞)
≤ y < 0 and 0 < y ≤
csc y = x ⇒ sin y =
)0, − π2 )
π
π⎞
⎛
x = tan ⎜ y − ⎟
2⎠
⎝
1
, x ≥1
x
Let y = arccsc x.
x + (1 x)
(which is undefined).
0
So, y = π 2. Therefore,
arctan x + arctan (1 x) = π 2.
130. (a) arcsin ( − x) = − arcsin x, x ≤ 1
Let y = arcsin ( − x).
Then − x = sin y ⇒ x = −sin y ⇒ x = sin (− y ).
So, − y = arcsin x ⇒ y = −arcsin x.
Therefore, arcsin ( − x) = −arcsin x.
2
shifted right one unit.
132. f ( x) = arctan x +
x
2
−
f ( x) is the graph of arcsin x
θ
π
1
Range: [−π 2, π 2]
6
Then for −
2
Domain: [0, 2]
36 − x 2
x
= arccos
6
6
129. (a) arccsc x = arcsin
)2, π2 )
π
x = 1 + sin y
x
51
y
x − 1 = sin y
θ
128. arcsin
Inverse Functions
Range: [0, π ]
−4
x
−2
2
4
f ( x) is the graph of arctan x
shifted π 4 unit upward.
133. f ( x ) = arcsec 2 x
y
2 x = sec y
(− 12 , π(
1
x = sec y
2
Domain: ( −∞, −1 2], [1 2, ∞)
Range: [0, π 2), (π 2, π ]
x
− 32
⎛ x⎞
134. f ( x ) = arccos⎜ ⎟
⎝ 4⎠
x
= cos y
4
x = 4 cos y
3
2
y
π
3π
4
Domain: [− 4, 4]
Range: [0, π ]
( 12 , 0(
− 12
π
4
−4
−2
x
2
4
135. Because f ( −3) = 8 and f is one-to-one, you have
f
−1
(8)
= −3.
136. Because f (0) = 5 + arccos(0) = 5 + π 2, and f is
one-to-one, f
−1
(π
2 + 5) = 0.
(b) arccos( − x) = π − arccos x, x ≤ 1
Let y = arccos( − x). Then
− x = cos y ⇒ x = −cos y ⇒ x = cos(π − y ).
So, π − y = arccos x ⇒ y = π − arccos x.
Therefore, arccos( − x) = π − arccos x.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
52
Chapter 1
Preparation for Calculus
137. Let f and g be one-to-one functions.
Let ( f
(f
g )( x) = y, then x =
(f
145. True
g)
−1
( y).
Also:
g )( x ) = y
147. (a) arccot x = y if and only if cot y = x,
0 < y < π.
f ( g ( x )) = y
g ( x ) = f −1 ( y )
For x > 0, cot y > 0 and 0 < y <
x = g −1 ( f −1 ( y ))
x = ( g −1
So, ( f
(f
g)
g)
−1
−1
( y)
= g
−1
f −1 ( y ))
= ( g −1
So, tan y =
f −1 )( y ) and
−1
So, ( f −1 )
−1
( x)
−1
= y then x = f −1 ( y ) and f ( x) = y.
For x = 0, arccot (0) =
= f.
−1
139. Let y = sin x. Then sin y = x and
cos(sin −1 x) = cos( y ) =
1 − x 2 , as indicated in
2
.
π
2
< y < π.
Therefore, you need to add π to get
⎛1⎞
y = π + arctan ⎜ ⎟.
⎝ x⎠
x
So, cos y =
y
x2
140. Suppose g ( x) and h( x) are both inverses of f ( x). Then
the graph of f ( x) contains the point ( a, b) if and only if
the graphs of g ( x) and h( x) contain the point (b, a).
Because the graphs of g ( x) and h( x) are the same,
g ( x) = h( x). So, the inverse of f ( x) is unique.
141. False. Let f ( x ) = x 2 .
.
1
⎛1⎞
< 0 and arctan ⎜ ⎟ < 0.
x
⎝ x⎠
0 ≤ y ≤ π , y =/
1
π
2
.
1
⎛1⎞
and y = arccos⎜ ⎟.
x
⎝ x⎠
(c) y = arccsc x if and only if csc y = x,
−
π
2
≤ y ≤
So, sin y =
x ≥ 1,
π
2
x ≥ 1,
, y =/ 0.
1
⎛1⎞
and y = arcsin ⎜ ⎟.
x
⎝ x⎠
⎛ 1 ⎞
148. (a) arccot (0.5) = arctan ⎜
⎟ = arctan ( 2) ≈ 1.1071
⎝ 0.5 ⎠
⎛ 1 ⎞
(b) arcsec( 2.7) = arccos⎜
⎟ ≈ 1.1914
⎝ 2.7 ⎠
142. True; if f has a y-intercept.
⎛ −1 ⎞
(c) arccsc( −3.9) = arcsin ⎜
⎟ ≈ − 0.2593
⎝ 3.9 ⎠
143. False
arcsin 2 0 + arccos 2 0 = 0 +
2
(b) y = arcsec x if and only if sec y = x,
the figure.
1−
π
For x < 0, cot y < 0 and
So, tan y =
π
1
⎛1⎞
> 0 and y = arctan ⎜ ⎟.
x
⎝ x⎠
f .
138. If f has an inverse, then f and f −1 are both one-to-one.
Let ( f −1 )
146. False. Let f ( x) = x or g ( x) = 1 x.
π
2
2
=/ 1
(d) arccot ( − 0.5) = π + arctan ( −2.0) ≈ 2.0344
144. False
⎡ π π⎤
The range of y = arcsin x is ⎢− , ⎥.
⎣ 2 2⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.5
149. tan (arctan x + arctan y ) =
Inverse Functions
53
tan (arctan x + arctan y )
x + y
=
, xy =/ 1
1 − tan (arctan x ) tan (arctan y )
1 − xy
So,
⎛x + y⎞
arctan x + arctan y = arctan ⎜
⎟, xy =/ 1.
⎝ 1 − xy ⎠
1
1
and y = .
Let x =
2
3
5
1 + 1
⎛1⎞
⎛1⎞
2
3
arctan ⎜ ⎟ + arctan ⎜ ⎟ = arctan
= arctan 6
1−
1 − 12 ⋅ 13
⎝ 2⎠
⎝ 3⎠
(
)
1
6
150. arcsin (sin x) =/ x for many values of x outside
⎡ π π⎤
⎢− 2 , 2 ⎥.
⎣
⎦
= arctan
5
6
5
6
= arctan 1 =
π
4
152. f will be symmetric about the line y = x if f is one-toone, and equals its inverse. So assume
f ( x1 ) = f ( x2 )
ax1 + b
ax2 + b
=
cx1 − a
cx2 − a
For example, arcsin (sin 2π ) = arcsin (0) = 0 =/ 2π .
3
acx1x2 − a 2 x1 + bcx2 − ab = acx1 x2 + bcx1 − ab
g
(a 2 + bc) x2 = (a 2 + bc) x1.
f
2π
−2 π
So, x1 = x2 if a 2 + bc =/ 0.
−3
To show that f = f −1 , solve for x as follows:
151. y = ax 2 + bx + c. Interchange x and y, and solve for y
using the quadratic formula.
ay 2 + by + c − x = 0
y =
Because x ≤
f −1 ( x) =
−b ±
b 2 − 4a (c − x )
2a
−b
, use the negative sign.
2a
−b −
b 2 − 4ac + 4ax
2a
ax + b
cx − a
ycx − ay = ax + b
y =
( yc
− a) x = b + ay
x =
f −1 ( x ) =
ay + b
yc − a
ax + b
= f ( x)
cx − a
So, f is symmetric about the line y = x is and only if
a 2 + bc =/ 0.
153. f is one-to-one if f ( x1 ) = f ( x2 ) implies x1 = x2 . So assume
f ( x1 ) = f ( x2 )
ax1 + b
ax2 + b
=
cx1 + d
cx2 + d
acx1 x2 + adx1 + bcx2 + bd = acx1 x2 + adx2 + bcx1 + bd
adx1 + bcx2 = adx2 + bcx1
(ad
− bc ) x1 = ( ad − bc) x2 .
So, x1 = x2 if ad − bc =/ 0. To find f −1 , solve for x as follows.
ax + b
cx + d
ycx + yd = ax + b
y =
( yc
− a) x = b − yd
x =
b − yd
yc − a
f −1 ( x) =
b − dx
cx − a
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
54
Chapter 1
Preparation for Calculus
Section 1.6 Exponential and Logarithmic Functions
1. (a) 253 2 = 53 = 125
1
= e3
e −3
(d)
(b) 811 2 = 9
2. (a) 64
8. 4 x = 64 ⇒ x = 3
1
1
=
271 3
3
(d) 27 −1 3 =
13
7. 3x = 81 ⇒ x = 4
1
1
=
32
9
(c) 3−2 =
9. 6 x − 2 = 36 ⇒ x − 2 = 2 ⇒ x = 4
= 4
10. 5 x +1 = 125 ⇒ x + 1 = 3 ⇒ x = 2
1
1
=
54
625
(b) 5− 4 =
13
⎛1⎞
(c) ⎜ ⎟
⎝8⎠
1
=
2
3
1
⎛1⎞
(d) ⎜ ⎟ =
4
64
⎝ ⎠
3. (a)
(5 )(5 )
2
3
= 5
2+3
(b)
(52 )(5−3 )
(c)
53
53
1
= 4 =
2
25
5
5
= 5 = 3125
= 5 2 − 3 = 5 −1 =
1
5
(b)
(54 )
3
12.
( 14 )
x
= 16 ⇒ 4− x = 16 ⇒ − x = 2 ⇒ x = −2
13.
( 13 )
x −1
14.
( 15 )
2x
= 27 ⇒ 31− x = 27 ⇒ 1 − x = 3 ⇒ x = −2
= 625 ⇒ 5−2 x = 54 ⇒ −2 x = 4 ⇒ x = −2
16. 182 = (5 x − 7) ⇒ ±18 = 5 x − 7 ⇒ x = 5, − 11
5
2
17. x 3 4 = 8 ⇒ x = 84 3 = 24 = 16
18.
(x
+ 3)
43
= 16 ⇒ x + 3 = ±163 4
⇒ x + 3 = ± 8 ⇒ x = 5, −11
= 52 = 25
3
1
−1
23 3
(c) ⎡( 27) ( 27) ⎤ = ⎡⎣27 −1 3 ⎤⎦ = 27 −1 =
⎣
⎦
27
(d)
= 32 ⇒ 2− x = 32 ⇒ − x = 5 ⇒ x = −5
3
= 26 = 64
12
x
15. 43 = ( x + 2) ⇒ 4 = x + 2 ⇒ x = 2
2
( 22 )
( 12 )
5
26
⎛1⎞
(d) ⎜ ⎟ 26 − 4 = 22 = 4
2
⎝4⎠
4. (a)
11.
(25)3 2 32
= 5332 = (125)9 = 1125
19. e x = 5 ⇒ x = ln 5 ≈ 1.609
20. e x = 1 = e0 ⇒ x = 0
5. (a) e 2 (e 4 ) = e6
21. e −2 x = e5 ⇒ −2 x = 5 ⇒ x = − 52
(b)
(e 3 )
4
22. e3 x = e − 4 ⇒ 3x = − 4 ⇒ x = − 43
(c)
(e 3 )
−2
(d)
e5
= e2
e3
⎛1⎞
6. (a) ⎜ ⎟
⎝e⎠
(b)
= e12
= e −6 =
−2
(e 3 )
= e2
4
1
e6
1,000,000
1
⎛
⎞
23. ⎜1 +
⎟
1,000,000
⎝
⎠
≈ 2.718280469
e ≈ 2.718281828
1,000,000
1
⎛
⎞
e > ⎜1 +
⎟
1,000,000
⎝
⎠
= e12
(c) e0 = 1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.6
24. 1 + 1 +
+
1
2
1
6
+
1
24
+
1
120
+
1
720
+
1
5040
Exponential and Logarithmic Functions
55
= 2.71825396
e ≈ 2.718281828
e >1+1+
25. y = 3x
1
2
+
1
6
+
1
24
+
+
1
120
+
1
720
28. y = 2− x
1
5040
2
x
–2
−1
0
1
2
x
–2
−1
0
1
y
1
9
1
3
1
3
9
y
1
16
1
2
1
1
2
y
2
3
1
16
0.002
y
4
2
3
2
−3
−2
1
1
0
1
9
3
−1
29. f ( x) = 3− x
–1
2
2
26. y = 3x −1
y
x
−1
x
−1
x
−2
1
1
3
1
2
3
x
3
y
9
2
0
±1
±2
1
1
3
0.0123
y
y
12
2
10
8
6
4
2
−2
−1
27. y =
x
−1
1
x
1
( 13 )
x
2
3
4
30. f ( x ) = 3
= 3− x
x
–2
−1
0
1
2
y
9
3
1
1
3
1
9
x
x
0
±1
±2
y
1
9
9
y
y
4
4
3
3
2
2
1
−2
−1
x
1
−2
−1
x
1
2
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
56
Chapter 1
Preparation for Calculus
31. y = e − x
x
35. h( x) = e x − 2
y
–1
4
0
1
1
1
e
x
0
1
2
3
4
1
e
e2
3
y
e
y
e
−2
e
−1
y
−1
32. y =
x
1
2
3
3
2
1 ex
2
1
x
–1
0
1
2
y
1
2e
1
2
e
2
e2
2
x
1
2
3
36. g ( x) = −e x 2
y
4
3
x
–2
y
−
2
1
e
0
2
4
−1
−e
− e2
y
1
x
x
−1
1
2
1
2
3
3
−2
33. y = e x + 2
−3
−4
x
–2
–1
y
1
+ 2
e2
1
+ 2
e
y
6
5
0
1
2
3
e + 2
e + 2
−5
2
2
37. y = e − x
Symmetric with respect to the y-axis
Horizontal asymptote
y = 0
4
y
3
2
1
2
x
−3 −2 −1
1
2
3
34. y = e x −1
x
−1
x
–1
0
1
2
y
1
e2
1
e
1
e
1
38. y = e − x 4
y
12
y
10
3
8
2
6
4
1
−3 −2 −1
−1
−2
x
1
2
3
x
−8 −6 −4 −2
2
4
6
8
−4
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.6
39. f ( x ) =
Exponential and Logarithmic Functions
46. (a)
1
3 + ex
57
10
Because e x > 0, 3 + e x > 0.
−8
Domain: all real numbers
10
−2
40. f ( x ) =
1
2 − ex
The graph approaches 8 as x → ∞. The graph
approaches 0 as x → −∞.
2 − e x = 0 ⇒ x = ln 2
(b)
10
Domain: all x ≠ ln 2
41. f ( x ) =
1 − 4x
−8
10
1 − 4 ≥ 0 ⇒ 4 ≤ 1 ⇒ x ln 4 ≤ ln 1 = 0
x
x
As x → ±∞, the graph approaches 4.
Domain: x ≤ 0
42. f ( x ) =
1 + 3− x
47. y = Ce ax
Because 1 + 3− x > 0 for all x, the domain is all real
numbers.
43. f ( x ) = sin e − x
Horizontal asymptote: y = 0
Matches (c)
48. y = Ce − ax
Horizontal asymptote: y = 0
Domain: all real numbers
Reflection in the y-axis
44. f ( x ) = cos e − x
Matches (d)
Domain: all real numbers
45. (a)
−2
49. y = C (1 − e − ax )
7
f
Vertical shift C units
g
Reflection in both the x- and y-axes
Matches (a)
−5
7
−1
Horizontal shift 2 units to the right.
(b)
3
f
−2
4
h
Horizontal asymptotes: y = C and y = 0
−3
A reflection in the x-axis and a vertical shrink.
(c)
C
1 + e − ax
C
lim
= C
x → ∞ 1 + e − ax
C
= 0
lim
x → −∞ 1 + e − ax
50. y =
Matches (b)
51. y = Ca x
7
(0, 2): 2
f
q
= Ca 0 = C
(3, 54): 54
−4
8
−1
Vertical shift 3 units upward and a reflection in the
y-axis.
= 2a 3
27 = a 3
3 = a
y = 2(3x )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
58
Chapter 1
Preparation for Calculus
62. f ( x) = −2 ln x
52. y = Ca x
(1, 2): 2
= Ca
(2, 1): 1
= Ca 2
y
2
1
Dividing eliminates C:
2
Ca
1
=
=
1
Ca 2
a
1
and C = 4.
2
So, a =
x
⎛1⎞
y = 4⎜ ⎟ = 4( 2− x )
⎝ 2⎠
x
1
2
3
4
−1
−2
Domain: x > 0
63. f ( x) = ln 2 x
53. f ( x) = ln x + 1
y
Vertical shift 1 unit upward
2
Matches (b)
1
54. f ( x) = −ln x
x
1
Reflection in the x-axis
2
−1
Matches (d)
55. f ( x) = ln ( x − 1)
Domain: x > 0
Horizontal shift 1 unit to the right
Matches (a)
64. f ( x) = ln x
y
56. f ( x) = −ln ( − x)
3
2
Reflection in the y-axis and the x-axis
1
Matches (c)
−3
−2
ln 0.1353… = −2
e
3
Domain: x ≠ 0
65. f ( x) = ln ( x − 3)
ln 2 = 0.6931…
0.6931…
2
−3
e −2 = 0.1353…
59.
1
−2
ln 1 = 0
58.
x
−1
e0 = 1
57.
3
y
= 2
4
3
ln 0.5 = − 0.6931…
60.
e − 0.6931… =
2
1
x
1
2
−1
1
2
3
4
5
6
7
−2
61. f ( x) = 3 ln x
−3
−4
Domain: x > 3
y
3
2
1
x
−1
1
2
3
4
5
−2
−3
Domain: x > 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.6
66. f ( x) = ln x − 4
Exponential and Logarithmic Functions
59
72. 3 units upward: ln x + 3
Reflected in x-axis: ln ( − x) + 3
y
x
1
−1
2
3
4
5
6
−2
y = ln ( − x) + 3
73. f ( x ) = e 2 x
−3
−4
g ( x) = ln
−5
1
ln x
2
x =
−6
5
Domain: x > 0
f
67. h( x) = ln ( x + 2)
g
−3
6
y
−1
3
74. f ( x ) = e x 3
2
1
x
−3 −2
−1
1
2
3
g ( x) = ln x 3 = 3 ln x
−2
8
f
−3
g
Domain: x > −2
−4
68. f ( x) = ln ( x − 2) + 1
y
−4
75. f ( x ) = e x − 1
4
g ( x) = ln ( x + 1)
2
5
x
−2
8
4
6
f
−2
g
−4
−3
Domain: x > 2
69. 8 units upward: e x + 8
Reflected in x-axis: − (e x + 8)
76. f ( x ) = e x −1
g ( x) = 1 + ln x
y = − (e x + 8) = − e x − 8
70. 2 units to the left: e x + 2
6
−1
4
f
g
−2
4
6 units downward: e x + 2 − 6
y = ex+ 2 − 6
−2
71. 5 units to the right: ln ( x − 5)
1 unit downward: ln ( x − 5) − 1
y = ln ( x − 5) − 1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
60
Chapter 1
Preparation for Calculus
y = e 4 x −1
77. (a)
(b)
1.25
f
ln y = 4 x − 1
f −1
ln y + 1 = 4 x
x =
f −1 ( x ) =
(ln y
1 ln x
4(
1
4
+ 1)
− 0.75
1
4
(ln e4 x −1 + 1) = 14 (4 x − 1 + 1) =
( 14 (ln x + 1)) = e(
ln x +1) −1
x
= eln x = x
y = 3e − x
78. (a)
(b)
y
= e− x
3
y
ln = − x
3
x = −ln
f −1 ( x) = ln
1.5
+ 1)
(c) f −1 ( f ( x)) = f −1 (e 4 x −1 ) =
f ( f −1 ( x)) = f
− 1.5
5
f
−1
5
f −1
−1
3
y
= ln
3
y
3
= ln 3 − ln x
x
(c) f −1 ( f ( x)) = f −1 (3e − x ) = ln 3 − ln (3e − x ) = ln 3 − ln 3 − ln e − x = x
⎛ 3⎞
⎛ x⎞
f ( f −1 ( x)) = f ⎜ ln ⎟ = 3e −ln(3 x) = 3eln(3 x) = 3⎜ ⎟ = x
x
⎝
⎠
⎝ 3⎠
y = 2 ln ( x − 1)
79. (a)
(b)
y
= ln ( x − 1)
2
ey 2 = x − 1
x = 1 + ey
f
−1
( x)
=1+e
6
f −1
f
−3
9
−2
2
x2
(c) f −1 ( f ( x )) = f −1 ( 2 ln ( x − 1)) = 1 + eln( x −1) = 1 + x − 1 = x
⎛ x⎞
f ( f −1 ( x )) = f (1 + x x 2 ) = 2 ln ⎡⎣(1 + e x 2 ) − 1⎤⎦ = 2⎜ ⎟ = x
⎝ 2⎠
y = 3 + ln ( 2 x)
80. (a)
(b)
6
f
y − 3 = ln 2 x
e y −2 = 2x
x =
f −1 ( x ) =
f −1
−2
1 e y −3
2
1 ex −3
2
(c) f −1 ( f ( x )) = f −1 (3 + ln ( 2 x)) =
f ( f −1 ( x )) = f
2
7
−1
1 e3 + ln( 2 x) − 3
2
=
1
2
( 2 x)
= x
( 12 e ) = 3 + ln(e ) = 3 + ( x − 3) =
x −3
x −3
x
=
81. ln e x = x 2
84. eln
82. ln e 2 x −1 = 2 x − 1
85. −1 + ln e 2 x = −1 + 2 x
83. eln(5 x + 2) = 5 x + 2
86. −8 + eln x = −8 + x3
x
x
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 1.6
87. (a) ln 6 = ln 2 + ln 3 ≈ 1.7917
103.
(b) ln 23 = ln 2 − ln 3 ≈ − 0.4055
1⎡
1 x( x + 3)
2 ln ( x + 3) + ln x − ln ( x 2 − 1)⎤⎦ = ln 2
⎣
x −1
3
3
(c) ln 81 = 4 ln 3 ≈ 4.3944
3 =
(d) ln
= ln
88. (a) ln 0.25 = ln
1
4
1
3
(2 ln
2 + ln 3) ≈ 0.8283
105. 2 ln 3 −
x
= ln x − ln 4
4
x5 = ln x5 2 =
90. ln
106.
5
2
ln x
(
93. ln x
)
3
z + 1 = ln ( z + 1)
13
1
2
=
(b) ln e 2 x = 3
ln ( x 2 + 5)
2x = 3
1
3
x =
ln ( z + 1)
96. ln z ( z − 1) = ln z + ln ( z − 1)
2
3
3
2
x = 6
(b) ln e − x = 0
−x = 0
x = 0
109. (a) ln x = 2
2
x = e 2 ≈ 7.389
(b) e x = 4
x = ln 4 ≈ 1.386
97. ln 3e 2 = ln 3 + 2 ln e = 2 + ln 3
110. (a) ln x 2 = 8
1
98. ln = ln 1 − ln e = −1
e
x 2 = e8
x = ± e 4 ≈ ± 54.598
99. ln x + ln 7 = ln ( x ⋅ 7) = ln (7 x)
100. ln y + ln x = ln ( yx
⎛ x2 + 1⎞
⎜ 2
⎟
⎝ x − 1⎠
2 x = 12
= ln z + 2 ln ( z − 1)
2
x +1
108. (a) eln 2x = 12
x −1
1 ⎛ x − 1⎞
⎛ x − 1⎞
= ln ⎜
⎟ = ln ⎜
⎟
x
x
2 ⎝ x ⎠
⎝
⎠
1
= ⎡⎣ln ( x − 1) − ln x⎤⎦
2
1
1
= ln ( x − 1) − ln x
2
2
2
9
2
x = 4
12
95. ln
2
3⎡
x2 + 1
2
⎤ = 3 ln
ln
x
+
1
−
ln
x
+
1
−
ln
x
−
1
(
)
(
)
(
)
⎦ 2 ( x + 1)( x − 1)
2⎣
12
2
= ln x +
x −1
107. (a) eln x = 4
x + 5 = ln x + ln ( x + 5)
2
(b) e −2 x = 5
)
−2 x = ln 5
x = − 12 ln 5 ≈ − 0.805
x − 2
101. ln ( x − 2) − ln ( x + 2) = ln
x + 2
111.
102. 3 ln x + 2 ln y − 4 ln z = ln x + ln y − ln z
3
= ln
x3 y 2
z4
2
2
2
x 2 + 1 = ln
= ln
92. ln ( xyz ) = ln x + ln y + ln z
94. ln
1
ln( x 2 + 1) = ln 9 − ln
2
xy
= ln x + ln y − ln z
z
91. ln
x( x + 3)
⎛ x ⎞
= ln ⎜ 2
⎟
⎝ x − 1⎠
1 = ln 1 − 3 ln 2 + 2 ln 3 ≈ − 4.2765
(d) ln 72
(
)
89. ln
2
x
104. 2 ⎡⎣ln x − ln ( x + 1) − ln ( x − 1)⎤⎦ = 2 ln
( x + 1)( x − 1)
= ln 1 − 2 ln 2 ≈ −1.3862
(b) ln 24 = 3 ln 2 + ln 3 ≈ 3.1779
(c) ln 3 12 =
3
ln 3 ≈ 0.5493
1
2
61
Exponential and Logarithmic Functions
4
ex > 5
ln e x > ln 5
x > ln 5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
62
Chapter 1
Preparation for Calculus
e1− x < 6
112.
121.
y
ln e1− x < ln 6
4
1 − x < ln 6
3
x > 1 − ln 6
(8, 3)
2
(2, 1)
1
113. −2 < ln x < 0
(1, 0)
x
e −2 < x < e0 = 1
1
< x <1
e2
114. 1 < ln x < 100
e1 < x < e100
4
6
x
1
2
8
y
0
1
3
8
(a) y is an exponential function of x: False
e < x < e100
115.
2
(b) y is a logarithmic function of x:
True; y = log 2 x
(c) x is an exponential function of y: True; 2 y = x
3
(d) y is a linear function of x:
f=g
False
9
0
122. The graph is that of y2 = eln x .
The domain of y1 = ln (e x ) is ( − ∞, ∞).
−3
116.
The domain of y2 = eln x is x > 0.
3
No, ln e x ≠ eln x for all real values of x. They are equal
for x > 0.
f=g
−1
5
10
⎛ I ⎞
ln ⎜
⎟
ln 10 ⎝ 10−16 ⎠
10
⎡ln I − ln 10−16 ⎤⎦
=
ln 10 ⎣
123. (a) β =
−1
117. The domain of the natural logarithmic function
is (0, ∞) and the range is ( −∞, ∞). The function is
continuous, increasing, and one-to-one, and its
graph is concave downward. In addition, if a and b
are positive numbers and n is rational, then ln (1) = 0,
=
10
[ln I + 16ln 10]
ln 10
ln ( a ⋅ b) = ln a + ln b, ln ( a n ) = n ln a, and
=
10
ln I + 160
ln 10
ln ( a b) = ln a − ln b.
= 10 log10 I + 160
118. The functions f ( x ) = e x and g ( x) = ln x are inverses
124. β (10− 5 ) =
of each other. So, ln e x = g ( f ( x)) = x.
= − 50 + 160 = 110 decibels
119. f ( x ) = e x . Domain is ( −∞, ∞) and range is (0, ∞). f is
continuous, increasing, one-to-one, and concave upwards
on its entire domain.
lim e x = 0 and lim e x = ∞
x →−∞
125. False
ln x + ln 25 = ln ( 25 x) ≠ ln ( x + 25)
126. False. The property is
x →∞
120. The graphs of f ( x) = ln x and g ( x) = e are mirror
x
images in the line y = x.
10
ln 10− 5 + 160
ln 10
ln xy = ln x + ln y (for x, y > 0 ). As a counter
example, let x = y = e. Then
ln xy = ln e 2 = 2
and
ln x ln y = 1 ⋅ 1 = 1.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 1
127.
45
g
(
y = ln x +
(c)
f
ey = x +
−3
(e y
3
−5
128.
2 xe y = e 2 y − 1
130. p( x ) =
f
p′( x ) =
g
10,000
x
ln x
⎛1⎞
x⎜ ⎟
⎝ x ⎠ = ln x − 1
2
(ln x)
(ln x)2
(ln x)(1) −
(a) p′(1000) =
0
The graphs intersect twice: ( 4.1771, 1.4296) and
(a)
(ln 1000)2
(b) p′(1,000,000) =
g ( x) = x1 4 grows more rapidly.
(
ln 1000 − 1
≈ 0.1238
About 12.4 primes per 100 integers
(5503.647, 8.6132).
129. f ( x ) = ln x +
e2 y − 1
2e x
x =
10
0
x2 + 1
2
and (6, 46,656).
The function f ( x ) = 6 x grows more rapidly.
)
− x) = x 2 + 1
The graphs intersect three times: ( − 0.7899, 0.2429),
(1.6242, 18.3615)
x2 + 1
63
x2 + 1
)
ln (1,000,000) − 1
(ln 1,000,000)2
≈ 0.0671
About 6.7 primes per 100 integers
(c) p′(1,000,000,000) =
6
ln (1,000,000,000) − 1
(ln 1,000,000,000)2
≈ 0.0459
−9
9
About 4.6 primes per 100 integers
131.
−6
Domain: −∞ < x < ∞
(
(b) f ( − x) = ln − x +
(
x2 + 1
⎡ −x +
= ln ⎢⎢
⎣⎢
(
)
(− x −
( x x + 1)
⎡
= ln ⎢⎢
⎢⎣ x +
(
= −ln ( x +
3 ⋅ 2 ⋅ 1 = 479,001,600
Stirlings Formula:
12
)(
)
⎛ 12 ⎞
12! ≈ ⎜ ⎟
⎝e⎠
)
x2 + 1 ⎤
⎥
⎥
2
x +1
⎦⎥
x2 + 1 − x −
2
⎡
= ln ⎢⎢
⎢⎣ − x −
n = 12
12! = 12 ⋅ 11 ⋅ 10
⎤
⎥
⎥
x2 + 1 ⎥
⎦
132.
2π (12) ≈ 475,687,487
n = 15
15! = 15 ⋅ 14
3 ⋅ 2 ⋅ 1 = 1,307,674,368,000
2
Stirlings Formula:
)
15
⎛ 15 ⎞
15! ≈ ⎜ ⎟
⎝e⎠
⎤
⎥
⎥
x2 + 1 ⎥
⎦
1
x2
)
+ 1) = − f ( x )
2π (15) ≈ 1,300,430,722,200
≈ 1.3004 × 1012
Review Exercises for Chapter 1
1. y = 5 x − 8
x = 0: y = 5(0) − 8 = −8 ⇒ (0, − 8), y-intercept
y = 0: 0 = 5 x − 8 ⇒ x =
8
5
⇒
( 85 , 0), x-intercept
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64
Chapter 1
Preparation for Calculus
2. y = x 2 − 8 x + 12
x = 0: y = (0) − 8(0) + 12 = 12 ⇒ (0, 12), y -intercept
2
y = 0: x 2 − 8 x + 12 = ( x − 6)( x − 2) = 0 ⇒ x = 2, 6 ⇒ ( 2, 0), (6, 0), x -intercepts
x −3
x−4
3. y =
x = 0: y =
0−3
3
⎛ 3⎞
=
⇒ ⎜ 0, ⎟, y-intercept
0 − 4
4
⎝ 4⎠
y = 0: 0 =
x −3
⇒ x = 3 ⇒ (3, 0), x-intercept
x − 4
4. y = ( x − 3)
x + 4
x = 0: y = (0 − 3) 0 + 4 = − 3 4 = − 3( 2) = − 6 ⇒ (0, − 6), y -intercept
y = 0: ( x − 3)
x + 4 = 0 ⇒ x = 3, − 4 ⇒ (3, 0), ( − 4, 0), x-intercepts
5. y = x 2 + 4 x does not have symmetry with respect to either axis or the origin.
6. Symmetric with respect to y-axis because
y = ( − x) − (− x) + 3
4
2
y = x 4 − x 2 + 3.
7. Symmetric with respect to both axes and the origin because:
y 2 = (− x2 ) − 5
( − y )2
y2 = x2 − 5
y2 = x2 − 5
= x2 − 5
8. Symmetric with respect to the origin because:
(− x)(− y )
= −2
( − y )2
= ( − x) − 5
2
y 2 = x2 − 5
10. y = − x 2 + 4
y-intercept: y = − (0) + 4 = 4
2
xy = − 2.
(0, 4)
1
9. y = − x + 3
2
−x + 4 = 0
2
x-intercepts:
y-intercept: y = −
(2
1
( 0) + 3 = 3
2
− x)( 2 + x) = 0
x = ±2
(2, 0), (− 2, 0)
(0, 3)
1
x-intercept: − x + 3 = 0
2
1
− x = −3
2
x = 6
Symmetric with respect to the y-axis because
− ( − x) + 4 = − x 2 + 4.
2
y
5
(0, 4)
(6, 0)
Symmetry: none
3
2
1
(−2, 0)
y
−3
(2, 0)
−1
−1
1
x
3
6
4
(0, 3)
2
(6, 0)
−2
2
4
x
6
−2
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 1
11. y = x 3 − 4 x
13. y = 2 4 − x
y-intercept: y = 0 − 4(0) = 0
3
y-intercept: y = 2 4 − 0 = 2 4 = 4
(0, 4)
(0, 0)
x3 − 4 x = 0
x-intercept: 2 4 − x = 0
2
x ( x − 4) = 0
4− x = 0
x( x − 2)( x + 2) = 0
4− x = 0
x-intercepts:
x = 4
x = 0, 2, − 2
(4, 0)
(0, 0), (2, 0), (− 2, 0)
Symmetric with respect to the origin because
( − x)
3
65
Symmetry: none
− 4( − x) = − x3 + 4 x = − ( x3 − 4 x).
y
5
(0, 4)
y
4
3
3
1
(− 2, 0)
−4 −3
−1
2
1
(0, 0) (2, 0)
1
3
x
4
−1
(4, 0)
x
1
−1
2
3
4
5
−2
−3
14. y = x − 4 − 4
−4
y-intercept: y = 0 − 4 − 4 = − 4 − 4 = 4 − 4 = 0
y2 = 9 − x
12.
(0, 0)
y2 + x − 9 = 0
y-intercept: y 2 = 9 − 0 = 9 ⇒ y = ± 3
x-intercepts: x − 4 − 4 = 0
x − 4 = 4
(0, 3), (0, − 3)
x − 4 = 4 or x − 4 = − 4
x-intercept: 02 = 9 − x ⇒ x = 9
x = 8
(9, 0)
(0, 0), (8, 0)
Symmetric with respect to the x-axis because
(− y )2
+ x − 9 = y 2 + x − 9 = 0.
Symmetry: none
y
6
y
5
4
4
2
(0, 3)
(9, 0)
2
1
−4
−5
1 2 3 4 5 6 7
(0, 0)
−2
−2
x
−1
−2
x = 0
2
(8, 0)
4
6
8
x
10
−4
9
−6
(0, −3)
15. 5 x + 3 y = −1 ⇒ y =
1
3
( −5 x
− 1)
x − y = −5 ⇒ y = x + 5
1
3
( −5 x
− 1) = x + 5
−5 x − 1 = 3 x + 15
−16 = 8 x
−2 = x
For x = −2, y = x + 5 = −2 + 5 = 3.
Point of intersection is: ( − 2, 3)
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66
Chapter 1
Preparation for Calculus
− 2x + 9
4
6x − 7
6x − 4 y = 7 ⇒ y =
4
− 2x + 9
6x − 7
=
4
4
− 2x + 9 = 6x − 7
20. The line is horizontal and has slope 0.
16. 2 x + 4 y = 9 ⇒ y =
y
(−7, 8)
(−1, 8)
6
4
2
−8
− 8 x = −16
−6
−4
2
−2
x = 2
For x = 2, y =
6( 2) − 7
4
=
5
4
21. y − ( −5) =
7
4
y + 5 =
⎛ 5⎞
Point of intersection: ⎜ 2, ⎟
⎝ 4⎠
(x
− 3)
−
7
x
4
21
4
4 y + 20 = 7 x − 21
0 = 7 x − 4 y − 41
y
x − y = −5 ⇒ y = x + 5
17.
x
−2
2
x2 − y = 1 ⇒ y = x2 − 1
−8 −6 −4 −2
x + 5 = x2 − 1
x
2
−4
0 = x − x −6
−8
−10
0 = ( x − 3)( x + 2)
6
8
(3, −5)
−6
2
4
(0, − 414(
x = 3 or x = − 2
For x = 3, y = 3 + 5 = 8.
22. Because m is undefined the line is vertical.
x = −8 or x + 8 = 0
For x = − 2, y = − 2 + 5 = 3.
y
Points of intersection: (3, 8), ( − 2, 3)
6
4
18. x 2 + y 2 = 1 ⇒ y 2 = 1 − x 2
2
−x + y = 1 ⇒ y = x + 1
1 − x = ( x + 1)
2
(−8, 1)
−6
2
−4
x
−2
2
−2
−4
1 − x2 = x2 + 2x + 1
0 = 2x2 + 2x
0 = 2 x( x + 1)
23.
x = 0 or x = −1
For x = 0, y = 0 + 1 = 1.
y
y − 0 = − 23 ( x − ( −3))
3
y = − 23 x − 2
2x + 3y + 6 = 0
2
1
(− 3, 0)
−4 −3
x
−1
1
2
3
For x = −1, y = −1 + 1 = 0.
−3
Points of intersection: (0, 1), ( −1, 0)
y
19.
−4
24. Because m = 0, the line is horizontal.
y − 4 = 0( x − 5)
5
4
3
y
y = 4 or y − 4 = 0
( 5, )
5
2
8
6
2
1
(5, 4)
( 32 , 1 )
1
2
3
2
x
4
5
⎛5⎞
3
⎜ ⎟ −1
3
2⎠
⎝
2
Slope =
=
=
7
⎛ 3⎞
7
5−⎜ ⎟
2
⎝ 2⎠
−4
x
−2
2
4
6
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 1
25. y = 6
y
31. (a)
7
Slope: 0
y -intercept: (0, 6)
y −5 =
7
16
(x
67
+ 3)
16 y − 80 = 7 x + 21
5
0 = 7 x − 16 y + 101
4
3
(b) 5 x − 3 y = 3 has slope 53 .
2
1
y −5 =
x
−4 −3 −2 −1
1
2
3
4
5
3
(x
+ 3)
3 y − 15 = 5 x + 15
0 = 5 x − 3 y + 30
y
26. x = − 3
3
Slope: undefined
(c) 3x + 4 y = 8
2
Line is vertical.
4 y = − 3x + 8
1
−5 −4
x
−2 −1
−1
1
y =
−2
−3
27. y = 4 x − 2
−3
x + 2
4
4
Perpendicular line has slope .
3
y
Slope: 4
4
y -intercept: (0, − 2)
2
4
( x − (− 3))
3
3 y − 15 = 4 x + 12
y −5 =
3
1
x
−4 −3 −2 −1
1
2
3
4 x − 3 y + 27 = 0 or y =
4
−2
−3
4
x + 9
3
(d) Slope is undefined so the line is vertical.
x = −3
28. 3 x + 2 y = 12
x +3 = 0
y
2 y = − 3x + 12
−3
x + 6
2
3
Slope : −
2
y -intercept: (0, 6)
6
y =
29.
m =
y −0 =
y =
4y − x =
30.
2−0
1
=
8−0
4
1
( x − 0)
4
1
x
4
0
m =
2
x
−4 −2
−2
2
4
6
2 x + 3 y − 16 = 0
8
(b) x + y = 0 has slope −1. Slope of the perpendicular
line is 1.
−4
y − 4 = 1( x − 2)
y
y = x + 2
4
0 = x − y + 2
3
2
1
−4
(c) m =
x
−1
1
2
3
4
−2
3
( x − 2)
4
4 y − 16 = −3 x + 6
−4
−2
( x − (− 5))
5
5 y − 25 = − 2 x − 10
4 −1
3
= −
2−6
4
y −4 = −
−3
3x + 4 y − 22 = 0
2
−1 − 5
−6
=
= −
10 − ( − 5)
15
5
(d) Because the line is horizontal the slope is 0.
y = 4
y −5 =
5 y + 2 x − 15 = 0
2
( x − 2)
3
3 y − 12 = −2 x + 4
y − 4 = −
32. (a)
4
y −4 = 0
y
8
33. The slope is −850.
6
V = −850t + 12,500.
4
2
−2
−2
x
2
4
6
V (3) = −850(3) + 12,500 = $9950
8
−4
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68
Chapter 1
Preparation for Calculus
34. (a) C = 9.25t + 13.50t + 36,500 = 22.75t + 36,500
39.
Domain: ( −∞, ∞)
(b) R = 30t
(c)
f ( x) = x 2 + 3
30t = 22.75t + 36,500
Range: [3, ∞)
7.25t = 36,500
40. g ( x) =
t ≈ 5034.48 hours to break even
6− x
Domain: 6 − x ≥ 0
35. f ( x) = 5 x + 4
6 ≥ x
(a) f (0) = 5(0) + 4 = 4
(−∞, 6]
Range: [0, ∞)
(b) f (5) = 5(5) + 4 = 29
(c) f ( − 3) = 5( − 3) + 4 = −11
41.
(d) f (t + 1) = 5(t + 1) + 4 = 5t + 9
f ( x) = − x + 1
Domain: ( − ∞, ∞)
Range: ( −∞, 0]
36. f ( x ) = x 3 − 2 x
(a) f ( − 3) = ( − 3) − 2(− 3) = − 27 + 6 = − 21
3
42.
(b) f ( 2) = 23 − 2( 2) = 8 − 4 = 4
2
x +1
Domain: all x ≠ −1; ( −∞, −1) ∪ (−1, ∞)
h( x ) =
Range: all y ≠ 0; ( −∞, 0) ∪ (0, ∞)
(c) f ( −1) = ( −1) − 2( −1) = −1 + 2 = 1
3
(d) f (c − 1) = (c − 1) − 2(c − 1)
3
43. x − y 2 = 6
= c 3 − 3c 2 + 3c − 1 − 2c + 2
y = ±
∆x
4( x + ∆x) − 4 x
2
(
2
) − 4x
8 x∆x + 4( ∆x)
∆x
= 8 x + 4∆x,
f (1) = 2(1) − 6 = − 4
=
44. x 2 − y = 0
(2 x
y
6
Function of x because
there is one value for
y for each x.
5
4
3
2
2
1
x
−3
− 6) − ( − 4)
x −1
2x − 6 + 4
=
x −1
2x − 2
=
x −1
2( x − 1)
=
x −1
= 2, x ≠ 1
−2
−1
1
2
3
∆x ≠ 0
45. y =
38. f ( x ) = 2 x − 6
8 10 12 14
−2
∆x
=
x −1
2
4 x 2 + 8 x∆x + 4( ∆x) − 4 x 2
2
f ( x) − f ( −1)
4
−4
∆x
=
x
2
−1
−3
∆x
2
1
2
4 x + 2 x∆x + ( ∆x)
=
3
Not a function because
there are two values of
y for some x.
37. f ( x ) = 4 x 2
=
4
x−6
2
= c 3 − 3c 2 + c + 1
f ( x + ∆x) − f ( x)
y
y
x −2
x −2
y is a function of x
because there is one
value of y for each x.
4
3
2
1
x
−2 −1
1
3
4
5
6
−2
−3
−4
46. x = 9 − y 2
Not a function of x
since there are two
values of y for some x.
y
4
2
1
x
− 12 − 9 − 6 − 3
−1
3
6
12
−2
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 1
47. f ( x ) = x 3 − 3x 2
49. (a) f ( x ) = x 2 ( x − 6)
69
2
100
The leading coefficient is
positive and the degree is
even so the graph will rise
to the left and to the right.
6
(0, 0)
−6
6
−4
(2, −4)
10
− 25
6
(a) The graph of g is obtained from f by a vertical shift
down 1 unit, followed by a reflection in the x-axis:
(b) g ( x) = x 3 ( x − 6)
(b) The graph of g is obtained from f by a vertical shift
upwards of 1 and a horizontal shift of 2 to the right.
g ( x) = f ( x − 2) + 1 = ( x − 2) − 3( x − 2) + 1
2
300
The leading coefficient is
positive and the degree is
odd so the graph will rise
to the right and fall to
the left.
g ( x) = −⎡⎣ f ( x ) − 1⎤⎦ = − x 3 + 3 x 2 + 1
3
2
(c) h( x) = x 3 ( x − 6)
−2
10
− 100
3
200
48. (a) Odd powers: f ( x) = x, g ( x) = x , h( x ) = x
3
The leading coefficient is
positive and the degree is
even so the graph will rise
to the left and to the right.
5
g
2
h
−4
10
− 800
−3
3
f
−2
The graphs of f, g, and h all rise to the right and fall
to the left. As the degree increases, the graph rises
and falls more steeply. All three graphs pass through
the points (0, 0), (1, 1), and ( −1, −1) and are
50. (a)
(b)
(c)
(d)
3 (cubic), negative leading coefficient
4 (quartic), positive leading coefficient
2 (quadratic), negative leading coefficient
5, positive leading coefficient
symmetric with respect to the origin.
51. For company (a) the profit rose rapidly for the first year,
and then leveled off. For the second company (b), the
profit dropped, and then rose again later.
Even powers: f ( x ) = x 2 , g ( x) = x 4 , h( x) = x 6
52. (a)
y
g
4
h
x
f
x
y
−3
2 x + 2 y = 24
3
y = 12 − x
0
The graphs of f, g, and h all rise to the left and to the
right. As the degree increases, the graph rises more
steeply. All three graphs pass through the points
(0, 0), (1, 1), and (−1, 1) and are symmetric with
A = xy = x(12 − x )
(b) Domain: 0 < x < 12 or (0, 12)
40
respect to the y-axis.
All of the graphs, even and odd, pass through the
origin. As the powers increase, the graphs become
flatter in the interval −1 < x < 1.
(b) y = x will look like h( x ) = x , but rise and fall
7
5
even more steeply. y = x8 will look like
h( x) = x 6 , but rise even more steeply.
0
12
0
(c) Maximum area is A = 36 in.2 . In general, the
maximum area is attained when the rectangle is
a square. In this case, x = 6.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
70
Chapter 1
Preparation for Calculus
53. (a) y = −1.204 x + 64.2667
(b)
57. (a)
f ( x) =
1x
2
−3
y =
1x
2
−3
70
2( y + 3) = x
2( x + 3) = y
0
f −1 ( x) = 2 x + 6
33
0
(c) The data point (27, 44) is probably an error.
Without this point, the new model is
y = −1.4344 x + 66.4387.
(b)
7
f
−1
− 11
10
f
54. (a) Using a graphing utility, you obtain
y = − 0.043x 2 + 4.19 x − 56.2.
(b)
−7
(c) f −1 ( f ( x )) = f −1
50
f(f
10
80
( x))
= f ( 2 x + 6) =
+ 6) − 3 = x
(5 x
− 7) + 7
y + 7
= x
5
x + 7
= y
5
x + 7
f −1 ( x) =
5
y = − 0.043( 26) + 4.19( 26) − 56.2
2
≈ $23.7 thousand
(d) For x = 34 :
y = − 0.043(34) + 4.19(34) − 56.2
2
(b)
55. (a) Using a graphing utility,
6
f
− 10
−1
6
f
y = 0.61t 2 + 11.0t + 172
− 10
1200
(c) f
4
−1
( f ( x))
f ( f −1 ( x ))
32
0
The model fits the data well.
59. (a)
56. (a) Yes, y is a function of t. At each time t, there is one
and only one displacement y.
(b) The amplitude is approximately
(0.25 − (−0.25)) 2 = 0.25. The period is
(c) One model is y =
1
1
⎛ 2π ⎞
cos⎜ t ⎟ ≈
cos(5.7t )
4
1.1
4
⎝
⎠
= f −1 (5 x − 7) =
= x
5
⎛ x + 7⎞
⎛ x + 7⎞
= f⎜
⎟ = 5⎜
⎟ −7 = x
⎝ 5 ⎠
⎝ 5 ⎠
f ( x) =
x +1
y =
x +1
y −1 = x
2
x2 − 1 = y
f −1 ( x) = x 2 − 1, x ≥ 0
approximately 1.1.
(d)
(2 x
x
y = 5x − 7
(c) For x = 26 :
(b)
1
2
f ( x) = 5 x − 7
58. (a)
0
≈ $36.6 thousand
−1
( 12 x − 3) = 2( 12 x − 3) + 6 =
(b)
4
f −1
f
0.5
−3
(1.1, 0.25)
0
2.2
(0.5, −0.25)
−0.5
The model appears to fit the data.
6
−2
(c) f −1 ( f ( x)) = f −1
(
)
x +1 =
f ( f −1 ( x)) = f ( x 2 − 1) =
=
( x 2 − 1)
( x 2 − 1) + 1
2
−1 = x
x 2 = x for x ≥ 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 1
63. f ( x) = 2 arctan ( x + 3)
f ( x) = x3 + 2
60. (a)
y = x3 + 2
3
y − 2 = x
3
x − 2 = y
f −1 ( x) =
71
y
4
x − 2
3
−6
−4
x
−2
2
−2
(b)
3
f
f
−4
−1
−4
64. h( x) = −3 arcsin ( 2 x)
5
y
−3
4
(c) f −1 ( f ( x)) = f −1 ( x3 + 2) =
f ( f −1 ( x)) = f
61. (a)
(
3
f ( x) =
3
x +1
y =
3
x +1
3
) (
x − 2 =
3
( x 3 + 2) − 2
x − 2
)
3
= x
+ 2 = x
−π
2
−π
4
π
2
x
−2
−4
65. Let θ = arcsin 12 .
y −1 = x
3
2
sin θ =
x −1 = y
3
(
1
2
)
1
sin arcsin 12 = sin θ =
f −1 ( x ) = x 3 − 1
(b)
π
4
4
1
2
θ
3
66. Let θ = arccot 2.
f −1
5
cot θ = 2
f
−4
1
tan (arccot ) = tan θ =
5
1
2
θ
2
−2
(c) f −1 ( f ( x)) = f −1
=
(
(
x +1
3
x +1
3
)
3
)
67. f ( x) = e x matches (d).
The graph is increasing and the domain is all real x.
−1= x
f ( f −1 ( x )) = f ( x3 − 1) =
3
( x3 − 1) + 1
= x
68. f ( x ) = e − x matches (a).
The graph is decreasing and the domain is all real x.
f ( x) = x 2 − 5,
62. (a)
x ≥ 0
69. f ( x) = ln ( x + 1) + 1 matches (c).
y = x2 − 5
The graph is increasing and the domain is x > −1.
y +5 = x
70. f ( x) = − ln ( x + 1) + 1 matches (b).
x +5 = y
f −1 ( x) =
(b)
The graph is decreasing and the domain is x > −1.
x + 5
71. f ( x) = ln x + 3
4
f −1
−6
Vertical shift three units upward
6
Vertical asymptote: x = 0
f
y
−6
(c) f −1 ( f ( x)) = f −1 ( x 2 − 5)
=
f ( f −1 ( x)) = f
( x2
(
5
4
− 5) + 5 = x for x ≥ 0.
) (
x + 5 =
x + 5
)
2
−5 = x
3
2
1
x
1
2
3
4
5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
72
Chapter 1
Preparation for Calculus
72. f ( x ) = ln ( x − 1)
73. ln
Horizontal shift one unit to the right
Vertical asymptote: x = 1
y
5
4 x 2 − 1 1 ( 2 x − 1)( 2 x + 1)
= ln
4x2 + 1 5
4x2 + 1
1
= ⎡⎣ln ( 2 x − 1) + ln ( 2 x + 1) − ln ( 4 x 2 + 1)⎤⎦
5
74. ln ⎡⎣( x 2 + 1)( x − 1)⎤⎦ = ln ( x 2 + 1) + ln ( x − 1)
2
1
x
75. ln 3 +
4
3
−1
1
ln ( 4 − x 2 ) − ln x = ln 3 + ln
3
3
4 − x 2 − ln x
⎛ 3 3 4 − x2 ⎞
⎟
= ln ⎜
⎜
⎟
x
⎝
⎠
−2
76. 3⎡⎣ln x − 2 ln ( x 2 + 1)⎤⎦ + 2 ln 5 = 3 ln x − 6 ln ( x 2 + 1) + ln 52
⎡ 25 x 3 ⎤
6
⎥
= ln x 3 − ln ( x 2 + 1) + ln 25 = ln ⎢
6
⎢ ( x 2 + 1) ⎥
⎣
⎦
77. ln
x +1 = 2
x + 1 = e2
x + 1 = e4
x = e 4 − 1 ≈ 53.598
78. ln x + ln ( x − 3) = 0
ln x( x − 3) = 0
x( x − 3) = e0
x 2 − 3x − 1 = 0
3±
x =
3+
x =
79. (a)
13
2
f ( x ) = ln
x
y = ln
x
ey =
13
2
only because
3−
13
2
< 0.
f ( x ) = e1 − x
80. (a)
y = e1 − x
ln y = 1 − x
x
x = 1 − ln y
e2 y = x
y = 1 − ln x
e2 x = y
f
−1
( x)
f −1 ( x ) = 1 − ln x
= e2 x
(b)
(b)
2
f
4
−1
f
−3
3
−1
( f ( x))
= f
f −1
5
−2
−2
(c) f
f
−4
−1
(ln
)
x = e 2 ln
f ( f −1 ( x )) = f (e 2 x ) = ln
x
= eln x = x
e 2 x = ln e x = x
(c) f −1 ( f ( x )) = f −1 (e1 − x ) = 1 − ln (e1 − x )
= 1 − (1 − x ) = x
f ( f −1 ( x )) = f (1 − ln x) = e1 − (1 − ln x) = eln x = x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 1
81. f = e − x 2
82. f = 4e − x
y
73
2
y
6
5
4
4
1
2
−5 −4 −3 −2 −1
−2
2
x
1 2 3 4 5
−2
−3
−4
−5
x
4
−2
Problem Solving for Chapter 1
1. (a)
x2 − 6 x + y 2 − 8 y = 0
( x2 − 6 x + 9) + ( y 2 − 8 y + 16)
(x
= 9 + 16
− 3) + ( y − 4) = 25
2
2
Center: (3, 4); Radius: 5
4
3
3
3
. Slope of tangent line is − . So, y − 0 = − ( x − 0) ⇒ y = − x Tangent line
3
4
4
4
4−0
4
(c) Slope of line from (6, 0) to (3, 4) is
= − .
3−6
3
3
3
3
9
Slope of tangent line is . So, y − 0 = ( x − 6) ⇒ y = x − Tangent line
4
4
4
2
3
3
9
(d) − x = x −
4
4
2
3
9
x =
2
2
x = 3
(b) Slope of line from (0, 0) to (3, 4) is
9⎞
⎛
Intersection: ⎜ 3, − ⎟
4⎠
⎝
2. Let y = mx + 1 be a tangent line to the circle from the point (0, 1). Because the center of the circle is at (0, −1)
and the radius is 1 you have the following.
x 2 + ( y + 1) = 1
2
x 2 + ( mx + 1 + 1) = 1
2
(m2 + 1) x 2 + 4mx + 3 = 0
Setting the discriminant b 2 − 4ac equal to zero,
16m 2 − 4( m 2 + 1)(3) = 0
16m 2 − 12m 2 = 12
4m 2 = 12
m = ±
Tangent lines: y =
3
3x + 1 and y = − 3 x + 1.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
74
Chapter 1
Preparation for Calculus
⎧1, x ≥ 0
3. H ( x) = ⎨
⎩0, x < 0
(e)
y
⎧1
1
⎪ , x ≥ 0
H ( x) = ⎨ 2
2
⎪0, x < 0
⎩
y
4
3
4
2
3
1
2
x
−4 −3 −2 −1
−1
1
2
3
1
4
−2
−4 −3 −2 −1
−1
−3
−2
−4
−3
⎧−1, x ≥ 0
(a) H ( x) − 2 = ⎨
⎩−2, x < 0
y
x
1
2
3
4
−4
⎧1, x ≥ 2
(f ) − H ( x − 2) + 2 = ⎨
⎩2, x < 2
4
3
y
2
1
4
x
−4 −3 −2 −1
−1
1
2
3
3
4
1
−3
−4 −3 −2 −1
−1
−4
x
1
2
3
4
−2
⎧1, x ≥ 2
(b) H ( x − 2) = ⎨
⎩0, x < 2
−3
−4
y
4
3
2
1
x
−4 −3 −2 −1
−1
1
2
3
4
−2
−3
−4
⎧−1, x ≥ 0
(c) − H ( x) = ⎨
⎩0, x < 0
y
4
3
2
1
x
−4 −3 −2 −1
−1
1
2
3
4
−2
−3
−4
⎧1, x ≤ 0
(d) H ( − x ) = ⎨
⎩0, x > 0
y
4
3
2
−4 −3 −2 −1
−1
x
1
2
3
4
−2
−3
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 1
4. (a) f ( x + 1)
75
f ( x)
(f )
y
y
4
4
2
−3
x
−1
1
−4
3
x
−2
−2
−2
−4
−4
(b) f ( x ) + 1
(g) f ( x
2
4
2
4
)
y
y
4
4
2
x
−4
4
−4
−2
x
−2
−2
−4
−4
(c) 2 f ( x)
5. (a) x + 2 y = 100 ⇒ y =
y
x2
⎛ 100 − x ⎞
A( x) = xy = x⎜
+ 50 x
⎟ = −
2 ⎠
2
⎝
4
−4
100 − x
2
Domain: 0 < x < 100 or (0, 100)
x
−2
2
4
(b)
−2
1600
−4
(d) f ( − x)
0
110
0
Maximum of 1250 m 2 at x = 50 m, y = 25 m.
y
(c) A( x) = − 12 ( x 2 − 100 x)
4
2
−4
x
−2
2
4
= − 12 ( x 2 − 100 x + 2500) + 1250
−2
= − 12 ( x − 50) + 1250
−4
A(50) = 1250 m 2 is the maximum.
2
x = 50 m, y = 25 m
(e) − f ( x)
y
4
2
−4
x
−2
2
4
−2
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
76
Chapter 1
Preparation for Calculus
6. (a) 4 y + 3 x = 300 ⇒ y =
300 − 3 x
4
−3 x 2 + 300 x
⎛ 300 − 3 x ⎞
A( x) = x( 2 y ) = x⎜
⎟ =
2
2
⎝
⎠
Domain: 0 < x < 100
y
(b)
4000
3500
3000
2500
2000
1500
1000
500
x
25
50
75
100
Maximum of 3750 ft 2 at x = 50 ft, y = 37.5 ft.
(c) A( x) = − 32 ( x 2 − 100 x)
= − 32 ( x 2 − 100 x + 2500) + 3750
= − 32 ( x − 50) + 3750
2
A(50) = 3750 square feet is the maximum area, where x = 50 ft and y = 37.5 ft.
7. The length of the trip in the water is
4 + x2
+
2
So, the total time is T =
22 + x 2 , and the length of the trip over land is
1 + (3 − x )
4
1 + (3 − x ) .
2
2
hours.
8. f ( x ) = e x = e − x
y
8
6
4
2
−8 −6 −4 −2
x
2
4
6
8
y = e x − e− x
ye x = e 2 x − 1
(e x )
2
(Quadratic in e x )
− ye x − 1 = 0
ex =
ex =
ey =
y ±
y2 + 4
2
x +
y2 + 4
2
x +
x2 + 4
2
⎡x +
f −1 ( x) = y = ln ⎢
⎢⎣
x2 + 4 ⎤
⎥
2
⎥⎦
( Use positive solution.)
Inverse
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 1
9. (a) Slope =
9−4
= 5. Slope of tangent line is less than 5.
3−2
(b) Slope =
4 −1
= 3. Slope of tangent line is greater than 3.
2 −1
(c) Slope =
4.41 − 4
= 4.1. Slope of tangent line is less than 4.1.
2.1 − 2
(d) Slope =
f ( 2 + h ) − f ( 2)
( 2 + h) − 2
=
77
( 2 + h)2
−4
h
4h + h 2
=
h
= 4 + h, h ≠ 0
(e) Letting h get closer and closer to 0, the slope approaches 4. So, the slope at (2, 4) is 4.
10.
y
4
3
2
(4, 2)
1
x
1
2
3
4
5
−1
(a) Slope =
3−2
1
1
= . Slope of tangent line is greater than .
9−4
5
5
(b) Slope =
2 −1
1
1
= . Slope of tangent line is less than .
4 −1
3
3
(c) Slope =
2.1 − 2
10
10
=
. Slope of tangent line is greater than .
4.41 − 4
41
41
(d) Slope =
f ( 4 + h ) − f ( 4)
=
( 4 + h) − 4
(e)
4+ h − 2
=
h
4+ h − 2
⋅
h
4+ h −2
h
( 4 + h) − 4 =
4+ h + 2
=
4 + h + 2
h 4 + h + 2
As h gets closer to 0, the slope gets closer to
11. f ( x ) = y =
(
)
1
, h ≠ 0
4+ h + 2
1
1
. The slope is at the point (4, 2).
4
4
1
1− x
(a) Domain: all x ≠ 1 or ( −∞, 1) ∪ (1, ∞)
Range: all y ≠ 0 or ( −∞, 0) ∪ (0, ∞)
x −1
1
1
1− x
⎛ 1 ⎞
(b) f ( f ( x)) = f ⎜
=
=
=
⎟ =
1− x −1
⎛ 1 ⎞
−x
x
⎝1 − x ⎠
1−⎜
⎟
1− x
⎝1 − x ⎠
Domain: all x ≠ 0, 1 or ( −∞, 0) ∪ (0, 1) ∪ (1, ∞)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78
Chapter 1
Preparation for Calculus
1
1
⎛ x − 1⎞
(c) f f ( f ( x)) = f ⎜
=
= x
⎟ =
1
⎝ x ⎠ 1 − ⎛ x − 1⎞
⎜
⎟
x
⎝ x ⎠
(
)
Domain: all x ≠ 0, 1 or ( −∞, 0) ∪ (0, 1) ∪ (1, ∞)
(d) The graph is not a line. It has holes at (0, 0) and (1, 1).
y
2
1
x
−2
1
2
−2
y
12. Using the definition of absolute value, you can rewrite the equation.
4
y + y = x+ x
3
2
⎧2 y ,
⎨
⎩0,
y > 0
⎧2 x, x > 0
=⎨
.
y ≤ 0 ⎩0,
x ≤ 0
1
−4 −3 −2 −1
(x
x2 + 6x − 9 = 0
x =
(x
14. (a)
2
8
2
+ 3) + y 2 = 18
6
2
−8
−4 −2
−2
x
2
4
2
Circle of radius
x
1
y
− 3) + y 2 = 2( x 2 + y 2 )
x2 + y2 + 6x − 9 = 0
= −3 ± 18
0
−4
x2 − 6x + 9 + y 2 = 2x2 + 2 y 2
36 + 36
2
≈ 1.2426, − 7.2426
4
2I
I
=
x2 + y 2
( x − 3)2 + y 2
(b)
x2 − 6x + 9 = 2x2
−6 ±
3
−3
For any x ≤ 0, y is any y ≤ 0. So, the graph of y + y = x + x is as follows.
2I
I
=
2
x2
x
( − 3)
2
−2
For x > 0 and y > 0, you have 2 y = 2 x ⇒ y = x.
13. (a)
x
1
−6
18 and center ( −3, 0).
3
I
kI
=
2
x2 + y2
x
−
4
(
) + y2
(x
− 4) + y 2 = k ( x 2 + y 2 )
(k
− 1) x 2 + 8 x + ( k − 1) y 2 = 16
2
If k = 1, then x = 2 is a vertical line. Assume k ≠ 1.
8x
16
+ y2 =
k −1
k −1
8x
16
16
16
x2 +
+
+ y2 =
+
k − 1 ( k − 1)2
k − 1 ( k − 1)2
x2 +
2
4 ⎞
16k
⎛
2
, Circle
⎜x +
⎟ + y =
2
k − 1⎠
⎝
k
( − 1)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 1
79
(b) If k = 3, ( x + 2) + y 2 = 12
2
y
6
4
2
−6
−4
x
−2
2
4
−2
−4
(c) As k becomes very large,
16k
4
→ 0.
→ 0 and
2
k −1
(k − 1)
The center of the circle gets closer to (0, 0), and its radius approaches 0.
d1d 2 = 1
15.
⎡( x + 1) + y 2 ⎤⎡( x − 1) + y 2 ⎤ = 1
⎣
⎦⎣
⎦
2
(x
2
2
2
2
2
+ 1) ( x − 1) + y 2 ⎡( x + 1) + ( x − 1) ⎤ + y 4 = 1
⎣
⎦
( x2
− 1) + y 2 ⎡⎣2 x 2 + 2⎤⎦ + y 4 = 1
2
y
x4 − 2x2 + 1 + 2x2 y 2 + 2 y 2 + y 4 = 1
( x4
(x + y
2
2
)
2
Let y = 0. Then x 4 = 2 x 2 ⇒ x = 0 or x 2 = 2.
So, (0, 0),
(
)
2
+ 2x2 y2 + y 4 ) − 2x2 + 2 y 2 = 0
(
2, 0 and −
)
= 2( x − y
2
2
)
(− 2 , 0)
1
( 2 , 0)
x
−2
2
−1
(0, 0)
−2
2, 0 are on the curve.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 2
Limits and Their Properties
Section 2.1
A Preview of Calculus..........................................................................81
Section 2.2
Finding Limits Graphically and Numerically .....................................82
Section 2.3
Evaluating Limits Analytically ............................................................93
Section 2.4
Continuity and One-Sided Limits......................................................105
Section 2.5
Infinite Limits .....................................................................................117
Review Exercises ........................................................................................................125
Problem Solving .........................................................................................................133
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 2
Limits and Their Properties
Section 2.1 A Preview of Calculus
1. Precalculus: ( 20 ft/sec)(15 sec) = 300 ft
7. f ( x ) = 6 x − x 2
2. Calculus required: Velocity is not constant.
(a)
y
10
Distance ≈ (20 ft/sec)(15 sec) = 300 ft
8
P
6
3. Calculus required: Slope of the tangent line at x = 2 is
the rate of change, and equals about 0.16.
4. Precalculus: rate of change = slope = 0.08
5. (a) Precalculus: Area =
1 bh
2
=
1
2
(5)( 4)
= 10 sq. units
≈ 2( 2.5)
= 5 sq. units
(a)
x
2
4
(b) slope = m =
8
(6 x − x 2 ) − 8
x−2
= ( 4 − x), x ≠ 2
(b) Calculus required: Area = bh
6. f ( x) =
−2
x
=
(x
For x = 3, m = 4 − 3 = 1
For x = 2.5, m = 4 − 2.5 = 1.5 =
3
2
For x = 1.5, m = 4 − 1.5 = 2.5 =
5
2
y
P(4, 2)
2
− 2)( 4 − x)
x−2
(c) At P( 2, 8), the slope is 2. You can improve your
approximation by considering values of x close to 2.
8. Answers will vary. Sample answer:
x
1
2
3
4
=
The instantaneous rate of change of an automobile’s
position is the velocity of the automobile, and can be
determined by the speedometer.
x −2
x −4
(b) slope = m =
=
5
(
x −2
x + 2
)(
x −2
)
1
,x ≠ 4
x + 2
1
1
=
3
1 + 2
1
x = 3: m =
≈ 0.2679
3 + 2
1
x = 5: m =
≈ 0.2361
5 + 2
1
1
(c) At P( 4, 2) the slope is
=
= 0.25.
4
4 + 2
x = 1: m =
You can improve your approximation of the slope at
x = 4 by considering x-values very close to 4.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
81
82
Chapter 2
Limits and Their Properties
9. (a) Area ≈ 5 +
Area ≈
1
2
5
2
+
5
3
+
5
4
≈ 10.417
(5 + 1.55 + 52 + 2.55 + 53 + 3.55 + 54 + 4.55 ) ≈ 9.145
(b) You could improve the approximation by using more rectangles.
10. (a) D1 =
(5 − 1)
(b) D2 =
1+
2
( 52 )
2
+ (1 − 5)
+
1+
2
=
16 + 16 ≈ 5.66
( 52 − 53 )
2
+
( 53 − 54 )
1+
2
+
1+
( 54 − 1)
2
≈ 2.693 + 1.302 + 1.083 + 1.031 ≈ 6.11
(c) Increase the number of line segments.
Section 2.2 Finding Limits Graphically and Numerically
1.
x
3.9
3.99
3.999
4.001
4.01
4.1
f (x)
0.2041
0.2004
0.2000
0.2000
0.1996
0.1961
lim
x → 4 x2
2.
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.5132
0.5013
0.5001
?
0.4999
0.4988
0.4881
x +1 −1
1⎞
⎛
≈ 0.5000 ⎜ Actual limit is .⎟
x
2⎠
⎝
lim
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.9983
0.99998
1.0000
1.0000
0.99998
0.9983
lim
x→0
4.
( Actual limit is 1.) ( Make sure you use radian mode.)
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.0500
0.0050
0.0005
–0.0005
–0.0050
–0.0500
lim
cos x − 1
≈ 0.0000
x
( Actual limit is 0.) ( Make sure you use radian mode.)
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.9516
0.9950
0.9995
1.0005
1.0050
1.0517
lim
x→0
6.
sin x
≈ 1.0000
x
x
x→0
5.
1⎞
⎛
⎜ Actual limit is .⎟
5⎠
⎝
x
x→0
3.
x − 4
≈ 0.2000
− 3x − 4
ex − 1
≈ 1.0000
x
(Actual limit is 1.)
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
1.0536
1.0050
1.0005
0.9995
0.9950
0.9531
lim
x→0
ln ( x + 1)
x
≈ 1.0000
(Actual limit is 1.)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.2
7.
x
0.9
0.99
0.999
1.001
1.01
1.1
f (x)
0.2564
0.2506
0.2501
0.2499
0.2494
0.2439
lim
x →1
8.
x − 2
≈ 0.2500
x + x −6
x
–4.1
–4.01
–4.001
–4
–3.999
–3.99
–3.9
f (x)
1.1111
1.0101
1.0010
?
0.9990
0.9901
0.9091
x + 4
≈ 1.0000
x 2 + 9 x + 20
lim
0.9
0.99
0.999
1.001
1.01
1.1
f (x)
0.7340
0.6733
0.6673
0.6660
0.6600
0.6015
x →1
10.
x4 − 1
≈ 0.6666
x6 − 1
–3.1
–3.01
–3.001
–3
–2.999
–2.99
–2.9
f (x)
27.91
27.0901
27.0090
?
26.9910
26.9101
26.11
x 3 + 27
≈ 27.0000
x + 3
x → −3
–6.1
–6.01
–6.001
–6
–5.999
–5.99
–5.9
f (x)
–0.1248
–0.1250
–0.1250
?
–0.1250
–0.1250
–0.1252
10 − x − 4
≈ − 0.1250
x + 6
lim
x
f (x)
lim
x→2
13.
1⎞
⎛
⎜ Actual limit is − .⎟
8⎠
⎝
1.9
1.99
1.999
2
2.001
2.01
2.1
0.1149
0.115
0.1111
?
0.1111
0.1107
0.1075
x ( x + 1) − 2 3
≈ 0.1111
x − 2
1⎞
⎛
⎜ Actual limit is .⎟
9⎠
⎝
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
1.9867
1.9999
2.0000
2.0000
1.9999
1.9867
lim
x→0
14.
( Actual limit is 27.)
x
x → −6
12.
2⎞
⎛
⎜ Actual limit is .⎟
3⎠
⎝
x
lim
11.
(Actual limit is 1.)
x
lim
sin 2 x
≈ 2.0000
x
(Actual limit is 2.) (Make sure you use radian mode.)
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.4950
0.5000
0.5000
0.5000
0.5000
0.4950
lim
x→0
tan x
≈ 0.5000
tan 2 x
83
1⎞
⎛
⎜ Actual limit is .⎟
4⎠
⎝
2
x → −4
9.
Finding Limits Graphically and Numerically
1⎞
⎛
⎜ Actual limit is .⎟
2⎠
⎝
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
84
15.
Chapter 2
x
1.9
1.99
1.999
2.001
2.01
2.1
f (x)
0.5129
0.5013
0.5001
0.4999
0.4988
0.4879
lim
x→2
16.
Limits and Their Properties
ln x − ln 2
≈ 0.5000
x − 2
1⎞
⎛
⎜ Actual limit is .⎟
2⎠
⎝
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
3.99982
4
4
0
0
0.00018
lim
x→0
4
does not exist.
1 + e1 x
25. (a) f (1) exists. The black dot at (1, 2) indicates that
17. lim ( 4 − x) = 1
x →3
f (1) = 2.
18. lim sec x = 1
(b) lim f ( x) does not exist. As x approaches 1 from the
x→0
x →1
19. lim f ( x) = lim ( 4 − x) = 2
x→2
left, f (x) approaches 3.5, whereas as x approaches 1
from the right, f (x) approaches 1.
x→2
20. lim f ( x) = lim ( x 2 + 3) = 4
x →1
21. lim
x→2
(c) f ( 4) does not exist. The hollow circle at
x →1
x − 2
x − 2
(4, 2) indicates that
(d) lim f ( x) exists. As x approaches 4, f ( x) approaches
does not exist.
For values of x to the left of 2,
for values of x to the right of 2,
x→4
x − 2
(x
− 2)
2: lim f ( x) = 2.
= −1, whereas
x − 2
(x
− 2)
= 1.
x→0
oscillates between –1 and 1 as x approaches 0.
lim tan x does not exist because the function increases
x →π 2
π
2
from the left and
decreases without bound as x approaches
the right.
indicates that f is not defined at –2.
(b) lim f ( x) does not exist. As x approaches –2, the
23. lim cos(1 x) does not exist because the function
without bound as x approaches
x→4
26. (a) f ( −2) does not exist. The vertical dotted line
4
22. lim
does not exist. The function approaches
x → 0 2 + e1 x
2 from the left side of 0 by it approaches 0 from the left
side of 0.
24.
f is not defined at 4.
π
2
from
x → −2
values of f ( x) do not approach a specific number.
(c) f (0) exists. The black dot at (0, 4) indicates that
f (0) = 4.
(d) lim f ( x ) does not exist. As x approaches 0 from the
x→0
left, f ( x) approaches 12 , whereas as x approaches 0
from the right, f ( x) approaches 4.
(e) f ( 2) does not exist. The hollow circle at
(2, 12 ) indicates that
f ( 2) is not defined.
(f ) lim f ( x) exists. As x approaches 2, f ( x) approaches
x→2
1 : lim f
2 x→2
( x)
= 12 .
(g) f ( 4) exists. The black dot at ( 4, 2) indicates that
f ( 4) = 2.
(h) lim f ( x) does not exist. As x approaches 4, the
x→4
values of f ( x) do not approach a specific number.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.2
27.
y
Finding Limits Graphically and Numerically
32. You need f ( x) − 1 =
6
85
1
2 − x
−1 =
< 0.01.
x −1
x −1
5
4
3
Let δ =
f
2
1
−
x
−2 −1
−1
1
2
3
4
5
−2
lim f ( x ) exists for all values of c ≠ 4.
x→c
y
28.
and you have
1
f ( x) − 1 =
π
2
33. You need to find δ such that 0 < x − 1 < δ implies
lim f ( x ) exists for all values of c ≠ π .
x→c
29. One possible answer is
5
f
2
1
x
1
2
3
4
5
30. One possible answer is
y
4
3
So take δ =
2
1
1
2
−1
31. You need f ( x ) − 3 =
(x
+ 1) − 3 = x − 2 < 0.4.
So, take δ = 0.4. If 0 < x − 2 < 0.4, then
x − 2 =
(x
1
. Then 0 < x − 1 < δ implies
11
1
1
< x −1<
11
11
1
1
−
< x −1< .
11
9
−
x
−1
1
− 1 < 0.1. That is,
x
1
− 1 < 0.1
x
1
1 − 0.1 <
< 1 + 0.1
x
9
1
11
<
<
10
10
x
10
10
x >
>
9
11
10
10
−1 > x −1 >
−1
9
11
1
1
> x −1 > − .
9
11
6
4
f ( x) − 1 =
−0.1 <
y
−2
1
2− x
1 101
1
−1 =
<
=
100 101 100
x −1
x −1
= 0.01.
x
π
−1
−3
1
1
1
1
< x − 2 <
⇒1−
< x −1<1+
101
101
101
101
100
102
⇒
< x −1<
101
101
100
⇒ x −1 >
101
2
−π
2
−2 −1
−1
1
1
. If 0 < x − 2 <
, then
101
101
+ 1) − 3 = f ( x) − 3 < 0.4, as desired.
Using the first series of equivalent inequalities, you
obtain
f ( x) − 1 =
1
− 1 < 0.1.
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
86
Chapter 2
Limits and Their Properties
34. You need to find δ such that 0 < x − 2 < δ implies
f ( x) − 3 = x 2 − 1 − 3 = x 2 − 4 < 0.2. That is,
37. lim ( x 2 − 3) = 22 − 3 = 1 = L
x→2
( x2
−0.2 < x − 4 < 0.2
2
4 − 0.2 <
x2
< 4 + 0.2
3.8 <
x2
< 4.2
3.8 <
x
<
So take δ =
(
(x
+ 2)( x − 2) < 0.01
4.2 − 2
4.2 − 2 ≈ 0.0494.
Then 0 < x − 2 < δ implies
−
x 2 − 4 < 0.01
x + 2 x − 2 < 0.01
4.2
3.8 − 2 < x − 2 <
)
x − 2 <
4.2 − 2
3.8 − 2 < x − 2 <
4.2 − 2.
Using the first series of equivalent inequalities, you obtain
If you assume 1 < x < 3, then δ ≈ 0.01 5 = 0.002.
x − 2 < 0.002 =
35. lim (3 x + 2) = 3(2) + 2 = 8 = L
x + 2 x − 2 < 0.01
( x2
+ 2) − 8 < 0.01
38. lim ( x 2 + 6) = 42 + 6 = 22 = L
3 x − 6 < 0.01
x→4
3 x − 2 < 0.01
≈ 0.0033 = δ
So, if 0 < x − 2 < δ =
0.01
,
3
you have
( x2
+ 6) − 22 < 0.01
x 2 − 16 < 0.01
(x
+ 4)( x − 4) < 0.01
3 x − 2 < 0.01
x − 4 <
3x − 6 < 0.01
(3 x
− 3) − 1 < 0.01
f ( x) − L < 0.01.
x→2
0 < x − 2 <
1
1
(0.01) <
(0.01)
5
x + 2
x 2 − 4 < 0.01
f ( x ) − 3 = x 2 − 4 < 0.2.
0.01
3
0.01
x + 2
So, if 0 < x − 2 < δ ≈ 0.002, you have
4.2 − 2 < x − 2 <
(3 x
− 3) − 1 < 0.01
0.01
x + 4
+ 2) − 8 < 0.01
If you assume 3 < x < 5, then δ =
f ( x) − L < 0.01.
So, if 0 < x − 4 < δ ≈
2−
x
< 0.01
3
− 13 ( x − 6) < 0.01
≈ 0.00111.
0.01
, you have
9
0.01
0.01
x − 4 <
<
9
x + 4
x⎞
6
⎛
= 4 = L
36. lim ⎜ 6 − ⎟ = 6 −
x → 6⎝
3⎠
3
x⎞
⎛
⎜ 6 − ⎟ − 4 < 0.01
3⎠
⎝
0.01
9
( x + 4)( x − 4) < 0.01
x 2 − 16 < 0.01
( x2
+ 6) − 22 < 0.01
f ( x ) − L < 0.01.
x − 6 < 0.03
0 < x − 6 < 0.03 = δ
So, if 0 < x − 6 < δ = 0.03, you have
− 13 ( x − 6) < 0.01
2−
x
< 0.01
3
x⎞
⎛
⎜ 6 − ⎟ − 4 < 0.01
3⎠
⎝
f ( x) − L < 0.01.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.2
39. lim ( x + 2) = 4 + 2 = 6
42. lim
x→4
( 3 x + 1) =
x →3 4
Given ε > 0:
(x
Finding Limits Graphically and Numerically
+ 2) − 6 < ε
( 34 x + 1) − 134
3
x
4
So, let δ = ε . So, if 0 < x − 4 < δ = ε , you have
3
4
x −4 < ε
40. lim ( 4 x + 5) = 4(− 2) + 5 = − 3
3
4
+ 5) − (− 3) < ε
4 x + 2 < ε
x + 2 <
ε
4
4ε .
3
4ε ,
3
you have
4
ε
3
x −3 < ε
< ε
( 34 x + 1) − 134
< ε
−
f ( x) − L < ε .
43. lim 3 = 3
x→6
.
Given ε > 0:
x + 2 <
ε
4
, you have
ε
3−3 < ε
f ( x) − L < ε .
f ( x) − L < ε .
( 1 x − 1) =
1
2
( − 4)
− 1 = −3
44. lim ( −1) = −1
x→2
Given ε > 0: −1 − ( −1) < ε
Given ε > 0:
( 12 x − 1) − (−3)
0 < ε
So, for any δ > 0, you have
(4 x + 5) − ( −3) < ε
x → −4 2
3−3 < ε
So, any δ > 0 will work.
4
4x + 8 < ε
1
2
4ε
3
9
4
= δ
So, if 0 < x + 2 < δ =
1x
2
x −3 < ε
3
x
4
4x + 8 < ε
41. lim
13
4
< ε
x −3 <
Given ε > 0:
4
9
4
So, if 0 < x − 3 < δ =
x → −2
ε
−
So, let δ =
f ( x) − L < ε .
So, let δ =
=
< ε
x −3 <
+ 2) − 6 < ε
(4 x
(3) + 1
Given ε > 0:
x −4 < ε = δ
(x
3
4
87
< ε
+ 2 < ε
x − ( −4) < ε
x − ( −4) < 2ε
0 < ε
So, any δ > 0 will work.
So, for any δ > 0, you have
(−1) − (−1)
f ( x) − L
< ε
< ε.
So, let δ = 2ε .
So, if 0 < x − ( −4) < δ = 2ε , you have
x − ( −4) < 2ε
1x
2
+ 2 < ε
( 12 x − 1) + 3
< ε
f ( x) − L < ε .
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
88
Chapter 2
3
45. lim
x→0
Limits and Their Properties
48. lim x − 3 = 3 − 3 = 0
x = 0
x →3
Given ε > 0:
3
Given ε > 0:
x −0 < ε
3
x −3 < ε
x < ε
So, let δ = ε .
x < ε3 = δ
So, for 0 < x − 3 < δ = ε , you have
So, let δ = ε 3.
x −3 < ε
So, for 0 x − 0 δ = ε 3 , you have
x −3 −0 < ε
x < ε3
3
3
f ( x) − L < ε .
x < ε
x →1
x→4
( x2
+ 1) − 2 < ε
x2 − 1 < ε
4 = 2
(x
Given ε > 0:
+ 1)( x − 1) < ε
x − 2 < ε
x − 2
x + 2 < ε
x + 2
x − 4 < ε
x + 2
Assuming 1 < x < 9, you can choose δ = 3ε . Then,
0 < x − 4 < δ = 3ε ⇒ x − 4 < ε
⇒
x −1 <
So for 0 < x − 1 < δ =
x2 − 1 < ε
( x2
+ 1) − 2 < ε
f ( x) − 2 < ε .
x − 5 − 10 < ε
−( x − 5) − 10 < ε
(x
− 5 < 0)
−x − 5 < ε
50. lim ( x 2 + 4 x) = (− 4) 2 + 4(− 4) = 0
x → −4
Given ε > 0:
x − ( −5) < ε
( x2
+ 4 x) − 0 < ε
So, let δ = ε .
x( x + 4) < ε
So for x − ( −5) < δ = ε , you have
x + 4 <
−( x + 5) < ε
−( x − 5) − 10 < ε
x − 5 − 10 < ε
f ( x) − L < ε .
ε
, you have
3
1
1
ε
x −1 < ε <
3
x +1
x − 2 < ε.
x → −5
ε
x +1
If you assume 0 < x < 2, then δ = ε 3.
x + 2
47. lim x − 5 = ( −5) − 5 = −10 = 10
Given ε > 0:
)
Given ε > 0:
f ( x) − L < ε .
x =
(
49. lim x 2 + 1 = 12 + 1 = 2
x −0 < ε
46. lim
x −3 −0 < ε
ε
x
If you assume − 5 < x < − 3, then δ =
(because x
− 5 < 0)
So for 0 < x − (− 4) < δ =
x + 4 <
ε
5
<
ε
5
ε
5
.
, you have
1
ε
x
x ( x + 4) < ε
(x
2
+ 4 x) − 0 < ε
f ( x) − L < ε .
51. lim f ( x) = lim 4 = 4
x →π
x →π
52. lim f ( x) = lim x = π
x →π
x →π
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.2
Finding Limits Graphically and Numerically
x +5 −3
x − 4
1
lim f ( x) =
x→4
6
55. f ( x ) =
53. f ( x) =
89
x −9
x −3
lim f ( x) = 6
x →9
10
The domain is
[−5, 4) ∪ (4, ∞).
0.5
−6
The graphing utility
does not show the hole
⎛ 1⎞
at ⎜ 4, ⎟.
⎝ 6⎠
6
− 0.1667
0
The domain is all x ≥ 0 except x = 9. The graphing
utility does not show the hole at (9, 6).
x −3
x − 4x + 3
1
lim f ( x) =
x →3
2
54. f ( x) =
10
0
2
ex 2 − 1
x
1
lim f ( x) =
x→0
2
56. f ( x) =
The domain is all
x ≠ 1, 3. The graphing
4
2
−3
utility does not show the
⎛ 1⎞
hole at ⎜ 3, ⎟.
⎝ 2⎠
5
−4
−2
2
−1
The domain is all x ≠ 0. The graphing utility does not
⎛ 1⎞
show the hole at ⎜ 0, ⎟.
⎝ 2⎠
57. C (t ) = 9.99 − 0.79 ⎡⎣⎡−
⎣ (t − 1)⎤⎦ ⎤⎦
(a)
16
0
6
8
(b)
t
3
3.3
3.4
3.5
3.6
3.7
4
C
11.57
12.36
12.36
12.36
12.36
12.36
12.36
lim C (t ) = 12.36
t → 3.5
(c)
t
2
2.5
2.9
3
3.1
3.5
4
C
10.78
11.57
11.57
11.57
12.36
12.36
12.36
The lim C (t ) does not exist because the values of C approach different values as t approaches 3 from both sides.
t →3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
90
Chapter 2
Limits and Their Properties
58. C (t ) = 5.79 − 0.99 ⎡⎣⎡−
⎣ (t − 1)⎤⎦ ⎤⎦
(a)
12
0
6
4
(b)
t
3
3.3
3.4
3.5
3.6
3.7
4
C
7.77
8.76
8.76
8.76
8.76
8.76
8.76
lim C (t ) = 8.76
t → 3.5
(c)
t
2
2.5
2.9
3
3.1
3.5
4
C
6.78
7.77
7.77
7.77
8.76
8.76
8.76
The limit lim C (t ) does not exist because the values of C approach different values as t approaches 3 from both sides.
t →3
59. lim f ( x) = 25 means that the values of f approach 25
x →8
62. (a) No. The fact that f ( 2) = 4 has no bearing on the
existence of the limit of f ( x ) as x approaches 2.
as x gets closer and closer to 8.
60. In the definition of lim f ( x), f must be defined on both
x→c
sides of c, but does not have to be defined at c itself. The
value of f at c has no bearing on the limit as x approaches c.
61. (i) The values of f approach different numbers as x
approaches c from different sides of c:
y
4
3
2
1
x
−4 −3 −2 −1
−1
1
2
3
4
(b) No. The fact that lim f ( x) = 4 has no bearing on
x→2
the value of f at 2.
63. (a) C = 2π r
C
6
3
=
=
≈ 0.9549 cm
2π
2π
π
5.5
(b) When C = 5.5: r =
≈ 0.87535 cm
2π
6.5
When C = 6.5: r =
≈ 1.03451 cm
2π
So 0.87535 < r < 1.03451.
r =
(c)
−3
−4
(ii) The values of f increase without bound as x
approaches c:
y
6
5
4
lim
x →3 π
( 2π r )
= 6; ε = 0.5; δ ≈ 0.0796
4 3
π r , V = 2.48
3
4
(a) 2.48 = π r 3
3
1.86
3
r =
64. V =
π
3
2
r ≈ 0.8397 in.
1
x
−3 −2 −1
−1
2
3
4
5
(b)
2.45 ≤
V
−2
y
4
0.5849 ≤
r 3 ≤ 0.5992
0.8363 ≤
r
x
2
3
4
65. f ( x) = (1 + x )
1x
lim (1 + x)
1x
−3
≤ 0.8431
(c) For ε = 2.51 − 2.48 = 0.03, δ ≈ 0.003
3
−4 −3 −2
4 3
π r ≤ 2.51
3
2.45 ≤
(iii) The values of f oscillate between two fixed
numbers as x approaches c:
≤ 2.51
x→0
= e ≈ 2.71828
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.2
Finding Limits Graphically and Numerically
91
68. (a) lim f ( x ) exists for all c ≠ −3.
y
x→c
7
(b) lim f ( x ) exists for all c ≠ −2, 0.
x→c
3
69. False. The existence or nonexistence of f ( x ) at
(0, 2.7183)
2
x = c has no bearing on the existence of the limit
of f ( x ) as x → c.
1
x
−3 −2 −1
−1
1
2
3
4
5
70. True
x
f (x)
x
f (x)
–0.1
2.867972
0.1
2.593742
–0.01
2.731999
0.01
2.704814
⎧x − 4, x ≠ 2
f ( x) = ⎨
x = 2
⎩0,
–0.001
2.719642
0.001
2.716942
f ( 2) = 0
–0.0001
2.718418
0.0001
2.718146
lim f ( x) = lim ( x − 4) = 2 ≠ 0
–0.00001
2.718295
0.00001
2.718268
–0.000001
2.718283
0.000001
2.718280
x +1 − x −1
66. f ( x) =
71. False. Let
x→2
x→2
72. False. Let
⎧x − 4, x ≠ 2
f ( x) = ⎨
x = 2
⎩0,
lim f ( x) = lim ( x − 4) = 2 and f ( 2) = 0 ≠ 2
x
x→2
x
–1
–0.5
–0.1
0
0.1
0.5
1.0
f(x)
2
2
2
Undef.
2
2
2
x→2
73. f ( x) =
x = 0.5 is true.
lim
x → 0.25
lim f ( x) = 2
As x approaches 0.25 =
x→0
Note that for
−1 < x < 1, x ≠ 0, f ( x) =
(x
+ 1) + ( x − 1)
= 2.
x
f ( x) =
74. f ( x) =
y
lim
x→0
3
1
2
from either side,
= 0.5.
x
x is not defined on an open interval
containing 0 because the domain of f is x ≥ 0.
1
75. Using a graphing utility, you see that
x
−1
x approaches
1
4
x = 0 is false.
f ( x) =
−2
x
1
2
sin x
=1
x
sin 2 x
lim
= 2, etc.
x→0
x
−1
lim
x→0
67.
0.002
(1.999, 0.001)
(2.001, 0.001)
So, lim
x→0
1.998
2.002
0
Using the zoom and trace feature, δ = 0.001. So
(2 − δ , 2 + δ ) = (1.999, 2.001).
Note:
x2 − 4
= x + 2 for x ≠ 2.
x − 2
sin nx
= n.
x
76. Using a graphing utility, you see that
tan x
=1
x
tan 2 x
lim
= 2,
x→0
x
lim
x→0
So, lim
x→0
tan ( nx )
x
etc.
= n.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
92
Chapter 2
Limits and Their Properties
77. If lim f ( x) = L1 and lim f ( x) = L2 , then for every ε > 0, there exists δ1 > 0 and δ 2 > 0 such that
x→c
x→c
x − c < δ1 ⇒ f ( x) − L1 < ε and x − c < δ 2 ⇒ f ( x) − L2 < ε . Let δ equal the smaller of δ1 and δ 2 .
Then for x − c < δ , you have L1 − L2 = L1 − f ( x) + f ( x) − L2 ≤ L1 − f ( x ) + f ( x) − L2 < ε + ε .
Therefore, L1 − L2 < 2ε . Since ε > 0 is arbitrary, it follows that L1 = L2 .
78. f ( x) = mx + b, m ≠ 0. Let ε > 0 be given. Take
ε
δ =
m
(b) You are given lim g ( x) = L > 0. Let
x→c
1
ε = L. There exists δ > 0 such that
2
0 < x − c < δ implies that
.
If 0 < x − c < δ =
ε
m
, then
g ( x) − L < ε =
m x −c < ε
mx − mc < ε
( mx
−
+ b) − ( mc + b) < ε
which shows that lim ( mx + b) = mc + b.
x→c
have g ( x) >
x→c
exists δ > 0 such that if
< ε.
This means the same as f ( x) − L < ε when
0 < x − c < δ.
So, lim f ( x) = L.
1 ⎛
h⎞
b⎜1 − ⎟
2 ⎝
2⎠
Area rectangle = bh
Area triangle =
Because these are equal,
x→c
80. (a)
L
> 0, as desired.
2
81. The radius OP has a length equal to the altitude z of the
h
h
triangle plus . So, z = 1 − .
2
2
then
( f ( x) − L) − 0
L
L
< g ( x) − L <
2
2
3L
L
<
g ( x) <
2
2
For x in the interval (c − δ , c + δ ), x ≠ c, you
79. lim ⎣⎡ f ( x) − L⎤⎦ = 0 means that for every ε > 0 there
0 < x − c < δ,
L
. That is,
2
(3x + 1)(3x − 1) x 2 + 0.01 = (9 x 2 − 1) x 2 +
1
100
1
= 9 x4 − x2 +
100
1
=
(10 x 2 − 1)(90 x 2 − 1)
100
1 ⎛
h⎞
b⎜1 − ⎟ = bh
2 ⎝
2⎠
h
= 2h
1−
2
5
h =1
2
2
h = .
5
P
So, (3x + 1)(3x − 1) x + 0.01 > 0 if
2
10 x 2 − 1 < 0 and 90 x 2 − 1 < 0.
1
⎛
,
Let ( a, b) = ⎜ −
90
⎝
1 ⎞
⎟.
90 ⎠
For all x ≠ 0 in ( a, b), the graph is positive.
h
O
b
You can verify this with a graphing utility.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.3
Evaluating Limits Analytically
93
82. Consider a cross section of the cone, where EF is a diagonal of the inscribed cube. AD = 3, BC = 2.
Let x be the length of a side of the cube. Then EF = x 2.
By similar triangles,
A
EF
AG
=
BC
AD
x 2
3− x
=
2
3
G
E
F
3 2x = 6 − 2x
Solving for x,
(3
)
B
2 + 2 x = 6
x =
D
C
6
9 2 −6
=
≈ 0.96.
7
3 2 + 2
Section 2.3 Evaluating Limits Analytically
1.
4.
6
−4
8
10
−5
10
−6
− 10
f (t ) = t t − 4
(a) lim h( x) = 0
x→4
(b) lim h( x ) = −5
(a) lim f (t ) = 0
x → −1
2.
t →4
(b) lim f (t ) = −5
10
t → −1
5. lim x3 = 23 = 8
0
10
x→2
6. lim x 4 = (− 3) 4 = 81
−5
g ( x) =
(
x → −3
)
x −3
12
7. lim ( 2 x − 1) = 2(0) − 1 = −1
x −9
x→0
(a) lim g ( x) = 2.4
8. lim ( 2 x + 3) = 2(− 4) + 3 = − 8 + 3 = − 5
x→4
x → −4
(b) lim g ( x ) = 4
9. lim ( x 2 + 3 x) = ( −3) + 3( −3) = 9 − 9 = 0
x→0
2
3.
x → −3
4
10. lim ( − x 3 + 1) = ( −2) + 1 = − 8 + 1 = − 7
3
−␲
␲
x→2
11. lim ( 2 x 2 + 4 x + 1) = 2( −3) + 4(−3) + 1
2
−4
x →−3
= 18 − 12 + 1 = 7
f ( x) = x cos x
(a) lim f ( x) = 0
x→0
(b)
12. lim ( 2 x 3 − 6 x + 5) = 2(1) − 6(1) + 5
3
x →1
= 2−6 + 5 =1
lim f ( x ) ≈ 0.524
x →π 3
⎛ π⎞
⎜= ⎟
⎝ 6⎠
x +1 =
13. lim
x →3
14. lim
x→2
3
3+1 = 2
12 x + 3 =
3
12(2) + 3
=
3
24 + 3 =
3
27 = 3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
94
Chapter 2
Limits and Their Properties
15. lim ( x + 3) = ( −4 + 3) = 1
2
2
30.
x → −4
16. lim (3 x − 2) = (3(0) − 2) = (− 2) 4 = 16
4
4
x→0
17. lim
x→2
1
1
=
2
x
18. lim
x → −5
lim cos x = cos
x → 5π 3
5π
1
=
3
2
3π
⎛π x ⎞
31. lim tan ⎜ ⎟ = tan
= −1
x →3
4
4
⎝ ⎠
7π
−2 3
⎛π x ⎞
=
32. lim sec⎜ ⎟ = sec
x→7
6
3
⎝ 6 ⎠
5
5
5
=
= −
x + 3
2
−5 + 3
33. lim e x cos 2x = e0 cos 0 = 1
x→0
x
1
1
19. lim 2
= 2
=
x →1 x + 4
1 + 4
5
20. lim
x →1
x→0
3x + 5
3(1) + 5
3+5
8
=
=
=
= 4
x +1
1+1
2
2
21. lim
3x
=
x + 2
22. lim
x + 6
=
x + 2
x→7
x →3
23.
34. lim e − x sin π x = e0 sin 0 = 0
3(7)
7 + 2
=
21
= 7
3
3+ 6
=
3+ 2
lim sin x = sin
x →π 2
π
9
3
=
5
5
x →1
)
⎛x⎞
⎛1⎞
36. lim ln ⎜ x ⎟ = ln ⎜ ⎟ = ln e −1 = −1
x →1
⎝e ⎠
⎝e⎠
37. (a) lim f ( x) = 5 − 1 = 4
x →1
(b) lim g ( x) = 43 = 64
x→4
=1
2
(
35. lim ln 3x + e x = ln 3 + e
(c) lim g ( f ( x)) = g ( f (1)) = g ( 4) = 64
x →1
24. lim tan x = tan π = 0
38. (a) lim f ( x ) = ( −3) + 7 = 4
x →π
x → −3
πx
π
(b) lim g ( x) = 42 = 16
1
25. lim cos
= cos
=
x →1
3
3
2
26. lim sin
x→2
πx
2
= sin
π ( 2)
2
x→4
(c) lim g ( f ( x)) = g ( 4) = 16
x → −3
= 0
39. (a) lim f ( x) = 4 − 1 = 3
x →1
27. lim sec 2 x = sec 0 = 1
(b) lim g ( x) =
x→0
(c) lim g ( f ( x)) = g (3) = 2
28. lim cos 3 x = cos 3π = −1
x →π
29.
3+1 = 2
x →3
lim sin x = sin
x → 5π 6
x →1
5π
1
=
6
2
( )
40. (a) lim f ( x) = 2 42 − 3( 4) + 1 = 21
x→4
(b) lim g ( x) =
x → 21
3
21 + 6 = 3
(c) lim g ( f ( x)) = g ( 21) = 3
x→4
41. (a) lim ⎡⎣5 g ( x)⎤⎦ = 5 lim g ( x) = 5( 2) = 10
x→c
x→c
(b) lim ⎡⎣ f ( x) + g ( x)⎤⎦ = lim f ( x) + lim g ( x) = 3 + 2 = 5
x→c
x→c
x→c
(c) lim ⎡⎣ f ( x) g ( x)⎤⎦ = ⎡⎢ lim f ( x)⎤⎥
x→c
⎣x → c
⎦
(d) lim
x→c
f ( x)
g ( x)
=
lim f ( x)
x→c
lim g ( x)
x→c
=
⎡ lim g ( x)⎤ = (3)( 2) = 6
⎢⎣x → c
⎥⎦
3
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.3
Evaluating Limits Analytically
95
42. (a) lim ⎡⎣4 f ( x)⎤⎦ = 4 lim f ( x) = 4(2) = 8
x→c
x→c
3
11
(b) lim ⎡⎣ f ( x) + g ( x)⎤⎦ = lim f ( x) + lim g ( x) = 2 +
=
x→c
x→c
x→c
4
4
3
⎛ 3⎞
(c) lim ⎣⎡ f ( x) g ( x)⎦⎤ = ⎡⎢ lim f ( x)⎤⎡
g ( x)⎤⎥ = 2⎜ ⎟ =
⎥⎢ xlim
x→c
→c
⎣x → c
⎦⎣
⎦
4
2
⎝ ⎠
(d) lim
x→c
f ( x)
g ( x)
=
lim f ( x)
x→c
lim g ( x)
x→c
=
2
8
=
( 3 4) 3
x3 − 8
and g ( x) = x 2 + 2 x + 4 agree except
x − 2
at x = 2.
3
3
3
43. (a) lim ⎡⎣ f ( x)⎤⎦ = ⎡⎢ lim f ( x)⎤⎥ = ( 4) = 64
x→c
⎣x → c
⎦
(b) lim
x→c
f ( x) =
lim f ( x) =
47. f ( x ) =
4 = 2
x→c
lim f ( x) = lim g ( x) = lim ( x 2 + 2 x + 4)
(c) lim ⎡⎣3 f ( x)⎤⎦ = 3 lim f ( x) = 3( 4) = 12
x→c
x→c
(d) lim ⎣⎡ f ( x)⎦⎤
= ⎡⎢ lim f ( x)⎤⎥
⎣x → c
⎦
32
x→c
44. (a) lim 3 f ( x) =
x→c
(b) lim
x→c
f ( x)
18
=
3
32
lim f ( x) =
x→c
= ( 4)
3
32
x→2
= 8
12
lim f ( x)
−9
9
0
27
3
=
=
lim 18
18
2
x→c
2
2
2
(c) lim ⎣⎡ f ( x)⎦⎤ = ⎡⎢ lim f ( x)⎤⎥ = ( 27) = 729
x→c
⎣x → c
⎦
23
= ⎡⎢ lim f ( x)⎤⎥
⎣x → c
⎦
23
= ( 27)
23
= 9
48. f ( x ) =
x3 + 1
and g ( x) = x 2 − x + 1 agree except at
x +1
x = −1.
lim f ( x) = lim g ( x ) = lim ( x 2 − x + 1)
x → −1
x → −1
x2 − 1
( x + 1)( x − 1)
and
=
x +1
x +1
g ( x) = x − 1 agree except at x = −1.
lim f ( x) = lim g ( x) = lim ( x − 1) = −1 − 1 = −2
x → −1
x → −1
x → −1
= ( −1) 2 − (−1) + 1 = 3
45. f ( x ) =
x → −1
x→2
27 = 3
x→c
(d) lim ⎡⎣ f ( x)⎤⎦
x→c
x→2
= 22 + 2(2) + 4 = 12
7
−4
4
−1
3
−3
49. f ( x) =
4
+ 4) ln ( x + 6)
x − 16
agree except at x = − 4.
−4
2
lim f ( x) = lim g ( x) =
3x + 5 x − 2
( x + 2)(3 x − 1)
=
and
x + 2
x + 2
g ( x) = 3 x − 1 agree except at x = −2.
46. f ( x ) =
(x
2
x → −4
x → −4
and g ( x) =
ln ( x + 6)
x − 4
ln 2
≈ − 0.0866
−8
1
−7
3
lim f ( x) = lim g ( x) = lim (3 x − 1)
x → −2
x → −2
x → −2
= 3(− 2) − 1 = − 7
−2
3
−4
5
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
96
Chapter 2
50. f ( x ) =
Limits and Their Properties
e2 x − 1
and g ( x) = e x + 1 agree except at
ex − 1
53. lim
x − 4
x − 4
= lim
4
x
→
− 16
( x + 4)( x − 4)
x → 4 x2
x = 0.
= lim
x→4 x
lim f ( x) = lim g ( x) = e0 + 1 = 2
x→0
x→0
54. lim
3
1
1
1
=
=
4+ 4
8
+ 4
− ( x − 5)
5− x
= lim
x → 5 ( x − 5)( x + 5)
− 25
x →5 x2
= lim
−2
x →5 x
2
−1
−1
1
=
= −
5+ 5
10
+5
( x + 3)( x − 2)
x2 + x − 6
= lim
x → −3
x → −3 ( x + 3)( x − 3)
x2 − 9
−1
55. lim
51. lim
x → 0 x2
x
x
1
1
= lim
= lim
=
= −1
x → 0 x( x − 1)
x→0 x − 1
0 −1
− x
= lim
−3 − 2
−5
5
x − 2
=
=
=
6
−3
−3 − 3
−6
x → −3 x
52. lim
x → 0 x2
2x
2x
2
= lim
= lim
0
0
x
→
x
→
+ 4x
x ( x + 4)
x + 4
=
( x − 2)( x + 4)
x2 + 2 x − 8
= lim
x → 2 x2 − x − 2
x → 2 ( x − 2)( x + 1)
56. lim
2
2
1
=
=
0 + 4
4
2
= lim
x→2
57. lim
x→4
x +5 −3
= lim
x→4
x − 4
= lim
x→4
58. lim
x →3
x +1 − 2
= lim
x →3
x −3
= lim
x →3
59. lim
x→0
x + 5 −
x
5
x +5 −3
⋅
x − 4
2+ x −
x
x→4
3
1
1
=
6
9 +3
x +1+ 2
x −3
= lim
x →3 x − 3 ⎡
x +1 + 2
(
)⎣ x + 1 + 2⎤⎦
1
1
1
=
=
4
x +1+ 2
4 + 2
x + 5 −
x
= lim
x→0
2
)
1
=
x + 5 + 3
= lim
x +1− 2
⋅
x −3
x→0 x
x→0
x + 5 +3
x + 5 +3
( x + 5) − 9
( x − 4)( x + 5 +
= lim
60. lim
x + 4
2+ 4
6
=
=
= 2
x +1
2+1
3
x→0
x→0
+ 5) − 5
(
5
2
2+ x − 2
2 + x +
x + 5 +
x + 5 +
⋅
x + 5 +
2+ x −
x
= lim
= lim
(
(x
5
)
⋅
)
= lim
x→0
5
5
1
x + 5 +
2+ x +
2+ x +
= lim
2 x
x→0
5
=
1
5 +
=
1
2 +
5
=
1
5
=
10
2 5
2
2
1
2+ x +
2
2
=
1
2 2
=
2
4
1
1
−
−x
−1
−1
1
+
x
3
3 = lim 3 − (3 + x) = lim
61. lim
= lim
=
= −
x→0
x → 0 (3 + x )3( x )
x → 0 (3 + x )(3)( x )
x → 0 (3 + x )3
x
(3)3
9
4 − ( x + 4)
1
1
−
4 = lim 4( x + 4)
62. lim x + 4
x→0
x→0
x
x
−1
−1
1
= lim
=
= −
x → 0 4( x + 4)
4(4)
16
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.3
2( x + ∆x ) − 2 x
63. lim
∆x
∆x → 0
(x
64. lim
2
∆x
∆x → 0
(x
65. lim
x 2 + 2 x∆x + ( ∆x) − x 2
∆x( 2 x + ∆ x )
2
= lim
∆x
∆x → 0
= lim
∆x → 0
x 2 + 2 x∆x + ( ∆x) − 2 x − 2∆x + 1 − x 2 + 2 x − 1
2
∆x
= lim ( 2 x + ∆x ) = 2 x
∆x
∆x → 0
+ ∆x) − 2( x + ∆x) + 1 − ( x 2 − 2 x + 1)
∆x → 0
97
2 x + 2∆x − 2 x
2 ∆x
= lim
= lim 2 = 2
∆x → 0 ∆x
∆x → 0
∆x
= lim
∆x → 0
+ ∆x ) − x 2
Evaluating Limits Analytically
2
= lim
∆x → 0
∆x
= lim ( 2 x + ∆x − 2) = 2 x − 2
∆x → 0
(x
66. lim
+ ∆x) − x3
3
∆x
∆x → 0
x 3 + 3 x 2 ∆x + 3 x ( ∆ x ) + ( ∆ x ) − x 3
2
= lim
∆x → 0
= lim
(
67. lim
x→0
68. lim
2
∆x
2
)=
⎡⎛ sin x ⎞⎛ 1 ⎞⎤
sin x
1
⎛1⎞
= lim ⎢⎜
⎟⎜ ⎟⎥ = (1)⎜ ⎟ =
x → 0 ⎝ x ⎠⎝ 5 ⎠
5x
5
5
⎝ ⎠
⎣
⎦
3(1 − cos x)
x→0
69. lim
∆x
∆x 3 x + 3 x∆x + ( ∆x)
∆x → 0
3
x
x2
76.
2
) = 3x
⎡ sin x 1 − cos
= lim ⎢
⋅
x → 0⎣ x
x
1 − tan x
cos x − sin x
= lim
x → π 4 sin x cos x − cos 2 x
x − cos x
lim
cos θ tan θ
θ
θ →0
= lim
sin θ
x → π 4 cos
θ →0
θ
x → π 4 cos
⎡ sin x sin x ⎤
tan 2 x
sin 2 x
= lim
= lim ⎢
⋅
⎥
x→0
x → 0 x cos 2 x
x→0
cos 2 x ⎦
x
⎣ x
x →π 4
= − 2
(1 − e− x )e− x
1 − e− x
1 − e− x e− x
=
lim
⋅
=
lim
x →0 ex − 1
x → 0 ex − 1
x→0
1 − e− x
e− x
77. lim
= lim e − x = 1
x→0
78. lim
4(e2 x − 1)
x→0
ex − 1
73.
h→0
h
⎡1 − cos h
= lim ⎢
(1 − cos h)⎤⎥
h → 0⎣
h
⎦
= (0)(0) = 0
74. lim φ sec φ = π ( −1) = −π
φ →π
75.
lim
= lim
x→0
4(e x − 1)(e x + 1)
ex − 1
= lim 4(e x + 1) = 4( 2) = 8
x→0
= (1)(0) = 0
(1 − cos h)2
lim
x
= lim ( −sec x )
sin 2 x
⎡ sin x
⎤
= lim ⎢
71. lim
sin x⎥ = (1) sin 0 = 0
x→0
x → 0⎣ x
x
⎦
72. lim
x(sin x − cos x)
−1
= lim
x⎤
⎥
⎦
=1
−(sin x − cos x)
= lim
= (1)(0) = 0
70. lim
2
x → π 4 sin
⎡ ⎛ (1 − cos x) ⎞⎤
= lim ⎢3⎜
⎟⎥ = (3)(0) = 0
x→0
x
⎢⎣ ⎝
⎠⎥⎦
sin x(1 − cos x)
x→0
(
lim 3x 2 + 3 x∆x + ( ∆x)
∆x → 0
79. lim
sin 3t
3
⎛ sin 3t ⎞⎛ 3 ⎞
⎛ 3⎞
= lim⎜
⎟⎜ ⎟ = (1)⎜ ⎟ =
t → 0⎝ 3t ⎠⎝ 2 ⎠
2t
2
⎝ 2⎠
80. lim
⎡ ⎛ sin 2 x ⎞⎛ 1 ⎞⎛ 3x ⎞⎤
sin 2 x
= lim ⎢2⎜
⎟⎥
⎟⎜ ⎟⎜
x → 0 ⎝ 2 x ⎠⎝ 3 ⎠ sin 3 x
3x
⎝
⎠⎦
⎣
t →0
x → 0 sin
2
⎛1⎞
= 2(1)⎜ ⎟(1) =
3
⎝ 3⎠
cos x
= lim sin x = 1
x →π 2
x
x → π 2 cot
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
98
Chapter 2
81. f ( x ) =
Limits and Their Properties
x + 2 −
x
2
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.358
0.354
0.354
?
0.354
0.353
0.349
It appears that the limit is 0.354.
2
The graph has a hole at x = 0.
−3
3
−2
x + 2 −
x
Analytically, lim
x→0
2
x→0
= lim
x→0 x
82. f ( x ) =
x + 2 −
x
= lim
(
2
x + 2 − 2
x + 2 +
x + 2 +
x + 2 +
⋅
2
)
= lim
x→0
2
2
1
x + 2 +
2
=
1
2 2
=
2
≈ 0.354.
4
4 − x
x − 16
x
15.9
15.99
15.999
16
16.001
16.01
16.1
f (x)
–0.1252
–0.125
–0.125
?
–0.125
–0.125
–0.1248
It appears that the limit is –0.125.
1
The graph has a hole at x = 16.
0
20
−1
4− x
= lim
x →16 x − 16
x →16
Analytically, lim
(4 − x )
( x + 4)( x − 4)
= lim
x →16
−1
1
= − .
8
x + 4
1
1
−
2
+
x
2
83. f ( x) =
x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
–0.263
–0.251
–0.250
?
–0.250
–0.249
–0.238
It appears that the limit is –0.250.
3
The graph has a hole at x = 0.
−5
1
−2
1
1
−
1
1
−1
2
2 = lim 2 − ( 2 + x) ⋅ 1 = lim − x
x
+
Analytically, lim
⋅
= lim
= − .
x→0
x
→
0
x
→
0
x
→
0
2( 2 + x)
2( 2 + x) x
2( 2 + x )
4
x
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.3
84. f ( x ) =
Evaluating Limits Analytically
99
x 5 − 32
x − 2
x
1.9
1.99
1.999
1.9999
2.0
2.0001
2.001
2.01
2.1
f (x)
72.39
79.20
79.92
79.99
?
80.01
80.08
80.80
88.41
It appears that the limit is 80.
100
The graph has a hole at x = 2.
−4
3
−25
( x − 2)( x 4 + 2 x3 + 4 x 2 + 8 x + 16)
x5 − 32
= lim
= lim ( x 4 + 2 x3 + 4 x 2 + 8 x + 16) = 80.
x→2 x − 2
x→2
x→2
x −2
Analytically, lim
(Hint: Use long division to factor x 5 − 32. )
85. f (t ) =
sin 3t
t
t
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (t)
2.96
2.9996
3
?
3
2.9996
2.96
It appears that the limit is 3.
4
The graph has a hole at t = 0.
−2␲
2␲
−1
Analytically, lim
t →0
86. f ( x) =
sin 3t
⎛ sin 3t ⎞
= lim 3⎜
⎟ = 3(1) = 3.
t → 0 ⎝ 3t ⎠
t
cos x − 1
2x2
x
–1
–0.1
–0.01
0.01
0.1
1
f (x)
–0.2298
–0.2498
–0.25
–0.25
–0.2498
–0.2298
It appears that the limit is –0.25.
1
The graph has a hole at x = 0.
−␲
␲
−1
Analytically,
cos x − 1 cos x + 1
cos 2 x − 1
−sin 2 x
sin 2 x
−1
⋅
=
=
=
⋅
2
2
2
2x
cos x + 1
2 x (cos x + 1)
2 x (cos x + 1)
x2
2(cos x + 1)
⎡ sin 2 x
⎤
−1
1
⎛ −1 ⎞
lim ⎢ 2 ⋅
⎥ = 1⎜ ⎟ = − = −0.25
x → 0⎢ x
2(cos x + 1) ⎦⎥
4
⎝ 4⎠
⎣
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
100
Chapter 2
Limits and Their Properties
87. f ( x ) =
sin x 2
x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
–0.099998
–0.01
–0.001
?
0.001
0.01
0.099998
It appears that the limit is 0.
1
The graph has a hole at x = 0.
−2␲
2␲
−1
⎛ sin x 2 ⎞
sin x 2
= lim x⎜
⎟ = 0(1) = 0.
x→0
x→0
x
⎝ x ⎠
Analytically, lim
88. f ( x ) =
sin x
3
x
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.215
0.0464
0.01
?
0.01
0.0464
0.215
It appears that the limit is 0.
2
The graph has a hole at x = 0.
−3
3
−2
Analytically, lim
sin x
= lim
x→0
x
x→0 3
89. f ( x) =
3
⎛ sin x ⎞
x2 ⎜
⎟ = (0)(1) = 0.
⎝ x ⎠
ln x
x −1
4
x
0.5
0.9
0.99
1.01
1.1
1.5
f (x)
1.3863
1.0536
1.0050
0.9950
0.9531
0.8109
−1
It appears that the limit is 1.
Analytically, lim
x →1
90. f ( x ) =
ln x
= 1.
x −1
e3 x − 8
e2 x − 4
5
x
0.5
0.6
0.69
0.70
0.8
0.9
f (x)
2.7450
2.8687
2.9953
3.0103
3.1722
3.3565
It appears that the limit is 3.
Analytically, lim
6
−1
x → ln 2
−1
2
0
(e x − 2)(e2 x + 2e x + 4) = lim e2 x + 2e x + 4 = 4 + 4 + 4 = 3.
e3 x − 8
= lim
2x
x → ln 2
x → ln 2
2+ 2
e − 4
ex + 2
(e x − 2)(e x + 2)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.3
91. lim
∆x → 0
Evaluating Limits Analytically
101
f ( x + ∆x ) − f ( x)
3( x + ∆x) − 2 − (3 x − 2)
3 x + 3∆x − 2 − 3x + 2
3∆x
= lim
= lim
= lim
= 3
∆x → 0
∆x → 0
∆x → 0 ∆x
∆x
∆x
∆x
( x + ∆x) − 4( x + ∆x) − ( x 2 − 4 x)
f ( x + ∆x) − f ( x)
x 2 + 2 x∆x + ∆x 2 − 4 x − 4∆x − x 2 + 4 x
= lim
= lim
∆x → 0
∆x → 0
∆x
∆x
∆x
∆x( 2 x + ∆x − 4)
= lim
= lim ( 2 x + ∆x − 4) = 2 x − 4
∆x → 0
∆x → 0
∆x
2
92. lim
∆x → 0
93. lim
∆x → 0
f ( x + ∆x) − f ( x)
∆x
1
1
−
3
+
∆
+
+ 3
x
x
x
= lim
∆x → 0
∆x
x + 3 − ( x + ∆x + 3) 1
= lim
⋅
∆x → 0 ( x + ∆x + 3)( x + 3)
∆x
= lim
∆x → 0
= lim
∆x → 0
94. lim
∆x → 0
f ( x + ∆x) − f ( x)
∆x
(x
−∆x
+ ∆x + 3)( x + 3)∆x
(x
−1
−1
=
2
+ ∆x + 3)( x + 3)
( x + 3)
= lim
∆x → 0
= lim
x + ∆x −
∆x
∆x → 0 ∆x
(
x
x→0
∆x → 0
x + ∆x − x
x + ∆x +
95. lim ( 4 − x 2 ) ≤ lim f ( x) ≤ lim ( 4 + x 2 )
x→0
= lim
x→0
x
)
x + ∆x −
∆x
= lim
∆x → 0
x
1
x + ∆x +
⋅
x + ∆x +
x + ∆x +
1
=
2 x
x
99. f ( x) = x sin
4 ≤ lim f ( x) ≤ 4
x→0
x
x
1
x
0.5
Therefore, lim f ( x ) = 4.
x→0
− 0.5
0.5
96. lim ⎡⎣b − x − a ⎤⎦ ≤ lim f ( x) ≤ lim ⎡⎣b + x − a ⎤⎦
x→a
x→a
x→a
b ≤ lim f ( x) ≤ b
x→a
Therefore, lim f ( x) = b.
x→a
− 0.5
1⎞
⎛
lim ⎜ x sin ⎟ = 0
x⎠
x → 0⎝
100. h( x) = x cos
97. f ( x) = x sin x
1
x
6
0.5
− 2␲
2␲
− 0.5
0.5
−6
1⎞
⎛
lim ⎜ x cos ⎟ = 0
x⎠
x → 0⎝
− 0.5
lim x sin x = 0
x→0
101. (a) Two functions f and g agree at all but one point
(on an open interval) if f ( x) = g ( x) for all x in the
98. f ( x) = x cos x
interval except for x = c, where c is in the interval.
6
− 2␲
2␲
−6
x2 − 1
( x + 1)( x − 1)
and
=
x −1
x −1
g ( x) = x + 1 agree at all points except x = 1.
(b) f ( x) =
(Other answers possible.)
lim x cos x = 0
x→0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
102
Chapter 2
Limits and Their Properties
102. An indeterminant form is obtained when evaluating a
limit using direct substitution produces a meaningless
fractional expression such as 0 0. That is,
lim
x→c
103. If a function f is squeezed between two functions h and
g, h( x) ≤ f ( x) ≤ g ( x), and h and g have the same limit
L as x → c, then lim f ( x ) exists and equals L
x→c
f ( x)
g ( x)
for which lim f ( x) = lim g ( x) = 0
x→c
x→c
104. (a) Use the dividing out technique because the numerator and denominator have a common factor.
x2 + x − 2
( x + 2)( x − 1)
= lim
x → −2
x → −2
x + 2
x + 2
= lim ( x − 1) = − 2 − 1 = − 3
lim
x → −2
(b) Use the rationalizing technique because the numerator involves a radical expression.
x + 4 − 2
= lim
x→0
x
lim
x→0
= lim
x→0 x
= lim
x→0
x + 4 − 2
−
x
( x + 4) − 4
(
x + 4 + 2
)
1
=
x + 4 + 2
105. f ( x ) = x, g ( x) = sin x, h( x) =
x + 4 + 2
x + 4 + 2
1
1
=
4
4 + 2
sin x
x
3
lim
t →2
f
g
h
−5
107. s(t ) = −16t 2 + 500
s ( 2) − s ( t )
2 −t
2
= lim
t →2
−3
= lim
When the x-values are “close to” 0 the magnitude of f is
approximately equal to the magnitude of g. So,
g
f ≈ 1 when x is “close to” 0.
106. f ( x ) = x, g ( x) = sin 2 x, h( x) =
sin 2 x
x
2 −t
t →2
= lim
5
−16( 2) + 500 − ( −16t 2 + 500)
t →2
= lim
436 + 16t − 500
2−t
2
16(t 2 − 4)
2 −t
16(t − 2)(t + 2)
t →2
2−t
= lim −16(t + 2) = −64 ft/sec
t →2
The paint can is falling at about 64 feet/second.
2
g
−3
3
h
f
−2
When the x-values are “close to” 0 the magnitude of g is
“smaller” than the magnitude of f and the magnitude of
g is approaching zero “faster” than the magnitude of f.
f ≈ 0 when x is “close to” 0.
So, g
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.3
103
500
5 5
5 5
=
sec. The velocity at time a =
is
16
2
2
108. s(t ) = −16t 2 + 500 = 0 when t =
⎛5 5 ⎞
s⎜⎜
⎟ − s (t )
2 ⎟⎠
⎝
lim
=
⎛5 5 ⎞
5 5
t → ⎜⎜
⎟⎟
t
−
2
⎝
⎠
2
Evaluating Limits Analytically
0 − ( −16t 2 + 500)
lim
⎛5 5 ⎞
t → ⎜⎜
⎟⎟
⎝ 2 ⎠
5 5
−t
2
125 ⎞
⎛
16⎜ t 2 −
⎟
4 ⎠
⎝
= lim
⎛5 5 ⎞
5 5
t → ⎜⎜
⎟⎟
−t
⎝ 2 ⎠
2
⎛
5 5 ⎞⎛
5 5⎞
16⎜⎜ t +
⎟⎜ t −
⎟
⎟⎜
2
2 ⎟⎠
⎝
⎠⎝
= lim
⎛5 5 ⎞
5 5
t → ⎜⎜
⎟⎟
−t
⎝ 2 ⎠
2
=
⎡
⎛
5 5 ⎞⎤
lim ⎢−16⎜⎜ t +
⎟⎥ = −80 5 ft/sec
5 5⎢
2 ⎟⎠⎥⎦
t→
⎝
⎣
2
≈ −178.9 ft/sec.
The velocity of the paint can when it hits the ground is about 178.9 ft/sec.
109. s(t ) = −4.9t 2 + 200
lim
s(3) − s(t )
3−t
t →3
−4.9(3) + 200 − ( −4.9t 2 + 200)
2
= lim
3−t
t →3
= lim
t →3
= lim
4.9(t 2 − 9)
3−t
4.9(t − 3)(t + 3)
t →3
3−t
= lim ⎡−
4.9(t + 3)⎤⎦
t → 3⎣
= −29.4 m/sec
The object is falling about 29.4 m/sec.
110. −4.9t 2 + 200 = 0 when t =
lim
s ( a ) − s (t )
t →a
a −t
200
20 5
20 5
=
sec. The velocity at time a =
is
4.9
7
7
0 − ⎡⎣−4.9t 2 + 200⎤⎦
t →a
a −t
4.9(t + a )(t − a )
= lim
t →a
a −t
= lim
=
⎡
⎛
20 5 ⎞⎤
lim ⎢−4.9⎜⎜ t +
⎟⎥ = −28 5 m/sec
20 5 ⎢
7 ⎟⎠⎥⎦
t→
⎝
⎣
7
≈ −62.6 m/sec.
The velocity of the object when it hits the ground is about 62.6 m/sec.
111. Let f ( x) = 1 x and g ( x) = −1/ x. lim f ( x) and lim g ( x) do not exist. However,
x →0
x →0
⎡ 1 ⎛ 1 ⎞⎤
lim ⎡ f ( x) + g ( x)⎤⎦ = lim ⎢ + ⎜ − ⎟⎥ = lim [0] = 0
x→0 x
x→0
⎝ x ⎠⎦
⎣
and therefore does not exist.
x → 0⎣
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
104
Chapter 2
Limits and Their Properties
112. Suppose, on the contrary, that lim g ( x) exists. Then,
x→c
If lim f ( x) = 0, then lim ⎡⎣− f ( x) ⎤⎦ = 0.
x→c
x→c
118. (a)
because lim f ( x ) exists, so would lim ⎡⎣ f ( x) + g ( x )⎤⎦ ,
x→c
x→c
− f ( x) ≤ f ( x) ≤ f ( x)
which is a contradiction. So, lim g ( x) does not exist.
lim ⎡− f ( x) ⎤⎦ ≤ lim f ( x) ≤ lim f ( x)
x →c⎣
x→c
x→c
113. Given f ( x) = b, show that for every ε > 0 there exists
x→c
a δ > 0 such that f ( x) − b < ε whenever
Therefore, lim f ( x) = 0.
x − c < δ . Because f ( x) − b = b − b = 0 < ε for
every ε > 0, any value of δ > 0 will work.
x→c
(b) Given lim f ( x ) = L:
x→c
For every ε > 0, there exists δ > 0 such that
f ( x) − L < ε whenever 0 < x − c < δ . Since
114. Given f ( x ) = x , n is a positive integer, then
n
f ( x) − L ≤ f ( x ) − L < ε for
lim x n = lim ( xx n −1 )
x→c
x→c
x − c < δ , then lim f ( x) = L .
= ⎡⎢ lim x⎤⎥ ⎡⎢ lim x n −1 ⎤⎥ = c ⎡⎢ lim ( xx n − 2 )⎤⎥
⎣x → c ⎦ ⎣ x → c
⎦
⎣x → c
⎦
x→c
= c ⎡⎢ lim x⎤⎥ ⎡⎢ lim x n − 2 ⎤⎥ = c(c) lim ( xx n − 3 )
x→c
⎣x → c ⎦ ⎣ x → c
⎦
=
119. Let
⎧ 4, if x ≥ 0
f ( x) = ⎨
⎩−4, if x < 0
= cn.
115. If b = 0, the property is true because both sides are
equal to 0. If b ≠ 0, let ε > 0 be given. Because
lim f ( x ) = L, there exists δ > 0 such that
lim f ( x) = lim 4 = 4.
x→0
x→0
lim f ( x ) does not exist because for
x→0
x→c
f ( x) − L < ε b whenever 0 < x − c < δ . So,
whenever 0 < x − c < δ , we have
b f ( x ) − L < ε or
x→c
0 ≤ lim f ( x) ≤ 0
x < 0, f ( x) = −4 and for x ≥ 0, f ( x) = 4.
120. The graphing utility was set in degree mode, instead of
radian mode.
bf ( x) − bL < ε
121. The limit does not exist because the function approaches
1 from the right side of 0 and approaches −1 from the
left side of 0.
which implies that lim ⎣⎡bf ( x)⎦⎤ = bL.
x→c
116. Given lim f ( x) = 0:
2
x→c
For every ε > 0, there exists δ > 0 such that
−3
f ( x) − 0 < ε whenever 0 < x − c < δ .
3
Now f ( x) − 0 = f ( x ) = f ( x) − 0 < ε for
−2
x − c < δ . Therefore, lim f ( x) = 0.
x→c
122. False. lim
x →π
− M f ( x ) ≤ f ( x ) g ( x) ≤ M f ( x)
117.
(
lim − M f ( x)
x→c
)≤
(
lim f ( x) g ( x ) ≤ lim M f ( x)
x→c
x→c
− M (0) ≤ lim f ( x) g ( x) ≤ M (0)
x→c
0 ≤ lim f ( x) g ( x ) ≤ 0
x→c
Therefore, lim f ( x) g ( x) = 0.
x→c
)
sin x
0
=
= 0
x
π
123. True.
124. False. Let
⎧x x ≠ 1
f ( x) = ⎨
,
⎩3 x = 1
c = 1.
Then lim f ( x ) = 1 but f (1) ≠ 1.
x →1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.4
125. False. The limit does not exist because f ( x ) approaches
Continuity and One-Sided Limits
sec x − 1
x2
(a) The domain of f is all x ≠ 0, π /2 + nπ .
129. f ( x ) =
3 from the left side of 2 and approaches 0 from the right
side of 2.
(b)
4
−3
2
− 3␲
2
6
3␲
2
−2
1 x2
2
and g ( x) = x 2 .
(c) lim f ( x ) =
x→0
Then f ( x ) < g ( x) for all x ≠ 0. But
(d)
lim f ( x) = lim g ( x) = 0.
127. lim
x →0
The domain is not
obvious. The hole at
x = 0 is not apparent.
−2
126. False. Let f ( x ) =
x→0
105
x→0
1
2
sec x − 1 sec x − 1 sec x + 1
sec 2 x − 1
=
⋅
=
sec x + 1 x 2 (sec x + 1)
x2
x2
=
1 − cos x
1 − cos x 1 + cos x
= lim
⋅
x→0
1 + cos x
x
x
tan 2 x
1 ⎛ sin 2 x ⎞
1
=
⎜
⎟
x (sec x + 1) cos 2 x ⎝ x 2 ⎠ sec x + 1
2
sec x − 1
1 ⎛ sin 2 x ⎞
1
= lim
⎜ 2 ⎟
2
x →0
x → 0 cos 2 x
x
x
sec
x
+1
⎝
⎠
1 − cos 2 x
sin 2 x
= lim
= lim
x → 0 x(1 + cos x )
x → 0 x(1 + cos x )
So, lim
1
⎛1⎞
= 1(1)⎜ ⎟ = .
2
⎝ 2⎠
sin x
sin x
⋅
x →0 x
1 + cos x
= lim
sin x ⎤
⎡ sin x ⎤ ⎡
= ⎢lim
⎥
⎥ ⎢ xlim
x
→
0
→
0
1
+
cos x ⎦
x
⎣
⎦⎣
130. (a) lim
x→0
= (1)(0) = 0
1 − cos x
1 − cos x 1 + cos x
= lim
⋅
x→0
x2
x2
1 + cos x
1 − cos 2 x
x → 0 x (1 + cos x )
= lim
⎧0, if x is rational
128. f ( x) = ⎨
⎩1, if x is irrational
2
sin 2 x
1
⋅
x → 0 x2
1 + cos x
= lim
⎧0, if x is rational
g ( x) = ⎨
⎩x, if x is irrational
1
⎛1⎞
= (1)⎜ ⎟ =
2
⎝ 2⎠
lim f ( x) does not exist.
(b) From part (a),
x→0
1 − cos x
1
≈
⇒ 1 − cos x
2
x2
1
≈ x 2 ⇒ cos x
2
1
≈ 1 − x 2 for x
2
≈ 0.
No matter how “close to” 0 x is, there are still an infinite
number of rational and irrational numbers so that
lim f ( x) does not exist.
x→0
lim g ( x) = 0
x→0
when x is “close to” 0, both parts of the function are
“close to” 0.
(c) cos(0.1) ≈ 1 −
1
2
(0.1) = 0.995
2
(d) cos(0.1) ≈ 0.9950, which agrees with part (c).
Section 2.4 Continuity and One-Sided Limits
lim f ( x ) = 3
2. (a)
(b) lim f ( x) = 3
(b)
(c) lim f ( x) = 3
(c) lim f ( x) = −2
The function is continuous at x = 4 and is continuous
on ( −∞, ∞ ).
The function is continuous at x = −2.
1. (a)
x → 4+
x → 4−
x→4
lim f ( x) = −2
x → −2+
lim f ( x) = −2
x → −2−
x → −2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
106
Chapter 2
3. (a)
Limits and Their Properties
lim f ( x) = 0
x → 3+
7. lim
1
1
1
=
=
8+8
16
x +8
8. lim
2
2
1
=
=
2 + 2 2
x + 2
x → 8+
(b) lim f ( x) = 0
x → 3−
(c) lim f ( x) = 0
x → 2−
x →3
The function is NOT continuous at x = 3.
4. (a)
(b)
9. lim
x → 5+
lim f ( x ) = 3
x −5
x −5
= lim
2
+
x − 25
x → 5 ( x + 5)( x − 5)
x → −3+
= lim
x → 5+
lim f ( x) = 3
x → −3−
(c) lim f ( x) = 3
10. lim
x → 4+
x → −3
4− x
− ( x − 4)
−1
= lim
= lim
x 2 − 16
x → 4+ ( x + 4)( x − 4)
x → 4+ x + 4
The function is NOT continuous at x = −3 because
f ( −3) = 4 ≠ lim f ( x).
=
x → −3
5. (a)
lim f ( x) = −3
11.
x → 2+
(b) lim f ( x) = 3
(c) lim f ( x) does not exist
x→2
12. lim
x → 4−
The function is NOT continuous at x = 2.
(b)
x
lim
x → −3−
x2 − 9
1
−1
= −
4+ 4
8
x
does not exist because
x2 − 9
decreases without bound as x → −3−.
x → 2−
6. (a)
1
1
=
10
x +5
x − 2
x − 2
= lim
⋅
x − 4
x → 4− x − 4
= lim
lim f ( x) = 0
x → 4−
x → −1+
lim f ( x ) = 2
x − 4
x → −1
13. lim
The function is NOT continuous at x = −1.
x → 0−
14.
lim
x →10+
x
x
= lim
x → 0−
x − 10
x − 10
− 4)
1
x + 2
x → 4−
(c) lim f ( x) does not exist.
(
(x
= lim
x → −1−
x + 2
x + 2
x + 2
=
)
1
1
=
4
4 + 2
−x
= −1
x
= lim
x →10+
x − 10
=1
x − 10
1
1
−
−∆x
1
x
x
x = lim x − ( x + ∆x) ⋅ 1 = lim
+
∆
15. lim
⋅
−
∆x
∆x
∆x
∆x → 0
∆x → 0− x( x + ∆x )
∆x → 0− x( x + ∆x )
= lim
∆x → 0−
=
16.
lim
∆x → 0+
(x
+ ∆x) + ( x + ∆x ) − ( x 2 + x)
2
∆x
−1
x( x + ∆x)
−1
1
= − 2
x ( x + 0)
x
x 2 + 2 x( ∆x) + ( ∆x ) + x + ∆x − x 2 − x
2
= lim
∆x
∆x → 0+
2 x( ∆x) + ( ∆x ) + ∆x
2
= lim
∆x
= lim ( 2 x + ∆x + 1)
∆x → 0+
∆x → 0+
= 2x + 0 + 1 = 2x + 1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.4
17. lim f ( x) = lim
x → 3−
x → 3−
x + 2
5
=
2
2
30. f ( x ) =
18. lim f ( x) = lim ( x 2 − 4 x + 6) = 9 − 12 + 6 = 3
x → 3−
x → 3−
x → 3+
x → 3+
lim f ( x) = lim ( − x 2 + 4 x − 2) = − 9 + 12 − 2 = 1
Since these one-sided limits disagree, lim f ( x)
x →3
does not exist.
19. lim cot x does not exist because
x →π
lim cot x and lim cot x do not exist.
x →π +
x →π −
20. lim sec x does not exist because
x →π 2
lim sec x and
x → (π 2)+
x → (π 2)−
x → 4−
= 3 for 3 ≤ x < 4)
22. lim ( 2 x − x
x → 2+
)
= 2( 2) − 2 = 2
23. lim ( 2 − − x ) does not exist because
x →3
lim ( 2 − − x
x → 3−
)
= 2 − ( −3) = 5
has a discontinuity at x = −1 because f ( −1) is not
defined.
31. f ( x ) =
x
+ x
2
has discontinuities at each integer k because
lim f ( x ) ≠ lim f ( x).
x →k−
x → k+
x <1
⎧x,
⎪
x = 1 has a discontinuity at
32. f ( x) = ⎨2,
⎪2 x − 1, x > 1
⎩
x = 1 because f (1) = 2 ≠ lim f ( x) = 1.
x →1
33. g ( x) =
49 − x 2 is continuous on [−7, 7].
9 − t 2 is continuous on [−3, 3].
34. f (t ) = 3 −
35. lim f ( x) = 3 = lim f ( x ). f is continuous on [−1, 4].
x → 0−
x → 0+
36. g ( 2) is not defined. g is continuous on [−1, 2).
6
has a nonremovable discontinuity at x = 0
x
because lim f ( x ) does not exist.
37. f ( x) =
x→0
and
lim ( 2 − − x
x → 3+
)
= 2 − ( −4) = 6.
4
has a nonremovable discontinuity at
x −6
x = 6 because lim f ( x ) does not exist.
38. f ( x ) =
⎛
x ⎞
24. lim⎜1 − − ⎟ = 1 − ( −1) = 2
x →1
2
⎝
⎠
39. f ( x) = 3x − cos x is continuous for all real x.
25. lim ln ( x − 3) = ln 0
40. f ( x) = x 2 − 4 x + 4 is continuous for all real x.
x → 3+
does not exist.
26. lim ln (6 − x ) = ln 0
x → 6−
does not exist.
27. lim ln ⎡⎣ x 2 (3 − x)⎤⎦ = ln ⎡⎣4(1)⎤⎦ = ln 4
x → 2−
28. lim ln
x → 5+
29. f ( x ) =
107
x2 − 1
x +1
lim sec x do not exist.
21. lim (5 x − 7) = 5(3) − 7 = 8
(x
Continuity and One-Sided Limits
x
5
= ln = ln 5
1
x − 4
1
x − 4
2
has discontinuities at x = −2 and x = 2
because f ( −2) and f ( 2) are not defined.
x→6
41. f ( x) =
1
1
has nonremovable
=
2
4− x
( 2 − x)( 2 + x)
discontinuities at x = ±2 because lim f ( x) and
lim f ( x) do not exist.
x→2
x → −2
42. f ( x) = cos
πx
2
is continuous for all real x.
x
is not continuous at x = 0, 1.
x − x
x
1
for x ≠ 0, x = 0 is
Because 2
=
x − x
x −1
a removable discontinuity, whereas x = 1 is a
nonremovable discontinuity.
43. f ( x ) =
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
108
Chapter 2
Limits and Their Properties
x
has nonremovable discontinuities at
x − 4
x = 2 and x = −2 because lim f ( x) and lim f ( x)
44. f ( x ) =
2
x→2
x → −2
do not exist.
x −5
x −5
=
x 2 − 25
( x + 5)( x − 5)
has a nonremovable discontinuity at x = − 5 because
lim f ( x ) does not exist, and has a removable
x → −5
discontinuity at x = 5 because
1
1
.
=
x →5 x + 5
10
lim f ( x) = lim
x →5
47. f ( x) =
x + 2
x + 2
=
x − 3 x − 10
( x + 2)( x − 5)
has a nonremovable discontinuity at x = 5 because
lim f ( x ) does not exist, and has a removable
x →5
discontinuity at x = −2 because
lim f ( x) = lim
x → −2 x
1
1
= − .
7
−5
x + 2
x + 2
=
48. f ( x ) = 2
x − x −6
( x − 3)( x + 2)
50. f ( x ) =
⎧ x, x ≤ 1
51. f ( x) = ⎨ 2
⎩x , x > 1
has a possible discontinuity at x = 1.
1.
2.
3.
lim f ( x ) = lim x = 1 ⎫
⎪
x →1−
⎬ lim f ( x ) = 1
lim f ( x) = lim x 2 = 1⎪ x →1
+
+
x →1
x →1
⎭
x →1−
f ( −1) = lim f ( x)
x →1
⎧−2 x + 3, x < 1
52. f ( x ) = ⎨ 2
x ≥1
⎩x ,
has a possible discontinuity at x = 1.
2.
x → −2 x
f (1) = 1
f is continuous at x = 1, therefore, f is continuous for
all real x.
discontinuity at x = − 2 because
lim f ( x) = lim
x −5
x →5
1.
x →3
x −5
has a nonremovable discontinuity at x = 5 because
lim f ( x ) does not exist.
has a nonremovable discontinuity at x = 3 because
lim f ( x) does not exist, and has a removable
x → −2
x + 7
has a nonremovable discontinuity at x = −7 because
lim f ( x ) does not exist.
2
x → −2
x + 7
x → −7
x
45. f ( x) = 2
is continuous for all real x.
x +1
46. f ( x ) =
49. f ( x) =
1
1
= − .
5
−3
3.
f (1) = 12 = 1
lim f ( x ) = lim ( −2 x + 3) = 1⎫
⎪
x →1−
f ( x) = 1
⎬ lim
lim f ( x) = lim x 2 = 1
⎪ x →1
+
+
x →1
x →1
⎭
x →1−
f (1) = lim f ( x)
x →1
f is continuous at x = 1, therefore, f is continuous for
all real x.
⎧x
⎪ + 1, x ≤ 2
53. f ( x ) = ⎨ 2
⎪3 − x, x > 2
⎩
has a possible discontinuity at x = 2.
1.
2.
f ( 2) =
2
+1 = 2
2
⎫
⎛x
⎞
lim f ( x) = lim ⎜ + 1⎟ = 2⎪
⎪
−
−
x→2
x→2 ⎝ 2
⎠
f ( x) does not exist.
⎬ xlim
→2
lim f ( x ) = lim (3 − x) = 1 ⎪
⎪⎭
x → 2+
x → 2+
Therefore, f has a nonremovable discontinuity at x = 2.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.4
Continuity and One-Sided Limits
109
x ≤ 2
⎧−2 x,
54. f ( x ) = ⎨ 2
x
−
4
x
+
1,
x > 2
⎩
has a possible discontinuity at x = 2.
1.
2.
f ( 2) = −2( 2) = −4
lim f ( x) = lim ( −2 x) = −4
⎫
⎪ lim f x does not exist.
( )
⎬
lim f ( x ) = lim ( x 2 − 4 x + 1) = −3⎪ x → 2
+
+
x→2
x→2
⎭
x → 2−
x → 2−
Therefore, f has a nonremovable discontinuity at x = 2.
⎧ πx
⎪tan ,
4
55. f ( x ) = ⎨
⎪ x,
⎩
⎪⎧ln ( x + 1), x ≥ 0
57. f ( x) = ⎨
x < 0
⎪⎩1 − x 2 ,
x <1
x ≥1
has a possible discontinuity at x = 0.
⎧ πx
⎪tan , −1 < x < 1
= ⎨
4
⎪ x,
x ≤ −1 or x ≥ 1
⎩
1.
has possible discontinuities at x = −1, x = 1.
1.
f ( −1) = −1
f (1) = 1
2.
lim f ( x ) = −1
lim f ( x ) = 1
f ( −1) = lim f ( x)
f (1) = lim f ( x)
3.
x → −1
x →−1
x →1
x →1
f is continuous at x = ±1, therefore, f is continuous for
all real x.
2.
⎧⎪10 − 3e5 − x , x > 5
58. f ( x) = ⎨
3
x ≤ 5
⎪⎩10 − 5 x,
has a possible discontinuity at x = 5.
x −3 ≤ 2
2.
x −3 > 2
⎧ πx
⎪csc , 1 ≤ x ≤ 5
= ⎨
6
⎪2,
x < 1 or x > 5
⎩
3.
π
f (1) = csc
2.
lim f ( x) = 2
3.
f (1) = lim f ( x)
6
= 2
x →1
x →1
f (5) = csc
f (5) = 7
lim f ( x) = 10 − 3e5 − 5 = 7⎫
⎪
⎬ lim f ( x ) = 7
lim f ( x ) = 10 − 53 (5) = 7 ⎪ x → 5
−
x →5
⎭
x → 5+
f (5) = lim f ( x)
x →5
f is continuous at x = 5, so, f is continuous for all
real x.
has possible discontinuities at x = 1, x = 5.
1.
lim f ( x) = 1 − 0 = 1⎫
⎪
f ( x) does not exist.
⎬ xlim
lim f ( x) = 0
⎪ →0
+
x→0
⎭
x → 0−
So, f has a nonremovable discontinuity at x = 0.
1.
⎧ πx
⎪csc ,
6
56. f ( x) = ⎨
⎪2,
⎩
f (0) = ln (0 + 1) = ln 1 = 0
5π
= 2
6
lim f ( x ) = 2
x →5
f (5) = lim f ( x)
x →5
f is continuous at x = 1 and x = 5, therefore, f is
continuous for all real x.
59. f ( x) = csc 2 x has nonremovable discontinuities at
integer multiples of π 2.
60. f ( x) = tan
πx
has nonremovable discontinuities at each
2
2k + 1, k is an integer.
61. f ( x) = x − 8 has nonremovable discontinuities at
each integer k.
62. f ( x ) = 5 − x has nonremovable discontinuities at
each integer k.
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110
Chapter 2
Limits and Their Properties
63. f (1) = 3
4 sin x
= 4
x
lim g ( x) = lim ( a − 2 x) = a
64. lim g ( x) = lim
Find a so that lim ( ax − 4) = 3
x →1−
a(1) − 4 = 3
x → 0−
x → 0−
x → 0+
x → 0+
Let a = 4.
a = 7.
65. Find a and b such that lim ( ax + b) = − a + b = 2 and lim ( ax + b) = 3a + b = −2.
x → −1+
x → 3−
a − b = −2
(+)3a
+ b = −2
= −4
4a
a = −1
2 + ( −1) = 1
b =
x ≤ −1
⎧ 2,
⎪
f ( x) = ⎨− x + 1, −1 < x < 3
⎪−2,
x ≥ 3
⎩
x2 − a2
x→a x − a
= lim ( x + a ) = 2a
66. lim g ( x) = lim
x→a
x→a
Find a such 2a = 8 ⇒ a = 4.
67. f (1) = arctan (1 − 1) + 2 = 2
Find a such that lim ( ae
x −1
x →1−
+ 3) = 2
ae 1 −1 + 3 = 2
71. f ( g ( x)) =
( x2
1
+ 5) − 6
=
1
x2 − 1
Nonremovable discontinuities at x = ±1
72. f ( g ( x )) = sin x 2
Continuous for all real x
73. y = x − x
Nonremovable discontinuity at each integer
a +3 = 2
0.5
a = −1.
−3
3
68. f ( 4) = 2e 4 a − 2
Find a such that lim ln ( x − 3) + x 2 = 2e 4 a − 2
− 1.5
x → 4+
ln ( 4 − 3) + 42 = 2e 4 a − 2
16 = 2e 4 a − 2
9 = e4a
74. h( x) =
Nonremovable discontinuities at x = −5 and x = 3
2
ln 9 = 4a
a =
69. f ( g ( x)) = ( x − 1)
ln 9
ln 32
ln 3
=
=
.
4
4
2
70. f ( g ( x )) =
−8
7
−2
2
Continuous for all real x
1
x −1
1
1
=
x 2 + 2 x − 15
( x + 5)( x − 3)
2
⎪⎧x − 3 x, x > 4
75. g ( x) = ⎨
⎪⎩2 x − 5, x ≤ 4
Nonremovable discontinuity at x = 4
Nonremovable discontinuity at x = 1; continuous for
all x > 1
10
−2
8
−2
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Section 2.4
Continuity and One-Sided Limits
111
⎧ cos x − 1
, x < 0
⎪
76. f ( x) = ⎨
x
⎪5 x,
x ≥ 0
⎩
f (0) = 5(0) = 0
lim f ( x) = lim
x → 0−
3
(cos x
x → 0−
− 1)
x
= 0
−7
lim f ( x) = lim (5 x) = 0
x → 0+
x → 0+
2
−3
Therefore, lim f ( x) = 0 = f (0) and f is continuous on the entire real line.
(x
x→0
= 0 was the only possible discontinuity.)
77. f ( x) =
x
x2 + x + 2
Continuous on ( −∞, ∞)
⎧2 x − 4, x ≠ 3
84. f ( x ) = ⎨
x = 3
⎩1,
Since lim f ( x) = lim ( 2 x − 4) = 2 ≠ 1,
x →3
x +1
78. f ( x) =
x
Continuous on (0, ∞)
79. f ( x) = 3 −
85. f ( x ) =
−4
x + 3
The graph appears to be continuous on the interval
[−4, 4]. Because f (0) is not defined, you know that
πx
f has a discontinuity at x = 0. This discontinuity is
removable so it does not show up on the graph.
4
Continuous on:
…, ( −6, − 2),( −2, 2),( 2, 6), (6, 10), …
82. f ( x) = cos
4
−2
Continuous on [−3, ∞)
81. f ( x ) = sec
sin x
x
3
x
Continuous on [0, ∞)
80. f ( x) = x
x →3
f is continuous on (− ∞, 3) and (3, ∞ ).
86. f ( x) =
1
x
x3 − 8
x − 2
14
Continuous on (−∞, 0) and (0, ∞)
⎧ x2 − 1
, x ≠1
⎪
83. f ( x) = ⎨ x − 1
⎪2,
x =1
⎩
Since lim f ( x) = lim
x →1
x →1
x2 − 1
( x − 1)( x + 1)
= lim
x →1
x −1
x −1
= lim ( x + 1) = 2,
−4
4
0
The graph appears to be continuous on the interval
[−4, 4]. Because f (2) is not defined, you know that
f has a discontinuity at x = 2. This discontinuity is
removable so it does not show up on the graph.
x →1
f is continuous on (−∞, ∞ ).
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112
Chapter 2
87. f ( x) =
Limits and Their Properties
ln ( x 2 + 1)
88. f ( x) =
x
− e− x + 1
ex − 1
5
3
−4
4
−4
−2
−3
The graph appears to be continuous on the interval
[−4, 4]. Because f (0) is not defined, you know that f
The graph appears to be continuous on the interval
[−4, 4]. Because f (0) is not defined, you know that f
has a discontinuity at x = 0. This discontinuity is
removable so it does not show up on the graph.
has a discontinuity at x = 0. This discontinuity is
removable so it does not show up on the graph.
89. f ( x) =
1 x4
12
− x3 + 4 is continuous on the interval [1, 2]. f (1) =
there exists a number c in [1, 2] such that f (c) = 0.
90. f ( x ) = −
4
37
12
and f ( 2) = − 83 . By the Intermediate Value Theorem,
5
⎛π x ⎞
+ tan ⎜ ⎟ is continuous on the interval [1, 4].
x
⎝ 10 ⎠
5
⎛π ⎞
⎛ 2π ⎞
f (1) = −5 + tan ⎜ ⎟ ≈ −4.7 and f ( 4) = − + tan ⎜ ⎟ ≈ 1.8. By the Intermediate Value Theorem, there exists a number
4
⎝ 10 ⎠
⎝ 5 ⎠
c in [1, 4] such that f (c) = 0.
⎡ π⎤
⎛π ⎞
91. h is continuous on the interval ⎢0, ⎥. h(0) = − 2 < 0 and h⎜ ⎟ ≈ 0.91 > 0. By the Intermediate Value Theorem,
⎝2⎠
⎣ 2⎦
⎡ π⎤
there exists a number c in ⎢0, ⎥ such that h(c) = 0.
⎣ 2⎦
92. g is continuous on the interval [0, 1]. g (0) ≈ − 2.77 < 0 and g (1) ≈ 1.61 > 0. By the Intermediate Value Theorem,
there exists a number c in [0, 1] such that g (c) = 0.
93. f ( x) = x3 + x − 1
95. g (t ) = 2 cos t − 3t
f ( x ) is continuous on [0, 1].
g is continuous on [0, 1].
f (0) = −1 and f (1) = 1
g (0) = 2 > 0 and g (1) ≈ −1.9 < 0.
By the Intermediate Value Theorem, f (c) = 0 for at
By the Intermediate Value Theorem, g (c) = 0 for at
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of f ( x), you find that
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of g (t ), you find that
x ≈ 0.68. Using the root feature, you find that
x ≈ 0.6823.
t ≈ 0.56. Using the root feature, you find that
t ≈ 0.5636.
94. f ( x) = x 4 − x 2 + 3 x − 1
f ( x) is continuous on [0, 1].
f (0) = −1 and f (1) = 2
By the Intermediate Value Theorem, f (c) = 0 for at
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of f ( x), you find that
x ≈ 0.37. Using the root feature, you find that
x ≈ 0.3733.
96. h(θ ) = tanθ + 3θ − 4 is continuous on [0, 1].
h(0) = − 4 and h(1) = tan(1) −1 ≈ 0.557.
By the Intermediate Value Theorem, h(c) = 0 for at
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of h(θ ), you find that
θ ≈ 0.91. Using the root feature, you obtain
θ ≈ 0.9071.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.4
97. f ( x ) = x + e x − 3
f is continuous on [0, 1].
f (0) = e0 − 3 = − 2 < 0 and
Continuity and One-Sided Limits
101. f ( x ) = x 3 − x 2 + x − 2
f is continuous on [0, 3].
f (0) = −2 and f (3) = 19
−2 < 4 < 19
f (1) = 1 + e − 3 = e − 2 > 0.
By the Intermediate Value Theorem, f (c) = 0 for at
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of f ( x), you find that
x ≈ 0.79. Using the root feature, you find that
x ≈ 0.7921.
98. g ( x) = 5 ln ( x + 1) − 2
g is continuous on [0, 1].
g (0) = 5 ln (0 + 1) − 2 = − 2 and
g (1) = 5 ln ( 2) − 2 > 0.
By the Intermediate Value Theorem, g (c) = 0 for at
least one value of c between 0 and 1. Using a graphing
utility to zoom in on the graph of g ( x), you find that
x ≈ 0.49. Using the root feature, you find that
x ≈ 0.4918.
99. f ( x ) = x 2 + x − 1
f is continuous on [0, 5].
The Intermediate Value Theorem applies.
x3 − x 2 + x − 2 = 4
x3 − x 2 + x − 6 = 0
− 2)( x 2 + x + 3) = 0
(x
x = 2
(x
+ x + 3 has no real solution.)
2
c = 2
So, f ( 2) = 4.
102. f ( x) =
35
20
⎛5⎞
f⎜ ⎟ =
and f ( 4) =
6
3
⎝ 2⎠
35
20
< 6 <
6
3
The Intermediate Value Theorem applies.
x2 + x
= 6
x −1
x2 + x = 6x − 6
−1 < 11 < 29
The Intermediate Value Theorem applies.
x 2 + x − 12 = 0
(x
+ 4)( x − 3) = 0
x = −4 or x = 3
c = 3( x = −4 is not in the interval.)
So, f (3) = 11.
100. f ( x) = x 2 − 6 x + 8
f is continuous on [0, 3].
f (0) = 8 and f (3) = −1
x2 − 5x + 6 = 0
(x
− 2)( x − 3) = 0
x = 2 or x = 3
c = 3 ( x = 2 is not in the interval.)
So, f (3) = 6.
103. (a) The limit does not exist at x = c.
(b) The function is not defined at x = c.
(c) The limit exists at x = c, but it is not equal to the
value of the function at x = c.
(d) The limit does not exist at x = c.
104. Answers will vary. Sample answer:
−1 < 0 < 8
y
5
4
3
2
1
The Intermediate Value Theorem applies.
x2 − 6x + 8 = 0
(x
− 2)( x − 4) = 0
x = 2 or x = 4
c = 2 ( x = 4 is not in the interval.)
So, f ( 2) = 0.
x2 + x
x −1
⎡5 ⎤
f is continuous on ⎢ , 4⎥. The nonremovable
⎣2 ⎦
discontinuity, x = 1, lies outside the interval.
f (0) = −1 and f (5) = 29
x 2 + x − 1 = 11
113
−2 −1
x
1
3 4 5 6 7
−2
−3
The function is not continuous at x = 3 because
lim f ( x) = 1 ≠ 0 = lim f ( x).
x → 3+
x → 3−
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
114
Chapter 2
Limits and Their Properties
105. If f and g are continuous for all real x, then so is f + g
(Theorem 2.11, part 2). However, f g might not be
continuous if g ( x) = 0. For example, let f ( x) = x and
g ( x) = x 2 − 1. Then f and g are continuous for all real
x, but f g is not continuous at x = ±1.
106. A discontinuity at c is removable if the function f can
be made continuous at c by appropriately defining (or
redefining) f (c). Otherwise, the discontinuity is
111. The functions agree for integer values of x:
g ( x) = 3 − − x = 3 − ( − x ) = 3 + x⎫⎪
⎬ for x an integer
f ( x) = 3 + x = 3 + x
⎪⎭
However, for non-integer values of x, the functions
differ by 1.
f ( x ) = 3 + x = g ( x) − 1 = 2 − − x .
For example,
f
( 12 ) = 3 + 0 = 3, g ( 12 ) = 3 − (−1) = 4.
nonremovable.
(b) f ( x ) =
112. lim f (t ) ≈ 28
x − 4
(a) f ( x ) =
t → 4−
x − 4
lim f (t ) ≈ 56
t → 4+
sin ( x + 4)
x + 4
x ≥ 4
⎧1,
⎪
⎪0,
(c) f ( x) = ⎨
⎪1,
⎪0,
⎩
−4 < x < 4
x = −4
x < −4
x = 4 is nonremovable, x = −4 is removable
At the end of day 3, the amount of chlorine in the pool
has decreased to about 28 oz. At the beginning of day 4,
more chlorine was added, and the amount is now about
56 oz.
⎧0.40,
0 < t ≤ 10
⎪
113. C (t ) = ⎨0.40 + 0.05 t − 9 , t > 10, t not an integer
⎪0.40 + 0.05 t − 10 , t > 10, t an integer
(
)
⎩
y
C
4
0.7
3
0.6
2
0.5
0.4
1
−6 −4 −2
0.3
x
−1
2
4
0.2
6
0.1
t
−2
2
107. True
1.
2.
3.
6
8 10 12 14
f (c) = L is defined.
There is a nonremovable discontinuity at each integer
greater than or equal to 10.
lim f ( x) = L exists.
Note: You could also express C as
x→c
0 < t ≤ 10
⎧⎪0.40,
C (t ) = ⎨
⎪⎩0.40 − 0.05 10 − t , t > 10
f (c) = lim f ( x)
x→c
All of the conditions for continuity are met.
108. True. If f ( x) = g ( x), x ≠ c, then
lim f ( x) = lim g ( x ) (if they exist) and at least one of
x→c
4
x→c
⎛ t + 2
⎞
114. N (t ) = 25⎜ 2
− t⎟
2
⎝
⎠
t
0
1
1.8
2
3
3.8
N (t )
50
25
5
50
25
5
these limits then does not equal the corresponding
function value at x = c.
109. False. A rational function can be written as
P( x) Q( x) where P and Q are polynomials of degree m
Discontinuous at every positive even integer. The
company replenishes its inventory every two months.
and n, respectively. It can have, at most, n
discontinuities.
x →1
Number of units
110. False. f (1) is not defined and lim f ( x) does not exist.
N
50
40
30
20
10
t
2
4
6
8
10 12
Time (in months)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.4
115. Let s(t ) be the position function for the run up to the
campsite. s(0) = 0 (t = 0 corresponds to 8:00 A.M.,
s( 20) = k (distance to campsite)). Let r (t ) be the
position function for the run back down the mountain:
r (0) = k , r (10) = 0. Let f (t ) = s(t ) − r (t ).
When t = 0 (8:00 A.M.),
f (0) = s(0) − r (0) = 0 − k < 0.
When t = 10 (8:00 A.M.), f (10) = s(10) − r (10) > 0.
Continuity and One-Sided Limits
⎧−1, if x < 0
⎪
120. sgn ( x) = ⎨0, if x = 0
⎪1,
if x > 0
⎩
(a)
lim sgn ( x) = −1
x → 0−
(b) lim sgn ( x) = 1
x → 0+
(c) limsgn ( x) does not exist.
x→0
Because f (0) < 0 and f (10) > 0, then there must be a
y
4
value t in the interval [0, 10] such that f (t ) = 0. If
3
2
f (t ) = 0, then s(t ) − r (t ) = 0, which gives us
s(t ) = r (t ). Therefore, at some time t, where
1
4 3
π r be the volume of a sphere with radius r.
3
500π
V is continuous on [5, 8]. V (5) =
≈ 523.6 and
3
2048π
V (8) =
≈ 2144.7. Because
3
523.6 < 1500 < 2144.7, the Intermediate Value
Theorem guarantees that there is at least one value r
between 5 and 8 such that V ( r ) = 1500. (In fact,
r ≈ 7.1012.)
117. Suppose there exists x1 in [a, b] such that
f ( x1 ) > 0 and there exists x2 in [a, b] such that
[ x1, x2 ] (or [ x2 , x1] if x2 < x1 ). So, f would have a zero in
[a, b], which is a contradiction. Therefore, f ( x) > 0 for
all x in [a, b] or f ( x) < 0 for all x in [a, b].
118. Let c be any real number. Then lim f ( x ) does not exist
x→c
because there are both rational and irrational numbers
arbitrarily close to c. Therefore, f is not continuous at c.
119. If x = 0, then f (0) = 0 and lim f ( x ) = 0. So, f is
x→0
t→x
2
3
4
−3
−4
121. (a)
S
60
50
40
30
20
10
t
5
10 15 20 25 30
(b) There appears to be a limiting speed and a possible
cause is air resistance.
⎧0, 0 ≤ x < b
122. (a) f ( x ) = ⎨
⎩b, b < x ≤ 2b
y
2b
b
x
b
2b
NOT continuous at x = b.
⎧x
0 ≤ x ≤ b
⎪⎪ 2 ,
(b) g ( x) = ⎨
⎪b − x , b < x ≤ 2b
⎪⎩
2
continuous at x = 0.
If x ≠ 0, then lim f (t ) = 0 for x rational, whereas
1
−2
f ( x2 ) < 0. Then by the Intermediate Value Theorem,
f ( x) must equal zero for some value of x in
x
−4 −3 −2 −1
0 ≤ t ≤ 10, the position functions for the run up and the
run down are equal.
116. Let V =
115
y
2b
lim f (t ) = lim kt = kx ≠ 0 for x irrational. So, f is not
t→x
t→x
continuous for all x ≠ 0.
b
x
b
2b
Continuous on [0, 2b].
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
116
Chapter 2
Limits and Their Properties
124. Let y be a real number. If y = 0, then x = 0. If
y > 0, then let 0 < x0 < π 2 such that
2
⎪⎧1 − x , x ≤ c
123. f ( x) = ⎨
x, x > c
⎪⎩
M = tan x0 > y ( this is possible since the tangent
f is continuous for x < c and for x > c. At x = c, you
function increases without bound on [0, π 2)). By the
need 1 − c 2 = c. Solving c 2 + c − 1, you obtain
c =
−1 ±
1+ 4
=
2
−1 ±
2
5
Intermediate Value Theorem, f ( x) = tan x is
continuous on [0, x0 ] and 0 < y < M , which implies
.
that there exists x between 0 and x0 such that
tan x = y. The argument is similar if y < 0.
125. f ( x) =
x + c2 − c
,c > 0
x
Domain: x + c 2 ≥ 0 ⇒ x ≥ −c 2 and x ≠ 0, ⎡⎣−c 2 , 0) ∪ (0, ∞)
lim
x →0
x + c2 − c
= lim
x →0
x
x + c2 − c
⋅
x
x + c2 + c
x +c +c
2
= lim
x→0
( x + c2 ) − c2
1
= lim
x ⎡ x + c + c⎤
⎣
⎦
x +c +c
x→0
2
2
=
1
2c
Define f (0) = 1 ( 2c) to make f continuous at x = 0.
126. 1.
2.
f (c) is defined.
127. h( x) = x x
lim f ( x) = lim f (c + ∆x) = f (c) exists.
x→c
15
∆x → 0
[Let x = c + ∆x. As x → c, ∆x → 0]
3.
lim f ( x) = f (c).
−3
x→c
3
Therefore, f is continuous at x = c.
−3
h has nonremovable discontinuities at
x = ±1, ± 2, ± 3, ….
128. (a) Define f ( x) = f 2 ( x) − f1 ( x). Because f1 and f 2 are continuous on [a, b], so is f.
f ( a) = f 2 ( a ) − f1 ( a) > 0 and f (b) = f 2 (b) − f1 (b) < 0
By the Intermediate Value Theorem, there exists c in [a, b] such that f (c) = 0.
f (c) = f 2 (c) − f1 (c) = 0 ⇒ f1 (c) = f 2 (c)
(b) Let f1 ( x) = x and f 2 ( x) = cos x, continuous on [0, π 2], f1 (0) < f 2 (0) and f1 (π 2) > f 2 (π 2).
So by part (a), there exists c in [0, π 2] such that c = cos(c).
Using a graphing utility, c ≈ 0.739.
129. The statement is true.
If y ≥ 0 and y ≤ 1, then y ( y − 1) ≤ 0 ≤ x 2 , as desired. So assume y > 1. There are now two cases.
Case l: If x ≤ y − 12 , then 2 x + 1 ≤ 2 y and
y ( y − 1) = y ( y + 1) − 2 y
Case 2: If x ≥ y −
(
x2 ≥ y −
1
2
1
2
)
2
≤ ( x + 1) − 2 y
= y2 − y +
= x + 2x + 1 − 2 y
> y2 − y
≤ x + 2y − 2y
= y ( y − 1)
2
2
2
1
4
= x2
In both cases, y ( y − 1) ≤ x 2 .
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.5
117
Infinite Limits
130. P(1) = P(02 + 1) = P(0) + 1 = 1
2
P( 2) = P(12 + 1) = P(1) + 1 = 2
2
P(5) = P( 22 + 1) = P( 2) + 1 = 5
2
Continuing this pattern, you see that P( x) = x for infinitely many values of x.
So, the finite degree polynomial must be constant: P( x) = x for all x.
Section 2.5 Infinite Limits
1.
lim 2
x
= ∞
x2 − 4
lim 2
x
= ∞
x − 4
x → −2+
x → −2−
6. f ( x ) =
As x approaches 4 from the left, x − 4 is a small
negative number. So,
2
lim f ( x) = ∞.
x → 4−
1
2. lim
= ∞
x → −2+ x + 2
1
= −∞
lim
x → −2− x + 2
3.
4.
As x approaches 4 from the right, x − 4 is a small
positive number. So,
lim f ( x) = −∞.
x → 4+
πx
= −∞
4
πx
lim tan
= ∞
4
x → −2−
lim tan
7. f ( x ) =
x → −2+
lim sec
x → −2+
lim sec
x → −2−
−1
x − 4
πx
4
πx
1
(x
− 4)
2
As x approaches 4 from the left or right, ( x − 4) is a
2
small positive number. So,
= ∞
lim f ( x ) = lim f ( x) = ∞.
x → 4+
= −∞
4
1
5. f ( x) =
x − 4
8. f ( x ) =
x → 4−
−1
( x − 4)
2
As x approaches 4 from the left or right, ( x − 4) is a
2
As x approaches 4 from the left, x − 4 is a small
negative number. So,
small positive number. So,
lim f ( x) = −∞
lim f ( x) = lim f ( x ) = −∞.
x → 4−
x → 4−
x → 4+
As x approaches 4 from the right, x − 4 is a small
positive number. So,
lim f ( x) = ∞
x → 4+
9. f ( x) =
1
x2 − 9
x
–3.5
–3.1
–3.01
–3.001
−2.999
–2.99
–2.9
–2.5
0.308
1.639
16.64
166.6
−166.7
−16.69
−1.695
−0.364
f ( x)
lim f ( x) = ∞
2
x → −3−
lim f ( x ) = −∞
x → −3+
−6
6
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
118
Chapter 2
Limits and Their Properties
10. f ( x ) =
x
x −9
x
–3.5
–3.1
–3.01
–3.001
−2.999
–2.99
–2.9
–2.5
−1.077
−5.082
−50.08
−500.1
499.9
49.92
4.915
0.9091
2
f ( x)
lim f ( x) = −∞
2
x → −3−
lim f ( x ) = ∞
x → −3+
−6
6
−2
11. f ( x) =
x2
x −9
x
–3.5
–3.1
–3.01
–3.001
−2.999
–2.99
–2.9
–2.5
3.769
15.75
150.8
1501
−1499
−149.3
−14.25
−2.273
2
f ( x)
lim f ( x) = ∞
4
x → −3−
lim f ( x ) = −∞
x → −3+
−6
6
−4
12. f ( x ) = cot
x
f ( x)
πx
3
–3.5
–3.1
–3.01
–3.001
−2.999
–2.99
–2.9
–2.5
−1.7321
−9.514
−95.49
−954.9
954.9
95.49
9.514
1.7321
lim f ( x) = −∞
4
x → −3−
lim f ( x ) = ∞
x → −3+
−6
6
−4
13. f ( x) =
lim
x → 0+
1
x2
15. f ( x ) =
1
1
= ∞ = lim 2
x2
x → 0− x
Therefore, x = 0 is a vertical asymptote.
2
( x − 3)3
2
lim
= −∞
3
−
x → 3 ( x − 3)
14. f ( x ) =
lim
x → 3+
2
(x
− 3)
3
= ∞
Therefore, x = 3 is a vertical asymptote.
lim
x → −2−
x2
x2
=
x − 4
( x + 2)( x − 2)
2
x2
x2
= ∞ and lim 2
= −∞
+
x − 4
x → −2 x − 4
2
Therefore, x = −2 is a vertical asymptote.
lim
x → 2−
x2
x2
= −∞ and lim 2
= ∞
+
x − 4
x→2 x − 4
2
Therefore, x = 2 is a vertical asymptote.
16. f ( x) =
3x
x2 + 9
No vertical asymptotes because the denominator
is never zero.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.5
17. g (t ) =
t −1
t2 + 1
22. h( x) =
No vertical asymptotes because the denominator
is never zero.
=
3s + 4
3s + 4
=
s 2 − 16
( s − 4)( s + 4)
3s + 4
3s + 4
lim 2
= −∞ and lim 2
= ∞
s → 4− s − 16
s → 4+ s − 16
=
18. h( s ) =
Therefore, s = 4 is a vertical asymptote.
3s + 4
3s + 4
= −∞ and lim 2
= ∞
2
−
+
s
−
16
s
− 16
s → −4
s → −4
lim
Therefore, s = − 4 is a vertical asymptote.
3
3
=
x2 + x − 2
( x + 2)( x − 1)
3
3
lim 2
= ∞ and lim 2
= −∞
−
+
2
x
x
x
x − 2
+
−
+
x → −2
x → −2
Therefore, x = − 2 is a vertical asymptote.
lim
x →1−
3
3
= −∞ and lim 2
= ∞
x2 + x − 2
x →1+ x + x − 2
Therefore, x = 1 is a vertical asymptote.
20.
x3 − 8
( x − 2)( x 2 + 2 x + 4)
=
x − 2
x − 2
= x 2 + 2 x + 4, x ≠ 2
lim g ( x ) = 4 + 4 + 4 = 12
g ( x) =
x→2
There are no vertical asymptotes. The graph has a
hole at x = 2.
21. f ( x ) =
=
=
x 2 − 2 x − 15
3
x − 5x2 + x − 5
( x − 5)( x + 3)
(x
x + 3
,x ≠ 5
x2 + 1
lim f ( x) =
x →5
− 5)( x 2 + 1)
5+ 3
15
=
52 + 1
26
There are no vertical asymptotes. The graph has a hole
at x = 5.
119
x2 − 9
x + 3x 2 − x − 3
( x − 3)( x + 3)
3
(x
− 1)( x + 1)( x + 3)
x −3
, x ≠ −3
( x + 1)( x − 1)
lim h( x) = −∞ and lim h( x) = ∞
x → −1−
x → −1+
Therefore, x = −1 is a vertical asymptote.
lim h( x) = ∞ and lim h( x) = −∞
x →1−
x →1+
Therefore, x = 1 is a vertical asymptote.
lim h( x ) =
19. f ( x) =
Infinite Limits
x → −3
−3 − 3
3
= −
(− 3 + 1)(− 3 − 1)
4
Therefore, the graph has a hole at x = − 3.
23. f ( x ) =
e −2x
x −1
lim f ( x) = −∞ and lim = ∞
x →1−
x →1+
Therefore, x = 1 is a vertical asymptote.
24. g ( x) = xe − 2x
The function is continuous for all x. Therefore, there are
no vertical asymptotes.
25. h(t ) =
ln (t 2 + 1)
t + 2
lim h(t ) = − ∞ and lim = ∞
t → −2 −
t → −2 +
Therefore, t = − 2 is a vertical asymptote.
26. f ( z ) = ln ( z 2 − 4) = ln ⎡⎣( z + 2)( z − 2)⎤⎦
= ln ( z + 2) + ln ( z − 2)
The function is undefined for − 2 < z < 2.
Therefore, the graph has holes at z = ± 2.
27. f ( x ) =
1
ex − 1
lim f ( x) = −∞ and lim f ( x) = ∞
x → 0−
x → 0+
Therefore, x = 0 is a vertical asymptote.
28. f ( x) = ln ( x + 3)
lim f ( x) = − ∞
x → −3
Therefore, x = − 3 is a vertical asymptote.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
120
Chapter 2
Limits and Their Properties
29. f ( x) = cscπ x =
1
sinπ x
34.
lim
x → −1−
x2 − 2x − 8
= ∞
x +1
4
− 10
8
x2 − 2 x − 8
lim
= −∞
x +1
x → −1+
Let n be any integer.
lim f ( x ) = − ∞ or ∞
Vertical asymptote at x = −1
x→n
−8
Therefore, the graph has vertical asymptotes at x = n.
sinπ x
cosπ x
2n + 1
cosπ x = 0 for x =
, where n is an integer.
2
30. f ( x ) = tanπ x =
lim
x → 2 n +1
2
35.
t
31. s(t ) =
sin t
sin t = 0 for t = nπ , where n is an integer.
x2 + 1
= −∞
x → −1− x + 1
36.
ln ( x 2 + 1)
lim
x → −1+
lim
x +1
ln ( x 2 + 1)
x +1
x → −1−
37.
lim
x → −1+
38. lim
x →1−
lim s(t ) = 1
Therefore, the graph has a hole at t = 0.
32. g (θ ) =
tan θ
θ
=
sin θ
θ cos θ
cos θ = 0 for θ =
lim
θ → π + nπ
π
2
+ nπ , where n is an integer.
2
3
−5
39. lim
x → 2+
−1
− 1)
2
= −∞
x
= ∞
x − 2
x2
4
1
=
=
4+ 4
2
x → 2− x + 4
41.
lim
x → −3−
2
x +3
x +3
= lim
−
x
+
( x + x − 6) x → −3 ( 3)( x − 2)
2
= lim
x → −3−
42.
lim
x →−(1 2)+
6x2 + x − 1
=
4x2 − 4 x − 3
lim g (θ ) = 1
=
θ →0
Therefore, the graph has a hole at θ = 0.
x2 − 1
= lim ( x − 1) = −2
x → −1 x + 1
x → −1
1⎞
⎛
43. lim ⎜1 + ⎟ = −∞
x⎠
x → 0− ⎝
Removable discontinuity at x = −1
1⎞
⎛
44. lim ⎜ 6 − 3 ⎟ = −∞
+
x
x→0 ⎝
⎠
33. lim
2
−3
3
45.
−5
−5
1
= ∞
x +1
(x
40. lim
g (θ ) = ∞ or − ∞
+ nπ .
3
= −∞
Therefore, the graph has vertical asymptotes at
π
−8
= ∞
2
θ =
3
Vertical asymptote at x = −1
lim s (t ) = ∞ or − ∞ (for n ≠ 0)
t →0
−3
Vertical asymptote at x = −1
t → nπ
Therefore, the graph has vertical asymptotes at
t = nπ , for n ≠ 0.
8
lim
f ( x) = ∞ or −∞
Therefore, the graph has vertical asymptotes at
2n + 1
x =
.
2
x2 + 1
= ∞
x → −1+ x + 1
lim
⎛
lim ⎜ x 2 +
x → −4− ⎝
1
1
= −
x − 2
5
lim
(3x − 1)(2 x + 1)
(2 x − 3)(2 x + 1)
lim
3x − 1
5
=
2x − 3
8
x → −(1 2)+
x → −(1 2)+
2 ⎞
⎟ = −∞
x + 4⎠
πx⎞
⎛x
46. lim ⎜ + cot
⎟ = ∞
+
3
2 ⎠
x →3 ⎝
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.5
2
47. lim
x → 0+ sin
1
x − 25
lim f ( x) = −∞
57. f ( x ) =
= ∞
x
−2
48. lim
= ∞
+
x → (π 2) cos x
e
49. lim
x
( x − 8)
x → 8−
3
121
Infinite Limits
0.3
2
−8
x → 5−
8
− 0.3
πx
58. f ( x ) = sec
= −∞
6
8
lim f ( x) = −∞
−9
x → 4+
50. lim ln ( x 2 − 16) = − ∞
9
x → 4+
−6
51.
lim
x → (π 2)−
52. lim e
ln cos x = ln cos
π
2
59. A limit in which f ( x ) increases or decreases without
= ln 0 = − ∞
bound as x approaches c is called an infinite limit. ∞ is
not a number. Rather, the symbol
sin x = 1(0) = 0
− 0.5 x
x → 0+
lim f ( x) = ∞
x→c
53.
54.
lim
x → (1 2)
−
x sec π x =
lim
x → (1 2)
−
x
= ∞
cos π x
says how the limit fails to exist.
60. The line x = c is a vertical asymptote if the graph of f
approaches ± ∞ as x approaches c.
lim x tan π x = −∞
2
x → (1 2)+
61. One answer is
x2 + x + 1
x2 + x + 1
55. f ( x ) =
=
3
x −1
( x − 1)( x 2 + x + 1)
lim f ( x) = lim
x →1+
x →1+
f ( x) =
1
= ∞
x −1
x −3
x −3
= 2
.
x
−
6
x
+
2
x
−
4 x − 12
(
)(
)
62. No. For example, f ( x ) =
3
1
has no vertical
x2 + 1
asymptote.
−4
y
63.
5
3
2
−3
1
( x − 1)( x + x + 1)
x −1
=
2
x + x +1
x2 + x + 1
lim f ( x) = lim ( x − 1) = 0
56. f ( x ) =
2
3
x →1−
−2
x
−1
1
−2
x →1−
4
64. m =
−8
8
m0
1 − (v 2 c 2 )
lim m = lim
v → c−
−4
65. (a)
3
−1
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.0411
0.0067
0.0017
≈ 0
≈ 0
≈ 0
0.5
lim
x → 0+
−1.5
v → c−
m0
1 − (v 2 c 2 )
= ∞
x − sin x
= 0
x
1.5
−0.25
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
122
Chapter 2
(b)
Limits and Their Properties
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.0823
0.0333
0.0167
0.0017
≈ 0
≈ 0
0.25
lim
x → 0+
−1.5
x − sin x
= 0
x2
1.5
−0.25
(c)
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.1646
0.1663
0.1666
0.1667
0.1667
0.1667
0.25
lim
x → 0+
− 1.5
x − sin x
= 0.1667 (1 6)
x3
1.5
− 0.25
(d)
x
1
0.5
0.2
0.1
0.01
0.001
0.0001
f (x)
0.1585
0.3292
0.8317
1.6658
16.67
166.7
1667.0
1.5
lim
x → 0+
−1.5
x − sin x
x − sin x
= ∞ or n > 3, lim
= ∞.
x4
xn
x → 0+
1.5
−1.5
66. lim P = ∞
Total distance
Total time
2d
50 =
(d x) + (d y)
68. (a) Average speed =
V → 0+
As the volume of the gas decreases, the pressure
increases.
67. (a) r =
(b) r =
(c)
lim
x → 25−
2(7)
625 − 49
2(15)
625 − 225
2x
625 − x
2
=
7
ft sec
12
=
50 =
2 xy
y + x
50 y + 50 x = 2 xy
3
ft sec
2
50 x = 2 xy − 50 y
50 x = 2 y ( x − 25)
= ∞
25 x
= y
x − 25
Domain: x > 25
(b)
(c)
x
30
40
50
60
y
150
66.667
50
42.857
lim
x → 25+
25 x
= ∞
x − 25
As x gets close to 25 mi/h, y becomes larger and
larger.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 2.5
Infinite Limits
123
1
1
1
1
2
bh − r 2θ = (10)(10 tan θ ) − (10) θ = 50 tan θ − 50 θ
2
2
2
2
69. (a) A =
⎛ π⎞
Domain: ⎜ 0, ⎟
⎝ 2⎠
(b)
θ
0.3
0.6
0.9
1.2
1.5
f (θ )
0.47
4.21
18.0
68.6
630.1
100
0
1.5
0
(c)
lim A = ∞
θ → π 2−
70. (a) Because the circumference of the motor is half that of the saw arbor, the saw makes 1700 2 = 850 revolutions per minute.
(b) The direction of rotation is reversed.
⎛ ⎛π
⎞⎞
(c) 2( 20 cot φ ) + 2(10 cot φ ): straight sections. The angle subtended in each circle is 2π − ⎜ 2⎜ − φ ⎟ ⎟ = π + 2φ .
2
⎠⎠
⎝ ⎝
So, the length of the belt around the pulleys is 20(π + 2φ ) + 10(π + 2φ ) = 30(π + 2φ ).
Total length = 60 cot φ + 30(π + 2φ )
⎛ π⎞
Domain: ⎜ 0, ⎟
⎝ 2⎠
(d)
(e)
φ
0.3
0.6
0.9
1.2
1.5
L
306.2
217.9
195.9
189.6
188.5
450
␲
2
0
0
(f )
lim L = 60π ≈ 188.5
φ → (π 2)−
(All the belts are around pulleys.)
(g)
lim L = ∞
φ → 0+
71. False. For instance, let
f ( x) =
x −1
or
x −1
g ( x) =
x
.
x2 + 1
72. True
2
73. False. The graphs of y = tan x, y = cot x, y = sec x
and y = csc x have vertical asymptotes.
74. False. Let
⎧1
⎪ , x ≠ 0
f ( x) = ⎨ x
⎪3, x = 0.
⎩
The graph of f has a vertical asymptote at x = 0, but
f (0) = 3.
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124
Chapter 2
Limits and Their Properties
1
1
and g ( x ) = 4 , and c = 0.
2
x
x
75. Let f ( x ) =
⎛ x2 − 1⎞
1⎞
1
1
⎛1
and
but
lim
lim
−
=
lim
,
=
∞
=
∞
⎜
⎟ = −∞ ≠ 0.
⎜
⎟
4
x → 0⎝ x 2
x→0
x → 0 x2
x → 0 x4
x4 ⎠
⎝ x ⎠
lim
76. Given lim f ( x) = ∞ and lim g ( x) = L :
x→c
x→c
(1) Difference:
Let h( x) = − g ( x). Then lim h( x ) = − L, and lim ⎡⎣ f ( x) − g ( x)⎤⎦ = lim ⎡⎣ f ( x) + h( x)⎤⎦ = ∞, by the Sum Property.
x→c
x→c
x→c
(2) Product:
If L > 0, then for ε = L 2 > 0 there exists δ1 > 0 such that g ( x) − L < L 2 whenever 0 < x − c < δ1.
So, L 2 < g ( x) < 3L 2. Because lim f ( x) = ∞ then for M > 0, there exists δ 2 > 0 such that
x→c
f ( x) > M ( 2 L) whenever x − c < δ 2 . Let δ be the smaller of δ1 and δ 2 . Then for 0 < x − c < δ ,
you have f ( x) g ( x) > M ( 2 L)( L 2) = M . Therefore lim f ( x) g ( x) = ∞. The proof is similar for L < 0.
x→c
(3) Quotient: Let ε > 0 be given.
There exists δ1 > 0 such that f ( x ) > 3L 2ε whenever 0 < x − c < δ1 and there exists δ 2 > 0 such that
g ( x) − L < L 2 whenever 0 < x − c < δ 2 . This inequality gives us L 2 < g ( x) < 3L 2. Let δ be the
smaller of δ1 and δ 2 . Then for 0 < x − c < δ , you have
g ( x)
f ( x)
<
3L 2
= ε.
3L 2ε
Therefore, lim
x→c
g ( x)
= 0.
f ( x)
77. Given lim f ( x) = ∞, let g ( x) = 1. Then
x→c
lim
x→c
g ( x)
f ( x)
78. Given lim
x →c
Then, lim
x→c
= 0 by Theorem 1.15.
1
= 0. Suppose lim f ( x ) exists and equals L.
x→c
f ( x)
lim 1
1
1
= x→c
=
= 0.
lim f ( x)
f ( x)
L
x→c
This is not possible. So, lim f ( x ) does not exist.
x→c
1
is defined for all x > 3.
x −3
Let M > 0 be given. You need δ > 0 such that
1
f ( x) =
> M whenever 3 < x < 3 + δ .
x −3
79. f ( x) =
Equivalently, x − 3 <
1
whenever
M
x − 3 < δ , x > 3.
1
. Then for x > 3 and
M
1
1
x − 3 < δ,
>
= M and so f ( x ) > M .
8
x −3
So take δ =
1
1
is defined for all x < 5. Let N < 0 be given. You need δ > 0 such that f ( x ) =
< N whenever
x −5
x −5
1
1
1
whenever x − 5 < δ , x < 5. Equivalently,
< − whenever
5 − δ < x < 5. Equivalently, x − 5 >
N
x −5
N
80. f ( x ) =
1
. Note that δ > 0 because N < 0. For x − 5 < δ and
N
1
1
1
1
x < 5,
>
= − N , and
= −
< N.
x −5
δ
x −5
x −5
x − 5 < δ , x < 5. So take δ = −
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 2
125
Review Exercises for Chapter 2
1. Calculus required. Using a graphing utility, you can estimate the length to be 8.3. Or, the length is slightly longer
than the distance between the two points, approximately 8.25.
11
−9
9
−1
2. Precalculus. L =
(9
− 1) + (3 − 1)
2
2
≈ 8.25
x −3
x 2 − 7 x + 12
3. f ( x) =
x
2.9
2.99
2.999
3
3.001
3.01
3.1
f (x)
–0.9091
–0.9901
–0.9990
?
–1.0010
–1.0101
–1.1111
lim f ( x ) ≈ −1.0000 (Actual limit is −1.)
x →3
6
−6
12
−6
x + 4 − 2
x
4. f ( x ) =
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
0.2516
0.2502
0.2500
?
0.2500
0.2498
0.2485
lim f ( x ) ≈ 0.2500
x→0
(Actual limit is 14 .)
0.5
−5
5
0
5. h( x) =
x( 4 − x )
4x − x2
=
= 4 − x, x ≠ 0
x
x
(a) lim h( x) = 4 − 0 = 4
x→0
(b) lim h( x) = 4 − ( −1) = 5
x → −1
6. f (t ) =
ln (t + 2)
t
(a) lim f (t ) does not exist because lim f (t ) = − ∞
t →0−
t →0
and lim f (t ) = ∞.
t →0+
(b) lim f (t ) =
t → −1
ln 1
= 0
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
126
Chapter 2
Limits and Their Properties
7. lim( x + 4) = 1 + 4 = 5
x →1
Let ε > 0 be given. Choose δ = ε . Then for 0 < x − 1 < δ = ε , you have
x −1 < ε
(x
+ 4) − 5 < ε
f ( x) − L < ε .
x =
8. lim
x →9
9 = 3
Let ε > 0 be given. You need
x −3 < ε ⇒
x −3 < ε
x +3
x +3 ⇒ x −9 < ε
x + 3.
Assuming 4 < x < 16, you can choose δ = 5ε .
So, for 0 < x − 9 < δ = 5ε , you have
x − 9 < 5ε <
x +3ε
x −3 < ε
f ( x) − L < ε .
9. lim (1 − x 2 ) = 1 − 22 = −3
x→2
Let ε > 0 be given. You need
1 − x 2 − ( −3) < ε ⇒ x 2 − 4 = x − 2 x + 2 < ε ⇒ x − 2 <
Assuming 1 < x < 3, you can choose δ =
So, for 0 < x − 2 < δ =
x − 2 <
ε
5
<
ε
5
ε
5
1
ε
x + 2
.
, you have
ε
x + 2
x − 2 x + 2 < ε
x2 − 4 < ε
4 − x2 < ε
(1 − x 2 ) − (−3)
< ε
f ( x) − L < ε .
10. lim 9 = 9. Let ε > 0 be given. δ can be any positive
x →5
number. So, for 0 < x − 5 < δ , you have
9−9 < ε
f ( x) − L < ε .
11. lim x 2 = (− 6) 2 = 36
13. lim ( x − 2) = (6 − 2) = 16
2
14. lim
x −3 =
3
x → −5
3
(− 5) − 3 =
3
−8 = − 2
15. lim
4
4
4
=
=
4 −1
3
−1
16. lim
x
2
2
2
= 2
=
=
2 +1
4+1
5
+1
x→4 x
x → −6
12. lim (5 x − 3) = 5(0) − 3 = − 3
2
x→6
x → 2 x2
x→0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 2
t + 2
1
1
= lim
= −
2
t
→
−
− 4
t − 2
4
17. lim
19. lim
t → −2 t 2
x→4
x −3 −1
= lim
x→4
x − 4
t 2 − 16
(t − 4)(t + 4)
= lim
t →4 t − 4
t →4
t − 4
= lim(t + 4) = 4 + 4 = 8
= lim
18. lim
x→4
4+ x − 2
⋅
x
4+ x + 2
= lim
x→0
4+ x + 2
x −3 +1
x −3 +1
− 3) − 1
(
)
x −3 +1
1
1
=
2
x −3 +1
x→4
4+ x − 2
= lim
x→0
x
x→0
(x
( x − 4)
= lim
t →4
20. lim
x −3 −1
⋅
x − 4
127
1
1
=
4
4+ x + 2
⎡1 ( x + 1)⎤⎦ − 1
1 − ( x + 1)
−1
21. lim ⎣
= lim
= lim
= −1
x→0
x → 0 x( x + 1)
x→0 x + 1
x
22. lim
(1
s →0
)
1+ s −1
s
(
⎡1
= lim ⎢
s → 0⎢
⎣
)
1+ s −1
s
⋅
(1
(1
)
)
1 + s + 1⎤
⎥
1 + s + 1⎥
⎦
⎡1 (1 + s )⎤⎦ − 1
−1
1
= lim ⎣
= lim
= −
s →0 ⎡
s
→
0
2
s 1 1 + s + 1⎤
(1 + s)⎡⎣ 1 1 + s + 1⎤⎦
⎣
⎦
(
23. lim
x→0
24.
)
(
⎛ x ⎞⎛ 1 − cos x ⎞
1 − cos x
= lim ⎜
⎟⎜
⎟ = (1)(0) = 0
x → 0 sin x ⎝
sin x
x
⎠
⎝
⎠
25. lim e x −1 sin
x →1
4(π 4)
4x
=
= π
x → (π 4) tan x
1
26. lim
lim
27. lim
∆x → 0
x→2
1
∆x → 0 2
∆x → 0
πx
2
ln ( x − 1)
= e0 sin
2
ln ( x − 1)
= lim
x→2
π
2
=1
2 ln ( x − 1)
ln ( x − 1)
= lim 2 = 2
x→2
sin ⎡⎣(π 6) + ∆x⎤⎦ − (1 2)
sin (π 6)cos ∆x + cos(π 6)sin ∆x − (1 2)
= lim
∆
x
→
0
∆x
∆x
= lim
28. lim
)
cos(π + ∆x) + 1
∆x
= lim
∆x → 0
⋅
(cos ∆x
∆x
− 1)
+ lim
∆x → 0
3 sin ∆x
3
⋅
= 0+
(1) =
2
∆x
2
3
2
cos π cos ∆x − sin π sin ∆x + 1
∆x
⎡ (cos ∆x − 1) ⎤
sin ∆x ⎤
⎡
= lim ⎢−
sin π
⎥ − ∆lim
∆x → 0
x → 0⎢
∆x
∆x ⎥⎦
⎣
⎣
⎦
= −0 − (0)(1) = 0
29. lim ⎡⎣ f ( x) g ( x )⎤⎦ = ⎡⎢ lim f ( x )⎤⎡
g ( x)⎤⎥
⎥⎢xlim
x→c
→c
⎣x → c
⎦⎣
⎦
= (− 6)
( 12 ) = − 5
( 12 ) = − 3
lim f ( x)
−6
f ( x)
30. lim
= x→c
=
= −12
1
x → c g ( x)
lim g ( x)
x→c
31. lim ⎡⎣ f ( x) + 2 g ( x)⎤⎦ = lim f ( x) + 2 lim g ( x)
x→c
x→c
x→c
( 2)
= −6 + 2
2
32. lim ⎡⎣ f ( x)⎤⎦ = ⎡⎢ lim f ( x )⎤⎥
x→c
⎣x → c
⎦
2
= ( − 6) = 36
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
128
Chapter 2
33. f ( x) =
Limits and Their Properties
2x + 9 − 3
x
1
The limit appears to be
−1
1
.
3
1
0
x
–0.01
–0.001
0
0.001
0.01
f (x)
0.3335
0.3333
?
0.3333
0.331
lim f ( x) ≈ 0.3333
x→0
2x + 9 − 3
⋅
x
lim
x→0
2x + 9 + 3
(2 x + 9) − 9
= lim
= lim
x → 0 x ⎡ 2 x + 9 + 3⎤
x→0
2x + 9 + 3
⎣
⎦
2
=
2x + 9 + 3
2
1
=
3
9 + 3
⎡1 ( x + 4)⎤⎦ − (1 4)
34. f ( x) = ⎣
x
3
The limit appears to be −
−8
1
1
16
−3
x
–0.01
–0.001
0
0.001
0.01
f (x)
–0.0627
–0.0625
?
–0.0625
–0.0623
lim f ( x) ≈ − 0.0625 = −
x→0
1
16
1
1
−
−1
1
+
4
4 = lim 4 − ( x + 4) = lim
x
= −
lim
x→0
x → 0 ( x + 4)4( x )
x → 0 ( x + 4)4
x
16
35. f ( x ) = lim
20(e x 2 − 1)
x→0
x −1
3
The limit appears to be 0.
−3
3
−3
x
–0.1
–0.01
–0.001
0.001
0.01
0.1
f (x)
0.8867
0.0988
0.0100
−0.0100
−0.1013
−1.1394
lim f ( x) ≈ 0.0000
x→0
lim
x→0
20(e x 2 − 1)
x −1
=
20(e0 − 1)
0 −1
=
0
= 0
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 2
129
ln ( x + 1)
36. f ( x ) =
x +1
1
The limit appears to be 0.
−1
2
−1
x
–0.1
–0.01
–0.001
0
0.001
0.01
0.1
f (x)
−0.1171
−0.0102
−0.0010
?
0.0010
0.0099
0.0866
lim f ( x) ≈ 0.0000
x→0
ln ( x + 1)
lim
x +1
x→0
37. v = lim
s ( 4) − s ( t )
4 −t
t →4
= lim
t →4
= lim
t →4
= lim
ln 1
0
=
= 0
1
1
=
2
⎣⎡−4.9(16) + 250⎤⎦ − ⎡⎣−4.9t + 250⎤⎦
4 −t
4.9(t 2 − 16)
4 −t
4.9(t − 4)(t + 4)
4−t
t →4
= lim ⎡−
4.9(t + 4)⎤⎦ = −39.2 m/sec
t →4 ⎣
The object is falling at about 39.2 m/sec.
38. −4.9t 2 + 250 = 0 ⇒ t =
When a =
lim
50
, the velocity is
7
s ( a ) − s (t )
t →a
50
sec
7
a −t
⎡−4.9a 2 + 250⎤⎦ − ⎡⎣−4.9t 2 + 250⎤⎦
= lim ⎣
t →a
a −t
2
2
4.9(t − a )
= lim
t →a
a −t
4.9(t − a )(t + a )
= lim
t →a
a −t
= lim ⎡−
4.9
t + a )⎤⎦
(
⎣
t →a
= −4.9( 2a )
50 ⎞
⎛
⎜a =
⎟
7⎠
⎝
= −70 m/sec.
The velocity of the object when it hits the ground is about 70 m/sec.
39. lim
x → 3+
1
1
1
=
=
3+ 3
6
x + 3
40. lim
x → 6−
x −6
x −6
= lim
x 2 − 36
x → 6− ( x − 6)( x + 6)
= lim
x → 6−
=
1
x + 6
1
12
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
130
Chapter 2
41. lim
x → 4−
Limits and Their Properties
x − 2
x − 2
x + 2
= lim
⋅
x − 4
x → 4− x − 4
x + 2
x − 4
= lim
−
x → 4 ( x − 4)
x + 2
(
4
has a nonremovable discontinuity at
x −5
x = 5 because lim f ( x) does not exist.
51. f ( x ) =
x →5
)
52. f ( x ) =
1
= lim
x + 2
x → 4−
has nonremovable discontinuities at x = ± 3
1
=
4
x −3
42. lim
x → 3−
x −3
because lim f ( x ) and lim f ( x) do not exist.
x →3
−( x − 3)
= lim
x → 3−
x −3
53. f ( x) =
= −1
x → 2−
x
x
1
=
=
,x ≠ 0
x3 − x
x( x 2 − 1)
( x − 1)( x + 1)
x → −1
x→4
at x = 4.
45. lim f ( x) = 0
x→2
x +3
x − 3 x − 18
x +3
=
( x + 3)( x − 6)
54. f ( x ) =
46. lim g ( x) = 1 + 1 = 2
x →1+
47. lim h(t ) does not exist because lim h(t ) = 1 + 1 = 2
t →1−
t →1
1
2
(1 + 1)
x →1
and has a removable discontinuity at x = 0 because
1
= −1.
lim f ( x) = lim
x→0
x → 0 ( x − 1)( x + 1)
44. lim x − 1 does not exist. There is a break in the graph
t →1+
x → −3
has nonremovable discontinuities at x = ±1
because lim f ( x) and lim f ( x) do not exist,
43. lim ( 2 x + 1) = 2(1) + 1 = 3
and lim h(t ) =
1
1
=
x2 − 9
( x − 3)( x + 3)
=
= 1.
2
1
, x ≠ −3
x −6
has a nonremovable discontinuity at x = 6
because lim f ( x) does not exist, and has a
48. lim f ( s ) = 2
x→6
s → −2
removable discontinuity at x = − 3 because
49. f ( x ) = x 2 − 4 is continuous for all real x.
lim f ( x ) = lim
x → −3
50. f ( x) = x 2 − x + 20 is continuous for all real x.
x → −3
1
1
= − .
x −6
9
55. f ( 2) = 5
Find c so that lim (cx + 6) = 5.
x → 2+
c ( 2) + 6 = 5
2c = −1
c = −
1
2
56. lim ( x + 1) = 2
x →1+
lim ( x + 1) = 4
x → 3−
Find b and c so that lim ( x 2 + bx + c ) = 2 and lim ( x 2 + bx + c ) = 4.
x →1−
Consequently you get
Solving simultaneously,
57. f ( x ) = −3 x 2 + 7
Continuous on ( −∞, ∞)
x → 3+
1+ b + c = 2
b
and 9 + 3b + c = 4.
= −3 and
c = 4.
(4 x − 1)( x + 2)
4 x2 + 7 x − 2
=
x + 2
x + 2
Continuous on (− ∞, − 2) ∪ (− 2, ∞). There is a
58. f ( x ) =
removable discontinuity at x = − 2.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 2
59. f ( x) =
x − 4
(a)
60. f ( x) = x + 3
lim x + 3 = k + 3 where k is an integer.
x → k+
lim x + 3 = k + 2 where k is an integer.
(b)
Nonremovable discontinuity at each integer k
Continuous on ( k , k + 1) for all integers k
61. g ( x) = 2e
is continuous on all intervals ( n, n + 1),
where n is an integer. g has nonremovable discontinuities
at each n.
62. h( x) = −2 ln 5 − x
Because 5 − x > 0 except for x = 5, h is continuous
on ( −∞, 5) ∪ (5, ∞).
lim f ( x) = 4
x → 2+
x→2
68. f ( x ) =
(x
− 1) x
(a) Domain: ( −∞, 0] ∪ [1, ∞)
(b) lim f ( x) = 0
x → 0−
(c) lim f ( x) = 0
x →1+
69. f ( x) =
3
x
3
= −∞
x
3
lim = ∞
x → 0+ x
lim
x → 0−
(3x + 2)( x − 1)
3x 2 − x − 2
=
x −1
x −1
lim f ( x) = lim (3 x + 2) = 5
63. f ( x ) =
x →1
lim f ( x ) = −4
x → 2−
(c) lim f ( x) does not exist.
x → k−
x 4
⎡ x − 2⎤
x2 − 4
= ( x + 2) ⎢
⎥
x − 2
⎢⎣ x − 2 ⎦⎥
67. f ( x) =
Continuous on [4, ∞)
131
Therefore, x = 0 is a vertical asymptote.
x →1
Removable discontinuity at x = 1
Continuous on ( −∞, 1) ∪ (1, ∞)
⎧5 − x, x ≤ 2
64. f ( x) = ⎨
⎩2 x − 3, x > 2
lim (5 − x) = 3
x → 2−
lim ( 2 x − 3) = 1
Nonremovable discontinuity at x = 2
Continuous on ( −∞, 2) ∪ ( 2, ∞)
65. f is continuous on [1, 2]. f (1) = −1 < 0 and
f ( 2) = 13 > 0. Therefore by the Intermediate Value
Theorem, there is at least one value c in (1, 2) such that
2c 3 − 3 = 0.
Therefore, x = 2 is a vertical asymptote.
x3
x3
=
x −9
( x + 3)( x − 3)
2
x3
x3
= − ∞ and lim 2
= ∞
x → −3− x − 9
x → −3+ x − 9
lim
2
Therefore, x = − 3 is a vertical asymptote.
lim
x → −3−
x3
x3
= − ∞ and lim 2
= ∞
+
x −9
x→3 x − 9
2
Therefore, x = 3 is a vertical asymptote.
72. f ( x) =
2t
lim
Nonremovable discontinuity every 6 months
5
( x − 2) 4
5
5
= ∞ = lim
4
− 2) 4
x → 2+ ( x − 2)
lim
x → 2− ( x
71. f ( x ) =
x → 2+
66. A = 5000(1.06)
70. f ( x ) =
x → −6−
6x
6x
= −
36 − x 2
( x + 6)( x − 6)
6x
6x
= ∞ and lim
= −∞
2
36 − x 2
x → −6+ 36 − x
Therefore, x = − 6 is a vertical asymptote.
9000
lim
x → 6−
6x
6x
= ∞ and lim
= −∞
2
36 − x 2
x → 6+ 36 − x
Therefore, x = 6 is a vertical asymptote.
0
4000
5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
132
Chapter 2
2x + 1
2x + 1
=
2
x − 64
( x + 8)( x − 8)
73. g ( x) =
lim
x → −8−
Limits and Their Properties
79.
2x + 1
2x + 1
= − ∞ and lim 2
= ∞
x 2 − 64
x → −8+ x − 64
80.
Therefore, x = − 8 is a vertical asymptote.
lim
x → 8−
lim
x +1
1
1
= lim 2
=
3
+
x +1
3
x → −1 x − x + 1
lim
x +1
1
1
= lim 2
= −
−
x4 − 1
4
x → −1 ( x + 1)( x − 1)
x → −1+
x → −1−
1⎞
⎛
81. lim ⎜ x − 3 ⎟ = −∞
x ⎠
x → 0+ ⎝
2x + 1
2x + 1
= − ∞ and lim 2
= ∞
+
x 2 − 64
x
− 64
x →8
Therefore, x = 8 is a vertical asymptote.
1
82. lim
x → 2− 3
1
sin π x
sin π x = 0 for x = n, where n is an integer.
74. f ( x) = csc π x =
x − 4
2
= −∞
83. lim
sin 4 x
⎡ 4 ⎛ sin 4 x ⎞⎤
4
= lim ⎢ ⎜
⎟⎥ =
5x
5
x → 0+ ⎣ 5 ⎝ 4 x ⎠⎦
84. lim
sec x
= ∞
x
85. lim
csc 2 x
1
= lim
= ∞
x
x → 0+ x sin 2 x
x → 0+
lim f ( x) = ∞ or −∞
x→n
Therefore, the graph has vertical asymptotes at x = n.
x → 0+
75. g ( x) = ln ( 25 − x 2 ) = ln ⎡⎣(5 + x)(5 − x)⎤⎦
lim ln ( 25 − x 2 ) = 0
x → 0+
x →5
lim ln ( 25 − x 2 ) = 0
cos 2 x
= −∞
x
x → 0−
x → −5
86. lim
Therefore, the graph has holes at x = ± 5. The graph
does not have any vertical asymptotes.
87. lim ln (sin x) = −∞
x → 0+
76. f ( x ) = 7e − 3 x
lim 7e − 3 x = ∞
88. lim 12e − 2
x → 0−
x → 0−
Therefore, x = 0 is a vertical asymptote.
89. C =
x2 + 2 x + 1
77. lim
= −∞
x −1
x →1−
78.
lim
x → (1 2)+
(c) C (90) = $720,000
–0.1
–0.01
–0.001
0.001
0.01
0.1
f(x)
2.0271
2.0003
2.0000
2.0000
2.0003
2.0271
lim
lim
80,000 p
= ∞
− p
p →100− 100
tan 2x
x
x
x→0
80,000 p
, 0 ≤ p < 100
100 − p
(b) C (50) = $80.000
(d)
(a)
= ∞
(a) C (15) ≈ $14,117.65
x
= ∞
2x − 1
90. f ( x ) =
x
tan 2 x
= 2
x
⎧ tan 2 x
, x ≠ 0
⎪
(b) Yes, define f ( x ) = ⎨ x
.
⎪2,
x
0
=
⎩
Now f ( x) is continuous at x = 0.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 2
133
Problem Solving for Chapter 2
x 2 + ( y − 1) +
2
1. (a) Perimeter ∆PAO =
x 2 + ( x 2 − 1) +
2
=
x2 + x4 + 1
Perimeter ∆PBO =
(x
− 1) + y 2 +
x2 + y2 + 1
=
(x
− 1) + x 4 +
x2 + x4 + 1
2
2
x 2 + ( x 2 − 1) +
x2 + x4 + 1
(x
x2 + x4 + 1
2
(b) r ( x ) =
(c)
x2 + y2 + 1
− 1) + x 4 +
2
x
4
2
1
0.1
0.01
Perimeter ∆PAO
33.02
9.08
3.41
2.10
2.01
Perimeter ∆PBO
33.77
9.60
3.41
2.00
2.00
r ( x)
0.98
0.95
1
1.05
1.005
lim r ( x) =
x → 0+
1+ 0+1
2
=
=1
1+ 0 +1
2
x
1
1
bh = (1)( x) =
2
2
2
y
x2
1
1
=
Area ∆PBO = bh = (1)( y ) =
2
2
2
2
2. (a) Area ∆PAO =
(b) a( x) =
(c)
Area ∆PBO
x2 2
=
= x
Area ∆PAO
x 2
x
4
2
1
0.1
0.01
Area ∆PAO
2
1
12
1 20
1 200
Area ∆PBO
8
2
12
1 200
1 20,000
a( x)
4
2
1
1 10
1 100
lim a( x ) = lim x = 0
x → 0+
x → 0+
3. (a) There are 6 triangles, each with a central angle of 60° = π 3. So,
π⎤
3 3
⎡1 ⎤
⎡1
Area hexagon = 6⎢ bh⎥ = 6⎢ (1) sin ⎥ =
≈ 2.598.
3⎦
2
⎣2 ⎦
⎣2
h = sin θ
h = sin 60°
1
1
60°
θ
Error = Area (Circle) − Area (Hexagon) = π −
3 3
≈ 0.5435
2
(b) There are n triangles, each with central angle of θ = 2π n. So,
n sin ( 2π n)
2π ⎤
⎡1 ⎤
⎡1
An = n ⎢ bh⎥ = n ⎢ (1) sin ⎥ =
.
n⎦
2
⎣2 ⎦
⎣2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
134
Chapter 2
(c)
Limits and Their Properties
n
6
12
24
48
96
An
2.598
3
3.106
3.133
3.139
(d) As n gets larger and larger, 2π n approaches 0. Letting x = 2π n, An =
sin ( 2π n)
sin ( 2π n)
=
2n
(2π n)
π =
sin x
π
x
which approaches (1)π = π .
4. (a) Slope =
4− 0
4
=
3−0
3
(b) Slope = −
5. (a) Slope = −
3
3
Tangent line: y − 4 = − ( x − 3)
4
4
3
25
y = − x +
4
4
(
(c) Let Q = ( x, y ) = x,
25 − x 2
(b) Slope of tangent line is
5
( x − 5)
12
5
169
y =
x −
Tangent line
12
12
)
(
(c) Q = ( x, y ) = x, − 169 − x 2
25 − x 2 − 4
⋅
x −3
(d) lim mx = lim
x →3
x →3
= lim
x →3
= lim
x →3
mx =
25 − x 2 + 4
25 − x 2 + 4
25 − x − 16
(
− 3)
25 − x 2 + 4
(3 − x)(3 + x)
( x − 3)( 25 − x 2 +
= lim
x →3
x →5
)
12 −
x→5
169 − x 2 12 +
⋅
x −5
12 +
144 − (169 − x 2 )
= lim
x→5
)
(x
(
− 5) 12 +
4
−6
3
=
= −
4 + 4
4
25 − x 2 + 4
x→5
= lim
This is the slope of the tangent line at P.
169 − x 2
x − 25
169 − x 2
169 − x 2
)
2
= lim
−( 3 + x )
)
− 169 − x 2 + 12
x −5
(d) lim mx = lim
2
(x
5
.
12
y + 12 =
25 − x 2 − 4
x −3
mx =
12
5
(x
(
− 5) 12 +
(x
12 +
x→5
169 − x 2
+ 5)
169 − x
2
=
)
10
5
=
12 + 12
12
This is the same slope as part (b).
6.
a + bx −
x
3
a + bx −
x
=
3
⋅
a + bx +
a + bx +
3
=
3
x
(
(a + bx) − 3
)
a + bx +
3
b
3 + bx +
3
Letting a = 3 simplifies the numerator.
So, lim
x→0
Setting
3 + bx −
x
b
3 +
3
3
= lim
x→0 x
=
(
bx
3 + bx +
3
)
= lim
x→0
3, you obtain b = 6. So, a = 3 and b = 6.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 2
3 + x1 3 − 2
⋅
x −1
(d) lim f ( x) = lim
7. (a) 3 + x1 3 ≥ 0
x →1
x1 3 ≥ −3
x ≥ −27
x →1
= lim
x →1
Domain: x ≥ −27, x ≠ 1 or [−27, 1) ∪ (1, ∞)
(b)
= lim
0.5
x →1
= lim
− 30
x →1
12
(
− 1)
(c)
13
lim f ( x ) =
− 2
−27 − 1
≈ 0.0714
x → −27 +
=
−2
1
=
−28
14
8. lim f ( x) = lim ( a 2 − 2) = a 2 − 2
x → 0−
x → 0−
( x1 3 − 1)( x 2 3 + x1 3 + 1)(
(x
+ x
13
a2 − 2 = a
a = −1, 2
1
1
2
3
4
−2
−3
−4
(a)
f (1) = 1 + −1 = 1 + ( −1) = 0
f ( 0) = 0
x→2
f
(b) f continuous at 2: g1
( 12 ) = 0 + (−1) = −1
f ( −2.7) = −3 + 2 = −1
lim f ( x ) = 3: g1 , g3 , g 4
x → 2−
x
−4 −3 −2 −1
9. (a) lim f ( x) = 3: g1 , g 4
(b)
lim f ( x) = −1
x →1−
lim f ( x) = −1
y
x →1+
3
lim f ( x) = −1
2
x →1 2
1
x
−1
1
12
2
(a − 2)(a + 1) = 0
10.
=
)
3
a2 − a − 2 = 0
(c)
+ 2)
3 + x1 3 + 2
y
ax
tan x
⎛
⎞
= a⎜ because lim
= 1⎟
x→0 x
x → 0+ tan x
⎝
⎠
Thus,
(
+ 1)
4
lim f ( x) = lim
x → 0+
3 + x1 3 + 2
1
23
(1 + 1 + 1)(2
11.
)
x1 3 − 1
1
=
3 + x1 3 + 2
3 + x1 3 + 2
− 0.1
3 + ( −27)
3 + x1 3 + 2
3 + x1 3 − 4
(x
135
1
−1
(c) f is continuous for all real numbers except
x = 0, ±1, ± 2, ± 3, …
−2
(a) f
( 14 ) =
f (3) =
4 = 4
1
3
= 0
f (1) = 1 = 1
(b)
lim f ( x) = 1
x →1−
lim f ( x) = 0
x →1+
lim f ( x) = −∞
x → 0−
lim f ( x) = ∞
x → 0+
(c) f is continuous for all real numbers except
x = 0, ±1, ± 12 , ± 13 , …
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
)
136
Chapter 2
Limits and Their Properties
v2 =
12. (a)
192,000
+ v0 2 − 48
r
2
192,000
= v 2 − v0 2 + 48
r
192,000
r = 2
v − v0 2 + 48
1
x
a
192,000
lim r =
v→0
48 − v0 2
Let v0 =
(b)
v2 =
x → a+
(ii) lim Pa , b ( x ) = 0
x → a−
1920
+ v0 2 − 2.17
r
(iii) lim Pa , b ( x) = 0
x → b+
(iv) lim Pa , b ( x) = 1
x → b−
(c) Pa , b is continuous for all positive real numbers
except x = a, b.
1920
lim r =
− v0 2
2.17
v→0
2.17 mi sec
(≈
1.47 mi/sec).
10,600
r = 2
v − v0 2 + 6.99
(c)
lim r =
v→0
Let v0 =
b
(b) (i) lim Pa , b ( x) = 1
48 = 4 3 mi sec.
1920
= v 2 − v0 2 + 2.17
r
1920
r = 2
v − v0 2 + 2.17
Let v0 =
y
13. (a)
10,600
6.99 − v0 2
(d) The area under the graph of U, and above the x-axis,
is 1.
14. Let a ≠ 0 and let ε > 0 be given. There exists
δ1 > 0 such that if 0 < x − 0 < δ1 then
f ( x) − L < ε . Let δ = δ1 a . Then for
0 < x − 0 < δ = δ1 a , you have
x <
6.99 ≈ 2.64 mi sec.
Because this is smaller than the escape velocity for
Earth, the mass is less.
δ1
a
ax < δ1
f ( ax) − L < ε .
As a counterexample, let
⎧1, x ≠ 0
.
a = 0 and f ( x ) = ⎨
⎩2, x = 0
Then lim f ( x ) = 1 = L, but
x→0
lim f ( ax) = lim f (0) = lim 2 = 2.
x→0
x→0
x→0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 3
Differentiation
Section 3.1
The Derivative and the Tangent Line Problem.................................138
Section 3.2
Basic Differentiation Rules and Rates of Change.............................154
Section 3.3
Product and Quotient Rules and Higher-Order Derivatives.............167
Section 3.4
The Chain Rule...................................................................................182
Section 3.5
Implicit Differentiation.......................................................................199
Section 3.6
Derivatives of Inverse Functions .......................................................214
Section 3.7
Related Rates ......................................................................................225
Section 3.8
Newton’s Method ...............................................................................235
Review Exercises ........................................................................................................245
Problem Solving .........................................................................................................260
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R
Differentiation
3
Section 3.1 The Derivative and the Tangent Line Problem
1. At ( x1 , y1 ), slope = 0.
2
2
∆x
∆x → 0
− 6( ∆x) − ( ∆x)
= lim
2
∆x
= lim ( − 6 − ∆x ) = − 6
∆x → 0
y
∆x → 0
6
f(4) = 5
5
f (0 + ∆t ) − f (0)
(4, 5)
4
9. Slope at (0, 0) = lim
f(4) − f(1) = 3
3
3( ∆t ) − ( ∆t ) − 0
2
(1, 2)
= lim
1
∆t → 0
∆t
= lim (3 − ∆t ) = 3
x
1
2
3
f ( 4) − f (1)
4
5
(x
− 1) + f (1)
4 −1
3
= ( x − 1) + 2
3
= 1( x − 1) + 2
6
∆t → 0
10. Slope at (1, 5) = lim
h(1 + ∆t ) − h(1)
∆t
∆t → 0
= lim
(1 + ∆t )2
+ 4(1 + ∆t ) − 5
∆t
∆t → 0
= x +1
1 + 2( ∆t ) + ( ∆t ) + 4 + 4(∆t ) − 5
2
f ( 4) − f (1)
4 −1
f ( 4) − f (3)
4−3
4 −1
= lim
=
5− 2
=1
3
≈
5 − 4.75
= 0.25
1
f ( 4) − f (1)
>
f ( 4) − f (3)
4−3
6(∆t ) + (∆t ) 2
= lim
∆t → 0
∆t
= lim (6 + ∆t ) = 6
∆t → 0
.
11.
This slope is steeper than the slope of the line
f ( 4) − f (1)
through (1, 2) and ( 4, 5). So,
< f ′(1).
4 −1
5. f ( x) = 3 − 5 x is a line. Slope = −5
+ 1 is a line. Slope =
7. Slope at ( 2, − 5) = lim
= lim
∆x → 0
+ ∆x ) − 9 − ( −5)
2
∆x
4 + 4( ∆x) + ( ∆x) − 4
2
= lim
f ′( x) = lim
f ( x + ∆x ) − f ( x)
∆x
∆x → 0
7 −7
= lim
∆x → 0
∆x
= lim 0 = 0
12. g ( x) = −3
3
2
∆x
(2
f ( x) = 7
∆x → 0
g ( 2 + ∆x) − g ( 2)
∆x → 0
∆t
∆t → 0
(b) The slope of the tangent line at (1, 2) equals f ′(1).
3
x
2
∆t
∆t → 0
f(1) = 2
2
6. g ( x) =
5 − 9 − 6( ∆x) − ( ∆x) + 4
= lim
f(4) − f(1)
y=
(x − 1) + f(1) = x + 1
4−1
So,
∆x
∆x → 0
At ( x2 , y2 ), slope = − 52 .
4. (a)
5 − (3 + ∆x) − (− 4)
= lim
2. At ( x1 , y1 ), slope = 23 .
(c) y =
∆x
∆x → 0
At ( x2 , y2 ), slope = 52 .
3. (a), (b)
f (3 + ∆x) − f (3)
8. Slope at (3, − 4) = lim
g ′( x) = lim
∆x → 0
= lim
g ( x + ∆x ) − g ( x )
∆x
−3 − ( −3)
∆x
0
= lim
= 0
∆x → 0 ∆x
∆x → 0
∆x → 0
∆x
= lim ( 4 + ∆x ) = 4
∆x → 0
138
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
13.
f ( x) = −10 x
f ′( x) = lim
∆x → 0
= lim
The Derivative and the Tangent Line Problem
15. h( s ) = 3 +
f ( x + ∆x ) − f ( x)
h′( s ) = lim
∆x
−10( x + ∆x) − ( −10 x)
∆s
2
2 ⎞
⎛
3 + ( s + ∆s ) − ⎜ 3 + s ⎟
3
3 ⎠
⎝
= lim
∆s → 0
∆s
2
2
2
3 + s + ∆s − 3 − s
3
3
3
= lim
∆s → 0
∆s
2
∆s
2
= lim 3
=
∆s → 0 ∆s
3
∆x
−10 x − 10∆x + 10 x
= lim
∆x → 0
∆x
−10∆x
= lim
∆x → 0
∆x
= lim ( −10) = −10
∆x → 0
f ( x) = 7 x − 3
f ′( x) = lim
∆x → 0
= lim
f ( x + ∆x ) − f ( x )
∆x
7( x + ∆x ) − 3 − (7 x − 3)
16.
∆x
7 x + 7 ∆x − 3 − 7 x + 3
= lim
∆x → 0
∆x
7 ( ∆x )
= lim
∆x → 0 ∆x
= lim 7 = 7
∆x → 0
∆x → 0
f ( x) = 5 −
f ′( x) = lim
2
x
3
f ( x + ∆x) − f ( x)
∆x
∆x → 0
2
2
( x + ∆x) − ⎛⎜ 5 − x ⎞⎟
3
3 ⎠
⎝
lim
∆x → 0
∆x
2
2
2
5 − x − ∆x − 5 + x
3
3
3
lim
∆x → 0
∆x
2
− ( ∆x)
lim 3
∆x → 0
∆x
2
⎛ 2⎞
lim ⎜ − ⎟ = −
∆x → 0 ⎝ 3 ⎠
3
5−
=
=
=
=
17.
2
s
3
h( s + ∆s ) − h( s )
∆s → 0
∆x → 0
14.
139
f ( x) = x 2 + x − 3
f ′( x) = lim
f ( x + ∆x ) − f ( x )
∆x
∆x → 0
= lim
(x
+ ∆x) + ( x + ∆x) − 3 − ( x 2 + x − 3)
2
∆x
∆x → 0
= lim
x + 2 x( ∆x) + ( ∆x) + x + ∆x − 3 − x 2 − x + 3
2
2
∆x
∆x → 0
2 x( ∆x) + ( ∆x) + ∆x
2
= lim
∆x
= lim ( 2 x + ∆x + 1) = 2 x + 1
∆x → 0
∆x → 0
18.
f ( x) = x 2 − 5
f ′( x) = lim
f ( x + ∆x) − f ( x)
∆x
∆x → 0
= lim
(x
2
∆x
∆x → 0
= lim
+ ∆x) − 5 − ( x 2 − 5)
x + 2 x( ∆x) + ( ∆x) − 5 − x 2 + 5
2
2
∆x
∆x → 0
= lim
2 x( ∆x) + ( ∆x)
2
∆x
= lim ( 2 x + ∆x) = 2 x
∆x → 0
∆x → 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
140
19.
Chapter 3
Differentiation
f ( x) = x3 − 12 x
f ( x + ∆x) − f ( x)
f ′( x) = lim
∆x
∆x → 0
⎡( x + ∆x)3 − 12( x + ∆x)⎤ − ⎡ x3 − 12 x⎤
⎦
⎦ ⎣
= lim ⎣
∆x → 0
∆x
x3 + 3 x 2 ∆x + 3 x( ∆x) + (∆x) − 12 x − 12 ∆x − x3 + 12 x
2
= lim
3
∆x
∆x → 0
= lim
∆x → 0
3 x ∆x + 3x( ∆x) + ( ∆x) − 12 ∆x
2
2
3
∆x
(
)
= lim 3 x + 3 x ∆x + ( ∆x) − 12 = 3 x 2 − 12
∆x → 0
20.
2
2
f ( x) = x3 + x 2
f ′( x) = lim
f ( x + ∆x) − f ( x)
∆x
∆x → 0
⎡( x + ∆x)3 + ( x + ∆x) 2 ⎤ − ⎡ x 3 + x 2 ⎤
⎦
⎦ ⎣
= lim ⎣
∆x → 0
∆x
x 3 + 3x 2 ∆x + 3 x(∆x ) + ( ∆x) + x 2 + 2 x ∆x + (∆x) − x3 − x 2
2
= lim
3
∆x
∆x → 0
= lim
∆x → 0
2
3 x ∆x + 3 x( ∆x) + (∆x) + 2 x ∆x + ( ∆x)
2
2
3
∆x
(
2
)
= lim 3x + 3x ∆x + ( ∆x) + 2 x + ( ∆x) = 3 x 2 + 2 x
∆x → 0
21.
f ( x) =
1
x −1
f ′( x) = lim
2
2
22.
f ( x + ∆x ) − f ( x )
∆x
∆x → 0
f ( x) =
1
x2
f ′( x) = lim
= lim
−∆x
= lim
∆x → 0 ∆x( x + ∆x − 1)( x − 1)
= lim
∆x → 0
= −
−1
( x + ∆x − 1)( x − 1)
= lim
− 1)
∆x → 0
∆x → 0
= lim
∆x → 0
2
(x
+ ∆x)
2
−
1
x2
∆x
∆x → 0
1
(x
∆x
1
1
1
−
+
∆
−
−1
x
x
x
1
= lim
∆x → 0
∆x
( x − 1) − ( x + ∆x − 1)
= lim
∆x → 0 ∆x( x + ∆x − 1)( x − 1)
= lim
f ( x + ∆x) − f ( x)
∆x → 0
x − ( x + ∆x )
2
2
∆x( x + ∆x) x 2
2
− 2 x ∆x − ( ∆ x )
2
∆ x ( x + ∆x ) x 2
2
−2 x − ∆x
( x + ∆x ) x 2
2
−2 x
x4
2
= − 3
x
=
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
23.
f ( x) =
f ( x + ∆x ) − f ( x)
∆x
∆x → 0
x + ∆x + 4 −
∆x
= lim
∆x → 0
(x
= lim
∆x → 0
x + 4
1
x + ∆x + 4 +
∆x → 0
⎛
⋅ ⎜⎜
⎝
x + ∆x + 4 +
x + ∆x + 4 +
x + 4⎞
⎟
x + 4 ⎟⎠
+ ∆x + 4) − ( x + 4)
∆x ⎡⎣ x + ∆x + 4 +
= lim
f ( x) =
141
x + 4
f ′( x) = lim
24.
The Derivative and the Tangent Line Problem
x + 4 ⎤⎦
x + 4
1
x + 4 +
=
x + 4
=
2
1
x + 4
4
x
f ( x + ∆x ) − f ( x )
f ′( x) = lim
∆x
4
−
x + ∆x
∆x
∆x → 0
= lim
∆x → 0
4
= lim
∆x → 0
x − 4 x + ∆x ⎛
⋅ ⎜⎜
∆x x x + ∆x
⎝
x + ∆x ⎞
⎟
x + ∆x ⎟⎠
x +
x +
4 x − 4( x + ∆x)
= lim
∆x → 0
∆x
x
= lim
∆x → 0
x
(
x + ∆x
x + ∆x
x
=
x
4
x
(
−4
x +
x
(
x +
x + ∆x
−4
x +
)
=
x
x + ∆x
)
)
−2
x
25. (a) f ( x) = x 2 + 3
f ′( x) = lim
(b)
f ( x + ∆x ) − f ( x)
∆x
∆x → 0
(− 1, 4)
⎡( x + ∆x)2 + 3⎤ − ( x 2 + 3)
⎦
= lim ⎣
∆x → 0
∆x
x 2 + 2 x∆x + ( ∆x) + 3 − x 2 − 3
2
= lim
∆x
∆x → 0
= lim
8
2 x∆x + ( ∆x)
−3
3
−1
(c) Graphing utility confirms
dy
= − 2 at ( −1, 4).
dx
2
∆x
= lim ( 2 x + ∆x) = 2 x
∆x → 0
∆x → 0
At ( −1, 4), the slope of the tangent line is
m = 2( −1) = − 2.
The equation of the tangent line is
y − 4 = − 2( x + 1)
y − 4 = − 2x − 2
y = − 2x + 2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
142
Chapter 3
Differentiation
26. (a) f ( x ) = x 2 + 2 x − 1
f ( x + ∆x ) − f ( x)
f ′( x) = lim
∆x
∆x → 0
⎡( x + ∆x)2 + 2( x + ∆x) − 1⎤ − ⎡ x 2 + 2 x − 1⎤
⎦
⎦ ⎣
= lim ⎣
∆x → 0
∆x
⎡ x 2 + 2 x∆x + ( ∆x)2 + 2 x + 2∆x − 1⎤ − ⎡ x 2 + 2 x − 1⎤
⎦
⎦ ⎣
= lim ⎣
∆x → 0
∆x
2 x∆x + ( ∆x) + 2∆x
2
= lim
∆x
= lim ( 2 x + ∆x + 2) = 2 x + 2
∆x → 0
∆x → 0
At (1, 2), the slope of the tangent line is m = 2(1) + 2 = 4.
The equation of the tangent line is
y − 2 = 4( x − 1)
y − 2 = 4x − 4
y = 4 x − 2.
(b)
8
(1, 2)
− 10
8
−4
(c) Graphing utility confirms
27. (a)
dy
= 4 at (1, 2).
dx
f ( x) = x3
f ′( x) = lim
f ( x + ∆x ) − f ( x )
∆x
∆x → 0
= lim
(x
+ ∆x) − x3
3
∆x
∆x → 0
x 3 + 3 x 2 ∆x + 3x( ∆x) + ( ∆x)
2
= lim
∆x
∆x → 0
= lim
∆x → 0
3x ∆x + 3 x( ∆x ) + ( ∆x)
2
2
(
∆x
= lim 3 x + 3 x ∆x + ( ∆x )
∆x → 0
3
2
2
3
) = 3x
2
At ( 2, 8), the slope of the tangent is m = 3( 2) = 12. The equation of the tangent line is
2
y − 8 = 12( x − 2)
y − 8 = 12 x − 24
y = 12 x − 16.
(b)
10
(2, 8)
−5
5
−4
(c) Graphing utility confirms
dy
= 12 at ( 2, 8).
dx
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
28. (a)
The Derivative and the Tangent Line Problem
143
f ( x) = x3 + 1
f ′( x) = lim
f ( x + ∆x ) − f ( x )
∆x
∆x → 0
⎡( x + ∆x)3 + 1⎤ − ( x 3 + 1)
⎦
= lim ⎣
∆x → 0
∆x
x 3 + 3 x 2 ( ∆x) + 3 x( ∆x) + ( ∆x) + 1 − x3 − 1
2
= lim
3
∆x
∆x → 0
= lim ⎡3 x 2 + 3x( ∆x) + ( ∆x) ⎤ = 3x 2
⎦
∆x → 0 ⎣
2
At ( −1, 0), the slope of the tangent line is m = 3( −1) = 3. The equation of the tangent line is
2
y − 0 = 3( x + 1)
y = 3x + 3.
(b)
6
−9
(−1, 0)
9
−6
(c) Graphing utility confirms
29. (a)
f ( x) =
dy
= 3 at ( −1, 0).
dx
x
f ′( x) = lim
f ( x + ∆x) − f ( x)
∆x
∆x → 0
= lim
∆x → 0
= lim
x + ∆x −
∆x
∆x → 0 ∆x
= lim
∆x → 0
(
(x
x
x + ∆x +
x + ∆x +
⋅
+ ∆x) − x
x + ∆x +
1
x + ∆x +
x
x
=
x
x
)
1
2
x
At (1, 1), the slope of the tangent line is m =
1
1
= .
2
2 1
The equation of the tangent line is
1
( x − 1)
2
1
1
y −1 = x −
2
2
1
1
y = x + .
2
2
y −1 =
(b)
3
(1, 1)
−1
5
−1
(c) Graphing utility confirms
dy
1
= at (1, 1).
dx
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
144
Chapter 3
30. (a)
Differentiation
f ( x) =
x −1
f ′( x) = lim
(b)
f ( x + ∆x ) − f ( x)
−2
x + ∆x − 1 −
∆x
= lim
∆x → 0
∆x → 0
(5, 2)
∆x
∆x → 0
= lim
4
(x
∆x
(
x −1
+ ∆x − 1) − ( x − 1)
x + ∆x − 1 +
1
x + ∆x − 1 +
= lim
∆x → 0
⎛
⋅ ⎜⎜
⎝
x −1
x −1
=
x − 1⎞
⎟
x − 1 ⎟⎠
x + ∆x − 1 +
x + ∆x − 1 +
)
2
At (5, 2), the slope of the tangent line is m =
10
−4
(c) Graphing utility confirms
1
x −1
dy
1
= at (5, 2).
dx
4
1
1
= .
4
2 5−1
The equation of the tangent line is
1
( x − 5)
4
1
5
y − 2 = x −
4
4
1
3
y = x + .
4
4
y − 2 =
31. (a)
f ( x) = x +
f ′( x) = lim
4
x
f ( x + ∆x ) − f ( x)
(b)
− 12
∆x
∆x → 0
12
(− 4, − 5)
4
4⎞
⎛
− ⎜x + ⎟
x + ∆x ⎝
x⎠
= lim
∆x → 0
∆x
x( x + ∆x )( x + ∆x ) + 4 x − x 2 ( x + ∆x) − 4( x + ∆x)
= lim
∆x → 0
x( ∆x)( x + ∆x)
(x
6
+ ∆x) +
x 3 + 2 x 2 ( ∆x) + x( ∆x) − x3 − x 2 ( ∆x) − 4( ∆x)
− 10
(c) Graphing utility confirms
dy
3
= at ( − 4, − 5).
dx
4
2
= lim
x( ∆x)( x + ∆x)
∆x → 0
x 2 ( ∆x) + x( ∆x) − 4( ∆x)
2
= lim
x( ∆x)( x + ∆x)
∆x → 0
= lim
∆x → 0
=
x + x( ∆x ) − 4
2
x( x + ∆x)
4
x − 4
=1− 2
x2
x
2
At ( − 4, − 5), the slope of the tangent line is m = 1 −
4
( − 4)
2
=
3
.
4
The equation of the tangent line is
3
( x + 4)
4
3
y +5 = x +3
4
3
y = x − 2.
4
y +5 =
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
32. (a)
6
x + 2
f ( x + ∆x ) − f ( x )
f ( x) = x +
f ′( x) = lim
∆x
∆x → 0
6
6
−
2
2
x
+
∆
x
+
x
+
(
)
= lim
∆x → 0
= lim
∆x
6 x + 12 − 6( x + ∆x + 2)
∆x( x + ∆x + 2)( x + 2)
∆x → 0
6 x + 12 − 6 x − 6∆x − 12
= lim
∆x → 0 ∆x( x + ∆x + 2)( x + 2)
= lim
∆x → 0
=
−6∆x
∆x( x + ∆x + 2)( x + 2)
−6
(x
+ 2)
The Derivative and the Tangent Line Problem
145
34. Using the limit definition of derivative, f ′( x) = 4 x.
Because the slope of the given line is –4, you have
4 x = −4
x = −1.
At the point ( −1, 2) the tangent line is parallel to
4 x + y + 3 = 0. The equation of this line is
y − 2 = −4( x + 1)
y = −4 x − 2.
35. From Exercise 27 we know that f ′( x) = 3 x 2 .
Because the slope of the given line is 3, you have
3x 2 = 3
x = ±1.
2
Therefore, at the points (1, 1) and ( −1, −1) the tangent
At (0, 3), the slope of the tangent line is
m = − 64 = − 32 .
lines are parallel to 3x − y + 1 = 0.
These lines have equations
y − 1 = 3( x − 1) and y + 1 = 3( x + 1)
The equation of the tangent line is
3
y − 3 = − ( x − 0)
2
3
y −3 = − x
2
3
y = − x + 3.
2
(b)
y = 3 x + 2.
36. Using the limit definition of derivative, f ′( x) = 3 x 2 .
Because the slope of the given line is 3, you have
3x 2 = 3
x 2 = 1 ⇒ x = ±1.
Therefore, at the points (1, 3) and ( −1, 1) the tangent
6
lines are parallel to 3x − y − 4 = 0. These lines have
equations
(0, 3)
− 10
y = 3x − 2
8
y − 3 = 3( x − 1) and y − 1 = 3( x + 1)
y = 3x
−6
(c) Graphing utility confirms
dy
3
= − at (0, 3).
dx
2
33. Using the limit definition of derivative, f ′( x) = 2 x.
Because the slope of the given line is 2, you have
2x = 2
x =1
At the point (1, 1) the tangent line is parallel to
y = 3x + 4.
37. Using the limit definition of derivative,
−1
f ′( x) =
.
2x x
1
Because the slope of the given line is − , you have
2
1
1
−
= −
2
2x x
x = 1.
2 x − y + 1 = 0. The equation of this line is
Therefore, at the point (1, 1) the tangent line is parallel to
y − 1 = 2( x − 1)
x + 2 y − 6 = 0. The equation of this line is
y = 2 x − 1.
1
( x − 1)
2
1
1
y −1 = − x +
2
2
1
3
y = − x + .
2
2
y −1 = −
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
146
Chapter 3
Differentiation
38. Using the limit definition of derivative,
−1
f ′( x) =
.
32
2( x − 1)
1
Because the slope of the given line is − , you have
2
−1
1
= −
32
2
2( x − 1)
1 = ( x − 1)
42. The slope of the graph of f is –1 for x < 4, 1 for
x > 4, and undefined at x = 4.
y
3
2
f′
1
x
1
2
3
4
5
6
32
1 = x − 1 ⇒ x = 2.
At the point ( 2, 1), the tangent line is parallel to
x + 2 y + 7 = 0. The equation of the tangent line is
43. The slope of the graph of f is negative for x < 0 and
positive for x > 0. The slope is undefined at x = 0.
y
1
( x − 2)
2
1
y = − x + 2.
2
y −1 = −
2
1
f′
x
−2 −1
1
2
3
4
39. The slope of the graph of f is 1 for all x-values.
y
−2
4
44. The slope is positive for −2 < x < 0 and negative for
0 < x < 2. The slope is undefined at x = ± 2, and 0 at
3
2
f′
x
−3 −2 −1
−1
1
x = 0.
3
2
y
−2
f′
2
1
40. The slope of the graph of f is 0 for all x-values.
−2
y
x
−1
1
2
−1
2
−2
1
f′
−2
−1
1
x
2
−1
45. Answers will vary.
Sample answer: y = − x
−2
y
4
41. The slope of the graph of f is negative for
x < 4, positive for x > 4, and 0 at x = 4.
y
3
2
1
f′
4
−4
−6
2
3
4
−2
−3
2
−6 −4 −2
−2
x
−4 −3 −2 −1
−1
−4
x
2
4
6
46. Answers will vary.
Sample answer: y = x
−8
y
4
3
2
1
x
−4 −3 −2
1
2
3
4
−2
−3
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
47. g ( 4) = 5 because the tangent line passes through ( 4, 5).
g ′( 4) =
5−0
5
= −
4−7
3
48. h( −1) = 4 because the tangent line passes through
The Derivative and the Tangent Line Problem
55. Let ( x0 , y0 ) be a point of tangency on the graph of f.
By the limit definition for the derivative,
f ′( x) = 4 − 2 x. The slope of the line through ( 2, 5) and
( x0 , y0 ) equals the derivative of f at
6−4
2
1
=
=
3 − ( −1)
4
2
5 − y0 = ( 2 − x0 )( 4 − 2 x0 )
5 − ( 4 x0 − x0 2 ) = 8 − 8 x0 + 2 x0 2
49. f ( x ) = 5 − 3 x and c = 1
50. f ( x ) = x and c = −2
3
51. f ( x ) = − x 2 and c = 6
x0 :
5 − y0
= 4 − 2 x0
2 − x0
(−1, 4).
h′( −1) =
147
0 = x0 2 − 4 x0 + 3
0 = ( x0 − 1)( x0 − 3) ⇒ x0 = 1, 3
Therefore, the points of tangency are (1, 3) and
(3, 3), and the corresponding slopes are 2 and –2. The
equations of the tangent lines are:
52. f ( x ) = 2
x and c = 9
y − 5 = 2( x − 2)
y − 5 = −2( x − 2)
y = 2x + 1
53. f (0) = 2 and f ′( x) = −3, −∞ < x < ∞
y = −2 x + 9
y
f ( x) = −3 x + 2
7
6
(2, 5)
5
y
4
3
2
2
1
1
−3 −2 −1
x
2
−1
1
2
3
6
56. Let ( x0 , y0 ) be a point of tangency on the graph of f. By
f
−3
x
−2
3
−2
the limit definition for the derivative, f ′( x) = 2 x. The
54. f (0) = 4, f ′(0) = 0; f ′( x) < 0 for x < 0, f ′( x) > 0
for x > 0
slope of the line through (1, − 3) and ( x0 , y0 ) equals the
derivative of f at x0 :
Answers will vary: Sample answer: f ( x ) = x 2 + 4
−3 − y0
= 2 x0
1 − x0
y
−3 − y0 = (1 − x0 )2 x0
f
12
−3 − x0 2 = 2 x0 − 2 x0 2
10
x0 2 − 2 x0 − 3 = 0
8
6
( x0
4
2
−6 −4 −2
(3, 3)
(1, 3)
x
2
4
6
− 3)( x0 + 1) = 0 ⇒ x0 = 3, −1
Therefore, the points of tangency are (3, 9) and
(−1, 1), and the corresponding slopes are 6 and –2. The
equations of the tangent lines are:
y + 3 = 6( x − 1)
y + 3 = −2( x − 1)
y = 6x − 9
y = −2 x − 1
y
10
(3, 9)
8
6
4
(−1, 1)
−8 −6 −4 −2
−2
−4
x
2
4
6
(1, −3)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
148
Chapter 3
57. (a)
Differentiation
f ( x) = x 2
f ′( x) = lim
f ( x + ∆x) − f ( x )
∆x → 0
∆x
∆x → 0
= lim
(b) g ′( x) = lim
(x
= lim
2
∆x
= lim
= lim
∆x( 2 x + ∆x)
∆x → 0
∆x → 0
∆x
= lim ( 2 x + ∆x) = 2 x
∆x → 0
∆x
(
2
3
2
)
) = 3x
2
At x = −1, g ′( −1) = 3 and the tangent line is
At x = −1, f ′( −1) = −2 and the tangent line is
or
(
∆x 3 x 2 + 3x( ∆x) + (∆x )
= lim 3 x 2 + 3 x( ∆x) + ( ∆x)
∆x → 0
y − 1 = −2( x + 1)
2
3
= lim
x 2 + 2 x( ∆x) + (∆x) − x 2
∆x → 0
∆x
x + 3 x 2 ( ∆x) + 3 x(∆x) + ( ∆x) − x3
∆x → 0
∆x
∆x
2
= lim
( x + ∆x)3 − x3
∆x → 0
+ ∆x) − x 2
∆x → 0
g ( x + ∆x) − g ( x)
∆x
y + 1 = 3( x + 1)
y = −2 x − 1.
or
y = 3 x + 2.
At x = 0, g ′(0) = 0 and the tangent line is y = 0.
At x = 0, f ′(0) = 0 and the tangent line is y = 0.
At x = 1, g ′(1) = 3 and the tangent line is
At x = 1, f ′(1) = 2 and the tangent line is
y − 1 = 3( x − 1)
y = 2 x − 1.
or
y = 3 x − 2.
2
2
−3
−3
3
3
−2
−3
For this function, the slopes of the tangent lines are
always distinct for different values of x.
For this function, the slopes of the tangent lines are
sometimes the same.
58. (a) g ′(0) = −3
(b) g ′(3) = 0
(c) Because g ′(1) = − 83 , g is decreasing (falling) at x = 1.
(d) Because g ′( −4) =
7
,
3
g is increasing (rising) at x = −4.
(e) Because g ′( 4) and g ′(6) are both positive, g (6) is greater than g ( 4), and g (6) − g ( 4) > 0.
(f ) No, it is not possible. All you can say is that g is decreasing (falling) at x = 2.
59. f ( x ) =
(a)
1 2
x
2
6
−6
6
−2
f ′(0) = 0, f ′(1 2) = 1 2, f ′(1) = 1, f ′( 2) = 2
(b) By symmetry: f ′( −1 2) = −1 2, f ′( −1) = −1, f ′( −2) = −2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
(c)
The Derivative and the Tangent Line Problem
149
y
4
f′
3
2
1
x
−4 −3 −2
1
2
3
4
−2
−3
−4
(
(d) f ′( x) = lim
∆x → 0
60. f ( x ) =
)
1
1
1 2
1
2
x + 2 x ( ∆x ) + ( ∆ x ) − x 2
( x + ∆x)2 − x 2
f ( x + ∆x) − f ( x)
∆x ⎞
⎛
2
2
2
2
= lim
= lim
= lim ⎜ x +
⎟ = x
∆x → 0
∆x → 0
∆x → 0 ⎝
∆x
∆x
∆x
2 ⎠
1 3
x
3
(a)
6
−9
9
−6
f ′(0) = 0, f ′(1 2) = 1 4, f ′(1) = 1, f ′( 2) = 4, f ′(3) = 9
(b) By symmetry: f ′( −1 2) = 1 4, f ′(−1) = 1, f ′( −2) = 4, f ′( −3) = 9
y
(c)
f′
5
4
3
2
1
−3 −2 −1
x
1
−1
(d) f ′( x) = lim
2
3
f ( x + ∆x) − f ( x)
∆x
1
1
( x + ∆x)3 − x3
3
3
= lim
∆x → 0
∆x
1 3
1
2
3
x + 3 x 2 ( ∆x) + 3 x( ∆x) + (∆x) − x3
3
3
= lim
∆x → 0
∆x
1
2⎤
⎡
= lim ⎢ x 2 + x( ∆x) + (∆x ) ⎥ = x 2
∆x → 0 ⎣
3
⎦
∆x → 0
(
61. g ( x) =
)
f ( x + 0.01) − f ( x)
0.01
62. g ( x) =
f ( x + 0.01) − f ( x)
0.01
(
2
= ⎡2( x + 0.01) − ( x + 0.01) − 2 x + x 2 ⎤100
⎣
⎦
= 2 − 2 x − 0.01
)
= 3 x + 0.01 − 3 x 100
8
3
f
g
g
f
−2
4
−1
8
−1
−1
The graph of g ( x) is approximately the graph of
f ′( x) = 2 − 2 x.
The graph of g ( x) is approximately the graph of
f ′( x) =
3
2
x
.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
150
Chapter 3
Differentiation
63. f ( 2) = 2( 4 − 2) = 4, f ( 2.1) = 2.1( 4 − 2.1) = 3.99
3.99 − 4
f ′( 2) ≈
= −0.1
2.1 − 2
⎡⎣Exact: f ′( 2) = 0⎤⎦
1 3
(2 ) = 2, f (2.1) = 2.31525
4
2.31525 − 2
f ′( 2) ≈
= 3.1525 ⎡⎣Exact: f ′( 2) = 3⎤⎦
2.1 − 2
64. f ( 2) =
65. f ( x ) = x 2 − 5, c = 3
f ′(3) = lim
f ( x ) − f (3)
x −3
x →3
= lim
x 2 − 5 − (9 − 5)
x →3
(x
lim
=
x −3
− 3)( x + 3)
x −3
= lim ( x + 3) = 6
x →3
x →3
66. g ( x) = x 2 − x, c = 1
g ′(1) = lim
g ( x ) − g (1)
x −1
x →1
= lim
x →1
= lim
x2 − x − 0
x −1
x( x − 1)
x −1
= lim x = 1
x →1
f ( x ) − f ( −2 )
x →−2
x+ 2
f ′( −2) = lim
= lim
(x
3
+ 2 x + 1) − 1
2
x + 2
x →−2
x ( x + 2)
= lim x 2 = 4
x →−2
x →−2
x+ 2
= lim
2
68. f ( x) = x3 + 6 x, c = 2
f ( x ) − f ( 2)
x − 2
x→2
( x3 + 6 x) − 20
lim
x − 2
x→2
= lim
x→2
(x
− 2)( x 2 + 2 x + 10)
x − 2
= lim ( x 2 + 2 x + 10) = 18
x→2
g ′(0) = lim
g ( x ) − g ( 0)
= lim
x − 0
x→0
x
As x → 0− ,
x→0
−1
=
x
x
x
. Does not exist.
→ −∞.
x
x
As x → 0+ ,
1
→ ∞.
x
=
x
Therefore g ( x) is not differentiable at x = 0.
70. f ( x) =
3
,c = 4
x
f ( x ) − f ( 4)
f ′( 4) = lim
x→4
= lim
3
x
x − 4
− 43
x − 4
12 − 3 x
= lim
x → 4 4 x ( x − 4)
x→4
= lim
x→4
− 3( x − 4)
4 x ( x − 4)
= lim −
x→4
3
3
= −
4x
16
71. f ( x ) = ( x − 6) , c = 6
23
x→6
67. f ( x ) = x 3 + 2 x 2 + 1, c = −2
=
x,c = 0
f ′(6) = lim
x →1
f ′( 2) = lim
69. g ( x) =
= lim
f ( x ) − f ( 6)
x −6
(x
x→6
− 6) − 0
1
= lim
.
x →6 x − 6 1 3
x−6
(
)
23
Does not exist.
Therefore f ( x) is not differentiable at x = 6.
72. g ( x) = ( x + 3) , c = −3
13
g ′( −3) = lim
x → −3
= lim
x → −3
g ( x ) − g ( −3)
x − ( −3)
(x
+ 3) − 0
1
= lim
.
x → −3 x + 3 2 3
x+3
(
)
13
Does not exist.
Therefore g ( x) is not differentiable at x = −3.
73. h( x) = x + 7 , c = −7
h′( −7) = lim
x →−7
= lim
x →−7
h ( x ) − h ( −7 )
x − ( −7 )
x+7 −0
x+7
= lim
.
x →−7 x + 7
x +7
Does not exist.
Therefore h( x) is not differentiable at x = −7.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
74. f ( x) = x − 6 , c = 6
The Derivative and the Tangent Line Problem
151
84. f is differentiable for all x ≠ 1.
f ( x ) − f ( 6)
x→6
x −6
x −6 −0
x −6
= lim
= lim
.
x→6
x→6 x − 6
x −6
f ′(6) = lim
f is not continuous at x = 1.
3
−4
5
Does not exist.
Therefore f ( x) is not differentiable at x = 6.
75. f ( x ) is differentiable everywhere except at
−3
85. f ( x) = x − 1
The derivative from the left is
x = 3. (Discontinuity)
76. f ( x ) is differentiable everywhere except at
x = ±3. (Sharp turns in the graph)
77. f ( x) is differentiable everywhere except at
x = −4. (Sharp turn in the graph)
78. f ( x ) is differentiable everywhere except at
x = ±2. (Discontinuities)
79. f ( x ) is differentiable on the interval (1, ∞). (At
lim
x −1
= lim
x −1 − 0
x −1
x →1−
lim
x →1+
x −1 − 0
f ( x) − f (1)
= lim
= 1.
+
x −1
x −1
x →1
The one-sided limits are not equal. Therefore, f is not
differentiable at x = 1.
86. f ( x ) =
1 − x2
The derivative from the left does not exist because
lim
x →1−
f ( x) − f (1)
= lim
x −1
x →1−
x = 0. (Discontinuity)
1 − x2 − 0
x −1
1 − x2
⋅
x −1
= lim
x →1−
81. f ( x) = x − 5 is differentiable everywhere except at
= lim −
x = −5. There is a sharp corner at x = 5.
7
= −1.
The derivative from the right is
x = 1 the tangent line is vertical.)
80. f ( x) is differentiable everywhere except at
f ( x ) − f (1)
x →1−
x →1−
1+ x
1 − x2
1 − x2
1 − x2
= −∞.
(Vertical tangent)
−1
11
−1
4x
82. f ( x ) =
is differentiable everywhere except at
x −3
x = 3. f is not defined at x = 3. (Vertical asymptote)
The limit from the right does not exist since f is
undefined for x > 1. Therefore, f is not differentiable at
x = 1.
⎧( x − 1)3 , x ≤ 1
⎪
87. f ( x ) = ⎨
2
⎪⎩( x − 1) , x > 1
The derivative from the left is
15
lim
x →1−
f ( x) − f (1)
x −1
= lim
(x
− 1) − 0
3
x −1
x →1−
= lim ( x − 1) = 0.
2
−8
12
−6
The derivative from the right is
83. f ( x ) = x 2 5 is differentiable for all x ≠ 0. There is a
sharp corner at x = 0.
lim
x →1+
f ( x) − f (1)
x −1
= lim
(x
− 1) − 0
2
x −1
= lim ( x − 1) = 0.
x →1+
x →1+
5
−6
x →1−
6
The one-sided limits are equal. Therefore, f is
differentiable at x = 1. ( f ′(1) = 0)
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
152
Chapter 3
Differentiation
90. Note that f is continuous at x = 2.
⎧ x, x ≤ 1
88. f ( x ) = ⎨ 2
⎩x , x > 1
⎧1
⎪ x + 1, x < 2
f ( x) = ⎨ 2
⎪ 2x ,
x ≥ 2
⎩
The derivative from the left is
lim
f ( x ) − f (1)
x −1
x →1−
= lim
x →1−
x −1
= lim 1 = 1.
x −1
x →1−
The derivative from the left is
⎛1
⎞
⎜ x + 1⎟ − 2
f ( x ) − f ( 2)
2
⎝
⎠
= lim
lim
x−2
x −2
x → 2−
x → 2−
1
( x − 2) 1
= lim 2
= .
−
2
x−2
x→2
The derivative from the right is
lim =
x →1+
f ( x) − f (1)
x −1
= lim
= lim ( x + 1) = 2.
x −1
x →1+ x − 1
x →1+
2
The one-sided limits are not equal. Therefore, f is not
differentiable at x = 1.
89. Note that f is continuous at x = 2.
The derivative from the right is
2
⎪⎧x + 1, x ≤ 2
f ( x) = ⎨
⎪⎩4 x − 3, x > 2
lim
x → 2+
f ( x ) − f ( 2)
= lim
x −2
x → 2+
= lim
The derivative from the left is
lim
x → 2−
x → 2+
( x 2 + 1) − 5
f ( x) − f ( 2)
= lim
x−2
x−2
x → 2−
= lim ( x + 2) = 4.
= lim
x → 2+
x → 2−
= lim
The derivative from the right is
lim
x → 2+
f ( x ) − f ( 2)
x − 2
= lim
x → 2+
(4 x
− 3) − 5
x − 2
x → 2+
= lim 4 = 4.
2x + 2
2x + 2
2x − 4
(x
(
− 2)
2x + 2
2( x − 2)
(x
(
− 2)
2x + 2
)
)
2
1
= .
2
2x + 2
The one-sided limits are equal. Therefore, f is
1⎞
⎛
differentiable at x = 2. ⎜ f ′( 2) = ⎟
2⎠
⎝
x → 2+
The one-sided limits are equal. Therefore, f is
differentiable at x = 2. ( f ′( 2) = 4)
91.
2x − 2
⋅
x−2
4
−3
3
−2
Let g ( x) =
axb.
x
For f ( x) = a xb,
lim
x → 0−
f ( x ) − f ( 0)
x −0
= lim
x → 0−
axb − 0
x
= lim
x → 0−
axb
x
= lim a xb ⋅ lim
1
1
−1
= −1 ⋅ lim
= lim
= ∞.
x
x → 0− x
x → 0− x
= lim a xb ⋅ lim
1
1
= 0 ⋅ lim
= 0.
x
x → 0+ x
x → 0−
x → 0−
On the other hand,
lim
x → 0+
f ( x ) − f ( 0)
x −0
= lim
x → 0+
axb − 0
x
= lim
x → 0+
axb
x
So, f is not differentiable at x = 0 because lim
x→0
x → 0+
x → 0+
f ( x ) − f ( 0)
x −0
does not exist. f is differentiable for all x ≠ n,
n an integer.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.1
92. (a) f ( x) = x 2 and f ′( x) = 2 x
The Derivative and the Tangent Line Problem
153
g ( x) = x3 and g ′( x) = 3 x 2
(b)
y
y
5
3
4
f
g′
3
2
2
x
−4 −3 −2 −1
f'
g
1
1
1
2
3
4
x
−1
−2
1
2
−1
−3
(c) The derivative is a polynomial of degree 1 less than the original function. If h( x) = x n , then h′( x) = nx n −1.
(d) If f ( x ) = x 4 , then
f ′( x) = lim
∆x → 0
= lim
∆x → 0
f ( x + ∆x) − f ( x)
∆x
(x
+ ∆x) − x 4
∆x
4
x 4 + 4 x3 ( ∆x) + 6 x 2 ( ∆x) + 4 x( ∆x) + (∆x) − x 4
∆x → 0
∆x
2
3
4
= lim
= lim
(
∆x 4 x3 + 6 x 2 ( ∆x) + 4 x(∆x) + ( ∆x)
2
∆x
∆x → 0
3
)=
(
lim 4 x 3 + 6 x 2 ( ∆x ) + 4 x(∆x) + ( ∆x)
∆x → 0
2
3
) = 4x .
3
So, if f ( x ) = x 4 , then f ′( x) = 4 x 3 which is consistent with the conjecture. However, this is not a proof because you
must verify the conjecture for all integer values of n, n ≥ 2.
93. False. The slope is lim
∆x → 0
f ( 2 + ∆x) − f ( 2)
.
∆x
94. False. y = x − 2 is continuous at x = 2, but is not
differentiable at x = 2. (Sharp turn in the graph)
95. False. If the derivative from the left of a point does not
equal the derivative from the right of a point, then the
derivative does not exist at that point. For example, if
f ( x) = x , then the derivative from the left at x = 0 is
–1 and the derivative from the right at x = 0 is 1. At
x = 0, the derivative does not exist.
96. True—see Theorem 3.1.
⎧⎪x sin (1 x ), x ≠ 0
97. f ( x) = ⎨
x = 0
⎪⎩0,
Using the Squeeze Theorem, you have
− x ≤ x sin (1 x) ≤ x , x ≠ 0.
So, lim x sin (1 x ) = 0 = f (0) and f is continuous at
x→0
x = 0.
Using the alternative form of the derivative, you have
lim
f ( x ) − f ( 0)
x −0
x→0
= lim
x→0
x sin (1 x) − 0
x −0
1⎞
⎛
= lim ⎜ sin ⎟.
x → 0⎝
x⎠
Because this limit does not exist ( sin (1 x) oscillates
between –1 and 1), the function is not differentiable
at x = 0.
⎧⎪x 2 sin (1 x), x ≠ 0
g ( x) = ⎨
x = 0
⎪⎩0,
Using the Squeeze Theorem again, you have
− x 2 ≤ x 2 sin (1 x) ≤ x 2 , x ≠ 0.
So, lim x 2 sin (1 x) = 0 = g (0)
x→0
and g is continuous at x = 0. Using the alternative form
of the derivative again, you have
lim
g ( x ) − g ( 0)
x 2 sin (1 x) − 0
x −0
1
= lim x sin = 0.
x→0
x
Therefore, g is differentiable at x = 0, g ′(0) = 0.
x→0
x −0
= lim
x→0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
154
Chapter 3
98.
Differentiation
3
−3
3
−1
As you zoom in, the graph of y1 = x 2 + 1 appears to be locally the graph of a horizontal line, whereas the graph
of y2 = x + 1 always has a sharp corner at (0, 1). y2 is not differentiable at (0, 1).
Section 3.2 Basic Differentiation Rules and Rates of Change
y = x1 2
1. (a)
y′ =
y′(1) =
9. y =
1 −1 2
x
2
1
2
10. y =
y = x3
(b)
y′ =
y′ =
y′ = 3 x 2
y′(1) = 3
11.
y = x −1 2
2. (a)
y′ = − 12 x −3 2
y′(1) = − 12
x = x1 4
1 −3 4
1
x
=
4
4 x3 4
f ( x) = x + 11
f (t ) = −2t 2 + 3t − 6
f ′(t ) = −4t + 3
3. y = 12
14. y = t 2 − 3t + 1
y′ = 0
y′ = 2t − 3
f ( x ) = −9
15. g ( x) = x 2 + 4 x 3
f ′( x) = 0
g ′( x) = 2 x + 12 x 2
5. y = x 7
16. y = 4 x − 3 x3
y′ = 7 x 6
y′ = 4 − 9 x 2
6. y = x12
17. s(t ) = t 3 + 5t 2 − 3t + 8
y′ = 12 x11
s′(t ) = 3t 2 + 10t − 3
1
= x −5
x5
18. y = 2 x3 + 6 x 2 − 1
y′ = −5 x −6 = −
5
x6
3
8. y = 7 = 3x −7
x
y′ = 3( − 7 x
4
12. g ( x) = 6 x + 3
13.
y′(1) = −1
7. y =
1 −4 5
1
x
=
5
5x4 5
g ′( x) = 6
y′ = − x −2
4.
x = x1 5
f ′( x ) = 1
y = x −1
(b)
5
−8
)
21
= − 8
x
y′ = 6 x 2 + 12 x
19. y =
y′ =
π
2
π
2
sin θ − cos θ
cos θ + sin θ
20. g (t ) = π cos t
g ′(t ) = −π sin t
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.2
21. y = x 2 −
1
2
y′ = 2 x +
1
2
Basic Differentiation Rules and Rates of Change
cos x
23. y =
1 ex
2
− 3 sin x
sin x
y′ =
1 x
e
2
− 3 cos x
24. y =
3 x
e
4
+ 2 cos x
y′ =
3 x
e
4
− 2 sin x
22. y = 7 + sin x
y′ = cos x
Function
Rewrite
Differentiate
Simplify
25. y =
5
2x2
y =
5 −2
x
2
y′ = −5 x −3
y′ = −
5
x3
26. y =
3
2x4
y =
3 −4
x
2
y′ = − 6 x −5
y′ = −
6
x5
3
y =
6 −3
x
125
y′ = −
18 −4
x
125
y′ = −
18
125 x 4
2
y =
y′ = −
2π −3
x
9
y′ = −
2π
9 x3
y = x −1 2
1
y′ = − x −3 2
2
y′ = −
1
2 x3 2
y = 4 x3
y′ = 12 x 2
y′ = 12 x 2
27. y =
28. y =
6
(5 x )
π
(3 x )
29. y =
x
x
30. y =
4
x −3
π
9
x −2
8
= 8 x −2 , ( 2, 2)
x2
16
f ′( x) = −16 x −3 = − 3
x
f ′( 2) = −2
31. f ( x) =
36.
f (t ) = 2 −
f ′(t ) = 4t − 2
f ′( 4) =
33.
g ′(π ) = 0
4
= 2 − 4t −1 , ( 4, 1)
t
4
= 2
t
3 t
e,
4
f ′(t ) =
3 t
e
4
f ( 0) =
3 0
e
4
(0, 34 )
=
3
4
g ′( x) = − 4e x
g ′(1) = − 4e
y = 2 x 4 − 3, (1, −1)
4
= t 2 − 4t −3
t3
12
g ′(t ) = 2t + 12t −4 = 2t + 4
t
39. g (t ) = t 2 −
y′(1) = 8
f ( x) = 2( x − 4) , ( 2, 8)
2
= 2 x 2 − 16 x + 32
f ′( x) = 4 x − 16
40.
f ′( 2) = 8 − 16 = − 8
35.
f (t ) =
38. g ( x) = − 4e x , (1, − 4e)
1
4
y′ = 8 x 3
34.
g (t ) = −2 cos t + 5, (π , 7)
g ′(t ) = 2 sin t
37.
32.
155
f (θ ) = 4 sin θ − θ , (0, 0)
f ′(θ ) = 4 cos θ − 1
f ′(0) = 4(1) − 1 = 3
41.
3
= 8 x + 3x − 2
x2
6
f ′( x) = 8 − 6 x − 3 = 8 − 3
x
f ( x) = 8 x +
4 x3 + 3x 2
= 4 x 2 + 3x
x
f ′( x) = 8 x + 3
f ( x) =
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
156
42.
43.
Chapter 3
Differentiation
2x4 − x
= 2 x − x −2
x3
2
f ′( x) = 2 + 2 x −3 = 2 + 3
x
f ( x) =
f ( x) =
53. (a)
y′ = 4 x 3 − 6 x
At (1, 0): y′ = 4(1) − 6(1) = −2
3
y − 0 = −2( x − 1)
Tangent line:
x3 − 3x 2 + 4
= x − 3 + 4 x −2
x2
f ′( x) = 1 −
y = x 4 − 3x2 + 2
y = − 2x + 2
2x + y − 2 = 0
x −8
8
=
x3
x3
3
(b)
4 x3 + 2 x + 5
= 4 x 2 + 2 + 5 x −1
x
5
h′( x) = 8 x − 5 x − 2 = 8 x − 2
x
3
44. h( x) =
45. y = x( x 2 + 1) = x 3 + x
−2
−1
54. (a)
y′ = 3 x 2 + 1
46. y = x 2 ( 2 x 2 − 3 x) = 2 x 4 − 3x3
f ( x) =
f ′( x) =
48.
49.
x − 6 3 x = x1 2 − 6 x1 3
2 −1 3 1 −2 3
2
1
t
− t
= 13 − 2 3
3
3
3t
3t
f ( x) = 6
(b)
3
− 5 sin x
x
3
( x − 1)
2
3
7
y = − x +
2
2
3x + 2 y − 7 = 0
y − 2 = −
5
−2
7
−1
55. (a)
g ( x) = x + e x
g ′( x) = 1 + e x
3
At (0, 1): g ′(0) = 1 + 1 = 2
Tangent line: y − 1 = 2( x − 0)
y = 2x + 1
51. f ( x ) = x − 2 − 2e x
f ′( x) = − 2 x − 3 − 2e x =
3
2
(1, 2)
2
+ 3 cos x = 2 x −1 3 + 3 cos x
x
2
2
f ′( x) = − x −4 3 − 3 sin x = − 4 3 − 3 sin x
3
3x
f ( x) =
= 2 x −3 4
x3
3
3
f ′( x) = − x −7 4 = − 7 4
2
2x
Tangent line:
x + 5 cos x = 6 x1 2 + 5 cos x
f ′( x) = 3 x −1 2 − 5 sin x =
50.
2
4
1 −1 2
1
2
x
− 2 x −2 3 =
− 23
x
2
2 x
f (t ) = t 2 3 − t 1 3 + 4
f ′(t ) =
f ( x) =
At (1, 2): f ′(1) = −
y′ = 8 x3 − 9 x 2 = x 2 (8 x − 9)
47.
2
(1, 0)
−2
− 2e x
x3
(b)
8
(0, 1)
52. g ( x) =
g ′( x) =
x − 3e
1
2
x
x
− 3e x
−4
4
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.2
56. (a) h(t ) = sin t + 12 et
y′ = 2 x = 0 ⇒ x = 0
)
At π , 12 et : h′(π ) = −1 + 12 eπ
Tangent line:
(
)
= ( −1 + 12 e )t + 12 e
y − 12 eπ = −1 + 12 eπ (t − π )
y
(b)
157
60. y = x 2 + 9
h′(t ) = cos t + 12 et
(
Basic Differentiation Rules and Rates of Change
π
π
At x = 0, y = 1.
Horizontal tangent: (0, 9)
61. y = − 4 x + e x
+ π − 12 π eπ
20
y′ = − 4 + e x = 0
ex = 4
x = ln 4
Horizontal tangent: (ln 4, − 4 ln 4 + 4)
−1
6
−2
y′ = 1 + 4e x cannot equal 0.
57. y = x 4 − 2 x 2 + 3
So, there are no horizontal tangents.
y′ = 4 x 3 − 4 x
63. y = x + sin x, 0 ≤ x < 2π
= 4 x( x 2 − 1)
y′ = 1 + cos x = 0
= 4 x( x − 1)( x + 1)
cos x = −1 ⇒ x = π
y′ = 0 ⇒ x = 0, ±1
At x = π : y = π
Horizontal tangents: (0, 3), (1, 2), ( −1, 2)
58. y = x3 + x
Therefore, there are no horizontal tangents.
1
= x −2
x2
y′ = −2 x −3
Horizontal tangent: (π , π )
64. y =
y′ = 3 x 2 + 1 > 0 for all x.
59. y =
62. y = x + 4e x
2
= − 3 cannot equal zero.
x
Therefore, there are no horizontal tangents.
y′ =
3 x + 2 cos x, 0 ≤ x < 2π
3 − 2 sin x = 0
π
3
2π
⇒ x =
or
2
3
3
sin x =
At x =
At x =
π
3
: y =
3π + 3
3
2π
2 3π − 3
: y =
3
3
⎛π
Horizontal tangents: ⎜⎜ ,
⎝3
3π +
3
3 ⎞ ⎛ 2π 2 3π − 3 ⎞
⎟⎟, ⎜⎜ ,
⎟⎟
3
⎠ ⎝ 3
⎠
65. k − x 2 = −6 x + 1 Equate functions.
−2 x = −6
Equate derivatives.
So, x = 3 and k − 9 = −18 + 1 ⇒ k = −8.
66. kx 2 = − 2 x + 3 Equate functions.
2kx = − 2
So, k = −
Equate derivatives.
2
1
= − , and
x
2x
1
⎛ 1⎞ 2
⎜− ⎟x = − 2x + 3 ⇒ − x = − 2x + 3 ⇒ x = 3 ⇒ k = − .
3
⎝ x⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
158
67.
Chapter 3
Differentiation
3
k
= − x + 3 Equate functions.
4
x
3
k
Equate derivatives.
− 2 = −
4
x
So, k =
x = x + 4 Equate functions.
68. k
k
2
3 2
x
3
3
3
3 2
4
and
= − x + 3⇒ x = − x +3
x
x
4
4
4
4
3
⇒ x = 3 ⇒ x = 2 ⇒ k = 3.
2
x
=1
So, k = 2
(2 x )
72. (a) The slope appears to be steepest between A and B.
(b) The average rate of change between A and B is
greater than the instantaneous rate of change at B.
Equate derivatives.
(c)
y
x and
f
x = x + 4 ⇒ 2 x = x + 4 ⇒ x = 4 ⇒ k = 4.
B C
A
69.
3kx 2 = 1
So, k =
Equate derivatives.
1
and
3x 2
⎛ 1 ⎞ 3
⎜ 2 ⎟x = x + 1
⎝ 3x ⎠
1
x = x +1
3
3
4
x = − ,k =
.
2
27
E
D
kx 3 = x + 1 Equate equations.
x
73. g ( x) = f ( x) + 6 ⇒ g ′( x) = f ′( x)
74. g ( x) = 3 f ( x) − 1 ⇒ g ′( x) = 3 f ′( x)
y
75.
3
f′
f
1
x
−3 −2 −1
1
2
3
−2
70.
kx = 4 x − 1 Equate equations.
4
4kx3 = 4
Equate derivatives.
If f is linear then its derivative is a constant function.
1
So, k = 3 and
x
f ( x) = ax + b
f ′( x) = a
⎛1⎞ 4
⎜ 3 ⎟x = 4x − 1
⎝x ⎠
x = 4x − 1
76.
y
2
1
x =
and k = 27.
3
71. The graph of a function f such that f ′ > 0 for all x and
the rate of change of the function is decreasing
(i.e., f ′′ < 0 ) would, in general, look like the graph
below.
y
f
1
−2
x
−1
1
3
4
f′
−3
−4
If f is quadratic, then its derivative is a linear function.
f ( x) = ax 2 + bx + c
f ′( x) = 2ax + b
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.2
Basic Differentiation Rules and Rates of Change
159
77. Let ( x1 , y1 ) and ( x2 , y2 ) be the points of tangency on y = x 2 and y = − x 2 + 6 x − 5, respectively.
The derivatives of these functions are:
y′ = 2 x ⇒ m = 2 x1 and y′ = −2 x + 6 ⇒ m = −2 x2 + 6
m = 2 x1 = −2 x2 + 6
x1 = − x2 + 3
Because y1 = x12 and y2 = − x2 2 + 6 x2 − 5:
(− x22 + 6 x2 − 5) − ( x12 ) = −2 x + 6
y2 − y1
=
2
x2 − x1
x2 − x1
m =
(− x22
(− x22
+ 6 x2 − 5) − ( − x2 + 3)
x2 − ( − x2 + 3)
2
= −2 x2 + 6
+ 6 x2 − 5) − ( x2 2 − 6 x2 + 9) = ( −2 x2 + 6)( 2 x2 − 3)
−2 x2 2 + 12 x2 − 14 = −4 x2 2 + 18 x2 − 18
2 x2 2 − 6 x2 + 4 = 0
2( x2 − 2)( x2 − 1) = 0
x2 = 1 or 2
x2 = 1 ⇒ y2 = 0, x1 = 2 and y1 = 4
So, the tangent line through (1, 0) and ( 2, 4) is
⎛ 4 − 0⎞
y −0 = ⎜
⎟( x − 1) ⇒ y = 4 x − 4.
⎝ 2 − 1⎠
So, the tangent line through ( 2, 3) and (1, 1) is
⎛ 3 − 1⎞
y −1 = ⎜
⎟( x − 1) ⇒ y = 2 x − 1.
⎝ 2 − 1⎠
y
y
5
5
(2, 4)
4
4
3
3
2
2
1
1
(1, 0) 2
−1
(2, 3)
(1, 1)
x
x
3
−1
2
3
−2
x2 = 2 ⇒ y2 = 3, x1 = 1 and y1 = 1
78. m1 is the slope of the line tangent to y = x. m2 is the slope of the line tangent to y = 1 x. Because
y = x ⇒ y′ = 1 ⇒ m1 = 1 and y =
1
1
1
⇒ y′ = − 2 ⇒ m2 = − 2 .
x
x
x
The points of intersection of y = x and y = 1 x are
x =
1
⇒ x 2 = 1 ⇒ x = ±1.
x
At x = ±1, m2 = −1. Because m2 = −1 m1 , these tangent lines are perpendicular at the points of intersection.
79.
f ( x) = 3x + sin x + 2
f ′( x) = 3 + cos x
Because cos x ≤ 1, f ′( x) ≠ 0 for all x and f does not have a horizontal tangent line.
80.
f ( x) = x5 + 3 x3 + 5 x
f ′( x) = 5 x 4 + 9 x 2 + 5
Because 5 x 4 + 9 x 2 ≥ 0, f ′( x) ≥ 5. So, f does not have a tangent line with a slope of 3.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
160
81.
Chapter 3
f ( x) =
Differentiation
x , ( −4, 0)
1 −1 2
1
x
=
2
2 x
1
0− y
=
−
4− x
2 x
f ′( x) =
4 + x = 2
xy
4 + x = 2
x
x
4 + x = 2x
x = 4, y = 2
The point ( 4, 2) is on the graph of f.
0− 2
( x − 4)
−4 − 4
4y − 8 = x − 4
Tangent line: y − 2 =
0 = x − 4y + 4
82.
2
, (5, 0)
x
2
f ′( x) = − 2
x
f ( x) =
−
2
0 − y
=
2
x
5− x
−10 + 2 x = − x 2 y
⎛2⎞
−10 + 2 x = − x 2 ⎜ ⎟
⎝ x⎠
−10 + 2 x = −2 x
4 x = 10
x =
5
4
,y =
2
5
8
⎛5 4⎞
⎛5⎞
The point ⎜ , ⎟ is on the graph of f. The slope of the tangent line is f ′⎜ ⎟ = − .
25
⎝2 5⎠
⎝ 2⎠
Tangent line:
y −
4
8⎛
5⎞
= − ⎜x − ⎟
5
25 ⎝
2⎠
25 y − 20 = −8 x + 20
8 x + 25 y − 40 = 0
83. f ′(1) appears to be close to −1.
f ′(1) = −1
84. f ′( 4) appears to be close to 1.
f ′( 4) = 1
3.64
0.77
3.33
16
1.24
−10
19
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.2
Basic Differentiation Rules and Rates of Change
161
85. (a) One possible secant is between (3.9, 7.7019) and ( 4, 8) :
8 − 7.7019
( x − 4)
4 − 3.9
y − 8 = 2.981( x − 4)
20
y −8 =
(4, 8)
y = S ( x) = 2.981x − 3.924
−2
12
−2
3 12
3
x ⇒ f ′( 4) = ( 2) = 3
2
2
T ( x) = 3( x − 4) + 8 = 3x − 4
(b) f ′( x) =
The slope (and equation) of the secant line approaches that of the tangent line at
(4, 8) as you choose points closer and closer to (4, 8).
(c) As you move further away from ( 4, 8), the accuracy of the approximation T gets worse.
20
f
T
−2
12
−2
(d)
∆x
–3
–2
–1
–0.5
–0.1
0
0.1
0.5
1
2
3
f ( 4 + ∆x)
1
2.828
5.196
6.548
7.702
8
8.302
9.546
11.180
14.697
18.520
T ( 4 + ∆x)
–1
2
5
6.5
7.7
8
8.3
9.5
11
14
17
86. (a) Nearby point: (1.0073138, 1.0221024)
1.0221024 − 1
( x − 1)
1.0073138 − 1
y = 3.022( x − 1) + 1
2
Secant line: y − 1 =
(1, 1)
−3
3
(Answers will vary.)
−2
(b) f ′( x) = 3 x 2
T ( x) = 3( x − 1) + 1 = 3 x − 2
(c) The accuracy worsens as you move away from (1, 1).
2
(1, 1)
−3
3
f
T
−2
(d)
∆x
–3
–2
–1
–0.5
–0.1
0
0.1
0.5
1
2
3
f ( x)
–8
–1
0
0.125
0.729
1
1.331
3.375
8
27
64
T ( x)
–8
–5
–2
–0.5
0.7
1
1.3
2.5
4
7
10
The accuracy decreases more rapidly than in Exercise 85 because y = x3 is less "linear" than y = x 3 2 .
87. False. Let f ( x) = x and g ( x) = x + 1. Then
f ′( x) = g ′( x) = x, but f ( x) ≠ g ( x).
88. True. If f ( x) = g ( x) + c, then
f ′( x) = g ′( x) + 0 = g ′( x).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
162
Chapter 3
Differentiation
89. False. If y = π 2 , then dy dx = 0. ( π 2 is a constant.)
90. True. If y = x π = (1 π ) ⋅ x, then
dy dx = (1 π )(1) = 1 π .
Instantaneous rate of change:
91. True. If g ( x) = 3 f ( x), then g ′( x) = 3 f ′( x).
1
92. False. If f ( x ) =
= x − n , then
xn
f ′( x) = − nx
− n −1
93. f (t ) = 4t + 5,
2 −1
[1, 2]
13 − 9
= 4
1
97. g ( x) = x 2 + e x ,
3
≈ 0.866
2
(1 2) − 0
(π 6) − 0
=
3
π
≈ 0.955
[0, 1]
g ′( x) = 2 x + e x
Instantaneous rate of change:
(0, 1): g ′(0) = 1
(1, 1 + e): g ′(1) =
2 + e ≈ 4.718
Average rate of change:
1−0
=
(1 + e) − (1)
1
= e ≈ 2.718
1 x
e , [0, 2]
2
1
h′( x) = 3 x 2 − e x
2
98. h( x) = x3 −
Instantaneous rate of change:
At (3, 2): f ′(3) = 6
At (3.1, 2.61) : f ′(3.1) = 6.2
Instantaneous rate of change:
Average rate of change:
f (3.1) − f (3)
2.61 − 2
=
= 6.1
3.1 − 3
0.1
[1, 2]
1⎞
1
⎛
⎜ 0, − ⎟ : h′(0) = −
2⎠
2
⎝
1 2⎞
1 2
⎛
⎜ 2, 8 − e ⎟ : h′( 2) = 12 − e ≈ 8.305
2 ⎠
2
⎝
Average rate of change:
h ( 2) − h ( 0 )
2 −0
Instantaneous rate of change:
(1, −1)
=
−0
g (1) − g (0)
[3, 3.1]
f ′(t ) = 2t
1
95. f ( x ) = − ,
x
1
f ′( x) = 2
x
⎛π 1 ⎞
⎛π ⎞
⎜ , ⎟ ⇒ f ′⎜ ⎟ =
⎝ 6 2⎠
⎝6⎠
(π 6)
(These are the same because f is a line of slope 4.)
94. f (t ) = t 2 − 7,
⇒ f ′(0) = 1
f (π 6) − f (0)
Instantaneous rate of change is the constant 4. Average
rate of change:
=
(0, 0)
Average rate of change:
−n
= n +1 .
x
f ′(t ) = 4. So, f ′(1) = f ′( 2) = 4.
f ( 2) − f (1)
⎡ π⎤
96. f ( x ) = sin x, ⎢0, ⎥
⎣ 6⎦
f ′( x) = cos x
⎡8 − (1 2)e 2 ⎤⎦ − ( −1 2)
= ⎣
2
17 − e 2
4
≈ 2.403
⇒ f ′(1) = 1
=
1⎞
1
⎛
⎜ 2, − ⎟ ⇒ f ′( 2) =
2⎠
4
⎝
Average rate of change:
f ( 2) − f (1)
2 −1
=
(−1 2)
− ( −1)
2 −1
=
1
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.2
99. (a) s(t ) = −16t 2 + 1362
103. From (0, 0) to ( 4, 2), s(t ) =
v(t ) = −32t
2 −1
v (t ) =
= 1298 − 1346 = −48 ft/sec
mi/min.
= 30 mi/h for 0 < t < 4
v
When t = 2: v( 2) = −64 ft/sec
(d) −16t 2 + 1362 = 0
1362
⇒t =
16
⎛
1362 ⎞
⎟ = −32⎜⎜
4 ⎟⎠
⎝
1362
≈ 9.226 sec
4
1362 ⎞
⎟
4 ⎟⎠
= −8 1362 ≈ −295.242 ft/sec
s(t ) = −16t − 22t + 220
2
100.
1
2
s(t ) = t − 4 ⇒ v(t ) = 1 mi/min. = 60 mi/h.
When t = 1: v(1) = −32 ft/sec
⎛
(e) v⎜⎜
⎝
(60)
⇒ v (t ) =
(6, 2) to (10, 6),
(c) v(t ) = s′(t ) = −32t
t2 =
1
2
1t
2
163
Similarly, v(t ) = 0 for 4 < t < 6. Finally, from
v(t ) = −32t − 22
v(3) = −118 ft/sec
s(t ) = −16t 2 − 22t + 220
= 112 ( height after falling 108 ft )
Velocity (in mi/h)
(b)
s( 2) − s(1)
Basic Differentiation Rules and Rates of Change
60
50
40
30
20
10
t
2
4
6
8
10
Time (in minutes)
(The velocity has been converted to miles per hour.)
104. From (0, 0) to (6, 5), s(t ) =
v (t ) =
5
6
(60)
5
t
6
⇒ v (t ) =
5
6
mi/min.
= 50 mi/h for 0 < t < 6
Similarly, v(t ) = 0 for 6 < t < 8.
Finally, from (8, 5) to (10, 6),
s (t ) =
1t
2
−16t − 22t + 108 = 0
2
+ 1 ⇒ v (t ) =
1
2
mi/min = 30 mi h.
v
−2(t − 2)(8t + 27) = 0
Velocity (in mph)
50
t = 2
v( 2) = −32( 2) − 22
= −86 ft/sec
40
30
20
10
101.
s(t ) = −4.9t 2 + v0t + s0
t
2
4
6
8
10
Time (in minutes)
= −4.9t 2 + 120t
(The velocity has been converted to miles per hour.)
v(t ) = −9.8t + 120
v(5) = −9.8(5) + 120 = 71 m/sec
v(10) = −9.8(10) + 120 = 22 m/sec
102. s(t ) = −4.9t 2 + v0t + s0
= −4.9t 2 + s0 = 0 when t = 5.6.
s0 = 4.9t 2 = 4.9(5.6) ≈ 153.7 m
2
105. v = 40 mi/h =
2
3
mi/min
( 23 mi/min )(6 min) = 4 mi
v = 0 mi/h = 0 mi/min
(0 mi/min )( 2 min )
= 0 mi
v = 60 mi/h = 1 mi/min
(1 mi/min )(2 min )
= 2 mi
Distance (in miles)
s
10
8
(10, 6)
6
(6, 4)
4
(8, 4)
2
(0, 0)
t
2
4
6
8
10
Time (in minutes)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
164
Chapter 3
Differentiation
106. This graph corresponds with Exercise 103.
109. (a) Using a graphing utility,
R(v) = 0.417v − 0.02.
s
Distance (in miles)
10
(b) Using a graphing utility,
B(v ) = 0.0056v 2 + 0.001v + 0.04.
8
(10, 6)
6
(c) T (v) = R(v) + B(v) = 0.0056v 2 + 0.418v + 0.02
4
(4, 2)
2
(6, 2)
(d)
80
t
(0, 0)
2
4
6
8
T
10
Time (in minutes)
107. V = s 3 ,
dV
= 3s 2
ds
When s = 6 cm,
0
dV
= 108 cm3 per cm change in s.
ds
dT
= 0.0112v + 0.418
dv
(e)
For v = 40, T ′( 40) ≈ 0.866
dA
108. A = s ,
= 2s
ds
110.
120
0
For v = 80, T ′(80) ≈ 1.314
2
When s = 6 m,
B
R
For v = 100, T ′(100) ≈ 1.538
dA
= 12 m 2 per m change in s.
ds
(f ) For increasing speeds, the total stopping distance
increases.
C = (gallons of fuel used)( cost per gallon )
52,200
⎛ 15,000 ⎞
= ⎜
⎟(3.48) =
x
⎝ x ⎠
dC
52,200
= −
dx
x2
x
10
15
20
25
30
35
40
C
5220
3480
2610
2088
1740
1491.4
1305
dC dx
–522
–232
–130.5
–83.52
–58
–42.61
–32.63
The driver who gets 15 miles per gallon would benefit more. The rate of change at x = 15 is larger in absolute value than that
at x = 35.
1
111. s(t ) = − at 2 + c and s′(t ) = − at
2
⎡−(1 2)a(t0 + ∆t )2 + c⎤ − ⎡−(1 2)a(t0 − ∆t )2 + c)⎤
s(t0 + ∆t ) − s(t0 − ∆t )
⎦ ⎣
⎦
Average velocity:
= ⎣
2 ∆t
(t0 + ∆t ) − (t0 − ∆t )
=
(
−(1 2)a t0 2 + 2t0 ∆t + ( ∆t )
−2at0 ∆t
=
= − at0 = s′(t0 )
2 ∆t
112.
2
) + (1 2)a(t
2
0
− 2t0 ∆t + ( ∆t )
2
)
2 ∆t
instantaneous velocity at t = t0
1,008,000
+ 6.3Q
Q
dC
1,008,000
= −
+ 6.3
dQ
Q
C =
C (351) − C (350) ≈ 5083.095 − 5085 ≈ −$1.91
When Q = 350,
dC
≈ −$1.93.
dQ
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.2
Basic Differentiation Rules and Rates of Change
113. y = ax 2 + bx + c
1
,x > 0
x
1
y′ = − 2
x
114. y =
Because the parabola passes through (0, 1) and (1, 0),
you have:
(0, 1): 1
(1, 0):
165
At ( a, b), the equation of the tangent line is
= a ( 0) + b ( 0) + c ⇒ c = 1
2
0 = a(1) + b(1) + 1 ⇒ b = − a − 1
2
y −
So, y = ax 2 + ( − a − 1) x + 1.
1
1
= − 2 ( x − a)
a
a
or
y = −
x
2
+ .
a2
a
The x-intercept is ( 2a, 0).
From the tangent line y = x − 1, you know that the
⎛ 2⎞
The y-intercept is ⎜ 0, ⎟.
⎝ a⎠
derivative is 1 at the point (1, 0).
y′ = 2ax + ( −a − 1)
The area of the triangle is A =
1 = 2a(1) + (− a − 1)
1 = a −1
1
1
⎛2⎞
bh = ( 2a)⎜ ⎟ = 2.
2
2
⎝a⎠
y
a = 2
b = − a − 1 = −3
2
( )
(a, b) = a, a1
Therefore, y = 2 x 2 − 3 x + 1.
1
x
1
2
3
115. y = x 3 − 9 x
y′ = 3 x 2 − 9
y + 9 = (3 x 2 − 9)( x − 1)
Tangent lines through (1, − 9):
( x3 − 9 x) + 9
= 3x3 − 3x 2 − 9 x + 9
0 = 2 x 3 − 3 x 2 = x 2 ( 2 x − 3)
x = 0 or x =
3
2
( 32 , − 818 ). At (0, 0), the slope is y′(0) = −9. At ( 32 , − 818 ), the slope is y′( 32 ) = − 94 .
y − 0 = −9( x − 0) and
y + 81
= − 94 ( x − 32 )
8
The points of tangency are (0, 0) and
Tangent Lines:
y = −9 x
y = − 94 x −
9x + y = 0
9 x + 4 y + 27 = 0
27
4
116. y = x 2
y′ = 2 x
(a) Tangent lines through (0, a):
y − a = 2 x ( x − 0)
x2 − a = 2x2
−a = x 2
±
(
The points of tangency are ±
(
At −
)
− a , − a , the slope is
Tangent lines: y + a = 2
−a = x
) ( −a , − a), the slope is y′( −a ) = 2
y′( − − a ) = −2 − a .
− a ( x − − a ) and y + a = −2 − a ( x + − a )
− a , − a . At
y = 2 − ax + a
−a .
y = −2 − ax + a
Restriction: a must be negative.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
166
Chapter 3
Differentiation
(b) Tangent lines through ( a, 0): y − 0 = 2 x( x − a)
x 2 = 2 x 2 − 2ax
0 = x 2 − 2ax = x( x − 2a )
The points of tangency are (0, 0) and ( 2a, 4a 2 ). At (0, 0), the slope is y′(0) = 0. At ( 2a, 4a 2 ), the slope is y′( 2a) = 4a.
Tangent lines: y − 0 = 0( x − 0) and y − 4a 2 = 4a( x − 2a )
y = 0
y = 4ax − 4a 2
Restriction: None, a can be any real number.
3
x ≤ 2
⎪⎧ax ,
117. f ( x) = ⎨ 2
x
b
,
x > 2
+
⎪⎩
f must be continuous at x = 2 to be differentiable at x = 2.
lim f ( x) = lim ax3 = 8a
⎫
8a = 4 + b
⎪
⎬
2
−
8
a
4 = b
lim f ( x) = lim ( x + b) = 4 + b⎪
x → 2+
x → 2+
⎭
x → 2−
x → 2−
2
⎪⎧3ax , x < 2
f ′( x) = ⎨
x > 2
⎪⎩2 x,
For f to be differentiable at x = 2, the left derivative must equal the right derivative.
3a( 2) = 2( 2)
2
12a = 4
a =
1
3
b = 8a − 4 = − 43
x < 0
⎧cos x,
118. f ( x ) = ⎨
⎩ax + b, x ≥ 0
f (0) = b = cos(0) = 1 ⇒ b = 1
⎧−sin x, x < 0
f ′( x) = ⎨
x > 0
⎩a,
So, a = 0.
Answer: a = 0, b = 1
119. f1 ( x) = sin x is differentiable for all x ≠ nπ , n an
integer.
f 2 ( x) = sin x is differentiable for all x ≠ 0.
You can verify this by graphing f1 and f 2 and observing
the locations of the sharp turns.
120. Let f ( x) = cos x.
f ′( x) = lim
f ( x + ∆x) − f ( x)
∆x
cos x cos ∆x − sin x sin ∆x − cos x
= lim
∆x → 0
∆x
cos x(cos ∆x − 1)
⎛ sin ∆x ⎞
= lim
− lim sin x⎜
⎟
∆x → 0
∆x → 0
∆x
⎝ ∆x ⎠
∆x → 0
= 0 − sin x(1) = −sin x
121. You are given f : R → R satisfying
(*) f ′( x)
=
f ( x + n) − f ( x )
for all real numbers x and
n
all positive integers n. You claim that
f ( x) = mx + b, m, b ∈ R.
For this case,
⎡m( x + n) + b⎤⎦ − [mx + b]
= m.
f ′( x) = m = ⎣
n
Furthermore, these are the only solutions:
Note first that f ′( x + 1) =
f ( x + 2) − f ( x + 1)
,
1
and f ′( x) = f ( x + 1) − f ( x). From (*) you have
2 f ′( x ) = f ( x + 2) − f ( x)
= ⎡⎣ f ( x + 2) − f ( x + 1)⎤⎦ + ⎡⎣ f ( x + 1) − f ( x)⎤⎦
= f ′( x + 1) + f ′( x).
Thus, f ′( x) = f ′( x + 1).
Let g ( x) = f ( x + 1) − f ( x).
Let m = g (0) = f (1) − f (0).
Let b = f (0). Then
g ′( x) = f ′( x + 1) − f ′( x) = 0
g ( x) = constant = g (0) = m
f ′( x) = f ( x + 1) − f ( x) = g ( x) = m ⇒ f ( x) = mx + b.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
Product and Quotient Rules and Higher-Order Derivatives
167
Section 3.3 Product and Quotient Rules and Higher-Order Derivatives
1. g ( x ) = ( x 2 + 3)( x 2 − 4 x)
g ′( x ) = ( x 2 + 3)( 2 x − 4) + ( x 2 − 4 x )( 2 x)
= 2 x − 4 x + 6 x − 12 + 2 x − 8 x
3
2
3
2
7.
f ( x) =
f ′( x) =
x
x2 + 1
( x2
+ 1)(1) − x( 2 x)
( x2
= 4 x3 − 12 x 2 + 6 x − 12
= 2( 2 x 3 − 6 x 2 + 3 x − 6)
2. y = (3 x − 4)( x 3 + 5)
y′ = (3 x − 4)(3 x 2 ) + ( x3 + 5)(3)
= 9 x3 − 12 x 2 + 3 x3 + 15
8. g (t ) =
g ′(t ) =
=
t (1 − t 2 ) = t1 2 (1 − t 2 )
1
h′(t ) = t1 2 ( −2t ) + (1 − t 2 ) t −1 2
2
1
1
= −2t 3 2 + 1 2 − t 3 2
2t
2
5 32
1
= − t + 12
2
2t
=
1 − 5t 2
1 − 5t 2
=
2t1 2
2 t
s ( s 2 + 8) = s1 2 ( s 2 + 8)
4. g ( s ) =
1
g ′( s ) = s1 2 ( 2 s ) + ( s 2 + 8) s −1 2
2
1 32
32
−1 2
= 2s + s + 4s
2
5 32
4
= s + 12
2
s
=
5.
5s + 8
2 s
2
f ( x) = e x cos x
f ′( x) = e x ( −sin x) + e x cos x
= e (cos x − sin x )
x
6. g ( x) =
g ′( x) =
=
x sin x
⎛ 1 ⎞
x cos x + sin x⎜
⎟
⎝2 x ⎠
1
x cos x +
sin x
2 x
=
2
1 − x2
( x2
+ 1)
2
3t 2 − 1
2t + 5
(2t
+ 5)(6t ) − (3t 2 − 1)( 2)
(2t
+ 5)
2
12t 2 + 30t − 6t 2 + 2
( 2t
= 12 x3 − 12 x 2 + 15
3. h(t ) =
+ 1)
=
+ 5)
2
6t 2 + 30t + 2
(2t
+ 5)
2
x
x1 2
= 3
x +1
x +1
1 −1 2
3
( x + 1) 2 x − x1 2 (3x 2 )
h′( x) =
2
( x3 + 1)
9. h( x) =
=
3
x3 + 1 − 6 x3
2 x1 2 ( x3 + 1)
1 − 5 x3
=
2
10. f ( x ) =
f ′( x) =
=
=
=
11. g ( x ) =
g ′( x ) =
=
2
x ( x 3 + 1)
2
x2
2 x +1
(2
)
x + 1 ( 2 x) − x 2 ( x −1 2 )
(2
)
x +1
2
4 x3 2 + 2 x − x3 2
(2
)
x +1
2
3x3 2 + 2 x
(2
x (3
(2
)
x +1
2
)
x + 2
)
x +1
2
sin x
ex
e x cos x − sin x(e x )
(e x )
2
cos x − sin x
ex
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
168
12.
Chapter 3
f (t ) =
f ′(t ) =
13.
Differentiation
cos t
t3
t 3 ( −sin t ) − cos t (3t 2 )
(t )
3
2
= −
t sin t + 3 cos t
t4
f ( x) = ( x 3 + 4 x)(3x 2 + 2 x − 5)
f ′( x) = ( x 3 + 4 x)(6 x + 2) + (3x 2 + 2 x − 5)(3 x 2 + 4)
= 6 x 4 + 24 x 2 + 2 x 3 + 8 x + 9 x 4 + 6 x 3 − 15 x 2 + 12 x 2 + 8 x − 20
= 15 x 4 + 8 x 3 + 21x 2 + 16 x − 20
f ′(0) = −20
14.
y = ( x 2 − 3 x + 2)( x 3 + 1)
y′ = ( x 2 − 3 x + 2)(3 x 2 ) + ( x3 + 1)( 2 x − 3)
18.
f ′( x) =
= 3x 4 − 9 x3 + 6 x 2 + 2 x 4 − 3x3 + 2 x − 3
x2
x cos x − sin x
=
x2
= 5 x 4 − 12 x3 + 6 x 2 + 2 x − 3
y′( 2) = 5( 24 ) − 12( 23 ) + 6( 22 ) + 2( 2) − 3 = 9
15.
x − 4
x −3
(x
− 3)
19.
x − 6x + 4
(x
− 3)
f ( x) =
x − 4
x + 4
( x + 4)(1) − ( x − 4)(1)
f ′( x) =
=
=
f ′(3) =
(x
1
= −
4
+ 4)
+ 4)
2
+ 4)
8
f ( x) =
f ′( x) =
f ′(0) =
2
(3 + 4) 2
20.
=
2
8
(x
f ( x) = e x sin x
f ′(0) = 1
x + 4− x + 4
(x
=
π
)
2
= e x (cos x + sin x)
1−6+ 4
2
3π − 6
f ′( x) = e x cos x + e x sin x
2
f ′(1) =
(1 − 3)
π2
2
2
=
3
=
2
(x
(
2
2x − 6x − x + 4
2
=
17.
− 3)
)
3 3π − 18
=
− 3)( 2 x) − ( x 2 − 4)(1)
(x
f ′( x) =
16.
(
(π 6) 3 2 − (1 2)
⎛π ⎞
f ′⎜ ⎟ =
π 2 36
⎝6⎠
2
f ( x) =
sin x
x
( x)(cos x) − (sin x)(1)
f ( x) =
cos x
ex
e x ( − sin x) − cos x(e x )
(e x )
2
− sin x − cos x
ex
0 −1
= −1
1
8
49
f ( x) = x cos x
f ′( x) = ( x)( −sin x) + (cos x)(1) = cos x − x sin x
⎛π ⎞
f ′⎜ ⎟ =
⎝4⎠
π⎛ 2⎞
2
− ⎜⎜
⎟ =
2
4 ⎝ 2 ⎟⎠
2
(4 − π )
8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
Function
x 2 + 3x
21. y =
7
Product and Quotient Rules and Higher-Order Derivatives
Rewrite
1
3
y = x2 + x
7
7
Differentiate
2
3
y′ = x +
7
7
Simplify
2x + 3
y′ =
7
22. y =
5x2 − 3
4
y =
5 2
3
x −
4
4
y′ =
23. y =
6
7 x2
y =
6 −2
x
7
y′ = −
12 −3
x
7
y′ = −
12
7 x3
24. y =
10
3x3
y =
10 −3
x
3
y′ = −
30 −4
x
3
y′ = −
10
x4
25. y =
4x 3 2
x
y = 4 x1 2 , x > 0
y′ = 2 x −1 2
y′ =
2
,x > 0
x
26. y =
2x
x1 3
y = 2x2 3
y′ =
4 −1 3
x
3
y′ =
27.
f ( x) =
f ′( x) =
=
=
=
=
4 − 3x − x 2
x2 − 1
( x2
− 1)( −3 − 2 x ) − ( 4 − 3 x − x 2 )( 2 x)
( x2
− 1)
2
−3 x 2 + 3 − 2 x 3 + 2 x − 8 x + 6 x 2 + 2 x 3
(x
2
− 1)
2
3x 2 − 6 x + 3
( x2
− 1)
− 1)
(x
2
− 1) ( x + 1)
2
f ( x) =
f ′( x) =
=
=
2
3( x − 1)
y′ =
=
2
3( x 2 − 2 x + 1)
( x2
10
x
4
28.
2
=
3
(x
+ 1)
2
,x ≠ 1
169
=
5x
2
4
3 x1 3
x2 + 5x + 6
x2 − 4
2
( x − 4)(2 x + 5) − ( x 2 + 5 x + 6)(2 x)
( x2
− 4)
2
2 x3 + 5 x 2 − 8 x − 20 − 2 x3 − 10 x 2 − 12 x
( x2
− 4)
2
− 5 x 2 − 20 x − 20
( x2
− 4)
2
− 5( x 2 + 4 x + 4)
(x
− 2) ( x + 2)
2
− 5( x + 2)
(x
= −
2
− 2) ( x + 2)
2
5
(x
− 2)
2
2
2
, x ≠ 2, − 2
Alternate solution:
f ( x) =
=
=
f ′( x) =
x2 + 5x + 6
x2 − 4
( x + 3)( x + 2)
(x
+ 2)( x − 2)
x + 3
, x ≠ −2
x − 2
(x
= −
− 2)(1) − ( x + 3)(1)
(x
− 2)
2
5
(x
− 2)
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
170
29.
Chapter 3
4 ⎞
4x
⎛
f ( x) = x⎜1 −
⎟ = x −
x + 3⎠
x +3
⎝
f ′( x) = 1 −
=
=
30.
Differentiation
(x
( x2
32.
+ 3)
2
+ 6 x + 9) − 12
(x
+ 3)
2
x2 + 6x − 3
(x
+ 3)
f ( x) =
2 ⎤
⎡
⎡ x − 1⎤
= x4 ⎢
f ( x) = x 4 ⎢1 −
⎥
x + 1⎥⎦
⎣
⎣ x + 1⎦
)
x + 3 = x1 3 ( x1 2 + 3)
3
x
= x
56
(
)
x + 3
+ 3x
13
5 −1 6
x
+ x −2 3
6
5
1
=
+ 23
16
6x
x
⎡ 2 ⎤ ⎡ x2 − 1 ⎤
⎥ + ⎢
⎥ 4 x3 )
= x4 ⎢
2
2 (
⎣⎢ ( x + 1) ⎦⎥ ⎢⎣ ( x + 1) ⎦⎥
33. h( s ) = ( s 3 − 2) = s 6 − 4s 3 + 4
2
h′( s ) = 6s 5 − 12 s 2 = 6s 2 ( s 3 − 2)
2⎤
⎥
⎥⎦
34. h( x) = ( x 2 + 3) = x 6 + 9 x 4 + 27 x 2 + 27
3
3x − 1
= 3 x1 2 − x −1 2
x
3
1
3x + 1
f ′( x) = x −1 2 + x −3 2 =
2
2
2 x3 2
f ( x) =
h′( x) = 6 x5 + 36 x 3 + 54 x
= 6 x ( x 4 + 6 x 2 + 9)
= 6 x( x 2 + 3)
Alternate solution:
f ( x) =
(
f ′( x) =
⎡ ( x + 1) − ( x − 1) ⎤ ⎡ x − 1⎤
3
⎥ + ⎢
f ′( x) = x 4 ⎢
⎥(4 x )
⎢⎣
⎥⎦ ⎣ x + 1⎦
( x + 1)2
31.
x
Alternate solution:
2
⎡ 2 x2 + x −
= 2 x3 ⎢
2
⎢⎣ ( x + 1)
3
⎛1
⎞
⎛1
⎞
f ′( x) = x1 3 ⎜ x −1 2 ⎟ + ( x1 2 + 3)⎜ x −2 3 ⎟
⎝2
⎠
⎝3
⎠
5 −1 6
= x
+ x −2 3
6
5
1
=
+ 23
6 x1 6
x
+ 3)4 − 4 x(1)
(x
f ( x) =
3x − 1
3x − 1
=
x1 2
x
35.
⎛1⎞
x1 2 (3) − (3 x − 1)⎜ ⎟( x −1 2 )
⎝ 2⎠
f ′( x) =
x
1 −1 2
x (3 x + 1)
= 2
x
3x + 1
=
2 x3 2
f ( x) =
f ′( x) =
=
=
2 − (1 x)
x −3
( x2
2
2x − 1
2x − 1
= 2
x( x − 3)
x − 3x
=
− 3 x)2 − ( 2 x − 1)( 2 x − 3)
( x2
− 3 x)
2
2x2 − 6x − 4x2 + 8x − 3
( x2
− 3x)
− 2x2 + 2 x − 3
( x2
− 3 x)
2
2
=
2x2 − 2x + 3
x 2 ( x − 3)
2
x2
1 ⎞
⎛2
36. g ( x) = x 2 ⎜ −
⎟ = 2x −
x + 1⎠
x +1
⎝x
g ′( x) = 2 −
37.
(x
+ 1)2 x − x 2 (1)
(x
+ 1)
2
=
2( x 2 + 2 x + 1) − x 2 − 2 x
(x
+ 1)
2
=
x2 + 2 x + 2
(x
+ 1)
2
f ( x) = ( 2 x 3 + 5 x)( x − 3)( x + 2)
f ′( x) = (6 x 2 + 5)( x − 3)( x + 2) + ( 2 x 3 + 5 x)(1)( x + 2) + ( 2 x 3 + 5 x)( x − 3)(1)
= (6 x 2 + 5)( x 2 − x − 6) + ( 2 x3 + 5 x)( x + 2) + ( 2 x 3 + 5 x)( x − 3)
= (6 x 4 + 5 x 2 − 6 x 3 − 5 x − 36 x 2 − 30) + ( 2 x 4 + 4 x 3 + 5 x 2 + 10 x) + ( 2 x 4 + 5 x 2 − 6 x3 − 15 x)
= 10 x 4 − 8 x 3 − 21x 2 − 10 x − 30
Note: You could simplify first: f ( x ) = ( 2 x 3 + 5 x)( x 2 − x − 6)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
38.
Product and Quotient Rules and Higher-Order Derivatives
171
f ( x) = ( x3 − x)( x 2 + 2)( x 2 + x − 1)
f ′( x) = (3 x 2 − 1)( x 2 + 2)( x 2 + x − 1) + ( x3 − x)( 2 x)( x 2 + x − 1) + ( x3 − x)( x 2 + 2)( 2 x + 1)
= (3 x 4 + 5 x 2 − 2)( x 2 + x − 1) + ( 2 x 4 − 2 x 2 )( x 2 + x − 1) + ( x5 + x3 − 2 x)( 2 x + 1)
= (3 x 6 + 5 x 4 − 2 x 2 + 3x5 + 5 x3 − 2 x − 3 x 4 − 5 x 2 + 2) + ( 2 x 6 − 2 x 4 + 2 x5 − 2 x3 − 2 x 4 + 2 x 2 )
+ ( 2 x 6 + 2 x 4 − 4 x 2 + x 5 + x3 − 2 x )
= 7 x6 + 6 x5 + 4 x3 − 9 x 2 − 4 x + 2
39.
40.
f ( x) =
49. y =
f ′( x) =
( x 2 − c 2 )(2 x) − ( x 2 + c 2 )(2 x)
2
( x2 − c2 )
f ( x) =
c −x
c2 + x2
f ′( x) =
41.
x2 + c2
x2 − c2
2
= −
4 xc 2
(x
2
−c
2
)
f (t ) = t 2 sin t
50. y =
y′ =
f (θ ) = (θ + 1) cos θ
= cos θ − (θ + 1) sin θ
45.
=
sin x
x3
= cos x cot 2 x
52. y = x sin x + cos x
x3 cos x − sin x(3 x 2 )
( x3 )
2
=
x cos x − 3 sin x
x4
f ′( x) = − e x + sec 2 x
46. y = e x − cot x
y′ = e x + csc 2 x
g ′(t ) =
4
cos x
− cos x
sin 2 x
= cos x(csc 2 x − 1)
y′ = x cos x + sin x − sin x = x cos x
53.
t + 6 csc t = t1 4 + 6 csc t
1 −3 4
1
t
− 6 csc t cot t = 3 4 − 6 csc t cot t
4
4t
1
48. h( x) =
− 12 sec x = x −1 − 12 sec x
x
−1
h′( x) = − x −2 − 12 sec x tan x = 2 − 12 sec x tan x
x
f ( x) = x 2 tan x
f ′( x) = x 2 sec 2 x + 2 x tan x = x( x sec 2 x + 2 tan x)
f ( x) = − e x + tan x
47. g (t ) =
sec x( x tan x − 1)
x sec x tan x − sec x
=
2
x
x2
y′ = csc x cot x − cos x
cos t
t
−t sin t − cos t
t sin t + cos t
f ′(t ) =
= −
t2
t2
f ′( x) =
sec x
x
51. y = −csc x − sin x
43. f (t ) =
f ( x) =
(2 cos x)2
−6 cos 2 x + 6 sin x − 6 sin 2 x
4 cos 2 x
3
= ( −1 + tan x sec x − tan 2 x)
2
3
= sec x( tan x − sec x)
2
(c 2 + x 2 )(−2 x) − (c 2 − x 2 )(2 x) = − 4 xc 2
2
2
(c 2 + x 2 )
(c 2 + x 2 )
f ′(θ ) = (θ + 1)( −sin θ ) + (cos θ )(1)
44.
3 − 3 sin x
2 cos x
2 cos x
(−3 cos x)( 2 cos x) − (3 − 3 sin x)(−2 sin x)
=
=
2
f ′(t ) = t 2 cos t + 2t sin t = t (t cos t + 2 sin t )
42.
y′ =
2
3(1 − sin x )
54.
f ( x) = sin x cos x
f ′( x) = sin x( −sin x) + cos x(cos x) = cos 2 x
55. y = 2 x sin x + x 2e x
y′ = 2 x (cos x) + 2 sin x + x 2e x + 2 xe x
= 2 x cos x + 2 sin x + xe x ( x + 2)
56. h( x) = 2e x cos x
h′( x) = 2(e x cos x − e x sin x) = 2e x (cos x − sin x)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
172
Chapter 3
57. y =
y′ =
58. y =
y′ =
Differentiation
ex
4
x
4
xe x − e x 4 2
(
(4 x )
x
)
2
=
2e x
x +1
e x ⎡4
⎣
(
x − 2
16 x
)
x
x
x ⎤
⎦ = e ( 4 x − 2) = e ( 2 x − 1)
32
16 x
8 x3 2
2
( x2
+ 1)2e x − 2e x ( 2 x )
( x2
+ 1)
2
=
2e x ( x 2 − 2 x + 1)
( x2
+ 1)
2
⎛ x + 1⎞
59. g ( x) = ⎜
⎟( 2 x − 5)
⎝ x + 2⎠
⎡ ( x + 2)(1) − ( x + 1)(1) ⎤ 2 x 2 + 8 x − 1
⎛ x + 1⎞
⎥ =
g ′( x) = ⎜
⎟( 2) + ( 2 x − 5) ⎢
⎝ x + 2⎠
⎢⎣
⎥⎦
( x + 2) 2
( x + 2) 2
(Form of answer may vary.)
60.
⎛ x2 − x − 3 ⎞ 2
f ( x) = ⎜
⎟( x + x + 1)
2
⎝ x +1 ⎠
f ′( x) = 2
61. g (θ ) =
g ′(θ ) =
x5 + 2 x3 + 2 x 2 − 2
( x 2 + 1)
2
θ
1 − sin θ
1 − sin θ + θ cos θ
(1 − sin θ )
2
(Form of answer may vary.)
(Form of answer may vary.)
62.
f (θ ) =
sin θ
1 − cos θ
f ′(θ ) =
1
cos θ − 1
=
2
cos θ − 1
(1 − cos θ )
(Form of answer may vary.)
63.
y =
y′ =
1 + csc x
1 − csc x
(1 −
csc x)( −csc x cot x ) − (1 + csc x)(csc x cot x )
(1 −
csc x)
2
=
−2 csc x cot x
(1 −
csc x)
2
( )
−2( 2) 3
⎛π ⎞
y′⎜ ⎟ =
= −4 3
2
⎝6⎠
(1 − 2)
64.
f ( x) = tan x cot x = 1
66.
f ( x) = sin x(sin x + cos x)
f ′( x) = 0
f ′( x) = sin x(cos x − sin x ) + (sin x + cos x)cos x
f ′(1) = 0
= sin x cos x − sin 2 x + sin x cos x + cos 2 x
= sin 2 x + cos 2 x
sec t
65. h(t ) =
t
sec t (t tan t − 1)
t (sec t tan t ) − (sec t )(1)
=
h′(t ) =
t2
t2
sec π (π tan π − 1)
1
= 2
h′(π ) =
2
π
π
π
⎛π ⎞
f ′⎜ ⎟ = sin
+ cos
=1
2
2
⎝4⎠
π
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
67. (a)
f ( x) = ( x3 + 4 x − 1)( x − 2),
Product and Quotient Rules and Higher-Order Derivatives
(1, − 4)
f ′( x) = ( x3 + 4 x − 1)(1) + ( x − 2)(3 x 2 + 4)
70. (a)
f ′( x) =
= x3 + 4 x − 1 + 3x3 − 6 x 2 + 4 x − 8
= 4 x3 − 6 x 2 + 8 x − 9
( x − 3)
2
= −
6
( x − 3)
2
−6
= − 6; Slope at ( 4, 7)
1
Tangent line:
y − 7 = − 6( x − 4) ⇒ y = − 6 x + 31
Tangent line: y + 4 = −3( x − 1) ⇒ y = −3x − 1
3
8
(b)
−1
x + 3
,
(4, 7)
x −3
( x − 3)(1) − ( x + 3)(1)
f ′( 4) =
f ′(1) = −3; Slope at (1, − 4)
(b)
f ( x) =
173
(4, 7)
3
−5
10
(1, − 4)
−6
(c) Graphing utility confirms
68. (a)
dy
= −3 at (1, − 4).
dx
f ( x) = ( x − 2)( x 2 + 4), (1, − 5)
−8
(c) Graphing utility confirms
f ( x) = tan x,
71. (a)
f ′( x) = ( x − 2)( 2 x) + ( x + 4)(1)
2
f ′( x) = sec 2 x
= 2x2 − 4x + x2 + 4
⎛π ⎞
f ′⎜ ⎟ = 2;
⎝4⎠
= 3x 2 − 4 x + 4
f ′(1) = −3; Slope at (1, − 5)
(b)
⎛π ⎞
⎜ , 1⎟
⎝4 ⎠
⎛π ⎞
Slope at ⎜ , 1⎟
⎝4 ⎠
π⎞
⎛
y − 1 = 2⎜ x − ⎟
4⎠
⎝
Tangent line:
Tangent line: y − ( − 5) = 3( x − 1) ⇒ y = 3 x − 8
y − 1 = 2x −
3
−3
4x − 2 y − π + 2 = 0
6
(1, −5)
(b)
f ′( x) =
f ′( −5) =
( (
−␲
x
,
( −5, 5)
x + 4
( x + 4)(1) − x(1)
(x
+ 4)
4
( −5 + 4)
2
2
= 4;
␲
−4
(c) Graphing utility confirms
=
4
(x
+ 4)
2
Slope at ( −5, 5)
⎛π ⎞
⎜ , 2⎟
⎝3 ⎠
′
f ( x ) = sec x tan x
⎛π ⎞
f ′⎜ ⎟ = 2 3;
⎝3⎠
Tangent line:
⎛π ⎞
Slope at ⎜ , 2 ⎟
⎝3 ⎠
8
π⎞
⎛
y − 2 = 2 3⎜ x − ⎟
3⎠
⎝
(− 5, 5)
−8
dy
⎛π ⎞
= 2 at ⎜ , 1⎟.
dx
⎝4 ⎠
f ( x ) = sec x,
72. (a)
Tangent line: y − 5 = 4( x + 5) ⇒ y = 4 x + 25
(b)
2
4
dy
(c) Graphing utility confirms
= 3 at (1, − 5).
dx
f ( x) =
π
π
,1
4
− 15
69. (a)
dy
= − 6 at ( 4, 7).
dx
6 3 x − 3 y + 6 − 2 3π = 0
1
(b)
−6
(c) Graphing utility confirms
dy
= 4 at ( −5, 5).
dx
6
( π3 , 2(
−␲
␲
−2
(c) Graphing utility confirms
dy
⎛π ⎞
= 2 3 at ⎜ , 2 ⎟.
dx
⎝3 ⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
174
Chapter 3
73. (a)
Differentiation
f ( x) = ( x − 1)e x ,
(1, 0)
f ′( x) = ( x − 1)e + e = e
x
x
76.
f ′(1) = e
( x2
f ′( x) =
(b)
+ 9)(0) − 27( 2 x)
(x
Tangent line: y − 0 = e( x − 1)
y = e( x − 1)
3⎞
⎛
⎜ −3, ⎟
2⎠
⎝
27
;
x2 + 9
f ( x) =
x
−54( −3)
f ′( −3) =
(9
3
+ 9)
+ 9)
2
2
2
−54 x
=
(x
2
+ 9)
2
1
2
=
3
1
= ( x + 3)
2
2
1
y = x +3
2
2y − x − 6 = 0
y −
−3
3
(1, 0)
−3
(c) Graphing utility confirms
dy
= e at (1, 0).
dx
77.
74. (a)
f ( x) =
f ′( x) =
⎛ 1⎞
⎜ 0, ⎟
⎝ 4⎠
ex
,
x + 4
(x
+ 4) e x − e x
(x
+ 4)
2
=
f ′( x ) =
e x ( x + 3)
(x
+ 4)
2
f ′( −2) =
3
f ′(0) =
16
( x2
+ 16)(16) − 16 x( 2 x )
( x2
+ 16)
256 − 16( 4)
20
=
2
2
=
256 − 16 x 2
( x2
+ 16)
2
12
25
8
12
=
( x + 2)
5
25
12
16
y =
x −
25
25
25 y − 12 x + 16 = 0
1.5
78.
−2
2
(c) Graphing utility confirms
f ( x) =
f ′( x) =
f ′( 2) =
8
;
x + 4
2
( x2
(x
2
−16( 2)
(4
+ 4)
2
+ 4)
= −
2
1
2
1
( x − 2)
2
1
y = − x + 2
2
2y + x − 4 = 0
y −1 = −
dy
3
⎛
at ⎜ 0,
=
dx
16
⎝
1⎞
⎟.
4⎠
f ′( 2) =
4x ⎛ 4 ⎞
; ⎜ 2, ⎟
x2 + 6 ⎝ 5 ⎠
( x2
+ 6)( 4) − 4 x( 2 x)
( x2
+ 6)
2
=
24 − 4 x 2
( x2
+ 6)
2
24 − 16
2
=
102
25
4
2
=
( x − 2)
5
25
2
16
y =
x +
25
25
25 y − 2 x − 16 = 0
y −
(2, 1)
+ 4)(0) − 8( 2 x)
f ( x) =
f ′( x) =
−0.5
75.
8⎞
⎛
⎜ −2, − ⎟
5⎠
⎝
16 x
;
x + 16
2
y +
1
3
Tangent line: y −
=
( x − 0)
4
16
3
1
y =
x +
16
4
(b)
f ( x) =
=
−16 x
(x
2
+ 4)
2
79.
2x − 1
= 2 x −1 − x −2
x2
2( − x + 1)
f ′( x) = −2 x −2 + 2 x −3 =
x3
f ( x) =
f ′( x) = 0 when x = 1, and f (1) = 1.
Horizontal tangent at (1, 1).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
80.
Product and Quotient Rules and Higher-Order Derivatives
x2
x +1
f ( x) =
84. f ( x) =
2
( x2
f ′( x) =
+ 1)( 2 x) − ( x 2 )( 2 x)
( x2
+ 1)
2
=
2x
( x2
+ 1)
2
f ′( x) = 0 when x = 0.
f ′( x ) =
e x (8) − 8( x − 2)e x
e2 x
f ( x) = e x sin x,
=
−4
x
x
−2
4
3π
.
4
⎞
⎟⎟.
⎠
x +1
83. f ( x) =
x −1
( x − 1) − ( x + 1) = −2
f ′( x) =
2
2
( x − 1)
( x − 1)
1
1
2 y + x = 6 ⇒ y = − x + 3; Slope: −
2
2
−2
1
= −
2
2
x
−
1
(
)
− 1) = 4
2
(x
(4 x
x = −1, 3; f ( −1) = 0, f (3) = 2
1
1
1
( x + 1) ⇒ y = − x −
2
2
2
1
1
7
y − 2 = − ( x − 3) ⇒ y = − x +
2
2
2
f(x) =
x+1
x−1
(3, 2)
x
−2
2
4
)
=
−1
(x
− 1)
2
− 5)( x − 1) = x + 1
4 x − 10 x + 4 = 0
2
(x
− 2)( 2 x − 1) = 0 ⇒ x =
1
,2
2
⎛1⎞
⎛1⎞
f ⎜ ⎟ = −1, f ( 2) = 2; f ′⎜ ⎟ = −4, f ′( 2) = −1
⎝ 2⎠
⎝ 2⎠
Two tangent lines:
1⎞
⎛
y + 1 = −4⎜ x − ⎟ ⇒ y = −4 x + 1
2⎠
⎝
y − 2 = −1( x − 2)
85. f ′( x) =
x − 1 = ±2
y
(
4x − 5
1
=
− 1)( x + 1)
x
−
( 1)2
g ′( x) =
6
4
1
, −1
2
−1 − x
f ′( x) = 0 when cos x = − sin x ⇒ x =
⎛ 3π
2 3π
Horizontal tangent is at ⎜ ,
⎜ 4 2 e
⎝
2
5 − ( x ( x − 1))
x
y −0 = −
(2, 2)
−2
f ′( x) = e cos x + e sin x = e (cos x + sin x )
−6 −4 −2
2
x
x−1
f(x) =
y = −4x + 1
24 − 8 x
ex
0 ≤ x ≤ π
x
(−1, 0)
− 1)
y = −x + 4
ex
2y + x = 7
(x
6
Horizontal tangent is at (3, 8e − 3 ).
(x
y
(−1, 5)
g ′( x) = 0 when x = 3.
82.
− 1)
Let ( x, y ) = ( x, x ( x − 1)) be a point of tangency on the
8( x − 2)
g ′( x) =
(x
−1
=
2
graph of f.
Horizontal tangent is at (0, 0).
81. g ( x) =
x
x −1
( x − 1) − x
175
g ( x) =
(x
+ 2)3 − 3x(1)
(x
(x
⇒ y = −x + 4
+ 2)
2
+ 2)5 − (5 x +
(x
+ 2)
2
=
6
( x + 2)
4)(1)
=
(x
2
6
+ 2)
2
5x + 4
3x
2x + 4
=
+
= f ( x) + 2
x
x
2
2
+
+
) ( x + 2)
(
) (
f and g differ by a constant.
x(cos x − 3) − (sin x − 3x )(1) x cos x − sin x
=
x2
x2
x(cos x + 2) − (sin x + 2 x)(1) x cos x − sin x
=
g ′( x) =
x2
x2
sin x + 2 x sin x − 3x + 5 x
=
= f ( x) + 5
g ( x) =
x
x
86. f ′( x) =
6
f and g differ by a constant.
−4
−6
2y + x = −1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
176
Chapter 3
Differentiation
87. (a) p′( x ) = f ′( x ) g ( x) + f ( x ) g ′( x)
⎛ 1⎞
p′(1) = f ′(1) g (1) + f (1) g ′(1) = 1( 4) + 6⎜ − ⎟ = 1
⎝ 2⎠
(b) q′( x) =
q′( 4) =
g ( x) f ′( x) − f ( x) g ′( x)
g ( x)
2
3( −1) − 7(0)
2
3
= −
1
3
88. (a) p′( x) = f ′( x) g ( x) + f ( x) g ′( x)
p′( 4) =
(b) q′( x) =
q′(7) =
1
(8) + 1(0) = 4
2
g ( x) f ′( x) − f ( x) g ′( x)
g ( x)
4( 2) − 4( −1)
4
2
2
=
12
3
=
16
4
89. Area = A(t ) = (6t + 5) t = 6t 3 2 + 5t1 2
A′(t ) = 9t1 2 +
5 −1 2
18t + 5
t
=
cm 2 /sec
2
2 t
⎛1
90. V = π r 2 h = π (t + 2)⎜
⎝2
V ′(t ) =
91.
1
⎞
t ⎟ = (t 3 2 + 2t1 2 )π
2
⎠
1⎛ 3 1 2
3t + 2
−1 2 ⎞
π in.3 /sec
⎜ t + t ⎟π =
2⎝ 2
4t1 2
⎠
x ⎞
⎛ 200
C = 100⎜ 2 +
⎟, 1 ≤ x
x + 30 ⎠
⎝ x
⎛ 400
⎞
dC
30
⎟
= 100⎜ − 3 +
2
⎜ x
dx
( x + 30) ⎟⎠
⎝
dC
= −$38.13 thousand 100 components
dx
dC
(b) When x = 15:
= −$10.37 thousand 100 components
dx
dC
(c) When x = 20:
= −$3.80 thousand 100 components
dx
As the order size increases, the cost per item decreases.
(a) When x = 10:
92.
4t ⎤
⎡
P(t ) = 500 ⎢1 +
50 + t 2 ⎥⎦
⎣
⎡ 50 + t 2 ( 4) − ( 4t )( 2t ) ⎤
⎡
⎡
2⎤
2 ⎤
(
)
⎥ = 500 ⎢ 200 − 4t ⎥ = 2000 ⎢ 50 − t ⎥
P′(t ) = 500 ⎢
2
2
⎢
⎥
⎢ 50 + t 2 ⎥
⎢ 50 + t 2 2 ⎥
(50 + t 2 )
)⎦
)⎦
⎣
⎦
⎣(
⎣(
P′( 2) ≈ 31.55 bacteria/h
93. (a)
cot x =
cos x
sin x
sin x( −sin x) − (cos x)(cos x )
d
d ⎡ cos x ⎤
sin 2 x + cos 2 x
1
= −
= − 2 = −csc 2 x
[cot x] = ⎢
⎥ =
2
sin 2 x
sin x
dx
dx ⎣ sin x ⎦
(sin x)
(b)
sec x =
1
cos x
d
d ⎡ 1
[sec x] = ⎢
dx
dx ⎣ cos
(c)
csc x =
⎤
(cos x)(0) − (1)(−sin x) = sin x = 1 ⋅ sin x = sec x tan x
⎥ =
cos x cos x
cos x cos x
x⎦
(cos x)2
1
sin x
d
d ⎡ 1 ⎤
(sin x)(0) − (1)(cos x) = − cos x = − 1 ⋅ cos x = −csc x cot x
[csc x] = ⎢
⎥ =
sin x sin x
sin x sin x
dx
dx ⎣ sin x ⎦
(sin x)2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
94.
Product and Quotient Rules and Higher-Order Derivatives
177
f ( x) = sec x
g ( x) = csc x, [0, 2π )
f ′( x) = g ′( x )
1
sin x
⋅
sec x tan x
sin 3 x
cos x cos x
sec x tan x = −csc x cot x ⇒
= −1 ⇒
= −1 ⇒
= −1 ⇒ tan 3 x = −1 ⇒ tan x = −1
3
1
cos x
csc x cot x
cos
x
⋅
sin x sin x
3π 7π
,
4 4
x =
95. (a) h(t ) = 112.4t + 1332
98.
p(t ) = 2.9t + 282
(b)
f ( x) = 4 x5 − 2 x3 + 5 x 2
f ′( x) = 20 x 4 − 6 x 2 + 10 x
f ′′( x) = 80 x 3 − 12 x + 10
400
3000
p(t)
h(t)
99.
f ( x) = 4 x3 2
f ′( x) = 6 x1 2
2
2
10
10
f ′′( x) = 3 x −1 2 =
0
0
(c) A =
112.4t + 1332
2.9t + 282
100.
3
x
f ( x) = x 2 + 3 x −3
f ′( x) = 2 x − 9 x − 4
10
f ′′( x) = 2 + 36 x − 5 = 2 +
2
10
101.
0
A represents the average health care expenses per
person (in thousands of dollars).
(d) A′(t ) ≈
3407.5
(t + 98.53)2
f ( x) =
f ′( x) =
27,834
≈
8.41t 2 + 1635.6t + 79,524
f ′′( x) =
A′(t ) represents the rate of change of the average
health care expenses per person per year t.
96. (a)
r
r + h
r + h = r csc θ
sin θ =
=
h′(θ ) = r ( −csc θ ⋅ cot θ )
⎛π ⎞
h′(30°) = h′⎜ ⎟
⎝6⎠
f ′′( x) =
(
= −3960 2 ⋅
97.
f ( x) =
f ′( x) =
h = r csc θ − r = r (csc θ − 1)
(b)
102.
)
3 = −7920 3 mi/rad
=
f ( x) = x 4 + 2 x3 − 3 x 2 − x
f ′( x) = 4 x 3 + 6 x 2 − 6 x − 1
f ′′( x) = 12 x 2 + 12 x − 6
=
=
=
x
x −1
( x − 1)(1) − x(1)
(x
− 1)
36
x5
=
2
−1
(x
− 1)
2
2
(x
− 1)
3
x 2 + 3x
x − 4
(x
− 4)( 2 x + 3) − ( x 2 + 3 x)(1)
(x
− 4)
2
2 x 2 − 5 x − 12 − x 2 − 3 x
−
(x
−
(x
−
x 2 − 8 x − 12
x 2 − 8 x + 16
( x − 4)
2
4) ( 2 x − 8) − ( x 2 − 8 x − 12)( 2 x − 8)
( x − 4)4
4) ⎡⎣( x − 4)( 2 x − 8) − 2( x 2 − 8 x − 12)⎤⎦
( x − 4) 4
4)( 2 x − 8) − 2( x 2 − 8 x − 12)
3
( x − 4)
2
(x
=
2 x 2 − 16 x + 32 − 2 x 2 + 16 x + 24
(x
− 4)
3
56
(x
− 4)
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
178
103.
Chapter 3
Differentiation
f ( x) = x sin x
112.
f ′( x) = − h′( x )
f ′′( x) = x( −sin x) + cos x + cos x
f ′( 2) = − h′( 2) = −4
= − x sin x + 2 cos x
113.
104.
f ( x ) = 4 − h( x )
f ′( x) = x cos x + sin x
f ( x) = sec x
f ′( x) =
f ′( x) = sec x tan x
f ′′( x) = sec x(sec 2 x) + tan x(sec x tan x)
f ′( 2) =
= sec x(sec 2 x + tan 2 x)
105. g ( x) =
g ′( x) =
=
ex
x
g ( x)
h( x )
h( x) g ′( x) − g ( x)h′( x )
⎡⎣h( x )⎤⎦
h( 2) g ′( 2) − g ( 2)h′( 2)
2
⎡⎣h( 2)⎤⎦
(−1)(−2) − (3)(4)
2
( −1)
2
= −10
xe x − e x
x2
x 2 ( xe x + e x − e x ) − 2 x( xe x − e x )
g ′′( x) =
=
f ( x) =
x4
ex 2
( x − 2 x + 2)
x3
106. h(t ) = e sin t
114.
f ( x ) = g ( x ) h( x )
f ′( x) = g ( x )h′( x) + h( x) g ′( x)
f ′( 2) = g ( 2)h′( 2) + h( 2) g ′( 2)
= (3)( 4) + ( −1)( −2)
= 14
t
h′(t ) = et cos t + et sin t = et (cos t + sin t )
h′′(t ) = et ( − sin t + cos t ) + et (cos t + sin t )
= 2e cos t
t
115. The graph of a differentiable function f such that
f ( 2) = 0, f ′ < 0 for −∞ < x < 2, and f ′ > 0 for
2 < x < ∞ would, in general, look like the graph
below.
y
107.
f ′( x) = x
2
4
f ′′( x) = 2 x
3
2
108.
f ′′( x) = 2 − 2 x
−1
f ′′′( x) = 2 x −2 =
109.
f ′′′( x) = 2
f ( 4) ( x ) =
1
2
x2
x
1
3
2
4
One such function is f ( x ) = ( x − 2) .
2
x
1
(2) x −1 2 =
2
110. f (4) ( x) = 2 x + 1
1
x
116. The graph of a differentiable function f such that
f > 0 and f ′ < 0 for all real numbers x would, in
general, look like the graph below.
y
f (5) ( x) = 2
f ( 6) ( x ) = 0
111.
f ( x ) = 2 g ( x ) + h( x )
f
x
f ′( x) = 2 g ′( x) + h′( x)
f ′( 2) = 2 g ′( 2) + h′( 2)
= 2( −2) + 4
= 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
y
117.
It appears that f is cubic, so
f ′ would be quadratic and
f ′′ would be linear.
f′
2
f
1
Product and Quotient Rules and Higher-Order Derivatives
121.
179
y
f′
f″
1
−2
x
−1
1
2
π
2
−1
−2
x
2π
−3
−4
f″
y
118.
f′
3
f
f ′′
−2
−1
y
122.
It appears that f is quadratic
so f ′ would be linear and
f ′′ would be constant.
2
1
−1
3
f′
π
2
x
2
f ′′
4
x
π
−2
123. v(t ) = 36 − t 2 , 0 ≤ t ≤ 6
y
119.
a(t ) = v′(t ) = −2t
4
3
2
1
f′
−3 −2 −1
v(3) = 27 m/sec
a(3) = −6 m/sec2
x
1 2 3 4 5
The speed of the object is decreasing.
f″
−3
−4
−5
124. v(t ) =
120.
y
f ′′
a(t ) = v′(t ) =
3
f′
2
(a) a(5) =
1
−4
−3
100t
2t + 15
(2t + 15)(100) − (100t )(2)
2
(2t + 15)
1500
⎡⎣2(5) + 15⎤⎦
=
1500
(2t + 15)
2
= 2.4 ft/sec 2
2
x
−1
(b) a(10) =
−1
(c) a( 20) =
125. s(t ) = −8.25t 2 + 66t
v(t ) = s′(t ) = 16.50t + 66
a(t ) = v′(t ) = −16.50
1500
⎡⎣2(10) + 15⎤⎦
≈ 1.2 ft/sec 2
2
1500
⎡⎣2( 20) + 15⎤⎦
2
≈ 0.5 ft/sec 2
t(sec)
0
1
2
3
4
s(t) (ft)
0
57.75
99
123.75
132
66
49.5
33
16.5
0
–16.5
–16.5
–16.5
–16.5
–16.5
v(t ) = s′(t ) (ft/sec)
a(t ) = v′(t ) (ft/sec
2
)
Average velocity on:
57.75 − 0
= 57.75
[0, 1] is
1−0
99 − 57.75
= 41.25
[1, 2] is
2 −1
123.75 − 99
= 24.75
[2, 3] is
3− 2
132 − 123.75
= 8.25
[3, 4] is
4−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
180
Chapter 3
126. (a)
Differentiation
s position function
y
v velocity function
16
12
a acceleration function
s
8
v
4
t
−1
1
4
5
6
7
a
(b) The speed of the particle is the absolute value of its velocity. So, the particle’s speed is slowing
down on the intervals (0, 4 3) and (8 3, 4) and it speeds up on the intervals ( 4 3, 8 3) and ( 4, 6).
16
t= 8
3
12
8
v
speed
4
t
−4
1
3
5
6
7
−8
−12
−16
t=4
t= 4
3
f ( x) = x n
127.
f ( x) =
128.
f (n) ( x) = n( n − 1)( n − 2) " ( 2)(1) = n!
f ( n) ( x ) =
Note: n! = n( n − 1) " 3 ⋅ 2 ⋅ 1 ( read “ n factorial”)
1
x
(−1)n (n)(n
− 1)( n − 2)"( 2)(1)
(−1) n!
=
n +1
x
x n +1
n
129. f ( x) = g ( x)h( x)
(a)
f ′( x) = g ( x )h′( x) + h( x) g ′( x)
f ′′( x) = g ( x)h′′( x) + g ′( x)h′( x) + h( x) g ′′( x) + h′( x) g ′( x)
= g ( x)h′′( x) + 2 g ′( x)h′( x) + h( x) g ′′( x)
f ′′′( x) = g ( x)h′′′( x) + g ′( x)h′′( x) + 2 g ′( x)h′′( x) + 2 g ′′( x)h′( x) + h( x) g ′′′( x) + h′( x) g ′′( x)
= g ( x)h′′′( x) + 3 g ′( x)h′′( x) + 3 g ′′( x)h′( x) + g ′′′( x)h( x)
f (4) ( x) = g ( x )h(4) ( x) + g ′( x)h′′′( x) + 3 g ′( x)h′′′( x) + 3 g ′′( x)h′′( x) + 3 g ′′( x )h′′( x) + 3 g ′′′( x )h′( x)
+ g ′′′( x)h′( x ) + g (4) ( x)h( x )
= g ( x )h(4) ( x) + 4 g ′( x)h′′′( x) + 6 g ′′( x)h′′( x) + 4 g ′′′( x )h′( x) + g (4) ( x)h( x)
(b) f (n) ( x) = g ( x)h(n) ( x) +
+
n( n − 1)( n − 2) " ( 2)(1)
1⎡⎣( n − 1)( n − 2) " ( 2)(1)⎤⎦
g ′( x)h(n −1) ( x) +
n( n − 1)( n − 2) " ( 2)(1)
(3)( 2)(1) ⎡⎣( n − 3)( n − 4) " (2)(1)⎤⎦
n( n − 1)( n − 2) " ( 2)(1)
(2)(1)⎡⎣(n
− 2)( n − 3) " ( 2)(1)⎤⎦
g ′′( x)h(n − 2) ( x)
g ′′′( x )h(n − 3) ( x) + "
n( n − 1)( n − 2) " ( 2)(1) (n −1)
g
( x)h′( x) + g (n) ( x)h( x)
⎡⎣( n − 1)( n − 2) " ( 2)(1)⎤⎦ (1)
n!
n!
= g ( x ) h ( n) ( x ) +
g ′( x )h(n −1) ( x) +
g ′′( x)h(n − 2) ( x) + "
1!( n − 1)!
2!( n − 2)!
+
+
(n
n!
g (n −1) ( x )h′( x) + g (n) ( x)h( x )
− 1)!1!
Note: n! = n( n − 1)"3 ⋅ 2 ⋅ 1 (read “n factorial”)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.3
130.
Product and Quotient Rules and Higher-Order Derivatives
181
⎡⎣ xf ( x )⎤⎦′ = xf ′( x) + f ( x)
⎡⎣ xf ( x)⎤⎦′′ = xf ′′( x ) + f ′( x ) + f ′( x) = xf ′′( x) + 2 f ′( x)
⎡⎣ xf ( x)⎤⎦′′′ = xf ′′′( x) + f ′′( x) + 2 f ′′( x) = xf ′′′( x) + 3 f ′′( x)
In general, ⎡⎣ xf ( x)⎤⎦
131.
( n)
= xf (n) ( x) + nf (n −1) ( x).
f ( x) = x n sin x
f ′( x) = x cos x + nx
n
y = 2 sin x + 3
135.
n −1
y′ = 2 cos x
sin x
y′′ = −2 sin x
When n = 1: f ′( x) = x cos x + sin x
y′′ + y = −2 sin x + ( 2 sin x + 3) = 3
When n = 2: f ′( x) = x 2 cos x + 2 sin x
When n = 3: f ′( x) = x 3 cos x + 3 x 2 sin x
y = 3 cos x + sin x
136.
y′ = −3 sin x + cos x
When n = 4: f ′( x) = x 4 cos x + 4 x 3 sin x
y′′ = −3 cos x − sin x
For general n, f ′( x) = x n cos x + nx n −1 sin x.
132.
f ( x) =
cos x
= x − n cos x
xn
y′′ + y = ( −3 cos x − sin x) + (3 cos x + sin x) = 0
137. False. If y = f ( x) g ( x), then
dy
= f ( x ) g ′( x) + g ( x) f ′( x ).
dx
f ′( x) = − x − n sin x − nx − n −1 cos x
= − x − n −1 ( x sin x + n cos x)
= −
x sin x + n cos x
x n +1
When n = 1: f ′( x ) = −
x sin x + cos x
x2
138. True. y is a fourth-degree polynomial.
dny
= 0 when n > 4.
dx n
139. True
x sin x + 2 cos x
When n = 2: f ′( x ) = −
x3
When n = 3: f ′( x) = −
x sin x + 3 cos x
x4
When n = 4: f ′( x ) = −
x sin x + 4 cos x
x5
For general n, f ′( x ) = −
x sin x + n cos x
.
x n +1
1
1
2
, y′ = − 2 , y′′ = 3
x
x
x
⎡2⎤
⎡ 1⎤
x3 y′′ + 2 x 2 y′ = x3 ⎢ 3 ⎥ + 2 x 2 ⎢− 2 ⎥ = 2 − 2 = 0
⎣x ⎦
⎣ x ⎦
133. y =
134.
y = 2 x3 − 6 x + 10
y′ = 6 x 2 − 6
y′′ = 12 x
y′′′ = 12
− y′′′ − xy′′ − 2 y′ = –12 − x(12 x) − 2(6 x 2 − 6) = − 24 x 2
h′(c) = f (c) g ′(c) + g (c ) f ′(c)
= f (c)(0) + g (c)(0)
= 0
140. True
141. True
142. True. If v(t ) = c then a(t ) = v′(t ) = 0.
143.
⎧⎪x 2 , x ≥ 0
f ( x) = x x = ⎨ 2
⎪⎩− x , x < 0
⎧2 x, x > 0
f ′( x) = ⎨
= 2 x
⎩− 2 x, x < 0
⎧2, x > 0
f ′′( x) = ⎨
⎩− 2, x < 0
f ′′(0) does not exist because the left and right
derivatives do not agree at x = 0.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
182
Chapter 3
f ′g )′ = fg ′′ + f ′g ′ − f ′g ′ − f ′′g
( fg ′ −
144. (a)
Differentiation
= fg ′′ − f ′′g
(b)
( fg )′′
=
( fg ′ +
f ′g )′
= fg ′′ + f ′g ′ + f ′g ′ + f ′′g
True
= fg ′′ + 2 f ′g ′ + f ′′g
≠ fg ′′ + f ′′g
145.
False
d
d
⎡( f ( x) g ( x))h( x)⎤⎦
⎡⎣ f ( x) g ( x)h( x)⎤⎦ =
dx
dx ⎣
d
=
⎡ f ( x) g ( x)⎤⎦ h( x) + f ( x) g ( x)h′( x)
dx ⎣
= ⎡⎣ f ( x) g ′( x) + g ( x) f ′( x)⎤⎦ h( x) + f ( x) g ( x)h′( x)
= f ′( x ) g ( x)h( x) + f ( x) g ′( x)h( x) + f ( x) g ( x)h′( x)
Section 3.4 The Chain Rule
y = f ( g ( x))
u = g ( x)
y = f (u )
u = 5x − 8
y = u4
u = x +1
y = u −1 2
3. y = csc3 x
u = csc x
y = u3
4. y = 3 tan (π x 2 )
u = π x2
y = 3 tan u
5. y = e − 2 x
u = − 2x
y = eu
u = ln x
y = u3
1. y = (5 x − 8)
2. y =
4
1
x +1
6. y = (ln x)
3
7. y = ( 4 x − 1)
12. g ( x) =
3
y′ = 3( 4 x − 1) ( 4) = 12( 4 x − 1)
2
8. y = 5( 2 − x3 )
2
g ′( x) =
) ( −3 x )
3
= 60 x 2 ( x3 − 2)
3
3
9. g ( x) = 3( 4 − 9 x)
= −60 x ( 2 − x
2
2
g ′( x) = 12( 4 − 9 x ) ( −9) = −108( 4 − 9 x )
f (t ) = (9t + 2)
23
f ′(t ) =
2
−1 3
(9t + 2) (9) =
3
f (t ) =
5 − t = (5 − t )
f ′(t ) =
3
)
3
13. y =
y′ =
4
3
11.
12
−1 2
1
4 − 3 x 2 ) ( −6 x) = −
(
2
3x
4 − 3x 2
4
y′ = 5( 4)( 2 − x
10.
4 − 3x 2 = (4 − 3x 2 )
3
6
9t + 2
12
1
−1
(5 − t )−1 2 (−1) =
2
2 5−t
3
14.
3
6 x 2 + 1 = (6 x 2 + 1)
13
−2 3
1 2
(6 x + 1) (12 x) = 2 4 x 2 3 =
3
(6 x + 1)
f ( x) =
f ′( x) =
4x
3
(6 x 2 + 1)
x 2 − 4 x + 2 = ( x 2 − 4 x + 2)
12
−1 2
1 2
( x − 4 x + 2) ( 2 x − 4) =
2
x−2
x − 4x + 2
2
15. y = 2 4 9 − x 2 = 2(9 − x 2 )
14
−3 4
⎛1⎞
y′ = 2⎜ ⎟(9 − x 2 ) ( −2 x)
⎝ 4⎠
−x
−x
=
=
34
3
2
4 9 − x2
(9 − x )
(
)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
2
Section 3.4
16.
f ( x) =
3
12 x − 5 = (12 x − 5)
13
24.
1
4
(12 x − 5)− 2 3 (12) =
23
3
−
12
x
5)
(
f ′( x) =
17. y = ( x − 2)
−2
(1)
=
2
−1
( x − 2)
(4 − 5t − t 2 )
f (t ) = (t − 3)
3
−1 2
12
⎡1
⎤
y′ = x ⎢ (1 − x 2 ) ( −2 x)⎥ + (1 − x 2 ) (1)
2
⎣
⎦
= − x 2 (1 − x 2 )
2
=
= (1 − x 2 )
2t + 5
(t 2
+ 5t − 4)
2
(1)
−3
=
(t
− 3)
−5
3
−4
22. g (t ) =
12
=
(t
− 2)
27. y =
+ 5)
3
−1 2
t
(t 2 − 2)
=
1 − 2x2
1 − x2
f ( x ) = x 2 ( x − 2)
− x3
+ x 16 − x 2 =
2 16 − x 2
− x(3 x 2 − 32)
2 16 − x 2
3
4
x
x +1
( x2
+ 1)
( x2
+ 1)
( x2
+ 1)
x
=
2
12
(x
2
+ 1)
12
−1 2
⎛1⎞
− x⎜ ⎟( x 2 + 1) ( 2 x)
2
⎝ ⎠
⎡ x2 + 1 1 2 ⎤ 2
) ⎥⎦
⎢⎣(
(1)
− x 2 ( x 2 + 1)
12
−1 2
x2 + 1
−1 2
⎡⎣ x 2 + 1 − x 2 ⎤⎦
x +1
1
1
=
=
32
3
2
( x + 1)
( x 2 + 1)
=
= (t 2 − 2)
1
2
y′ =
28. y =
y′ =
=
23.
⎡− x 2 + (1 − x 2 )⎤
⎣
⎦
−1 2
5
t − 2
−3 2
1 2
g ′(t ) = − (t − 2) ( 2t )
2
−t
=
32
2
( t − 2)
= −
12
1 2
x 16 − x 2
2
12
−1 2
1 ⎛1
⎞
y′ = x 2 ⎜ (16 − x 2 ) ( −2 x) ⎟ + x(16 − x 2 )
2 ⎝2
⎠
=
3
(3 x
=
+ (1 − x 2 )
−1 2
26. y =
−2
1
−1 2
21. y =
= (3 x + 5)
3x + 5
1
−3 2
y′ = − (3 x + 5) (3)
2
−3
=
32
2(3 x + 5)
2
2
12
−2
= − 3(t − 2)
4
y′ = 12(t − 2)
= −
= ( 2 x − 5) ⎡⎣6 x + ( 2 x − 5)⎤⎦
−2
f ′(t ) = −2(t − 3)
− 2)
3
25. y = x 1 − x 2 = x(1 − x 2 )
2
) (− 5 − 2t )
5 + 2t
(t
2
2
s′(t ) = − ( 4 − 5t − t
20. y = −
3
= ( 2 x − 5) (8 x − 5)
−1
1
18. s(t ) =
= ( 4 − 5t − t 2 )
2
4 − 5t − t
19.
183
f ′( x) = x(3)( 2 x − 5) ( 2) + ( 2 x − 5) (1)
−1
y′ = −1( x − 2)
=
f ( x) = x( 2 x − 5)
The Chain Rule
2
x
x + 4
4
( x4
+ 4)
12
(1)
x4 + 4 − 2 x4
( x4
+ 4)
32
−1 2
1 4
( x + 4) ( 4 x 3 )
2
x4 + 4
− x
=
4 − x4
( x4
+ 4)
32
=
4 − x4
( x4
+ 4)
3
3
4
f ′( x) = x 2 ⎡4( x − 2) (1)⎤ + ( x − 2) ( 2 x)
⎣
⎦
= 2 x( x − 2) ⎡⎣2 x + ( x − 2)⎤⎦
3
= 2 x ( x − 2) ( 3 x − 2)
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
184
Chapter 3
Differentiation
⎛ x +5⎞
29. g ( x) = ⎜ 2
⎟
⎝ x + 2⎠
2
31.
⎛ 2
⎞
⎛ x + 5 ⎞⎜ ( x + 2) − ( x + 5)( 2 x) ⎟
g ′( x) = 2⎜ 2
⎟
2
⎟⎟
⎝ x + 2 ⎠⎜⎜
( x 2 + 2)
⎝
⎠
=
=
⎛ 1 − 2v ⎞
f (v ) = ⎜
⎟
⎝1+ v⎠
⎛ 1 − 2v ⎞
f ′(v ) = 3⎜
⎟
⎝1+ v ⎠
( x2 + 2)
−2( x + 5)( x 2 + 10 x − 2)
3
( x 2 + 2)
⎛ 3x 2 − 2 ⎞
32. g ( x) = ⎜
⎟
⎝ 2x + 3 ⎠
(1 + v)
3
2v ) ⎞
⎟
⎟
⎠
4
3
2
⎛ ( 2 x + 3)(6 x) − (3 x 2 − 2)( 2) ⎞
⎜
⎟
2
⎜
⎟
( 2 x + 3)
⎝
⎠
3(3x 2 − 2) (6 x 2 + 18 x + 4)
2
2
2 ⎞
⎛ 3
⎛ t 2 ⎞⎜ (t + 2)( 2t ) − t (3t ) ⎟
h′(t ) = 2⎜ 3
⎟
2
⎟⎟
⎝ t + 2 ⎠⎜⎜
( t 3 + 2)
⎝
⎠
2t 2 ( 4t − t 4 )
3
( t 3 + 2)
2
(2 x
+ 3)
4
6(3 x 2 − 2) (3 x 2 + 9 x + 2)
2
=
(2 x
3
+ 3)
4
2
5
2
=
2t 3 ( 4 − t 3 )
=
(( x + 3) + x)
f ′( x) = 2(( x + 3) + x)(5( x
f ( x) =
(1 + v)(−2) − (1 −
2
(1 + v)
2
⎛ 3x 2 − 2 ⎞
g ′( x) = 3⎜
⎟
⎝ 2x + 3 ⎠
2
(t 3 + 2)
⎜
⎜
⎝
−9(1 − 2v)
=
⎛ t
⎞
30. h(t ) = ⎜ 3
⎟
+
t
2
⎝
⎠
33.
2⎛
2( x + 5)( 2 − 10 x − x 2 )
2
=
3
5
)
+ 3) ( 2 x) + 1
4
2
9
5
4
9
5
4
= 2 ⎡10 x( x 2 + 3) + ( x 2 + 3) + 10 x 2 ( x 2 + 3) + x⎤ = 20 x( x 2 + 3) + 2( x 2 + 3) + 20 x 2 ( x 2 + 3) + 2 x
⎥⎦
⎣⎢
(
34. g ( x) = 2 + ( x 2 + 1)
(
)
4 3
g ′( x) = 3 2 + ( x 2 + 1)
35. y =
y′ =
) (4( x
4 2
2
)
3
3
x +1
2
x +1
)
4 2
x +1
x
37. y =
1 − 3x 2 − 4 x3 2
2
(
+ 1) ( 2 x) = 24 x( x 2 + 1) 2 + ( x 2 + 1)
x ( x 2 + 1)
2
y′ = −
2
y
The zero of y′ corresponds
to the point on the graph of
y where the tangent line
is horizontal.
−1
5
y′ =
y′ has no zeros.
−5
4
y′
−2
y′
38. g ( x ) =
−2
2x
x +1
1
2 x ( x + 1)
y
( x + 1) x
2 x( x + 1)
y′ has no zeros.
g ′( x ) =
36. y =
4
x −1+
2
x +1
1
+
x −1
2
1
x +1
g ′ has no zeros.
7
y
6
32
y′
−6
g
6
−1
g′
−2
10
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.4
39.
cos π x + 1
x
dy
−π x sin π x − cos π x − 1
=
dx
x2
π x sin π x + cos π x + 1
= −
x2
The Chain Rule
185
43. y = e3 x
y =
y′ = 3e3 x
At (0, 1), y′ = 3.
44. y = e − 3 x
y′ = − 3e − 3 x
The zeros of y′ correspond to the points on the graph of
y where the tangent lines are horizontal.
At (0, 1), y′ = − 3.
3
y
45. y = ln x3 = 3 ln x
−5
5
y′ =
y′
At (1, 0), y′ = 3.
−3
40.
1
x
dy
1
1
= 2 x tan − sec 2
dx
x
x
y = x 2 tan
The zeros of y′ correspond
to the points on the graph of
y where the tangent lines
are horizontal.
41. (a)
3
ln x
2
3⎛ 1 ⎞
3
y′ = ⎜ ⎟ =
2⎝ x ⎠
2x
46. y = ln x 3 2 =
6
y
−4
5
At (1, 0), y′ =
y′
−6
47.
y = sin x
y′(0) = 1
1 cycle in [0, 2π ]
y = sin 2 x
y′ = 2 cos 2 x
y′(0) = 2
2 cycles in [0, 2π ]
The slope of sin ax at the origin is a.
42. (a)
y = sin 3 x
y′ = 3 cos 3 x
y′(0) = 3
3 cycles in [0, 2π ]
(b)
⎛ x⎞
y = sin ⎜ ⎟
⎝ 2⎠
⎛1⎞
⎛ x⎞
y′ = ⎜ ⎟ cos⎜ ⎟
⎝ 2⎠
⎝ 2⎠
1
y′(0) =
2
3
.
2
y = cos 4 x
dy
= −4 sin 4 x
dx
y′ = cos x
(b)
3
x
48.
y = sin π x
dy
= π cos π x
dx
49. g ( x ) = 5 tan 3 x
g ′( x ) = 15 sec 2 3 x
50. h( x) = sec( x 2 )
h′( x) = 2 x sec( x 2 ) tan ( x 2 )
51. y = sin (π x ) = sin (π 2 x 2 )
2
y′ = cos(π 2 x 2 ) ⎡⎣2π 2 x⎤⎦ = 2π 2 x cos(π 2 x 2 )
= 2π 2 x cos(π x)
2
(
52. y = cos(1 − 2 x ) = cos (1 − 2 x)
2
2
)
y′ = −sin (1 − 2 x) ( 2(1 − 2 x)( −2))
2
= 4(1 − 2 x ) sin (1 − 2 x )
2
Half cycle in [0, 2π ]
The slope of sin ax at the origin is a.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
186
Chapter 3
Differentiation
53. h( x) = sin 2 x cos 2 x
60. g (θ ) = cos 2 8θ = (cos 8θ )
h′( x) = sin 2 x( −2 sin 2 x ) + cos 2 x( 2 cos 2 x )
g ′(θ ) = 2(cos 8θ )( −sin 8θ )8 = −16 cos 8θ sin 8θ
= 2 cos 2 x − 2 sin 2 x
2
2
= 2 cos 4 x
61.
Alternate solution: h( x) =
1
2
sin 4 x
h′( x) =
1
2
cos 4 x( 4) = 2 cos 4 x
( ) sec ( 2θ ) + tan( ) sec( ) tan( )
= 12 sec( 12θ ) ⎡sec ( 12θ ) + tan ( 12θ )⎤
⎣
⎦
g ′(θ ) = sec
2 1
1θ
2
1
2
2
55.
1θ 1
2 2
63.
2
( 14 )(sin 2θ )(cos 2θ )(2)
1
2
sin 4θ
f (t ) = 3sec 2 (π t − 1)
f ′(t ) = 6 sec(π t − 1) sec(π t − 1) tan (π t − 1)(π )
= 6π sec 2 (π t − 1) tan (π t − 1) =
−sin 2 x − 2 cos 2 x
−1 − cos 2 x
=
3
sin x
sin 3 x
64.
cos v
= cos v ⋅ sin v
csc v
= cos 2 v − sin 2 v = cos 2v
65.
y′ = 8 sec x ⋅ sec x tan x = 8 sec x tan x
2
y =
x +
1
1
2
sin ( 2 x ) =
x + sin ( 4 x 2 )
4
4
1
1
2
2
+ cos( 4 x )(8 x) =
+ 2 x cos( 2 x )
4
2 x
66. y = sin x1/3 + (sin x)
1/3
g ′(t ) = 10 cos π t ( −sin π t )(π )
−2/3
⎛1
⎞ 1
y′ = cos x1/3 ⎜ x −2/3 ⎟ + (sin x) cos x
3
3
⎝
⎠
= −10π (sin π t )(cos π t )
= −5π sin 2π t
=
2
f ′(θ ) = 2( tan 5θ )(sec 2 5θ )5 = 10 tan 5θ sec 2 5θ
cos3 (π t − 1)
y = 3x − 5 cos(π x) = 3x − 5 cos(π 2 x 2 )
dy
1
= x −1/2
2
dx
2
2
6π sin (π t − 1)
dy
2
= 3 + 5 sin (π 2 x 2 )( 2π 2 x) = 3 + 10π 2 x sin (π x)
dx
57. y = 4 sec 2 x
f (θ ) = tan 2 5θ = ( tan 5θ )
(sin 2θ )
= −4π cot (π t + 2) csc 2 (π t + 2)
g ′(v) = cos v(cos v) + sin v( −sin v)
59.
1
4
h′(t ) = 4 cot (π t + 2)( −csc 2 (π t + 2)(π ))
2
58. g (t ) = 5 cos 2 π t = 5(cos π t )
sin 2 2θ =
62. h(t ) = 2 cot 2 (π t + 2)
sin 4 x
=
1
4
= sin 2θ cos 2θ =
sin 2 x( −sin x) − cos x( 2 sin x cos x)
f ′( x) =
56. g (v) =
1θ
2
cot x
cos x
=
sin x
sin 2 x
f ( x) =
f (θ ) =
f ′(θ ) = 2
54. g (θ ) = sec 12θ tan 12θ
1θ
2
2
1 ⎡ cos x1/3
cos x ⎤
⎢ 2/3 +
⎥
3⎢ x
(sin x)2/3 ⎦⎥
⎣
67. y = sin ( tan 2 x)
y′ = cos( tan 2 x)(sec 2 2 x)( 2) = 2 cos( tan 2 x) sec2 2 x
68. y = cos sin ( tan π x)
y′ = −sin sin ( tan π x) ⋅
69. f ( x ) = e 2 x
−π sin sin ( tan π x) cos( tan π x) sec 2 π x
−1/ 2
1
sin ( tan π x)) cos( tan π x) sec 2 π x(π ) =
(
2
2 sin ( tan π x)
71. y = e
f ′( x) = 2e 2 x
70. y = e − x
x
dy
e x
=
dx
2 x
2
2
dy
= − 2 xe − x
dx
72.
y = x 2e − x
dy
= − x 2e − x + 2 xe − x
dx
= xe − x ( 2 − x)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.4
73. g (t ) = (e − t + et )
3
83.
2
74. g (t ) = e
g ′(t ) = e − 3
75. y = ln e x
2
t2
= e
=
6
t 3e 3 t
2
=
6e − 3
t3
t2
= x2
84.
dy
=1
dx
g ′( x) =
⎛1 + ex ⎞
y = ln ⎜
x⎟
⎝1 − e ⎠
dy
ex
ex
=
+
x
dx
1+ e
1 − ex
77.
y =
h′( x) =
87.
x
−1
2
= 2(e x + e − x )
e x + e− x
− 2(e x − e − x )
(e x + e − x )
2
e x − e− x
78. y =
2
dy
e x + e− x
=
dx
2
79.
y = x 2e x − 2 xe x + 2e x = e x ( x 2 − 2 x + 2)
dy
= e x ( 2 x − 2) + e x ( x 2 − 2 x + 2) = x 2e x
dx
80.
y = xe x − e x = e x ( x − 1)
dy
= e x + e x ( x − 1) = xe x
dx
81.
82.
4x
2x2 + 3
y = (ln x)
4
4(ln x)
dy
3⎛ 1 ⎞
= 4(ln x) ⎜ ⎟ =
dx
x
⎝ x⎠
2e
1 − e2 x
88.
3
y = x ln x
dy
⎛1⎞
= x⎜ ⎟ + ln x = 1 + ln x
dx
⎝ x⎠
−2
dy
= − 2(e x + e − x ) (e x − e − x )
dx
=
2
x
86. h( x) = ln ( 2 x 2 + 3)
= ln (1 + e x ) − ln (1 − e x )
=
y = ln e x = x
85. g ( x) = ln x 2 = 2 ln x
dy
= 2x
dx
76.
y = e x (sin x + cos x)
= e x ( 2 cos x) = 2e x cos x
− 3t − 2
(6t −3 )
89.
1
ln ( x 2 − 1)
2
2x2 − 1
y = ln x
x 2 − 1 = ln x +
dy
1
1 ⎛ 2x ⎞
=
+ ⎜ 2
⎟ =
dx
x
2 ⎝ x − 1⎠
x( x 2 − 1)
1
ln ( x 2 − 9)
2
1 1
x
y′ =
(2 x) = 2
2 x2 − 9
x −9
90. y = ln
x2 − 9 =
91. f ( x) = ln
x
= ln x − ln ( x 2 + 1)
x2 + 1
f ′( x) =
1
2x
1 − x2
− 2
=
x
x +1
x( x 2 + 1)
⎛1⎞
⎛1
⎞
f ′( x) = e − x ⎜ ⎟ − e− x ln x = e − x ⎜ − ln x ⎟
⎝ x⎠
⎝x
⎠
⎛ 2x ⎞
92. f ( x ) = ln ⎜
⎟ = ln ( 2 x) − ln ( x + 3)
⎝ x + 3⎠
1
1
1
1
=
−
f ′( x) =
( 2) −
2x
x + 3
x
x +3
f ( x) = e3 ln x
93. g (t ) =
f ( x) = e − x ln x
f ′( x) =
e3
x
187
dy
= e x (cos x − sin x) + (sin x + cos x)(e x )
dx
g ′(t ) = 3(e − t + et ) (et − e − t )
−3 t 2
The Chain Rule
g ′(t ) =
ln t
t2
t 2 (1 t ) − 2t ln t
t4
=
1 − 2 ln t
t3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
188
Chapter 3
94. h(t ) =
h′(t ) =
Differentiation
ln t
t
t (1 t ) − ln t
x +1
1
= ⎡⎣ln ( x + 1) − ln ( x − 1)⎤⎦
x −1
2
95. y = ln
1 − ln t
t2
=
t2
dy
1⎡ 1
1 ⎤
1
= ⎢
−
=
⎥
dx
2 ⎣ x + 1 x − 1⎦
1 − x2
96. y = ln 3
y′ =
97.
y =
(
−
(
x2 + 1
+ ln x +
x
)
−x x
dy
=
dx
=
=
98.
y =
x2 + 1 +
⎛
+ ⎜
2
x + 1 ⎝x +
1
x2
−
x2 + 1
x2 + 1
x2 + 4
1 ⎛2 +
− ln ⎜
2
2x
4 ⎜⎝
)
(
Note that
x2 + 4 + 4x
4x
1
2 +
=
x2 + 4
(
−1 + (1 2) 2 −
2x
=
−
x + 4
4x
x2 + 4
2
=
x2
x2 + 1
=
x2 + 1
x2
(
x2 + 4
−
x2 + 4
1⎛
⎜
4⎝ 2 +
⋅
⎞⎛
⎟⎜
x + 4 ⎠⎝
1
2
2 −
x2 + 4
2−
x2 + 4
)
(
2
x2 + 4
1 2− x + 4 ⎛
−
⎜
x3
− x2
4
⎝
x2 + 4
x + 4
)+
dy
cos x
=
= cot x
sin x
dx
100. y = ln csc x
1
(− csc x cot x) = − cot x
csc x
=
2 −
)
x2 + 4 +
1
ln x
4
⎞
1
⎟ +
x + 4 ⎠ 4x
x
2
x2 + 4
.
− x2
⎞
1
⎟ +
x + 4 ⎠ 4x
x
2
x2 + 4
1
+
x3
4x
x2 + 4
1
+
=
3
x
4x
99. y = ln sin x
y′ =
2
− x2 + 4
x2 + 4 ⎞
1
⎟ =
− ln 2 +
2
⎟
x
2x
4
⎠
2
+
⎞
⎟
x + 1⎠
x
2
1 + x2
1
2 +
⎞⎛
⎟⎜1 +
x + 1 ⎠⎝
1
x2 + 1 + x ⎞
⎟
x 2 + 1 ⎠⎟
4
−1
dy
=
+
dx
2 x x2 + 4
=
⎞⎛
⎟⎜
x 2 + 1 ⎠⎝⎜
1
1
+
1⎡ 1
1 ⎤
4
−
=
3 ⎢⎣ x − 2
x + 2 ⎥⎦
3( x 2 − 4)
)
⎛
+ ⎜
⎝x +
x2
− 2 x2 x
dy
=
dx
So,
x2 + 1
1
x2
x2 + 1
1
x − 2
= ⎡⎣ln ( x − 2) − ln ( x + 2)⎤⎦
3
x + 2
x2 + 4
.
x3
101. y = ln
cos x
cos x − 1
= ln cos x − ln cos x − 1
dy
− sin x
− sin x
sin x
=
−
= − tan x +
dx
cos x
cos x − 1
cos x − 1
102. y = ln sec x + tan x
dy
sec x tan x + sec 2 x
=
dx
sec x + tan x
=
sec x(sec x + tan x)
sec x + tan x
= sec x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.4
103. y = ln
The Chain Rule
189
−1 + sin x
= ln −1 + sin x − ln 2 + sin x
2 + sin x
dy
cos x
cos x
3 cos x
=
−
=
dx
−1 + sin x
2 + sin x
(sin x − 1)(sin x + 2)
1
ln (1 + sin 2 x)
2
sin x cos x
dy
⎛ 1 ⎞ 2 sin x cos x
= ⎜ ⎟
=
2
2
1
sin
1 + sin 2 x
dx
x
+
⎝ ⎠
104. y = ln 1 + sin 2 x =
105.
x 2 + 8 x = ( x 2 + 8 x) , (1, 3)
12
y =
y′ =
−1/ 2
2( x + 4)
1 2
x + 8 x) ( 2 x + 8) =
=
(
12
2
2( x 2 + 8 x)
1+ 4
y′(1) =
106.
12 + 8(1)
(2, 2)
1/5
=
x2 + 8x
5
5
=
3
9
=
y = (3 x 3 + 4 x ) ,
y′ =
x + 4
109.
−4/5
1 3
3 x + 4 x) (9 x 2 + 4)
(
5
f ′(t ) =
9x + 4
5(3 x3 + 4 x)
4/5
=
=
f ′( −2) = −
f ( x) =
(t
(3x 2 )
=
−15 x 2
( x 3 − 2)
110.
2
( x2
− 3x)
=
⎛ 1⎞
= ( x 2 − 3 x) , ⎜ 4, ⎟
⎝ 16 ⎠
−2
2
f ′( x) = −2( x 2 − 3 x)
−3
(2 x − 3) =
− 3 x)
2
3t − 3 − 3t − 2
(t
− 1)
2
−5
(t − 1)
2
x + 4
, (9, 1)
2x − 5
(2 x − 5)(1) − ( x + 4)(2)
(2 x
3
f ′(9) = −
5
32
111.
− 5)
2
2x − 5 − 2x − 8
= −
−2( 2 x − 3)
( x2
f ( x) =
f ′( x) =
60
3
= −
100
5
1
f ′( 4) = −
−2
− 1)
f ′(0) = −5
−1
5
1⎞
⎛
= 5( x 3 − 2) , ⎜ −2, − ⎟
2⎠
x3 − 2
⎝
f ( x) =
f ′( x ) = −5( x 3 − 2)
108.
3t + 2
, (0, − 2)
t −1
(t − 1)(3) − (3t + 2)(1)
2
1
y′( 2) =
2
107.
f (t ) =
(2 x
− 5)
2
13
(2 x − 5)
2
13
(18 − 5)
2
= −
y = 26 − sec3 4 x,
1
13
(0, 25)
y′ = −3 sec 4 x sec 4 x tan 4 x 4
2
= −12 sec3 4 x tan 4 x
y′(0) = 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
190
Chapter 3
1
+
x
112. y =
y′ = − x − 2
Differentiation
12
⎛π 2 ⎞
cos x = x −1 + (cos x) , ⎜ , ⎟
⎝2 π⎠
1
1
sin x
−1 2
+ (cos x) (− sin x) = − 2 −
x
2
2 cos x
−3 2
⎛π ⎞
⎛ 3π ⎞
y′⎜ ⎟ = −3 sin ⎜ ⎟ =
2
⎝4⎠
⎝ 4 ⎠
y′(π /2) is undefined.
113. (a) f ( x ) = ( 2 x 2 − 7) ,
Tangent line: y +
(4, 5)
1/ 2
f ′( x) =
−1/ 2
1
2 x 2 − 7) ( 4 x) =
(
2
f ′( 4) =
8
5
⎛π
2⎞
y = cos 3 x, ⎜⎜ , −
⎟⎟
4
2
⎝
⎠
y′ = −3 sin 3 x
116. (a)
2x
y =
2 x2 − 7
(b)
Tangent line:
8
y − 5 = ( x − 4) ⇒ 8 x − 5 y − 7 = 0
5
2
3 2π
−3 2
x +
−
2
8
2
2
( π4 , − 22 (
−π
2
π
2
−2
6
(b)
2
−3 2 ⎛
π⎞
=
⎜x − ⎟
2
2 ⎝
4⎠
(4, 5)
⎛π ⎞
f ( x) = tan 2 x, ⎜ , 1⎟
⎝4 ⎠
117. (a)
−6
6
f ′( x) = 2 tan x sec 2 x
−2
f ( x) = (9 − x 2 )
114. (a)
f ′( x) =
f ′(1) =
2/3
⎛π ⎞
f ′⎜ ⎟ = 2(1)( 2) = 4
⎝4⎠
, (1, 4)
Tangent line:
−1/3
2
−4 x
9 − x 2 ) ( −2 x ) =
(
1/3
3
3(9 − x 2 )
−4
3(8)
1/3
= −
2
3
π⎞
⎛
y − 1 = 4⎜ x − ⎟ ⇒ 4 x − y + (1 − π ) = 0
4⎠
⎝
(b)
−␲
Tangent line:
2
y − 4 = − ( x − 1) ⇒ 2 x + 3 y − 14 = 0
3
(b)
4
␲
( (
π
,1
4
−4
6
⎛π ⎞
118. (a) y = 2 tan 3 x, ⎜ , 2 ⎟
⎝4 ⎠
(1, 4)
−2
y′ = 6 tan 2 x ⋅ sec 2 x
5
⎛π ⎞
y′⎜ ⎟ = 6(1)( 2) = 12
⎝4⎠
−1
f ( x) = sin 2 x,
115. (a)
(π , 0)
Tangent line:
f ′( x) = 2 cos 2 x
π⎞
⎛
y − 2 = 12⎜ x − ⎟ ⇒ 12 x − y + ( 2 − 3π ) = 0
4⎠
⎝
f ′(π ) = 2
Tangent line:
y = 2( x − π ) ⇒ 2 x − y − 2π = 0
(b)
2
0
(b)
3
( π4 , 2(
−π
2
(π , 0)
2␲
π
2
−1
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.4
119. (a)
⎛1
⎞
y = 4 − x 2 − ln ⎜ x + 1⎟,
⎝2
⎠
(0, 4)
f ′( x) =
1
dy
⎛1⎞
= − 2x −
dx
(1 2) x + 1⎜⎝ 2 ⎟⎠
25 − x 2 = ( 25 − x 2 ) ,
f ( x) =
121.
1
(25 − x 2 )(− 2 x) =
2
−x
25 − x 2
3
4
Tangent line:
dy
1
When x = 0,
= − .
dx
2
y − 4 = −
1
( x − 0)
2
1
y = − x + 4
2
Tangent line: y − 4 = −
(b)
191
(3, 4)
12
f ′(3) = −
1
= − 2x −
x + 2
The Chain Rule
3
( x − 3) ⇒ 3x + 4 y − 25 = 0
4
8
(3, 4)
−9
9
8
−4
(0, 4)
−4
4
−4
120. (a)
1 − x2
y = 2e
1 − x2
y′ = 2e
,
f ′( x) =
(1, 2)
( − 2 x)
= − 4 xe1 − x
x
f ( x) =
122.
2 − x
2
(2 − x 2 )
= x (2 − x 2 )
32
−1 2
,
(1, 1)
for x > 0
f ′(1) = 2
2
Tangent line: y − 1 = 2( x − 1) ⇒ 2 x − y − 1 = 0
y′(1) = − 4
3
Tangent line: y − 2 = − 4( x − 1)
y = − 4x + 6
(b)
2
(1, 1)
−2
7
2
−1
−4
4
−1
f ( x) = 2 cos x + sin 2 x,
123.
0 < x < 2π
f ′( x) = −2 sin x + 2 cos 2 x
= −2 sin x + 2 − 4 sin 2 x = 0
2 sin 2 x + sin x − 1 = 0
(sin x
+ 1)( 2 sin x − 1) = 0
sin x = −1 ⇒ x =
3π
2
1
π 5π
⇒ x = ,
2
6 6
π 3π 5π
Horizontal tangents at x = ,
,
6 2 6
sin x =
⎛ π 3 3 ⎞ ⎛ 3π ⎞
Horizontal tangent at the points ⎜ ,
⎜ 6 2 ⎟⎟, ⎜⎝ 2 , 0 ⎟⎠, and
⎝
⎠
⎛ 5π 3 3 ⎞
⎜⎜ , −
⎟
2 ⎟⎠
⎝ 6
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
192
Chapter 3
Differentiation
x
2x − 1
f ( x) =
124.
f ′( x) =
=
=
x −1
(2 x
− 1)
32
(2 x
− 1)
12
126.
− x( 2 x − 1)
− 1)
2
− 1)
= 432 x( x 3 + 4)( x 3 + 1)
32
127.
= 0 ⇒ x =1
f ( x) = 5( 2 − 7 x)
2
= 108 x( x3 + 4) ⎡⎣3 x 3 + x 3 + 4⎤⎦
32
f ( x) =
1
−1
= ( x − 6)
x −6
f ′( x) = −( x − 6)
Horizontal tangent at (1, 1)
125.
2
f ′′( x) = 54 x 2 ( 2)( x 3 + 4)(3x 2 ) + 108 x( x 3 + 4)
x −1
(2 x
3
f ′( x) = 18( x 3 + 4) (3 x 2 ) = 54 x 2 ( x 3 + 4)
−1 2
2x − 1
2x − 1 − x
(2 x
f ( x) = 6( x 3 + 4)
f ′′( x) = 2( x − 6)
−2
−3
=
2
(x
− 6)
3
4
f ′( x) = 20( 2 − 7 x) ( −7) = −140( 2 − 7 x)
3
128.
3
f ′′( x) = −420( 2 − 7 x) ( −7) = 2940( 2 − 7 x)
2
f ( x) =
8
(x
− 2)
= 8( x − 2)
2
f ′( x) = −16( x − 2)
2
f ′′( x) = 48( x − 2)
129.
−2
−3
−4
=
48
(x
− 2)
4
f ( x) = sin x 2
f ′( x) = 2 x cos x 2
f ′′( x) = 2 x ⎡⎣2 x( −sin x 2 )⎤⎦ + 2 cos x 2
= 2(cos x 2 − 2 x 2 sin x 2 )
130.
f ( x) = sec 2 π x
f ′( x) = 2 sec π x(π sec π x tan π x )
= 2π sec 2 π x tan π x
f ′′( x) = 2π sec 2 π x(sec 2 π x)(π ) + 2π tanπ x( 2π sec 2 π x tan π x)
= 2π 2 sec 4 π x + 4π 2 sec 2 π x tan 2 π x
= 2π 2 sec 2 π x(sec 2 π x + 2 tan 2 π x )
= 2π 2 sec 2 π x(3 sec 2 π x − 2)
131.
f ( x ) = (3 + 2 x )e − 3 x
f ′( x ) = (3 + 2 x)( − 3e − 3 x ) + 2e − 3 x
= ( − 7 − 6 x )e − 3 x
f ′′( x ) = ( − 7 − 6 x )( − 3e − 3 x ) − 6e − 3 x
= 3(6 x + 5)e − 3 x
132.
g ( x) =
x + e x ln x
g ′( x) =
1
2
g ′′( x) = −
= −
x
+
ex
+ e x ln x
x
1
xe x − e x
ex
+
+
+ e x ln x
32
2
4x
x
x
1
4x
x
+
e x ( 2 x − 1)
x2
+ e x ln x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.4
133.
h( x ) =
1
9
(3 x
h′( x ) =
13
9
(1, 649 )
+ 1) ,
(3 x
3
+ 1) (3) = (3x + 1)
2
g ′(t ) = t 2 (ln 2)2t + ( 2t )2t
2
= t 2t (t ln 2 + 2)
= 2t t ( 2 + t ln 2)
h′′(1) = 24
1
−1 2
⎛ 1⎞
134. f ( x) =
= ( x + 4) , ⎜ 0, ⎟
x + 4
⎝ 2⎠
1
−3 2
f ′( x) = − ( x + 4)
2
3
3
−5 2
f ′′( x) = ( x + 4)
=
52
4
4( x + 4)
135.
3
128
142.
=
143.
(0, 1)
2
−sin ( x )( 2 x) =
−2 x sin ( x 2 )
f ′′(0) = 0
144.
⎛π
⎜ ,
⎝6
146.
⎛π ⎞
g ′′⎜ ⎟ = 32 3
⎝6⎠
147.
f ( x) = 4 x
h(θ ) = 2−θ cos πθ
g (α ) = 5−α 2 sin 2α
x2
x −1
= 2 log 2 x − log 2 ( x − 1)
f ( x) = log 2
g ′( x ) = − (ln 5)5− x
2
1
−
x ln 2 ( x − 1) ln 2
x − 2
(ln 2) x( x − 1)
x −1
2
1
= log 3 x + log 3 ( x − 1) − log 3 2
2
1
1
1
h′( x) =
+ ⋅
−0
x ln 3 2 ( x − 1) ln 3
148. h( x) = log 3
y = 5x − 2
dy
= (ln 5)5 x − 2
dx
140. y = x(6− 2 x )
y′ = x( − 2 ln 6)6− 2 x + 6− 2 x
(− 2 x ln 6
(ln 5)5−α 2 sin 2α
y = log10 ( 2 x) = log10 2 + log10 x
=
138. g ( x ) = 5− x
1
2
y = log 3 x
f ′( x) =
f ′( x) = (ln 4)4 x
= 6
t2
dy
1
1
= 0+
=
dx
x ln 10
x ln 10
= 8 sec 2 ( 2t ) tan ( 2t )
−2x
32t ( 2t ln 3 − 1)
dy
1
=
dx
x ln 3
⎞
3⎟
⎠
g ′′(t ) = 4 sec( 2t ) ⋅ sec ( 2t ) tan ( 2t )2
139.
t2
g ′(α ) = 5−α 2 2 cos 2α −
145.
g ′(t ) = 2 sec 2 ( 2t )
137.
t ( 2 ln 3)32t − 32t
= − 2−θ ⎡⎣(ln 2) cos πθ + π sin πθ ⎤⎦
= −4 x 2 cos( x 2 ) − 2 sin ( x 2 )
g (t ) = tan 2t ,
32t
t
h′(θ ) = 2−θ ( − π sin πθ ) − (ln 2)2−θ cos πθ
f ′′( x) = −2 x cos( x 2 )( 2 x) − 2 sin ( x 2 )
136.
f (t ) =
f ′(t ) =
f ( x) = cos x 2 ,
f ′( x) =
193
141. g (t ) = t 2 2t
h′′( x ) = 2(3x + 1)(3) = 18 x + 6
f ′′(0) =
The Chain Rule
+ 1)
x
=
1 ⎡1
1
⎤
+
ln 3 ⎢⎣ x
2( x − 1) ⎥⎦
=
1 ⎡ 3x − 2 ⎤
ln 3 ⎢⎣ 2 x( x − 1) ⎥⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
194
149.
Chapter 3
Differentiation
1
log 5 ( x 2 − 1)
2
dy
x
1
2x
= ⋅ 2
= 2
dx
2 ( x − 1) ln 5
( x − 1) ln 5
y = log 5
x2 − 1 =
y
155.
3
−3
y = log10
150.
x2 − 1
x
x
−1
2
f′
= log10 ( x − 1) − log10 x
2
The zeros of f ′ correspond to the points where the graph
of f has horizontal tangents.
2x
1
dy
= 2
−
dx
( x − 1) ln 10 x ln 10
156.
y
1 ⎡ 2x
1⎤
1 ⎡ x2 + 1 ⎤
⎢
⎥
=
−
=
ln 10 ⎢⎣ x 2 − 1 x ⎥⎦
ln 10 ⎢ x( x 2 − 1) ⎥
⎣
⎦
4
3
2
−3
10 ⎡ t (1 t ) − ln t ⎤
⎢
⎥
ln 4 ⎣
t2
⎦
10
5
= 2
[1 − ln t] = 2 (1 − ln t )
t ln 4
t ln 2
t + 1 = t3 2
1 ln (t + 1)
2 ln 2
1 ⎡ 32 1
3
⎤
t
+ t 1 2 ln (t + 1)⎥
2 ln 2 ⎢⎣
t +1 2
⎦
f ′(t ) =
153.
y
f′
−4
The zeros of f ′ correspond to the points where the graph
of f has horizontal tangents.
157. g ( x) = f (3x)
g ′( x) = f ′(3 x)(3) ⇒ g ′( x ) = 3 f ′(3 x)
158. g ( x ) = f ( x 2 )
g ′( x ) = f ′( x 2 )( 2 x) ⇒ g ′( x ) = 2 xf ′( x 2 )
159.
f ( x ) = g ( x ) h( x )
f ′( x) = g ( x )h′( x) + g ′( x )h( x)
3
2
f ′(5) = ( −3)( −2) + (6)(3) = 24
1
x
−2
f
x
4
f′
−2
g ′(t ) =
f (t ) = t 3 2 log 2
f
−1
10 log 4 t
10 ⎛ ln t ⎞
151. g (t ) =
=
⎜
⎟
t
ln 4 ⎝ t ⎠
152.
f
2
2
3
160.
−2
f ′( x) = g ′( h( x))h′( x)
−3
f ′(5) = g ′(3)( −2) = −2 g ′(3)
The zeros of f ′ correspond to the points where the graph
of f has horizontal tangents.
154.
f ( x) = g ( h( x ))
Not possible, you need g ′(3) to find f ′(5).
y
f
3
f′
161.
f
f ( x) =
2
1
−3
f′
−2
−1
f ′( x) =
x
−1
1
2
3
−2
f ′(5) =
−3
g ( x)
h( x )
h( x) g ′( x) − g ( x )h′( x)
⎡⎣h( x)⎤⎦
(3)(6)
f is increasing on (1, ∞) so f ′ must be positive there.
162.
− ( −3)(−2)
(3)
f is decreasing on ( −∞, −1) so f ′ must be negative there.
f ( x) = ⎡⎣ g ( x)⎤⎦
2
2
=
12
4
=
9
3
3
f ′( x) = 3⎡⎣ g ( x )⎤⎦ g ′( x)
2
f ′(5) = 3( −3) (6) = 162
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.4
The Chain Rule
195
163. (a) h( x) = f ( g ( x)), g (1) = 4, g ′(1) = − 12 , f ′( 4) = −1
h′( x) = f ′( g ( x)) g ′( x)
( )
h′(1) = f ′( g (1)) g ′(1) = f ′( 4) g ′(1) = ( −1) − 12 =
1
2
(b) s( x) = g ( f ( x )), f (5) = 6, f ′(5) = −1, g ′(6) does not exist.
s′( x) = g ′( f ( x)) f ′( x)
s′(5) = g ′( f (5)) f ′(5) = g ′(6)( −1)
s′(5) does not exist because g is not differentiable at 6.
164. (a) h( x) = f ( g ( x))
168. y = A cos ω t
h′( x) = f ′( g ( x)) g ′( x)
3.5
= 1.75
2
y = 1.75 cos ω t
(a) Amplitude: A =
h′(3) = f ′( g (3)) g ′(3) = f ′(5)(1) =
1
2
(b) s( x) = g ( f ( x))
2π
π
=
10
5
πt
y = 1.75 cos
5
Period: 10 ⇒ ω =
s′( x) = g ′( f ( x)) f ′( x)
s′(9) = g ′( f (9)) f ′(9) = g ′(8)( 2) = (−1)( 2) = −2
F = 132,400(331 − v)
165. (a)
πt ⎤
πt
⎡ π
(b) v = y′ = 1.75⎢− sin ⎥ = −0.35π sin
5
5
5
⎣
⎦
−1
F ′ = ( −1)(132,400)(331 − v)
−2
(−1)
=
132,400
(331 − v)2
When v = 30, F ′ ≈ 1.461.
(b) F = 132,400(331 + v )
169. (a) Using a graphing utility, you obtain a model similar to
T (t ) = 56.1 + 27.6 sin (0.48t − 1.86).
(a)
100
−1
F ′ = ( −1)(132,400)(331 + v)
−2
(−1)
=
−132,400
(331 + v)2
0
When v = 30, F ′ ≈ −1.016.
(b)
166. y =
1
3
cos 12t −
v = y′ =
1
3
1
4
13
0
100
sin 12t
[−12 sin 12t] − 14 [12 cos 12t]
= −4 sin 12t − 3 cos 12t
0
13
0
When t = π 8, y = 0.25 ft and v = 4 ft/sec.
The model is a good fit.
167. θ = 0.2 cos 8t
The maximum angular displacement is θ = 0.2 (because
−1 ≤ cos 8t ≤ 1).
dθ
= 0.2[−8 sin 8t ] = −1.6 sin 8t
dt
When t = 3, dθ dt = −1.6 sin 24 ≈ 1.4489 rad/sec.
(c)
20
0
13
− 20
T ′(t ) ≈ 13.25 cos (0.48t − 1.86)
(d) The temperature changes most rapidly around spring
(March–May), and fall (Oct–Nov).
170. (a) According to the graph C ′( 4) > C ′(1).
(b) Answers will vary.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
196
Chapter 3
171. (a)
Differentiation
(b)
350
T ′( p ) =
34.96
3.955
+
p
p
T ′(10) ≈ 4.75 deg lb in.2
T ′(70) ≈ 0.97 deg lb in.2
0
100
0
172. (a) g ( x) = f ( x) − 2 ⇒ g ′( x) = f ′( x)
(b) h( x) = 2 f ( x) ⇒ h′( x) = 2 f ′( x)
(c) r ( x) = f ( −3 x) ⇒ r ′( x) = f ′( −3x)(−3) = −3 f ′( −3 x)
So, you need to know f ′( −3 x).
−2
−1
0
1
2
3
f ′( x)
4
2
3
− 13
−1
−2
−4
g ′( x)
4
2
3
− 13
−1
−2
−4
h′( x)
8
4
3
− 23
−2
−4
−8
12
1
−1
−2
x
( )
r ′(0) = −3 f ′(0) = ( −3) − 13 = 1
r ′( −1) = −3 f ′(3) = ( −3)( −4) = 12
(d) s( x) = f ( x + 2) ⇒ s′( x) = f ′( x + 2)
So, you need to know f ′( x + 2).
s′( −2) = f ′(0) = − 13 , etc.
173.
r ′( x)
s′( x)
S = C(R2 − r 2 )
dS
dr ⎞
⎛ dR
= C⎜ 2R
− 2r ⎟
dt
dt
dt ⎠
⎝
Because r is constant, you have dr dt = 0 and
dS
= (1.76 × 105 )( 2)(1.2 × 10−2 )(10−5 )
dt
= 4.224 × 10 −2 = 0.04224 cm/sec 2 .
174. C (t ) = P(1.05)
10
(c)
≈ $48.79
dC
t
= P ln (1.05)(1.05)
dt
–4
⎡
⎤
3
⎥ = 400 − 1200(t 2 + 2)− 2
N = 400 ⎢1 −
2
⎢
2
(t + 2) ⎥⎦
⎣
−3
4800t
N ′(t ) = 2400(t 2 + 2) ( 2t ) =
3
2
( t + 2)
(a) N ′(0) = 0 bacteria/day
(b) N ′(1) =
t
(a) C (10) = 29.95(1.05)
(b)
175.
− 13
(c)
(d)
4800(1)
=
4800
≈ 177.8 bacteria/day
27
(1 + 2)
4800( 2)
9600
N ′( 2) =
=
≈ 44.4 bacteria/day
3
216
4
+
2
(
)
4800(3)
14,400
N ′(3) =
=
≈ 10.8 bacteria/day
1331
( 9 + 2) 3
4800( 4)
19,200
N ′( 4) =
=
≈ 3.3 bacteria/day
3
5832
(16 + 2)
3
When t = 1,
dC
≈ 0.051P.
dt
(e)
When t = 8,
dC
≈ 0.072 P.
dt
(f ) The rate of change of the population is decreasing as
t → ∞.
dC
t
= ln (1.05) ⎡P(1.05) ⎤
⎣
⎦
dt
= ln (1.05)C (t )
176. (a)
V =
k
t +1
V (0) = 10,000 =
The constant of proportionality is ln 1.05.
V =
(b)
k
= k
0+1
10,000
−1 2
= 10,000(t + 1)
t +1
− 5000
dV
−3 2
⎛ 1⎞
= 10,000⎜ − ⎟(t + 1)
=
dt
⎝ 2⎠
t
( + 1)3 2
V ′(1) =
(c) V ′(3) =
− 5000
≈ −1767.77 dollars/year
23 2
− 5000
− 5000
=
= − 625 dollars/year
43 2
8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.4
f ′( x) = β cos β x
d
d
⎡ f ( − x )⎤⎦ =
⎡− f ( x)⎤⎦
dx ⎣
dx ⎣
f ′( − x)( −1) = − f ′( x)
f ′′( x) = − β sin β x
2
f ′′′( x) = − β 3 cos β x
f ′( − x) = f ′( x).
f (4) = β 4 sin β x
So, f ′( x) is even.
(b) f ′′( x) + β 2 f ( x) = − β 2 sin β x + β 2 (sin β x) = 0
(c)
(b) If f ( − x) = f ( x), then
f (2 k ) ( x ) = ( −1) β 2 k sin β x
k
f (2 k −1) ( x ) = ( −1)
k +1
d
d
⎡ f ( − x)⎤⎦ =
⎡ f ( x)⎤⎦
dx ⎣
dx ⎣
f ′( − x)( −1) = f ′( x)
β 2 k −1 cos β x
178. (a) Yes, if f ( x + p ) = f ( x) for all x, then
f ′( − x) = − f ′( x).
f ′( x + p ) = f ′( x), which shows that f ′ is
periodic as well.
(b) Yes, if g ( x) = f ( 2 x), then g ′( x) = 2 f ′( 2 x).
Because f ′ is periodic, so is g ′.
So, f ′ is odd.
182.
u2
( ) (2uu′)
r ′(1) = f ′( g (1)) g ′(1)
Note that g (1) = 4 and f ′( 4) =
u =
d
d ⎡ 2⎤
1
⎡u ⎤ =
u = u2
⎦
dx ⎣ ⎦
dx ⎣
2
uu′
u
=
= u′ , u ≠ 0
u
u2
179. (a) r ′( x) = f ′( g ( x)) g ′( x)
5−0
5
= .
6−2
4
183.
Also, g ′(1) = 0. So, r ′(1) = 0.
s′( 4) = g ′( f ( 4)) f ′( 4)
184.
5 ⎛5⎞
6−4
1
Note that f ( 4) = , g ′⎜ ⎟ =
= and
2 ⎝ 2⎠
6−2
2
1⎛ 5 ⎞
5
5
f ′( 4) = . So, s′( 4) = ⎜ ⎟ = .
2⎝ 4 ⎠
8
4
g ( x) = sin 2 x + cos 2 x = 1 ⇒ g ′( x ) = 0
g ( x) + 1 = f ( x)
Taking derivatives of both sides, g ′( x) = f ′( x).
Equivalently,
f ′( x) = 2 sec x ⋅ sec x tan x = 2 sec 2 x tan x and
x ≠
5
3
f ( x) = x 2 − 9
⎛ x2 − 9 ⎞
⎟,
f ′( x) = 2 x⎜ 2
⎜ x −9 ⎟
⎝
⎠
x ≠ ±3
185. h( x) = x cos x
h′( x) = − x sin x +
g ′( x) = 2 sin x cos x + 2 cos x( −sin x) = 0
(b) tan 2 x + 1 = sec 2 x
−1 2
g ( x) = 3 x − 5
⎛ 3x − 5 ⎞
g ′( x) = 3⎜
,
⎜ 3x − 5 ⎟⎟
⎝
⎠
(b) s′( x) = g ′( f ( x)) f ′( x)
180. (a)
197
181. (a) If f ( − x) = − f ( x), then
177. f ( x) = sin β x
(a)
The Chain Rule
186.
x
cos x,
x
x ≠ 0
f ( x) = sin x
⎛ sin x ⎞
f ′( x) = cos x⎜
, x ≠ kπ
⎜ sin x ⎟⎟
⎝
⎠
g ′( x) = 2 tan x ⋅ sec 2 x = 2 sec 2 x tan x, which
are the same.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
198
Chapter 3
187. (a)
Differentiation
f ( x) = tan x
f (π 4) = 1
f ′( x) = sec 2 x
f ′(π 4) = 2
f ′′( x) = 2 sec 2 x tan x
f ′′(π 4) = 4
(b)
f
P2
P1
−1
1
(4)( x − π 4)2 + 2( x − π 4) + 1
2
(c) P2 is a better approximation than P1.
(d) The accuracy worsens as you move away from
x = π 4.
= 2( x − π 4) + 2( x − π 4) + 1
2
188. (a)
p
2
0
P1 ( x) = 2( x − π 4) + 1
P2 ( x) =
5
2
3
2
f ′(π 6) =
3
f ( x) = sec x
f (π 6) =
f ′( x) = sec x tan x
f ′′( x) = sec x(sec 2 x) + tan x(sec x tan x)
f ′′(π 6) =
10 3
9
= sec3 x + sec x tan 2 x
P1 ( x) =
2
2
( x − π 6) +
3
3
P2 ( x) =
1 ⎛ 10 ⎞⎛
2⎛
π⎞
π⎞
⋅⎜
⎟⎜ x − ⎟ + ⎜ x − ⎟ +
2 ⎝ 3 3 ⎠⎝
6⎠
3⎝
6⎠
2
2
3
2⎛
2
π⎞
π⎞
⎛ 5 ⎞⎛
= ⎜
⎟⎜ x − ⎟ + ⎜ x − ⎟ +
6⎠
3⎝
6⎠
3
⎝ 3 3 ⎠⎝
2
(b)
3
f
P2
− 1.5
1.5
P1
−1
(c) P2 is a better approximation than P1.
(d) The accuracy worsens as you move away from x = π 6.
189. (a)
f ( x) = e x
f ( 0) = 1
f ′( x) = e
x
f ′(0) = 1
f ′′( x) = e x
f ′′(0) = 1
190. (a)
1
x
f ′( x) =
f (1) = ln (1) = 0
f ′(1) = 1
f ′′( x) = −1 x 2
P1 ( x) = 1( x − 0) + 1 = x + 1
f ′′(1) = −1
P1 ( x) = 1( x − 1) + 0 = x − 1
1
(1)( x − 0)2 + 1( x − 0) + 1
2
1
= x2 + x + 1
2
P2 ( x) =
(b)
f ( x) = ln x
1
(−1)( x − 1)2 + 1( x − 1) + 0
2
1
2
= − ( x + 1) + x − 1
2
P2 ( x) =
5
(b)
f
P2
3
P1
P1
f
−3
−4
6
4
P2
−1
−5
(c) P2 is a better approximation than P1.
(d) The accuracy worsens as you move away from
x = 0.
(c) P2 is a better approximation than P1.
(d) The accuracy worsens as you move away from
x = 0.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
191. False. If y = (1 − x) , then y′ =
12
1
2
Implicit Differentiation
199
(1 − x)−1 2 (−1).
192. False. If f ( x ) = sin 2 2 x, then f ′( x) = 2(sin 2 x)( 2 cos 2 x).
193. True
194. True
f ( x ) = a1 sin x + a2 sin 2 x + " + an sin nx
195.
f ′( x ) = a1 cos x + 2a2 cos 2 x + " + nan cos nx
f ′(0) = a1 + 2a2 + " + nan
a1 + 2a2 + " + nan = f ′(0) = lim
f ( x ) − f ( 0)
x −0
x→0
(
)
= lim
x→0
(
f ( x)
sin x
)
⋅
f ( x)
sin x
= lim
≤1
x
→
0
x
sin x
(
)
n +1
n
⎡
⎤
x k − 1 Pn′ ( x) − Pn ( x)( n + 1) x k − 1 kx k −1
x k − 1 Pn′ ( x ) − ( n + 1)kx k −1Pn ( x)
d ⎢ Pn ( x) ⎥
=
=
196.
n+2
2n + 2
dx ⎢ x k − 1 n +1 ⎥
xk − 1
xk − 1
⎣
⎦
(
)
(
)
(
)
n +1
d ⎡ 1
dx n ⎢⎣ x k −
(
)
n+2
d ⎡ d n ⎡ 1 ⎤⎤
k
k −1
⎢
⎥ = x − 1 Pn′ ( x ) − ( n + 1)kx Pn ( x)
dx ⎣ dx n ⎢⎣ x k − 1⎥⎦ ⎦
Pn ( x) = x k − 1
Pn + 1 ( x) = x k − 1
n
(
)
⎤
⇒
1⎥⎦
(
)
Pn + 1 (1) = −( n + 1)kPn (1)
For n = 1,
d ⎡ 1
dx ⎢⎣ x k −
P1 ( x)
− kx k −1
⎤
=
=
⇒ P1 (1) = − k . Also, P0 (1) = 1.
2
2
⎥
k
k
1⎦
x −1
x −1
(
)
(
)
You now use mathematical induction to verify that Pn (1) = ( − k ) n! for n ≥ 0. Assume true for n. Then
n
Pn +1 (1) = −( n + 1)k Pn (1) = −( n + 1)k (− k ) n! = ( − k )
n
n +1
( n + 1)!.
Section 3.5 Implicit Differentiation
1.
x2 + y 2 = 9
4.
2 x + 2 yy′ = 0
y′ = −
2.
6 x 2 + 9 y 2 y′ = 0
9 y 2 y′ = − 6 x 2
x
y
− 6x2
2x2
= − 2
2
9y
3y
y′ =
x 2 − y 2 = 25
2 x − 2 yy′ = 0
x 3 − xy + y 2 = 7
5.
x
y′ =
y
3.
2 x3 + 3 y 3 = 64
3x − xy′ − y + 2 yy′ = 0
2
(2 y
− x ) y′ = y − 3 x 2
x1 2 + y1 2 = 16
y′ =
1 −1 2
1
x
+ y −1 2 y′ = 0
2
2
y′ = −
x −1 2
y −1 2
= −
y
x
6.
y − 3x 2
2y − x
x 2 y + y 2 x = −2
x 2 y′ + 2 xy + y 2 + 2 yxy′ = 0
( x2
)
(
+ 2 xy y′ = − y 2 + 2 xy
y′ =
)
− y( y + 2 x)
x( x + 2 y )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
200
Chapter 3
Differentiation
13. sin x = x(1 + tan y )
x3 y 3 − y − x = 0
7.
cos x = x(sec 2 y ) y′ + (1 + tan y )(1)
3 x 3 y 2 y′ + 3 x 2 y 3 − y′ − 1 = 0
(3 x 3 y 2
)
− 1 y′ = 1 − 3 x 2 y 3
y′ =
y′ =
1 − 3x y
3x3 y 2 − 1
2 3
(−csc y) y′ = 1 −
2
1
−1 2
( xy ) ( xy′ + y ) = 2 xy + x 2 y′
2
x
y
y′ +
= 2 xy + x 2 y′
2 xy
2 xy
y′ =
4 xy
y′ − x cos( xy ) y′ = y cos( xy )
y′ =
16. x = sec
1 = −
xy
y′ =
xe y − 10 x + 3 y = 0
9.
xe y
dy
dy
+ e y − 10 + 3
= 0
dx
dx
dy y
( xe + 3) = 10 − e y
dx
17.
1
y
y′
1
1
sec tan
2
y
y
y
⎛1⎞ ⎛1⎞
− y2
= − y 2 cos⎜ ⎟ cot ⎜ ⎟
sec(1 y ) tan (1 y )
⎝ y⎠ ⎝ y⎠
3 dy
dy
+ 2y
= 0
y dx
dx
⎞
dy ⎛ 3
⎜ − 2y⎟
dx ⎝ y
⎠
dy
2x
2 xy
=
=
dx
(3 y ) − 2 y 3 − 2 y 2
e xy + x 2 − y 2 = 10
18.
ln ( xy ) + 5 x = 30
ln x + ln y + 5 x = 30
dy
ye + 2 x
= − xy
dx
xe − 2 y
xy
1
1 dy
+
+5 = 0
x
y dx
1 dy
1
= − −5
y dx
x
sin x + 2 cos 2 y = 1
cos x − 4(sin 2 y ) y′ = 0
dy
y
⎛ y + 5 xy ⎞
= − − 5y = −⎜
⎟
dx
x
x
⎝
⎠
cos x
y′ =
4 sin 2 y
12.
1 − x cos( xy )
2x =
dy
⎛ dy
⎞
+ y ⎟e xy + 2 x − 2 y
= 0
⎜x
dx
dx
⎝
⎠
dy xy
( xe − 2 y) = − ye xy − 2 x
dx
11.
y cos( xy )
x 2 − 3 ln y + y 2 = 10
2x −
10 − e y
dy
=
dx
xe y + 3
10.
1
1
=
= − tan 2 y
1 − csc 2 y
−cot 2 y
y′ = [ xy′ + y] cos( xy )
xy − y
x − 2x2
y′
y = sin xy
15.
⎛ x
⎞
y
− x 2 ⎟ y′ = 2 xy −
⎜⎜
⎟
2 xy
⎝ 2 xy
⎠
y
2 xy −
2 xy
y′ =
x
− x2
2 xy
y′ =
cot y = x − y
14.
xy = x 2 y + 1
8.
cos x − tan y − 1
x sec 2 y
(sin π x + cos π y )2 = 2
2(sin π x + cos π y ) ⎡⎣π cos π x − π (sin π y ) y′⎤⎦ = 0
π cos π x − π (sin π y ) y′ = 0
y′ =
cos π x
sin π y
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
19. 4 x 3 + ln y 2 + 2 y = 2 x
12 x 2 +
20.
2
y′ + 2 y′ = 2
y
⎛2
⎞
2
⎜ + 2 ⎟ y′ = 2 − 12 x
⎝y
⎠
y′ =
2 − 12 x 2
2 y + 2
y′ =
y (1 − 6 x 2 )
y − 6 yx 2
=
1+ y
1+ y
Implicit Differentiation
4 xy + ln x 2 y = 7
4 xy + 2 ln x + ln y = 7
2
1
4 xy′ + 4 y + + y′ = 0
x
y
⎛
1⎞
2
⎜ 4 x + ⎟ y ′ = −4 y −
y⎠
x
⎝
2
−4 y −
x
y′ =
1
4x +
y
−4 xy 2 − 2 y
4x2 y + x
y′ =
21. (a) x 2 + y 2 = 64
y
(b)
y 2 = 64 − x 2
y = ±
12
64 − x 2
y1 =
64 − x 2
4
12
4
− 12
x
−4
− 12
−1 2
dy
1
= ± (64 − x 2 ) ( −2 x) =
dx
2
(c) Explicitly:
201
∓x
64 − x
2
=
y2 = −
64 − x 2
−x
±
64 − x
2
= −
x
y
(d) Implicitly: 2 x + 2 yy′ = 0
y′ = −
x
y
22. (a) 25 x 2 + 36 y 2 = 300
23. (a) 16 y 2 − x 2 = 16
36 y 2 = 300 − 25 x 2 = 25(12 − x 2 )
16 y 2 = x 2 + 16
25
(12 − x 2 )
36
5
y=±
12 − x 2
6
y2 =
y2 =
y =
y
(b)
6
4
6
12 − x 2
6
4
2
−6 −4 −2
±
x 2 + 16
4
y
(b)
y1 = 5
x2
x 2 + 16
+1 =
16
16
y1 = 1
x 2 + 16
4
2
x
−2
−4
−6
(c) Explicitly:
2
4
y2 = − 5
6
−6
6
x
−4
12 − x 2
−6
−1 2
5⎛1⎞
dy
= ± ⎜ ⎟(12 − x 2 ) ( − 2 x)
6⎝ 2⎠
dx
5x
= ∓
6 12 − x 2
25 x
= −
36 y
(d) Implicitly: 50 x + 72 y ⋅ y′ = 0
y′ =
− 50 x
25 x
= −
72 y
36 y
6
−2
y2 = −
1
4
x 2 + 16
−1 2
1 2
( x + 16) (−2 x)
2
4
±x
±x
x
=
=
= −
2
±
4
4
16
y
y
(
)
4 x + 16
dy
=
(c) Explicitly:
dx
±
(d) Implicitly: 16 y 2 − x 2 = 16
32 yy′ − 2 x = 0
32 yy′ = 2 x
y′ =
2x
x
=
32 y
16 y
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
202
Chapter 3
Differentiation
x2 + y 2 − 4x + 6 y + 9 = 0
24. (a)
(x
(c) Explicitly:
− 4 x + 4) + ( y + 6 y + 9) = − 9 + 4 + 9
2
dy
1
2 −1 2
2( x − 2)⎤⎦
= ± ⎡4 − ( x − 2) ⎤ ⎡−
⎣
⎦ ⎣
2
dx
x − 2
= ∓
2
4 − ( x − 2)
2
(x
− 2) + ( y + 3) = 4
2
2
(y
+ 3) = 4 − ( x − 2)
2
y + 3 = ±
2
4 − ( x − 2)
y = −3 ±
2
4 − ( x − 2)
= −
2
y
(b)
1
−1
−1
y1 = − 3 +
x − 2
y +3
(d) Implicitly:
2 x + 2 yy′ − 4 + 6 y′ = 0
4 − (x − 2)2
x
1
2
3
4
5
2 yy′ + 6 y′ = − 2 x + 4
6
y′( 2 y + 6) = − 2( x − 2)
−2
−3
−4
y′ =
−5
−6
y2 = − 3 −
4 − (x −
2)2
xy = 6
25.
xy′ + y (1) = 0
+ y ) = x3 + y 3
3
x 2 y′ + 2 xy + 2 xyy′ + y 2 = 0
1
6
( x2
+ 2 xy ) y′ = −( y 2 + 2 xy )
y′ = −
y3 − x2 = 4
3 y 2 y′ − 2 x = 0
y2 =
2 yy′ =
2( 2)
3( 22 )
29.
1
=
3
tan ( x + y ) = x
(1 + y′) sec2 ( x
+ y) = 1
2 yy′ =
y′ =
x 2 − 49
x 2 + 49
( x2
=
+ 49)( 2 x) − ( x 2 − 49)( 2 x )
( x2
+ 49)
+ 49)
1 − sec 2 ( x + y )
sec 2 ( x + y )
tan 2 ( x + y ) + 1
= −
2
x2
x +1
2
At (0, 0): y′ = 0
98 x
y ( x 2 + 49)
− tan 2 ( x + y )
= −sin 2 ( x + y )
2
196 x
( x2
y( y + 2 x)
x( x + 2 y )
At ( −1, 1): y′ = −1
2x
3y2
y′ =
27.
x − 2
y + 3
x 2 y + xy 2 = 0
At ( −6, −1): y′ = −
At ( 2, 2): y′ =
= −
3x 2 y + 3 xy 2 = 0
y
y′ = −
x
y′ =
2( y + 3)
x3 + 3x 2 y + 3xy 2 + y 3 = x3 + y 3
xy′ = − y
26.
(x
28.
− 2( x − 2)
2
At (7, 0) : y′ is undefined.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
x cos y = 1
30.
( x2
35.
x[− y′ sin y] + cos y = 0
4 y′( x 2 y + y 3 − x 2 ) = 4( 2 xy − x3 − xy 2 )
y′ =
1
⎛ π⎞
At ⎜ 2, ⎟ : y′ =
2 3
⎝ 3⎠
x 3 + y 3 − 6 xy = 0
36.
3e xy − x = 0
3x 2 + 3 y 2 y′ − 6 xy′ − 6 y = 0
y′(3 y 2 − 6 x) = 6 y − 3x 2
3e xy xy′ = 1 − 3 ye xy
1 − 3 ye xy
3xe xy
y′ =
37.
1
x
1
y′ =
2 xy
2 yy′ =
( x2
(y
− 3) = 4( x − 5),
2
2
y −3
y′ =
At (6, 1): y′ =
1
2e
y = −x + 7
+ 4) y ′ + y ( 2 x ) = 0
38.
(x
+ 2) + ( y − 3) = 37,
2
2
=
−2 x ⎡⎣8 ( x + 4)⎤⎦
x2 + 4
−16 x
At ( 2, 1): y′ =
( x2
+ 4)
(y
(x
+ 2)
y −3
2
At ( 4, 4): y′ = −
6
= −6
1
Tangent line: y − 4 = −6( x − 4)
y = −6 x + 28
8 ⎞
⎛
⎜ Or, you could just solve for y: y = 2
⎟
+ 4⎠
x
⎝
39.
(4 − x) y 2 = x3
(4 − x)(2 yy′) + y 2 (−1) = 3x 2
At ( 2, 2): y′ = 2
− 3) y′ = −( x + 2)
y′ = −
−32
1
= −
64
2
y′ =
(4, 4)
2( x + 2) + 2( y − 3) y′ = 0
2
=
2
= −1
1−3
Tangent line: y − 1 = −1( x − 6)
+ 4) y = 8
−2 xy
y′ = 2
x + 4
34.
(6, 1)
2( y − 3) y′ = 4
At (e, 1): y′ =
( x2
6 y − 3x 2
2 y − x2
= 2
2
y − 2x
3y − 6x
(16 3) − (16 9) = 32 = 4
⎛ 4 8⎞
At ⎜ , ⎟ : y′ =
(64 9) − (8 3) 40 5
⎝ 3 3⎠
1
9
y 2 = ln x
33.
2 xy − x3 − xy 2
x2 y + y3 − x2
At (1, 1) : y′ = 0
3e xy [ xy′ + y] − 1 = 0
32.
2
4 x 2 yy′ + 4 y 3 y′ − 4 x 2 y′ = 8 xy − 4 x3 − 4 xy 2
1
cot y
x
cot y
=
x
At (3, 0): y′ =
+ y2 ) = 4 x2 y
4 x3 + 4 x 2 yy′ + 4 xy 2 + 4 y 3 y′ = 4 x 2 y′ + 8 xy
=
y′ =
203
2( x 2 + y 2 )( 2 x + 2 yy′) = 4 x 2 y′ + y(8 x)
cos y
y′ =
x sin y
31.
Implicit Differentiation
3x 2 + y 2
2 y(4 − x)
xy = 1,
(1, 1)
xy′ + y = 0
y′ =
−y
x
At (1, 1): y′ = −1
Tangent line: y − 1 = −1( x − 1)
y = −x + 2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
204
Chapter 3
Differentiation
7 x 2 − 6 3xy + 13 y 2 − 16 = 0,
40.
(
)
3, 1
y 2 ( x2 + y2 ) = 2 x2 ,
44.
y x + y = 2x
2 2
14 x − 6 3 xy′ − 6 3 y + 26 yy′ = 0
At
(
)
3, 1 : y′ =
(
3 x −
y = −
3
y′ =
)
(
(
−4, 2 3
)
18 x − 2 xy 2
2x2 y − 8 y
y =
y′ =
⎛
At ⎜1,
⎝
3
6
3
( x + 4)
6
46.
3( x + y
6( x + y
2
2
(x
+ 2 y ) y′ = − 2 x − y
y′ =
)
At ( 2, 0), y′ =
− 2x − y
x + 2y
−4
= −2
2
Tangent line: y − 0 = − 2( x − 2)
y = − 2x + 4
= 100( x − y ),
2
(2, 0)
2 x + xy′ + y + 2 yy′ = 0
13
(4, 2)
2
)(2 x + 2 yy′) = 100(2 x − 2 yy′)
At ( 4, 2): 6(16 + 4)(8 + 4 y′) = 100(8 − 4 y′)
960 + 480 y′ = 800 − 400 y′
880 y′ = −160
47.
x + y − 1 = ln ( x 2 + y 2 ),
1 + y′ =
(1, 0)
2 x + 2 yy′
x2 + y 2
x 2 + y 2 + ( x 2 + y 2 ) y′ = 2 x + 2 yy′
At (1, 0): 1 + y′ = 2
2
y′ = − 11
Tangent line:
x 2 + xy + y 2 = 4,
1
2
2
9
−9
x +
4
2
y =
1
( x − 8)
2
1
y = − x + 5
2
43.
9
−9
=
( x − 1)
4
4
4 y − 9 = − 9x + 9
y −
4 y + 9 x = 18
−x
⎛ y⎞
= −⎜ ⎟
y −1 3
⎝ x⎠
2
2
3
−y
x
Tangent line:
Tangent line: y − 1 = −
2
+
1x
3
9⎞
−9 4
−9
=
⎟, y′ =
4⎠
1
4
3
8
x +
3
6
3
−1 3
− 1)
xy′ = − y
)
2 −1 3
2
x
+ y −1 3 y′ = 0
3
3
y′ =
(x
4 xy′ + 4 y = 0
(8, 1)
x 2 3 + y 2 3 = 5,
1
3
⎛ 9⎞
⎜1, ⎟
⎝ 4⎠
4 xy = 9,
45.
24
1
=
=
48 3
2 3
Tangent line: y − 2 3 =
At (8, 1): y′ = −
y =
2(16) 2 3 − 16 3
=
1
3
Tangent line: y − 1 =
18( −4) − 2( −4)(12)
At −4, 2 3 : y′ =
42.
6 y′ = 2
y′ =
2
(
2 y′ + 2 + 4 y′ = 4
)
x 2 yy′ + 2 xy − 18 x − 8 yy′ = 0
2
At (1, 1):
3x + 4
x 2 y 2 − 9 x 2 − 4 y 2 = 0,
41.
2 yy′x + 2 xy 2 + 4 y 3 y′ = 4 x
6 3 − 14 3
−8 3
=
= − 3
8
26 − 6 3 3
Tangent line: y − 1 = −
(1, 1)
2
2
6 3 y − 14 x
26 y − 6 3x
y′ =
4
y′ = 1
2 x − 4
y − 2 = − 11
(
)
Tangent line: y = x − 1
11y + 2 x − 30 = 0
2
y = − 11
x +
30
11
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
y 2 + ln ( xy ) = 2,
48.
2 yy′ +
(e, 1)
(b)
xy′ + y
= 0
xy
−1
3e
Because
−1
( x − e)
3e
4
−1
y =
x +
3e
3
Tangent line: y − 1 =
x2
y2
+
= 1,
2
8
yy′
= 0
x +
4
y
3 x ( −2 ) y
1
−
= 1 ⇒ x + = 1 ⇒ y = −2 x + 4,
6
8
2
4
Tangent line.
(1, 2)
51.
y′ =
y′ =
y = −2 x + 4
52.
b
−1
,
sin y
0 < y < π
sin 2 y + cos 2 y = 1
sin 2 y = 1 − cos 2 y
y0 y y02
− x0 x x02
− 2 =
+ 2
2
b
b
a2
a
+
cos y = x
y′ =
−b 2 x
y − y0 = 2 0 ( x − x0 ), Tangent line at ( x0 , y0 )
a y0
a
1
1 + x2
−sin y ⋅ y′ = 1
−b 2 x
x2
y2
2 x 2 yy′
+ 2 = 1 ⇒ 2 + 2 = 0 ⇒ y′ = 2
2
a
b
a
b
a y
y02
2
1
π
π
= cos 2 y , − < y <
sec 2 y
2
2
sec 2 y = 1 + tan 2 y = 1 + x 2
Tangent line: y − 2 = −2( x − 1)
Because
tan y = x
y′ sec 2 y = 1
At (1, 2): y′ = −2
x02
2
xx
yy
x02
y2
− 02 = 1, you have 02 − 20 = 1.
2
a
b
a
b
Note: From part (a),
4x
y′ = −
y
(b)
x0b 2
( x − x0 ), Tangent line at ( x0 , y0 )
y0 a 2
yy0
y2
xx
x2
− 02 = 02 − 02
2
b
b
a
a
At (e, 1): 2ey′ + ey′ + 1 = 0
y′ =
205
x2
y2
2x
2 yy′
xb 2
− 2 = 1 ⇒ 2 − 2 = 0 ⇒ y′ =
2
a
b
a
b
ya 2
y − y0 =
2 xy 2 y′ + xy′ + y = 0
49. (a)
Implicit Differentiation
sin y =
= 1, you have
y0 y
xx
+ 02 = 1.
b2
a
y′ =
1 − cos 2 y =
−1
1 − x2
1 − x2
, −1 < x < 1
Note: From part (a),
1( x) 2( y )
1
1
+
= 1 ⇒ y = − x + 1 ⇒ y = −2 x + 4,
2
8
4
2
Tangent line.
50. (a)
x2
y2
−
= 1, (3, − 2)
6
8
x
y
− y′ = 0
3
4
y
x
y′ =
4
3
4x
y′ =
3y
At (3, − 2): y′ =
4(3)
3( −2)
53.
x2 + y2 = 4
2 x + 2 yy′ = 0
−x
y′ =
y
y′′ =
=
y ( −1) + xy′
y2
− y + x( − x y )
y2
− y − x2
y3
4
= − 3
y
=
2
= −2
Tangent line: y + 2 = −2( x − 3)
y = −2 x + 4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
206
Chapter 3
Differentiation
x2 y − 4 x = 5
54.
x 2 − y 2 = 36
2 x − 2 yy′ = 0
x
y′ =
y
x − yy′ = 0
55.
x 2 y′ + 2 xy − 4 = 0
4 − 2 xy
y′ =
x2
2
x y′′ + 2 xy′ + 2 xy′ + 2 y = 0
1 − yy′′ − ( y′) = 0
⎡ 4 − 2 xy ⎤
x 2 y′′ + 4 x ⎢
⎥ + 2y = 0
2
⎣ x
⎦
4
′′
x y + 4 x( 4 − 2 xy ) + 2 x 2 y = 0
2
2
⎛ x⎞
1 − yy′′ − ⎜ ⎟ = 0
⎝ y⎠
2
y − y 3 y′′ = x 2
x 4 y′′ + 16 x − 8 x 2 y + 2 x 2 y = 0
x 4 y′′ = 6 x 2 y − 16 x
6 xy − 16
y′′ =
x3
56.
y′′ =
y 2 − x2
36
= − 3
3
y
y
xy − 1 = 2 x + y 2
xy′ + y = 2 + 2 yy′
xy′ − 2 yy′ = 2 − y
(x
− 2 y ) y′ = 2 − y
y′ =
2− y
x − 2y
xy′′ + y′ + y′ = 2 yy′′ + 2( y′)
2
xy′′ − 2 yy′′ = 2( y′) − 2 y′
2
2
⎛ 2− y ⎞
( x − 2 y ) y′′ = 2( y′)2 − 2 y′ = 2⎜
⎟ −
⎝ x − 2y ⎠
2( 2 − y )⎡⎣( 2 − y ) − ( x − 2 y )⎤⎦
y′′ =
( x − 2 y )3
=
=
57.
⎛ 2− y ⎞
2⎜
⎟
⎝ x − 2y ⎠
=
2( 4 − 2 x + 2 y − 2 y + xy − y 2 )
(x
2( − 5)
(2 y
− x)
3
=
− 2 y)
3
2( 2 − y )( 2 − x + y )
(x
=
− 2 y)
3
2( y 2 − xy + 2 x − 4)
(2 y
− x)
3
10
(x
− 2 y)
3
y 2 = x3
y3 = 4 x
58.
3 y 2 y′ = 4
2 yy′ = 3x 2
y′ =
3x 2
3 x 2 xy
3 y x3
3y
=
⋅
=
⋅
=
2y
2 y xy
2x y2
2x
y′′ =
2 x(3 y′) − 3 y( 2)
4 x2
=
2 x ⎣⎡3 ⋅ (3 y 2 x)⎦⎤ − 6 y
4x2
y′ =
4
3y2
3 y 2 y′′ + 6 y( y′) = 0
2
yy′′ + 2( y′) = 0
2
=
3y
3x
=
4 x2
4y
y′′ =
− 2( y′)
y
2
=
−2 ⎛ 4 ⎞
⎜
⎟
y ⎝ 3y2 ⎠
2
32
y′′ = − 5
9y
Note: y = ( 4 x)
13
4
−2 3
( 4 x)
3
8
32
32
−5 3
= −
= − 5
y′′ = − ( 4)( 4 x)
53
9
9y
9( 4 x)
y′ =
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
x 2 + y 2 = 25
59.
60.
2 x + 2 yy′ = 0
y′ =
Implicit Differentiation
207
x 2 + y 2 = 36
2 x + 2 yy′ = 0
−x
y
y′ = −
x
y
At ( 4, 3):
At (6, 0) ; slope is undefined.
Tangent line:
−4
y −3 =
( x − 4) ⇒ 4 x + 3 y − 25 = 0
3
Tangent line: x = 6
3
( x − 4) ⇒ 3 x − 4 y = 0
4
Normal line: y − 3 =
Normal line: y = 0
8
(6, 0)
− 12
12
6
(4, 3)
−9
−8
9
(
At 5,
)
11 , slope is
−6
At ( −3, 4):
y −
Tangent line:
Tangent line:
−4
( x + 3) ⇒ 4 x + 3 y = 0
3
Normal line: y − 4 =
11 =
5x +
11 y − 36 = 0
y −
Normal line:
11 =
5 y − 5 11 =
6
(− 3, 4)
5y −
−9
−5
( x − 5)
11
11 y − 11 = −5 x + 25
3
( x + 3) ⇒ 3x − 4 y + 25 = 0
4
y − 4 =
−5
11
11
( x − 5)
5
11x − 5 11
11x = 0
9
8
(5, 11)
−6
− 12
12
−8
61.
x2 + y2 = r 2
2 x + 2 yy′ = 0
y′ =
−x
= slope of tangent line
y
y
= slope of normal line
x
Let ( x0 , y0 ) be a point on the circle. If x0 = 0, then the tangent line is horizontal, the normal line is
vertical and, hence, passes through the origin. If x0 ≠ 0, then the equation of the normal line is
y − y0 =
y =
y0
( x − x0 )
x0
y0
x
x0
which passes through the origin.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
208
62.
Chapter 3
Differentiation
y2 = 4x
63. 25 x 2 + 16 y 2 + 200 x − 160 y + 400 = 0
2 yy′ = 4
50 x + 32 yy′ + 200 − 160 y′ = 0
2
= 1 at (1, 2)
y
y′ =
y′ =
Equation of normal line at (1, 2) is
Horizontal tangents occur when x = −4:
y − 2 = −1( x − 1), y = 3 − x.
25(16) + 16 y 2 + 200(−4) − 160 y + 400 = 0
The centers of the circles must be on the normal line
and at a distance of 4 units from (1, 2).
Therefore,
(x
2
2
25 x( x + 8) = 0 ⇒ x = 0, − 8
x = 1 ± 2 2.
(
2)
)
Centers of the circles: 1 + 2 2, 2 − 2 2 and
2, 2 + 2
Horizontal tangents: ( −4, 0), ( −4, 10)
25 x 2 + 400 + 200 x − 800 + 400 = 0
2( x − 1) = 16
(1 − 2
y( y − 10) = 0 ⇒ y = 0,10
Vertical tangents occur when y = 5:
− 1) + ⎡⎣(3 − x) − 2⎤⎦ = 16
2
Vertical tangents: (0, 5), ( −8, 5)
y
(− 4, 10)
(x − 1 − 2 2) + ( y − 2 + 2 2)
(x − 1 + 2 2) + ( y − 2 − 2 2)
2
2
2
2
10
6
(− 8, 5)
Equations:
(0, 5)
4
= 16
(− 4, 0)
−10 − 8 − 6 − 4
= 16
64. 4 x 2 + y 2 − 8 x + 4 y + 4 = 0
8 x + 2 yy′ − 8 + 4 y′ = 0
8 − 8x
4 − 4x
=
y′ =
y + 2
2y + 4
Horizontal tangents occur when x = 1:
4(1) + y − 8(1) + 4 y + 4 = 0
2
200 + 50 x
160 − 32 y
2
x
−2
2
y
(1, 0)
−1
1
x
2
3
4
−1
(2, − 2)
(0, − 2)
−3
−4
(1, − 4)
−5
y 2 + 4 y = y( y + 4) = 0 ⇒ y = 0, − 4
Horizontal tangents: (1, 0), (1, − 4)
Vertical tangents occur when y = −2:
4 x 2 + ( −2) − 8 x + 4( −2) + 4 = 0
2
4 x 2 − 8 x = 4 x( x − 2) = 0 ⇒ x = 0, 2
Vertical tangents: (0, − 2), ( 2, − 2)
65.
y = x
x2 + 1
1
ln ( x 2 + 1)
2
1 ⎛ dy ⎞
1
x
+ 2
⎜ ⎟ =
y ⎝ dx ⎠
x
x +1
ln y = ln x +
⎡ 2 x2 + 1 ⎤
dy
2 x2 + 1
= y⎢ 2
⎥ =
dx
x2 + 1
⎣⎢ x( x + 1) ⎥⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
66.
Implicit Differentiation
209
x 2 ( x + 1)( x + 2) , x > 0
y =
y 2 = x 2 ( x + 1)( x + 2)
2 ln y = 2 ln x + ln ( x + 1) + ln ( x + 2)
2 dy
2
1
1
=
+
+
y dx
x
x +1 x + 2
dy
y ⎡2
1
1 ⎤
= ⎢ +
+
dx
x + 1 x + 2 ⎥⎦
2 ⎣x
x 2 ( x + 1)( x + 2) ⎡ 2( x + 1)( x + 2) + x ( x + 2) + x ( x + 1) ⎤
4 x2 + 9 x + 4
⎢
⎥ =
2
x ( x + 1)( x + 2)
⎥⎦
2 ( x + 1)( x + 2)
⎣⎢
dy
=
dx
67.
y =
3x − 2
x2
(x
+ 1)
70.
2
ln y
1
ln (3 x − 2) − 2 ln ( x + 1)
2
1 ⎛ dy ⎞
2
3
2
+
−
⎜ ⎟ =
y ⎝ dx ⎠
x
2(3 x − 2)
x +1
ln y = 2 ln x +
1 ⎛ dy ⎞
1
1
1
1
+
−
−
⎜ ⎟=
y ⎝ dx ⎠ x + 1 x − 2 x − 1 x + 2
⎡
⎤
4 ⎤
2 x2 + 4
dy
⎡ −2
⎢
⎥
= y⎢ 2
+ 2
=
y
⎥
2
2
−
−
1 x
4⎦
dx
⎢⎣ ( x − 1)( x − 4) ⎥⎦
⎣x
( x + 1)( x + 2) ⋅
2x2 + 4
=
( x − 1)( x − 2) ( x + 1)( x − 1)( x + 2)( x − 2)
⎡ 3x 2 + 15 x − 8 ⎤
dy
= y⎢
⎥
dx
⎢⎣ 2 x(3 x − 2)( x + 1) ⎥⎦
=
68.
y =
3 x 3 + 15 x 2 − 8 x
2( x + 1)
3
3x − 2
=
x −1
x2 + 1
2
71.
1⎡
ln ( x 2 − 1) − ln ( x 2 + 1)⎤⎦
2⎣
1 ⎛ dy ⎞
1 ⎡ 2x
2x ⎤
−
⎜ ⎟ = ⎢ 2
y ⎝ dx ⎠
2 ⎣ x − 1 x 2 + 1⎦⎥
ln y =
dy
=
dx
( x + 1)( x − 2)
( x − 1)( x + 2)
= ln ( x + 1) + ln ( x − 2) − ln ( x − 1) − ln ( x + 2)
y =
x2 − 1 ⎡ 2x
x 2 + 1 ⎢⎣ x 4 −
=
=
2x
( x − 1) ( x − 2)
2
2
y = x2 x
2
ln x
x
1 ⎛ dy ⎞
2⎛ 1 ⎞
2
⎛ 2⎞
⎜ ⎟ = ⎜ ⎟ + ln x⎜ − 2 ⎟ = 2 (1 − ln x)
y ⎝ dx ⎠
x⎝ x ⎠
x
⎝ x ⎠
2y
dy
= 2 (1 − ln x) = 2 x(2 x) − 2 (1 − ln x)
dx
x
ln y =
⎤
1⎥⎦
( x 2 − 1) 2 x
12
( x 2 + 1) ( x 2 − 1)( x 2
2( x 2 + 2)
12
69.
y =
( x 2 + 1)
x( x − 1)
32
+ 1)
1 ⎛ dy ⎞
⎛1⎞
⎜ ⎟ = ( x − 1)⎜ ⎟ + ln x
y ⎝ dx ⎠
⎝ x⎠
12
dy
⎡x − 1
⎤
= y⎢
+ ln x⎥
dx
⎣ x
⎦
32
y ⎡ 4x2 + 4x −
⎢
2 ⎢ x( x 2 − 1)
⎣
y = x x −1
ln y = ( x − 1)(ln x)
( x 2 − 1)
x +1
3
1
ln y = ln x + ln ( x − 1) − ln ( x + 1)
2
2
1 ⎛ dy ⎞ 1 3 ⎛ 1 ⎞ 1 ⎛ 1 ⎞
⎜ ⎟= + ⎜
⎟− ⎜
⎟
y ⎝ dx ⎠ x 2 ⎝ x − 1 ⎠ 2 ⎝ x + 1 ⎠
3
1 ⎤
dy
y ⎡2
= ⎢ +
−
2 ⎣ x x − 1 x + 1⎥⎦
dx
=
72.
2
2 ⎤ ( 2 x + 2 x − 1) x − 1
⎥ =
⎥⎦
( x + 1)3 2
= x x − 2( x − 1 + x ln x)
73.
y = ( x − 2)
x +1
ln y = ( x + 1) ln ( x − 2)
1 ⎛ dy ⎞
⎛ 1 ⎞
⎜ ⎟ = ( x + 1)⎜
⎟ + ln ( x − 2)
y ⎝ dx ⎠
⎝ x − 2⎠
dy
⎡x + 1
⎤
= y⎢
+ ln ( x − 2)⎥
dx
⎣x − 2
⎦
= ( x − 2)
x +1
⎤
⎢ x − 2 + ln ( x − 2)⎥
⎣
⎦
x +1 ⎡
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
210
74.
Chapter 3
Differentiation
y = (1 + x)
1x
75.
ln y = ln x
1
ln (1 + x)
x
1 ⎛ dy ⎞
1⎛ 1 ⎞
⎛ 1⎞
⎜ ⎟ = ⎜
⎟ + ln (1 + x)⎜ − 2 ⎟
y ⎝ dx ⎠
x ⎝1 + x ⎠
⎝ x ⎠
ln y =
y′ =
76.
(1 + x)1 x ⎡
ln ( x + 1) ⎤
1
−
⎢
⎥
x
⎣x + 1
⎦
x
x > 0
ln x
= (ln x)(ln x) = (ln x)
2
y′
= 2 ln x(1 x )
y
ln ( x + 1) ⎤
dy
y⎡ 1
= ⎢
−
⎥
dx
x ⎣x + 1
x
⎦
=
y = x ln x,
2 y ln x
2 x ln x ⋅ ln x
=
x
x
y = (ln x )
ln x
,
x >1
ln y = ln ⎡(ln x) ⎤ = (ln x) ln (ln x )
⎣
⎦
y′
1
1
1
= (ln x )
⋅ + ln (ln x)
y
x
ln x x
ln x
1
(1 + ln(ln x))
x
y
y′ = (1 + ln (ln x ))
x
=
= (ln x )
ln x
(1 + ln(ln x))
x
77. Find the points of intersection by letting y 2 = 4 x in the equation 2 x 2 + y 2 = 6.
( x + 3)( x
The curves intersect at (1, ± 2).
2 x 2 + 4 x = 6 and
− 1) = 0
2x 2 + y 2 = 6 4
Ellipse:
Parabola:
4 x + 2 yy′ = 0
2 yy′ = 4
y′ = −
2x
y
y′ =
y 2 = 4x
(1, 2)
−6
6
(1, − 2)
2
y
−4
At (1, 2), the slopes are:
y′ = −1
y′ = 1
At (1, − 2), the slopes are:
y′ = 1
y′ = −1
Tangents are perpendicular.
78. Find the points of intersection by letting y 2 = x 3 in the equation 2 x 2 + 3 y 2 = 5.
2 x 2 + 3 x3 = 5 and 3x3 + 2 x 2 − 5 = 0
Intersect when x = 1.
2
Points of intersection: (1, ±1)
y 2 = x 3:
y′ =
(1, 1)
2 x 2 + 3 y 2 = 5:
2 yy′ = 3 x 2
2
3x
2y
2x2 + 3y2 = 5
4 x + 6 yy′ = 0
y′ = −
−2
4
(1, − 1)
−2
y 2= x 3
2x
3y
At (1, 1), the slopes are:
y′ =
3
2
y′ = −
2
3
At (1, −1), the slopes are:
3
2
y′ =
2
3
Tangents are perpendicular.
y′ = −
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
79. y = − x and x = sin y
82.
Point of intersection: (0, 0)
y = − x:
x2 + y2 = C 2
y′ = −
1 = y′cos y
are orthogonal.
y′ = 1
2
2
K=1
Tangents are perpendicular.
−3
4
−4
80. Rewriting each equation and differentiating:
x(3 y − 29) = 3
1⎛ 3
⎞
y = ⎜ + 29 ⎟
3⎝ x
⎠
1
y′ = − 2
x
3
x
y =
+1
3
2
x (3y − 29) = 3 15
x 3 = 3y − 3
−15
−2
For each value of x, the derivatives are negative
reciprocals of each other. So, the tangent lines are
orthogonal at both points of intersection.
xy = C
x2 − y2 = K
xy′ + y = 0
2 x − 2 yy′ = 0
y
y′ = −
x
(
y = 5 − x2
)
x.
84. Answers will vary. Sample answer: Given an implicit
equation, first differentiate both sides with respect to x.
Collect all terms involving y′ on the left, and all other
terms to the right. Factor out y′ on the left side. Finally,
divide both sides by the left-hand factor that does not
contain y′.
85. (a) True
12
−3
81.
3
83. Answers will vary. Sample answer: In the explicit form
of a function, the variable is explicitly written as a
function of x. In an implicit equation, the function is only
implied by an equation. An example of an implicit
function is x 2 + xy = 5. In explicit form it would be
x+y=0
x 3 = 3( y − 1)
C=1
−3
C=2
−2
6
(0, 0)
K = −1
3
x = sin y
y′ = x
x
y
slopes is ( − x y )( K ) = ( − x Kx)( K ) = −1. The curves
At (0, 0), the slopes are:
−6
y′ = K
At the point of intersection ( x, y ), the product of the
y′ = sec y
y′ = −1
211
y = Kx
2 x + 2 yy′ = 0
x = sin y:
y′ = −1
Implicit Differentiation
( )
( )
(b) False.
d
cos y 2 = −2 y sin y 2 .
dy
(c) False.
d
cos( y 2 ) = −2 yy′ sin ( y 2 ).
dx
86. (a) The slope is greater at x = − 3.
(b) The graph has vertical tangent lines at about
(− 2, 3) and (2, 3).
x
y′ =
y
At any point of intersection ( x, y ) the product of the
(c) The graph has a horizontal tangent line at about
(0, 6).
slopes is ( − y x)( x y ) = −1. The curves are orthogonal.
2
2
C=4
−3
3
C=1
K = −1
−2
−3
3
K=2
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
212
Chapter 3
87. (a)
Differentiation
x 4 = 4( 4 x 2 − y 2 )
10
4 y 2 = 16 x 2 − x 4
y 2 = 4x2 −
y = ±
(b)
− 10
10
1 4
x
4
− 10
1 4
x
4
4x2 −
1 4
x
4
y = 3 ⇒ 9 = 4 x2 −
10
36 = 16 x 2 − x 4
− 10
10
x 4 − 16 x 2 + 36 = 0
x2 =
7, 1 −
16 ±
y1 =
1
3
(
1
3
y2 =
(
7, −1 +
7, 1 +
7, y′ =
1
3
1
3
1
3
(
7, y′ = −
)
(
2(3)
.
(
1⎡
3⎣
)
7 + 7 x + 8 7 + 23⎤.
⎦
)
7 − 7 , and the line is
)
(
1⎡
3⎣
7 + 3 =
1
3
(
)
7 − 7 x + 23 − 8 7 ⎤.
⎦
)
7 − 7 , and the line is
)(
1
3
x(8 − x 2 )
)
)
(
1
7 + 3 = − ⎡
3⎣
7 −7 x +1−
7, y′ = −
2
7
7 + 3 =
(
)
7 . So, there are four values of x:
7 + 7 , and the line is
)(
For x = 1 +
y4 = −
(
7 − 7 x −1+
For x = −1 +
y3 = −
1
3
7, y′ =
)(
(
28
28 = 8 ± 2 7 = 1 ±
7 + 7 x +1+
For x = 1 −
− 10
256 − 144
=8±
2
To find the slope, 2 yy′ = 8 x − x 3 ⇒ y′ =
For x = −1 −
y2
y3
Note that x 2 = 8 ±
−1 −
y4
y1
(
)
(
)
)
(
)
7 − 7 x − 23 − 8 7 ⎤.
⎦
)
7 + 7 , and the line is
)(
)
(
1
7 + 3 = − ⎡
3⎣
7 + 7 x −1−
7 + 7 x − 8 7 + 23 ⎤.
⎦
(c) Equating y3 and y4 :
−
1
3
(
7x +
)(
7 −7 x +1−
(
)(
)
7 + 3 = −
7 −7 x +1−
) (
7 =
7 − 7 − 7x − 7 + 7 7 =
1
3
(
)(
)
7 + 7 x −1−
)(
7 + 7 x −1−
7x −
7 + 3
7
)
7 − 7 + 7x − 7 − 7 7
16 7 = 14 x
x =
If x =
8 7
7
⎛8 7 ⎞
8 7
, 5 ⎟⎟.
, then y = 5 and the lines intersect at ⎜⎜
7
⎝ 7
⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.5
x +
88.
1
x
2
+
1
2
y =
Implicit Differentiation
213
c
dy
= 0
y dx
y
dy
= −
dx
x
y0
Tangent line at ( x0 , y0 ): y − y0 = −
(
y-intercept: (0, y
x-intercept: x0 +
+
(x
− x0 )
)
y )
y0 , 0
x0
0
x0
x0
0
Sum of intercepts:
(x
0
+
) (
y0 + y0 +
x0
x 2 + y 2 = 100, slope =
89.
)
y0 = x0 + 2
x0
3
4
3
4
x
=
⇒ y = − x
4
3
y
⎛ 16 ⎞
x 2 + ⎜ x 2 ⎟ = 100
⎝9 ⎠
25 2
x = 100
9
x = ±6
2
=
( c)
2
= c
x2
y2
+
= 1, ( 4, 0)
4
9
2x
2 yy′
+
= 0
4
9
−9 x
y′ =
4y
3
⎛
Points on ellipse: ⎜1, ±
2
⎝
⎛ 3
At ⎜1,
⎝ 2
p x p −1
p x p −1 y
⋅ q −1 =
⋅
q y
q
yq
p x p −1 p
⋅ p x
q
x
q
=
p p
x
q
r real
ln y = ln ( x r ) = r ln x
3
⎛
At ⎜1, −
2
⎝
q −1
So, if y = x , n = p q, then y′ = nx
n
y = xr ,
)
So, −9 x 2 + 36 x = 4 y 2 = 36 − 9 x 2 ⇒ x = 1.
qy q −1 y′ = px p −1
(b)
y0
But, 9 x 2 + 4 y 2 = 36 ⇒ 4 y 2 = 36 − 9 x 2 .
yq = x p
=
x0 +
−9 x( x − 4) = 4 y 2
y = x p q ; p, q integers and q > 0
y′ =
(
−9 x
y −0
=
4y
x − 4
Points: (6, − 8) and ( −6, 8)
90. (a)
y 0 + y0 =
91.
2 x + 2 yy′ = 0
y′ = −
x0
n −1
⎞
3⎟
⎠
−9 x
−9
3
⎞
=
= −
3 ⎟: y′ =
4y
2
⎠
4 ⎡⎣(3 2) 3 ⎤⎦
⎞
3 ⎟: y′ =
⎠
Tangent lines: y = −
.
y =
3
2
3
3
( x − 4) = − x + 2 3
2
2
3
( x − 4) =
2
3
x − 2 3
2
y′
r
=
y
x
y′ =
yr
xr ⋅ r
=
= rx r −1.
x
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
214
Chapter 3
Differentiation
92. x = y 2
93. (a)
1 = 2 yy′
y′ =
1
,
2y
slope of tangent line
Consider the slope of the normal line joining ( x0 , 0) and
At ( 4, 2): y′ =
= ( y 2 , y ) on the parabola.
( x, y )
(a) If x0 =
y − 2 = 2( x − 4)
y = 2x − 6
1
2
y 2 = x0 −
1,
4
(b)
1
2
1,
2
4
−6
then y 2 =
−
1
4
1
2
−4
(c) If x0 = 1, then y =
1
2
x2
( 2 x − 6)
+
32
8
(c)
then y 2 = 0 ⇒ y = 0. Same as part (a).
2
6
= − 14 , which is
impossible. So, the only normal line is the x-axis
( y = 0).
(b) If x0 =
−4
1
= −
4( 2)
2
Slope of normal line is 2.
y −0
−2 y = 2
y − x0
y 2 − x0 = −
x2
y2
+
=1
32
8
−x
2x
2 yy′
+
= 0 ⇒ y′ =
32
8
4y
2
=1
x 2 + 4( 4 x 2 − 24 x + 36) = 32
= x and there are three
17 x 2 − 96 x + 112 = 0
normal lines.
(17 x
⎛1 1 ⎞
The x-axis, the line joining ( x0 , 0) and ⎜ ,
⎟,
2⎠
⎝2
− 28)( x − 4) = 0 ⇒ x = 4,
28
17
⎛ 28 46 ⎞
Second point: ⎜ , − ⎟
⎝ 17 17 ⎠
1 ⎞
⎛1
and the line joining ( x0 , 0) and ⎜ , −
⎟
2
2⎠
⎝
If two normals are perpendicular, then their slopes are –1
and 1. So,
−2 y = − 1 =
y −0
1
⇒ y =
y 2 − x0
2
and
12
(1 4) −
= −1 ⇒
x0
1
1
3
− x0 = − ⇒ x0 = .
4
2
4
The perpendicular normal lines are y = − x +
y = x −
3
and
4
3
.
4
Section 3.6 Derivatives of Inverse Functions
1.
f ( x) = x3 − 1,
f ′( x) = 3 x
a = 26
2.
f ( x) = 5 − 2 x3 ,
f ′( x) = −6 x
2
a = 7
2
f is monotonic (increasing) on ( −∞, ∞) therefore f has
f is monotonic (decreasing) on ( −∞, ∞) therefore f has
an inverse.
an inverse.
f (3) = 26 ⇒ f
( f −1 )′ (26)
=
−1
(26)
= 3
1
1
1
1
=
=
=
f ′(3)
27
f ′( f −1 ( 26))
3(32 )
f ( −1) = 7 ⇒ f −1 (7) = −1
( f −1 )′ (7)
=
−1
1
1
1
=
=
=
2
f ′( −1)
6
f ′( f −1 (7))
−6( −1)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.6
3.
f ( x) = x3 + 2 x − 1,
a = 2
Derivatives of Inverse Functions
7. f ( x ) =
f ′( x) = 3 x 2 + 2 > 0
f is monotonic (increasing) on ( −∞, ∞) therefore f has
f ′( x) =
x +6
,
x −2
(x
=
f (1) = 2 ⇒ f −1 ( 2) = 1
( f −1 )′ (2)
4.
1
1
1
1
=
=
=
f ′(1)
5
f ′( f −1 ( 2))
3(12 ) + 2
=
an inverse.
1
(−243 − 54) = −11 ⇒ f −1(−11) = −3
27
( f −1)′(−11) = f ′ f −11(−11) = f ′(1−3)
(
)
f (−3) =
1
1
1
=
=
1
1
4
2
17
5(−3) + 6( −3)
(459)
27
27
5.
f ( x) = sin x,
)
a = 1 2, −
π
2
≤ x ≤
π
2
⎛ π π⎞
f ′( x) = cos x > 0 on ⎜ − , ⎟
⎝ 2 2⎠
⎡ π π⎤
f is monotonic (increasing) on ⎢− , ⎥ therefore f has
⎣ 2 2⎦
an inverse.
π
1
π
⎛π ⎞
⎛1⎞
f ⎜ ⎟ = sin
=
⇒ f −1 ⎜ ⎟ =
6
2
6
⎝6⎠
⎝ 2⎠
( f )′ ⎛⎜⎝ 12 ⎞⎟⎠ =
1
−1
=
6.
1
1
=
=
π⎞
⎛
⎛π ⎞
f ′⎜ ⎟
cos⎜ ⎟
⎝6⎠
⎝6⎠
2
2 3
=
3
3
a = 1, 0 ≤ x ≤ π 2
f is monotonic (decreasing) on [0, π 2] therefore f has
an inverse.
f (0) = 1 ⇒ f −1 (1) = 0
=
1
f ′( f −1 (1))
=
< 0 on ( 2, ∞ )
2
an inverse.
8. f ( x ) =
=
f ′( f
1
−1
=
(3))
1
1
=
= −2
2
′
f ( 6)
−8 (6 − 2)
x +3
,
x +1
x > −1, a = 2
(x
+ 1)(1) − ( x + 3)(1)
f ′( x) =
(x
−2
=
(x
+ 1)
2
+ 1)
2
< 0 on ( −1, ∞)
f is monotonic (decreasing) on ( −1, ∞) therefore f has
an inverse.
f (1) = 2 ⇒ f −1 ( 2) = 1
( f −1 )′ (2) =
1
1
1
=
=
= −2
f ′(1) ( −2) (1 + 1)2
f ′( f −1 ( 2))
9. f ( x ) = x 3 −
4
,
x
f ′( x) = 3 x 2 +
a = 6, x > 0
4
> 0
x2
f is monotonic (increasing) on (0, ∞) therefore f has an
inverse.
( f −1 )′ (6) =
f ′( x) = −2 sin 2 x < 0 on (0, π 2)
( f −1 )′ (1)
2
f ( 2) = 6 ⇒ f −1 (6) = 2
⎛
⎛ 1 ⎞⎞
f ′⎜ f −1 ⎜ ⎟ ⎟
⎝ 2 ⎠⎠
⎝
f ( x) = cos 2 x,
−8
( x − 2)
− 2)
f is monotonic (decreasing) on ( 2, ∞) therefore f has
( f −1 )′ (3)
f is monotonic (increasing) on ( −∞, ∞) therefore f has
(
− 2)(1) − ( x + 6)(1)
f (6) = 3 ⇒ f −1 (3) = 6
1 5
( x + 2 x3 ), a = −11
27
1
f ′( x) =
(5 x 4 + 6 x 2 )
27
f ( x) =
=
x > 0, a = 3
(x
an inverse.
215
1
1
1
=
=
−2 sin 0
f ′(0)
0
10. f ( x ) =
f ′( x) =
1
1
1
1
=
=
=
2
f ′( 2)
13
f ′( f −1 (6))
3( 22 ) + 4 2
x − 4,
2
a = 2,
x ≥ 4
1
> 0 on ( 4, ∞)
x − 4
f is monotonic (increasing) on [4, ∞) therefore f has an
inverse.
f (8) = 2 ⇒ f −1 ( 2) = 8
1
1
=
4
2 8− 4
( f −1 )′ (2) = f ′ f 1−1(2) = f ′1(8) = 114 = 4
(
)
f ′(8) =
So, ( f −1 )′ (1) is undefined.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
216
11.
Chapter 3
Differentiation
⎛ 1 1⎞
⎜ , ⎟
⎝ 2 8⎠
f ( x) = x3 ,
f ( x) = arccos( x 2 )
15. (a)
⎛1
f −1 ( x ) = 3 x , ⎜ ,
⎝8
( f −1 )′ ( x) = 3 13 x
12.
=
y =
(1, −1)
3− x
f ( x) =
,
4
( f −1 )′ ( x) = − 14
( f −1 )′ (−1) = − 14
f ( x) =
x − 4,
f ′( x) =
1
x − 4
( f −1 )′ ( x)
= 2x
( f −1 )′ (1)
= 2
(5, 1)
,
1.5
1
1 + x2
1
f ′( −1) =
2
1
π
y +
= ( x + 1)
4
2
1
1 π
y = x + − , tangent line
2
2
4
f ′( x) =
(b)
2
(
−1, − π
4
)
−3
3
−2
(1, 5)
1
f ′( x) =
(
f′
y −
2
1 − (3 x )
)
π
4
4 − x
x
−2
x2
(4
− x) x
(b)
2
(3)
3
2 6 =
1 − 9(1 18)
(
= 3 2 x −
y = 3 2x +
( f )′ (2) = − 12
−1
, tangent line
17. (a) f ( x) = arcsin 3 x
+ 1)
f −1 ( x ) =
=
2
f ( x) = arctan x
f ′(1) = −2
( f −1 )′ ( x)
1 − x4
0
4
1 + x2
−8 x
( x2
π
(0, π2 (
−1.5
(−1, 1)
f −1 ( x) = x 2 + 4,
f ′( x) =
− 2x
=
2
16. (a)
−1
f ( x) =
( 2 x)
= 0( x − 0)
(b)
f ′(1) = −4
14.
2
4
3
f ( x ) = 3 − 4 x,
2
1
f ′(5) =
2
π
y −
1⎞
⎟
2⎠
f ′( x) = −4
13.
1− x
4
f ′(0) = 0
3
⎛1⎞
f ′⎜ ⎟ =
4
⎝ 2⎠
( f −1 )′ ⎛⎜⎝ 18 ⎞⎟⎠
−1
f ′( x) =
f ′( x) = 3 x 2
2 6
π
=
=
3
1 − 9 x2
3
= 3 2
12
)
− 1, Tangent line
4
2
−1
1
(
2 π
,
6 4
(
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.6
f ( x) = arcsec x
18. (a)
( 2) =
1
=
2
f′
y −
4
1 − ( xy )
2
2
(
2
x −
2
=
2
)
π
2, 4
⎛ 3
⎞
2 3 −6
y′ = ⎜⎜
− 1⎟⎟ 2 =
3
⎝ 3
⎠
)
23.
−4
5
x = y3 − 7 y 2 + 2
dy
dy
1 = 3y
− 14 y
dx
dx
dy
1
=
dx
3 y 2 − 14 y
24.
2
dy
1
−1
At ( −4, 1):
=
=
.
dx
3 − 14
11
− x2 − 2x
2t
1 − t4
x
2
−3(1 2)
g ′( x) =
−3
=
1 − ( x 4)
4 − x2
2
2
f ′( x) = 3x 2 − 14 x and f ′(1) = −11. So,
26.
x = 2 ln ( y 2 − 3)
27.
dy
1
2y
y − 3 dx
2
4x − 1
2x
1
=
2
x
4x2 − 1
f ( x) = arctan (e x )
f ′( x) =
2
dy
y −3
=
dx
4y
f ( x) = arcsec 2 x
f ′( x) =
dy
1
−1
=
=
.
dx
−11
11
1 = 2
1 − ( x + 1)
1
=
2
25. g ( x) = 3 arccos
Let f ( x) = x − 7 x + 2. Then
20.
1
f (t ) = arcsin t 2
f ′(t ) =
Alternate Solution:
3
f ( x) = arcsin ( x + 1)
f ′( x) =
−2
19.
1 ⎛1
2
⎞
⎜ y′ + 1⎟ =
3
3 4⎝2
⎠
2 ⎛1
2
⎞
⎜ y′ + 1⎟ =
3
3⎝ 2
⎠
4
(
2
⎛1 ⎞
At ⎜ , 1⎟:
⎝2 ⎠
2
π
x +
− 1, tangent line
2
4
y =
(b)
2
arctan ( 2 x)
3
2
1
⎛ dy
⎞
+ y⎟ =
(2)
⎜x
3 1 + 4x2
⎝ dx
⎠
1
x2 − 1
x
217
arctan ( xy ) =
22.
1
f ′( x) =
π
Derivatives of Inverse Functions
1
1 + (e x )
2
ex
1 + e2 x
ex =
2
At (0, 2):
21.
dy
4−3
1
=
=
dx
8
8
x arctan x = e y
1
dy
+ arctan x = e y ⋅
x
dx
1 + x2
π⎞ 1 π
π dy
⎛
At ⎜1, ln ⎟: +
=
4⎠ 2
4
4 dx
⎝
dy
π + 2
=
dx
π
28.
f ( x) = arctan
x
⎛ 1 ⎞⎛ 1 ⎞
f ′( x) = ⎜
⎟⎜
⎟ =
2
⎝ 1 + x ⎠⎝ 2 x ⎠
29. g ( x) =
g ′( x ) =
=
arcsin 3x
x
(
x3
)
1 − 9 x 2 − arcsin 3 x
x
3x −
1
x (1 + x)
2
1 − 9 x 2 arcsin 3x
x2 1 − 9 x2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
218
Chapter 3
30. g ( x) =
g ′( x) =
Differentiation
arccos x
x +1
(x
= −
34.
− arccos x
1 − x2
2
( x + 1)
x +1+
(x
2
h′(t ) =
1
+ 2e 2 x arcsin x
1 − x2
=
⎤
⎥
1− x ⎦
1
36.
2
h′( x) = 2 x arctan (5 x) + x 2
1
1 + (5 x )
2
(5)
−1 2
1
(1 − t 2 ) (−2t )
2
−t
1 − t2
f ( x) = arcsin x + arccos x =
π
2
37. y = 2 x arccos x − 2 1 − x 2
−1 2
⎛1⎞
− 2⎜ ⎟(1 − x 2 ) ( −2 x)
2
⎝
⎠
1− x
2x
2x
+
= 2 arccos x
2
1− x
1 − x2
1
y′ = 2 arccos x − 2 x
5x2
= 2 x arctan (5 x) +
1 + 25 x 2
= 2 arccos x −
33. h( x) = arccot 6 x
−6
1 + 36 x 2
2
38. y = ln (t 2 + 4) −
1
t
arctan
2
2
2t
1
1
⎛1⎞
y′ = 2
− ⋅
2⎜ ⎟
t + 4
2 1 + ( t 2) ⎝ 2 ⎠
=
y =
1 − t2
f ′( x) = 0
32. h( x) = x 2 arctan (5 x)
39.
9 x2 − 1
35. h(t ) = sin (arccos t ) =
1 − x2
⎡
= e 2 x ⎢2 arcsin x +
⎣
h′( x) =
−1
x
1 − x 2 arccos x
+ 1)
9x2 − 1
3x
=
31. g ( x) = e 2 x arcsin x
g ′( x) = e 2 x
−3
f ′( x) =
−1
+ 1)
f ( x) = arccsc 3 x
2t
1
2t − 1
−
= 2
t2 + 4 t2 + 4
t + 4
1⎛ 1 x + 1
1
1
⎞
+ arctan x ⎟ = ⎡⎣ln ( x + 1) − ln ( x − 1)⎤⎦ + arctan x
⎜ ln
2⎝ 2 x − 1
4
2
⎠
dy
1⎛ 1
1 ⎞
12
1
= ⎜
−
=
⎟+
dx
4 ⎝ x + 1 x − 1⎠ 1 + x2
1 − x4
40. y =
y′ =
1⎡
⎛ x ⎞⎤
x 4 − x 2 + 4 arcsin ⎜ ⎟⎥
2 ⎢⎣
⎝ 2 ⎠⎦
⎡
−1 2
1⎢ 1
x ( 4 − x 2 ) ( −2 x) +
2⎢ 2
⎣
41. g (t ) = tan (arcsin t ) =
g ′(t ) =
(
1 − t − t −t
2
4 − x2 + 2
⎤
⎥ = 1⎡
⎢
2⎥
2⎣
1 − ( x 2) ⎦
1
t
1−t
43.
2
1 − t2
1 − t2
)=
f ′( x) = 0
π
2
+
4 − x2 +
y = x arcsin x +
⎤
4
⎥ =
4 − x2 ⎦
1 − x2
⎞
⎟ + arcsin x −
1− x ⎠
1
2
4 − x2
x
1 − x2
= arcsin x
32
44.
42. f ( x ) = arcsec x + arccsc x =
4 − x2
⎛
dy
= x⎜
dx
⎝
1
(1 − t 2 )
− x2
1
ln (1 + 4 x 2 )
4
dy
2x
1 ⎛ 8x ⎞
=
+ arctan ( 2 x) − ⎜
⎟ = arctan ( 2 x)
4 ⎝ 1 + 4 x2 ⎠
dx 1 + 4 x 2
y = x arctan 2 x −
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.6
45. y = 8 arcsin
y′ = 2
1
1 − ( x 4)
46. y = 25 arcsin
=
=
1 − ( x 2)
=
−
2
25 − x 2 − x
−1 2
1
25 − x 2 ) ( −2 x) =
(
2
x
1 + x2
16 − x 2
25
25 − x 2
50. y =
−
(25 − x 2 )
x2
+
25 − x 2
2 x2
=
25 − x 2
25 − x 2
⎛
2 3π ⎞
,
⎜⎜ −
⎟⎟
⎝ 2 8 ⎠
1
arccos x,
2
y′ =
(1 + x 2 ) + (1 − x 2 )
2
(1 + x 2 )
⎛
2 3π ⎞
2
−1
At ⎜⎜ −
,
.
= −
⎟⎟, y′ =
2
8
2
2
1
2
⎝
⎠
2
(1 + x )
3π
2⎛
= −
⎜x +
8
2 ⎜⎝
2
1
2
⎛1⎞
⎜ ⎟ =
2
4 + x2
1 + ( x 4) ⎝ 2 ⎠
Tangent line: y −
π
1
( x − 2)
4
π
1
1
−
y = x +
4
4
2
4
⎛1 π ⎞
⎜ , ⎟
⎝2 3⎠
2
1 − x2
Tangent line: y −
2
1 − (1 4)
π
3
=
y =
y =
2
3π
1
−
x +
2
8
2
2
1
⎛ π⎞
At ⎜ 2, ⎟, y′ =
= .
4 + 4
4
⎝ 4⎠
2x2 + 8 + x
⎛1 π ⎞
At ⎜ , ⎟, y′ =
⎝2 3⎠
2⎞
⎟
2 ⎟⎠
⎛ π⎞
⎜ 2, ⎟
⎝ 4⎠
⎛ x⎞
51. y = arcsin ⎜ ⎟,
⎝ 2⎠
y′ =
2
x
+
2
2
x2 + 4
( x + 4)
+ 4)
2 1 − x2
y = −
x
1
−
2
2( x 2 + 4)
−2
1
1
1
+ ( x 2 + 4) ( 2 x )
2
2 1 + ( x 2)
2
( x2
−1
Tangent line: y −
2
49. y = 2 arcsin x,
y′ =
x2
(1 + x2 ) − x(2 x)
1
+
2
1 + x2
(1 + x 2 )
48. y = arctan
=
16 − (16 − x 2 ) + x 2
16 − x 2
x2
+
=
=
2
2 16 − x 2
2 16 − x 2
1
2
y′ =
2
x
− x 25 − x 2
5
47. y = arctan x +
y′ =
−1 2
16 − x 2
x
− (16 − x 2 ) ( −2 x)
2
4
−
−
16 − x 2
y′ = 5
219
x
x 16 − x 2
−
4
2
8
=
Derivatives of Inverse Functions
=
4
.
3
4 ⎛
1⎞
⎜x − ⎟
2⎠
3⎝
4
2
π
x +
−
3
3
3
4 3
2 3
π
x +
−
3
3
3
=
⎛ 2 π⎞
52. y = arcsec( 4 x), ⎜
⎜ 4 , 4 ⎟⎟
⎝
⎠
4
1
y′ =
for x > 0
=
x 16 x 2 − 1
4 x 16 x 2 − 1
⎛ 2 π⎞
At ⎜⎜
, ⎟⎟, y′ =
⎝ 4 4⎠
Tangent line: y −
(
1
)
2 4
2−1
⎛
= 2 2 ⎜⎜ x −
4
⎝
2⎞
⎟
4 ⎟⎠
π
y = 2 2x +
= 2 2.
π
4
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
220
Chapter 3
Differentiation
53. y = 4 x arccos( x − 1),
−1
y′ = 4 x
1 − ( x − 1)
2
(1, 2π )
⎛1 π ⎞
54. y = 3 x arcsin x, ⎜ , ⎟
⎝2 4⎠
1
y′ = 3 x
+ 3 arcsin x
1 − x2
+ 4 arccos( x − 1)
At (1, 2π ), y′ = − 4 + 2π .
3
⎛1 π ⎞
At ⎜ , ⎟, y′ =
2
4
2
⎝
⎠
Tangent line: y − 2π = ( 2π − 4)( x − 1)
y = ( 2π − 4) x + 4
55.
π
Tangent line: y −
⎛
= ⎜
⎝
3 +
⎛
y = ⎜
⎝
3 +
4
3 +
π
2
.
π ⎞⎛
1⎞
⎟⎜ x − ⎟
2 ⎠⎝
2⎠
π⎞
3
⎟x −
2⎠
2
f ( x) = arccos x
−1
f ′( x) =
= − 2 when x = ±
1 − x2
When x =
3 2, f
Tangent lines:
y −
(
3
.
2
)
3 2 = π 6. When x = −
(
3 2, f −
⎛
3⎞
⎛π
= − 2⎜⎜ x −
⎟⎟ ⇒ y = − 2 x + ⎜ +
6
2
⎝6
⎝
⎠
π
56. g ( x) = arctan x, g ′( x) =
1
1
, g ′(1) =
1 + x2
2
π
1
Tangent line: y −
= ( x − 1)
4
2
π
1
1
y = x +
−
2
4
2
f ( x) = arctan x, a = 0
f ( 0) = 0
)
3 2 = 5π 6.
⎛
5π
3⎞
⎛ 5π
= − 2⎜⎜ x +
−
⎟⎟ ⇒ y = − 2 x + ⎜
6
2 ⎠
⎝ 6
⎝
y −
57.
1
⎛π ⎞
+ 3⎜ ⎟ =
34
⎝6⎠
⎞
3⎟
⎠
⎞
3⎟
⎠
58.
f ( x ) = arccos x, a = 0
−1
f ′( x ) =
1 − x2
−x
f ′′( x ) =
(1 − x 2 )
f ′(0) = −1
,
32
f ′′(0) = 0
,
f ′(0) = 1
f ′′(0) = 0
P2 ( x) = f (0) + f ′(0) x +
y
f 3
1
f ′′(0) x 2 = x
2
P1 = P2
1
−2
y
π
− x
2
π
1
− x
P2 ( x ) = f (0) + f ′(0) x + f ′′(0) x 2 =
2
2
P1 ( x) = f (0) + f ′(0) x = x
1.0
2
P1 ( x ) = f (0) + f ′(0) x =
1
f ′( x) =
,
1 + x2
− 2x
f ′′( x) =
,
2
(1 + x 2 )
1.5
π
f ( 0) =
x
−1
1
−1
P1 = P 2
f
0.5
x
− 1.0
0.5 1.0 1.5
− 1.0
− 1.5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.6
59.
f ( x) = arcsin x, a =
Derivatives of Inverse Functions
y
1
2
1.5
0.5
1 − x2
x
f ′′( x) =
P1
1.0
1
f ′( x) =
(1 − x 2 )
221
x
0.5 1.0 1.5
P2
− 1.0
32
f
− 1.5
1⎞
2 3⎛
1⎞
π
⎛1⎞
⎛ 1 ⎞⎛
P1 ( x) = f ⎜ ⎟ + f ′⎜ ⎟⎜ x − ⎟ =
+
⎜x − ⎟
2
2
2
6
3
2⎠
⎝ ⎠
⎝ ⎠⎝
⎠
⎝
π
1 ⎞ 1 ⎛ 1 ⎞⎛
1⎞
2 3⎛
1 ⎞ 2 3⎛
1⎞
⎛1⎞
⎛ 1 ⎞⎛
P2 ( x) = f ⎜ ⎟ + f ′⎜ ⎟⎜ x − ⎟ + f ′′⎜ ⎟⎜ x − ⎟ =
+
⎜x − ⎟ +
⎜x − ⎟
2 ⎠ 2 ⎝ 2 ⎠⎝
2⎠
6
3 ⎝
2⎠
9 ⎝
2⎠
⎝ 2⎠
⎝ 2 ⎠⎝
2
60.
f ( x) = arcsin x, a = 1
f ′( x) =
f ′′( x) =
2
y
1
1 + x2
− 2x
P1 (x)
π
2
f
π
4
(1 + x 2 )
x
2
−4
P1 ( x) = f (1) + f ′(1)( x − 1) =
π
−2
2
P2 (x)
1
( x − 1)
2
+
4
4
1
1
1
π
2
2
P2 ( x) = f (1) + f ′(1)( x − 1) + f ′′(1)( x − 1) =
+ ( x − 1) − ( x − 1)
2
4
2
4
⎛ π ⎞
⎜ − , 1⎟
⎝ 4 ⎠
x 2 + x arctan y = y − 1,
61.
2 x + arctan y +
63.
x
y ′ = y′
1 + y2
1
1 − x2
⎛
x ⎞
⎜1 −
⎟ y′ = 2 x + arctan y
+
y2 ⎠
1
⎝
2 x + arctan y
y′ =
x
1−
1 + y2
π
π
y =
1
1 + ( xy )
2
[y
+ xy′] =
Tangent line: y −
(0, 0)
1
1 − ( x + y)
2
[1 + y′]
⎛ 2
2⎞
⎜⎜ 2 , 2 ⎟⎟
⎝
⎠
,
y′ = 0
−1
y′ =
1 − x2
⎛
2
= −1⎜⎜ x −
2
⎝
y = −x +
arctan ( x + y ) = y 2 +
1
1 + ( x + y)
At (1, 0):
2
[1 + y′]
π
4
2⎞
⎟
2 ⎟⎠
2
,
(1, 0)
= 2 yy′
1
[1 + y′] = 0 ⇒ y′ = −1
2
Tangent line: y − 0 = −1( x − 1)
y = −x + 1
At (0, 0): 0 = 1 + y′ ⇒ y′ = −1
Tangent line: y = − x
2
2
arctan ( xy ) = arcsin ( x + y ),
62.
2
π
⎛ 2
2⎞
′
At ⎜
⎜ 2 , 2 ⎟⎟ : y = −1
⎝
⎠
64.
−2π
π
x +1−
8+π
16 + 2π
1 − y2
1− y
π
−2π ⎛
π⎞
⎜x + ⎟
8 + π⎝
4⎠
1
+
1
− +
−
⎛ π ⎞
2
4 =
2 = −2π
At ⎜ − , 1⎟ : y′ =
4
π
π
−
8+π
⎝ 4 ⎠
1−
2+
2
4
Tangent line: y − 1 =
arcsin x + arcsin y =
65. f is not one-to-one because many different x-values yield
the same y-value.
Example: f (0) = f (π ) = 0
Not continuous at
( 2n
− 1)π
2
, where n is an integer.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
222
Chapter 3
Differentiation
66. f is not one-to-one because different x-values yield the
same y-value.
73. (a) cot θ =
⎛ x⎞
3
⎛ 4⎞
Example: f (3) = f ⎜ − ⎟ =
5
⎝ 3⎠
θ = arccot ⎜ ⎟
⎝5⎠
Not continuous at ± 2.
(b)
67. Because you know that f
−1
−1 5
−5 dx
dθ
dx
=
= 2
2
+ 25 dt
dt
dt
x
1 + ( x 5)
exists and that
y1 = f ( x1 ) by Theorem 3.17, then ( f −1 )′ ( y1 ) =
1
,
f ′( x1 )
provided that f ′( x1 ) ≠ 0.
68. Theorem 3.17: Let f be a function that is differentiable
on an interval I. If f has an inverse function g, then g is
If
dx
dθ
= − 400 and x = 10,
= 16 rad/h.
dt
dt
If
dx
dθ
= − 400 and x = 3,
≈ 58.824 rad/h.
dt
dt
74. (a) cot θ =
differentiable at any x for which f ′( g ( x)) ≠ 0.
Moreover, g ′( x ) =
⎛ x⎞
1
, f ′( g ( x)) ≠ 0.
f ′( g ( x))
(b)
dθ
−3 dx
= 2
dt
x + 9 dt
If x = 10,
(
)
at ( − 12 , 1)
70. (a) Since the slope of the tangent line to f at −1, − 12
1 , the
2
is m =
slope of the tangent line to f −1
1
(1 2)
If x = 3,
dθ
≈ 11.001 rad/h.
dt
dθ
≈ 66.667 rad/h.
dt
A lower altitude results in a greater rate of change
of θ .
= 2.
(b) Since the slope of the tangent line to f at ( 2, 1) is 2,
the slope of the tangent line to f
m =
x
3
θ = arccot ⎜ ⎟
⎝ 3⎠
69. The derivatives are algebraic. See Theorem 3.18.
is
x
5
−1
at (1, 2) is
h(t ) = −16t 2 + 256
75. (a)
−16t 2 + 256 = 0 when t = 4 sec
1
.
2
h
71. Because the slope of f at (1, 3) is m = 2, the slope of
θ
500
f −1 at (3, 1) is 1 2.
h
−16t 2 + 256
=
500
500
16
⎡
θ = arctan ⎢
(−t 2 + 16)⎤⎥⎦
⎣ 500
(b) tan θ =
72. From Example 5, you have y′ = 2 1 − x 2 . At the
point (0, 0), m = 2 1 − 0 = 2, and the equation of the
tangent line is y = 2 x. On the interval ( − 0.266, 0.266),
the tangent line is within 0.01 unit of the graph of the
original function. A person saying that the original
function is “locally linear” means that a linear function
is a good approximation of the original function near a
point (in this case, the origin).
dθ
−8t 125
=
2
dt
1 + ⎡⎣( 4 125)( −t 2 + 16)⎤⎦
−1000t
=
2
15,625 + 16(16 − t 2 )
When t = 1, dθ dt ≈ −0.0520 rad/sec.
When t = 2, dθ dt ≈ −0.1116 rad/sec.
76. cos θ =
800
s
⎛ 800 ⎞
θ = arccos⎜
⎟
⎝ s ⎠
dθ
dθ ds
=
⋅
=
dt
ds dt
s
−1
⎛ −800 ⎞ ds
=
⎜ 2 ⎟
s ⎠ dt
⎝
s
1 − (800 s )
2
ds
800
,
s 2 − 8002 dt
s > 800
θ
800
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.6
77. tan θ =
h
300
y
78.
300
dh
= 5 ft sec
dt
200
⎛ h ⎞
⎟
⎝ 300 ⎠
θ = arctan ⎜
100
θ
dθ
1 300
⎛ dh ⎞
=
⎜ ⎟
dt
1 + ( h 2 3002 ) ⎝ dt ⎠
h
200
300
θ
50
x
dx
x 2 + 2500 dθ
=
dt
50
dt
300
=
(5)
3002 + h 2
1500
3
rad sec when h = 100
=
=
3002 + h 2
200
79. (a)
dθ
= 30( 2π ) = 60π rad/min
dt
x
tan θ =
50
⎛ x⎞
θ = arctan ⎜ ⎟
⎝ 50 ⎠
223
dθ
dθ dx
50
dx
=
= 2
dt
dx dt
x + 2500 dt
x
100
Derivatives of Inverse Functions
When θ = 45° =
π
4
, x = 50:
dx
(50) + 2500 60π = 6000π ft/min
=
( )
dt
50
2
Let y = arccos u. Then
cos y = u
dy
= u′
dx
dy
u′
= −
= −
dx
sin y
− sin y
(b)
1
u′
1 − u2
1 − u2
y
.
u
Let y = arctan u. Then
tan y = u
sec 2 y
1 + u2
dy
= u′
dx
dy
u′
u′
=
=
.
2
dx
sec y
1 − u2
u
y
1
Let y = arcsec u. Then
(c)
sec y = u
sec y tan y
dy
= u′
dx
dy
u′
=
=
dx
sec y tan y
u
u
u′
u2 − 1
u2 − 1
y
.
1
Note: The absolute value sign in the formula for the derivative of arcsec u is necessary because the inverse secant
function has a positive slope at every value in its domain.
(d)
Let y = arccot u. Then
cot y = u
− csc 2 y
(e)
1 + u2
dy
= u′
dx
dy
u′
u′
=
= −
.
− csc 2 y
dx
1 + u2
1
y
u
Let y = arccsc u. Then
csc y = u
− csc y cot y
dy
= u′
dx
dy
u′
=
= −
− csc y cot y
dx
u
u
1
u′
u −1
2
.
y
u2 − 1
Note: The absolute value sign in the formula for the derivative of arccsc u is necessary because the inverse cosecant
function has a negative slope at every value in its domain.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
224
Chapter 3
Differentiation
80. f ( x) = kx + sin x
87. f ( x ) = sec x,
For k ≥ 1, f is one-to-one, and for k ≤ −1, f is
0 ≤ x <
π
2
,π ≤ x <
3π
2
y
one-to-one. Therefore, f has an inverse for k ≥ 1
and k ≤ −1.
4
2
81. True
u′
d
[arcsec u] =
dx
u
−π
2
u2 − 1
x
π
2
−2
−4
d
− u′
[arccsc u] =
dx
u u2 − 1
(a) y = arcsec x,
π
0 ≤ y <
82. True
d
1
> 0 for all x.
[arctan x] =
dx
1 + x2
x ≤ −1
3π
2
π ≤ y <
or
2
x ≥1
or
y
3π
2
83. True
d
sec 2 x
sec 2 x
⎡arctan ( tan x)⎦⎤ =
=
=1
⎣
1 + tan 2 x
sec2 x
dx
π
2
⎛
⎞
⎟,
1− x ⎠
x
θ = arctan ⎜
85. Let
⎝
2
1− x
x
= x
sin θ =
1
arcsin x = θ .
y′ =
tan y + 1 = sec y
⎞
⎟ for −1 < x < 1.
1− x ⎠
2
tan y = ±
x
2
x
⎛ x − 2⎞
,
f ( x) = arcsin ⎜
⎟ − 2 arcsin
2
2
⎝
⎠
x
,
x < 1.
=
1
1
2 1 − (1 4)( x − 4 x + 4)
2
, as indicated in the figure.
π
⎛π
⎞
So, cos⎜ − θ ⎟ = x and
− θ = arccos x which
2
2
⎝
⎠
π
x
− arctan
.
gives arccos x =
2
1 − x2
(
0≤ x ≤ 4
)
⎡ 1 4 x ⎤
⎥
− 2⎢
2
⎢1 − ( x 2)2 ⎥
1 − ⎡⎣( x − 2) 2⎤⎦
⎣
⎦
12
f ′( x) =
1 − x2
sec 2 y − 1
On 0 ≤ y < π 2 and π ≤ y < 3π 2, tan y ≥ 0.
x2
π−
θ
2
x2 − 1
x
88.
1 − x2
1
2
x
Then tan θ =
1
sec y tan y
=
θ
x
6
1 = sec y tan y ⋅ y′
1
86. Let θ = arctan
4
x = sec y
−1 < x < 1
2
⎛
So, arcsin x = arctan ⎜
⎝
1−
2
y = arcsec x
(b)
x
tan θ =
x
−6 −4 −2
dy
84. False. The derivative
is undefined when x = ± 1.
dx
1
=
2
x − ( x 4)
2
1
−
2
x 1 − ( x 4)
1
−
2
x − ( x 2 4)
=0
Because the derivative is zero, you can conclude that
the function is constant. (By letting x = 0 in f ( x),
you can see that the constant is − π 2.)
x
θ
1 − x2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.7
Related Rates
225
Section 3.7 Related Rates
y =
1.
x 2 + y 2 = 25
4.
x
dy
⎛ 1 ⎞ dx
= ⎜
⎟
dt
⎝ 2 x ⎠ dt
dx
dy
= 2 x
dt
dt
2x
dx
dy
+ 2y
= 0
dt
dt
⎛ x ⎞ dx
dy
= ⎜− ⎟
dt
⎝ y ⎠ dt
(a) When x = 4 and dx dt = 3:
dx
⎛ y ⎞ dy
= ⎜− ⎟
dt
⎝ x ⎠ dt
dy
1
3
=
(3) =
dt
4
2 4
(a) When x = 3, y = 4, and dx dt = 8:
(b) When x = 25 and dy dt = 2:
dy
3
= − (8) = − 6
dt
4
dx
= 2 25 ( 2) = 20
dt
(b) When x = 4, y = 3, and dy dt = −2:
dx
3
3
= − ( −2 ) =
dt
4
2
y = 3x 2 − 5 x
2.
dy
dx
= (6 x − 5)
dt
dt
dx
dy
1
=
dt
6 x − 5 dt
(a) When x = 3 and
5.
dx
= 2
dt
dy
dx
= 4x
dt
dt
(a) When x = −1:
dx
= 2:
dt
dy
= ⎡⎣6(3) − 5⎤⎦ 2 = 26
dt
(b) When x = 2 and
dy
= 4( −1)( 2) = −8 cm/sec
dt
dy
= 4:
dt
(b) When x = 0:
dx
1
4
=
(4) =
6( 2) − 5
7
dt
dy
= 4(0)( 2) = 0 cm/sec
dt
(c) When x = 1:
xy = 4
3.
x
dy
dx
+ y
= 0
dt
dt
dy
⎛ y ⎞ dx
= ⎜− ⎟
dt
⎝ x ⎠ dt
⎛ x ⎞ dy
dx
= ⎜− ⎟
dt
⎝ y ⎠ dt
(a) When x = 8, y = 1/2, and dx dt = 10:
dy
1/2
5
= − (10) = −
dt
8
8
(b) When x = 1, y = 4, and dy dt = −6:
dx
1
3
= − ( −6) =
dt
4
2
y = 2x2 + 1
dy
= 4(1)( 2) = 8 cm/sec
dt
6.
dx
1
,
= 6
1 + x 2 dt
dy
dx
− 2x
− 2x
−12 x
6 =
=
⋅
=
2
2( )
2
2
2
dt
dt
1
1
1
+
+
x
x
(
)
(
)
( + x2 )
y =
(a) When x = − 2:
( −12)(− 2) = 24 in./sec
dy
=
2
dt
25
⎡1 + ( − 2)2 ⎤
⎣
⎦
(b) When x = 0:
−12(0)
dy
=
= 0 in./sec
dt
(1 + 0)2
(c) When x = 2:
( −12)(2) = − 24 in./sec
dy
=
2
dt
25
(1 + 22 )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
226
7.
Chapter 3
Differentiation
dx
= 3
dt
dy
dx
= sec 2 x ⋅
= sec 2 x(3) = 3 sec 2 x
dt
dt
y = tan x,
12. (a) sin
θ
2
dy
2
⎛ π⎞
= 3 sec 2 ⎜ − ⎟ = 3( 2) = 12 ft/sec
dt
⎝ 3⎠
=
π
(b) When x = − :
4
(b)
2
s
= 6 ft/sec
dA
s2
dθ
dθ
1
cos θ
where
=
= rad/min.
dt
2
dt
dt
2
When θ =
When θ =
dy
= 3 sec 2 (0) = 3 ft/sec
dt
(a) When x =
6
π
4
13.
s 2 ⎛ 1 ⎞⎛ 1 ⎞
s2
.
⎜ ⎟⎜ ⎟ =
2 ⎝ 2 ⎠⎝ 2 ⎠
8
s
V =
4 3
πr
3
(a) When r = 9,
:
dV
2
= 4π (9) (3) = 972π in.3 /min.
dt
When r = 36,
dV
2
= 4π (36) (3) = 15,552π in.3 /min.
dt
9. Yes, y changes at a constant rate.
dy
dx
= a⋅
dt
dt
(b) If dr dt is constant, dV dt is proportional to r 2 .
No, the rate dy dt is a multiple of dx dt .
10. Answers will vary. See page 149.
A = πr2
dr
= 4
dt
dA
dr
= 2π r
dt
dt
(a) When r = 8,
3 dt
=
3s 2
.
8
dr
= 3
dt
dV
dr
= 4π r 2
dt
dt
⎛ 3⎞
dy
⎛π ⎞
= − 4 sin ⎜ ⎟ = − 4⎜⎜
⎟⎟ = − 2 3 cm/sec
dt
⎝3⎠
⎝ 2 ⎠
11.
,
s 2 ⎛ 3 ⎞⎛ 1 ⎞
⎜
⎟⎜ ⎟ =
2 ⎜⎝ 2 ⎟⎠⎝ 2 ⎠
b
:
3
π dA
=
h
:
π
6 dt
s
⎛ 2⎞
dy
⎛π ⎞
= − 4 sin ⎜ ⎟ = − 4⎜⎜
⎟⎟ = − 2 2 cm/sec
dt
⎝4⎠
⎝ 2 ⎠
(c) When x =
,
dθ
dA
(c) If s and
is constant,
is proportional to cos θ .
dt
dt
dy
⎛π ⎞
⎛1⎞
= − 4 sin ⎜ ⎟ = − 4⎜ ⎟ = − 2 cm/sec
dt
⎝6⎠
⎝ 2⎠
(b) When x =
π dA
θ
dx
= 4
dt
dy
dx
= − sin x ⋅
= − sin x( 4) = − 4 sin x
dt
dt
y = cos x,
π
θ
s2 ⎛
s2
θ
θ⎞
sin θ
⎜ 2 sin cos ⎟ =
2⎝
2
2⎠
2
(c) When x = 0:
8.
⇒ b = 2 s sin
2
h
θ
cos =
⇒ h = s cos
s
2
2
1
1⎛
θ ⎞⎛
θ⎞
A = bh = ⎜ 2s sin ⎟⎜ s cos ⎟
2
2⎝
2 ⎠⎝
2⎠
π
( 2)
(1 2)b
θ
(a) When x = − :
3
dy
⎛ π⎞
= 3 sec 2 ⎜ − ⎟ = 3
dt
⎝ 4⎠
=
14.
4 3 dV
πr ,
= 800
3
dt
dV
dr
= 4π r 2
dt
dt
dr
1 ⎛ dV ⎞
1
=
(800)
⎜
⎟ =
dt
4π r 2 ⎝ dt ⎠
4π r 2
V =
(a) When r = 30,
dA
= 2π (8)( 4) = 64π cm 2 /min.
dt
(b) When r = 32,
dA
= 2π (32)( 4) = 256π cm 2 /min.
dt
dr
1
2
800) =
cm/min.
=
2(
dt
9
π
4π (30)
(b) When r = 60,
dr
1
1
800) =
cm/min.
=
2(
dt
18
π
4π (60)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.7
V = x3
15.
17.
dx
= 6
dt
dV
dx
= 3x 2
dt
dt
V =
=
Related Rates
227
[because 2r
= 3h]
1 2
1 ⎛9 ⎞
π r h = π ⎜ h 2 ⎟h
3
3 ⎝4 ⎠
3π 3
h
4
dV
= 10
dt
(a) When x = 2,
4( dV dt )
dV
dh
9π 2 dh
h
=
⇒
=
dt
dt
dt
4
9π h 2
dV
2
= 3( 2) (6) = 72 cm3/sec.
dt
When h = 15,
(b) When x = 10,
4(10)
dh
8
ft/min.
=
=
2
dt
405π
9π (15)
dV
2
= 3(10) (6) = 1800 cm 3/sec.
dt
s = 6x2
16.
dx
= 6
dt
ds
dx
= 12 x
dt
dt
h
r
(a) When x = 2,
ds
= 12( 2)(6) = 144 cm 2 /sec.
dt
(b) When x = 10,
ds
= 12(10)(6) = 720 cm 2 /sec.
dt
V =
18.
1 2
1 25 3
25π 3
h =
h
πr h = π
3
3 144
3(144)
r
h
5 ⎞
⎛
h.⎟
⇒ r =
⎜ By similar triangles, =
5
12
12 ⎠
⎝
5
dV
= 10
dt
dV
25π 2 dh
dh
⎛ 144 ⎞ dV
h
=
⇒
= ⎜
2⎟
dt
dt
dt
144
⎝ 25π h ⎠ dt
When h = 8,
19.
r
12
h
dh
144
9
ft/min.
=
(10) =
25π (64)
10π
dt
12
6
1
3
1
(a) Total volume of pool =
1
(2)(12)(6) + (1)(6)(12) = 144 m3
2
Volume of 1 m of water =
1
(1)(6)(6) = 18 m3
2
2
h=1
(see similar triangle diagram)
18
% pool filled =
(100%) = 12.5%
144
12
b=6
(b) Because for 0 ≤ h ≤ 2, b = 6h, you have
V =
1
bh(6) = 3bh = 3(6h)h = 18h 2
2
dV
dh
1
dh
1
1
1
m/min.
= 36h
=
⇒
=
=
=
4
144h
144(1)
144
dt
dt
dt
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
228
Chapter 3
1
bh(12) = 6bh = 6h 2
2
20. V =
(a)
Differentiation
(since b
= h)
dV
dh
dh
1 dV
= 12h
⇒
=
dt
dt
dt
12h dt
When h = 1 and
dV
dh
1
1
= 2,
=
(2) = ft/min.
12(1)
6
dt
dt
12 ft
3 ft
h ft
3 ft
(b) If
dh
3
1
dV
3
⎛1⎞
ft/min and h = 2 ft, then
= in./min =
= (12)( 2)⎜ ⎟ = ft 3 /min.
dt
8
32
dt
4
⎝ 32 ⎠
x 2 + y 2 = 252
21.
2x
dx
dy
+ 2y
= 0
dt
dt
dy
− x dx
−2 x
=
⋅
=
dt
y dt
y
(a) When x = 7, y =
(b)
576 = 24,
because
25
y
dx
= 2.
dt
−2(7)
dy
7
=
= − ft/sec.
dt
24
12
x
−2(15)
dy
3
=
= − ft/sec.
dt
20
2
When x = 15, y =
400 = 20,
When x = 24, y = 7,
−2( 24)
dy
48
=
= −
ft/sec.
dt
7
7
1
xy
2
dA
dx ⎞
1 ⎛ dy
= ⎜x
+ y ⎟
dt
dt ⎠
2 ⎝ dt
A =
From part (a) you have x = 7, y = 24,
dx
dy
7
= 2, and
= − . So,
dt
dt
12
dA
⎤
1⎡ ⎛ 7 ⎞
527 2
= ⎢7⎜ − ⎟ + 24( 2)⎥ =
ft /sec.
dt
2 ⎣ ⎝ 12 ⎠
24
⎦
(c)
tan θ =
sec2 θ
x
y
θ
25
y
dθ
1 dx
x dy
=
⋅
− 2 ⋅
dt
y dt
y
dt
⎡ 1 dx
dθ
x dy ⎤
= cos 2θ ⎢ ⋅
− 2 ⋅ ⎥
dt
y
dt ⎦
⎣ y dt
x
dx
dy
7
24
= 2,
= − and cos θ =
, you have
dt
dt
12
25
2
dθ
7 ⎛ 7 ⎞⎤
1
⎛ 24 ⎞ ⎡ 1
rad/sec.
= ⎜ ⎟ ⎢ ( 2) −
− ⎟⎥ =
2⎜
dt
⎝ 25 ⎠ ⎢⎣ 24
( 24) ⎝ 12 ⎠⎥⎦ 12
Using x = 7, y = 24,
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.7
Related Rates
229
x 2 + y 2 = 25
22.
2x
dx
dy
+ 2y
= 0
dt
dt
dx
y dy
0.15 y
= − ⋅
= −
dt
x dt
x
When x = 2.5, y =
23. When y = 6, x =
18.75,
dx
= −
dt
dy
⎛
⎞
= 0.15 ⎟
⎜ because
dt
⎝
⎠
18.75
0.15 ≈ −0.26 m/sec.
2.5
122 − 62 = 6 3, and s =
x 2 + (12 − y )
5
y
2
x
=
108 + 36 = 12.
s
12 − y
( x, y )
x
y
12
x 2 + (12 − y ) = s 2
2
2x
dx
dy
ds
+ 2(12 − y )( −1)
= 2s
dt
dt
dt
dx
dy
ds
x
+ ( y − 12)
= s
dt
dt
dt
Also, x 2 + y 2 = 122.
2x
− x dx
dx
dy
dy
+ 2y
= 0 ⇒
=
dt
dt
dt
y dt
So, x
⎛ − x dx ⎞
dx
ds
+ ( y − 12)⎜
⎟ = s .
dt
y
dt
dt
⎝
⎠
(12)(6) −0.2 = −1 = − 3 m/sec (horizontal)
dx ⎡
ds
dx
sy ds
12 x ⎤
⇒
=
⋅
=
( )
⎢x − x +
⎥ = s
dt ⎣
y ⎦
dt
dt
12 x dt
15
5 3
12
( )6 3
(
(
)
)
− 3
dy
− x dx
−6 3
1
=
=
⋅
= m/sec (vertical)
dt
y dt
6
15
5
24. Let L be the length of the rope.
L2 = 144 + x 2
(a)
2L
dL
dx
= 2x
dt
dt
dx
L dL
4L
=
⋅
= −
dt
x dt
x
dL
⎛
⎞
= −4 ft/sec ⎟
⎜ since
dt
⎝
⎠
4 ft/sec
13 ft
12 ft
When L = 13:
x =
L2 − 144 =
169 − 144 = 5
4(13)
dx
52
= −
= −
= −10.4 ft/sec
dt
5
5
Speed of the boat increases as it approaches the dock.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
230
Chapter 3
(b) If
Differentiation
dx
= −4, and L = 13:
dt
dL
x dx
5
−20
=
=
ft/sec
( − 4) =
dt
L dt
13
13
L2 − 144
( −4)
L
dL
x dx
=
=
dt
L dt
dL
−4
lim
= lim
+
+
L →12 dt
L →12 L
L2 − 144 = 0
s 2 = x2 + y 2
25. (a)
s 2 = 902 + x 2
27.
x = 20
dx
= −450
dt
dy
= −600
dt
ds
dx
dy
= 2x
+ 2y
2s
dt
dt
dt
x( dx dt ) + y( dy dt )
ds
=
dt
s
dx
= −25
dt
ds
dx
ds
x dx
= 2x
⇒
=
⋅
2s
dt
dt
dt
s dt
When x = 20, s =
902 + 202 = 10 85,
ds
−50
20
=
≈ −5.42 ft/sec.
(−25) =
dt
10 85
85
y
2nd
300
200
y
s
20 ft
x
3rd
100
x
−100
100
1st
s
x
200
300
90 ft
Home
When x = 225 and y = 300, s = 375 and
225( −450) + 300( −600)
ds
=
= −750 mi/h.
dt
375
375
1
(b) t =
= h = 30 min
750
2
26.
x2 + y2 = s2
2x
dy
⎛
⎞
= 0⎟
⎜ because
dt
⎝
⎠
dx
ds
+ 0 = 2s
dt
dt
dx
s ds
=
dt
x dt
When s = 10, x =
28. s 2 = 902 + x 2
x = 90 − 20 = 70
dx
= 25
dt
ds
x dx
=
⋅
dt
s dt
When x = 70, s =
902 + 702 = 10 130,
ds
70
175
=
≈ 15.35 ft/sec.
( 25) =
dt
10 130
130
100 − 25 =
75 = 5 3,
2nd
20 ft
dx
10
480
=
= 160 3 ≈ 277.13 mi/h.
(240) =
dt
5 3
3
x
3rd
1st
s
y
90 ft
x
Home
5 mi
s
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.7
29. (a)
y
15
=
⇒ 15 y − 15 x = 6 y
y − x
6
y =
5
x
3
31. x(t ) =
2π
= 12 seconds
π 6
(b) When x =
1
,y =
2
⎛1⎞
12 − ⎜ ⎟
⎝2⎠
2
=
3
m.
2
⎛
3⎞
Lowest point: ⎜⎜ 0,
⎟⎟
2
⎝
⎠
(c) When x =
15
231
1
πt
sin , x 2 + y 2 = 1
2
6
(a) Period:
dx
= 5
dt
dy
5 dx
5
25
= ⋅
= (5) =
ft/sec
dt
3 dt
3
3
Related Rates
1
, y =
4
⎛1⎞
1−⎜ ⎟
⎝ 4⎠
2
=
15
and t = 1:
4
dx
1⎛π ⎞
πt
π
πt
cos
= ⎜ ⎟ cos
=
dt
2⎝ 6 ⎠
6
12
6
6
x
x2 + y 2 = 1
y
(b)
d ( y − x)
dt
=
dy
dx
25
10
−
=
−5 =
ft/sec
dt
dt
3
3
y
20
=
y − x
6
30. (a)
2x
So,
− x dx
dx
dy
dy
+ 2y
= 0 ⇒
=
dt
dt
dt
y dt
14
π
dy
⎛π ⎞
= −
⋅
cos⎜ ⎟
dt
15 4 12
⎝6⎠
20 y − 20 x = 6 y
−π ⎛ 1 ⎞ 3
−π
=
⎜ ⎟
12
2
24
15 ⎝ ⎠
=
14 y = 20 x
10
x
7
y =
dx
= −5
dt
−50
dy
10 dx
10
=
=
ft/sec
(−5) =
dt
7 dt
7
7
Speed =
32. x(t ) =
− 5π
5π
=
m/sec
120
120
3
sin π t , x 2 + y 2 = 1
5
(a) Period:
2π
π
(b) When x =
20
x
= 2 seconds
3
,y =
5
⎛
Lowest point: ⎜ 0,
⎝
6
1
− 5π
=
.
120
5
⎛ 3⎞
1−⎜ ⎟
⎝5⎠
2
=
4
m.
5
4⎞
⎟
5⎠
y
d ( y − x)
dy
dx
(b)
=
−
dt
dt
dt
−50
=
− ( −5)
7
−50 35
−15
=
+
=
ft/sec
7
7
7
(c) When x =
3
,y =
10
⎛1⎞
1−⎜ ⎟
⎝4⎠
2
=
15
and
4
3
3
1
1
= sin π t ⇒ sin π t =
⇒ t = :
10
5
2
6
dx
3
= π cos π t
dt
5
x2 + y 2 = 1
2x
So,
− x dx
dx
dy
dy
+ 2y
= 0 ⇒
=
dt
dt
dt
y dt
dy
−3 10 3
−9π
−9 5π
⎛π ⎞
=
⋅ π cos⎜ ⎟ =
=
.
dt
125
15 4 5
25 5
⎝6⎠
Speed =
−9 5π
≈ 0.5058 m/sec
125
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
232
Chapter 3
Differentiation
33. Because the evaporation rate is proportional to the
surface area, dV dt = k ( 4π r 2 ). However, because
y
50
dy
= 4 m/sec
dt
1 dy
dθ
sec 2θ ⋅
=
50 dt
dt
1
dθ
dy
cos 2θ ⋅
=
50
dt
dt
V = ( 4 3)π r 3 , you have
dV
dr
= 4π r 2 .
dt
dt
Therefore, k ( 4π r 2 ) = 4π r 2
tan θ =
37.
dr
dr
⇒ k =
.
dt
dt
y
34.
1
1
1
=
+
R
R1
R2
dR1
=1
dt
dR2
= 1.5
dt
1 dR
1
1
dR
dR2
⋅
= 2 ⋅ 1 + 2 ⋅
2
R
dt
R1
dt
R2
dt
y
θ
When y = 50, θ =
When R1 = 50 and R2 = 75:
R = 30
pV 1.3 = k
35.
dV
dp
+ V 1.3
= 0
dt
dt
dV
dp ⎞
⎛
V 0.3 ⎜1.3 p
+V ⎟ = 0
dt
dt ⎠
⎝
1.3 pV 0.3
1.3 p
36.
32r tan θ = v ,
32r sec 2θ
38.
Likewise,
2
.
2
dx
= −600 mi/h
dt
(sec2θ ) ddtθ = − x52 ⋅ dx
dt
dθ
x 2 ⎛ 5 ⎞ dx
⎛ 5 ⎞ dx
= cos 2θ ⎜ − 2 ⎟
= 2 ⎜− 2 ⎟
dt
L ⎝ x ⎠ dt
⎝ x ⎠ dt
⎛ 52 ⎞⎛ 1 ⎞ dx
= ⎜ − 2 ⎟⎜ ⎟
⎝ L ⎠⎝ 5 ⎠ dt
⎛1⎞
= ( −sin 2θ )⎜ ⎟( −600) = 120 sin 2θ
⎝ 5⎠
r is a constant.
dθ
v
dv
=
cos 2θ .
dt
16r
dt
, and cos θ =
y
,y = 5
x
tan θ =
dV
dp
= −V
dt
dt
dθ
dv
= 2v
dt
dt
dv
16r
dθ
=
sec 2θ
dt
v
dt
4
1 ⎛ 2⎞
1
dθ
=
rad/sec.
⎜
⎟ ( 4) =
dt
50 ⎜⎝ 2 ⎟⎠
25
rg tan θ = v 2
2
π
2
So,
⎡ 1
⎤
dR
1
2
= (30) ⎢
1 +
1.5)⎥ = 0.6 ohm/sec
2( )
2(
dt
⎢⎣ (50)
⎥⎦
(75)
x
50
L
y=5
θ
x
(a) When θ = 30°,
dθ
120
1
=
= 30 rad/h = rad/min.
dt
4
2
(b) When θ = 60°,
dθ
3
⎛ 3⎞
= 120⎜ ⎟ = 90 rad/h = rad/min.
dt
2
⎝ 4⎠
(c) When θ = 75°,
dθ
= 120 sin 2 75° ≈ 111.96 rad/h ≈ 1.87 rad/min.
dt
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.7
39. H =
4347
e369,444 (50t + 19,793)
400,000,000
(b)
(a) t = 65° ⇒ H ≈ 99.79%
Related Rates
233
2000
4␲
0
t = 80° ⇒ H ≈ 60.20%
⎛ − 369,444(50) ⎞
⎟t ′
(b) H ′ = H ⋅ ⎜
⎜ (50t + 19,793)2 ⎟
⎝
⎠
− 2000
(c)
At t = 75 and t ′ = 2, H ′ ≈= − 4.7% h.
tan θ =
40.
dx dt = −600π sin θ is greatest when
sin θ = 1 ⇒ θ =
π
2
+ nπ
(or 90° +
dx dt is least when θ = nπ
x
50
dθ
= 30( 2π ) = 60π rad/min = π rad/sec
dt
1 ⎛ dx ⎞
⎛ dθ ⎞
sec 2θ ⎜ ⎟ =
⎜ ⎟
50 ⎝ dt ⎠
⎝ dt ⎠
(or n ⋅ 180°).
(d) For θ = 30°,
dx
1
= −600π sin (30°) = −600π = −300π cm/sec.
dt
2
For θ = 60°,
dx
= −600π sin (60°)
dt
dx
⎛ dθ ⎞
= 50 sec 2θ ⎜ ⎟
dt
⎝ dt ⎠
Police
= −600π
42. sin 18° =
θ
n ⋅ 180°).
50 ft
3
= −300 3π cm/sec.
2
x
y
x dy
1 dx
⋅
+ ⋅
y 2 dt
y dt
dx
x dy
=
⋅
= (sin 18°)( 275) ≈ 84.9797 mi/hr
dt
y dt
0 = −
x
(a) When θ = 30°,
dx
200π
=
ft/sec.
dt
3
y
x
41.
(b) When θ = 60°,
dx
= 200π ft/sec.
dt
(c) When θ = 70°,
dx
≈ 427.43π ft/sec.
dt
18°
43. tan θ =
dθ
= (10 rev/sec)( 2π rad/rev) = 20π rad/sec
dt
(a)
x
30
dθ
1 dx
−sin θ
=
dt
30 dt
dx
dθ
= −30 sin θ
dt
dt
= −30 sin θ ( 20π )
cos θ =
= −600π sin θ
P
44. (i) (a)
dx
dy
negative ⇒
positive
dt
dt
(b)
dy
dx
positive ⇒
negative
dt
dt
(ii) (a)
dx
dy
negative ⇒
negative
dt
dt
30
θ
x
x
x
⇒ x = 50 tan θ
50
dx
dθ
= 50 sec 2 θ
dt
dt
θ
d
2 = 50 sec 2 θ
dt
π
π
1
dθ
2
=
cos θ , − ≤ θ ≤
25
4
4
dt
(b)
dy
dx
positive ⇒
positive
dt
dt
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
234
Chapter 3
Differentiation
45. x 2 + y 2 = 25; acceleration of the top of the ladder =
First derivative: 2 x
d2y
dt 2
dx
dy
+ 2y
= 0
dt
dt
dx
dy
x
+ y
= 0
dt
dt
Second derivative: x
d 2x
dx dx
d2y
dy dy
+
⋅
+ y 2 +
⋅
= 0
2
dt
dt dt
dt
dt dt
⎛ 1 ⎞ ⎡ d 2 x ⎛ dx ⎞
d2y
⎛ dy ⎞
= ⎜ ⎟ ⎢− x 2 − ⎜ ⎟ − ⎜ ⎟
2
dt
⎝ dt ⎠
⎝ dt ⎠
⎝ y ⎠ ⎢⎣ dt
2
When x = 7, y = 24,
2
⎤
⎥
⎥⎦
dy
7
dx
dx
d 2x
is constant,
= − , and
= 2 (see Exercise 25). Because
= 0.
dt
12
dt
dt
dt 2
2
d2y
1⎡
1⎡
49 ⎤
1 ⎡ 625 ⎤
2
⎛ 7⎞ ⎤
=
−
7
0
−
2
−
−
−4 −
=
−
≈ −0.1808 ft/sec 2
⎢
⎥ =
(
)
(
)
⎜
⎟
⎢
⎥
2
dt
24 ⎢⎣
24 ⎣
144 ⎦
24 ⎣⎢ 144 ⎦⎥
⎝ 12 ⎠ ⎥⎦
46. L2 = 144 + x 2 ; acceleration of the boat =
dL
dx
= 2x
dt
dt
dL
dx
L
= x
dt
dt
First derivative: 2 L
d 2x
dt 2
Second derivative: L
d 2L
dL dL
d 2x
dx
+
⋅
= x 2 +
⋅
2
dt
dt dt
dt
dt
d 2x
⎛ 1 ⎞⎡ d 2 L
= ⎜ ⎟ ⎢L 2 +
2
dt
⎝ x ⎠ ⎢⎣ dt
dx
dt
2
⎛ dL ⎞
⎛ dx ⎞
⎜ ⎟ −⎜ ⎟
⎝ dt ⎠
⎝ dt ⎠
2
⎤
⎥
⎥⎦
d 2L
dL
dx
dL
= 0.
is constant,
= −10.4, and
= −4 (see Exercise 28). Because
dt 2
dt
dt
dt
1
1
1
d 2x
2
2
= ⎡13(0) + ( −4) − ( −10.4) ⎤ = [16 − 108.16] = [−92.16] = −18.432 ft/sec 2
⎦
dt 2
5⎣
5
5
When L = 13, x = 5,
47. (a) dy dt = 3( dx dt ) means that y changes three times as fast as x changes.
(b) y changes slowly when x ≈ 0 or x ≈ L. y changes more rapidly when x is near the middle of the interval.
48. y(t ) = −4.9t 2 + 20
49. (a) A = ( base)( height ) = 2 xe − x
y
dy
= −9.8t
dt
y(1) = −4.9 + 20 = 15.1
y′(1) = −9.8
(b)
) + 2e
2 2
y
12
(0, 0)
When y = 15.1:
2
= ( − 2 x 2 + 2) e − x
20
By similar triangles:
(
2
dA
= ⎡2 x − xe − x
⎢⎣
dt
x
x
20
y
=
x
x − 12
20 x − 240 = xy
For x = 2 and
2 2
− x2 2 ⎤
dx
⎥⎦ dt
dx
dt
dx
= 4,
dt
dA
− 24
= − 6e − 2 ( 4) = 2 ≈ − 3.25 cm 2 min.
dt
e
20 x − 240 = x(15.1)
(20
− 15.1) x = 240
x =
240
4.9
20 x − 240 = xy
dx
dy
dx
= x
+ y
20
dt
dt
dt
dx
x
dy
=
dt
20 − y dt
At t = 1,
dx
240 4.9
=
(−9.8) ≈ −97.96 m/sec.
dt
20 − 15.1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.8
Newton’s Method
235
Section 3.8 Newton’s Method
The following solutions may vary depending on the software or calculator used, and on rounding.
1.
f ( x) = x 2 − 5
f ′( x) = 2 x
x1 = 2.2
2.
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
2.2000
–0.1600
4.4000
–0.0364
2.2364
2
2.2364
0.0013
4.4727
0.0003
2.2361
f ′( xn )
xn −
f ′( xn )
f ( x) = x3 − 3
f ′( x ) = 3 x 2
x1 = 1.4
3.
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
1.4000
–0.2560
5.8800
–0.0435
1.4435
2
1.4435
0.0080
6.2514
0.0013
1.4423
f ′( xn )
xn −
f ′( xn )
f ( x) = cos x
f ′( x) = −sin x
x1 = 1.6
4.
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
1.6000
–0.0292
–0.9996
0.0292
1.5708
2
1.5708
0.0000
–1.0000
0.0000
1.5708
f ′( xn )
xn −
f ′( xn )
f ( x) = tan x
f ′( x) = sec2 x
x1 = 0.1
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
0.1000
0.1003
1.0101
0.0993
0.0007
2
0.0007
0.0007
1.0000
0.0007
0.0000
f ′( xn )
xn −
f ′( xn )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
236
5.
Chapter 3
Differentiation
f ( x) = x3 + 4
f ′( x ) = 3 x 2
x1 = −2
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
–2.0000
–4.0000
12.0000
–0.3333
–1.6667
2
–1.6667
–0.6296
8.3333
–0.0756
–1.5911
3
–1.5911
–0.0281
7.5949
–0.0037
–1.5874
4
–1.5874
–0.0000
7.5596
0.0000
–1.5874
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is –1.587.
6.
f ( x) = 2 − x3
f ′( x ) = −3 x 2
x1 = 1.0
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
1.0000
1.0000
–3.0000
–0.3333
1.3333
2
1.3333
–0.3704
–5.3333
0.0694
1.2639
3
1.2639
–0.0190
–4.7922
0.0040
1.2599
4
1.2599
0.0001
–4.7623
0.0000
1.2599
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is 1.260.
7.
f ( x) = x3 + x − 1
f ′( x) = 3x 2 + 1
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
0.5000
–0.3750
1.7500
–0.2143
0.7143
2
0.7143
0.0788
2.5307
0.0311
0.6832
3
0.6832
0.0021
2.4003
0.0009
0.6823
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is 0.682.
8.
f ( x) = x5 + x − 1
f ′( x) = 5 x 4 + 1
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
0.5000
–0.4688
1.3125
–0.3571
0.8571
2
0.8571
0.3196
3.6983
0.0864
0.7707
3
0.7707
0.0426
2.7641
0.0154
0.7553
4
0.7553
0.0011
2.6272
0.0004
0.7549
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is 0.755.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.8
9.
f ( x) = 5
Newton’s Method
237
x − 1 − 2x
5
− 2
2 x −1
From the graph you see that these are two zeros. Begin with x = 1.2.
f ′( x) =
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
1.2000
–0.1639
3.5902
–0.0457
1.2457
2
1.2457
–0.0131
3.0440
–0.0043
1.2500
3
1.2500
–0.0001
3.0003
–0.0003
1.2500
f ′( xn )
xn −
f ′( xn )
Approximation of the zero of f is 1.250.
Similarly, the other zero is approximately 5.000.
(Note: These answers are exact)
10.
f ( x) = x − 2
x +1
1
x +1
f ′( x) = 1 −
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
5.0000
0.1010
0.5918
0.1707
4.8293
2
4.8293
0.0005
0.5858
0.00085
4.8284
f ′( xn )
xn −
f ′( xn )
Approximation of the zero of f is 4.8284.
11.
f ( x) = x − e − x
f ′( x) = 1 + e − x
x1 = 0.5
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
xn −
f ( xn )
f ′( xn )
1
0.5
–0.1065
1.6065
–0.0663
0.5663
2
0.5663
0.0013
1.5676
0.0008
0.5671
3
0.5671
0.0001
1.5672
–0.0000
0.5671
Approximation of the zero of f is 0.567.
12.
f ( x) = x − 3 + ln x
f ′( x) = 1 +
1
x
x1 = 2.0
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
2.0
–0.3069
1.5
–0.2046
2.2046
2
2.2046
–0.0049
1.4536
–0.0033
2.2079
3
2.2079
–0.0001
1.4529
–0.0000
2.2079
f ′( xn )
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is 2.208.
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238
13.
Chapter 3
Differentiation
f ( x) = x 3 − 3.9 x 2 + 4.79 x − 1.881
f ′( x ) = 3x 2 − 7.8 x + 4.79
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
0.5000
–0.3360
1.6400
–0.2049
0.7049
2
0.7049
–0.0921
0.7824
–0.1177
0.8226
3
0.8226
–0.0231
0.4037
–0.0573
0.8799
4
0.8799
–0.0045
0.2495
–0.0181
0.8980
5
0.8980
–0.0004
0.2048
–0.0020
0.9000
6
0.9000
0.0000
0.2000
0.0000
0.9000
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is 0.900.
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
1.1
0.0000
–0.1600
–0.0000
xn −
f ( xn )
f ′( xn )
1.1000
Approximation of the zero of f is 1.100.
n
xn
f ( xn )
f ′( xn )
1
1.9
0.0000
0.8000
f ( xn )
f ′( xn )
xn −
0.0000
f ( xn )
f ′( xn )
1.9000
Approximation of the zero of f is 1.900.
14.
f ( x) = x 4 + x3 − 1
f ′( x) = 4 x 3 + 3 x 2
From the graph you see that these are two zeros. Begin with x1 = 1.0
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
1.0000
1.0000
7.0000
0.1429
0.8571
2
0.8571
0.1695
4.7230
0.0359
0.8213
3
0.8213
0.0088
4.2390
0.0021
0.8192
4
0.8192
0.0003
4.2120
0.0000
0.8192
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is 0.819.
Similarly, the other zero is approximately –1.380.
15.
f ( x ) = 1 − x + sin x
f ′( x ) = −1 + cos x
x1 = 2
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
2.0000
–0.0907
–1.4161
0.0640
1.9360
2
1.9360
–0.0019
–1.3571
0.0014
1.9346
3
1.9346
0.0000
–1.3558
0.0000
1.9346
f ′( xn )
xn −
f ′( xn )
Approximate zero: x ≈ 1.935
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.8
16.
Newton’s Method
239
f ( x) = x 3 − cos x
f ′( x) = 3x 2 + sin x
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
0.9000
0.1074
3.2133
0.0334
0.8666
2
0.8666
0.0034
3.0151
0.0011
0.8655
3
0.8655
0.0000
3.0087
0.0000
0.8655
xn −
f ( xn )
f ′( xn )
Approximation of the zero of f is 0.866.
17. h( x) = f ( x) − g ( x) = 2 x + 1 −
h′( x) = 2 −
2
x + 4
1
x + 4
n
xn
h( xn )
h′( xn )
h( xn )
h′( xn )
1
0.6000
0.0552
1.7669
0.0313
0.5687
2
0.5687
0.0000
1.7661
0.0000
0.5687
xn −
h( xn )
h′( xn )
Point of intersection of the graphs of f and g occurs when x ≈ 0.569.
18. h( x) = e x 2 − 2 + x 2
h′( x) =
1 x2
e + 2x
2
Two points of intersection
n
xn
h( xn )
h′( xn )
h( xn )
h′( xn )
xn −
h( xn )
h′( xn )
1
–1
–0.3935
–1.6967
0.2319
–1.2319
2
–1.2319
0.0577
–2.1937
–0.0263
–1.2056
3
–1.2056
0.0007
–2.1376
–0.0004
–1.2052
One point of intersection of the graphs of f and g occurs when x ≈ −1.205.
n
xn
h( xn )
h′( xn )
h( xn )
h′( xn )
xn −
h( xn )
h′( xn )
1
1
0.6487
2.8244
0.2297
0.7703
2
0.7703
0.0632
2.2755
0.0277
0.7425
3
0.7425
0.0009
2.2098
0.0004
0.7421
Another point of intersection of the graphs of f and g occurs when x ≈ 0.742.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
240
Chapter 3
Differentiation
19. h( x) = f ( x) − g ( x) = x − tan x
h′( x) = 1 − sec 2 x
n
xn
h( xn )
h′( xn )
h( xn )
h′( xn )
1
4.5000
–0.1373
–21.5048
0.0064
4.4936
2
4.4936
–0.0039
–20.2271
0.0002
4.4934
xn −
h( xn )
h′( xn )
Point of intersection of the graphs of f and g occurs when x ≈ 4.493.
Note: f ( x) = x and g ( x) = tan x intersect infinitely often.
20. h( x) = arctan x − arccos x
h′( x) =
1
+
1 + x2
1
1 − x2
n
xn
h( xn )
h′( xn )
h( xn )
h′( xn )
xn −
h( xn )
h′( xn )
1
0.5
–0.5835
1.9547
–0.2985
0.7985
2
0.7985
0.0278
2.2718
0.0122
0.7863
3
0.7863
0.0003
2.2365
0.0001
0.7862
Point of intersection of the graphs of f and g occurs when x ≈ 0.786.
21. (a)
22. (a) f ( x) = x n − a, a > 0
f ( x ) = x 2 − a, a > 0
f ′( x) = nx n −1
f ′( x) = 2 x
xn +1 = xn −
(b)
f ( xn )
x 2 − a 1⎛
a⎞
= xn − n
= ⎜ xn + ⎟
f ′( xn )
xn ⎠
2 xn
2⎝
xi +1 = xi −
f ( xi )
xn − a
(n − 1) xi n + a
= xi − i n −1 =
f ′( xi )
nxi
nxi n −1
5: xn + 1 =
1⎛
5⎞
⎜ xn +
⎟, x1 = 2
xn ⎠
2⎝
n
1
2
3
4
i
1
2
3
4
xn
2
2.25
2.2361
2.2361
xi
1.5
1.5694
1.5651
1.5651
(b)
For example, given x1 = 2,
x2 =
1⎛
5⎞
9
= 2.25.
⎜2 + ⎟ =
2⎝
2⎠
4
4
6 : xi + 1 =
4
6 ≈ 1.565
3
15: xi + 1 =
3 xi 4 + 6
, x1 = 1.5
4 xi 3
2 xi 3 + 15
, x1 = 2.5
3 xi 2
5 ≈ 2.236
7 : xn + 1 =
1⎛
7⎞
⎜ xn +
⎟, x1 = 2
2⎝
xn ⎠
n
1
2
3
4
5
xn
2
2.75
2.6477
2.6458
2.6458
3
i
1
2
3
4
xi
2.5
2.4667
2.4662
2.4662
15 ≈ 2.466
7 ≈ 2.646
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.8
23. y = 2 x 3 − 6 x 2 + 6 x − 1 = f ( x)
y′ = 3 x 2 − 2
x1 = 1
x1 = 0
f ′( x) = 0; therefore, the method fails.
x2 = −1
xn
f ( xn )
f ′( xn )
1
1
1
0
241
24. y = x3 − 2 x − 2, x1 = 0
y′ = 6 x 2 − 12 x + 6 = f ′( x)
n
Newton’s Method
x3 = 0
x4 = −1
and so on.
Fails to converge
25. Let g ( x) = f ( x) − x = cos x − x
g ′( x) = −sin x − 1.
g ( xn )
g ( xn )
n
xn
g ( xn )
g ′( xn )
1
1.0000
–0.4597
–1.8415
0.2496
0.7504
2
0.7504
–0.0190
–1.6819
0.0113
0.7391
3
0.7391
0.0000
–1.6736
0.0000
0.7391
g ′( xn )
xn −
g ′( xn )
The fixed point is approximately 0.74.
26. Let g ( x) = f ( x) − x = cot x − x
g ′( x) = −csc 2 x − 1.
g ( xn )
g ( xn )
n
xn
g ( xn )
g ′( xn )
1
1.0000
–0.3579
–2.4123
0.1484
0.8516
2
0.8516
0.0240
–2.7668
–0.0087
0.8603
3
0.8603
0.0001
–2.7403
0.0000
0.8603
g ′( xn )
xn −
g ′( xn )
The fixed point is approximately 0.86.
27. Let g ( x ) = e x 10 − x
g ′( x ) =
1 x 10
e
−1
10
n
xn
g ( xn )
g ′( xn )
1
1.0
0.1052
–0.8895
–0.1182
1.1182
2
1.1182
0.0001
–0.8882
–0.0001
1.1183
g ( xn )
g ′( xn )
xn −
g ( xn )
g ′( xn )
The fixed point is approximately 1.12.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
242
Chapter 3
Differentiation
28. Let g ( x ) = x + ln x
g ′( x) =
1
+1
x
g ( xn )
g ( xn )
n
xn
g ( xn )
g ′( xn )
1
0.5
–0.1931
3
–0.0644
0.5644
2
0.5644
–0.0076
2.7718
–0.0027
0.5671
xn −
g ′( xn )
g ′( xn )
The fixed point is approximately 0.57.
29.
1
−a = 0
x
1
f ′( x) = − 2
x
(1 xn ) − a = x + x 2 ⎛ 1 − a ⎞ = x + x − x 2a = 2 x − x 2a = x 2 − ax
xn +1 = xn −
n
n ⎜
n
n
n
n
n
n(
n)
⎟
−1 xn 2
⎝ xn
⎠
f ( x) =
30. (a) xn + 1 = xn ( 2 − 3xn )
31. f ( x ) = x 3 − 3x 2 + 3, f ′( x) = 3x 2 − 6 x
i
1
2
3
4
xi
0.3000
0.3300
0.3333
0.3333
4
(a)
−4
5
≈ 0.333
1
3
−2
(b) xn +1 = xn ( 2 − 11xn )
(b) x1 = 1
i
1
2
3
4
xi
0.1000
0.0900
0.0909
0.0909
1
11
≈ 0.091
x2 = x1 −
f ( x1 )
f ′( x1 )
≈ 1.333
Continuing, the zero is 1.347.
1
(c) x1 =
4
f ( x1 )
x2 = x1 −
≈ 2.405
f ′( x1 )
Continuing, the zero is 2.532.
(d)
y = −3x + 4
y
f
3
−2
x
1
4
5
y = −1.313x + 3.156
The x-intercept of y = −3x + 4 is 43 . The x-intercept
of y = 1.313 x + 3.156 is approximately 2.405.
The x-intercepts correspond to the values resulting
from the first iteration of Newton's Method.
(e) If the initial guess x1 is not "close to" the desired
zero of the function, the x-intercept of the tangent
line may approximate another zero of the function.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 3.8
32. f ( x) = sin x, f ′( x) = cos x
(a)
Newton’s Method
243
33. Answers will vary. See page 229.
If f is a function continuous on [a, b] and differentiable
2
on ( a, b) where c ∈ [a, b] and f (c) = 0, Newton’s
−␲
Method uses tangent lines to approximate c such that
f (c) = 0.
␲
First, estimate an initial x1 close to c (see graph).
−2
(b) x1 = 1.8
y
f ( x1 )
x2 = x1 −
f ′( x1 )
≈ 6.086
1
(c) x1 = 3
−1
f ( x1 )
≈ 3.143
f ′( x1 )
x2 = x1 −
(d)
y
x
a 3
f(x)
x1
x2
c
2 b
x
−1
−2
y = 0.99x + 3.111
2
Then determine x2 by x2 = x1 −
y = − 0.227x + 1.383
1
π
2
x
π
f ( x1 )
f ′( x1 )
Calculate a third estimate by x3 = x2 −
−1
.
f ( x2 )
.
f ′( x2 )
Continue this process until xn − xn +1 is within the
−2
desired accuracy.
The x-intercept of y = −0.227 x + 1.383 is
approximately 6.086. The x-intercept of
y = 0.99 x + 3.111 is approximately 3.143.
Let xn +1 be the final approximation of c.
34. At x = − 3 and x = 2, the tangent lines to the curve are
The x-intercepts correspond to the values resulting
from the first iteration of Newton’s Method.
horizontal. Hence, Newton’s Method will not converge
for these initial approximations.
(e) If the initial guess x1 is not "close to" the desired
zero of the function, the x-intercept of the tangent
line may approximate another zero of the function.
35. y = f ( x) = 4 − x 2 , (1, 0)
d =
(x
− 1) + ( y − 0)
2
2
=
(x
− 1) + ( 4 − x 2 ) =
2
2
x 4 − 7 x 2 − 2 x + 17
d is minimized when D = x 4 − 7 x 2 − 2 x + 17 is a minimum.
g ( x) = D′ = 4 x3 − 14 x − 2
g ′( x) = 12 x 2 − 14
n
xn
g ( xn )
g ′( xn )
g ( xn )
g ′( xn )
1
2.0000
2.0000
34.0000
0.0588
xn −
g ( xn )
g ′( xn )
y
5
1.9412
(1.939, 0.240)
3
2
2
1.9412
0.0830
31.2191
0.0027
1.9385
3
1.9385
–0.0012
31.0934
0.0000
1.9385
1
(1, 0)
x
−3
−1
−1
1
3
x ≈ 1.939
Point closest to (1, 0) is ≈ (1.939, 0.240).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
244
Chapter 3
Differentiation
36. Maximize: C =
C′ =
3t 2 + t
50 + t 3
−3t 4 − 2t 3 + 300t + 50
(50 + t 3 )
f ( xn )
2
= 0
f ( xn )
f ′( xn )
xn −
n
xn
1
4.5000
12.4375
915.0000
0.0136
4.4864
2
4.4864
0.0658
904.3822
0.0001
4.4863
f ′( xn )
f ( xn )
f ′( xn )
Let f ( x) = 3t 4 + 2t 3 − 300t − 50
f ′( x) = 12t 3 + 6t 2 − 300.
Because f ( 4) = −354 and f (5) = 575, the solution is in the interval ( 4, 5).
Approximation: t ≈ 4.486 hours
Minimize: T =
37.
Distance rowed
Distance walked
+
Rate rowed
Rate walked
x2 + 4
x 2 − 6 x + 10
+
3
4
x
x −3
T′ =
+
= 0
3 x2 + 4
4 x 2 − 6 x + 10
T =
4x
x 2 − 6 x + 10 = −3( x − 3)
x2 + 4
16 x 2 ( x 2 − 6 x + 10) = 9( x − 3) ( x 2 + 4)
2
7 x 4 − 42 x 3 + 43x 2 + 216 x − 324 = 0
Let f ( x) = 7 x 4 − 42 x3 + 43 x 2 + 216 x − 324 and f ′( x) = 28 x3 − 126 x 2 + 86 x + 216. Becasuse f (1) = −100 and
f ( 2) = 56, the solution is in the interval (1, 2).
f ( xn )
f ′( xn )
f ( xn )
xn −
n
xn
1
1.7000
19.5887
135.6240
0.1444
1.5556
2
1.5556
–1.0480
150.2780
–0.0070
1.5626
3
1.5626
0.0014
49.5591
0.0000
1.5626
f ′( xn )
f ( xn )
f ′( xn )
Approximation: x ≈ 1.563 mi
38. Set T = 300 and obtain the following equation.
0.2988 x 4 − 22.625 x3 + 628.49 x 2 − 7565.9 x + 33,478 = 300
0.2988 x 4 − 22.625 x3 + 628.49 x 2 − 7565.9 x + 33,178 = 0
From the graph, T = 300 when x ≈ 17,and x ≈ 22.
Using Newton’s Method with x1 = 17, you obtain x = 17.2 years.
Using Newton’s Method with x1 ≈ 22, you obtain x ≈ 22.1 years.
39. False. Let f ( x) = ( x 2 − 1) ( x − 1). x = 1 is a discontinuity. It is not a zero of f ( x ). This statement would be true if
f ( x) = p( x) q( x) was given in reduced form.
40. True
41. True
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
245
44. Let ( x1 , y1 ) be the point of tangency.
42. True
f ( x) = cos x, f ′( x) = −sin x, f ′( x1 ) = −sin ( x1 ).
43. f ( x ) = −sin x
f ′( x ) = −cos x
At the point of tangency,
y −0
f ′( x1 ) = 1
x1 − 0
Let ( x0 , y1 ) = ( x0 , − sin ( x0 )) be a point on the graph of
f. If ( x0 , y0 ) is a point of tangency, then
−cos( x0 )
−sin ( x1 ) = cos( x1 ) x1
−sin ( x0 )
y −0
y
.
= 0
= 0 =
x0 − 0
x0
x0
cos( x1 ) + x1 sin ( x1 ) = 0
Using Newton’s method with initial guess 3, you obtain
x1 ≈ 2.798 and y1 ≈ −0.942.
So, x0 = tan ( x0 ).
x0 ≈ 4.4934
Slope = −cos( x0 ) ≈ 0.217
You can verify this answer by graphing y1 = −sin x and
the tangent line y2 = 0.217 x.
2
−1
5
−2
Review Exercises for Chapter 3
1.
f ( x) = 12
f ′( x) = lim
2.
f ( x + ∆x ) − f ( x)
f ′( x) = lim
∆x
∆x → 0
f ( x) = 5 x − 4
∆x → 0
∆x
⎡5( x + ∆x) − 4⎤⎦ − (5 x − 4)
= lim ⎣
∆x → 0
∆x
5 x + 5∆ x − 4 − 5 x + 4
= lim
∆x → 0
∆x
5∆x
= lim
= 5
∆x → 0 ∆x
12 − 12
= lim
∆x → 0
∆x
0
= lim
= 0
∆x → 0 ∆x
3.
f ( x + ∆x) − f ( x)
f ( x) = x 2 − 4 x + 5
f ′( x) = lim
f ( x + ∆x) − f ( x)
∆x
∆x → 0
⎡( x + ∆x)2 − 4( x + ∆x) + 5⎤ − ⎡ x 2 − 4 x + 5⎤
⎦
⎦ ⎣
= lim ⎣
∆x → 0
∆x
=
(x
lim
2
2
∆x
∆x → 0
2 x( ∆x) + ( ∆x ) − 4( ∆x)
2
= lim
∆x → 0
)
+ 2 x( ∆x ) + ( ∆x) − 4 x − 4( ∆x) + 5 − ( x 2 − 4 x + 5)
∆x
= lim ( 2 x + ∆x − 4) = 2 x − 4
∆x → 0
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246
4.
Chapter 3
f ( x) =
Differentiation
6
x
f ( x + ∆x) − f ( x)
∆x
6
6
−
−6 ∆x
−6
−6
+
∆
x
x
x = lim 6 x − (6 x + 6 ∆x) = lim
= lim
= lim
= 2
∆x → 0
∆x → 0
∆x → 0 ∆x( x + ∆x ) x
∆x → 0 ( x + ∆ x ) x
x
∆x
∆x( x + ∆x) x
f ′( x) = lim
∆x → 0
5. g ( x) = 2 x 2 − 3 x, c = 2
14.
g ( x ) − g ( 2)
g ′( 2) = lim
x→2
f ′( x) =
x − 2
(2 x 2 − 3x) − 2
= lim
x→2
= lim
2 −2
t
3
−4 −3
4
g ′(t ) =
t = − 3
3
3t
x − 2
= lim ( 2 x + 1) = 2( 2) + 1 = 5
x→2
8
8
= x−4
4
5x
5
32 −5
32
h′( x) = − x = − 5
5
5x
16. h( x) =
1
,c = 3
x + 4
f ( x) − f (3)
f ′(3) = lim
17.
x −3
1
1
−
x
+
4
7
= lim
x→3
x −3
7 − x − 4
= lim
x → 3 ( x − 3)( x + 4)7
x→3
= lim
x→3
1 −1 2
1
x +1
+ x −3 2 =
x
2
2
2 x3 2
15. g (t ) =
x − 2
( x − 2)(2 x + 1)
x→2
6. f ( x) =
f ( x) = x1 2 − x −1 2
f (θ ) = 4θ − 5 sin θ
f ′(θ ) = 4 − 5 cos θ
18. g (α ) = 4 cos α + 6
g ′(α ) = −4 sin α
−1
1
= −
49
( x + 4) 7
19.
f (t ) = 3 cos t − 4et
f ′(t ) = −3 sin t − 4et
7. f is differentiable for all x ≠ 3.
20. g ( s ) =
5
3
sin s − 2e s
g ′( s ) =
5
3
cos s − 2e s
8. f is differentiable for all x ≠ −1.
9. y = 25
y′ = 0
10.
21.
f (t ) = 4t 4
f ′(t ) = 16t 3
11.
f ( x) = x 3 − 11x 2
f ′( x) = 3 x 2 − 22 x
22.
f ′(1) = 6 − 4 = 2
g '( s ) = 15s 4 − 8s 3
23.
x + 3 3 x = 6 x1 2 + 3 x1 3
h′( x) = 3 x −1 2 + x −2 3 =
f ( x) = 3x 2 − 4 x, (1, −1)
f ′( x) = 6 x − 4
12. g ( s ) = 3s 5 − 2s 4
13. h( x) = 6
27
= 27 x − 3 , (3, 1)
x3
81
f ′( x) = 27( − 3) x − 4 = − 4
x
81
f ′(3) = − 4 = −1
3
f ( x) =
3
+
x
1
3
x2
f ( x) = 2 x 4 − 8, (0, − 8)
f ′( x) = 8 x3
f ′(0) = 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
24.
f (θ ) = 3 cos θ − 2θ , (0, 3)
f ′(θ ) = − 3 sin θ − 2
30. g ( x) = ( 2 x3 + 5 x)(3x − 4)
g ′( x) = ( 2 x3 + 5 x)(3) + (3 x − 4)(6 x 2 + 5)
f ′(0) = − 3 sin (0) − 2 = − 2
25.
= 6 x 3 + 15 x + 18 x3 − 24 x 2 + 15 x − 20
= 24 x 3 − 24 x 2 + 30 x − 20
F = 200 T
F ′(t ) =
31. h( x) =
100
T
h′( x) =
(a) When T = 4, F ′( 4) = 50 vibrations/sec/lb.
(b) When T = 9, F′(9) = 33 13 vibrations/sec/lb.
26.
32.
x sin x = x1 2 sin x
1
2
x
sin x +
f ′(t ) = 2t 5 ( −sin t ) + cos t (10t 4 )
S = 6s 2
= −2t 5 sin t + 10t 4 cos t
dS
(a) When s = 3,
= 12(3) = 36 in.2 /in.
ds
33.
x2 + x − 1
x2 − 1
f ( x) =
f ′( x) =
dS
(b) When s = 5,
= 12(5) = 60 in.2 /in.
ds
27. s(t ) = −16t 2 + v0t + s0 ; s0 = 600, v0 = − 30
=
( x2
s′(t ) = v(t ) = − 32t − 30
(b) Average velocity =
s(3) − s(1)
3−1
366 − 554
=
2
= − 94 ft/sec
34.
f ( x) =
f ′( x) =
=
− 1)( 2 x + 1) − ( x 2 + x − 1)( 2 x)
( x2
( x2
− 1)
=
( x2
(e) When
t ≈ 5.258, v(t ) ≈ − 32(5.258) − 30 ≈ −198.3 ft/sec.
s(t ) = −16t 2 + s0
35. y =
y′ =
=
s(9.2) = −16(9.2) + s0 = 0
2
s0 = 1354.24
36. y =
The building is approximately 1354 feet high (or 415 m).
f ( x) = (5 x 2 + 8)( x 2 − 4 x − 6)
f ′( x) = (5 x 2 + 8)( 2 x − 4) + ( x 2 − 4 x − 6)(10 x )
= 10 x3 + 16 x − 20 x 2 − 32 + 10 x3 − 40 x 2 − 60 x
= 20 x3 − 60 x 2 − 44 x − 32
y′ =
2
+ 4)( 2) − ( 2 x + 7)( 2 x)
( x2
+ 4)
2
2 x 2 + 8 − 4 x 2 − 14 x
( x2
+ 4)
2
− 2 x 2 − 14 x + 8
( x2
(d) s(t ) = 0 = −16t 2 − 30t + 600
Using a graphing utility or the Quadratic Formula,
t ≈ 5.258 seconds.
2
2x + 7
x2 + 4
(c) v(1) = − 32(1) − 30 = − 62 ft/sec
v(3) = − 32(3) − 30 = −126 ft/sec
− 1)
−( x 2 + 1)
(a) s(t ) = −16t 2 − 30t + 600
29.
x cos x
f (t ) = 2t 5 cos t
dS
= 12 s
ds
28.
247
+ 4)
2
=
− 2( x 2 + 7 x − 4)
( x2
+ 4)
2
x4
cos x
(cos x) 4 x3
− x 4 ( −sin x)
cos 2 x
4 x cos x + x 4 sin x
cos 2 x
3
sin x
x4
( x 4 ) cos x − (sin x)(4 x3 )
2
( x4 )
=
x cos x − 4 sin x
x5
37. y = 3 x 2 sec x
y′ = 3 x 2 sec x tan x + 6 x sec x
= 4(5 x3 − 15 x 2 − 11x − 8)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
248
Chapter 3
Differentiation
38. y = 2 x − x 2 tan x
44.
y′ = 2 − x 2 sec 2 x − 2 x tan x
f ( x) =
f ′( x) =
39. y = 4 xe x − cot x
y′ = 4 xe x + 4e x + csc2 x
=
40. g ( x) = 3 x sin x + x cos x
2
= 5 x cos x + (3 − x 2 )sin x
f ( x) = ( x + 2)( x 2 + 5), ( −1, 6)
= 2 x 2 + 4 x + x 2 + 5 = 3x 2 + 4 x + 5
45.
y = 4 x + 10
46.
= 2 x 2 − 2 x − 24 + x 2 + 6 x − 1
= 3 x 2 + 4 x − 25
47.
f ′(0) = 0 + 0 − 25 = − 25
Tangent line: y − 4 = − 25( x − 0)
y = − 25 x + 4
48.
(x
− 1)
2
(1 −
cos x)
2
g (t ) = −8t 3 − 5t + 12
h( x ) = 6 x − 2 + 7 x 2
36
+ 14
x4
f ( x) = 15 x5 2
f ′( x) =
75 3 2
x
2
f ′′( x) =
225 1 2
x
4
=
225
4
x
f ( x) = 20 5 x = 20 x1 5
f ′( x) = 4 x −4 5
x + 1 ⎛1
⎞
, ⎜ , − 3⎟
x − 1 ⎝2
⎠
− 1) − ( x + 1)
−2 sin x
h′′( x ) = 36 x − 4 + 14 =
f ′( x) = ( x − 4)( 2 x + 6) + ( x 2 + 6 x − 1)(1)
(x
2
h′( x) = −12 x − 3 + 14 x
f ( x) = ( x − 4)( x 2 + 6 x − 1), (0, 4)
f ′( x) =
cos x )
g ′′(t ) = −48t
Tangent line: y − 6 = 4( x + 1)
f ( x) =
(1 −
g ′(t ) = −24t 2 − 5
f ′( −1) = 3 − 4 + 5 = 4
43.
cos x )( −sin x) − (1 + cos x)(sin x )
π⎞
⎛
Tangent line: y − 1 = −2⎜ x − ⎟
2⎠
⎝
y = −2 x + 1 + π
f ′( x) = ( x + 2)( 2 x) + ( x 2 + 5)(1)
42.
(1 −
−2
⎛π ⎞
= −2
f ′⎜ ⎟ =
1
⎝2⎠
g ′( x) = 3 x cos x + 3 sin x − x 2 sin x + 2 x cos x
41.
1 + cos x ⎛ π ⎞
, ⎜ , 1⎟
1 − cos x ⎝ 2 ⎠
=
f ′′( x) =
−2
(x
− 1)
2
49.
−2
⎛1⎞
= −8
f ′⎜ ⎟ =
(1 4)
⎝ 2⎠
1⎞
⎛
Tangent line: y + 3 = −8⎜ x − ⎟
2⎠
⎝
y = −8 x + 1
−16 −9 5
16
x
= − 95
5
5x
f (θ ) = 3 tan θ
f ′(θ ) = 3 sec 2 θ
f ′′(θ ) = 6 sec θ (sec θ tan θ ) = 6 sec 2 θ tan θ
50.
h(t ) = 10 cos t − 15 sin t
h′(t ) = −10 sin t − 15 cos t
h′′(t ) = −10 cos t + 15 sin t
51. v(t ) = 20 − t 2 , 0 ≤ t ≤ 6
a(t ) = v′(t ) = − 2t
v(3) = 20 − 32 = 11 m/sec
a(3) = − 2(3) = − 6 m/sec 2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
52. v(t ) =
a (t ) =
=
90t
4t + 10
( 4t + 10)90 − 90t ( 4)
( 4t
+ 10)
900
(4t
+ 10)
=
2
60. y =
= sec5 x tan x(sec 2 x − 1)
225
(2t
+ 5)
= sec5 x tan 3 x
2
61. y = x(6 x + 1)
90
≈ 6.43 ft/sec
14
225
a(1) =
≈ 4.59 ft/sec 2
49
90(5)
4
4
4
= (6 x + 1) (36 x + 1)
4
62.
f ( s ) = ( s 2 − 1)
(s3 + 5)
52
32
f ′( s ) = ( s 2 − 1) (3s 2 ) + ( s 3 + 5)( 52 )( s 2 − 1) ( 2 s )
32
= s( s 2 − 1) ⎡⎣3s( s 2 − 1) + 5( s 3 + 5)⎤⎦
3
3
63.
f ′( x) =
2
=
−1
1
55. y = 2
= ( x 2 + 4)
x + 4
y′ = −1( x 2 + 4)
−2
(2 x)
1
(5 x
+ 1)
2
= −
x2 + 1
3( x 2 + 1)
−3
(5)
−1 2
1 2
( x + 1) (2 x)
2
x2 + 1
− 3x
3( x 2 + 1) − 3 x 2
(x
2
+ 1)
=
32
3
(x
2
+ 1)
32
2x
(x
= (5 x + 1)
f ′( x) = − 2(5 x + 1)
= −
2
+ 4)
⎛ x + 5⎞
64. h( x) = ⎜ 2
⎟
⎝ x + 3⎠
2
10
(5 x
+ 1)
2
⎛ 2
⎞
⎛ x + 5 ⎞⎜ ( x + 3)(1) − ( x + 5)( 2 x ) ⎟
h′( x) = 2⎜ 2
⎟
2
⎟⎟
⎝ x + 3 ⎠⎜⎜
( x 2 + 3)
⎝
⎠
−2
3
=
2( x + 5)( − x 2 − 10 x + 3)
( x2
57. y = 5 cos(9 x + 1)
y′ = −5 sin (9 x + 1)(9) = −45 sin (9 x + 1)
58. y = 1 − cos 2 x + 2 cos 2 x
= 2[2 sin x cos x] − 4 sin x cos x
= 0
x
sin 2 x
−
2
4
1
1
1
y′ =
− cos 2 x( 2) = (1 − cos 2 x) = sin 2 x
2
4
2
+ 3)
3
65. g (t ) = t 2e t 4
g ′(t ) =
=
y′ = 2 sin 2 x − 4 cos x sin x
59. y =
(8s3 − 3s + 25)
12
2
f ( x) =
32
3x
f ( x) =
3
y′ = 3( x 2 − 6) ( 2 x) = 6 x( x 2 − 6)
56.
52
= s( s 2 − 1)
4
y′ = 4(7 x + 3) (7) = 28(7 x + 3)
54. y = ( x − 6)
5
= (6 x + 1) (30 x + 6 x + 1)
= 15 ft/sec
= 18 ft/sec
50
225
a(10) =
= 0.36 ft/sec2
252
2
5
= 30 x(6 x + 1) + (6 x + 1)
90(10)
53. y = (7 x + 3)
5
y′ = x 5(6 x + 1) (6) + (6 x + 1) (1)
30
225
a(5) =
= 1 ft/sec 2
152
(c) v(10) =
sec 7 x
sec5 x
−
7
5
y′ = sec6 x(sec x tan x) − sec 4 x(sec x tan x )
2
(a) v(1) =
(b) v(5) =
249
1 2 t4
t e
4
1 t4
te
4
66. h( z ) = e − z
[t
y′ =
+ 8]
2 2
h′( z ) = − ze − z
67. y =
+ 2te t 4
2 2
e 2 x + e − 2 x = (e 2 x + e − 2 x )
12
−1 2
1 2x
( e + e − 2 x ) ( 2e 2 x − 2e − 2 x ) =
2
e2 x − e− 2 x
e2 x + e− 2 x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
250
Chapter 3
Differentiation
68. y = 3e − 3 t
78.
y′ = 3e − 3 t (3t − 2 ) =
69. g ( x) =
g ′( x) =
70.
f (θ ) =
9e − 3 t
t2
x2
ex
e x ( 2 x) − x 2e x
e2 x
=
x( 2 − x )
b
a + bx
1
+ 2 ln
ax
a
x
b
1
= −
+ 2 ⎡⎣ln ( a + bx ) − ln x⎤⎦
ax
a
dy
b⎡ b
1⎛ 1 ⎞
1⎤
= − ⎜− 2 ⎟ + 2 ⎢
− ⎥
dx
a ⎝ x ⎠ a ⎣ a + bx
x⎦
y = −
=
ex
1 sin 2θ
e
2
=
⎤
1
b ⎡ −a
1
b
+ 2⎢
−
⎥ =
2
2
ax
a ⎣⎢ x( a + bx) ⎦⎥
ax
ax( a + bx)
(a
f ′(θ ) = cos 2θ esin 2θ
71. g ( x) = ln
1
x = ln x
2
72. h( x) = ln
x( x − 1)
x − 2
= ln x + ln ( x − 1) − ln ( x − 2)
1
1
1
x2 − 4 x + 2
h′( x) =
+
−
= 3
x
x −1
x − 2
x − 3x 2 + 2 x
73.
75.
y =
77.
x 2 − 1, (3, 2)
−2 3
1 2
( x − 1) (2 x) = 2 2 x 2 3
3
3( x − 1)
2(3)
f ′(3) =
=
3( 4)
1
2
−1
4
= 4( x 2 + 1) , ( −1, 2)
x2 + 1
−2
8x
f ′( x) = − 4( x 2 + 1) ( 2 x) = −
2
2
x
( + 1)
81. f ( x) =
f ′( −1) = −
82. f ( x ) =
=
2
2
=
8
= 2
4
− 3)(3) − (3 x + 1)( 4)
(4 x
− 3)
2
12 x − 9 − 12 x − 4
f ′( 4) = −
83.
⎡( −1) + 1⎤
⎣
⎦
(4 x
= −
1 ⎛ a + bx ⎞
1
y = − ln ⎜
⎟ = − ⎡⎣ln ( a + bx) − ln x⎤⎦
a ⎝ x ⎠
a
8( −1)
3x + 1
, ( 4, 1)
4x − 3
f ′( x) =
1
y = 2 ⎡⎣a + bx − a ln ( a + bx)⎤⎦
b
dy
1⎛
ab ⎞
x
= 2 ⎜b −
⎟ =
dx
b ⎝
a + bx ⎠
a + bx
dy
1⎛ b
1⎞
1
= − ⎜
− ⎟ =
dx
a ⎝ a + bx
x⎠
x( a + bx)
3
f ′( x) =
⎤
dy
ab
x
1⎡ b
⎥ =
= 2⎢
−
2
2
dx
b ⎢ a + bx
(a + bx) ⎥⎦ ( a + bx)
⎣
76.
1 − x 3 , ( − 2, 3)
80. f ( x ) =
1
2⎛ 2x ⎞
7 x2 − 6
+ ⎜ 2
⎟ =
x
3⎝ x − 2 ⎠
3 x3 − 6 x
1⎡
a ⎤
ln ( a + bx) +
b 2 ⎢⎣
a + bx ⎥⎦
1
x 2 ( a + bx)
−1 2
1
−3 x 2
1 − x 3 ) ( −3 x 2 ) =
(
2
2 1 − x3
−12
= −2
f ′( −2) =
2(3)
23
2
f ( x) = ln ⎡ x( x 2 − 2) ⎤ = ln x + ln ( x 2 − 2)
⎢⎣
⎥⎦
3
f ′( x) =
=
f ′( x) =
f ( x) = x ln x
−1 2 ⎛ 1 ⎞
⎛ x⎞
f ′( x) = ⎜ ⎟(ln x) ⎜ ⎟ + ln x
⎝ 2⎠
⎝ x⎠
1
1 + 2 ln x
=
+ ln x =
2 ln x
2 ln x
74.
79. f ( x ) =
1
2x
g ′( x) =
+ bx) − bx
ax 2 ( a + bx)
(4 x
− 3)
2
13
(4 x
− 3)
2
13
(16
− 3)
2
= −
1
13
1
⎛π 1 ⎞
csc 2 x, ⎜ , ⎟
2
⎝ 4 2⎠
y′ = −csc 2 x cot 2 x
y =
⎛π ⎞
y′⎜ ⎟ = 0
⎝4⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
⎛π ⎞
y = csc 3 x + cot 3 x, ⎜ , 1⎟
⎝6 ⎠
84.
1
1
cos 8t − sin 8t
4
4
1
1
y′ = ( − sin 8t )8 − (cos 8t )8
4
4
= − 2 sin 8t − 2 cos 8t
y =
90.
y′ = −3 csc 3x cot 3x − 3 csc 2 3x
⎛π ⎞
y′⎜ ⎟ = 0 − 3 = −3
⎝6⎠
85.
y = (8 x + 5)
At time t =
3
y′ = 3(8 x + 5) (8) = 24(8 x + 5)
2
y =
1
−1
> (5 x + 1)
5x + 1
y′ = ( −1)(5 x + 1)
−2
(5)
y′′ = ( − 5)( − 2)(5 x + 1)
87.
= − 5(5 x + 1)
−3
(5)
=
π
4
,
⎡ ⎛ π ⎞⎤ 1
⎡ ⎛ π ⎞⎤
1
⎛π ⎞
y ⎜ ⎟ = cos ⎢8 ⎜ ⎟⎥ − sin ⎢8 ⎜ ⎟⎥
4
4
4
4
⎝ ⎠
⎣ ⎝ ⎠⎦
⎣ ⎝ 4 ⎠⎦
1
1
= (1) = ft.
4
4
2
y′′ = 24( 2)(8 x + 5)(8) = 384(8 x + 5)
86.
251
⎡ ⎛ π ⎞⎤
⎡ ⎛ π ⎞⎤
⎛π ⎞
v(t ) = y′⎜ ⎟ = − 2 sin ⎢8 ⎜ ⎟⎥ − 2cos ⎢8 ⎜ ⎟⎥
⎝4⎠
⎣ ⎝ 4 ⎠⎦
⎣ ⎝ 4 ⎠⎦
−2
= − 2(0) − 2(1) = − 2 ft/sec
50
(5 x
+ 1)
3
91. (a) You get an error message because ln h does not
exist for h = 0.
f ( x) = cot x
f ′( x) = −csc 2 x
(b) Reversing the data, you obtain
h = 0.8627 − 6.4474 ln p.
f ′′( x) = −2 csc x( −csc x ⋅ cot x)
(c)
25
= 2 csc x cot x
2
88.
y = sin 2 x
0
y′ = 2 sin x cos x = sin 2 x
(d) If p = 0.75, h ≈ 2.72 km.
y′′ = 2 cos 2 x
89. T =
(e) If h = 13 km, p ≈ 0.15 atmosphere.
700
2
t + 4t + 10
(f ) h = 0.8627 − 6.4474 ln p
T = 700(t 2 + 4t + 10)
T′ =
−1
1 = − 6.4474
−1400(t + 2)
(t 2
+ 4t + 10)
4 + 10)
≈ −18.667 deg/h.
2
p = 0.0514 and
−1400(3 + 2)
(9
+ 12 + 10)
≈ −7.284 deg/h.
2
−1400(5 + 2)
(25 +
dp
= − 0.0816 atm km
dh
20 + 10)
dp
= − 0.0080 atm km
dh
As the altitude increases, the rate of change of
pressure decreases.
(c) When t = 5,
T′ =
p = 0.5264 and
For h = 20,
(b) When t = 3,
T′ =
(implicit differentiation )
For h = 5,
−1400(1 + 2)
(1 +
1 dp
p dh
dp
p
=
dh
− 6.4474
2
(a) When t = 1,
T′ =
1
0
≈ −3.240 deg/h.
2
(d) When t = 10,
T′ =
−1400(10 + 2)
(100
+ 40 + 10)
2
≈ −0.747 deg/h.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
252
Chapter 3
Differentiation
⎛ 10 +
92. y = 10 ln ⎜
⎜
⎝
(a)
100 − x 2 ⎞
⎟ −
⎟
x
⎠
)
(
100 − x 2 = 10 ⎡ln 10 +
⎣⎢
100 − x 2 − ln x⎤ −
⎦⎥
100 − x 2
20
0
10
0
(b)
−x
dy
⎡
= 10 ⎢
2
dx
⎢⎣ 100 − x 10 +
(
100 − x
2
)
−
1⎤
+
x⎥
⎥⎦
=
⎡
⎤ 10
−10
+
⎢
⎥ −
2
x
100 − x ⎣10 + 100 − x ⎦
=
⎡
⎤ 10
−10
+ 1⎥ −
⎢
2
x
100 − x ⎣10 + 100 − x
⎦
x
2
x
100 − x 2
x
100 − x 2
x
2
⎡
100 − x 2 ⎤ 10
⎢
⎥ −
x
100 − x 2 ⎢⎣10 + 100 − x 2 ⎥⎦
x
10
=
−
x
10 + 100 − x 2
x
=
=
(
x 10 −
100 − x 2
x
2
) − 10 = −
x
When x = 5, dy dx = −
(c)
93.
lim
x → 10 −
100 − x 2
x
3. When x = 9, dy dx = − 19 9.
dy
= 0
dx
x 2 + y 2 = 64
2 x + 2 yy′ = 0
y
x
= 1 − 4 y′
y′ +
2 y
2 x
2 yy′ = − 2 x
y′ = −
94.
x
y
xy′ + y = 2
xy − 8
xy y′ = 2
xy − y
2
xy − y
x +8
x + 4 xy − y = 6
2
3
y′ =
2 x + 4 xy′ + 4 y − 3 y y′ = 0
2
( 4 x − 3 y ) y′ =
− 2x − 4 y
y′ =
2x + 4 y
3 y2 − 4x
2
95.
xy = x − 4 y
96.
=
=
x 3 y − xy 3 = 4
x 3 y′ + 3 x 2 y − x3 y 2 y′ − y 3 = 0
x 3 y′ − 3 xy 2 y′ = y 3 − 3 x 2 y
y′( x 3 − 3 xy 2 ) = y 3 − 3 x 2 y
y 3 − 3x 2 y
y′ = 3
x − 3xy 2
y′ =
97.
x +8
xy y′
xy
2( x − 4 y ) − y
x + 8( x − 4 y )
2x − 9 y
9 x − 32 y
x sin y = y cos x
( x cos y ) y′ + sin y = − y sin x + y′ cos x
y′( x cos y − cos x) = − y sin x − sin y
y′ =
y sin x + sin y
cos x − x cos y
y( y 2 − 3x 2 )
x( x 2 − 3 y 2 )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
253
cos( x + y ) = x
98.
−(1 + y′) sin ( x + y ) = 1
− y′ sin ( x + y ) = 1 + sin ( x + y)
y′ = −
1 + sin ( x + y )
sin ( x + y)
= −csc( x + 1) − 1
99. x 2 + y 2 = 10
2 x + 2 yy′ = 0
−x
y
y′ =
4
At (3, 1), y′ = −3
(3, 1)
Tangent line:
y − 1 = −3( x − 3) ⇒ 3x + y − 10 = 0
Normal line:
y −1 =
−4
y ln x + y 2 = 0,
101.
(e, −1)
y
+ 2 yy′ = 0
x
−y
y′(ln x + 2 y ) =
x
y′ ln x +
3
2
y′ =
3
y − 4 = ( x − 6)
Tangent line:
2
3
y = x −5
2
2 y − 3x + 10 = 0
Normal line:
6
1
( x − 3) ⇒ x − 3 y = 0
3
100. x 2 − y 2 = 20
2 x − 2 yy′ = 0
x
y′ =
y
At (6, 4), y′ =
−6
−y
x(ln x + 2 y )
At (e, −1): y′ =
−1
e
−1
( x − e)
e
−1
y =
x
e
Tangent line: y + 1 =
2
( x − 6)
3
2
y = − x +8
3
3 y + 2 x − 24 = 0
y − 4 = −
Normal line: y + 1 = e( x − e)
y = ex − e 2 − 1
2
4
P1
P2
−6
6
f
0
(e, −1)
3
0
102.
−4
ln ( x + y ) = x, (0, 1)
1
(1 + y′) = 1
x + y
1 + y′ = x + y
y′ = x + y − 1
8
At (0, 1): y′ = 0
Tangent line: y − 1 = 0 ⇒ y = 1
Normal line: x = 0
(0, 1)
−7
5
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
254
103.
Chapter 3
x2 + 1
x + 4
1
ln y = ln x + ln ( x 2 + 1) − ln ( x + 4)
2
y′
x
1
1
=
+ 2
−
y
x
x +1 x + 4
y =
y′ =
=
104.
Differentiation
y =
x
107.
⎡ π π⎤
f is monotonic (increasing) on ⎢− , ⎥ therefore f has
⎣ 4 4⎦
an inverse.
+ 4)
2
x2 + 1
(2 x + 1)3 ( x 2
− 1)
4
⎛π ⎞
f ′⎜ ⎟ =
3
⎝6⎠
2
⎛
3⎞
⎟⎟ =
⎝ 3 ⎠
( f −1 )⎜⎜
x+3
ln y = 3 ln ( 2 x + 1) + 2 ln ( x 2 − 1) − ln ( x + 3)
y′
6
4x
1
=
+ 2
−
y
2x + 1 x − 1 x + 3
y′ =
=
105.
− 1) ⎛ 6
4x
1 ⎞
+ 2
−
⎜
⎟
x+3
⎝ 2x + 1 x − 1 x + 3 ⎠
(2 x + 1)
3
(x
2
2
(2 x + 1)2 ( x 2
⎛ 3⎞
3
π
⇒ f −1 ⎜⎜
⎟⎟ =
3
3
6
⎝
⎠
⎛π ⎞
f⎜ ⎟ =
⎝6⎠
x3 + 8 x 2 + 4
(x
3 π
π
,− ≤ x ≤
3
4
4
a =
⎛ π π⎞
f ′( x) = sec 2 x > 0 on ⎜ − , ⎟
⎝ 4 4⎠
x2 + 1 ⎛ 1
x
1 ⎞
−
⎜ + 2
⎟
x + 4 ⎝x
x + 1 x + 4⎠
x
f ( x) = tan x,
− 1)(12 x3 + 45 x 2 + 8 x − 17)
( x + 3)
2
f ( x) = x 3 + 2, a = −1
108.
1
⎛
⎛ 3 ⎞⎞
f ′⎜ f −1 ⎜⎜
⎟⎟ ⎟⎟
⎜
⎝ 3 ⎠⎠
⎝
f is monotonic (increasing) on ( −∞, ∞) therefore f has
f ( x) = cos x, a = 0, 0 ≤ x ≤ π
f ′( x) = −sin x < 0 on (0, π )
f is monotonic (decreasing) on [0, π ] therefore f has
an inverse.
π
⎛π ⎞
f ⎜ ⎟ = 0 ⇒ f −1 (0) =
2
⎝2⎠
( f −1 )(0)
=
f ′( f
an inverse.
1
−1
( 0) )
=
f ( −31 3 ) = −1 ⇒ f −1 (−1) = −31 3
f ′( −31 3 ) = 32 3
( f )′ (−1) =
106.
f ( x) = x
1
f ′( x) = x
2
f ′( f −1 ( −1))
x − 3,
1
1
1
=
=
=
f ′( −31 3 ) 3(32 3 ) 35 3
a = 4
1
+
x −3
x −3 > 0
f ( 4) = 4 ⇒ f −1 ( 4) = 4
f ′( 4) = 2 + 1 = 3
( f −1 )′ (4)
=
f ′( f
1
−1
( 4))
=
1
1
=
′
f ( 4)
3
1 − x2
+ x 2 (1 − x 2 )
1 − x2
−1 2
= (1 − x 2 )
−3 2
110. y = arctan ( 2 x 2 − 3)
y′ =
f is monotonic (increasing) on [3, ∞) therefore f has
an inverse.
(1 − x 2 )
12
y′ =
1
1
=
= −1
π
−1
⎛ ⎞
f ′⎜ ⎟
⎝2⎠
x
109. y = tan (arcsin x ) =
1
1
1
3
=
=
4
π
4
⎛ ⎞
⎛ ⎞
f ′⎜ ⎟
⎜ ⎟
⎝6⎠
⎝ 3⎠
⎛π ⎞
f ′⎜ ⎟ = −1
⎝2⎠
f ′( x) = 3 x 2 > 0
−1
=
1
(2 x
2
− 3) + 1
2
( 4 x)
4x
4 x 4 − 12 x 2 + 10
2x
=
2x4 − 6 x2 + 5
=
111. y = x arcsec x
x
y′ =
x
x2 − 1
+ arcsec x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
255
1
arctan e 2 x
2
1⎛ 1 ⎞ 2x
e2 x
2e ) =
y′ = ⎜
4 x ⎟(
2⎝1 + e ⎠
1 + e4 x
112. y =
113. y = x(arcsin x) − 2 x + 2 1 − x 2 arcsin x
2
2 x arcsin x
y′ =
1− x
2
2
x
x 2 − 4 − 2 arcsec ,
2
114. y =
x
y′ =
115.
+ (arcsin x) − 2 +
x2 − 4
y =
2 1 − x2
(
x 2)
2x
−
1− x
2
arcsin x = (arcsin x)
( x 2)
2
−1
=
x
x2 − 4
4
−
x
(a) When x =
x
dy
= 4
dt
x2 − 4
=
x2 − 4
x2 − 4
x
dx
= 4 units/sec.
dt
(c) When x = 4,
dx
= 8 units/sec.
dt
x2 − 4
x
dθ
= 3( 2π ) rad/min
dt
dx
⎛ dθ ⎞
sec 2θ ⎜ ⎟ =
dt
dt
⎝ ⎠
x
dx
= ( tan 2θ + 1)(6π ) = 6π ( x 2 + 1)
dt
1 dx
,
= 2 2 units/sec.
2 dt
(b) When x = 1,
=
tanθ = x
117.
x
dy
= 2 units/sec
dt
dy
dx
1 dx
=
⇒
= 2
dt
dt
2 x dt
2
2 < x < 4
1
−
1− x
2
1
,
2
dx
⎛1
⎞ 15π
km/min = 450π km/h.
= 6π ⎜ + 1⎟ =
dt
2
⎝4
⎠
When x =
116. Surface area = A = 6 x 2 , x = length of edge
1
dx
= 8
dt
θ
x
dA
dx
= 12 x
= 12(6.5)(8) = 624 cm 2 /sec
dt
dt
118.
s(t ) = 60 − 4.9t 2
s′(t ) = −9.8t
s = 35 = 60 − 4.9t 2
4.9t 2 = 25
t =
tan 30 =
x (t ) =
dx
=
dt
5
4.9
s (t )
1
=
x (t )
3
3s ( t )
ds
=
3
dt
s (t)
3 ( −9.8)
5
≈ −38.34 m/sec
4.9
30°
x(t )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
256
Chapter 3
Differentiation
119. f ( x) = x3 − 3 x − 1
From the graph you can see that f ( x) has three real zeros.
f ′( x) = 3x 2 − 3
f ( xn )
f ( xn )
f ( xn )
f ′( xn )
–1.5000
0.1250
3.7500
0.0333
–1.5333
–1.5333
–0.0049
4.0530
–0.0012
–1.5321
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
n
xn
1
2
xn −
f ′( xn )
f ′( xn )
f ( xn )
f ′( xn )
xn −
n
xn
1
–0.5000
0.3750
–2.2500
–0.1667
–0.3333
2
–0.3333
–0.0371
–2.6667
0.0139
–0.3472
3
–0.3472
–0.0003
–2.6384
0.0001
–0.3473
n
xn
f ( xn )
f ′( xn )
1
1.9000
0.1590
7.8300
0.0203
1.8797
2
1.8797
0.0024
7.5998
0.0003
1.8794
f ( xn )
f ′( xn )
xn −
f ( xn )
f ′( xn )
The three real zeros of f ( x) are x ≈ −1.532, x ≈ −0.347, and x ≈ 1.879.
120. f ( x ) = x3 + 2 x + 1
From the graph, you can see that f ( x) has one real zero.
f ′( x) = 3 x 2 + 2
f changes sign in [−1, 0].
f ( xn )
f ( xn )
f ( xn )
f ′( xn )
–0.5000
–0.1250
2.7500
–0.0455
–0.4545
–0.4545
–0.0029
2.6197
–0.0011
–0.4534
n
xn
1
2
f ′( xn )
xn −
f ′( xn )
On the interval [−1, 0]: x ≈ −0.453.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
257
121. g ( x) = xe x − 4
g ′( x) = ( x + 1)e x
From the graph, there is one zero near 1.
g ( xn )
g ( xn )
n
xn
g ( xn )
g ′( xn )
1
1.0
–1.2817
5.4366
–0.2358
1.2358
2
1.2358
0.2525
7.6937
0.0328
1.2030
3
1.2030
0.0059
7.3359
0.0008
1.2022
xn −
g ′( xn )
g ′( xn )
To three decimal places, x = 1.202.
122.
f ( x) = 3 − x ln x
f ′( x) = −1 − ln x
From the graph, there is one zero near 3.
f ( xn )
f ( xn )
f ′( xn )
3
–0.2958
–2.0986
0.1410
2.8590
2
2.8590
–0.0034
–2.0505
0.0016
2.8574
3
2.8574
–0.0000
–2.0499
0.0000
2.8574
n
xn
1
xn −
f ′( xn )
f ( xn )
f ′( xn )
To three decimal places, x = 2.857.
123. f ( x) = x 4 + x3 − 3 x 2 + 2
From the graph you can see that f ( x) has two real zeros.
f ′( x) = 4 x 3 + 3 x 2 − 6 x
f ( xn )
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
− 2.0
− 2.0
− 8.0
0.25
− 2.25
2
− 2.25
1.0508
−16.875
− 0.0623
− 2.1877
3
− 2.1877
0.0776
−14.3973
− 0.0054
− 2.1823
4
− 2.1823
0.0004
−14.3911
− 0.00003
− 2.1873
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
−1.0
−1.0
5.0
− 0.2
− 0.8
2
− 0.8
− 0.0224
4.6720
− 0.0048
− 0.7952
3
− 0.7952
− 0.00001
4.6569
− 0.0000
− 0.7952
f ′( xn )
xn −
xn −
f ′( xn )
f ( xn )
f ′( xn )
The two zeros of f ( x) are x ≈ − 2.1823 and x ≈ − 0.7952.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
258
Chapter 3
Differentiation
124. f ( x ) = 3 x − 1 − x
From the graph you can see that f ( x) has two real zeros.
2
3
−1
x −1
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
1
1.1
− 0.1513
3.7434
− 0.0404
1.1404
2
1.1404
− 0.0163
3.0032
− 0.0054
1.1458
3
1.1458
− 0.0003
2.9284
− 0.0000
1.1459
n
xn
f ( xn )
f ′( xn )
1
8.0
− 0.0627
− 0.4331
0.1449
7.8551
2
7.8551
− 0.0004
− 0.4271
0.0010
7.8541
f ′( x) =
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
xn −
xn −
f ( xn )
f ′( xn )
The two zeros of f ( x) are x ≈ 1.1459 and x ≈ 7.8541.
125. Find the zeros of f ( x ) = x 4 − x − 3.
f ′( x) = 4 x3 − 1
From the graph you can see that f ( x) has two real zeros.
f changes sign in [−2, −1].
f ( xn )
f ( xn )
f ( xn )
f ′( xn )
–1.2000
0.2736
–7.9120
–0.0346
–1.1654
–1.1654
0.0100
–7.3312
–0.0014
–1.1640
n
xn
1
2
f ′( xn )
xn −
f ′( xn )
On the interval [−2, −1]: x ≈ −1.164.
f changes sign in [1, 2].
f ( xn )
f ( xn )
f ′( xn )
1.5000
0.5625
12.5000
0.0450
1.4550
2
1.4550
0.0268
11.3211
0.0024
1.4526
3
1.4526
–0.0003
11.2602
0.0000
1.4526
n
xn
1
f ′( xn )
xn −
f ( xn )
f ′( xn )
On the interval [1, 2]: x ≈ 1.453.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 3
259
126. Find the zeros of f ( x) = sin π x + x − 1.
f ′( x) = π cos π x + 1
From the graph you can see that f ( x) has three real zeros.
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
xn −
0.2000
–0.2122
3.5416
–0.0599
0.2599
2
0.2599
–0.0113
3.1513
–0.0036
0.2635
3
0.2635
0.0000
3.1253
0.0000
0.2635
n
xn
f ( xn )
f ′( xn )
1
1.0000
0.0000
–2.1416
n
xn
f ( xn )
f ′( xn )
1
1.8000
0.2122
3.5416
0.0599
1.7401
2
1.7401
0.0113
3.1513
0.0036
1.7365
3
1.7365
0.0000
3.1253
0.0000
1.7365
n
xn
1
f ( xn )
f ′( xn )
0.0000
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
f ( xn )
xn −
f ′( xn )
1.0000
xn −
f ( xn )
f ′( xn )
The three real zeros of f ( x) are x ≈ 0.264, x = 1, and x ≈ 1.737.
127. Find the zeros of f ( x) = ln x + x.
f ′( x) =
1
+1
x
From the graph you can see that f ( x) has one real zero.
n
xn
f ( xn )
f ′( xn )
f ( xn )
f ′( xn )
xn −
f ( xn )
f ′( xn )
1
0.5
–0.1931
3.0000
–0.0644
0.5644
2
0.5644
–0.0076
2.7718
–0.0027
0.5671
3
0.5671
0.0001
2.7634
–0.0000
0.5671
The real zero of f ( x) is x ≈ 0.567.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
260
Chapter 3
Differentiation
128. Find the zeros of f ( x) = arcsin x − 1 + x.
f ′( x) =
1
1 − x2
+1
From the graph you can see that f ( x) has one real zero.
f ( xn )
n
xn
f ( xn )
f ′( xn )
1
0.5
0.0236
2.1547
0.0110
0.4890
2
0.4890
0.0001
2.1465
0.0000
0.4890
xn −
f ′( xn )
f ( xn )
f ′( xn )
The real zero of f ( x) is x ≈ 0.489.
Problem Solving for Chapter 3
1. (a) x 2 + ( y − r ) = r 2 , Circle
2
x 2 = y, Parabola
Substituting:
(y
3
− r) = r 2 − y
2
y 2 − 2ry + r 2 = r 2 − y
−3
3
y 2 − 2ry + y = 0
−1
y ( y − 2r + 1) = 0
2
Because you want only one solution, let 1 − 2r = 0 ⇒ r =
1
1⎞
1
⎛
. Graph y = x 2 and x 2 + ⎜ y − ⎟ = .
2⎠
4
2
⎝
(b) Let ( x, y ) be a point of tangency:
x 2 + ( y − b) = 1 ⇒ 2 x + 2( y − b) y′ = 0 ⇒ y′ =
2
x
, Circle
b − y
y = x 2 ⇒ y′ = 2 x, Parabola
Equating:
3
x
2x =
b − y
2(b − y ) = 1
b − y =
−3
1
1
⇒ b = y +
2
2
3
−1
Also, x 2 + ( y − b) = 1 and y = x 2 imply:
2
2
⎡
1 ⎞⎤
1
3
5
2
⎛
y + ( y − b) = 1 ⇒ y + ⎢ y − ⎜ y + ⎟⎥ = 1 ⇒ y +
= 1 ⇒ y = and b =
2
4
4
4
⎝
⎠⎦
⎣
⎛ 5⎞
Center: ⎜ 0, ⎟
⎝ 4⎠
2
5⎞
⎛
Graph y = x 2 and x 2 + ⎜ y − ⎟ = 1.
4⎠
⎝
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 3
261
2. Let ( a, a 2 ) and (b, −b 2 + 2b − 5) be the points of tangency. For y = x 2 , y′ = 2 x and for y = − x 2 + 2 x − 5,
y′ = −2 x + 2. So, 2a = −2b + 2 ⇒ a + b = 1, or a = 1 − b. Furthermore, the slope of the common tangent line is
a 2 − ( −b 2 + 2b − 5)
a −b
=
(1 − b)2 + b 2 − 2b + 5
(1 − b) − b
= −2b + 2
1 − 2b + b 2 + b 2 − 2b + 5
= −2b + 2
1 − 2b
⇒ 2b 2 − 4b + 6 = 4b 2 − 6b + 2
⇒
⇒ 2b 2 − 2b − 4 = 0
⇒ b2 − b − 2 = 0
⇒ (b − 2)(b + 1) = 0
b = 2, −1
For b = 2, a = 1 − b = −1 and the points of tangency are ( −1, 1) and ( 2, −5). The tangent line has slope
−2: y − 1 = −2( x = 1) ⇒ y = −2 x − 1
For b = −1, a = 1 − b = 2 and the points of tangency are ( 2, 4) and ( −1, − 8). The tangent line has slope
4: y − 4 = 4( x − 2) ⇒ y = 4 x − 4
y
10
8
6
4
−8 −6 −4 −2
x
2 4 6 8 10
−4
−6
3. (a) f ( x ) = cos x
P1 ( x) = a0 + a1 x
f ( 0) = 1
P1 (0) = a0 ⇒ a0 = 1
f ′(0) = 0
P1′(0) = a1 ⇒ a1 = 0
P1 ( x) = 1
(b) f ( x ) = cos x
f ( 0) = 1
P2 (0) = a0 ⇒ a0 = 1
f ′(0) = 0
P2′(0) = a1 ⇒ a1 = 0
f ′′(0) = −1
P2′′ (0) = 2a2 ⇒ a2 = − 12
P2 ( x) = 1 −
(c)
P2 ( x) = a0 + a1 x + a2 x 2
1 x2
2
x
−1.0
−0.1
−0.001
0
0.001
0.1
1.0
cos x
0.5403
0.9950
≈1
1
≈1
0.9950
0.5403
P2 ( x)
0.5
0.9950
≈1
1
≈1
0.9950
0.5
P2 ( x) is a good approximation of f ( x) = cos x when x is near 0.
(d) f ( x ) = sin x
P3 ( x) = a0 + a1 x + a2 x 2 + a3 x3
f ( 0) = 0
P3 (0) = a0 ⇒ a0 = 0
f ′(0) = 1
P3′(0) = a1 ⇒ a1 = 1
f ′′(0) = 0
P3′′(0) = 2a2 ⇒ a2 = 0
f ′′′(0) = −1
P3′′′(0) = 6a3 ⇒ a3 = − 16
P3 ( x) = x −
1 x3
6
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262
Chapter 3
Differentiation
(d) Let ( a, a 2 ), a ≠ 0, be a point on the parabola y = x 2 .
4. (a) y = x 2 , y′ = 2 x, Slope = 4 at ( 2, 4)
Tangent line: y − 4 = 4( x − 2)
Tangent line at ( a, a 2 ) is y = 2a( x − a) + a 2 .
y = 4x − 4
(b) Slope of normal line: −
Normal line at ( a, a 2 ) is y = −(1 2a )( x − a ) + a 2 .
1
4
To find points of intersection, solve:
1
( x − a) + a 2
2a
1
1
x2 +
x = a2 +
2a
2
1
1
1
1
x2 +
x +
= a2 + +
2a
16a 2
2 16a 2
x2 = −
1
( x − 2)
4
1
9
y = − x +
4
2
1
9
y = − x +
= x2
4
2
⇒ 4 x 2 + x − 18 = 0
Normal line: y − 4 = −
2
1 ⎞
1 ⎞
⎛
⎛
⎜x +
⎟ = ⎜a +
⎟
4
4
a
a⎠
⎝
⎠
⎝
⇒ ( 4 x + 9)( x − 2) = 0
2
1
1 ⎞
⎛
= ±⎜ a +
⎟
4a
4
a⎠
⎝
1
1
x +
= a +
⇒ x = a
4a
4a
x +
9
x = 2, −
4
⎛ 9 81 ⎞
Second intersection point: ⎜ − , ⎟
⎝ 4 16 ⎠
x +
(c) Tangent line: y = 0
Normal line: x = 0
(Point of tangency)
1
1 ⎞
1
2a 2 + 1
⎛
= −⎜ a +
= −
⎟ ⇒ x = −a −
4a
4a ⎠
2a
2a
⎝
The normal line intersects a second time at x = −
2a 2 + 1
.
2a
5. Let p( x) = Ax3 + Bx 2 + Cx + D
p′( x) = 3 Ax 2 + 2 Bx + C.
At (1, 1):
A+
At ( −1, − 3):
B + C + D =
3A + 2B + C
1
Equation 1
= 14
Equation 2
A+
B − C + D = −3
3 A + 2B + C
= −2
Equation 3
Equation 4
Adding Equations 1 and 3: 2 B + 2 D = −2
Subtracting Equations 1 and 3: 2 A + 2C = 4
Adding Equations 2 and 4: 6 A + 2C = 12
Subtracting Equations 2 and 4: 4 B = 16
So, B = 4 and D =
1
2
(−2
− 2 B ) = −5. Subtracting 2 A + 2C = 4 and 6 A + 2C = 12,
you obtain 4 A = 8 ⇒ A = 2. Finally, C =
1
2
(4
− 2 A) = 0. So, p( x) = 2 x3 + 4 x 2 − 5.
6. f ( x ) = a + b cos cx
From Equation 3, b =
f ′( x) = −bc sin cx
At (0, 1): a + b = 1
Equation 1
3
⎛π 3⎞
⎛ cπ ⎞
At ⎜ , ⎟: a + b cos⎜ ⎟ =
2
⎝ 4 2⎠
⎝ 4 ⎠
Equation 2
⎛ cπ ⎞
−bc sin ⎜ ⎟ = 1
⎝ 4 ⎠
Equation 3
From Equation 1, a = 1 − b. Equation 2 becomes
(1 − b)
3
cπ
1
⎛ cπ ⎞
+ b cos⎜ ⎟ =
⇒ −b + b cos
= .
4
2
4
2
⎝ ⎠
−1
. So:
c sin (cπ 4)
1
−1
1
⎛ cπ ⎞
+
cos⎜ ⎟ =
2
c sin (cπ 4)
c sin (cπ 4) ⎝ 4 ⎠
1
⎛ cπ ⎞
⎛ cπ ⎞
1 − cos⎜ ⎟ = c sin ⎜ ⎟
4
2
⎝ ⎠
⎝ 4⎠
Graphing the equation
g (c) =
1
⎛ cπ ⎞
⎛ cπ ⎞
c sin ⎜ ⎟ + cos⎜ ⎟ − 1,
2
4
⎝ ⎠
⎝ 4⎠
you see that many values of c will work. One answer:
1
3
3
1
c = 2, b = − , a =
⇒ f ( x) =
− cos 2 x
2
2
2
2
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Problem Solving for Chapter 3
8. (a) b 2 y 2 = x 3 ( a − x ); a, b > 0
x4 = a2 x2 − a2 y 2
7. (a)
a2 y 2 = a2 x2 − x4
y =
±
a2 x2 − x4
and y2 = −
a
a2 x2 − x4
.
a
2
(b)
a = 12
−3
y2 = −
b
x3 ( a − x)
b
and
.
(c) Differentiating implicitly:
−2
2b 2 yy′ = 3 x 2 ( a − x) − x 3 = 3ax 2 − 4 x3
(± a, 0) are the x-intercepts, along with (0, 0).
(c) Differentiating implicitly:
y′ =
4 x3 = 2a 2 x − 2a 2 yy′
x(a 2 − 2 x 2 )
a2 y
(3ax 2
− 4 x3 )
2b 2 y
= 0
⇒ 3ax 2 = 4 x 3
2a 2 x − 4 x 3
2a 2 y
3a = 4 x
= 0 ⇒ 2x2 = a2 ⇒ x =
x =
±a
2
3
y2 =
4
a
a
=
− a2 y2
4
2
a4
a2 y2 =
4
a2
y2 =
4
a
y=±
2
3a
4
3a ⎞
27 a 3 ⎛ 1 ⎞
⎛ 3a ⎞ ⎛
b2 y 2 = ⎜ ⎟ ⎜ a −
⎟ =
⎜ a⎟
4⎠
64 ⎝ 4 ⎠
⎝ 4⎠ ⎝
2
2
⎛ a2 ⎞
2⎛ a ⎞
2 2
⎜ ⎟ =a ⎜ ⎟−a y
⎝2⎠
⎝ 2⎠
4
x3 ( a − x)
Graph y1 =
b affects the height.
a=2
a=1
=
b2
(b) a determines the x-intercept on the right: ( a, 0).
3
y′ =
x3 ( a − x)
y2 =
a2 x2 − x4
a
Graph: y1 =
263
27 a 4
3 3a 2
y
⇒
=
±
256b 2
16b
⎛ 3a 3 3a 2 ⎞ ⎛ 3a −3 3a 2 ⎞
Two points: ⎜⎜ ,
⎟, ⎜ ,
⎟
16b ⎟⎠ ⎜⎝ 4
16b ⎟⎠
⎝ 4
a⎞
⎛ a a⎞ ⎛ a
Four points: ⎜
, ⎟, ⎜
, − ⎟,
2
2⎠
2
2
⎝
⎠ ⎝
⎛ a a⎞
, ⎟,
⎜−
2 2⎠
⎝
a⎞
⎛ −a
,− ⎟
⎜
2⎠
2
⎝
Line determined by (0, 30) and (90, 6):
y
9. (a)
(0, 30)
30
x
90
30 − 6
24
4
4
( x − 0) = − x = − x ⇒ y = − x + 30
0 − 90
90
15
15
4
10
When x = 100: y = − (100) + 30 =
> 3
15
3
y − 30 =
(90, 6)
(100, 3)
100
Not drawn to scale
As you can see from the figure, the shadow determined by the man extends beyond the shadow determined by the child.
Line determined by (0, 30) and (60, 6):
y
(b)
30
(0, 30)
30 − 6
2
2
( x − 0) = − x ⇒ y = − x + 30
0 − 60
5
5
2
When x = 70: y = − (70) + 30 = 2 < 3
5
y − 30 =
(60, 6)
(70, 3)
x
60
70
Not drawn to scale
As you can see from the figure, the shadow determined by the child extends beyond the shadow determined by the man.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
264
Chapter 3
Differentiation
(c) Need (0, 30), ( d , 6), ( d + 10, 3) collinear.
30 − 6
6−3
24
3
=
⇒
=
⇒ d = 80 feet
0−d
d − ( d + 10)
d
10
(d) Let y be the distance from the base of the street light to the tip of the shadow. You know that dx / dt = −5.
For x > 80, the shadow is determined by the man.
y
y − x
5
dy
5 dx
−25
=
⇒ y = x and
=
=
30
6
4
dt
4 dt
4
For x < 80, the shadow is determined by the child.
y
y − x − 10
10
100
dy
10 dx
50
and
=
⇒ y =
x +
=
= −
30
3
9
9
dt
9 dt
9
Therefore:
⎧ 25
⎪⎪− 4 , x > 80
dy
= ⎨
dt
⎪− 50 , 0 < x < 80
⎪⎩ 9
dy / dt is not continuous at x = 80.
ALTERNATE SOLUTION for parts (a) and (b):
(a) As before, the line determined by the man’s shadow is
ym = −
4
x + 30
15
The line determined by the child’s shadow is obtained by finding the line through (0, 30) and (100, 3):
y − 30 =
30 − 3
27
( x − 0) ⇒ yc = − x + 30
0 − 100
100
By setting ym = yc = 0, you can determine how far the shadows extend:
4
1
x = 30 ⇒ x = 112.5 = 112
15
2
27
1
Child: yc = 0 ⇒
x = 30 ⇒ x = 111.11 = 111
100
9
Man: ym = 0 ⇒
The man’s shadow is 112
1
1
7
− 111 = 1 ft beyond the child’s shadow.
2
9
18
(b) As before, the line determined by the man’s shadow is
2
ym = − x + 30
5
For the child’s shadow,
y − 30 =
30 − 3
27
( x − 0) ⇒ yc = − x + 30
0 − 70
70
2
x = 30 ⇒ x = 75
5
27
700
7
Child: yc = 0 ⇒
x = 30 ⇒ x =
= 77
70
9
9
Man: ym = 0 ⇒
So the child’s shadow is 77
7
7
− 75 = 2 ft beyond the man’s shadow.
9
9
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 3
265
1
dy
dx
= x −2 3
3
dt
dt
1 −2 3 dx
1 = (8)
3
dt
dx
= 12 cm/sec
dt
10. (a) y = x1 3 ⇒
(b) D =
x2 + y 2 ⇒
dD
1
dy ⎞ x( dx / dt ) + y( dy / dt )
⎛ dx
= ( x 2 + y 2 )⎜ 2 x
+ 2y ⎟ =
dt
2
dt
dt ⎠
⎝
x2 + y 2
=
(c) tan θ =
8(12) + 2(1)
98
=
68
=
64 + 4
49
cm/sec
17
x( dy / dt ) − y ( dx / dt )
y
dθ
⇒ sec 2θ ⋅
=
x
dt
x2
68
2
θ
8
From the triangle, sec θ =
68 8. So
8(1) − 2(12)
dθ
−16
4
=
=
= − rad/sec.
dt
64(68 64)
68
17
11. (a) v(t ) = − 27
t + 27 ft/sec
5
a(t ) = − 27
ft/sec 2
5
t + 27 = 0 ⇒
(b) v(t ) = − 27
5
27 t
5
= 27 ⇒ t = 5 seconds
27 5 + 27 5 + 6 = 73.5 feet
S (5) = − 10
()
()
2
(c) The acceleration due to gravity on Earth is greater in magnitude than that on the moon.
12. E ′( x) = lim
∆x → 0
E ( x + ∆x) − E ( x)
E ( x ) E ( ∆x ) − E ( x )
⎛ E ( ∆x ) − 1 ⎞
E ( ∆x ) − 1
= lim
= lim E ( x )⎜
⎟ = E ( x) ∆lim
∆x → 0
∆x → 0
x→0
∆x
∆x
∆
x
∆x
⎝
⎠
E ( ∆x ) − E (0)
But, E ′(0) = lim
∆x → 0
∆x
= lim
E ( ∆x) − 1
∆x
∆x → 0
= 1. So, E ′( x) = E ( x) E ′(0) = E ( x) exists for all x.
For example: E ( x) = e x .
13.
a + bx
1 + cx
f ( 0) = a = e 0 = 1 ⇒ a = 1
f ( x) =
f ′( x) =
(1 + cx)(b) − (a + bx)c
2
(1 + cx)
=
b − ac
(1 + cx)
2
f ′(0) = b − ac = 1 ⇒ b = 1 + c
f ′′( x) =
(1 + cx) (0) − (b − ac)2c(1 + cx)
4
(1 + cx)
2
=
2c( ac − b)
(1 + cx)
3
f ′′(0) = 2c( ac − b) = 2c(c − (1 + c)) = 2c( −1) = 1 ⇒ c = −
So, b = 1 + c = 1 −
1
1+ x
2
f ( x) =
1
1− x
2
1
1
= .
2
2
1
2
6
f
ex
−5
2
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
266
Chapter 3
14. (a)
Differentiation
z (degrees)
0.1
0.01
0.0001
sin z
z
0.0174524
0.0174533
0.0174533
sin z
≈ 0.0174533
z
sin z
π
In fact, lim
.
=
z →0 z
180
(b) lim
z →0
(c)
sin ( z + ∆z ) − sin z
d
(sin z ) = ∆lim
z →0
dz
∆z
sin z ⋅ cos ∆z + sin ∆z ⋅ cos z − sin z
= lim
∆z → 0
∆z
⎡
= lim ⎢sin
∆z → 0
⎣
⎡
⎛ cos ∆z − 1 ⎞⎤
z⎜
⎟⎥ + lim ⎢cos
∆z
⎝
⎠⎦ ∆z → 0 ⎣
⎛ sin ∆z ⎞⎤
z⎜
⎟⎥
⎝ ∆z ⎠⎦
π
⎛ π ⎞
= (sin z )(0) + (cos z )⎜
cos z
⎟ =
180
⎝ 180 ⎠
π
⎛ π
⎞
(d) S (90) = sin ⎜
90 ⎟ = sin = 1
2
⎝ 180 ⎠
⎛ π
⎞
180 ⎟ = −1
C (180) = cos⎜
180
⎝
⎠
π
d
d
sin (cz ) = c ⋅ cos(cz ) =
S ( z) =
C( z)
180
dz
dz
(e) The formulas for the derivatives are more complicated in degrees.
15. j (t ) = a′(t )
(a) j (t ) is the rate of change of acceleration.
(b)
s(t ) = −8.25t 2 + 66t
v(t ) = −16.5t + 66
a(t ) = −16.5
a′(t ) = j (t ) = 0
The acceleration is constant, so j (t ) = 0.
16.
y = ln x
1
x
1
y − b = ( x − a)
a
1
y = x + b − 1, Tangent line
a
y′ =
If x = 0, c = b − 1. So, b − c = b − (b − 1) = 1.
(c) a is position.
b is acceleration.
c is jerk.
d is velocity.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 4
Applications of Differentiation
Section 4.1
Extrema on an Interval .......................................................................268
Section 4.2
Rolle’s Theorem and the Mean Value Theorem...............................278
Section 4.3
Increasing and Decreasing Functions
and the First Derivative Test ..............................................................289
Section 4.4
Concavity and the Second Derivative Test .......................................317
Section 4.5
Limits at Infinity .................................................................................339
Section 4.6
A Summary of Curve Sketching........................................................354
Section 4.7
Optimization Problems.......................................................................380
Section 4.8
Differentials ........................................................................................397
Review Exercises ........................................................................................................403
Problem Solving .........................................................................................................419
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 4
Applications of Differentiation
Section 4.1 Extrema on an Interval
1.
f ( x) =
f ′( x) =
8. Critical number: x = 0
x2
x + 4
x = 0: neither
2
(x
2
+ 4)( 2 x) − ( x
(x
2
+ 4)
2
)(2 x)
2
=
8x
(x
2
+ 4)
2
9. Critical numbers: x = 1, 2, 3
x = 1, 3: absolute maxima (and relative maxima)
f ′(0) = 0
2.
x = 2: absolute minimum (and relative minimum)
f ( x) = cos
f ′( x) = −
f ′(0) = 0
π
2
πx
10. Critical numbers: x = 2, 5
2
sin
x = 2: neither
πx
2
x = 5: absolute maximum (and relative maximum)
f ′( 2) = 0
11.
f ′( x) = 3 x 2 − 6 x = 3 x( x − 2)
4
3. f ( x) = x + 2 = x + 4 x −2
x
8
f ′( x) = 1 − 8 x −3 = 1 − 3
x
f ′( 2) = 0
4.
f ( x ) = −3 x
Critical numbers: x = 0, 2
12. g ( x) = x 4 − 8 x 2
g ′( x) = 4 x3 − 16 x = 4 x( x 2 − 4)
Critical numbers: x = 0, − 2, 2
x +1
f ′( x) = −3 x ⎡ 12 ( x + 1)
⎣
( )
−1 2
x + 1( −3)
⎤ +
⎦
= − 32 ( x + 1)
−1 2
⎡⎣ x + 2( x + 1)⎤⎦
= − 32 ( x + 1)
−1 2
(3 x
13. g (t ) = t
f ( x) = ( x + 2)
f ′( x) =
2
3
(x
+ 2)
23
+ 2)
Critical number: t =
−1 3
f ′( −2) is undefined.
14.
f ( x) =
6. Using the limit definition of the derivative,
lim
x → 0−
lim
x → 0+
f ( x ) − f ( 0)
x −0
f ( x ) − f ( 0)
x −0
=
4 − t, t < 3
−1 2
12
⎡1
⎤
g ′(t ) = t ⎢ ( 4 − t ) ( −1)⎥ + ( 4 − t )
2
⎣
⎦
1
−1 2
= ( 4 − t ) ⎡−
⎣ t + 2( 4 − t )⎤⎦
2
8 − 3t
=
2 4 −t
f ′ − 23 = 0
5.
f ( x) = x3 − 3 x 2
(4 − x ) − 4
lim
x → 0−
= lim
x → 0+
x
(4 − x ) − 4
x −0
=1
= −1
f ′(0) does not exist, because the one-sided derivatives
are not equal.
7. Critical number: x = 2
f ′( x) =
8
3
4x
x2 + 1
( x2
+ 1)( 4) − ( 4 x )( 2 x)
( x2
+ 1)
2
=
4(1 − x 2 )
( x2
+ 1)
2
Critical numbers: x = ±1
15. h( x) = sin 2 x + cos x, 0 < x < 2π
h′( x) = 2 sin x cos x − sin x = sin x( 2 cos x − 1)
Critical numbers in (0, 2π ) : x =
π
3
, π,
5π
3
x = 2: absolute maximum (and relative maximum)
268
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Section 4.1
16.
f (θ ) = 2 sec θ + tan θ , 0 < θ < 2π
g ′( x) = 4 x − 8 = 4( x − 2)
= sec θ ( 2 tan θ + sec θ )
Critical number: x = 2
Left endpoint: (0, 0)
⎡ ⎛ sin θ ⎞
1 ⎤
= sec θ ⎢2⎜
⎥
⎟ +
cos
θ
cos
θ⎦
⎠
⎣ ⎝
Critical number: ( 2, − 8) Minimum
= sec 2 θ ( 2 sin θ + 1)
7π 11π
,
6 6
Critical numbers in (0, 2π ) : θ =
Right endpoint: (6, 24) Maximum
24. h( x) = 5 − x 2 , [−3, 1]
f (t ) = te −2t
h′( x) = −2 x
f ′(t ) = e −2t − 2te −2t = e −2t (1 − 2t )
Critical number: x = 0
Critical number: t =
Left endpoint: ( −3, − 4) Minimum
1
2
Critical number: (0, 5) Maximum
18. g ( x) = 4 x 2 (3x )
Right endpoint: (1, 4)
g ′( x) = 8 x(3x ) + 4 x 2 3x ln 3 = 4 x(3x )( 2 + x ln 3)
25.
Critical numbers: x = 0, −1.82
f ( x) = x3 −
3 2
x ,
2
f ′( x) = 3 x − 3x
2
19.
f ( x) = x 2 log 2 ( x 2 + 1) = x
f ′( x) = 2 x
=
ln ( x + 1)
ln ( x 2 + 1)
2
+ x2
(
− 1)
)
Left endpoint: −1, − 52 Minimum
2x
Right endpoint: ( 2, 2) Maximum
(ln 2)( x
2x ⎡
x
2
⎢ln ( x + 1) + 2
ln 2 ⎣
x +
2
2
+ 1)
Critical number: (0, 0)
⎤
⎥
1⎦
Critical number: x = 0
(
Critical number: 1, − 12
26.
)
f ( x) = 2 x3 − 6 x, [0, 3]
f ′( x) = 6 x 2 − 6 = 6( x 2 − 1)
20. g (t ) = 2t ln t
21.
[−1, 2]
= 3 x( x
ln 2
2
ln 2
⎛1⎞
g ′(t ) = 2ln t + 2t ⎜ ⎟ = 2ln t + 2
⎝t ⎠
Critical number: x = 1 ( x = −1 not in interval.)
1
Critical number: t =
e
Critical number: (1, − 4) Minimum
f ( x ) = 3 − x,
269
23. g ( x) = 2 x 2 − 8 x, [0, 6]
f ′(θ ) = 2 sec θ tan θ + sec 2 θ
17.
Extrema on an Interval
Left endpoint: (0, 0)
Right endpoint: (3, 36) Maximum
[−1, 2]
f ′( x) = −1 ⇒ no critical numbers
Left endpoint: ( −1, 4) Maximum
Right endpoint: ( 2, 1) Minimum
3
22. f ( x) = x + 2, [0, 4]
4
3
f ′( x) =
⇒ no critical numbers
4
27.
f ( x ) = 3 x 2 3 − 2 x,
f ′( x) = 2 x −1 3 − 2 =
[−1, 1]
(
21 −
3
3
x
)
x
Left endpoint: ( −1, 5) Maximum
Critical number: (0, 0) Minimum
Right endpoint: (1, 1)
Left endpoint: (0, 2) Minimum
Right endpoint: ( 4, 5) Maximum
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
270
Chapter 4
28. g ( x) =
g ′( x) =
3
Applications of Differentiation
[−2, 2]
33. f ( x) = a xb,
x = x1 3 , [− 8, 8]
1
3x 2 3
y
From the graph of f, you see
that the maximum value of f is
2 for x = 2, and the minimum
value is –2 for −2 ≤ x < −1.
Critical number: x = 0
Left endpoint: ( − 8, − 2) Minimum
2
1
−2
Critical number: (0, 0)
h′( s ) =
− 2)
2
[−2, 2]
34. h( x) = a2 − xb,
From the graph you see that the maximum value of h is 4
at x = −2, and the minimum value is 0 for 1 < x ≤ 2.
1
−1
= ( s − 2) , [0, 1]
s − 2
−1
(s
1
−2
Right endpoint: (8, 2) Maximum
29. h( s ) =
x
−1
y
4
2
3
1⎞
⎛
Left endpoint: ⎜ 0, − ⎟ Maximum
2⎠
⎝
1
Right endpoint: (1, −1) Minimum
−2
30. h(t ) =
h′(t ) =
t
, [−1, 6]
t + 3
(t + 3)(1) − t (1)
(t
+ 3)
2
35.
=
3
(t
+ 3)
2
x
1
2
⎡ 5π 11π ⎤
f ( x) = sin x, ⎢ ,
⎥
⎣6 6 ⎦
f ′( x) = cos x
Critical number: x =
No critical numbers
1⎞
⎛
Left endpoint: ⎜ −1, − ⎟ Minimum
2⎠
⎝
⎛ 2⎞
Right endpoint: ⎜ 6, ⎟ Maximum
⎝ 3⎠
31. y = 3 − t − 3 ,
−1
[−1, 5]
For x < 3, y = 3 + (t − 3) = t
and y′ = 1 ≠ 0 on [−1, 3)
For x > 3, y = 3 − (t − 3) = 6 − t
and y′ = −1 ≠ 0 on (3, 5]
So, x = 3 is the only critical number.
Left endpoint: ( −1, −1) Minimum
3π
2
⎛ 5π 1 ⎞
Left endpoint: ⎜ , ⎟ Maximum
⎝ 6 2⎠
⎛ 3π
⎞
Critical number: ⎜ , −1⎟ Minimum
⎝ 2
⎠
⎛ 11π 1 ⎞
,− ⎟
Right endpoint: ⎜
2⎠
⎝ 6
36. g ( x) = sec x,
⎡ π π⎤
⎢− 6 , 3 ⎥
⎣
⎦
g ′( x) = sec x tan x
⎛ π 2 ⎞ ⎛ π
⎞
Left endpoint: ⎜ − ,
⎟ ≈ ⎜ − , 1.1547 ⎟
6
6
3⎠ ⎝
⎠
⎝
Right endpoint: (5, 1)
⎛π ⎞
Right endpoint: ⎜ , 2 ⎟ Maximum
⎝3 ⎠
Critical number: (3, 3) Maximum
Critical number: (0, 1) Minimum
32. g ( x) = x + 4 , [−7, 1]
37. y = 3 cos x,
[0, 2π ]
y′ = −3 sin x
g is the absolute value function shifted 4 units to the
left. So, the critical number is x = − 4.
Critical number in (0, 2π ): x = π
Left endpoint: ( − 7, 3)
Left endpoint: (0, 3) Maximum
Critical number: ( − 4, 0) Minimum
Critical number: (π , − 3) Minimum
Right endpoint: (1, 5) Maximum
Right endpoint: ( 2π , 3) Maximum
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.1
⎛π x ⎞
38. y = tan ⎜ ⎟,
⎝ 8 ⎠
π
y′ = 2 x −
2
8
= 0 ⇒ 2 x2 = 8 ⇒ x = 2
x
( x = −2 not in domain )
Right endpoint: ( 2, 1) Maximum
Critical number: x = 2
f ( x) = arctan x , [−2, 1]
Left endpoint: (1, 1)
2
Right endpoint: (5, 25 − 8 ln5) ≈ (5, 12.124) Maximum
2x
1 + x4
Critical number: ( 2, 4 − 8 ln2) ≈ ( 2, −1.545) Minimum
Critical number: x = 0
Left endpoint: ( −2, arctan4) ≈ ( −2, 1.326) Maximum
⎛ π⎞
Right endpoint: (1, arctan1) = ⎜1, ⎟ ≈ (1, 0.785)
⎝ 4⎠
Critical number: (0, 0) Minimum
⎛ 1 ⎞
y′ = x⎜
⎟ + ln ( x + 3)
⎝ x + 3⎠
⎛ ln4 ⎞
Right endpoint: ⎜ 4,
⎟ ≈ ( 4, 0.347)
4 ⎠
⎝
⎛ 1⎞
Critical number: ⎜ e, ⎟ ≈ ( 2.718, 0.368) Maximum
⎝ e⎠
Left endpoint: (0, 0) Minimum
Right endpoint: (3, 3 ln 6) ≈ (3, 5.375) Maximum
45. f ( x) = 2 x − 3
(a) Minimum: (0, − 3)
41. h( x) = 5e x − e 2 x , [−1, 2]
= e ( 5 − 2e
5 − 2e x = 0 ⇒ e x =
Left endpoint: (0, 0) Minimum
44. y = x ln ( x + 3), [0, 3]
Left endpoint: (1, 0) Minimum
h′( x) = 5e − 2e
y′ = e x sin x + e x cos x = e x (sin x + cos x)
Right endpoint: (π , 0) Minimum
Critical number: x = e
2x
43. y = e x sin x, [0, π ]
Critical number:
⎛ 3π
2 3π 4 ⎞ ⎛ 3π
⎞
e ⎟⎟ ≈ ⎜ , 7.46 ⎟ Maximum
⎜⎜ ,
4
2
4
⎠
⎝
⎠ ⎝
ln x
40. g ( x) =
, [1, 4]
x
⎛1⎞
x⎜ ⎟ − ln x
1 − ln x
x
g ′( x) = ⎝ ⎠ 2
=
x
x2
x
8
x
2x −
Left endpoint: (0, 0) Minimum
f ′( x) =
271
42. y = x 2 − 8 ln x, [1, 5]
[0, 2]
⎛π x ⎞
sec ⎜ ⎟ ≠ 0
y′ =
8
⎝ 8 ⎠
39.
Extrema on an Interval
x
x
)
5
⎛5⎞
⇒ x = ln ⎜ ⎟ ≈ 0.916
2
⎝ 2⎠
⎛5⎞
Critical number: x = ln ⎜ ⎟
⎝ 2⎠
Maximum: ( 2, 1)
(b) Minimum: (0, − 3)
(c) Maximum: ( 2, 1)
(d) No extrema
5
1⎞
⎛
Left endpoint: ⎜ −1, − 2 ⎟ ≈ ( −1, 1.704)
e
e
⎝
⎠
Right endpoint: ( 2, 5e 2 − e 4 ) ≈ ( 2, −17.653) Minimum
⎛ ⎛ 5 ⎞ 25 ⎞
Critical number: ⎜ ln ⎜ ⎟,
⎟ Maximum
⎝ ⎝ 2⎠ 4 ⎠
⎛ ⎛ 5 ⎞⎞
Note: h⎜ ln ⎜ ⎟ ⎟ = 5eln(5 2) − e2 ln(5 2)
⎝ ⎝ 2 ⎠⎠
46. f ( x ) =
4 − x2
(a) Minima: ( −2, 0) and ( 2, 0)
Maximum: (0, 2)
(b) Minimum: ( −2, 0)
(c) Maximum: (0, 2)
(
(d) Maximum: 1,
3
)
2
25
⎛5⎞ ⎛5⎞
= 5⎜ ⎟ − ⎜ ⎟ =
4
⎝ 2⎠ ⎝ 2⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
272
Chapter 4
Applications of Differentiation
3
,
x −1
47. f ( x ) =
49. f ( x) =
(1, 4]
x + 4e x
Right endpoint: ( 4, 1) Minimum
2 10
[− 2, 2]
,
(2, 3.6542)
4
(−2, 2.1098)
8
−3
3
(−0.7753, 1.9070)
0
4
0
2
,
2 − x
48. f ( x ) =
f ′( x) =
[0, 2)
0
(2 x 2
+ 8 x + 5)e x
10
2 10
x + 4
Right endpoint: ( 2, 3.6542) Maximum
Left endpoint: (0, 1) Minimum
Critical point: ( − 0.7753, 1.9070) Minimum
3
x
x + cos ,
2
50. f ( x ) =
(0, 1)
−1
5
[0, 2π ]
3
−1
(1.729, 1.964)
2␲
0
0
f ′( x) =
1
2
x
−
1
x
sin
2
2
Left endpoint: (0, 1) Minimum
Critical point: (1.729, 1.964) Maximum
51. (a)
5
(1, 4.7)
0
Minimum: (0.4398, −1.0613)
1
(0.4398, −1.0613)
−2
f ( x) = 3.2 x5 + 5 x3 − 3.5 x,
(b)
[0, 1]
f ′( x) = 16 x + 15 x − 3.5
4
2
16 x + 15 x − 3.5 = 0
4
2
x2 =
x =
−15 ±
(15) − 4(16)(−3.5)
2(16)
2
=
−15 ± 449
32
−15 + 449
≈ 0.4398
32
Left endpoint: (0, 0)
Critical point: (0.4398, −1.0613) Minimum
Right endpoint: (1, 4.7) Maximum
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.1
52. (a)
Extrema on an Interval
273
3
⎛ 8⎞
Maximum: ⎜ 2, ⎟
⎝ 3⎠
(2, 83 )
0
3
0
(b)
f ( x) =
4
x 3 − x,
3
f ′( x) =
2(6 − 3x )
6( 2 − x )
2( 2 − x)
⎤
4⎡ ⎛ 1⎞
4
−1 2
12
−1 2 ⎛ 1 ⎞
=
=
x⎜ ⎟(3 − x) ( −1) + (3 − x ) (1)⎥ = (3 − x) ⎜ ⎟ ⎡−
⎣ x + 2(3 − x)⎤⎦ =
3 ⎣⎢ ⎝ 2 ⎠
3
3 3−x
3 3− x
3− x
⎝ 2⎠
⎦
[0, 3]
Left endpoint: (0, 0) Minimum
⎛
Critical point: ⎜ 2,
⎝
8⎞
⎟ Maximum
3⎠
Right endpoint: (3, 0) Minimum
53. (a)
6
(3, 5.3753)
Minimum: (1.0863, −1.3972)
−1
4
(1.0863, −1.3972)
−3
(b) f ( x ) = ( x 2 − 2 x) ln ( x + 3),
f ′( x) = ( x 2 − 2 x) ⋅
[0, 3]
x 2 − 2 x + ( 2 x 2 + 4 x − 6) ln ( x + 3)
1
+ ( 2 x − 2) ln ( x + 3) =
x + 3
x + 3
Left endpoint: (0, 0)
Critical point: (1.0863, −1.3972) Minimum
Right endpoint: (3, 5.3753) Maximum
54. (a)
5
(−2, π )
Minimum: ( 2.1111, −1.0502)
(4, 0)
−2
4
(2.1111, −1.0502)
−3
x
(b) f ( x) = ( x − 4) arcsin ,
4
f ′( x) = ( x − 4)
1
4
1−
[− 2, 4]
+ arcsin
2
x
16
x
=
4
x − 4
4 1−
+ arcsin
2
x
16
x
4
Left endpoint: ( − 2, π ) Maximum
Critical point: ( 2.1111, −1.0502) Minimum
Right endpoint: ( 4, 0)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
274
Chapter 4
Applications of Differentiation
55. f ( x ) = (1 + x 3 ) ,
[0, 2]
12
58.
−1 2
3 2
x (1 + x 3 )
2
−3 2
3
f ′′( x) = ( x 4 + 4 x)(1 + x3 )
4
−5 2
3
f ′′′( x) = − ( x 6 + 20 x 3 − 8)(1 + x3 )
8
f ′( x) =
f ( x) = x ln ( x + 1), [0, 2]
f ′( x) =
x
+ ln ( x + 1)
( x + 1)
f ′′( x) =
x +1− x
=
Setting f ′′′ = 0, you have x 6 + 20 x 3 − 8 = 0.
400 − 4(1)( −8)
−20 ±
x =
3
f ′′′( x) =
2
x =
3
−10 ±
108 =
x =
(
f ′′
56.
3
−10 ±
108 =
−10 +
108
f ( x) =
f ′( x) =
f ′′( x) =
f ′′′( x) =
(x
2
+ 1)
3 − 1 ≈ 0.732.
+ 1)
+ 1)
(x
+ 1) − ( x + 2)2( x + 1)
2 2
f ′( x) = − xe − x
)
3
60.
= e− x
2 2
( x2
f ′′′( x) = e − x
2 2
( 2 x)
= xe
− x2 2
)−e
− 1)
8
27
+ 1)
(x
f ( 4 ) ( 0) =
56
81
f ( x) =
f (5) ( x) =
2 2
(
(x
560
243
f ( 4) ( x ) =
, [0, 1]
f ′′( x) = − x − xe
2
3
f (5) ( x) =
f ′′′( x) =
4
− x2 2
(x
+ 1)
4
=
−x − 3
(x
+ 1)
3
[0, 2]
−1 3
+ 1)
−4 3
−7 3
f (4) ( x) = − 56
x + 1)
81 (
1
f ′′(1) = is the maximum value.
2
f ( x) = e − x
2
23
Setting f ′′′ = 0, you have x = 0, ±1.
57.
2
f ( x ) = ( x + 1) ,
f ′′′( x ) =
24 x − 24 x3
( x2
1
x + 2
=
2
x +1
( x + 1)
+
+ 1)
f ′′( x) = − 92 ( x + 1)
2
2
1
(x
f ′( x) =
⎡1 ⎤
⎢ 2 , 3⎥
⎣
⎦
−2(1 − 3x
( x2
59.
) ≈ 1.47 is the maximum value.
1
,
x2 + 1
−2 x
+ 1)
f ′′(0) = 2 is the maximum value.
3 −1
In the interval [0, 2], choose
3
(x
1
x +1
+
2
(x
+ 1)
−10 3
−13 3
is the maximum value.
1
,
x +1
[−1, 1]
2
24 x − 24 x3
( x 2 + 1)
24(5 x 4 − 10 x 2 + 1)
5
( x 2 + 1)
−240 x(3 x 4 − 10 x 2 + 3)
6
( x 2 + 1)
4
f (4) (0) = 24 is the maximum value.
− x2 2
61. f ( x ) = tan x
(
+ ( x 2 − 1) − xe − x
(3 − x )
2
f ′′(0) = 1 is the maximum value.
2 2
)
f is continuous on [0, π 4] but not on [0, π ].
lim
x → (π 2)−
tan x = ∞.
62. A: absolute minimum
B: relative maximum
C: neither
D: relative minimum
E: relative maximum
F: relative minimum
G: neither
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.1
5
(b) No
4
3
66. (a) No
f
2
1
(b) Yes
x
−2 −1
1
3
4
5
6
67. (a) No
−2
−3
64.
(b) Yes
68. (a) No
y
(b) Yes
5
4
f
3
69. P = VI − RI 2 = 12 I − 0.5 I 2 , 0 ≤ I ≤ 15
2
−2 −1
275
65. (a) Yes
y
63.
Extrema on an Interval
P = 0 when I = 0.
x
1
2
3
4
5
6
P = 67.5 when I = 15.
−2
P′ = 12 − I = 0
−3
Critical number: I = 12 amps
When I = 12 amps, P = 72, the maximum output.
No, a 20-amp fuse would not increase the power
output. P is decreasing for I > 12.
70. x =
v 2 sin 2θ π
3π
≤ θ ≤
,
32
4
4
dθ
is constant.
dt
dx
dx dθ
v 2 cos 2θ dθ
=
(by the Chain Rule) =
dt
dθ dt
16
dt
In the interval [π 4, 3π 4], θ = π 4, 3π 4 indicate minimums for dx dt and θ = π 2 indicates a maximum for dx dt.
This implies that the sprinkler waters longest when θ = π 4 and 3π 4. So, the lawn farthest from the sprinkler gets the
most water.
S = 6hs +
71.
3s 2 ⎛
⎜
2 ⎜⎝
(
dS
3s 2
=
−
dθ
2
=
3 − cos θ ⎞ π
π
⎟⎟, ≤ θ ≤
sin θ
2
⎠ 6
3csc θ cot θ + csc 2 θ
(
3s 2
csc θ −
2
csc θ =
3cot θ
sec θ =
3
)
)
3cot θ + csc θ = 0
θ = arcsec 3 ≈ 0.9553 radians
⎛π ⎞
(
3s 2
S ⎜ ⎟ = 6hs +
2
⎝6⎠
( 3)
3s 2
⎛π ⎞
S ⎜ ⎟ = 6hs +
2
⎝6⎠
( 3)
)
S arcsec 3 = 6hs +
3s 2
2
( 2)
S is minimum when θ = arcsec 3 ≈ 0.9553 radian.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
276
Chapter 4
Applications of Differentiation
72. (a) Because the grade at A is 9%, A( −500, 45)
The grade at B is 6%, B (500, 30).
y
A
B
9%
6%
−500
x
500
y = ax 2 + bx + c
(b)
y′ = 2ax + b
At A: 2a( −500) + b = −0.09
At B : 2a(500) + b = 0.06
Solving these two equations, you obtain
a =
3
40,000
b = −
and
3
.
200
Using the points A( −500, 45) and B(500, 30), you obtain
45 =
3
3
2
(−500) + ⎛⎜ − ⎞⎟(−500) + C
40,000
⎝ 200 ⎠
30 =
3
3
2
(500) + ⎛⎜ − ⎞⎟(500) + C.
40,000
⎝ 200 ⎠
In both cases, C = 18.75 =
(c)
3
3
75
75
x2 −
x +
. So, y =
40,000
200
4
4
x
–500
–400
–300
–200
–100
0
100
200
300
400
500
d
0
0.75
3
6.75
12
18.75
12
6.75
3
0.75
0
For −500 ≤ x ≤ 0, d = ( ax 2 + bx + c) − ( −0.09 x).
For 0 ≤ x ≤ 500, d = ( ax 2 + bx + c) − (0.06 x).
(d) y′ =
3
3
x −
= 0
20,000
200
x =
3
20,000
⋅
= 100
200
3
The lowest point on the highway is (100, 18), is not directly over the origin.
73. True. See Exercise 37.
74. True. This is stated in the Extreme Value Theorem.
77. If f has a maximum value at
x = c, then f (c) ≥ f ( x) for all x in I. So,
− f (c) ≤ − f ( x) for all x in I. So, − f has a minimum
value at x = c.
75. True
76. False. Let f ( x ) = x 2 . x = 0 is a critical number of f.
g ( x) = f ( x − k ) = ( x − k )
2
x = k is a critical number of g.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.1
78.
f ( x) = ax3 + bx 2 + cx + d ,
Extrema on an Interval
277
a ≠ 0
f ′( x) = 3ax 2 + 2bx + c
The quadratic polynomial can have zero, one, or two zeros.
x =
−2b ±
4b 2 − 12ac
−b ±
=
6a
b 2 − 3ac
3a
Zero critical numbers: b 2 < 3ac.
Example: ( a = b = c = 1, d = 0) f ( x) = x 3 + x 2 + x has no critical numbers.
One critical number: b 2 = 3ac.
Example: ( a = 1, b = c = d = 0) f ( x) = x 3 has one critical number, x = 0.
Two critical numbers: b 2 > 3ac.
1
Example: ( a = c = 1, b = 2, d = 0) f ( x) = x3 + 2 x 2 + x has two critical numbers: x = −1, − .
3
79. First do an example: Let a = 4 and f ( x) = 4.
Then R is the square 0 ≤ x ≤ 4, 0 ≤ y ≤ 4.
Its area and perimeter are both k = 16.
Claim that all real numbers a > 2 work. On the one hand, if a > 2 is given, then let f ( x) = 2a ( a − 2).
Then the rectangle
2a ⎫
2a 2
⎧
R = ⎨( x, y ): 0 ≤ x ≤ a, 0 ≤ y ≤
:
⎬ has k =
a − 2⎭
a − 2
⎩
2a 2
⎛ 2a ⎞
Area = a⎜
⎟ =
a −2
⎝ a − 2⎠
2a( a − 2) + 2( 2a)
2a 2
⎛ 2a ⎞
Perimeter = 2a + 2⎜
.
=
⎟ =
a − 2
a − 2
⎝ a − 2⎠
To see that a must be greater than 2, consider
R =
{( x, y ): 0
≤ x ≤ a, 0 ≤ y ≤ f ( x )}.
f attains its maximum value on [0, a] at some point P( x0 , y0 ), as indicated in the figure.
y
P ( x0 , y0 )
y0
f
R
O
A
x0
x
a
Draw segments OP and PA. The region R is bounded by the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ y0 , so area ( R) = k ≤ ay0 .
Furthermore, from the figure, y0 < OP and y0 < PA. So, k = Perimeter ( R ) > OP + PA > 2 y0 . Combining,
2 y0 < k ≤ ay0 ⇒ a > 2.
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278
Chapter 4
Applications of Differentiation
Section 4.2 Rolle’s Theorem and the Mean Value Theorem
1. f ( x ) =
9. f ( x ) = − x 2 + 3x,
1
x
f (0) = − (0) + 3(0)
2
f ( −1) = f (1) = 1. But, f is not continuous on [−1, 1].
2. Rolle’s Theorem does not apply to f ( x) = cot ( x 2) over
[π , 3π ] because f is not continuous at
x = 2π .
3. Rolle’s Theorem does not apply to f ( x ) = 1 − x − 1
f (3) = − (3) + 3(3) = 0
2
f is continuous on [0, 3] and differentiable on (0, 3).
Rolle’s Theorem applies.
f ′( x) = −2 x + 3 = 0
−2 x = − 3 ⇒ x =
over [0, 2] because f is not differentiable at x = 1.
c-value:
(2 − x 2 3 )
4. f ( x) =
x1 3
f is not differentiable at x = 0.
5. f ( x ) = x 2 − x − 2 = ( x − 2)( x + 1)
x-intercepts: ( −1, 0), ( 2, 0)
f ′( x) = 2 x − 1 = 0 at x = 12 .
6. f ( x ) = x 2 + 6 x = x( x + 6)
x-intercepts: (0, 0), ( − 6, 0)
f ′( x) = 2 x + 6 = 0 at x = − 3.
7. f ( x ) = x
f ( 2) = 4 − 16 + 5 = − 7
f (6) = 36 − 48 + 5 = − 7
f is continuous on [2, 6] and differentiable on ( 2, 6).
Rolle’s Theorem applies.
f ′( x) = 2 x − 8 = 0
2x = 8 ⇒ x = 4
c-value: 4
11. f ( x ) = ( x − 1)( x − 2)( x − 3), [1, 3]
f (1) = (1 − 1)(1 − 2)(1 − 3) = 0
f (3) = (3 − 1)(3 − 2)(3 − 3) = 0
f is continuous on [1, 3]. f is differentiable on (1, 3).
Rolle’s Theorem applies.
x + 4
x-intercepts: ( −4, 0), (0, 0)
1
( x + 4)−1 2 + ( x + 4)1 2
2
−1 2 ⎛ x
⎞
= ( x + 4) ⎜ + ( x + 4) ⎟
2
⎝
⎠
f ( x) = x3 − 6 x 2 + 11x − 6
f ′( x) = 3 x 2 − 12 x + 11 = 0
f ′( x) = x
8
−1 2
⎛3
⎞
= 0 at x = −
f ′( x) = ⎜ x + 4 ⎟( x + 4)
3
⎝2
⎠
8. f ( x ) = −3 x
3
2
10. f ( x) = x 2 − 8 x + 5, [2, 6]
(2 − x2 3 )
−
3
2
3
f ( −1) = f (1) = 1
f ′( x) =
[0, 3]
x =
c-values:
6 ± 3
3
6− 3 6 + 3
,
3
3
x +1
x-intercepts: ( −1, 0), (0, 0)
1
( x + 1)−1 2 − 3( x + 1)1 2
2
−1 2 ⎛ x
⎞
= −3( x + 1) ⎜ + ( x + 1) ⎟
⎝2
⎠
f ′( x) = −3 x
f ′( x) = −3( x + 1)
−1 2 ⎛ 3
2
⎞
⎜ x + 1⎟ = 0 at x = −
2
3
⎝
⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.2
Rolle’s Theorem and the Mean Value Theorem
12. f ( x ) = ( x − 4)( x + 2) , [− 2, 4]
2
13.
2
f (8) = (8)
f ( 4) = ( 4 − 4)( 4 + 2) = 0
2
23
23
−1 = 3
−1 = 3
f is continuous on [− 2, 4]. f is differentiable on
f is continuous on [−8, 8]. f is not differentiable
(− 2, 4). Rolle’s Theorem applies.
on ( −8, 8) because f ′(0) does not exist. Rolle’s
f ( x) = ( x − 4)( x 2 + 4 x + 4) = x3 − 12 x − 16
Theorem does not apply.
14. f ( x ) = 3 − x − 3 , [0, 6]
f ′( x) = 3 x 2 − 12 = 0
f ( 0) = f ( 6) = 0
3 x 2 = 12
f is continuous on [0, 6]. f is not differentiable on
x 2 = 4 ⇒ x = ±2
( Note: x
(0, 6) because
= − 2 is not in the interval.)
f ′(3) does not exist. Rolle’s Theorem
does not apply.
c-value: 2
15.
f ( x) = x 2 3 − 1, [−8, 8]
f ( −8) = ( − 8)
f ( − 2) = ( − 2 − 4)( − 2 + 2) = 0
279
x2 − 2x
, [−1, 6]
x + 2
1+ 2
f ( −1) =
= 3
1
36 − 12
f ( 6) =
= 3
8
f ( x) =
f is continuous on [−1, 6]. f is differentiable on ( −1, 6). Rolle’s Theorem applies.
f ′( x) =
(x
+ 2)( 2 x − 2) − ( x 2 − 2 x)(1)
(x
+ 2)
2
f ′( x) = x 2 + 4 x − 4 = 0 ⇒ x =
( Note: − 2 − 2
f ( x) =
f ( −1) =
f (1) =
2x2 + 4x − 2 x − 4 − x2 + 2 x
2
−1
+ 2)
− 4 ± 16 + 16
= −2 + 2
2
2
(x
+ 2)
2
2
)
2
x2 − 1
, [−1, 1]
x
(−1)
(x
x2 + 4x − 4
=
2 is not in the interval.
c-value: − 2 + 2
16.
=
−1
= 0
12 − 1
= 0
1
f is not continuous on [−1, 1] because f (0) does not exist.
17. f ( x ) = sin x, [0, 2π ]
f (0) = sin 0 = 0
f ( 2π ) = sin ( 2π ) = 0
f is continuous on [0, 2π ]. f is differentiable on
(0, 2π ). Rolle’s Theorem applies.
f ′( x) = cos x = 0 ⇒ x =
Rolle’s Theorem does not apply.
c-values:
π 3π
2
,
2
π 3π
2
,
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
280
Chapter 4
Applications of Differentiation
18. f ( x ) = cos 2 x, [−π , π ]
f ( x) = x − 1, [−1, 1]
23.
f ( −1) = f (1) = 0
f ( −π ) = cos( − 2π ) = 1
f (π ) = cos 2π = 1
f is continuous on [−1, 1]. f is not differentiable on
f is continuous on [−π , π ] and differentiable on
(−1, 1) because
(−π , π ). Rolle’s Theorem applies.
does not apply.
f ′( x) = −2 sin 2 x
1
−2 sin 2 x = 0
−1
sin 2 x = 0
π
1
π
x = −π , − , 0, , π
2
2
π
−1
24. f ( x ) = x − x −1 3 , [0, 1]
π
c-values: − , 0,
2
2
f (0) = f (1) = 0
19. f ( x) = tan x, [0, π ]
f is continuous on [0, 1]. f is differentiable on
f (0) = tan 0 = 0
(0, 1). (Note: f is not differentiable at
f (π ) = tan π = 0
1
f ′( x) = 1 −
3 x2
1
1 =
20. f ( x) = sec x, [π , 2π ]
f is not continuous on
[π , 2π ] because f (3π 2) = sec(3π 2) does not exist.
3
Rolle’s Theorem does not apply.
21. f ( x) = ( x 2 − 2 x)e x , [0, 2]
3
3 x2
1
x2 =
3
1
x2 =
27
1
=
27
x =
f ( 0) = f ( 2 ) = 0
f is continuous on [0, 2] and differentiable on (0, 2), so
Rolle’s Theorem applies.
2
1
0
1
2 ≈ 1.414
22. f ( x ) = x − 2 ln x, [1, 3]
f (1) = 1
f (3) = 3 − 2 ln 3 =/ 1
Because f (1) =/ f (3), Rolle’s Theorem does not apply
on [1, 3].
3
9
3
≈ 0.1925
9
c-value:
f ′( x) = ( x 2 − 2 x)e x + ( 2 x − 2)e x = e x ( x 2 − 2)
= 0
3
exist. Rolle’s Theorem does not apply.
= 0 ⇒ x =
x = 0. )
Rolle’s Theorem applies.
f is not continuous on [0, π ] because f (π 2) does not
c-value:
f ′(0) does not exist. Rolle’s Theorem
−1
25. f ( x) = x − tan π x, ⎡⎣− 14 , 14 ⎤⎦
( )
f ( 14 ) =
f − 14 = − 14 + 1 =
1
4
3
4
− 1 = − 34
Rolle’s Theorem does not apply.
0.75
− 0.25
0.25
− 0.75
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.2
26.
πx
x
− sin
, [−1, 0]
2
6
f ( −1) = f (0) = 0
Rolle’s Theorem and the Mean Value Theorem
29. f (t ) = −16t 2 + 48t + 6
f ( x) =
(a) f (1) = f ( 2) = 38
(b) v = f ′(t ) must be 0 at some time in (1, 2).
f is continuous on [−1, 0]. f is differentiable on
f ′(t ) = −32t + 48 = 0
(−1, 0). Rolle’s Theorem applies.
f ′( x) =
cos
πx
6
=
t =
1 π
πx
− cos
= 0
2
6
6
3
6
π
arccos
3
⎡Value needed in ( −1, 0).⎤⎦
π ⎣
sec
⎛ 1
⎞
3
⎟ = 0
C ′( x) = 10⎜ − 2 +
2
⎜ x
( x + 3) ⎟⎠
⎝
3
1
= 2
x2 + 6x + 9
x
(b)
c-value: –0.5756
0.02
2 x2 − 6x − 9 = 0
0
f ( x) = 2 + arcsin ( x 2 − 1), [−1, 1]
2x
1 − ( x 2 − 1)
2
=
4
In the interval
3+3 3
≈ 4.098 ≈ 410 components
(3, 6): c =
2
2x
2x2 − x4
f ′(0) does not exist. Rolle’s Theorem does not apply.
108
6 ± 6 3
3±3 3
=
=
4
2
f ( −1) = f (1) = 2
f ′( x) =
6 ±
x =
−0.01
27.
25
3
(a) C (3) = C (6) =
≈ −0.5756 radian
−1
3
2
x ⎞
⎛1
30. C ( x) = 10⎜ +
⎟
x + 3⎠
⎝x
π
x = −
281
y
31.
Tangent line
3
(c2, f (c2))
(a, f (a))
f
Secan
−1
t line
1
(b, f (b))
(c1, f (c1))
−1
a
x
b
Tangent line
28. f ( x) = 2 + ( x 2 − 4 x)( 2− x 4 ), [0, 4]
f (0) = f ( 4) = 2
32.
y
f is continuous on [0, 4]. f is differentiable on (0, 4).
Rolle’s Theorem applies.
⎛ 1⎞
f ′( x) = ( 2 x − 4)2− x 4 + ( x 2 − 4 x) ln 2 ⋅ 2− x 4 ⎜ − ⎟
⎝ 4⎠
⎡
⎛ x2
⎞⎤
= 2− x 4 ⎢2 x − 4 − (ln 2)⎜
− x ⎟⎥
⎝4
⎠⎦
⎣
= 0 ⇒ x ≈ 1.6633
c-value: 1.6633
−3
x
a
b
33. f is not continuous on the interval [0, 6]. ( f is not
continuous at x = 2.)
34. f is not differentiable at x = 2. The graph of f is not
smooth at x = 2.
3
0
f
4
35. f ( x) =
1
, [0, 6]
x −3
f has a discontinuity at x = 3.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
282
Chapter 4
Applications of Differentiation
36. f ( x) = x − 3 , [0, 6]
40. f ( x ) = 2 x 3 is continuous on [0, 6] and differentiable on
(0, 6).
f is not differentiable at x = 3.
f ( 6) − f ( 0)
37. f ( x) = − x 2 + 5
(a) Slope =
1− 4
= −1
2 +1
f ′( x) = 6 x 2 = 72
x 2 = 12
y − 4 = −( x + 1)
Secant line:
x = ±2 3
y = −x + 3
x + y −3 = 0
(b) f ′( x) = −2 x = −1
⇒
x = c =
In the interval (0, 6) : c = 2 3.
1
2
1
19
⎛1⎞
(c) f (c) = f ⎜ ⎟ = − + 5 =
4
4
⎝ 2⎠
19
1⎞
⎛
y −
= −⎜ x − ⎟
4
2⎠
⎝
4 y − 19 = −4 x + 2
Tangent line:
41. f ( x ) = x 3 + 2 x is continuous on [−1, 1] and
differentiable on ( −1, 1).
f (1) − f ( −1)
1 − ( −1)
3 − ( −3)
=
2
3x 2 = 1
1
3
x = ±
7
Secant
Tangent
−6
3
3
c = ±
f
6
−1
42. f ( x ) = x 4 − 8 x is continuous on [0, 2] and
38. f ( x ) = x 2 − x − 12
differentiable on (0, 2).
−6 − 0
=1
−2 − 4
(a) Slope =
Secant line:
f ( 2) − f ( 0)
y −0 = x − 4
x − y − 4 = 0
(b) f ′( x) = 2 x − 1 = 1
⇒
y + 12 = x − 1
f
15
Secant
Tangent
f ′( x) = 4 x 3 − 8 = 4( x3 − 2) = 0
x3 = 2
c =
3
f (1) − f (0)
1−0
1− 4
= −1
3
f ′( x) = 2 x = −1
1
x = −
2
c = −
=1
2 −1 3
x
=1
3
3
8
⎛ 2⎞
x = ⎜ ⎟ =
27
⎝ 3⎠
on ( −2, 1).
=
2
43. f ( x ) = x 2 3 is continuous on [0, 1] and differentiable on
39. f ( x) = x 2 is continuous on [−2, 1] and differentiable
1 − ( −2 )
3
2
f ′( x) =
−15
f (1) − f ( −2)
0−0
= 0
2
(0, 1).
5
−15
=
x =
x − y − 13 = 0
(d)
2− 0
x = c =1
(c) f (c) = f (1) = −12
Tangent line:
= 3
f ′( x ) = 3 x 2 + 2 = 3
4 x + 4 y − 21 = 0
(d)
432 − 0
= 72
6− 0
=
6 −0
c =
8
27
x +1
is not continuous at x = 0.
x
The Mean Value Theorem does not apply.
44. f ( x ) =
1
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.2
45. f ( x) = 2 x + 1 is not differentiable at x = −1 2.
The Mean Value Theorem does not apply.
2 − x is continuous on [−7, 2] and
46. f ( x ) =
differentiable on ( −7, 2).
f ( 2) − f ( −7)
0−3
1
= −
9
3
=
2 − ( −7)
−1
1
f ′( x) =
= −
3
2 2− x
50. f ( x ) = ( x + 3) ln ( x + 3) is continuous on [−2, −1]
and differentiable on ( −2, −1).
f ( −1) − f ( −2)
2 ln 2 − 0
=
= ln 4
−1 − ( −2)
1
f ′( x) = ( x + 3)
1
+ ln ( x + 3) = 1 + ln ( x + 3)
x + 3
1 + ln ( x + 3) = ln 4
ln ( x + 3) = ln 4 − 1 = ln 4 − ln e = ln
2 2 − x = 3
4
e
4
x =
− 3 ≈ 1.386
e
c =
1
x = −
4
4 − 3e
e
51. f ( x ) = x log 2 x = x
1
4
47. f ( x) = sin x is continuous on [0, π ] and differentiable
on (0, π ).
f (π ) − f (0)
π −0
0− 0
π
= 0
ln x
ln 2
f is continuous on [1, 2] and differentiable on (1, 2).
f ( 2) − f (1)
2 −1
=
f ′( x) = x
2−0
= 2
2 −1
=
1
ln x
1 + ln x
+
=
= 2
x ln 2
ln 2
ln 2
f ′( x) = cos x = 0
1 + ln x = 2 ln 2 = ln 4
x = π 2
c =
xe = 4
π
x =
2
48. f ( x ) = e −3x is continuous on [0, 2] and differentiable
on (0, 2).
f ( 2) − f ( 0)
e −1
=
2−0
2
e −6 − 1
2
e −6 − 1 1 − e −6
=
=
6
−6
⎛ 1 − e −6 ⎞
−3 x = ln ⎜
⎟
⎝ 6 ⎠
1
c = ln
3
⎛ 1 − e −6 ⎞
1
1 ⎛ 6 ⎞
ln ⎜
⎟ = ln ⎜
⎟
3
6
3 ⎝ 1 − e −6 ⎠
⎝
⎠
⎛ 6 ⎞
= ln
⎜
−6 ⎟
⎝1 − e ⎠
4
e
f is continuous on [0, 1] and differentiable on (0, 1).
f ′( x) = −3e −3 x =
x = −
c =
4
e
52. f ( x) = arctan (1 − x)
−6
e −3 x
4
e
x +3 =
3
2 − x =
2
9
2 − x =
4
c = −
283
Rolle’s Theorem and the Mean Value Theorem
3
6
1 − e −6
49. f ( x) = cos x + tan x is not continuous at
f (1) − f (0)
1− 0
f ′( x) =
=
0 − (π 4)
=
1−0
= −
π
4
−1
1 + (1 − x)
2
−1
π
= −
x − 2x + 2
4
4
2
x − 2x + 2 =
2
π
x2 − 2 x −
4
π
+ 2 = 0
x ≈ 1.5227, 0.4773
c = 0.4773
x = π 2. The Mean Value Theorem does not apply.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
284
Chapter 4
53. f ( x) =
Applications of Differentiation
54. f ( x) = x − 2 sin x, [−π , π ]
x ⎡ 1 ⎤
, − ,2
x + 1 ⎢⎣ 2 ⎥⎦
(a) – (c)
2␲
(a) – (c)
1
tangent
f
Tangent
− 2␲
− 0.5
tangent
2
Secant
− 2␲
−1
(b) Secant line:
(b) Secant line:
slope =
f ( 2) − f ( −1 2)
2 − ( −1 2)
=
2 3 − ( −1)
52
1
=
2
3
y − π = 1( x − π )
cos x = 0
( x + 1) =
3
2
2
2
x = c = ±
3
= −1 ±
2
In the interval [−1 2, 2]: c = −1 +
=
=
(
−1 +
⎡−1 +
⎣
−2 +
(
6 2
)
(
6 2
)
Tangent lines:
)
π⎞
⎛π
⎞
⎛
y − ⎜ − 2 ⎟ = 1⎜ x − ⎟
2
2⎠
⎝
⎠
⎝
y = x − 2
π⎞
⎛ π
⎞
⎛
y − ⎜ − + 2 ⎟ = 1⎜ x + ⎟
2⎠
⎝ 2
⎠
⎝
y = x + 2
6
6
x , [1, 9]
55. f ( x) =
−2
+1
6
y −1+
2
π
⎛ π⎞
f ⎜− ⎟ = − + 2
2
⎝ 2⎠
6
2
6 2⎤ +1
⎦
Tangent line: y − 1 +
π
π
⎛π ⎞
− 2
f⎜ ⎟ =
2
⎝2⎠
x = −1 ±
f (c) =
π − ( −π )
=1
2π
(c) f ′( x) = 1 − 2 cos x = 1
2
3
+ 1)
π − ( −π )
=
y = x
=
(x
f (π ) − f ( −π )
slope =
2
2
y −
= ( x − 2)
3
3
2
y = ( x − 1)
3
(c) f ′( x) =
secant 2␲
f
(a) – (c)
2
2⎛
= ⎜⎜ x −
3⎝
6
Tangent
⎞
6
+ 1⎟⎟
2
⎠
6
2
6
2
= x −
+
3
3
3
3
1
y = 2x + 5 − 2 6
3
(
3
Secant
f
1
9
1
)
(b) Secant line:
f (9) − f (1)
slope =
9 −1
=
3−1
1
=
8
4
1
( x − 1)
4
1
3
y = x +
4
4
y −1 =
(c) f ′( x) =
1
2
=
1
4
x
x = c = 4
f ( 4) = 2
Tangent line: y − 2 =
y =
1
( x − 4)
4
1
x +1
4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.2
56. f ( x ) = x 4 − 2 x 3 + x 2 , [0, 6]
(a) – (c)
Rolle’s Theorem and the Mean Value Theorem
57. f ( x) = 2e x 4 cos
πx
4
285
,0 ≤ x ≤ 2
1000
(a) – (c)
3
Tangent
Secant
f
Tangent
0
f
Secant
6
−100
0
2
0
(b) Secant line:
slope =
f ( 6) − f ( 0)
=
6 −0
y − 0 = 150( x − 0)
(b) Secant line:
900 − 0
= 150
6
slope =
f ( 2) − f ( 0)
2−0
y − 2 = −1( x − 0)
y = 150 x
=
0− 2
= −1
2 −0
y = −x + 2
(c) f ′( x) = 4 x3 − 6 x 2 + 2 x = 150
πx⎞
π x ⎞π
⎛1
x 4⎛
(c) f ′( x) = 2⎜ e x 4 cos
⎟ + 2e ⎜ −sin
⎟
4 ⎠
4 ⎠4
⎝4
⎝
Using a graphing utility, there is one solution in
(0, 6), x = c ≈ 3.8721 and f (c) ≈ 123.6721
πx π
π x⎤
⎡1
= e x 4 ⎢ cos
− sin
4
2
4 ⎥⎦
⎣2
Tangent line: y − 123.6721 = 150( x − 3.8721)
f ′(c) = −1 ⇒ c ≈ 1.0161, f (c) ≈ 1.8
y = 150 x − 457.143
Tangent line: y − 1.8 = −1( x − 1.0161)
y = − x + 2.8161
58. f ( x) = ln sec π x
(a) – (c)
0. 5
f
secant
0
0.25
tangent
−0. 1
(b) Secant line: slope =
y −0 =
y =
(c) f ′( x) =
f (1 4) − f (0)
(1 4) − 0
(2 ln 2)( x − 0)
(ln 4) x
= 4 ln
2 = 2 ln 2 ≈ 1.3863
1
⋅ sec π x ⋅ tan π x ⋅ π = π tan π x
sec π x
f ′(c ) = π tan π c = ln 4
c =
1
π
tan −1
ln4
π
≈ 0.1323
f (c) ≈ 0.0889
Tangent line: y − 0.0889 = 1.3863( x − 0.1323)
y = 1.3863 x − 0.0945
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
286
Chapter 4
Applications of Differentiation
59. s(t ) = −4.9t 2 + 300
(a) vavg =
s(3) − s(0)
255.9 − 300
=
= −14.7 m/sec
3−0
3
62. f ( a) = f (b) and f ′(c) = 0 where c is in the interval
(a, b).
(a)
(b) s(t ) is continuous on [0, 3] and differentiable on
g ( a ) = g (b ) = f ( a ) + k
(0, 3). Therefore, the Mean Value Theorem applies.
g ′( x) = f ′( x ) ⇒ g ′(c) = 0
v(t ) = s′(t ) = −9.8t = −14.7 m/sec
Interval: [a, b]
−14.7
= 1.5 sec
t =
−9.8
Critical number of g: c
(b)
g ( a + k ) = g (b + k ) = f ( a )
g ′( x) = f ′( x − k )
S (12) − S (0) 200 ⎡⎣5 − (9 14)⎤⎦ − 200 ⎡⎣5 − (9 2)⎤⎦
=
12 − 0
12
450
=
7
⎛
⎞
9
450
⎟ =
S ′(t ) = 200⎜
⎜ ( 2 + t )2 ⎟
7
⎝
⎠
1
1
=
2
28
+
t
2
(
)
g ′(c + k ) = f ′(c) = 0
Interval: [a + k , b + k ]
Critical number of g : c + k
(c)
⎛c⎞
g ′⎜ ⎟ = kf ′(c) = 0
⎝k⎠
t = 2 7 − 2 ≈ 3.2915 months
⎡a b ⎤
Interval: ⎢ , ⎥
⎣k k ⎦
S ′(t ) is equal to the average value in April.
61. No. Let f ( x) = x 2 on [−1, 2].
f ′(0) = 0 and zero is in the interval ( −1, 2) but
f ( −1) ≠ f ( 2).
g ( x) = f ( kx)
⎛a⎞
⎛b⎞
g ⎜ ⎟ = g ⎜ ⎟ = f ( a)
⎝k⎠
⎝k⎠
g ′( x) = kf ′( kx)
2+ t = 2 7
f ′( x) = 2 x
g ( x) = f ( x − k )
(b)
9 ⎞
⎛
60. S (t ) = 200⎜ 5 −
⎟
2
+ t⎠
⎝
(a)
g ( x) = f ( x) + k
Critical number of g :
c
k
x = 0
⎧0,
63. f ( x ) = ⎨
⎩1 − x, 0 < x ≤ 1
No, this does not contradict Rolle’s Theorem. f is not
continuous on [0, 1].
64. No. If such a function existed, then the Mean Value
Theorem would say that there exists c ∈ ( −2, 2) such
that
f ′(c ) =
f ( 2) − f ( −2)
2 − ( −2)
=
6 + 2
= 2.
4
But, f ′( x) < 1 for all x.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.2
65. Let S (t ) be the position function of the plane. If
Rolle’s Theorem and the Mean Value Theorem
287
70. (a) f is continuous on [−10, 4] and changes sign,
t = 0 corresponds to 2 P.M., S (0) = 0, S (5.5) = 2500
( f (−8)
and the Mean Value Theorem says that there exists a
time t0 , 0 < t0 < 5.5, such that
Theorem, there exists at least one value of x in
[−10, 4] satisfying f ( x) = 0.
2500 − 0
≈ 454.54.
5.5 − 0
S ′(t0 ) = v(t0 ) =
Applying the Intermediate Value Theorem to the
velocity function on the intervals [0, t0 ] and [t0 , 5.5],
you see that there are at least two times during the
flight when the speed was 400 miles per hour.
(0 < 400 < 454.54)
> 0, f (3) < 0). By the Intermediate Value
(b) There exist real numbers a and b such that
−10 < a < b < 4 and f ( a) = f (b) = 2.
Therefore, by Rolle’s Theorem there exists at least
one number c in ( −10, 4) such that f ′(c) = 0.
This is called a critical number.
y
(c)
8
66. Let T (t ) be the temperature of the object. Then
4
T (0) = 1500° and T (5) = 390°. The average
x
−8
temperature over the interval [0, 5] is
390 − 1500
= −222° F/h.
5−0
By the Mean Value Theorem, there exist a time t0 ,
0 < t0 < 5, such that T ′(t0 ) = −222°F/h.
67. Let S (t ) be the difference in the positions of the
−4
−4
−8
71. f is continuous on [−5, 5] and does not satisfy the
conditions of the Mean Value Theorem. ⇒ f is not
differentiable on ( −5, 5). Example: f ( x) = x
2 bicyclists, S (t ) = S1 (t ) − S2 (t ). Because
y
S (0) = S ( 2.25) = 0, there must exist a time
t0 ∈ (0, 2.25) such that S ′(t0 ) = v(t0 ) = 0.
8
(−5, 5)
f(x) = ⏐x⏐
6
(5, 5)
4
At this time, v1 (t0 ) = v2 (t0 ).
2
68. Let t = 0 correspond to 9:13 A.M. By the Mean Value
(
Theorem, there exists t0 in 0,
v′(t0 ) = a(t0 ) =
4
1
30
) such that
85 − 35
= 1500 mi/h 2 .
1 30
⎛π x ⎞
⎛ π x ⎞⎛
⎛ π x ⎞ ⎞⎛ π ⎞
69. f ( x) = 3 cos 2 ⎜ ⎟, f ′( x) = 6 cos⎜ ⎟⎜ −sin ⎜ ⎟ ⎟⎜ ⎟
⎝ 2 ⎠
⎝ 2 ⎠⎝
⎝ 2 ⎠ ⎠⎝ 2 ⎠
−4
x
−2
2
72. f is not continuous on [−5, 5].
⎧1 x, x ≠ 0
Example: f ( x ) = ⎨
x = 0
⎩0,
y
⎛π x ⎞ ⎛π x ⎞
= −3π cos⎜ ⎟ sin ⎜ ⎟
⎝ 2 ⎠ ⎝ 2 ⎠
f (x) = 1x
4
2
(5, 15)
7
(a)
f′
x
f
2
(− 5, − 15)
−2␲
4
−2
4
2␲
−5
−7
(b) f and f ′ are both continuous on the entire real line.
(c) Because f ( −1) = f (1) = 0, Rolle’s Theorem
applies on [−1, 1]. Because f (1) = 0 and f ( 2) = 3,
Rolle’s Theorem does not apply on [1, 2].
(d) lim f ′( x) = 0
x → 3−
lim f ′( x ) = 0
x → 3+
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
288
Chapter 4
Applications of Differentiation
73. f ( x ) = x 5 + x 3 + x + 1
77. f ′( x) = 0
f is differentiable for all x.
f ( x) = c
f ( −1) = −2 and f (0) = 1, so the Intermediate Value
Theorem implies that f has at least one zero c in
[−1, 0], f (c) = 0.
Suppose f had 2 zeros, f (c1 ) = f (c2 ) = 0. Then
Rolle’s Theorem would guarantee the existence of a
number a such that
f ′( a ) = f (c2 ) − f (c1 ) = 0.
But, f ′( x) = 5 x 4 + 3x 2 + 1 > 0 for all x. So, f has
exactly one real solution.
74. f ( x ) = 2 x 5 + 7 x − 1
f is differentiable for all x.
f (0) = −1 and f (1) = 8, so the Intermediate Value
Theorem implies that f has at least one zero c in
[0, 1], f (c) = 0.
Suppose f had 2 zeros, f (c1 ) = f (c2 ) = 0. Then
Rolle’s Theorem would guarantee the existence of a
number a such that
f ′( a ) = f (c2 ) − f (c1 ) = 0.
But f ′( x) = 10 x + 7 > 0 for all x. So, f ( x) = 0 has
f ( 2) = 5
So, f ( x) = 5.
78. f ′( x) = 4
f ( x) = 4 x + c
f ( 0) = 1 ⇒ c = 1
So, f ( x) = 4 x + 1.
79. f ′( x) = 2 x
f ( x) = x 2 + c
f (1) = 0 ⇒ 0 = 1 + c ⇒ c = −1
So, f ( x ) = x 2 − 1.
80. f ′( x) = 6 x − 1
f ( x) = 3 x 2 − x + c
f ( 2) = 7 ⇒ 7 = 3( 22 ) − 2 + c
= 10 + c ⇒ c = − 3
So, f ( x) = 3 x 2 − x − 3.
4
exactly one real solution.
75. f ( x) = 3 x + 1 − sin x
f is differentiable for all x.
f ( −π ) = −3π + 1 < 0 and f (0) = 1 > 0, so the
Intermediate Value Theorem implies that f has at least
one zero c in [−π , 0], f (c) = 0.
Suppose f had 2 zeros, f (c1 ) = f (c2 ) = 0. Then
Rolle’s Theorem would guarantee the existence of a
number a such that
f ′( a ) = f (c2 ) − f (c1 ) = 0.
But f ′( x) = 3 − cos x > 0 for all x. So, f ( x) = 0 has
exactly one real solution.
76. f ( x) = 2 x − 2 − cos x
f (0) = −3, f (π ) = 2π − 2 + 1 = 2π − 1 > 0. By the
Intermediate Value Theorem, f has at least one zero.
Suppose f had 2 zeros, f (c1 ) = f (c2 ) = 0. Then
Rolle’s Theorem would guarantee the existence of a
number a such that
f ′( a ) = f (c2 ) − f (c1 ) = 0.
But, f ′( x) = 2 + sin x ≥ 1 for all x. So, f has exactly
one real solution.
81. False. f ( x) = 1 x has a discontinuity at x = 0.
82. False. f must also be continuous and differentiable on
each interval. Let
f ( x) =
x3 − 4 x
.
x2 − 1
83. True. A polynomial is continuous and differentiable
everywhere.
84. True
85. Suppose that p( x) = x 2 n + 1 + ax + b has two real roots
x1 and x2 . Then by Rolle’s Theorem, because
p( x1 ) = p( x2 ) = 0, there exists c in ( x1 , x2 ) such that
p′(c) = 0. But p′( x) = ( 2n + 1) x 2 n + a ≠ 0, because
n > 0, a > 0. Therefore, p( x) cannot have two real
roots.
86. Suppose f ( x) is not constant on ( a, b). Then there exists
x1 and x2 in ( a, b) such that f ( x1 ) ≠ f ( x2 ). Then by the
Mean Value Theorem, there exists c in ( a, b) such that
f ′(c ) =
f ( x2 ) − f ( x1 )
x2 − x1
≠ 0.
This contradicts the fact that f ′( x) = 0 for all x in ( a, b).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
Increasing and Decreasing Functions and the First Derivative Test
87. If p( x) = Ax 2 + Bx + C , then
p′( x ) = 2 Ax + B =
=
=
f (b ) − f ( a )
b − a
( Ab 2 + Bb + C ) − ( Aa 2 + Ba + C )
b − a
A(b 2 − a 2 ) + B(b − a )
b − a
(b
− a ) ⎡⎣ A(b + a) + B⎤⎦
=
b − a
= A(b + a) + B.
So, 2 Ax = A(b + a ) and x = (b + a ) 2 which is the
midpoint of [a, b].
88. (a) f ( x ) = x 2 , g ( x ) = − x 3 + x 2 + 3 x + 2
f ( −1) = g ( −1) = 1, f ( 2) = g ( 2) = 4
Let h( x) = f ( x) − g ( x). Then, h( −1) = h( 2) = 0.
So, by Rolle’s Theorem these exists c ∈ ( −1, 2)
such that h′(c) = f ′(c) − g ′(c) = 0.
So, at x = c, the tangent line to f is parallel to the
tangent line to g.
h( x) = x3 − 3 x − 2, h′( x )
= 3x 2 − 3 = 0 ⇒ x = c = 1
(b) Let h( x) = f ( x) − g ( x). Then h( a ) = h(b) = 0 by
90. f ( x ) =
So, at x = c, the tangent line to f is parallel to the
tangent line to g.
89. Suppose f ( x) has two fixed points c1 and c2 . Then, by
the Mean Value Theorem, there exists c such that
f ′(c ) =
f (c2 ) − f (c1 )
c2 − c1
=
c2 − c1
= 1.
c2 − c1
This contradicts the fact that f ′( x) < 1 for all x.
cos x differentiable on ( −∞, ∞).
f ′( x) = − 12 sin x
− 12 ≤ f ′( x ) ≤
1
2
⇒ f ′( x ) < 1
for all real numbers.
So, from Exercise 70, f has, at most, one fixed point.
( x ≈ 0.4502)
91. Let f ( x) = cos x. f is continuous and differentiable for
all real numbers. By the Mean Value Theorem, for any
interval [a, b], there exists c in ( a, b) such that
f (b ) − f ( a )
= f ′(c)
b − a
cos b − cos a
= −sin c
b − a
cos b − cos a = ( −sin c)(b − a)
cos b − cos a = −sin c b − a
cos b − cos a ≤ b − a since −sin c ≤ 1.
92. Let f ( x) = sin x. f is continuous and differentiable for
all real numbers. By the Mean Value Theorem, for any
interval [a, b], there exists c in ( a, b) such that
f (b ) − f ( a )
= f ′(c )
b − a
sin(b) − sin ( a ) = (b − a ) cos(c)
sin (b) − sin ( a ) = b − a cos(c)
Rolle’s Theorem, there exists c in ( a, b) such that
h′(c) = f ′(c) − g ′(c) = 0.
1
2
289
sin a − sin b ≤ a − b .
93. Let 0 < a < b. f ( x) =
x satisfies the hypotheses of
the Mean Value Theorem on [a, b]. Hence, there exists c
in ( a, b) such that
f ′(c) =
So,
f (b ) − f ( a )
1
=
=
b − a
2 c
b −
a = (b − a )
b − a
.
b − a
1
b − a
<
.
2 c
2 a
Section 4.3 Increasing and Decreasing Functions and the First Derivative Test
1. (a) Increasing: (0, 6) and (8, 9). Largest: (0, 6)
(b) Decreasing: (6, 8) and (9, 10). Largest: (6, 8)
2. (a) Increasing: ( 4, 5), (6, 7). Largest: ( 4, 5), (6, 7)
(b) Decreasing: ( −3, 1), (1, 4), (5, 6). Largest: ( −3, 1)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
290
Chapter 4
Applications of Differentiation
3. f ( x ) = x 2 − 6 x + 8
4. y = −( x + 1)
2
From the graph, f is decreasing on ( −∞, 3) and
From the graph, f is increasing on ( −∞, −1) and
increasing on (3, ∞).
decreasing on ( −1, ∞).
Analytically, f ′( x) = 2 x − 6.
Analytically, y′ = − 2( x + 1).
Critical number: x = 3
Critical number: x = −1
Test intervals:
−∞ < x < 3
3 < x < ∞
Test intervals:
− ∞ < x < −1
−1 < x < ∞
Sign of f ′( x):
f′ < 0
f′ > 0
Sign of y′:
y′ > 0
y′ < 0
Conclusion:
Increasing
Decreasing
Conclusion:
5. y =
Decreasing
Increasing
x3
− 3x
4
From the graph, y is increasing on ( −∞, − 2) and ( 2, ∞), and decreasing on ( −2, 2).
Analytically, y′ =
3x 2
3
3
− 3 = ( x 2 − 4) = ( x − 2)( x + 2)
4
4
4
Critical numbers: x = ± 2
Test intervals:
− ∞ < x < −2
−2 < x < 2
2 < x < ∞
Sign of y′:
y′ > 0
y′ < 0
y′ > 0
Conclusion:
Increasing
Decreasing
Increasing
6. f ( x ) = x 4 − 2 x 2
From the graph, f is decreasing on ( −∞, −1) and (0, 1), and increasing on ( −1, 0) and (1, ∞).
Analytically, f ′( x) = 4 x 3 − 4 x = 4 x( x − 1)( x + 1).
Critical numbers: x = 0, ±1.
Test intervals:
− ∞ < x < −1
−1 < x < 0
0 < x <1
1< x < ∞
Sign of f ′:
f′ < 0
f′ > 0
f′ < 0
f′ > 0
Conclusion:
7. f ( x ) =
Decreasing
Increasing
Decreasing
Increasing
1
(x
+ 1)
2
From the graph, f is increasing on ( −∞, −1) and decreasing on ( −1, ∞).
Analytically, f ′( x) =
−2
( x + 1)
3
.
No critical numbers. Discontinuity: x = −1
Test intervals:
− ∞ < x < −1
−1 < x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
Conclusion:
Increasing
Decreasing
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
8. y =
Increasing and Decreasing Functions and the First Derivative Test
291
x2
2x − 1
From the graph, y is increasing on ( −∞, 0) and (1, ∞), and decreasing on (0, 1 2) and (1 2, 1).
Analytically, y′ =
(2 x
− 1)2 x − x 2 ( 2)
( 2 x − 1)
2
=
2 x2 − 2 x
( 2 x − 1)
2
=
2 x( x − 1)
(2 x
− 1)
2
Critical numbers: x = 0, 1
Discontinuity: x = 1 2
Test intervals:
−∞ < x < 0
0 < x <12
12 < x <1
1< x < ∞
Sign of y′:
y′ > 0
y′ < 0
y′ < 0
y′ > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
9. g ( x ) = x 2 − 2 x − 8
g ′( x ) = 2 x − 2
Critical number: x = 1
Test intervals:
−∞ < x < 1
1< x < ∞
Sign of g ′( x) :
g′ < 0
g′ > 0
Conclusion:
Decreasing
Increasing
Increasing on: (1, ∞)
Decreasing on: ( −∞, 1)
10. h( x) = 12 x − x 3
h′( x) = 12 − 3 x 2 = 3( 4 − x 2 ) = 3( 2 − x)( 2 + x)
Critical numbers: x = ± 2
Test intervals:
−∞ < x < −2
−2 < x < 2
2 < x < ∞
Sign of h′( x):
h′ < 0
h′ > 0
h′ < 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on: ( − 2, 2)
Decreasing on: ( −∞, − 2), ( 2, ∞ )
Domain: [−4, 4]
11. y = x 16 − x 2
y′ =
−2( x 2 − 8)
16 − x
2
=
−2
16 − x 2
( x − 2 2 )( x + 2 2 )
Critical numbers: x = ±2 2
Test intervals:
− 4 < x < −2 2
−2 2 < x < 2 2
2 2 < x < 4
Sign of y′:
y′ < 0
y′ > 0
y′ < 0
Conclusion:
Decreasing
Increasing
Decreasing
(
Increasing on: −2 2, 2 2
(
)
)(
Decreasing on: − 4, − 2 2 , 2 2, 4
)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
292
Chapter 4
12. y = x +
y′ =
Applications of Differentiation
9
x
( x − 3)( x + 3)
1−9
x2 − 9
=
=
2
x
x2
x2
Critical numbers: x = ± 3
Discontinuity: x = 0
Test intervals:
−∞ < x < −3
−3 < x < 0
0 < x < 3
3 < x < ∞
Sign of y′:
y′ > 0
y′ < 0
y′ < 0
y′ > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on: ( − ∞, − 3), (3, ∞ )
Decreasing on: ( − 3, 0), (0, 3)
13. f ( x ) = sin x − 1,
0 < x < 2π
f ′( x) = cos x
Critical numbers: x =
π 3π
2
,
2
Test intervals:
0 < x <
Sign of f ′( x) :
f′ > 0
Conclusion:
π
π
2
2
Increasing
< x <
f′ < 0
Decreasing
3π
2
3π
< x < 2π
2
f′ > 0
Increasing
⎛ π ⎞ ⎛ 3π
⎞
Increasing on: ⎜ 0, ⎟, ⎜ , 2π ⎟
⎝ 2⎠ ⎝ 2
⎠
⎛ π 3π ⎞
Decreasing on: ⎜ ,
⎟
⎝2 2 ⎠
x
14. h( x) = cos ,
0 < x < 2π
2
1
x
h′( x) = − sin
2
2
Critical numbers: none
Test interval:
0 < x < 2π
Sign of h′( x):
h′ < 0
Conclusion:
Decreasing
Decreasing on 0 < x < 2π
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
Increasing and Decreasing Functions and the First Derivative Test
293
15. y = x − 2 cos x, 0 < x < 2π
y′ = 1 + 2 sin x
y′ = 0: sin x = −
1
2
7π 11π
,
6 6
Critical numbers: x =
7π
11π
< x <
6
6
11π
< x < 2π
6
y′ > 0
y′ < 0
y′ > 0
Increasing
Decreasing
Increasing
Test intervals:
0 < x <
Sign of y′ :
Conclusion:
7π
6
⎛ 7π ⎞ ⎛ 11π
⎞
Increasing on: ⎜ 0,
, 2π ⎟
⎟, ⎜
⎝ 6 ⎠ ⎝ 6
⎠
⎛ 7π 11π ⎞
Decreasing on: ⎜ ,
⎟
⎝ 6 6 ⎠
16. f ( x ) = sin 2 x + sin x, 0 < x < 2π
f ′( x) = 2 sin x cos x + cos x = cos x( 2 sin x + 1)
2 sin x + 1 = 0 ⇒ sin x = −
π 3π
cos x = 0 ⇒ x =
Critical numbers:
1
7π 11π
⇒ x =
,
2
6 6
2
,
2
π 7π 3π 11π
2
,
6
,
2
Test intervals:
0 < x <
Sign of f ′( x) :
f′ > 0
Conclusion:
,
6
π
π
2
2
< x <
7π
6
f′ < 0
Increasing
Decreasing
7π
3π
< x <
6
2
3π
11π
< x <
2
6
11π
< x < 2π
6
f′ > 0
f′ < 0
f′ > 0
Increasing
Decreasing
Increasing
⎛ π ⎞ ⎛ 7π 3π ⎞ ⎛ 11π
⎞
Increasing on: ⎜ 0, ⎟, ⎜ ,
, 2π ⎟
⎟, ⎜
2
6
2
6
⎝
⎠ ⎝
⎠ ⎝
⎠
⎛ π 7π ⎞ ⎛ 3π 11π ⎞
Decreasing on: ⎜ ,
⎟, ⎜ ,
⎟
⎝2 6 ⎠ ⎝ 2 6 ⎠
17. g ( x) = e − x + e3 x
g ′( x) = −e − x + 3e3 x
Critical number: x = − 14 ln 3
Test intervals:
−∞ < x < − 14 ln 3
− 14 ln 3 < x < ∞
Sign of g ′( x) :
g′ < 0
g′ > 0
Conclusion:
Decreasing
Increasing
(
Increasing on: − 14 ln 3, ∞
(
)
)
Decreasing on: −∞, − 14 ln 3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
294
Chapter 4
18. h( x) =
h′( x) = −
Applications of Differentiation
x e− x ,
x ≥ 0
1
x e− x +
2
x
1
2
Critical number: x =
⎛ 1
−
e− x = e− x ⎜
⎝2 x
(x
1 − 2x
⎞
x ⎟ = e− x −
2 x
⎠
= 0 is an endpoint )
1
2
1
< x < ∞
2
Test intervals:
0 < x <
Sign of h′( x) :
h′ > 0
h′ < 0
Conclusion:
Increasing
Decreasing
⎛ 1⎞
Increasing on: ⎜ 0, ⎟
⎝ 2⎠
⎛1 ⎞
Decreasing on: ⎜ , ∞ ⎟
⎝2 ⎠
⎛ x⎞
19. f ( x) = x 2 ln ⎜ ⎟,
⎝ 2⎠
21. (a)
x > 0
f ′( x) = 2 x − 4
⎛ x ⎞ x2
⎛ x⎞
= 2 x ln ⎜ ⎟ + x
f ′( x) = 2 x ln ⎜ ⎟ +
2
x
⎝ ⎠
⎝ 2⎠
Test intervals:
0 < x <
Sign of f ′( x) :
f′ < 0
Conclusion:
2
e
Decreasing
2
< x < ∞
e
f ′( x) =
Sign of f ′:
f′ < 0
f′ > 0
Decreasing
Increasing
Increasing on: ( 2, ∞)
22. (a)
f ( x) = x 2 + 6 x + 10
f ′( x) = 2 x + 6
Critical number: x = −3
x > 0
(b)
Test intervals:
− ∞ < x < −3
−3 < x < ∞
Sign of f ′:
f′ < 0
f′ > 0
Conclusion:
Decreasing
Increasing
Decreasing on: ( −∞, − 3)
Critical number: x = e 2
Test intervals:
0 < x < e2
e2 < x < ∞
Sign of f ′( x) :
f′ > 0
f′ < 0
Increasing
2 < x < ∞
(c) Relative minimum: ( 2, − 4)
x
1
− ln x
x
2 x = 2 − ln x
2 x3 2
x
Conclusion:
−∞ < x < 2
Decreasing on: ( −∞, 2)
Increasing
2 ⎞
⎛
Decreasing on: ⎜ 0,
⎟
e⎠
⎝
20.
Test intervals:
Conclusion:
f′ > 0
⎛ 2
⎞
, ∞⎟
Increasing on: ⎜
⎝ e ⎠
ln x
f ( x) =
,
x
Critical number: x = 2
(b)
2
e
Critical number: x =
f ( x) = x 2 − 4 x
Increasing on: ( −3, ∞)
(c) Relative minimum: ( −3, 1)
Decreasing
Increasing on: (0, e 2 )
Decreasing on: (e 2 , ∞)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
23. (a)
(b)
Increasing and Decreasing Functions and the First Derivative Test
f ( x) = −2 x 2 + 4 x + 3
24. (a)
f ( x) = − 3 x 2 − 4 x − 2
f ′( x) = −4 x + 4 = 0
f ′( x) = − 6 x − 4 = 0
Critical number: x = 1
Critical number: x = − 23
Test intervals:
−∞ < x < 1
1< x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
Conclusion:
Increasing
(b)
Decreasing
Test intervals:
− ∞ < x < − 23
− 23 < x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
Conclusion:
Increasing on: ( −∞, 1)
Increasing
Decreasing
(
)
Decreasing on: ( − 23 , ∞)
Relative maximum: ( − 23 , − 23 )
Increasing on: −∞, − 23
Decreasing on: (1, ∞)
(c) Relative maximum: (1, 5)
(c)
25. (a)
295
f ( x) = 2 x3 + 3 x 2 − 12 x
f ′( x) = 6 x 2 + 6 x − 12 = 6( x + 2)( x − 1) = 0
Critical numbers: x = −2, 1
(b)
Test intervals:
− ∞ < x < −2
−2 < x < 1
1< x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
f′ > 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: ( −∞, − 2), (1, ∞ )
Decreasing on: ( −2, 1)
(c) Relative maximum: ( −2, 20)
Relative minimum: (1, − 7)
26. (a)
f ( x) = x3 − 6 x 2 + 15
f ′( x) = 3 x 2 − 12 x = 3 x( x − 4)
Critical numbers: x = 0, 4
(b)
Test intervals:
−∞ < x < 0
0 < x < 4
4 < x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
f′ > 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: ( −∞, 0), ( 4, ∞)
Decreasing on: (0, 4)
(c) Relative maximum: (0, 15)
Relative minimum: ( 4, −17)
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296
Chapter 4
27. (a)
Applications of Differentiation
f ( x) = ( x − 1) ( x + 3) = x3 + x 2 − 5 x + 3
2
f ′( x) = 3 x 2 + 2 x − 5 = ( x − 1)(3 x + 5)
Critical numbers: x = 1, − 53
(b)
Test intervals:
−∞ < x < − 53
−5 3 < x < 1
1< x < ∞
Sign of f ′:
f′ > 0
f′ < 0
f′ > 0
Conclusion:
Increasing
(
Decreasing
Increasing
)
Increasing on: −∞, − 53 and (1, ∞ )
(
)
Decreasing on: − 53 , 1
(
(c) Relative maximum: − 53 ,
256
27
)
Relative minimum: (1, 0)
28. (a)
f ( x) = ( x + 2) ( x − 1)
2
f ′( x) = 3 x( x + 2)
Critical numbers: x = −2, 0
(b)
Test intervals:
− ∞ < x < −2
−2 < x < 0
0 < x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
f′ > 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: ( −∞, − 2), (0, ∞)
Decreasing on: ( −2, 0)
(c) Relative maximum: ( −2, 0)
Relative minimum: (0, − 4)
29. (a)
f ( x) =
x5 − 5 x
5
f ′( x) = x 4 − 1
Critical numbers: x = −1, 1
(b)
Test intervals:
− ∞ < x < −1
−1 < x < 1
1< x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
f′ > 0
Conclusion:
Increasing
Decreasing
Increasing
Increasing on: ( −∞, −1), (1, ∞)
Decreasing on: ( −1, 1)
4⎞
⎛
(c) Relative maximum: ⎜ −1, ⎟
5⎠
⎝
4⎞
⎛
Relative minimum: ⎜1, − ⎟
5
⎝
⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
30. (a)
Increasing and Decreasing Functions and the First Derivative Test
f ( x) = x 4 − 32 x + 4
33. (a)
f ′( x) = 4 x3 − 32 = 4( x3 − 8)
f ( x ) = ( x + 2)
f ′( x) =
Critical number: x = 2
(b)
Test intervals:
−∞ < x < 2
2 < x < ∞
Sign of f ′( x):
f′ < 0
f′ > 0
Conclusion:
Decreasing
Increasing
f ′( x) =
34. (a)
1 −2 3
1
x
=
3
3x 2 3
−∞ < x < 0
0 < x < ∞
Sign of f ′( x):
f′ > 0
f′ > 0
(b)
Increasing
Increasing
f ( x) = ( x − 3)
13
1
1
( x − 3)−2 3 =
23
3
3( x − 3)
Test intervals:
−∞ < x < 3
3 < x < ∞
Sign of f ′:
f′ > 0
f′ > 0
Increasing
Increasing
(c) No relative extrema
f ( x) = x 2 3 − 4
2 −1 3
2
x
= 13
3
3x
35. (a)
−∞ < x < 0
0 < x < ∞
Sign of f ′( x):
f′ < 0
f′ > 0
Decreasing
Increasing on: (0, ∞)
f ( x) = 5 − x − 5
f ′( x) = −
Test intervals:
Conclusion:
Decreasing
Increasing on: ( −∞, ∞)
Critical number: x = 0
(b)
f′ > 0
Conclusion:
(c) No relative extrema
f ′( x) =
f′ < 0
Critical number: x = 3
Increasing on: ( −∞, ∞)
32. (a)
Sign of f ′:
f ′( x) =
Test intervals:
Increasing
−2 < x < ∞
(c) Relative minimum: ( −2, 0)
+1
Conclusion:
− ∞ < x < −2
Increasing on: ( −2, ∞)
Critical number: x = 0
(b)
Test intervals:
Decreasing on: ( −∞, − 2)
(c) Relative minimum: ( 2, − 44)
f ( x) = x
2
2
( x + 2)−1 3 =
13
3
3( x + 2)
Conclusion:
Decreasing on: ( −∞, 2)
31. (a)
23
Critical number: x = −2
(b)
Increasing on: ( 2, ∞)
13
297
Increasing
⎧ 1, x < 5
x −5
= ⎨
x −5
⎩−1, x > 5
Critical number: x = 5
(b)
Test intervals:
−∞ < x < 5
5 < x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
Conclusion:
Increasing
Decreasing on: ( −∞, 0)
Increasing on: ( −∞, 5)
(c) Relative minimum: (0, − 4)
Decreasing on: (5, ∞)
Decreasing
(c) Relative maximum: (5, 5)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
298
Chapter 4
36. (a)
Applications of Differentiation
f ( x) = x + 3 − 1
f ′( x) =
⎧ 1, x > −3
x +3
= ⎨
x +3
⎩−1, x < −3
Critical number: x = −3
(b)
Test intervals:
− ∞ < x < −3
−3 < x < ∞
Sign of f ′( x):
f′ < 0
f′ > 0
Conclusion:
Decreasing
Increasing
Increasing on: ( −3, ∞)
Decreasing on: ( −∞, − 3)
(c) Relative minimum: ( −3, −1)
37. (a)
f ( x) = 2 x +
f ′( x) = 2 −
1
x
1
2x2 − 1
=
2
x
x2
2
2
Critical numbers: x = ±
Discontinuity: x = 0
(b)
Test intervals:
−∞ < x < −
Sign of f ′:
f′ > 0
Conclusion:
2
2
−
2
< x < 0
2
f′ < 0
Increasing
Decreasing
⎛
2⎞
Increasing on: ⎜⎜ −∞, −
⎟
2 ⎟⎠
⎝
and
⎛
2 ⎞
Decreasing on: ⎜⎜ −
, 0 ⎟⎟
2
⎝
⎠
and
0 < x <
f′ < 0
Decreasing
2
2
2
< x < ∞
2
f′ > 0
Increasing
⎛ 2 ⎞
, ∞ ⎟⎟
⎜⎜
⎝ 2
⎠
⎛
2⎞
⎜⎜ 0,
⎟⎟
2
⎝
⎠
⎛
⎞
2
, − 2 2 ⎟⎟
(c) Relative maximum: ⎜⎜ −
⎝ 2
⎠
⎛ 2
⎞
Relative minimum: ⎜
⎜ 2 , 2 2 ⎟⎟
⎝
⎠
38. (a)
f ( x) =
f ′( x) =
x
x −5
( x − 5) − x
(x
− 5)
2
=
−5
(x
− 5)
2
No critical numbers
Discontinuity: x = 5
(b)
Test intervals:
−∞ < x < 5
5 < x < ∞
Sign of f ′( x):
f′ < 0
f′ < 0
Conclusion:
Decreasing
Decreasing
Decreasing on: ( −∞, 5), (5, ∞)
(c) No relative extrema
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
39. (a)
f ( x) =
f ′( x) =
Increasing and Decreasing Functions and the First Derivative Test
299
x2
x −9
2
( x2
− 9)( 2 x) − ( x 2 )( 2 x)
(x
2
− 9)
2
=
−18 x
(x
2
− 9)
2
Critical number: x = 0
Discontinuities: x = −3, 3
(b)
Test intervals:
−∞ < x < −3
−3 < x < 0
0 < x < 3
3 < x < ∞
Sign of f ′( x):
f′ > 0
f′ > 0
f′ < 0
f′ < 0
Conclusion:
Increasing
Increasing
Decreasing
Decreasing
Increasing on: ( −∞, − 3), ( −3, 0)
Decreasing on: (0, 3), (3, ∞)
(c) Relative maximum: (0, 0)
40. (a)
f ( x) =
f ′( x) =
x2 − 2x + 1
x +1
(x
+ 1)( 2 x − 2) − ( x 2 − 2 x + 1)(1)
(x
+ 1)
2
=
x2 + 2x − 3
(x
+ 1)
2
=
(x
+ 3)( x − 1)
(x
+ 1)
2
Critical numbers: x = −3, 1
Discontinuity: x = −1
(b)
Test intervals:
−∞ < x < −3
−3 < x < − 1
−1 < x < 1
1< x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
f′ < 0
f′ > 0
Conclusion:
Increasing
Decreasing
Decreasing
Increasing
Increasing on: ( −∞, − 3), (1, ∞ )
Decreasing on: ( −3, −1), ( −1, 1)
(c) Relative maximum: ( −3, − 8)
Relative minimum: (1, 0)
41. (a)
2
⎪⎧4 − x , x ≤ 0
f ( x) = ⎨
x > 0
⎪⎩−2 x,
⎧−2 x, x < 0
f ′( x) = ⎨
x > 0
⎩−2,
Critical number: x = 0
(b)
Test intervals:
−∞ < x < 0
0 < x < ∞
Sign of f ′:
f′ > 0
f′ < 0
Conclusion:
Increasing
Decreasing
Increasing on: ( −∞, 0)
Decreasing on: (0, ∞)
(c) Relative maximum: (0, 4)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
300
Chapter 4
42. (a)
Applications of Differentiation
⎧2 x + 1, x ≤ −1
f ( x) = ⎨ 2
⎩x − 2, x > −1
x < −1
⎧2,
f ′( x) = ⎨
⎩2 x, x > −1
Critical numbers: x = −1, 0
(b)
Test intervals:
−∞ < x < −1
−1 < x < 0
0 < x < ∞
Sign of f ′:
f′ > 0
f′ < 0
f′ > 0
Conclusion:
Increasing
Increasing on: ( −∞, −1)
Decreasing
Increasing
(0, ∞)
and
Decreasing on: ( −1, 0)
(c) Relative maximum: ( −1, −1)
Relative minimum: (0, − 2)
43. (a)
⎧3 x + 1, x ≤ 1
f ( x) = ⎨
2
⎩5 − x , x > 1
x <1
⎧3,
f ′( x) = ⎨
−
x
x >1
2
,
⎩
Critical number: x = 1
(b)
Test intervals:
−∞ < x < 1
1< x < ∞
Sign of f ′:
f′ > 0
f′ < 0
Conclusion:
Increasing
Decreasing
Increasing on: ( −∞, 1)
Decreasing on: (1, ∞)
(c) Relative maximum: (1, 4)
44. (a)
3
x ≤ 0
⎪⎧− x + 1,
f ( x) = ⎨ 2
⎪⎩− x + 2 x, x > 0
2
x < 0
⎪⎧−3 x ,
f ′( x) = ⎨
⎪⎩−2 x + 2, x > 0
Critical numbers: x = 0, 1
(b)
Test intervals:
−∞ < x < 0
0 < x <1
1< x < ∞
Sign of f ′:
f′ < 0
f′ > 0
f′ < 0
Conclusion:
Decreasing
Increasing
Decreasing
Increasing on: (0, 1)
Decreasing on: ( −∞, 0)
and
(1, ∞)
(c) Relative maximum: (1, 1)
Note: (0, 1) is not a relative minimum
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
45.
Increasing and Decreasing Functions and the First Derivative Test
f ( x ) = (3 − x )e x − 3
f ′( x) = (3 − x)e
48.
x −3
−e
x −3
x
+ arctan x
1 + x2
f ′( x) = 0
Critical number: x = 2
Critical number: x = 0
Test intervals:
−∞ < x < 2
2 < x < ∞
Sign of f ′( x):
f′ > 0
f′ < 0
Increasing
Decreasing
Increasing on: ( − ∞, 2)
4
f′ > 0
Decreasing
Increasing
6
−6
6
49. g ( x) = ( x)3− x
g ′( x) = (1 − x ln 3)3− x
Critical number: x = 0
Test intervals:
−∞ < x < 0
0 < x < ∞
Sign of f ′( x):
f′ < 0
f′ > 0
Decreasing
Critical number: x =
Increasing
Increasing on: (0, ∞)
4
−5
Relative minimum: (0, −1)
1
≈ 0.9102
ln 3
Test intervals:
−∞ < x <
Sign of f ′( x):
f′ > 0
Conclusion:
Decreasing on: ( −∞, 0)
1
ln 3
Decreasing
1
−2
⎛ 1
⎞
, ∞⎟
Decreasing on: ⎜
ln
3
⎝
⎠
f ( x) = 4( x − arcsin x), −1 ≤ x ≤ 1
1
< x < ∞
ln 3
f′ < 0
Increasing
1 ⎞
⎛
Increasing on: ⎜ −∞,
⎟
ln 3 ⎠
⎝
5
−2
4
−3
1 ⎞
⎛ 1
,
Relative maximum: ⎜
⎟ ≈ (0.9102, 0.3349)
⎝ ln 3 e ln 3 ⎠
4
1 − x2
Critical number: x = 0
50.
Test intervals:
−1 ≤ x < 0
0 < x ≤1
Sign of f ′( x):
f′ < 0
f′ < 0
Decreasing
(0, 0)
−1.5
2 −3
2 −3
( 2 x)
Critical number: x = 0
3
No relative extrema
f ( x) = 2 x
f ′( x) = (ln 2)2 x
Decreasing
Decreasing on: [−1, 1]
(Absolute maximum at
x = −1, absolute minimum
at x = 1 )
f′ < 0
−3
f ′( x) = ( x − 1)e x + e x = xe x
Conclusion:
Sign of f ′( x):
Relative minimum: (0, 0)
f ( x) = ( x − 1)e x
f ′( x) = 4 −
0 < x < ∞
Decreasing on: ( −∞, 0)
−1
47.
−∞ < x < 0
Increasing on: (0, ∞)
−2
Relative minimum: ( 2, e −1 )
Conclusion:
Test intervals:
Conclusion:
0.5
Decreasing on: ( 2, ∞)
46.
f ( x) = x arctan x
f ′( x) =
= e x − 3 (2 − x)
Conclusion:
301
1.5
Test intervals:
−∞ < x < 0
0 < x < ∞
Sign of f ′( x):
f′ < 0
f′ > 0
Conclusion:
Decreasing
Increasing
Increasing on: (0, ∞)
−3
3
Decreasing on: ( −∞, 0)
( 18 )
Relative minimum: 0,
−3
3
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
302
51.
Chapter 4
Applications of Differentiation
ln x
ln 4
f ( x) = x − log 4 x = x −
f ′( x) = 1 −
53. g ( x) =
1
1
= 0 ⇒ x ln 4 = 1 ⇒ x =
x ln 4
ln 4
Critical number: x =
1
ln 4
(e 2 x
+ 1)2e 2 x − e 2 x ( 2e 2 x )
(e 2 x
+ 1)
2
0 < x <
Sign of f ′( x):
f′ < 0
1
ln 4
Decreasing
No relative extrema.
Increasing
−3
3
54. h( x) = ln ( 2 − ln x)
Domain: x > 0 and 2 − ln x > 0 ⇒ 0 < x < e 2
h′( x) =
−1
2
−2
(0.7213, 0.9570)
1 ⎞
⎛
Decreasing on: ⎜ 0,
⎟
⎝ ln 4 ⎠
5
−1
1
1
1
⎛ 1⎞
=
⎜− ⎟ =
2 − ln x ⎝ x ⎠
x ln x − 2 x
x(ln x − 2)
No critical numbers.
Relative maximum:
h′( x) < 0 on entire domain.
⎛ 1
1
⎛ 1 ⎞ ⎞ ⎛ 1 ln (ln 4) + 1 ⎞
− log 4 ⎜
,
,
⎟
⎜
⎟⎟ = ⎜
ln
4
ln
4
ln 4
⎝ ln 4 ⎠ ⎠ ⎝ ln 4
⎝
⎠
Decreasing on: (0, e 2 )
4
−2
No relative extrema.
≈ (0.7213, 0.9570)
x3
− ln x
3
55.
10
−6
f ( x) = e −1 ( x − 2) = e1 (2 − x) , x =/ 2
⎛
⎞
1
⎟
f ′( x) = e1 (2 − x) ⎜
2
⎜ ( 2 − x) ⎟
⎝
⎠
Domain: x > 0
1
x2 − 1
=
x
x
No critical numbers.
x = 2 is a vertical asymptote.
Critical number: x = 1
Test intervals:
0 < x <1
1< x < ∞
Sign of f ′( x):
f′ < 0
f′ > 0
Conclusion:
+ 1)
2
f′ > 0
⎛ 1
⎞
Increasing on: ⎜
, ∞⎟
⎝ ln 4 ⎠
f ′( x) = x 2 −
(e 2 x
Increasing on: ( −∞, ∞)
1
< x < ∞
ln 4
5
52. f ( x ) =
2e 2 x
=
No critical numbers.
Test intervals:
Conclusion:
g ′( x) =
e2 x
e +1
2x
Decreasing
2 < x < ∞
Sign of f ′( x):
f′ > 0
f′ > 0
Increasing
Increasing
Increasing on: ( −∞, 2), ( 2, ∞)
3
4
No relative extrema.
Decreasing on: (0, 1)
⎛
Relative minimum: ⎜1,
⎝
−∞ < x < 2
Conclusion:
Increasing
Increasing on: (1, ∞)
Test intervals:
1⎞
⎟
3⎠
−1
−1
3
4
−1
−1
56.
f ( x) = earctan x
6
⎛ 1 ⎞
f ′( x) = earctan x ⎜
= 0
2⎟ /
⎝1 + x ⎠
−4
8
No critical numbers.
Increasing on: ( −∞, ∞)
−2
No relative extrema.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
57. (a)
Increasing and Decreasing Functions and the First Derivative Test
303
x
+ cos x, 0 < x < 2π
2
1
f ′( x) =
− sin x = 0
2
f ( x) =
π 5π
Critical numbers: x =
6
,
6
Test intervals:
0 < x <
Sign of f ′( x):
f′ > 0
Conclusion:
π
π
4
4
5π
4
< x <
f′ < 0
Increasing
5π
< x < 2π
4
f′ > 0
Decreasing
Increasing
⎛ π ⎞ ⎛ 5π
⎞
Increasing on: ⎜ 0, ⎟, ⎜ , 2π ⎟
6
6
⎝
⎠ ⎝
⎠
⎛ π 5π ⎞
Decreasing on: ⎜ ,
⎟
⎝6 6 ⎠
⎛π π + 6 3 ⎞
(b) Relative maximum: ⎜ ,
⎟⎟
⎜6
12
⎝
⎠
⎛ 5π 5π − 6 3 ⎞
Relative minimum: ⎜⎜ ,
⎟⎟
12
⎝ 6
⎠
5
(c)
2␲
0
0
58. (a)
f ( x) = sin x cos x + 5 =
f ′( x) = cos 2 x
Critical numbers:
1
sin 2 x + 5, 0 < x < 2π
2
π 3π 5π 7π
4
,
4
,
4
Test intervals:
0 < x <
Sign of f ′:
f′ > 0
Conclusion:
,
4
π
π
4
4
Increasing
< x <
f′ < 0
Decreasing
3π
4
3π
5π
< x <
4
4
5π
7π
< x <
4
4
7π
< x < 2π
4
f′ > 0
f′ < 0
f′ > 0
Increasing
Decreasing
Increasing
⎛ π ⎞ ⎛ 3π 5π ⎞ ⎛ 7π
⎞
Increasing on: ⎜ 0, ⎟, ⎜ ,
⎟, ⎜ , 2π ⎟
⎝ 4⎠ ⎝ 4 4 ⎠ ⎝ 4
⎠
⎛ π 3π ⎞ ⎛ 5π 7π ⎞
Decreasing on: ⎜ ,
⎟, ⎜ ,
⎟
⎝4 4 ⎠ ⎝ 4 4 ⎠
⎛ π 11 ⎞ ⎛ 5π 11 ⎞
(b) Relative maxima: ⎜ , ⎟, ⎜ , ⎟
⎝4 2⎠ ⎝ 4 2⎠
(c)
7
⎛ 3π 9 ⎞ ⎛ 7π 9 ⎞
Relative minima: ⎜ , ⎟, ⎜ , ⎟
⎝ 4 2⎠ ⎝ 4 2⎠
2␲
0
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
304
Chapter 4
Applications of Differentiation
f ( x) = sin x + cos x,
59. (a)
0 < x < 2π
f ′( x) = cos x − sin x = 0 ⇒ sin x = cos x
Critical numbers: x =
π 5π
4
,
4
Test intervals:
0 < x <
Sign of f ′( x):
f′ > 0
Conclusion:
π
π
4
4
< x <
5π
4
f′ < 0
Increasing
Decreasing
5π
< x < 2π
4
f′ > 0
Increasing
⎛ π ⎞ ⎛ 5π
⎞
Increasing on: ⎜ 0, ⎟, ⎜ , 2π ⎟
⎝ 4⎠ ⎝ 4
⎠
⎛ π 5π ⎞
Decreasing on: ⎜ ,
⎟
⎝4 4 ⎠
⎛π
(b) Relative maximum: ⎜ ,
⎝4
⎞
2⎟
⎠
⎛ 5π
Relative minimum: ⎜ , −
⎝ 4
(c)
⎞
2⎟
⎠
3
2␲
0
−3
f ( x) = x + 2 sin x,
60. (a)
0 < x < 2π
f ′( x) = 1 + 2 cos x = 0 ⇒ cos x = −
Critical numbers:
2π 4π
,
3 3
Test intervals:
0 < x <
Sign of f ′( x):
f′ > 0
Conclusion:
1
2
2π
3
Increasing
2π
4π
< x <
3
3
4π
< x < 2π
3
f′ < 0
f′ > 0
Decreasing
Increasing
⎛ 2π ⎞ ⎛ 4π
⎞
Increasing on: ⎜ 0,
⎟, ⎜ , 2π ⎟
⎝ 3 ⎠ ⎝ 3
⎠
⎛ 2π 4π ⎞
Decreasing on: ⎜ ,
⎟
⎝ 3 3 ⎠
⎛ 2π 2π
(b) Relative maximum: ⎜ ,
+
⎝ 3 3
⎞ ⎛ 2π
⎞
3 ⎟ ≈ ⎜ , 3.826 ⎟
⎠ ⎝ 3
⎠
⎛ 4π 4π
−
Relative minimum: ⎜ ,
⎝ 3 3
⎞
⎛ 4π
⎞
3 ⎟ ≈ ⎜ , 2.457 ⎟
⎠
⎝ 3
⎠
(c)
7
0
2␲
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
61. (a)
f ( x) = cos 2 ( 2 x),
Increasing and Decreasing Functions and the First Derivative Test
305
0 < x < 2π
f ′( x) = − 4 cos 2 x sin 2 x = 0 ⇒ cos 2 x = 0 or sin 2 x = 0
Critical numbers: x =
π 3π 5π 7π π
4
,
4
,
Test intervals:
0 < x <
Sign of f ′( x):
f′ < 0
4
4
,
π
π
4
4
2
, π,
3π
2
< x <
π
π
2
2
f′ > 0
Conclusion:
Decreasing
Test intervals:
π < x <
Sign of f ′( x):
f′ < 0
Conclusion:
,
5π
4
Decreasing
< x <
3π
4
3π
< x < π
4
f′ < 0
Increasing
f′ > 0
Decreasing
Increasing
5π
3π
< x <
4
2
3π
7π
< x <
2
4
7π
< x < 2π
4
f′ > 0
f′ < 0
f′ > 0
Increasing
Decreasing
Increasing
⎛ π π ⎞ ⎛ 3π ⎞ ⎛ 5π 3π ⎞ ⎛ 7π
⎞
Increasing on: ⎜ , ⎟, ⎜ , π ⎟, ⎜ ,
⎟, ⎜ , 2π ⎟
4
2
4
4
2
4
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
⎛ π ⎞ ⎛ π 3π ⎞ ⎛ 5π ⎞ ⎛ 3π 7π ⎞
Decreasing on: ⎜ 0, ⎟, ⎜ ,
⎟, ⎜ π ,
⎟, ⎜ ,
⎟
4 ⎠ ⎝ 2 4 ⎠
⎝ 4⎠ ⎝2 4 ⎠ ⎝
⎛π ⎞
⎛ 3π ⎞
(b) Relative maxima: ⎜ , 1⎟, (π , 1), ⎜ , 1⎟
2
⎝
⎠
⎝ 2 ⎠
(c)
⎛ π ⎞ ⎛ 3π ⎞ ⎛ 5π ⎞ ⎛ 7π ⎞
Relative minima: ⎜ , 0 ⎟, ⎜ , 0 ⎟, ⎜ , 0 ⎟, ⎜ , 0 ⎟
⎝4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠ ⎝ 4 ⎠
3
2␲
0
−1
62. (a)
f ( x) = sin x −
3 cos x, 0 < x < 2π
f ′( x) = cos x +
3 sin x = 0 ⇒
tan x =
3 sin x = − cos x
−1
− 3
=
3
3
Critical numbers: x =
5π 11π
,
6 6
Test intervals:
0 < x <
Sign of f ′( x):
f′ > 0
Conclusion:
5π
6
Increasing
5π
11π
< x <
6
6
11π
< x < 2π
6
f′ < 0
f′ > 0
Decreasing
Increasing
⎛ 5π ⎞ ⎛ 11π
⎞
, 2π ⎟
Increasing on: ⎜ 0,
⎟, ⎜
⎝ 6 ⎠ ⎝ 6
⎠
⎛ 5π 11π ⎞
Decreasing on: ⎜ ,
⎟
⎝ 6 6 ⎠
⎛ 5π ⎞
(b) Relative maximum: ⎜ , 2 ⎟
⎝ 6 ⎠
⎛ 11π
⎞
, − 2⎟
Relative minimum: ⎜
⎝ 6
⎠
(c)
3
0
2␲
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
306
Chapter 4
Applications of Differentiation
f ( x) = sin 2 x + sin x,
63. (a)
0 < x < 2π
f ′( x) = 2 sin x cos x + cos x = cos x( 2 sin x + 1) = 0
Critical numbers: x =
π 7π 3π 11π
2
,
6
Test intervals:
0 < x <
Sign of f ′( x):
f′ > 0
Conclusion:
,
2
,
6
π
π
2
2
< x <
7π
6
f′ < 0
Increasing
Decreasing
7π
3π
< x <
6
2
3π
11π
< x <
2
6
11π
< x < 2π
6
f′ > 0
f′ < 0
f′ > 0
Increasing
Decreasing
Increasing
⎛ π ⎞ ⎛ 7π 3π ⎞ ⎛ 11π
⎞
, 2π ⎟
Increasing on: ⎜ 0, ⎟, ⎜ ,
⎟, ⎜
⎝ 2⎠ ⎝ 6 2 ⎠ ⎝ 6
⎠
⎛ π 7π ⎞ ⎛ 3π 11π ⎞
Decreasing on: ⎜ ,
⎟, ⎜ ,
⎟
⎝2 6 ⎠ ⎝ 2 6 ⎠
⎛ 7π 1 ⎞ ⎛ 11π 1 ⎞
(b) Relative minima: ⎜ , − ⎟, ⎜
,− ⎟
4⎠ ⎝ 6
4⎠
⎝ 6
⎛ π ⎞ ⎛ 3π ⎞
Relative maxima: ⎜ , 2 ⎟, ⎜ , 0 ⎟
⎝2 ⎠ ⎝ 2 ⎠
(c)
3
2␲
0
−1
f ( x) =
64. (a)
f ′( x) =
sin x
, 0 < x < 2π
1 + cos 2 x
cos x( 2 + sin 2 x)
(1 + cos2 x)
Critical numbers: x =
2
= 0
π 3π
2
,
2
Test intervals:
0 < x <
Sign of f ′( x ):
f′ > 0
Conclusion:
π
π
2
2
Increasing
< x <
f′ < 0
Decreasing
3π
2
3π
< x < 2π
2
f′ > 0
Increasing
⎛ π ⎞ ⎛ 3π
⎞
Increasing on: ⎜ 0, ⎟, ⎜ , 2π ⎟
2
2
⎝
⎠ ⎝
⎠
⎛ π 3π ⎞
Decreasing on: ⎜ ,
⎟
⎝2 2 ⎠
⎛π ⎞
(b) Relative maximum: ⎜ , 1⎟
⎝2 ⎠
⎛ 3π
⎞
Relative minimum: ⎜ , −1⎟
2
⎝
⎠
2
(c)
0
2␲
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
Increasing and Decreasing Functions and the First Derivative Test
67. f (t ) = t 2 sin t , [0, 2π ]
65. f ( x ) = 2 x 9 − x 2 , [−3, 3]
2(9 − 2 x 2 )
(a) f ′( x) =
(a) f ′(t ) = t 2 cos t + 2t sin t = t (t cos t + 2 sin t )
9 − x2
(b)
y
(b)
40
y
f′
30
f
10
8
f′
20
10
4
2
x
−1
1
π
2
− 10
2
−8
− 10
2(9 − 2 x
9− x
2
)
2
t = 0 or t = −2 tan t
= 0
t cot t = −2
t ≈ 2.2889, 5.0870 (graphing utility)
3
3 2
= ±
2
2
Critical numbers: t = 2.2889, 5.0870
(d) Intervals:
(d) Intervals:
⎛
3 2⎞
⎜⎜ −3,−
⎟
2 ⎟⎠
⎝
f ′( x) < 0
⎛ 3 2 3 2⎞
,
⎜⎜ −
⎟
2
2 ⎟⎠
⎝
f ′( x) > 0
⎛3 2 ⎞
, 3⎟⎟
⎜⎜
⎝ 2
⎠
f ′( x) < 0
Decreasing
Increasing
Decreasing
f is increasing when f ′ is positive and decreasing
when f ′ is negative.
)
(
x 2 − 3 x + 16 , [0, 5]
66. f ( x) = 10 5 −
5( 2 x − 3)
(a) f ′( x) = −
x 2 − 3 x + 16
(0, 2.2889) (2.2889, 5.0870) (5.0870, 2π )
f ′(t ) > 0
f ′(t ) < 0
f ′(t ) > 0
Increasing
68. f ( x) =
(b)
y
6
f
f
4
2
6
(c) −
1
1
x
− sin
2
2
2
8
12
f′
f′
π
x
−3
Increasing
x
x
+ cos , [0, 4π ]
2
2
(a) f ′( x) =
15
−1
Decreasing
f is increasing when f ′ is positive and decreasing
when f ′ is negative.
y
3
f
(c) t (t cos t + 2 sin t ) = 0
Critical numbers: x = ±
(b)
t
2π
− 20
(c)
307
1
3
2π
3π
4π
x
4
5( 2 x − 3)
x 2 − 3x + 16
(c)
= 0
Critical number: x =
3
2
(d) Intervals:
⎛ 3⎞
⎛3 ⎞
⎜ 0, ⎟
⎜ , 5⎟
2
⎝
⎠
⎝2 ⎠
f ′( x) > 0 f ′( x) < 0
1
1
x
− sin = 0
2
2
2
x
sin = 1
2
π
x
=
2
2
Critical number: x = π
(d) Intervals:
(0, π )
f ′( x) >
0
(π , 4π )
f ′( x) >
0
Increasing Decreasing
Increasing Increasing
f is increasing when f ′ is positive and decreasing
when f ′ is negative.
f is increasing when f ′ is positive.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
308
Chapter 4
69. (a)
A pplications of Differentiation
x
f ( x) = −3 sin , [0, 6π ]
3
x
f ′( x) = −cos
3
(b)
(c) Critical numbers: x =
(d) Intervals:
⎛ 3π ⎞
⎜ 0,
⎟
⎝ 2 ⎠
f′ < 0
y
4
f
2
x
4π
⎛ 3π 9π ⎞
⎜ ,
⎟
⎝ 2 2 ⎠
f′ > 0
⎛ 9π
⎞
⎜ , 6π ⎟
⎝ 2
⎠
′
f < 0
Decreasing Increasing Decreasing
f is increasing when f ′ is positive and decreasing
when f ′ is negative.
f′
2π
3π 9π
,
2 2
−2
−4
70. (a)
f ( x) = 2 sin 3 x + 4 cos 3 x, [0, π ]
(c) f ′( x ) = 0 ⇒ tan 3 x =
f ′( x) = 6 cos 3 x − 12 sin 3x
(b)
Critical numbers: x ≈ 0.1545, 1.2017, 2.2489
(d) Intervals:
y
12
(0, 0.1545) (0.1545, 1.2017) (1.2017, 2.2489) ( 2.2489, π )
f'
f′ > 0
8
f
4
−8
(a) f ′( x) = ( 4 − 2 x − x 2 )e x
2x2 − 1
2x
y
(b)
10
y
f
x
4
1
f
3
2
−10
f′
2
−20
f
1
−30
x
−1
f′ < 0
72. f ( x ) = ( 4 − x 2 )e x , [0, 2]
1 2
( x − ln x), (0, 3]
2
(b)
f′ > 0
Decreasing
Increasing
Decreasing
f is increasing when f ′ is positive and decreasing when f ′ is
negative.
−12
(a) f ′( x) =
f′ < 0
Increasing
x
π
−4
71. f ( x ) =
1
2
1
2
3
4
−1
(c)
2x − 1
= 0
2x
( 4 − 2 x − x 2 )e x
= 0
2
(c)
Critical number: x =
⎛
2⎞
(d) Intervals: ⎜⎜ 0,
⎟
2 ⎟⎠
⎝
f ′( x) < 0
Decreasing
Critical number: x ≈ 1.2361
1
=
2
2
2
⎛ 2 ⎞
, 3⎟⎟
⎜⎜
⎝ 2
⎠
f ′( x) > 0
(d) Intervals: (0, 1.2361)
( x = −1 +
f ′( x) > 0
(1.2361, 2)
f ′( x) < 0
Increasing
Decreasing
5
)
(e) f is increasing when f ′ is positive, and decreasing
when f ′ is negative.
Increasing
(e) f is increasing when f ′ is positive, and decreasing
when f ′ is negative.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
73. f ( x ) =
(
Increasing and Decreasing Functions and the First Derivative Test
)(
)
x 2 − 1 x3 − 3x
x5 − 4 x3 + 3x
=
= x 3 − 3 x, x ≠ ± 1
x2 − 1
x2 − 1
y
f ( x ) = g ( x) = x3 − 3 x for all x ≠ ± 1.
(
−4 −3
3, 0
x
−1
1 2 3 4 5
−2
−3
−4
−5
f symmetric about origin
(
5
4
3
(−1, 2)
)
f ′( x) = 3 x 2 − 3 = 3 x 2 − 1 , x ≠ ±1 ⇒ f ′( x ) ≠ 0
zeros of f : (0, 0), ±
309
)
(1, −2)
g ( x) is continuous on ( −∞, ∞) and f ( x) has holes at ( −1, 2) and (1, − 2).
74.
f (t ) = cos 2 t − sin 2 t = 1 − 2 sin 2 t = g (t )
77. f is quadratic ⇒ f ′ is a line.
f ′(t ) = −4 sin t cos t = −2 sin 2t
y
4
f symmetric with respect to y-axis
zeros of f : ±
π
f′
2
4
−4
y
2
4
−2
Relative maximum: (0, 1)
⎛ π
⎞ ⎛π
⎞
Relative minimum: ⎜ − , −1⎟, ⎜ , −1⎟
⎝ 2
⎠ ⎝2
⎠
x
−2
−4
78. f is a 4th degree polynomial ⇒ f ′ is a cubic polynomial.
y
2
6
f′
1
−π
π
x
−1
x
−6 −4 −2
2
6
4
−2
The graphs of f ( x ) and g ( x) are the same.
75. f ( x) = c is constant ⇒ f ′( x) = 0.
79. f has positive, but decreasing slope.
y
y
4
4
2
f′
2
f′
−4
−2
2
x
−4
x
−2
4
2
4
−2
−2
−4
−4
76. f ( x) is a line of slope ≈ 2 ⇒ f ′( x) = 2.
80. f has positive slope.
y
4
6
3
2
−6
f′
6
x
−3 −2 −1
1
2
3
−2
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
310
Chapter 4
Applications of Differentiation
In Exercises 81 – 86, f ′( x ) > 0 on ( −∞ , − 4), f ′( x ) < 0 on
(– 4, 6) and f ′( x ) > 0 on (6, ∞ ).
81. g ( x) = f ( x ) + 5
< 0 on ( − ∞, 2))
(c) f has a relative minimum at x = 2.
g ′( x ) = 3 f ′( x )
(ii) (a) Critical numbers:
x = 0, 1 ( Because f ′(1) = 0)
g ′( −5) = 3 f ′( −5) > 0
g ( x) = − f ( x)
(b) f increasing on ( − ∞, 0) and (1, ∞)
g ′( x) = − f ′( x )
(Because f ′
g ′( −6) = − f ′( −6) < 0
> 0 on these intervals)
f decreasing on
84. g ( x) = − f ( x)
(0, 1) (Because f ′
g ′( x) = − f ′( x )
< 0 on (0, 1))
(c) f has a relative maximum at x = 0, and a
relative minimum at x = 1.
g ′(0) = − f ′(0) > 0
85. g ( x ) = f ( x − 10)
(iii) (a) Critical numbers: x = −1, 0, 1
g ′( x ) = f ′( x − 10)
(Because f ′(−1)
g ′(0) = f ′( −10) > 0
= f ′(0) = f ′(1) = 0)
(b) f increasing on ( − ∞, −1) and (0, 1)
86. g ( x) = f ( x − 10)
(Because f ′
g ′( x) = f ′( x − 10)
> 0 on these intervals)
f decreasing on ( −1, 0) and (1, ∞)
g ′(8) = f ′( −2) < 0
87. No. f does have a horizontal tangent line at x = c, but
f could be increasing (or decreasing) on both sides of the
point. For example, f ( x ) = x 3 at x = 0.
88. Yes. An example is f ( x ) = e − x , f ′( x) = − e − x .
⎧> 0,
x < 4 ⇒ f is increasing on ( −∞, 4).
⎪
′
89. f ( x)⎨undefined, x = 4
⎪< 0,
x > 4 ⇒ f is decreasing on ( 4 ∞).
⎩
Two possibilities for f ( x) are given below.
y
(Because f ′
< 0 on these intervals)
(c) f has a relative maximum at x = −1 and
x = 1. f has a relative minimum at x = 0.
(iv) (a) Critical numbers: x = − 3, 1, 5
(Because f ′(− 3)
= f ′(1) = f ′( s ) = 0)
(b) f increasing on ( − 3, 1) and (1, 5)
(Because f ′
> 0 on these intervals). In fact,
f is increasing on ( − 3, 5).
f decreasing on ( − ∞, − 3) and (5, ∞)
(Because f ′
6
< 0 on these intervals)
(c) f has a relative minimum at x = − 3, and a
4
relative maximum at x = 5.
2
x
2
6
8
−2
(b)
> 0 on ( 2, ∞ ))
(− ∞, 2) (Because f ′
g ( x) = 3 f ( x) − 3
(a)
(b) f increasing on
f decreasing on
g ′(0) = f ′(0) < 0
83.
(i) (a) Critical number: x = 2 ( Because f ′( 2) = 0)
(2, ∞) (Because f ′
g ′( x) = f ′( x )
82.
90.
x = 1 is not a relative extremum.
91. Critical number: x = 5
f ′( 4) = −2.5 ⇒ f is decreasing at x = 4.
y
f ′(6) = 3 ⇒ f is increasing at x = 6.
2
1
x
1
3
4
5
(5, f (5)) is a relative minimum.
−1
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
Increasing and Decreasing Functions and the First Derivative Test
311
92. Critical number: x = 2
f ′(1) = 2 ⇒ f is decreasing at x = 1.
f ′(3) = 6 ⇒ f is increasing at x = 3.
(2, f (2)) is not a relative extremum.
In Exercises 93 and 94, answers will vary.
Sample answers:
93. (a)
94. (a)
y
y
2
1
f
1
x
−1
π
2
1
x
−2
−1
(b) The critical numbers are in intervals ( −0.50, − 0.25)
(b) The critical numbers are in the intervals
⎛ π ⎞ ⎛π π ⎞
⎛ 3π 5π ⎞
⎜ 0, ⎟, ⎜ , ⎟, and ⎜ ,
⎟ because the sign of
⎝ 6⎠ ⎝3 2⎠
⎝ 4 6 ⎠
f ′ changes in these intervals. f is increasing on
and (0.25, 0.50) because the sign of f ′ changes in
these intervals. f is decreasing on approximately
(−1, − 0.40), (0.48, 1), and increasing on
(−0.40, 0.48).
⎛ π⎞
⎛ 3π 6π ⎞
approximately ⎜ 0, ⎟ and ⎜ ,
⎟ and decreasing
⎝ 7⎠
⎝ 7 7 ⎠
⎛ π 3π ⎞
⎛ 6π ⎞
on ⎜ ,
⎟ and ⎜ , π ⎟.
7
7
⎝
⎠
⎝ 7
⎠
(c) Relative minimum when x ≈ −0.40: ( −0.40, 0.75)
Relative maximum when x ≈ 0.48: (0.48, 1.25)
(c) Relative minima when x ≈
Relative maxima when x ≈
3π
,π
7
π 6π
7
,
7
95. s(t ) = 4.9(sin θ )t 2
(a) s′(t ) = 4.9(sin θ )( 2t ) = 9.8(sin θ )t
speed = s′(t ) = 9.8(sin θ )t
(b)
θ
s′(t )
0
0
π
π
π
4
3
2
4.9 2t
4.9 3t
The speed is maximum for θ =
π
2
9.8t
2π
3
3π
4
π
4.9 3t
4.9 2t
0
.
96. (a) M = − 0.06803t 4 + 3.7162t 3 − 76.281t 2 + 716.56t − 2393.0
(b)
350
9
100
20
(c) Using a graphing utility, the maximum is approximately (17.7, 322.0), which compares well with the actual maximum in
2007: (17, 326.0).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
312
Chapter 4
Applications of Differentiation
3t
,t ≥ 0
27 + t 3
97. C =
(a)
t
0
0.5
1
1.5
2
2.5
3
C(t)
0
0.055
0.107
0.148
0.171
0.176
0.167
The concentration seems greatest near t = 2.5 hours.
(b)
0.25
0
3
0
The concentration is greatest when t ≈ 2.38 hours.
(27 + t 3 )(3) − (3t )(3t 2 )
2
(27 + t 3 )
(c) C ′ =
C ′ = 0 when t = 3
3
=
3( 27 − 2t 3 )
(27 + t 3 )
2
2 ≈ 2.38 hours.
By the First Derivative Test, this is a maximum.
98. f ( x) = x, g ( x) = sin x, 0 < x < π
(a)
x
0.5
1
1.5
2
2.5
3
f(x)
0.5
1
1.5
2
2.5
3
g(x)
0.479
0.841
0.997
0.909
0.598
0.141
f ( x) seems greater than g ( x) on (0, π ).
(b)
5
f
g
␲
0
−2
x > sin x on (0, π ) so, f ( x) > g ( x).
(c) Let h( x) = f ( x ) − g ( x) = x − sin x
h′( x ) = 1 − cos x > 0 on (0, π ).
Therefore, h( x) is increasing on (0, π ). Because h(0) = 0 and h′( x) > 0 on (0, π ),
h( x ) > 0
x − sin x > 0
x > sin x
f ( x) > g ( x ) on (0, π )
99. v = k ( R − r )r 2 = k ( Rr 2 − r 3 )
v′ = k ( 2 Rr − 3r 2 )
= kr ( 2 R − 3r ) = 0
r = 0 or 23 R
Maximum when r =
2 R.
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
100. R =
101. (a) s(t ) = 6t − t 2 , t ≥ 0
0.001T 4 − 4T + 100
(a) R′ =
0.004T 3 − 4
2 0.001T − 4T + 100
4
v(t ) = 6 − 2t
= 0
(b) v(t ) = 0 when t = 3.
Critical number: T = 10°
Moving in positive direction for 0 ≤ t < 3 because
v(t ) > 0 on 0 ≤ t < 3.
Minimum resistance: R ≈ 8.3666 ohms
125
(b)
313
Increasing and Decreasing Functions and the First Derivative Test
(c) Moving in negative direction when t > 3.
(d) The particle changes direction at t = 3.
− 100
102. (a) s(t ) = t 2 − 7t + 10, t ≥ 0
100
v(t ) = 2t − 7
− 25
The minimum resistance is approximately
R ≈ 8.37 ohms at T = 10°.
(b) v(t ) = 0 when t =
7
2
Particle moving in positive direction for
t >
7
2
because v′(t ) > 0 on
( 72 , ∞).
(c) Particle moving in negative direction on ⎡⎣0,
(d) The particle changes direction at t =
7
2
).
7
.
2
103. (a) s(t ) = t 3 − 5t 2 + 4t , t ≥ 0
v(t ) = 3t 2 − 10t + 4
(b) v(t ) = 0 for t =
10 ± 100 − 48
5 ± 13
=
6
3
Particle is moving in a positive direction on
⎡ 5 − 13 ⎞
⎟⎟ ≈ [0, 0.4648) and
⎢0,
3
⎢⎣
⎠
⎛ 5 + 13 ⎞
, ∞ ⎟⎟ ≈ ( 2.8685, ∞) because v > 0 on these intervals.
⎜⎜
3
⎝
⎠
(c) Particle is moving in a negative direction on
⎛ 5 − 13 5 + 13 ⎞
,
⎜⎜
⎟⎟ ≈ (0.4648, 2.8685)
3
3
⎝
⎠
(d) The particle changes direction at t =
5 ± 13
.
3
104. (a) s(t ) = t 3 − 20t 2 + 128t − 280
105. Answers will vary.
v(t ) = 3t 2 − 40t + 128
106. Answers will vary.
(b) v(t ) = (3t − 16)(t − 8)
v(t ) = 0 when t =
16
,8
3
)
v(t ) < 0 for ( 16
, 8)
3
v(t ) > 0 for ⎡⎣0, 16
and (8, ∞)
3
(c)
(d) The particle changes direction at t =
16
3
and 8.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
314
Chapter 4
Applications of Differentiation
107. (a) Use a cubic polynomial
(d)
4
f ( x ) = a3 x + a2 x + a1 x + a0
3
(2, 2)
2
−2
(b) f ′( x) = 3a3 x 2 + 2a2 x + a1.
f (0) = 0: a3 (0) + a2 (0) + a1(0) + a0 = 0 ⇒
a0 = 0
f ′(0) = 0:
3a3 (0) + 2a2 (0) + a1 = 0 ⇒
a1 = 0
f ( 2) = 2: a3 ( 2) + a2 ( 2) + a1( 2) + a0 = 2 ⇒
8a3 + 4a2 = 2
3
2
2
3
2
f ′( 2) = 0:
4
(0, 0)
−4
3a3 ( 2) + 2a2 ( 2) + a1 = 0 ⇒ 12a3 + 4a2 = 0
2
(c) The solution is a0 = a1 = 0, a2 =
f ( x ) = − 12 x 3 +
3
,a
2 3
= − 12 :
3 2
x .
2
108. (a) Use a cubic polynomial
(d)
1200
(4, 1000)
f ( x ) = 3a3 x + a2 x + a1 x + a0
3
2
(b) f ′( x) = 3a3 x 2 + 2a2 x + a1
−3
f (0) = 0:
a3 (0) + a2 (0) + a1(0) + a0 = 0 ⇒
a0 = 0
f ′(0) = 0:
3a3 (0) + 2a2 (0) + a1 = 0 ⇒
a1 = 0
3
2
2
(0, 0)
8
−400
f ( 4) = 1000: a3 ( 4) + a2 ( 4) + a1( 4) + a0 = 1000 ⇒ 64a3 + 16a2 = 100
3
2
f ′( 4) = 0:
3a3 ( 4) + 2a2 ( 4) + a1 = 0 ⇒
2
(c) The solution is a0 = a1 = 0, a2 =
f ( x ) = − 125
x3 +
4
375
, a3
2
48a3 + 8a2 = 0
= − 125
4
375 2
x .
2
109. (a) Use a fourth degree polynomial
f ( x ) = a4 x 4 + a3 x 3 + a2 x 2 + a1 x + a0 .
(b) f ′( x) = 4a4 x 3 + 3a3 x 2 + 2a 2 x + a1
f (0) = 0: a4 (0) + a3 (0) + a2 (0) + a1(0) + a0 = 0 ⇒
a0 = 0
f ′(0) = 0:
a1 = 0
4
3
2
4a4 (0) + 3a3 (0) + 2a2 (0) + a1 = 0 ⇒
3
2
f ( 4) = 0: a4 ( 4) + a3 ( 4) + a2 ( 4) + a1( 4) + a0 = 0 ⇒ 256a4 + 64a3 + 16a2 = 0
4
f ′( 4) = 0:
3
2
4a4 ( 4) + 3a3 ( 4) + 2a2 ( 4) + a1 = 0 ⇒
256a4 + 48a3 + 8a2 = 0
f ( 2) = 4: a4 ( 2) + a3 ( 2) + a2 ( 2) + a1( 2) + a0 = 4 ⇒
16a4 + 8a3 + 4a2 = 4
3
4
f ′( 2) = 0:
2
3
2
4a4 ( 2) + 3a3 ( 2) + 2a2 ( 2) + a1 = 0 ⇒
3
2
(c) The solution is a0 = a1 = 0, a2 = 4, a3 = −2,
f ( x) =
(d)
1 x4
4
a4 =
32a4 + 12a3 + 4a2 = 0
1.
4
− 2 x3 + 4 x 2
5
(2, 4)
−2
(0, 0)
(4, 0)
5
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.3
Increasing and Decreasing Functions and the First Derivative Test
315
110. (a) Use a fourth-degree polynomial
f ( x ) = a4 x 4 + a3 x 3 + a2 x 2 + a1 x + a0 .
(b) f ′( x) = 4a4 x 3 + 3a3 x 2 + 2a 2 x + a1
f (1) = 2:
a4 (1) + a3 (1) + a2 (1) + a1(1) + a0 = 2 ⇒
a4 + a3 + a2 + a1 + a0 = 2
f ′(1) = 0:
4a4 (1) + 3a3 (1) + 2a2 (1) + a1 = 0 ⇒
4a4 + 3a3 + 2a2 + a1 = 0
f (−1) = 4: a4 ( −1) + a3 ( −1) + a2 ( −1) + a1(−1) + a0 = 4 ⇒
a4 − a3 + a2 − a1 + a0 = 4
f ′( −1) = 0:
− 4a4 + 3a3 − 2a2 + a1 = 0
4
3
3
4
2
3
2
4a4 (−1) + 3a3 ( −1) + 2a2 ( −1) + a1 = 0 ⇒
3
f (3) = 4:
2
a4 (3) + a3 (3) + a2 (3) + a1(3) + a0 = 4 ⇒ 81a4 + 27 a3 + 9a2 + a1 + a0 = 4
4
f ′(3) = 0:
3
2
4a4 (3) + 3a3 (3) + 2a2 (3) + a1 = 0 ⇒
3
(c) The solution is a0 =
f ( x ) = − 18 x 4 +
(d)
2
23
, a1
8
1 x3
2
+
1 x2
4
2
= − 32 , a2 =
−
3
x
2
+
1,
4
a3 =
1,
2
108a4 + 27 a3 + 6a2 + a1 = 0
a4 = − 18
23
.
8
6
(−1 , 4)
(3, 4)
(1, 2)
−4
6
−2
111. True.
Let h( x) = f ( x) + g ( x) where f and g are increasing.
Then h′( x ) = f ′( x) + g ′( x) > 0 because
f ′( x) > 0 and g ′( x) > 0.
117. Assume that f ′( x) < 0 for all x in the interval (a, b) and
let x1 < x2 be any two points in the interval. By the
Mean Value Theorem, you know there exists a number c
such that x1 < c < x2 , and
f ′(c ) =
112. False.
Let h( x) = f ( x) g ( x) where f ( x) = g ( x) = x. Then
h( x) = x is decreasing on ( −∞, 0).
2
f ( x2 ) − f ( x1 )
x2 − x1
Because f ′(c) < 0 and x2 − x1 > 0, then
f ( x2 ) − f ( x1 ) < 0, which implies that
f ( x2 ) < f ( x1 ). So, f is decreasing on the interval.
113. False.
Let f ( x ) = x 3 , then f ′( x) = 3 x 2 and f only has one
critical number. Or, let f ( x ) = x + 3 x + 1, then
3
f ′( x) = 3( x + 1) has no critical numbers.
2
114. True.
If f ( x ) is an nth-degree polynomial, then the degree of
f ′( x ) is n − 1.
115. False. For example, f ( x ) = x 3 does not have a relative
extrema at the critical number x = 0.
116. False. The function might not be continuous on the
interval.
118. Suppose f ′( x) changes from positive to negative at c.
Then there exists a and b in I such that f ′( x) > 0 for all
x in (a, c) and f ′( x) < 0 for all x in (c, b). By Theorem
4.5, f is increasing on (a, c) and decreasing on (c, b).
Therefore, f (c) is a maximum of f on (a, b) and so, a
relative maximum of f.
119. Let f ( x ) = (1 + x ) − nx − 1. Then
n
f ′( x) = n(1 + x)
− n = n ⎡(1 + x)
⎣
because x > 0 and n > 1.
n −1
n −1
− 1⎤ > 0
⎦
So, f ( x ) is increasing on (0, ∞). Because
f (0) = 0 ⇒ f ( x) > 0 on (0, ∞)
(1 + x)n
− nx − 1 > 0 ⇒ (1 + x) > 1 + nx.
n
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
316
Chapter 4
Applications of Differentiation
120. Let x1 and x2 be two real numbers, x1 < x2 . Then
x13
122. f ( x) = axebx
< x2 ⇒ f ( x1 ) < f ( x2 ). So f is increasing on
3
2
f ′( x) = ax( 2bx)ebx + aebx = aebx (1 + 2bx 2 )
2
(−∞, ∞).
2
2
f ( 4) = 2: 2 = 4ae16b ⇒ 2a =
121. Let x1 and x2 be two positive real numbers,
0 < x1 < x2 . Then
1
1
⇒ a = e −16b
2
e16b
Relative maximum at x = 4:
f ′( 4) = 0 ⇒ 1 + 2b(16) = 0 ⇒ b = −
1
1
>
x1
x2
f ( x1 ) > f ( x2 )
1 12
e =
2
So, a =
So, f is decreasing on (0, ∞).
e − x2
xe
2
f ( x) =
1
32
e
,
2
32
.
Notice the f is increasing on (0, 4) and decreasing on
(4, ∞),
so ( 4, 2) is a relative maximum.
123. First observe that
tan x + cot x + sec x + csc x =
sin x
cos x
1
1
+
+
+
cos x
sin x
cos x sin x
=
sin 2 x + cos 2 x + sin x + cos x
sin x cos x
=
1 + sin x + cos x ⎛ sin x + cos x − 1 ⎞
⎜
⎟
sin x cos x ⎝ sin x + cos x − 1 ⎠
=
(sin x
+ cos x) − 1
sin x cos x(sin x + cos x − 1)
=
2 sin x cos x
sin x cos x(sin x + cos x − 1)
=
2
sin x + cos x − 1
2
Let t = sin x + cos x − 1. The expression inside the absolute value sign is
f (t ) = sin x + cos x +
2
2
2
= (sin x + cos x − 1) + 1 +
= t +1+
t
sin x + cos x − 1
sin x + cos x − 1
π⎞
π
π
⎛
Because sin ⎜ x + ⎟ = sin x cos + cos x sin =
4⎠
4
4
⎝
2, 2 ⎤⎦ and t = sin x + cos x − 1 ∈ ⎡⎣−1 −
sin x + cos x ∈ ⎡⎣−
f ′(t ) = 1 −
(
f −1 +
)
(
)(
t +
t2 − 2
2
=
=
2
2
t
t
2 = −1 +
=
2
(sin x + cos x),
2
2
2 +1+
−1 +
4 − 2⎛
⎜
2 − 1 ⎜⎝
2 t −
t
2
2
)
2
=
2 +
2 + 1⎞
4 2 − 2 + 4−
⎟ =
1
2 + 1 ⎟⎠
(
For t > 0, f is decreasing and f (t ) > f −1 +
(
For t < 0, f is increasing on −
2 − 1, −
2 ⎤⎦.
2, −1 +
)
2
2 −1
2
= 2+ 3 2
)
2 = 2+3 2
(
2 , then decreasing on −
)
(
2, 0 . So f (t ) < f −
)
2 = 1 − 2 2.
Finally, f (t ) ≥ 2 2 − 1.
(You can verify this easily with a graphing utility.)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
Concavity and the Second Derivative Test
317
Section 4.4 Concavity and the Second Derivative Test
1.
y = x2 − x − 2
y′ = 2 x − 1
y′′ = 2
y′′ > 0 for all x.
Concave upward: ( −∞, ∞)
2.
g ( x) = 3x 2 − x3
g ′( x ) = 6 x − 3 x 2
g ′′( x ) = 6 − 6 x
g ′′( x ) = 0 when x = 1.
Intervals:
−∞ < x < 1
1< x < ∞
Sign of g ′′ :
g ′′ > 0
g ′′ < 0
Conclusion:
Concave upward
Concave downward
Concave upward: ( −∞, 1)
Concave downward: (1, ∞)
3.
f ( x) = − x3 + 6 x 2 − 9 x − 1
f ′( x) = −3x 2 + 12 x − 9
f ′′( x) = −6 x + 12 = −6( x − 2)
Intervals:
−∞ < x < 2
2 < x < ∞
Sign of f ′′ :
f ′′ > 0
f ′′ < 0
Conclusion:
f ′′( x) = 0 when x = 2.
Concave upward
Concave downward
Concave upward: ( −∞, 2)
Concave downward: ( 2, ∞)
4.
h( x ) = x 5 − 5 x + 2
h′( x) = 5 x 4 − 5
h′′( x) = 20 x 3
h′′( x) = 0 when x = 0.
Intervals:
−∞ < x < 0
0 < x < ∞
Sign of h′′ :
h′′ < 0
h′′ > 0
Conclusion:
Concave downward
Concave upward
Concave upward: (0, ∞)
Concave downward: ( −∞, 0)
5.
f ( x) =
f ′( x) =
f ′′( x) =
24
x 2 + 12
− 48 x
( x2
+ 12)
2
−144( 4 − x 2 )
( x2
+ 12)
Intervals:
−∞ < x < − 2
−2 < x < 2
2 < x < ∞
Sign of f ′′ :
f ′′ > 0
f ′′ < 0
f ′′ > 0
Conclusion:
Concave upward
Concave downward
Concave upward
3
f ′′( x) = 0 when x = ± 2.
Concave upward: ( −∞, − 2), ( 2, ∞ )
Concave downward: ( −2, 2)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
318
6.
Chapter 4
Applications of Differentiation
2x2
3x 2 + 1
4x
f ( x) =
f ′( x) =
(3x 2
+ 1)
2
− 4(3 x − 1)(3 x + 1)
f ′′( x) =
(3x 2
+ 1)
3
1
f ′′( x) = 0 when x = ± .
3
Intervals:
−∞ < x < − 13
− 13 < x <
Sign of f ′′ :
f ′′ < 0
f ′′ > 0
Conclusion:
Concave downward
1
3
Concave upward
1
3
< x < ∞
f ′′ < 0
Concave downward
⎛ 1 1⎞
Concave upward: ⎜ − , ⎟
⎝ 3 3⎠
1 ⎞⎛ 1 ⎞
⎛
Concave downward: ⎜ − ∞, − ⎟⎜ , ∞ ⎟
3 ⎠⎝ 3 ⎠
⎝
7. f ( x) =
f′ =
f ′′ =
x2 + 1
x2 − 1
−4 x
( x2 − 1)
4(3 x 2 + 1)
3
( x 2 − 1)
2
f is not continuous at x = ± 1.
Intervals:
− ∞ < x < −1
−1 < x < 1
1< x < ∞
Sign of f ′′ :
f ′′ > 0
f ′′ < 0
f ′′ > 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave upward: ( −∞, −1), (1, ∞ )
Concave downward: ( −1, 1)
8.
y =
y′ =
(−3x5 + 40 x3 + 135 x)
1 −15 x 4 + 120 x 2 + 135
)
270 (
1
270
y′′ = − 92 x( x − 2)( x + 2)
y′′ = 0 when x = 0, ± 2.
Intervals:
−∞ < x < −2
−2 < x < 0
0 < x < 2
2 < x < ∞
Sign of y′′ :
y′′ > 0
y′′ < 0
y′′ > 0
y′′ < 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave downward
Concave upward: ( −∞, − 2), (0, 2)
Concave downward: ( −2, 0), ( 2, ∞)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
9.
g ( x) =
g ′( x ) =
Concavity and the Second Derivative Test
319
x2 + 4
4 − x2
16 x
(4 − x 2 )
16(3x 2 + 4)
g ′′( x ) =
3
(4 − x 2 )
2
=
16(3x 2 + 4)
(2
− x) ( 2 + x)
3
3
f is not continuous at x = ± 2.
Intervals:
−∞ < x < −2
−2 < x < 2
2 < x < ∞
Sign of g ′′ :
g ′′ < 0
g ′′ > 0
g ′′ < 0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward: ( −2, 2)
Concave downward: ( −∞, − 2), ( 2, ∞)
10.
h( x ) =
h′( x) =
h′′( x) =
x2 − 1
2x − 1
2( x − x + 1)
2
(2 x
− 1)
2
−6
(2 x
− 1)
Intervals:
−∞ < x <
Sign of h′′ :
h′′ > 0
h′′ < 0
Conclusion:
Concave upward
Concave downward
1
2
1
2
< x < ∞
3
f ′′ is not continuous at x =
1
.
2
1⎞
⎛
Concave upward: ⎜ −∞, ⎟
2⎠
⎝
⎛1 ⎞
Concave downward: ⎜ , ∞ ⎟
⎝2 ⎠
11.
⎛ π π⎞
y = 2 x − tan x, ⎜ − , ⎟
⎝ 2 2⎠
y′ = 2 − sec x
y′′ = −2 sec 2 x tan x
y′′ = 0 when x = 0.
π
Intervals:
−
Sign of y′′ :
y′′ > 0
2
Conclusion:
2
< x < 0
Concave upward
0 < x <
π
2
y′′ < 0
Concave downward
⎛ π ⎞
Concave upward: ⎜ − , 0 ⎟
⎝ 2 ⎠
⎛ π⎞
Concave downward: ⎜ 0, ⎟
⎝ 2⎠
12.
y = x + 2 csc x,
(−π , π )
y′ = 1 − 2 csc x cot x
y′′ = −2 csc x( −csc 2 x) − 2 cot x(−csc x cot x)
= 2(csc3 x + csc x cot 2 x)
Intervals:
−π < x < 0
0 < x < π
Sign of y′′ :
y′′ < 0
y′′ > 0
Conclusion:
Concave downward
Concave upward
y′′ = 0 when x = 0.
Concave upward: (0, π )
Concave downward: ( −π , 0)
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320
13.
Chapter 4
Applications of Differentiation
f ( x) = x3 − 6 x 2 + 12 x
f ′( x) = 3 x 2 − 12 x + 12
f ′′( x) = 6( x − 2) = 0 when x = 2.
Intervals:
−∞ < x < 2
2 < x < ∞
Sign of f ′′ :
f ′′ < 0
f ′′ > 0
Conclusion:
Concave upward: ( 2, ∞)
Concave downward
Concave upward
Concave downward: ( −∞, 2)
Point of inflection: ( 2, 8)
14.
f ( x) = − x3 + 6 x 2 − 5
f ′( x) = − 3 x 2 + 12 x
f ′′( x) = − 6 x + 12 = − 6( x − 2) = 0 when x = 2.
Intervals:
−∞ < x < 2
2 < x < ∞
Sign of f ′′ :
f ′′ > 0
f ′′ < 0
Conclusion:
Concave upward: ( − ∞, 2)
Concave upward
Concave downward
Concave downward: ( 2, ∞)
Point of inflection: ( 2, 11)
15.
f ( x) =
1 4
x
2
+ 2 x3
Intervals:
−∞ < x < −2
−2 < x < 2
0 < x < ∞
f ′′( x) = 6 x 2 + 12 x = 6 x( x + 2)
Sign of f ′′ :
f ′′ > 0
f ′′ < 0
f ′′ > 0
f ′′( x) = 0 when x = 0, − 2
Conclusion:
Concave upward
f ′( x) = −1 − 12 x3
Intervals:
−∞ < x < 0
0 < x < ∞
f ′′( x) = − 36 x 2 = 0 when x = 0.
Sign of f ′′ :
f ′′ < 0
f ′′ < 0
f ′( x) = 2 x3 + 6 x 2
Concave downward
Concave upward
Concave upward: ( −∞, − 2), (0, ∞)
Concave downward: ( −2, 0)
Points of inflection: ( −2, −8) and (0, 0)
16.
f ( x) = 4 − x − 3 x 4
Concave downward: ( − ∞, ∞)
Conclusion:
Concave downward
Concave downward
No points of inflection
17.
f ( x ) = x ( x − 4)
3
2
3
2
f ′( x) = x ⎡3( x − 4) ⎤ + ( x − 4) = ( x − 4) ( 4 x − 4)
⎣
⎦
f ′′( x) = 4( x − 1)⎡⎣2( x − 4)⎤⎦ + 4( x − 4) = 4( x − 4)⎡⎣2( x − 1) + ( x − 4)⎤⎦ = 4( x − 4)(3 x − 6) = 12( x − 4)( x − 2)
2
f ′′( x) = 12( x − 4)( x − 2) = 0 when x = 2, 4.
Intervals:
−∞ < x < 2
2 < x < 4
4 < x < ∞
Sign of f ′′( x):
f ′′( x) > 0
f ′′( x) < 0
f ′′( x) > 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave upward: ( − ∞, 2), ( 4, ∞)
Concave downward: ( 2, 4)
Points of inflection: ( 2, −16), ( 4, 0)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
18.
Concavity and the Second Derivative Test
321
f ( x) = ( x − 2) ( x − 1)
3
f ′( x) = ( x − 2) ( 4 x − 5)
2
f ′′( x) = 6( x − 2)( 2 x − 3)
f ′′( x) = 0 when x =
3
, 2.
2
Intervals:
−∞ < x <
Sign of f ′′:
f ′′ > 0
Conclusion:
3
2
< x < 2
3
2
f ′′ < 0
Concave upward
Concave downward
2 < x < ∞
f ′′ > 0
Concave upward
3⎞
⎛
Concave upward: ⎜ −∞, ⎟, ( 2, ∞)
2⎠
⎝
⎛3
Concave downward: ⎜ ,
⎝2
⎞
2⎟
⎠
1⎞
⎛3
Points of inflection: ⎜ , − ⎟, ( 2, 0)
16 ⎠
⎝2
19.
x + 3, Domain: [−3, ∞)
f ( x) = x
−1 2
⎛1⎞
f ′( x) = x⎜ ⎟( x + 3)
+
⎝ 2⎠
f ′′( x) =
=
6
20.
x +3 =
x + 3 − 3( x + 2)( x + 3)
3( x + 2)
2
x + 3
−1 2
4( x + 3)
3( x + 4)
4( x + 3)
32
= 0 when x = − 4.
f ( x) = x 9 − x , Domain: x ≤ 9
f ′( x) =
f ′′( x) =
3(6 − x)
2 9− x
3( x − 12)
4(9 − x)
32
= 0 when x = 12.
x = 12 is not in the domain. f ′′ is not continuous at
x = 9.
x = − 4 is not in the domain. f ′′ is not continuous at
Interval:
−∞ < x < 9
x = − 3.
Sign of f ′′:
f ′′ < 0
Interval:
−3 < x < ∞
Sign of f ′′:
f ′′ > 0
Conclusion:
Concave upward
Conclusion:
Concave downward
Concave downward: ( −∞, 9)
No point of inflection
Concave upward: ( −3, ∞)
There are no points of inflection.
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322
21.
Chapter 4
f ( x) =
f ′( x) =
Applications of Differentiation
4
x +1
−8 x
2
( x 2 + 1)
8(3 x 2 − 1)
f ′′( x) =
3
( x 2 + 1)
2
f ′′( x) = 0 for x = ±
3
3
Intervals:
−∞ < x < −
Sign of f ′′:
f ′′ > 0
Conclusion:
3
3
−
3
< x <
3
3
3
f ′′ < 0
Concave upward
3
< x < ∞
3
f ′′ > 0
Concave downward
Concave upward
⎛
3⎞ ⎛ 3 ⎞
, ∞ ⎟⎟
Concave upward: ⎜⎜ − ∞, −
⎟, ⎜
3 ⎟⎠ ⎜⎝ 3
⎝
⎠
⎛
3
3⎞
,
Concave downward: ⎜⎜ −
⎟⎟
3
3
⎝
⎠
⎛
3
Points of inflection: ⎜⎜ −
,
3
⎝
22.
⎞
3⎟⎟ and
⎠
⎛ 3
,
⎜⎜
⎝ 3
⎞
3⎟⎟
⎠
x +3
, Domain: x > 0
x
x −3
f ′( x) =
2 x3 2
9− x
f ′′( x) =
= 0 when x = 9
4 x5 2
f ( x) =
Intervals:
0 < x < 9
9 < x < ∞
Sign of f ′′:
f ′′ > 0
f ′′ < 0
Conclusion:
Concave upward
Concave downward
Concave upward: (0, 9)
Concave downward: (9, ∞)
Points of inflection: (9, 4)
23.
x
f ( x) = sin , 0 ≤ x ≤ 4π
2
1
⎛ x⎞
f ′( x) = cos⎜ ⎟
2
⎝2⎠
f ′′( x) = −
1
⎛ x⎞
sin ⎜ ⎟
4
⎝ 2⎠
Intervals:
0 < x < 2π
2π < x < 4π
Sign of f ′′:
f ′′ < 0
f ′′ > 0
Conclusion:
Concave downward
Concave upward
f ′′( x) = 0 when x = 0, 2π , 4π .
Concave upward: ( 2π , 4π )
Concave downward: (0, 2π )
Point of inflection: ( 2π , 0)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
24.
Concavity and the Second Derivative Test
323
3x
, 0 < x < 2π
2
3x
3x
cot
f ′( x) = −3 csc
2
2
9 ⎛ 3 3x
3x
3x ⎞
+ csc
cot 2 ⎟ ≠ 0 for any x in the domain of f .
f ′′( x) = ⎜ csc
2⎝
2
2
2⎠
f ( x) = 2 csc
f ′′ is not continuous at x =
2π
4π
and x =
.
3
3
Intervals:
0 < x <
Sign of f ′′( x):
f ′′ > 0
Conclusion:
2π
3
Concave upward
2π
4π
< x <
3
3
4π
< x < 2π
3
f ′′ < 0
f ′′ > 0
Concave downward
Concave upward
⎛ 2π ⎞ ⎛ 4π
⎞
Concave upward: ⎜ 0,
⎟, ⎜ , 2π ⎟
⎝ 3 ⎠ ⎝ 3
⎠
⎛ 2π 4π ⎞
Concave downward: ⎜ ,
⎟
⎝ 3 3 ⎠
No point of inflection
25.
π⎞
⎛
f ( x) = sec⎜ x − ⎟, 0 < x < 4π
2⎠
⎝
π⎞ ⎛
π⎞
⎛
f ′( x) = sec⎜ x − ⎟ tan ⎜ x − ⎟
2
2⎠
⎝
⎠
⎝
π⎞
π⎞
π⎞
⎛
⎛
⎛
f ′′( x) = sec3 ⎜ x − ⎟ + sec⎜ x − ⎟ tan 2 ⎜ x − ⎟ ≠ 0 for any x in the domain of f .
2
2
2⎠
⎝
⎠
⎝
⎠
⎝
f ′′ is not continuous at x = π , x = 2π , and x = 3π .
Intervals:
0 < x < π
π < x < 2π
2π < x < 3π
3π < x < 4π
Sign of f ′′:
f ′′ > 0
f ′′ < 0
f ′′ > 0
f ′′ < 0
Conclusion:
Concave upward
Concave downward
Concave upward
Concave upward
Concave upward: (0, π ), ( 2π , 3π )
Concave downward: (π , 2π ), (3π , 4π )
No point of inflection
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
324
26.
Chapter 4
Applications of Differentiation
f ( x) = sin x + cos x, 0 ≤ x ≤ 2π
f ′( x) = cos x − sin x
f ′′( x) = sin x − cos x
f ′′( x) = 0 when x =
3π 7π
,
.
4 4
3π
4
Intervals:
0 < x <
Sign of f ′′:
f ′′( x) < 0
Conclusion:
Concave downward
3π
7π
< x <
4
4
7π
< x < 2π
4
f ′′( x) > 0
f ′′( x) < 0
Concave upward
Concave downward
⎛ 3π 7π ⎞
Concave upward: ⎜ ,
⎟
⎝ 4 4 ⎠
⎛ 3π ⎞ ⎛ 7π
⎞
Concave downward: ⎜ 0,
⎟, ⎜ , 2π ⎟
⎝ 4 ⎠ ⎝ 4
⎠
⎛ 3π ⎞ ⎛ 7π ⎞
Points of inflection: ⎜ , 0 ⎟, ⎜ , 0 ⎟
⎝ 4 ⎠ ⎝ 4 ⎠
27.
f ( x) = 2 sin x + sin 2 x, 0 ≤ x ≤ 2π
f ′( x) = 2 cos x + 2 cos 2 x
f ′′( x) = −2 sin x − 4 sin 2 x = −2 sin x(1 + 4 cos x)
f ′′( x) = 0 when x = 0, 1.823, π , 4.460.
Intervals:
0 < x < 1.823
1.823 < x < π
π < x < 4.460
4.460 < x < 2π
Sign of f ′′:
f ′′ < 0
f ′′ > 0
f ′′ < 0
f ′′ > 0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward
Concave upward: (1.823, π ), ( 4.460, 2π )
Concave downward: (0, 1.823), (π , 4.460)
Points of inflection: (1.823, 1.452), (π , 0), ( 4.46, −1.452)
28.
f ( x) = x + 2 cos x, [0, 2π ]
f ′( x) = 1 − 2 sin x
f ′′( x) = −2 cos x
f ′′( x) = 0 when x =
π 3π
2
,
Intervals:
0 < x <
Sign of f ′′:
f ′′ < 0
Conclusion:
2
.
π
π
2
2
Concave downward
< x <
3π
2
f ′′ > 0
Concave upward
3π
< x < 2π
2
f ′′ < 0
Concave downward
⎛ π 3π ⎞
Concave upward: ⎜ ,
⎟
⎝2 2 ⎠
⎛ π ⎞ ⎛ 3π
⎞
Concave downward: ⎜ 0, ⎟, ⎜ , 2π ⎟
⎝ 2⎠ ⎝ 2
⎠
⎛ π π ⎞ ⎛ 3π 3π ⎞
Points of inflection: ⎜ , ⎟, ⎜ ,
⎟
⎝2 2⎠ ⎝ 2 2 ⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
29.
Concavity and the Second Derivative Test
325
y = e −3 x
y′ =
y′′ =
3 −3 x
e
x2
e −3 x (9 − 6 x)
x4
y′′ = 0 when x =
3
. y is not defined at x = 0.
2
Test intervals:
−∞ < x < 0
0 < x <
Sign of y′′:
y′′ > 0
y′′ > 0
Conclusion:
Concave upward
3
2
Concave upward
3
< x < ∞
2
y′′ < 0
Concave downward
⎛3
⎞
Point of inflection: ⎜ , e −2 ⎟
⎝2
⎠
⎛
Concave upward: ( − ∞, 0), ⎜ 0,
⎝
3⎞
⎟
2⎠
⎛3 ⎞
Concave downward: ⎜ , ∞ ⎟
⎝2 ⎠
30.
1 x
(e − e − x )
2
1
y ′ = (e x + e − x )
2
1 x
y′′ = (e − e − x )
2
y′′ = 0 when x = 0.
y =
Test interval:
−∞ < x < 0
0 < x < ∞
Sign of y′′:
y′′ < 0
y′′ > 0
Conclusion:
Concave downward
Concave upward
Point of inflection: (0, 0)
Concave upward: (0, ∞)
Concave downward: ( − ∞, 0)
31.
f ( x) = x − ln x, Domain: x > 0
f ′( x) = 1 −
f ′′( x) =
1
x
1
x2
f ′′( x) > 0 on the entire domain of f. There are no points of inflection.
Concave upward: (0, ∞)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
326
32.
Chapter 4
y = ln
y′ =
y′′ =
Applications of Differentiation
x2 + 9 =
1
ln ( x 2 + 9)
2
x
x2 + 9
9 − x2
( x 2 + 9)
2
y′′ = 0 when x = ± 3.
Test interval:
−∞ < x < −3
−3 < x < 3
3 < x < ∞
Sign of y′′:
y′′ < 0
y′′ > 0
y′′ < 0
Conclusion:
Concave downward
Concave upward
Concave downward
1
⎛
⎞
Points of inflection: ⎜ ± 3, ln 18 ⎟
2
⎝
⎠
Concave upward: ( − 3, 3)
Concave downward: ( − ∞, − 3), (3, ∞)
33.
f ( x) = arcsin x 4 5 ,
f ′( x) =
f ′′( x) =
−1 ≤ x ≤ 1
4
5 x1 5 1 − x8 5
20 x8 5 − 4
25 x 6 5 (1 − x8 5 )
32
f ′′( x) = 0 when 20 x8 5 = 4 ⇒ x8 5 =
1
⎛1⎞
⇒ x = ±⎜ ⎟
5
⎝5⎠
58
≈ ± 0.3657.
f ′′ is undefined at x = 0.
Test intervals:
⎛1⎞
−1 < x < − ⎜ ⎟
⎝5⎠
Sign of f ′′:
f ′′ > 0
Conclusion:
58
Concave upward
⎛ ⎛ 1 ⎞5 8
Points of inflection: ⎜ ± ⎜ ⎟ , arcsin
⎜ ⎝5⎠
⎝
⎛1⎞
−⎜ ⎟
⎝ 5⎠
58
⎛1⎞
0 < x < ⎜ ⎟
⎝5⎠
< x < 0
f ′′ < 0
Concave downward
58
f ′′ < 0
Concave downward
⎛1⎞
⎜ ⎟
⎝5⎠
58
< x <1
f ′′ > 0
Concave upward
1⎞
⎟ ≈ ( ± 0.3657, 0.4636)
5 ⎟⎠
58
58
⎛
⎞
⎛ 1 ⎞ ⎞ ⎛⎛ 1 ⎞
Concave upward: ⎜ −1, − ⎜ ⎟ ⎟, ⎜ ⎜ ⎟ , 1⎟
⎜
⎟
⎝ 5 ⎠ ⎟⎠ ⎜⎝ ⎝ 5 ⎠
⎝
⎠
⎛ ⎛ 1 ⎞ 5 8 ⎞ ⎛ ⎛ 1 ⎞5 8 ⎞
Concave downward: ⎜ − ⎜ ⎟ , 0 ⎟, ⎜ 0, ⎜ ⎟ ⎟
⎜ ⎝5⎠
⎟ ⎜ ⎝5⎠ ⎟
⎝
⎠ ⎝
⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
34.
Concavity and the Second Derivative Test
327
f ( x) = arctan ( x 2 )
f ′( x) =
f ′′( x) =
2x
x4 + 1
2(1 − 3x 4 )
( x4
+ 1)
2
f ′′( x) = 0 when 3x 4 = 1 ⇒ x = ± 4
Test interval:
−∞ < x < − 4
Sign of f ′′( x):
f ′′ < 0
Conclusion:
1
3
⎛
1
Points of inflection: ⎜⎜ ± 4 , arctan
3
⎝
4
−4
1
< x <
3
4
1
3
f ′′ > 0
Concave downward
⎛
1
Concave upward: ⎜ − 4 ,
⎜
3
⎝
1
≈ ± 0.7598.
3
Concave upward
4
1
< x < ∞
3
f ′′ < 0
Concave downward
1⎞
⎟ ≈ ( ± 0.7598, 0.5236)
3 ⎟⎠
1⎞
⎟
3 ⎟⎠
⎛
1⎞ ⎛ 1 ⎞
Concave downward: ⎜⎜ − ∞, − 4 ⎟⎟, ⎜⎜ 4 , ∞ ⎟⎟
3⎠ ⎝ 3 ⎠
⎝
35.
f ( x) = 6 x − x 2
38.
f ′( x) = 6 − 2 x
f ′( x) = − 3 x 2 + 14 x − 15 = − ( x − 3)(3 x − 5)
f ′′( x) = −2
f ′′( x) = − 6 x + 14 = − 2(3x − 7)
Critical number: x = 3
Critical numbers: x = 3,
f ′′(3) = − 2 < 0
Therefore, (3, 9) is a relative maximum.
f ( x) = x 2 + 3 x − 8
( 53 ) = 4 > 0
is a relative minimum.
Therefore, ( 53 , − 275
27 )
f ′′
f ′( x) = 2 x + 3
f ′′( x) = 2
Critical number: x = − 32
( )
f ′′ − 32 = 2 > 0
(
)
is a relative minimum.
Therefore, − 32 , − 41
4
37.
f ( x) = x3 − 3 x 2 + 3
f ′( x) = 3 x 2 − 6 x = 3x( x − 2)
f ′′( x) = 6 x − 6 = 6( x − 1)
Critical numbers: x = 0, x = 2
5
3
f ′′(3) = − 4 < 0
Therefore, (3, 9) is a relative maximum.
36.
f ( x) = − x3 + 7 x 2 − 15 x
39.
f ( x) = x 4 − 4 x3 + 2
f ′( x) = 4 x 3 − 12 x 2 = 4 x 2 ( x − 3)
f ′′( x) = 12 x 2 − 24 x = 12 x( x − 2)
Critical numbers: x = 0, x = 3
However, f ′′(0) = 0, so you must use the First
Derivative Test. f ′( x) < 0 on the intervals ( −∞, 0)
and (0, 3); so, (0, 2) is not an extremum. f ′′(3) > 0
so (3, − 25) is a relative minimum.
f ′′(0) = −6 < 0
Therefore, (0, 3) is a relative maximum.
f ′′( 2) = 6 > 0
Therefore, ( 2, −1) is a relative minimum.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
328
40.
Chapter 4
Applications of Differentiation
f ( x) = − x 4 + 4 x3 + 8 x 2
f ′( x) = −4 x3 + 12 x 2 + 16 x = −4 x( x − 4)( x + 1)
44.
f ′( x) =
f ′′( x) = −12 x 2 + 24 x + 16 = −4(3x 2 − 6 x − 4)
f ′′( −1) = −20 < 0
f ′′(0) = 16 > 0
Therefore, (0, 0) is a relative minimum.
f ′′( 4) = −80 < 0
Therefore, ( 4, 128) is a relative maximum.
41.
Critical number: x = 0
However, f ′′(0) is undefined, so you must use the First
Derivative Test. Because f ′( x) < 0 on ( −∞, 0) and
f ′( x) > 0 on (0, ∞), (0, −3) is a relative minimum.
x2 + 1
x
f ′( x) =
f ′′( x) =
x2 + 1
1
( x2
+ 1)
32
Critical number: x = 0
f ′′(0) = 1 > 0
Therefore, (0, 1) is a relative minimum.
43.
4
f ( x) = x +
x
f ′( x) = 1 −
− 1)
2
45.
f ( x) = cos x − x, 0 ≤ x ≤ 4π
f ′( x) = −sin x − 1 ≤ 0
Therefore, f is non-increasing and there are no relative
extrema.
46. f ( x) = 2 sin x + cos 2 x, 0 ≤ x ≤ 2π
f ′( x) = 2 cos x − 2 sin 2 x = 2 cos x − 4 sin x cos x
π π 5π 3π
f ′′( x) = −2 sin x − 4 cos 2 x
2
3 x1 3
2
f ′′( x) = − 4 3
9x
f ( x) =
(x
= 2 cos x(1 − 2 sin x) = 0 when x =
f ( x) = x 2 3 − 3
f ′( x) =
42.
x
x −1
−1
There are no critical numbers and x = 1 is not in the
domain. There are no relative extrema.
Critical numbers: x = −1, 0, 4
Therefore ( −1, 3) is a relative maximum.
f ( x) =
4
x2 − 4
=
2
x
x2
8
f ′′( x) = 3
x
Critical numbers: x = ±2
f ′′( −2) = −1 < 0
Therefore, ( −2, − 4) is a relative maximum.
f ′′( 2) = 1 > 0
Therefore, ( 2, 4) is a relative minimum.
, , , .
6 2 6 2
⎛π ⎞
f ′′⎜ ⎟ = − 3 < 0
⎝6⎠
⎛π 3⎞
Therefore, ⎜ , ⎟ is a relative maximum.
⎝ 6 2⎠
⎛π ⎞
f ′′⎜ ⎟ = 2 > 0
⎝2⎠
⎛π ⎞
Therefore, ⎜ , 1⎟ is a relative minimum.
⎝2 ⎠
⎛ 5π ⎞
f ′′⎜ ⎟ = − 3 < 0
⎝ 6 ⎠
⎛ 5π 3 ⎞
Therefore, ⎜ , ⎟ is a relative maximum.
⎝ 6 2⎠
⎛ 3π ⎞
f ′′⎜ ⎟ = 6 > 0
⎝ 2 ⎠
⎛ 3π
⎞
Therefore, ⎜ , − 3⎟ is a relative minimum.
2
⎝
⎠
47. y = f ( x) = 8 x 2 − ln x
1
x
1
f ′′( x) = 16 + 2
x
f ′( x) = 16 x −
f ′( x) = 0 ⇒ 16 x =
1
1
⇒ 16 x 2 = 1 ⇒ x = ±
x
4
Critical number:
1
1
⎛
⎞
x =
⎜ x = − is not in the domain.⎟
4
4
⎝
⎠
⎛1⎞
f ′′⎜ ⎟ > 0
⎝4⎠
1⎞ ⎛1 1
⎛1 1
⎞
Therefore, ⎜ , − ln ⎟ = ⎜ , + ln 4 ⎟ is a relative
4⎠ ⎝ 4 2
⎝4 2
⎠
minimum.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
48.
y = f ( x) = x ln x
52.
f ′( x) = ln x + 1
f ′′( x) =
Concavity and the Second Derivative Test
g ′( x ) =
1
x
g ′′( x ) =
Critical number: ln x + 1 = 0 ⇒ ln x = −1
⇒ x = e −1 =
1
e
2
g ′′(3) < 0
⎛
Therefore, ⎜ 3,
⎝
minimum.
⎛1 1⎞
Therefore, ⎜ , − ⎟ is a relative minimum.
⎝e e⎠
53.
x
ln x
1 −( x − 3)2 2
e
2π
2
−1
( x − 3)e−( x − 3) 2
2π
2
1
( x − 2)( x − 4)e−( x − 3)
2π
Critical number: x = 3
⎛1⎞
f ′′⎜ ⎟ > 0
⎝e⎠
49. y = f ( x) =
g ( x) =
329
1 ⎞
⎟ ≈ (3, 0.399) is a relative
2π ⎠
f ( x ) = x 2e − x
f ′( x) = − x 2e − x + 2 xe − x = xe − x ( 2 − x)
f ′′( x) = − e − x ( 2 x − x 2 ) + e− x ( 2 − 2 x)
Domain: 0 < x < 1, x > 1
f ′( x) =
(ln x)(1) − ( x)(1 x)
2
(ln x)
f ′′( x) =
2 − ln x
x(ln x)
=
ln x − 1
(ln x)
= e − x ( x 2 − 4 x + 2)
2
Critical numbers: x = 0, 2
f ′′(0) > 0
Therefore, (0, 0) is a relative minimum.
Critical number: x = e
f ′′(e) > 0
f ′′( 2) < 0
Therefore, (e, e) is a relative minimum.
50. y = f ( x) = x 2 ln
x
, Domain: x > 0
4
Therefore, ( 2, 4e −2 ) is a relative maximum.
54.
f ′( x) = − xe − x + e − x = e − x (1 − x)
2⎛ 1 ⎞
x
x⎞
⎛
= x⎜1 + 2 ln ⎟
f ′( x) = x ⎜ ⎟ + 2 x ln
4
4⎠
⎝ x⎠
⎝
f ′′( x) = 1 + 2 ln
f ′′( x) = −e − x + ( −e − x )(1 − x) = e − x ( x − 2)
x
x
⎛1⎞
+ 2 x⎜ ⎟ = 3 + 2 ln
x
4
4
⎝ ⎠
Critical number: x = 1
f ′′(1) < 0
Critical number: x = 4e −1 2
Therefore, (1, e −1 ) is a relative maximum.
f ′′( 4e −1 2 ) > 0
Therefore, ( 4e −1 2 , −8e −1 ) is a relative minimum.
e x + e− x
51. f ( x) =
2
x
e − e− x
f ′( x) =
2
e x + e− x
f ′′( x) =
2
Critical number: x = 0
f ′′(0) > 0
Therefore, (0, 1) is a relative minimum.
f ( x) = xe − x
55.
f ( x ) = 8 x( 4 − x )
f ′( x) = −8( 4− x )( x ln 4 − 1)
f ′′( x) = 8( 4− x ) ln 4( x ln 4 − 2)
Critical number: x =
1
1
=
ln 4
2 ln 2
⎛ 1 ⎞
f ′′⎜
⎟ < 0
⎝ 2 ln 2 ⎠
⎛ 1
4e −1 ⎞
,
Therefore, ⎜
⎟ is a relative maximum.
⎝ 2 ln 2 ln 2 ⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
330
56.
Chapter 4
Applications of Differentiation
y = f ( x) = x 2 log 3 x = x 2
f ′( x) =
x( 2 ln x + 1)
59. f ( x ) = 0.2 x 2 ( x − 3) , [−1, 4]
3
ln x
ln 3
(a)
2
f ′′( x) = ( x − 3)( 4 x 2 − 9.6 x + 3.6)
ln 3
2 ln x + 3
f ′′( x) =
ln 3
= 0.4( x − 3)(10 x 2 − 24 x + 9)
(b) f ′′(0) < 0 ⇒ (0, 0) is a relative maximum.
Critical number: ln x = −
( 56 ) > 0 ⇒ (1.2, –1.6796) is a relative minimum.
1
⇒ x = e −1 2
2
f ′′
Points of inflection:
(3, 0), (0.4652, − 0.7048), (1.9348, − 0.9049)
f ′′(e −1 2 ) > 0
Therefore, (e −1 2 , − 0.1674) is a relative minimum.
57.
f ′( x) = 0.2 x(5 x − 6)( x − 3)
(c)
y
f ( x) = arcsec x − x
f″
f′
2
f ′( x) =
1
x2 − 1
x
− 1 = 0 when x
x 2 − 1 = 1.
1
x 2 ( x 2 − 1) = 1
1+
5
1
x −1 x
−
2
x
1+
f is increasing when f ′ > 0 and decreasing when
f ′ < 0. f is concave upward when f ′′ > 0 and
5
2
concave downward when f ′′ < 0.
= ± 1.272.
2
f ′′( x) = −
60. f ( x ) = x 2
x
(x
2
− 1)
32
x
(a)
f ′′(1.272) < 0
f ′′( x) =
f ′′( −1.272) > 0
Therefore, ( −1.272, 3.747) is a relative minimum.
f ( x) = arcsin x − 2 x
f ′( x) =
f ′′( x) =
1
1 − x2
x
(1 − x 2 )
−2
f ′( x) =
6 − x 2 , ⎡⎣−
3 x( 4 − x 2 )
6 − x2
6( x 4 − 9 x 2 + 12)
(6 − x 2 )
⎛
⎞
3
, 0.68 ⎟⎟ is a relative maximum.
⎜⎜ −
⎝ 2
⎠
33
2
.
(
)
f ′′( ±2) < 0 ⇒ ±2, 4 2 are relative maxima.
Points of inflection: ( ±1.2758, 3.4035)
(c)
⎛
3⎞
f ′′⎜⎜ −
⎟⎟ < 0
2
⎝
⎠
9−
(b) f ′′(0) > 0 ⇒ (0, 0) is a relative minimum.
y
3
Critical numbers: x = ±
2
⎛ 3
⎞
, − 0.68 ⎟⎟ is a relative minimum.
⎜⎜
⎝ 2
⎠
32
f ′′( x) = 0 when x = ±
32
⎛ 3⎞
f ′′⎜⎜
⎟⎟ > 0
⎝ 2 ⎠
6 ⎤⎦
6,
f ′( x) = 0 when x = 0, x = ± 2.
Therefore, (1.272, − 0.606) is a relative maximum.
58.
4
f
x 4 − x 2 − 1 = 0 when x 2 =
or x = ±
x
−2 −1
6
f
x
−3
3
f ''
f'
−6
The graph of f is increasing when f ′ > 0 and
decreasing when f ′ < 0. f is concave upward
when f ′′ > 0 and concave downward when
f ′′ < 0.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
61. f ( x ) = sin x −
(a)
1
1
sin 3x + sin 5 x,
3
5
[0, π ]
π
,x =
π
5π
.
6
,x =
6
2
f ′′( x) = −sin x + 3 sin 3 x − 5 sin 5 x
f ′′( x) = 0 when x =
π
6
x ≈ 1.1731, x ≈ 1.9685
,x =
2 x cos x +
sin x
2x
Critical numbers: x ≈ 1.84, 4.82
5π
,
6
=
=
5π
⎞
, 0.2667 ⎟
6
⎝
⎠
(1.9685, 0.9637), ⎛⎜
(4 x 2 + 1) sin x
2 cos x
−
2x
2x 2x
4 x cos x − ( 4 x 2 + 1) sin x
2x 2x
(b) Relative maximum: (1.84, 1.85)
⎛π
⎞
Points of inflection: ⎜ , 0.2667 ⎟, (1.1731, 0.9638),
⎝6
⎠
Relative minimum: ( 4.82, − 3.09)
Points of inflection: (0.75, 0.83), (3.42, − 0.72)
(c)
y
Note: (0, 0) and (π , 0) are not points of inflection
4
f′
because they are endpoints.
2
f
(c)
y
x
π
2
4
−2
π
4
π
2
f′
π
x
−4
f is increasing when f ′ > 0 and decreasing when
f ′ < 0. f is concave upward
when f ′′ > 0 and concave downward when
f ′′ < 0.
−4
−6
−8
f ''
−2
f
2
cos x
cos x
sin x
+
−
2x
2x
2x 2x
f ′′( x) = − 2 x sin x +
⎛π ⎞
⎛π
⎞
(b) f ′′⎜ ⎟ < 0 ⇒ ⎜ , 1.53333⎟ is a relative
2
2
⎝ ⎠
⎝
⎠
maximum.
f″
The graph of f is increasing when f ′ > 0 and
decreasing when f ′ < 0. f is concave upward
when f ′′ > 0 and concave downward when
f ′′ < 0.
331
2 x sin x, [0, 2π ]
62. f ( x ) =
(a) f ′( x) =
f ′( x) = cos x − cos 3x + cos 5 x
f ′( x) = 0 when x =
Concavity and the Second Derivative Test
63. (a)
y
4
3
2
1
x
1
2
3
4
f ′ < 0 means f decreasing
f ′ increasing means concave upward
(b)
y
4
3
2
1
x
1
2
3
4
f ′ > 0 means f increasing
f ′ increasing means concave upward
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
332
Chapter 4
64. (a)
Applications of Differentiation
67. (a)
y
y
f
f'
3
4
f ''
3
2
−2
1
x
−1
3
−1
x
1
2
3
4
f ′ < 0 means f decreasing
y
(b)
f ′ decreasing means concave downward
f″
f′
f
4
(b)
y
4
x
−2
3
2
−2
2
−4
1
x
1
2
3
4
f ′ > 0 means f increasing
68. (a) The graph of f is increasing and concave downward:
f ′ > 0, f ′′ < 0.
(b) The graph of f is decreasing and concave upward:
f ′ < 0, f ′′ > 0.
f ′ decreasing means concave downward
65. Answers will vary. Sample answer:
f ( x) = x 4 .
Let
69.
f ′′( x ) = 12 x 2
y
4
f ′′(0) = 0, but (0, 0) is not a point of inflection.
2
(2, 0) (4, 0)
x
y
2
4
6
6
5
4
3
70.
y
2
2
1
−3
−2
−1
x
1
2
1
3
(0, 0)
(2, 0)
x
−1
1
3
66. (a) The rate of change of sales is increasing.
S ′′ > 0
(b) The rate of change of sales is decreasing.
S ′ > 0, S ′′ < 0
(c) The rate of change of sales is constant.
S ′ = C , S ′′ = 0
(d) Sales are steady.
S = C , S ′ = 0, S ′′ = 0
71.
y
3
2
1
(2, 0)
(4, 0)
x
1
2
3
4
5
(e) Sales are declining, but at a lower rate.
S ′ < 0, S ′′ > 0
(f ) Sales have bottomed out and have started to rise.
S ′ > 0, S ′′ > 0 Answers will vary.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
72.
y
73.
y
333
Concavity and the Second Derivative Test
3
f
2
(0, 0)
(2, 0)
−1
1
x
−4
x
8
3
−1
12
f″
−8
f ′′ is linear.
f ′ is quadratic.
f is cubic.
f concave upward on ( −∞, 3), downward on (3, ∞).
74. (a)
d
12
t
10
(b) Because the depth d is always increasing, there are no relative extrema. f ′( x) > 0
(c) The rate of change of d is decreasing until you reach the widest point of the jug, then the rate increases until you reach the
narrowest part of the jug’s neck, then the rate decreases until you reach the top of the jug.
75. (a) n = 1:
n = 2:
n = 3:
f ( x) = x − 2
f ( x ) = ( x − 2)
f ′( x) = 1
f ′( x) = 2( x − 2)
f ′′( x) = 0
f ′′( x) = 2
No point of inflection
n = 4:
f ( x ) = ( x − 2)
2
f ( x ) = ( x − 2)
3
f ′( x) = 3( x − 2)
f ′( x) = 4( x − 2)
2
f ′′( x) = 6( x − 2)
No point of inflection
3
f ′′( x) = 12( x − 2)
Point of inflection: ( 2, 0)
2
No point of inflection
Relative minimum: ( 2, 0)
Relative minimum: ( 2, 0)
6
6
6
4
6
f(x) = (x − 2)3
−9
9
−9
9
−9
9
f(x) = (x − 2)2
f(x) = x − 2
Point of
inflection
−6
−6
−6
−9
9
f(x) = (x − 2)4
−6
Conclusion: If n ≥ 3 and n is odd, then ( 2, 0) is point of inflection. If n ≥ 2 and n is even, then ( 2, 0) is a relative minimum.
(b) Let f ( x ) = ( x − 2) , f ′( x) = n( x − 2)
n
n −1
, f ′′( x) = n( n − 1)( x − 2)
n−2
.
For n ≥ 3 and odd, n − 2 is also odd and the concavity changes at x = 2.
For n ≥ 4 and even, n − 2 is also even and the concavity does not change at x = 2.
So, x = 2 is point of inflection if and only if n ≥ 3 is odd.
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334
Chapter 4
f ( x) =
76. (a)
f ′( x) =
3
Applications of Differentiation
(b) f ′′( x) does not exist at x = 0.
x
1 −2 3
x
3
y
3
f ′′( x) = − 92 x −5 3
2
Point of inflection: (0, 0)
1
(0, 0)
−6
−4
x
−2
2
4
6
−2
−3
77. f ( x ) = ax 3 + bx 2 + cx + d
Relative maximum: (3, 3)
Relative minimum: (5, 1)
Point of inflection: ( 4, 2)
f ′( x) = 3ax 2 + 2bx + c, f ′′( x) = 6ax + 2b
f (3) = 27 a + 9b + 3c + d = 3 ⎫⎪
⎬ 98a + 16b + 2c = −2 ⇒ 49a + 8b + c = −1
f (5) = 125a + 25b + 5c + d = 1⎪⎭
f ′(3) = 27 a + 6b + c = 0, f ′′( 4) = 24a + 2b = 0
49a + 8b + c = −1
27 a + 6b + c =
22a + 2b
a =
1,
2
f ( x) =
2a
45
,
2
− 6 x2 +
0
22a + 2b = −1
0
= −1
b = −6, c =
1 x3
2
24a + 2b =
=
1
d = −24
45
x
2
− 24
78. f ( x ) = ax 3 + bx 2 + cx + d
Relative maximum: ( 2, 4)
Relative minimum: ( 4, 2)
Point of inflection: (3, 3)
f ′( x) = 3ax 2 + 2bx + c, f ′′( x ) = 6ax + 2b
f ( 2) = 8a + 4b + 2c + d = 4 ⎪⎫
⎬ 56a + 12b + 2c = −2 ⇒ 28a + 6b + c = −1
f ( 4) = 64a + 16b + 4c + d = 2⎪⎭
f ′( 2) = 12a + 4b + c = 0, f ′( 4) = 48a + 8b + c = 0, f ′′(3) = 18a + 2b = 0
28a + 6b + c = −1
18a + 2b =
12a + 4b + c = 0
16a + 2b
= −1
16a + 2b = −1
2a
= 1
a =
1,
2
f ( x) =
0
b = − 92 , c = 12, d = −6
1 x3
2
−
9 2
x
2
+ 12 x − 6
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
Concavity and the Second Derivative Test
335
79. f ( x ) = ax 3 + bx 2 + cx + d
Maximum: ( −4, 1)
Minimum: (0, 0)
(a) f ′( x) = 3ax 2 + 2bx + c,
f ′′( x) = 6ax + 2b
f ( 0) = 0 ⇒ d = 0
f ( −4) = 1 ⇒
f ′( −4) = 0 ⇒
f ′(0) = 0 ⇒
−64a + 16b − 4c = 1
48a − 8b + c = 0
c = 0
Solving this system yields a =
f ( x) =
+
1 x3
32
1
32
and b = 6a =
3
.
16
3 2
x
16
(b) The plane would be descending at the greatest rate at the point of inflection.
f ′′( x) = 6ax + 2b =
3
x
16
+
3
8
= 0 ⇒ x = −2.
Two miles from touchdown.
80. (a) line OA : y = −0.06 x
slope: − 0.06
line CB : y = 0.04 x + 50
slope: 0.04
y
f ( x ) = ax 3 + bx 2 + cx + d
150
f ′( x) = 3ax + 2bx + c
2
(−1000, 60):
60 = ( −1000) a + (1000) b − 1000c + d
2
(−1000, 60)
A
−0.06 = (1000) 3a − 2000b + c
90 = (1000) a + (1000) b + 1000c + d
3
2
C
(0, 50)
2
(1000, 90):
(1000, 90)
B
100
3
−1000
O
x
1000
0.04 = (1000) 3a + 2000b + c
2
The solution to this system of four equations is a = −1.25 × 10−8 , b = 0.000025, c = 0.0275, and d = 50.
(b) y = −1.25 × 10−8 x 3 + 0.000025 x 2 + 0.0275 x + 50
100
−1100
1100
−10
(c)
0.1
−1100
1100
− 0.1
(d) The steepest part of the road is 6% at the point A.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
336
81.
Chapter 4
Applications of Differentiation
C = 0.5 x 2 + 15 x + 5000
C =
C
5000
= 0.5 x + 15 +
x
x
C = average cost per unit
dC
5000
= 0.5 −
= 0 when x = 100
dx
x2
By the First Derivative Test, C is minimized when x = 100 units.
5.755 3 8.521 2
6.540
T −
T +
T + 0.99987, 0 < T < 25
108
106
105
82. S =
17.265 2 17.042
6.540
T −
T +
108
106
105
34.53
17.042
S ′′ =
T −
= 0 when T ≈ 49.4, which is not in the domain
108
106
S ′′ < 0 for 0 < T < 25 ⇒ Concave downward.
(a) S ′ =
(b) The maximum is approximately ( 4, 1).
(c)
1.001
0
0.996
25
(d) When t = 20, S ≈ 0.998.
5000t 2
,0 ≤ t ≤ 3
8 + t2
83. S =
(a)
t
0.5
1
1.5
2
2.5
3
S
151.5
555.6
1097.6
1666.7
2193.0
2647.1
Increasing at greatest rate when 1.5 < t < 2
(b)
3000
0
3
0
Increasing at greatest rate when t ≈ 1.5.
(c)
S =
S ′(t ) =
5000t 2
8 + t2
80,000t
(8 + t 2 )
80,000(8 − 3t 2 )
S ′′(t ) =
3
(8 + t 2 )
2
S ′′(t ) = 0 for t = ±
8
2 6
≈ 1.633 yrs.
. So, t =
3
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.4
337
100t 2
,t > 0
65 + t 2
84. S =
(a)
Concavity and the Second Derivative Test
100
0
35
0
(b) S ′(t ) =
13,000t
(65 + t 2 )
13,000(65 − 3t 2 )
S ′′(t ) =
3
(65 + t 2 )
2
= 0 ⇒ t = 4.65
S is concave upwards on (0, 4.65), concave downwards on ( 4.65, 30).
(c) S ′(t ) > 0 for t > 0.
As t increases, the speed increases, but at a slower rate.
85.
f ( x) = 2(sin x + cos x),
⎛π ⎞
f⎜ ⎟ = 2 2
⎝4⎠
f ′( x) = 2(cos x − sin x),
⎛π ⎞
f ′⎜ ⎟ = 0
⎝4⎠
⎛π ⎞
f ′′( x) = 2( −sin x − cos x), f ′′⎜ ⎟ = −2 2
⎝4⎠
π⎞
⎛
P1 ( x) = 2 2 + 0⎜ x − ⎟ = 2 2
4⎠
⎝
P1′ ( x) = 0
π⎞ 1
π⎞
⎛
⎛
P2 ( x) = 2 2 + 0⎜ x − ⎟ + −2 2 ⎜ x − ⎟ = 2 2 −
4⎠ 2
4⎠
⎝
⎝
(
)
2
π⎞
⎛
2⎜ x − ⎟
4⎠
⎝
4
2
P1
− 2␲
π⎞
⎛
P2′ ( x) = −2 2 ⎜ x − ⎟
4⎠
⎝
2␲
f
P2
−4
P2′′ ( x) = −2 2
The values of f , P1 , P2 , and their first derivatives are equal at x = π 4. The values of the second derivatives of f and P2 are
equal at x = π 4. The approximations worsen as you move away from x = π 4.
86.
f ( x) = 2(sin x + cos x),
f ( 0) = 2
f ′( x) = 2(cos x − sin x),
f ′(0) = 2
f ′′( x) = 2( −sin x − cos x),
f ′′(0) = −2
P1 ( x) = 2 + 2( x − 0) = 2(1 + x)
P1′ ( x) = 2
P2 ( x) = 2 + 2( x − 0) +
4
P2
f
1
2
(−2)( x
− 0) = 2 + 2 x − x
2
2
P2′ ( x) = 2 − 2 x
P2′′ ( x) = −2
−6
6
P1
−4
The values of f , P1 , P2 , and their first derivatives are equal at x = 0. The values of the second derivatives of f and P2 are
equal at x = 0. The approximations worsen as you move away from x = 0.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
338
Chapter 4
Applications of Differentiation
f ( x) = arctan x, a = −1,
87.
1
,
f ′( x) =
1 + x2
2x
,
f ′′( x) = −
2
(1 + x 2 )
f (−1) = −
4
1
f ′( −1) =
2
1
f ′′( −1) =
2
P1 ( x) = f ( −1) + f ′( −1)( x + 1) = −
P1′ ( x) =
π
π
4
+
1
( x + 1)
2
1
2
P2 ( x) = f ( −1) + f ′( −1)( x + 1) +
1
1
1
π
2
2
f ′′(−1)( x + 1) = − + ( x + 1) + ( x + 1)
2
4
2
4
1
1
+ ( x + 1)
2
2
1
P2′′ ( x) =
2
P2′ ( x) =
The values of f , P1 , P2 , and their first derivatives are equal when x = −1. The approximations worsen as
you move away from x = −1.
4
P2
−6
6
f
P1
−4
f ( x) =
88.
f ′( x) =
f ′′( x) =
x
,
x −1
f ( 2) =
−( x + 1)
2
x ( x − 1)
2
,
3x 2 + 6 x − 1
,
3
4 x3 2 ( x − 1)
2
f ′( 2) = −
f ′′( 2) =
3
3 2
= −
4
2 2
23
23 2
=
16
8 2
P1 ( x) =
⎛ 3 2⎞
3 2
5 2
x +
2 + ⎜⎜ −
⎟⎟( x − 2) = −
4
4
2
⎝
⎠
P1′ ( x) = −
3 2
4
P2 ( x) =
⎛ 3 2⎞
1 ⎛ 23 2 ⎞
2
2 + ⎜⎜ −
⎟⎟( x − 2) + ⎜⎜
⎟( x − 2) =
4 ⎠
2 ⎝ 16 ⎟⎠
⎝
P2′ ( x) = −
3 2
23 2
+
( x − 2)
4
16
P2′′ ( x) =
2 −
3 2
23 2
( x − 2) +
( x − 2)2
4
32
23 2
16
The values of f , P1 , P2 and their first derivatives are equal at x = 2. The values of the second derivatives of f and P2 are equal
at x = 2. The approximations worsen as you move away from x = 2.
3
P1
P2
f
−1
5
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
89.
Limits at Infinity
339
⎛1⎞
f ( x) = x sin ⎜ ⎟
⎝ x⎠
⎡ 1
1
⎛ 1 ⎞⎤
⎛1⎞
⎛1⎞
⎛1⎞
f ′( x) = x ⎢− 2 cos⎜ ⎟⎥ + sin ⎜ ⎟ = − cos⎜ ⎟ + sin ⎜ ⎟
x
⎝ x ⎠⎦
⎝ x⎠
⎝ x⎠
⎝ x⎠
⎣ x
1⎡ 1
1
1
1
⎛ 1 ⎞⎤
⎛1⎞
⎛1⎞
⎛1⎞
f ′′( x) = − ⎢ 2 sin ⎜ ⎟⎥ + 2 cos⎜ ⎟ − 2 cos⎜ ⎟ = − 3 sin ⎜ ⎟ = 0
x⎣x
x
x
⎝ x ⎠⎦
⎝ x⎠ x
⎝ x⎠
⎝ x⎠
1
x =
1
π
⎛1 ⎞
Point of inflection: ⎜ , 0 ⎟
⎝π ⎠
−1
( π1 , 0(
When x > 1 π , f ′′ < 0, so the graph is concave downward.
90.
f ( x) = x( x − 6) = x 3 − 12 x 2 + 36 x
2
f ′( x) = 3 x 2 − 24 x + 36 = 3( x − 2)( x − 6) = 0
f ′′( x) = 6 x − 24 = 6( x − 4) = 0
Relative extrema: ( 2, 32) and (6, 0)
Point of inflection ( 4, 16) is midway between the relative
extrema of f.
91. True. Let y = ax 3 + bx 2 + cx + d , a ≠ 0. Then
y′′ = 6ax + 2b = 0 when x = −(b 3a), and the
concavity changes at this point.
92. False. f ( x ) = 1 x has a discontinuity at x = 0.
93. False. Concavity is determined by f ′′. For example, let
f ( x ) = x and c = 2. f ′(c) = f ′( 2) > 0, but f is not
concave upward at c = 2.
1
−1
94. False. For example, let f ( x) = ( x − 2) .
4
95. f and g are concave upward on ( a, b) implies that f ′ and
g ′ are increasing on ( a, b), and f ′′ > 0 and g ′′ > 0 .
So, ( f + g )′′ > 0 ⇒ f + g is concave upward on
(a, b) by Theorem 4.7.
96. f, g are positive, increasing, and concave upward on
(a, b) ⇒ f ( x) > 0, f ′( x) ≥ 0 and f ′′( x) > 0, and
g ( x) > 0, g ′( x) ≥ 0 and g ′′( x) > 0 on ( a, b). For
x ∈ ( a, b),
( fg )′ ( x)
= f ′( x) g ( x ) + f ( x) g ′( x)
( fg )′′ ( x)
= f ′′( x) g ( x) + 2 f ′( x) g ′( x) + f ( x) g ′′( x) > 0
So, fg is concave upward on ( a, b).
Section 4.5 Limits at Infinity
1. f ( x) =
2x2
x + 2
2
4. f ( x) = 2 +
x2
x +1
4
No vertical asymptotes
Horizontal asymptote: y = 2
No vertical asymptotes
Horizontal asymptote: y = 2
Matches (f ).
Matches (a).
2x
2. f ( x) =
x2 + 2
No vertical asymptotes
Horizontal asymptotes: y = ±2
Matches (c).
3. f ( x) =
x
x2 + 2
No vertical asymptotes
Horizontal asymptote: y = 0
f (1) < 1
Matches (d).
5. f ( x ) =
4 sin x
x2 + 1
No vertical asymptotes
Horizontal asymptote: y = 0
f (1) > 1
Matches (b).
6. f ( x) =
2 x 2 − 3x + 5
x2 + 1
No vertical asymptotes
Horizontal asymptote: y = 2
Matches (e).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
340
Chapter 4
7. f ( x) =
Applications of Differentiation
4x + 3
2x − 1
x
100
101
102
103
104
105
106
f(x)
7
2.26
2.025
2.0025
2.0003
2
2
lim f ( x ) = 2
x→∞
10
− 10
10
− 10
8. f ( x ) =
2x2
x +1
x
100
101
102
103
104
105
106
f(x)
1
18.18
198.02
1998.02
19,998
199,998
1,999,998
lim f ( x) = ∞
(Limit does not exist )
x→∞
20
0
10
−2
−6 x
9. f ( x ) =
4x2 + 5
x
100
101
102
103
104
105
106
f(x)
–2
–2.98
–2.9998
–3
–3
–3
–3
lim f ( x) = −3
x→∞
10
− 10
10
− 10
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
Limits at Infinity
341
10
10. f ( x) =
2x2 − 1
x
100
101
102
103
104
105
106
f(x)
10.0
0.7089
0.0707
0.0071
0.0007
0.00007
0.000007
lim f ( x ) = 0
x→∞
10
−9
9
−2
11. f ( x ) = 5 −
1
x2 + 1
x
100
101
102
103
104
105
106
f(x)
4.5
4.99
4.9999
4.999999
5
5
5
lim f ( x ) = 5
x→∞
6
−1
8
0
12. f ( x) = 4 +
3
x2 + 2
x
100
101
102
103
104
105
106
f(x)
5
4.03
4.0003
4.0
4.0
4
4
lim f ( x ) = 4
x→∞
10
0
15
0
13. (a) h( x) =
f ( x)
2
=
x
lim h( x) = ∞
x→∞
(b) h( x) =
f ( x)
3
x
lim h( x) = 5
=
5 x 3 − 3 x 2 + 10 x
10
= 5x − 3 +
x2
x
(Limit does not exist )
14. (a) h( x) =
5 x − 3x + 10 x
3 10
= 5− + 2
x3
x
x
(b) h( x) =
3
2
x→∞
x →∞
−4 x 2 + 2 x − 5
5
= −4 x + 2 −
x
x
x
lim h( x) = −∞
(Limit does not exist )
=
x→∞
f ( x)
2
=
−4 x 2 + 2 x − 5
2
5
= −4 + − 2
x2
x
x
=
−4 x 2 + 2 x − 5
4
2
5
= − + 2 − 3
x3
x
x
x
x
lim h( x) = −4
x→∞
f ( x)
5 x − 3x + 10 x
5
3
10
=
= − 2 + 3
x4
x4
x x
x
lim h( x) = 0
(c) h( x) =
f ( x)
3
2
(c) h( x) =
f ( x)
3
x
lim h( x) = 0
x →∞
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
342
Chapter 4
Applications of Differentiation
15. (a) lim
x2 + 2
= 0
x3 − 1
(b) lim
x + 2
=1
x2 − 1
(c) lim
x2 + 2
= ∞
x −1
16. (a) lim
3 − 2x
= 0
3x3 − 1
x→∞
x→∞
x→∞
( Limit does not exist )
(
(c) lim
3 − 2x
= −∞
3x − 1
17. (a) lim
5 − 2 x3 2
= 0
3x 2 − 4
( Limit does not exist )
(b) lim
5 − 2 x3 2
2
= −
3x3 2 − 4
3
(c) lim
5 − 2 x3 2
= −∞
3x − 4
x→∞
x→∞
(Limit does not exist )
x→∞
x→∞
32
5x
= ∞
(c) lim
x→∞ 4
x +1
x⎞
⎛5
20. lim ⎜ − ⎟ = ∞
x → −∞ ⎝ x
3⎠
(
= −1, for x < 0 we have x = −
x2 − x
4 + (5 x 2 )
4x2 + 5
22. lim
=
lim
= 4
x → −∞ x 2 + 3
x → −∞ 1 + 3 x 2
( )
⎛1⎞
−2 − ⎜ ⎟
⎝ x⎠
= lim
x → −∞
1
1−
x
(
= −2, for x < 0, x = −
28. lim
( )
x →∞
5 x3 + 1
5 + 1 x3
= lim
3
2
x
→∞
10 x − 3 x + 7
10 − 3 x + 7 x
3
=
5+0
1
=
10 − 0
2
x2
)
5x2 + 2
x2 + 3
x→∞
= lim
x→∞
= lim
x→∞
x
1x
0
23. lim 2
= lim
=
= 0
x→∞ x − 1
x → ∞ 1 − 1 x2
1
24. lim
⎛
⎜
⎜
⎝
(Limit does not exist )
2 − (1 x)
2x − 1
2 −0
2
21. lim
= lim
=
=
x → ∞ 3x + 2
x → ∞ 3 + (2 x)
3+ 0
3
1
x
x2 − x ⎞
⎟
− x 2 ⎟⎠
2+
x → −∞
( Limit does not exist )
)
2x + 1
= lim
3⎞
⎛
19. lim ⎜ 4 + ⎟ = 4 + 0 = 4
x → ∞⎝
x⎠
x2
x2 + 1
1
= lim
x → −∞ ⎛
2
x + 1⎞
⎜
⎟
⎜ − x2 ⎟
⎝
⎠
−1
= lim
x → −∞
1 + (1 x 2 )
x → −∞
5 x3 2
5
=
4 x3 2 + 1
4
(b) lim
)
x → −∞
27. lim
5 x3 2
= 0
4x2 + 1
18. (a) lim
x2
x
26. lim
2
x→∞
2
= −1, for x < 0 we have x = −
3 − 2x
2
(b) lim
= −
x → ∞ 3x − 1
3
x→∞
x − x
1
= lim
x → −∞ ⎛
2
x − x⎞
⎜
⎟
⎜ − x2 ⎟
⎝
⎠
−1
= lim
x → −∞
1 − (1 x )
x → −∞
2
x→∞
x
25. lim
5x2 + 2
x 1 + (3 x 2 )
5 x 2 + (2 x)
1 + 3 x2
= ∞
Limit does not exist.
29. lim
x→∞
x2 − 1
= lim
x→∞
2x − 1
= lim
x→∞
x2 − 1
x2
2 −1x
1 − 1 x2
1
=
2 −1x
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
(
6
⎛
x4 − 1
x4 − 1 ⎜1 − x
= lim
3
x → −∞ x 3 − 1 ⎜
x3 − 1
⎜ 1x
⎝
30. lim
x → −∞
1 x2 − 1 x
−1 + 1 x3
= lim
x → −∞
(
for x < 0, we have −
31. lim
x→∞
⎟
⎟
⎠
6
= 0,
+ 1)
13
= lim
x→∞
x1 3 + 1 x
23
(1 + 1 x 2 )
13
39. lim log10 (1 + 10− x ) = 0
x →∞
⎛5
x2 + 1⎞
5
40. lim ⎜ + ln
⎟ =
2
x→∞ 2
x
2
⎝
⎠
= ∞
= 0−
42. lim arcsec(u + 1) =
u →∞
2x
( x6 − 1)
13
x → −∞
(1 x )
2x
lim
13
13
x → −∞
( x6 − 1) (1 ( x6 ) )
2
=
= lim
x → −∞
33. lim
x→∞
2 x
(1 − 1 x )
6
13
= 0
1
= 0
2 x + sin x
lim
π
2
π
2
= −1
x +1
Therefore, y = 1 and y = −1 are both horizontal
asymptotes.
4
y=1
y = −1
−6
6
⎛1⎞
34. lim cos⎜ ⎟ = cos 0 = 1
x→∞
⎝ x⎠
35. Because ( −1 x) ≤ (sin 2 x) x ≤ (1 x) for all x ≠ 0, you
= −
=1
x +1
x
x → −∞
2
x +1
x
lim
x→∞
π
x
43. f ( x) =
Limit does not exist.
32. lim
−4
44. f ( x) =
3x + 2
x − 2
have by the Squeeze Theorem,
y = 3 is a horizontal asymptote (to the right).
1
sin 2 x
1
lim − ≤ lim
≤ lim
x→∞
x→∞
x→∞ x
x
x
sin 2 x
0 ≤ lim
≤ 0.
x→∞
x
y = −3 is a horizontal asymptote (to the left).
Therefore, lim
x→∞
343
⎛8⎞
41. lim (8t −1 − arctan t ) = lim ⎜ ⎟ − lim arctan t
t →∞
t →∞ ⎝ t ⎠
t →∞
)
x 6 = x3
⎛
⎞
x + 1 ⎜ 1 x2 3 ⎟
= lim
1
3
1
3
x→∞
( x 2 + 1) ⎜⎜⎝ 1 ( x 2 ) ⎟⎟⎠
x +1
( x2
) ⎞⎟
Limits at Infinity
sin 2 x
= 0.
x
8
y=3
− 10
10
y = −3
−6
36. lim
x→∞
x − cos x
cos x ⎞
⎛
= lim ⎜1 −
⎟
x → ∞⎝
x
x ⎠
=1−0 =1
Note:
x2 + 2
lim f ( x) = 3
x→∞
cos x
= 0 by the Squeeze Theorem because
x
1
cos x
1
− ≤
≤ .
x
x
x
lim
x→∞
(
x →∞
3x
45. f ( x ) =
lim f ( x ) = −3
x → −∞
Therefore, y = 3 and y = −3 are both horizontal
asymptotes.
6
)
37. lim 2 − 5e − x = 2
y=3
−9
8
38. lim
= 2
x →∞ 4 − 10 − x 2
9
y = −3
−6
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
344
Chapter 4
Applications of Differentiation
9 x2 − 2
2x + 1
46. f ( x) =
47. lim x sin
x→∞
(Let x = 1 t. )
3
is a horizontal asymptote (to the right).
2
y =
y = −
1
sin t
= lim
=1
+
x
t
t →0
48. lim x tan
3
is a horizontal asymptote (to the left).
2
x→∞
⎡ sin t
1
tan t
1 ⎤
= lim
= lim ⎢
⋅
⎥
+
+
cos t ⎦
x
t
x→0
x→0 ⎣ t
= (1)(1) = 1
6
y=
y = −3
2
(Let x = 1 t. )
3
2
−9
9
−6
(
49. lim x +
x → −∞
(
50. lim x −
x →∞
(
51. lim 3x +
x → −∞
⎡
x 2 + 3 = lim ⎢ x +
x → −∞
⎢⎣
)
)
(
⎡
x 2 + x = lim ⎢ x −
x →∞
⎢⎣
)
x2 + 3 ⋅
)
(
⎡
9 x 2 − x = lim ⎢ 3 x +
x → −∞
⎢⎣
)
x → −∞
3x −
= lim
x → −∞
x → −∞
(
x→∞
16 x 2 − x
) 44xx ++
x +
9x2 − x ⋅
−1
= −
1 + (1 x)
1
2
9x2 − x ⎤
⎥
9 x 2 − x ⎥⎦
3x −
3x −
x
= lim
52. lim 4 x −
−x
x2 + x ⎤
= lim
⎥ = lim
x →∞
x →∞ 1 +
x 2 + x ⎥⎦
x + x2 + x
)
(
= lim
x −
x +
x2 + x ⋅
x2 + 3 ⎤
−3
= 0
⎥ = lim
2
x → −∞
x + 3 ⎥⎦
x − x2 + 3
x −
3−
3+
9 x2 − x
1
9x
2
(for x < 0 you have x = −
− x
− x2
1
9 − (1 x)
16 x 2 − x
16 x 2 − x
= lim
x→∞
= lim
x→∞
= lim
x→∞
=
x2
)
1
6
16 x 2 − (16 x 2 − x)
4x +
(16 x 2
− x)
x
4x +
4+
16 x 2 − x
1
16 − 1 x
1
1
=
=
4+ 4
8
53.
x
100
101
102
103
104
105
106
f(x)
1
0.513
0.501
0.500
0.500
0.500
0.500
(
lim x −
x →∞
)
x( x − 1) = lim
x →∞
x −
x2 − x x +
⋅
1
x +
x2 − x
x − x
2
= lim
x →∞
x
x+
x − x
2
= lim
x →∞
1
1+
1 − (1 x)
=
1
2
2
−1
8
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
54.
x
100
101
102
103
104
105
106
f(x)
1.0
5.1
50.1
500.1
5000.1
50,000.1
500,000.1
x2 − x
lim
x→∞
x2 − x
⋅
1
x2 + x
x2 − x
x2 + x
x2 − x
= lim
x→∞
x3
x2 + x
x2 − x
345
Limits at Infinity
= ∞
Limit does not exist.
30
0
50
0
55.
x
100
101
102
103
104
105
106
f(x)
0.479
0.500
0.500
0.500
0.500
0.500
0.500
Let x = 1 t.
sin (t 2)
1 sin (t 2)
1
⎛ 1 ⎞
= lim
=
lim x sin ⎜ ⎟ = lim
+
+
t
t 2
2
t →0
t →0 2
⎝ 2x ⎠
x→∞
1
−2
2
−1
56.
x
100
101
102
103
104
105
106
f(x)
2.000
0.348
0.101
0.032
0.010
0.003
0.001
lim
x→∞
x +1
= 0
x x
3
0
25
−1
57. (a) lim f ( x) = 4 means that f ( x) approaches 4 as x
x→∞
becomes large.
(b)
lim f ( x ) = 2 means that f ( x) approaches 2 as x
x → −∞
becomes very large (in absolute value) and negative.
f ′( x) < 0 for x < 2.
8
f ′( x) > 0 for x > 2.
4
lim f ( x ) = lim f ( x) = 6
x → −∞
58. Answers will vary.
y
59. x = 2 is a critical number.
x→∞
For example, let
−6
f ( x) =
+ 6.
2
0.1( x − 2) + 1
−2
x
2
4
6
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
346
Chapter 4
Applications of Differentiation
60. Yes. For example, let f ( x) =
6 x − 2
(x
− 2) + 1
2
64. y =
.
x − 4
x −3
y
5
4
Intercepts: (0, 4 3), ( 4, 0)
y
3
2
Symmetry: none
8
Horizontal asymptote: y = 1
Vertical asymptote: x = 3
4
−1
x
1
3
2
4
5
6
7
2
3
4
−2
−3
2
x
−4
−2
2
4
65. y =
6
−2
61. (a)
x +1
x2 − 4
y
4
3
Intercepts: (0, −1 4), ( −1, 0)
4
Symmetry: none
−4
5
2
1
−4
x
−1
Horizontal asymptote: y = 0
−2
Vertical asymptotes: x = ± 2
−4
−3
−2
66. y =
(b) When x increases without bound, 1 x approaches
zero and e1 x approaches 1. Therefore, f ( x)
Symmetry: origin
horizontal asymptote at y = 1. As x approaches
approaches zero from the left, 1 x approaches −∞,
67. y =
approaches zero, and f ( x) approaches 2. The
1 2
6
x2
x + 16
2
Intercept: (0, 0)
limit does not exist because the left limit does not
equal the right limit. Therefore, x = 0 is a
nonremovable discontinuity.
x
−1
−2
−3
−4
−5
−6
Vertical asymptotes: x = ± 3
approaches ∞, and f ( x) approaches zero. As x
62. (a)
−5−4
Horizontal asymptote: y = 0
zero from the right, 1 x approaches ∞, e1 x
e
6
5
4
3
2
1
Intercept: (0, 0)
approaches 2 (1 + 1) = 1. So, f ( x) has a
1x
y
2x
9 − x2
y
Symmetry: y-axis
2
Horizontal asymptote: y = 1
1
y
y′ =
4
f
3
2
f′
1
2
3
−3
x→∞
68. y =
lim f ′( x) = 0
x
−8 −6 −4 −2
2
2
4
6
8
−1
−2
Symmetry: y-axis
x→∞
Horizontal asymptote: y = 2
of a horizontal line, lim f ′( x) = 0.
x→∞
x
1− x
Intercept: (0, 0)
Vertical asymptotes: x = ± 2
y
Horizontal asymptote: y = −1
2
Vertical asymptote: x = 1
x
−2
−3
−4
1
2
3
4
5
y
( x − 4)
16( x 2 + 4)
y′′ =
3
( x 2 − 4)
2
3
−1
4x
y′ = −
4
1
Symmetry: none
2x2
x − 4
2
Intercept: (0, 0)
x→∞
(c) Because lim f ( x ) = 3, the graph approaches that
63. y =
+ 16)
4
−4
(b) lim f ( x) = 3
( x2
Relative minimum: (0, 0)
x
−4
32 x
2
y′′ = (0) < 0
8
6
4
2
−4
−2
x
2
4
6
Relative maximum: (0, 0)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
y
69. xy 2 = 9
Domain: x > 0
3
Intercepts: none
1
2
Intercept:
x
Symmetry: x-axis
y = ±
−1
1
2
3
4
5
6
y = 0 = 3+
7
−2
3
x
−3
Horizontal asymptote: y = 3
Horizontal asymptote: y = 0
Vertical asymptote: x = 0
Vertical asymptote: x = 0
y
9
x2
8
7
6
5
4
3
2
y
Intercepts: none
Symmetry: y-axis
Horizontal asymptote: y = 0
Vertical asymptote: x = 0
71. y =
2
−8 −6 −4
4
6
76. y =
7
6
5
4
3
2
(0, 0)
Horizontal asymptote: y = 3
Vertical asymptote: x = 1
Symmetry: none
Horizontal asymptote: y = 1
x
Vertical asymptote: x = 0
1 2 3 4 5 6
y
−2
y
4
8
2
6
Intercept: (0, 0)
4
2
Symmetry: origin
−4
x
−4 −3 −2 −1
Vertical asymptotes: x = ± 1
77. y =
3
2x2 − 3
73. y = 2 − 2 =
x
x2
Symmetry: y-axis
Vertical asymptote: x = 0
x2 − 4
1
x
−4 −3 −2
2
3
4
Horizontal asymptote: none
Vertical asymptotes: x = ± 2 (discontinuities)
y
20
16
12
8
4
y
6
5
−5 −4 −3 −2 −1
Intercept: (1, 0)
Horizontal asymptote: y = 1
x3
Symmetry: origin
2
Vertical asymptote: x = 0
Symmetry: none
4
Intercepts: none
3
Horizontal asymptote: y = 2
1
x −1
=
x
x
2
Domain: ( −∞, − 2), ( 2, ∞)
y
4
3 ⎞
, 0⎟
2 ⎟⎠
x
−2
−2
1
Horizontal asymptote: y = 0
74. y = 1 −
4
4 + x2
+1=
2
x
x2
Intercept: none
−4 −3 −2 −1
3x
1 − x2
⎛
Intercepts: ⎜⎜ ±
⎝
1 2 3 4 5
8
y
Symmetry: none
72. y =
2
−2
x
−4 −3 −2 −1
x
−4
3x
x −1
Intercept:
2
2
2 ⎛ 2 ⎞
⇒
= − 3 ⇒ x = − ; ⎜ − , 0⎟
x
x
3 ⎝ 3 ⎠
Symmetry: none
−4
70. x 2 y = 9 ⇒ y =
347
2
x
75. y = 3 +
4
Limits at Infinity
−4 −3 −2 −1
−2
−3
−4
−5
−6
x
1
2
3
4
x
1 2 3 4 5
−8
−12
−16
−20
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
348
Chapter 4
78. y =
Applications of Differentiation
x
x − 4
Domain: all x ≠ 0
Domain: ( −∞, − 2), ( 2, ∞)
f ′( x) =
Intercepts: none
Symmetry: origin
lim
x
x2 − 4
x
= 1, lim
30
⇒ No points of inflection
x4
Vertical asymptote: x = 0
= −1.
x2 − 4
x → −∞
10
⇒ No relative extrema
x3
f ′′( x) = −
Horizontal asymptotes: y = ± 1 because
x→∞
5
x2
79. f ( x ) = 9 −
2
Horizontal asymptote: y = 9
Vertical asymptotes: x = ± 2 (discontinuities)
y=9
12
x=0
y
5
4
3
2
−5 −4 −3
−1
−6
6
−2
x
1
3 4 5
−2
−3
−4
−5
80.
f ( x) =
f ′( x) =
1
1
=
x2 − x − 2
( x + 1)( x − 2)
−( 2 x − 1)
(x
( x2
f ′′( x) =
2
− x − 2)
= 0 when x =
2
1
.
2
− x − 2) ( −2) + ( 2 x − 1)( 2)( x 2 − x − 2)( 2 x − 1)
2
( x2
− x − 2)
4
=
⎛1⎞
⎛1 4⎞
Because f ′′⎜ ⎟ < 0, ⎜ , − ⎟ is a relative maximum.
⎝ 2⎠
⎝2 9⎠
6( x 2 − x + 1)
( x2
− x − 2)
x = −1
Because f ′′( x) ≠ 0, and it is undefined in the domain of f,
−3
Vertical asymptotes: x = −1, x = 2
3
y=0
Horizontal asymptote: y = 0
f ( x) =
f ′( x) =
f ′′( x) =
=
x=2
2
( 12 , − 94(
there are no points of inflection.
81.
3
−2
x − 2
x − 2
=
x2 − 4 x + 3
x
( − 1)( x − 3)
( x2
− 4 x + 3) − ( x − 2)( 2 x − 4)
( x2
( x2
− 4 x + 3)
2
=
− x2 + 4x − 5
( x2
− 4 x + 3)
2
≠ 0
− 4 x + 3) ( −2 x + 4) − ( − x 2 + 4 x − 5)( 2)( x 2 − 4 x + 3)( 2 x − 4)
2
( x2
2( x 3 − 6 x 2 + 15 x − 14)
( x2
− 4 x + 3)
3
=
− 4 x + 3)
4
2( x − 2)( x 2 − 4 x + 7)
( x2
− 4 x + 3)
3
= 0 when x = 2.
Because f ′′( x) > 0 on (1, 2) and f ′′( x) < 0 on ( 2, 3), then
(2, 0) is a point of inflection.
2
x=3
−1
5
Vertical asymptotes: x = 1, x = 3
Horizontal asymptote: y = 0
y=0
x=1
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
f ( x) =
82.
f ′( x) =
x +1
x2 + x + 1
− x ( x + 2)
2
3x 2 + 1
2
g ′( x) =
= 0 when x = 0, − 2.
( x 2 + x + 1)
2( x3 + 3 x 2 − 1)
f ′′( x) =
3
( x 2 + x + 1)
(3x 2
+ 1)
32
−18 x
g ′′( x) =
= 0
349
2x
g ( x) =
84.
Limits at Infinity
(3x
2
+ 1)
52
No relative extrema. Point of inflection: (0, 0).
when x ≈ 0.5321, − 0.6527, − 2.8794.
f ′′(0) < 0
Horizontal asymptotes: y = ±
Therefore, (0, 1) is a relative maximum.
2
3
No vertical asymptotes
f ′′( −2) > 0
4
y= 2
3
Therefore,
−6
1⎞
⎛
⎜ −2, − ⎟
3⎠
⎝
6
y=− 2
3 −4
is a relative minimum.
Points of inflection:
(0.5321, 0.8440), (−0.6527, 0.4491) and
⎛ x ⎞
85. g ( x) = sin ⎜
⎟, 3 < x < ∞
⎝ x − 2⎠
⎛ x ⎞
−2cos⎜
⎟
⎝ x − 2⎠
g ′( x) =
2
( x − 2)
(−2.8794, − 0.2931)
Horizontal asymptote: y = 0
(− 0.6527, 0.4491)
2
Horizontal asymptote: y = sin (1)
(0.5321, 0.8440)
−3
Relative maximum:
x
π
2π
=
⇒ x =
≈ 5.5039
x − 2
2
π −2
3
(0, 1)
(−2, − 13(
y=0
No vertical asymptotes
−2
(− 2.8794, − 0.2931)
1.2
f ( x) =
83.
f ′( x) =
f ′′( x) =
3x
4x2 + 1
3
(4 x 2
+ 1)
y = sin(1)
⇒ No relative extrema
32
−36 x
(4 x
2
+ 1)
3
12
0
= 0 when x = 0.
52
Point of inflection: (0, 0)
Horizontal asymptotes: y = ±
No vertical asymptotes
2
3
2
86.
2 sin 2 x
; Hole at (0, 4)
x
4 x cos 2 x − 2 sin 2 x
f ′( x) =
x2
f ( x) =
There are an infinite number of relative extrema. In the
interval ( −2π , 2π ), you obtain the following.
Relative minima: ( ± 2.25, − 0.869), ( ± 5.45, − 0.365)
Relative maxima: ( ± 3.87, 0.513)
y= 3
−3
( π 2−π 2 , 1(
2
3
y= −3
2
Horizontal asymptote: y = 0
6
No vertical asymptotes
y=0
−2
(−3.87, 0.513)
− 2␲
−2
(−5.45, −0.365)
(−2.25, −0.869)
(3.87, 0.513)
2␲
(5.45, −0.365)
(2.25, −0.869)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
350
87.
Chapter 4
Applications of Differentiation
f ( x) = 2 + ( x 2 − 3)e − x
f ′( x) = −e − x ( x + 1)( x − 3)
90. f ( x ) = −
(a)
Critical numbers: x = −1, x = 3
x3 − 2 x 2 + 2
1
1
, g ( x) = − x + 1 − 2
2x2
2
x
4
f=g
Relative minimum: ( −1, 2 − 2e) ≈ (−1, − 3.4366)
−6
6
Relative maximum: (3, 2 + 6e −3 ) ≈ (3, 2.2987)
−4
Horizontal asymptote: y = 2
4
x3 − 2 x 2 + 2
2x2
3
⎡x
2 x2
2 ⎤
= −⎢ 2 −
+
⎥
2
2
2
2
x
x
x2 ⎦
⎣
(b) f ( x ) = −
y=2
−2
6
(−1, 2 − 2e)
1
1
= − x + 1 − 2 = g ( x)
2
x
−4
70
(c)
88.
10 ln x
10 ln x
=
, Domain: x > 0
2
x5 2
x x
5
f ′( x) = 7 2 ( 2 − 5 ln x)
x
f ( x) =
Critical number: x = e
25
−80
−70
The graph appears as the slant asymptote
1
y = − x + 1.
2
≈ 1.4918
Relative maximum: (1.4918, 1.4715)
Horizontal asymptote: y = 0
3
91.
(1.4918, 1.4715)
−1
10
y=0
C =
x3 − 3x 2 + 2
2
, g ( x) = x +
x( x − 3)
x( x − 3)
8
(a)
C
x
500
x
500
⎛
⎞
lim ⎜ 0.5 +
⎟ = 0.5
x → ∞⎝
x ⎠
C = 0.5 +
93. lim N (t ) = ∞
f=g
t →∞
−4
8
(b) f ( x) =
lim E (t ) = c
t →∞
−2
x3 − 3x 2 + 2
x( x − 3)
94. (a)
2
=
+
x( x − 3)
x( x − 3)
= x +
lim T = 1700°
t → 0+
This is the temperature of the kiln.
x 2 ( x − 3)
(c)
v1
⎡
⎤
1
⎥ = 100[1 − 0] = 100%
lim 100 ⎢1 −
c
v2 → ∞
⎢
(v1 v2 ) ⎥⎦
⎣
92. C = 0.5 x + 500
−6
89. f ( x ) =
80
2
= g ( x)
x( x − 3)
70
(b) lim T = 72°
t →∞
This is the temperature of the room.
(c) No. y = 72 is the horizontal asymptote.
95. (a) lim
n →∞
− 80
80
(b)
− 70
The graph appears as the slant asymptote y = x.
0.83
= 0.83 = 83%
1 + e −0.2 n
P′ =
0.166e −0.2 n
(1 + e−0.2n )
2
P′(3) ≈ 0.038
P′(10) ≈ 0.017
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
96. (a) T1 (t ) = −0.003t 2 + 0.68t + 26.6
(b)
x2 + 2
lim f ( x) = 6 = L
(a)
T1
351
6x
98. f ( x ) =
90
Limits at Infinity
x→∞
lim f ( x) = −6 = K
x → −∞
− 10
130
(c)
6 x1
(b) f ( x1 ) + ε =
− 10
+ε = 6
x12 + 2
90
6 x1 = (6 − ε )
T2
x12 + 2
36 x12 = ( x12 + 2)(6 − ε )
− 10
120
36 x12 − (6 − ε ) x12 = 2(6 − ε )
2
x12 ⎡⎣36 − 36 + 12ε − ε 2 ⎤⎦ = 2(6 − ε )
2
2
− 10
1451 + 86t
58 + t
(d) T1 (0) ≈ 26.6°
T2 =
T2 (0) ≈ 25.0°
(e) lim T2 =
t →∞
T1 has no horizontal asymptote.
2x2
97. f ( x ) = 2
x + 2
(a) lim f ( x ) = 2 = L
2
12ε − ε 2
(d) N = x2 = (ε − 6)
2
12ε − ε 2
3x
x→∞
2 x12
+ε = 2
+ 2
+ 2ε =
x2 + 3
f ( x1 ) + ε =
x12
+ε
(c) M = x1 = (6 − ε )
99. lim
x→∞
x12
2 x12
+ 4
= 3
3 x1
x12 + 3
x1 =
x2 = − x1 by symmetry
4 − 2ε
ε
2
9 x12 − (3 − ε ) x12 = 3(3 − ε )
2
x12 (9 − 9 + 6ε − ε 2 ) = 3(3 − ε )
2
2
> 0. For x > M :
x >
x12 =
4 − 2ε
2 x 2 + x 2ε + 2ε > 2 x 2 + 4
2x2
+ε > 2
x + 2
3(3 − ε )
2
6ε − ε 2
x1 = (3 − ε )
ε
x 2ε > 4 − 2ε
x12 + 3
9 x12 = (3 − ε ) ( x12 + 3)
4 − 2ε
ε
+ε = 3
3 x1 = (3 − ε )
ε = 4 − 2ε
x12
(c) Let M =
12ε − ε 2
2
12ε − ε 2
x2 = − x1 by symmetry
(f ) No. The limiting temperature is 86°.
2 x12
2
x1 = (6 − ε )
86
= 86
1
(b) f ( x1 ) + ε =
2(6 − ε )
x12 =
2
3
6ε − ε 2
3
6ε − ε 2
Let M = x1 = (3 − ε )
(a) When ε = 0.5:
2
2x2
− 2 > −ε = ε
2
x + 2
f ( x) − L > ε
(d) Similarly, N = −
4 − 2ε
ε
.
M = (3 − 0.5)
3
6(0.5) − (0.5)
2
=
5 33
11
(b) When ε = 0.1:
M = (3 − 0.1)
3
6(0.1) − (0.1)
2
=
29 177
59
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
352
Chapter 4
3x
100. lim
2
= 0. Let ε > 0 be given. You need
x
M > 0 such that
= −3
x +3
x → −∞
Applications of Differentiation
2
102. lim
x→∞
3 x1
f ( x1 ) − ε =
x12 + 3
− ε = −3
3 x1 = (ε − 3)
x12
+ 3
9 x12 = (ε − 3) ( x12 + 3)
2
9 x12 − (ε − 3) x12 = 3(ε − 3)
2
x12 (9 − ε 2 + 6ε − 9) = 3(ε − 3)
2
2
x12 =
3(ε − 3)
6ε − ε
x1 = (ε − 3)
Let x1 = N = (ε − 3)
2
2
3
6ε − ε 2
=
−5 33
11
6(0.1) − (0.1)
1
= 0. Let ε > 0 be given. You need
x2
M > 0 such that
x→∞
1
1
1
− 0 = 2 < ε whenever x > M .
x2
x
ε
Let M =
⇒ x >
1
ε
1
1
ε
ε
x → −∞
.
1
ε
⇒
1
< ε ⇒ f ( x) − L < ε .
x2
.
−1
3
ε
,
⇒ f ( x) − L < ε .
104. lim
⇒ x2 >
−1
3
1
> −3 ε
x
1
− < 3 ε
x
1
− 3 < ε
x
ε
For x > M , you have
x >
−1
1
−1
< ε ⇒ − x3 >
⇒ x < 13
ε
ε
x3
For x < N =
101. lim
1
−1
− 0 = 3 < ε whenever x < N .
x3
x
f ( x) − L =
2
−29 177
=
59
x2 >
1
= 0. Let ε > 0. You need N < 0 such that
x3
Let N =
3
2
< ε ⇒ f ( x) − L < ε .
x
x > 2ε ⇒
x → −∞
2
x
1
4
>
⇒ x > 2
2
ε
ε
For x > M = 4 ε , you have
2
(b) When ε = 0.1:
f ( x) − L =
2
< ε ⇒
x
103. lim
3
N = (0.1 − 3)
x > M.
2
3
6ε − ε 2
6(0.5) − (0.5)
2
< ε whenever
x
Let M = 4 ε .
(a) When ε = 0.5:
N = (0.5 − 3)
2
−0 =
x
f ( x) − L =
1
= 0. Let ε > 0 be given.
x − 2
You need N < 0 such that
1
−1
− 0 =
< ε
f ( x) − L =
x − 2
x − 2
whenever x < N .
−1
−1
1
< ε ⇒ x − 2 <
⇒ x < 2−
ε
ε
x − 2
Let N = 2 −
x − 2 <
1
ε
. For x < N = 2 −
1
ε
,
−1
ε
−1
< ε
x − 2
⇒ f ( x) − L < ε .
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.5
Limits at Infinity
353
106. line: y + 2 = m( x − 0) ⇒ mx − y − 2 = 0
105. line: mx − y + 4 = 0
y
y
5
4
(4, 2)
y = mx + 4
2
3
d
2
x
−4
(3, 1)
1
(0, − 2)
x
−2 −1
−1
1
2
3
4
4
−4
(a) d =
=
(b)
Ax1 + By1 + C
A + B
2
2
m(3) − 1(1) + 4
=
m2 + 1
3m + 3
(a) d =
m2 + 1
=
6
(b)
− 12
m→∞
m → −∞
The line approaches the vertical line x = 0. So, the
distance from (3, 1) approaches 3.
x→∞
q( x)
=
2
m( 4) − 1( 2) − 2
m2 + 1
4m − 4
m2 + 1
10
−5
(c) lim d ( m) = 3 = lim d ( m)
p( x )
A + B
2
12
−2
107. lim
Ax1 + By1 + C
= lim
x→∞
5
−3
(c) lim d ( m) = 4; lim d ( m) = 4
m→∞
m → −∞
The line approaches the vertical line x = 0. So, the
distance from ( 4, 2) approaches 4.
an x n + " + a1 x + a0
bm x m + " + b1 x + b0
Divide p( x) and q( x) by x m .
an
a
a
+ " + m1−1 + m0
m−n
p( x)
0+"+ 0+ 0
0
x
x
x
= lim
=
=
= 0.
Case 1: If n < m: lim
b1
b0
x →∞ q ( x )
x →∞
+
+
+
0
0
"
b
b
m
m
bm + " + m −1 + m
x
x
a1
a0
an + " + m − 1 + m
p( x)
x
x = an + " + 0 + 0 = a n .
= lim
Case 2: If m = n: lim
b1
b
x →∞ q ( x )
x →∞
bm + " + 0 + 0
bm
bm + " + m −1 + m0
x
x
a
a
an x n − m + " + m1−1 + m0
p( x)
±∞ + " + 0
x
x
= lim
Case 3: If n > m: lim
=
= ±∞.
b1
b0
x →∞
x →∞ q ( x )
bm + " + 0
bm + " + m −1 + m
x
x
108. lim x 3 = ∞. Let M > 0 be given. You need N > 0 such that f ( x ) = x 3 > M whenever x > N .
x→∞
x 3 > M ⇒ x > M 1 3. Let N = M 1 3 . For x > N = M 1 3 , x > M 1 3 ⇒ x 3 > M ⇒ f ( x) > M .
109. False. Let f ( x ) =
2x
x2 + 2
. (See Exercise 57(b).)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
354
Chapter 4
Applications of Differentiation
110. False. Let y1 =
x + 1, then y1 (0) = 1. So y1′ = 1 2
(
)
x + 1 and y1′ (0) = 1 2. Finally, y1′′ = −
1
4( x + 1)
32
1
1
1
and y1′′ (0) = − . Let p = ax 2 + bx + 1, then p(0) = 1. So, p′ = 2ax + b and p′(0) =
⇒ b = .
4
2
2
1
1
Finally, p′′ = 2a and p′′(0) = − ⇒ a = − . Therefore,
4
8
⎧⎪( −1 8) x 2 + (1 2) x + 1, x < 0
f ( x) = ⎨
and f (0) = 1,
x ≥ 0
⎪⎩ x + 1,
⎧⎪(1 2) − (1 4) x, x < 0
1
f ′( x) = ⎨
and f ′(0) = , and
2
x
x
1
2
+
1
,
>
0
⎪⎩
(
)
⎧
(−1 4), x < 0
1
⎪
f ′′( x) = ⎨
and f ′′(0) = − .
32
4
⎪⎩−1 4( x + 1) , x > 0
(
)
f ′′( x) < 0 for all real x, but f ( x) increases without bound.
Section 4.6 A Summary of Curve Sketching
1.
1
−3
x − 2
1
y′ = −
⇒ undefined when x = 2
( x − 2) 2
y =
y′′ =
2
( x − 2)
3
⇒ undefined when x = 2
7⎞
⎛7 ⎞ ⎛
Intercepts: ⎜ , 0 ⎟, ⎜ 0, − ⎟
3
2⎠
⎝
⎠ ⎝
Vertical asymptote: x = 2
Horizontal asymptote: y = −3
y′
y′′
Conclusion
−∞ < x < 2
−
−
Decreasing, concave down
2 < x < ∞
−
+
Decreasing, concave up
y
No relative extrema, no points of inflection
y
x=2
( 73 , 0 (
x
4
−2
−4
( 0, − 72 (
y = −3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
2.
y =
y′ =
x
x +1
A Summary of Curve Sketching
355
2
1 − x2
(x
y′′ = −
2
+ 1)
2
=
2 x (3 − x 2 )
( x2
+ 1)
3
(1 − x)( x
(x
2
+ 1)
+ 1)
2
= 0 when x = ± 1.
= 0 when x = 0, ±
3.
Horizontal asymptote: y = 0
y′
y′′
Conclusion
–
–
Decreasing, concave down
–
0
Point of inflection
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
–
0
Point of inflection
y
−∞ < x < −
x = −
−
3
3
4
−
3
3 < x < −1
1
−
2
x = −1
−1 < x < 0
x = 0
0
0 < x <1
1
2
x =1
1< x <
x =
y
3
4
3
3 < x < ∞
3.
y =
y′ =
x2
x2 + 3
6x
( x 2 + 3)
18(1 − x 2 )
y′′ =
3
( x2 + 3)
2
(
1
3
–
+
(1, 12 )
3, 3
4
)
(−1, − 12 )
x
1
2
(0, 0)
(−
Decreasing, concave up
3, − 3
4
y=0
)
= 0 when x = 0.
= 0 when x = ±1.
Horizontal asymptote: y = 1
y
−∞ < x < −1
x = −1
1
4
−1 < x < 0
x = 0
0
0 < x <1
x =1
1< x < ∞
1
4
y′
y′′
Conclusion
–
–
Decreasing, concave down
–
0
Point of inflection
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
y
y=1
1
(− 1, 14 (
−4
(1, 14 (
(0, 0))
2
x
4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
356
4.
Chapter 4
y =
y′ =
Applications of Differentiation
x2 + 1
x2 − 4
−10 x
= 0 when x = 0 and undefined when x = ± 2.
( x 2 − 4)
10(3x 2 + 4)
y′′ =
3
( x 2 − 4)
2
< 0 when x = 0.
Intercept: (0, − 1 4)
Symmetric about y-axis
Vertical asymptotes: x = ±2
Horizontal asymptote: y = 1
y′
y′′
Conclusion
−∞ < x < −2
+
+
Increasing, concave up
−2 < x < 0
+
−
Increasing, concave down
y
x = 0
− 14
Relative maximum
0 < x < 2
−
−
Decreasing, concave down
2 < x < ∞
−
+
Decreasing, concave up
y
8
6
(0, − 14 )
4
2
x
−8 −6 −4 −2
2
x = −2
5.
y =
y′ =
y′′ =
4
y=1
x=2
3x
x2 − 1
−3( x 2 + 1)
( x 2 − 1)
6 x( x 2 + 3)
3
( x 2 − 1)
2
undefined when x = ±1
y′
y′′
Conclusion
− ∞ < x < −1
−
−
Decreasing, concave down
−1 < x < 0
−
+
Decreasing, concave up
−3
0
Point of inflection
y
x = 0
Intercept: (0, 0)
0
Symmetry with respect to origin
0 < x <1
−
−
Decreasing, concave down
Vertical asymptotes: x = ±1
1< x < ∞
−
+
Decreasing, concave up
Horizontal asymptote: y = 0
x = −1 y x = 1
y=0
−3 −2 −1
1
(0, 0)
x
1
2
3
4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
6.
A Summary of Curve Sketching
357
x −3
3
=1−
x
x
3
f ′( x) = 2 undefined when x = 0
x
6
f ′′( x) = − 3 ≠ 0
x
f ( x) =
Vertical asymptote: x = 0
Intercept: (3, 0)
Horizontal asymptote: y = 1
y′
y′′
Conclusion
−∞ < x < 0
+
+
Increasing, concave up
0 < x < ∞
+
−
Increasing, concave down
y
y
4
3
y=1
2
1
x
−4 −3 −2 −1
x=0
1
3
−2
4
(3, 0)
−3
−4
7.
32
x2
f ( x) = x +
f ′( x) = 1 −
( x − 4)( x 2 + 4 x + 16)
64
=
= 0 when x = 4 and undefined when x = 0.
x3
x3
192
x4
f ′′( x) =
(
Intercept: −2 3 4, 0
)
Vertical asymptote: x = 0
Slant asymptote: y = x
y′
y′′
Conclusion
−∞ < x < 0
+
+
Increasing, concave up
0 < x < 4
−
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
y
x = 4
6
4 < x < ∞
y
(−2 4, 0)
3
(4, 6)
8
6
4
y=x
2
x
−8 −6
2
−4
4
6
8
x=0
−6
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
358
Chapter 4
Applications of Differentiation
x3
9x
= x + 2
x −9
x −9
8. f ( x ) =
2
x 2 ( x 2 − 27)
f ′( x) =
= 0 when x = 0, ± 3 3 and is undefined when x = ± 3.
( x 2 − 9)
18 x( x 2 + 27)
f ′′( x) =
3
( x 2 − 9)
2
= 0 when x = 0
Intercept: (0, 0)
Symmetry: origin
Vertical asymptotes: x = ±3
Slant asymptote: y = x
y′
y′′
Conclusion
+
−
Increasing, concave down
0
−
Relative maximum
−3 3 < x < −3
−
−
Decreasing, concave down
−3 < x < 0
−
+
Decreasing, concave up
0
0
Point of inflection
0 < x < 3
−
−
Decreasing, concave down
3 < x < 3 3
−
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
y
−∞ < x < −3 3
x = −3 3
−
9 3
2
x = 0
0
3 3 < x < ∞
9.
3 3, 9 3
2
)
9
x = −3
y=x
x
−12 −9 −6 −3
3
−6
9 3
2
x = 3 3
(
y
(
6
9 12
x=3
−3 3, − 9 3
2
)
x 2 − 6 x + 12
4
= x − 2+
x − 4
x − 4
( x − 2)( x − 6) = 0 when x = 2, 6 and is undefined when x = 4.
4
y′ = 1 −
=
2
( x − 4)
( x − 4) 2
y =
y′′ =
8
(x
− 4)
3
Vertical asymptote: x = 4
Slant asymptote: y = x − 2
y′
y′′
Conclusion
+
−
Increasing, concave down
0
−
Relative maximum
2 < x < 4
−
−
Decreasing, concave down
4 < x < 6
−
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
y
−∞ < x < 2
x = 2
x = 6
6 < x < ∞
−2
6
y
x=4
8
6
(6, 6)
4
y=x−2
2
(0, − 3)
x
6
8
10
(2, − 2)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
10.
A Summary of Curve Sketching
359
4
− x2 − 4x − 7
= −x − 1 −
x +3
x + 3
x2 + 6x + 5
( x + 1)( x + 5) = 0 when x = −1, − 5 and is undefined when x = − 3.
y′ = −
= −
2
2
( x + 3)
( x + 3)
y =
y′′ =
−8
(x
+ 3)
3
(
Intercept: 0, − 73
)
No symmetry
Vertical asymptote: x = − 3
Slant asymptote: y = − x − 1
y
−∞ < x < −5
x = −5
6
y′
y′′
Conclusion
−
+
Decreasing, concave up
0
+
Relative minimum
−5 < x < −3
+
+
Increasing, concave up
− 3 < x < −1
+
−
Increasing, concave down
0
−
Relative maximum
−
−
Decreasing, concave down
x = −1
−2
−1 < x < ∞
y
12
10
8
6
4
2
(−5, 6)
−10 −8 −6 −4 −2
−4
−6
−8
−10
(−1, −2)
x
2
(0, − 73)
11. y = x 4 − x , Domain: ( −∞, 4]
y′ =
y′′ =
8 − 3x
8
= 0 when x = and undefined when x = 4.
3
2 4− x
3x − 16
4( 4 − x )
Note: x =
32
= 0 when x =
16
and undefined when x = 4.
3
16
is not in the domain.
3
y
−∞ < x <
8
3
y′
+
y′′
–
Conclusion
y
Increasing, concave down
4
8
x =
3
8
< x < 4
3
16
3 3
x = 4
0
0
–
Relative maximum
(
8 16 3
,
9
3
2
(0, 0)
–
–
Decreasing, concave down
Undefined
Undefined
Endpoint
(
−2
(4, 0)
x
2
4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
360
12.
Chapter 4
Applications of Differentiation
h( x) = x 9 − x 2 , Domain: − 3 ≤ x ≤ 3
9 − 2 x2
h′( x) =
h′′( x) =
9− x
x( 2 x 2 − 27)
(9 − x 2 )
3
3 2
= ±
and undefined when x = ± 3.
2
2
= 0 when x = ±
2
32
= 0 when x = 0 and undefined when x = ± 3.
Intercepts: (0, 0), ( ±3, 0)
Symmetric with respect to the origin
x = −3
y′
y′′
Conclusion
0
Undefined
Undefined
Endpoint
−
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
3
0
Point of inflection
+
−
Increasing, concave down
3
2
−3 < x < −
3
2
x = −
y
−
9
2
3
< x < 0
2
−
x = 0
3
2
0 < x <
x =
0
3
2
9
2
−
−
−
5
4
3
2
1
Relative maximum
(− 3, 0)
3
< x < 3
2
−2 −1
−5 −4
Decreasing, concave down
(
(0, 0)
0
Undefined
Undefined
)
(3, 0)
x
(
Endpoint
3 2, 9
2
2
1 2 3 4 5
−
x = 3
13.
0
y
3 2, 9
−
2
2
)
−5
y = 3x 2 3 − 2 x
y′ = 2 x −1 3 − 2 =
y′′ =
2(1 − x1 3 )
x1 3
= 0 when x = 1 and undefined when x = 0.
−2
< 0 when x ≠ 0.
3x 4 3
y
−∞ < x < 0
x = 0
0
0 < x <1
x =1
1< x < ∞
y′
y′′
Conclusion
–
–
Decreasing, concave down
Undefined
Undefined
Relative minimum
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
y
5
(1, 1)
1
( 278 , 0 )
x
(0, 0)
1
2
3
5
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
14.
y = ( x + 1) − 3( x + 1)
2
A Summary of Curve Sketching
361
23
y′ = 2( x + 1) − 2( x + 1)
−1 3
=
2( x + 1)
(x
43
− 2
+ 1)
13
= 0 when x = 0, − 2 and undefined when x = −1.
6( x + 1) + 2
2
( x + 1)− 4 3 =
43
3
3( x + 1)
43
y′′ = 2 +
Intercepts: ( −1, 0), ( ± 33 4 − 1, 0)
y′
y
−∞ < x < − 2
−2
x = −2
− 2 < x < −1
x = −1
0
−1 < x < 0
x = 0
0
0 < x < ∞
y′′
Conclusion
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
Undefined
+
Relative maximum
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
y
(−33/4
− 1, 0) 8
6
(−2, 0) 4
(−1, 0) (0, 0)
2
−5 −4
x
−2
2
3
(33/4 − 1, 0)
−4
15. y = 2 − x − x3
y′ = −1 − 3 x 2
No critical numbers
y′′ = −6 x = 0 when x = 0.
y
−∞ < x < 0
x = 0
2
0 < x < ∞
y′
y′′
Conclusion
–
+
Decreasing, concave up
–
0
Point of inflection
–
–
Decreasing, concave down
y
5
4
(0, 2)
1
−3 −2 −1
(1, 0)
2
x
3
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362
16.
Chapter 4
Applications of Differentiation
y = − 13 ( x 3 − 3 x + 2)
y′ = − x 2 + 1 = 0 when x = ±1.
y′′ = −2 x = 0 when x = 0.
y
−∞ < x < −1
− 43
x = −1
−1 < x < 0
− 23
x = 0
0 < x <1
x =1
0
1< x < ∞
y′
y′′
Conclusion
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
y
2
1
(−2, 0)
(1, 0)
x
−1
1
−2
3
2
(0, (
(−1, − 43 (
17.
−2
y = 3x 4 + 4 x3
y′ = 12 x 3 + 12 x 2 = 12 x 2 ( x + 1) = 0 when x = 0, x = −1.
y′′ = 36 x 2 + 24 x = 12 x(3x + 2) = 0 when x = 0, x = − 23 .
y′
y
−∞ < x < −1
x = −1
–1
− 23
−1 < x <
x = − 23
− 16
27
− 23 < x < 0
x = 0
0
0 < x < ∞
y′′
Conclusion
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
0
Point of inflection
+
+
Increasing, concave up
y
2
1
(− 43 , 0 (
(0, 0)
x
−2
(− 1, − 1)
(
1
16
− 2 , − 27
3
(
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
18.
y = − 2 x 4 + 3x 2
y
3
y′ = − 8 x + 6 x = 0 when x = 0, ±
.
2
1
y′′ = − 24 x 2 + 6 = 0 when x = ± .
2
3
−
y
(− 12, 58) (12, 58)
2
( 23 , 98 )
(− )
1
(− 26 , 0)
( 26 , 0)
3 9
,
2 8
(0, 0)1
−2
3
2
x = −
Symmetry: y-axis
⎛
6 ⎞
Intercepts: ⎜⎜ ±
, 0 ⎟⎟
⎝ 2
⎠
3
2
−∞ < x < −
3
1
< x < −
2
2
1
x = −
2
1
− < x < 0
2
x = 0
5
8
0
y′
y′′
Conclusion
+
−
Increasing, concave down
0
−
Relative maximum
−
−
Decreasing, concave down
−2
0
Point of inflection
−
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
2
0
Point of inflection
+
−
Increasing, concave down
0
−
Relative maximum
−
−
Decreasing, concave down
363
x
2
0 < x <
−1
−2
x =
x =
1
2
5
8
1
2
1
< x <
2
3
2
3
2
3
< x < ∞
2
19.
9
8
A Summary of Curve Sketching
9
8
y = x5 − 5 x
y′ = 5 x 4 − 5 = 5( x 4 − 1) = 0 when x = ± 1.
y′′ = 20 x3 = 0 when x = 0.
y′
y
−∞ < x < −1
x = −1
4
−1 < x < 0
x = 0
0
0 < x <1
x =1
–4
1< x < ∞
y′′
Conclusion
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
–
0
Point of inflection
–
+
Decreasing, concave up
0
+
Relative minimum
+
+
Increasing, concave up
y
)− 4 5, 0 )
6
(−1, 4)
4
(0, 0)
−2
−1
−2
−4
−6
1
) 4 5, 0 )
x
2
(1, −4)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
364
Chapter 4
20. y = ( x − 1)
Applications of Differentiation
5
y′ = 5( x − 1) = 0 when x = 1.
4
y′′ = 20( x − 1) = 0 when x = 1.
3
y
−∞ < x < 1
y
2
y′
y′′
Conclusion
+
−
Increasing, concave down
1
(1, 0)
x
−1
x =1
0
1< x < ∞
0
0
Point of inflection
+
+
Increasing, concave up
1
2
3
−1
21. y = 2 x − 3
y′ =
2( 2 x − 3)
2x − 3
3
.
2
undefined at x =
y′′ = 0
y
y
y′
Conclusion
3
−∞ < x <
x =
3
2
22.
3
2
3
2
0
< x < ∞
−
Decreasing
Undefined
Relative minimum
+
Increasing
(0, 3)
2
1
x
( 32 , 0(
3
4
y = x2 − 6x + 5
y′ =
2( x − 3)( x 2 − 6 x + 5)
x2 − 6 x + 5
=
2( x − 3)( x − 5)( x − 1)
(x
− 5)( x − 1)
= 0 when x = 3 and undefined when x = 1, x = 5.
y′′ =
2( x 2 − 6 x + 5)
x − 6x + 5
2
y
−∞ < x < 1
x =1
0
1< x < 3
x = 3
4
3 < x < 5
x = 5
5 < x < ∞
0
=
2( x − 5)( x − 1)
(x
− 5)( x − 1)
undefined when x = 1, x = 5.
y′
y′′
Conclusion
−
+
Decreasing, concave up
Undefined
Undefined
Relative minimum, point of inflection
+
−
Increasing, concave down
0
−
Relative maximum
−
−
Decreasing, concave down
Undefined
Undefined
Relative minimum, point of inflection
+
+
Increasing, concave up
y
5
(3, 4)
4
3
2
1
(1, 0)
1
2
(5, 0)
3
4
5
6
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
23.
A Summary of Curve Sketching
365
f ( x ) = e3 x ( 2 − x )
f ′( x) = −e3 x + 2( 2 − x)e3 x = e3 x (5 − 3 x) = 0 when x =
f ′′( x) = −3e3 x ( − 4 + 3 x) = 0 when x =
f ( x)
4
3
−∞ < x <
2e 4
3
4
3
x =
4
5
< x <
3
3
5
x =
3
5
< x < ∞
3
y
e5
3
f ′( x)
5
.
3
4
.
3
f ′′( x)
Conclusion
+
+
Increasing, concave up
54.6
0
Point of inflection
+
–
Increasing, concave down
0
– 445.2
Relative maximum
–
–
Decreasing, concave down
( (
5, e5
3 3
50
40
(
30
20
4 , 2e 4
3 3
(
10
(0, 2)
(2, 0)
1
24.
x
3
4
5
f ( x) = −2 + e3 x ( 4 − 2 x)
f ′( x) = −2e3 x (3x − 5) = 0 when x = 53.
f ′′( x) = −6e3 x (3x − 4) = 0 when x =
4.
3
Horizontal asymptote (to left): y = −2
f ( x)
−∞ < x <
x =
4
3
4
3
< x <
x =
5
3
4
3
70.7975
5
3
5
3
96.9421
< x < ∞
f ′( x)
f ′′( x)
Conclusion
+
+
Increasing, concave up
109.1963
0
Point of inflection
+
–
Increasing, concave down
0
– 890.4790
Relative maximum
–
–
Decreasing, concave down
y
100
( 53 , 96.9421)
80
60
( 43 , 70.7975)
40
20
1
3
4
5
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
366
25.
Chapter 4
g (t ) =
10
1 + 4e − t
g ′(t ) =
40e − t
Applications of Differentiation
> 0 for all t.
(1 + 4e−t )
40e − t ( 4e −t − 1)
g ′′(t ) =
3
(1 + 4e−t )
2
= 0 at t ≈ 1.386.
lim g (t ) = 10 ⇒ t = 10 is a horizontal asymptote.
t →∞
lim g (t ) = 0 ⇒ t = 0 is a horizontal asymptote.
t → −∞
g (t )
y
g ′(t )
g ′′(t )
Conclusion
+
+
Increasing, concave up
2.5
0
Point of inflection
+
−
Increasing, concave down
y = 10
8
−∞ < t < 1.386
t = 1.386
5
1.386 < t < ∞
26.
h( x ) =
8
2 + 3e− x 2
h′( x) =
12e x 2
6
(1.386, 5)
4
(0, 2)
2
−4
y=0
t
−2
2
4
6
(2e x 2 + 3)
6e x 2 ( 3 − 2e x 2 )
h′′( x) =
3
(2e x 2 + 3)
2
No critical numbers, no relative extrema
lim h( x) =
x→∞
8
= 4 ⇒ x = 4 is a horizontal asymptote.
2
lim h( x) = 0 ⇒ x = 0 is a horizontal asymptote.
x → −∞
h′′( x ) = 0: 3 = 2e x 2 ⇒ e x 2 =
3
⎛ 3⎞
⇒ x = 2 ln ⎜ ⎟
2
⎝ 2⎠
⎛ 8⎞
Intercept: ⎜ 0, ⎟
⎝ 5⎠
h( x )
−∞ < x < 2 ln
x = 2 ln
3
2
3
2 ln < x < ∞
2
h′( x)
h′′( x)
Conclusion
y
6
3
2
2
+
+
Increasing, concave up
1
2
0
Point of inflection
+
–
Increasing, concave down
y=4
5
4
( (
3
0, 8
(
2
5
−3 −2 −1
−2
1
2 ln 3 , 2
2
2
3
4
(
5
x
y=0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
27.
A Summary of Curve Sketching
367
y = ( x − 1) ln ( x − 1), Domain: x > 1
y′ = 1 + ln ( x − 1) = 0 when ln ( x − 1) = −1 ⇒ ( x − 1) = e −1 ⇒ x = 1 + e −1
y′′ =
1
x −1
y
y′
y′′
Conclusion
−
+
Decreasing, concave up
0
e
Relative minimum
y
1 < x < 1 + e −1
3
2
1
− e −1
x = 1 + e −1
1+ e
28.
y =
y′ =
y′′ =
−1
< x < ∞
+
(2, 0)
x
−1
+
Increasing, concave up
2
−1
3
(1.368, − 0.368)
1 3
x − ln x, Domain: x > 0
24
(x
− 2)( x 2 + 2 x + 4)
8x
= 0 when x = 2.
x3 + 4
4x2
y
y
0 < x < 2
x = 2
− 0.3598
2 < x < ∞
5
y′
y′′
Conclusion
−
+
Decreasing, concave up
0
3
Relative minimum
+
+
Increasing, concave down
4
3
2
1
(2, 13 − ln2)
1
3
4
x
5
2
29.
⎛ x − 2⎞
g ( x ) = 6 arcsin ⎜
⎟ , Domain: [0, 4]
⎝ 2 ⎠
g ′( x ) =
12( x − 2)
= 0 when x = 2.
(4 x − x 2 )( x 2 − 4 x + 8)
12( x 4 − 8 x3 + 24 x 2 − 32 x + 32)
g ′′( x ) =
32
⎡( 4 x − x 2 )( x 2 − 4 x + 8)⎤
⎣
⎦
g ( x)
0 < x < 2
x = 2
2 < x < 4
0
y
12
g ′( x)
g ′′( x)
Conclusion
−
+
Decreasing, concave up
0
+
Relative minimum
+
+
10
8
6
4
2
Increasing, concave down
−3 −2 −1
(2, 0)
x
1
2
3
4
5
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
368
30.
Chapter 4
Applications of Differentiation
h( x) = 7 arctan ( x + 1) − ln ( x 2 + 2 x + 2)
h′( x ) =
h′′( x ) =
5 − 2x
5
= 0 when x = .
x2 + 2x + 2
2
2( x 2 − 5 x − 7)
(x
2
+ 2 x + 2)
2
= 0 when x =
h( x )
−∞ < x < −1.1401
x = −1.1401
–0.9935
−1.1401 < x <
x =
5
2
5
2
5
< x < 6.1401
2
f ( x) =
h′( x)
h′′( x)
Conclusion
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
–
Relative maximum
–
–
Decreasing, concave down
8
4
−8
6.0707
6.1401 < x < ∞
31.
53
2
y
6.4635
x = 6.1401
5±
–
0
Point of inflection
–
+
Decreasing, concave up
( 25 , 6.4635)
(6.1401, 6.0707)
−4
4
(−1.1401, −0.9935)
x
8
x
27 x
= x
3x − 3
3
27(1 − x ln 3)
1
= 0 ⇒ x =
≈ 0.910
3x
ln 3
27 ln 3( x ln 3 − 2)
2
f ′′( x) =
= 0 ⇒ x =
≈ 1.820
3x
ln 3
f ′( x) =
lim f ( x) = 0, lim f ( x) = −∞
x →∞
x → −∞
Horizontal symptote: y = 0
Intercept: (0, 0)
f ( x)
−∞ < x < 0.910
f ′( x)
f ′′( x)
Conclusion
+
–
Increasing, concave down
0
–
Relative maximum
0.910 < x < 1.820
–
–
Decreasing, concave down
x = 1.820
–
0
Point of inflection
x = 0.910
1.820 < x < ∞
y
10
9.041
(0.91, 9.04)
8
(1.82, 6.65)
6
4
6.652
–
+
Decreasing, concave up
2
y=0
(0, 0)
−1
x
1
2
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
32.
A Summary of Curve Sketching
369
g (t ) = (5 − t )5t
5 ln 5 − 1
1
= 5−
≈ 4.379
ln 5
ln 5
5 ln 5 − 2
2
g ′′(t ) = 5t ln 5(5 ln 5 − 2 − t ln 5) = 0 ⇒ t =
= 5−
≈ 3.757
ln 5
ln 5
g ′(t ) = 5t (5 ln 5 − 1 − t ln 5) = 0 ⇒ t ln 5 = 5 ln 5 − 1 ⇒ t =
lim g (t ) = −∞ and lim g (t ) = 0
t →∞
t →−∞
Horizontal asymptote: y = 0
Intercepts: (5, 0), (0, 5)
g (t )
−∞ < t < 3.757
t = 3.757
525.553
3.757 < t < 4.379
t = 4.379
714.301
4.379 < t < ∞
33.
g ′′( x) =
Conclusion
+
+
Increasing, concave up
+
0
Point of inflection
+
–
Increasing, concave down
0
–
Relative maximum
ln ( x − x 2 )
ln 4
,
–
y
800
700
600
500
400
300
200
100
Decreasing, concave down
(3.757, 525.553)
(4.379, 714.301)
(0, 5)
−100
1
(5, 0)
2
3
4
y=0
6
x
Domain: 0 < x < 1
2x − 1
1
= 0 when x = .
ln 4 ⋅ x( x − 1)
2
−2 x 2 + 2 x − 1
ln 4 ⋅ x 2 ( x − 1)
g ( x)
0 < x <
x =
g ′′(t )
–
g ( x) = log 4 ( x − x 2 ) =
g ′( x) =
g ′(t )
1
< x <1
2
g ′( x)
g ′′( x)
Conclusion
+
–
Increasing, concave down
y
1
1
2
1
2
2
0.5
−1
−1
0
–
Relative maximum
x=0
x
−0.5
−2
(0.5, −1)
1.0
x=1
−3
–
–
Decreasing, concave down
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
370
34.
Chapter 4
Applications of Differentiation
f ( x) = log 2 x 2 − 4 x =
f ′( x) =
f ′′( x) =
2( x − 2)
ln x 2 − 4 x
ln 2
= 0 when x = 2 and undefined when x = 0 and x = 4.
x( x − 4) ln 2
−2( x 2 − 4 x + 8)
x 2 ( x − 4) ln 2
2
f ( x)
f ′( x)
−∞ < x < 0
x = 0
Undefined
0 < x < 2
2
2 < x < 4
Undefined
4 < x < ∞
f ′( x) =
f ′′( x) =
–
–
Decreasing, concave down
Undefined
Undefined
Undefined
+
–
Increasing, concave down
y
0
–
Relative maximum
–
–
Decreasing, concave down
4
−4
x = 4
f ( x) =
Conclusion
6
x = 2
35.
f ′′( x)
Undefined
Undefined
Undefined
+
–
Increasing, concave down
20 x
1
19 x 2 − 1
− =
2
x +1 x
x( x 2 + 1)
− (19 x 4 − 22 x 2 − 1)
x 2 ( x 2 + 1)
2
3
2
−2
8
x
−6
4
x3 + x + 4
=
= 0 for x ≈ −1.379
x +1
x2 + 1
f ( x) = x +
2
x4 + 2x2 − 8x + 1
= 0 for x ≈ 1.608, x ≈ 0.129
( x 2 + 1)
8(3 x 2 − 1)
= 0 for x = ±
f ′′( x) =
3
( x 2 + 1)
= 0 for x ≈ ±1.84
6
−4
f ′( x) =
= 0 for x ≈ ±1.10
2(19 x 6 − 63 x9 − 3 x 2 − 1)
x3 ( x 2 + 1)
36.
−2
(2, 2)
2
1
≈ ± 0.577
3
Slant asymptote: y = x
Vertical asymptote: x = 0
Points of inflection: ( −0.577, 2.423), (0.577, 3.577)
Horizontal asymptote: y = 0
Relative maximum: (0.129, 4.064)
Minimum: ( −1.10, −9.05)
Relative minimum: (1.608, 2.724)
Maximum: (1.10, 9.05)
Points of inflection: ( −1.84, −7.86), (1.84, 7.86)
5
10
−6
−15
15
−10
6
−3
37.
f ( x) =
f ′( x) =
f ′′( x) =
−2 x
4
x + 7
−14
2
( x2
+ 7)
32
+ 7)
6
−4
42 x
( x2
−6
< 0
52
= 0 at x = 0
Horizontal asymptotes: y = ± 2
Point of inflection: (0, 0)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
f ( x) =
38.
f ′( x) =
f ′′( x) =
4x
f ′( x) = 2 − 4 cos x
2
( x2
+ 15)
32
−180 x
( x2
+ 15)
371
f ( x) = 2 x − 4 sin x, 0 ≤ x ≤ 2π
41.
x + 15
60
A Summary of Curve Sketching
52
> 0
f ′′( x) = 4 sin x
= 0 at x = 0
1
π 5π
⇒ x = ,
2
3 3
f ′′( x) = 0 ⇒ x = 0, π , 2π
f ′( x) = 0 ⇒ cos x =
Horizontal asymptotes: y = ± 4
⎛ π 2π
⎞
Relative minimum: ⎜ ,
− 2 3⎟
⎝3 3
⎠
Point of inflection: (0, 0)
⎛ 5π 10π
⎞
+ 2 3⎟
Relative maximum: ⎜ ,
⎝ 3 3
⎠
6
−8
Points of inflection: (0, 0), (π , 2π ), ( 2π , 4π )
8
y
−6
16
x
⎛ x ⎞
+ ln ⎜
⎟
2
⎝ x + 3⎠
1
3
+
y′ =
x( x + 3)
2
12
y =
39.
y′′ =
8
4
π
2
−3( 2 x + 3)
x 2 ( x + 3)
2
π
3π
2
x
2π
f ( x) = − x + 2 cos x, 0 ≤ x ≤ 2π
42.
f ′( x) = −1 − 2 sin x
5
f ′′( x) = −2 cos x
−8
f ( x) = 0 at x ≈ 1.030
6
f ′( x) = 0 ⇒ sin x = −
−5
Vertical asymptotes: x = −3, x = 0
x
Slant asymptote: y =
2
3x
40. y =
(1 + 4e− x 3 )
2
3e x 3 − 4( x − 3)
y′ =
2e x 3
2( x − 6)
y′′ =
3e x 3
Slant asymptote: y =
3
x
2
f ′′( x) = 0 ⇒ x =
π 3π
,
2 2
Relative minimum:
⎛ 7π
⎜ ,−
⎝ 6
3 −
7π ⎞
⎟ ≈ (3.665, − 5.397)
6 ⎠
Relative maximum:
⎛ 11π
,
⎜
⎝ 6
3 −
11π ⎞
⎟ ≈ (5.760, − 4.028)
6 ⎠
y
4
2
π
35
1
7π 11π
⇒ x =
,
2
6 6
−2
3π
2
2π
x
−4
−6
0
0
20
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
372
Chapter 4
Applications of Differentiation
1
sin 3 x, 0 ≤ x ≤ 2π
18
1
cos x − cos 3x
6
1
cos x − [cos 2 x cos x − sin 2 x sin x]
6
1⎡
cos x − ⎣(1 − 2 sin 2 x) cos x − 2 sin 2 x cos x⎤⎦
6
1
⎡
⎤
⎡5 2
cos x ⎢1 − (1 − 2 sin 2 x − 2 sin 2 x)⎥ = cos x ⎢ + sin 2
6
⎣
⎦
⎣6 3
43. y = sin x −
y′ =
=
=
=
y′ = 0:
⎤
x⎥
⎦
cos x = 0 ⇒ x = π 2, 3π 2
5 2
+ sin 2 x = 0 ⇒ sin 2 x = −5 4, impossible
6 3
1
y′′ = −sin x + sin 3 x = 0 ⇒ 2 sin x = sin 3x
2
= sin 2 x cos x + cos 2 x sin x
= 2 sin x cos 2 x + ( 2 cos 2 x − 1) sin x
= sin x( 2 cos 2 x + 2cos 2 x − 1)
= sin x( 4 cos 2 x − 1)
sin x = 0 ⇒ x = 0, π , 2π
2 = 4 cos 2 x − 1 ⇒ cos x = ±
3 2 ⇒ x =
π 5π 7π 11π
6
,
6
,
,
6
6
y
⎛ π 19 ⎞
Relative maximum: ⎜ , ⎟
⎝ 2 18 ⎠
2
1
⎛ 3π 19 ⎞
Relative minimum: ⎜ , − ⎟
18 ⎠
⎝ 2
−1
4⎞
⎛ π 4 ⎞ ⎛ 5π 4 ⎞
⎛ 7π 4 ⎞ ⎛ 11π
,− ⎟
Points of inflection: ⎜ , ⎟, ⎜ , ⎟, (π , 0), ⎜ , − ⎟, ⎜
9⎠ ⎝ 6
9⎠
⎝ 6 9⎠ ⎝ 6 9⎠
⎝ 6
π
2
π
3π
2
x
−2
1
cos 2 x, 0 ≤ x ≤ 2π
4
1
y′ = −sin x + sin 2 x = −sin x + sin x cos x
2
= sin x(cos x − 1)
44. y = cos x −
y′ = 0:
sin x = 0 ⇒ x = 0, π , 2π
cos x − 1 = 0 ⇒ x = 0, 2π
y′′ = −cos x + cos 2 x
= −cos x + 2 cos 2 x − 1
= ( 2 cos x + 1)(cos x − 1)
2π 4π
,
3 3
cos x − 1 = 0 ⇒ x = 0, 2π
y
y′′ = 0: 2 cos x + 1 = 0 ⇒ x =
5⎞
⎛
Relative minimum: ⎜π , − ⎟
4⎠
⎝
3 ⎞ ⎛ 4π
3⎞
⎛ 2π
Points of inflection: ⎜ , − ⎟, ⎜ , − ⎟
8⎠ ⎝ 3
8⎠
⎝ 3
2
1
−1
π
2
π
3π
2
2π
x
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
45.
y = 2 x − tan x, −
π
< x <
2
π
46.
2
y′ = 2 − sec x = 0 when x = ± .
4
y′′ = −2 sec 2 x tan x = 0 when x = 0.
π 3π
4
y′′ = 2 csc 2 x cot x = 0 when x =
⎛π π
⎞
Relative maximum: ⎜ , − 1⎟
⎝4 2
⎠
⎛ 3π 3π
⎞
Relative maximum: ⎜ ,
− 5⎟
⎝ 4 2
⎠
π⎞
⎛ π
Relative minimum: ⎜ − , 1 − ⎟
4
2⎠
⎝
⎛π π
⎞
Relative minimum: ⎜ , − 3⎟
⎝4 2
⎠
Point of inflection: (0, 0)
π
⎛π
⎞
Point of inflection: ⎜ , π − 4 ⎟
⎝2
⎠
2
Vertical asymptotes: x = 0, π
Vertical asymptotes: x = ±
5
4
3
2
1
2
1
π
4
−1
,
4
π
2
.
.
y
y
−π
2
373
y = 2( x − 2) + cot x, 0 < x < π
y′ = 2 − csc 2 x = 0 when x =
π
2
A Summary of Curve Sketching
π
2
x
π
−1
−2
−3
−4
−5
−2
47. y = 2(csc x + sec x), 0 < x <
π
y
2
y′ = 2(sec x tan x − csc x cot x) = 0 ⇒ x =
⎛π
⎞
Relative minimum: ⎜ , 4 2 ⎟
4
⎝
⎠
Vertical asymptotes: x = 0,
x
π
π
4
16
12
8
4
−4
π
4
π
2
x
2
⎛π x ⎞
⎛π x ⎞
48. y = sec 2 ⎜ ⎟ − 2 tan ⎜ ⎟ − 1, − 3 < x < 3
⎝ 8 ⎠
⎝ 8 ⎠
⎛π x ⎞
⎛ π x ⎞⎛ π ⎞
⎛ π x ⎞⎛ π ⎞
y′ = 2 sec 2 ⎜ ⎟ tan ⎜ ⎟⎜ ⎟ − 2 sec2 ⎜ ⎟⎜ ⎟ = 0 ⇒ x = 2
⎝ 8 ⎠
⎝ 8 ⎠⎝ 8 ⎠
⎝ 8 ⎠⎝ 8 ⎠
Relative minimum: ( 2, −1)
y
5
4
3
2
−5 − 4 −3 −2 −1
−2
−3
−4
−5
3 4 5
x
(2, −1)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
374
49.
Chapter 4
Applications of Differentiation
3π
3π
< x <
2
2
x + sin x cos x
g ′( x) =
= 0 when x = 0.
cos 2 x
g ( x) = x tan x, −
2(cos x + x sin x)
g ′′( x) =
cos3 x
3π π π 3π
Vertical asymptotes: x = − , − , ,
2
2 2 2
Intercepts: ( −π , 0), (0, 0), (π , 0)
Symmetric with respect to y-axis.
⎛ π⎞
⎛ π 3π ⎞
Increasing on ⎜ 0, ⎟ and ⎜ ,
⎟
⎝ 2⎠
⎝2 2 ⎠
Points of inflection: ( ±2.80, −1)
y
10
8
6
4
2
−π
π
4
−2
−4
−6
−8
π
3π
2
x
sin x cos x − x
sin 2 x
y
f ′ is quadratic.
f″
f
f ′′ is linear.
The zeros of f ′ correspond
to the points where the graph
of f has horizontal tangents.
The zero of f ′′ corresponds to
the point where the graph of f ′
has a horizontal tangent.
54. f ′′ is constant.
x
−2
2
−1
f′
−2
y
f
f ′ is linear.
f ''
f is quadratic.
The zero of f ′ corresponds
to the points where the graph
of f has a horizontal tangent.
There are no zeros on of f ′′,
which means the graph of f ′
has no horizontal tangent.
55. f ( x ) =
50. g ( x ) = x cot x, − 2π < x < 2π
g ′( x ) =
53. f is cubic.
4( x − 1)
x
f'
2
x2 − 4 x + 5
Vertical asymptote: none
Horizontal asymptote: y = 4
9
x
g ′(0) does not exist. But lim x cot x = lim
= 1.
x→0
x → 0 tan x
Vertical asymptotes: x = ±2π , ± π
⎛ 3π ⎞ ⎛ π ⎞ ⎛ π ⎞ ⎛ 3π ⎞
Intercepts: ⎜ − , 0 ⎟, ⎜ − , 0 ⎟, ⎜ , 0 ⎟, ⎜ , 0 ⎟
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝2 ⎠ ⎝ 2 ⎠
Symmetric with respect to y-axis.
Decreasing on (0, π ) and (π , 2π )
Points of inflection: ( ±4.49, 1)
−6
The graph crosses the horizontal asymptote y = 4.
If a function has a vertical asymptote at x = c, the
graph would not cross it because f (c) is undefined.
56. g ( x) =
y
−π
3x 4 − 5 x + 3
x4 + 1
4
Vertical asymptote: none
3
Horizontal asymptote: y = 3
2
−2π
9
−1
−1
π
2π
7
x
−2
−3
−4
−6
51. Because the slope is negative, the function is decreasing
on ( 2, 8), and so f (3) > f (5).
52. If f ′( x) = 2 in [−5, 5], then f ( x) = 2 x + 3 and
f ( 2) = 7 is the least possible value of f ( 2). If
6
−1
The graph crosses the horizontal asymptote y = 3.
If a function has a vertical asymptote at x = c, the
graph would not cross it because f (c) is undefined.
f ′( x) = 4 in [−5, 5], then f ( x) = 4 x + 3 and
f ( 2) = 11 is the greatest possible value of f ( 2).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
sin 2 x
x
Vertical asymptote: none
Horizontal asymptote: y = 0
57. h( x) =
A Summary of Curve Sketching
61. f ( x ) = −
375
3
x 2 − 3x − 1
= −x + 1 +
x − 2
x − 2
3
3
−3
−2␲
6
−3
2␲
The graph appears to approach the slant asymptote
y = − x + 1.
−1
Yes, it is possible for a graph to cross its horizontal
asymptote.
It is not possible to cross a vertical asymptote because
the function is not continuous there.
62. g ( x) =
2 x 2 − 8 x − 15
5
= 2x + 2 −
x −5
x −5
18
cos 3 x
58. f ( x) =
4x
Vertical asymptote: x = 0
Horizontal asymptote: y = 0
2
−10
20
−2
The graph appears to approach the slant asymptote
y = 2 x + 2.
−2␲
2␲
63. f ( x ) =
−2
Yes, it is possible for a graph to cross its horizontal
asymptote.
It is not possible to cross a vertical asymptote because
the function is not continuous there.
59. h( x) =
6 − 2x
3− x
2 x3
2x
= 2x − 2
x +1
x +1
2
4
−6
6
−4
2(3 − x)
if x ≠ 3
⎧2,
=
= ⎨
3− x
Undefined,
if x = 3
⎩
The rational function is not reduced to lowest terms.
The graph appears to approach the slant asymptote
y = 2 x.
64. h( x) =
3
4
− x3 + x 2 + 4
= −x + 1 + 2
x2
x
10
−2
−10
4
10
−1
There is a hole at (3, 2).
−10
The graph appears to approach the slant asymptote
y = − x + 1.
x2 + x − 2
60. g ( x) =
x −1
=
(x
+ 2)( x − 1)
x −1
if x ≠ 1
⎧x + 2,
= ⎨
Undefined,
if x = 1
⎩
The rational function is not reduced to lowest terms.
4
−8
4
−4
There is a hole at (1, 3).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
376
Chapter 4
65.
Applications of Differentiation
y
4
x
−2
2
−4
4
x
−2
2
4
−4
(or any vertical translation of f )
66.
−2
−4
−4
y
y
2
100
10
f
80
1
8
60
x
−2
2
x
3
6
9
2
4
6
8
10
(or any vertical translation of f )
x2 + 1
1
2
x
−2
−1
1
−1
−1
−2
−2
2
x
−4 −2
−2
12 15
cos 2 π x
1
f ′′
6
4
(a)
x
4
−2
68.
y
120
69. f ( x) =
f″
−8
8
(or any vertical translation of f )
y
−6 −3
2
x
−4
−2
−4
f
2
2
−4
4
4
f″
4
f
y
y
67.
y
(or any vertical translation of the 3 segments of f )
, (0, 4)
1.5
4
0
−0.5
On (0, 4) there seem to be 7 critical numbers: 0.5, 1.0, 1.5, 2.0, 2.5, 3.0, 3.5
(b) f ′( x) =
(
−cos π x x cos π x + 2π ( x 2 + 1) sin π x
(x
Critical numbers ≈
2
+ 1)
32
)
= 0
1
3
5
7
, 0.97, , 1.98, , 2.98, .
2
2
2
2
The critical numbers where maxima occur appear to be integers in part (a), but approximating them using f ′ shows that
they are not integers.
70. f ( x ) = tan (sin π x)
(a)
3
−2␲
2␲
−3
(b) f ( − x ) = tan (sin ( −π x)) = tan ( −sin π x) = − tan (sin π x) = − f ( x)
Symmetry with respect to the origin
(c) Periodic with period 2
(d) On ( −1, 1), there is a relative maximum at
( 12 , tan 1) and a relative minimum at (− 12 , − tan 1).
(e) On (0, 1), the graph of f is concave downward.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
Horizontal asymptote: y = 0
1
x −3
x
x + 5
of f is increasing.
(b) No. Let f ( x) = x 2 + 1 (positive and concave up).
(
Slant asymptote: y = 3 x + 2
y = 3x + 2 +
1
3x 2 − 7 x − 5
=
x −3
x −3
Slant asymptote: y = − x
(b) f ′′( x) = 0 at x2 and x3 (point of inflection).
(d) f has a relative maximum at x1 .
1
−x + 2x + 1
=
x − 2
x − 2
2
f ( x) = ln x, g ( x) =
x
1
1
, g ′( x) =
x
2 x
For x > 4, g ′( x) > f ′( x). g is increasing at
a higher rate than f for “large” values of x.
25
(e) f has a point of inflection at x2 and x3 (change in
concavity).
78. (a) f ′( x) = 0 for x = −2 (relative maximum) and
x = 2 (relative minimum).
f ′ is negative for −2 < x < 2 (decreasing).
f ′ is positive for x > 2 and x < −2 (increasing).
(b) f ′′( x) = 0 at x = 0 (point of inflection).
f ′′ is positive for x > 0 (concave upward).
g
f ′′ is negative for x < 0 (concave downward).
f
(c) f ′ is increasing on (0, ∞). ( f ′′ > 0)
0
500
(d) f ′( x ) is minimum at x = 0. The rate of change of f
0
f ( x) = ln x, g ( x) =
f ′( x) =
77. (a) f ′( x) = 0 at x0 , x2 and x4 (horizontal tangent).
(c) f ′( x ) does not exist at x1 (sharp corner).
74. Vertical asymptote: x = 2
f ′( x) =
)
g ( x) = ln x 2 + 1 is not concave up.
73. Vertical asymptote: x = 3
(b)
f ( x)
f ′( x) = g ′( x) f ( x) and f ′( x) > 0. So, the graph
2
75. (a)
f ′( x )
Because f ( x) > 0, you know that
Horizontal asymptote: none
y = −x +
g ′( x) =
(a) Yes. If the graph of g is increasing, then g ′( x) > 0.
72. Vertical asymptote: x = −5
y =
377
76. g ( x) = ln f ( x), f ( x ) > 0
71. Vertical asymptote: x = 3
y =
A Summary of Curve Sketching
4
x
1
1
, g ′( x) =
4
x
4 x3
at x = 0 is less than the rate of change of f for all
other values of x.
For x > 256, g ′( x) > f ′( x). g is increasing at
a higher rate than f for “large” values of x.
f ( x) = ln x increases very slowly for “large”
values of x.
15
g
f
0
20,000
0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
378
Chapter 4
A pplications of Differentiation
79. Tangent line at P : y − y0 = f ′( x0 )( x − x0 )
(a) Let y = 0: − y0 = f ′( x0 )( x − x0 )
(d) Let x = 0: y − y0 =
f ′( x0 ) x = x0 f ′( x0 ) − y0
y = y0 +
f ( x0 )
y0
x = x0 −
= x0 −
f ′( x0 )
f ′( x0 )
(b) Let x = 0: y − y0 = f ′( x0 )( − x0 )
(e)
BC = x0 −
y = y0 − x0 f ′( x0 )
(f ) PC
1
( x − x0 )
f ′( x0 )
80. f ( x ) =
f ′( x0 )
f ( x0 ) 1 + ⎡⎣ f ′( x0 )⎤⎦
f ′( x0 )
2
2
AB = x0 − ( x0 + f ( x0 ) f ′( x0 )) = f ( x0 ) f ′( x0 )
(h)
AP
2
= f ( x0 ) f ′( x0 ) + y02
2
AP = f ( x0 )
2
1 + ⎡⎣ f ′( x0 )⎤⎦
2
+ f ( x0 ) f ′( x0 ), 0)
2 xn
x +1
4
(a) For n even, f is symmetric about the y-axis. For n odd, f is symmetric about the origin.
(b) The x-axis will be the horizontal asymptote if the degree of the numerator is less than 4.
That is, n = 0, 1, 2, 3.
(c) n = 4 gives y = 2 as the horizontal asymptote.
(d) There is a slant asymptote y = 2 x if n = 5:
(e)
2.5
2 x5
2x
= 2x − 4
.
4
x +1
x +1
2.5
n=4
n=0
n=5
n=2
−3
−3
3
3
n=3
n=1
− 1.5
− 1.5
n
0
1
2
3
4
5
M
1
2
3
2
1
0
N
2
3
4
5
2
3
81. f ( x ) =
ax
( x − b)
2
(g)
x = x0 + y0 f ′( x0 ) = x0 + f ( x0 ) f ′( x0 )
( x0
f ( x0 )
− x0 =
⎛ f ( x0 ) ⎞
f ( x0 ) f ′( x0 ) + f ( x0 )
= y02 + ⎜⎜
⎟⎟ =
2
f ′( x0 )
⎝ f ′( x0 ) ⎠
PC =
− y0 f ′( x0 ) = − x + x0
x-intercept:
2
x0 f ′( x0 ))
1
(c) Normal line: y − y0 = −
( x − x0 )
′
f ( x0 )
Let y = 0: − y0 = −
f ( x0 )
f ′( x0 )
2
y = f ( x0 ) − x0 f ′( x0 )
(0, f ( x0 ) −
x0
f ′( x0 )
⎛
x0 ⎞
y-intercept: ⎜ 0, y0 +
⎟
⎜
′
f ( x0 ) ⎟⎠
⎝
⎛
f ( x0 ) ⎞
x-intercept: ⎜ x0 −
, 0⎟
⎜
f ′( x0 ) ⎟⎠
⎝
y-intercept:
−1
(− x0 )
f ′( x0 )
2
Answers will vary. Sample answer: The graph has a vertical asymptote at x = b. If a and b are both positive,
or both negative, then the graph of f approaches ∞ as x approaches b, and the graph has a minimum at x = – b.
If a and b have opposite signs, then the graph of f approaches −∞ as x approaches b, and the graph has a
maximum at x = – b.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.6
82.
A Summary of Curve Sketching
379
1
1
( ax)2 − ( ax) = ( ax)( ax − 2), a ≠ 0
2
2
1
2
f ′( x) = a x − a = a( ax − 1) = 0 when x = .
a
f ′′( x) = a 2 > 0 for all x.
f ( x) =
⎛2 ⎞
(a) Intercepts: (0, 0), ⎜ , 0 ⎟
⎝a ⎠
y
a=2
(b)
a = −2
5
⎛1 1⎞
Relative minimum: ⎜ , − ⎟
⎝a 2⎠
4
a=1
Points of inflection: none
−3
83. y =
4 + 16 x 2
x
2
a = −1 −1
3
84. y =
(x
x2 + 6x =
+ 3) − 9
2
As x → ∞, y → 4 x. As x → −∞, y → −4 x.
y → x + 3 as x → ∞, and y → − x − 3 as x → −∞.
Slant asymptotes: y = ± 4 x
Slant asymptotes: y = x + 3, y = − x − 3
y
y
12
15
10
12
8
9
6
2
−8 −6 −4 −2
3
x
2
4
6
8
−9 −6 −3
x
−3
3
6
f ( x) − f ( a)
f (b ) − f ( a )
−
x − a
b − a
85. Let λ =
, a < x < b.
x −b
λ ( x − b) =
f ( x) − f ( a )
x − a
−
λ ( x − b)( x − a) = f ( x) − f ( a ) −
f (b ) − f ( a )
b − a
f (b ) − f ( a )
( x − a)
b − a
f (b ) − f ( a )
f ( x) = f ( a ) +
( x − a) + λ ( x − b)( x − a)
b − a
f (b ) − f ( a )
⎫
⎪⎧
Let h(t ) = f (t ) − ⎨ f ( a) +
(t − a) + λ (t − a)(t − b)⎪⎬.
b
a
−
⎪⎭
⎩⎪
h( a ) = 0, h(b) = 0, h( x) = 0
By Rolle’s Theorem, there exist numbers α1 and α 2 such that a < α1 < x < α 2 < b and h′(α1 ) = h′(α 2 ) = 0.
By Rolle’s Theorem, there exists β in ( a, b) such that h′′( β ) = 0.
Finally,
0 = h′′( β ) = f ′′( β ) − {2λ} ⇒ λ =
1
2
f ′′( β ).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
380
Chapter 4
Applications of Differentiation
Section 4.7 Optimization Problems
1. (a)
(b)
First Number, x
Second Number
Product, P
10
110 − 10
10(110 − 10) = 1000
20
110 − 20
20(110 − 20) = 1800
30
110 − 30
30(110 − 30) = 2400
40
110 − 40
40(110 − 40) = 2800
50
110 − 50
50(110 − 50) = 3000
60
110 − 60
60(110 − 60) = 3000
First Number, x
Second Number
Product, P
10
110 − 10
10(110 − 10) = 1000
20
110 − 20
20(110 − 20) = 1800
30
110 − 30
30(110 − 30) = 2400
40
110 − 40
40(110 − 40) = 2800
50
110 − 50
50(110 − 50) = 3000
60
110 − 60
60(110 − 60) = 3000
70
110 − 70
70(110 − 70) = 2800
80
110 − 80
80(110 − 80) = 2400
90
110 − 90
90(110 − 90) = 1800
100
110 − 100
100(110 − 100) = 1000
The maximum is attained near x = 50 and 60.
(c) P = x(110 − x) = 110 x − x 2
(d)
3500
(55, 3025)
0
120
0
The solution appears to be x = 55.
(e)
dP
= 110 − 2 x = 0 when x = 55.
dx
d 2P
= −2 < 0
dx 2
P is a maximum when x = 110 − x = 55. The two numbers are 55 and 55.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
2. (a)
Height, x
Length & Width
Volume
1
24 − 2(1)
1[24 − 2(1)]2 = 484
2
24 − 2(2)
2[24 − 2(2)]2 = 800
3
24 − 2(3)
3[24 − 2(3)]2 = 972
4
24 − 2(4)
4[24 − 2(4)]2 = 1024
5
24 − 2(5)
5[24 − 2(5)]2 = 980
6
24 − 2(6)
6[24 − 2(6)]2 = 864
Optimization Problems
381
The maximum is attained near x = 4.
(b) V = x( 24 − 2 x ) , 0 < x < 12
2
dV
2
= 2 x( 24 − 2 x )(−2) + ( 24 − 2 x) = ( 24 − 2 x )( 24 − 6 x)
dx
= 12(12 − x)( 4 − x) = 0 when x = 12, 4 (12 is not in the domain ).
(c)
d 2V
= 12( 2 x − 16)
dx 2
d 2V
< 0 when x = 4.
dx 2
When x = 4, V = 1024 is maximum.
(d)
1200
0
12
0
The maximum volume seems to be 1024.
3. Let x and y be two positive numbers such that
x + y = S.
P = xy = x( S − x) = Sx − x 2
dP
S
= S − 2 x = 0 when x = .
dx
2
5. Let x and y be two positive numbers such that
xy = 147.
S = x + 3y =
147
+ 3y
y
dS
147
= 3 − 2 = 0 when y = 7.
dy
y
2
d P
S
= −2 < 0 when x = .
dx 2
2
P is a maximum when x = y = S 2.
d 2S
294
= 3 > 0 when y = 7.
dy 2
y
S is minimum when y = 7 and x = 21.
4. Let x and y be two positive numbers such that
xy = 185.
185
S = x + y = x +
x
185
dS
= 1 − 2 = 0 when x =
dx
x
370
d 2S
= 3 > 0 when x =
dx 2
x
S is a minimum when x = y =
6. Let x be a positive number.
1
x
dS
1
= 1 − 2 = 0 when x = 1.
dx
x
S = x +
185.
185
185.
d 2S
2
= 3 > 0 when x = 1.
dx 2
x
The sum is a minimum when x = 1 and 1 x = 1.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
382
Chapter 4
Applications of Differentiation
7. Let x and y be two positive numbers such that
x + 2 y = 108.
11. Let x be the length and y the width of the rectangle.
xy = 32
P = xy = y (108 − 2 y ) = 108 y − 2 y 2
dP
= 108 − 4 y = 0 when y = 27.
dy
2
d P
= −4 < 0 when y = 27.
dy 2
P is a maximum when x = 54 and y = 27.
8. Let x and y be two positive numbers such that
x 2 + y = 54.
P = xy = x(54 − x 2 ) = 54 x − x3
dP
= 54 − 3 x 2 = 0 when x = 3 2.
dx
y =
64
⎛ 32 ⎞
P = 2 x + 2 y = 2 x + 2⎜ ⎟ = 2 x +
x
⎝ x⎠
dP
64
= 2 − 2 = 0 when x = 4 2.
dx
x
d 2P
128
= 3 > 0 when x = 4 2.
dx 2
x
P is minimum when x = y = 4 2 ft.
12. Let x be the length and y the width of the rectangle.
xy = A
y =
2
d P
= −6 x < 0 when x = 3 2.
dx 2
The product is a maximum when x = 3 2 and y = 36.
9. Let x be the length and y the width of the rectangle.
2 x + 2 y = 80
y = 40 − x
A = xy = x( 40 − x) = 40 x − x 2
32
x
A
x
2A
⎛ A⎞
P = 2 x + 2 y = 2 x + 2⎜ ⎟ = 2 x +
x
⎝x⎠
dP
2A
A.
= 2 − 2 = 0 when x =
dx
x
d 2P
4A
A.
= 3 > 0 when x =
dx 2
x
P is minimum when x = y =
A cm. (A square!)
dA
= 40 − 2 x = 0 when x = 20.
dx
d2A
= −2 < 0 when x = 20.
dx 2
y
A is maximum when x = y = 20 m.
x
10. Let x be the length and y the width of the rectangle.
2x + 2 y = P
P − 2x
P
=
− x
2
2
P
⎛P
⎞
A = xy = x⎜ − x ⎟ = x − x 2
2
2
⎝
⎠
y =
dA
P
P
=
− 2 x = 0 when x = .
dx
2
4
d2A
P
= −2 < 0 when x = .
2
dx
4
(x
13. d =
− 2) + ⎡⎣ x 2 − (1 2)⎤⎦
2
2
x 4 − 4 x + (17 4)
=
Because d is smallest when the expression inside the
radical is smallest, you need only find the critical
numbers of
f ( x) = x 4 − 4 x +
17
.
4
f ′( x) = 4 x 3 − 4 = 0
x =1
A is maximum when x = y = P 4 units. (A square!)
( 12 ) is
By the First Derivative Test, the point nearest to 2,
(1, 1).
y
y
4
3
2
x
( x, x 2 )
(2, 12(
d
1
x
−2
−1
1
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
16. f ( x ) =
14. f ( x ) = ( x − 1) , ( −5, 3)
2
(x
d =
2
2
+ 5) + ⎡( x − 1) − 3⎤
⎣
⎦
2
=
( x2
( x2
=
x 4 − 4 x 3 + x 2 + 18 x + 29
=
+ 10 x + 25) + ( x 2 − 2 x − 2)
+ 10 x + 25) + ( x 4 − 4 x3 + 8 x + 4)
Because d is smallest when the expression inside the radical
is smallest, you need to find the critical numbers of
g ( x) = x 4 − 4 x3 + x 2 + 18 x + 29
g ′( x) = 4 x3 − 12 x 2 + 2 x + 18
= 2( x + 1)( 2 x − 8 x + 9) = 0
(x
− 12) +
2
By the First Derivative Test, x = −1 yields a minimum.
So, ( −1, 4) is closest to ( −5, 3).
(x
15. d =
=
− 4) +
2
(
x −0
)
2
x 2 − 7 x + 16
Because d is smallest when the expression inside the
radical is smallest, you need only find the critical
numbers of
f ( x) = x 2 − 7 x + 16.
f ′( x) = 2 x − 7 = 0
x =
By the First Derivative Test, the point nearest to ( 4, 0) is
(7 2,
)
x 2 − 24 x + 144 + x − 8
=
x 2 − 23x + 136
)
2
g ( x) = x 2 − 23 x + 136
g ′( x) = 2 x − 23 = 0 when x =
g ′′( x ) = 2 > 0 at x =
23
2
23
2
The point nearest to (12, 0) is
⎛ 23
⎜ ,
⎝2
⎛ 23 ⎞ ⎞ ⎛ 23
f ⎜ ⎟ ⎟ = ⎜⎜ ,
⎝ 2 ⎠⎠ ⎝ 2
17. xy = 30 ⇒ y =
14 ⎞
⎟
2 ⎟⎠
30
x
⎛ 30
⎞
+ 2 ⎟ (see figure)
A = ( x + 2)⎜
⎝ x
⎠
dA
⎛ −30 ⎞ ⎛ 30
⎞
= ( x + 2)⎜ 2 ⎟ + ⎜
+ 2⎟
dx
⎝ x ⎠ ⎝ x
⎠
=
2( x 2 − 30)
x2
30
=
30
= 0 when x =
30.
30
By the First Derivative Test, the dimensions ( x + 2) by
(y
7 2.
x −8 −0
Because d is smallest when the expression inside the radical
is smallest, you need to find the critical numbers of
y =
7
2
(
=
2
x = −1
383
x − 8, (12, 0)
d =
2
Optimization Problems
(
+ 2) are 2 +
) (
30 by 2 +
)
30 (approximately
7.477 by 7.477). These dimensions yield a minimum
area.
y
4
x+2
3
x
( x, x )
2
d
1
y
y+2
x
1
2
3
(4, 0)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
384
Chapter 4
Applications of Differentiation
18. xy = 36 ⇒ y =
36
x
⎛ 36
⎞
+ 3⎟
A = ( x + 3)( y + 3) = ( x + 3)⎜
⎝ x
⎠
108
= 36 +
+ 3x + 9
x
−108
dA
=
+ 3 = 0 ⇒ 3 x 2 = 108 ⇒ x = 6, y = 6
dx
x2
Dimensions: 9 × 9
xy = 245,000 (see figure)
19.
S = x + 2y
490,000 ⎞
⎛
= ⎜x +
⎟ where S is the length
x
⎝
⎠
of fence needed.
dS
490,000
=1−
= 0 when x = 700.
dx
x2
d 2S
980,000
=
> 0 when x = 700.
dx 2
x3
S is a minimum when x = 700 m and y = 350 m.
x+3
x
y
y+3
y
x
20.
S = 2 x 2 + 4 xy = 337.5
y =
337.5 − 2 x 2
4x
⎡ 337.5 − 2 x 2 ⎤
1 3
V = x2 y = x2 ⎢
⎥ = 84.375 x − x
4x
2
⎣
⎦
3 2
dV
= 84.375 − x = 0 ⇒ x 2 = 56.25 ⇒ x = 7.5 and y = 7.5.
2
dx
y
d 2V
= −3x < 0 for x = 7.5.
dx 2
The maximum value occurs when x = y = 7.5 cm.
21.
x
x
⎛ x⎞
16 = 2 y + x + π ⎜ ⎟
⎝ 2⎠
32 = 4 y + 2 x + π x
y =
32 − 2 x − π x
4
A = xy +
π ⎛ x⎞
π x2
π
π
1
⎛ 32 − 2 x − π x ⎞
= 8x − x2 − x2 + x2
⎜ ⎟ = ⎜
⎟x +
2⎝ 2⎠
4
8
2
4
8
⎝
⎠
2
π
π
π⎞
dA
8
32
⎛
.
= 8 − x − x + x = 8 − x⎜1 + ⎟ = 0 when x =
=
π
dx
2
4
4
1
4
4
+
+π
(
)
⎝
⎠
π⎞
32
d2A
⎛
.
= −⎜1 + ⎟ < 0 when x =
2
dx
4⎠
4+π
⎝
y =
32 − 2 ⎡⎣32 ( 4 + π )⎤⎦ − π ⎡⎣32 ( 4 + π )⎦⎤
16
=
4
4+π
16
32
The area is maximum when y =
ft and x =
ft.
4+π
4+π
x
2
y
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
22. You can see from the figure that A = xy and y =
Optimization Problems
385
6− x
.
2
y
6
5
y=
4
6−x
2
3
2
( x, y )
1
x
1
2
4
3
5
6
1
⎛6 − x⎞
2
A = x⎜
⎟ = (6 x − x ).
2
⎝ 2 ⎠
dA
1
= (6 − 2 x ) = 0 when x = 3.
dx
2
d2A
= −1 < 0 when x = 3.
dx 2
A is a maximum when x = 3 and y = 3 2.
23. (a)
y −2
0−2
=
0 −1
x −1
y = 2+
L =
2
x −1
x2 + y2 =
2 ⎞
⎛
x2 + ⎜ 2 +
⎟
x − 1⎠
⎝
2
=
x2 + 4 +
8
4
,
+
x − 1 ( x − 1)2
x >1
10
(b)
(2.587, 4.162)
0
10
0
L is minimum when x ≈ 2.587 and L ≈ 4.162.
(c) Area = A( x) =
A′( x) = 1 +
(x
1
1 ⎛
2 ⎞
x
xy = x⎜ 2 +
⎟ = x +
2
2 ⎝
x − 1⎠
x −1
( x − 1) − x
( x − 1)2
=1−
1
(x
− 1)
2
= 0
− 1) = 1
2
x − 1 = ±1
x = 0, 2 (select x = 2)
They y = 4 and A = 4.
Vertices: (0, 0), ( 2, 0), (0, 4)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
386
Chapter 4
24. (a)
Applications of Differentiation
)
(
1
1
base × height =
2 36 − h 2 (6 + h) =
2
2
1/2
−1/2
dA
1
= (36 − h 2 ) ( −2h)(6 + h) + (36 − h 2 )
dh
2
A =
= (36 − h 2 )
−1/2
36 − h 2 (6 + h)
−2( h 2 + 3h − 18)
⎡− h(6 + h) + (36 − h 2 )⎤ =
⎣
⎦
36 − h 2
=
−2( h + 6)( h − 3)
36 − h 2
dA
= 0 when h = 3, which is a maximum by the First Derivative Test. So, the sides are 2 36 − h 2 = 6 3, an
dh
equilateral triangle. Area = 27 3 sq. units.
6
6
h
36 − h 2
6 + h
=
2 3 6+ h
(b) cos α =
6 + h
2 3
36 − h 2
6 + h
1
2
⎛ ⎞
Area = 2⎜ ⎟ 36 − h 2 (6 + h) = (6 + h) tan α = 144 cos 4 α tan α
⎝ 2⎠
tan α =
)
(
A′(α ) = 144 ⎡⎣cos 4 α sec 2 α + 4 cos3 ( −sin α ) tan α ⎤⎦ = 0
⇒ cos 4 α sec 2 α = 4 cos3 α sin α tan α
1 = 4 cos α sin α tan α
1
= sin 2 α
4
1
sin α =
⇒ α = 30° and A = 27 3.
2
2
α
3
6+h
6
h
6
36 − h 2
(c) Equilateral triangle
25.
A = 2 xy = 2 x 25 − x 2 (see figure)
dA
⎛ 1 ⎞⎛
= 2 x⎜ ⎟⎜
dx
⎝ 2 ⎠⎝
⎞
⎛ 25 − 2 x 2 ⎞
5 2
2
≈ 3.54.
⎟ + 2 25 − x = 2⎜
⎟ = 0 when x = y =
2
25 − x ⎠
⎝ 25 − x 2 ⎠
−2 x
2
By the First Derivative Test, the inscribed rectangle of maximum area has vertices
⎛ 5 2 ⎞ ⎛ 5 2 5 2⎞
, 0 ⎟⎟, ⎜⎜ ±
,
⎜⎜ ±
⎟.
2
2
2 ⎟⎠
⎝
⎠ ⎝
Width:
y
8
6
( x,
25 − x 2
5 2
; Length: 5 2
2
(
x
−6 −4 −2
−2
2
4
6
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
26.
Optimization Problems
387
A = 2 xy = 2 x r 2 − x 2 (see figure)
2( r 2 − 2 x 2 )
dA
=
= 0 when x =
dx
r 2 − x2
2r
.
2
2r by
By the First Derivative Test, A is maximum when the rectangle has dimensions
(
)
2r 2.
y
(−x,
r 2 − x2
( ( x,
r 2 − x2
(
x
(−r, 0)
(r, 0)
200 − 2 x
2
⎛ y⎞
27. (a) P = 2 x + 2π r = 2 x + 2π ⎜ ⎟ = 2 x + π y = 200 ⇒ y =
= (100 − x)
π
π
⎝ 2⎠
y
2
y
x
(b)
Width, y
Length, x
2
10
π
20
π
30
π
40
π
2
2
2
2
50
π
2
60
π
Area, xy
2
(100
− 10)
(10) (100
(100
− 20)
(20) (100
− 20) ≈ 1019
(100
− 30)
(30) (100
2
− 30) ≈ 1337
(100
− 40)
(40) (100
2
− 40) ≈ 1528
(100
− 50)
(50) (100
2
− 50) ≈ 1592
(100
− 60)
(60) (100
2
− 60) ≈ 1528
π
2
π
π
π
π
π
− 10) ≈ 573
The maximum area of the rectangle is approximately 1592 m2.
(c) A = xy = x
(d) A′ =
2
π
(100
2
π
(100
− x) =
2
π
(100 x − x 2 )
− 2 x ). A′ = 0 when x = 50.
Maximum value is approximately 1592 when length = 50 m and width =
(e)
100
π
.
2000
(50, 1591.6)
0
100
0
Maximum area is approximately
1591.55 m 2 ( x = 50 m).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
388
Chapter 4
Applications of Differentiation
28. V = π r 2 h = 22 cubic inches or h =
(a)
Radius, r
Height
Surface Area
2
⎡
22 ⎤
⎥ ≈ 220.3
2π (0.2) ⎢0.2 +
2
⎢⎣
π (0.2) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 111.0
2π (0.4) ⎢0.4 +
2
⎢⎣
π (0.4) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 75.6
2π (0.6) ⎢0.6 +
2
⎢⎣
π (0.6) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 59.0
2π (0.8) ⎢0.8 +
2
⎢⎣
π (0.8) ⎥⎦
22
0.2
π (0.2)
22
0.4
π (0.4)
22
0.6
π (0.6)
22
(b)
0.8
π (0.8)
Radius, r
Height
Surface Area
2
⎡
22 ⎤
⎥ ≈ 220.3
2π (0.2) ⎢0.2 +
2
⎢⎣
π (0.2) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 111.0
2π (0.4) ⎢0.4 +
2
⎢⎣
π (0.4) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 75.6
2π (0.6) ⎢0.6 +
2
π (0.6) ⎦⎥
⎣⎢
2
⎡
22 ⎤
⎥ ≈ 59.0
2π (0.8) ⎢0.8 +
2
⎢⎣
π (0.8) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 50.3
2π (1.0) ⎢1.0 +
2
⎢⎣
π (1.0) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 45.7
2π (1.2) ⎢1.2 +
2
⎢⎣
π (1.2) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 43.7
2π (1.4) ⎢1.4 +
2
⎢⎣
π (1.4) ⎥⎦
2
⎡
22 ⎤
⎥ ≈ 43.6
2π (1.6) ⎢1.6 +
2
π (1.6) ⎦⎥
⎣⎢
2
⎡
22 ⎤
⎥ ≈ 44.8
2π (1.8) ⎢1.8 +
2
⎢⎣
π (1.8) ⎥⎦
22
0.2
π (0.2)
22
0.4
π (0.4)
22
0.6
π (0.6)
22
0.8
π (0.8)
1.0
π (1.0)
22
22
1.2
π (1.2)
22
1.4
π (1.4)
22
1.6
π (1.6)
22
1.8
π (1.8)
22
2.0
22
π r2
π ( 2.0)
2
⎡
22 ⎤
⎥ ≈ 47.1
2π ( 2.0) ⎢2.0 +
2
⎢⎣
π ( 2.0) ⎥⎦
The minimum seems to the about 43.6 for r = 1.6.
(c) S = 2π r 2 + 2π rh
22 ⎤
44
⎡
= 2π r ( r + h) = 2π r ⎢r +
= 2π r 2 +
π r 2 ⎥⎦
r
⎣
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
(d)
Optimization Problems
389
100
(1.52, 43.46)
−1
4
−10
The minimum seems to be 43.46 for r ≈ 1.52.
(e)
dS
44
= 4π r − 2 = 0 when r =
dr
r
22
h =
≈ 3.04 in.
π r2
Note: Notice that h =
3
11 π ≈ 1.52 in.
⎛ 111 3 ⎞
22
22
=
= 2⎜ 1 3 ⎟ = 2r.
23
2
πr
π (11 π )
⎝π ⎠
29. Let x be the sides of the square ends and y the length of
the package.
V = π r2x
30.
x + 2π r = 108 ⇒ x = 108 − 2π r (see figure)
P = 4 x + y = 108 ⇒ y = 108 − 4 x
V = π r 2 (108 − 2π r ) = π (108r 2 − 2π r 3 )
V = x 2 y = x 2 (108 − 4 x) = 108 x 2 − 4 x3
dV
= π ( 216r − 6π r 2 ) = 6π r (36 − π r )
dr
36
= 0 when r =
and x = 36.
dV
= 216 x − 12 x 2
dx
= 12 x(18 − x) = 0 when x = 18.
π
d 2V
= 216 − 24 x = −216 < 0 when x = 18.
dx 2
2
d V
36
= π ( 216 − 12π r ) < 0 when r =
.
π
dr 2
The volume is maximum when x = 18 in. and
y = 108 − 4(18) = 36 in.
r
x
Volume is maximum when x = 36 in. and
r = 36 π ≈ 11.459 in.
31. No. The volume will change because the shape of the container changes when squeezed.
32. No, there is no minimum area. If the sides are x and y, then 2 x + 2 y = 20 ⇒ y = 10 − x.
The area is A( x) = x(10 − x) = 10 x − x 2 . This can be made arbitrarily small by selecting x ≈ 0.
33.
V = 14 =
h =
4 3
π r + π r 2h
3
14 − ( 4 3)π r 3
πr2
=
14
πr2
−
4
r
3
4 ⎞
28 8 2
4
28
⎛ 14
− πr = πr2 +
S = 4π r 2 + 2π rh = 4π r 2 + 2π r ⎜ 2 − r ⎟ = 4π r 2 +
π
3
3
3
r
r
r
⎝
⎠
8
28
dS
= π r − 2 = 0 when r =
3
dr
r
8
56
d 2S
= π + 3 > 0 when r =
3
dr 2
r
3
3
21
≈ 1.495 cm.
2π
21
.
2π
The surface area is minimum when r =
r
3
21
cm and h = 0.
2π
h
The resulting solid is a sphere of radius r ≈ 1.495 cm.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
390
Chapter 4
Applications of Differentiation
4 3
π r + π r 2h
3
4000
4
h =
− r
3
πr2
34. V = 4000 =
Let k = cost per square foot of the surface area of the sides, then 2k = cost per square foot of the hemispherical ends.
⎡
4 ⎞⎤
8000 ⎤
⎛ 4000
⎡16
C = 2k ( 4π r 2 ) + k ( 2π rh) = k ⎢8π r 2 + 2π r ⎜ 2 − r ⎟⎥ = k ⎢ π r 2 +
3 ⎠⎦
r ⎦⎥
⎝ πr
⎣3
⎣
dC
8000 ⎤
⎡ 32
= k ⎢ π r − 2 ⎥ = 0 when r =
dr
3
r ⎦
⎣
By the Second Derivative Test, you have
The cost is minimum when r =
3
750
π
3
750
π
≈ 6.204 ft and h ≈ 24.814 ft.
d 2C
12,000 ⎤
⎡ 32
= k⎢ π +
> 0 when r =
dr 2
r 3 ⎥⎦
3
⎣
3
750
π
.
ft and h ≈ 24.814 ft.
35. Let x be the length of a side of the square and y the length of a side of the triangle.
4 x + 3 y = 10
A = x2 +
=
(10
1 ⎛ 3 ⎞
y⎜
y⎟
2 ⎜⎝ 2 ⎟⎠
− 3 y)
16
2
+
3 2
y
4
dA
1
3
= (10 − 3 y )( −3) +
y = 0
dy
8
2
−30 + 9 y + 4 3 y = 0
y =
30
9 + 4 3
d2A
9 + 4 3
=
> 0
dy 2
8
A is minimum when y =
30
10 3
and x =
.
9 + 4 3
9+ 4 3
36. (a) Let x be the side of the triangle and y the side of the square.
A =
3⎛
π ⎞ 2 4⎛
π⎞ 2
⎜ cot ⎟ x + ⎜ cot ⎟ y where 3 x + 4 y = 20
4⎝
3⎠
4⎝
4⎠
2
=
3 2 ⎛
3 ⎞
20
x + ⎜5 − x⎟ , 0 ≤ x ≤
.
4
4 ⎠
3
⎝
3
3 ⎞⎛ 3 ⎞
⎛
x + 2⎜ 5 − x ⎟⎜ − ⎟ = 0
2
4 ⎠⎝ 4 ⎠
⎝
60
x =
4 3 +9
A′ =
(
)
When x = 0, A = 25, when x = 60 4 3 + 9 , A ≈ 10.847, and when x = 20 3, A ≈ 19.245. Area is maximum
when all 20 feet are used on the square.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
Optimization Problems
391
(b) Let x be the side of the square and y the side of the pentagon.
A =
4⎛
π ⎞ 2 5⎛
π⎞ 2
⎜ cot ⎟ x + ⎜ cot ⎟ y where 4 x + 5 y = 20
4⎝
4⎠
4⎝
5⎠
2
4 ⎞
⎛
= x 2 + 1.7204774⎜ 4 − x ⎟ , 0 ≤ x ≤ 5.
5 ⎠
⎝
4 ⎞
⎛
A′ = 2 x − 2.75276384⎜ 4 − x ⎟ = 0
5 ⎠
⎝
x ≈ 2.62
When x = 0, A ≈ 27.528, when x ≈ 2.62, A ≈ 13.102, and when x = 5, A ≈ 25. Area is maximum when all 20 feet
are used on the pentagon.
(c) Let x be the side of the pentagon and y the side of the hexagon.
A =
=
5⎛
π ⎞ 2 6⎛
π⎞ 2
⎜ cot ⎟ x + ⎜ cot ⎟ y where 5 x + 6 y = 20
4⎝
5⎠
4⎝
6⎠
5⎛
π⎞ 2 3
⎜ cot ⎟ x +
4⎝
5⎠
2
2
( 3)⎛⎜⎝ 20 −6 5x ⎞⎟⎠ , 0 ≤
x ≤ 4.
5⎛
π⎞
⎛ 5 ⎞⎛ 20 − 5 x ⎞
⎜ cot ⎟ x + 3 3 ⎜ − ⎟⎜
⎟ = 0
2⎝
5⎠
6
⎝ 6 ⎠⎝
⎠
x ≈ 2.0475
A′ =
When x = 0, A ≈ 28.868, when x ≈ 2.0475, A ≈ 14.091, and when x = 4, A ≈ 27.528. Area is maximum when all
20 feet are used on the hexagon
(d) Let x be the side of the hexagon and r the radius of the circle.
A =
6⎛
π⎞ 2
2
⎜ cot ⎟ x + π r where 6 x + 2π r = 20
4⎝
6⎠
2
=
3 3 2
10
⎛ 10 3x ⎞
x + π⎜
.
−
⎟ ,0 ≤ x ≤
2
π ⎠
3
⎝π
⎛ 10 3x ⎞
A′ = 3 3 − 6⎜
−
= 0
π ⎟⎠
⎝π
x ≈ 1.748
When x = 0, A ≈ 31.831, when x ≈ 1.748, A ≈ 15.138, and when x = 10 3, A ≈ 28.868. Area is maximum when all
20 feet are used on the circle.
In general, using all of the wire for the figure with more sides will enclose the most area.
37. Let S be the strength and k the constant of
proportionality. Given
h 2 + w2 = 202 , h 2 = 202 − w2 ,
S = kwh 2
S = kw( 400 − w2 ) = k ( 400 w − w3 )
dS
20 3
= k ( 400 − 3w2 ) = 0 when w =
in.
dw
3
and h =
20 6
in.
3
d 2S
20 3
= −6kw < 0 when w =
.
dw2
3
38. Let A be the amount of the power line.
A = h − y + 2
x2 + y 2
2y
dA
= −1 +
dy
x2 + y2
= 0 when y =
d 2A
2x2
=
> 0 for y =
32
dy 2
( x2 + y 2 )
x
.
3
x
.
3
The amount of power line is minimum when
y = x 3.
y
(0, h)
These values yield a maximum.
h−y
y
(−x, 0)
x
(x, 0)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
392
Chapter 4
39.
Applications of Differentiation
x2 + 4
θ
40. sin α =
2
4−x
x
C ( x) = 2k
x + 4 + k ( 4 − x)
2 xk
C ′( x) =
I =
− k = 0
x2 + 4
2x =
tan α =
2
3x 2 = 4
2
3
Oil well
4
3
2
h
2 tan α
⇒ h = 2 tan α ⇒ s =
= 2 sec α
2
sin α
k sin α
k sin α
k
=
= sin α cos 2 α
s2
4 sec 2 α
4
dI
k
= ⎡⎣sin α ( −2 sin α cos α ) + cos 2 α (cos α )⎤⎦
dα
4
k
= cos α ⎡⎣cos 2 α − 2 sin 2 α ⎤⎦
4
k
= cos α ⎡⎣1 − 3 sin 2 α ⎤⎦
4
1
π 3π
, or when sin α = ±
.
= 0 when α = ,
2 2
3
x2 + 4
4x2 = x2 + 4
x =
h
h
π
⇒ s =
,0 < α <
s
sin α
2
Because α is acute, you have
2
3
4− 2
3
sin α =
Refinery
1
⎛ 1 ⎞
⇒ h = 2 tan α = 2⎜
⎟ =
3
⎝ 2⎠
2 ft.
Because ( d 2 I ) ( dα 2 ) = ( k 4) sin α (9 sin 2 α − 7) < 0
The path of the pipe should go underwater from the
oil well to the coast following the hypotenuse of a
right triangle with leg lengths of 2 kilometers and
2 3 kilometers for a distance of 4 3 kilometers.
when sin α = 1
3, this yields a maximum.
Then the pipe should go down the coast to the refinery
(
)
for a distance of 4 − 2
3 kilometers.
h
α
s
α
4 ft
S =
41. (a)
x 2 + 4, L =
x2 + 4
+
2
Time = T =
dT
=
dx
2
x
x2 + 4
+
4
1 + (3 − x )
2
x 2 − 6 x + 10
4
x −3
= 0
x 2 − 6 x + 10
9 − 6 x + x2
x2
=
2
x + 4
4( x 2 − 6 x + 10)
x 4 − 6 x3 + 9 x 2 + 8 x − 12 = 0
S=
2
x
x2 + 4
3−x
1
L=
1 + (3 − x( 2
Q
You need to find the roots of this equation in the interval [0, 3]. By using a computer or graphing utility you can
determine that this equation has only one root in this interval ( x = 1). Testing at this value and at the endpoints,
you see that x = 1 yields the minimum time. So, the man should row to a point 1 mile from the nearest point on
the coast.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
(b)
x2 + 4
+
v1
T =
dT
=
dx
v1
x
x2 + 4
x
Because
x2 + 4
x −3
+
v2
= 0
x 2 − 6 x + 10
= sin θ1 and
x −3
x 2 − 6 x + 10
= −sin θ 2
θ1
2
3−x
x
d 2T
4
1
=
+
> 0
32
32
dx 2
v1 ( x 2 + 4)
v2 ( x 2 − 6 x + 10)
Because
393
x 2 − 6 x + 10
v2
sin θ1
sin θ 2
sin θ1
sin θ 2
.
−
= 0 ⇒
=
v1
v2
v1
v2
you have
Optimization Problems
1
θ2
Q
this condition yields a minimum time.
42.
p (t ) =
250
1 + 4e − t 3
p′(t ) =
1000
e−t 3
; p′( 2) ≈ 18.35 elk month
3 (1 + 4e − t 3 )2
p′′(t ) =
−t 3
−t 3
− 1)
1000 e ( 4e
= 0 when t ≈ 4.16 months.
3
9
(1 + 4e−t 3 )
43. f ( x) = 2 − 2 sin x
44.
y
2
1
−1
x
x +
2
d12
2
v2
x − a
+
v2
d 2 + (a − x)
2
2
= 0
Because
π
4
π
2
x
x
x 2 + d12
(a) Distance from origin to y-intercept is 2.
Distance from origin to x-intercept is π 2 ≈ 1.57.
(b) d =
d 2 2 + ( a − x)
x 2 + d12
+
v1
dT
=
dx
v1
3
−π
4
T =
x2 + y 2 =
x 2 + ( 2 − 2 sin x)
2
= sin θ1 and
x − a
d 2 + ( a − x)
2
2
= −sin θ 2
you have
sin θ1
sin θ 2
sin θ1
sin θ 2
.
−
= 0 ⇒
=
v1
v2
v1
v2
Because
3
(0.7967, 0.9795)
−␲
4
d 2T
d12
d22
=
+
> 0
32
2
2 32
dx
v1 ( x 2 + d12 )
v2 ⎡d 2 2 + ( a − x) ⎤
⎣
⎦
␲
2
this condition yields a minimum time.
−1
Minimum distance = 0.9795 at x = 0.7967.
(c) Let f ( x ) = d 2 ( x) = x 2 + ( 2 − 2 sin x) .
2
f ′( x) = 2 x + 2( 2 − 2 sin x)(−2 cos x)
Setting f ′( x) = 0, you obtain x ≈ 0.7967, which
corresponds to d = 0.9795.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
394
45.
Chapter 4
V =
Applications of Differentiation
1 2
1
π r h = π r 2 144 − r 2
3
3
f (c ) = f (c + x)
46. (a)
10ce − c = 10(c + x)e −(c + x)
−1 2
⎤
dV
1 ⎡ ⎛1⎞
= π ⎢r 2 ⎜ ⎟(144 − r 2 ) ( −2r ) + 2r 144 − r 2 ⎥
dr
3 ⎣ ⎝ 2⎠
⎦
=
c
c + x
= c+ x
ec
e
c+x
ce
= (c + x )e c
1 ⎡ 288r − 3r 3 ⎤
π⎢
⎥
3 ⎣ 144 − r 2 ⎦
⎡ r (96 − r ) ⎤
⎥ = 0 when r = 0, 4 6.
= π⎢
⎢⎣ 144 − r 2 ⎥⎦
2
By the First Derivative Test, V is maximum when
r = 4 6 and h = 4 3.
ce x = c + x
ce − c = x
x
c =
(b) A( x) = xf (c)
⎡ ⎛ x ⎞ −x
= x ⎢10⎜ x
⎟e
⎣ ⎝ e − 1⎠
10 x 2 x (1 − e x )
= x
e
e −1
Area of circle: A = π (12) = 144π
2
Lateral surface area of cone:
(
S = π 4 6
) (4 6 )
2
(
+ 4 3
)
2
= 48 6π
(c) A( x) =
Area of sector:
144π − 48 6π =
1 2
θ r = 72θ
2
144π − 48 6π
72
2π
3 − 6 ≈ 1.153 radians or 66°
=
3
x
ex − 1
10 x 2 x
e
ex − 1
(ex −1) ⎤
⎥
⎦
(1− ex )
6
θ =
(
)
0
9
0
The maximum area is 4.591 for x = 2.118 and
f ( x) = 2.547.
(d) c =
x
ex − 1
2
lim c = 1
x → 0+
lim c = 0
x →∞
0
0
4
47. Let d be the amount deposited in the bank, i be the interest rate paid by the bank,
and P be the profit.
P = (0.12)d − id
d = ki 2 ( because d is proportional to i 2 )
P = (0.12)( ki 2 ) − i( ki 2 ) = k (0.12i 2 − i 3 )
dP
0.24
= k (0.24i − 3i 2 ) = 0 when i =
= 0.08.
di
3
d 2P
= k (0.24 − 6i ) < 0 when i = 0.08 ( Note: k > 0).
di 2
The profit is a maximum when i = 8%.
48. (a) The profit is increasing on (0, 40).
(b) The profit is decreasing on ( 40, 60).
(c) In order to yield a maximum profit, the company should spend about $40 thousand.
(d) The point of diminishing returns is the point where the concavity changes, which in this case is x = 20 thousand dollars.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.7
49.
Optimization Problems
395
L
, a > 0, b > 0, L > 0
1 + ae − x b
⎛ a
⎞
aL − x b
− L⎜ − e − x b ⎟
e
b
⎝
⎠
b
y′ =
=
2
2
(1 + ae− x b ) (1 + ae− x b )
y =
(1 + ae− x b ) ⎛⎜⎝ −baL2 e− x b ⎞⎟⎠ − ⎛⎜⎝ aLb e− x b ⎞⎟⎠2(1 + ae− x b )⎛⎜⎝ −ba e− x b ⎞⎟⎠
2
y′′ =
(1 + ae− x b )
4
(1 + ae− x b )⎛⎜⎝ −baL2 e− x b ⎞⎟⎠ + 2⎛⎜⎝ aLb e− x b ⎞⎛⎟⎜⎠⎝ ba e− x b ⎞⎟⎠
=
(1 + ae− x b )
y′′ = 0 if ae − x b = 1 ⇒
y (b ln a ) =
=
Lae − x b ( ae − x b − 1)
(1 + ae− x b ) b2
3
−x
⎛1⎞
= ln ⎜ ⎟ ⇒ x = b ln a
b
⎝a⎠
L
1 + ae
3
−(b ln a) b
L
L
=
1 + a(1 a )
2
=
Therefore, the y-coordinate of the inflection point is L 2.
50.
A = ( base)( height ) = 2 xe − x
52. S 2 = 4m − 1 + 5m − 6 + 10m − 3
2
Using a graphing utility, you can see that the minimum
occurs when m = 0.3.
2
2
dA
= − 4 x 2 e − x + 2e − x
dx
= 2e − x (1 − 2 x 2 ) = 0 when x =
2
Line y = 0.3x
2
.
2
S 2 = 4(0.3) − 1 + 5(0.3) − 6 + 10(0.3) − 3 = 4.7 mi.
2e −1 2
A =
S2
y
30
3
20
2
(
−2
−1
2
2,
1
−1
e 2
)
2
10
(0.3, 4.7)
x
m
1
2
3
−1
51.
S1 = ( 4m − 1) + (5m − 6) + (10m − 3)
2
2
2
dS1
= 2( 4m − 1)( 4) + 2(5m − 6)(5) + 2(10m − 3)(10)
dm
64
= 282m − 128 = 0 when m =
.
141
64
Line: y =
x
141
53. S3 =
256
320
640
858
−1 +
−6 +
−3 =
≈ 6.1 mi
141
141
141
141
m2 + 1
+
5m − 6
m2 + 1
+
10m − 3
m2 + 1
Using a graphing utility, you can see that the minimum
occurs when x ≈ 0.3.
Line: y ≈ 0.3x
S3 =
⎛ 64 ⎞
⎛ 64 ⎞
⎛ 64 ⎞
S = 4⎜
⎟ − 1 + 5⎜
⎟ − 6 + 10⎜
⎟−3
141
141
⎝
⎠
⎝
⎠
⎝ 141 ⎠
=
4m − 1
4(0.3) − 1 + 5(0.3) − 6 + 10(0.3) − 3
(0.3)2
+1
≈ 4.5 mi.
S3
30
20
10
(0.3, 4.5)
m
1
2
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
396
Chapter 4
Applications of Differentiation
54. (a) Label the figure so that r 2 = x 2 + h 2 .
Then, the area A is 8 times the area of the region given by OPQR:
(
⎡1
⎤
⎡1
A = 8 ⎢ h 2 + ( x − h ) h⎥ = 8 ⎢ ( r 2 − x 2 ) + x −
2
⎣
⎦
⎣2
8x2
A′( x) = 8 r 2 − x 2 −
8x2
r 2 − x2
r 2 − x2
)
⎤
r 2 − x 2 ⎥ = 8 x r 2 − x 2 + 4 x 2 − 4r 2
⎦
+ 8x = 0
= 8x + 8 r 2 − x2
r 2 − x2
x 2 = x r 2 − x 2 + (r 2 − x 2 )
2x2 − r 2 = x r 2 − x2
4 x 4 − 4 x 2 r 2 + r 4 = x 2 (r 2 − x 2 )
5x4 − 5x2r 2 + r 4 = 0
5r 2 ±
25r 4 − 20r 4
r2 ⎡
=
5±
10
10 ⎣
Take positive value.
x2 =
x = r
h
Quadratic in x 2 .
θ
2
5 ⎦⎤.
R
O
x
h
Q
P r
5+ 5
≈ 0.85065r Critical number
10
θ
x
h
= and cos = . The area A of the cross equals the sum of two large rectangles minus the common
2
r
2
r
square in the middle.
(b) Note that sin
θ
A = 2( 2 x)( 2h) − 4h 2 = 8 xh − 4h 2 = 8r 2 sin
θ
2
cos
θ
2
− 4r 2 sin 2
θ
2
θ⎞
⎛
= 4r 2 ⎜ sin θ − sin 2 ⎟
2⎠
⎝
θ
θ⎞
⎛
A′(θ ) = 4r 2 ⎜ cos θ − sin cos ⎟ = 0
2
2⎠
⎝
θ
θ
1
cos θ = sin cos = sin θ
2
2
2
tan θ = 2
θ = arctan( 2) ≈ 1.10715
(c) Note that x 2 =
(
63.4°
or
)
r2
5+
10
5 and r 2 − x 2 =
(
)
r2
5−
10
5.
A( x) = 8 x r 2 − x 2 + 4 x 2 − 4r 2
⎡r 2
= 8⎢ 5 +
⎣10
(
5
) (
12
⎡r 4
⎤
= 8⎢ ( 20)⎥
10
⎣
⎦
=
8 2
r
5
⎡4
= 2r 2 ⎢
⎣⎢ 5
+ 2r 2 +
5 − 2r 2 +
5 −1+
12
⎤
5⎥
⎦
)
r2
5−
10
2
5
(
r2
+ 4 5+
10
)
5 − 4r 2
5r 2 − 4 r 2
2 5 2
r
5
5⎤
2
⎥ = 2r
5 ⎦⎥
(
)
5 −1
Using the angle approach, note that tan θ = 2, sin θ =
1
1⎛
1 ⎞
2
⎛θ ⎞
and sin 2 ⎜ ⎟ = (1 − cos θ ) = ⎜1 −
⎟.
2
2⎝
5
5⎠
⎝ 2⎠
4r 2
⎛ 2
θ⎞
1⎛
1 ⎞⎞
⎛
So, A(θ ) = 4r 2 ⎜ sin θ − sin 2 ⎟ = 4r 2 ⎜
− ⎜1 −
⎟ =
⎟
2⎠
2⎝
5 ⎠⎠
⎝
⎝ 5
(
)
5 −1
2
= 2r 2
(
)
5 −1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.8
55. f ( x ) = x 3 − 3 x; x 4 + 36 ≤ 13 x 2
Differentials
397
3
1⎞
1
⎛
56. Let a = ⎜ x + ⎟ and b = x 3 + 3 , x > 0.
x
x
⎝
⎠
x 4 − 13 x 2 + 36 = ( x 2 − 9)( x 2 − 4)
6
1⎞
1⎞
⎛
⎛
a 2 − b 2 = ⎜ x + ⎟ − ⎜ x3 + 3 ⎟
x
x
⎝
⎠
⎝
⎠
= ( x − 3)( x − 2)( x + 2)( x + 3) ≤ 0
So, −3 ≤ x ≤ −2 or 2 ≤ x ≤ 3.
2
6
1⎞
1
⎛
⎛
⎞
= ⎜ x + ⎟ − ⎜ x6 + 6 + 2⎟
x⎠
x
⎝
⎝
⎠
f ′( x) = 3 x − 3 = 3( x + 1)( x − 1)
2
f is increasing on ( −∞, −1) and (1, ∞).
Let f ( x ) =
So, f is increasing on [−3, − 2] and [2, 3].
f ( −2) = −2, f (3) = 18. The maximum value of f is 18.
=
(x
(
+ 1 x) − x 6 + 1 x + 2
6
(x
6
(
+ 1 x) + x + 1 x
3
a −b
a + b
2
2
3
3
)
)
= a −b
3
1⎞ ⎛
1⎞
⎛
= ⎜ x3 + 3x +
+ 3 ⎟ − ⎜ x3 + 3 ⎟
x
x
x
⎝
⎠ ⎝
⎠
= 3x +
Let g ( x) = x +
3
1⎞
⎛
= 3⎜ x + ⎟.
x
x⎠
⎝
1
1
, g ′( x) = 1 − 2 = 0 ⇒ x = 1.
x
x
2
and g ′′(1) = 2 > 0. So g is a minimum at
x3
x = 1: g (1) = 2.
g ′′( x ) =
Finally, f is a minimum of 3( 2) = 6.
Section 4.8 Differentials
1.
f ( x) = x 2
f ′( x) = 2 x
Tangent line at (2, 4): y − f ( 2) = f ′( 2)( x − 2)
y − 4 = 4( x − 2)
y = 4x − 4
x
2.
1.9
1.99
2
2.01
2.1
f ( x) = x 2
3.6100
3.9601
4
4.0401
4.4100
T ( x) = 4 x − 4
3.6000
3.9600
4
4.0400
4.4000
f ( x) =
6
= 6 x −2
x2
f ′( x) = −12 x −3 =
⎛
Tangent line at ⎜ 2,
⎝
−12
x3
3⎞
⎟:
2⎠
3
−12
−3
y −
=
( x − 2) = ( x − 2)
2
8
2
3
9
y = − x +
2
2
x
6
x2
3
9
T ( x) = − x +
2
2
f ( x) =
1.9
1.99
2
2.01
2.1
1.6620
1.5151
1.5
1.4851
1.3605
1.65
1.515
1.5
1.485
1.35
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
398
3.
Chapter 4
A pplications of Differentiation
f ( x) = x5
f ′( x) = 5 x 4
Tangent line at ( 2, 32):
y − f ( 2) = f ′( 2)( x − 2)
y − 32 = 80( x − 2)
y = 80 x − 128
x
1.9
1.99
2
2.01
2.1
24.7610
31.2080
32
32.8080
40.8410
24.0000
31.2000
32
32.8000
40.0000
1.9
1.99
2
2.01
2.1
1.3784
1.4107
1.4142
1.4177
1.4491
1.3789
1.4107
1.4142
1.4177
1.4496
1.9
1.99
2
2.01
2.1
f ( x) = sin x
0.9463
0.9134
0.9093
0.9051
0.8632
T ( x) = (cos 2)( x − 2) + sin 2
0.9509
0.9135
0.9093
0.9051
0.8677
f ( x) = x
5
T ( x) = 80 x − 128
4.
f ( x) =
x
1
f ′( x) =
2
x
(
Tangent line at 2,
)
2:
y − f ( 2) = f ′( 2)( x − 2)
y −
1
( x − 2)
2 2
1
x
y =
+
2 2
2
2 =
x
f ( x) =
T ( x) =
5.
x
x
2
2
+
1
2
f ( x) = sin x
f ′( x) = cos x
Tangent line at ( 2, sin 2):
y − f ( 2) = f ′( 2)( x − 2)
y − sin 2 = (cos 2)( x − 2)
y = (cos 2)( x − 2) + sin 2
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.8
6.
f ( x) = log 2 x =
ln x
,
ln 2
Differentials
399
(2, 1)
1
x ln 2
1
f ′( 2) =
2 ln 2
f ′( x) =
1
( x − 2)
2 ln 2
1
1
y =
x +1−
2 ln 2
ln 2
Tangent line at ( 2, 1): y − 1 =
x
f ( x ) = log 2 x
T ( x) =
1
1
x +1−
2 ln 2
ln 2
1.9
1.99
2
2.01
2.1
0.9260
0.9928
1
1.0072
1.0704
0.9279
0.9928
1
1.0072
1.0721
7. y = f ( x) = x3 , f ′( x) = 3x 2 , x = 1, ∆x = dx = 0.1
dy = f ′( x) dx
∆y = f ( x + ∆x) − f ( x)
= f (1.1) − f (1)
= f ′(1)(0.1)
= 0.331
= 3(0.1)
= 0.3
8. y = f ( x) = 6 − 2 x 2 , f ′( x) = − 4 x, x = − 2, ∆x = dx = 0.1
∆y = f ( x + ∆x) − f ( x)
dy = f ′( x) dx
= f ( −1.9) − f (− 2)
(
= 6 − 2( −1.9) − 6 − 2( − 2)
2
2
)
= − 4( − 2)(0.1)
= 0.8
= −1.22 − (− 2) = 0.78
9. y = f ( x) = x 4 + 1, f ′( x) = 4 x3 , x = −1, ∆x = dx = 0.01
∆y = f ( x + ∆x) − f ( x)
dy = f ′( x) dx
= f ( −0.99) − f ( −1)
= f ′( −1)(0.01)
= ( −4)(0.01) = −0.04
= ⎡( −0.99) + 1⎤ − ⎡(−1) + 1⎤ ≈ −0.0394
⎣
⎦ ⎣
⎦
4
4
10. y = f ( x) = 2 − x 4 , f ′( x) = −4 x3 , x = 2, ∆x = dx = 0.01
∆y = f ( x + ∆x) − f ( x)
= f ( 2.01) − f ( 2)
≈ −14.3224 − ( −14) = − 0.3224
dy = f ′( x) dx
= ( −4 x 3 ) dx
= −4( 2) (0.01)
3
= −0.32
11.
y = 3x 2 − 4
dy = 6 x dx
12.
y = 3x 2 3
dy = 2 x −1 3 dx =
2
dx
x1 3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
400
13.
Chapter 4
Applications of Differentiation
y = x tan x
24.
dy = ( x sec 2 x + tan x)dx
14.
dy =
y = csc 2 x
x +
y =
26. (a) f (1.9) = f ( 2 − 0.1) ≈ f ( 2) + f ′( 2)( −0.1)
( )
1
x
≈ 1 + − 12 (−0.1) = 1.05
(b) f ( 2.04) = f ( 2 + 0.04) ≈ f ( 2) + f ′( 2)(0.04)
( )
≈ 1 + − 12 (0.04) = 0.98
9 − x2
−x
9 − x2
27. (a) g ( 2.93) = g (3 − 0.07) ≈ g (3) + g ′(3)( −0.07)
(b) g (3.1) = g (3 + 0.1) ≈ g (3) + g ′(3)(0.1)
⎞
1 − 2x2
dx
1 − x ⎟ dx =
1 − x2
⎠
( )
≈ 8 + − 12 (0.1) = 7.95
2
28. (a) g ( 2.93) = g (3 − 0.07) ≈ g (3) + g ′(3)( −0.07)
≈ 8 + (3)( −0.07) = 7.79
y = 3x − sin 2 x
dy = (3 − 2 sin x cos x) dx = (3 − sin 2 x) dx
20.
( )
≈ 8 + − 12 ( −0.07) = 8.035
dx
y = x 1 − x2
⎛
−x
+
dy = ⎜ x
− x2
1
⎝
19.
dx
≈ 1 + (1)(0.04) = 1.04
−1 2
1
dy = (9 − x 2 ) ( −2 x ) dx =
2
18.
2
(b) f ( 2.04) = f ( 2 + 0.04) ≈ f ( 2) + f ′( 2)(0.04)
1 ⎞
x −1
⎛ 1
dy = ⎜
−
dx
⎟ dx =
2x x ⎠
2x x
⎝2 x
17.
1 + ( x − 2)
≈ 1 + (1)( −0.1) = 0.9
x +1
15. y =
2x − 1
3
dy = −
dx
( 2 x − 1)2
y =
1
25. (a) f (1.9) = f ( 2 − 0.1) ≈ f ( 2) + f ′(2)( −0.1)
dy = ( − 2csc 2 x cot 2 x)dx
16.
y = arctan ( x − 2)
(b) g (3.1) = g (3 + 0.1) ≈ g (3) + g ′(3)(0.1)
≈ 8 + (3)(0.1) = 8.3
sec 2 x
y = 2
x +1
⎡ x 2 + 1 2 sec 2 x tan x − sec 2 x( 2 x) ⎤
(
)
⎥ dx
dy = ⎢
2
⎢
⎥
2
1
x
+
(
)
⎣
⎦
⎡ 2 sec 2 x x 2 tan x + tan x − x ⎤
(
) ⎥ dx
= ⎢
2
⎢
⎥
2
( x + 1)
⎣
⎦
29. x = 10 in., ∆x = dx = ±
1
in.
32
(a) A = x 2
dA = 2 xdx
5
⎛ 1⎞
∆A ≈ dA = 2(10)⎜ ± ⎟ = ± in.2
8
⎝ 32 ⎠
(b) Percent error:
1
21. y = ln 4 − x = ln ( 4 − x 2 )
2
1 ⎛ −2 x ⎞
−x
dy = ⎜
dx
⎟ dx =
2 ⎝ 4 − x2 ⎠
4 − x2
2
22.
y = e −0.5 x cos 4 x
dy = ⎡⎣e −0.5 x ( − 4 sin 4 x) + (− 0.5)e −0.5 x cos 4 x⎤⎦ dx
= e
23.
−0.5 x
[− 4 sin 4 x
− 0.5 cos 4 x] dx
y = x arcsin x
⎛
dy = ⎜
⎝
x
1 − x2
⎞
+ arcsin x ⎟ dx
⎠
dA
58
5
1
=
=
=
= 0.00625 = 0.625%
A
100
800
100
30. r = 16 in., ∆r = dr = ±
1
in.
4
(a) A = π r 2
dA = 2π r dr
⎛ 1⎞
∆A ≈ dA = 2π (16)⎜ ± ⎟ = ± 8π in.2
⎝ 4⎠
(b) Percent error:
dA
8π
1
=
=
= 0.03125 = 3.125%
2
32
A
π (16)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 4.8
31. b = 36 cm, h = 50 cm,
∆b = ∆h = db = dh = ± 0.25 cm
(a)
2
(b)
∆S ≈ dS = 8π (8)( ± 0.02) = ± 1.28π in.2
(c) Percent error of volume:
dV
5.12π
=
= 0.0075 or 0.75%
4π 8 2
V
()
3
Percent error of surface area:
dS
1.28π
=
= 0.005 or 0.5%
2
S
4π (8)
∆C = dC = ±0.9 cm
2
1 2
⎛C ⎞
A = πr2 = π ⎜ ⎟ =
C
π
π
2
4
⎝ ⎠
1
1
± 28.8
dA =
C dC =
(64)(±0.9) =
2π
2π
π
dA
28.8 π
=
≈ 0.028125 = 2.8%
2
A
⎡⎣1 ( 4π )⎤⎦ (64)
(b)
⎡1 ( 2π )⎤⎦ C dC
dA
2 dC
= ⎣
=
≤ 0.03
2
A
⎡⎣1 ( 4π )⎦⎤ C
C
dC
0.03
≤
= 0.015 = 1.5%
2
C
35.
T = 2.5 x + 0.5 x 2 , ∆x = dx = 26 − 25 = 1, x = 25
dT = ( 2.5 + x)dx = ( 2.5 + 25)(1) = 27.5 mi
36. Because the slope of the tangent line is greater at
x = 900 than at x = 400, the change in profit is
greater at x = 900 units.
37. (a)
T = 2π
dT =
L g
π
g
L g
dL
Relative error:
(
(π dL) g L g
dT
=
T
2π L g
V = x3
dV = 3 x 2 dx
2
=
S = 6x2
dS = 12 x dx
∆S ≈ dS = 12(15)( ± 0.03) = ± 5.4 in.2
(c) Percent error of volume:
dV
20.25
=
= 0.006 or 0.6%
V
153
Percent error of surface area:
dS
5.4
=
= 0.004 or 0.4%
2
S
6(15)
)
dL
2L
1
= ( relative error in L)
2
1
= (0.005) = 0.0025
2
∆V ≈ dV = 3(15) ( ± 0.03) = ± 20.25 in.3
(b)
dT
27.5
=
≈ 7.3%
T
375
Percentage change =
33. x = 15 in., ∆x = dx = ± 0.03 in.
(a)
S = 4π r 2
dS = 8π r dr
C = 64 cm
C
2π
4 3
πr
3
∆V ≈ dV = 4π (8) ( ± 0.02) = ± 5.12π in.3
10.75
≈ 0.011944 = 1.19%
1 36 50
(
)( )
2
C = 2π r ⇒ r =
V =
dV = 4π r 2 dr
(b) Percent error:
32. (a)
401
34. r = 8 in., dr = ∆r = ± 0.02 in.
1
(a) A = bh
2
1
1
dA = b dh + h db
2
2
1
1
∆A ≈ dA = (36)( ± 0.25) + (50)( ± 0.25)
2
2
= ±10.75 cm 2
dA
=
A
Differentials
Percentage error:
(b)
dT
1
(100) = 0.25% = %
T
4
(0.0025)(3600)( 24)
= 216 sec = 3.6 min
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
402
38.
Chapter 4
Applications of Differentiation
43. Let f ( x ) =
E = IR
E
R =
I
f ( x + ∆x) ≈ f ( x) + f ′( x) dx =
E
dR = − 2 dI
I
(
f ( x + ∆x ) =
)
− E I dI
dR
=
R
2
= −
E I
4
624 ≈
4
4
dI
I
625 +
(
1
4
625
)
4
3
x3
dx
(−1)
1
= 4.998
500
= 5−
624 ≈ 4.9980.
4
Using a calculator,
401,493,267 e369,444 (50t +19,793)
dt
2,000,000 (50t + 19,793)2
1
x +
4
4
dR
dI
dI
= −
=
R
I
I
39. dH = −
x , x = 625, dx = −1.
4
44. Let f ( x ) = x 3 , x = 3, dx = −0.01.
f ( x + ∆x) ≈ f ( x) + f ′( x) dx = x 3 + 3x 2 dx
At t = 72 and dt = 1, dH ≈ −2.65.
f ( x + ∆x) = ( 2.99) ≈ 33 + 3(3) (−0.01)
3
2
= 27 − 0.27 = 26.73
40. h = 50 tan θ
Using a calculator: ( 2.99) ≈ 26.7309
3
45.
h
f ( x) =
f ′( x) =
θ
x + 4
1
x + 4
2
At (0, 2), f (0) = 2, f ′(0) =
50 ft
θ = 71.5° = 1.2479 radians
1
4
1
( x − 0)
4
1
y = x + 2
4
Tangent line: y − 2 =
dh = 50 sec θ ⋅ dθ
2
50 sec2 (1.2479)
dh
=
dθ ≤ 0.06
50 tan (1.2479)
h
6
9.9316
dθ ≤ 0.06
2.9886
y
(0, 2)
f
−6
dθ ≤ 0.018
6
−2
41. Let f ( x) =
x , x = 100, dx = −0.6.
46.
f ( x + ∆x) ≈ f ( x ) + f ′( x ) dx
=
x +
f ( x + ∆x ) =
99.4
≈
1
2
100 +
x
f ( x) = tan x
f ′( x) = sec 2 x
f ( 0) = 0
dx
f ′(0) = 1
1
(−0.6) = 9.97
2 100
Tangent line at (0, 0): y − 0 = ( x − 0)
y = x
99.4 ≈ 9.96995
Using a calculator:
4
f
42. Let f ( x ) =
x , x = 27, dx = −1.
3
1
dx
3 3 x2
1
1
27 +
(−1) = 3 − ≈ 2.9630
3
2
27
3 27
f ( x + ∆x ) ≈ f ( x) + f ′( x) dx =
3
26 ≈
3
Using a calculator,
3
3
−␲
(0, 0)
y
␲
x+
−4
47. In general, when ∆x → 0, dy approaches ∆y.
26 ≈ 2.9625
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 4
48. Propagated error = f ( x + ∆x) − f ( x),
403
50. Yes. y = x is the tangent line approximation to
f ( x) = sin x at (0, 0).
dy
, and the percent error
relative error =
y
f ′( x) = cos x
f ′(0) = 1
dy
=
× 100.
y
Tangent line: y − 0 = 1( x − 0)
y = x
49. (a) Let f ( x ) =
x , x = 4, dx = 0.02,
(
)
f ′( x) = 1 2
51. True
x.
Then
52. True,
f ( 4.02) ≈ f ( 4) + f ′( 4) dx
4.02 ≈
4 +
1
2 4
(0.02)
= 2+
1
(0.02).
4
dy
∆y
=
= a
∆x
dx
53. True
54. False
(b) Let
f ( x ) = tan x, x = 0, dx = 0.05, f ′( x ) = sec 2 x.
Let f ( x ) =
x , x = 1, and ∆x = dx = 3. Then
∆y = f ( x + ∆x) − f ( x) = f ( 4) − f (1) = 1
Then
f (0.05) ≈ f (0) + f ′(0) dx
and dy = f ′( x) dx =
tan 0.05 ≈ tan 0 + sec 2 0(0.05) = 0 + 1(0.05).
1
2 1
(3)
=
3
.
2
So, dy > ∆y in this example.
Review Exercises for Chapter 4
1.
f ( x ) = x 2 + 5 x,
[−4, 0]
4. h( x) = 3 x − x,
f ′( x) = 2 x + 5 = 0 when x = −5 2
h′( x) =
Critical number: x = −5 2
2.
Critical number: ( −5 2, −25 4)
Minimum
Right endpoint: (0, 0)
Maximum
f ( x) = x + 6 x , [− 6, 1]
Critical numbers: x = 0, − 4
Left endpoint: ( − 6, 0)
Minimum
Critical number: (0, 0)
Minimum
Critical number: ( − 4, 32)
Maximum
Right endpoint: (1, 7)
f ( x) =
x − 2, [0, 4]
f ′( x) =
1
2
x
−1 = 0 ⇒ 2
x = 3 ⇒ x = 94
Left endpoint: (0, 0)
Minimum
Critical number: (9 4, 9 4)
Maximum
Right endpoint: (9, 0)
Minimum
2
f ′( x) = 3x 2 + 12 x = 3 x( x + 4) = 0 when x = 0, − 4
3.
2
Critical number: x = 9 4
Left endpoint: ( −4, − 4)
3
3
[0, 9]
5.
f ( x) =
f ′( x) =
4x
, [− 4, 4]
x2 + 9
( x2
+ 9) 4 − 4 x ( 2 x )
( x2
+ 9)
2
(
)
Critical number: (3, 23 )
Right endpoint: ( 4, 16
25 )
Minimum
Right endpoint: ( 4, 0)
Maximum
+ 9)
2
)
Critical number: − 3, − 23
Left endpoint: (0, − 2)
( x2
Critical numbers: x = ± 3
(
No critical numbers on (0, 4)
36 − 4 x 2
= 0 ⇒ 36 − 4 x 2 = 0 ⇒ x = ± 3
Left endpoint: − 4, − 16
25
x
=
Minimum
Maximum
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
404
6.
Chapter 4
Applications of Differentiation
x
f ( x) =
x +1
2
12. f ( x) = sin 2 x, [− π , π ]
, [0, 2]
Yes. f ( − π ) = f (π ) = 0. f is continuous on [− π , π ]
−3 2
−1 2
⎡ 1
⎤
f ′( x) = x ⎢− ( x 2 + 1) ( 2 x )⎥ + ( x 2 + 1)
⎣ 2
⎦
=
and differentiable on ( − π , π ).
f ′( x) = 2 cos 2 x = 0 for x = ±
1
( x 2 + 1)
32
π 3π
c-values: ± , ±
4
4
No critical numbers
Left endpoint: (0, 0)
13. f ( x ) = x 2 3 , 1 ≤ x ≤ 8
Minimum
(
Right endpoint: 2, 2
5
)
2 −1/3
x
3
f (b ) − f ( a )
4 −1
3
=
=
8−1
7
b − a
2
3
f ′(c ) = c −1 3 =
3
7
f ′( x) =
Maximum
7. g ( x) = 2 x + 5 cos x, [0, 2π ]
g ′( x) = 2 − 5 sin x = 0 when sin x =
2.
5
Critical numbers: x ≈ 0.41, x ≈ 2.73
3
2744
⎛ 14 ⎞
≈ 3.764
c = ⎜ ⎟ =
729
⎝9⎠
Left endpoint: (0, 5)
Critical number: (0.41, 5.41)
8.
Critical number: ( 2.73, 0.88)
Minimum
Right endpoint: ( 2π , 17.57)
Maximum
f ′( x) = 2 cos 2 x = 0 when x =
π 3π 5π 7π
4
,
4
,
4
,
4
.
Left endpoint: (0, 0)
⎛π ⎞
Critical number: ⎜ , 1⎟
⎝4 ⎠
Maximum
⎛ 3π
⎞
Critical number: ⎜ , −1⎟
4
⎝
⎠
Minimum
⎛ 5π ⎞
Critical number: ⎜ , 1⎟
⎝ 4 ⎠
Maximum
⎛ 7π
⎞
Critical number: ⎜ , −1⎟
⎝ 4
⎠
Minimum
15. The Mean Value Theorem cannot be applied. f is not
differentiable at x = 5 in [2, 6].
16. The Mean Value Theorem cannot be applied. f is not
defined for x < 0.
17. f ( x) = x − cos x, −
Right endpoint: ( 2π , 0)
9. No, Rolle’s Theorem cannot be applied.
f (0) = −7 ≠ 25 = f ( 4)
10. Yes. f ( −3) = f ( 2) = 0. f is continuous on [−3, 2],
differentiable on ( −3, 2).
f ′( x) = ( x + 3)(3 x − 1) = 0 for x = 13.
1
3
x2
is not continuous on [−2, 2]. f ( −1)
1 − x2
is not defined.
11. No. f ( x ) =
1
,1 ≤ x ≤ 4
x
1
f ′( x) = − 2
x
f (b ) − f ( a )
(1 4) − 1 = −3 4 = − 1
=
4 −1
3
4
b − a
1
−1
f ′(c ) = 2 = −
4
c
c = 2
14. f ( x ) =
f ( x) = sin 2 x, [0, 2π ]
c-value:
3π
π
,± .
4
4
f ′( x) = 1 + sin x
f (b ) − f ( a )
b − a
=
π
2
≤ x ≤
π
2
(π 2) − (−π 2)
(π 2) − (−π 2)
=1
f ′(c) = 1 + sin c = 1
c = 0
18. f ( x ) = x log 2 x = x ⋅
ln x
, [1, 2]
ln 2
1
[ln x + 1]
ln 2
f (b ) − f ( a )
2−0
=
= 2
b − a
2−1
f ′( x) =
f ′(c ) =
1
[ln c + 1] = 2
ln 2
ln c = 2 ln 2 − 1
c = e 2 ln
2 −1
=
4
≈ 1.4715
e
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 4
19. No; the function is discontinuous at x = 0 which is in
the interval [−2, 1].
21.
20. (a) f ( x) = Ax + Bx + C
Critical number: x = − 32
f ′( x) = 2 Ax + B
A( x22 − x12 ) + B( x2 − x1 )
=
x2 − x1
f ( x) = x 2 + 3x − 12
f ′( x) = 2 x + 3
2
f ( x2 ) − f ( x1 )
405
x2 − x1
= A( x1 + x2 ) + B
Intervals:
−∞ < x < − 32
− 32 < x < ∞
Sign of f ′( x) :
f ′( x) < 0
f ′( x) > 0
Decreasing
Increasing
Conclusion:
f ′(c ) = 2 Ac + B = A( x1 + x2 ) + B
2 Ac = A( x1 + x2 )
22. h( x) = ( x + 2)
13
x + x2
c = 1
2
= Midpoint of [ x1, x2 ]
h′( x) =
f ′( x) = 4 x − 3
21 − 1
= 5
4−0
=
b − a
f ′(c) = 4c − 3 = 5
c = 2 = Midpoint of [0, 4]
23.
1
1
−2 3
( x + 2) =
23
3
3( x + 2)
Critical number: x = −2
(b) f ( x) = 2 x 2 − 3x + 1
f (b ) − f ( a )
+8
Intervals:
(−∞, −2)
(−2, ∞)
Sign of h′( x) :
h′( x) > 0
h′( x) > 0
Conclusion:
Increasing
Increasing
h is increasing on ( −∞, ∞).
f ( x) = ( x − 1) ( x − 3)
2
f ′( x) = ( x − 1) (1) + ( x − 3)( 2)( x − 1)
2
= ( x − 1)(3 x − 7)
Critical numbers: x = 1 and x =
24. g ( x) = ( x + 1)
Intervals:
−∞ < x < 1
1< x <
Sign of f ′( x) :
f ′( x) > 0
f ′( x) < 0
f ′( x) > 0
Increasing
Decreasing
Increasing
Conclusion:
7
3
25. h( x) =
3
g ′( x) = 3( x + 1)
7
3
7
3
< x < ∞
x ( x − 3) = x3 2 − 3 x1 2
Domain: (0, ∞)
2
Critical number: x = −1
h′( x) =
Intervals:
−∞ < x < −1
−1 < x < ∞
Sign of g ′( x) :
g ′( x ) > 0
g ′( x ) > 0
Conclusion:
Increasing
Increasing
26. f ( x ) = sin x + cos x,
3( x − 1)
3 3 2 3 −1 2
3
x − x
= x −1 2 ( x − 1) =
2
2
2
2 x
Critical number: x = 1
Intervals:
0 < x <1
1< x < ∞
Sign of h′( x):
h′( x) < 0
h′( x) > 0
Conclusion:
Decreasing
Increasing
0 ≤ x ≤ 2π
f ′( x) = cos x − sin x
Critical numbers: x =
π 5π
4
,
4
π
π
4
4
5π
< x < 2π
4
0 < x <
Sign of f ′( x) :
f ′( x) > 0
f ′( x) < 0
f ′( x) > 0
Increasing
Decreasing
Increasing
Conclusion:
< x <
5π
4
Intervals:
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
406
27.
Applications of Differentiation
f (t ) = ( 2 − t ) 2t
1
ln 2
Intervals:
−∞ < t < 2 −
f ′(t ) = 0: ( 2 − t )ln2 = 1
Sign of f ′(t ) :
f ′(t ) > 0
f ′(t ) < 0
Increasing
Decreasing
2−t =
1
ln 2
t = 2−
Conclusion:
2−
1
< t < ∞
ln 2
f ′(t ) = ( 2 − t )2t ln 2 − 2t = 2t ⎡⎣( 2 − t ) ln 2 − 1⎤⎦
1
≈ 0.5573, Critical number
ln 2
⎛
1 ⎞
Increasing on: ⎜ − ∞, 2 −
⎟
ln 2 ⎠
⎝
⎛
⎞
1
Decreasing on: ⎜ 2 −
, ∞⎟
ln 2 ⎠
⎝
28. g ( x) = 2 x ln x
29. (a) f ( x ) = x 2 − 6 x + 5
f ′( x) = 2 x − 6 = 0 when x = 3.
⎛1⎞
g ′( x) = 2 x⎜ ⎟ + 2 x ln x = 2 + 2 x ln x = 0
⎝ x⎠
(b)
ln x = −1
Critical number: x =
1
e
Intervals:
−∞ < x < 3
3 < x < ∞
Sign of f ′( x) :
f ′( x) < 0
f ′( x) > 0
Decreasing
Increasing
Conclusion:
⎛1 ⎞
Increasing on: ⎜ , ∞ ⎟
⎝e ⎠
(c) Relative minimum: (3, − 4)
⎛ 1⎞
Decreasing on: ⎜ 0, ⎟
⎝ e⎠
(d)
1
e
−3
1
< x < ∞
e
Intervals:
0 < x <
Sign of g ′( x) :
g ′( x) < 0
g ′( x) > 0
Conclusion:
Decreasing
Increasing
3
9
−5
30. (a) f ( x ) = 4 x 3 − 5 x
f ′( x) = 12 x 2 − 5 = 0 when x = ±
(b)
15
6
5
15
.
= ±
12
6
15
6
15
< x < ∞
6
−∞ < x <
Sign of f ′( x) :
f ′( x) > 0
f ′( x) < 0
f ′( x) > 0
Increasing
Decreasing
Increasing
Conclusion:
−
15
< x <
6
Intervals:
⎛
15 5 15 ⎞
(c) Relative maximum: ⎜⎜ −
,
⎟
6
9 ⎟⎠
⎝
⎛ 15
5 15 ⎞
Relative minimum: ⎜⎜
, −
⎟
6
9 ⎟⎠
⎝
(d)
4
−6
6
−4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 4
31. (a) h(t ) =
1t4
4
407
(c) Relative minimum: ( 2, −12)
− 8t
h′(t ) = t − 8 = 0 when t = 2.
3
(b)
Intervals:
−∞ < t < 2
2 < t < ∞
Sign of h′(t ):
h′(t ) < 0
h′(t ) > 0
Conclusion:
Decreasing
Increasing
(d)
10
−2
6
− 15
32. (a) g ( x) =
g ′( x) =
1 3
( x − 8 x)
4
3 2
8
2 6
x − 2 = 0 ⇒ x2 =
⇒ x = ±
4
3
3
(b)
2 6
3
−
2 6
2 6
< x <
3
3
2 6
< x < ∞
3
Intervals:
−∞ < x < −
Sign of g ′( x) :
g ′( x) > 0
g ′( x) < 0
g ′( x) > 0
Conclusion:
Increasing
Decreasing
Increasing
⎛ 2 6 8 6⎞
(c) Relative maximum: ⎜⎜ −
,
⎟
3
9 ⎟⎠
⎝
⎛2 6
8 6⎞
Relative minimum: ⎜⎜
, −
⎟
3
9 ⎟⎠
⎝
(d)
6
−9
9
−6
33. (a) f ( x ) =
f ′( x) =
x + 4
x2
x 2 (1) − ( x + 4)( 2 x )
4
x
f ′( x) = 0 when x = − 8.
= −
x2 + 8x
x +8
= − 3
x4
x
Discontinuity at: x = 0
(b)
Intervals:
−∞ < x < − 8
−8 < x < 0
0 < x < ∞
Sign of f ′( x) :
f ′( x) < 0
f ′( x) > 0
f ′( x) < 0
Decreasing
Increasing
Decreasing
Conclusion:
(
1
(c) Relative minimum: − 8, − 16
(d)
)
8
− 10
5
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
408
f ( x) =
34. (a)
f ′( x) =
=
=
Applications of Differentiation
x 2 − 3x − 4
x − 2
(x
− 2)( 2 x − 3) − ( x 2 − 3 x − 4)(1)
(x
− 2)
2
2 x 2 − 7 x + 6 − x 2 + 3x + 4
(x
− 2)
2
x 2 − 4 x + 10
(x
− 2)
2
f ′( x) ≠ 0 since x 2 − 4 x + 10 = 0 has no real roots.
Discountinuity at: x = 2
(b)
Intervals:
−∞ < x < 2
2 < x < ∞
Sign of f ′( x) :
f ′( x) > 0
f ′( x) > 0
Increasing
Increasing
Conclusion:
(c) No relative extrema
(d)
6
−8
10
−6
f ( x) = cos x − sin x, (0, 2π )
35. (a)
f ′( x) = − sin x − cos x = 0 ⇒ − cos x = sin x ⇒ tan x = −1
Critical numbers: x =
(b)
3π 7π
,
4 4
3π
7π
< x <
4
4
7π
< x < 2π
4
f ′( x) < 0
f ′( x) > 0
f ′( x) < 0
Decreasing
Increasing
Decreasing
Intervals:
0 < x <
Sign of f ′( x) :
Conclusion:
⎛ 3π
(c) Relative minimum: ⎜ , −
⎝ 4
⎛ 7π
Relative maximum: ⎜ ,
⎝ 4
(d)
3π
4
⎞
2⎟
⎠
⎞
2⎟
⎠
2
0
2p
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 4
g ( x) =
3
⎛π x
⎞
− 1⎟,
sin ⎜
2
⎝ 2
⎠
g ′( x) =
3⎛ π ⎞
2
2
⎛π x
⎞
− 1⎟ = 0 when x = 1 + , 3 + .
⎜ ⎟ cos⎜
π
π
2⎝ 2 ⎠
⎝ 2
⎠
36. (a)
(b)
[0, 4]
0 < x <1+
Intervals:
409
2
1+
π
2
π
2
< x < 3+
3+
π
2
π
< x < 4
Sign of g ′( x) :
g ′( x) > 0
g ′( x) < 0
g ′( x) > 0
Conclusion:
Increasing
Decreasing
Increasing
2 3⎞
⎛
(c) Relative maximum: ⎜1 + , ⎟
π 2⎠
⎝
2 3⎞
⎛
Relative minimum: ⎜ 3 + , − ⎟
π 2⎠
⎝
2
(d)
0
4
−2
37.
f ( x) = x3 − 9 x 2
f ′( x) = 3 x 2 − 18 x
f ′′( x) = 6 x − 18 = 0 when x = 3.
Intervals:
−∞ < x < 3
3 < x < ∞
Sign of f ′′( x) :
f ′′( x) < 0
f ′′( x) > 0
Conclusion:
Concave downward
Concave upward
Point of inflection: (3, −54)
38.
f ( x) = 6 x 4 − x 2
f ′( x) = 24 x3 − 2 x
f ′′( x) = 72 x 2 − 2 = 0 ⇒ x 2 =
1
36
⇒ x = ± 16
Intervals:
−∞ < x < − 16
− 16 < x <
Sign of f ′′( x) :
f ′′( x) > 0
f ′′( x) < 0
Conclusion:
Concave upward
(
1
6
1
6
< x < ∞
f ′′( x) > 0
Concave downward
Concave upward
) ( 16 , − 2165 )
5
,
Points of inflection: − 16 , − 216
39.
g ( x) = x
x + 5, Domain: x ≥ −5
1
3x + 10
−1 2
12
−1 2
⎛1⎞
+ ( x + 5) = ( x + 5) ( x + 2( x + 5)) =
g ′( x) = x⎜ ⎟( x + 5)
2
2 x +5
⎝ 2⎠
g ′′( x) =
2
x + 5 (3) − (3 x + 10)( x + 5)
4( x + 5)
−1 2
=
6( x + 5) − (3 x + 10)
4( x + 5)
32
=
3x + 20
4( x + 5)
32
> 0 on ( −5, ∞).
Concave upward on ( −5, ∞)
No point of inflection
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
410
40.
Applications of Differentiation
f ( x) = 3 x − 5 x3
f ′( x) = 3 − 15 x 2
f ′′( x) = − 30 x = 0 when x = 0.
Intervals:
−∞ < x < 0
0 < x < ∞
Sign of f ′′( x) :
f ′′( x) > 0
f ′′( x) < 0
Conclusion:
Concave upward
Concave downward
Point of inflection: (0, 0)
41.
f ( x) = x + cos x, 0 ≤ x ≤ 2π
f ′( x) = 1 − sin x
f ′′( x) = −cos x = 0 when x =
, .
2 2
π
π
2
2
Intervals:
0 < x <
Sign of f ′′( x):
f ′′( x) < 0
Conclusion:
π 3π
Concave downward
< x <
3π
2
f ′′( x) > 0
Concave upward
3π
< x < 2π
2
f ′′( x) < 0
Concave downward
⎛ π π ⎞ ⎛ 3π 3π ⎞
Points of inflection: ⎜ , ⎟, ⎜ ,
⎟
⎝2 2⎠ ⎝ 2 2 ⎠
42.
x
f ( x) = tan , (0, 2π )
4
1
x
f ′( x) = sec 2
4
4
1 ⎛
x
x ⎞⎛ 1 ⎞
f ′′( x) = ( 2)⎜ sec 2 tan ⎟⎜ ⎟
4 ⎝
4
4 ⎠⎝ 4 ⎠
1
x
x
= sec 2 tan > 0 on (0, 2π ).
8
4
4
45.
g ′( x ) = −4 x( 2 x 2 − 1) = 0 ⇒ x = 0, ±
43.
f ( x) = ( x + 9)
g ′′(0) = 4 > 0
46.
h′′(t ) =
f ′′( x) = 2 > 0 ⇒ ( −9, 0) is a relative minimum.
h′′(3) =
f ( x) = 2 x3 + 11x 2 − 8 x − 12
f ′( x) = 6 x 2 = 22 x − 8 = 2( x + 4)(3x − 1)
1
3
f ′′( x) = 12 x + 22
f ′′( − 4) < 0 ⇒ ( − 4, 68) is a relative maximum.
is a relative minimum.
( 13 ) > 0 ⇒ ( 13 , − 361
27 )
f ′′
h(t ) = t − 4 t + 1, Domain: [−1, ∞]
2
= 0 ⇒ t = 3
t +1
h′(t ) = 1 −
2
Critical numbers: x = − 4,
(0, 0) is a relative minimum.
⎛ 1 ⎞
⎛ 1 1⎞
, ⎟ are relative maxima.
g ′′⎜ ±
⎟ = −8 < 0 ⎜ ±
2⎠
2 2⎠
⎝
⎝
f ′( x) = 2( x + 9) = 0 ⇒ x = −9
44.
1
2
g ′′( x ) = 4 − 24 x 2
Concave upward on (0, 2π )
No point of inflection
g ( x ) = 2 x 2 (1 − x 2 )
47.
1
(t
+ 1)
1
> 0
8
3/ 2
(3, −5) is a relative minimum.
18
x
18
f ′( x) = 2 − 2 = 0 ⇒ 2 x 2 = 18 ⇒ x = ± 3
x
f ( x) = 2 x +
Critical numbers: x = ± 3
36
x3
f ′′( − 3) < 0 ⇒ ( − 3, −12) is a relative maximum.
f ′′( x) =
f ′′(3) > 0 ⇒ (3, 12) is a relative minimum.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 4
411
[0, 4π ]
48. h( x) = x − 2 cos x,
1
2
7π 11π 19π 23π
Critical numbers: x =
,
,
,
6 6
6
6
h′′( x) = 2 cos x
h′( x) = 1 + 2 sin x = 0 ⇒ sin x = −
⎛ 7π ⎞
⎛ 7π 7π
h′′⎜ ⎟ = − 3 < 0 ⇒ ⎜ ,
+
6
⎝ ⎠
⎝ 6 6
⎛ 11π ⎞
⎛ 11π 11π
h′′⎜
,
−
⎟ = 3 > 0 ⇒ ⎜
6
⎝ 6 ⎠
⎝ 6
⎞
3 ⎟ ≈ (3.665, 5.397) is a relative maximum.
⎠
⎞
3 ⎟ ≈ (5.760, 4.028) is a relative minimum.
⎠
⎛ 19π ⎞
⎛ 19π 19π
⎞
h′′⎜
,
+ 3 ⎟ ≈ (9.948, 11.680) is a relative maximum.
⎟ = − 3 < 0 ⇒ ⎜
6
⎝ 6 ⎠
⎝ 6
⎠
⎛ 23π ⎞
⎛ 23π 23π
⎞
− 3 ⎟ ≈ (12.043, 10.311) is a relative minimum.
h′′⎜
,
⎟ = 3 > 0 ⇒ ⎜
6
⎝ 6 ⎠
⎝ 6
⎠
49.
53. (a)
y
D = 0.00188t 4 − 0.1273t 3 + 2.672t 2 − 7.81t + 77.1,
7
6
(5, f(5))
5
0 ≤ t ≤ 40
4
(3, f(3))
3
(b)
2
1
800
(6, 0)
x
−1
(0, 0) 2 3 4 5
7
y
50.
0
40
0
7
6
(c) Maximum occurs at t = 40 (2010).
5
Minimum occurs at t ≈ 1.6 (1970).
4
3
(d) D′(t ) is greatest at t = 40 (2010).
2
1
−1
x
1
2
3
4
5
6
7
51. The first derivative is positive and the second derivative
is negative. The graph is increasing and is concave
down.
52.
⎛Q⎞
⎛ x⎞
C = ⎜ ⎟ s + ⎜ ⎟r
x
⎝ ⎠
⎝ 2⎠
dC
Qs
r
= − 2 +
= 0
dx
x
2
Qs
r
=
x2
2
2Qs
x2 =
r
2Qs
x =
r
⎛ 18,000 ⎞
54. t = 50 log10 ⎜
⎟
⎝ 18,000 − h ⎠
(a) Domain: 0 ≤ h < 18,000
(b)
Vertical asymptote:
h = 18,000
t
100
80
60
40
20
h
4,000
(c)
12,000
t = 50 log10 18,000 − 50 log10 (18,000 − h)
dt
50
=
dh
(ln 10)(18,000 − h)
d 2t
50
=
2
dh 2
ln
10
18,000
− h)
(
)(
No critical numbers
As t increases, the rate of change of the altitude is
increasing.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
412
Applications of Differentiation
1⎞
⎛
55. lim ⎜ 8 + ⎟ = 8 + 0 = 8
x → ∞⎝
x⎠
5x2
x + 2
66. g ( x) =
2
5x2
5
= lim
= 5
x→∞ x + 2
x → ∞1 + 2 x 2
( )
1 − 4x
1x − 4
56. lim
= lim
= −4
x → −∞ x + 1
x → −∞ 1 + 4 x
lim
2
Horizontal asymptote: y = 5
2 x2
2
2
57. lim 2
= lim
=
x → ∞ 3x + 5
x → ∞ 3 + 5 x2
3
10
y=5
3
4x
4 x
= lim
= 0
4
x
→
∞
+3
1+3x
58. lim
−9
x → ∞ x4
9
−2
3x 2
= −∞
+ 5
59. lim
60. lim
x → −∞
61. lim
x→∞
Discontinuity: x = 4
x2 + x
=12
−2 x
lim
5 cos x
= 0, because 5 cos x ≤ 5.
x
62. lim
x→∞
x3
x2 + 2
= lim
x→∞
= lim
x→∞
x→∞
2 + (3 x )
2x + 3
= lim
= 2
x
→
∞
1 − (4 x)
x − 4
Vertical asymptote: x = 4
Horizontal asymptote: y = 2
x3
8
x 1 + 2 x2
x2
1 + 2 x2
y=2
= ∞
−6
12
Limit does not exist.
63. lim
x → −∞ x
−4
6x
= 6
+ cos x
68. f ( x ) =
lim
x
does not exist.
x → −∞ 2 sin x
3x
x2 + 2
3x
x + 2
x→∞
64. lim
65. f ( x ) =
2x + 3
x − 4
67. h( x) =
x → −∞ x
2
= lim
x→∞
= lim
x→∞
3
− 2
x
Discontinuity: x = 0
3x
lim
x2 + 2
x → −∞
⎛3
⎞
lim ⎜ − 2 ⎟ = −2
x → ∞⎝ x
⎠
= lim
x → −∞
= lim
x → −∞
Vertical asymptote: x = 0
Horizontal asymptote: y = −2
3x x
x + 2
2
x2
3
1 + (2 x 2 )
= 3
3x x
(
x2 + 2 −
3
− 1 + (2 x 2 )
x2
)
= −3
Horizontal asymptotes: y = ±3
4
3
−5
y=3
5
−6
y = −3
6
y = −2
−7
−4
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Review Exercises for Chapter 4
413
73. f ( x ) = 4 x − x 2 = x( 4 − x)
5
3 + 2e − x
5
5
lim
=
x → ∞ 3 + 2e − x
3
5
lim
= 0
x → −∞ 3 + 2e − x
69. f ( x ) =
Domain: ( −∞, ∞); Range: (−∞, 4]
f ′( x) = 4 − 2 x = 0 when x = 2.
f ′′( x) = −2
Horizontal asymptotes: y = 0, y =
5
3
3
Therefore, ( 2, 4) is a relative maximum.
Intercepts: (0, 0), ( 4, 0)
y=5
y
3
5
−3
3
y=0
(2, 4)
4
3
−1
2
70. g ( x) = 30 xe − 2 x
lim 30 xe
−2 x
x→∞
1
(0, 0)
(4, 0)
x
= 0
1
Horizontal asymptote: y = 0
8
2
3
5
74. f ( x ) = 4 x 3 − x 4 = x 3 ( 4 − x)
Domain: ( −∞, ∞); Range: ( −∞, 27)
f ′( x) = 12 x 2 − 4 x 3 = 4 x 2 (3 − x) = 0 when
y=0
−2
10
−1
x = 0, 3.
f ′′( x) = 24 x − 12 x 2 = 12 x( 2 − x) = 0 when
71. g ( x) = 3 ln (1 + e − x
lim 3 ln (1 + e − x
x→∞
4
)
4
)
= 0
x = 0, 2.
f ′′(3) < 0
Therefore, (3, 27) is a relative maximum.
Horizontal asymptote: y = 0
9
Points of inflection: (0, 0), ( 2, 16)
Intercepts: (0, 0), ( 4, 0)
y
−9
9
y=0
30
(3, 27)
25
−3
20
(2, 16)
15
⎛ x ⎞
72. h( x) = 10 ln ⎜
⎟
⎝ x + 1⎠
10
(0, 0) 5
(4, 0)
x
Discontinuities: x = 0, x = −1
−2
1
2
3
5
⎛ x ⎞
⎛ x ⎞
lim 10 ln ⎜
10 ln ⎜
⎟ = xlim
⎟ = 0
→
−∞
1
+
x
⎝
⎠
⎝ x + 1⎠
x→∞
Vertical asymptotes: x = 0, x = −1
Horizontal asymptote: y = 0
6
x=0
y=0
−9
9
x = −1
−6
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414
Chapter 4
Applications of Differentiation
77. f ( x ) = x1 3 ( x + 3)
75. f ( x ) = x 16 − x 2
Domain: [−4, 4]; Range: [−8, 8]
16 − 2 x
f ′( x) =
2
Domain: ( −∞, ∞); Range: ( −∞, ∞)
= 0 when x = ± 2 2 and
16 − x
undefined when x = ± 4.
2
2 x( x − 24)
(16 − x )
2
(
(x
f ′′( x) =
32
= 0 when x = −1 and
+ 3) x 2 3
13
−2
x 5 3 ( x + 3)
43
is undefined when x = 0, − 3.
By the First Derivative Test ( −3, 0) is a relative
)
f ′′ −2 2 > 0
(
)
Therefore, −2 2, − 8 is a relative minimum.
(
x +1
f ′( x) =
undefined when x = −3, 0.
2
f ′′( x) =
23
3
)
4 is a relative minimum. (0, 0)
is a point of inflection.
)
f ′′ 2 2 < 0
(
(
maximum and −1, −
Intercepts: ( −3, 0), (0, 0)
)
Therefore, 2 2, 8 is a relative maximum.
y
Point of inflection: (0, 0)
4
3
Intercepts: ( −4, 0), (0, 0), ( 4, 0)
2
1
(− 3, 0)
Symmetry with respect to origin
−5 −4
(0, 0)
x
−2 −1
1
2
y
)2
8
(− 1, − 1.59)
2, 8 )
−3
6
4
2
(−4, 0)
−8 −6
(4, 0)
−2
2
4
6
78. f ( x) = ( x − 3)( x + 2)
x
8
16,875
Domain: ( −∞, ∞); Range: ⎡− 256 , ∞
⎣
(0, 0)
)−2
3
)
f ′( x) = ( x − 3)(3)( x + 2) + ( x + 2)
2
−8
2, −8 )
3
= ( 4 x − 7)( x + 2) = 0 when x = −2, 74 .
2
76. f ( x ) = ( x − 4)
2
2
f ′′( x) = ( 4 x − 7)( 2)( x + 2) + ( x + 2) ( 4)
2
Domain: ( −∞, ∞); Range: [0, ∞)
f ′( x ) = 4 x( x − 4) = 0 when x = 0, ± 2.
2
2 3
f ′′( x) = 4(3 x 2 − 4) = 0 when x = ±
.
3
f ′′(0) < 0
Therefore, (0, 16) are relative maximum.
f ′′( ± 2) > 0
= 6( 2 x − 1)( x + 2) = 0 when x = −2, 12 .
( 74 ) > 0
f ′′
( ,−
7
4
Therefore,
16,875
256
) is a relative minimum.
Points of inflection: ( −2, 0),
( 12 , − 625
16 )
Intercepts: ( −2, 0), (0, − 24,) (3, 0)
y
Therefore, ( ± 2, 0) are relative minima.
(
Points of inflection: ± 2 3 3, 64 9
)
(3, 0)
(−2, 0)
−4
−2
2
x
4
− 20
(0,−24)
Intercepts: ( −2, 0), (0, 16), ( 2, 0)
− 40
Symmetry with respect to y-axis
− 60
( 74 , − 16.875
256 )
y
24
20
(0, 16)
(−2, 0)
(2, 0)
8
4
x
−3 −2 −1
1
2
3
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415
Review Exercises for Chapter 4
79.
5 − 3x
x − 2
1
f ( x) =
f ′( x) =
(x
− 2)
Domain: ( −∞, 0), (0, ∞); Range: (−∞, − 6], [6, ∞ )
> 0 for all x ≠ 2
2
f ′( x) = 3 x 2 + 1 −
−2
f ′′( x) =
(x
− 2)
3
when x = ±1.
Concave downward on ( 2, ∞)
Vertical asymptote: x = 2
f ′′( x) = 6 x +
Horizontal asymptote: y = −3
Therefore, ( −1, − 6) is a relative maximum.
f ′′(1) > 0
x=2
( 53 , 0 (
1
1
2
3
(
−2
4
5
0, − 5
2
(
6
y = −3
Vertical asymptote: x = 0
−4
Symmetric with respect to
origin
−5
−6
82. f ( x ) = x 2 +
2x
1 + x2
80. f ( x ) =
2(1 − x)(1 + x)
f ′( x) =
(1 + x 2 )
−4 x(3 − x 2 )
3
(1 + x 2 )
2
f ′′( x) =
= 0 when x = 0, ±
f ′′( −1) > 0
Therefore, ( −1, −1) is a relative minimum.
)
3 2 , (0, 0),
5
−2
−1
(−1, −6) − 5
(1, 6)
1
2
x=0
1
x3 + 1
=
x
x
f ′′( x) = 2 +
3.
1
2 x3 − 1
=
= 0 when x =
2
x
x2
3
1
.
2
2( x 3 + 1)
2
=
= 0 when x = −1.
x3
x3
⎛ 1 ⎞
f ′′⎜ 3 ⎟ > 0
⎝ 2⎠
Therefore, (1, 1) is a relative maximum.
3, −
f ′( x) = 2 x −
= 0 when x = ±1.
f ′′(1) < 0
Points of inflection:
10
Domain: ( −∞, 0), (0, ∞); Range: ( −∞, ∞)
Domain: ( −∞, ∞); Range: [−1, 1]
(−
y
Therefore, (1, 6) is a relative
minimum.
x
−2 −1
8
6x4 + 8
=
≠ 0
3
x
x3
f ′′( −1) < 0
5⎞
⎛5 ⎞ ⎛
Intercepts: ⎜ , 0 ⎟, ⎜ 0, − ⎟
2⎠
⎝3 ⎠ ⎝
2
4
x2
(3x 2 + 4)( x 2 − 1) = 0
3x 4 + x 2 − 4
=
x2
x2
=
Concave upward on ( −∞, 2)
y
4
x
81. f ( x ) = x 3 + x +
(
3,
3 2
)
3 ⎞
⎛ 1
Therefore, ⎜ 3 , 3 ⎟ is a relative minimum.
4⎠
⎝ 2
Point of inflection: ( −1, 0)
Intercept: ( −1, 0)
Vertical asymptote: x = 0
y
Intercept: (0, 0)
3
Symmetric with respect to the origin
2
Horizontal asymptote: y = 0
1
(−1, 0)
−3 −2
y
( 12 , 34 )
3
3
x
1
2
3
3
2
(1, 1)
1
x
−1
1
2
3
(−1, −1)
−2
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
x
416
Chapter 4
Applications of Differentiation
83. 4 x + 3 y = 400 is the perimeter.
8
⎛ 400 − 4 x ⎞
2
A = 2 xy = 2 x⎜
⎟ = (100 x − x )
3
3
⎝
⎠
8
dA
= (100 − 2 x ) = 0 when x = 50.
3
dx
16
d 2A
= −
< 0 when x = 50.
2
3
dx
y
x
200
ft.
3
A is a maximum when x = 50 ft and y =
84. Ellipse:
85. You have points (0, y ), ( x, 0), and (1, 8). So,
x2
y2
1
+
= 1, y =
144 − x 2
144 16
3
=
2
⎛ 8x ⎞
Let f ( x) = L2 = x 2 + ⎜
⎟ .
⎝ x − 1⎠
⎤
144 − x 2 ⎥
⎦
4 ⎡ 144 − 2 x 2 ⎤
⎢
⎥ = 0 when x =
3 ⎣ 144 − x 2 ⎦
y −8
0 −8
8x
or y =
=
.
x −1
x −1
0 −1
m =
4
⎛2
⎞
A = ( 2 x)⎜
144 − x 2 ⎟ = x 144 − x 2
3
3
⎝
⎠
− x2
dA
4⎡
= ⎢
+
dx
3 ⎣ 144 − x 2
x
⎛ x ⎞ ⎡ ( x − 1) − x ⎤
⎥ =0
f ′( x ) = 2 x + 128⎜
⎟⎢
2
⎝ x − 1 ⎠⎢⎣ ( x − 1) ⎥⎦
72 = 6 2.
x−
The dimensions of the rectangle are 2 x = 12 2 by
2
y =
144 − 72 = 4 2.
3
64 x
( x − 1)
3
=0
x ⎡( x − 1) − 64⎤ = 0 when x = 0, 5 (minimum).
⎣
⎦
3
y
Vertices of triangle: (0, 0), (5, 0), (0, 10)
12
8
( x,
1
3
144 − x 2
(
y
10
x
−12
8
12
−8
6
−12
4
(0, y)
(1, 8)
2
(x, 0)
x
2
4
6
8
10
86. You have points (0, y ), ( x, 0), and (4, 5). So,
m =
y −5
5−0
5x
or y =
=
.
x − 4
0− 4
4− x
⎛ 5x ⎞
Let f ( x ) = L2 = x 2 + ⎜
⎟
⎝ x − 4⎠
2
(0, y)
⎛ x ⎞⎡ x − 4 − x ⎤
f ′( x ) = 2 x + 50⎜
⎥ = 0
⎟⎢
2
⎝ x − 4 ⎠⎢⎣ ( x − 4) ⎥⎦
100 x
x −
= 0
x
( − 4) 3
3
x ⎡( x − 4) − 100⎤ = 0 when x = 0 or x = 4 +
⎣
⎦
L =
x2 +
25 x 2
(x
− 4)
2
=
x
x −4
(x
− 4) + 25 =
2
3
3
L
(4, 5)
5
(x, 0)
4
100.
100 + 4
100 2 3 + 25 ≈ 12.7 ft
3
100
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
417
Review Exercises for Chapter 4
87.
h
0
5
10
15
20
P
10,332
5,583
2,376
1,240
517
ln P
9.243
8.627
7.773
7.123
6.248
89. You can form a right triangle with vertices (0, 0), ( x, 0)
and (0, y ). Assume that the hypotenuse of length L
passes through (4, 6).
(0, y)
12
(a)
(4, 6)
L
−2
22
(0, 0)
0
y = −0.1499h + 9.3018 is the regression line for
m =
data ( h, ln P ).
(b) ln P = ah + b
2
⎛ x ⎞ ⎡ −4 ⎤
⎥ = 0
f ′( x) = 2 x + 72⎜
⎟⎢
2
⎝ x − 4 ⎠⎢⎣ ( x − 4) ⎦⎥
P = Ce ah , C = eb
For our data, a = −0.1499 and
C = e9.3018 = 10,957.7.
(c)
3
x ⎡( x − 4) − 144⎤ = 0 when x = 0 or x = 4 +
⎣
⎦
L ≈ 14.05 ft
−0.1499 h
3
144.
12,000
90.
0
22
0
(d)
y −6
6−0
6x
or y =
=
0− 4
4− x
x − 4
⎛ 6x ⎞
Let f ( x ) = L2 = x 2 + y 2 = x 2 + ⎜
⎟ .
⎝ x − 4⎠
P = e ah + b = eb e ah
P = 10,957.7e
(x, 0)
dP
= (10,957.71)( −0.1499)e −0.1499 h
dh
= −1.642.56e
−0.1499 h
dP
For h = 5,
≈ −776.3. For h = 18,
dh
dP
≈ −110.6.
dh
88. f ( x ) = x n , n is a positive integer.
L1
or L1 = 6 csc θ (see figure)
6
L
sec θ = 2 or L2 = 9 sec θ
9
L = L1 + L2 = 6 csc θ + 9 sec θ
csc θ =
dL
= −6 csc θ cot θ + 9 sec θ tan θ = 0
dθ
tan 3 θ =
sec θ =
csc θ =
(a) f ′( x) = nx n −1
1 + tan 2 θ =
sec θ
=
tan θ
n−2
The function has a point of inflection at (0, 0) when
n is odd and n ≥ 3.
L =
(32 3 + 22 3 )
6
21 3
= 3(32 3 + 22 3 )
L1
θ
L2
θ
9
3
3
2
3
⎛ 2⎞
1+ ⎜ ⎟
⎝ 3⎠
23
=
32 3 + 22 3
31 3
32 3 + 22 3
21 3
12
The function has a relative minimum at (0, 0) when
n is even.
(b) f ′′( x) = n( n − 1) x
2
⇒ tan θ =
3
32
(32 3 + 22 3 )
+9
12
31 3
ft ≈ 21.07 ft
6
(π2 − θ(
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
418
91.
Chapter 4
V =
Applications of Differentiation
(
1 2
1
π x h = π x2 r +
3
3
dV
1 ⎡
= π⎢
dx
3 ⎣
2r 2 + 2r
(
− x3
r − x
2
r 2 − x2
+ 2x r +
2
r2
) (see figure)
⎤
− x )⎥ =
3
⎦
πx
2
r 2 − x2
( 2r
2
+ 2r
)
r 2 − x 2 − 3x 2 = 0
r 2 − x 2 − 3x 2 = 0
2r
r 2 − x 2 = 3 x 2 − 2r 2
4r 2 (r 2 − x 2 ) = 9 x 4 − 12 x 2 r 2 + 4r 4
0 = 9 x 4 − 8 x 2 r 2 = x 2 (9 x 2 − 8r 2 )
x = 0,
2 2r
3
(0, r)
h
r
x
(x, −
r 2 − x2
(
By the First Derivative Test, the volume is a maximum when
2 2r
and h = r +
3
x =
r 2 − x2 =
Thus, the maximum volume is V =
92.
4r
.
3
1 ⎛ 8r 2 ⎞⎛ 4r ⎞
32π r 3
π⎜
cubic units.
⎟⎜ ⎟ =
3 ⎝ 9 ⎠⎝ 3 ⎠
81
)
(
V = π x 2 h = π x 2 2 r 2 − x 2 = 2π x 2
r 2 − x 2 (see figure)
−1 2
⎡ ⎛1⎞
⎤
dV
= 2π ⎢ x 2 ⎜ ⎟( r 2 − x 2 ) ( −2 x) + 2 x r 2 − x 2 ⎥ =
dx
2
⎝
⎠
⎣
⎦
(x,
r2 − x2
(
(x, −
r2 − x2
(
2π x
r2 − x
(2r 2 − 3x 2 ) = 0 when x = 0 and x 2
2
=
2r 2
⇒ x =
3
6r
.
3
x
r
h
By the First Derivative Test, the volume is a maximum when x =
6r
2r
and h =
.
3
3
4π r 3
⎛ 2 ⎞⎛ 2r ⎞
.
Thus, the maximum volume is V = π ⎜ r 2 ⎟⎜
⎟ =
3 3
⎝ 3 ⎠⎝ 3 ⎠
93. y = f ( x) = 0.5 x 2 , f ′( x) = x, x = 3, ∆x = dx = 0.01
∆y = f ( x + ∆x) − f ( x)
= f (3.01) − f (3)
dy = f ′( x)dx
= f ′(3)dx
= 4.53005 − 4.5
= 3(0.01)
= 0.03005
= 0.03
94. y = f ( x) = x3 − 6 x, f ′( x) = 3 x 2 − 6, x = 2,
∆x = dx = 0.1
∆y = f ( x + ∆x) − f ( x)
= f ( 2.1) − f ( 2)
dy = f ′( x)dx
= f ′( 2)dx
= − 3.339 − (− 4)
= 6(0.01)
= 0.661
= 0.06
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 4
95.
y = x(1 − cos x) = x − x cos x
98.
y =
[∆p
−x
dy =
36 − x 2
−x
V =
8
4
) − (75 − 74 ) = − 14
= dp because p is linear.]
99. P = 100 xe − x 400 , x changes from 115 to 120.
36 − x 2
x − x 400 ⎞
⎛
dP = 100⎜ e − x 400 −
e
⎟ dx
400
⎝
⎠
dx
115 ⎞
⎛
= e −115 400 ⎜100 −
⎟(120 − 115)
4 ⎠
⎝
≈ 267.24
97. r = 9 cm, dr = ∆r = ± 0.025
(a)
(
dp = − 14 dx = − 14 (1) = − 14
36 − x 2
−1 2
dy
1
= (36 − x 2 ) ( −2 x ) =
dx
2
1x
4
∆p = p(8) − p(7) = 75 −
dy
= 1 + x sin x − cos x
dx
dy = (1 + x sin x − cos x) dx
96.
p = 75 −
419
4 3
πr
3
Approximate percentage change:
dP
267.24
(100) =
(100) ≈ 3.1%
P
8626.57
dV = 4π r 2 dr
∆V ≈ dV = 4π (9) ( ± 0.025) = ± 8.1π cm 3
2
(b)
S = 4π r 2
dS = 8π r dr
∆S ≈ dS = 8π (9)( ± 0.025) = ±1.8π cm 2
(c) Percent error of volume:
dV
8.1π
=
= 0.0083, or 0.83%
4π 9 3
V
()
3
Percent error of surface area:
dS
1.8π
=
= 0.0056, or 0.56%
2
S
4π (9)
Problem Solving for Chapter 4
1. p( x) = x 4 + ax 2 + 1
(a)
p′( x) = 4 x3 + 2ax = 2 x( 2 x 2 + a )
p′′( x) = 12 x 2 + 2a
For a ≥ 0, there is one relative minimum at (0, 1).
(b) For a < 0, there is a relative maximum at (0, 1).
(c) For a < 0, there are two relative minima at x = ±
a
− .
2
(d) If a < 0, there are three critical points; if a > 0, there is only one critical point.
a=1 a=3 y
a=2
a=0
8
7
6
5
4
3
2
a = −1
a = −2
a = −3
−2
−1
−2
x
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
420
Applications of Differentiation
2. (a) For a = −3, − 2, −1, 0, p has a relative maximum at (0, 0).
For a = 1, 2, 3, p has a relative maximum at (0, 0) and 2 relative minima.
(b)
p′( x) = 4ax3 − 12 x = 4 x( ax 2 − 3) = 0 ⇒ x = 0, ±
3
a
p′′( x) = 12ax 2 − 12 = 12( ax 2 − 1)
For x = 0, p′′(0) = −12 < 0 ⇒ p has a relative maximum at (0, 0).
(c) If a > 0, x = ±
⎛
p′′⎜⎜ ±
⎝
(d)
3
are the remaining critical numbers.
a
3⎞
⎛ 3⎞
⎟⎟ = 12a⎜ ⎟ − 12 = 24 > 0 ⇒ p has relative minima for a > 0.
a⎠
⎝a⎠
(0, 0) lies on
y = −3 x 2 .
Let x = ±
9 18
9
3
⎛ 3⎞
⎛3⎞
. Then p( x) = a⎜ ⎟ − 6⎜ ⎟ =
−
= − .
a
a
a
a
⎝a⎠
⎝a⎠
2
So, y = −
y
a=2
a=1
x
−3
1
a = −1
a = −3
f ( x) =
3⎞
2
⎟ = −3 x is satisfied by all the relative extrema of p.
a ⎟⎠
a=3
2
3.
2
⎛
9
= −3⎜⎜ ±
a
⎝
−8
2
3
a = −2
a=0
c
+ x2
x
c
c
c
+ 2 x = 0 ⇒ 2 = 2 x ⇒ x3 =
⇒ x =
x2
x
2
2c
f ′′( x) = 3 + 2
x
f ′( x) = −
3
c
2
If c = 0, f ( x) = x 2 has a relative minimum, but no relative maximum.
If c > 0, x =
3
⎛ c⎞
c
is a relative minimum, because f ′′⎜ 3 ⎟ > 0.
⎜ 2⎟
2
⎝
⎠
If c < 0, x =
3
c
is a relative minimum, too.
2
Answer: All c.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 4
421
f ( x) = ax 2 + bx + c, a ≠ 0
4. (a)
f ′( x) = 2ax
f ′′( x) = 2a ≠ 0
No point of inflection
f ( x) = ax 3 + bx 2 + cx + d , a ≠ 0
(b)
f ′( x) = 3ax 2 + 2bx + c
f ′′( x) = 6ax + 2b = 0 ⇒ x =
−b
3a
One point of inflection
y⎞
k
⎛
y′ = ky⎜1 − ⎟ = ky − y 2
L⎠
L
⎝
(c)
y′′ = ky′ −
If y =
5. Set
+ + + + + + − − − − −
y″
2k
2 ⎞
⎛
yy′ = ky′⎜1 − y ⎟
L
L ⎠
⎝
y=
L
2
L
, then y′′ = 0, and this is a point of inflection because of the analysis above.
2
f (b) − f ( a) − f ′( a)(b − a )
(b
− a)
2
= k.
Define F ( x) = f ( x) − f ( a ) − f ′( a )( x − a) − k ( x − a) .
2
F ( a ) = 0, F (b) = f (b) − f ( a ) − f ′( a )(b − a ) − k (b − a) = 0
2
F is continuous on [a, b] and differentiable on ( a, b).
There exists c1 , a < c1 < b, satisfying F ′(c1 ) = 0.
F ′( x) = f ′( x) − f ′( a ) − 2k ( x − a) satisfies the hypothesis of Rolle’s Theorem on [a, c1]:
F ′( a) = 0, F ′(c1 ) = 0.
There exists c2 , a < c2 < c1 satisfying F ′′(c2 ) = 0.
Finally, F ′′( x) = f ′′( x) − 2k and F ′′(c2 ) = 0 implies that
k =
f ′′(c2 )
So, k =
2
.
f (b) − f ( a ) − f ′( a )(b − a )
(b
− a)
2
=
f ′′(c2 )
2
⇒ f (b) = f ( a ) + f ′( a )(b − a ) +
1
2
f ′′(c2 )(b − a ) .
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
422
Applications of Differentiation
6.
d
5
13
θ
12
132 + x 2 , sin θ =
d =
x
.
d
Let A be the amount of illumination at one of the corners, as indicated in the figure. Then
A =
(132
A′( x) = kI
kI
kIx
sin θ =
32
2
+ x2 )
(13 + x 2 )
( x2
⇒ ( x 2 + 169)
+ 169)
32
12
⎛ 3⎞
− x⎜ ⎟( x 2 + 169) ( 2 x)
⎝ 2⎠
= 0
3
(169 + x 2 )
(1)
= 3x 2 ( x 2 + 169)
32
12
x 2 + 169 = 3x 2
2 x 2 = 169
13
≈ 9.19 ft
2
x =
By the First Derivative Test, this is a maximum.
7. Distance =
f ′( x) =
x
(4
42 + x 2 +
x
4 + x
2
2
−
(4
− x) + 42 = f ( x)
2
4− x
(4
− x) + 42
2
= 0
− x ) + 4 2 = − ( x − 4) 4 2 + x 2
2
x 2 ⎡⎣16 − 8 x + x 2 + 16⎤⎦ = ( x 2 − 8 x + 16)(16 + x 2 )
32 x 2 − 8 x3 + x 4 = x 4 − 8 x3 + 32 x 2 − 128 x + 256
128 x = 256
x = 2
The bug should head towards the midpoint of the opposite side.
Without Calculus: Imagine opening up the cube:
P
x
Q
The shortest distance is the line PQ, passing through the midpoint.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 4
423
8. Let T be the intersection of PQ and RS. Let MN be the perpendicular to SQ and PR passing through T.
Let TM = x and TN = b − x.
SN
MR
b − x
=
⇒ SN =
MR
b − x
x
x
NQ
PM
b − x
=
⇒ NQ =
PM
b − x
x
x
b − x
b − x
SQ =
d
( MR + PM ) =
x
x
2
1
1⎛ b − x ⎞
1 ⎡
(b − x) ⎥⎤ = 1 d ⎡ 2 x 2 − 2bx + b 2 ⎤
dx + ⎜
d ⎟(b − x) = d ⎢ x +
⎢
⎥
x
x
2
2⎝ x
2 ⎢
2 ⎣
⎠
⎥⎦
⎦
⎣
2
2
1 ⎡ x( 4 x − 2b) − ( 2 x − 2bx + b ) ⎤
⎥
A′( x) = d ⎢
x2
2 ⎢
⎥⎦
⎣
A′( x) = 0 ⇒ 4 x 2 − 2 xb = 2 x 2 − 2bx + b 2
A( x) = Area =
2 x2 = b2
b
x =
2
So, you have SQ =
(
S
Q
N
b−x
)
b − b 2
b − x
d =
d =
x
b 2
(
)
x
P
d
Using the Second Derivative Test, this is a minimum. There is no maximum.
9. f continuous at x = 0: 1 = b
b
T
2 − 1 d.
M
R
10. f continuous at x = −1: a = 2
f continuous at x = 1: a + 1 = 5 + c
f continuous at x = 0: 2 = c
f differentiable at x = 1: a = 2 + 4 = 6. So, c = 2.
f continuous at x = 1: b + 2 = d + 4 ⇒ b = d + 2
⎧1,
⎪
f ( x) = ⎨6 x
⎪x 2
⎩
⎧6 x
= ⎨ 2
⎩x
+ 1,
x = 0
f differentiable at x = 0: 0 = 0
0 < x ≤1
f differentiable at x = 1: 2b = d
+ 4 x + 2, 1 < x ≤ 3
+ 1,
So, b = −2 and d = −4.
0 ≤ x ≤1
+ 4 x + 2, 1 < x ≤ 3
11. Let h( x) = g ( x) − f ( x), which is continuous on [a, b] and
y
differentiable on ( a, b). h( a ) = 0 and h(b) = g (b) − f (b).
g
By the Mean Value Theorem, there exists c in ( a, b) such that
h (b ) − h ( a )
g (b ) − f (b )
=
.
b − a
b − a
Because h′(c) = g ′(c) − f ′(c) > 0 and b − a > 0,
f
h′(c) =
x
a
b
g (b) − f (b) > 0 ⇒ g (b) > f (b).
12. (a) Let M > 0 be given. Take N =
(b) Let ε > 0 be given. Let M =
M , you have f ( x) = x 2 > M .
M . Then whenever x > N =
1
ε
. Then whenever x > M =
1
ε
, you have x 2 >
1
ε
⇒
1
1
< ε ⇒ 2 − 0 < ε.
x2
x
(c) Let ε > 0 be given. There exists N > 0 such that f ( x) − L < ε whenever x > N .
Let δ =
1
1
. Let x = .
y
N
If 0 < y < δ =
⎛1⎞
1
1
1
⇒ x > N and f ( x ) − L = f ⎜ ⎟ − L < ε .
, then <
N
x
N
⎝ y⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
424
13.
y = (1 + x 2 )
15. Assume y1 < d < y2 . Let g ( x) = f ( x) − d ( x − a).
−1
g is continuous on [a, b] and therefore has a minimum
−2 x
y′ =
(1 + x 2 )
2(3 x 2 − 1)
3
( x 2 + 1)
(c, g (c)) on [a, b]. The point c cannot be an endpoint
2
y′′ =
y :
Applications of Differentiation
= 0 ⇒ x = ±
1
3
= ±
3
3
0
3
3
So, a < c < b and g ′(c) = 0 ⇒ f ′(c) = d .
3
3
⎛
3
,
The tangent line has greatest slope at ⎜⎜ −
3
⎝
⎛ 3
least slope at ⎜
⎜ 3 ,
⎝
v
14. (a) s =
3⎞
⎟ and
4 ⎟⎠
3⎞
⎟.
4 ⎟⎠
16. The line has equation
x
y
4
+
= 1 or y = − x + 4.
3
4
3
Rectangle:
4
⎛ 4
⎞
Area = A = xy = x⎜ − x + 4 ⎟ = − x 2 + 4 x.
3
⎝ 3
⎠
8
8
3
A′( x) = − x + 4 = 0 ⇒ x = 4 ⇒ x =
3
3
2
km ⎛
m⎞
⎜1000
⎟
5
h ⎝
km ⎠
=
v
sec
18
⎛
⎞
⎜ 3600 ⎟
h ⎠
⎝
Dimensions:
3
×2
2
Calculus was helpful.
v
20
40
60
80
100
Circle: The distance from the center ( r , r ) to the line
s
5.56
11.11
16.67
22.22
27.78
x
y
+
− 1 = 0 must be r:
3
4
d
5.1
13.7
27.2
44.2
66.4
d ( s ) = 0.071s 2 + 0.389 s + 0.727
(b) The distance between the back of the first vehicle
and the front of the second vehicle is d ( s ), the safe
stopping distance. The first vehicle passes the given
point in 5.5/s seconds, and the second vehicle takes
d ( s ) s more seconds. So,
T =
(c)
g ′( a ) = f ′( a ) − d = y1 − d < 0
g ′(b) = f ′(b) − d = y2 − d > 0.
+++ −−−− −−−− +++
−
of [a, b] because
d ( s)
s
+
5.5
.
s
r =
r
r
+ −1
3
4
=
1
1
+
9 16
7 r − 12
12 7 r − 12
=
5
12
5
5r = 7 r − 12 ⇒ r = 1 or r = 6.
Clearly, r = 1.
Semicircle: The center lies on the line
x
y
+
=1
3
4
and satisfies x = y = r.
10
So
r
r
7
12
+
=1⇒
r =1⇒ r =
.
3
4
12
7
No calculus necessary.
0
30
0
1
(0.071s 2 + 0.389s + 0.727) + 5.5
s
s
The minimum is attained when s ≈ 9.365 m/sec.
T =
(d) T ( s ) = 0.071s + 0.389 +
T ′( s ) = 0.071 −
=
6.227
s
6.227
⇒ s2
s2
6.227
⇒ s ≈ 9.365 m/sec
0.071
T (9.365) ≈ 1.719 seconds
9.365 m/sec ⋅
3600
= 33.7 km/h
1000
(e) d (9.365) = 10.597 m
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 4
425
p( x) = ax 3 + bx 2 + cx + d
17.
p′( x) = 3ax 2 + 2bx + c
p′′( x) = 6ax + 2b
6ax + 2b = 0
x = −
b
3a
The sign of p′′( x) changes at x = −b 3a. Therefore, ( −b 3a, p( −b 3a)) is a point of inflection.
⎛ b3 ⎞
⎛ b2 ⎞
2b3
bc
⎛ b⎞
⎛ b⎞
p⎜ − ⎟ = a⎜ −
+ b⎜ 2 ⎟ + c ⎜ − ⎟ + d =
−
+ d
3⎟
2
a
a
a
a
a
27
9
3
27
3
⎝ 3a ⎠
⎝
⎠
⎝
⎠
⎝
⎠
When p( x) = x 3 − 3 x 2 + 2, a = 1, b = −3, c = 0, and d = 2.
x0 =
y0 =
−( −3)
3(1)
2( −3)
27(1)
=1
3
−
2
(−3)(0)
3(1)
+ 2 = −2 − 0 + 2 = 0
The point of inflection of p( x) = x 3 − 3 x 2 + 2 is ( x0 , y0 ) = (1, 0).
18. (a)
T
R
PQ − TR
x
8.5 − x
PQ
8.5 in.
x
C
P
Q
x 2 + PQ 2 = C 2 ⇒ PQ 2 = C 2 − x 2
TR 2 + (8.5 − x) = x 2 ⇒ TR 2 = 17 x − 8.52
2
( PQ
− TR) + 8.52 = PQ 2 ⇒ 2( PQ)(TR ) = TR 2 + 8.52
2
So, 2( PQ )(TR ) = 17 x − 8.52 + 8.52.
8.5 x = ( PQ)(TR ) =
(8.5 x)2
17 x − 8.52
C 2 − x2
17 x − 8.52
= C 2 − x2
C 2 = x2 +
(8.5 x)
2
17 x − 8.5
=
2
17 x3
17 x − 8.52
2 x3
C2 =
2 x − 8.5
(b) Domain: 4.25 < x < 8.5
(c) To minimize C, minimize f ( x ) = C 2 :
f ′( x) =
x =
(2 x
− 8.5)(6 x 2 ) − 2 x3 ( 2)
(2 x
− 8.5)
2
=
8 x3 − 51x 2
(2 x
− 8.5)
2
= 0
51
= 6.375
8
By the First Derivative Test, x = 6.375 is a minimum.
(d) For x = 6.375, C ≈ 11.0418 in.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4
426
Applications of Differentiation
19. f ( x ) = sin (ln x)
(a) Domain: x > 0
(0, ∞)
or
(b) f ( x ) = 1 = sin (ln x) ⇒ ln x =
π
2
+ 2kπ .
Two values are x = eπ 2 , e(π 2) + 2π .
(c) f ( x ) = −1 = sin (ln x) ⇒ ln x =
3π
+ 2kπ .
2
Two values are x = e −π 2 , e3π 2 .
(d) Because the range of the sine function is [−1, 1], parts (b) and (c) show that the range of f is [−1, 1].
(e) f ′( x) =
1
cos(ln x)
x
f ′( x) = 0 ⇒ cos(ln x) = 0 ⇒ ln x =
f (eπ
)
π
2
⎫
⎪⎪
π
f (1) = 0
⎬ Maximum is 1 at x = e
⎪
f (10) ≈ 0.7440⎪
⎭
(f )
2
+ kπ ⇒ x = eπ
2
on [1, 10].
=1
2
≈ 4.8105.
2
5
0
−2
1
lim f ( x) seems to be − . (This is incorrect.)
+
2
x →0
(g) For the points x = eπ 2 , x = e −3π 2 , x = e −7π 2 , … you have f ( x) = 1.
For the points x = e −π 2 , x = e −5π 2 , x = e −9π 2 , … you have f ( x) = −1.
That is, as x → 0+ , there is an infinite number of points where f ( x) = 1, and an infinite number where f ( x) = −1.
So, lim sin (ln x) does not exist.
x → 0+
You can verify this by graphing f ( x ) on small intervals close to the origin.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R
Integration
5
Section 5.1
Antiderivatives and Indefinite Integration.........................................428
Section 5.2
Area .....................................................................................................436
Section 5.3
Riemann Sums and Definite Integrals...............................................451
Section 5.4
The Fundamental Theorem of Calculus ............................................460
Section 5.5
Integration by Substitution.................................................................473
Section 5.6
Numerical Integration.........................................................................487
Section 5.7
The Natural Logarithmic Function: Integration................................496
Section 5.8
Inverse Trigonometric Functions: Integration...................................506
Section 5.9
Hyperbolic Functions .........................................................................516
Review Exercises ........................................................................................................527
Problem Solving .........................................................................................................535
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R
Integration
5
Section 5.1 Antiderivatives and Indefinite Integration
1.
2.
3.
d⎛2
d
⎞
+ C⎟ =
(2 x −3 + C ) = −6 x −4 = −x64
⎜
dx ⎝ x3
dx
⎠
1
d⎛ 4
d ⎛ 4 1 −1
⎞
⎞
+ C⎟ =
⎜ 2x −
⎜ 2x − x + C ⎟
2x
2
dx ⎝
dx ⎝
⎠
⎠
1
1
= 8 x 3 + x −2 = 8 x 3 +
2
2x2
4.
dy
= 5
dt
y = 5t + C
Check:
5.
dy
= 9t 2
dt
y = 3t 3 + C
dy
= x3 2
dx
2
y = x5 2 + C
5
Check:
d 3
⎡3t + C ⎤⎦ = 9t 2
Check:
dt ⎣
6.
2 x −2
−1
+C = 2 +C
−2
x
Check:
Rewrite
7.
∫
8.
∫ 4x2
9.
∫
3
1
1
x
dx
x
1
10.
∫ (3x)2 dx
11.
∫ ( x + 7) dx
∫ (8 x
3
∫ (x
32
Check:
428
1 x −1
+C
4 −1
−
∫x
x −1 2
+C
−1 2
−
1 ⎛ x −1 ⎞
⎜
⎟+C
9 ⎝ −1 ⎠
−1
+C
9x
−3 2
dx
x2
+ 7x + C
2
14.
1
+C
4x
2
+C
x
⎛
∫ ⎜⎝
x +
d
(2 x 4 − 3x3 + 4 x + C ) = 8 x3 − 9 x 2 + 4
dx
+ 2 x + 1) dx =
2 52
x + x2 + x + C
5
1 ⎞
⎟ dx =
2 x⎠
⎛
∫ ⎜⎝ x
12
+
1 −1 2 ⎞
x ⎟ dx
2
⎠
12
x3 2
1⎛ x ⎞
⎟ + C
+ ⎜
32
2 ⎜⎝ 1 2 ⎟⎠
2
= x3 2 + x1 2 + C
3
=
− 9 x 2 + 4) dx = 2 x 4 − 3x 3 + 4 x + C
Check:
13.
1 −2
x dx
4∫
⎤
d ⎡ x2
⎢ + 7 x + C⎥ = x + 7
dx ⎣ 2
⎦
Check:
12.
3 43
x +C
4
dx
1 −2
x dx
9∫
=
Simplify
43
13
dx
d ⎡ −1
⎤
+ C ⎥ = 2 x −3
dx ⎢⎣ x 2
⎦
x
+C
43
∫x
x dx
Integrate
d ⎡2 5 2
⎤
x + C ⎥ = x3 2
dx ⎢⎣ 5
⎦
dy
= 2 x −3
dx
y =
Given
d
[5t + C ] = 5
dt
Check:
d ⎛2 32
1 −1 2
⎞
12
12
⎜ x + x + C⎟ = x + x
dx ⎝ 3
2
⎠
1
= x +
2 x
d ⎛2 5 2
⎞
2
32
⎜ x + x + x + C ⎟ = x + 2x + 1
dx ⎝ 5
⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.1
15.
∫
x 2 dx =
3
Check:
16.
∫(
4
∫x
x5 3
3
+ C = x5 3 + C
53
5
dx =
23
d ⎛3 5 3
⎞
23
⎜ x + C⎟ = x =
dx ⎝ 5
⎠
)
x + 1 dx =
3
∫ (x
34
3
17.
∫
1
dx =
x5
Check:
18.
∫
4
+ 1) dx = x 7 4 + x + C
7
4
∫ 3x
∫
=
∫ (1 − 3x
−7
dx =
∫ (x
12
3x −1
5x−3
+
+ C
−1
−3
= x +
3
5
− 3 +C
3x
x
=1−
=
21.
∫ ( x + 1)(3x − 2) dx
=
Check:
22.
2
+ x − 2) dx
1 2
x − 2x + C
2
d⎛ 3 1 2
⎞
2
⎜ x + x − 2 x + C ⎟ = 3x + x − 2
dx ⎝
2
⎠
∫ (4t
2
+ 3) dt =
2
=
x3 2
x1 2
+ 6
+ C
32
12
Check:
∫ (16t
4
+ 24t 2 + 9) dt
16t 5
+ 8t 3 + 9t + C
5
⎞
d ⎛ 16t 5
+ 8t 3 + 9t + C ⎟ = 16t 4 + 24t 2 + 9
⎜
dt ⎝ 5
⎠
= ( 4t 2 + 3)
23.
d ⎛2 3 2
⎞
12
⎜ x + 12 x + C ⎟
dx ⎝ 3
⎠
2⎛ 3
⎞
⎛1
⎞
= ⎜ x1 2 ⎟ + 12⎜ x −1 2 ⎟
3⎝ 2
⎠
⎝2
⎠
x + 6
x
∫ (3x
= ( x + 1)(3 x − 2)
+ 6 x −1 2 ) dx
= x1 2 + 6 x −1 2 =
3
5
+ 4
x2
x
x 4 − 3x 2 + 5
x4
= x3 +
=
Check:
+ 5 x − 4 ) dx
= x −
3 x −6
1
+C = − 6 + C
2x
−6
2 32
x + 12 x1 2 + C
3
2
= x1 2 ( x + 18) + C
3
−2
= 1 − 3x − 2 + 5 x − 4
d⎛ 1
d ⎛ 1 −6
⎞
⎞
⎜− 6 + C⎟ =
⎜− x + C ⎟
dx ⎝ 2 x
dx ⎝ 2
⎠
⎠
x + 6
dx =
x
x 4 − 3x2 + 5
dx =
x4
x −4
−1
+C =
+C
dx =
4x4
−4
3
⎛ 1⎞
= ⎜ − ⎟( − 6) x − 7 = 7
x
2
⎝ ⎠
19.
∫
Check:
d ⎡
3
5
d ⎡
5
⎤
⎤
x + − 3 + C⎥ =
x + 3 x −1 − x − 3 + C ⎥
3x
3
dx ⎢⎣
x
dx ⎢⎣
⎦
⎦
x3 + 1
d ⎛ −1
d ⎛ 1 −4
⎞
⎞
⎜ 4 + C⎟ =
⎜− x + C⎟
dx ⎝ 4 x
dx ⎝ 4
⎠
⎠
1
1
= − ( − 4 x −5 ) = 5
x
4
3
dx =
x7
Check:
∫x
−5
20.
x2
d ⎛4 7 4
⎞
34
Check:
⎜ x + x + C⎟ = x + 1 =
dx ⎝ 7
⎠
429
Antiderivatives and Indefinite Integration
∫ (5 cos x + 4 sin x) dx
2
= 5 sin x − 4 cos x + C
Check:
d
(5 sin x − 4 cos x + C ) = 5 cos x + 4 sin x
dx
24.
∫ (θ
2
+ sec 2 θ ) dθ =
Check:
25.
26.
d ⎛1 3
⎞
2
2
⎜ θ + tan θ + C ⎟ = θ + sec θ
dθ ⎝ 3
⎠
x
∫ (2 sin x − 5e ) dx
Check:
1 3
θ + tan θ + C
3
= − 2 cos x − 5e x + C
d
(− 2 cos x − 5e x + C ) = 2 sin x − 5e x
dx
2
∫ sec y( tan y − sec y) dy = ∫ (sec y tan y − sec y) dy
= sec y − tan y + C
Check:
d
(sec y − tan y + C ) = sec y tan y − sec2 y
dy
= sec y ( tan y − sec y )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
430
27.
Chapter 5
∫ ( tan
2
Check:
28.
31.
35. f ′( x ) = 6 x, f (0) = 8
y dy = tan y + C
f ( x) =
f ( x) = 3x 2 + 8
36. f ′( s) = 10 s − 12 s 3 , f (3) = 2
f ( s) =
⎛4
∫ ⎜⎝ x
f ( 2) = 10
f ′( x) =
f ( x) =
38. f ′′( x) = x 2
f ′(0) = 8
f ( 0) = 4
y
f(x) = 4x + 2
dx =
f ( x) =
1 4
x + 8x + 4
12
f ′( 4) = 2
x
+C
2
f ( x) =
f ( 0) = 0
x2
+
2
f ′( x) =
2
∫x
−3 2
dx = −2 x −1 2 + C1 = −
x2
f ( x) = 2
6
f′
4
−2
3
2
39. f ′′( x) = x −3 2
2
8
−4
x
2
∫x
1 3
x +8
3
1 4
⎛1
⎞
f ( x ) = ∫ ⎜ x 3 + 8⎟ dx =
x + 8 x + C2
12
⎝3
⎠
f ( 0) = 0 + 0 + C 2 = 4 ⇒ C 2 = 4
f(x) = 4x
1
1 3
x + C1
3
f ′(0) = 0 + C1 = 8 ⇒ C1 = 8
f ′( x) =
f ′( x) =
5
2
y
= x 2 + x + C2
f ( x) = x 2 + x + 4
3
f ( x) =
∫ (2 x + 1) dx
f ( 2) = 6 + C2 = 10 ⇒ C2 = 4
d
4
Check:
(4 ln x + tan x + C ) = x + sec2 x
dx
34. f ′( x) = x
= 2 x + C1
f ′( x) = 2 x + 1
⎞
+ sec 2 x ⎟ dx = 4 ln x + tan x + C
⎠
33. f ′( x) = 4
∫ 2 dx
f ′( 2) = 4 + C1 = 5 ⇒ C1 = 1
⎞
d ⎛x
5
− 5 ln x + C ⎟ = x −
⎜
dx ⎝ 2
x
⎠
−3 −2 −1
4
f ′( 2) = 5
3x
+C
ln 3
f′
− 3s 4 + C
37. f ′′( x) = 2
⎞
d ⎛
3x
+ C ⎟ = cos x + 3x
⎜ sin x +
dx ⎝
ln 3
⎠
Answers will vary.
2
f ( s) = 5s 2 − 3s 4 + 200
⎞
d ⎛ 2
4x
+ C ⎟ = 2x − 4x
⎜x −
dx ⎝
ln 4
⎠
f ( x) = 4 x + C
3
2
2
32.
∫ (10s − 12s ) ds = 5s
f (3) = 2 = 5(3) − 3(3) + C = 45 − 243 + C ⇒ C = 200
5⎞
x2
⎛
− 5 ln x + C
⎜ x − ⎟ dx =
x⎠
2
⎝
Check:
= 3x2 + C
2
4x
+C
ln 4
= sin x +
∫ 6 x dx
f (0) = 8 = 3(0) + C ⇒ C = 8
= 2 x 2 + cot x + C
= x2 −
x
∫
2
d
(2 x 2 + cot x + C ) = 4 x − csc2 x
dx
∫ (cos x + 3 ) dx
Check:
∫ sec
d
( tan y + C ) = sec2 y = tan 2 y + 1
dy
x
∫ (2 x − 4 ) dx
Check:
30.
y + 1) dy =
2
∫ (4 x − csc x) dx
Check:
29.
Integration
x
2
4
Answers will vary.
2
+ C1
x
2
+ C1 = 2 ⇒ C1 = 3
2
2
f ′( x) = −
+3
x
f ′( 4) = −
f ( x) =
∫ (−2 x
−1 2
+ 3) dx = −4 x1 2 + 3 x + C2
f ( 0) = 0 + 0 + C 2 = 0 ⇒ C 2 = 0
f ( x) = −4 x1 2 + 3x = −4
x + 3x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.1
40. f ′′( x) = sin x
Antiderivatives and Indefinite Integration
(b)
f ′(0) = 1
dy
= x 2 − 1, ( −1, 3)
dx
f ( 0) = 6
f ′( x) =
y =
∫ sin
x dx = −cos x + C1
3 =
f ′(0) = −1 + C1 = 1 ⇒ C1 = 2
f ′( x) = −cos x + 2
f ( x) =
∫ (−cos
C =
x + 2) dx = −sin x + 2 x + C2
y =
f ( 0) = 0 + 0 + C 2 = 6 ⇒ C 2 = 6
f ( x) = −sin x + 2 x + 6
431
x3
− x+C
3
(−1)3
3
− ( −1) + C
5
(−1, 3)
7
3
−4
4
3
7
x
− x+
3
3
−5
44. (a) Answers will vary. Sample answer:
41. f ′′( x) = e x
y
f ′(0) = 2
(1, 3)
f ( 0) = 5
f ′( x) =
∫e
dx = e x + C1
x
x
1
7
f ′(0) = 2 = e 0 + C1 ⇒ C1 = 1
f ′( x) = e x + 1
f ( x) =
∫ (e
x
+ 1) dx = e x + x + C2
(b)
dy
−1
= 2 , x > 0, (1, 3)
dx
x
f ( 0) = 5 = e 0 + 0 + C 2 ⇒ C 2 = 4
y =
f ( x) = e x + x + 4
42. f ′′( x) =
f ′(1) = 4
dx =
− x −1
1
+C = +C
x
−1
1
+C ⇒C = 2
1
1
y = +2
x
2
x2
5
2
2
dx = ∫ 2 x − 2 dx = − + C1
x2
x
f ′(1) = 4 = − 2 + C1 ⇒ C1 = 6
∫
f ′( x) = −
f ( x) =
−2
3=
f (1) = 3
f ′( x) =
1
∫ − x 2 dx = ∫ − x
∫
2
+ 6
x
⎛ 2
⎞
⎜ − + 6 ⎟ dx = − 2 ln x + 6 x + C2
⎝ x
⎠
−1
8
−1
45. (a)
9
−3
f (1) = 3 = 6 + C2 ⇒ C2 = − 3
f ( x) = − 2 ln x + 6 x − 3
43. (a) Answers will vary. Sample answer.
3
−9
dy
(b)
= 2 x, ( −2, − 2)
dx
y
y =
5
∫ 2 x dx
= x2 + C
−2 = ( − 2 ) + C = 4 + C ⇒ C = − 6
2
y = x2 − 6
x
−4
4
12
(c)
−5
− 15
15
−8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
432
Chapter 5
46. (a)
Integration
50. f (0) = − 4. Graph of f ′ is given.
20
(a) f ′( 4) ≈ −1.0
0
(b) No. The slopes of the tangent lines are greater than
2 on [0, 2]. Therefore, f must increase more than
6
0
dy
=2
dx
(b)
4 units on [0, 4].
x , ( 4, 12)
(c) No, f (5) < f ( 4) because f is decreasing on [4, 5].
4
y = ∫ 2 x1 2 dx = x3 2 + C
3
4 32
4
32
4
+C ⇒C =
12 = ( 4) + C = (8) + C =
3
3
3
3
4
4
y = x3 2 +
3
3
(c)
(d) f is a maximum at x = 3.5 because f ′(3.5) ≈ 0
and the First Derivative Test.
(e) f is concave upward when f ′ is increasing on
(−∞, 1) and (5, ∞).
f is concave downward on
(1, 5). Points of inflection at x = 1, 5.
20
51. (a) h(t ) =
∫ (1.5t + 5) dt
= 0.75t 2 + 5t + C
h(0) = 0 + 0 + C = 12 ⇒ C = 12
0
h(t ) = 0.75t 2 + 5t + 12
6
0
(b) h(6) = 0.75(6) + 5(6) + 12 = 69 cm
2
47. They are the same. In both cases you are finding a
function F ( x) such that F ′( x) = f ( x).
48. f ( x ) = tan x ⇒ f ′( x) = 2 tan x ⋅ sec x
2
2
g ( x) = sec x ⇒ g ′( x ) = 2 sec x ⋅ sec x tan x = f ′( x)
2
The derivatives are the same, so f and g differ by a
constant. In fact, tan 2 x + 1 = sec 2 x.
49. Because f ′′ is negative on ( −∞, 0), f ′ is decreasing on
(−∞, 0). Because f ′′ is positive on (0, ∞), f ′ is
increasing on (0, ∞). f ′ has a relative minimum at
(0, 0). Because f ′ is positive on ( −∞, ∞ ), f is
increasing on ( −∞, ∞).
y
f″
x
−2
t , 0 ≤ t ≤ 10
2 32
kt + C
3
P(0) = 0 + C = 500 ⇒ C = 500
P (t ) =
∫ kt
12
dt =
2
k + 500 = 600 ⇒ k = 150
3
2
P(t ) = (150)t 3 2 + 500 = 100t 3 2 + 500
3
P(1) =
P(7) = 100(7)
32
+ 500 ≈ 2352 bacteria
53. a(t ) = −32 ft sec 2
v (t ) =
s (t ) =
2
1
−3
dP
= k
dt
∫ − 32 dt
= −32t + C1
v(0) = 60 = C1
3
f′
52.
1
2
3
∫ (−32t + 60) dt
= −16t 2 + 60t + C2
s ( 0) = 6 = C 2
s(t ) = −16t 2 + 60t + 6, Position function
−2
f
−3
The ball reaches its maximum height when
v(t ) = −32t + 60 = 0
32t = 60
t =
s
15
8
seconds.
(158 ) = −16(158 )
2
(158 ) + 6 = 62.25 feet
+ 60
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.1
54. a(t ) = −32 ft sec2
v(t ) =
∫ − 32 dt
Antiderivatives and Indefinite Integration
57. From Exercise 56, f (t ) = −4.9t 2 + v0t + 2. If
f (t ) = 200 = −4.9t 2 + v0t + 2,
= −32t + C1
v(0) = 0 + C1 = V0 ⇒ C1 = V0
Then v(t ) = −9.8t + v0 = 0
s′(t ) = − 32t + V0
for this t value. So, t = v0 9.8 and you solve
s (t ) =
⎛v ⎞
⎛v ⎞
−4.9⎜ 0 ⎟ + v0 ⎜ 0 ⎟ + 2 = 200
⎝ 9.8 ⎠
⎝ 9.8 ⎠
∫ (− 32t
+ V0 )dt = −16t 2 + V0t + C2
2
s(0) = 0 + 0 + C2 = S0 ⇒ C2 = S0
s′(t ) = −32t + v0 = 0 when t =
⎛v 2 ⎞
+ ⎜ 0 ⎟ = 198
⎝ 9.8 ⎠
−4.9v0 2
s(t ) = −16t 2 + V0t + S0
(9.8)
v0
= time to reach
32
2
−4.9v0 2 + 9.8v0 2 = (9.8) 198
2
4.9v0 2 = (9.8) 198
maximum height.
2
2
⎛v ⎞
⎛v ⎞
⎛v ⎞
s⎜ 0 ⎟ = −16⎜ 0 ⎟ + v0 ⎜ 0 ⎟ = 550
32
32
⎝ ⎠
⎝ ⎠
⎝ 32 ⎠
v2
v2
− 0 + 0 = 550
64
32
v0 2 = 3880.8
v0 ≈ 62.3 m sec.
58. From Exercise 56, f (t ) = −4.9t 2 + 1800. (Using the
v0 = 35,200
2
canyon floor as position 0.)
v0 ≈ 187.617 ft sec
f (t ) = 0 = −4.9t 2 + 1800
4.9t 2 = 1800
55. v0 = 16 ft sec
s0 = 64 ft
t2 =
s(t ) = −16t 2 + 16t + 64 = 0
(a)
−16(t 2 − t − 4) = 0
t =
1±
1+
17
2
v (t ) =
17
s (t ) =
= −0.8t 2 + s0
s0 = 320
So, the height of the cliff is 320 meters.
v(t ) = −1.6t
v( 20) = −32 m sec
60.
56. a(t ) = −9.8
= −9.8t + C1
v(0) = v0 = C1 ⇒ v(t ) = −9.8t + v0
f (t ) =
∫ (−1.6t ) dt
2
= −16 17 ≈ −65.970 ft sec
∫ − 9.8 dt
= −1.6t + v0 = −1.6t , because the
s( 20) = 0 ⇒ −0.8( 20) + s0 = 0
≈ 2.562 seconds.
⎛ 1 + 17 ⎞
⎛ 1 + 17 ⎞
v⎜⎜
⎟⎟ = −32⎜⎜
⎟⎟ + 16
2
2
⎝
⎠
⎝
⎠
v (t ) =
∫ −1.6 dt
stone was dropped, v0 = 0.
2
v(t ) = s′(t ) = −32t + 16
(b)
1800
⇒ t ≈ 9.2 sec
4.9
59. a = −1.6
Choosing the positive value,
t =
433
∫ (−9.8t + v0 ) dt
= −4.9t + v0t + C2
2
f (0) = s0 = C2 ⇒ f (t ) = −4.9t 2 + v0t + s0
So, f (t ) = −4.9t 2 + 10t + 2.
v(t ) = −9.8t + 10 = 0 ( Maximum height when v = 0.)
9.8t = 10
t =
10
9.8
⎛ 10 ⎞
f ⎜ ⎟ ≈ 7.1 m
⎝ 9.8 ⎠
∫ v dv
= −GM ∫
1
dy
y2
GM
1 2
v =
+C
y
2
When y = R, v = v0 .
GM
1 2
+C
v0 =
R
2
GM
1
C = v0 2 −
R
2
GM
1 2
1 2 GM
+ v0 −
v =
y
R
2
2
2GM
2GM
+ v0 2 −
v2 =
y
R
⎛1
1⎞
v 2 = v0 2 + 2GM ⎜ − ⎟
R⎠
⎝y
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Integration
66. v(0) = 45 mi h = 66 ft sec
61. x(t ) = t 3 − 6t 2 + 9t − 2, 0 ≤ t ≤ 5
30 mi h = 44 ft sec
(a) v(t ) = x′(t ) = 3t 2 − 12t + 9
= 3(t 2 − 4t + 3) = 3(t − 1)(t − 3)
a(t ) = v′(t ) = 6t − 12 = 6(t − 2)
(b) v(t ) > 0 when 0 < t < 1 or 3 < t < 5.
(c) a(t ) = 6(t − 2) = 0 when t = 2.
v( 2) = 3(1)( −1) = −3
62. x(t ) = (t − 1)(t − 3)
2
0 ≤ t ≤ 5
15 mi h = 22 ft sec
a (t ) = − a
v(t ) = − at + 66
a
s(t ) = − t 2 + 66t ( Let s(0) = 0.)
2
v(t ) = 0 after car moves 132 ft.
− at + 66 = 0 when t =
= t 3 − 7t 2 + 15t − 9
(a) v(t ) = x′(t ) = 3t 2 − 14t + 15 = (3t − 5)(t − 3)
a(t ) = v′(t ) = 6t − 14
and 3 < t < 5.
5
3
(c) a(t ) = 6t − 14 = 0 when t =
x (t ) =
)
a(t ) = −16.5
( )=
−3 = 2
− 23
v(t ) = −16.5t + 66
− 43
s(t ) = −8.25t 2 + 66t
−1 2
t > 0
= t
t
v(t )dt = 2t1 2 + C
∫
(a) −16.5t + 66 = 44
t =
x(1) = 4 = 2(1) + C ⇒ C = 2
⎛ 22 ⎞
s⎜
⎟ ≈ 73.33 ft
⎝ 16.5 ⎠
Position function: x(t ) = 2t1 2 + 2
1
−1
Acceleration function: a(t ) = v′(t ) = − t −3 2 = 3 2
2
2t
(b) −16.5t + 66 = 22
t =
64. (a) a(t ) = cos t
f (t ) =
∫ v(t ) dt
=
∫ sin t dt
= −cos t + C2
f (0) = 3 = −cos(0) + C2 = −1 + C2 ⇒ C2 = 4
f (t ) = −cos t + 4
(b) v(t ) = 0 = sin t for t = kπ , k = 0, 1, 2, …
65. (a)
v(0) = 25 km h = 25 ⋅
v(13) = 80 km h = 80 ⋅
1000
3600
1000
3600
=
=
250
36
800
36
m sec
v(t ) = at + C
v(13) =
550
36
⇒ v(t ) = at +
= 13a +
t/se
0
132
73.33
feet
117.33
feet
It takes 1.333 seconds to reduce the speed from 45
from 30 mi h to 15 mi h, and 1.333 seconds to
reduce the speed from 15 mi h to 0 mi h. Each
time, less distance is needed to reach the next speed
reduction.
250
36
250
36
= 13a
a =
(b)
250
36
800
36
≈ 2.667
mi h to 30 mi h, 1.333 seconds to reduce the speed
m sec
a(t ) = a (constant acceleration )
v ( 0) =
6f
= 0)
=6
(because v0
/h
= sin t + C1 = sin t
44
16.5
44
(16.5
) ≈ 117.33 ft
c
(c)
mi
=
s
∫ a(t ) dt
∫ cos t dt
45
v (t ) =
22
≈ 1.333
16.5
=4
63. v(t ) =
1
7
3
/h
( ) = (3( ) − 5)(
7
3
7.
3
mi
v
7
3
2
a ⎛ 66 ⎞
⎛ 66 ⎞
⎛ 66 ⎞
s⎜ ⎟ = − ⎜ ⎟ + 66⎜ ⎟
a
2
a
⎝ ⎠
⎝ ⎠
⎝a⎠
33
= 132 when a =
= 16.5.
2
30
(b) v(t ) > 0 when 0 < t <
66
.
a
4f
t/se
c
mi
0 m /h =
i/h 22 f
t/se
c
Chapter 5
15
434
550
468
s (t ) = a
s(13) =
=
275
234
≈ 1.175 m sec 2
t2
250
+
t ( s ( 0) = 0)
2
36
275 (13)
250
+
(13) ≈ 189.58 m
234 2
36
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.1
67. Truck: v(t ) = 30
s(t ) = 30t ( Let s(0) = 0.)
Automobile: a(t ) = 6
Antiderivatives and Indefinite Integration
435
⎧−1, 0 ≤ x < 2
⎪
72. f ′( x) = ⎨2, 2 < x < 3
⎪
⎩0, 3 < x ≤ 4
⎧− x + C1 , 0 ≤ x < 2
⎪
f ( x ) = ⎨2 x + C2 , 2 < x < 3
⎪C ,
3 < x ≤ 4
⎩ 3
v(t ) = 6t ( Let v(0) = 0.)
s(t ) = 3t 2 ( Let s(0) = 0.)
At the point where the automobile overtakes the truck:
f (0) = 1 ⇒ C1 = 1
30t = 3t 2
f continuous at
0 = 3t 2 − 30t
x = 2 ⇒ −2 + 1 = 4 + C2 ⇒ C2 = −5
0 = 3t (t − 10) when t = 10 sec.
f continuous at x = 3 ⇒ 6 − 5 = C3 = 1
(a) s(10) = 3(10) = 300 ft
2
⎧− x + 1, 0 ≤ x < 2
⎪
f ( x ) = ⎨2 x − 5, 2 ≤ x < 3
⎪1,
3 ≤ x ≤ 4
⎩
(b) v(10) = 6(10) = 60 ft sec ≈ 41 mi h
68. a(t ) = k
y
v(t ) = kt
2
k
s(t ) = t 2 because v(0) = s(0) = 0.
2
1
x
At the time of lift-off, kt = 160 and ( k 2)t 2 = 0.7.
1
2
3
4
Because ( k 2)t 2 = 0.7,
t =
⎛
v⎜⎜
⎝
1.4
k
1.4 ⎞
⎟ = k
k ⎟⎠
73.
1.4
= 160
k
1.4k = 1602 ⇒ k =
So, ⎣⎡s( x)⎦⎤ + ⎣⎡c( x)⎦⎤ = k for some constant k.
Because, s(0) = 0 and c(0) = 1, k = 1.
2
1602
1.4
≈ 18,285.714 mi h 2
2
properties.]
74.
d
(ln Cx
dx
75.
d
1
1
(ln x + C ) = x + 0 = x
dx
71. f ′′( x ) = 2 x
f ′( 2) = 0 ⇒ 4 + C = 0 ⇒ C = − 4
2
[Note that s( x) = sin x and c( x) = cos x satisfy these
69. False. f has an infinite number of antiderivatives, each
differing by a constant.
f ′( x) = x 2 + C
2
Therefore, ⎡⎣s( x)⎦⎤ + ⎡⎣c( x)⎤⎦ = 1.
≈ 7.45 ft sec2 .
70. True
2
2
d ⎡
⎡s( x)⎦⎤ + ⎡⎣c( x )⎤⎦ ⎤⎥ = 2 s( x ) s′( x) + 2c( x)c′( x)
⎦
dx ⎣⎢⎣
= 2 s ( x ) c ( x ) − 2c ( x ) s ( x ) = 0
)
=
d
(ln C + ln x
dx
)
= 0+
1
1
=
x
x
x3
− 4 x + C1
3
8
16
f ( 2) = 0 ⇒ − 8 + C1 = 0 ⇒ C1 =
3
3
f ( x) =
f ( x) =
x3
16
− 4x +
3
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
436
Chapter 5
Integration
76. f ( x + y ) = f ( x ) f ( y ) − g ( x) g ( y )
g ( x + y ) = f ( x) g ( y ) + g ( x) f ( y )
f ′(0) = 0
[Note: f ( x) = cos x and g ( x) = sin x satisfy these conditions]
f ′( x + y ) = f ( x) f ′( y ) − g ( x) g ′( y ) (Differentiate with respect to y)
g ′( x + y ) = f ( x) g ′( y ) + g ( x) f ′( y ) (Differentiate with respect to y)
Letting y = 0, f ′( x) = f ( x) f ′(0) − g ( x ) g ′(0) = − g ( x) g ′(0)
g ′( x ) = f ( x) g ′(0) + g ( x) f ′(0) = f ( x) g ′(0)
So, 2 f ( x) f ′( x) = −2 f ( x) g ( x) g ′(0)
2 g ( x) g ′( x ) = 2 g ( x) f ( x) g ′(0).
Adding, 2 f ( x) f ′( x) + 2 g ( x) g ′( x) = 0.
Integrating, f ( x ) + g ( x ) = C.
2
2
Clearly C ≠ 0, for if C = 0, then f ( x ) = − g ( x) ⇒ f ( x) = g ( x) = 0, which contradicts that f, g are nonconstant.
2
Now, C = f ( x + y ) + g ( x + y ) =
2
2
2
( f ( x ) f ( y ) − g ( x ) g ( y ))
2
+ ( f ( x) g ( y ) + g ( x) f ( y ))
2
= f ( x) f ( y ) + g ( x) g ( y ) + f ( x) g ( y ) + g ( x) f ( y )
2
2
2
2
2
2
2
2
2
2
2
2
= ⎡ f ( x) + g ( x) ⎤⎡ f ( y ) + g ( y ) ⎤ = C 2
⎣
⎦⎣
⎦
So, C = 1 and you have f ( x) + g ( x ) = 1.
2
2
Section 5.2 Area
6
1.
∑(3i + 2)
i =1
9
2.
∑ (k 2
k =3
4
3.
∑ k2
k =0
6
4.
i =1
6
∑2
= 3(1 + 2 + 3 + 4 + 5 + 6) + 12 = 75
i =1
+ 1) = (32 + 1) + ( 42 + 1) + … + (92 + 1) = 287
1
1 1
1
1
158
= 1+ + +
+
=
2 5 10 17
85
+1
3
3
3
3
37
=
+ +
=
j
4
5
6
20
∑
j=4
4
5.
6
= 3∑i +
∑c
= c + c + c + c = 4c
k =1
4
6.
∑ ⎡⎣(i − 1)
i =1
11
7.
2
3
+ (i + 1) ⎤ = (0 + 8) + (1 + 27) + ( 4 + 64) + (9 + 125) = 238
⎦
6
1
∑ 5i
9.
i =1
14
8.
⎤
j =1
9
∑1+i
i =1
⎡ ⎛ j⎞
∑ ⎢⎣7⎜⎝ 6 ⎟⎠ + 5⎥⎦
10.
4
⎡
2
⎛ j⎞ ⎤
j =1
⎣⎢
⎦⎥
∑ ⎢1 − ⎜⎝ 4 ⎟⎠ ⎥
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2
11.
3
2 n ⎡⎛ 2i ⎞
⎛ 2i ⎞⎤
⎢⎜ ⎟ − ⎜ ⎟⎥
∑
n i =1 ⎣⎢⎝ n ⎠
⎝ n ⎠⎦⎥
12.
2
3 n ⎡ ⎛
3i ⎞ ⎤
⎢2⎜1 + ⎟ ⎥
∑
n i =1 ⎣⎢ ⎝
n ⎠ ⎥⎦
13.
∑7
12
20
17.
) ∑ i2 − ∑1 =
i =1
i =1
15
= 7(12) = 84
19.
30
∑ (−18) = (−18)(30)
i =1
−1 =
i =1
∑ i(i − 1)
2
24
∑ 4i
i =1
16.
16
i =1
i =1
10
15
⎡10(11)( 21) ⎤
⎢
⎥ − 10 = 375
6
⎣
⎦
15
15
i =1
i =1
∑ i 3 − 2∑ i 2 + ∑ i
15 (16)
15(16)(31) 15(16)
−2
+
4
6
2
= 14,400 − 2480 + 120 = 12,040
24
⎡ 24( 25) ⎤
= 4 ∑i = 4 ⎢
⎥ = 1200
i =1
⎣ 2 ⎦
16
10
i =1
=
= −540
i =1
15.
=
i =1
i =1
14.
10
∑ (i 2
∑ i2
437
⎡19( 20)(39) ⎤
= ⎢
⎥ = 2470
6
⎣
⎦
2
i =1
18.
19
∑ (i − 1)
=
Area
25
20.
∑ (i3
i =1
− 2i ) =
⎡16(17) ⎤
⎥ − 64 = 616
⎣ 2 ⎦
∑ (5i − 4) = 5∑ i − 4(16) = 5⎢
=
2
25
∑ i3
i =1
2
25
− 2∑ i
i =1
(25)2 (26)2
− 2
4
= 105,625 − 650
25( 26)
2
= 104,975
⎤
2i + 1
1 n
1 ⎡ n( n + 1)
n+ 2
2
= 2 ∑ ( 2i + 1) = 2 ⎢2
+ n⎥ =
=1+
= S ( n)
2
n
n i =1
n ⎣
2
n
n
i =1
⎦
12
S (10) =
= 1.2
10
S (100) = 1.02
n
21.
∑
S (1000) = 1.002
S (10,000) = 1.0002
7j + 4
1 n
=
∑ ( 7 j + 4)
n2
n 2 j =1
j =1
n
22.
∑
=
⎤
1 ⎡ n( n + 1)
+ 4 n⎥
⎢7
n2 ⎣
2
⎦
=
7n 2 + 7n
4n
7 n + 15
+ 2 =
= S ( n)
2
n
2n
2n
S (10) =
17
4
= 4.25
S (100) = 3.575
S (1000) = 3.5075
S (10,000) = 3.50075
6k ( k − 1)
6 n
6 ⎡ n( n + 1)( 2n + 1) n( n + 1) ⎤
= 3 ∑ (k 2 − k ) = 3 ⎢
−
⎥
3
6
2
n
n
n
k =1
k =1
⎣
⎦
n
23.
∑
=
6 ⎡ 2n 2 + 3n + 1 − 3n − 3 ⎤
1
2
2
⎢
⎥ = 2 ⎡⎣2n − 2⎤⎦ = 2 − 2 = S ( n)
6
n2 ⎣
n
n
⎦
S (10) = 1.98
S (100) = 1.9998
S (1000) = 1.999998
S (10,000) = 1.99999998
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
438
Chapter 5
Integration
2i 3 − 3i
1 n
= 4 ∑ ( 2i 3 − 3i)
4
n
n i =1
i =1
n
24.
∑
=
=
2
n( n + 1) ⎤
1 ⎡ n 2 ( n + 1)
⎢
⎥
2
−3
4
4
2
n ⎢⎣
⎥⎦
(n
+ 1)
2
−
2
2n
S (10) = 0.5885
3( n + 1)
2n
3
=
n3 + 2n 2 − 2n − 3
= S ( n)
2n3
S (100) = 0.5098985
S (1000) = 0.5009989985
S (10,000) = 0.50009999
25.
y
y
10
10
8
8
6
6
4
4
2
2
x
1
∆x =
x
2
1
2
2−0
1
=
4
2
Left endpoints: Area ≈
1
2
Right endpoints: Area ≈
[5 + 6 + 7 + 8]
1
2
=
[6 + 7 + 8 + 9]
26
2
=
= 13
= 15
30
2
13 < Area < 15
26.
y
y
10
10
8
8
6
6
4
4
2
2
x
1
∆x =
2
3
x
4
1
2
3
4
4−2
1
=
6
3
Left endpoints: Area ≈
1⎡
20 19
17 16 ⎤
37
7+
+
+6+
+ ⎥ =
≈ 12.333
3 ⎢⎣
3
3
3
3⎦
3
Right endpoints: Area ≈
1 ⎡ 20 19
17 16 15 ⎤
35
+
+6+
+
+ ⎥ =
≈ 11.667
3 ⎢⎣ 3
3
3
3
3⎦
3
35
37
< Area <
3
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2
27.
y
439
y
50
50
40
40
30
30
20
20
10
10
x
1
∆x =
Area
2
3
4
x
5
1
2
3
4
5
5−2
1
=
6
2
Left endpoints: Area ≈
1
2
Right endpoints: Area ≈
[5 + 9 + 14 +
1
2
[9 + 14 +
20 + 27 + 35] = 55
20 + 27 + 35 + 44] =
149
2
= 74.5
55 < Area < 74.5
28.
y
y
10
10
8
8
6
6
4
4
2
2
x
1
∆x =
2
x
3
1
2
3
3−1
1
=
8
4
Left endpoints: Area ≈
1 ⎡2
4⎣
Right endpoint: Area ≈
1 ⎡ 41
4 ⎣16
+
41
16
+
+
13
4
13
4
+
+
65
16
65
16
+5+
+5+
97
16
+
97
16
+
29
4
29
4
+
+
137
16
137 ⎤
16 ⎦
=
155
16
= 9.6875
+ 10⎤⎦ = 11.6875
9.6875 < Area < 11.6875
29.
y
y
1
1
π
4
π
∆x = 2
−0
4
x
π
2
=
π
4
π
2
x
π
8
Left endpoints: Area ≈
π⎡
Right endpoints: Area ≈
⎛π ⎞
⎛π ⎞
⎛ 3π ⎞⎤
cos(0) + cos⎜ ⎟ + cos⎜ ⎟ + cos⎜ ⎟⎥ ≈ 1.1835
8 ⎢⎣
8
4
⎝ ⎠
⎝ ⎠
⎝ 8 ⎠⎦
π⎡
⎛π ⎞
⎛π ⎞
⎛ 3π ⎞
⎛ π ⎞⎤
cos⎜ ⎟ + cos⎜ ⎟ + cos⎜ ⎟ + cos⎜ ⎟⎥ ≈ 0.7908
8 ⎢⎣ ⎝ 8 ⎠
⎝4⎠
⎝ 8 ⎠
⎝ 2 ⎠⎦
0.7908 < Area < 1.1835
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
440
Chapter 5
Integration
y
30.
y
1
1
x
π
2
π −0
∆x =
6
=
x
π
2
π
6
Left endpoints: Area ≈
Right endpoints: Area ≈
π⎡
π
π
π
2π
5π ⎤
sin 0 + sin
+ sin
+ sin
+ sin
+ sin
≈ 1.9541
6 ⎢⎣
6
3
2
3
6 ⎥⎦
π⎡
π
π
π
2π
5π
⎤
sin
+ sin
+ sin
+ sin
+ sin
+ sin π ⎥ ≈ 1.9541
6 ⎢⎣
6
3
2
3
6
⎦
By symmetry, the answers are the same. The exact area (2) is larger.
31. S = ⎡⎣3 + 4 +
9
2
+ 5⎤⎦ (1) =
s = ⎡⎣1 + 3 + 4 + 92 ⎤⎦ (1) =
= 16.5
33
2
25
2
= 12.5
32. S = [5 + 5 + 4 + 2](1) = 16
s = [4 + 4 + 2 + 0](1) = 10
33. S ( 4) =
1⎛1⎞
⎜ ⎟+
4⎝ 4⎠
⎛1⎞
s( 4) = 0⎜ ⎟ +
⎝ 4⎠
1⎛1⎞
⎜ ⎟+
2⎝ 4⎠
1⎛1⎞
⎜ ⎟+
4⎝ 4⎠
3⎛1⎞
⎜ ⎟+
4⎝ 4⎠
1⎛1⎞
⎜ ⎟+
2⎝ 4⎠
1+
⎛1⎞
1⎜ ⎟ =
⎝ 4⎠
3⎛1⎞
1+
⎜ ⎟ =
4⎝ 4⎠
2 +
8
2 +
8
3 + 2
3
≈ 0.768
≈ 0.518
34. S ( 4) = 4(e − 0 + e − 0.5 + e −1 + e −1.5 ) 12 ≈ 4.395
s( 4) = 4(e − 0.5 + e −1 + e −1.5 + e − 2 ) 12 ≈ 2.666
1 ⎛1⎞
1 ⎛1⎞
1 ⎛1⎞
1 ⎛1⎞
1 1 1 1 1
⎛1⎞
≈ 0.746
35. S (5) = 1⎜ ⎟ +
⎜ ⎟+
⎜ ⎟+
⎜ ⎟+
⎜ ⎟ = + + + +
5 6 7 8 9
⎝ 5 ⎠ 6 5⎝ 5 ⎠ 7 5⎝ 5 ⎠ 8 5⎝ 5 ⎠ 9 5⎝ 5 ⎠
s(5) =
1 ⎛1⎞
1 ⎛1⎞
1 ⎛1⎞
1 ⎛ 1 ⎞ 1⎛ 1 ⎞
1 1 1 1
1
+ + + +
≈ 0.646
⎜ ⎟+
⎜ ⎟+
⎜ ⎟+
⎜ ⎟+ ⎜ ⎟ =
6 5⎝ 5 ⎠ 7 5⎝ 5 ⎠ 8 5⎝ 5 ⎠ 9 5⎝ 5 ⎠ 2 ⎝ 5 ⎠
6 7 8 9 10
⎛1⎞
36. S (5) = 1⎜ ⎟ +
⎝5⎠
=
s(5) =
n
1⎡
⎢1 +
5 ⎢⎣
i =1
24
+
5
2
21
+
5
⎛1⎞ ⎛1⎞
1−⎜ ⎟ ⎜ ⎟ +
⎝ 5⎠ ⎝ 5⎠
⎛ 24i ⎞
∑ ⎜⎝ n2 ⎟⎠
n →∞
37. lim
2
⎛1⎞ ⎛1⎞
1−⎜ ⎟ ⎜ ⎟ +
⎝5⎠ ⎝ 5⎠
= lim
n →∞
2
⎛ 2⎞ ⎛1⎞
1−⎜ ⎟ ⎜ ⎟ +
⎝ 5⎠ ⎝5⎠
16
+
5
2
2
2
⎛ 3⎞ ⎛ 1⎞
1−⎜ ⎟ ⎜ ⎟ +
⎝ 5⎠ ⎝ 5⎠
⎛ 4⎞ ⎛1⎞
1−⎜ ⎟ ⎜ ⎟
⎝ 5⎠ ⎝ 5⎠
2
2
9⎤
⎥ ≈ 0.859
5 ⎥⎦
⎛ 2⎞ ⎛1⎞
1−⎜ ⎟ ⎜ ⎟ +
⎝ 5⎠ ⎝ 5⎠
⎛ 3⎞ ⎛ 1⎞
1−⎜ ⎟ ⎜ ⎟ +
⎝ 5⎠ ⎝ 5⎠
⎛ 4⎞ ⎛1⎞
1 − ⎜ ⎟ ⎜ ⎟ + 0 ≈ 0.659
⎝ 5⎠ ⎝ 5⎠
⎡ ⎛ n 2 + n ⎞⎤
24 n
24 ⎛ n( n + 1) ⎞
1⎞
⎛
i = lim 2 ⎜
= lim ⎢12⎜
⎟
⎟⎥ = 12 nlim
⎜1 + ⎟ = 12
2 ∑
2
→∞ ⎝
n →∞ n
n →∞
n i =1
2
n
n⎠
⎠⎦
⎝
⎠
⎣ ⎝
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2
⎛ 3i ⎞⎛ 3 ⎞
n
∑ ⎜⎝ n ⎟⎜
⎟
n→∞
⎠⎝ n ⎠
38. lim
= lim
n→∞
i =1
n
39. lim ∑
n →∞
i =1
Area
441
9 n
9 ⎡ n( n + 1) ⎤
9 ⎛ n + 1⎞
9
⎢
⎥ = nlim
∑ i = nlim
⎜
⎟ =
→ ∞ n2
→∞ 2⎝
n 2 i =1
2
n
2
⎠
⎣
⎦
1 n −1
1 ⎡ ( n − 1)( n)( 2n − 1) ⎤
1
2
i − 1) = lim 3 ∑ i 2 = lim 3 ⎢
⎥
3(
n
→∞
n
→∞
n i =1
n ⎣
6
n
⎦
= lim
n →∞
⎡ 1 ⎛ 2 − (3 n) + (1 n 2 ) ⎞⎤
1 ⎡ 2n3 − 3n 2 + n ⎤
1
⎢ ⎜
⎟⎥ =
⎢
⎥ = nlim
3
→∞ ⎢ 6 ⎜
⎟
6⎣
n
1
3
⎥
⎦
⎠⎦
⎣ ⎝
2
n
2i ⎞ ⎛ 2 ⎞
2 n
2
⎛
40. lim ∑ ⎜1 + ⎟ ⎜ ⎟ = lim 3 ∑ ( n + 2i )
n →∞
n
→∞
n ⎠ ⎝n⎠
n i =1
i =1 ⎝
n
n
⎤
2⎡n 2
n + 4n ∑ i + 4 ∑ i 2 ⎥
∑
⎢
3
n →∞ n
i =1
i =1 ⎦
⎣ i =1
= lim
= lim
n →∞
⎛ n( n + 1) ⎞ 4( n)( n + 1)( 2n + 1) ⎤
2⎡ 3
n + ( 4n)⎜
⎥
⎟+
3⎢
n ⎣⎢
2
6
⎝
⎠
⎦⎥
2
4 2
2 ⎤
4⎞
26
⎡
⎛
= 2 lim ⎢1 + 2 + + + + 2 ⎥ = 2⎜1 + 2 + ⎟ =
n →∞ ⎣
n
3 n 3n ⎦
3⎠
3
⎝
n
⎡
i ⎞⎛ 2 ⎞
1⎡ n
1 n ⎤
1⎡
1 ⎛ n( n + 1) ⎞⎤
n2 + n ⎤
1⎞
⎛
⎛
41. lim ∑ ⎜1 + ⎟⎜ ⎟ = 2 lim ⎢∑ 1 + ∑ i⎥ = 2 lim ⎢n + ⎜
= 2 lim ⎢1 +
= 2⎜1 + ⎟ = 3
⎥
⎟
2 ⎥
n →∞
n
n
n
→∞
→∞
→∞
n ⎠⎝ n ⎠
n ⎣ i =1
n i =1 ⎦
n ⎢⎣
n⎝
2
2n ⎦
2⎠
⎝
i =1 ⎝
⎣
⎠⎥⎦
3
n
3i ⎞ ⎛ 3 ⎞
3 n ⎡ 2n + 3i ⎤
⎛
42. lim ∑ ⎜ 2 + ⎟ ⎜ ⎟ = lim ∑ ⎢
n→∞
n→∞ n
n ⎠ ⎝n⎠
n ⎦⎥
i =1 ⎝
i =1 ⎣
= lim
n→∞
= lim
n→∞
3
n4
n
∑ (8n3
i =1
3
+ 36n 2i + 54ni 2 + 27i 3 )
2
n( n + 1)( 2n + 1)
n 2 ( n + 1) ⎞
3 ⎛ 4
2 n( n + 1)
⎜
⎟
n
n
n
8
36
54
27
+
+
+
⎟
n4 ⎜
2
6
4
⎝
⎠
⎛
(n + 1) + 9(n + 1)(2n + 1) + 27 ⋅ (n + 1)2 ⎞⎟
= lim 3 ⎜ 8 + 18
n→∞ ⎜
n
n2
4
n2 ⎟
⎝
⎠
27 ⎞
609
⎛
= 152.25
= 3 ⎜ 8 + 18 + 18 +
⎟ =
4⎠
4
⎝
43. (a)
y
3
2
1
x
1
(b) ∆x =
3
2−0
2
=
n
n
⎛ 2⎞
⎛ 2⎞
⎛ 2⎞
⎛ 2⎞
Endpoints: 0 < 1⎜ ⎟ < 2⎜ ⎟ < … < ( n − 1)⎜ ⎟ < n⎜ ⎟ = 2
⎝n⎠
⎝n⎠
⎝n⎠
⎝ n⎠
(c) Because y = x is increasing, f ( mi ) = f ( xi −1 ) on [ xi −1 , xi ].
s( n) =
n
∑ f ( xi −1 )∆x
i =1
=
n
⎛ 2i − 2 ⎞⎛ 2 ⎞
⎟⎜ ⎟ =
n ⎠⎝ n ⎠
∑ f ⎜⎝
i =1
n
⎡
⎛ 2 ⎞⎤⎛ 2 ⎞
∑ ⎢⎣(i − 1)⎜⎝ n ⎟⎠⎥⎦⎜⎝ n ⎟⎠
i =1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
442
Chapter 5
Integration
(d) f ( M i ) = f ( xi ) on [ xi −1 , xi ]
n
∑ f ( xi ) ∆x
S ( n) =
i =1
(e)
⎛ 2i ⎞ 2
n
∑ f ⎜⎝ n ⎟⎠ n
=
=
i =1
i =1
x
5
10
50
100
s( n)
1.6
1.8
1.96
1.98
S ( n)
2.4
2.2
2.04
2.02
⎡
n
(f ) lim
n →∞
⎛ 2 ⎞⎤ ⎛ 2 ⎞
∑ ⎢⎣(i − 1)⎜⎝ n ⎟⎠⎥⎦⎜⎝ n ⎟⎠
= lim
n
4
∑ (i − 1)
n →∞ n 2
i =1
⎡ ⎛ 2 ⎞⎤⎛ 2 ⎞
n
∑ ⎢⎣i⎜⎝ n ⎟⎠⎥⎦⎜⎝ n ⎟⎠
= lim
⎤
⎡ 2( n + 1) 4 ⎤
4 ⎡ n( n + 1)
− n⎥ = lim ⎢
− ⎥ = 2
⎢
n
→∞
2
n⎦
⎣
⎦
⎣ n
n →∞ n 2
i =1
n
2( n + 1)
⎡ ⎛ 2 ⎞⎤⎛ 2 ⎞
4 n
⎛ 4 ⎞ n( n + 1)
lim ∑ ⎢i⎜ ⎟⎥⎜ ⎟ = lim 2 ∑ i = lim ⎜ 2 ⎟
= lim
= 2
n
n
n
→∞
→∞
→∞
n i =1
2
n
⎝n ⎠
i =1 ⎣ ⎝ n ⎠⎦ ⎝ n ⎠
n →∞
44. (a)
y
4
3
2
1
x
2
4
3−1
2
=
n
n
(b) ∆ x =
Endpoints:
2
4
<1+
<
n
n
⎛ 2⎞
< 1 + 2⎜ ⎟ <
⎝n⎠
1 < 1+
⎛2⎞
1 < 1 + 1⎜ ⎟
⎝n⎠
< 1+
2n
= 3
n
⎛ 2⎞
⎛ 2⎞
< 1 + ( n − 1)⎜ ⎟ < 1 + n⎜ ⎟
n
⎝ ⎠
⎝n⎠
(c) Because y = x is increasing, f ( mi ) = f ( xi −1 ) on [ xi −1 , xi ].
s( n) =
n
∑ f ( xi −1 ) ∆x
n
∑
=
i =1
i =1
⎡
⎛ 2 ⎞⎤ ⎛ 2 ⎞
f ⎢1 + (i − 1)⎜ ⎟⎥ ⎜ ⎟ =
⎝ n ⎠⎦ ⎝ n ⎠
⎣
n
⎡
⎛ 2 ⎞⎤⎛ 2 ⎞
∑ ⎢⎣1 + (i − 1)⎜⎝ n ⎟⎠⎥⎦⎜⎝ n ⎟⎠
i =1
(d) f ( M i ) = f ( xi ) on [ xi −1 , xi ]
n
∑ f ( xi ) ∆x
S ( n) =
=
i =1
(e)
n
⎡
⎛ 2 ⎞⎤⎛ 2 ⎞
∑ f ⎢⎣1 + i⎜⎝ n ⎟⎠⎥⎦⎜⎝ n ⎟⎠
i =1
x
5
10
50
100
s( n)
3.6
3.8
3.96
3.98
S ( n)
4.4
4.2
4.04
4.02
n
⎡
⎛ 2 ⎞⎤ ⎛ 2 ⎞
∑ ⎢1 + (i − 1)⎜⎝ n ⎟⎠⎥⎜⎝ n ⎟⎠
n→∞
(f ) lim
i =1 ⎣
⎦
=
n
⎡
⎛ 2 ⎞⎤⎛ 2 ⎞
∑ ⎢⎣1 + i⎜⎝ n ⎟⎠⎥⎦⎜⎝ n ⎟⎠
i =1
⎞⎤
2 ⎛ n( n + 1)
2n + 2
4⎤
2⎤
⎛ 2 ⎞⎡
⎡
⎡
= lim ⎜ ⎟ ⎢n + ⎜
− n ⎟⎥ = lim ⎢2 +
− ⎥ = lim ⎢4 − ⎥ = 4
n → ∞⎝ n ⎠ ⎢
n
n
→
∞
→
∞
n
2
n
n
n
⎣
⎦
⎣
⎦
⎝
⎠⎥⎦
⎣
⎡
2( n + 1) ⎤
⎡
2⎡
2⎤
⎛ 2 ⎞⎤ ⎛ 2 ⎞
⎛ 2 ⎞ n( n + 1) ⎤
⎡
lim ∑ ⎢1 + i⎜ ⎟⎥ ⎜ ⎟ = lim ⎢n + ⎜ ⎟
4+ ⎥ = 4
⎥ = nlim
⎢2 +
⎥ = nlim
⎢
∞
n
→
∞
→
∞
→
n⎣
2
n
n⎦
⎝ n ⎠⎦ ⎝ n ⎠
⎝n⎠
⎣
i =1 ⎣
⎦
⎣
⎦
n
n→∞
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2
1⎞
⎛
45. y = −4 x + 5 on [0, 1]. ⎜ Note: ∆x = ⎟
n⎠
⎝
⎛ i ⎞⎛ 1 ⎞
f ⎜ ⎟⎜ ⎟ =
⎝ n ⎠⎝ n ⎠
n
∑
s( n) =
i =1
n
⎡
⎛i⎞
⎤⎛ 1 ⎞
∑ ⎢⎣−4⎜⎝ n ⎟⎠ + 5⎥⎦⎜⎝ n ⎟⎠
⎛ i ⎞⎛ 1 ⎞
n
∑ f ⎜⎝ n ⎟⎜
⎟
⎠⎝ n ⎠
S ( n) =
i =1
4 n
= − 2∑ i + 5
n i =1
⎡⎛ i ⎞ 2
⎤⎛ 1 ⎞
∑ ⎢⎜⎝ n ⎟⎠ + 2⎥⎜⎝ n ⎟⎠
⎥⎦
i =1 ⎢
⎣
n
=
4 n( n + 1)
+5
2
n2
1⎞
⎛
= −2⎜1 + ⎟ + 5
n⎠
⎝
⎡1 n ⎤
= ⎢ 3 ∑ i2 ⎥ + 2
⎣ n i =1 ⎦
= −
n→∞
n( n + 1)( 2n + 1)
=
6n 3
Area = lim S ( n) =
n →∞
y
443
1⎞
⎛
47. y = x 2 + 2 on [0, 1]. ⎜ Note: ∆x = ⎟
n⎠
⎝
i =1
Area = lim s( n) = 3
Area
5
+ 2 =
1⎛
3
1⎞
⎜2 + + 2 ⎟ + 2
6⎝
n
n ⎠
7
3
y
4
3
3
2
1
−2
1
x
−1
1
2
3
x
1
5−2
3⎞
⎛
46. y = 3 x − 2 on [2, 5]. ⎜ Note: ∆x =
= ⎟
n
n⎠
⎝
3i ⎞⎛ 3 ⎞
⎛
f ⎜ 2 + ⎟⎜ ⎟
n ⎠⎝ n ⎠
⎝
n
∑
S ( n) =
i =1
⎡ ⎛
n
⎤⎛ 3 ⎞
3i ⎞
⎟ − 2⎥ ⎜ ⎟
n⎠
⎦⎝ n ⎠
∑ ⎢⎣3⎜⎝ 2 +
=
i =1
⎛ 3⎞
= 18 + 3⎜ ⎟
⎝n⎠
n
=
n
⎡ ⎛ 2i ⎞ 2
∑ ⎢3⎜⎝ n ⎟⎠
i =1
⎣⎢
⎤⎛ 2 ⎞
+ 1⎥⎜ ⎟
⎦⎥⎝ n ⎠
n
24
2
∑i 2 + n ∑ 1
n3 i =1
i =1
=
24 ⎛ n( n + 1)( 2n + 1) ⎞ 2
⎜
⎟ + ( n)
n3 ⎝
6
⎠ n
4( n + 1)( 2n + 1)
=
+ 2
n2
i =1
27
51
=
2
2
⎛ 2i ⎞⎛ 2 ⎞
i =1
∑i −6
n →∞
n
∑ f ⎜⎝ n ⎟⎜
⎟
⎠⎝ n ⎠
S ( n) =
2 n
Area = lim S ( n) = 12 +
3
2 −0
2⎞
⎛
= ⎟
48. y = 3 x 2 + 1 on [0, 2]. ⎜ Note: ∆x =
n
n⎠
⎝
=
27 ⎛ ( n + 1)n ⎞
27 ⎛
1⎞
⎜
⎟ = 12 +
⎜1 + ⎟
2
2
n2 ⎝
n⎠
⎝
⎠
= 12 +
2
Area = lim S ( n) = 8 + 2 = 10
n→∞
y
y
15
3
10
2
5
x
1
2
3
4
5
6
x
1
2
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
444
Chapter 5
Integration
50. y = 4 − x 2 on [−2, 2]. Find area of region over the
3⎞
⎛
49. y = 25 − x 2 on [1, 4]. ⎜ Note: ∆x = ⎟
n⎠
⎝
2
⎡
3i ⎞⎛ 3 ⎞
3i ⎞ ⎤⎛ 3 ⎞
⎛
⎛
s( n) = ∑ f ⎜1 + ⎟⎜ ⎟ = ∑ ⎢25 − ⎜1 + ⎟ ⎥⎜ ⎟
n ⎠⎝ n ⎠
n ⎠ ⎥⎦⎝ n ⎠
⎝
⎢
i =1 ⎝
i =1 ⎣
3 n ⎡
9i 2
6i ⎤
= ∑ ⎢24 − 2 − ⎥
n i =1 ⎣
n
n⎦
n
n
2⎞
⎛
interval [0, 2]. ⎜ Note: ∆x = ⎟
n⎠
⎝
s( n) =
3⎡
9 n( n + 1)( 2n + 1) 6 n( n + 1) ⎤
−
⎢24n − 2
⎥
n⎣
n
n
6
2 ⎦
9
9
= 72 −
(n + 1)(2n + 1) − (n + 1)
n
2n 2
i =1
=
=
Area = lim s( n) = 72 − 9 − 9 = 54
⎛ 2i ⎞⎛ 2 ⎞
f ⎜ ⎟⎜ ⎟
⎝ n ⎠⎝ n ⎠
n
∑
n
⎡
i =1
⎣⎢
2
⎛ 2i ⎞ ⎤ ⎛ 2 ⎞
∑ ⎢4 − ⎜⎝ n ⎟⎠ ⎥⎜⎝ n ⎟⎠
⎦⎥
n
= 8−
8
∑ i2
n 3 i =1
= 8−
8n( n + 1)( 2n + 1)
4⎛
3
1⎞
= 8 − ⎜2 + + 2 ⎟
n
n ⎠
6n3
3⎝
n →∞
1
8
16
Area = lim s( n) = 8 − =
n →∞
2
3
3
32
Area =
3
y
20
15
10
y
5
−1
x
1
−5
2
3
4
5
3
2
1
x
−1
1
3−1
2⎞
⎛
= ⎟
51. y = 27 − x3 on [1, 3]. ⎜ Note: ∆x =
n
n
⎝
⎠
s( n) =
⎛
n
∑ f ⎜⎝1 +
i =1
=
2i ⎞⎛ 2 ⎞
⎟⎜ ⎟ =
n ⎠⎝ n ⎠
n
⎡
i =1
⎣⎢
⎛
∑ ⎢27 − ⎜⎝1 +
2i ⎞
⎟
n⎠
3
⎤⎛ 2 ⎞
⎥⎜ ⎟
⎦⎥⎝ n ⎠
2 ⎡
8i
12i
6i ⎤
∑ ⎢26 − n3 − n2 − n ⎥
n i =1 ⎣
⎦
n
3
y
2
2
2⎡
8 n 2 ( n + 1)
12 n( n + 1)( 2n + 1) 6 n( n + 1) ⎤
= ⎢26n − 3
− 2
−
⎥
n ⎢⎣
n
4
n
6
n
2 ⎥⎦
4
4
6n + 1
2
= 52 − 2 ( n + 1) − 2 ( n + 1)( 2n + 1) −
n
n
n
Area = lim s( n) = 52 − 4 − 8 − 6 = 34
30
24
18
12
6
−2 −1
−6
x
1
2
4
5
n →∞
1−0
1⎞
⎛
52. y = 2 x − x 3 on [0, 1]. ⎜ Note: ∆x =
= ⎟
n
n⎠
⎝
Because y both increases and decreases on [0, 1], T(n) is neither an upper nor lower sum.
T ( n) =
n
∑
i =1
⎡ ⎛ i ⎞ ⎛ i ⎞3 ⎤⎛ 1 ⎞
⎛ i ⎞⎛ 1 ⎞
f ⎜ ⎟⎜ ⎟ = ∑ ⎢2⎜ ⎟ − ⎜ ⎟ ⎥⎜ ⎟
⎝ n ⎠⎝ n ⎠
i =1 ⎢
⎣ ⎝ n ⎠ ⎝ n ⎠ ⎥⎦⎝ n ⎠
2
n( n + 1)
2
1
1 ⎡ n 2 ( n + 1) ⎤
1 1
2
1
= 2 ∑ i − 4 ∑ i3 =
− 4⎢
⎥ = 1+ − −
−
2
4
4
4
4
n i =1
n i =1
n
n ⎣⎢
n
n
n2
⎦⎥
n
n
Area = lim T ( n) = 1 −
n →∞
y
n
1
3
=
4
4
2.0
1.5
1.0
0.5
x
0.5
1.0
2.0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2
Area
445
⎛
1 − ( −1)
2⎞
= ⎟
53. y = x 2 − x 3 on [−1, 1]. ⎜ Note: ∆x =
n
n
⎝
⎠
Because y both increases and decreases on [−1, 1], T(n) is neither an upper nor a lower sum.
⎛
n
2i ⎞⎛ 2 ⎞
⎟⎜ ⎟ =
n ⎠⎝ n ⎠
∑ f ⎜⎝ −1 +
T ( n) =
i =1
n
⎡⎛
i =1
⎣⎝
n
⎡
i =1
⎣
∑ ⎢2 −
=
⎡⎛
i =1
⎣⎢
2
3
2i ⎞
2i ⎞ ⎤⎛ 2 ⎞
⎛
⎟ − ⎜ −1 + ⎟ ⎥ ⎜ ⎟
n⎠
n ⎠ ⎦⎥⎝ n ⎠
⎝
4i
4i ⎞ ⎛
6i 12i 2
8i 3 ⎞⎤ ⎛ 2 ⎞
+ 2 ⎟ − ⎜ −1 +
− 2 + 3 ⎟⎥ ⎜ ⎟
n
n ⎠ ⎝
n
n
n ⎠⎦ ⎝ n ⎠
2
∑ ⎢⎜1 −
=
n
∑ ⎢⎜⎝ −1 +
10i 16i 2
8i 3 ⎤ ⎛ 2 ⎞
4 n
20 n
32 n
16 n
+ 2 − 3 ⎥⎜ ⎟ = ∑ 1 − 2 ∑ i + 3 ∑i 2 − 4 ∑ i 3
n i =1
n i =1
n
n
n ⎦⎝ n ⎠
n i =1
n i =1
4
20 n( n + 1) 32 n( n + 1)( 2n + 1) 16 n 2 ( n + 1)
+ 3 ⋅
− 4 ⋅
( n) − 2 ⋅
n
n
n
n
2
6
4
1 ⎞ 16 ⎛
3
1⎞
2
1⎞
⎛
⎛
= 4 − 10⎜1 + ⎟ + ⎜ 2 + + 2 ⎟ − 4⎜1 + + 2 ⎟
3⎝
n⎠
n
n ⎠
n
n ⎠
⎝
⎝
2
=
Area = lim T ( n) = 4 − 10 +
n →∞
32
2
−4 =
3
3
y
2
1
x
−1
1
2 −1
1⎞
⎛
= ⎟
54. y = 2 x 3 − x 2 on [1, 2]. ⎜ Note: ∆x =
n
n⎠
⎝
s( n) =
n
∑
i =1
=
i ⎞⎛ 1 ⎞
⎛
f ⎜1 + ⎟⎜ ⎟ =
n ⎠⎝ n ⎠
⎝
⎛ 2i 3
n
∑ ⎜ n3
i =1
+
⎝
3
2
⎡ ⎛
i⎞
i ⎞ ⎤⎛ 1 ⎞
⎛
+
−
+
2
1
1
⎢
∑ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠ ⎥⎜⎝ n ⎟⎠
⎥⎦
i =1 ⎢
⎣
n
⎞⎛ 1 ⎞
5i 2
4i
+
+ 1⎟⎜ ⎟
n2
n
⎠⎝ n ⎠
2 n 2 ( n + 1)
5 n( n + 1)( 2n + 1)
4 n( n + 1)
⋅
+ 3 ⋅
+ 2 ⋅
+1
4
4
6
2
n
n
n
2
=
Area = lim sn =
n→∞
1
5
31
+ + 2+1=
2
3
6
y
12
10
8
6
4
2
−3
−2
−1
x
1
2
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
446
Chapter 5
Integration
2−0
2⎞
⎛
= ⎟
55. f ( y ) = 4 y, 0 ≤ y ≤ 2 ⎜ Note: ∆y =
n
n⎠
⎝
S ( n) =
n
∑ f (mi ) ∆y
5−0
5⎞
⎛
= ⎟
57. f ( y ) = y 2 , 0 ≤ y ≤ 5 ⎜ Note: ∆y =
n
n⎠
⎝
S ( n) =
i =1
i =1
=
∑
⎛ 2i ⎞⎛ 2 ⎞
f ⎜ ⎟⎜ ⎟
⎝ n ⎠⎝ n ⎠
n
⎛ 2i ⎞⎛ 2 ⎞
n
i =1
=
∑ 4⎜⎝ n ⎟⎜
⎟
⎠⎝ n ⎠
i =1
16
= 2
n
⎛ 5i ⎞⎛ 5 ⎞
n
∑ f ⎜⎝ n ⎟⎜
⎟
⎠⎝ n ⎠
⎛ 5i ⎞ ⎛ 5 ⎞
∑ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠
i =1
=
125 n 2
∑i
n 3 i =1
=
125 n( n + 1)( 2n + 1)
⋅
n3
6
=
125 ⎛ 2n 2 + 3n + 1 ⎞ 125 125 125
+
+
⎜
⎟ =
n2 ⎝
6
3
2n
6n 2
⎠
n
∑i
i =1
8( n + 1)
8
⎛ 16 ⎞ n( n + 1)
= ⎜ 2⎟ ⋅
=
= 8+
2
n
n
⎝n ⎠
8⎞
⎛
Area = lim S ( n) = lim ⎜ 8 + ⎟ = 8
n →∞
n →∞ ⎝
n⎠
2
=
n
⎛ 125 125 125 ⎞ 125
+
+
Area lim S ( n) = lim ⎜
⎟ =
n →∞
n →∞ ⎝ 3
2n
6n 2 ⎠
3
y
y
6
4
4
3
2
2
x
−5
−2
1
x
2
4
6
5
10
15
20
25
−4
8
−1
−6
56. g ( y ) =
S ( n) =
4−2
2⎞
1
⎛
= ⎟
y, 2 ≤ y ≤ 4. ⎜ Note: ∆y =
n
n⎠
2
⎝
⎛
n
2i ⎞⎛ 2 ⎞
⎟⎜ ⎟
n ⎠⎝ n ⎠
∑ g ⎜⎝ 2 +
i =1
=
n
∑
i =1
=
2 − 1 1⎞
⎛
= ⎟
58. f ( y ) = 4 y − y 2 , 1 ≤ y ≤ 2. ⎜ Note: ∆y =
n
n⎠
⎝
S ( n) =
⎛
n
∑ f ⎜⎝1 +
i =1
i⎞
1⎛
2i ⎞⎛ 2 ⎞
2 n ⎛
⎜ 2 + ⎟⎜ ⎟ = ∑ ⎜1 + ⎟
n ⎠⎝ n ⎠
n i =1 ⎝
n⎠
2⎝
n +1
2⎡
1 n( n + 1) ⎤
⎢n +
⎥ = 2+
n⎣
n
n
2 ⎦
=
2
1 n ⎡ ⎛
i⎞ ⎛
i⎞ ⎤
4
1
1
+
−
+
⎢
∑ ⎜ n ⎟⎠ ⎜⎝ n ⎟⎠ ⎥
n i =1 ⎢⎣ ⎝
⎥⎦
=
1 n ⎛
4i
2i
i2 ⎞
−1−
− 2⎟
⎜4 +
∑
n i =1 ⎝
n
n
n ⎠
=
1 n ⎛
2i
i2 ⎞
− 2⎟
⎜3 +
∑
n i =1 ⎝
n
n ⎠
=
1⎡
2 n( n + 1)
1 n( n + 1)( 2n + 1) ⎤
− 2
⎢3n +
⎥
n⎣
n
n
2
6
⎦
Area = lim S ( n) = 2 + 1 = 3
n →∞
y
5
4
3
n + 1 ( n + 1)( 2n + 1)
−
n
6
= 3+
2
1
x
1
2
3
4
5
i ⎞⎛ 1 ⎞
⎟⎜ ⎟
n ⎠⎝ n ⎠
Area = lim S ( n) = 3 + 1 −
n →∞
1
11
=
3
3
y
5
3
2
1
x
1
2
3
4
5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2
Area
447
3−1
2⎞
⎛
= ⎟
59. g ( y ) = 4 y 2 − y 3 , 1 ≤ y ≤ 3. ⎜ Note: ∆y =
n
n⎠
⎝
⎛
n
∑ g ⎜⎝1 +
S ( n) =
i =1
2i ⎞⎛ 2 ⎞
⎟⎜ ⎟
n ⎠⎝ n ⎠
=
2
3
⎡ ⎛
2i ⎞
2i ⎞ ⎤ 2
⎛
+
−
+
4
1
1
⎢
∑ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠ ⎥ n
⎥⎦
i =1 ⎢
⎣
=
2 n ⎡
4i
4i 2 ⎤ ⎡
6i 12i 2
8i 3 ⎤
+ 2 ⎥ − ⎢1 +
+ 2 + 3⎥
4 ⎢1 +
∑
n i =1 ⎣
n
n ⎦ ⎣
n
n
n ⎦
=
2 n ⎡
10i
4i 2
8i 3 ⎤
+ 2 − 3⎥
∑
⎢3 +
n i =1 ⎣
n
n
n ⎦
=
2
2⎡
10 n( n + 1)
4 n( n + 1)( 2n + 1)
8 n 2 ( n + 1) ⎤
⎢3n +
⎥
+ 2
− 2
2
6
4
n⎢
n
n
n
⎥⎦
⎣
n
8
44
−4 =
3
3
Area = lim S ( n) = 6 + 10 +
n →∞
y
10
8
6
2
x
−4 −2
−2
−4
1⎞
⎛
60. h( y ) = y 3 + 1, 1 ≤ y ≤ 2 ⎜ Note: ∆y = ⎟
n⎠
⎝
⎛
n
∑ h⎜⎝1 +
S ( n) =
i =1
n
⎡⎛
∑ ⎢⎜⎝1 +
=
i =1
⎢⎣
i ⎞⎛ 1 ⎞
⎟⎜ ⎟
n ⎠⎝ n ⎠
3
⎤1
i⎞
⎟ + 1⎥
n⎠
⎥⎦ n
1 ⎛
i3
3i 2
3i ⎞
= ∑ ⎜2 + 3 + 2 + ⎟
n i =1 ⎝
n
n
n⎠
y
n
1⎡
1 n 2 ( n + 1)
3 n( n + 1)( 2n + 1) 3 3n( n + 1) ⎤
= ⎢2n + 3
+ 2
+
⎥
4
6
2n ⎥⎦
n ⎢⎣
n
n
n
2
= 2+
(n
+ 1)
1 ( n + 1)( 2n + 1) 3( n + 1)
+
+
2
2n
n2 4
n2
5
4
3
2
1
2
Area = lim S ( n) = 2 +
n →∞
x
2
4
6
8
10
1
3
19
+1+
=
4
2
4
61. f ( x) = x 2 + 3, 0 ≤ x ≤ 2, n = 4
xi + xi −1
.
2
Let ci =
∆x =
1
1
3
5
7
, c1 = , c2 = , c3 = , c4 =
2
4
4
4
4
Area ≈
n
∑ f (ci ) ∆x
i =1
=
4
∑ ⎡⎣ci2
i =1
1 ⎡⎛ 1
69
⎛1⎞
⎞ ⎛9
⎞ ⎛ 25
⎞ ⎛ 49
⎞⎤
+ 3⎤⎦ ⎜ ⎟ = ⎢⎜
+ 3⎟ + ⎜
+ 3⎟ + ⎜
+ 3⎟ + ⎜
+ 3⎟⎥ =
2 ⎣⎝ 16
8
⎝ 2⎠
⎠ ⎝ 16
⎠ ⎝ 16
⎠ ⎝ 16
⎠⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
448
Chapter 5
Integration
62. f ( x) = x 2 + 4 x, 0 ≤ x ≤ 4, n = 4
xi + xi −1
.
2
1
3
5
7
∆x = 1, c1 = , c2 = , c3 = , c4 =
2
2
2
2
Let ci =
n
∑ f (ci ) ∆x
Area ≈
=
i =1
4
∑ ⎡⎣ci2
i =1
π
63. f ( x ) = tan x, 0 ≤ x ≤
π
16
, c1 =
π
32
n
⎛π ⎞
4
∑ ( tan ci ) ⎜⎝ 16 ⎟⎠
=
i =1
∆x =
π
π
Area ≈
2
π⎛
π
3π
5π
7π ⎞
+ tan
+ tan
+ tan ⎟ ≈ 0.345
⎜ tan
16 ⎝
32
32
32
32 ⎠
, n = 4
xi + xi + 1
.
2
, c1 =
8
=
i =1
64. f ( x ) = cos x, 0 ≤ x ≤
Let ci =
3π
5π
7π
, c3 =
, c4 =
32
32
32
, c2 =
∑ f (ci ) ∆x
Area ≈
,n = 4
xi + xi −1
.
2
Let ci =
∆x =
4
⎡⎛ 1
⎞ ⎛9
⎞ ⎛ 25
⎞ ⎛ 49
⎞⎤
+ 4ci ⎤⎦ (1) = ⎢⎜ + 2 ⎟ + ⎜ + 6 ⎟ + ⎜
+ 10 ⎟ + ⎜
+ 14 ⎟⎥ = 53
4
4
4
4
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
π
16
3π
5π
7π
, c3 =
, c4 =
16
16
16
, c2 =
n
∑ f (ci ) ∆x
=
i =1
⎛π ⎞
4
∑ cos (ci) ⎜⎝ 8 ⎟⎠
i =1
=
π⎛
π
3π
5π
7π ⎞
+ cos
+ cos
+ cos
⎜ cos
⎟ ≈ 1.006
8⎝
16
16
16
16 ⎠
65. f ( x) = ln x, 1 ≤ x ≤ 5, n = 4
xi + xi −1
, ∆x = 1
2
3
5
7
9
c1 = , c2 = , c3 = , c4 =
2
2
2
2
Let ci =
Area ≈
n
∑ f (ci ) ∆x
=
i =1
4
∑ ⎡⎣ln(ci )⎤⎦(1)
≈ 0.40547 + 0.91629 + 1.25276 + 1.50408 ≈ 4.0786
i =1
66. f ( x ) = xe x , 0 ≤ x ≤ 2, n = 4
xi + xi −1
1
, ∆x =
2
2
1
3
5
7
c1 = , c2 = , c3 = , c4 =
4
4
4
4
Let ci =
Area ≈
n
∑ f (ci ) ∆ x
⎛1⎞
⎛1⎞
⎛1⎞
∑ ⎡⎣ci eci ⎤⎦ ⎜⎝ 2 ⎟⎠ ≈ [0.32101 + 1.58775 + 4.36293 + 10.07055]⎜⎝ 2 ⎟⎠ ≈ (16.34224) ⎜⎝ 2 ⎟⎠
4
=
i =1
67.
≈ 8.1711
i =1
y
68.
y
4
4
3
3
2
2
1
1
1
2
3
4
x
(b) A ≈ 6 square units
1
2
3
4
x
(a) A ≈ 3 square units
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.2
69. You can use the line y = x bounded by x = a and
x = b. The sum of the areas of these inscribed
rectangles is the lower sum.
449
Area
y
(c)
8
6
y
4
2
x
1
3
2
4
Midpoint Rule:
M ( 4) = 2 23 + 4 54 + 5 75 + 6 92 =
x
a
b
The sum of the areas of these circumscribed rectangles is
the upper sum.
6112
315
≈ 19.403
(d) In each case, ∆x = 4 n. The lower sum uses left
end-points, (i − 1)( 4 n). The upper sum uses right
endpoints, (i)( 4 n). The Midpoint Rule uses
y
(
midpoints, i −
(e)
1
2
)(4 n).
N
4
8
20
100
200
s(n)
15.333 17.368 18.459 18.995 19.06
S(n)
21.733 20.568 19.739 19.251 19.188
x
a
b
You can see that the rectangles do not contain all of the
area in the first graph and the rectangles in the second
graph cover more than the area of the region. The exact
value of the area lies between these two sums.
70. See the definition of area, page 296.
71. (a)
y
8
M(n) 19.403 19.201 19.137 19.125 19.125
(f ) s (n) increases because the lower sum approaches
the exact value as n increases. S (n) decreases
because the upper sum approaches the exact value
as n increases. Because of the shape of the graph,
the lower sum is always smaller than the exact
value, whereas the upper sum is always larger.
72. (a) Left endpoint of first subinterval is 1.
6
Left endpoint of last subinterval is 4 −
4
2
(b) Right endpoint of first subinterval is 1 +
x
1
2
3
4
=
1
4
15 .
4
=
5
4
1
2
=
Right endpoint of second subinterval is 1 +
Lower sum:
s( 4) = 0 + 4 + 5 13 + 6 = 15 13 =
(b)
1
4
46
3
≈ 15.333
.
3
.
2
(c) The rectangles lie above the graph.
(d) The heights would be equal to that constant.
y
73. True. (Theorem 5.2 (2))
8
6
74. True. (Theorem 5.3)
4
75. Suppose there are n rows and n + 1 columns in the
+ n, as do
figure. The stars on the left total 1 + 2 +
2
the stars on the right. There are n( n + 1) stars in total, so
x
1
2
3
4
2[1 + 2 +
Upper sum:
11 =
S ( 4) = 4 + 5 13 + 6 + 6 52 = 2115
326
15
≈ 21.733
1+ 2+
+ n] = n( n + 1)
+ n =
1
2
(n)(n + 1).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
450
Chapter 5
76. (a) θ =
Integration
2π
n
77. For n odd,
r
h
(b) sin θ =
r
h = r sin θ
A =
h
θ
r
1 block
n = 3, 2 rows,
4 blocks
n = 5, 3 rows,
9 blocks
2
n,
1
1
1
bh = r ( r sin θ ) = r 2 sin θ
2
2
2
n +1
⎛ n + 1⎞
rows, ⎜
⎟ blocks,
2
⎝ 2 ⎠
For n even,
2π ⎞
⎛1
(c) An = n⎜ r 2 sin
⎟
n ⎠
⎝2
=
n = 1, 1 row,
n = 2, 1 row,
⎛ sin ( 2π n) ⎞
2π
r 2n
sin
= πr2⎜
⎟
2
n
⎝ 2π n ⎠
Let x = 2π n. As n → ∞, x → 0.
2 block
n = 4, 2 rows,
6 blocks
n = 6, 3 rows,
12 blocks
n,
n
rows,
2
n 2 + 2n
blocks,
4
⎛ sin x ⎞
2
2
lim An = lim π r 2 ⎜
⎟ = π r (1) = π r
x→0
⎝ x ⎠
n →∞
n
78. (a)
∑ 2i
= n( n + 1)
i =1
The formula is true for n = 1: 2 = 1(1 + 1) = 2.
Assume that the formula is true for n = k :
k
∑ 2i
= k ( k + 1).
i =1
k +1
Then you have
∑ 2i
=
i =1
k
∑ 2i + 2(k
+ 1) = k ( k + 1) + 2( k + 1) = ( k + 1)( k + 2)
i =1
which shows that the formula is true for n = k + 1.
n
(b)
∑ i3
i =1
=
n 2 ( n + 1)
4
2
The formula is true for n = 1 because 13 =
12 (1 + 1)
4
Assume that the formula is true for n = k :
∑ i3
k
k +1
∑ i3
i =1
=
k
∑ i 3 + (k
i =1
+ 1) =
3
=
4
= 1.
4
k 2 ( k + 1)
.
4
2
=
i =1
Then you have
2
k 2 ( k + 1)
(k + 1) ⎡k 2 + 4 k + 1 ⎤ = (k + 1) k + 2 2
3
+ ( k + 1) =
(
)⎦
(
)
⎣
4
4
4
2
2
2
which shows that the formula is true for n = k + 1.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.3
451
Riemann Sums and Definite Integrals
79. Assume that the dartboard has corners at ( ±1, ±1).
A point (x, y) in the square is closer to the center than the top edge if
x + y
2
1
x + y ≤ 1 − 2y + y
2
y
≤1− y
2
2
y ≤
1
2
(x, y)
(1 − x 2 ).
−1
(0, 0)
By symmetry, a point (x, y) in the square is closer to the center than the right edge if
x ≤
1
2
(x, 1)
2
1
2
(1 − x 2 ) and
x =
1
2
(1 − y 2 ) intersect at (
x
−1
(1 − y 2 ).
In the first quadrant, the parabolas y =
1
2 − 1,
)
2 − 1 . There are 8 equal
regions that make up the total region, as indicated in the figure.
y
1
(
2 − 1,
2 − 1(
(1, 1)
−1
x
1
−1
Area of shaded region S =
Probability =
∫0
2 −1
2 2
5
⎡1
⎤
2
⎢ 2 (1 − x ) − x⎥ dx = 3 − 6
⎣
⎦
⎡2 2
8S
5⎤
4 2
5
= 2⎢
− ⎥ =
−
Area square
3
6
3
3
⎣
⎦
Section 5.3 Riemann Sums and Definite Integrals
1. f ( x ) =
∆xi =
n
x , y = 0, x = 0, x = 3, ci =
3(i − 1)
3i 2
−
2
n
n2
∑ f (ci )∆xi
n →∞
lim
i =1
2
=
3i 2
n2
3
( 2i − 1)
n2
n
= lim ∑
3i 2 3
( 2i − 1)
n2 n2
3 3
= lim 3
n →∞ n
∑ (2i 2
n →∞
i =1
n
i =1
−i
)
3 3 ⎡ n( n + 1)( 2n + 1) n( n + 1) ⎤
−
⎢2
⎥
n →∞ n 3
6
2
⎣
⎦
= lim
⎡ ( n + 1)( 2n + 1) n + 1⎤
= lim 3 3 ⎢
−
⎥
n →∞
3n 2
2n 2 ⎦
⎣
⎡2
⎤
= 3 3 ⎢ − 0⎥ = 2 3 ≈ 3.464
⎣3
⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
452
Chapter 5
2. f ( x ) =
3
Integration
x , y = 0, x = 0, x = 1, ci =
i3
n3
(i − 1) = 3i 2 − 3i + 1
i3
−
3
n
n3
n3
3
∆xi =
n
n →∞
i 3 ⎡ 3i 2 − 3i + 1⎤
⎢
⎥
n3 ⎣
n3
⎦
n
lim ∑ f (ci ) ∆xi = lim ∑
n →∞
i =1
3
i =1
= lim
1 n
∑ (3i3 − 3i 2 + i)
n 4 i =1
= lim
2
⎛ n( n + 1)( 2n + 1) ⎞ n( n + 1) ⎤
1 ⎡ ⎛ n 2 ( n + 1) ⎞
⎢
⎥
⎜
⎟ − 3⎜
3
⎟+
4
⎟
4
6
2
n ⎢ ⎜⎝
⎥
⎝
⎠
⎠
⎣
⎦
= lim
1 ⎡ 3n 4 + 6n3 + 3n 2
2n3 + 3n 2 + n
n2 + n ⎤
−
+
⎥
4⎢
4
2
2 ⎦
n ⎣
= lim
1 ⎡ 3n 4
n3
n2 ⎤
1
1 ⎤
3
⎡3
+
−
+
−
=
⎢
⎥ = nlim
2⎥
→∞ ⎢
2
4⎦
n
n
n4 ⎣ 4
4
2
4
4
⎣
⎦
n →∞
n →∞
n →∞
n →∞
6−2
4
⎛
⎞
3. y = 8 on [2, 6]. ⎜ Note: ∆x =
= , ∆ → 0 as n → ∞ ⎟
n
n
⎝
⎠
n
∑ f (ci ) ∆xi
=
i =1
6
∫2
4i ⎞⎛ 4 ⎞
⎛
f ⎜ 2 + ⎟⎜ ⎟ =
n ⎠⎝ n ⎠
⎝
n
∑
i =1
n
⎛ 4⎞
∑ 8⎜⎝ n ⎟⎠
=
i =1
n
∑
i =1
32
1 n
1
= ∑ 32 = (32n) = 32
n
n i =1
n
8 dx = lim 32 = 32
n →∞
⎛
3 − ( −2)
⎞
5
4. y = x on [−2, 3]. ⎜ Note: ∆x =
= , ∆ → 0 as n → ∞ ⎟
n
n
⎝
⎠
n
∑ f (ci ) ∆xi
=
i =1
5i ⎞⎛ 5 ⎞
⎛
f ⎜ −2 + ⎟⎜ ⎟
n ⎠⎝ n ⎠
⎝
n
∑
i =1
=
5i ⎞⎛ 5 ⎞
25 n
25 ⎛
1⎞
5
25
⎛ 25 ⎞ n( n + 1)
= −10 +
+
⎟⎜ ⎟ = −10 + 2 ∑ i = −10 + ⎜ 2 ⎟
⎜1 + ⎟ =
n ⎠⎝ n ⎠
n i =1
n⎠
2
2⎝
2 2n
⎝n ⎠
⎛
n
∑ ⎝⎜ −2 +
i =1
3
∫ − 2 x dx
25 ⎞
5
⎛5
= lim ⎜ +
⎟ =
n →∞ ⎝ 2
2n ⎠
2
⎛
1 − ( −1)
⎞
2
5. y = x3 on [−1, 1]. ⎜ Note: ∆x =
= , ∆ → 0 as n → ∞ ⎟
n
n
⎝
⎠
n
∑ f (ci ) ∆xi
=
i =1
⎛
n
∑ f ⎜⎝ −1 +
i =1
=
2i ⎞
⎟
n⎠
n
⎡
i =1
⎣
6i 12i 2
8i 3 ⎤ ⎛ 2 ⎞
− 2 + 3 ⎥⎜ ⎟
n
n
n ⎦⎝ n ⎠
∑ ⎢−1 +
= −2 +
1
∫ −1
3
⎛
n
∑ ⎜⎝ −1 +
i =1
=
2i ⎞⎛ 2 ⎞
⎟⎜ ⎟
n ⎠⎝ n ⎠
⎛2⎞
⎜ ⎟
⎝n⎠
12 n
24 n
16 n
i − 3 ∑ i 2 + 4 ∑ i3
2∑
n i =1
n i =1
n i =1
1⎞
3
1⎞
2
1⎞
2
⎛
⎛
⎛
= −2 + 6⎜1 + ⎟ − 4⎜ 2 + + 2 ⎟ + 4⎜1 + + 2 ⎟ =
n⎠
n
n ⎠
n
n ⎠
n
⎝
⎝
⎝
2
= 0
x3 dx = lim
n →∞ n
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.3
Riemann Sums and Definite Integrals
453
4 −1
3
⎛
⎞
6. y = 4 x 2 on [1, 4]. ⎜ Note: ∆x =
= , ∆ → 0 as n → ∞ ⎟
n
n
⎝
⎠
n
∑ f (ci )∆xi
=
i =1
⎛
n
∑ f ⎜⎝1 +
i =1
=
2
3i ⎞ ⎛ 3 ⎞
⎛
4⎜1 + ⎟ ⎜ ⎟
n ⎠ ⎝n⎠
⎝
n
∑
i =1
=
12 n ⎛
6i 9i 2 ⎞
+ 2⎟
⎜1 +
∑
n i =1 ⎝
n
n ⎠
=
12 ⎡
6 n( n + 1)
9 n( n + 1)( 2n + 1) ⎤
+ 2
⎢n +
⎥
n⎣
n
n
2
6
⎦
= 12 + 36
4
∫1
3i ⎞⎛ 3 ⎞
⎟⎜ ⎟
n ⎠⎝ n ⎠
n +1
( n + 1)(2n + 1)
+ 18
n
n2
36( n + 1) 18( n + 1)( 2n + 1) ⎤
⎡
4 x 2 dx = lim ⎢12 +
+
⎥
n →∞
n
n2
⎣
⎦
= 12 + 36 + 36 = 84
2 −1
1
⎛
⎞
7. y = x 2 + 1 on [1, 2]. ⎜ Note: ∆x =
= , ∆ → 0 as n → ∞ ⎟
n
n
⎝
⎠
n
∑ f (ci ) ∆xi
=
i =1
⎛
n
∑ f ⎜⎝1 +
i =1
=
=
n
⎡⎛
i =1
⎣⎢
n
⎡
i =1
⎣
∑ ⎢⎜⎝1 +
∑ ⎢1 +
= 2+
2
∫1 ( x
2
i ⎞⎛ 1 ⎞
⎟⎜ ⎟
n ⎠⎝ n ⎠
2
⎤⎛ 1 ⎞
i⎞
⎟ + 1⎥ ⎜ ⎟
n⎠
⎥⎦ ⎝ n ⎠
⎤⎛ 1 ⎞
i2
2i
+ 2 + 1⎥⎜ ⎟
n
n
⎦⎝ n ⎠
2 n
1 n 2
1 ⎞ 1⎛
3
1⎞
10
3
1
⎛
i
i = 2 + ⎜1 + ⎟ + ⎜ 2 + + 2 ⎟ =
+
+
+
∑
2
3∑
n i =1
n i =1
n ⎠ 6⎝
n
n ⎠
3
2n 6n 2
⎝
10
3
1 ⎞
⎛ 10
+ 1) dx = lim ⎜
+
+
⎟ =
n →∞ ⎝ 3
3
2n 6n 2 ⎠
⎛
1 − ( −2)
⎞
3
8. y = 2 x 2 + 3 on [−2, 1]. ⎜ Note: ∆x =
= , ∆ → 0 as n → ∞ ⎟
n
n
⎝
⎠
n
∑ f (ci ) ∆xi
=
i =1
3i ⎞⎛ 3 ⎞
⎛
f ⎜ −2 + ⎟⎜ ⎟
n ⎠⎝ n ⎠
⎝
n
∑
i =1
=
⎡ ⎛
n
∑ ⎢2⎜⎝ −2 +
⎣⎢
i =1
∫ − 2 (2 x
1
2
2
⎤⎛ 3 ⎞
3i ⎞
⎟ + 3⎥ ⎜ ⎟
n⎠
⎥⎦ ⎝ n ⎠
=
⎡ ⎛
⎤
3
12i
9i 2 ⎞
+ 2 ⎟ + 3⎥
⎢2⎜ 4 −
∑
n i =1 ⎣ ⎝
n
n ⎠
⎦
=
3 n
∑
n i =1
=
(n + 1)( 2n + 1)
3⎡
24 n( n + 1) 18 n( n + 1)( 2n + 1) ⎤
n +1
+ 2
+9
⎢11n −
⎥ = 33 − 36
2
6
n⎣
n
n
n
n2
⎦
n
⎡
24i 18i 2 ⎤
+ 2 ⎥
⎢11 −
n
n ⎦
⎣
⎡
n +1
(n + 1)(2n + 1) ⎤ = 33 − 36 + 18 = 15
+ 3) dx = lim ⎢33 − 36
+9
⎥
n →∞
n
n2
⎣
⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
454
Chapter 5
n
9. lim
∆ →0
∑ (3ci
Integration
+ 10) ∆xi =
∫ −1 (3x + 10) dx
5
i =1
on the interval [−1, 5].
n
10. lim
∆ →0
∑ 6ci (4 − ci )
2
26.
2
∫0
2e − x dx
27. Rectangle
A = bh = 3( 4)
2
∫ 0 6 x(4 − x) dx
4
∆xi =
i =1
3
∫0
A =
4 dx = 12
y
on the interval [0, 4].
5
n
11. lim
∆ →0
∑
ci2 + 4 ∆xi =
i =1
3
∫0
x 2 + 4 dx
3
Rectangle
2
on the interval [0, 3].
1
⎛3⎞
n
∑ ⎜ c 2 ⎟ ∆xi
∆ →0
12. lim
i =1 ⎝ i
⎠
=
3
∫1
x
3
dx
x2
1
⎛
3⎞
13. lim ∑ ⎜1 + ⎟ ∆xi =
∆ →0
ci ⎠
i =1 ⎝
3
4
5
28. Rectangle
on the interval [1, 3].
n
2
A = bh = 10(6) = 60
5
∫1
3⎞
⎛
⎜1 + ⎟ dx
x⎠
⎝
6
∫ − 4 6 dx
A =
= 60
y
on the interval [1, 5].
n
∑ (2−ci sin ci ) ∆xi
∆ →0
14. lim
i =1
=
π
∫0
2− x sin x dx
4
Rectangle
on the interval [0, π ].
15.
2
−4
4
∫ 0 5 dx
x
−2
2
4
6
29. Triangle
16.
17.
2
∫0
(6 − 3x) dx
∫ − 4 (4 − x ) dx
4
A =
1 bh
2
A =
4
∫0
=
1
2
(4)(4)
= 8
x dx = 8
y
2
18.
∫0
19.
2
∫ − 5 (25 − x ) dx
20.
∫ −1 x 2
21.
∫0
22.
∫0
23.
∫0
24.
∫ 0 ( y − 2)
x 2 dx
4
Triangle
5
1
π 2
π 4
2
2
4
dx
+ 2
x
2
4
30. Triangle
cos x dx
tan x dx
1
1
bh = (8)( 2) = 8
2
2
8 x
A = ∫
dx = 8
0 4
A =
y
y 3 dy
4
25.
2
4
∫1
2
dx
x
2
dy
2
Triangle
x
2
4
6
8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.3
31. Trapezoid
Riemann Sums and Definite Integrals
455
35. Semicircle
A =
b1 + b2
⎛ 4 + 10 ⎞
h = ⎜
⎟ 2 = 14
2
⎝ 2 ⎠
A =
∫ 0 (3x + 4) dx
2
1 2
1
49π
2
π r = π ( 7) =
2
2
2
7
49
π
A = ∫
49 − x 2 dx =
−7
2
A =
= 14
y
y
12
12
10
Semicircle
8
8
6
Trapezoid
4
4
2
x
−1
1
2
3
4
6
8
36. Semicircle
32. Trapezoid
A =
2
−4
−4
A =
x
−8 −6 −4 −2
b1 + b 2
2
3
∫0
A = 12 π r 2
8+ 2
(3) = 15
2
h =
A =
(8 − 2 x) dx = 15
r
∫ −r
r 2 − x 2 dx = 12 π r 2
y
y
r
Semicircle
8
6
Trapezoid
−r
x
r
4
2
−r
x
1
2
3
4
In Exercises 37 – 44,
33. Triangle
A =
1 bh
2
(2)(1)
=1
A =
∫ −1 (1 − x ) dx
=1
=
1
2
1
4
∫2
Triangle
x
−1
1
34. Triangle
=
( 2a ) a = a 2
A =
1 bh
2
A =
∫ − a (a − x ) dx
1
2
a
= a2
x 3 dx = 60,
4
∫2
x dx = 6,
dx = 2
2
4
37.
∫4
38.
∫2
39.
∫2
40.
∫2
41.
∫ 2 ( x − 9) dx
42.
3
∫ 2 ( x + 4) dx = ∫ 2
43.
∫ 2 ( 12 x
y
1
4
∫2
x dx = − ∫ x dx = −6
2
2
x 3 dx = 0
4
8 x dx = 8∫
4
x dx = 8(6) = 48
4
2
25 dx = 25 ∫ dx = 25( 2) = 50
4
2
4
4
∫2
=
x dx − 9 ∫ dx = 6 − 9( 2) = −12
4
2
y
a
−a
4
4
x 3 dx + 4 ∫ dx = 60 + 4( 2) = 68
4
2
Triangle
a
x
4
3
)
− 3 x + 2 dx =
=
∫
4
1
2 2
1
2
4
4
2
2
x3 dx − 3 ∫ x dx + 2 ∫ dx
(60) − 3(6) + 2(2) = 16
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
456
44.
Chapter 5
Integration
∫ 2 (10 + 4 x − 3x ) dx = 10∫ 2
4
4
3
4
4
2
2
dx + 4 ∫ x dx − 3∫ x3 dx
51. (a) Quarter circle below x-axis:
− 14 π r 2 = − 14 π ( 2) = −π
2
= 10( 2) + 4(6) − 3(60) = −136
45. (a)
7
∫0
f ( x) dx =
5
∫0
f ( x) dx +
(b)
∫ 5 f ( x) dx
= −∫
(c)
∫ 5 f ( x) dx
= 0
(d)
∫ 0 3 f ( x) dx
46. (a)
0
5
5
∫5
f ( x) dx = 10 + 3 = 13
0
0
3
6
(b)
∫ 6 f ( x) dx
= −∫
(c)
∫ 3 f ( x) dx
= 0
(d)
∫3
47. (a)
∫2
3
3
6
6
f ( x ) dx = −(−1) = 1
6
3
− 5 f ( x ) dx = −5∫
6
3
⎣⎡ f ( x) + g ( x)⎤⎦ dx =
(b)
∫ 2 ⎡⎣g ( x) − f ( x)⎤⎦ dx
=
(c) Triangle + Semicircle below x -axis:
(e) Sum of absolute values of (b) and (c):
4 + (1 + 2π ) = 5 + 2π
(f ) Answers to (d) plus
2(10) = 20: (3 − 2π ) + 20 = 23 − 2π
52. (a)
6
(d)
∫ 2 3 f ( x) dx
0
∫ −1
4
= 3∫
f ( x ) dx =
∫ 2 f ( x) dx + ∫ 2 g ( x) dx
(d)
∫ 5 f ( x) dx
=
(e)
∫ 0 f ( x) dx
= − 12 + 2 + 2 + 2 − 4 +
(f )
∫ 4 f ( x) dx
= 2 − 4 = −2
6
6
∫ 2 g ( x) dx − ∫ 2 f ( x) dx
6
7
11
11
∫ −1
−
1
2
1
2
(4)(2) +
1
2
=5
= −3
1
2
= 2
6
10
y
2
1
∫0
(3, 2) (4, 2)
(11, 1)
1
x
−2
f ( x) dx −
1
3 f ( x) dx = 3( 2) = 6
∫ 0 f ( x) dx = − 12 + 12 (2)(2) + 2 + 12 (2)(2) − 12
f ( x) dx = 3(10) = 30
2
1
2
(c)
2
6
1
0
f ( x) dx = −5( −1) = 5
6
6
= − ∫ f ( x) dx =
1
2 g ( x) dx = 2∫ g ( x ) dx = 2( −2) = −4
∫2
48. (a)
∫ 0 − f ( x) dx
∫3
= −2 − 10 = −12
(c)
= 4
(b)
= 10 + ( −2) = 8
6
(4)(2)
(d) Sum of parts (b) and (c): 4 − (1 + 2π ) = 3 − 2π
∫ 0 f ( x) dx = ∫ 0 f ( x) dx + ∫ 3 f ( x) dx = 4 + (−1) = 3
6
1
2
2
f ( x ) dx = 3(10) = 30
5
(b) Triangle: 12 bh =
− 12 ( 2)(1) − 12 π ( 2) = −(1 + 2π )
f ( x ) dx = −10
5
= 3∫
7
2
−1
f ( x ) dx
4
8
12
(0, ⫺1)
−2
(8, ⫺2)
5
⎡⎣ f ( x) + 2⎤⎦ dx =
= 0 − 5 = −5
(b)
∫ 0 f ( x) dx − ∫ 1 f ( x) dx = 5 − (−5) = 10
(c)
∫ −1 3 f ( x) dx
= 3∫
∫ 0 3 f ( x) dx
= 3∫ f ( x ) dx = 3(5) = 15
(d)
1
1
1
0
1
−1
Upper estimate: [32 + 24 + 12 − 4 − 20]( 2) = 88
[−6 + 8 + 30](2)
= 64 left endpoint estimate
(b) [8 + 30 + 80]( 2) = 236 right endpoint estimate
(c)
[0 + 18 + 50](2)
= 136 midpoint estimate
If f is increasing, then (a) is below the actual value and
(b) is above.
∫ 0 f ( x) dx + ∫ 0
5
5
2 dx
(b)
∫ − 2 f ( x + 2) dx = ∫ 0 f ( x) dx = 4 (Let u
(c)
∫ − 5 f ( x) dx
= 2∫
0
(d)
∫ − 5 f ( x) dx
= 0
(f
1
0
∫0
= 4 + 10 = 14
f ( x ) dx = 3(0) = 0
49. Lower estimate: [24 + 12 − 4 − 20 − 36]( 2) = −48
50. (a)
53. (a)
3
5
5
5
5
= x + 2.)
f ( x) dx = 2( 4) = 8 ( f even )
odd)
54. (a) The left endpoint approximation will be greater than
the actual area so,
n
∑ f ( xi )∆x
>
i =1
∫ 1 f ( x) dx.
5
(b) The right endpoint approximation will be less than
the actual area so,
n
∑ f ( xi )∆x
i =1
<
∫ 1 f ( x) dx.
5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.3
⎧4, x < 4
55. f ( x) = ⎨
⎩ x, x ≥ 4
60.
f ( x ) dx = 4( 4) + 4( 4) +
8
∫0
1
2
(4)( 4)
Riemann Sums and Definite Integrals
2
∫1
ln x dx
y
= 40
1
y
1
(8, 8)
8
457
x
2
−1
4
(a) A ≈
1
square units
3
x
4
8
61. f ( x ) =
x > 6
⎧⎪6,
56. f ( x) = ⎨ 1
⎪⎩− 2 x + 9, x ≤ 6
1
x −4
is not integrable on the interval [3, 5] because f has a
discontinuity at x = 4.
y
62. f ( x) = x x is integrable on [−1, 1], but is not
12
10
continuous on [−1, 1]. There is discontinuity at
(0, 9)
8
x = 0. To see that
6
4
1
∫ −1
2
x
2
4
6
8
10
x
dx
x
12
is integrable, sketch a graph of the region bounded by
f ( x) = x x and the x-axis for −1 ≤ x ≤ 1. You see
∫ 0 f ( x) dx = 6(6) + (3) + 6(6) = 36 + 9 + 36 = 81
12
16
2
that the integral equals 0.
y
57.
y
2
4
1
3
2
x
−2
1
2
1
1
2
3
x
4
−2
(a) A ≈ 5 square units
58.
63.
y
∫ − 2 f ( x) dx + ∫ 1 f ( x) dx
1
5
∫ − 2 f ( x) dx
5
=
a = −2, b = 5
4
64.
3
∫ −3 f ( x) dx + ∫ 3 f ( x) dx − ∫ a f ( x) dx = ∫ −1 f ( x) dx
3
b
6
2
6
∫ −3 f ( x) dx + ∫ b f ( x) dx = ∫ −1 f ( x) dx
a
6
1
x
1
4
(b) A ≈
1
2
3
4
4
3
a = −3, b = −1
1
square units
65. Answers will vary. Sample answer: a = π , b = 2π
2π
y
59.
6
∫π
sin x dx < 0
y
1
1
1
2
3
(c) A ≈ 2 square units
x
π
2
3π
2
x
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
458
Chapter 5
Integration
66. Answers will vary. Sample answer: a = 0, b = π
π
∫0
cos x dx = 0
67. True
68. False
1
∫0 x
y
x dx ≠
(∫
1
0
x dx
)(∫
1
0
x dx
)
69. True
π
4
π
2
3π
4
70. True
x
π
71. False
−1
∫ 0 (− x) dx
2
= −2
72. True. The limits of integration are the same.
73. f ( x ) = x 2 + 3x, [0, 8]
x0 = 0, x1 = 1, x2 = 3, x3 = 7, x4 = 8
∆x1 = 1, ∆x2 = 2, ∆x3 = 4, ∆x4 = 1
c1 = 1, c2 = 2, c3 = 5, c4 = 8
4
∑ f (ci ) ∆x = f (1) ∆x1 + f (2) ∆x2 + f (5) ∆x3 + f (8) ∆x4
i =1
= ( 4)(1) + (10)( 2) + ( 40)( 4) + (88)(1) = 272
74. f ( x) = sin x, [0, 2π ]
x0 = 0, x1 =
π
∆x1 =
c1 =
4
4
π
6
4
, x2 =
, ∆x2 =
, c2 =
∑ f (ci ) ∆xi
i =1
π
π
3
π
12
π
3
, x3 = π , x4 = 2π
, ∆x3 =
, c3 =
2π
, ∆x4 = π
3
2π
3π
, c4 =
3
2
⎛π ⎞
⎛π ⎞
⎛ 2π ⎞
⎛ 3π ⎞
= f ⎜ ⎟ ∆x1 + f ⎜ ⎟ ∆x2 + f ⎜ ⎟ ∆x3 + f ⎜ ⎟ ∆x4
⎝6⎠
⎝3⎠
⎝ 3 ⎠
⎝ 2 ⎠
⎛ 1 ⎞⎛ π ⎞ ⎛ 3 ⎞⎛ π ⎞ ⎛ 3 ⎞⎛ 2π ⎞
= ⎜ ⎟⎜ ⎟ + ⎜⎜
⎟⎜ ⎟ + ⎜
⎟⎜ ⎟ + ( −1)(π ) ≈ −0.708
⎝ 2 ⎠⎝ 4 ⎠ ⎝ 2 ⎟⎠⎝ 12 ⎠ ⎜⎝ 2 ⎟⎠⎝ 3 ⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.3
b
459
b−a
⎛b − a⎞
, ci = a + i(∆x) = a + i⎜
⎟
n
⎝ n ⎠
75. ∆x =
∫0
Riemann Sums and Definite Integrals
n
∑ f (ci )∆x
∆ →0
x dx = lim
i =1
n
⎡
i =1
⎣
⎛ b − a ⎞⎤ ⎛ b − a ⎞
⎟ ⎜
⎟
n ⎠⎥⎦ ⎝ n ⎠
∑ ⎢a + i⎜⎝
n→∞
= lim
2
⎡⎛ b − a ⎞ n
⎛b − a⎞ n ⎤
= lim ⎢⎜
a
+
⎟∑
⎜
⎟ ∑ i⎥
n→∞ ⎝
⎝ n ⎠ i =1 ⎥⎦
⎢⎣ n ⎠ i =1
2
⎡b − a
⎛ b − a ⎞ n( n + 1) ⎤
an) + ⎜
= lim ⎢
⎥
(
⎟
n→∞
2
⎝ n ⎠
⎥⎦
⎣⎢ n
⎡
(b − a)2 n + 1⎤⎥
= lim ⎢a(b − a) +
n →∞⎢
2 ⎥
n
⎣
⎦
= a (b − a ) +
(b
− a)
2
2
b − a⎤
⎡
= (b − a ) ⎢a +
2 ⎥⎦
⎣
=
(b
− a )( a + b)
2
=
b2 − a 2
2
b−a
⎛b − a⎞
, ci = a + i( ∆x) = a + i⎜
⎟
n
⎝ n ⎠
76. ∆x =
b
∫a
n
∑ f (ci )∆x
∆ →0
x 2 dx = lim
i =1
= lim
n →∞
n
∑
i =1
2
⎡
⎛ b − a ⎞⎤ ⎛ b − a ⎞
⎢a + i⎜ n ⎟⎥ ⎜ n ⎟
⎝
⎠⎦ ⎝
⎠
⎣
2 ⎤
⎡⎛ b − a ⎞ n ⎛
⎞
2ai(b − a)
2
2⎛ b − a ⎞
a
i
= lim ⎢⎜
+
+
⎜
⎟⎥
∑
⎟
⎜
⎟
⎜
n →∞ ⎢⎝
n ⎠ i =1 ⎝
n
⎝ n ⎠ ⎟⎠⎥⎦
⎣
2
2a(b − a ) n( n + 1) ⎛ b − a ⎞ n( n + 1)( 2n + 1) ⎤
⎛ b − a ⎞⎡ 2
= lim ⎜
+⎜
⎢na +
⎥
⎟
⎟
n →∞ ⎝
n ⎠ ⎢⎣
n
2
6
⎝ n ⎠
⎦⎥
2
3
⎡
a(b − a ) ( n + 1) (b − a) ( n + 1)( 2n + 1) ⎤
= lim ⎢a 2 (b − a ) +
+
⎥
n →∞ ⎢
6
n
n2
⎥⎦
⎣
1
1
2
3
= a 2 (b − a) + a(b − a ) + (b − a ) = (b3 − a 3 )
3
3
⎧1, x is rational
77. f ( x) = ⎨
⎩0, x is irrational
is not integrable on the interval [0, 1]. As
∆ → 0, f (ci ) = 1 or f (ci ) = 0 in each subinterval
because there are an infinite number of both rational and
irrational numbers in any interval, no matter how small.
⎧0, x = 0
⎪
78. f ( x ) = ⎨ 1
⎪ , 0 < x ≤1
⎩x
y
1
2
f(x) = x
The limit
1
n
lim
∆ →0
∑
f (ci )∆xi
i =1
x
1
2
does not exist.
This does not contradict Theorem 5.4
because f is not continuous on [0, 1].
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
460
Chapter 5
Integration
79. The function f is nonnegative between x = −1 and
x = 1.
y
2
∫ a (1 − x ) dx
b
So,
2
∫0
80. To find
x dx, use a geometric approach.
y
2
f(x) = 1 − x 2
3
2
is a maximum for
x
−2
1
a = −1 and b = 1.
1
2
−1
x
−1
−2
1
2
3
−1
So,
x dx = 1( 2 − 1) = 1.
2
∫0
81. Let f ( x ) = x 2 , 0 ≤ x ≤ 1, and ∆xi = 1 n. The appropriate Riemann Sum is
n
∑ f (ci ) ∆xi
=
i =1
lim
n →∞
2
1 n
⎛i⎞ 1
= 3 ∑ i2.
n
n i =1
i =1
n
∑ ⎜⎝ n ⎟⎠
1 2
⎡1 + 22 + 32 +
n3 ⎣
82. I ( f ) − J ( f ) =
1 n( 2n + 1)( n + 1)
2n 2 + 3n + 1
1
1 ⎞
1
⎛1
+ n 2 ⎤⎦ = lim 3 ⋅
= lim
= lim ⎜ +
+
⎟ =
n →∞ n
n →∞
n →∞ ⎝ 3
6
6n 2
2n 6n 2 ⎠
3
2
2
∫ 0 x f ( x) dx − ∫ 0 xf ( x) dx.
1
1
Observe that
⎛
x3
x⎞
x3
x2 ⎞
x3
x3
2
2
2
⎛
− x⎜ f ( x ) − ⎟ =
− x⎜ f ( x) − xf ( x ) +
− xf ( x) + x 2 f ( x) −
= x 2 f ( x) − xf ( x)
⎟ =
4
2⎠
4
4⎠
4
4
⎝
⎝
2
So, I ( f ) − J ( f ) =
2
2
∫ 0 ⎡⎣x f ( x) − xf ( x) ⎤⎦ dx
1
Furthermore, 6 + f ( x) =
So I ( f ) − J ( f ) =
x
. Then I ( f ) =
2
=
1
∫0 x
1
⎡ x3
∫ 0 ⎢⎢ 4
⎣
2
x⎞ ⎤
⎛
− x⎜ f ( x) − ⎟ ⎥ dx ≤
2 ⎠ ⎦⎥
⎝
2⎛
x⎞
1
⎜ ⎟ dx = and J ( f ) =
8
⎝ 2⎠
1
1
∫0
x3
1
dx =
4
16
⎛ x2 ⎞
∫ 0 x⎜⎝ 4 ⎟⎠
=
1
16
1
1
1
−
=
8 16
16
The maximum value is
1
.
16
Section 5.4 The Fundamental Theorem of Calculus
1. f ( x ) =
π
∫0
4
x2 + 1
4
dx is positive.
x2 + 1
3. f ( x) = x
5
2
∫ −2 x
−5
x2 + 1
x 2 + 1 dx = 0
5
−5
5
5
−5
−2
4. f ( x ) = x 2 − x
5
2. f ( x ) = cos x
π
∫0
cos x dx = 0
2
∫ −2 x
2
2 − x dx is negative.
−2
0
2
␲
−5
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.4
5.
6.
7.
6 x dx = ⎡⎣3x 2 ⎤⎦ = 3( 2) − 0
0
2
2
∫0
2
∫ −3 8 dt
0
10.
2
∫ 1 (6 x
11.
∫ 0 (2t − 1)
12.
∫ 1 (4 x
13.
∫ 1 ⎜⎝ x 2
1
∫1
dt =
= 14 − 6 + 7 +
((−1)
∫ 0 (4t
1
2
)
9.
− (−1) = −(1 + 1) = −2
2
3
2
=
3
2
(−1)2 ⎤⎦
33
2
= ⎡⎣2(8) −
2
3 2⎤
x
2 ⎦1
2
∫ −1 (t − 2) dt
1
1
⎡t 3
⎤
= ⎢ − 2t ⎥
⎣3
⎦ −1
10
⎛1
⎞ ⎛ 1
⎞
= ⎜ − 2⎟ − ⎜ − + 2⎟ = −
3
⎝3
⎠ ⎝ 3
⎠
3
2
( 4)⎤⎦
− ⎡⎣2(1) −
− 4t + 1) dt = ⎡⎣ 43 t 3 − 2t 2 + t ⎤⎦ =
0
1
3
2
4
3
(1)⎤⎦
(
= (16 − 6) − 2 −
− 2+1=
3
2
) = 192
1
3
2
1
⎞
⎡ 3
⎤
⎛ 3
⎞
− 1⎟ dx = ⎢− − x⎥ = ⎜ − − 2 ⎟ − ( −3 − 1) =
x
2
2
⎠
⎣
⎦1
⎝
⎠
−1
12
∫ 1 (u
4
4
⎡2
⎤
⎡2
− 2u −1 2 ) du = ⎢ u 3 2 − 4u1 2 ⎥ = ⎢
3
⎣
⎦1
⎣3
( 4)
3
2
⎤ ⎡2
⎤
− 4 4 ⎥ − ⎢ − 4⎥ =
3
3
⎦ ⎣
⎦
8
16.
∫ −8
3
3
43
⎡3
⎤
x1 3 dx = ⎢ x 4 3 ⎥ = ⎡84 3 − ( − 8) ⎤ = (16 − 16) = 0
⎦
4⎣
4
⎣4
⎦ −8
17.
∫ −1(
3
18.
∫1
8
1
)
x −
x
19.
∫0
20.
∫ 0 (2 − t )
21.
∫ −1 (t
22.
∫ −8
3
2
0
−1
( 34 − 2) − ( 34 + 2) = −4
1
t − 2 dt = ⎡⎣ 34 t 4 3 − 2t ⎤⎦ =
−1
2
dx =
x
8
1
− ⎡7( −1) −
⎣
3
u −2
du =
u
4
( 4)⎤⎦
⎡u 2
1⎞
1⎤
1⎞
⎛
⎛1
⎞ ⎛
⎜ u − 2 ⎟ du = ⎢ + ⎥ = ⎜ − 1⎟ − ⎜ 2 − ⎟ = −2
u ⎠
2
u
2
2⎠
⎝
⎝
⎠
⎝
⎣
⎦ −2
−1
15.
2
3
2
− 3 x 2 ) dx = ⎡⎣ x 4 − x 3 ⎤⎦ = (81 − 27) − (1 − 1) = 54
1
3
⎛3
∫ −2
= ⎡⎣7t − 32 t 2 ⎤⎦
−1
0
− 3x ) dx = ⎡⎣2 x 3 −
2
14.
2
2
= ⎡⎣7( 2) −
= ⎡⎣ x 2 − x⎤⎦
−1
= 0−
2
∫ −1 (7 − 3t ) dt
1
∫ −1 (2 x − 1) dx
3
8.
= [8t ]− 3 = 8(1) − 8( − 3) = 32
1
461
The Fundamental Theorem of Calculus
13
8
8
2 ∫ x −1 2 dx = ⎡⎣ 2 ( 2) x1 2 ⎤⎦ = ⎡⎣2 2 x ⎤⎦ = 8 − 2 2
1
1
1
8
1
dx =
⎤
1 1
1 ⎡ x2
2
1⎛ 1 2 ⎞
1
x − x1 2 ) dx = ⎢ − x3 2 ⎥ = ⎜ − ⎟ = −
(
∫
0
3
3⎣ 2
3
3
2
3
18
⎝
⎠
⎦0
t dt =
12
32
∫ 0 (2t − t ) dt
2
− t 2 3 ) dt = ⎡⎣ 34 t 4 3 − 53 t 5 3 ⎤⎦ = 0 −
−1
0
2
2
⎡t t
⎤
2 ⎤
2 2
16 2
⎡4
= ⎢ t3 2 − t5 2⎥ = ⎢
(20 − 6t )⎥ =
(20 − 12) =
3
5
15
15
15
⎣
⎦0
⎣
⎦0
( 34 + 53 ) = − 2720
x − x2
1 −1
dx = ∫ ( x 2 3 − x 5 3 ) dx
2 −8
23 x
−1
=
−1
⎡ x5 3
⎤
1 ⎡3 5 3 3 8 3⎤
1
32
4569
x − x ⎥ = ⎢
(24 − 15 x)⎥ = − (39) + (144) =
⎢
2 ⎣5
8
80
80
80
⎦ −8
⎣ 80
⎦ −8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
462
23.
Chapter 5
5
∫0
Integration
2 x − 5 dx =
∫ 0 (5 − 2 x) dx + ∫ 5 2 (2 x − 5) dx (split up the integral at the zero x
52
5
52
= ⎡⎣5 x − x 2 ⎤⎦
0
24.
5
Note: By Symmetry,
∫0
∫ 1 (3 − 1x − 31) dx
=
∫1
=
∫1
4
5
52
(2 x
3
⎡⎣3 + ( x − 3)⎤⎦ dx +
3
x dx +
5
2
)
( 252 − 254 ) − 0 + (25 − 25) − ( 254 − 252 ) = 2( 252 − 254 ) =
5
+ ⎡⎣ x 2 − 5 x⎤⎦
=
52
2 x − 5 dx = 2 ∫
=
25
2
− 5) dx.
4
∫3
⎡⎣3 − ( x − 3)⎤⎦ dx
∫ 3 (6 − x)dx
4
3
4
⎡ x2 ⎤
⎡
x2 ⎤
= ⎢ ⎥ + ⎢6 x − ⎥
2 ⎦3
⎣ 2 ⎦1 ⎣
9 ⎞⎤
⎛9 1⎞ ⎡
⎛
= ⎜ − ⎟ + ⎢( 24 − 8) − ⎜18 − ⎟⎥
2 ⎠⎦
⎝ 2 2⎠ ⎣
⎝
9
13
= 4 + 16 − 18 +
=
2
2
25.
4
∫0
x 2 − 9 dx =
2
2
∫ 0 (9 − x ) dx + ∫ 3 ( x
3
4
− 9) dx (split up integral at the zero x = 3)
3
4
− 4 x + 3) dx −
∫1 ( x
⎡
⎡ x3
⎤
x3 ⎤
64
⎛ 64
⎞
= ⎢9 x − ⎥ + ⎢ − 9 x⎥ = ( 27 − 9) + ⎜
− 36 ⎟ − (9 − 27) =
3 ⎦0 ⎣ 3
3
⎝ 3
⎠
⎣
⎦3
26.
4
∫0
x 2 − 4 x + 3 dx =
∫ 0 (x
1
2
3
2
− 4 x + 3) dx +
1
3
∫ 3 (x
4
2
− 4 x + 3) dx (split up the integral at the zeros x = 1, 3)
4
⎡ x3
⎤
⎡ x3
⎤
⎡ x3
⎤
= ⎢ − 2 x 2 + 3x⎥ − ⎢ − 2 x 2 + 3 x⎥ + ⎢ − 2 x 2 + 3 x⎥
⎣3
⎦0 ⎣ 3
⎦1 ⎣ 3
⎦3
⎛1
⎞
⎛1
⎞ ⎛ 64
⎞
= ⎜ − 2 + 3⎟ − (9 − 18 + 9) + ⎜ − 2 + 3⎟ + ⎜ − 32 + 12 ⎟ − (9 − 18 + 9)
⎝3
⎠
⎝3
⎠ ⎝ 3
⎠
4
4 4
= −0+ + −0 = 4
3
3 3
π
27.
π
∫ 0 (1 + sin x) dx = [ x − cos x]0
28.
π
∫ 0 (2 + cos x) dx = [2 x + sin x]0
29.
∫0
= (π + 1) − (0 − 1) = 2 + π
π
π 4
π 4
1 − sin 2 θ
dθ =
cos 2 θ
sec 2 θ
dθ =
tan 2 θ + 1
π 4
∫0
= ( 2π + 0) − 0 = 2π
π 4
dθ = [θ ]0
π 4
sec 2 θ
dθ =
sec 2 θ
30.
∫0
31.
∫ −π 6 sec
32.
π 2
2
∫ π 4 (2 − csc x) dx = [2 x + cot x]π 4
33.
∫ −π 3 4 sec θ tan θ dθ
π 6
2
∫0
π 6
x dx = [tan x] −π
6
=
π
4
π 4
∫0
π 4
dθ = [θ ]0
=
π
4
3 ⎛
3⎞
2 3
− ⎜⎜ −
⎟⎟ =
3
3
3
⎝
⎠
π 2
π 3
=
π 3
= [4 sec θ ]−π
3
π
π −2
⎛π
⎞
= (π + 0) − ⎜ + 1⎟ =
−1 =
2
2
⎝2
⎠
= 4( 2) − 4( 2) = 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.4
34.
π 2
∫ −π 2 (2t + cos t ) dt
463
⎛π 2
⎞ ⎛π 2
⎞
= ⎜
+ 1⎟ − ⎜
− 1⎟ = 2
4
4
⎝
⎠ ⎝
⎠
π 2
= ⎡⎣t 2 + sin t ⎤⎦
−π
The Fundamental Theorem of Calculus
2
2
35.
⎡ x
⎤
⎛
⎞ ⎛
⎞
(2 x + 6) dx = ⎢ ln2 2 + 6 x⎥ = ⎜ ln42 + 12 ⎟ − ⎜ ln1 2 + 0⎟ = ln32 + 12
⎝
⎠ ⎝
⎠
⎣
⎦0
2
∫0
3
3
t
⎡ 2
⎤
⎛
⎞ ⎛
1 ⎞
9 124
−
≈ − 72.546
(t − 5t ) dt = ⎢t2 − ln5 5 ⎥ = ⎜ 92 − 125
⎟ − ⎜0 −
⎟ =
ln
5
ln
5
2 ln 5
⎝
⎠ ⎝
⎠
⎣
⎦0
36.
∫0
37.
∫ −1 (e
38.
∫e
1
2e
1
1
+ sin θ ) dθ = ⎡⎣eθ − cos θ ⎤⎦ = (e − cos 1) − ⎡⎣e −1 − cos( −1)⎤⎦ = e −
−1
e
θ
1⎞
2e
⎛
⎜ cos x − ⎟ dx = [sin x − ln x]e = ⎡⎣sin ( 2e) − ln ( 2e)⎤⎦ − ⎡⎣sin (e) − ln (e)⎤⎦ = sin ( 2e) − sin (e) − ln 2
x
⎝
⎠
1
1
39. A =
∫0
40. A =
2
∫1
41. A =
∫0
42. A =
3
⎡ 2
⎤
( x − x 2 ) dx = ⎢ x2 − x3 ⎥ = 16
⎣
⎦0
2
1
1
⎡ 1⎤
dx = ⎢− ⎥ =
x2
2
⎣ x ⎦1
π 2
π 2
cos x dx = [sin x]0
π
∫ 0 (x
=1
π
⎡ x2
⎤
π2
π2 + 4
+ sin x ) dx = ⎢ − cos x⎥ =
+ 2 =
2
2
⎣2
⎦0
43. Because y > 0 on [0, 2],
Area =
2
∫ 0 (5 x
2
+ 2) dx = ⎡⎣ 53 x 3 + 2 x⎤⎦ =
0
2
40
3
+ 4 =
52 .
3
44. Because y > 0 on [0, 2],
Area =
3
∫ 0 ( x + x) dx
2
2
⎡ x4
x2 ⎤
= ⎢ +
⎥ = 4 + 2 = 6.
2 ⎦0
⎣4
45. Because y > 0 on [0, 8],
Area =
13
∫ 0 (1 + x ) dx
8
8
3
3
⎡
⎤
= ⎢ x + x 4 3 ⎥ = 8 + (16) = 20.
4
4
⎣
⎦0
46. Because y > 0 on [0, 4],
Area =
2
∫ 0 (− x + 4 x) dx
4
4
⎡ x3
⎤
64
32
= ⎢−
+ 2x2 ⎥ = −
+ 32 =
.
3
3
⎣ 3
⎦0
47. Because y > 0 on [1, e],
Area =
e
∫1
4
e
dx = [4 ln x] 1 = 4 ln e − 4 ln 1 = 4.
x
48. Because y > 0 on [0, 2],
Area =
2
∫0
2
e x dx = ⎣⎡e x ⎦⎤ = e 2 − e 0 = e 2 − 1.
0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
464
Chapter 5
Integration
3
49.
3
∫0
⎡ x4 ⎤
81
x 3 dx = ⎢ ⎥ =
4
4
⎣ ⎦0
53.
π 4
∫ −π 4 2 sec
51.
9
∫4
9
x dx = ⎡⎣ 23 x 3 2 ⎤⎦ =
4
f (c)(9 − 4) =
38
3
f (c) =
38
15
c =
38
15
c =
1444
225
4
∫1
2
3
(27
− 8) =
π
54.
π 3
π 3
∫ −π 3 cos x dx = [sin x]−π 3
⎡π
⎛ π ⎞⎤
f (c) ⎢ − ⎜ − ⎟⎥ =
⎣ 3 ⎝ 3 ⎠⎦
≈ 6.4178
≈ ± 0.4817
3
3
3 3
2π
c ≈ ±0.5971
55.
(9 − x 2 ) dx
3 − ( −3) ∫ − 3
1
3
3
=
1⎡
1 ⎤
9 x − x3 ⎥
6 ⎢⎣
3 ⎦ −3
1
⎡( 27 − 9) − ( −27 + 9)⎤⎦
6⎣
= 6
=
Average value = 6
9 − x 2 = 6 when x 2 = 9 − 6 or x = ± 3 ≈ ±1.7321.
10
(−
3, 6 )
(
3, 6 )
3
⎡
2x ⎤
= ⎢10 x −
⎥
ln 2 ⎦ 0
⎣
f (c)(3 − 0) = 30 −
7
ln 2
(10 − 2c )(3) = 30 −
7
ln 2
)
=
π
2
cos c =
⎛
8 ⎞ ⎛
1 ⎞
= ⎜ 30 −
⎟ − ⎜0 −
⎟
ln
2
ln
2⎠
⎝
⎠ ⎝
7
= 30 −
ln 2
3( 2
π
= ± arccos
1⎞
⎛
⎜ 5 − ⎟(3) = 15 − ln 4
c⎠
⎝
3
= 15 − ln 4
15 −
c
3
= ln 4
c
3
≈ 2.1640
c =
ln 4
c
2
⎛ 2 ⎞
c = ± arcsec⎜
⎟
⎝ π⎠
38
3
f (c)( 4 − 1) = 15 − ln 4
x
∫ 0 (10 − 2 ) dx
π
sec c = ±
1⎞
4
⎛
⎜ 5 − ⎟ dx = [5 x − ln x]1
x
⎝
⎠
3
4
sec 2 c =
= ( 20 − ln 4) − (5 − 0) = 15 − ln 4
52.
= 2(1) − 2( −1) = 4
4
⎡π
⎛ π ⎞⎤
f (c) ⎢ − ⎜ − ⎟⎥ = 4
⎣ 4 ⎝ 4 ⎠⎦
8
2 sec 2 c =
81
4
27
f (c) =
4
27
c3 =
4
3
3
c = 3
= 3 2 ≈ 1.8899
2
4
f (c)(3 − 0) =
50.
π 4
x dx = [2 tan x]−π
2
−4
4
0
56.
2
3 4( x + 1)
3
1
dx = 2 ∫ (1 + x −2 ) dx
∫
2
1
1
3−1
x
3
1⎤
⎡
= 2⎢x − ⎥
x ⎦1
⎣
1 ⎞ 16
⎛
= 2⎜ 3 − ⎟ =
3⎠
3
⎝
10
(
7
=
ln 2
2c =
7
3 ln 2
⎛ 7 ⎞
c = log 2 ⎜
⎟ ≈ 1.7512
⎝ 3 ln 2 ⎠
3, 16
3
)
4
0
0
Average value =
4( x 2 + 1)
x2
=
16
3
16
⇒ x =
3
3 (on [1, 3] )
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.4
57.
1
1
2e
1 − ( −1) ∫ −1
x
dx =
1
∫ −1 e
x
5
curve from 0 ≤ t ≤ 5 is approximately
(29 squares) (5) = 145 ft.
1
= ⎡⎣e x ⎤⎦ = e − e −1 ≈ 2.3504
−1
Average value = e − e −1 ≈ 2.3504
63. (a)
2e x = e − e −1
ex =
∫ 0 v(t ) dt. The area under the
62. The distance traveled is
dx
465
The Fundamental Theorem of Calculus
∫ 1 f ( x) dx = Sum of the areas
7
= A1 + A2 + A3 + A4
1
(e − e−1 )
2
=
1
2
(3 + 1) + 12 (1 + 2) + 12 (2 + 1) + (3)(1)
=8
⎛ e − e −1 ⎞
x = ln ⎜
⎟ ≈ 0.1614
⎝ 2 ⎠
y
5
4
A1
3
(0.16, 2.35)
A2
A3
A4
2
1
−1
1
−1
x
1
2
3
4
5
7
6
4
58.
4 1
1
1
⎡1
⎤
dx = ⎢ ln x⎥ = ln 4 ≈ 0.2310
∫
1
4 −1
2x
6
⎣6
⎦1
Average value =
7
(b) Average value =
1
ln 4 ≈ 0.2310
6
1
1
= ln 4
2x
6
6
2x =
ln 4
∫ 1 f ( x) dx
7 −1
=
8
4
=
6
3
(c) A = 8 + (6)( 2) = 20
20
10
=
6
3
Average value =
1
y
7
(2.16, 0.23)
3
x =
≈ 2.1640
ln 4
0
6
5
4
0
4
3
π
59.
π
1
2
⎡ 1
⎤
sin x dx = ⎢− cos x⎥ =
π − 0 ∫0
π
⎣ π
⎦0
2
2
Average value =
(0.690, π2 (
π
sin x =
2
2
=
2
0
x ≈ 0.881
5
6
7
dog from year 2 to year 6.
65. (a) F ( x) = k sec 2 x
2
F (0) = k = 500
π
F ( x) = 500 sec 2 x
(b)
2.71
−0.5
1
π 3
π 3 − 0∫0
500 sec 2 x dx =
1500
=
1500
π
π
[tan x]π0 3
(
3 −0
)
≈ 826.99 newtons
∫ 0 v(t ) dt. The area under the
8
curve from 0 ≤ t ≤ 8 is approximately
(18 squares) (30) ≈ 540 ft.
4
6
(0.881, π2 (
π
3
∫ 2 r′(t ) dt represents the net change in the weight of the
1.5
π
2
64. r (t ) represents the weight in pounds of the dog at time t.
(2.451, π2 (
−1
π 2
61. The distance traveled is
x
1
3␲
2
π 2
1
⎡2
⎤
cos x dx = ⎢ sin x⎥
60.
∫
0
2
0
π
π
−
( )
⎣
⎦0
cos x =
1
−␲
2
π
x ≈ 0.690, 2.451
Average value =
2
≈ 827 newtons
R
66.
R
1
k⎡ 2
r3 ⎤
2kR 2
2
2
k
R
−
r
dr
=
R
r
−
=
(
)
⎢
⎥
R − 0∫0
R⎣
3 ⎦0
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
466
67.
Chapter 5
Integration
5
5
1
(0.1729t + 0.1522t 2 − 0.0374t 3 ) dt ≈ 15 ⎡⎣0.08645t 2 + 0.05073t 3 − 0.00935t 4 ⎤⎦ 0 ≈ 0.5318 liter
5 − 0∫0
68. (a)
1
0
24
−1
The area above the x-axis equals the area below the x-axis. So, the average value is zero.
(b)
10
0
24
0
The average value of S appears to be g.
69. (a) v = −0.00086t 3 + 0.0782t 2 − 0.208t + 0.10
(b)
90
− 10
70
− 10
60
(c)
60
∫0
⎡ −0.00086t 4
⎤
0.0782t 3
0.208t 2
+
−
+ 0.10t ⎥ ≈ 2476 meters
v(t ) dt = ⎢
4
3
2
⎣
⎦0
70. (a) Because y < 0 on [0, 2],
∫ 0 f ( x) dx
2
(b)
∫ 2 f ( x) dx = (area of region B)
=
(c)
∫ 0 f ( x)
6
(d)
∫0
(e)
∫0
6
6
2
6
dx = − ∫
− 2 f ( x) dx = −2 ∫
⎡⎣2 + f ( x)⎤⎦ dx =
(f ) Average value =
71. F ( x) =
f ( x) dx +
2
0
6
∫0
6
1
6 0
∫ 0 (4t − 7) dt
x
∫
2
0
= −(area of region A) = −1.5.
∫ 0 f ( x) dx − ∫ 0 f ( x) dx
6
2
∫ 2 f ( x) dx
= 3.5 − (−1.5) = 5.0
= 1.5 + 5.0 = 6.5
f ( x) dx = −2( −1.5) = 3.0
2 dx +
∫ 0 f ( x) dx
f ( x) dx =
6
1
6
(3.5)
= 12 + 3.5 = 15.5
= 0.5833
x
= ⎡⎣2t 2 − 7t ⎤⎦ = 2 x 2 − 7 x
0
F ( 2) = 2( 22 ) − 7( 2) = −6
F (5) = 2(52 ) − 7(5) = 15
F (8) = 2(82 ) − 7(8) = 72
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.4
72. F ( x) =
3
∫ 2 (t + 2t − 2) dt
x
x
F (5) =
625
+ 25 − 10 − 4 = 167.25
4
F (8) =
84
+ 64 − 16 − 4 = 1068
4
x
∫1
20
dv =
v2
x
∫1
3
∫ 2 (t
2
x
20v −2 dv = −
20 ⎤
v ⎥⎦1
20
1⎞
⎛
= −
+ 20 = 20⎜1 − ⎟
x
x⎠
⎝
+ 2t − 2) dt = 0⎤
⎥⎦
77. g ( x) =
74. F ( x) =
x
∫2
1
2
= 0
g ( 2) =
∫ 0 f (t ) dt
≈ 4+ 2+1 = 7
g ( 4) =
∫ 0 f (t ) dt
≈ 7+ 2 = 9
g ( 6) =
∫ 0 f (t ) dt
≈ 9 + ( −1) = 8
g (8) =
∫ 0 f (t ) dt
≈ 8−3 = 5
x
F ( 2) =
1 1
−
= 0
4 4
F (5) =
1
1
21
−
= −
= −0.21
25 4
100
F (8) =
1
1
15
−
= −
64 4
64
x
∫ 0 f (t ) dt
= 10
x
2
1⎤
1
1
− 3 dt = − ∫ 2t −3 dt = 2 ⎥ = 2 −
2
t
t ⎦2
x
4
∫ 0 f (t ) dt
(a) g (0) =
()
F (5) = 20( 54 ) = 16
F (8) = 20( 78 ) = 35
2
F ( 2) = 20
467
⎡t 4
⎤
⎛ x4
⎞
x4
= ⎢ + t 2 − 2t ⎥ = ⎜
+ x 2 − 2 x ⎟ − ( 4 + 4 − 4) =
+ x2 − 2 x − 4
4
⎣4
⎦2
⎝4
⎠
F ( 2) = 4 + 4 − 4 − 4 = 0 ⎡Note: F ( 2) =
⎢⎣
73. F ( x) =
The Fundamental Theorem of Calculus
0
2
4
6
8
(b) g increasing on (0, 4) and decreasing on (4, 8)
(c) g is a maximum of 9 at x = 4.
(d)
y
10
8
6
4
2
75 F ( x) =
x
∫1
x
x
cos θ dθ = sin θ ⎤ = sin x − sin 1
⎥⎦1
2
78. g ( x) =
F ( 2) = sin 2 − sin 1 ≈ 0.0678
F (5) = sin 5 − sin 1 ≈ −1.8004
x
∫0
∫ 0 f (t ) dt
x
= 0
g ( 2) =
∫ 0 f (t ) dt
= − 12 ( 2)( 4) = −4
g ( 4) =
∫ 0 f (t ) dt
= − 12 ( 4)( 4) = −8
g ( 6) =
∫ 0 f (t ) dt
= −8 + 2 + 4 = − 2
g (8) =
∫ 0 f (t ) dt
= −2 + 6 = 4
x
= −cos x + cos 0
= 1 − cos x
F ( 2) = 1 − cos 2 ≈ 1.4161
F (8) = 1 − cos 8 ≈ 1.1455
8
∫ 0 f (t ) dt
sin θ dθ = −cos θ ⎤
⎥⎦ 0
F (5) = 1 − cos 5 ≈ 0.7163
6
(a) g (0) =
F (8) = sin 8 − sin 1 ≈ 0.1479
76. F ( x) =
4
0
2
4
6
8
(b) g decreasing on (0, 4) and increasing on (4, 8)
(c) g is a minimum of −8 at x = 4.
(d)
y
4
2
−2
x
−2
2
4
8
10
−4
−6
−8
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
468
Chapter 5
79. (a)
∫ 0 (t
x
Integration
x
⎡t 2
⎤
1
+ 2) dt = ⎢ + 2t ⎥ = x 2 + 2 x
2
2
⎣
⎦0
F ′( x) =
d ⎡1 2
⎤
x + 2 x⎥ = x + 2
(b)
dx ⎢⎣ 2
⎦
80. (a)
x
∫0
t (t + 1)dt =
2
∫ 0 (t
x
89. F ( x) =
+ t ) dt
3
x2 2
1
1
= x4 + x2 =
( x + 2)
4
2
4
81. (a)
d ⎡1 4
1 ⎤
x + x 2 ⎥ = x 3 + x = x( x 2 + 1)
dx ⎢⎣ 4
2 ⎦
x
∫8
x
3
3
3
⎡3 ⎤
t dt = ⎢ t 4 3 ⎥ = ( x 4 3 − 16) = x 4 3 − 12
4
4
⎣ 4 ⎦8
d ⎡3 4 3
⎤
(b)
x − 12⎥ = x1 3 =
dx ⎢⎣ 4
⎦
82. (a)
x
∫4
3
x
x
(b)
84. (a)
(b)
∫π 4
85. (a) F ( x) =
(b)
4
= tan x − 1
x
∫1
= [sec t ]π
87. F ( x ) =
∫ − 2 (t
x
2
3
= sec x − 2
92. F ( x) =
t
F ( x) =
x
∫0
t cos t dt
x
∫0
sec3 t dt
x+2
∫ x (4t + 1) dt
x
x+2
∫ x (4t + 1) dt
x
−1
x+2
∫ x (4t + 1) dt + ∫ 0 (4t + 1) dt
0
x
0
( 4t
+ 1) dt +
x+2
∫ 0 (4t + 1) dt
F ′( x ) = −( 4 x + 1) + 4( x + 2) + 1 = 8
94. F ( x ) =
= e ⎤⎦ = e − e
−1
x
x
∫−x
⎡t 4 ⎤
t 3 dt = ⎢ ⎥ = 0
⎣ 4 ⎦−x
F ′( x ) = 0
Alternate solution:
F ( x) =
∫−x t
=
∫−x t
x
1
⎤
dt = ln t ⎥ = ln x
t
⎦1
− 2t ) dt
t dt
x
= −∫
x
d
1
(ln x) =
dx
x
4
4
F ′( x ) = x cos x
=
d x
(e − e−1 ) = e x
dx
86. (a) F ( x) =
(b)
∫ −1 e dt
x
∫1
Alternate solution:
d
[sec x − 2] = sec x tan x
dx
t
x4 + 1
F ′( x ) = 8
x
d
[tan x − 1] = sec2 x
dx
x
91. F ( x) =
t 4 + 1 dt
2
= ⎡2( x + 2) + ( x + 2)⎤ − ⎡⎣2 x 2 + x⎤⎦
⎣
⎦
= 8 x + 10
x
x
x
∫ −1
x+2
sec 2 t dt = [tan t ]π
∫ π 3 sec t tan t dt
x2
x +1
2
= ⎡⎣2t 2 + t ⎤⎦
x
d ⎡ 2 3 2 16 ⎤
x − ⎥ = x1 2 =
dx ⎢⎣ 3
3⎦
x
F ′( x) =
93. F ( x ) =
⎡2 ⎤
t dt = ⎢ t 3 2 ⎥
⎣3 ⎦4
2 3 2 16
x −
3
3
2 32
= ( x − 8)
3
83. (a)
2
F ′( x ) = sec3 x
=
(b)
90. F ( x) =
t2
dt
t +1
x
∫1
F ′( x ) =
x
1 ⎤
⎡1
= ⎢ t4 + t2⎥
2 ⎦0
⎣4
(b)
88. F ( x) =
x
0
= −∫
3
dt
3
dt +
−x
0
x
∫0
t 3 dt +
t 3 dt
x
∫0
t 3 dt
F ′( x) = −( − x) ( −1) + ( x3 ) = 0
3
F ′( x ) = x 2 − 2 x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.4
95. F ( x) =
sin x
sin x
∫0
t dt = ⎡⎣ 23 t 3 2 ⎤⎦
0
=
2
3
(sin x)3 2
The Fundamental Theorem of Calculus
97. F ( x) =
F ′( x ) = (sin x) cos x = cos x sin x
x3
∫0
469
sin t 2 dt
F ′( x) = sin ( x 3 ) ⋅ 3 x 2 = 3 x 2 sin x 6
12
2
Alternate solution:
F ( x) =
sin x
∫0
F ′( x ) =
96. F ( x) =
sin x
x2
∫2
98. F ( x) =
t dt
d
(sin x) =
dx
t −3
2
99. g ( x) =
x2
⎡ t −2 ⎤
−1
1
⎡ 1 ⎤
+
dt = ⎢ ⎥ = ⎢− 2 ⎥ =
4
2
2
t
2
x
8
−
⎣
⎦2
⎣ ⎦2
x2
1,
2
g ( 2) ≈ 1, g (3) ≈
1,
2
g ( 4) = 0
y
2
t −3 dt
1
F ′( x) = ( x 2 )
x
g has a relative maximum at x = 2.
Alternate solution:
∫2
∫ 0 f (t ) dt
g (0) = 0, g (1) ≈
F ′( x ) = 2 x −5
F ( x) =
sin θ 2 dθ
F ′( x) = sin ( x 2 ) ( 2 x) = 2 x sin x 4
sin x (cos x)
x2
x2
∫0
−3
( 2 x)
= 2 x −5
f
g
x
1
2
3
4
−1
−2
4
t2
lim g (t ) = 4
100. (a) g (t ) = 4 −
t →∞
Horizontal asymptote: y = 4
(b) A( x) =
x
∫1
4( x − 1)
4⎞
4⎤
4
4x2 − 8x + 4
⎛
⎡
=
⎜ 4 − 2 ⎟ dt = ⎢4t + ⎥ = 4 x + − 8 =
t ⎠
t ⎦1
x
x
x
⎝
⎣
x
2
4
⎛
⎞
lim A( x ) = lim ⎜ 4 x + − 8⎟ = ∞ + 0 − 8 = ∞
x →∞ ⎝
x
⎠
x →∞
The graph of A( x) does not have a horizontal asymptote.
101. (a) v(t ) = 5t − 7, 0 ≤ t ≤ 3
Displacement =
∫ 0 (5t
3
3
⎡ 5t 2
⎤
45
3
− 7) dt = ⎢
− 7t ⎥ =
− 21 =
ft to the right
2
2
⎣ 2
⎦0
(b) Total distance traveled =
=
3
∫0
5t − 7 dt
∫ 0 (7 − 5t ) dt + ∫ 7 5 (5t − 7) dt
75
3
75
⎡
5t 2 ⎤
= ⎢7t −
⎥
2 ⎦0
⎣
3
⎡ 5t 2
⎤
+ ⎢
− 7t ⎥
2
⎣
⎦7 5
2
2
⎛ 7 ⎞ 5⎛ 7 ⎞
⎛5
⎞ ⎛ 5⎛ 7 ⎞
⎛ 7 ⎞⎞
= 7⎜ ⎟ − ⎜ ⎟ + ⎜ (9) − 21⎟ − ⎜ ⎜ ⎟ − 7⎜ ⎟ ⎟
⎜
5
2
2
2
2
5
⎝ ⎠
⎝ ⎠
⎝
⎠ ⎝ ⎝ ⎠
⎝ 5 ⎠ ⎟⎠
49 49 45
49 49
113
ft
=
−
+
− 21 −
+
=
5
10
2
10
5
10
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
470
Chapter 5
Integration
102. (a) v(t ) = t 2 − t − 12 = (t − 4)(t + 3), 1 ≤ t ≤ 5
Displacement =
2
∫ 1 (t
− t − 12) dt
5
5
⎡t 3
⎤
t2
56 ⎛ 56
⎛ 125 25
⎞ ⎛1 1
⎞
⎞
= ⎢ −
− 12t ⎥ = ⎜
−
− 60 ⎟ − ⎜ − − 12 ⎟ = −
ft to the left ⎟
⎜
3
2
3
2
3
2
3
3
⎝
⎠ ⎝
⎠
⎝
⎠
⎣
⎦1
(b) Total distance traveled =
2
∫ 1 ( −t
+ t + 12) dt +
4
∫ 4 (t
5
2
− t − 12) dt
4
5
⎡ t3
⎤
⎡t 3
⎤
t2
t2
= ⎢− +
+ 12t ⎥ + ⎢ −
− 12t ⎥
2
2
⎣ 3
⎦1
⎣3
⎦4
⎛ 64
⎞ ⎛ 1 1
⎞ ⎛ 125 25
⎞ ⎛ 64
⎞
= ⎜−
+ 8 + 48 ⎟ − ⎜ − + + 12 ⎟ + ⎜
−
− 60 ⎟ − ⎜
− 8 − 48 ⎟
2
⎝ 3
⎠ ⎝ 3 2
⎠ ⎝ 3
⎠ ⎝ 3
⎠
=
104 73 ⎛ 185 ⎞ ⎛ 104 ⎞
79
−
+ ⎜−
ft
⎟ − ⎜−
⎟ =
3
6
3
⎝ 6 ⎠ ⎝ 3 ⎠
103. (a) v(t ) = t 3 − 10t 2 + 27t − 18 = (t − 1)(t − 3)(t − 6), 1 ≤ t ≤ 7
Displacement =
3
∫1 (t
7
− 10t 2 + 27t − 18) dt
7
⎡ t 4 10t 3
⎤
27t 2
= ⎢ −
+
− 18t ⎥
4
3
2
⎣
⎦1
⎡ 7 4 10(73 ) 27(7 2 )
⎤ ⎡ 1 10 27
⎤
= ⎢ −
+
− 18(7)⎥ − ⎢ −
+
− 18⎥
4
3
2
4
3
2
⎢⎣
⎥⎦ ⎣
⎦
= −
91 ⎛ 91 ⎞
− ⎜− ⎟ = 0
12 ⎝ 12 ⎠
(b) Total distance traveled =
=
∫ 1 v (t )
7
3
∫1 (t
3
dt
− 10t 2 + 27t − 18) dt −
3
∫ 3 (t
6
− 10t 2 + 27t − 18) dt +
3
∫ 6 (t
7
− 10t 2 + 27t − 18) dt
Evaluating each of these integrals, you obtain
Total distance =
16
3
( )
− − 63
+
4
125
12
=
63
2
ft
104. (a) v(t ) = t 3 − 8t 2 + 15t = t (t − 3)(t − 5), 0 ≤ t ≤ 5
Displacement =
3
∫ 0 (t
5
− 8t 2 + 15t ) dt
5
⎡t 4
8t 3 15t 2 ⎤
= ⎢ −
+
⎥
3
2 ⎦0
⎣4
=
625 8(125) 375
125
−
+
=
ft to the right
4
3
2
12
(b) Total distance traveled =
=
∫ 0 v (t )
5
3
∫ 0 (t
3
dt
− 8t 2 + 15t ) dt −
3
∫ 3 (t
5
− 8t 2 + 15t ) dt
Evaluating each of these integrals, you obtain
Total distance =
63 ⎛ 16 ⎞
253
− ⎜− ⎟ =
≈ 21.08 ft
4
12
⎝ 3⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.4
105. (a) v(t ) =
1
t
The Fundamental Theorem of Calculus
471
,1≤ t ≤ 4
Because v(t ) > 0,
Displacement = Total Distance
4
∫1
Displacement =
4
t −1 2 dt = ⎡⎣2t1 2 ⎤⎦ = 4 − 2 = 2 ft to the right
1
(b) Total distance = 2 ft
106. (a) v(t ) = cos t , 0 ≤ t ≤ 3π
3π
Displacement =
∫0
(b) Total distance =
∫0
3π
cos t dt = [sin t ]0 = 0 ft
π 2
cos t dt −
π 2
= [sin t ]0
3π 2
∫π 2
3π 2
− [sin t ]π
2
cos t dt +
5π 2
∫ 3π 2
5π 2
+ [sin t ]3π
2
cos t dt −
3π
− [sin t]5π
2
3π
∫ 5π 2 cos t dt
= 1 − ( −2) + 2 − (−1) = 6
107. x(t ) = t 3 − 6t 2 + 9t − 2
x′(t ) = 3t 2 − 12t + 9 = 3(t 2 − 4t + 3) = 3(t − 3)(t − 1)
Total distance =
=
∫ 0 x′(t )
5
dt
∫ 0 3 (t − 3)(t − 1)
5
= 3∫
1
0
dt
(t 2 − 4t + 3) dt − 3 ∫ 1 (t 2 − 4t + 3) dt + 3 ∫ 3 (t 2 − 4t + 3) dt
3
5
= 4 + 4 + 20 = 28 units
108. x(t ) = (t − 1)(t − 3) = t 3 − 7t 2 + 15t − 9
2
x′(t ) = 3t 2 − 14t + 15
Using a graphing utility,
Total distance =
∫ 0 x′(t )
5
dt ≈ 27.37 units.
109. Let c(t) be the amount of water that is flowing out of the tank. Then c′(t ) = 500 − 5t L min is the rate of flow.
18
∫0
c′(t )dt =
18
18
∫0
⎡
5t 2 ⎤
(500 − 5t ) dt = ⎢500t − ⎥ = 9000 − 810 = 8190 L
2 ⎦0
⎣
110. Let c(t ) be the amount of oil leaking and t = 0 represent 1 p.m. Then c′(t ) = 4 + 0.75t gal min is the rate of flow.
(a) From 1 p.m. to 4 p.m. (3 hours):
∫ 0 (4 +
3
3
0.75 2 ⎤
123
⎡
0.75t ) dt = ⎢4t +
t
=
= 15.375 gal
2 ⎥⎦ 0
8
⎣
(b) From 4 p.m. to 7 p.m. (3 hours)
∫ 3 (4 + 0.75t ) dt
6
6
0.75 2 ⎤
⎡
= ⎢4t +
t
= 22.125 gal
2 ⎥⎦ 3
⎣
(c) The second answer is larger because the rate of flow is increasing.
111. The function f ( x) = x −2 is not continuous on [−1, 1].
1
∫ −1 x
−2
dx =
0
∫ −1 x
−2
dx +
1
∫0 x
−2
dx
Each of these integrals is infinite. f ( x ) = x −2 has a nonremovable discontinuity at x = 0.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
472
Chapter 5
Integration
2
is not continuous on [−2, 1].
x3
1
0 2
1 2
2
∫ − 2 x3 dx = ∫ − 2 x3 dx + ∫ 0 x3 dx
2
Each of these integrals is infinite. f ( x) = 3 has a nonremovable discontinuity at x = 0.
x
112. The function f ( x ) =
⎡π 3π ⎤
113. The function f ( x ) = sec 2 x is not continuous on ⎢ ,
⎥.
⎣4 4 ⎦
3π 4
∫π 4
sec 2 x dx =
π 2
∫ π 4 sec
2
x dx +
3π 4
∫π 2
sec 2 x dx
Each of these integrals is infinite. f ( x ) = sec 2 x has a nonremovable discontinuity at x =
π
2
⎡π 3π ⎤
114. The function f ( x) = csc x cot x is not continuous on ⎢ ,
⎥.
⎣2 2 ⎦
3π 2
∫π 2
π
3π 2
∫ π 2 csc x cot x dx + ∫ π
csc x cot x dx =
csc x cot x dx
Each of these integrals is infinite f ( x) = csc x cot x has a nonremovable discontinuity at x = π .
115. P =
π 2
2
π ∫0
π 2
⎡ 2
⎤
sin θ dθ = ⎢− cos θ ⎥
⎣ π
⎦0
= −
2
π
(0 − 1)
=
2
π
≈ 63.7%
116. Let F (t ) be an antiderivative of f (t ). Then,
v( x)
∫ u( x) f (t ) dt
v( x)
= ⎡⎣F (t )⎤⎦ u x = F (v( x)) − F (u ( x))
( )
d ⎡ v( x)
d
⎡F (v( x)) − F (u ( x)) = F ′(v( x))v′( x) − F ′(u ( x))u′( x) = f (v( x))v′( x) − f (u ( x))u′( x).
f (t ) dt ⎥⎤ =
dx ⎣⎢∫ u( x)
dx ⎣
⎦
117. True
118. True
119. f ( x ) =
1x
∫0
1
dt +
t2 + 1
x
∫0
1
dt
t2 + 1
By the Second Fundamental Theorem of Calculus, you have f ′( x) =
1
(1 x)
2
1
1
1
⎛ 1⎞
= −
+ 2
= 0.
⎜− 2 ⎟ + 2
1 + x2
x +1
+ 1⎝ x ⎠ x + 1
Because f ′( x) = 0, f ( x) must be constant.
120.
∫ c f (t ) dt
x
121. G ( x) =
= x2 + x − 2
Let f (t ) = 2t + 1. Then
∫ c f (t )dt
x
=
∫ c (2t + 1)dt
x
2
x
= ⎡⎣t 2 + t ⎤⎦ =
c
2
− c − c = −2
2
c2 + c − 2 = 0
(c
x
(a) G (0) =
x + x−c −c = x + x−2
2
⎡
⎤
∫ 0 ⎢⎣s ∫ 0 f (t ) dt ⎥⎦ ds
+ 2)(c − 1) = 0 ⇒ c = 1, − 2.
So, f ( x) = 2 x + 1, and c = 1 or c = −2.
s
⎡
⎤
∫ 0 ⎣⎢s ∫ 0 f (t ) dt ⎦⎥ ds
s
0
(b) Let F ( s ) = s ∫
G ( x) =
s
0
= 0
f (t ) dt.
∫ 0 F ( s) ds
x
G′( x) = F ( x) = x ∫
G′(0) = 0 ∫
0
0
x
0
f (t ) dt
f (t ) dt = 0
(c) G′′( x) = x ⋅ f ( x) +
∫ 0 f (t ) dt
(d) G′′(0) = 0 ⋅ f (0) +
∫ 0 f (t ) dt
x
0
= 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.5
Integration by Substitution
473
Section 5.5 Integration by Substitution
∫ f ( g( x )) g′( x ) dx
u = g( x)
du = g′( x ) dx
1.
∫ (8 x
+ 1) (16 x)dx
8x2 + 1
16x dx
2.
∫x
x3 + 1 dx
x3 + 1
3x 2 dx
3.
∫ tan
x sec 2 x dx
tan x
sec 2 x dx
4.
∫ sin 2 x dx
sin x
cos x dx
5.
4
∫ (1 + 6 x) (6) dx
2
2
2
2
cos x
Check:
6.
∫
∫
+C
4
3
⎡ 2
⎤
4( x 2 − 9)
3
d ⎢ ( x − 9)
+ C⎥ =
( 2 x ) = ( x 2 − 9) ( 2 x )
⎥
4
4
dx ⎢
⎣
⎦
(25 − x 2 )
32
32
+ C =
32
2
25 − x 2 ) + C
(
3
32
12
d ⎡2
2⎛ 3 ⎞
⎤
25 − x 2 ) + C ⎥ = ⎜ ⎟( 25 − x 2 ) ( −2 x) =
(
⎢
dx ⎣ 3
3⎝ 2 ⎠
⎦
3 − 4 x ( −8 x) dx =
2
Check:
4
4
2
3
5
( x 2 − 9)
25 − x ( −2 x ) dx =
∫
6 x)
+C
5
5
⎤
d ⎡ (1 + 6 x)
4
⎢
+ C ⎥ = 6(1 + 6 x)
dx ⎢
5
⎥
⎣
⎦
3
Check:
8.
(1 +
( x 2 − 9) (2 x) dx =
Check:
7.
=
∫ (3 − 4 x ) (−8 x) dx
2 13
=
(3 − 4 x 2 )
25 − x 2 (−2 x )
43
+C =
43
43
13
13
d ⎡3
3⎛ 4 ⎞
⎤
3 − 4 x 2 ) + C ⎥ = ⎜ ⎟(3 − 4 x 2 ) (−8 x ) = (3 − 4 x 2 ) (−8 x )
(
⎢
dx ⎣ 4
4⎝ 3 ⎠
⎦
4
2
( x 4 + 3) + C
1
1 ( x + 3)
x ( x + 3) dx = ∫ ( x 4 + 3) ( 4 x3 ) dx =
+C =
4
4
3
12
3
9.
∫
43
3
3 − 4 x2 ) + C
(
4
3
4
3
2
3
2
⎡ 4
⎤
3( x 4 + 3)
2
d ⎢ ( x + 3)
⎥
Check:
4 x3 ) = ( x 4 + 3) ( x3 )
+C =
(
⎢
⎥
12
12
dx
⎣
⎦
10.
∫
x 2 (6 − x3 ) dx = −
Check:
3
5
(6 − x 3 ) + C
1
1 (6 − x )
6 − x3 ) ( − 3 x 2 ) dx = − ⋅
+ C = −
(
∫
3
3
6
18
6
6
6
5
3
⎡
⎤
− 6(6 − x 3 ) ( − 3 x 2 )
5
d ⎢ (6 − x )
−
+ C⎥ =
= x 2 (6 − x 3 )
⎥
dx ⎢
18
18
⎣
⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
474
11.
Chapter 5
∫
Integration
5
5
⎡ 3
⎤
x3 − 1)
4
(
1
1 ⎢ ( x − 1) ⎥
3
2
x ( x − 1) dx = ∫ ( x − 1) (3x ) dx =
+C =
+C
⎥
3
3⎢
5
15
⎣
⎦
2
4
3
5
4
⎡ 3
⎤
5( x 3 − 1) (3 x 2 )
4
d ⎢ ( x − 1)
⎥
Check:
+C =
= x 2 ( x 3 − 1)
⎢
⎥
15
15
dx
⎣
⎦
12.
∫
4
4
⎡ 2
⎤
5 x 2 + 4)
3
(
1
1 ⎢ ( 5 x + 4) ⎥
2
x(5 x + 4) dx =
(5 x + 4) (10 x) dx = 10 ⎢ 4 ⎥ + C = 40 + C
10 ∫
⎣
⎦
3
2
4
3
⎡ 2
⎤
4(5 x 2 + 4) (10 x)
3
d ⎢ (5 x + 4)
⎥
Check:
+C =
= x(5 x 2 + 4)
⎥
40
40
dx ⎢
⎣
⎦
13.
∫
t
t 2 + 2 dt =
2
12
1
1 (t + 2 )
2
t
+
2
2
t
dt
=
(
)
(
)
2∫
2
32
(t 2
32
+ C =
+ 2)
32
+ C
3
32
12
⎡ 2
⎤
3 2(t 2 + 2) ( 2t )
12
d ⎢ ( t + 2)
⎥
Check:
+C =
= ( t 2 + 2) t
⎥
3
3
dt ⎢
⎣
⎦
14.
∫
t3
2t 4 + 3 dt =
4
12
1
1 ( 2t + 3)
4
3
2
t
+
3
8
t
dt
=
⋅
(
) ( )
8∫
8
( 3 2)
12
3 4
32
⎡ 4
⎤
2t + 3) (8t 3 )
(
d ⎢ ( 2t + 3)
+ C⎥ = 2
= t3
Check:
⎥
dt ⎢
12
12
⎣
⎦
15.
∫
5 x(1 − x 2 )
13
Check:
16.
∫
u2
dx = −
32
+ C =
(2t 4
+ 3)
32
+ C
12
2t 4 + 3
2
13
5
5 (1 − x )
2
−
x
−
x
dx
=
−
⋅
1
2
(
)
(
)
2∫
2
43
43
+ C = −
43
15
(1 − x 2 ) + C
8
43
13
13
d ⎡ 15
15 4
⎤
− (1 − x 2 ) + C ⎥ = − ⋅ (1 − x 2 ) ( −2 x) = 5 x(1 − x 2 ) = 5 x 3 1 − x 2
⎢
dx ⎣ 8
8 3
⎦
u 3 + 2 du =
3
12
1
1 (u + 2)
3
2
2
3
u
+
u
du
=
(
) ( )
3∫
3
32
32
+C =
2(u 3 + 2)
32
9
+C
32
⎡ 3
⎤
12
12
2 3
d ⎢ 2(u + 2)
Check:
+ C ⎥ = ⋅ (u 3 + 2) (3u 2 ) = (u 3 + 2) (u 2 )
⎥
9
9 2
du ⎢
⎣
⎦
17.
∫
x
(1 − x )
Check:
2
3
dx = −
2
−3
1
1 (1 − x )
2
1
x
2
x
dx
−
−
=
−
(
)
(
)
2∫
2
−2
−2
+C =
1
4(1 − x 2 )
2
+C
⎡
⎤
d ⎢
1
x
⎥ = 1 ( −2)(1 − x 2 )−3 (−2 x) =
C
+
2
3
⎢
⎥
2
dx 4(1 − x )
4
(1 − x 2 )
⎣
⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.5
18.
∫
x3
(1 + x4 )
∫
20.
∫
(1 + x3 )
2
⎤
−2
d ⎡
1
1
x2
⎢−
+ C ⎥ = − (−1)(1 + x3 ) (3x 2 ) =
2
3
dx ⎢ 3(1 + x )
3
⎥⎦
(1 + x3 )
⎣
6x2
(4 x
− 9)
3
Check:
−2
−1
1
1
−1
1 + x 4 ) ( 4 x3 ) dx = − (1 + x 4 ) + C =
+C
(
∫
4
4
4(1 + x 4 )
−1
3
⎡
⎤
−2
1
1 ⎢ (1 + x ) ⎥
1
3
2
+C = −
+C
dx = ∫ (1 + x ) (3 x ) dx =
⎢
⎥
−1
3
3
3(1 + x 3 )
⎣
⎦
x2
Check:
475
⎤
−2
d ⎡ −1
1
x3
⎢
⎥ = (1 + x 4 ) ( 4 x3 ) =
C
+
2
dx ⎢ 4(1 + x 4 )
4
⎥⎦
(1 + x 4 )
⎣
Check:
19.
dx =
2
Integration by Substitution
3
3
−3
1
1 ( 4 x − 9)
3
2
4
x
−
9
12
x
dx
=
⋅
(
) ( )
2∫
2
−2
dx =
−2
+C = −
1
4( 4 x − 9)
3
−2
+ C
⎡
⎤
−2
−3
d ⎢
d ⎡ 1 3
1
6x2
−1
⎤
+ C⎥ =
− ( 4 x − 9) + C ⎥ = − ( − 2)( 4 x 3 − 9) (12 x 2 ) =
2
3
⎢
⎥
dx ⎢ 4( 4 x 3 − 9)
dx ⎣ 4
4
⎦
( 4 x 3 − 9)
⎣
⎦
2
−1 2
1
1 (1 − x )
2
x
x
dx
1
2
−
−
=
−
(
)
(
)
2∫
2
12
12
21.
x
∫
1 − x2
Check:
22.
dx = −
12
−1 2
d ⎡
1
−(1 − x 2 ) + C ⎤ = − (1 − x 2 ) ( −2 x) =
⎥⎦
dx ⎣⎢
2
x
1 − x2
4
−1 2
1
1 (1 + x )
dx = ∫ (1 + x 4 ) ( 4 x3 ) dx =
4
4
12
1 + x4
12
x3
∫
Check:
+ C = − 1 − x2 + C
+C =
⎤
−1 2
d ⎡ 1 + x4
1 1
+ C ⎥ = ⋅ (1 + x 4 ) ( 4 x 3 ) =
⎢
dx ⎢⎣
2
2 2
⎥⎦
1 + x4
+C
2
x3
1 + x4
4
23.
⎛
∫ ⎜⎝1 +
3
1⎞ ⎛ 1 ⎞
⎛
⎟ ⎜ ⎟ dt = − ∫ ⎜1 +
t ⎠ ⎝ t2 ⎠
⎝
Check:
⎡
24.
∫ ⎢⎢x
2
+
⎣
Check:
25.
∫
1⎞
⎟
t⎠
3
⎡
⎛ 1 ⎞⎤
⎢1 + ⎜ t ⎟⎥
1
⎛
⎞
⎝ ⎠⎦
⎣
+C
⎜ − 2 ⎟ dt = −
4
⎝ t ⎠
4
3
3
⎤
d ⎡⎢ ⎡⎣1 + (1 t )⎤⎦
1 ⎛
1⎞ ⎛ 1 ⎞
1⎛
1⎞
−
+ C ⎥ = − ( 4)⎜1 + ⎟ ⎜ − 2 ⎟ = 2 ⎜1 + ⎟
⎥
dt ⎢
4
4 ⎝
t⎠ ⎝ t ⎠
t ⎝
t⎠
⎣
⎦
1 ⎤
⎥ dx =
(3x)2 ⎥⎦
⎛
∫ ⎜⎝ x
2
+
1 −2 ⎞
x3
1 ⎛ x −1 ⎞
x3
1
3x 4 − 1
+ ⎜
−
+C =
+C
x ⎟ dx =
⎟+C =
9 ⎠
3
9 ⎝ −1 ⎠
3
9x
9x
d ⎡1 3 1 −1
1
1
⎤
x − x + C ⎥ = x 2 + x −2 = x 2 +
2
dx ⎢⎣ 3
9
9
⎦
(3 x )
12
1
1
1 ⎡ ( 2 x) ⎤
−1 2
⎥ + C =
dx = ∫ ( 2 x) 2 dx = ⎢
2
2 ⎢⎣ 1 2 ⎥⎦
2x
Alternate Solution:
Check:
∫
1
dx =
2x
1
−1 2
∫ x dx =
2
d ⎡
1
−1 2
2 x + C ⎤⎦ = ( 2 x) ( 2) =
⎣
dx
2
2x + C
1 x1 2
+ C =
2 (1 2)
2x + C
1
2x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
476
26.
Chapter 5
∫
x
3
5x
2
Integration
dx = ∫
1 13
x dx =
5
3
1
3 43
3 3
25 x 4 + C
⋅ x
+C =
20
5 4
3
Alternate Solution:
∫
x
3
5x
Check:
27. y =
28. y =
∫
∫
2
dx = ∫ (5 x 2 )
−1 3
x dx =
3
d ⎡ 1
⎤
⋅ x4 3 + C⎥ =
dx ⎢⎣ 3 5 4
⎦
1
10
2
∫(5 x )
(10 x)dx =
1
3 4
⋅ ⋅ x1 3 =
5 4 3
3
2
1 (5 x )
⋅
10
(2 3)
23
−1 3
x
3
5x2
12⎤
⎡
⎤
−1 2
16 − x 2 ) ⎥
⎛ x2 ⎞
(
2
⎢
dx
x
dx
x
x
dx
=
−
−
−
=
−
+ C = 2 x 2 − 4 16 − x 2 + C
4
2
16
2
4
2
(
)
)
⎜ ⎟
⎥
∫
∫(
⎢
⎥
2
16 − x 2 ⎦
⎝ ⎠
12
⎣
⎦
⎡
⎢4 x +
⎣
4x
10 x 2
dx
1+ x
−1 2
10
=
(1 + x3 ) (3x 2 ) dx
3∫
3
(b)
=
∫
x +1
( x2
+ 2 x − 3)
(2, 2):
32
1
(4 − 22 ) + C ⇒ C = 2
3
32
1
(4 − x2 ) + 2
3
2
x2 − 8x + 1
−2
2
−1
32. (a) Answers will vary. Sample answer:
y
6
4
2
dx
x
−1 2
1
( x 2 − 8 x + 1) (2 x − 8) dx
2∫
12⎤
⎡ 2
1 ( x − 8 x + 1) ⎥
= ⎢
+ C = x2 − 8x + 1 + C
⎥
2⎢
12
⎣
⎦
10
=
31. (a) Answers will vary. Sample answer:
−2
(b)
dy
= esin x cos x, (π , 2)
dx
y =
∫e
(π , 2): 2
sin x
cos x dx = esin x + C
= esin π + C = 1 + C ⇒ C = 1
y = esin x + 1
y
3
6
2
−1
2 = −
dx
2
x − 4
−2
12
1
(4 − x 2 ) (−2 x dx)
2∫
32
1
+ C = − (4 − x 2 ) + C
3
4 − x 2 dx = −
32
1 2
⋅ (4 − x 2 )
2 3
y = −
−2
1
( x 2 + 2 x − 3) (2 x + 2) dx
2∫
∫
∫x
= −
−1
⎡ 2
⎤
1 ⎢ ( x + 2 x − 3) ⎥
=
+C
⎥
2⎢
−1
⎣
⎦
1
= −
+C
2
2( x + 2 x − 3)
30. y =
dy
= x 4 − x 2 , ( 2, 2)
dx
y =
12⎤
⎡
10 ⎢ (1 + x 3 ) ⎥
=
+ C
⎥
3⎢ 12
⎣
⎦
20
1 + x3 + C
=
3
29. y =
23
3
(5x 2 ) + C = 43 ⋅ 315 x 4 3 + C
20
+C =
x
10
−1
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.5
33.
∫ π sin π x dx
= −cos π x + C
34.
∫ sin 4 x dx
=
∫ cos 8 x dx
=
35.
Integration by Substitution
477
2⎛
x⎞
⎛ x⎞
2 ⎛ x ⎞⎛ 1 ⎞
⎜ ⎟ dx = 2 ∫ csc ⎜ ⎟⎜ ⎟ dx = −2 cot ⎜ ⎟ + C
⎝ 2⎠
⎝ 2 ⎠⎝ 2 ⎠
⎝ 2⎠
36.
∫ csc
1
1
(sin 4 x)(4) dx = − cos 4 x + C
4∫
4
37.
∫ θ 2 cos θ
1
1
(cos 8 x)(8) dx = sin 8 x + C
8∫
8
38.
∫ x sin x
1
1
2
1⎛ 1 ⎞
1
dθ = − ∫ cos ⎜ − 2 ⎟ dθ = −sin + C
θ⎝ θ ⎠
θ
dx =
1
(sin x 2 )(2 x) dx = − 12 cos x 2 + C
2∫
1
1 (sin 2 x)
1
+ C = sin 2 2 x + C OR
(sin 2 x)(2 cos 2 x) dx =
2∫
2
2
4
2
39.
∫
sin 2 x cos 2 x dx =
∫
sin 2 x cos 2 x dx = −
2
∫ sin 2 x cos 2 x dx
40.
∫
41.
∫ cot 3 x dx
1
1
1
2 sin 2 x cos 2 x dx = ∫ sin 4 x dx = − cos 4 x + C2
2∫
2
8
=
tan x sec 2 x dx =
csc 2 x
sin x
( tan x)
∫ cos3 x dx
−3
(cot x)−2
+C =
= − ∫ (cos x)
−3
(cos x)−2
2
( tan x)3 2 + C
3
(− csc2 x) dx
+C =
−2
= −
32
32
= − ∫ (cot x)
= −
42.
1
1 (cos 2 x)
1
+ C1 = − cos 2 2 x + C1 OR
(cos 2 x)(−2 sin 2 x) dx = −
2∫
2
2
4
1
1
1
1
+ C = tan 2 x + C = (sec 2 x − 1) + C = sec 2 x + C1
2 cot 2 x
2
2
2
(−sin x) dx
47.
44.
45.
∫
e − x ( − 3 x 2 ) dx = e − x + C
∫
( x + 1)e x
3
∫ e (e
x
−2 x
48.
∫
e 2 x + 2e x + 1
dx =
ex
x
2 + 2x
49.
2
1
e x + 2 x ( 2 x + 2) dx
2∫
1 2
= ex + 2x + C
2
dx =
+ 1) dx =
2
(e x
+ 1)
3
+ e− x )
2
=
) (e
−2
+ C
e x + e− x
−2
x
− e
sin π x
50.
∫e
tan 2 x
51.
∫e
−x
+C
−x
∫e
cos π x dx =
=
3
dx = 2 ∫ (e + e
x
dx −
∫e
−x
dx
∫ (e
x
+ 2 + e − x ) dx
= e x + 2 x − e− x + C
3
2e x − 2e − x
(e x
∫ 5e
5
= − e− 2 x + e− x + C
2
46. Let u = e x + e − x , du = (e x − e − x ) dx.
∫
5 − ex
dx =
e2 x
+C
−2
1
1
=
+ C = sec 2 x + C
2
2 cos x
2
43.
∫
−x
) dx
52.
1
π
1
π
∫e
sin π x
(π
cos π x) dx
esin π x + C
1
e tan 2 x ( 2 sec 2 2 x) dx
2∫
1
= e tan 2 x + C
2
sec 2 2 x dx =
sec 2 (e − x ) dx = − ∫ sec 2 (e − x )( − e− x ) dx
= − tan (e − x ) + C
∫ ln (e
2 x −1
) dx = ∫ (2 x − 1) dx
= x2 − x + C
53.
∫3
x2
3x 2
2 x2
⎛1⎞
dx = 2 ∫ 3x 2 ⎜ ⎟ dx = 2
+C =
3 +C
ln 3
ln 3
⎝ 2⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
478
54.
Chapter 5
Integration
(3 − x )
∫ (3 − x )7
2
dx = −
= −
55. f ( x ) =
x
∫ − sin 2 dx
1
2
(3 − x )
∫ − 2(3 − x)7
2
dx
1 ⎡ (3 − x ) 2 ⎤
⎣7
⎦ + C
2 ln 7
= 2 cos
x
+C
2
56. f ( x ) =
=
∫ 0.4
x
+ 4.
2
x3
∫x
⎛1⎞
dx = 3 ∫ 0.4 x 3 ⎜ ⎟ dx
⎝ 3⎠
3
0.4 x 3 + C
ln 0.4
3
1
3
0.4 x 3 + −
ln 0.4
2
ln 0.4
−x 4
= − 8e − x
4
⎛ 1⎞
dx = − 8 ∫ e − x 4 ⎜ − ⎟ dx
⎝ 4⎠
+C
=
∫
4
32
+
3
5
3
x + 6 dx =
∫ (u − 6) u du
32
12
∫ (u − 6u ) du
2 52
u − 4u 3 2 + C
5
2u 3 2
(u − 10) + C
5
2
32
= ( x + 6) ⎡⎣( x + 6) − 10⎤⎦ + C
5
2
32
= ( x + 6) ( x − 4) + C
5
3x − 4 dx =
u + 4
1
, dx = du
3
3
∫
u+4
⋅
3
1
u ⋅ du
3
1
(u3 2 + 4u1 2 ) du
9∫
1⎛ 2
8
⎞
= ⎜ u5 2 + u3 2 ⎟ + C
9⎝ 5
3
⎠
− 0.2 x3
(− 0.6 x 2 ) dx
2
8
( 3 x − 4) 5 2 + ( 3 x − 4) 3 2 + C
45
27
2
32
=
(3x − 4) ⎡⎣3(3x − 4) + 20⎤⎦ + C
135
2
=
(3x − 4)3 2 (9 x + 8) + C
135
3
5
= − e − 0.2 x + C
3
f ( 0) =
2(8 − x 2 )
16
5
+C = 7 ⇒ C =
3
3
=
3
∫e
+C =
=
∫x
+ 9
x 2e − 0.2 x dx
1
− 0.6
+C
32
62. u = 3 x − 4, x =
f ( 0) = 1 = − 8 + C ⇒ C = 9
f ( x ) = − 8e − x
2( 4)
3
=
f ( x) =
∫ 2e
32
3
=
3
1
1
3
+ C =
⇒ C =
−
ln 0.4
2
2
ln 0.4
58. f ( x ) =
f ( x) =
2(8 − x 2 )
61. u = x + 6, x = u − 6, dx = du
f ( 0) =
57. f ( x ) =
f ( x) =
f ( 2) =
⎛0⎞
Because f (0) = 6 = 2 cos⎜ ⎟ + C , C = 4. So,
⎝ 2⎠
f ( x ) = 2 cos
60. f ′( x) = −2 x 8 − x 2 , ( 2, 7)
=
3
5
19
= − + C ⇒ C =
2
3
6
3
5
19
f ( x ) = − e − 0.2 x +
3
6
59. f ′( x) = 2 x( 4 x 2 − 10) , ( 2, 10)
2
f ( x) =
(4 x 2 − 10)
12
3
+C =
2( 2 x 2 − 5)
3
3
+C
f ( 2) =
2(8 − 5)
+ C = 18 + C = 10 ⇒ C = −8
3
f ( x) =
3
2 2
(2 x − 5) − 8
3
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.5
Integration by Substitution
479
63. u = 1 − x, x = 1 − u , dx = − du
∫x
2
1 − x dx = − ∫ (1 − u )
2
u du
= − ∫ (u1 2 − 2u 3 2 + u 5 2 ) du
4
2
⎛2
⎞
= −⎜ u 3 2 − u 5 2 + u 7 2 ⎟ + C
3
5
7
⎝
⎠
2u 3 2
(35 − 42u + 15u 2 ) + C
105
2
= −
(1 − x)3 2 ⎡⎣35 − 42(1 − x) + 15(1 − x)2 ⎤⎦ + C
105
2
32
= −
(1 − x) (15 x 2 + 12 x + 8) + C
105
= −
64. u = 2 − x, x = 2 − u , dx = − du
∫ ( x + 1)
2 − x dx = − ∫ (3 − u ) u du
= − ∫ (3u1 2 − u 3 2 ) du
2
⎛
⎞
= −⎜ 2u 3 2 − u 5 2 ⎟ + C
5
⎝
⎠
2u 3 2
(5 − u ) + C
5
2
32
= − ( 2 − x) ⎡⎣5 − ( 2 − x)⎤⎦ + C
5
2
32
= − ( 2 − x) ( x + 3) + C
5
= −
65. u = 2 x − 1, x =
∫
x2 − 1
dx =
2x − 1
1
1
(u + 1), dx = du
2
2
⎡⎣(1 2)(u + 1)⎤⎦ − 1 1
du
2
u
2
∫
1 −1 2 ⎡ 2
u ⎣(u + 2u + 1) − 4⎤⎦ du
8∫
1
= ∫ (u 3 2 + 2u1 2 − 3u −1 2 ) du
8
1⎛ 2
4
⎞
= ⎜ u 5 2 + u 3 2 − 6u1 2 ⎟ + C
8⎝ 5
3
⎠
=
=
=
u1 2
(3u 2 + 10u − 45) + C
60
2x − 1 ⎡
2
3( 2 x − 1) + 10( 2 x − 1) − 45⎤ + C
⎦
60 ⎣
1
2 x − 1(12 x 2 + 8 x − 52) + C
60
1
2 x − 1(3 x 2 + 2 x − 13) + C
=
15
=
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
480
Chapter 5
Integration
66. u = x + 4, x = u − 4, du = dx
2x + 1
dx =
x + 4
∫
∫
2(u − 4) + 1
u
67. u = x + 1, x = u − 1, dx = du
−x
∫ ( x + 1) −
du
=
∫ (2u
=
4 32
u − 14u1 2 + C
3
2 12
u ( 2u − 21) + C
3
2
x + 4 ⎡⎣2( x + 4) − 21⎤⎦ + C
3
2
x + 4 ( 2 x − 13) + C
3
=
=
=
12
− 7u −1 2 ) du
x +1
dx =
−(u − 1)
∫u−
= −∫
(
du
u
)(
u +1
u
(
) du
u −1
)
u −1
= − ∫ (1 + u −1 2 ) du
= −(u + 2u1 2 ) + C
= −u − 2 u + C
= −( x + 1) − 2
x +1+C
= −x − 2
x +1 −1+ C
(
x + 1 + C1
= − x + 2
)
where C1 = −1 + C.
68. u = t + 10, t = u − 10, du = dt
∫ t (t
+ 10)
13
dt =
=
∫ (u − 10)u du
43
13
∫ (u − 10u ) du
13
3 7 3 15 4 3
u − u +C
7
2
3 43
u ( 2u − 35) + C
=
14
3
=
(t + 10)4 3 ⎡⎣2(t + 10) − 35⎤⎦ + C
14
3
43
=
(t + 10) (2t − 15) + C
14
=
69. Let u = x 2 + 1, du = 2 x dx.
∫ − 1 x( x
1
2
+ 1) dx =
3
2
∫ (x
1 1
2 −1
1
3
4
+ 1) ( 2 x) dx = ⎡ 18 ( x 2 + 1) ⎤ = 0
⎢⎣
⎥⎦ −1
70. Let u = 2 x 4 + 1, du = 8 x3 dx.
1
∫ 0 x (2 x
1
3
4
+ 1) dx =
2
1 1
8 0
∫
3
⎡
2 x 4 + 1) ⎤
(
13
1
⎢
(2 x + 1) (8 x ) dx = ⎢8 ⋅ 3 ⎥⎥ = 241 (33 − 13 ) = 12
⎣
⎦0
4
2
3
71. Let u = x 3 + 1, du = 3x 2 dx.
3 2⎤
⎡ 3
12
32 2
x + 1) ⎥
1 2 3
4
4
8
(
2
⎢
x + 1 dx = 2 ⋅ ∫ ( x + 1) (3 x ) dx =
= ⎡( x3 + 1) ⎤ = ⎡⎣27 − 2 2 ⎤⎦ = 12 −
⎢
⎥
⎣⎢
⎦⎥1
3 1
9
9
9
32
⎣
⎦1
2
2
∫1
2x
2
3
2
72. Let u = 1 − x 2 , du = −2 x dx.
1
∫0
1
x 1 − x 2 dx = −
12
32
1 1
(1 − x 2 ) (−2 x) dx = ⎡⎢⎣− 13(1 − x 2 ) ⎤⎥⎦ = 0 + 13 = 13
2∫0
0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.5
Integration by Substitution
481
73. Let u = 2 x + 1, du = 2 dx.
4
1
1 4
−1 2
dx = ∫ ( 2 x + 1) ( 2) dx = ⎡⎣ 2 x + 1 ⎤⎦ =
0
0
2
2x + 1
4
∫0
9 −
1 = 2
74. Let u = 1 + 2 x 2 , du = 4 x dx.
2
x
2
∫0
1 + 2 x2
dx =
75. Let u = 1 +
∫1
x , du =
1
9
(
x1+
−1 2
1 2
(1 + 2 x 2 ) (4 x) dx = ⎡⎢⎣ 12 1 + 2 x2 ⎤⎥⎦ = 32 − 12 = 1
4∫0
0
x
)
2
1
2
dx = 2 ∫
dx.
x
9
1
(1 +
x
)
9
−2 ⎛
1 ⎞
2 ⎤
1
1
⎡
⎜
⎟ dx = ⎢−
⎥ = −2 + 1 = 2
⎝2 x ⎠
⎣ 1 + x ⎦1
1
(u + 1).
2
When x = 1, u = 1. When x = 5, u = 9.
76. Let u = 2 x − 1, du = 2 dx, x =
x
dx =
2x − 1
5
∫1
9
∫1
1 2(u + 1) 1
1 9
du = ∫ (u1 2 + u −1 2 ) du
2
4 1
u
9
=
1 ⎡2 3 2
⎤
u + 2u1 2 ⎥
4 ⎢⎣ 3
⎦1
1 ⎡⎛ 2
⎞ ⎛2
⎞⎤
⎜ ( 27) + 2(3) ⎟ − ⎜ + 2 ⎟⎥
4 ⎢⎣⎝ 3
3
⎠ ⎝
⎠⎦
16
=
3
=
e − 2 x dx = −
2
e1 − x dx = − ∫ e1 − x ( −1) dx = ⎡⎣− e1 − x ⎤⎦ = − e −1 + 1
1
1
∫0
78.
∫1
79.
1
1 1 −2 x
1
1
⎡ 1
⎤
e ( − 2) dx = ⎢− e − 2 x ⎥ = − e − 2 +
2 ∫0
2
2
2
⎣
⎦0
1
77.
3
∫1
2
3
e3 x
1 3
1
e
⎛ 3⎞
⎡ 1 ⎤
dx = − ∫ e3 x ⎜ − 2 ⎟ dx = ⎢− e3 x ⎥ = − (e − e3 ) = (e 2 − 1)
x2
3 1
x
3
3
3
⎝
⎠
⎣
⎦1
80. Let u =
∫0
2
2
− x2
, du = − x dx.
2
xe − x
2 2
dx = − ∫
2
0
e− x
2 2
(− x) dx
2
= ⎡⎢− e− x 2 ⎤⎥
⎣
⎦0
81. u = x + 1, x = u − 1, dx = du
When x = 0, u = 1. When x = 7, u = 8.
Area =
=
7
∫0
x 3 x + 1 dx =
43
∫ 1 (u
8
3
∫ 1 (u − 1) u du
8
− u1 3 ) du
8
3
⎡3
⎤
= ⎢ u7 3 − u4 3⎥
4
⎣7
⎦1
3⎞
⎛ 384
⎞ ⎛3
= ⎜
− 12 ⎟ − ⎜ − ⎟
4⎠
⎝ 7
⎠ ⎝7
1209
=
28
2
= 1 − e −1 =
e −1
e
82. u = x + 2, x = u − 2, dx = du
When x = −2, u = 0. When x = 6, u = 8.
Area =
6
∫ −2 x
2 3
x + 2 dx
=
∫ 0 ( u − 2)
=
∫ 0 (u
8
8
73
2 3
u du
− 4u 4 3 + 4u1 3 ) du
8
12
4752
⎡3
⎤
= ⎢ u10 3 − u 7 3 + 3u 4 3 ⎥ =
7
35
⎣10
⎦0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
482
Chapter 5
83. Area =
2π 3
∫π 2
= 2∫
Integration
⎛ x⎞
sec 2 ⎜ ⎟ dx
⎝ 2⎠
2π 3
π 2
88.
−2x
∫ 0 (e
2
2
⎡ 1
⎤
+ 2) dx = ⎢− e − 2 x + 2 x⎥
⎣ 2
⎦0
⎛ x ⎞⎛ 1 ⎞
sec 2 ⎜ ⎟⎜ ⎟ dx
⎝ 2 ⎠⎝ 2 ⎠
2π 3
⎡
⎛ x ⎞⎤
= ⎢2 tan ⎜ ⎟⎥
⎝ 2 ⎠⎦π
⎣
(
1
1
= − e− 4 + 4 +
≈ 4.491
2
2
)
= 2
4
3 −1
2
84. Let u = 2 x, du = 2 dx.
Area =
=
π 4
∫ π 12
∫
−2
π 4
1
2 π 12
π 4
85.
5
∫0
89. f ( x ) = x 2 ( x 2 + 1) is even.
csc 2 x cot 2 x( 2) dx
= ⎡⎣− 12 csc 2 x⎤⎦
=
π 12
4
0
csc 2 x cot 2 x dx
2
∫ −2
1
2
x 2 ( x 2 + 1) dx = 2 ∫
( x 4 + x 2 ) dx
0
2
272
⎡ 32 8 ⎤
= 2⎢ + ⎥ =
3⎦
15
⎣5
5
e x dx = ⎡⎣e x ⎤⎦ 0 = e5 − 1 ≈ 147.413
150
90. f ( x ) = x( x 2 + 1) is odd.
3
∫ − 2 x( x
2
0
3
6
0
86.
+ 1) dx = 0
2
b
∫a
91. f ( x ) = sin 2 x cos x is even.
e dx = ⎡⎣− e ⎤⎦ a = e
−x
−x
b
−a
−e
−b
π 2
∫ −π 2 sin
2
x cos x dx = 2 ∫
π 2
0
sin 2 x(cos x) dx
π 2
⎡ sin 3 x ⎤
= 2⎢
⎥
⎣ 3 ⎦0
a
87.
∫0
6
xe − x
b
2 4
2
3
=
2
dx = ⎡⎢− 2e − x 4 ⎤⎥
⎣
⎦0
92. f ( x) = sin x cos x is odd.
6
π 2
∫ −π 2 sin x cos x dx
= − 2e − 3 2 + 2 ≈ 1.554
= 0
3
4
93.
−4.5
4.5
−3
π 4
94. (a)
∫ −π 4 sin x dx
(b)
∫ −π 4 cos x dx
π 4
⎡ x3 ⎤
64
2
x
dx
=
; the function x 2 is an even
⎢ ⎥ =
∫0
3
⎣ 3 ⎦0
function.
4
0
(a)
∫ −4 x
(b)
∫ −4 x
(c)
2
∫ 0 (− x ) dx
(d)
∫ − 4 3x
4
dx =
2
dx = 2 ∫ x 2 dx =
x 2 dx =
4
0
4
0
4
∫0
2
2
64
3
128
3
4
= − ∫ x 2 dx = −
0
64
3
4
dx = 3∫ x 2 dx = 64
0
= 0 because sin x is symmetric to the origin.
= 2∫
π 4
0
π 4
cos x dx = [2 sin x]0
=
2
⎡ x5
x3 ⎤
= 2⎢ + ⎥
3 ⎦0
⎣5
2 because cos x is symmetric to the y-axis.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.5
95.
96.
π 2
(c)
∫ −π 2 cos x dx
= 2∫
(d)
∫ −π 2 sin x cos x dx
π 2
π 2
∫ −3 (x
3
3
π 2
cos x dx = [2 sin x]0
0
∫ −π 2 (sin 4 x
483
= 2
= 0 because sin ( − x)cos( − x ) = −sin x cos x and so, is symmetric to the origin.
+ 4 x 2 − 3 x − 6) dx =
π 2
Integration by Substitution
∫ −3 (x
3
π 2
+ cos 4 x) dx =
∫ −π 2
2
− 3 x) dx +
sin 4 x dx +
97. If u = 5 − x 2 , then du = −2 x dx and
98. f ( x ) = x( x 2 + 1) is odd. So,
3
∫ − 2 x( x
2
3
π 2
∫ −π 2
2
∫ x (5 − x )
2
∫ − 3 (4 x
3
2
− 6) dx = 0 + 2∫
cos 4 x dx = 0 + 2∫
π 2
0
3
0
(4 x2
− 6) dx = 2 ⎡⎣ 43 x3 − 6 x⎤⎦ = 36
0
3
π 2
⎡2
⎤
cos 4 x dx = ⎢ sin 4 x⎥
⎣4
⎦0
= 0
dx = − 12 ∫ (5 − x 2 ) ( −2 x) dx = − 12 ∫ u 3 du.
3
+ 1) dx = 0.
2
99. (a) The second integral is easier. Use substitution with u = x 3 + 1 and du = 3 x 2 dx. The answer is
∫x
2
x3 + 1 dx =
1
3
∫( x
+ 1) 3 x 2 dx
=
2
9
( x3 + 1)
12
3
32
+ C.
(b) The first integral is easier. Use substitution with u = tan 3 x and du = 3sec 2 (3 x)dx. The answer is
2
∫ tan (3x) sec (3x) dx
100. (a)
=
∫ (2 x − 1)
2
dx =
1
2
∫ (2 x − 1)
2
dx =
∫ (4 x
1
3
2
∫ tan (3x) 3sec (3x) dx
∫ (2 x − 1)
2
2
2 dx =
1
6
(2 x
− 4 x + 1) dx =
=
1 tan 2 3 x
6
− 1) + C1 =
4 x3
3
3
+ C.
4 x3
3
− 2 x2 + x −
1
6
+ C1
− 2 x 2 + x + C2
They differ by constant: C2 = C1 − 16 .
(b)
∫ tan x sec
∫
2
tan 2 x
+ C1
2
x dx =
∫
tan x sec 2 x dx =
sec x(sec x tan x ) dx =
sec 2 x
+ C2
2
tan x
sec x − 1
sec 2 x
1
+ C1 =
+ C1 =
− + C1
2
2
2
2
1
They differ by a constant: C 2 = C1 − .
2
2
101.
2
dV
k
=
dt
(t + 1)2
V (t ) =
k
∫ (t + 1)2 dt
= −
k
+C
t +1
V (0) = − k + C = 500,000
1
V (1) = − k + C = 400,000
2
Solving this system yields k = −200,000 and C = 300,000. So, V (t ) =
When t = 4, V ( 4) = $340,000.
200,000
+ 300,000.
t +1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
484
Chapter 5
Integration
102. (a) The maximum flow is approximately R ≈ 62 thousand gallons at 9:00 A.M. (t ≈ 9).
(b) The volume of water used during the day is the area under the curve for 0 ≤ t ≤ 24. That is, V =
24
∫0
R(t )dt.
(c) The least amount of water is used approximately from 1 A.M. to 3 A.M. (1 ≤ t ≤ 3).
103.
b
1
∫
a
b−a
πt ⎤
1 ⎡
262.5
πt ⎤
⎡
⎢74.50 + 43.75 sin 6 ⎥ dt = b − a ⎢74.50t − π cos 6 ⎥
⎣
⎦
⎣
⎦a
b
(a)
1⎡
262.5
πt ⎤
1⎛
262.5 ⎞
≈ 102.352 thousand units
74.50t −
cos ⎥ = ⎜ 223.5 +
3 ⎢⎣
6 ⎦0
3⎝
π
π ⎠⎟
(b)
1⎡
262.5
πt ⎤
1⎛
262.5
⎞
− 223.5⎟ ≈ 102.352 thousand units
74.50t −
cos ⎥ = ⎜ 447 +
3 ⎢⎣
6 ⎦3
3⎝
π
π
⎠
(c)
1⎡
262.5
πt ⎤
1⎛
262.5 262.5 ⎞
+
= 74.5 thousand units
74.50t −
cos ⎥ =
⎜ 894 −
12 ⎢⎣
6 ⎦0
12 ⎝
π
π
π ⎠⎟
3
6
12
b
104.
b
1
1 ⎡ 1
1
⎤
⎡2 sin (60π t ) + cos(120π t )⎤⎦ dt =
−
cos(60π t ) +
sin (120π t )⎥
b − a∫a ⎣
b − a ⎢⎣ 30π
120π
⎦a
1 60
(a)
1
1
⎡ 1
⎤
−
cos(60π t ) +
sin (120π t )⎥
120π
(1 60) − 0 ⎢⎣ 30π
⎦0
(b)
1
1
⎡ 1
⎤
−
cos(60π t ) +
sin (120π t )⎥
120π
(1 240) − 0 ⎢⎣ 30π
⎦0
⎡⎛ 1
4
⎞ ⎛ 1 ⎞⎤
= 60 ⎢⎜
+ 0⎟ − ⎜ −
≈ 1.273 amps
⎟⎥ =
π
π
π
30
30
⎠ ⎝
⎠⎦
⎣⎝
1 240
⎡⎛
1
1 ⎞ ⎛ 1 ⎞⎤
= 240 ⎢⎜ −
+
⎟⎥
⎟ − ⎜−
120π ⎠ ⎝ 30π ⎠⎦
⎣⎝ 30 2π
=
1 30
(c)
1
1
⎡ 1
⎤
−
cos(60π t ) +
sin (120π t )⎥
⎢
120π
(1 30) − 0 ⎣ 30π
⎦0
2
π
(5 − 2 2 ) ≈ 1.382 amps
⎡⎛ 1 ⎞ ⎛ 1 ⎞⎤
= 30 ⎢⎜ −
⎟ − ⎜−
⎟⎥ = 0 amp
⎣⎝ 30π ⎠ ⎝ 30π ⎠⎦
105. u = 1 − x, x = 1 − u , dx = − du
When x = a, u = 1 − a. When x = b, u = 1 − b.
Pa , b =
b
∫a
15
15 1− b
x 1 − x dx =
− (1 − u ) u du
4
4 ∫ 1− a
b
=
1− b
1− b
⎡ (1 − x)3 2
⎤
⎤
15 1− b 3 2
15 ⎡ 2 5 2 2 3 2 ⎤
15 ⎡ 2u 3 2
12
−
=
−
=
−
=
⎢−
3
5
3x + 2)⎥
u
u
du
u
u
u
(
)
(
(
)
⎢
⎥
∫
⎢
⎥
4 1− a
4 ⎣5
3
4 ⎣ 15
2
⎦1− a
⎢⎣
⎥⎦ a
⎦1− a
0.75
⎡ (1 − x )3 2
⎤
(a) P0.50, 0.75 = ⎢−
(3x + 2)⎥ = 0.353 = 35.3%
2
⎢⎣
⎥⎦ 0.50
b
(b) P0, b
32
⎡ (1 − x)3 2
⎤
(1 − b) 3b + 2 + 1 = 0.5
= ⎢−
(3x + 2)⎥ = −
(
)
2
2
⎢⎣
⎥⎦ 0
32
(1 − b) (3b + 2) = 1
b ≈ 0.586 = 58.6%
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.5
Integration by Substitution
485
106. u = 1 − x, x = 1 − u , dx = − du
When x = a, u = 1 − a. When x = b, u = 1 − b.
Pa , b =
b
∫a
1155 3
1155 1− b
32
3
x (1 − x) dx =
− (1 − u ) u 3 2 du
32
32 ∫ 1− a
1− b
=
1155 1− b 9 2
6
2
⎡ 2 11 2 2 9 2
⎤
u
− u + u7 2 − u5 2 ⎥
(u − 3u 7 2 + 3u 5 2 − u 3 2 ) du = 1155
32 ∫ 1− a
32 ⎢⎣11
3
7
5
⎦1− a
=
⎤
⎡u5 2
⎤
1155 ⎡ 2u 5 2
105u 3 − 385u 2 + 495u − 231)⎥
= ⎢
(
(105u 3 − 385u 2 + 495u − 231)⎥
⎢
32 ⎣ 1155
⎦1− a
⎣ 16
⎦1− a
1− b
1− b
0.75
⎡u5 2
⎤
(a) P0, 0.25 = ⎢
(105u 3 − 385u 2 + 495u − 231)⎥
16
⎣
⎦1
≈ 0.025 = 2.5%
0
⎡u5 2
⎤
(b) P0.5, 1 = ⎢
(105u 3 − 385u 2 + 495u − 231)⎥ ≈ 0.736 = 73.6%
16
⎣
⎦ 0.5
107. (a)
4
g
0
9.4
f
−4
(b) g is nonnegative because the graph of f is positive at the beginning, and generally has more positive sections than
negative ones.
(c) The points on g that correspond to the extrema of f are points of inflection of g.
(d) No, some zeros of f , like x = π 2, do not correspond to an extrema of g. The graph of g continues to increase after
x = π 2 because f remains above the x-axis.
(e) The graph of h is that of g shifted 2 units downward.
4
0
9.4
−4
g (t ) =
∫ 0 f ( x) dx
t
=
π 2
∫0
f ( x) dx +
108. Let f ( x) = sin π x, 0 ≤ x ≤ 1.
Let ∆x =
n
∑
n →∞
i =1
t
= 2 + h(t ).
109. (a) Let u = 1 − x, du = − dx, x = 1 − u
1
and use righthand endpoints
n
i
ci = , i = 1, 2, …, n.
n
lim
∫ π 2 f ( x) dx
x = 0 ⇒ u = 1, x = 1 ⇒ u = 0
5
2
∫ 0 x (1 − x) dx
1
n
sin (iπ n)
= lim ∑ f (ci ) ∆x
∆x → 0
n
i =1
=
= −
1
π
(−1 − 1)
=
2
5
∫ 0 u (1 − u ) du
=
∫ 0 x (1 − x)
0
1
1
5
2
dx
x = 0 ⇒ u = 1, x = 1 ⇒ u = 0
1
⎤
cos π x⎥
π
⎦0
1
2 5
∫ 1 (1 − u ) u (−du )
(b) Let u = 1 − x, du = − dx, x = 1 − u
1
∫ 0 sin π x dx
= −
=
b
a
∫ 0 x (1 − x) dx
1
=
2
π
=
a b
∫ 1 (1 − u ) u (−du )
=
a
b
∫ 0 u (1 − u ) du
=
a
b
∫ 0 x (1 − x) dx
0
1
1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
486
Chapter 5
Integration
⎛π
⎞
⎛π
⎞
110. (a) sin x = cos⎜ − x ⎟ and cos x = sin ⎜ − x ⎟
⎝2
⎠
⎝2
⎠
Let u =
π 2
∫0
π
2
− x, du = − dx, x =
π
π 2
∫0
− u:
2
sin 2 x dx =
∫0
π 2
⎛π
⎞
cos 2 ⎜ − x ⎟ dx
⎝2
⎠
=
0
∫π 2
cos u ( − du )
=
∫0
π 2
(b) Let u =
π
2
− x as in part (a):
2
cos 2 u du =
π 2
∫0
cos 2 x dx
(2 x
+ 1) + C
⎛π
⎞
cos n ⎜ − x ⎟ dx
⎝2
⎠
π 2
∫0
sin n x dx =
=
n
∫ π 2 cos u(−du )
=
∫0
0
π 2
cos n u du =
π 2
∫0
cos n x dx
111. False
∫ (2 x + 1)
2
∫ (2 x + 1)
dx =
1
2
+ 1) dx =
1
2
2
2 dx =
1
6
3
112. False
∫ x( x
2
∫ (x
2
+ 1)( 2 x) dx =
1
4
( x2
+ 1) + C
2
113. True
∫ −10 (ax
10
3
+ bx 2 + cx + d ) dx =
∫ −10 (ax
10
3
+ cx) dx +
Odd
∫ −10 (bx
10
2
+ d ) dx = 0 + 2∫
10
0
(bx2 + d ) dx
Even
114. True
b
∫ a sin x dx = [−cos x]a
b
= −cos b + cos a = −cos(b + 2π ) + cos a =
b + 2π
∫a
sin x dx
115. True
4 ∫ sin x cos x dx = 2 ∫ sin 2 x dx = −cos 2 x + C
116. False
∫ sin
118. (a)
2
2 x cos 2 x dx =
1
(sin 2 x)2 (2 cos 2 x) dx
2∫
1 (sin 2 x)
+C
2
3
1
= sin 3 2 x + C
6
3
=
117. Let u = cx, du = c dx :
c∫
b
a
f (cx) dx = c ∫
cb
ca
du
f (u )
c
=
∫ ca f (u ) du
=
∫ ca f ( x) dx
cb
cb
d
[sin u − u cos u + C ] = cos u − cos u + u sin u
du
= u sin u
So,
∫ u sin u du
(b) Let u =
π2
∫0
sin
= sin u − u cos u + C.
x , u 2 = x, 2u du = dx.
x dx =
π
∫0
= 2∫
sin u ( 2u du )
π
0
u sin u du
π
= 2[sin u − u cos u]0 (part (a))
= 2 ⎡−
⎣ π cos(π )⎤⎦
= 2π
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.6
119. Because f is odd, f ( − x ) = − f ( x). Then
∫ − a f ( x) dx
a
121. Let f ( x ) = a0 + a1x + a2 x 2 +
∫ − a f ( x) dx + ∫ 0 f ( x) dx
a
0
=
= −∫
−a
0
f ( x ) dx +
1
∫0
f ( x) dx.
a
∫0
Numerical Integration
+ an x n .
⎡
x2
x3
f ( x) dx = ⎢a0 x + a1
+ a2
+
2
3
⎣
= a0 +
Let x = −u , dx = − du in the first integral.
a1 a2
+
+
2
3
487
+
1
+ an
x n +1 ⎤
⎥
n + 1⎦ 0
an
= 0 (Given)
n +1
When x = 0, u = 0. When x = − a, u = a.
By the Mean Value Theorem for Integrals, there exists c
in [0, 1] such that
∫ − a f ( x) dx
∫ 0 f ( x) dx
1
= −∫
= −∫
a
0
a
0
f ( −u )( − du ) +
f (u ) du +
∫ 0 f ( x) dx
a
∫ 0 f ( x) dx
a
1
0 = f (c).
= 0
So the equation has at least one real zero.
120. Let u = x + h, then du = dx.
122. α 2 ∫ f ( x) dx = α 2 (1) = α 2
1
When x = a, u = a + h.
0
When x = b, u = b + h. So,
∫ a f ( x + h) dx
b
=
= f (c)(1 − 0)
−2α ∫ f ( x) x dx = −2α (α ) = −2α 2
1
b+h
∫ a + h f (u ) du
0
b+h
∫ a + h f ( x) dx.
=
2
∫ 0 f ( x) x dx
1
= α2
Adding,
2
∫ 0 ⎡⎣α f ( x) − 2α xf ( x) +
1
∫ 0 f ( x)(α
1
x 2 f ( x)⎤⎦ dx = 0
− x) dx = 0.
2
Because (α − x) ≥ 0, f = 0. So, there are no such
2
functions.
Section 5.6 Numerical Integration
1. Exact:
2
∫0
2
x 2 dx = ⎡⎣ 13 x 3 ⎤⎦ =
0
Trapezoidal:
Simpson’s:
2
∫0
2
∫0
x 2 dx ≈
x 2 dx ≈
8
3
1 ⎡0
4⎣
⎢
1 ⎡0
6⎣
⎢
≈ 2.6667
( 12 )
+ 2
( 12 )
+ 4
2
2
( 32 )
+ 2(1) + 2
2
( 32 )
+ 2(1) + 4
2
2
2
2
+ ( 2) ⎤ =
⎦⎥
2
+ ( 2) ⎤ =
⎦⎥
8
3
11
4
= 2.7500
≈ 2.6667
2
2. Exact:
2
∫1
⎛ x2
⎞
⎡ x3
⎤
19
+ 1⎟ dx = ⎢ + x⎥ =
≈ 1.5833
⎜
12
⎝4
⎠
⎣12
⎦1
Trapezoidal:
Simpson’s:
2
∫1
1
⎛ ( 5 4) 2
⎞
⎛ ( 3 2) 2
⎞
⎛ ( 7 4) 2
⎞ ⎛ 22
⎛ x2
⎞
⎞
⎞⎤ 203
1 ⎡⎛ 12
+ 1⎟ dx ≈ ⎢⎜ + 1⎟ + 2⎜
+ 1⎟ + 2⎜
+ 1⎟ + 2⎜
+ 1⎟ + ⎜
+ 1⎟⎥ =
≈ 1.5859
⎜
⎜
⎟
⎜
⎟
⎜
⎟
8 ⎢⎝ 4
4
4
4
4
⎝4
⎠
⎠
⎠⎥⎦ 128
⎝
⎠
⎝
⎠
⎝
⎠ ⎝
⎣
⎛ x2
∫ 0 ⎜⎝ 4
⎛ ( 5 4) 2
⎞
⎛ (3 2) 2
⎞
⎛ ( 7 4) 2
⎞ ⎛ 22
⎞
⎞
⎞⎤
1 ⎡⎛ 12
19
⎢⎜ + 1⎟ + 4⎜
+ 1⎟ dx ≈
+ 1⎟ + 2⎜
+ 1⎟ + 4⎜
+ 1⎟ + ⎜
+ 1⎟⎥ =
≈ 1.5833
⎜
⎟
⎜
⎟
⎜
⎟
12
4
4
4
4
4
12
⎢
⎥
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
488
Chapter 5
Integration
2
3. Exact:
2
∫0
⎡ x4 ⎤
x 3 dx = ⎢ ⎥ = 4.0000
⎣ 4 ⎦0
Trapezoidal:
Simpson’s:
4. Exact:
3
∫2
2
∫0
2
∫0
x3 dx ≈
3
3
1⎡
17
3
3⎤
⎛1⎞
⎛ 3⎞
= 4.2500
⎢0 + 2⎜ ⎟ + 2(1) + 2⎜ ⎟ + ( 2) ⎥ =
4 ⎣⎢
4
⎝ 2⎠
⎝2⎠
⎦⎥
3
3
1⎡
24
3
3⎤
⎛1⎞
⎛ 3⎞
= 4.0000
⎢0 + 4⎜ ⎟ + 2(1) + 4⎜ ⎟ + ( 2) ⎥ =
6 ⎣⎢
6
⎝ 2⎠
⎝ 2⎠
⎦⎥
x 3 dx ≈
3
2
2
2
1
⎡ 2⎤
dx = ⎢− ⎥ = − +
=
2
x
3
2
3
⎣ x ⎦2
Trapezoidal:
Simpson’s:
3
∫2
⎛ 2 ⎞
⎛ 2 ⎞
⎛ 2 ⎞
2
1⎡ 2
2⎤
⎟ + 2⎜
⎟ + 2⎜
⎟ + 2 ⎥ ≈ 0.3352
dx ≈ ⎢ 2 + 2⎜
2
2
2
2
⎜ ( 9 4) ⎟
⎜ (10 4) ⎟
⎜ (11 4) ⎟ 3 ⎥
x
8⎢2
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
⎛ 2 ⎞
⎛ 2 ⎞
⎛ 2 ⎞
2
1⎡2
2⎤
⎢ 2 + 4⎜
⎟ + 2⎜
⎟ + 4⎜
⎟ + 2 ⎥ ≈ 0.3334
dx ≈
2⎟
2⎟
2⎟
2
⎜
⎜
⎜
x
12 ⎢ 2
3 ⎥
⎝ ( 9 4) ⎠
⎝ (10 4) ⎠
⎝ (11 4) ⎠
⎣
⎦
3
∫2
3
5. Exact:
3
∫1
⎡ x4 ⎤
81 1
x 3 dx = ⎢ ⎥ =
−
= 20
4
4
4
⎣ ⎦1
Trapezoidal:
Simpson’s:
6. Exact:
8
∫0
3
∫1
8
3
3
1 ⎡0
2⎣
x dx ≈
x dx ≈
1 ⎡0
3⎣
+ 2 + 2 3 2 + 2 3 3 + 2 3 4 + 2 3 5 + 2 3 6 + 2 3 7 + 2⎤⎦ ≈ 11.7296
+ 4 + 2 3 2 + 4 3 3 + 2 3 4 + 4 3 5 + 2 3 6 + 4 3 7 + 2⎤⎦ ≈ 11.8632
x dx = ⎡⎣ 23 x3 2 ⎤⎦ = 18 −
4
9
∫4
9
∫4
2
∫ 1 (4 − x ) dx
Trapezoidal:
5⎡
⎢2 + 2
16 ⎣
x dx ≈
x dx ≈
4
Simpson’s:
3
3
3
3
⎤
1⎡
3
⎛ 4⎞
⎛5⎞
⎛7⎞
⎛8⎞
⎢1 + 4⎜ ⎟ + 2⎜ ⎟ + 4( 2) + 2⎜ ⎟ + 4⎜ ⎟ + 27⎥ = 20.0000
9 ⎣⎢
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
⎦⎥
9
9
Trapezoidal:
8. Exact:
8
∫0
∫0
∫4
Simpson’s:
x 3 dx ≈
3
3
3
3
⎤
1⎡
3
⎛ 4⎞
⎛5⎞
⎛7⎞
⎛8⎞
⎢1 + 2⎜ ⎟ + 2⎜ ⎟ + 2( 2) + 2⎜ ⎟ + 2⎜ ⎟ + 27⎥ ≈ 20.2222
6 ⎣⎢
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
⎦⎥
8
Trapezoidal:
7. Exact:
x3 dx ≈
x dx = ⎡⎣ 34 x 4 3 ⎤⎦ = 12.0000
0
3
Simpson’s:
3
∫1
5⎡
⎢2 + 4
24 ⎣
=
38
3
37
+ 2
8
37
+
8
≈ 12.6667
21
+ 2
4
21 + 4
47
+ 2
8
47
+
8
26
+ 2
4
26 + 4
57
+ 2
8
57
+
8
31
+ 2
4
31 + 4
⎤
67
+ 3⎥ ≈ 12.6640
8
⎦
⎤
67
+ 3⎥ ≈ 12.6667
8
⎦
4
⎡
x3 ⎤
16 11
= ⎢4 x − ⎥ = −
−
= −9
3 ⎦1
3
3
⎣
2
∫ 1 (4 − x ) dx
4
2
∫ 1 (4 − x ) dx
4
16
3
≈
≈
2
2
2
⎫⎪
⎡
⎡
⎡
1 ⎪⎧
⎛ 3⎞ ⎤
⎛5⎞ ⎤
⎛7⎞ ⎤
⎨3 + 2 ⎢4 − ⎜ ⎟ ⎥ + 2(0) + 2 ⎢4 − ⎜ ⎟ ⎥ + 2( −5) + 2 ⎢4 − ⎜ ⎟ ⎥ − 12⎬ ≈ −9.1250
4⎪
2
2
2
⎝ ⎠ ⎦⎥
⎝ ⎠ ⎦⎥
⎝ ⎠ ⎦⎥
⎢⎣
⎢⎣
⎢⎣
⎪⎭
⎩
⎤
1⎡
9⎞
25 ⎞
49 ⎞
⎛
⎛
⎛
3 + 4⎜ 4 − ⎟ + 0 + 4⎜ 4 −
⎟ − 10 + 4⎜ 4 −
⎟ − 12⎥ = −9
6 ⎢⎣
4⎠
4⎠
4⎠
⎝
⎝
⎝
⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.6
9. Exact:
∫ 0 ( x + 2) 2
Trapezoidal:
Simpson’s:
⎡ −2 ⎤
−2 2
1
dx = ⎢
+
=
⎥ =
3
2
3
⎢⎣ ( x + 2) ⎥⎦ 0
2
1
∫ 0 ( x + 2)2 dx
2
1
∫ 0 ( x + 2) 2
≈
⎡
⎛
⎞
⎛
⎞
⎛
⎞ 2⎤
1 ⎢1
2
2
2
⎟ + 2⎜
⎟ + 2⎜
⎟+ ⎥
+ 2⎜
2
2
2
⎜ ((1 4) + 2) ⎟
⎜ ((1 2) + 2) ⎟
⎜ ( ( 3 4) + 2) ⎟ 9 ⎥
8⎢2
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
=
1 ⎡1
⎛ 32 ⎞
⎛8⎞
⎛ 32 ⎞ 2 ⎤
+ 2⎜ ⎟ + 2⎜ ⎟ + 2⎜
⎟ + ⎥ ≈ 0.3352
8 ⎢⎣ 2
⎝ 81 ⎠
⎝ 25 ⎠
⎝ 121 ⎠ 9 ⎦
⎡
⎛
⎞
⎛
⎞
⎛
⎞ 2⎤
1 ⎢1
2
2
2
⎟ + 2⎜
⎟ + 4⎜
⎟+ ⎥
+ 4⎜
⎜ ((1 4) + 2)2 ⎟
⎜ ((1 2) + 2)2 ⎟
⎜ ( ( 3 4) + 2) 2 ⎟ 9 ⎥
12 ⎢ 2
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
dx ≈
1 ⎡1
⎛ 32 ⎞
⎛8⎞
⎛ 32 ⎞ 2 ⎤
+ 4⎜ ⎟ + 2⎜ ⎟ + 4⎜
⎟ + ⎥ ≈ 0.3334
12 ⎢⎣ 2
⎝ 81 ⎠
⎝ 25 ⎠
⎝ 121 ⎠ 9 ⎦
=
10. Exact:
2
∫0
x 2 + 1 dx =
x
Trapezoidal:
Simpson’s:
2
11. Trapezoidal:
Simpson’s:
2
∫0
∫0
489
1
2
1
Numerical Integration
x
( x 2 + 1)
1⎡
3⎣
⎢
1 ⎡0
4⎢
x 2 + 1 dx ≈
⎣
32 2
⎤ =
⎦⎥ 0
1
3
( 12 ) ( 12 )
+ 2
x
x 2 + 1 dx ≈
1 ⎡0
6⎢
+ 4
2
1 + x 3 dx ≈
1 ⎡1
4⎣
⎢
+ 2 1+
∫0
2
∫0
1 + x 3 dx ≈
⎣
1 ⎡1
6⎢
⎣
(53 2 − 1) ≈ 3.393
( 12 ) ( 12 )
2
2
( 32 ) ( 32 )
+ 1 + 2(1) 12 + 1 + 4
( 18 ) + 2
( 18 ) + 2
+ 4 1+
( 32 ) ( 23 )
+ 1 + 2(1) 12 + 1 + 2
2 + 2 1+
2 + 4 1+
( 278 ) + 3⎤⎦⎥
( 278 ) + 3⎤⎥⎦
2
2
⎤
+ 1 + 2 22 + 1⎥ ≈ 3.457
⎦
⎤
+ 1 + 2 22 + 1⎥ ≈ 3.392
⎦
≈ 3.283
≈ 3.240
Graphing utility: 3.241
12. Trapezoidal:
Simpson’s:
1
2
∫0
1 + x3
1
2
∫0
1 + x3
⎡
⎛
1⎢
1 + 2⎜
⎜⎜
4⎢
⎝
⎣⎢
dx ≈
dx ≈
⎡
⎛
1⎢
1 + 4⎜
⎜⎜
6⎢
⎝
⎣⎢
⎞
⎟ + 2⎛
⎜
3 ⎟
⎝
1 + (1 2) ⎟⎠
1
⎞
⎟ + 2⎛
⎜
3 ⎟
⎝
1 + (1 2) ⎠⎟
1
⎛
⎞
⎜
+
2
⎟
⎜⎜
1 + 13 ⎠
⎝
1
⎛
⎞
⎜
4
+
⎟
⎜⎜
1 + 13 ⎠
⎝
1
⎤
⎞
⎟ + 1 ⎥ ≈ 1.397
⎥
3 ⎟
1 + (3 2) ⎟⎠ 3 ⎥
⎦
1
⎤
⎞
⎟ + 1 ⎥ ≈ 1.405
⎥
3 ⎟
1 + (3 2) ⎠⎟ 3 ⎥
⎦
1
Graphing utility: 1.402
13.
1
∫0
x
1 − x dx =
Trapezoidal:
Simpson’s:
1
∫0
1
∫0
x(1 − x) dx
x(1 − x) dx ≈
1
∫0
(1 − 14 ) + 2 12 (1 − 12 ) + 2 34 (1 − 34 ) ⎤⎥⎦
≈ 0.342
(1 − 14 ) + 2 12 (1 − 12 ) + 4 34 (1 − 34 ) ⎤⎦⎥
≈ 0.372
1 ⎡0
8⎢
⎣
+ 2
1
4
1 ⎡0
12 ⎣
⎢
+ 4
1
4
x(1 − x) dx ≈
Graphing utility: 0.393
14. Trapezoidal:
Simpson’s:
4
∫0
4
∫0
xe x dx ≈
xe x dx ≈
1
⎡0 + 2e1 + 2
2⎣
1
⎡0 + 4e1 + 2
3⎣
2e 2 + 2
2e 2 + 4
3e3 + 2e 4 ⎤⎦ ≈ 102.555
3e3 + 2e 4 ⎤⎦ ≈ 93.375
Graphing utility: 92.744
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
490
Chapter 5
15. Trapezoidal:
Simpson’s:
Integration
π 2
∫0
∫0
π 2
⎛
π 2 ⎡⎢
sin 0 + 2 sin ⎜
sin ( x 2 ) dx ≈
⎜
⎝
⎢
⎣
8
⎛
π 2 ⎡⎢
sin 0 + 4 sin ⎜
sin ( x 2 ) dx ≈
2
⎛
⎟⎟ + 2 sin ⎜⎜
⎠
⎝
4
π 2⎞
⎜
⎝
12 ⎢
⎣
π 2⎞
2
⎛
⎟⎟ + 2 sin ⎜⎜
⎠
⎝
4
π 2⎞
2
2
⎛3 π 2 ⎞
⎛
⎟⎟ + 2 sin ⎜⎜
⎟⎟ + sin ⎜⎜
⎝
⎠
⎝ 4 ⎠
2
π 2⎞
2
2
2⎠ ⎥
⎦
π ⎞ ⎤⎥
⎟⎟ ≈ 0.548
2
2
⎛3 π 2 ⎞
⎛
⎟⎟ + 4 sin ⎜⎜
⎟⎟ + sin ⎜⎜
4
⎝
⎠
⎝
⎠
2
π ⎞ ⎤⎥
⎟⎟ ≈ 0.550
2⎠ ⎥
⎦
Graphing utility: 0.549
16. Trapezoidal:
Simpson’s:
π 4
∫0
∫0
π 4
⎛ π 4⎞
⎛
π 4 ⎡⎢
tan 0 + 2 tan ⎜
⎟⎟ + 2 tan ⎜⎜
⎜
2
tan ( x 2 ) dx ≈
⎢
⎣
8
⎝
4
⎠
⎝
π 4⎞
2
2
2
⎛3 π 4 ⎞
⎛
⎟⎟ + 2 tan ⎜⎜
⎟⎟ + tan ⎜⎜
4
⎝
⎠
⎝
⎠
π ⎞ ⎤⎥
⎟⎟ ≈ 0.271
2
4⎠ ⎥
⎦
⎛ π 4⎞
⎛ π 4⎞
⎛3 π 4 ⎞
⎛ π⎞ ⎤
π 4 ⎡⎢
tan 0 + 4 tan ⎜
+ 2 tan ⎜
+ 4 tan ⎜
+ tan ⎜⎜
⎟
⎟
⎟
⎟⎟ ⎥ ≈ 0.257
⎜
⎟
⎜
⎟
⎜
⎟
12 ⎢
⎝ 4⎠ ⎥
⎝ 4 ⎠
⎝ 2 ⎠
⎝ 4 ⎠
2
tan ( x 2 ) dx ≈
2
2
⎣
2
⎦
Graphing utility: 0.256
17. Trapezoidal:
Simpson’s:
3.1
∫3
3.1
∫3
cos x 2 dx ≈
cos x 2 dx ≈
0.1 ⎡
cos
8 ⎣
0.1 ⎡
cos
12 ⎣
(3)2
(3)2
2
2
2
2
+ 2 cos(3.025) + 2 cos(3.05) + 2 cos(3.075) + cos(3.1) ⎤ ≈ −0.098
⎦
2
2
2
2
+ 4 cos(3.025) + 2 cos(3.05) + 4 cos(3.075) + cos(3.1) ⎤ ≈ −0.098
⎦
Graphing utility: −0.098
18. Trapezoidal:
Simpson’s:
π 2
∫0
π 2
∫0
1 + sin 2 x dx ≈
1 + sin 2 x dx ≈
π⎡
⎛π ⎞
⎛π ⎞
⎛ 3π ⎞
⎢1 + 2 1 + sin 2 ⎜ ⎟ + 2 1 + sin 2 ⎜ ⎟ + 2 1 + sin 2 ⎜ ⎟ +
16 ⎣⎢
⎝8⎠
⎝4⎠
⎝ 8 ⎠
π ⎡
⎛π ⎞
⎛π ⎞
⎛ 3π ⎞
⎢1 + 4 1 + sin 2 ⎜ ⎟ + 2 1 + sin 2 ⎜ ⎟ + 4 1 + sin 2 ⎜ ⎟ +
24 ⎣⎢
⎝8⎠
⎝4⎠
⎝ 8 ⎠
⎤
2 ⎥ ≈ 1.910
⎦⎥
⎤
2 ⎥ ≈ 1.910
⎦⎥
Graphing utility: 1.910
19. Trapezoidal:
Simpson’s:
2
∫0
x ln ( x + 1) dx ≈
x ln ( x + 1) dx ≈
2
∫0
1
⎡0 + 2(0.5) ln (1.5) + 2 ln ( 2) + 2(1.5) ln ( 2.5) + 2 ln (3)⎤⎦ ≈ 1.684
4⎣
1
⎡0 + 4(0.5) ln (1.5) + 2 ln ( 2) + 4(1.5) ln ( 2.5) + 2 ln (3)⎤⎦ ≈ 1.649
6⎣
Graphing utility: 1.648
20. Trapezoidal:
Simpson’s:
3
∫1
3
∫1
ln x dx ≈
ln x dx ≈
1
5.1284
⎡0 + 2 ln (1.5) + 2 ln 2 + 2 ln ( 2.5) + ln 3⎤⎦ ≈
≈ 1.282
4⎣
4
1
7.7719
⎡0 + 4 ln (1.5) + 2 ln 2 + 4 ln ( 2.5) + ln 3⎤⎦ ≈
≈ 1.295
6⎣
6
Graphing utility: 1.296
21. Trapezoidal:
Simpson’s:
2
∫0
2
∫0
xe − x dx ≈
2 xe− x dx ≈
1
2.2824
⎡⎣0 + e −1 2 + 2e−1 + 3e− 3 2 + 2e − 2 ⎤⎦ ≈
≈ 0.5706
4
4
1
3.5583
⎡0 + 2e−1 2 + 2e −1 + 6e − 3 2 + 2e − 2 ⎤⎦ ≈
≈ 0.5930
6⎣
6
Graphing utility: 0.594
22. Trapezoidal:
Simpson’s:
π
∫0
π
∫0
2 sin (π 4) 2 sin (π 2) 2 sin (3π 4)
⎤
sin x
π⎡
dx ≈ ⎢1 +
+
+
+ 0⎥ ≈ 1.836
x
8⎣
3π 4
π 4
π 2
⎦
4 sin (π 4) 2 sin (π 2) 4 sin (3π 4)
⎤
sin x
π⎡
dx ≈
+
+
+ 0⎥ ≈ 1.852
⎢1 +
x
12 ⎣
3π 4
π 4
π 2
⎦
Graphing utility: 1.852
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.6
23. Trapezoidal: Linear polynomials
Simpson’s: Quadratic polynomials
25.
491
f ( x) = 2 x3
f ′( x) = 6 x 2
f ′′( x) = 12 x
24. For a linear function, the Trapezoidal Rule is exact. The
(b
− a)
⎡max f ′′( x) ⎤⎦
12n 2 ⎣
and f ′′( x) = 0 for a linear function. Geometrically, a
error formula says that E ≤
Numerical Integration
f ′′′( x) = 12
3
f ( 4) ( x ) = 0
(a) Trapezoidal: Error ≤
linear function is approximated exactly by trapezoids:
y
(3 − 1)3
12( 42 )
(36)
= 1.5 because
f ′′( x) is maximum in [1, 3] when x = 3.
(b) Simpson’s: Error ≤
5
180( 44 )
( 0)
= 0 because
f (4) ( x) = 0.
x
a
(3 − 1)
b
26.
f ( x) = 5 x + 2
f ′( x) = 5
f ′′( x) = 0
The error is 0 for both rules.
27.
f ( x ) = ( x − 1)
−2
f ′( x ) = −2( x − 1)
f ′′( x ) = 6( x − 1)
−3
−4
f ′′′( x ) = −24( x − 1)
f (4) ( x ) = 120( x − 1)
−5
−6
(a) Trapezoidal: Error ≤
(b) Simpson’s: Error ≤
28.
(4
− 2)
12( 4
)
2
( 4 − 2)
3
5
180( 44 )
( 6)
=
(120)
=
1
because f ′′( x) is a maximum of 6 at x = 2.
4
1
because f (4) ( x) is a maximum of 120 at x = 2.
12
f ( x) = cos x
f ′( x) = −sin x
f ′′( x) = −cos x
f ′′′( x) = sin x
f (4) ( x) = cos x
(a) Trapezoidal: Error ≤
(b) Simpson’s: Error ≤
(π
− 0)
12( 4
(π
2
− 0)
180( 4
4
)
5
)
3
(1)
(1)
=
=
π3
192
≈ 0.1615 because f ′′( x) is at most 1 on [0, π ].
π5
46,080
≈ 0.006641 because f (4) ( x) is at most 1 on [0, π ].
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
492
29.
Chapter 5
Integration
f ( x ) = x −1 ,
f ′( x) = − x
1≤ x ≤ 3
−2
f ′′( x) = 2 x −3
f ′′′( x) = −6 x −4
f (4) ( x) = 24 x −5
(a) Maximum of f ′′( x ) = 2 x −3 is 2.
Trapezoidal: Error ≤
23
(2) ≤ 0.00001, n 2 ≥ 133,333.33, n ≥ 365.15 Let n = 366.
12n 2
(b) Maximum of f (4) ( x ) = 24 x −5 is 24.
Simpson’s: Error ≤
30.
25
(24) ≤ 0.00001, n 4 ≥ 426,666.67, n ≥ 25.56 Let n = 26.
180n 4
f ( x) = (1 + x) , 0 ≤ x ≤ 1
−1
f ′( x) = −(1 + x )
f ′′( x) = 2(1 + x)
31.
−2
f (4) ( x) = 24(1 + x)
12
1
−1 2
( x + 2)
2
1
−3 2
f ′′( x) = − ( x + 2)
4
3
−5 2
f ′′′( x) = ( x + 2)
8
−15
−7 2
f ( 4) ( x ) =
( x + 2)
16
f ′( x) =
−3
f ′′′( x) = −6(1 + x)
f ( x ) = ( x + 2) , 0 ≤ x ≤ 2
−4
−5
(a) Maximum of f ′′( x ) = 2(1 + x)
−3
is 2.
Trapezoidal:
Error ≤
(a) Maximum of f ′′( x) =
1
( 2) ≤ 0.00001
12n 2
n 2 ≥ 16,666.67
n ≥ 129.10. Let n = 130.
(b) Maximum of f (4) ( x ) = 24(1 + x)
−5
4( x + 2)
is
32
2
≈ 0.0884.
16
Trapezoidal:
is 24.
Error ≤
Simpson’s:
Error ≤
−1
1
(24) ≤ 0.00001
180n 4
(2
− 0) ⎛ 2 ⎞
⎜
⎟ ≤ 0.00001
12n 2 ⎜⎝ 16 ⎟⎠
3
n2 ≥
n 4 ≥ 13,333.33
n ≥ 76.8. Let n = 77.
n ≥ 10.75
Let n = 12. (In Simpson’s Rule n must be even.)
8 2 5
2 5
10 =
10
12(16)
24
(b) Maximum of f (4) ( x) =
−15
16( x + 2)
72
is
15 2
≈ 0.0829.
256
Simpson’s:
Error
≤
25 ⎛ 15 2 ⎞
⎜
⎟ ≤ 0.00001
180n 4 ⎜⎝ 256 ⎟⎠
n4 ≥
32(15) 2
180( 256)
105 =
2 5
10
96
n ≥ 6.2. Let n = 8 (even ).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.6
f ( x) = sin x, 0 ≤ x ≤
32.
f ′( x) = cos x
π
33. f ( x ) =
2
f ′′′( x) = −cos x
(a) Trapezoidal:
(π 2)
3
12n 2
(1)
≤ 0.00001
n ≥
π3
10
(b) f (4) ( x) = −
5
96
n ≥ 179.7. Let n = 180.
f ( 4 ) ( 0) =
5
180n 4
(1)
≤ 0.00001
π
n4 ≥
15
72
in [0, 2]
15
.
16
Simpson’s: Error ≤
32 ⎛ 15 ⎞
⎜ ⎟ ≤ 0.00001,
180n 4 ⎝ 16 ⎠
n 4 ≥ 16,666.67, n ≥ 11.36; let n = 12.
23
2
9( x + 1)
43
in [0, 2].
f ′′( x) is maximum when x = 0 and f ′′(0) =
Trapezoidal: Error ≤
(b) f (4) ( x) = −
`
1
.
4
8 ⎛1⎞
⎜ ⎟ ≤ 0.00001,
12n 2 ⎝ 4 ⎠
16(1 + x)
5
105
5760
n ≥ 8.5. Let n = 10 (even ).
(a) f ′′( x) = −
in [0, 2].
f (4) ( x) is maximum when x = 0 and
(b) Simpson’s:
34. f ( x ) = ( x + 1)
32
n 2 ≥ 16,666.67, n ≥ 129.10; let n = 130.
2
(π 2)
4(1 + x)
Trapezoidal: Error ≤
All derivatives are bounded by 1.
Error ≤
1
f ′′( x) is maximum when x = 0 and f ′′(0) =
f (4) ( x) = sin x
Error ≤
493
1+ x
(a) f ′′( x) = −
f ′′( x) = −sin x
Numerical Integration
2
.
9
8 ⎛ 2⎞
2
⎜ ⎟ ≤ 0.00001, n ≥ 14,814.81, n ≥ 121.72; let n = 122.
12n 4 ⎝ 9 ⎠
56
81( x + 1)
10 3
in [0, 2].
f (4) ( x) is maximum when x = 0 and f (4) (0) =
Simpson’s: Error ≤
56
.
81
32 ⎛ 56 ⎞
4
⎜ ⎟ ≤ 0.00001, n ≥ 12,290.81, n ≥ 10.53; let n = 12. (In Simpson’s Rule n must be even.)
180n 4 ⎝ 81 ⎠
35. f ( x ) = tan ( x 2 )
(a) f ′′( x) = 2 sec 2 ( x 2 ) ⎡⎣1 + 4 x 2 tan ( x 2 )⎤⎦ in [0, 1].
f ′′( x) is maximum when x = 1 and f ′′(1) ≈ 49.5305.
Trapezoidal: Error ≤
(1 − 0)
12n 2
3
(49.5305)
≤ 0.00001, n 2 ≥ 412,754.17, n ≥ 642.46; let n = 643.
(b) f (4) ( x) = 8 sec 2 ( x 2 ) ⎣⎡12 x 2 + (3 + 32 x 4 ) tan ( x 2 ) + 36 x 2 tan 2 ( x 2 ) + 48 x 4 tan 3 ( x 2 )⎦⎤ in [0, 1]
f (4) ( x) is maximum when x = 1 and f (4) (1) ≈ 9184.4734.
Simpson’s: Error ≤
(1 − 0)
180n 4
5
(9184.4734)
≤ 0.00001, n 4 ≥ 5,102,485.22, n ≥ 47.53; let n = 48.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
494
Chapter 5
Integration
36. f ( x ) = sin ( x 2 )
(a) f ′′( x) = 2 ⎡⎣−2 x 2 sin ( x 2 ) + cos( x 2 )⎤⎦ in [0, 1]. f ′′( x) is maximum when x = 1 and f ′′(1) ≈ 2.2853.
Trapezoidal: Error ≤
(1 − 0)
12n 2
3
(2.2853)
≤ 0.00001, n 2 ≥ 19,044.17, n ≥ 138.00; let n = 139.
(b) f (4) ( x) = (16 x 4 − 12) sin ( x 2 ) − 48 x 2 cos( x 2 ) in [0, 1]
f (4) ( x) is maximum when x ≈ 0.852 and f (4) (0.852) ≈ 28.4285.
(1 − 0)
Simpson’s: Error ≤
180n 4
5
(28.4285)
≤ 0.00001, n 4 ≥ 15,793.61, n ≥ 11.21; Let n = 12.
37. n = 4, b − a = 4 − 0 = 4
(a)
∫ 0 f ( x) dx
≈
4 ⎡3
8⎣
(b)
∫ 0 f ( x) dx
≈
4 ⎡3
12 ⎣
4
4
+ 2(7) + 2(9) + 2(7) + 0⎤⎦ =
+ 4(7) + 2(9) + 4(7) + 0⎤⎦ =
1
2
(49)
77
3
=
= 24.5
49
2
≈ 25.67
38. n = 8, b − a = 8 − 0 = 8
(a)
∫ 0 f ( x) dx
≈
8 ⎡0
16 ⎣
+ 2(1.5) + 2(3) + 2(5.5) + 2(9) + 2(10) + 2(9) + 2(6) + 0⎤⎦ =
1
2
(88)
(b)
∫ 0 f ( x) dx
≈
8 ⎡0
24 ⎣
+ 4(1.5) + 2(3) + 4(5.5) + 2(9) + 4(10) + 2(9) + 4(6) + 0⎤⎦ =
1
3
(134)
8
8
39. A =
π 2
∫0
= 44
=
134
3
x cos x dx
Simpson’s Rule: n = 14
π 2
∫0
x cos x dx ≈
π ⎡
⎢ 0 cos 0 + 4
84 ⎣
π
28
cos
π
28
+ 2
π
14
cos
π
14
+ 4
3π
3π
cos
+"+
28
28
π
2
cos
π⎤
⎥ ≈ 0.701
2⎦
y
1
1
2
π
π
4
2
x
40. Simpson’s Rule: n = 8
8 3∫
π 2
0
1−
2
sin 2 θ dθ ≈
3
3π ⎡
2
2
2
π
2
2 π
+ 2 1 − sin 2
+"+
⎢ 1 − sin 0 + 4 1 − sin
6 ⎣
3
3
16
3
8
1−
2
π ⎤
sin 2 ⎥
3
2⎦
≈ 17.476
41. Area ≈
1000
⎡125 + 2(125) + 2(120) + 2(112) + 2(90) + 2(90) + 2(95) + 2(88) + 2(75) + 2(35)⎤⎦ = 89,250 m 2
2(10) ⎣
42. (a) The integral
∫ 0 f ( x)dx would be overestimated because the trapezoids would be above the curve. Similarly, the integral
2
∫ 0 g ( x)dx would be underestimated.
2
(b) Simpson’s Rule would be more accurate because it takes into account the curvature of the graph.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.6
5
∫ 0 100 x
43. W =
Numerical Integration
495
125 − x3 dx
Simpson’s Rule: n = 12
5
∫ 0 100 x
125 − x 3 dx ≈
3
3
5 ⎡⎢
⎛5⎞
⎛5⎞
⎛ 10 ⎞
⎛ 10 ⎞
0 + 400⎜ ⎟ 125 − ⎜ ⎟ + 200⎜ ⎟ 125 − ⎜ ⎟
3(12) ⎢
⎝ 12 ⎠
⎝ 12 ⎠
⎝ 12 ⎠
⎝ 12 ⎠
⎣
3
⎤
⎛ 15 ⎞
⎛ 15 ⎞
+ 400⎜ ⎟ 125 − ⎜ ⎟ + " + 0⎥ ≈ 10,233.58 ft-lb
⎥
⎝ 12 ⎠
⎝ 12 ⎠
⎦
44. (a) Trapezoidal:
∫ 0 f ( x) dx
2
≈
2
⎡4.32 + 2(4.36) + 2( 4.58) + 2(5.79) + 2(6.14) + 2(7.25) + 2(7.64) + 2(8.08) + 8.14⎤⎦ ≈ 12.518
2(8) ⎣
≈
2
⎡4.32 + 4( 4.36) + 2( 4.58) + 4(5.79) + 2(6.14) + 4(7.25) + 2(7.64) + 4(8.08) + 8.14⎤⎦ ≈ 12.592
3(8) ⎣
Simpson’s:
∫ 0 f ( x) dx
2
(b) Using a graphing utility,
y = −1.37266 x3 + 4.0092 x 2 − 0.620 x + 4.28. Integrating,
45.
6
12
∫0
1 − x2
2
∫0
y dx ≈ 12.521.
dx Simpson’s Rule, n = 6
⎛1
⎞
⎜ − 0⎟
2
⎝
⎠ ⎡6 + 4 6.0209 + 2 6.0851 + 4 6.1968 + 2 6.3640 + 4 6.6002 + 6.9282⎤ ≈ 1 113.098 ≈ 3.1416
π ≈
(
) (
) (
) (
) (
)
[
]
⎦
3(6) ⎣
36
46. Simpson’s Rule: n = 6
π = 4∫
47.
t
∫ 0 sin
1
0
1
4 ⎡
4
2
4
2
4
1⎤
⎢1 +
+
+
+
+
+ ⎥ ≈ 3.14159
dx ≈
2
2
2
2
2
2
1+ x
3(6) ⎢
2⎥
1 + (1 6)
1 + ( 2 6)
1 + ( 3 6)
1 + ( 4 6)
1 + ( 5 6)
⎣
⎦
x dx = 2, n = 10
By trial and error, you obtain t ≈ 2.477.
48. Let f ( x ) = Ax 3 + Bx 2 + Cx + D. Then f (4) ( x) = 0.
Simpson’s: Error ≤
(b
− a)
( 0) = 0
180n 4
5
So, Simpson’s Rule is exact when approximating the integral of a cubic polynomial.
Example:
1
∫0
x 3 dx =
3
⎤
1⎡
1
⎛1⎞
⎢0 + 4⎜ ⎟ + 1⎥ =
6 ⎣⎢
4
⎝ 2⎠
⎥⎦
This is the exact value of the integral.
49. The quadratic polynomial
p( x) =
(x
( x1
− x2 )( x − x3 )
( x − x1 )( x − x3 ) y + ( x − x1 )( x − x2 ) y
y1 +
− x2 )( x1 − x3 )
( x2 − x1 )( x2 − x3 ) 2 ( x3 − x1 )( x3 − x2 ) 3
passes through the three points.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
496
Chapter 5
Integration
Section 5.7 The Natural Logarithmic Function: Integration
1.
2.
5
∫ x dx
∫
= 5∫
8. u = 5 − x3 , du = −3x 2 dx
1
dx = 5 ln x + C
x
x2
∫5 −
10
1
dx = 10 ∫ dx = 10 ln x + C
x
x
3. u = x + 1, du = dx
∫
∫
1
1
( 2) dx
2 ∫ 2x + 5
1
= ln 2 x + 5 + C
2
=
11.
6. u = 5 − 4 x, du = −4 dx
9
∫ 5 − 4 x dx
∫
x2 − 2x
1
1
dx = ∫ 3
(3x 2 − 6 x) dx
3
2
x − 3x
3 x − 3x 2
1
= ln x3 − 3x 2 + C
3
∫
x2 − 4
dx =
x
⎛
4⎞
⎟ dx
x⎠
∫ ⎜⎝ x −
x2
− 4 ln x + C
2
x2
=
− ln ( x 4 ) + C
2
9
1
(− 4dx)
4 ∫ 5 − 4x
9
= − ln 5 − 4 x + C
4
= −
=
7. u = x 2 − 3, du = 2 x dx
∫
1
(4 x3 + 3) dx
+ 3x
∫ x4
10. u = x 3 − 3x 2 , du = (3 x 2 − 6 x ) dx = 3( x 2 − 2 x) dx
5. u = 2 x + 5, du = 2 dx
1
4 x3 + 3
dx =
x 4 + 3x
= ln x 4 + 3 x + C
1
dx = ln x − 5 + C
x −5
∫ 2 x + 5 dx
1
1
(−3x 2 ) dx
3 ∫ 5 − x3
1
= − ln 5 − x3 + C
3
dx = −
9. u = x 4 + 3 x, du = ( 4 x3 + 3) dx
1
dx = ln x + 1 + C
x +1
4. u = x − 5, du = dx
∫
x3
12.
x
1
1
dx = ∫ 2
(2 x) dx
x2 − 3
2 x −3
1
= ln x 2 − 3 + C
2
∫
x3 − 8 x
dx =
x2
=
⎛
∫ ⎜⎝ x −
8⎞
⎟ dx
x⎠
x2
− 8 ln x + C
2
13. u = x 3 + 3 x 2 + 9 x, du = 3( x 2 + 2 x + 3) dx
∫
2
x2 + 2x + 3
1 3( x + 2 x + 3)
dx
dx
=
x3 + 3x 2 + 9 x
3 ∫ x3 + 3x 2 + 9 x
1
= ln x 3 + 3 x 2 + 9 x + C
3
14. u = x3 + 6 x 2 + 5, du = (3 x 2 + 12 x) dx = 3( x 2 + 4 x) dx
∫ x3
15.
∫
x2 + 4 x
1
dx =
+ 6 x2 + 5
3
x 2 − 3x + 2
dx =
x +1
∫
1
1
3( x 2 + 4 x ) dx = ln x3 + 6 x 2 + 5 + C
+ 6 x2 + 5
3
6 ⎞
⎛
∫ ⎜⎝ x − 4 + x + 1 ⎟⎠ dx
=
16.
∫ x3
2 x2 + 7 x − 3
dx =
x − 2
17.
∫
x3 − 3x 2 + 5
dx =
x −3
x2
− 4 x + 6 ln x + 1 + C
2
⎛
∫ ⎜⎝ 2 x + 11 +
19 ⎞
⎟ dx
x − 2⎠
= x 2 + 11x + 19 ln x − 2 + C
=
18.
∫
x3 − 6 x − 20
dx =
x+5
=
⎛
∫ ⎜⎝ x
2
+
5 ⎞
⎟ dx
x − 3⎠
x3
+ 5 ln x − 3 + C
3
⎛
∫ ⎜⎝ x
2
− 5 x + 19 −
115 ⎞
⎟ dx
x + 5⎠
x3 5 x 2
−
+ 19 x − 115 ln x + 5 + C
3
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.7
19.
∫
x4 + x − 4
dx =
x2 + 2
⎛
∫ ⎜⎝ x
=
=
20.
∫
26.
2
x ( x − 2)
∫ ( x − 1)3
(ln x)
2
x2 − 2x + 1 − 1
(x
=
∫ x − 1 dx − ∫ ( x − 1)3 dx
x2 + 2 + C
x2
1
− 4 x + ln x 2 − 5 + C
2
2
1
dx
x
2
− 1)
3
∫
27. u = 1 +
∫1 +
2x
dx =
∫
∫
(
(
)
x
dx = −
= −
1
x1 3 )
2
3
dx = 3∫
+ C
2
1
dx ⇒ (u − 1) du = dx
2x
(u
− 1)
u
du =
⎛
∫ ⎜⎝1 −
1⎞
⎟ du
u⎠
)
2 x − ln (1 +
2 x − ln 1 +
2 x + C1
)
2x + C
where C = C1 + 1.
28. u = 1 +
1
∫1− 3
2
ln 1 − 3
3
⎛ −3 ⎞
⎜
⎟ dx
x ⎜⎝ 2 x ⎟⎠
x + C
∫1 +
3 x , du =
1
3x
dx =
=
3
2
dx ⇒ dx = (u − 1) du
3
2 3x
12
∫ u 3 (u − 1) du
2 ⎛
1⎞
⎜1 − ⎟ du
u⎠
3∫ ⎝
2
⎡u − ln u ⎤⎦ + C
3⎣
2
= ⎡1 + 3 x − ln 1 +
3⎣
=
(
1
dx
3x 2 3
24. u = 1 + x1 3 , du =
∫ x 2 3 (1 +
=
1
ln ln x + C
3
1
x1−3
1
2( x − 1)
= u − ln u + C1
−3
x , du =
2 x
23. u = 1 − 3
1
2 x , du =
1
1
∫ ( x − 1)3 dx
1
= ln x − 1 +
1
1 1
1
⋅ dx
dx = ∫
3 ln x x
x ln ( x3 )
=
dx −
497
dx
3
x3
− 2 x + ln
3
− 1)
= 1+
22.
− 1)
(x
∫ (x
1
3
(ln x) + C
3
dx =
x
∫
=
x3 − 4 x 2 − 4 x + 20
x ⎞
⎛
dx = ∫ ⎜ x − 4 + 2
⎟ dx
x2 − 5
x − 5⎠
⎝
21. u = ln x, du =
dx =
x3
1
− 2 x + ln ( x 2 + 2) + C
3
2
=
∫
x ⎞
⎟ dx
x + 2⎠
− 2+
2
The Natural Logarithmic Function: Integration
1 ⎛ 1 ⎞
⎜
⎟ dx
1 + x1 3 ⎝ 3x 2 3 ⎠
=
2
3
3x −
(
2
ln 1 +
3
)
3x ⎤ + C
⎦
)
3 x + C1
= 3 ln 1 + x1 3 + C
25.
2x
∫ ( x − 1)2 dx
2x − 2 + 2
=
∫ ( x − 1)2
=
∫ ( x − 1)2
2( x − 1)
= 2∫
dx
dx + 2 ∫
1
(x
− 1)
2
dx
1
1
dx + 2 ∫
dx
x −1
( x − 1)2
= 2 ln x − 1 −
(x
2
+ C
− 1)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
498
Chapter 5
29. u =
Integration
1
x − 3, du =
2
dx ⇒ 2(u + 3)du = dx
x
x
(u + 3) du
dx = 2 ∫
u
x −3
2
∫
= 2∫
u 2 + 6u + 9
9⎞
⎛
du = 2 ∫ ⎜ u + 6 + ⎟ du
u
u⎠
⎝
⎡u 2
⎤
= 2 ⎢ + 6u + 9 ln u ⎥ + C1
⎣2
⎦
= u 2 + 12u + 18 ln u + C1
=
(
)
x −3
= x +6
2
(
+ 12
x + 18 ln
)
x − 3 + 18 ln
x − 3 + C1
x −3 +C
where C = C1 − 27.
30. u = x1 3 − 1, du =
∫3
3
x
dx =
x −1
∫
1
2
dx ⇒ dx = 3(u + 1) du
3x 2 3
u +1
2
3(u + 1) du
u
u +1 2
(u + 2u + 1) du
u
1⎞
⎛
= 3∫ ⎜ u 2 + 3u + 3 + ⎟ du
u⎠
⎝
= 3∫
⎡u3
⎤
3u 2
= 3⎢ +
+ 3u + ln u ⎥ + C
3
2
⎣
⎦
2
⎡ x1 3 − 1 3
(
) + 3( x1 3 − 1) + 3 x1 3 − 1 + ln x1 3 − 1⎤⎥ + C
= 3⎢
(
)
⎢
⎥
3
2
⎣
⎦
= 3 ln x1 3 − 1 +
⎛θ ⎞
3x 2 3
+ 3 x1 3 + x + C1
2
θ
⎛ θ ⎞⎛ 1 ⎞
= 3∫ cot ⎜ ⎟⎜ ⎟ dθ = 3 ln sin
+ C
3
⎝ 3 ⎠⎝ 3 ⎠
31.
∫ cot⎜⎝ 3 ⎟⎠ dθ
32.
∫
tan 5θ dθ =
∫
1
csc 2 x dx = ∫ (csc 2 x)( 2) dx
2
1
= − ln csc 2 x + cot 2 x + C
2
36.
θ⎞
⎛
∫ ⎜⎝ 2 − tan 4 ⎟⎠ dθ
=
∫ 2dθ
− 4 ∫ tan
= 2θ + 4 ln cos
1 5 sin 5θ
1
dθ = − ln cos 5θ + C
5 ∫ cos 5θ
5
θ ⎛1⎞
⎜ ⎟ dθ
4⎝ 4⎠
θ
4
+ C
37. u = 1 + sin t , du = cos t dt
33.
x⎛ 1 ⎞
x
x
cos t
∫ 1 + sin t dt
38. u = cot t , du = −csc 2 t dt
x
34.
∫ sec 2 dx = 2∫ sec 2 ⎜⎝ 2 ⎟⎠ dx = 2 ln sec 2 + tan 2
35.
∫ (cos 3θ
40.
− 1) dθ =
1
3
∫ cos 3θ (3) dθ
=
1
3
sin 3θ − θ + C
∫ (sec 2 x + tan 2 x) dx
=
1
2
−
= ln 1 + sin t + C
+C
∫
∫ dθ
∫ (sec 2 x + tan 2 x)(2) dx
csc 2 t
dt = −ln cot t + C
cot t
39. u = sec x − 1, du = sec x tan x dx
∫
=
1
2
sec x tan x
dx = ln sec x − 1 + C
sec x − 1
ln sec 2 x + tan 2 x − ln cos 2 x + C
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.7
41.
∫e
−x
tan (e − x ) dx = − ∫ tan (e − x )( − e− x ) dx
(
= − − ln cos (e − x )
)+C
The Natural Logarithmic Function: Integration
2
= 2 x −2 , x > 0
x2
−2
f ′( x) =
+C
x
f ′(1) = 1 = −2 + C ⇒ C = 3
47. f ′′( x) =
= ln cos (e − x ) + C
42.
∫ sec t (sec t
+ tan t ) dt =
∫ sec
2
t dt +
−2
+3
x
f ( x) = −2 ln x + 3 x + C1
∫ sec t tan t dt
f ′( x) =
= tan t + sec t + C
43.
3
f (1) = 1 = −2(0) + 3 + C1 ⇒ C1 = −2
1
dx = −3 ln x − 2 + C
x−2
(1, 0): 0 = −3 ln 1 − 2 + C ⇒ C = 0
y =
∫ 2 − x dx
= −3∫
f ( x) = −2 ln x + 3 x − 2
48. f ′′( x) =
y = −3 ln x − 2
10
f ′( x) =
(1, 0)
− 10
10
y =
∫
(−1, 0): 0
−4
(x
− 1)
2
− 2 = −4( x − 1)
−2
− 2,
x >1
4
− 2x + C
( x − 1)
f ′( 2) = 0 = 4 − 4 + C ⇒ C = 0
4
− 2x
x −1
f ( x) = 4 ln ( x − 1) − x 2 + C1
f ′( x) =
− 10
44.
499
x−2
dx =
x
⎛
∫ ⎜⎝1 −
2⎞
⎟ dx = x − 2 ln x + C
x⎠
f ( 2) = 3 = 4(0) − 4 + C1 ⇒ C1 = 7
f ( x) = 4 ln ( x − 1) − x 2 + 7
= −1 − 2 ln −1 + C = −1 + C ⇒ C = 1
y = x − 2 ln x + 1
49.
8
dy
1
=
, (0, 1)
dx
x + 2
(a)
(− 1, 0)
y
(0, 1)
3
−9
9
−4
x
−2
45.
2x
dx = ln x 2 − 9 + C
−9
(0, 4): 4 = ln 0 − 9 + C ⇒ C = 4 − ln 9
y =
∫ x2
y = ln x − 9 + 4 − ln 9
2
8
(0, 4)
4
−3
y =
(b)
1
∫ x + 2 dx
= ln x + 2 + C
y (0) = 1 ⇒ 1 = ln 2 + C ⇒ C = 1 − ln 2
⎛ x + 2⎞
So, y = ln x + 2 + 1 − ln 2 = ln ⎜
⎟ + 1.
⎝ 2 ⎠
−9
9
3
−4
−3
46. r =
∫
sec 2 t
dt = ln tan t + 1 + C
tan t + 1
6
−3
(π , 4): 4 = ln 0 + 1 + C ⇒ C = 4
r = ln tan t + 1 + 4
10
−8
(π , 4)
8
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
500
50.
Chapter 5
Integration
dy
ln x
, (1, − 2)
=
dx
x
(a)
(ln x)
ln x
∫ x dx = 2 + C
2
y =
(b)
y
2
x
4
So, y =
−1
(ln x)
−2
51.
4
∫0
(ln 1)2
y(1) = −2 ⇒ −2 =
1
2
2
−1
+ C ⇒ C = −2
2
11
−4
− 2.
4
5
5
⎡5
⎤
dx = ⎢ ln 3 x + 1⎥ = ln 13 ≈ 4.275
3x + 1
3
3
⎣
⎦0
4
55.
2
∫0
x2 − 2
dx =
x +1
2⎛
∫ 0 ⎜⎝ x − 1 −
1 ⎞
⎟ dx
x + 1⎠
2
⎡1
⎤
= ⎢ x 2 − x − ln x + 1 ⎥ = −ln 3
2
⎣
⎦0
1
1
1
52. ∫
dx = ⎡⎣ln 2 x + 3 ⎤⎦ −1
−1 2 x + 3
2
1
1
= [ln 5 − ln 1] = ln 5 ≈ 0.805
2
2
1
∫1
(1 +
∫e
1
x −1
∫ 0 x + 1 dx
57.
1
dx
x
2
2
1 − cos θ
dθ = ⎡⎣ln θ − sin θ ⎤⎦1
− sin θ
∫1 θ
= ln
⎛ 1 ⎞1
⎜
⎟ dx = ⎡⎣ln ln x ⎤⎦ e = ln 2
⎝ ln x ⎠ x
≈ 0.693
π 4
∫ π 8 (csc 2θ
1
59.
∫1 +
60.
∫1 +
61.
∫ x − 1 dx
x
1−
e2
e2
∫e
π
8
⇒ u =
π
4
,θ =
π
4
⇒ u =
2 − sin 2
≈ 1.929
1 − sin 1
π
2
1 π2
(csc u − cot u ) du
2 ∫π 4
π
1
= ⎡⎣− ln csc u + cot u − ln sin u ⎤⎦π
2
− cot 2θ ) dθ =
dx = 2
x
dx = 4
x
x
−2
1
≈ −0.386
e
58. u = 2θ , du = 2 dθ , θ =
1
∫ 0 1 dx + ∫ 0 x + 1 dx
1
2
1
dx =
x ln x
=
= ⎡⎣ x − 2 ln x + 1 ⎤⎦ 0 = 1 − 2 ln 2
ln x)
7
3⎤
⎡1
dx = ⎢ (1 + ln x) ⎥ =
3
x
⎣3
⎦1
54. u = ln x, du =
e2
56.
1
dx
x
53. u = 1 + ln x, du =
e
≈ −1.099
⎛
= ln ⎜⎜
⎝
=
1⎡
⎢− ln (1 + 0) − ln (1) + ln
2 ⎣⎢
=
1⎡
⎢ln
2 ⎣⎢
=
1 ⎛
ln ⎜1 +
2 ⎜⎝
(
2⎤
⎥
2 ⎦⎥
x +C
62.
∫ x − 1 dx
)
63.
∫ π 4 (csc x − sin x) dx
64.
∫ −π 4
(
)
2 + 1 + ln
(
x − x − 4 ln 1 +
x − 1⎞
⎟ + 2
x + 1 ⎟⎠
4
)
x +C
(
2 + 1 + ln
2⎤
⎥
2 ⎦⎥
2⎞
⎟
2 ⎟⎠
)
x − 2 ln 1 +
2
x + C
x2
π 2
π 4
= ln x − 1 +
x2
+ x + C
2
= ln
(
)
2 +1 −
2
≈ 0.174
2
⎛ 2 + 1⎞
sin 2 x − cos 2 x
dx = ln ⎜⎜
⎟⎟ − 2 2
cos x
⎝ 2 − 1⎠
≈ −1.066
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.7
The Natural Logarithmic Function: Integration
Note: In Exercises 65–68, you can use the Second
Fundamental Theorem of Calculus or integrate the
function.
x1
65. F ( x ) =
∫1
F ′( x ) =
1
x
66. F ( x ) =
t
67. F ( x ) =
F ′( x) =
dt
x
∫0
1
1
(3) =
3x
x
3x 1
F ( x) =
∫1
68. F ( x ) =
∫1
3x
dt = ⎡⎣ln t ⎤⎦1 = ln 3 x
t
1
1
F ′( x) =
(3) =
x
3x
tan t dt
x2 1
t
dt
2x
2
=
x2
x
3
6
dx = ⎣⎡6 ln x ⎦⎤1 = 6 ln 3
x
69. A =
∫1
70. A =
∫2
71. A =
∫0
72. A =
∫π 4
73. A =
∫1
4 1 1
⎛ 2 ln 2 ⎞
4
2
dx = 2 ∫
dx = 2 ln ln x ⎤⎦ 2 = 2⎡⎣ln (ln 4) − ln (ln 2)⎤⎦ = 2 ln ⎜
⎟ = 2 ln 2
2 ln x x
x ln x
⎝ ln 2 ⎠
π 4
π 4
tan x dx = −ln cos x ⎤
⎥⎦ 0
3π 4
4
dt
Alternate Solution:
F ′( x ) =
4
t
(by Second Fundamental Theorem of Calculus)
F ′( x ) = tan x
3
3x 1
∫1
501
= −ln
2
+ 0 = ln
2
2 =
⎛
3π 4
sin x
dx = −ln 1 + cos x ⎤⎦π 4 = −ln ⎜⎜1 −
1 + cos x
⎝
ln 2
2
⎛
2⎞
⎟ + ln ⎜⎜1 +
2 ⎟⎠
⎝
⎛2 +
2⎞
⎟ = ln ⎜⎜
2 ⎟⎠
⎝2 −
2⎞
⎟ = ln 3 + 2 2
2 ⎟⎠
(
)
4
x2 + 4
dx =
x
4
∫1
⎡ x2
⎤
4⎞
1 15
⎛
+ 8 ln 2 ≈ 13.045
⎜ x + ⎟ dx = ⎢ + 4 ln x⎥ = (8 + 4 ln 4) − =
x
2
2
2
⎝
⎠
⎣
⎦1
10
0
6
0
74. A =
5
5x
∫ 1 x2 + 2
5
2
dx =
5
∫1
5
1
5
5
5
(2 x dx) = ⎡⎢ ln x 2 + 2 ⎤⎥ = (ln 27 − ln 3) = ln 9 = 5 ln 3 ≈ 5.4931
x +2
2
2
⎣2
⎦1
2
4
0
6
0
10
75.
2
∫ 0 2 sec
πx
6
dx =
=
12
π
2
∫0
2
12 ⎡
πx
πx ⎤
⎛ π x ⎞π
+ tan
sec ⎜ ⎟ dx =
ln sec
π ⎢⎣
6
6 ⎥⎦ 0
⎝ 6 ⎠6
12 ⎛
π
π
⎞ 12
ln 2 +
− ln 1 + 0 ⎟ =
⎜ ln sec + tan
π⎝
3
3
π
⎠
(
)
3 ≈ 5.03041
0
4
0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
502
76.
Chapter 5
Integration
4
10
10
10
⎡
⎤
⎡
⎤ ⎡
⎤
= ⎢x2 +
ln cos(0.3x) ⎥ = ⎢16 +
ln cos(1.2)⎥ − ⎢1 +
ln cos(0.3)⎥ ≈ 11.7686
3
3
3
⎣
⎦1
⎣
⎦ ⎣
⎦
∫ 1 (2 x − tan(0.3x)) dx
4
8
0
5
−2
77. f ( x ) =
12
, b − a = 5 − 1 = 4, n = 4
x
Trapezoid:
Simpson:
4
1
⎡ f (1) + 2 f ( 2) + 2 f (3) + 2 f ( 4) + f (5)⎤⎦ = [12 + 12 + 8 + 6 + 2.4] = 20.2
2( 4) ⎣
2
4
1
⎡ f (1) + 4 f ( 2) + 2 f (3) + 4 f ( 4) + f (5)⎤⎦ = [12 + 24 + 8 + 12 + 2.4] ≈ 19.4667
3( 4) ⎣
3
Calculator:
5 12
∫1
x
dx ≈ 19.3133
Exact: 12 ln 5
78. f ( x) =
8x
, b − a = 4 − 0 = 4, n = 4
x2 + 4
Trapezoid:
Simpson:
4
1
⎡ f (0) + 2 f (1) + 2 f ( 2) + 2 f (3) + f ( 4)⎤⎦ = [0 + 3.2 + 4 + 3.6923 + 1.6] ≈ 6.2462
2( 4) ⎣
2
4
⎡ f (0) + 4 f (1) + 2 f ( 2) + 4 f (3) + f ( 4)⎤⎦ ≈ 6.4615
3( 4) ⎣
Calculator:
4
∫ 0 x2
8x
dx ≈ 6.438
+ 4
Exact: 4 ln 5
79. f ( x) = ln x, b − a = 6 − 2 = 4, n = 4
Trapezoid:
Simpson:
4
1
⎡ f ( 2) + 2 f (3) + 2 f ( 4) + 2 f (5) + f (6)⎤⎦ = [0.6931 + 2.1972 + 2.7726 + 3.2189 + 1.7918] ≈ 5.3368
2( 4) ⎣
2
4
⎡ f ( 2) + 4 f (3) + 2 f ( 4) + 4 f (5) + f (6)⎤⎦ ≈ 5.3632
3( 4) ⎣
Calculator:
6
∫ 2 ln x dx
≈ 5.3643
80. f ( x) = sec x, b − a =
Trapezoid:
Simpson:
π
3
2π
⎛ π⎞
,n = 4
− ⎜− ⎟ =
3
⎝ 3⎠
2π 3 ⎡ ⎛ π ⎞
π
⎛ π⎞
⎛π ⎞
⎛ π ⎞⎤
f ⎜ − ⎟ + 2 f ⎜ − ⎟ + 2 f ( 0) + 2 f ⎜ ⎟ + f ⎜ ⎟ ⎥ ≈
[2 + 2.3094 + 2 + 2.3094 + 2] ≈ 2.780
⎢
2( 4) ⎣ ⎝ 3 ⎠
12
⎝ 6⎠
⎝6⎠
⎝ 3 ⎠⎦
2π 3 ⎡ ⎛ π ⎞
⎛ π⎞
⎛π ⎞
⎛ π ⎞⎤
f ⎜ − ⎟ + 4 f ⎜ − ⎟ + 2 f (0) + 4 f ⎜ ⎟ + f ⎜ ⎟⎥ ≈ 2.6595
⎢
3( 4) ⎣ ⎝ 3 ⎠
⎝ 6⎠
⎝6⎠
⎝ 3 ⎠⎦
Calculator:
π 3
∫ −π 3 sec x dx
≈ 2.6339
81. Power Rule
83. Substitution: (u = x 2 + 4) and Log Rule
82. Substitution: (u = x 2 + 4) and Power Rule
84. Substitution: (u = tan x) and Log Rule
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.7
The Natural Logarithmic Function: Integration
y
85.
x
∫1
87.
2
1
x
x
⎛1⎞
3 ln x = ln x − ln ⎜ ⎟
⎝ 4⎠
⎛1⎞
2 ln x = −ln ⎜ ⎟ = ln 4
⎝ 4⎠
1
ln x = ln 4 = ln 2
2
x = 2
−1
2
1
1
2
A ≈ 1.25; Matches (d)
y
86.
x
∫ 1 4 t dt
⎡⎣3 ln t ⎤⎦1 = ⎡⎣ln t ⎤⎦1 4
x
−1
3
dt =
t
503
2
1
88.
x
1
2
3
4
x1
∫1
t
x
dt = ⎡⎣ln t ⎤⎦1 = ln x
(assume x
> 0)
(a) ln x = ln 5 ⇒ x = 5
−1
(b) ln x = 1 ⇒ x = e
−2
A ≈ 3; Matches (a)
89.
∫ cot u du
=
cos u
∫ sin u
du = ln sin u + C
Alternate solution:
d
1
⎡ln sin u + C ⎤⎦ =
cos u + C = cot u + C
du ⎣
sin u
90.
⎛ csc u + cot u ⎞
2
∫ csc u du = ∫ csc u⎜⎝ csc u + cot u ⎟⎠ du = − ∫ csc u + cot u (−csc u cot u − csc u ) du = −ln csc u + cot u + C
1
Alternate solution:
csc u cot u + csc u
d
1
⎡−ln csc u + cot u + C ⎦⎤ = −
(−csc u cot u − csc2 u) = csc( u + cot u ) = csc u
du ⎣
csc u + cot u
91. −ln cos x + C = ln
92. ln sin x + C = ln
1
+ C = ln sec x + C
cos x
1
+ C = −ln csc x + C
csc x
93. ln sec x + tan x + C = ln
(sec x
+ tan x)(sec x − tan x)
(sec x
− tan x)
+C
= ln
sec 2 x − tan 2 x
+ C
sec x − tan x
= ln
1
+ C = −ln sec x − tan x + C
sec x − tan x
94. −ln csc x + cot x + C = −ln
(csc x
+ cot x)(csc x − cot x)
(csc x
− cot x)
+ C
= −ln
csc 2 x − cot 2 x
+C
csc x − cot x
= −ln
1
+ C = ln csc x − cot x + C
csc x − cot x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
504
Chapter 5
Integration
95. Average value =
4 8
1
dx
∫
2
4 − 2 x2
99. P(t ) =
= (3000)( 4) ∫
0.25
dt
1 + 0.25t
= 12,000 ln 1 + 0.25t + C
4
= 4 ∫ x −2 dx
2
P(0) = 12,000 ln 1 + 0.25(0) + C = 1000
4
⎡ 1⎤
= ⎢−4 ⎥
⎣ x ⎦2
C = 1000
P(t ) = 12,000 ln 1 + 0.25t + 1000
⎛1 1⎞
= −4⎜ − ⎟ = 1
⎝4 2⎠
= 1000 ⎡⎣12 ln 1 + 0.25t + 1⎤⎦
P(3) = 1000 ⎡⎣12(ln 1.75) + 1⎤⎦ ≈ 7715
4 4( x + 1)
1
dx
96. Average value =
x2
4 − 2∫2
4⎛ 1
1⎞
= 2 ∫ ⎜ + 2 ⎟ dx
2 x
x ⎠
⎝
100.
dS
k
=
dt
t
k
dt = k ln t + C = k ln t + C because t > 1.
t
S ( 2) = k ln 2 + C = 200
4
S (t ) =
1⎤
⎡
= 2 ⎢ln x − ⎥
x ⎦2
⎣
1
1⎤
⎡
= 2 ⎢ln 4 − − ln 2 + ⎥
4
2⎦
⎣
∫
S ( 4) = k ln 4 + C = 300
1⎤
1
⎡
= 2 ⎢ln 2 + ⎥ = ln 4 +
≈ 1.8863
4⎦
2
⎣
97. Average value =
3000
∫ 1 + 0.25t dt
Solving this system yields k = 100 ln 2 and
C = 100. So,
S (t ) =
e 2 ln x
1
dx
∫
1
e −1
x
⎡ ln t
⎤
100 ln t
+ 100 = 100 ⎢
+ 1⎥.
ln 2
ln
2
⎣
⎦
e
2
2 ⎡ (ln x) ⎤
=
⎢
⎥
e − 1⎢ 2 ⎥
⎣
⎦1
1
=
(1 − 0)
e −1
1
=
≈ 0.582
e −1
98. Average value =
101. t =
2
1
πx
dx
sec
2 − 0∫0
6
102.
2
⎡1 ⎛ 6 ⎞
πx
πx ⎤
= ⎢ ⎜ ⎟ ln sec
+ tan
2
6
6 ⎥⎦ 0
π
⎝
⎠
⎣
=
=
(
3 − ln (1 + 0)⎤
⎦
(
3
3⎡
ln 2 +
π⎣
3
π
ln 2 +
10 300
1
dT
ln 2 ∫ 250 T − 100
=
300
10
10
⎡⎣ln (T − 100)⎤⎦ 250 =
[ln 200 − ln 150]
ln 2
ln 2
=
10 ⎡ ⎛ 4 ⎞⎤
ln ⎜ ⎟ ≈ 4.1504 min
ln 2 ⎢⎣ ⎝ 3 ⎠⎥⎦
50 90,000
50
1
dx = ⎡⎣3000 ln 400 + 3 x ⎤⎦ 40
50 − 40 ∫ 40 400 + 3 x
≈ $168.27
)
)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.7
103. f ( x ) =
The Natural Logarithmic Function: Integration
505
x
1 + x2
y
1
0.5
x
5
(a) y =
10
1
x
x intersects f ( x) =
:
2
1 + x2
1
x
x =
2
1 + x2
1 + x2 = 2
x =1
A =
1
⎡1
x ⎤ 1 ⎞
x2 ⎤
1
1
2
∫ 0 ⎝⎜ ⎢⎣1 + x 2 ⎥⎦ − 2 x ⎠⎟ dx = ⎢⎣ 2 ln( x + 1) − 4 ⎥⎦ = 2 ln 2 − 4
0
1⎛⎡
(b) f ′( x) =
(1 + x 2 ) − x(2 x)
2
(1 + x 2 )
=
1 − x2
(1 + x2 )
2
f ′(0) = 1
So, for 0 < m < 1, the graphs of f and y = mx enclose a finite region.
y
(c)
f(x) = 2x
x +1
0.5
y = mx
x
1−m
m
f ( x) =
x
intersects y = mx :
x2 + 1
x
= mx
1 + x2
1 = m + mx 2
x2 =
1− m
m
x =
A =
∫0
(1− m)
1− m
m
m
⎛ x
⎞
− mx ⎟ dx,
⎜
2
x
1
+
⎝
⎠
⎡1
mx 2 ⎤
= ⎢ ln (1 + x 2 ) −
⎥
2 ⎦0
⎣2
(1− m)
0 < m <1
m
=
1 ⎛
1 − m ⎞ 1 ⎛1 − m ⎞
ln ⎜1 +
⎟ − m⎜
⎟
2 ⎝
m ⎠ 2 ⎝ m ⎠
=
1 ⎛1⎞ 1
ln ⎜ ⎟ − (1 − m)
2 ⎝m⎠ 2
=
1
⎡m − ln ( m) − 1⎤⎦
2⎣
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
506
Chapter 5
Integration
104. (a) At x = −1, f ′( −1) ≈
109. Let f (t ) = ln t on [ x, y],
1
.
2
By the Mean Value Theorem,
1
The slope of f at x = −1 is approximately .
2
f ( y ) − f ( x)
(−∞, − 2).
Because 0 < x < c < y,
105. False
1
1
1
>
> . So,
x
c
y
1
ln y − ln x
1
<
< .
y
y − x
x
= ln ( x1 2 ) ≠ (ln x)
12
106. False
110. F ( x ) =
d
1
[ln x] =
dx
x
F ′( x) =
107. True
∫
x < c < y,
ln y − ln x
1
= .
y − x
c
increasing on ( − 2, ∞). Similarly, f is decreasing on
(ln x)
= f ′(c),
y − x
(b) Since the slope is positive for x > − 2, f is
1
2
0 < x < y.
2x 1
∫x
t
dt ,
x > 0
1
1
( 2) − = 0 ⇒ F is constant on (0, ∞).
2x
x
Alternate Solution:
1
dx = ln x + C1 = ln x + ln C = ln Cx , C ≠ 0
x
F ( x) = [ln t ⎤⎦ x = ln ( 2 x) − ln x = ln 2 + ln x − ln x
2x
= ln 2
108. False; the integrand has a nonremovable discontinuity at
x = 0.
111.
d
1
ln x = implies that
dx
x
1
∫ x dx = ln x + C.
The second formula follows by the Chain Rule.
Section 5.8 Inverse Trigonometric Functions: Integration
1.
2.
3.
∫
∫
∫x
dx
9 − x
2
⎛ x⎞
= arcsin ⎜ ⎟ + C
⎝ 3⎠
dx
1 − 4x2
1
4x − 1
2
12
1
= ∫
2
dx =
4.
∫ 1 + 9x2
dx = 4 ∫
5.
∫
6.
∫ 4 + ( x − 3)2 dx
1
1 − ( x + 1)
2
8. Let u = x 2 , du = 2 x dx.
2
1
dx = arcsin ( 2 x) + C
2
2
1 − 4x
∫
2
( 2 x)2 − 1
2x
t
dt =
=
9.
∫ t4
3
dx = 4 arctan (3 x) + C
1 + 9 x2
=
1
2∫
dx =
1
2∫
1
x
2
(x )
2
2
− 22
(2 x) dx
1
x2
arcsec
+C
4
2
dx = arcsec 2 x + C
1
⎛ x − 3⎞
arctan ⎜
⎟ + C
2
⎝ 2 ⎠
t
1
1
dt = ∫
( 2) dt
+ 25
2 (t 2 )2 + 52
10.
∫
=
⎛ t2 ⎞
1 1
arctan ⎜ ⎟ + C
2 5
⎝5⎠
=
⎛ t2 ⎞
1
arctan ⎜ ⎟ + C
10
⎝5⎠
1
x 1 − (ln x)
2
dx =
∫
1
1 − (ln x)
2
⋅
1
dx
x
= arcsin (ln x ) + C
7. Let u = t 2 , du = 2t dt.
1 − t4
1
x4 − 4
dx = arcsin ( x + 1) + C
1
∫
∫x
1
1 − (t
2
)
2
(2t ) dt
=
1
arcsin t 2 + C
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.8
Inverse Trigonometric Functions: Integration
11. Let u = e 2 x , du = 2e 2 x dx.
e2 x
1
2e 2 x
1
e2 x
dx
=
dx
=
arctan
+ C
4 + e4 x
2 ∫ 4 + (e 2 x ) 2
4
2
∫
sec 2 x
∫
13.
25 − tan x
2
dx =
∫
sec 2 x
52 − ( tan x )
∫x
9 x 2 − 25
sin x
∫ 7 + cos2 x dx
14.
1
dx = 2 ∫
(3 x ) (3 x )
2
− 52
3 dx
3x
2
= arcsec
+C
5
5
15.
16.
1
dx, u = x , x = u 2 , dx = 2u du
x 1− x
1
du
∫ u 1 − u 2 (2u du ) = 2∫ 1 − u 2 = 2 arcsin u + C = 2 arcsin
∫2
∫
( 7)
−1
2
+ cos 2 x
(−sin x) dx
= −
1
⎛ cos x ⎞
arctan ⎜
⎟ + C
7
⎝ 7 ⎠
= −
⎛
7
arctan ⎜⎜
7
⎝
7 cos
7
x⎞
⎟⎟ + C
⎠
3
dx, u =
x (1 + x)
1
x , du =
2
x +C
dx, dx = 2u du
x
x +C
x −3
1
2x
1
1
dx = ∫ 2
dx − 3∫ 2
dx = ln ( x 2 + 1) − 3 arctan x + C
+1
x +1
2 x +1
2
17.
∫ x2
18.
∫x
x2 + 3
x − 4
2
dx =
=
x + 5
∫
9 − ( x − 3)
2
x2
∫x
1
2
=
x − 4
2
∫ (x
x − 2
∫ ( x + 1)2
+ 4
2
− 4)
x2 − 4 +
dx =
∫
= −
20.
=
∫
3
2u du
du
= 3∫
= 3 arctan u + C = 3 arctan
2 ∫ u (1 + u 2 )
1 + u2
19.
dx
⎛ tan x ⎞
= arcsin ⎜
⎟ + C
⎝ 5 ⎠
12. u = 3 x, du = 3 dx, a = 5
2
2
507
dx =
=
(x
dx +
−1 2
3
∫x
x2 − 4
2 x dx + 3 ∫
dx
1
x
x2 − 4
dx
x
3
arcsec
+ C
2
2
− 3)
9 − ( x − 3)
2
dx +
∫
8
9 − ( x − 3)
2
dx
2
⎛ x − 3⎞
9 − ( x − 3) + 8 arcsin ⎜
⎟ + C = −
⎝ 3 ⎠
1
2x + 2
dx −
2 ∫ ( x + 1)2 + 4
3
∫ ( x + 1)2
+ 4
dx
1
3
⎛ x + 1⎞
ln ( x 2 + 2 x + 5) − arctan ⎜
⎟ + C
2
2
⎝ 2 ⎠
21. Let u = 3 x, du = 3 dx.
16
∫0
3
1 − 9x
2
dx =
⎛x
⎞
6 x − x 2 + 8 arcsin ⎜ − 1⎟ + C
3
⎝
⎠
16
∫0
1
1 − (3 x )
= ⎡⎣arcsin (3 x)⎤⎦ 0
16
2
(3) dx
=
π
6
22.
∫0
2
x⎤
⎡
dx = ⎢arcsin ⎥
2
2 ⎦0
⎣
4 − x
1
= arcsin
=
2
2
− arcsin 0
2
π
4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
508
Chapter 5
Integration
23. Let u = 2 x, du = 2 dx.
∫0
3 2
1
1
dx = ∫
1 + 4x2
2 0
2
3 2
∫
24.
1 + ( 2 x)
⎡1
⎤
= ⎢ arctan ( 2 x)⎥
⎣2
⎦0
2
3
dx
3 2
=
⎡1
2x ⎤
dx = ⎢ arcsec
⎥
2
3 ⎦
x 4x − 9
⎣3
1
3
3
3
1
1
2 3
arcsec( 2) − arcsec
3
3
3
π
1⎛ π ⎞ 1⎛ π ⎞
= ⎜ ⎟ − ⎜ ⎟ =
3⎝ 3 ⎠ 3⎝ 6 ⎠
18
=
π
6
6
1
∫ 3 25 + ( x − 3)2
25.
6
⎡1
⎛ x − 3 ⎞⎤
dx = ⎢ arctan ⎜
⎟⎥
⎝ 5 ⎠⎦ 3
⎣5
1
arctan (3 5)
5
≈ 0.108
=
26.
1
4
∫1
x 16 x − 5
2
dx =
4 dx
4
∫1
( 4 x) ( 4 x)
2
−
( 5)
4
⎡⎛ 1 ⎞
4x ⎤
= ⎢⎜
⎥ =
⎟ arcsec
5
5 ⎦1
⎝
⎠
⎣
2
1
16
1
⎛ 4 ⎞
−
arcsec
arcsec⎜
⎟ ≈ 0.091
5
5
5
⎝ 5⎠
27. Let u = e x , du = e x dx
ln 5
∫0
ln 5
ex
π
dx = ⎣⎡arctan (e x )⎦⎤
= arctan 5 −
≈ 0.588
0
1 + e2 x
4
28. Let u = e − x , du = −e − x dx
e− x
ln 4
∫ ln 2
1−e
−2 x
ln 4
π
⎛1⎞
⎛1⎞
⎛1⎞
dx = ⎣⎡−arcsin (e− x )⎦⎤
= −arcsin ⎜ ⎟ + arcsin⎜ ⎟ =
− arcsin⎜ ⎟ ≈ 0.271
ln 2
6
⎝ 4⎠
⎝ 2⎠
⎝ 4⎠
29. Let u = cos x, du = −sin x dx.
π
sin x
∫ π 2 1 + cos2 x dx
30.
π 2
∫0
= −∫
π
cos x
dx = ⎡⎣arctan (sin x)⎤⎦ 0
1 + sin 2 x
1
31. Let u = arcsin x, du =
1
∫0
2
1 − x2
∫0
33.
2
∫0
2
arccos x
1
1− x
2
dx = − ∫
dx
=
x2 − 2x + 2
2
1
1
1 − x2
2
0
1
2
=
=
2
π
4
π
4
dx.
⎡1
⎤
dx = ⎢ arcsin 2 x⎥
2
2
⎣
⎦0
1− x
arcsin x
32. Let u = arccos x, du = −
1
π
− sin x
dx = ⎡−
⎣ arctan (cos x )⎤⎦π
+ cos 2 x
π
π 21
2
=
π2
32
≈ 0.308
dx.
1
−arccos x
2
⎡ 1
⎤
dx = ⎢− arccos 2 x⎥
⎣ 2
⎦0
1 − x2
∫ 0 1 + ( x − 1)2 dx
=
3π 2
≈ 0.925
32
π
= ⎡⎣arctan ( x − 1)⎤⎦ 0 =
2
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.8
dx
=
2
x + 4 x + 13
2
34.
∫ −2
35.
∫ x2
∫ − 2 ( x + 2)2
2x
dx =
+ 6 x + 13
∫
=
36.
∫ x2
37.
∫
38.
∫
39.
2x − 5
dx =
+ 2x + 2
1
− x2 − 4 x
2
−x + 4x
2
2x − 3
3
∫2
4x − x
2
∫ x2
dx =
∫
dx =
∫
dx =
2x +
+ 6x
2x +
x2 + 6x
∫ x2
⎡1
1
⎛ x + 2 ⎞⎤
⎛ 4⎞
= ⎢ arctan ⎜
⎟⎥ = arctan ⎜ ⎟
3
⎝ 3 ⎠⎦ −2
⎝ 3⎠
+9
⎣3
6
1
dx − 6 ∫ 2
dx
+ 13
x + 6 x + 13
6
1
⎛ x + 3⎞
dx − 6 ∫
dx = ln x 2 + 6 x + 13 − 3 arctan ⎜
⎟+C
2
+ 13
⎝ 2 ⎠
4 + ( x + 3)
2x + 2
1
dx − 7 ∫
dx = ln x 2 + 2 x + 2 − 7 arctan ( x + 1) + C
2
+ 2x + 2
1 + ( x + 1)
⎛ x + 2⎞
dx = arcsin ⎜
⎟+C
⎝ 2 ⎠
1
4 − ( x + 2)
2
2
4 − ( x − 4 x + 4)
dx =
2
2x − 4
3
∫2
= −∫
4x − x
3
2
509
2
dx
2
Inverse Trigonometric Functions: Integration
dx +
2
2
∫
4 − ( x − 2)
1
3
∫2
⎛ x − 2⎞
dx = 2 arcsin ⎜
⎟+C
⎝ 2 ⎠
dx
4x − x2
(4 x − x 2 ) (4 − 2 x) dx + ∫ 2
−1/2
2
1
3
4 − ( x − 2)
2
dx
3
⎡
π
⎛ x − 2 ⎞⎤
= ⎢−2 4 x − x 2 + arcsin ⎜
≈ 1.059
⎟⎥ = 4 − 2 3 +
2
6
⎝
⎠⎦ 2
⎣
40.
∫ ( x − 1)
1
x − 2x
2
dx =
∫
1
(x
− 1)
(x
− 1) − 1
2
dx = arcsec x − 1 + C
41. Let u = x 2 + 1, du = 2 x dx.
∫ x4
x
1
2x
1
dx = ∫
dx = arctan ( x 2 + 1) + C
+ 2x2 + 2
2 ( x 2 + 1)2 + 1
2
42. Let u = x 2 − 4, du = 2 x dx.
∫
x
9 + 8x2 − x4
43. Let u =
∫
dx =
1
2∫
2x
25 − ( x 2 − 4)
2
dx =
et − 3. Then u 2 + 3 = et , 2u du = et dt , and
et − 3 dt =
2u 2
du =
+3
∫ u2
∫ 2 du − ∫ 6 u 2
∫
2u du
= dt.
u2 + 3
1
du
+ 3
= 2u − 2 3 arctan
44. Let u =
⎛ x2 − 4 ⎞
1
arcsin ⎜
⎟ + C
2
⎝ 5 ⎠
u
+ C = 2 et − 3 − 2 3 arctan
3
et − 3
+ C
3
x − 2, u 2 + 2 = x, 2u du = dx.
x − 2
dx =
x +1
2u 2
du =
+3
∫ u2
∫
2u 2 + 6 − 6
1
du = 2 ∫ du − 6∫ 2
du
u2 + 3
u +3
= 2u −
6
u
+C = 2
arctan
3
3
x − 2 − 2 3 arctan
x − 2
+ C
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
510
45.
Chapter 5
dx
x (1 + x)
3
∫1
Let u =
3
∫1
Integration
49. (a)
x , u 2 = x, 2u du = dx, 1 + x = 1 + u 2 .
2u du
=
u (1 + u 2 )
∫1
3
∫x
x − 1 dx =
=
3
dx
3− x x +1
1
∫0 2
Let u =
x + 1, u 2 = x + 1, 2u du = dx,
3− x =
∫1
4 − u2 .
2u du
2
2 4−u u
2
=
∫1
du
2
2
∫
x
dx =
x −1
=
47. (a)
∫
(b)
∫
(c)
48. (a)
1
1 − x2
x
1 − x2
4
−
π
6
12
u = x
dx = − 1 − x 2 + C ,
u = 1 − x2
1
dx cannot be evaluated using the basic
1 − x2
integration rules.
∫x
x
∫ e dx cannot be evaluated using the basic
2
1
∫ x2 e
1/ x
Note: In (b) and (c), substitution was necessary
before the basic integration rules could be used.
50. (a)
1
∫1 +
x4
1
x
dx cannot be evaluated using the basic
integration rules.
x
1
2x
(b) ∫
dx = ∫
dx
1 + x4
2 1 + ( x 2 )2
u = x2
dx = −e1 x + C , u =
u2 + 1
(2u ) du
u
⎛ u3
⎞
= 2⎜
+ u⎟ + C
3
⎝
⎠
2
= u (u 2 + 3) + C
3
2
=
x − 1( x + 2) + C
3
π
dx = arcsin x + C ,
integration rules.
2
1 2
(b) ∫ xe x dx = e x + C ,
2
(c)
=
∫
= 2 ∫ (u 2 + 1) du
⎛ 2⎞
⎛1⎞
= arcsin ⎜⎜
⎟⎟ − arcsin ⎜ ⎟
2
⎝ 2⎠
⎝
⎠
π
2
(c) Let u = x − 1. Then x = u 2 + 1 and
dx = 2u du.
4 − u2
⎛ u ⎞⎤
= arcsin ⎜ ⎟⎥
⎝ 2 ⎠⎦1
∫ (u + 1)(u )(2u ) du
2 ∫ (u 4 + u 2 ) du
⎛ u5
u3 ⎞
= 2⎜
+
⎟+C
3⎠
⎝5
2 3 2
u (3u + 5) + C
=
15
2
=
( x − 1)3 2 ⎡⎣3( x − 1) + 5⎤⎦ + C
15
2
=
( x − 1)3 2 (3x + 2) + C
15
π⎞ π
⎛π
= 2⎜ − ⎟ =
4⎠
6
⎝3
46.
2
32
( x − 1) + C , u = x − 1
3
x − 1 dx =
(b) Let u = x − 1. Then x = u 2 + 1 and
dx = 2u du.
2
du
1 + u2
= ⎡⎣2 arctan (u )⎤⎦ 1
∫
=
(c)
x3
∫1 +
x
4
1
arctan ( x 2 ) + C , u = x 2
2
1 4 x3
dx
4 ∫ 1 + x4
1
= ln (1 + x 4 ) + C , u = 1 + x 4
4
dx =
51. No. This integral does not correspond to any of the basic
differentiation rules.
52. The area is approximately the area of a square of side 1.
So, (c) best approximates the area.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.8
1
53. y′ =
∫
y =
4− x
1
2
,
Inverse Trigonometric Functions: Integration
56. (a)
(0, π )
y
x
−4
4
⎛ x⎞
y = arcsin ⎜ ⎟ + π
⎝ 2⎠
−4
2
(b) y′ =
1
54. y′ =
, ( 2, π )
4 + x2
1
1
x
y = ∫
dx = arctan + C
4 + x2
2
2
y =
25 − x 2
(5, π )
,
⎛ x⎞
dx = 2 arcsin ⎜ ⎟ + C
⎝5⎠
25 − x
2
∫
2
π = 2 arcsin (1) + C ⇒ C = 0
When x = 2, y = π :
⎛ x⎞
y = 2 arcsin ⎜ ⎟
⎝5⎠
1
⎛ 2⎞
π = arctan ⎜ ⎟ + C
2
⎝ 2⎠
5
7π
+C ⇒ C =
8
8
1
⎛ x ⎞ 7π
y = arctan ⎜ ⎟ +
2
8
⎝ 2⎠
π =
(5, π )
4
⎛ x⎞
dx = arcsin ⎜ ⎟ + C
2
⎝2⎠
4− x
When x = 0, y = π ⇒ C = π
511
π
55. (a)
−5
5
−5
y
4
dy
=
dx
x
57.
10
x2 − 1
,
(3, 0)
4
x
4
−6
12
−4
2
, (0, 2)
9 + x2
2
2
⎛ x⎞
y = ∫
dx = arctan ⎜ ⎟ + C
9 + x2
3
⎝ 3⎠
2 = C
−8
(b) y′ =
y =
58.
dy
1
=
,
dx
12 + x 2
( 4, 2)
4
2
⎛ x⎞
arctan ⎜ ⎟ + 2
3
⎝ 3⎠
−6
6
5
−4
−4
4
59.
dy
=
dx
−1
2y
16 − x 2
,
(0, 2)
3
−3
3
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
512
Chapter 5
Integration
y
dy
=
,
dx
1 + x2
60.
(0, 4)
63. Area =
3
∫1
1
dx =
x − 2x + 5
1
3
∫1 ( x − 1)2 + 4 dx
2
3
⎡1
⎛ x − 1 ⎞⎤
= ⎢ arctan ⎜
⎟⎥
⎝ 2 ⎠⎦1
⎣2
1
1
= arctan (1) − arctan (0)
2
2
7
−1
6
0
π
=
61. Area =
2
1
∫0
dx
4 − x2
64. Area =
1
⎡
⎛ x ⎞⎤
= ⎢2 arcsin ⎜ ⎟⎥
⎝ 2 ⎠⎦ 0
⎣
0
∫ −2 x 2
2
dx =
+ 4x + 8
=
1
2
2
x2 − 1
x
= [arcsec x]2
65. Area =
dx
π 2
3
−
π
4
=
π
4
π 2
3 cos x
1
π 2
= ⎡⎣3 arctan (sin x)⎤⎦ −π
2
2
= 3 arctan (1) − 3 arctan (−1)
⎛ 2 ⎞
= arcsec( 2) − arcsec⎜
⎟
⎝ 2⎠
π
dx
∫ −π 2 1 + sin 2 x dx = 3∫ −π 2 1 + sin 2 x (cos x dx)
2
=
+ 4
0
π
⎛π ⎞
= 2⎜ ⎟ =
3
⎝6⎠
∫2
2
0
∫ −2 ( x + 2)2
⎡
⎛ x + 2 ⎞⎤
= ⎢arctan ⎜
⎟⎥
⎝ 2 ⎠⎦ −2
⎣
= arctan (1) − arctan (0)
⎛1⎞
= 2 arcsin ⎜ ⎟ − 2 arcsin (0)
⎝2⎠
62. Area =
8
=
π
12
66. Area =
ln
∫0
3
4 ex
dx,
1 + e2 x
= 4 ⎡⎣arctan (e x )⎤⎦
0
ln
= 4 ⎡arctan
⎣
3π 3π
3π
+
=
4
4
2
(u
= ex )
3
( 3) − arctan(1)⎤⎦
π⎞ π
⎛π
= 4⎜ − ⎟ =
4⎠
3
⎝3
67. (a)
y
Shaded area is given by
2
1
∫ 0 arcsin x dx.
1
x
1
(b)
2
1
∫ 0 arcsin x dx
≈ 0.5708
(c) Divide the rectangle into two regions.
π
⎛π ⎞
Area rectangle = ( base)( height ) = 1⎜ ⎟ =
2
⎝2⎠
Area rectangle =
π
2
So,
1
=
∫ 0 arcsin x dx
π 2
1
∫ 0 arcsin x dx + ∫ 0
sin y dy
π 2
∫ 0 arcsin x dx + (−cos y)⎤⎦ 0
1
=
π
2
− 1,
(≈
=
1
∫ 0 arcsin x dx + 1
0.5708).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.8
68. (a)
513
1
⎛π ⎞
dx = [4 arctan x]0 = 4 arctan 1 − 4 arctan 0 = 4⎜ ⎟ − 4(0) = π
⎝4⎠
4
1
∫0 1 +
Inverse Trigonometric Functions: Integration
x2
(b) Let n = 6.
4∫
1
01
1
4
2
4
2
4
1⎤
⎛ 1 ⎞⎡
dx ≈ 4⎜ ⎟ ⎢1 +
+
+
+
+
+ ⎥ ≈ 3.1415918
18
1
1
36
1
1
9
1
1
4
1
4
9
1
25
36
+ x2
+
+
+
+
+
( )
( )
( )
( )
(
) 2 ⎥⎦
⎝ ⎠ ⎣⎢
(c) 3.1415927
1 x+2 2
dt
2∫ x t2 + 1
69. F ( x) =
(a) F ( x) represents the average value of f ( x) over the interval [ x, x + 2]. Maximum at x = −1, because the graph is greatest
on [−1, 1].
(b) F ( x) = [arctan t ]x
x+2
F ′( x) =
70.
∫
1
6x − x2
= arctan ( x + 2) − arctan x
1
1 + ( x + 2)
−
2
(1 + x 2 ) − ( x 2 + 4 x + 5) =
−4( x + 1)
1
=
= 0 when x = −1.
2
2
2
2
1+ x
( x + 1)( x + 4 x + 5) ( x + 1)( x 2 + 4 x + 5)
dx
(a) 6 x − x 2 = 9 − ( x 2 − 6 x + 9) = 9 − ( x − 3)
1
∫
6x − x
(b) u =
dx =
2
9 − ( x − 3)
2
⎛ x − 3⎞
= arcsin ⎜
⎟ +C
⎝ 3 ⎠
x , u 2 = x, 2u du = dx
1
∫
dx
∫
2
6u − u
2
4
(2u du )
=
∫
⎛
⎛ u ⎞
du = 2 arcsin ⎜
⎟ + C = 2 arcsin ⎜⎜
⎝ 6⎠
6−u
⎝
2
2
x⎞
⎟+C
6 ⎟⎠
4
(c)
y2
y1
−1
7
−2
The antiderivatives differ by a constant, π 2.
Domain: [0, 6]
71. False,
72. False,
∫ 3x
∫
dx
9 x 2 − 16
=
So,
∫
d ⎡
x
⎤
−arccos + C ⎥ =
dx ⎢⎣
2
⎦
dx
x
1
dx = arctan + C
25 + x 2
5
5
⎤
d ⎡
⎛u⎞
arcsin ⎜ ⎟ + C ⎥ =
⎢
dx ⎣
⎝a⎠
⎦
75.
73. True
3x
1
arcsec
+ C
12
4
du
a −u
2
2
1 − ( x 2)
2
=
1
4 − x2
74. False. Use substitution: u = 9 − e 2 x , du = −2e 2 x dx
⎛ u′ ⎞
⎜ ⎟ =
1 − (u a ) ⎝ a ⎠
1
2
12
2
u′
2
a − u2
⎛u⎞
= arcsin ⎜ ⎟ + C.
⎝a⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
514
76.
Chapter 5
Integration
d ⎡1
u
1 ⎡ u′ a ⎤
⎤
⎥
arctan + C ⎥ = ⎢
⎢
dx ⎣ a
a
a ⎢1 + (u a )2 ⎥
⎦
⎣
⎦
⎤
u′
u′
1⎡
⎢
⎥ = 2
2
a ⎢ (a 2 + u 2 ) a 2 ⎥
a + u2
⎣
⎦
=
So,
∫ a2
du
=
+ u2
∫ a2
u′
1
u
dx = arctan + C.
2
+ u
a
a
77. Assume u > 0.
⎡
⎡
⎤
d ⎡1
u
1⎢
u′ a
⎤
⎥ = 1⎢
+
=
arcsec
C
⎥
dx ⎢⎣ a
a
a ⎢ (u a ) (u a ) 2 − 1 ⎥
a ⎢u
⎦
⎥⎦
⎢⎣
⎣⎢
u′
(u 2
− a2 )
⎤
u′
⎥ =
.
2
2⎥
u u − a2
a ⎥
⎦
The case u < 0 is handled in a similar manner.
So,
∫u
du
u − a
2
=
2
∫u
78. Let f ( x) = arctan x −
f ′( x) =
u′
u − a
2
2
u
1
+ C.
arcsec
a
a
dx =
x
1 + x2
y
1
1− x
2x
−
=
> 0 for x > 0.
2
2
1 + x2
(1 + x ) (1 + x 2 )
2
2
Because f (0) = 0 and f is increasing for x > 0,
arctan x −
x
x
.
> 0 for x > 0. So, arctan x >
1 + x2
1 + x2
5
y3
4
3
2
y2
1
y1
2
4
6
x
8
10
Let g ( x) = x − arctan x
g ′( x) = 1 −
x2
1
=
> 0 for x > 0.
2
1+ x
1 + x2
Because g (0) = 0 and g is increasing for x > 0, x − arctan x > 0 for x > 0. So, x > arctan x. Therefore,
x
< arctan x < x.
1 + x2
79. (a) Area =
1
1
∫0 1 +
x2
dx
(b) Trapezoidal Rule: n = 8, b − a = 1 − 0 = 1
Area ≈ 0.7847
(c) Because
1
1
∫0 1 +
dx = [arctan x]0 =
1
x2
π
4
,
you can use the Trapezoidal Rule to approximate π 4, and therefore, π . For example, using n = 200, you obtain
π ≈ 4(0.785397) = 3.141588.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.8
Inverse Trigonometric Functions: Integration
515
80. (a) v(t ) = −32t + 500
550
0
20
0
(b) s(t ) =
∫ v(t ) dt
∫ (−32t
=
+ 500) dt
= −16t 2 + 500t + C
s(0) = −16(0) + 500(0) + C = 0 ⇒ C = 0
s(t ) = −16t 2 + 500t
When the object reaches its maximum height, v(t ) = 0.
v(t ) = −32t + 500 = 0
−32t = −500
t = 15.625
s(15.625) = −16(15.625) + 500(15.625)
2
= 3906.25 ft ( Maximum height )
1
∫ 32 + kv2
(c)
⎛
1
arctan ⎜⎜
32k
⎝
⎛
arctan ⎜⎜
⎝
dv = − ∫ dt
k ⎞
v ⎟ = −t + C1
32 ⎟⎠
k ⎞
v ⎟ = − 32kt + C
32 ⎟⎠
(
k
v = tan C −
32
32kt
)
(
32kt
(
)
32
tan C −
k
v =
)
When t = 0, v = 500, C = arctan 500 k 32 , and you have
⎛
32 ⎡
tan ⎢arctan ⎜⎜ 500
k
⎝
⎣⎢
v (t ) =
k ⎞
⎟ −
32 ⎟⎠
⎤
32kt ⎥.
⎦⎥
(d) When k = 0.001:
(
v (t ) =
)
32,000 tan ⎡arctan 500 0.00003125 −
⎣
0.032t ⎤
⎦
500
0
7
0
v(t ) = 0 when t0 ≈ 6.86 sec.
(e) h =
6.86
∫0
(
)
32,000 tan ⎡arctan 500 0.00003125 −
⎣
0.032 t ⎤ dt
⎦
Simpson’s Rule: n = 10; h ≈ 1088 feet
(f ) Air resistance lowers the maximum height.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
516
Chapter 5
Integration
Section 5.9 Hyperbolic Functions
1. (a) sinh 3 =
e3 − e −3
≈ 10.018
2
(b) tanh ( −2) =
2. (a) cosh 0 =
(b) sech 1 =
sinh ( −2)
e −2 − e2
≈ −0.964
e −2 + e 2
=
cosh ( −2)
⎛1 + 5 ⎞
6. (a) csch −1 2 = ln ⎜⎜
⎟ ≈ 0.481
2 ⎟⎠
⎝
(b) coth −1 3 =
e0 + e0
=1
2
2
⎛ e x − e− x ⎞
2
⎛
⎞
+⎜ x
7. tanh 2 x + sech 2 x = ⎜ x
−x ⎟
−x ⎟
⎝e + e ⎠
⎝e + e ⎠
2
≈ 0.648
e + e −1
3. (a) csch (ln 2) =
(b) coth (ln 5) =
=
1 ⎛ 4⎞
ln ⎜ ⎟ ≈ 0.347
2 ⎝ 2⎠
=
2
2
4
=
=
eln 2 − e − ln 2
2 − (1 2)
3
cosh (ln 5)
sinh (ln 5)
5 + (1 5)
=
=
5 − (1 5)
=
eln 5 + e − ln 5
eln 5 − e − ln 5
8. coth 2 x − csch 2 x =
13
12
4. (a) sinh −1 0 = 0
(
5. (a) cosh −1 2 = ln 2 +
(b) sech −1
⎛1 +
2
= ln ⎜
⎜
3
⎝
)
3 ≈ 1.317
9.
1 − ( 4 9) ⎞
⎟ ≈ 0.962
⎟
23
⎠
e 2 x − 2 + e −2 x + 4
(e x
+ e− x )
2
e 2 x + 2 + e −2 x
=1
e 2 x + 2 + e −2 x
cosh 2 x
1
−
sinh 2 x
sinh 2 x
=
cosh 2 x − 1
sinh 2 x
=
sinh 2 x
=1
sinh 2 x
(b) tanh −1 0 = 0
2
1 + (e 2 x + e −2 x ) 2
1 + cosh 2 x
=
2
2
2x
e + 2 + e −2 x
=
4
2
⎛ e x + e− x ⎞
2
= ⎜
⎟ = cosh x
2
⎝
⎠
2
⎛ e x − e− x ⎞
e 2 x − 2 + e −2 x
10. sinh 2 x = ⎜
⎟ =
2
4
⎝
⎠
⎛ e 2 x + e −2 x ⎞
−1 + ⎜
⎟
2x
−2 x
2
−1 + cosh 2 x
⎝
⎠ = −2 + e + e
=
2
2
4
So, sinh 2 x =
−1 + cosh 2 x
.
2
⎛ e x − e − x ⎞⎛ e x + e − x ⎞
e 2 x − e −2 x
11. 2 sinh x cosh x = 2⎜
= sinh 2 x
⎟⎜
⎟ =
2
2
2
⎝
⎠⎝
⎠
12. sinh 2 x + cosh 2 x =
e 2 x − e −2 x
e 2 x + e −2 x
+
= e2 x
2
2
⎛ e x − e − x ⎞⎛ e y + e − y ⎞ ⎛ e x + e − x ⎞⎛ e y − e − y ⎞
13. sinh x cosh y + cosh x sinh y = ⎜
⎟⎜
⎟ +⎜
⎟⎜
⎟
2
2
2
2
⎝
⎠⎝
⎠ ⎝
⎠⎝
⎠
1⎡ x+ y
−
+
x
y
= ⎣e
− e − x + y + e x − y − e ( ) + e x + y + e − x + y − e x − y − e −( x + y) ⎤⎦
4
=
1 ⎡ x+ y
e( x + y) − e −( x + y)
2e
− e −( x + y) ⎤ =
= sinh ( x + y )
⎦
4⎣
2
(
)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.9
14. 2 cosh
⎡ e( x + y )
x + y
x − y
= 2⎢
cosh
2
2
⎢⎣
2
+ e −( x + y) 2 ⎤ ⎡ e( x − y)
⎥⎢
2
⎦⎥ ⎣⎢
2
Hyperbolic Functions
517
+ e−( x − y) 2 ⎤
⎥
2
⎦⎥
⎡ e x + e y + e− y + e− x ⎤
e x + e− x
e y + e− y
= 2⎢
+
⎥ =
4
2
2
⎣
⎦
= cosh x + cosh y
15. sinh x =
3
2
2
13
⎛ 3⎞
cosh 2 x − ⎜ ⎟ = 1 ⇒ cosh 2 x =
⇒ cosh x =
4
⎝ 2⎠
32
3 13
=
13
13 2
tanh x =
csch x =
1
2
=
32
3
1
2 13
=
13
13 2
sech x =
coth x =
16. tanh x =
13
2
1
3
13
13
3
=
sinh x
e x − e− x
= lim
=1
x →0
x→0
2x
x
1
2
21. lim
2
3
3
⎛1⎞
2
2
⎜ ⎟ + sech x = 1 ⇒ sech x = ⇒ sech x =
4
2
⎝ 2⎠
cosh x =
coth x =
(coth x
1
= 2
12
23.
1
=
3 3
3
3
2 3
3
1
tanh x =
2
cosh x =
f ( x) = sinh (3 x)
f ( x) = cosh (8 x + 1)
f ′( x) = 8 sinh (8 x + 1)
25. y = sech (5 x 2 )
csch x =
3
sech x =
3
2
17. lim sinh x = ∞
x →∞
x →−∞
24.
y′ = −sech (5 x 2 ) tanh (5 x 2 )(10 x )
= −10 x sech (5 x 2 ) tanh (5 x 2 )
26.
coth x = 2
e x − e− x
= −1
x →−∞ e x + e − x
18. lim tanh x = lim
→ ∞ for x → 0 + , coth x → −∞ for x → 0− )
f ′( x) = 3 cosh (3 x)
3
Putting these in order:
sinh x =
x→0
1
2 3
=
3
3 2
3
⎛ 1 ⎞⎛ 2 3 ⎞
sinh x = tanh x cosh x = ⎜ ⎟⎜⎜
⎟⎟ =
3
⎝ 2 ⎠⎝ 3 ⎠
csch x =
22. lim coth x does not exist.
f ( x) = tanh ( 4 x 2 + 3 x)
f ′( x) = (8 x + 3) sech 2 ( 4 x 2 + 3 x)
27.
f ( x) = ln (sinh x)
f ′( x) =
1
(cosh x) = coth x
sinh x
19. lim sech x = 0
x →∞
20. lim csch x = 0
x →−∞
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
518
Chapter 5
Integration
x⎞
⎛
28. y = ln ⎜ tanh ⎟
2⎠
⎝
y = sinh (1 − x 2 ), (1, 0)
33.
y′ = cosh (1 − x 2 )( −2 x)
12
⎛ x⎞
sech 2 ⎜ ⎟
y′ =
tanh ( x 2)
⎝ 2⎠
=
1
2 sinh ( x 2) cosh ( x 2)
=
1
= csch x
sinh x
y′(1) = −2
Tangent line: y − 0 = −2( x − 1)
y = −2 x + 2
y = x cosh x ,
34.
(1, 1)
ln y = cosh x ln x
1
x
29. h( x) = sinh 2 x −
4
2
h′( x) =
y′
cosh x
=
+ sinh x ln x
y
x
cosh ( 2 x) − 1
1
1
cosh ( 2 x) −
=
= sinh 2 x
2
2
2
At (1, 1), y′ = cosh (1).
Tangent line: y − 1 = cosh (1)( x − 1)
30. y = x cosh x − sinh x
y = cosh (1) x − cosh (1) + 1
y′ = x sinh x + cosh x − cosh x
Note: cosh (1) ≈ 1.5431
= x sinh x
31.
f (t ) = arctan (sinh t )
35. y = (cosh x − sinh x) ,
2
(0, 1)
y′ = 2(cosh x − sinh x)(sinh x − cosh x)
1
cosh t
cosh t ) =
= sech t
f ′(t ) =
2 (
1 + sinh t
cosh 2 t
At (0, 1), y′ = 2(1)( −1) = −2.
32. g ( x) = sech 3 x
2
Tangent line: y − 1 = −2( x − 0)
g ′( x) = −2 sech (3 x) sech (3 x) tanh (3 x)(3)
y = −2 x + 1
= −6 sech 2 3x tanh 3 x
y = esinh x ,
36.
y′ = e
sinh x
(0, 1)
cosh x
y′(0) = e (1) = 1
0
Tangent line: y − 1 = 1( x − 0)
y = x +1
37.
f ( x) = sin x sinh x − cos x cosh x,
−4 ≤ x ≤ 4
f ′( x) = sin x cosh x + cos x sinh x − cos x sinh x + sin x cosh x
= 2 sin x cosh x = 0 when x = 0, ± π .
Relative maxima: ( ±π , cosh π )
Relative minimum: (0, −1)
(−π , cosh π )
12
− 2␲
(π , cosh π )
(0, − 1)
2␲
−2
38.
f ( x) = x sinh ( x − 1) − cosh ( x − 1)
6
f ′( x) = x cosh ( x − 1) + sinh ( x − 1) − sinh ( x − 1) = x cosh ( x − 1)
f ′( x) = 0 for x = 0.
By the First Derivative Test, (0, − cosh ( −1)) ≈ (0, −1.543) is a relative minimum.
−6
6
(0, −1.543)
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.9
g ( x) = x sech x
39.
g ′( x) = sech x − x sech x tanh x
Hyperbolic Functions
x
,
25
42. (a) y = 18 + 25 cosh
519
−25 ≤ x ≤ 25
y
= sech x(1 − x tanh x) = 0
80
x tanh x = 1
60
Using a graphing utility, x ≈ ±1.1997.
By the First Derivative Test, (1.1997, 0.6627) is a
20
relative maximum and ( −1.1997, − 0.6627) is a relative
−20
minimum.
x
−10
10
20
(b) At x = ±25, y = 18 + 25 cosh (1) ≈ 56.577.
1
(1.20, 0.66)
−␲
At x = 0, y = 18 + 25 = 43.
␲
(c) y′ = sinh
x
. At x = 25, y′ = sinh (1) ≈ 1.175.
25
(− 1.20, − 0.66) − 1
43.
h( x) = 2 tanh x − x
40.
=
1
2
∫ cosh(2 x)(2) dx
=
1
2
sinh 2 x + C
2
∫ sech (3x) dx
=
1
3
2
∫ sech (3x) (3 dx)
=
1
3
tanh (3 x) + C
∫ cosh 2 x dx
h′( x) = 2 sech 2 x − 1 = 0
sech 2 x =
1
2
44.
Using a graphing utility, x ≈ 0.8814.
From the First Derivative Test, (0.8814, 0.5328) is a
relative maximum and ( −0.8814, − 0.5328) is a relative
minimum.
45. Let u = 1 − 2 x, du = −2 dx.
∫ sinh (1 − 2 x) dx
= − 12 ∫ sinh (1 − 2 x)( −2) dx
= − 12 cosh (1 − 2 x) + C
2
(0.88, 0.53)
−3
46. Let u =
3
(− 0.88, − 0.53)
−2
∫
41. (a) y = 10 + 15 cosh
x
, −15 ≤ x ≤ 15
15
cosh
x , du =
x
x
1
2
x
dx.
⎛ 1 ⎞
x⎜
⎟ dx
⎝2 x ⎠
dx = 2∫ cosh
= 2 sinh
x +C
47. Let u = cosh ( x − 1), du = sinh ( x − 1) dx.
y
2
∫ cosh ( x − 1) sinh( x − 1) dx
30
=
1
3
cosh 3 ( x − 1) + C
20
48. Let u = cosh x, du = sinh x dx.
10
−10
x
10
20
sinh
∫ 1 + sinh 2 x dx
x
(c) y′ = sinh . At x = 15, y′ = sinh (1) ≈ 1.175.
15
sinh x
∫ cosh 2 x dx
=
−1
+ C
cosh x
= −sech x + C
(b) At x = ±15, y = 10 + 15 cosh (1) ≈ 33.146.
At x = 0, y = 10 + 15 cosh (0) = 25.
=
49. Let u = sinh x, du = cosh x dx.
cosh x
∫ sinh x
dx = ln sinh x + C
50. Let u = 2 x − 1, du = 2 dx.
2
∫ sech (2 x − 1) dx
=
1
2
=
1
2
2
∫ sech (2 x − 1)(2) dx
tanh ( 2 x − 1) + C
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
520
Chapter 5
Integration
59. Let u = 2 x, du = 2 dx.
x2
, du = x dx.
2
51. Let u =
∫0
⎛
x
2x ⎞
∫ ⎜⎝ csch 2 ⎟⎠ x dx = −coth 2 + C
2
2
2x
∫ x csch 2 dx =
2
2
2 4
1 − 4x
2
dx =
x tanh x dx = − ∫ sech 2 x( −sech x tanh x) dx
3
∫
ln 2
∫0
1
1
, du = − 2 dx.
x
x
csch (1 x) coth (1 x)
x2
1 − (2 x)
2 4
2
(2) dx
=
π
4
⎡ e x + e− x ⎤
−2 x
2e − x cosh x = 2e − x ⎢
⎥ =1+e
2
⎣
⎦
60.
= − 13 sech 3 x + C
53. Let u =
1
2 4
= ⎡⎣arcsin ( 2 x)⎤⎦ 0
52. Let u = sech x, du = −sech x tanh x dx.
∫ sech
∫0
2e − x cosh x dx =
−2 x
∫ 0 (1 + e ) dx
ln 2
ln 2
1
1⎛ 1 ⎞
coth ⎜ − 2 ⎟ dx
x
x⎝ x ⎠
dx = − ∫ csch
= csch
1
⎡
⎤
= ⎢ x − e −2 x ⎥
2
⎣
⎦0
1 ⎛ 1 ⎞⎤ ⎡
1⎤
⎡
= ⎢ln 2 − ⎜ ⎟⎥ − ⎢0 − ⎥
2
4
2⎦
⎝ ⎠⎦ ⎣
⎣
3
= + ln 2
8
1
+ C
x
54. Let u = sinh x, du = cosh x dx.
⎛ sinh x ⎞
dx = arcsin ⎜
⎟ +C
⎝ 3 ⎠
9 − sinh 2 x
cosh x
∫
61. Answers will vary.
62. f ( x) = cosh x and f ( x) = sech x take on only positive
values. f ( x ) = sinh x and f ( x) = tanh x are
⎛ e x − e− x ⎞
= arcsin ⎜
⎟ +C
6
⎝
⎠
55.
ln 2
∫0
tanh x dx =
ln 2
∫0
sinh x
dx,
cosh x
(u
= cosh x)
= ⎡⎣ln (cosh x)⎤⎦ 0
increasing functions.
63. The derivatives of f ( x) = cosh x and
f ( x) = sech x differ by a minus sign.
ln 2
(
= ln cosh (ln 2) − ln (cosh (0))
64. (a) f ( x) = cosh x is decreasing on ( −∞, 0) and
increasing on (0, ∞).
⎛5⎞
⎛5⎞
= ln ⎜ ⎟ − 0 = ln ⎜ ⎟
⎝ 4⎠
⎝4⎠
1
∫ 0 cosh
2
x dx =
11
∫0
+ cosh ( 2 x)
dx
2
1
=
1⎡
1
⎤
x + sinh ( 2 x)⎥
2 ⎢⎣
2
⎦0
=
1⎡
1
⎤
1 + sinh ( 2)⎥
2 ⎢⎣
2
⎦
1 1
=
+ sinh (1) cosh (1)
2 2
57.
4
1
∫ 0 25 − x 2 dx
=
4
58.
∫0
concave downward on (0, ∞).
65. y = cosh −1 (3 x)
3
y′ =
9x2 − 1
66. y = tanh −1
y′ =
x
2
1
2
⎛1⎞
=
2⎜ ⎟
4 − x2
1 − ( x 2) ⎝ 2 ⎠
y = tanh −1
67.
y′ =
1
=
x
⎛ 1 −1 2 ⎞
x ⎟
2⎜
⎠
x ⎝2
1−
( )
2
1
x (1 − x )
4
x⎤
4
⎡
dx = ⎢arcsin ⎥ = arcsin
2
5
5
⎣
⎦
25 − x
0
1
g ( x) = tanh x is concave upward on ( −∞, 0) and
1
1
1
1
dx + ∫
dx
10 ∫ 5 − x
10 5 + x
5+ x⎤
1
1
⎡1
= ⎢ ln
⎥ = 10 ln 9 = 5 ln 3
10
5
−
x
⎣
⎦0
4
(b) f ( x) = cosh x is concave upward on ( − ∞, ∞).
2 + (1/2)
5
eln 2 + e − ln 2
=
=
2
2
4
Note: cosh (ln 2) =
56.
g ( x) = tanh x is increasing on ( − ∞, ∞).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.9
68.
f ( x) = coth −1 ( x 2 )
77.
2x
f ′( x) =
2 x) =
2(
2
−
x4
1
1 − (x )
1
∫
1
1+e
2x
dx =
1
tan 2 x + 1
(sec2 x)
= sec x
x)
= ln
⎞
−2 csch −1 x
⎟ =
1 + x 2 ⎟⎠
x 1 + x2
−1
72. y = sech −1 (cos 2 x),
=
0 < x <
−1
cos 2 x 1 − cos 2 2 x
78.
dx = −
= −
(−2 sin 2 x)
79. Let u =
∫
1 + 4 x2
⎞
−1
⎟ + 2 sinh ( 2 x) −
2
1 + 4x ⎠
= 2 sinh
x4
4
2
−1
74. y = x tanh x + ln 1 − x
= x tanh −1 x +
76.
1
∫ 3 − 9 x2
∫ 2x
dx =
1 + e2 x − 1 − x + C
1
3 − x2
+ C
ln
12
3 + x2
1
2
1 + 4x2
= 2 ln
80. Let u = x3 2 , du =
2
∫
1
1 − 4x2
=
3
1+
ln
18
1−
dx =
3 + 3x
+C
3 − 3x
3x
+C
3x
1
1
(2) dx
2 ∫ 2 x 1 − ( 2 x)2
= −
1 ⎡1 +
ln ⎢
2 ⎢⎣
1
( x)
1+
= 2 sinh −1
x
1 + x3
dx =
3
2
(
2
⎛ 1 ⎞
⎜
⎟ dx
⎝2 x ⎠
x +C
)
x +
1+ x +C
x dx.
2
3∫
1
1 + (x
32
)
2
⎛3
⎜
⎝2
⎞
x ⎟ dx
⎠
2
sinh −1 ( x 3 2 ) + C
3
2
= ln x 3 2 + 1 + x 3 + C
3
=
)
(
1
1
(3) dx
3 ∫ 3 − (3 x ) 2
1 1
ln
32 3
dx.
x
1
dx = 2 ∫
x 1+ x
4x
1
ln (1 − x 2 )
2
=
⎞
⎟+C
⎟
⎠
1
−2 x
dx
2 ∫ 9 − ( x 2 )2
x , du =
−x
⎛ 1 ⎞
y′ = x⎜
+ tanh −1 x +
= tan −1 x
2⎟
1 − x2
⎝1 − x ⎠
75.
⎞
⎟+C
2x
e
+
1
⎠
ex
)
(
(2 x)
−1
⎞
⎟+C
⎟
⎠
1 + e2 x
ex
1⎛ 1 ⎞
3 − x2
= − ⎜ ⎟ ln
+ C
2⎝ 6 ⎠
3 + x2
since sin 2 x ≥ 0 for 0 < x < π 4.
⎛
y ′ = 2 x⎜
⎝
x
∫9 −
π
2 sin 2 x
2
=
= 2 sec 2 x,
cos 2 x sin 2 x
cos 2 x
73. y = 2 x sinh −1 ( 2 x) −
dx
2
⎛ −e x + e x 1 + e 2 x
= ln ⎜
⎜
e2 x
⎝
2
⎛
y′ = 2 csch −1 x⎜
⎜x
⎝
y′ =
e x 1 + (e x )
⎛
= ln ⎜
⎝1 +
1
y′ =
(2 cos 2 x) = 2 sec 2 x
1 − sin 2 2 x
71. y = (csch
ex
⎛1 +
= − ln ⎜
⎜
⎝
70. y = tanh −1 (sin 2 x )
−1
∫
521
= −csch −1 (e x ) + C
69. y = sinh −1 ( tan x)
y′ =
Hyperbolic Functions
81.
−1
∫ 4 x − x 2 dx
1
=
∫ ( x − 2)2
=
( x − 2) − 2
1
ln
4 ( x − 2) + 2
=
1
x−4
ln
+C
4
x
−4
dx
1 − 4x2 ⎤
⎥ +C
2x
⎦⎥
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
522
Chapter 5
82.
∫ ( x + 2)
83.
∫3
84.
∫1
dx
=
x2 + 4 x + 8
∫
dx
(x
+ 2)
(x
)
(
dx = ⎡ln x +
⎣⎢
x2 − 4
1
7
3
Integration
+ 2)
2
⎛2 +
1
= − ln ⎜
2 ⎜⎜
+ 4
⎝
(
7
x 2 − 4 ⎤ = ln 7 +
⎦⎥ 3
⎡ 1 ⎛2 +
dx = ⎢− ln ⎜
2 ⎜⎝
x 4 + x2
⎣⎢
4 + x 2 ⎞⎤
⎟⎥
⎟⎥
x
⎠⎦
1
1
1
=
5
)
1 1
1
(3) dx
3 ∫ − 1 4 2 − (3 x ) 2
1
1
∫0
1
25 x 2 + 1
1 1
5∫0
1
(5 x ) 2
(
⎡1
= ⎢ ln 5 x +
⎣5
=
89. y =
(
1
ln 5 +
5
x3 − 21x
dx =
x2
1 − 2x
dx =
x2
∫ 4x −
5 +
2
3⎞
⎟⎟
⎠
80 + 8 x − 16 x 2
4
dx
1
4∫
=
1
⎛ 4x − 1⎞
arcsin ⎜
⎟ +C
4
⎝ 9 ⎠
81 − ( 4 x − 1)
2
dx
(5) dx
=
∫ ( x − 1)
∫
= −
)
1
−4 x 2 + 8 x − 1
dx
2
2( x − 1)
1
ln
3
( )
3
3 +
2
− ⎡⎣2( x − 1)⎤⎦
dx
2
−4 x 2 + 8 x − 1
+C
2( x − 1)
1
⎤
25 x 2 + 1 ⎥
⎦0
26
)
20
⎞
⎟ dx
5 + 4 x − x2 ⎠
1
dx
= ∫ ( − x − 4) dx + 20 ∫
2
32 − ( x − 2)
∫ 5 + 4x −
⎛
∫ ⎜⎝ − x − 4 +
= −
3 + ( x − 2)
20
x2
ln
− 4x +
+ C
2
6
3 − ( x − 2)
= −
10
1+ x
x2
ln
− 4x +
+ C
2
3
5− x
=
90. y =
+1
1
∫
=
y =
1
1
[ln 7 − ln 1 + ln 7] = ln 7
24
12
dx =
⎛
45 ⎞
⎟ = ln ⎜⎜
5 ⎠⎟
⎝
88. Let u = 2( x − 1), du = 2 dx.
1⎡
⎛ 1 ⎞⎤
=
ln (7) − ln ⎜ ⎟⎥
⎢
24 ⎣
⎝ 7 ⎠⎦
86.
)
y =
1
⎡1 1 1
4 + 3x ⎤
ln
= ⎢
3
4
2
4 − 3 x ⎥⎦ −1
⎣
=
⎛7 +
5 = ln ⎜⎜
⎝ 3+
87. Let u = 4 x − 1, du = 4 dx.
(
∫ −1 16 − 9 x 2 dx
(
3
1 ⎛ 2 + 13 ⎞ 1
= − ln ⎜⎜
⎟⎟ + ln 2 +
2 ⎝
3
⎠ 2
85.
)
45 − ln 3 +
2
+ 2) + 4 ⎞⎟
+ C
⎟⎟
x + 2
⎠
(x
10
5− x
− x2
ln
− 4x −
+ C
2
3
x +1
4 − 2x
1
dx + 3∫
dx
2
2
x
( x − 2) − 4
∫ 4x −
= ln 4 x − x 2 +
( x − 2) − 2 + C
3
ln
4
( x − 2) + 2
= ln 4 x − x 2 +
3
x − 4
ln
+C
4
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.9
4
91. A = 2 ∫ sech
0
= 2∫
= 4∫
4
0
ex 2
e
4
x
dx
2
2
dx
+ e−x 2
92. A =
2
∫0
=
dx
2
x2
tanh 2 x dx =
3k
2
3kt
=
16
∫ x2
2
∫0
dx
+1
)
2
(
17 ≈ 5.237
94. A =
e 2 x − e −2 x
dx
e 2 x + e −2 x
5
ln 4 +
2
)
6
5
∫3
x2 − 4
(
= ⎡6 ln x +
⎣⎢
e4 + e −4
≈ 1.654
2
=
( x2 )
⎤
x4 + 1 ⎥
⎦0
≈ 5.207
2
∫ 16 dt
5 2
2∫0
(
523
dx
⎡5
= ⎢ ln x 2 +
⎣2
=
1
1
⎡1
⎤
= ⎢ ln (e 2 x + e −2 x )⎥ = ln (e 4 + e − 4 ) − ln 2
2
2
⎣2
⎦0
95.
x4 + 1
2x
(
1 2
1
= ∫ 2x
(2)(e2 x − e−2 x ) dx
2 0 e + e −2 x
= ln
5x
2
∫0
x2
(e ) + 1
4
= ⎡⎣8 arctan (e x 2 )⎤⎦
0
= 8 arctan (e 2 ) − 2π
0
93. A =
Hyperbolic Functions
dx
)
5
x2 − 4 ⎤
⎦⎥ 3
)
(
= 6 ln 5 +
21 − 6 ln 3 +
⎛5 +
= 6 ln ⎜⎜
⎝3+
21 ⎞
⎟ ≈ 3.626
5 ⎟⎠
5
)
1
dx
− 12 x + 32
1
∫ ( x − 6) 2
− 4
dx =
( x − 6) − 2 + C = 1 ln x − 8 + C
1
ln
2( 2)
4
x − 4
( x − 6) + 2
When x = 0: t = 0
C = −
When x = 1:
1
ln ( 2)
4
t = 10
−7
30k
1
1
1 ⎛7⎞
= ln
− ln ( 2) = ln ⎜ ⎟
−3
16
4
4
4 ⎝6⎠
k =
2 ⎛7⎞
ln ⎜ ⎟
15 ⎝ 6 ⎠
x −8
1
⎛ 3 ⎞⎛ 2 ⎞ ⎛ 7 ⎞
When t = 20: ⎜ ⎟⎜ ⎟ ln ⎜ ⎟( 20) = ln
4 2x − 8
⎝ 16 ⎠⎝ 15 ⎠ ⎝ 6 ⎠
2
x −8
⎛7⎞
ln ⎜ ⎟ = ln
2x − 8
⎝6⎠
x −8
49
=
36
2x − 8
62 x = 104
x =
104
52
=
≈ 1.677 kg
62
31
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
524
Chapter 5
Integration
96. (a) v(t ) = −32t
(b) s(t ) =
∫ v(t ) dt
∫(−32t ) dt
=
= −16t 2 + C
s(0) = −16(0) + C = 400 ⇒ C = 400
2
s(t ) = −16t 2 + 400
dv
= −32 + kv 2
dt
(c)
dv
= ∫ dt
− 32
dv
∫ 32 − kv 2 = − ∫ dt
∫ kv 2
Let u =
k v, then du =
32 +
32 −
1
1
ln
⋅
k 2 32
k dv.
k v
= −t + C
k v
Because v(0) = 0, C = 0.
ln
v
(
32 +
32 −
kv
= −2 32k t
kv
32 +
32 −
kv
= e −2
kv
32 +
k v = e −2
k e −2
k +
32 k t
)=
32
tanh
k
(
32 k t
(
32 (e
k (e
(
32 e −2
v =
⎡
(d) lim ⎢−
t →∞⎢
⎣
32 k t
32 −
32 k t
−2 32 k t
−2 32 k t
⎤
32k t ⎥ = −
⎥⎦
)
The velocity is bounded by −
kv
)
− 1) e
⋅
+ 1) e
)
−1
32 k t
32 k t
(
⎡
32 ⎢ − e
k ⎢ e
⎣⎢
=
32 k t
32 k t
− e−
+e
−
32 k t
32 k t
) ⎤⎥ = −
⎥
⎦⎥
32
tanh
k
(
32k t
)
32
k
k.
32
(e) Because ∫ tanh (ct ) dt = (1 c ) ln cosh (ct ) (which can be verified by differentiation), then
s (t ) = ∫ −
32
tanh
k
(
)
32k t dt = −
32
k
1
ln ⎡cosh
32k ⎣
(
)
1
32k t ⎤ + C = − ln ⎡cosh
⎦
k ⎣
(
)
32k t ⎤ + C .
⎦
When t = 0,
s(0) = C = 400 ⇒ 400 − (1 k ) ln ⎡cosh
⎣
When k = 0.01:
(
s2 (t ) = 400 − 100 ln cosh
0.32 t
(
)
32k t ⎤.
⎦
)
s1 (t ) = −16t + 400
2
s1 (t ) = 0 when t = 5 seconds
s2 (t ) = 0 when t ≈ 8.3 seconds
When air resistance is not neglected, it takes approximately 3.3 more seconds to reach the ground.
(f ) As k increases, the time required for the object to reach the ground increases.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 5.9
97. (a) y = a sech −1
x
−
a
dy
=
dx
−1
( x a)
a2 − x2 ,
(
1 − x2 a
2
)
+
525
a > 0
x
a − x
2
2
=
−a 2
x a − x
2
2
x
+
a − x
2
(b) Equation of tangent line through P = ( x0 , y0 ): y − a sech −1
When x = 0, y = a sech −1
Hyperbolic Functions
x0
−
a
a 2 − x02 +
x0
+
a
2
x2 − a2
=
x a − x
2
a 2 − x02 = −
a 2 − x02 = a sech −1
2
−
=
a2 − x2
x
a 2 − x02
( x − x0 )
x0
x0
.
a
So, Q is the point ⎡⎣0, a sech −1 ( x0 a )⎤⎦.
Distance from P to Q: d =
( x0
− 0) + ( y0 − asech −1 ( x0 a )) =
2
(
x02 + − a 2 − x02
)
2
=
a2 = a
y
Q
a
P
(a, 0)
x
L
98. In Example 5, a = 20. From Exercise 97(a),
y′ =
−
202 − x 2
.
x
99. Let
u = tanh −1 x, −1 < x < 1
tanh u = x.
sinh u
eu − e − u
= u
= x
cosh u
e + e−u
y
eu − e − u = xeu + xe − u
(0, y1)
e 2u − 1 = xe 2u + x
e 2u (1 − x) = 1 + x
(x, y)
1+ x
1− x
⎛1 + x ⎞
2u = ln ⎜
⎟
⎝1 − x ⎠
e 2u =
x
10
20
u =
The slope of the line connecting ( x, y ) and (0, y1 ) can
be determined by analyzing the shaded triangle. From
Exercise 97(b), the hypotenuse is a.
100. Let u = sinh −1 t. Then
sinh u =
20
20 2 − x 2
1 ⎛1 + x ⎞
ln ⎜
⎟, −1 < x < 1
2 ⎝1 − x ⎠
eu − e − u
= t
2
eu − e −u = 2t
e 2u − 2teu − 1 = 0
eu =
2t ±
x
m = −
202 − x 2
= y′
x
Hence, the boat is always pointing toward the person.
4t 2 + 4
2
= t ±
t2 + 1
= t +
t 2 + 1 ( because eu > 0)
(
u = ln t +
t2 + 1
)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
526
Chapter 5
Integration
101. Let y = arcsin ( tanh x). Then, sin y = tanh x =
e x − e− x
e x − e− x
and tan y =
= sinh x.
x
−x
2
e + e
x
e +e
−x
x
e −e
So, y = arctan (sinh x ). Therefore, arctan (sinh x) = arcsin( tanh x).
−x
y
2
b
102.
b
∫ −b
⎡ e xt ⎤
e xb
e − xb
2 ⎡ e xb − e − xb ⎤
2
−
= ⎢
e xt dt = ⎢ ⎥ =
⎥ = sinh ( xb)
x
x
x⎣
2
x
⎣ x ⎦ −b
⎦
e x + e− x
2
103. y = cosh x =
y = cosh −1 x
106.
cosh y = x
e x − e− x
y′ =
= sinh x
2
(sinh y)( y′)
=1
1
=
sinh y
y′ =
104. y = coth x =
y′ =
cosh x
sinh x
−1
sinh 2 x − cosh 2 x
=
= −csch 2 x
sinh 2 x
sinh 2 x
−2
(e x
=
1
x −1
2
sinh y = x
(cosh y ) y′
=1
y′ =
− e− x )
⎛ −2 ⎞ ⎛ e x − e − x ⎞
= ⎜ x
= −sech x tanh x
− x ⎟⎜ x
−x ⎟
⎝ e + e ⎠⎝ e + e ⎠
cosh y − 1
2
y = sinh −1 x
107.
2
105. y = sech x = x
e + e− x
y′ = −2(e x + e − x )
1
1
=
cosh y
1
sinh y + 1
2
=
1
x +1
2
y = sech −1 x
108.
sech y = x
−(sech y )( tanh y ) y′ = 1
y′ =
=
=
109. y = c cosh
−1
(sech y )( tanh y )
−1
(sech y)
1 − sech 2 y
−1
x 1 − x2
x
c
Let P( x1 , y1 ) be a point on the catenary.
x
c
y′ = sinh
The slope at P is sinh ( x1 c). The equation of line L is y − c =
When y = 0, c =
−1
( x − 0).
sinh ( x1 c)
x
⎛x ⎞
⇒ x = c sinh ⎜ 1 ⎟. The length of L is
sinh ( x1 c)
⎝c⎠
x
⎛x ⎞
c 2 sinh 2 ⎜ 1 ⎟ + c 2 = c ⋅ cosh 1 = y1 , the ordinate y1 of the point P.
c
c
⎝ ⎠
y
P(x1, y1)
(0, c)
L
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 5
527
110. There is no such common normal. To see this, assume there is a common normal.
y = cosh x ⇒ y′ = sinh x
Normal line at ( a, cosh a) is y − cosh a =
−1
( x − a).
sinh a
y
−1
Similarly, y − sinh c =
( x − c) is normal at (c, sinh c).
cosh c
Also,
(a, cosh a)
−1
−1
=
⇒ cosh c = sinh a.
sinh a
cosh c
The slope between the points is
Therefore, −
(c, sinh c)
y = cosh x
x
y = sinh x
sinh c − cosh a
.
c − a
a −c
= cosh c = sinh a.
cosh a − sinh c
cosh c > 0 ⇒ a > 0
sinh x < cosh x for all x ⇒ sinh c < cosh c = sinh a < cosh a. So, c < a. But,
−
a −c
< 0, a contradiction.
cosh a − sinh c
Review Exercises for Chapter 5
1.
2.
3.
∫ (4 x
∫
∫
2
+ x + 3) dx =
6
dx =
3
x
∫
x4 + 8
dx =
x3
4 x3
3
−3
∫ ( x + 8 x ) dx
2
∫ (5 cos x − 2 sec x) dx
5.
x
∫ (5 − e ) dx
∫
1 x2
2
+ 3x + C
x2 3
6 x −1 3 dx = 6 ⋅
+ C = 9x2 3 + C
(2 3)
4.
6.
+
=
1 2
4
x − 2 +C
2
x
= 5 sin x − 2 tan x + C
= 5x − e x + C
9.
f ′′( x) = 24 x, f ′( −1) = 7, f (1) = − 4
f ′( x) = 12 x 2 + C1
f ′( −1) = 7 = 12( −1) + C1 ⇒ C1 = − 5
2
f ′( x) = 12 x 2 − 5
f ( x ) = 4 x 3 − 5 x + C2
f (1) = − 4 = 4(1) − 5(1) + C2 ⇒ C2 = − 3
3
f ( x) = 4 x3 − 5 x − 3
10. f ′′( x) = 2cos x, f ′(0) = 4, f (0) = − 5
f ′( x) = 2sin x + C1
10
dx = 10 ln x + C
x
7. f ′( x) = − 6 x, f (1) = − 2
f ( x) = − 3x + C
f ′(0) = 4 = 2sin 0 + C1 ⇒ C1 = 4
f ′( x) = 2sin x + 4
f ( x) = − 2cos x + 4 x + C2
f (0) = − 5 = − 2cos 0 + 4(0) + C2
2
f (1) = − 2 = − 3(1) + C ⇒ C = 1
2
f ( x) = − 3x + 1
2
8. f ′( x) = 9 x 2 + 1, f (0) = 7
f ( x) = 3 x3 + x + C
f (0) = 7 = 3(0) + 0 + C ⇒ C = 7
2
f ( x) = 3 x3 + x + 7
= − 2 + C2 ⇒ C2 = − 3
f ( x) = − 2cos x + 4 x − 3
11. a(t ) = −32
v(t ) = −32t + 96
s(t ) = −16t 2 + 96t
(a) v(t ) = −32t + 96 = 0 when t = 3 sec.
s(3) = −144 + 288 = 144 ft
(b) v(t ) = −32t + 96 =
(c) s
96
2
when t =
( 32 ) = −16( 94 ) + 96( 32 ) = 108 ft
3
2
sec.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
528
Chapter 5
Integration
12. 45 mi h = 66 ft sec
30 mi h = 44 ft sec
a (t ) = − a
v(t ) = − at + 66 because v(0) = 66 ft sec.
a
s(t ) = − t 2 + 66t because s(0) = 0.
2
Solving the system
v(t ) = −at + 66 = 44
a
s(t ) = − t 2 + 66t = 264
2
you obtain t = 24 5 and a = 55 12. Now solve − (55 12)t + 66 = 0 and get t = 72 5.
2
55 12 ⎛ 72 ⎞
⎛ 72 ⎞
⎛ 72 ⎞
So, s⎜ ⎟ = −
⎜ ⎟ + 66⎜ ⎟ ≈ 475.2 ft.
2 ⎝ 5⎠
⎝ 5⎠
⎝ 5⎠
Stopping distance from 30 mi h to rest is 475.2 − 264 = 211.2 ft.
5
13.
∑ (5i − 3)
= 2 + 7 + 12 + 17 + 22 = 60
i =1
3
14.
∑ (k 2
k =0
10
15.
1
∑ 3i
+ 1) = 1 + 2 + 5 + 10 = 18
=
i =1
1
1
1
+
+"+
3(1) 3( 2)
3(10)
2
∑ ⎜⎝ n ⎟⎜
⎠⎝
17.
∑ 2i
20
i =1
2
⎛ 20( 21) ⎞
= 2⎜
⎟ = 420
⎝ 2 ⎠
30
18.
2
3 ⎛ 1 + 1⎞
3 ⎛ 2 + 1⎞
3 ⎛ n + 1⎞
⎛ 3 ⎞⎛ i + 1 ⎞
⎟ = ⎜
⎟ + ⎜
⎟ +"+ ⎜
⎟
n
n
n
n
n
n
⎠
⎝
⎠
⎝
⎠
⎝ n ⎠
i =1
n
16.
∑ (3i − 4)
i =1
30
30
i =1
i =1
= 3∑ i − 4 ∑ 1
2
20
19.
∑ (i + 1)
2
=
i =1
20
i =1
n
⎛
⎛ 3i ⎞ ⎞ 3
= lim ∑ ⎜ 8 − 2⎜ ⎟ ⎟
n→∞
⎝ n ⎠⎠ n
i =1 ⎝
(30)(31)
∑ (i 2
i =1
+ 2i + 1)
12
i =1
− 1) =
= lim
3⎡
6 n( n + 1) ⎤
⎢8n −
⎥
n⎣
n
2
⎦
n + 1⎤
⎡
= lim ⎢24 − 9
= 24 − 9 = 15
n→∞⎣
n ⎥⎦
y
12
∑(i3 − i)
8
i =1
(122 )(132 )
6
12(13)
4
2
= 6084 − 78 = 6006
=
3 n ⎛
6i ⎞
∑ ⎜8 − n ⎟⎠
n i =1 ⎝
n→∞
20( 21)( 41)
20( 21)
= 2
+ 20
6
2
= 2870 + 420 + 20 = 3310
∑ i(i 2
= lim
n→∞
=
20.
n
∑ f (ci)∆x
n→∞
Area = lim
− 4(30)
2
= 1395 − 120 − 1275
= 3⋅
3
, right endpoints
n
21. y = 8 − 2 x, ∆x =
4
−
2
−1
x
−2
1
2
3
4
5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 5
22. y = x 2 + 3, ∆x =
2
, right endpoints
n
23. y = 5 − x 2 , ∆x =
n
n
n →∞
i =1
⎡⎛ 2i ⎞
⎤⎛ 2 ⎞
= lim ∑ ⎢⎜ ⎟ + 3⎥⎜ ⎟
n →∞
n
⎢⎝ ⎠
⎥⎦⎝ n ⎠
i =1 ⎣
= lim
n →∞
i =1
2
n ⎡
3i ⎞ ⎤ ⎛ 3 ⎞
⎛
= lim ∑ ⎢5 − ⎜ −2 + ⎟ ⎥ ⎜ ⎟
n →∞
n ⎠ ⎦⎥ ⎝ n ⎠
⎝
⎢
i =1 ⎣
2
n
3
n
Area = lim ∑ f (ci ) ∆x
Area lim ∑ f (ci ) ∆x
n →∞
529
3 n ⎡
12i 9i 2 ⎤
− 2⎥
1+
∑
⎢
n →∞ n
n
n ⎦
i =1 ⎣
⎤
2 n ⎡ 4i 2
∑ ⎢ + 3⎥
n i =1 ⎣ n 2
⎦
= lim
⎤
2 ⎡ 4 n( n + 1)( 2n + 1)
= lim ⎢ 2
+ 3n⎥
n →∞ n n
6
⎣
⎦
3⎡
12 n( n + 1)
9 n( n + 1)( 2n + 1) ⎤
= lim ⎢n +
− 2
⎥
n →∞ n
n
n
2
6
⎣
⎦
⎡ 4 ( n + 1)( 2n + 1)
⎤
8
26
= lim ⎢
+ 6⎥ = + 6 =
2
n →∞ 3
3
3
n
⎣
⎦
⎡
n + 1 9 ( n + 1)( 2n + 1) ⎤
= lim ⎢3 + 18
−
⎥
n →∞
n
2
n2
⎣
⎦
= 3 + 18 − 9 = 12
y
y
12
10
6
8
4
6
3
4
2
1
2
x
−4 −3
2
1
x
−1
1
2
3
4
−2
24. y =
1 3
2
x , ∆x =
4
n
y
n
Area = lim ∑ f (ci ) ∆x
n →∞
20
i =1
15
3
n
1⎛
2i ⎞ ⎛ 2 ⎞
= lim ∑ ⎜ 2 + ⎟ ⎜ ⎟
n →∞
4
n ⎠ ⎝n⎠
⎝
i =1
10
1 n ⎡
24i
24i 2
8i 3 ⎤
8+
+ 2 + 3⎥
∑
⎢
n →∞ 2 n
n
n
n ⎦
i =1 ⎣
5
= lim
x
1
2
3
4
4 n ⎡
3i 3i 2
i3 ⎤
1+
+ 2 + 3⎥
∑
⎢
n →∞ n
n
n
n ⎦
i =1 ⎣
= lim
4⎡
3 n( n + 1)
3 n( n + 1)( 2n + 1)
1 n 2 ( n + 1)
⎢n +
+ 2
+ 3
n →∞ n ⎢
2
6
4
n
n
n
⎣
2
= lim
25. x = 5 y − y 2 , 2 ≤ y ≤ 5, ∆y =
⎤
⎥ = 4 + 6 + 4 + 1 = 15
⎥⎦
3
n
2
⎡
3i
3i ⎤ 3
Area = lim ∑ ⎢5⎛⎜ 2 + ⎞⎟ − ⎛⎜ 2 + ⎞⎟ ⎥⎛⎜ ⎞⎟
n →∞
n⎠ ⎝
n ⎠ ⎦⎥⎝ n ⎠
⎢ ⎝
i =1 ⎣
y
n
3 n ⎡
15i
i
9i 2 ⎤
− 4 − 12 − 2 ⎥
10 +
∑
⎢
n →∞ n
n
n
n ⎦
i =1 ⎣
= lim
3 ⎡
3i 9i ⎤
∑ ⎢6 + n − n2 ⎥
n →∞ n
i =1 ⎣
⎦
n
= lim
= lim
2
6
4
3
2
1
x
1
2
3
4
5
6
3⎡
3 n( n + 1)
9 n( n + 1)( 2n + 1) ⎤
− 2
⎢6n +
⎥
n
n
2
6
⎣
⎦
n →∞ n
9
27
⎡
⎤
= ⎢18 + − 9⎥ =
2
2
⎣
⎦
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
530
Chapter 5
Integration
5mb 2
mb 2
⎛ b ⎞⎛ b ⎞
⎛ 2b ⎞⎛ b ⎞
⎛ 3b ⎞⎛ b ⎞
⎛ 4b ⎞⎛ b ⎞
26. (a) S = m⎜ ⎟⎜ ⎟ + m⎜ ⎟⎜ ⎟ + m⎜ ⎟⎜ ⎟ + m⎜ ⎟⎜ ⎟ =
(1 + 2 + 3 + 4) =
16
8
⎝ 4 ⎠⎝ 4 ⎠
⎝ 4 ⎠⎝ 4 ⎠
⎝ 4 ⎠⎝ 4 ⎠
⎝ 4 ⎠⎝ 4 ⎠
mb
3mb
⎛b⎞
⎛ b ⎞⎛ b ⎞
⎛ 2b ⎞⎛ b ⎞
⎛ 3b ⎞⎛ b ⎞
s = m(0)⎜ ⎟ + m⎜ ⎟⎜ ⎟ + m⎜ ⎟⎜ ⎟ + m⎜ ⎟⎜ ⎟ =
(1 + 2 + 3) =
16
8
⎝ 4⎠
⎝ 4 ⎠⎝ 4 ⎠
⎝ 4 ⎠⎝ 4 ⎠
⎝ 4 ⎠⎝ 4 ⎠
2
(b) S ( n) =
⎛ bi ⎞⎛ b ⎞
f ⎜ ⎟⎜ ⎟ =
⎝ n ⎠⎝ n ⎠
n
⎛ mbi ⎞⎛ b ⎞
⎛b⎞
∑ ⎜⎝ n ⎟⎜
⎟ = m⎜ ⎟
⎠⎝ n ⎠
⎝ n⎠
i =1
2 n
⎛ bi ⎞⎛ b ⎞
f ⎜ ⎟⎜ ⎟ =
⎝ n ⎠⎝ n ⎠
n −1
2 n −1
n
∑
i =1
s( n) =
n −1
∑
i =0
(c) Area = lim
n →∞
27.
⎛ bi ⎞⎛ b ⎞
⎛b⎞
∑ m⎜⎝ n ⎟⎜
⎟ = m⎜ ⎟
n
⎠⎝ ⎠
⎝ n⎠
i =0
∑i
=
i =1
y
y = mx
2
mb 2 ( n + 1)
mb 2 ⎛ n( n + 1) ⎞
⎟ =
2 ⎜
n ⎝
2
2n
⎠
mb 2 ( n − 1)
mb 2 ⎛ ( n − 1)n ⎞
∑ i = n2 ⎜ 2 ⎟ =
2n
i =0
⎝
⎠
x=b
mb 2 ( n + 1)
mb 2 ( n − 1)
1
1
1
= lim
= mb 2 = (b)( mb) = ( base)( height )
n
→∞
2n
2n
2
2
2
28.
y
y
8
12
9
4
6
Triangle
2
3
−6 −4 −2
x
−3
3
6
9
∫ 0 (5 −
5
x − 5 ) dx =
1
2
(5)(5)
=
6
∫ −6
25
2
29. (a)
∫ 4 ⎣⎡ f ( x) + g ( x)⎤⎦ dx
=
∫ 4 f ( x) dx + ∫ 4 g ( x) dx
= 12 + 5 = 17
(b)
∫ 4 ⎡⎣ f ( x) − g ( x)⎤⎦ dx
=
∫ 4 f ( x) dx − ∫ 4 g ( x) dx
= 12 − 5 = 7
(c)
∫4
(d)
∫ 4 7 f ( x) dx
8
8
8
8
8
8
= 7∫
8
4
36 − x 2 dx =
1
2
π (6) = 18π
2
=
(b)
∫ 6 f ( x) dx
= −∫
(c)
∫ 4 f ( x) dx
4
= 0
(d)
∫3
6
−10 f ( x) dx = −10∫
6
3
31.
∫ 0 (3 + x) dx
32.
4
∫ 2 (x
8
8
f ( x) dx = 7(12) = 84
∫ 0 f ( x) dx
∫4
6
⎣⎡2 f ( x) − 3g ( x)⎤⎦ dx = 2 ∫ 4 f ( x) dx − 3∫ 4 g ( x) dx = 2(12) − 3(5) = 9
8
x
8
8
30. (a)
9
4
(semicircle)
(triangle)
33.
2
−4
−3
3
x
−2
∫ 0 f ( x) dx + ∫ 3 f ( x) dx
3
6
6
3
= 4 + (−1) = 3
f ( x) dx = −(−1) = 1
6
3
f ( x) dx = −10( −1) = 10
8
⎡
x2 ⎤
64
= ⎢3x +
= 56
⎥ = 24 +
2
2
⎣
⎦0
3
⎡ x5
⎤
211
231
⎛ 243
⎞ ⎛ 32
⎞
+ 4 x − 6) dx = ⎢ + 2 x 2 − 6 x⎥ = ⎜
+ 18 − 18 ⎟ − ⎜
+ 8 − 12 ⎟ =
+ 4 =
5
5
⎝ 5
⎠ ⎝5
⎠
⎣5
⎦2
x dx =
9
∫4
9
( 9 ) − ( 4 ) ⎤⎥⎦ =
2
⎡2
⎤
x 3 2 dx = ⎢ x5 2 ⎥ = ⎡⎢
5⎣
⎣5
⎦4
5
5
2
422
( 243 − 32) =
5
5
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
x
Review Exercises for Chapter 5
π 4
π 4
t dt = [tan t ]−π
34.
∫ −π 4 sec
35.
x
∫ 0 ( x + e ) dx
36.
2
6
2
= 1 − ( −1) = 2
42. Average value:
⎡ x2
⎤
= ⎢
+ e x ⎥ = 2 + e2 − 1 = 1 + e2
⎣2
⎦0
2
2
∫1
4
531
2
⎡ x4 ⎤
1
3
x
dx
=
⎢ ⎥ = 2
2 − 0 ∫0
⎣ 8 ⎦0
x3 = 2
x =
3
2
y
3
6
dx = 3 ln x ⎤⎦ 1 = 3 ln 6
x
8
6
37. A =
6
∫0
(8 − x) dx
4
6
⎡
x2 ⎤
= ⎢8 x − ⎥
2 ⎦0
⎣
x
1
= ( 48 − 18) − 0
x (1 − x) dx =
1
∫0
2
43. F ′( x ) = x 2 1 + x3
= 30
38. A =
( 3 2 , 2)
2
∫ 0 (x
1
12
− x3 2 ) dx
44. F ′( x ) =
1
x2
1
2
⎡2
⎤
= ⎢ x 3 2− x 5 2 ⎥
3
5
⎣
⎦0
45. F ′( x ) = x 2 + 3 x + 2
⎛ 2 2⎞
= ⎜ − ⎟ − ( 0)
⎝ 3 5⎠
4
=
15
39. A =
3
∫1
46. F ′( x ) = csc 2 x
47. u = x 3 + 3, du = 3x 2 dx
∫
2
dx
x
x2
x +3
3
dx =
3
+ 3)
−1 2
x 2 dx
−1 2
1
x 3 + 3) 3 x 2 dx
(
∫
3
12
2
= ( x 3 + 3) + C
3
= [2 ln x] 1
=
3
= 2 ln 3 − 2 ln 1
= ln 9
40. A =
∫ (x
x
∫ 0 (1 + e ) dx
48. u = 3 x 4 + 2, du = 12 x3 dx
2
∫ 6x
2
= ⎡⎣ x + e x ⎤⎦
0
3
3x 4 + 2 dx =
= ( 2 + e 2 ) − (0 + 1)
12
1
(3x 4 + 2) (12 x3 ) dx
2∫
4
1 ( 3 x + 2)
= ⋅
2
( 3 2)
= 1 + e 2 ≈ 8.3891
=
9
9 1
1
⎡1
⎤
dx = ⎢ 2 x ⎥
41. Average value:
9 − 4∫4
5
x
⎣
⎦4
2
2
= ( 3 − 2) =
5
5
y
2
1
=
5
x
2
5
x =
2
1
) 254 , 25 )
25
x =
4
x
2
4
6
8
32
+C
32
1 4
3 x + 2) + C
(
3
49. u = 1 − 3 x 2 , du = −6 x dx
2
∫ x(1 − 3x )
4
dx = − 16 ∫ (1 − 3 x 2 ) ( −6 x dx)
4
1 1 − 3x 2
= − 30
(
) +C
5
=
1
30
(3x 2 − 1)
5
+C
10
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
532
Chapter 5
Integration
50. u = x 2 + 8 x − 7, du = ( 2 x + 8) dx
∫
x+4
(x
2
+ 8 x − 7)
2
54.
∫
−2
1
x 2 + 8 x − 7) ( 2 x + 8) dx
(
∫
2
dx =
sin x
−1 2
dx = − ∫ (cos x) ( −sin x) dx
cos x
= −2(cos x)
12
= −2 cos x + C
−1
1 2
x + 8 x − 7) + C
(
2
−1
=
+C
2( x 2 + 8 x − 7)
= −
51.
∫ sin
52.
∫ x sin 3x
53.
∫
x cos x dx =
3
2
dx =
1
6
1
4
sin x + C
4
2
∫ (sin 3x )(6 x) dx
= − 16 cos 3 x 2 + C
55.
∫ xe
56.
∫
57.
∫ ( x + 1)5
− 3 x2
dx = −
12
∫ (1 + sec π x)
60.
∫ sec 2 x tan 2 x dx
2
sec π x tan π x dx =
+C
=
1
2
1
π
58.
( x + 1)2
1
∫ t2 2
∫ (1 + sec π x) (π sec π x tan π x) dx
2
∫ (sec 2 x tan 2 x)(2) dx
=
∫e
dx =
=
= −2 1 − sin θ + C
59.
1
6
1
2
− 3 x2
(− 6 x) dx
2
1
= − e−3 x + C
6
e1 x
⎛ 1⎞
dx = − ∫ e1 x ⎜ − 2 ⎟ dx = − e1 x + C
x2
⎝ x ⎠
cos θ
−1 2
dθ = − ∫ (1 − sin θ ) ( −cos θ ) dθ
1 − sin θ
= −2(1 − sin θ )
+C
−1 t
=
dt =
∫
1
2
( x + 1) 2 2 x + 1 dx
(
)
∫5
2
1
5( x + 1) + C
2 ln 5
2−1 t (t − 2 ) dt =
1 −1 t
2
+C
ln 2
1
(1 + sec π x)3 + C
3π
sec 2 x + C
61. (a) Answers will vary. Sample answer:
y
2
x
−3
(b)
3
dy
= x 9 − x 2 , (0, − 4)
dx
y =
2
∫ (9 − x )
12
3
−1 (9 − x )
2
32
2
x dx =
32
+C = −
32
1
(9 − x 2 ) + C
3
1
1
32
(9 − 0) + C = − (27) + C ⇒ C = 5
3
3
32
1
y = − (9 − x 2 ) + 5
3
−4 = −
62.
2 3
∫ 0 x ( x − 2)
1
3
−6
6
−5
dx
u = x 3 − 2, du = 3 x 2 dx, x 2 dx =
1
3
du
When x = 0, u = −2. When x = 1, u = −1
−1
∫ −2 u
63.
3
∫0
−1
3
1
u4 ⎤
1
16
15
5
−
= −
= −
du =
⎥ =
3
12 ⎦ −2
12 12
12
4
1
dx =
1+ x
−1 2
∫ 0 (1 + x) dx
3
3
12
= ⎡2(1 + x) ⎤ = 4 − 2 = 2
⎣
⎦0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 5
64.
6
x
6
∫3
3 x2 − 8
dx =
(
533
)
−1 2
12
1 6 2
( x − 8) (2 x) dx = ⎡⎢⎣13( x 2 − 8) ⎤⎥⎦ = 13 2 7 − 1
6∫3
3
65. u = 1 − y , y = 1 − u , dy = − du
When y = 0, u = 1. When y = 1, u = 0.
2π ∫
(y
0
1
+ 1) 1 − y dy = 2π ∫ − ⎡⎣(1 − u ) + 1⎤⎦
1
0
u du = 2π ∫
0
1
(u 3 2 − 2u1 2 ) du
4
28π
⎡2
⎤
= 2π ⎢ u 5 2 − u 3 2 ⎥ =
5
3
15
⎣
⎦1
0
66. u = x + 1, x = u − 1, dx = du
When x = −1, u = 0. When x = 0, u = 1.
0
2π ∫ x 2
−1
π
x + 1 dx = 2π ∫
u du = 2π ∫
2
(u5 2 − 2u3 2 + u1 2 ) du
0
1
4
2
32π
⎡2
⎤
= 2π ⎢ u 7 2 − u 5 2 + u 3 2 ⎥ =
5
3
105
⎣7
⎦0
1
π
∫0
68.
∫ −π 4 sin 2 x dx
π 4
= 0 because sin 2x is an odd function.
69. Trapezoidal Rule ( n = 4):
2
3
∫2 1 +
x2
dx
⎛
⎞
⎛
⎞
⎛
⎞
1⎡ 2
2
2
2
2 ⎤
⎢
⎥ ≈ 0.285
⎟ + 2⎜
⎟ + 2⎜
⎟ +
+ 2⎜
2
2
2
2
⎜ 1 + ( 9 4) ⎟
⎜ 1 + ( 5 2) ⎟
⎜ 1 + (11 4) ⎟ 1 + 32 ⎥
8 ⎢1 + 2
⎝
⎠
⎝
⎠
⎝
⎠
⎣
⎦
Simpson’s Rule ( n = 4):
≈
− 1)
π
⎡
⎛ x⎞
⎛ x⎞1
⎛ x ⎞⎤
cos⎜ ⎟ dx = 2∫ cos⎜ ⎟ dx = ⎢2 sin ⎜ ⎟⎥ = 2
0
⎝ 2⎠
⎝ 2⎠ 2
⎝ 2 ⎠⎦ 0
⎣
67.
≈
(u
0
1
2
3
∫2 1 +
x2
dx
⎛
⎞
⎛
⎞
⎛
⎞
1⎡ 2
2
2
2
2 ⎤
⎢
⎥ ≈ 0.284
⎟ + 2⎜
⎟ + 4⎜
⎟ +
+ 4⎜
2
2
2
2
⎜
⎟
⎜
⎟
⎜
⎟ 1 + 32 ⎥
12 ⎢1 + 2
⎝ 1 + ( 9 4) ⎠
⎝ 1 + ( 5 2) ⎠
⎝ 1 + (11 4) ⎠
⎣
⎦
Graphing utility: 0.284
70. Trapezoidal Rule ( n = 4):
Simpson’s Rule ( n = 4):
1
x3 2
∫ 0 3 − x 2 dx
≈
32
32
32
2(1 4)
2(1 2)
2(3 4)
1⎡
1⎤
⎢0 +
+
+
+ ⎥ ≈ 0.172
2
2
2
8⎢
2⎥
3 − (1 4)
3 − (1 2)
3 − ( 3 4)
⎣
⎦
32
32
32
4(1 4)
2(1 2)
4(3 4)
x3 2
1⎡
1⎤
dx
0
≈
⎢
+
+
+
+ ⎥ ≈ 0.166
2
2
2
2
3− x
12 ⎢
2⎥
3 − (1 4)
3 − (1 2)
3 − (3 4)
⎣
⎦
1
∫0
Graphing utility: 0.166
71. Trapezoidal Rule ( n = 4):
≈
3 ⎡
⎢ 0 ln (0 + 1) + 2
2( 4) ⎣⎢
Simpson’s Rule ( n = 4):
≈
3
∫0
3 ⎛3
⎞
ln ⎜ + 1⎟ + 2
4 ⎝4
⎠
3
∫0
3 ⎡
⎢ 0 ln (0 + 1) + 4
3( 4) ⎣⎢
x ln ( x + 1) dx
3 ⎛3
⎞
ln ⎜ + 1⎟ + 2
2 ⎝2
⎠
9 ⎛9
⎞
ln ⎜ + 1⎟ +
4 ⎝4
⎠
⎤
3 ln (3 + 1)⎥ ≈ 3.432
⎦⎥
3 ⎛3
⎞
ln ⎜ + 1⎟ + 4
2 ⎝2
⎠
9 ⎛9
⎞
ln ⎜ + 1⎟ +
4 ⎝4
⎠
⎤
3 ln (3 + 1)⎥ ≈ 3.414
⎦⎥
x ln ( x + 1) dx
3 ⎛3
⎞
ln ⎜ + 1⎟ + 2
4 ⎝4
⎠
Graphing utility: 3.406
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
534
Chapter 5
Integration
72. Trapezoidal Rule ( n = 4):
≈
1 + sin 2 x dx
π ⎡
⎤
1 + sin 2 π ⎥ ≈ 3.820
⎦⎥
π
3π
2
2 π
+ 2 1 + sin 2
+ 2 1 + sin 2
+
⎢ 1 + sin 0 + 2 1 + sin
2( 4) ⎣⎢
4
2
4
Simpson’s Rule ( n = 4):
≈
π
∫0
π
∫0
1 + sin 2 x dx
π ⎡
⎤
1 + sin 2 π ⎥ ≈ 3.829
⎦⎥
π
3π
2
2 π
+ 2 1 + sin 2
+ 4 1 + sin 2
+
⎢ 1 + sin 0 + 4 1 + sin
3( 4) ⎣⎢
4
2
4
Graphing utility: 3.820
73. u = 7 x − 2, du = 7 dx
1
∫ 7 x − 2 dx
74.
75.
∫
1
1
1
(7) dx = ln 7 x − 2 + C
7 ∫ 7x − 2
7
=
x2
1
1
1
dx = ∫ 3
3 x 2 ) dx = ln x3 + 1 + C
(
3
x +1
3 x +1
3
−sin x
sin x
∫ 1 + cos x dx = − ∫ 1 + cos x dx = −ln 1 + cos x
∫
ln
x
x
dx =
77. Let u = e
2x
1
1
1
(ln x)⎛⎜ ⎞⎟ dx = (ln x)2 + C
2∫
4
⎝ x⎠
+ e
−2 x
, du = ( 2e
−e
2x
−2 x
∫
79.
∫1
2x + 1
dx =
2x
4
) dx.
e 2 x − e −2 x
1 2e 2 x − 2e −2 x
∫ e2 x + e−2 x dx = 2 ∫ e2 x + e−2 x dx
1
= ln (e 2 x + e −2 x ) + C
2
4⎛
∫ 1 ⎜⎝1 +
1 ⎞
⎟ dx
2x ⎠
1
⎡
= ⎢ x + ln
2
⎣
+C
1
dx
x
76. u = ln x, du =
e2 x
1
dx = ln (e 2 x + 1) + C
2x
e +1
2
78.
= 4+
e
80.
∫1
81.
∫0
82.
∫0
ln x
dx =
x
π 3
π
4
⎤
x⎥
⎦1
1
ln 4 − 1 = 3 + ln 2
2
e
1⎛
∫ 1 (ln x) ⎜
1⎞
1
2⎤
⎡1
⎟ dx = ⎢ (ln x) ⎥ =
x
2
2
⎝ ⎠
⎣
⎦1
e
π 3
sec θ dθ = ⎡⎣ln sec θ + tan θ ⎤⎦ 0
tan
θ
3
π
dθ = 3 ∫ tan
0
(
= ln 2 +
3
θ ⎛1⎞
⎜ ⎟ dθ
3 ⎝ 3⎠
π
⎡
θ ⎤
= ⎢− 3 ln cos
3 ⎥⎦ 0
⎣
⎛1⎞
= − 3 ln ⎜ ⎟ + 3 ln (1)
⎝ 2⎠
= 3 ln 2
83. Let u = e 2 x , du = 2e 2 x dx.
∫ e2 x
1
dx =
+ e −2 x
e2 x
1
1
1
2x
2x
∫ 1 + e4 x dx = 2 ∫ 1 + e2 x 2 (2e ) dx = 2 arctan (e ) + C
( )
84. Let u = 5 x, du = 5 dx.
1
∫ 3 + 25 x 2 dx
=
1
5∫
( 3)
1
2
+ (5 x )
2
(5) dx
=
1
5x
arctan
+C
5 3
3
85. Let u = x 2 , du = 2 x dx.
∫
86.
∫x
x
1− x
4
dx =
1
9 x − 49
2
1
2∫
dx =
1
1 − ( x2 )
∫
2
(2 x) dx
1
3x
(3 x )
2
−7
=
1
arcsin x 2 + C
2
3 dx =
2
3x
1
+C
arcsec
7
7
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
)
Problem Solving for Chapter 5
535
2
⎛ x⎞
dx.
87. Let u = arctan ⎜ ⎟, du =
4 + x2
⎝ 2⎠
∫
arctan ( x 2)
1 ⎛
dx = ∫ ⎜ arctan
4 + x2
2 ⎝
2
88. Let u = arcsin ( 2 x), du =
∫
x ⎞⎛ 2 ⎞
1⎛
⎟⎜
⎟ dx = ⎜ arctan
2 ⎠⎝ 4 + x 2 ⎠
4⎝
1 − 4x2
dx
(arcsin 2 x) + C
1 ⎡⎣arcsin ( 2 x )⎤⎦
+ C =
2
2
4
2
arcsin 2 x
dx =
1 − 4x2
2
x⎞
⎟ +C
2⎠
89. y = sech ( 4 x − 1)
2
94.
y′ = − sech ( 4 x − 1) tanh ( 4 x − 1)( 4)
= − 4 sech ( 4 x − 1) tanh ( 4 x − 1)
90. y = 2 x − cosh
1
y′ = 2 −
2
x
95. Let u =
)
x = 2−
sinh x
2 x
4
( 4 x)
2
+1
16 x 2 + 1
2x
⎛ 2 ⎞
y′ = x⎜
+ tanh −1 2 x =
+ tanh −1 2 x
2⎟
1 − 4x2
⎝1 − 4x ⎠
∫
x
x −1
4
93. Let u = x , du = 3x dx.
2
2
3
∫ x (sech x )
2
=
∫
19
1
⎛2 ⎞
dx = tanh −1 ⎜ x ⎟ + C
6
3 ⎠
⎛4 2⎞
⎝
1−⎜ x ⎟
⎝9 ⎠
96. Let u = x 2 , du = 2 x dx.
92. y = x tanh −1 2 x
3
cosh 6 x + C
Alternate solution:
1
1
3 + 2x
∫ 32 − (2 x)2 dx = 12 ln 3 − 2 x + C
4
=
1
6
2
2
x, du = dx.
3
3
1
(sinh
=
∫ 9 − 4 x 2 dx
x
91. y = sinh −1 ( 4 x)
y′ =
∫ sinh 6 x dx
dx =
=
1
2∫
1
( x2 ) − 1
2
(
1
ln x 2 +
2
( 2 x) dx
)
x4 − 1 + C
2
1
(sech x3 ) (3x 2 ) dx
3∫
1
= tanh x 3 + C
3
dx =
Problem Solving for Chapter 5
1. (a) L(1) =
(b) L′( x) =
1
∫1
1
dt = 0
t
1
by the Second Fundamental Theorem of Calculus.
x
L′(1) = 1
(c) L( x ) = 1 =
2.718
∫1
x
∫1
1
dt for x ≈ 2.718
t
1
dt = 0.999896
t
(Note: The exact value of x is e, the base of the natural logarithm function.)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
536
Chapter 5
Integration
(d) First show that
x1
∫1
1
dt =
t
t
1
and du =
dt.
x1
x1
To see this, let u =
Then
1
dt =
t
x1
∫1
1
1
∫ 1 x1 t dt.
1
∫ 1 x1 ux1 ( x1 du )
1
=
1
1
∫ 1 x1 u du
=
1
1
∫ 1 x1 t dt.
Now,
L( x1 x2 ) =
2. (a)
x1 x2
∫1
1
dt =
t
x2
∫ 1 x1
⎛
1
t ⎞
du ⎜ using u = ⎟ =
u
x
⎝
1⎠
1
du +
u
1
∫ 1 x1
x2
∫1
1
du =
u
x1
∫1
1
du +
u
x2
∫1
1
du = L( x1 ) + L( x2 ).
u
y
10
7
6
5
4
3
2
1
−4
x
−2 −1
1 2
3
∫ −3
Area =
4 5
(9 − x 2 ) dx = 2∫
3
3
0
(b) Base = 6, height = 9, Area =
3
⎡
⎤
(9 − x 2 ) dx = 2⎢9 x − x3 ⎥ = 2[27 − 9] = 36
⎣
⎦0
2
3
bh =
2
3
(6)(9)
= 36
(c) Let the parabola be given by y = b 2 − a 2 x 2 , a, b > 0.
Area = 2 ∫
ba
0
(b2 − a 2 x 2 ) dx
ba
⎡
x3 ⎤
= 2 ⎢b 2 x − a 2 ⎥
3 ⎦0
⎣
y
b2
⎡ ⎛ b ⎞ a 2 ⎛ b ⎞3 ⎤
= 2 ⎢b 2 ⎜ ⎟ −
⎜ ⎟ ⎥
3 ⎝ a ⎠ ⎦⎥
⎣⎢ ⎝ a ⎠
⎡ b3 1 b3 ⎤
4 b3
= 2⎢ −
⎥ =
3 a⎦
3 a
⎣a
−
x
b
a
b
a
2b
, height = b 2
a
Base =
Archimedes’ Formula: Area =
3. (a) Let A =
2 ⎛ 2b ⎞ 2
4 b3
⎜ ⎟(b ) =
3⎝ a ⎠
3 a
f ( x)
b
∫ 0 f ( x) + f (b − x) dx.
Let u = b − x, du = − dx.
A =
f (b − u )
∫ b f (b − u ) + f (u ) (−du )
0
Then, 2 A =
So, A =
b
f ( x)
=
f (b − u )
b
∫ 0 f (b − u ) + f (u ) du
b
f (b − x )
=
∫ 0 f ( x) + f (b − x) dx + ∫ 0 f (b − x) + f ( x) dx
f (b − x )
b
∫ 0 f (b − x) + f ( x) dx
=
b
∫ 0 1 dx
= b.
b
.
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 5
(b) b = 1 ⇒
sin x
1
∫ 0 sin(1 − x) + sin x dx
(c) b = 3, f ( x) =
x +
4. S ( x) =
x
∫0
1
2
x
x
3
∫0
=
537
3− x
3
2
dx =
⎛πt2 ⎞
sin ⎜
⎟ dt
⎝ 2 ⎠
y
(a)
2
1
x
1
3
−1
−2
(b)
y
1.00
0.75
0.50
0.25
x
1
2
3
2
5
6
72 23
−0.25
The zeros of y = sin
(c) S ′( x) = sin
π x2
2
π x2
2
= 0 ⇒
Relative maxima at x =
correspond to the relative extrema of S(x).
π x2
2
= nπ ⇒ x 2 = 2n ⇒ x =
2 ≈ 1.4142 and x =
2n , n integer
6 ≈ 2.4495
Relative minima at x = 2 and x = 2 2 ≈ 2.8284
⎛ π x2 ⎞
π x2
π
(d) S ′′( x) = cos⎜
=
+ nπ ⇒ x 2 = 1 + 2n ⇒ x =
⎟(π x) = 0 ⇒
2
2
2
⎝
⎠
Points of inflection at x = 1,
5. (a)
1
∫ −1 cos x dx
1
∫ −1 cos x dx
3,
5, and
1 + 2n , n integer
7.
⎛ 1 ⎞
⎛ 1 ⎞
⎛ 1 ⎞
≈ cos⎜ −
⎟ + cos⎜
⎟ = 2 cos⎜
⎟ ≈ 1.6758
3⎠
⎝
⎝ 3⎠
⎝ 3⎠
1
= sin x⎤ = 2 sin (1) ≈ 1.6829
⎥⎦ −1
Error: 1.6829 − 1.6758 = 0.0071
(b)
1
1
1
1
3
+
=
1 + (1 3) 1 + (1 3)
2
∫ −1 1 +
x2
(Note:
exact answer is π 2 ≈ 1.5708)
dx ≈
(c) Let p( x) = ax 3 + bx 2 + cx + d .
1
1
∫ −1
⎡ ax 4
⎤
bx 3
cx 2
2b
p( x) dx = ⎢
+
+
+ dx⎥ =
+ 2d
3
2
3
⎣ 4
⎦ −1
2b
⎛ 1 ⎞
⎛ 1 ⎞
⎛b
⎞ ⎛b
⎞
p⎜ −
⎟ + p⎜
⎟ = ⎜ 3 + d ⎟ + ⎜ 3 + d ⎟ = 3 + 2d
3⎠
⎝
⎠ ⎝
⎠
⎝
⎝ 3⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
538
Chapter 5
Integration
y
6. (a)
5
4
3
2
1
(8, 3)
(6, 2)
f
(0, 0)
x
−1
−2
−3
−4
−5
(b)
2
4 5 6 7 8 9
(2, − 2)
x
0
1
2
3
4
5
6
F ( x)
0
− 12
−2
− 72
−4
− 72
−2
7
8
1
4
3
⎧− x,
0 ≤ x < 2
⎪
(c) f ( x ) = ⎨x − 4, 2 ≤ x < 6
⎪ 1 x − 1, 6 ≤ x ≤ 8
⎩2
F ( x) =
x
∫0
⎧( − x 2 2),
0 ≤ x < 2
⎪
⎪ 2
f (t ) dt = ⎨( x 2) − 4 x + 4, 2 ≤ x < 6
⎪
2
⎪⎩(1 4) x − x − 5, 6 ≤ x ≤ 8
F ′( x ) = f ( x). F is decreasing on (0, 4) and increasing on (4, 8). Therefore, the minimum is − 4 at x = 4,
and the maximum is 3 at x = 8.
⎧−1, 0 < x < 2
⎪
(d) F ′′( x) = f ′( x) = ⎨1,
2 < x < 6
⎪1 , 6 < x < 8
⎩2
x = 2 is a point of inflection, whereas x = 6 is not.
7. Let d be the distance traversed and a be the uniform acceleration. You can assume that v(0) = 0 and s(0) = 0. Then
a (t ) = a
v(t ) = at
s (t ) =
1 2
at .
2
2d
.
a
s(t ) = d when t =
The highest speed is v = a
2d
=
a
2ad .
)
ad
.
2
The lowest speed is v = 0.
The mean speed is
1
2
(
2ad + 0 =
The time necessary to traverse the distance d at the mean speed is t =
d
=
ad 2
2d
a
which is the same as the time calculated above.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 5
8.
∫ 0 f (t )( x − t ) dt
x
So,
∫ 0 xf (t ) dt − ∫ 0 tf (t ) dt
x
=
x
d x
f (t )( x − t ) dt = x f ( x) +
dx ∫ 0
= x∫
x
0
f (t ) dt −
∫ 0 f (t ) dt − x f ( x)
x
=
539
∫ 0 tf (t ) dt
x
∫ 0 f (t ) dt
x
Differentiating the other integral,
d x
dx ∫ 0
(∫
)
f (v ) dv dt =
x
0
∫ 0 f (v) dv.
x
So, the two original integrals have equal derivatives,
∫ 0 f (t )( x − t ) dt
x
∫ 0 (∫ 0 f (v) dv) dt + C.
x
=
t
Letting x = 0, you see that C = 0.
9. Consider F ( x) = ⎡⎣ f ( x)⎤⎦ ⇒ F ′( x) = 2 f ( x) f ′( x). So,
2
∫ a f ( x) f ′( x) dx
b
10. Consider
S ( n) =
1
∫0
1⎡
⎢
n⎣
b
∫a
=
x dx =
1
+
n
( x) dx
1 F′
2
1
2 x3 2 ⎤
3
⎥⎦
=
0
2
+"+
n
= ⎡⎣ 12 F ( x)⎤⎦ =
a
b
2.
3
1 ⎡F
2⎣
(b )
− F ( a)⎤⎦ =
1⎡f
2⎣
(b ) 2
2
− f ( a ) ⎤.
⎦
The corresponding Riemann Sum using right-hand endpoints is
n⎤
1
⎥ = 3 2 ⎡⎣ 1 +
n⎦
n
n ⎤⎦. So, lim
2 +"+
1+
2 +"+
n3 2
n →∞
n
=
2
.
3
1
11. Consider
1
∫0
x5 dx =
x6 ⎤
1
⎥ = .
6 ⎦0
6
The corresponding Riemann Sum using right endpoints is
S ( n) =
5
5
5
1 ⎡⎛ 1 ⎞
1
15 + 25 + " + n5
1
⎛ 2⎞
⎛n⎞ ⎤
= .
⎢⎜ ⎟ + ⎜ ⎟ + " + ⎜ ⎟ ⎥ = 6 ⎡⎣15 + 25 + " + n5 ⎤⎦. So, lim S ( n) = lim
n →∞
n →∞
n ⎣⎢⎝ n ⎠
n
n6
6
⎝n⎠
⎝ n ⎠ ⎥⎦
12. By Theorem 5.8, 0 < f ( x ) ≤ M ⇒
Similarly, m ≤ f ( x ) ⇒ m(b − a) =
So, m(b − a) ≤
So, 1 ≤
13. (a)
1
∫0
∫ a f ( x) dx
b
1 + x 4 dx ≤
∫ a f ( x) dx
b
b
∫a
m dx ≤
M dx = M (b − a).
b
∫a
≤
∫ a f ( x) dx.
b
≤ M (b − a ). On the interval [0, 1], 1 ≤
2.
(Note: ∫
1
1 + x 4 dx ≈ 1.0894
0
1 + x4 ≤
2 and b − a = 1.
)
(b) v is increasing (positive acceleration) on (0, 0.4) and (0.7, 1.0).
v
100
(c) Average acceleration =
80
60
v(0.4) − v(0)
60 − 0
=
= 150 mi h 2
0.4 − 0
0.4
(d) This integral is the total distance traveled in miles.
40
∫ 0 v(t ) dt
1
20
t
0.2
0.4
0.6
0.8
1.0
≈
1 ⎡0
10 ⎣
+ 2( 20) + 2(60) + 2( 40) + 2( 40) + 65⎤⎦ =
385
10
= 38.5 miles
(e) One approximation is
v(0.9) − v(0.8)
50 − 40
2
=
= 100 mi h
0.9 − 0.8
0.1
(other answers possible)
a(0.8) ≈
14. Because − f ( x) ≤ f ( x) ≤ f ( x) , − ∫
b
a
f ( x) dx ≤
∫ a f ( x) dx
b
≤
∫ a f ( x)
b
dx ⇒
∫ a f ( x) dx
b
≤
∫ a f ( x)
b
dx.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
540
Chapter 5
Integration
15. (a) (1 + i ) = 1 + 3i + 3i 2 + i 3 ⇒ (1 + i ) − i 3 = 3i 2 + 3i + 1
3
3
3i 2 + 3i + 1 = (i + 1) − i 3
3
(b)
n
∑ (3i 2
i =1
+ 3i + 1) =
n
∑ ⎡⎣(i + 1)
i =1
n
∑ (3i 2
So, ( n + 1) =
3
i =1
(c)
(n + 1)3
n
∑ (3i 2
−1 =
3
i =1
(
)
3
3
− i 3 ⎤ = ( 23 − 13 ) + (33 − 23 ) + " + ⎡⎢ ( n + 1) − n3 ⎤⎥ = ( n + 1) − 1
⎦
⎣
⎦
+ 3i + 1) + 1.
n
+ 3i + 1) =
∑ 3i 2
+
i =1
3( n)( n + 1)
+ n
2
3n( n + 1)
⇒ ∑ 3i 2 = n3 + 3n 2 + 3n −
−n
2
i =1
n
2n3 + 6n 2 + 6n − 3n 2 − 3n − 2n
2
3
2
2n + 3n + n
=
2
n( n + 1)( 2n + 1)
=
2
=
n
∑ i2
⇒
n( n + 1)( 2n + 1)
6
=
i =1
16. (a) y = f ( x ) = arcsin x
sin y = x
Area A =
π 4
π 4
∫ π 6 sin y ⋅ dy = [−cos y] π 6
= −
2
3
+
=
2
2
3 −
2
2
≈ 0.1589
π
⎛ 1 ⎞⎛ π ⎞
≈ 0.2618
Area B = ⎜ ⎟⎜ ⎟ =
⎝ 2 ⎠⎝ 6 ⎠
12
(b)
2 2
∫1 2
⎛ π ⎞⎛ 2 ⎞
arcsin x dx = Area (C ) = ⎜ ⎟⎜
⎟− A− B
⎝ 4 ⎠⎝ 2 ⎠
=
(c) Area A =
ln 3
∫0
π
2
8
−
3 −
2
2
−
π
12
⎛ 2
1⎞
= π⎜
− ⎟+
12 ⎠
⎝ 8
2 −
2
e y dy
(d)
3
∫1 ln x dx
≈ 0.1346
y
y = ln x
ln 3
= ⎡⎣e y ⎤⎦ 0 = 3 − 1 = 2
Area B =
3
ln 3
= 3( ln 3) − A = 3 ln 3 − 2 = ln 27 − 2 ≈ 1.2958
ey = x
A
B
tan y = x
Area A =
x
1
π 3
2
3
∫ π 4 tan y dy
π 3
= ⎡⎣−ln cos y ⎤⎦ π / 4
1
2
= −ln + ln
= ln
2
2
y
1
2 = ln 2
2
1
⎛π ⎞
⎛π ⎞
arctan x dx = ⎜ ⎟( 3 ) − ln 2 − ⎜ ⎟(1)
2
⎝4⎠
⎝3⎠
1
π(
=
4 3 − 3) − ln 2 ≈ 0.6818
12
2
Area C =
∫1
3
y = arctan x
π
3
π
4
A
C
B
x
1
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 5
17. Let u = 1 +
Area =
4
∫1
x,
541
x = u − 1, x = u 2 − 2u + 1, dx = ( 2u − 2) du.
1
dx =
x + x
2u − 2
du
− 2u + 1)
3
∫ 2 (u − 1) + (u 2
2(u − 1)
du
u2 − u
32
3
= ∫
du = ⎡⎣2 ln u ⎤⎦ 2
2 u
3
∫2
=
⎛ 3⎞
= 2 ln 3 − 2 ln 2 = 2 ln ⎜ ⎟
⎝ 2⎠
≈ 0.8109
18. Let u = tan x, du = sec 2 x dx.
Area =
π 4
∫0
1
dx =
sin x + 4 cos 2 x
2
=
π 4
sec 2 x
dx
tan 2 x + 4
1
du
+ 4
∫0
∫ 0 u2
1
⎡1
⎛ u ⎞⎤
= ⎢ arctan ⎜ ⎟⎥
⎝ 2 ⎠⎦ 0
⎣2
=
1
⎛1⎞
arctan ⎜ ⎟
⎝ 2⎠
2
19. (a) (i) y = e x
y1 = 1 + x
4
y
y1
−2
2
−1
(ii) y = e x
⎛ x2 ⎞
y2 = 1 + x + ⎜ ⎟
⎝ 2⎠
4
y
y2
−2
2
−1
(iii) y = e x
y3 = 1 + x +
x2
x3
+
2
6
4
y
−2
y3
2
−1
(b) nth term is x n n! in polynomial: y4 = 1 + x +
(c) Conjecture: e x = 1 + x +
x2
x3
x4
+
+
2!
3!
4!
x2
x3
+
+"
2!
3!
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 6
Differential Equations
Section 6.1
Slope Fields and Euler’s Method.......................................................543
Section 6.2
Differential Equations: Growth and Decay.......................................555
Section 6.3
Differential Equations: Separation of Variables ...............................564
Section 6.4
The Logistic Equation ........................................................................579
Section 6.5
First-Order Linear Differential Equations .........................................585
Section 6.6
Predator-Prey Differential Equations ................................................595
Review Exercises ........................................................................................................599
Problem Solving .........................................................................................................610
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 6
Differential Equations
Section 6.1 Slope Fields and Euler’s Method
1. Differential equation: y′ = 4 y
Solution: y = Ce
4. Differential equation:
4x
Solution: y 2 − 2 ln y = x 2
Check: y′ = 4Ce 4 x = 4 y
2. Differential equation: 3 y′ + 5 y = −e −2 x
Check: 2 yy′ −
Solution: y = e −2 x
y′ = −2e
−2 x
Check: 3( −2e −2 x ) + 5(e −2 x ) = −e −2 x
3. Differential equation: y′ =
dy
xy
= 2
dx
y −1
2xy
x2 − y2
2
y′ = 2 x
y
⎛
1⎞
⎜ y − ⎟ y′ = x
y⎠
⎝
x
y′ =
1
y −
y
y′ =
xy
y2 − 1
Solution: x 2 + y 2 = Cy
Check: 2 x + 2 yy′ = Cy′
y′ =
y′ =
−2 x
− C)
(2 y
−2 xy
2 y 2 − Cy
=
−2 xy
2 y − ( x2 + y2 )
=
−2 xy
y 2 − x2
=
2 xy
x2 − y 2
2
5. Differential equation: y′′ + y = 0
Solution: y = C1 sin x − C2 cos x
y′ = C1 cos x + C2 sin x
y′′ = −C1 sin x + C2 cos x
Check: y′′ + y = ( −C1 sin x + C2 cos x) + (C1 sin x − C2 cos x) = 0
6. Differential equation: y′′ + 2 y′ + 2 y = 0
Solution: y = C1e− x cos x + C2e − x sin x
Check:
y′ = −(C1 + C2 )e − x sin x + ( −C1 + C2 )e − x cos x
y′′ = 2C1e − x sin x − 2C2e − x cos x
y′′ + 2 y′ + 2 y = 2C1e− x sin x − 2C2e − x cos x +
2( −(C1 + C2 )e − x sin x + ( −C1 + C2 )e − x cos x) + 2(C1e − x cos x + C2e − x sin x)
= ( 2C1 − 2C1 − 2C2 + 2C2 )e − x sin x + ( −2C2 − 2C1 + 2C2 + 2C1 )e − x cos x = 0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
543
544
Chapter 6
Differential Equations
7. Differential equation: y′′ + y = tan x
Solution: y = −cos x ln sec x + tan x
y′ = ( −cos x)
=
1
(sec x ⋅ tan x + sec2 x) + sin x ln sec x + tan x
sec x + tan x
(−cos x)
sec x + tan x
(sec x)( tan x
+ sec x) + sin x ln sec x + tan x
= −1 + sin x ln sec x + tan x
y′′ = (sin x)
1
(sec x ⋅ tan x + sec2 x) + cos x ln sec x + tan x
sec x + tan x
= (sin x)(sec x) + cos x ln sec x + tan x
Check: y′′ + y = (sin x)(sec x) + cos x ln sec x + tan x − cos x ln sec x + tan x = tan x.
8. Differential equation: y′′ + 4 y′ = 2e x
2 −4 x
(e + e x )
5
2
8
2
y′ = ( −4e −4 x + e x ) = − e −4 x + e x
5
5
5
32 −4 x
2
y′′ =
e
+ ex
5
5
Solution: y =
2 ⎞
2 ⎞
⎛ 32
⎛ 8
⎛ 2 8⎞
Check: y′′ + 4 y′ = ⎜ e −4 x + e x ⎟ + 4⎜ − e−4 x + e x ⎟ = ⎜ + ⎟e x = 2e x
5 ⎠
5 ⎠
5⎠
⎝5
⎝ 5
⎝5
9. y = sin x cos x − cos 2 x
y′ = −sin 2 x + cos 2 x + 2 cos x sin x
= −1 + 2 cos 2 x + sin 2 x
Differential equation:
2 y + y′ = 2(sin x cos x − cos 2 x) + (−1 + 2 cos 2 x + sin 2 x )
= 2 sin x cos x − 1 + sin 2 x
= 2 sin 2 x − 1
⎛π ⎞
Initial condition ⎜ , 0 ⎟ :
⎝4 ⎠
sin
π
4
cos
π
4
− cos 2
π
4
2
2
2 ⎛ 2⎞
⋅
− ⎜⎜
⎟⎟ = 0
2
2
⎝ 2 ⎠
=
10. y = 6 x − 4 sin x + 1
12. y = e − cos x
y′ = 6 − 4 cos x
y′ = e − cos x (sin x ) = sin x ⋅ e − cos x
Differential equation: y′ = 6 − 4 cos x
Initial condition (0, 1): 0 − 0 + 1 = 1
11. y = 4e −6 x
y′ = sin x ⋅ e − cos x = sin x( y ) = y sin x
⎛π ⎞
Initial condition ⎜ , 1⎟ : e − cos(π 2) = e 0 = 1
⎝2 ⎠
2
y′ = 4e −6 x ( −12 x) = − 48 xe −6 x
2
Differential equation:
(
y′ = −12 xy = −12 x 4e −6 x
2
2
) = − 48xe
Differential equation:
−6 x 2
Initial condition (0, 4): 4e0 = 4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
In Exercises 13–20, the differential equation
is y (4) − 16 y = 0.
y (4) = 48 sin 2 x
y (4) − 16 y = 48 sin 2 x − 16(3 sin 2 x) = 0
y (4) = 3 cos x
Yes
y (4) − 16 y = − 45 cos x ≠ 0,
y = e −2 x
17.
y (4) = 16e −2 x
No
y (4) − 16 y = 16e −2 x − 16e −2 x = 0,
y = 2 sin x
14.
y (4) = 2 sin x
y
( 4)
Yes
− 16 y = 2 sin x − 16( 2 sin x) ≠ 0
y = 5 ln x
18.
No
30
x4
30
y (4) − 16 y = − 4 − 80 ln x ≠ 0,
x
No
y ( 4) = −
y = 3 cos 2 x
15.
545
y = 3 sin 2 x
16.
y = 3 cos x
13.
Slope Fields and Euler’s Method
y (4) = 48 cos 2 x
y (4) − 16 y = 48 cos 2 x − 48 cos 2 x = 0,
Yes
y = C1e 2 x + C2e −2 x + C3 sin 2 x + C4 cos 2 x
19.
y (4) = 16C1e 2 x + 16C2e −2 x + 16C3 sin 2 x + 16C4 cos 2 x
y (4) − 16 y = 0,
Yes
y = 3e 2 x − 4 sin 2 x
20.
y (4) = 48e 2 x − 64 sin 2 x
(
)
(
)
y (4) − 16 y = 48e 2 x − 64 sin 2 x − 16 3e 2 x − 4 sin 2 x = 0,
Yes
In Exercises 21–28, the differential equation is
xy′ − 2 y = x 3e x .
xy′ − 2 y = x(cos x) − 2(sin x) ≠ x3e x ,
No
21. y = x 2 , y′ = 2 x
xy′ − 2 y = x( 2 x) − 2( x 2 ) = 0 ≠ x3e x ,
No
26. y = cos x, y′ = −sin x
xy′ − 2 y = x( −sin x) − 2 cos x ≠ x3e x
No
22. y = x3 , y′ = 3 x 2
xy′ − 2 y = x(3 x 2 ) − 2 x3 = x3 ≠ x3e x
No
23. y = x 2e x , y′ = x 2e x + 2 xe x = e x ( x 2 + 2 x)
(
25. y = sin x, y′ = cos x
)
xy′ − 2 y = x e x ( x 2 + 2 x) − 2( x 2e x ) = x3e x ,
1
x
⎛1⎞
xy′ − 2 y = x⎜ ⎟ − 2 ln x ≠ x3e x ,
⎝ x⎠
27. y = ln x, y′ =
No
28. y = x 2e x − 5 x 2 , y′ = x 2e x + 2 xe x − 10 x
Yes
24. y = x 2 ( 2 + e x ), y′ = x 2 (e x ) + 2 x( 2 + e x )
xy′ − 2 y = x ⎡⎣ x 2e x + 2 xe x + 4 x⎤⎦ − 2 ⎡⎣ x 2e x + 2 x 2 ⎤⎦
= x 3e x ,
xy′ − 2 y = x ⎡⎣ x 2e x + 2 xe x − 10 x⎤⎦ − 2 ⎡⎣ x 2e x − 5 x 2 ⎤⎦
= x 3e x ,
Yes
Yes
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
546
Chapter 6
Differential Equations
29. y = Ce − x 2 passes through (0, 3).
31. y 2 = Cx 3 passes through ( 4, 4).
3 = Ce0 = C ⇒ C = 3
16 = C (64) ⇒ C =
Particular solution: y = 3e − x 2
Particular solution: y 2 =
30. y ( x 2 + y ) = C passes through (0, 2).
1
4
1 x 3 or
4
4 y 2 = x3
32. 2x 2 − y 2 = C passes through (3, 4).
2(0 + 2) = C ⇒ C = 4
2(9) − 16 = C ⇒ C = 2
Particular solution: y ( x 2 + y ) = 4
Particular solution: 2 x 2 − y 2 = 2
33. Differential equation: 4 yy′ − x = 0
General solution: 4 y 2 − x 2 = C
Particular solutions: C = 0, Two intersecting lines
C = ±1, C = ±4, Hyperbolas
2
2
2
C = −1
C=1
C=0
−3
−3
3
3
−2
−2
2
2
C=4
−3
3
−2
C = −4
−3
3
−3
−2
3
−2
34. Differential equation: yy′ + x = 0
General solution: x + y = C
2
2
Particular solutions: C = 0, Point
C = 1, C = 4, Circles
y
35. Differential equation: y′ + 2 y = 0
General solution: y = Ce −2 x
y′ + 2 y = C ( −2)e −2 x + 2(Ce −2 x ) = 0
Initial condition (0, 3): 3 = Ce0 = C
Particular solution: y = 3e −2 x
2
1
36. Differential equation: 3x + 2 yy′ = 0
1
2
x
General solution: 3x 2 + 2 y 2 = C
6 x + 4 yy′ = 0
2(3x + 2 yy′) = 0
3 x + 2 yy′ = 0
Initial condition (1, 3):
3(1) + 2(3) = 3 + 18 = 21 = C
2
2
Particular solution: 3x 2 + 2 y 2 = 21
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
Slope Fields and Euler’s Method
547
37. Differential equation: y′′ + 9 y = 0
General solution: y = C1 sin 3 x + C2 cos 3x
y′ = 3C1 cos 3x − 3C2 sin 3x,
y′′ = −9C1 sin 3 x − 9C2 cos 3 x
y′′ + 9 y = ( −9C1 sin 3 x − 9C2 cos 3 x) + 9(C1 sin 3 x + C2 cos 3 x) = 0
π
⎛π ⎞
Initial conditions ⎜ , 2 ⎟ and y′ = 1 when x = :
6
⎝6 ⎠
⎛π ⎞
⎛π ⎞
2 = C1 sin ⎜ ⎟ + C2 cos⎜ ⎟ ⇒ C1 = 2
⎝2⎠
⎝2⎠
y′ = 3C1 cos 3 x − 3C2 sin 3 x
1
⎛π ⎞
⎛π ⎞
1 = 3C1 cos⎜ ⎟ − 3C2 sin ⎜ ⎟ = −3C2 ⇒ C2 = −
3
⎝2⎠
⎝2⎠
Particular solution: y = 2 sin 3 x −
1
cos 3x
3
38. Differential equation: xy′′ + y′ = 0
General solution: y = C1 + C2 ln x
⎛1⎞
⎛1⎞
y′ = C2 ⎜ ⎟, y′′ = −C2 ⎜ 2 ⎟
⎝ x⎠
⎝x ⎠
1⎞
1
⎛
xy′′ + y′ = x⎜ −C2 2 ⎟ + C2 = 0
x ⎠
x
⎝
Initial conditions ( 2, 0) and y′ =
1
when x = 2:
2
0 = C1 + C2 ln 2
C2
x
C2
1
=
⇒ C2 = 1, C1 = −ln 2
2
2
y′ =
Particular solution: y = −ln 2 + ln x = ln
x
2
39. Differential equation: x 2 y′′ − 3xy′ + 3 y = 0
General solution: y = C1 x + C2 x3
y′ = C1 + 3C2 x 2 , y′′ = 6C2 x
x 2 y′′ − 3xy′ + 3 y = x 2 (6C2 x) − 3x(C1 + 3C2 x 2 ) + 3(C1x + C2 x3 ) = 0
Initial conditions ( 2, 0) and y′ = 4 when x = 2:
0 = 2C1 + 8C2
y′ = C1 + 3C2 x 2
4 = C1 + 12C2
C1 + 4C2 = 0 ⎫
⎬ C2 =
C1 + 12C2 = 4⎭
1,
2
C1 = −2
Particular solution: y = −2 x +
1 x3
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
548
Chapter 6
Differential Equations
40. Differential equation: 9 y′′ − 12 y′ + 4 y = 0
General solution: y = e 2 x 3 (C1 + C2 x )
(C1
(
+ C2 x) + C2e 2 x 3 = e 2 x 3 23 C1 + C2 + 23 C2 x
y′ =
2 e2 x 3
3
y′′ =
2 e2 x 3 2 C
3
3 1
(
)
+ C2 + 23 C2 x + e2 x 3 23 C2 =
(
9 y′′ − 12 y′ + 4 y = 9 23 e2 x 3
)( 23 C
1
(
2 e2 x 3 2 C
3
3 1
)
)
+ 2C2 + 23 C2 x
)
(
)
+ 2C2 + 23 C2 x − 12(e 2 x 3 ) 23 C1 + C2 + 23 C2 x + 4(e 2 x 3 )(C1 + C2 x) = 0
Initial conditions (0, 4) and (3, 0):
0 = e 2 (C1 + 3C2 )
4 = (1)(C1 + 0) ⇒ C1 = 4
0 = e 2 ( 4 + 3C2 ) ⇒ C2 = − 43
(
Particular solution: y = e 2 x 3 4 −
41.
)
dy
= 6x2
dx
y =
42.
4x
3
∫ 6x
47.
2
dx = 2 x3 + C
y =
(u
dy
= 10 x 4 − 2 x 3
dx
y =
∫ (10 x
4
− 2 x 3 ) dx = 2 x 5 −
x4
+ C
2
dy
x
43.
=
1 + x2
dx
1
x
y = ∫
dx = ln (1 + x 2 ) + C
1 + x2
2
(u
44.
= 1 + x 2 , du = 2 x dx)
dy
ex
=
4 + ex
dx
y =
45.
∫
48.
= −
= 2 x, du = 2 dx)
1
cos 2 x + C
2
dy
= tan 2 x = sec 2 x − 1
dx
y =
49.
∫ sin 2 x dx
∫ (sec
dy
= x
dx
Let u =
y =
∫x
2
x − 1) dx = tan x − x + C
x −6
x − 6, then x = u 2 + 6 and dx = 2u du.
x − 6 dx =
=
ex
dx = ln ( 4 + e x ) + C
4 + ex
dy
x − 2
2
=
=1−
dx
x
x
2⎞
⎛
y = ∫ ⎜1 − ⎟ dx
x⎠
⎝
= x − 2 ln x + C = x − ln x 2 + C
46.
dy
= sin 2 x
dx
∫ (u + 6)(u)(2u ) du
2∫ (u 4 + 6u 2 ) du
2
⎛ u5
⎞
= 2⎜
+ 2u 3 ⎟ + C
⎝5
⎠
2
52
32
= ( x − 6) + 4( x − 6) + C
5
2
32
= ( x − 6) ( x − 6 + 10) + C
5
2
32
= ( x − 6) ( x + 4) + C
5
dy
= x cos x 2
dx
y =
(u
2
∫ x cos( x ) dx
=
= x 2 , du = 2 x dx)
1
sin ( x 2 ) + C
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
50.
dy
= 2 x 4 x2 + 1
dx
58.
∫ 2x
4 x 2 + 1 dx
=
1
4
4 x 2 + 1 (8 x ) dx
=
1 ( 4 x + 1)
4
( 3 2)
=
32
1 2
(4 x + 1) + C
6
y =
∫
2
59.
32
+ C
y =
(u
∫ xe
dy
= e −2 x
dx
dx =
1 x2
e + C
2
dy
→ 0. Matches (d).
dx
dy
1
=
dx
x
For x = 0,
x2
dy
1
= . Matches (c).
dx
2
As x → ∞,
2
dy
= xe x
dx
549
dy
1
= cos x
2
dx
For x = 0,
60.
51.
Slope Fields and Euler’s Method
dy
is undefined (vertical tangent). Matches (a).
dx
61. (a), (b)
= x 2 , du = 2 x dx)
y
(4, 2)
5
52.
dy
= 5e − x 2
dx
y =
53.
−x 2
−x 2 ⎛ 1 ⎞
−x 2
∫ 5e dx = 5(−2)∫ e ⎜⎝ − 2 ⎟⎠ dx = −10e + C
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
1
4
3
2
dy dx
–4
Undef.
0
x
−2
8
(c) As x → ∞, y → −∞
As x → − ∞, y → −∞
62. (a), (b)
y
54.
4
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
dy dx
6
2
4
2
2
0
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
dy dx
−2 2
–2
0
0
−2 2
–8
(1, 1)
x
55.
4
−4
(c) As x → ∞, y → ∞
As x → − ∞, y → −∞
63. (a), (b)
56.
y
x
–4
–2
0
2
4
8
y
2
0
4
4
6
8
0
−
3
dy dx
3
−
3
0
(2, 2)
5
3
x
−4
57.
dy
= sin 2 x
dx
For x = 0,
dy
= 0. Matches (b).
dx
4
−3
(c) As x → ∞, y → −∞
As x → −∞, y → −∞
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
550
Chapter 6
Differential Equations
64. (a), (b)
66. (a) y′ =
y
2
−4
(0, −4)
y
(0, 1)
x
−2
1
, (0, 1)
y
3
2
−2
−4
x
3
−6
−3
(c) As x → ∞, y → −∞
As x → ∞, y → ∞
As x → − ∞, y → −∞
65. (a) y′ =
(b) y′ =
1
, (1, 0)
x
1
, (1, 1)
y
y
y
(1, 1)
(1, 0)
3
3
2
1
x
3
x
6
−1
−2
−3
−3
As x → ∞, y → ∞
As x → ∞, y → ∞
[Note: The solution is y = ln x. ]
(b) y′ =
67.
1
, ( 2, −1)
x
dy
= 0.25 y , y(0) = 4
dx
(a), (b)
y
12
(2, − 1)
3
2
1
−6
6
x
−1
6
−4
−2
−3
68.
As x → ∞, y → ∞
dy
= 4 − y , y ( 0) = 6
dx
(a), (b)
10
−5
5
0
69.
dy
= 0.02 y (10 − y ), y(0) = 2
dx
(a), (b)
12
−12
48
−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
70.
dy
= 0.2 x( 2 − y ), y(0) = 9
dx
71.
(a), (b)
Slope Fields and Euler’s Method
551
dy
= 0.4 y (3 − x), y(0) = 1
dx
(a), (b)
8
10
−2
−5
5
8
−2
0
72.
dy
1
πy
, y ( 0) = 2
= e − x 8 sin
dx
2
4
(a), (b)
5
−3
3
−3
y (0) = 2,
73. y′ = x + y,
n = 10, h = 0.1
y1 = y0 + hF ( x0 , y0 ) = 2 + (0.1)(0 + 2) = 2.2
y2 = y1 + hF ( x1 , y1 ) = 2.2 + (0.1)(0.1 + 2.2) = 2.43, etc.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
2
2.2
2.43
2.693
2.992
3.332
3.715
4.146
4.631
5.174
5.781
y(0) = 2, n = 20,
74. y′ = x + y,
h = 0.05
y1 = y0 + hF ( x0 , y0 ) = 2 + (0.05)(0 + 2) = 2.1
y2 = y1 + hF ( x1 , y1 ) = 2.1 + (0.05)(0.05 + 2.1) = 2.2075, etc.
The table shows the values for n = 0, 2, 4, … , 20.
n
0
2
4
6
8
10
12
14
16
18
20
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
2
2.208
2.447
2.720
3.032
3.387
3.788
4.240
4.749
5.320
5.960
75. y′ = 3 x − 2 y,
y(0) = 3,
n = 10,
h = 0.05
y1 = y0 + hF ( x0 , y0 ) = 3 + (0.05)(3(0) − 2(3)) = 2.7
y2 = y1 + hF ( x1 , y1 ) = 2.7 + (0.05)(3(0.05) + 2( 2.7)) = 2.4375, etc.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
yn
3
2.7
2.438
2.209
2.010
1.839
1.693
1.569
1.464
1.378
1.308
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
552
Chapter 6
Differential Equations
76. y′ = 0.5 x(3 − y ),
y (0) = 1,
n = 5,
h = 0.4
y1 = y0 + hF ( x0 , y0 ) = 1 + (0.4)(0.5(0)(3 − 1)) = 1
y2 = y1 + hF ( x1 , y1 ) = 1 + (0.4)(0.5(0.4)(3 − 1)) = 1.16, etc.
n
0
1
2
3
4
5
xn
0
0.4
0.8
1.2
1.6
2.0
yn
1
1
1.16
1.454
1.825
2.201
77. y′ = e xy ,
y (0) = 1,
n = 10, h = 0.1
y1 = y0 + hF ( x0 , y0 ) = 1 + (0.1)e0(1) = 1.1
y2 = y1 + hF ( x1 , y1 ) = 1.1 + (0.1)e(0.1)(1.1) ≈ 1.2116, etc.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
1
1.1
1.212
1.339
1.488
1.670
1.900
2.213
2.684
3.540
5.958
78. y′ = cos x + sin y , y (0) = 5, n = 10, h = 0.1
y1 = y0 + hF ( x0 , y0 ) = 5 + (0.1)(cos 0 + sin 5) ≈ 5.0041
y2 = y1 + hF ( x1 , y1 ) = 5.0041 + (0.1)(cos(0.1) + sin (5.0041)) ≈ 5.0078, etc.
79.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
5
5.004
5.008
5.010
5.010
5.007
4.999
4.985
4.965
4.938
4.903
dy
= y , y = 3e x , (0, 3)
dx
x
80.
0
0.2
0.4
0.6
0.8
1
y( x) (exact)
3
3.6642
4.4755
5.4664
6.6766
8.1548
y ( x) ( h = 0.2)
3
3.6000
4.3200
5.1840
6.2208
7.4650
y( x) ( h = 0.1)
3
3.6300
4.3923
5.3147
6.4308
7.7812
dy
2x
=
,y =
dx
y
x
2 x 2 + 4,
(0, 2)
0
0.2
0.4
0.6
0.8
1
y( x) (exact)
2
2.0199
2.0785
2.1726
2.2978
2.4495
y( x) ( h = 0.2)
2
2.000
2.0400
2.1184
2.2317
2.3751
y( x) ( h = 0.1)
2
2.0100
2.0595
2.1460
2.2655
2.4131
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.1
81.
dy
1
= y + cos x, y = (sin x − cos x + e x ),
dx
2
x
Slope Fields and Euler’s Method
553
(0, 0)
0
0.2
0.4
0.6
0.8
1
y( x) (exact)
0
0.2200
0.4801
0.7807
1.1231
1.5097
y ( x) ( h = 0.2)
0
0.2000
0.4360
0.7074
1.0140
1.3561
y( x) ( h = 0.1)
0
0.2095
0.4568
0.7418
1.0649
1.4273
82. As h increases (from 0.1 to 0.2), the error increases.
83.
dy
1
= − ( y − 72),
dt
2
(a)
(0, 140), h
a point ( x0 , y0 ) that satisfies the initial condition,
y ( x0 ) = y0 . Then, using a step size of h, find
t
0
1
2
3
Euler
140
112.7
96.4
86.6
(b) y = 72 + 68e −t 2
(c)
87. Consider y′ = F ( x, y ), y( x0 ) = y0 . Begin with
= 0.1
the point ( x1 , y1 ) = ( x0 + h, y0 + hF ( x0 , y0 )).
Continue generating the sequence of points
( xn +1, yn +1 ) = ( xn + h, yn + hF ( xn , yn )).
exact
t
0
1
2
3
Exact
140
113.24
97.016
87.173
dy
1
= − ( y − 72),
dt
2
(0, 140), h
= 0.05
88.
y = Ce kx
dy
= Cke kx
dx
Because dy dx = 0.07 y, you have Cke kx = 0.07Ce kx .
So, k = 0.07.
t
0
1
2
3
Euler
140
112.98
96.7
86.9
The approximations are better using h = 0.05.
84. When x = 0, y′ = 0, therefore (d) is not possible.
When x, y > 0, y′ < 0 (decreasing function) therefore
C cannot be determined.
89. False. Consider Example 2. y = x3 is a solution to
xy′ − 3 y = 0, but y = x 3 + 1 is not a solution.
90. True
91. True
(c) is the equation.
85. The general solution is a family of curves that satisfies
the differential equation. A particular solution is one
member of the family that satisfies given conditions.
92. False. The slope field could represent many different
differential equations, such as y′ = 2 x + 4 y.
86. A slope field for the differential equation
y′ = F ( x, y ) consists of small line segments at various
points ( x, y ) in the plane. The line segment equals the
slope y′ = F ( x, y ) of the solution y at the point ( x, y ).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
554
93.
Chapter 6
Differential Equations
dy
= −2 y , y (0) = 4, y = 4e −2 x
dx
(a)
x
0
0.2
0.4
0.6
0.8
1
y
4
2.6813
1.7973
1.2048
0.8076
0.5413
y1
4
2.5600
1.6384
1.0486
0.6711
0.4295
y2
4
2.4000
1.4400
0.8640
0.5184
0.3110
e1
0
0.1213
0.1589
0.1562
0.1365
0.1118
e2
0
0.2813
0.3573
0.3408
0.2892
0.2303
0.4312
0.4447
0.4583
0.4720
0.4855
r
(b) If h is halved, then the error is approximately halved ( r ≈ 0.5).
(c) When h = 0.05, the errors will again be approximately halved.
94.
dy
= x − y , y (0) = 1, y = x − 1 + 2e− x
dx
(a)
x
0
0.2
0.4
0.6
0.8
1
y
1
0.8375
0.7406
0.6976
0.6987
0.7358
y1
1
0.8200
0.7122
0.6629
0.6609
0.6974
y2
1
0.8000
0.6800
0.6240
0.6192
0.6554
e1
0
0.0175
0.0284
0.0347
0.0378
0.0384
e2
0
0.0375
0.0606
0.0736
0.0795
0.0804
0.47
0.47
0.47
0.48
0.48
r
(b) If h is halved, then the error is halved ( r ≈ 0.5).
(c) When h = 0.05, the error will again be approximately halved.
dI
+ RI = E (t )
dt
dI
4
+ 12 I = 24
dt
1
dI
= ( 24 − 12 I ) = 6 − 3I
4
dt
95. (a) L
y′ = ke kt
y′′ = k 2e kt
y′′ − 16 y = 0
2 kt
k e
− 16e kt = 0
k 2 − 16 = 0
I
3
(because ekt
≠ 0)
k = ±4
97. y = A sin ω t
t
−3
y = e kt
96.
y′ = Aω cos ω t
3
y′′ = − Aω 2 sin ω t
−3
(b) As t → ∞, I → 2. That is, lim I (t ) = 2. In fact,
t →∞
I = 2 is a solution to the differential equation.
y′′ + 16 y = 0
− Aω 2 sin ω t + 16 A sin ω t = 0
A sin ω t ⎡⎣16 − ω 2 ⎤⎦ = 0
If A ≠ 0, then ω = ±4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
f ( x) + f ′′( x) = − xg ( x) f ′( x),
98.
2 f ( x) f ′( x) + 2 f ′( x) f ′′( x) = −2 xg ( x) ⎡⎣ f ′( x)⎤⎦
Differential Equations: Growth and Decay
555
g ( x) ≥ 0
2
2
d ⎡
2
2
f ( x) + f ′( x) ⎤ = −2 x g ( x) ⎡⎣ f ′( x)⎤⎦
⎦
dx ⎣
For x < 0, − 2 x g ( x) ⎡⎣ f ′( x)⎤⎦ ≥ 0
2
For x > 0, − 2 x g ( x) ⎡⎣ f ′( x)⎤⎦ ≤ 0
2
So, f ( x ) + f ′( x ) is increasing for x < 0 and decreasing for x > 0.
2
2
f ( x ) + f ′( x) has a maximum at x = 0. So, it is bounded by its value at x = 0, f (0) + f ′(0) . So, f (and f ′ ) is bounded.
2
2
2
2
99. Let the vertical line x = k cut the graph of the solution y = f ( x) at ( k , t ). The tangent line at ( k , t ) is
y − t = f ′( k )( x − k )
Because y′ + p( x) y = q( x), you have
y − t = ⎡⎣q( k ) − p( k )t ⎤⎦ ( x − k )
⎛
1 q( k ) ⎞
For any value of t, this line passes through the point ⎜ k +
,
⎟.
⎜
p
k ) p( k ) ⎟⎠
(
⎝
To see this, note that
?
⎛
⎞
q( k )
1
− t = ⎡⎣q( k ) − p( k )t ⎤⎦ ⎜⎜ k +
− k ⎟⎟
p(k )
p
k
(
)
⎝
⎠
?
= q( k )k − p( k )tk +
q( k )
q( k )
− t − kq( k ) + p( k )kt =
− t.
p( k )
p( k )
Section 6.2 Differential Equations: Growth and Decay
1.
dy
= x +3
dx
y =
2.
∫ ( x + 3) dx
=
x2
+ 3x + C
2
−1
∫ (5 − 8 x) dx
1
=
∫ dx
ln y + 3 = x + C1
y + 3 = e x + C1 = Ce x
y = Ce x − 3
y
dy =
∫ − dx
ln 6 − y dy = − x + C1
= 5x − 4 x2 + C
dy
= y +3
dx
dy
= dx
y + 3
∫ y + 3 dy
dy
= 6− y
dx
dy
= dx
6− y
∫6 −
dy
= 5 − 8x
dx
y =
3.
4.
6 − y = e − x + C1 = Ce − x
y = 6 − Ce − x
5.
y′ =
5x
y
yy′ = 5 x
∫ yy′ dx
∫ y dy
=
=
∫ 5 x dx
∫ 5 x dx
1 2
5
y = x 2 + C1
2
2
2
2
y − 5x = C
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
556
Chapter 6
Differential Equations
y′ = −
x
4y
4 y y′ = −
x
6.
∫ 4 y dy
∫−
=
9. (1 + x 2 ) y′ − 2 xy = 0
2 xy
1 + x2
y′
2x
=
+
y
1 x2
y′ =
x dx
2
2 y 2 = − x 3 2 + C1
3
∫
6 y 2 + 2 x3 2 = C
7.
∫
y′ =
xy
y′
=
y
x
dy
=
y
∫
x dx
y = C (1 + x 2 )
xy + y′ = 100 x
10.
y′ = 100 x + xy = x(100 − y )
32
y = e(2 3)x + C1
y′
= x
100 − y
32
eC1 e(2 3)x
y′
( 2 x3 2 ) 3
∫ 100 −
y′
= x
1+ y
y′
∫1 +
y
dx =
dy
∫1 +
y
=
ln (1 + y ) =
y
dx =
∫ x dx
dy =
∫ x dx
x2
+ C1
2
x2
ln (100 − y ) = −
− C1
2
−ln (100 − y ) =
∫ x dx
∫ x dx
100 − y = e
( )
= Ce x
2 2
2 2
− x2 2
− 100
y = 100 − Ce − x
x 2 2 + C1
y = eC1 e x
( )
− x 2 2 − C1
− y = e − C1 e
x2
+ C1
2
1+ y = e
y
1
∫ 100 −
y′ = x(1 + y )
8.
2x
dx
x2
ln y = ln ⎡⎣C (1 + x 2 )⎤⎦
2 32
x + C1
3
= Ce
∫1 +
ln y = ln (1 + x 2 ) + ln C
x dx
=
dy
=
y
2x
dx
x2
ln y = ln (1 + x 2 ) + C1
∫
ln y =
∫1 +
∫
y′
dx =
y
∫
y′
dx =
y
−1
11.
−1
∫
2 2
dQ
k
= 2
dt
t
dQ
k
dt = ∫ 2 dt
dt
t
k
∫ dQ = − t + C
k
Q = − +C
t
dP
= k ( 25 − t )
dt
12.
∫
dP
dt =
dt
∫ k (25 − t ) dt
k
2
(25 − t ) + C
2
k
2
P = − ( 25 − t ) + C
2
∫ dP
= −
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
13. (a)
Differential Equations: Growth and Decay
y
15.
9
−5
x
−1
5
(0, 0)
dy
= x(6 − y ),
dx
dy
= − x dx
y −6
(b)
ln y − 6 =
(0, 0)
dy
1
= t , (0, 10)
dt
2
1
∫ dy = ∫ 2 t dt
1
y = t2 + C
4
1 2
10 = (0) + C ⇒ C = 10
4
1
y = t 2 + 10
4
16
− x2
+C
2
y − 6 = e−x
2 2+C
(0, 10)
= C1e− x
y = 6 + C1e − x
−4
2 2
4
−1
2 2
(0, 0): 0 = 6 + C1 ⇒ C1 = − 6
y = 6 − 6e − x
557
16.
2 2
dy
= −9 t ,
dt
∫ dy
=
∫ −9
(0, 10)
t dt
y = − 6t 3 2 + C
7
10 = 0 + C ⇒ C = 10
y = − 6t 3 2 + 10
−6
6
12
−1
14. (a)
(0, 10)
y
4
−1
3
−2
(0, 12 )
x
−4
4
17.
−4
(b)
dy
= xy ,
dx
⎛ 1⎞
⎜ 0, ⎟
⎝ 2⎠
dy
1
= − y, (0, 10)
dt
2
dy
1
∫ y = ∫ − 2 dt
1
ln y = − t + C1
2
dy
= x dx
y
y = e −(t 2) + C1 = eC1 e−t 2 = Ce −t 2
10 = Ce0 ⇒ C = 10
x2
ln y =
+ C
2
y = ex
2 2+C
y = 10e−t 2
= C1e x
2 2
1
⎛ 1⎞ 1
= C1e0 ⇒ C1 =
⎜ 0, ⎟ :
2
⎝ 2⎠ 2
1 x2 2
y = e
2
16
(0, 10)
−1
10
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
558
Chapter 6
Differential Equations
dy
3
= y, (0, 10)
dt
4
dy
3
= ∫ dt
y
4
18.
∫
ln y =
C =
3
t + C1
4
y =
y = e(3 4)t + C1
5 =
= eC1 e(3 4)t = Ce3t 4
k =
10 = Ce0 ⇒ C = 10
y = 10e
⎛ 1⎞
⎜ 0, ⎟, (5, 5)
⎝ 2⎠
21. y = Ce kt ,
3t 4
y =
40
1
2
1 kt
e
2
1 5k
e
2
ln 10
5
1 ⎣⎡(ln 10) 5⎦⎤t
1
1
e
= (10t 5 ) or y ≈ e0.4605t
2
2
2
22. y = Ce kt ,
(0, 4), ⎛⎜ 5,
⎝
(0, 10)
−5
C = 4
5
−5
19.
y = 4e kt
1
= 4e 5 k
2
ln (1 8)
k =
≈ −0.4159
5
dN
= kN
dt
(Theorem 6.1)
N = Ce kt
(0, 250): C
1⎞
⎟
2⎠
= 250
y = 4e −0.4159t
400
8
= ln
(1, 400): 400 = 250e ⇒ k = ln
250
5
k
N = 250e
ln (8 5)t
≈ 250e
0.4700t
When t = 4, N = 250e 4 ln(8 5) = 250e
5 = Ce ⇒ 10 = 2Ce k
k
ln(8 5)4
4
8192
⎛8⎞
.
= 250⎜ ⎟ =
5
⎝5⎠
2 = Ce5 k ⇒ 10 = 5Ce k
2Ce k = 5Ce5k
2e k = 5e5k
2
= e4k
5
dP
= kP
20.
dt
(Theorem 6.1)
P = Ce kt
(0, 5000): C
= 5000
19
(1, 4750): 4750 = 5000ek ⇒ k = ln⎛⎜ ⎞⎟
⎝ 20 ⎠
P = 5000e
(1, 5), (5, 2)
23. y = Ce kt ,
ln(19 20)t
≈ 5000e
−0.0513t
When t = 5, P = 5000eln(19 20)(5)
14
k =
1 ⎛ 2⎞
⎛ 2⎞
ln ⎜ ⎟ = ln ⎜ ⎟
4 ⎝5⎠
⎝ 5⎠
⎛ 2⎞
C = 5e − k = 5e −1 4 ln(2 5) = 5⎜ ⎟
⎝5⎠
−1 4
14
⎛5⎞
= 5⎜ ⎟
⎝ 2⎠
14
⎛ 5 ⎞ ⎡1 4 ln 2 5 ⎤t
y = 5⎜ ⎟ e ⎣ ( )⎦ ≈ 6.2872 e −0.2291t
⎝ 2⎠
5
⎛ 19 ⎞
= 5000⎜ ⎟ ≈ 3868.905.
⎝ 20 ⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
⎛ 1⎞
⎜ 3, ⎟, ( 4, 5)
⎝ 2⎠
24. y = Ce kt ,
1
= Ce3k ⇒ 1 = 2Ce3k
2
1
5 = Ce 4 k ⇒ 1 = Ce 4 k
5
1 4k
Ce
5
2Ce3k =
10e3k = e 4 k
Differential Equations: Growth and Decay
30. Because the half-life is 1599 years,
1
2
= 1e k (1599)
k =
1
1599
( 12 ).
ln
Because there are 1.5 g after 1000 years,
⎡ln 1 2 1599⎦⎤(1000)
1.5 = Ce ⎣ ( )
C ≈ 2.314.
So, the initial quantity is approximately 2.314 g.
⎡ln 1 2 1599⎤⎦(10,000)
When t = 10,000, y = 2.314e ⎣ ( )
10 = e k
≈ 0.03 g.
k = ln 10 ≈ 2.3026
y = Ce 2.3026t
5 = Ce
559
2.3026( 4)
31. Because the half-life is 1599 years,
1
2
= 1e k (1599)
( 12 ).
C ≈ 0.0005
k =
y = 0.0005e 2.3026t
Because there are 0.1 gram after 10,000 years,
25. In the model y = Ce kt , C represents the initial value of
y (when t = 0 ). k is the proportionality constant.
1
1599
ln
⎡ln 1 2 1599⎤⎦(10,000)
0.1 = Ce ⎣ ( )
C ≈ 7.63.
So, the initial quantity is approximately 7.63 g.
26. y′ =
dy
= ky
dt
dy
1
27.
= xy
2
dx
≈ 4.95 g.
32. Because the half-life is 5715 years,
dy
> 0 when xy > 0. Quadrants I and III.
dx
28.
⎡ln 1 2 1599⎦⎤(1000)
When t = 1000, y = 7.63e ⎣ ( )
1
2
= 1e k (5715)
k =
1
5715
ln
( 12 ).
dy
1
= x2 y
dx
2
Because there are 3 grams after 10,000 years,
dy
> 0 when y > 0. Quadrants I and II.
dx
C ≈ 10.089.
29. Because the initial quantity is 20 grams,
⎡ln 1 2 5715⎦⎤(10,000)
3 = Ce ⎣ ( )
So, the initial quantity is approximately 10.09 g.
⎡ln 1 2 5715⎤⎦(1000)
When t = 1000, y = 10.09e ⎣ ( )
y = 20e .
kt
≈ 8.94 g.
Because the half-life is 1599 years,
10 = 20e k (1599)
k =
1
1599
ln
( 12 ).
⎡ln 1 2 1599⎦⎤t
So, y = 20e ⎣ ( )
.
⎡ln 1 2 1599⎦⎤(1000)
When t = 1000, y = 20e ⎣ ( )
≈ 12.96 g .
When t = 10,000, y ≈ 0.26g.
33. Because the initial quantity is 5 grams, C = 5.
Because the half-life is 5715 years,
2.5 = 5e k (5715)
k =
1
5715
ln
( 12 ).
⎡ln 1 2 5715⎦⎤(1000)
≈ 4.43 g.
When t = 1000 years, y = 5e ⎣ ( )
⎡ln 1 2 5715⎦⎤(10,000)
When t = 10,000 years, y = 5e ⎣ ( )
≈ 1.49 g.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
560
Chapter 6
Differential Equations
34. Because the half-life is 5715 years,
1
2
= 1e
k =
38.
k (5715)
1
5715
ln
( 12 ).
Because there are 1.6 grams when t = 1000 years,
C ≈ 1.806.
So, the initial quantity is approximately 1.806 g.
⎡ln 1 2 5715⎦⎤(10,000)
When t = 10,000, y = 1.806e ⎣ ( )
≈ 0.54 g.
8000 = 4000e0.06t
= 1e k (24,100)
1
24,100
ln
2 = e0.06t
( 12 ).
ln 2 = 0.06t
Because there are 2.1 grams after 1000 years,
2.1 =
⎡ln 1 2 24,100⎦⎤(1000)
Ce ⎣ ( )
So, the initial quantity is approximately 2.161 g.
⎡ln 1 2
When t = 10,000, y = 2.161e ⎣ ( )
24,100⎤⎦(10,000)
≈ 1.62 g.
k =
k ( 24,100)
1
24,100
ln
ln 2
≈ 11.55 years.
0.06
36,000 = 18,000e 0.055t
ln 2 = 0.055t
t =
( ).
1
2
ln 2
≈ 12.6 years.
0.055
Amount after 10 years:
Because there are 0.4 grams after 10,000 years,
⎡ln 1 2
0.4 = Ce ⎣ ( )
40. Because A = 18,000e 0.055t , the time to double is given by
2 = e0.055t
36. Because the half-life is 24,100 years,
= 1e
t =
Amount after 10 years: A = 4000e(0.06)(10) ≈ $7288.48
C ≈ 2.161.
1
2
⎛1⎞
ln ⎜ ⎟t
2
ln (0.15) = ⎝ ⎠
5715
t ≈ 15,641.8 years
39. Because A = 4000e0.06t , the time to double is given by
35. Because the half-life is 24,100 years,
k =
1
C = Ce k (5715)
2
1
⎛1⎞
ln ⎜ ⎟
k =
5715 ⎝ 2 ⎠
0.15C = Ce[ln(1 2) 5715]t
⎡ln 1 2 5715⎤⎦(1000)
1.6 = Ce ⎣ ( )
1
2
y = Ce kt
24,100⎤⎦(10,000)
C ≈ 0.533.
So, the initial quantity is approximately 0.533 g.
⎡ln 1 2
When t = 1000, y = 0.533e ⎣ ( )
24,100⎦⎤(1000)
≈ 0.52 g.
A = 18,000e(0.055)(10) ≈ $31,198.55
41. Because A = 750e rt and A = 1500 when
t = 7.75, you have the following.
1500 = 750e7.75r
2 = e7.75r
ln 2 = 7.75r
37.
y = Ce kt
1C
2
= Ce
k =
r =
k (1599)
1
1599
ln
ln 2
≈ 0.0894 = 8.94%
7.75
Amount after 10 years: A = 750e 0.0894(10) ≈ $1833.67
( 12 )
ln 1 2 1599⎦⎤(100)
When t = 100, y = Ce ⎣⎡ ( )
≈ 0.9576 C
Therefore, 95.76% remains after 100 years.
42. Because A = 12,500e rt and A = 25,000 when
t = 20, you have the following.
25,000 = 12,500e 20 r
2 = e20 r
ln 2 = 20r
r =
ln 2
≈ 0.03466 ≈ 3.47%
20
Amount after 10 years:
A = 12,500e0.03466(10) ≈ $17,678.14
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
43. Because A = 500e rt and A = 1292.85 when
t = 10, you have the following.
1292.85 = 500e10 r
0.09 ⎞
⎛
48. 1,000,000 = P⎜1 +
⎟
12 ⎠
⎝
ln ( 2.5857) = 10r
ln ( 2.5857)
10
≈ 0.0950 = 9.50%
561
(12)( 25)
0.09 ⎞
⎛
P = 1,000,000⎜1 +
⎟
12 ⎠
⎝
≈ $106,287.83
2.5857 = e10 r
r =
Differential Equations: Growth and Decay
49. (a) 2000 = 1000(1 + 0.07)
−300
t
2 = 1.07t
The time to double is given by
ln 2 = t ln 1.07
1000 = 500e0.0950t
t =
2 = e0.0950t
ln 2 = 0.0950t
ln 2
≈ 10.24 years
ln 1.07
12t
0.07 ⎞
⎛
(b) 2000 = 1000⎜1 +
⎟
12 ⎠
⎝
ln 2
t =
≈ 7.30 years.
0.095
12 t
44. Because A = 6000e and A = 8950.95 when
t = 10, you have the following.
rt
8950.95 = 6000e
10 r
8950.95
= e10 r
6000
⎛ 8950.95 ⎞
ln ⎜
⎟ = 10r
⎝ 6000 ⎠
r =
1
8950.95
ln
= 0.04 = 4%
10
6000
The time to double is given by
2 = e0.04t
ln 2 = 0.04t
ln 2
≈ 17.33 years.
0.04
0.075 ⎞
⎛
45. 1,000,000 = P⎜1 +
⎟
12 ⎠
⎝
0.07 ⎞
⎛
2 = ⎜1 +
⎟
365 ⎠
⎝
365t
365t
2 = e0.07t
(12)(20)
ln 2 = 0.07t
−240
t =
ln 2
≈ 9.90 years
0.07
(12)(40)
P = 1,000,000(1.005)
0.08 ⎞
⎛
47. 1,000,000 = P⎜1 +
⎟
12 ⎠
⎝
0.07 ⎞
⎛
(c) 2000 = 1000⎜1 +
⎟
365 ⎠
⎝
(d) 2000 = 1000e(0.07)t
0.075 ⎞
⎛
P = 1,000,000⎜1 +
⎟
12 ⎠
⎝
≈ $224,174.18
0.06 ⎞
⎛
46. 1,000,000 = P⎜1 +
⎟
12 ⎠
⎝
0.07 ⎞
⎛
ln 2 = 12t ln ⎜1 +
⎟
12 ⎠
⎝
ln 2
≈ 9.93 years
t =
12 ln (1 + (0.07 12))
0.07 ⎞
⎛
ln 2 = 365t ln ⎜1 +
⎟
365 ⎠
⎝
ln 2
t =
≈ 9.90 years
365 ln (1 + (0.07 365))
12,000 = 6000e0.04t
t =
0.007 ⎞
⎛
2 = ⎜1 +
⎟
12 ⎠
⎝
−480
≈ $91,262.08
(12)(35)
0.08 ⎞
⎛
P = 1,000,000⎜1 +
⎟
12 ⎠
⎝
= $61,377.75
−420
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
562
Chapter 6
Differential Equations
50. (a) 2000 = 1000(1 + 0.055)
53. (a) P = Ce kt = Ce0.036t
t
P(1) = 34.6 = Ce0.036(1) ⇒ C ≈ 33.38
2 = 1.055t
ln 2 = t ln 1.055
t =
P = 33.38e0.036t
ln 2
≈ 12.95 years
ln 1.055
12 t
0.055 ⎞
⎛
(b) 2000 = 1000⎜1 +
⎟
12 ⎠
⎝
P = 33.38e0.036(10) ≈ 47.84 million.
(c) Because k > 0, the population is increasing.
54. (a) P = Ce kt = Ce −0.002t
12t
0.055 ⎞
⎛
2 = ⎜1 +
⎟
12 ⎠
⎝
P(1) = 10.0 = Ce −0.002(1) ⇒ C ≈ 10.02
0.055 ⎞
⎛
ln 2 = 12t ln ⎜1 +
⎟
12 ⎠
⎝
1
ln 2
t =
≈ 12.63 years
0.055 ⎞
12 ⎛
ln ⎜1 +
⎟
12 ⎠
⎝
0.055 ⎞
⎛
(c) 2000 = 1000⎜1 +
⎟
365 ⎠
⎝
0.055 ⎞
⎛
2 = ⎜1 +
⎟
365 ⎠
⎝
(b) For 2020, t = 10 and
365t
365t
0.055 ⎞
⎛
ln 2 = 365t ln ⎜1 +
⎟
365 ⎠
⎝
1
ln 2
t =
≈ 12.60 years
0.055 ⎞
365 ⎛
ln ⎜1 +
⎟
365 ⎠
⎝
P = 10.02e −0.002t
(b) For 2020, t = 10 and
P = 10.02e − 0.002(10) ≈ 9.82 million.
(c) Because k < 0, the population is decreasing.
55. (a) N = 100.1596(1.2455)
(b) N = 400 when t = 6.3 hours (graphing utility)
Analytically,
400 = 100.1596(1.2455)
400
= 3.9936
100.1596
t ln 1.2455 = ln 3.9936
t =
ln 3.9936
≈ 6.3 hours
ln 1.2455
56. (a) Let y = Ce kt .
2 = e0.055t
At time 2: 125 = Ce k (2) ⇒ C = 125e −2 k
ln 2 = 0.055t
ln 2
≈ 12.60 years
0.055
At time 4:
350 = Ce k (4) ⇒ 350 = (125e −2 k )(e 4 k )
14
5
51. (a) P = Ce kt = Ce −0.006t
P(1) = 2.2 = Ce −0.006(1) ⇒ C ≈ 2.21
k =
(b) For 2020, t = 10 and
P = 2.21e−0.006(10) ≈ 2.08 million.
(c) Because k < 0, the population is decreasing.
52. (a) P = Ce
= Ce
0.020 t
P(1) = 82.1 = Ce0.020(1) ⇒ C ≈ 80.47
P = 80.47e
= e2k
2k = ln
P = 2.21e −0.006t
kt
t
1.2455t =
(d) 2000 = 1000e0.055t
t =
t
0.020 t
(b) For 2020, t = 10 and
P = 80.47e0.020(10) ≈ 98.29 million.
(c) Because k > 0, the population is increasing.
1
2
14
5
ln
14
5
≈ 0.5148
C = 125e −2 k
= 125e −2(1 2)ln(14 5)
(145 ) =
= 125
625
14
≈ 44.64
Approximately 45 bacteria at time 0.
(b) y =
625
14
e(1 2) ln(14 5)t ≈ 44.64e0.5148t
(c) When t = 8,
y =
625 e(1 2)ln(14 5)8
14
(d) 25,000 =
=
( )
625 14
14 5
625 (1 2) ln(14 5)t
e
14
4
= 2744.
⇒ t ≈ 12.29 hours
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.2
57. (a)
P1 = Ce kt = 181e kt
205 = 181e10 k ⇒ k =
1
ln
10
205
( 181
) ≈ 0.01245
123
= e10 k
106
1 ⎛ 123 ⎞
⇒ k =
ln ⎜
⎟ ≈ 0.01487
10 ⎝ 106 ⎠
123 = 106e k (10) ⇒
t
(b) Using a graphing utility, P2 ≈ 182.3248(1.01091)
563
P1 = Ce kt = 106e kt (t = 0 ↔ 1920)
61. (a)
P1 ≈ 181e0.01245t ≈ 181(1.01253)
(c)
Differential Equations: Growth and Decay
t
1
P1 = 106e0.01487t = 106e10
300
⎛ 123 ⎞
ln⎜
⎟t
⎝ 106 ⎠
= 106(1.01499)
(b) Using a graphing utility, P2 ≈ 107.2727(1.01215) .
P1
t
P2
(c)
350
P1
0
150
50
P2
The model P2 fits the data better.
0
(d) Using the model P2 ,
320 = 182.3248(1.01091)
t
320
t
= (1.01091)
182.3248
ln (320 182.3248)
t =
ln (1.01091)
≈ 51.8 years, or 2011.
58. (a)
20 = 30(1 − e30 k )
ln (1 3)
30
N ≈ 30(1 − e
(b)
=
−0.0366 t
−ln 3
≈ −0.0366
30
)
25 = 30(1 − e −0.0366t )
e −0.0366t =
t =
The model P2 fits the data better.
(d)
P2 = 400 = 107.2727(1.01215)
t
400
t
= (1.01215)
107.2727
ln ( 400 107.2727)
t =
ln (1.01215)
≈ 109, or 2029.
62. A(t ) = V (t )e − 0.10t
30e30 k = 10
k =
100
75
1
6
−ln 6
≈ 49 days
−0.0366
59. (a) Because the population increases by a constant each
month, the rate of change from month to month will
always be the same. So, the slope is constant, and the
model is linear.
(b) Although the percentage increase is constant each
month, the rate of growth is not constant. The rate of
change of y is given by
= 100,000e0.8 t e − 0.10t = 100,000e0.8
dA
⎛ 0.4
⎞
= 100,000⎜
− 0.10 ⎟e0.8 t − 0.10t
dt
t
⎝
⎠
0.4
dA
= 0 when
= 0.10 ⇒ t = 16.
dt
t
The timber should be harvested in the year 2026
(2010 + 16).
Note: You could also use a graphing utility to graph
A(t ) and find the maximum value. Use a viewing
window of 0 ≤ x ≤ 30, 0 ≤ y ≤ 600,000.
63. β ( I ) = 10 log10
I
, I 0 = 10−16
I0
(a) β (10−14 ) = 10 log10
10−14
= 20 decibels
10−16
dy
= ry
dt
(b) β (10−9 ) = 10 log10
which is an exponential model.
(c) β (10 −6.5 ) = 10 log10
60. (a) Both functions represent exponential growth because
the graphs are increasing.
(b) g has a greater k value because its graph is increasing
at a greater rate than the graph of f.
t − 0.10 t
(d) β (10 −4 ) = 10 log10
10−9
= 70 decibels
10−16
10−6.5
= 95 decibels
10−16
10−4
= 120 decibels
10−16
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
t
564
Chapter 6
Differential Equations
93 = 10 log10
64.
I
= 10(log10 I + 16)
10−16
66.
−6.7 = log10 I ⇒ I = 10−6.7
80 = 10 log10
(See Example 6.)
y = 20 + Ce kt
I
= 10(log10 I + 16)
10−16
160 = 20 + Ce
k (0)
60 = 20 + 140e
−8 = log10 I ⇒ I = 10−8
⇒ C = 140
k (5)
2
= e5 k
7
1 ⎛ 2⎞
k = ln ⎜ ⎟ ≈ −0.25055
5 ⎝7⎠
⎛ 10−6.7 − 10−8 ⎞
Percentage decrease: ⎜
⎟(100) ≈ 95%
10−6.7
⎝
⎠
65. Because
dy
= k ( y − 20)
dt
dy
= k ( y − 80)
dt
30 = 20 + 140e(1 5) ln(2 7)t
t 5
∫
1
dy =
y − 80
t 5
1
⎛ 2⎞
= eln(2 7) = ⎜ ⎟
14
⎝7⎠
1
2
t
ln
= ln
14
5 7
1
5 ln
5 ln 14
14
t =
=
≈ 10.53 minutes
2
7
ln
ln
7
2
∫ k dt
ln ( y − 80) = kt + C.
When t = 0, y = 1500. So, C = ln 1420.
When t = 1, y = 1120. So,
k (1) + ln 1420 = ln (1120 − 80)
k = ln 1040 − ln 1420 = ln
⎡ln 104 142)⎤⎦t
So, y = 1420e ⎣ (
+ 80.
104
.
142
It will take 10.53 − 5 = 5.53 minutes longer.
67. False. If y = Ce kt , y′ = Ckekt ≠ constant.
When t = 5, y ≈ 379.2°F.
68. True
69. False. The prices are rising at a rate of 6.2% per year.
70. True
Section 6.3 Differential Equations: Separation of Variables
dy
x
=
dx
y
1.
∫ y dy
=
dy
= 0
dx
dy
5y
= − x2
dx
3. x 2 + 5 y
∫ x dx
2
y
x2
=
+ C1
2
2
y2 − x2 = C
∫ 5 y dy
5y
2
∫y
2
dy =
∫ 3x
2
dx
y3
= x 3 + C1
3
y 3 − 3x3 = C
∫ −x
=
−x
+ C1
3
2
dx
3
15 y 2 + 2 x3 = C
3x 2
dy
= 2
dx
y
2.
2
=
dy
6 − x2
=
dx
2 y3
4.
∫ 2y
3
dy =
2
∫ (6 − x ) dx
y4
x3
= 6x −
+ C1
2
3
3 y 4 + 2 x3 − 36 x = C
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
5.
dr
= 0.75 r
ds
dr
∫ r = ∫ 0.75 ds
ln r = 0.75 s + C1
r = e
Differential Equations: Separation of Variables
11.
1 − 4 x 2 y′ = x
∫ dy
0.75 s + C1
=
12.
s2
+C
2
r = 0.375 s 2 + C
∫ dy
ln y = 3 ln 2 + x + ln C = ln C ( 2 + x )
∫
3
∫
∫
dx
x
ln y = ln x + ln C = ln Cx
y = Cx
dy
=
y
ln y =
y
=
∫ 4 sin x dx
y2
= −4 cos x + C1
2
y 2 = C − 8 cos x
10.
y
∫
ln x
dx
x
∫ 12 y dy
=
∫ 7e
x
dx
6 y 2 = 7e x + C
15. yy′ − 2e x = 0
y
dy
= 2e x
dx
∫ y dy
=
∫ 2e
x
dx
dy
= −8 cos(π x)
dx
Initial condition (0, 3):
∫ − 8 cos(π x) dx
−8 sin (π x)
y
=
+C
2
π
−16
y2 =
sin (π x) + C
2
2
dy
= 7e x
dx
y2
= 2e x + C
2
=
dx ⎞
⎛
⎜ u = ln x, du =
⎟
x⎠
⎝
1
2
(ln x) + C1
2
yy′ = −8 cos(π x)
∫ y dy
dx
14. 12 yy′ − 7e x = 0
dy
= 4 sin x
dx
∫ y dy
x 2 − 16
2
2
y = e(1 2)(ln x) + C1 = Ce(ln x)
12 y
yy′ = 4 sin x
9.
∫
13. y ln x − xy′ = 0
3
xy′ = y
dy
=
y
=
x 2 − 16
11x
y = 11 x 2 − 16 + C
3
dx
2+ x
y = C ( x + 2)
11x
dy
=
dx
7. ( 2 + x) y′ = 3 y
8.
dx
1 − 4x2
x 2 − 16 y′ = 11x
r = 0.75
∫
∫
dx
−1 2
1
1 − 4 x 2 ) ( −8 x dx)
(
∫
8
1
y = −
1 − 4 x2 + C
4
∫ 0.75 s ds
dy
∫y =
=
1 − 4x2
x
= −
dr
= 0.75 s
ds
∫ dr
x
dy =
r = Ce0.75 s
6.
565
Particular solution:
9
5
= 2 +C ⇒ C =
2
2
y2
5
= 2e x +
2
2
y 2 = 4e x + 5
π
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
566
16.
Chapter 6
Differential Equations
y y′ = 0
x +
∫y
12
dy = − ∫ x1 2 dx
2
∫ (1 − y )
2 32
2
y = − x 3 2 + C1
3
3
y
32
+ x
32
( 9)
+ (1)
32
(
= C
21.
17. y ( x + 1) + y′ = 0
dy
= − ∫ ( x + 1) dx
y
(x
+ 1)
+ C1
u = Ce
⎡1 − ( x + 1)2 ⎤ 2
⎥⎦
Particular solution: y = e ⎢⎣
= e
(
)
0 = −1 + C ⇒ C = 1
1 − y2 =
1 − x2 − 1
(
− cos v 2
)2
1
= e1 2
e −1 2
(1− cos v2 ) 2
−r
dr =
∫e
−2 s
ds
Initial condition:
r (0) = 0: −1 = −
+ C
1
2
(ln x) + 2
2
y(1 + x 2 ) y′ = x(1 + y 2 )
⎛ 1 + e −2 s ⎞
1⎞
⎛1
− r = ln ⎜ e −2 s + ⎟ = ln ⎜
⎟
2⎠
2 ⎠
⎝2
⎝
y
x
dy =
dx
1 + y2
1 + x2
2 ⎞
⎛
r = ln ⎜
−2 s ⎟
⎝1 + e ⎠
1
1
ln (1 + y 2 ) = ln (1 + x 2 ) + C1
2
2
ln (1 + y 2 ) = ln (1 + x 2 ) + ln C = ln ⎡⎣C (1 + x 2 )⎤⎦
1 + y 2 = C (1 + x 2 )
)
3: 1+ 3 = C ⇒ C = 4
(
Particular solution: 1 + y = 4 1 + x
2
y = 3 + 4x
2
1
1
+C ⇒ C = −
2
2
Particular solution:
1
1
−e − r = − e −2 s −
2
2
1 −2 s
1
−r
e = e
+
2
2
Initial condition (1, 2): 2 = C
Particular solution: y =
+ C
1
− e − r = − e −2 s + C
2
2
2
x dx
dr
= er − 2s
ds
∫e
dy
= 2 ln x
2x
dx
ln x
∫ dy = ∫ x dx
(ln x)
)
12
− x2 + 2 x 2
18. 2 xy′ − ln x 2 = 0
(
= − 1 − x2
Particular solution: u = e
22.
Initial condition 0,
(
12
−1 2
Initial condition: u (0) = 1: C =
2 2
Initial condition ( −2, 1): 1 = Ce −1 2 , C = e1 2
19.
)
2
∫ (1 − x )
du
= uv sin v 2
dv
du
2
∫ u = ∫ v sin v dv
1
ln u = − cos v 2 + C1
2
2
2
y = Ce −( x + 1)
y =
y dy =
Particular solution:
= 27 + 1 = 28 = C
ln y = −
dy
= x 1 − y2
dx
Initial condition (0, 1):
Particular solution: y 3 2 + x3 2 = 28
∫
−1 2
− 1 − y2
Initial condition (1, 9):
32
y 1 − x2
20.
2
2
)
23. dP − kP dt = 0
dP
= k ∫ dt
P
ln P = kt + C1
∫
P = Ce kt
Initial condition: P(0) = P0 , P0 = Ce0 = C
Particular solution: P = P0e kt
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
24. dT + k (T − 70) dt = 0
Differential Equations: Separation of Variables
28.
dT
= − k ∫ dt
− 70
ln (T − 70) = − kt + C1
∫T
∫
y 3 = Cx 2
Initial condition:
T (0) = 140: 140 − 70 = 70 = Ce 0 = C
∫ 4 y dy
=
2 y2 =
Particular solution: 8 y 3 = x 2 , y =
dy
x
=
dx
4y
∫ x dx
x
+C
2
Initial condition (0, 2): 2( 2
)
= 0+ C ⇒ C = 8
x2
+8
2
4 y 2 − x 2 = 16
∫ 16 y dy
1
8
1 23
x
2
dy
0− y
y
=
= −
dx
2
( x + 2) − x
1
∫ − 2 dx
1
ln y = − x + C1
2
y = Ce − x 2
2 y2 =
dy
y −0
y
=
=
dx
x −0
x
dy
dx
= ∫
y
x
30. m =
∫
dy
−9 x
=
dx
16 y
26.
dy
=
y
∫
2
Particular solution:
m =
29.
2
23 = C (82 ), C =
Initial condition (8, 2):
Particular solution:
T − 70 = 70e − kt , T = 70(1 + e − kt )
y′ =
dy
2y
=
dx
3x
3
2
dy = ∫ dx
y
x
ln y 3 = ln x 2 + ln C
T − 70 = Ce − kt
25.
567
ln y = ln x + C1 = ln x + ln C = ln Cx
= − ∫ 9 x dx
y = Cx
−9 2
8y =
x +C
2
2
Initial condition (1, 1):
Particular solution:
31.
8 = −
9
25
+ C, C =
2
2
dy
= x
dx
y
−9 2
25
x +
2
2
16 y 2 + 9 x 2 = 25
2
8 y2 =
x
−2
dy
y
=
dx
2x
2
1
dy = ∫ dx
y
x
2
y′ =
27.
∫
−2
y =
∫ x dx
=
2 ln y = ln x + C1 = ln x + ln C
y 2 = Cx
32.
Initial condition (9, 1): 1 = 9C ⇒ C =
Particular solution:
1 2
x + C
2
dy
x
= −
dx
y
1
9
y
∫ y dy
4
1
y = x
9
2
2
9 y2 − x = 0
y =
1
3
−4
x
−2
2
−2
x
4
=
∫−
x dx
y2
− x2
=
+ C1
2
2
y2 + x2 = C
−4
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568
33.
Chapter 6
Differential Equations
36. (a) Euler’s Method gives y ≈ 0.2622 when x = 1.
dy
= 4− y
dx
(b)
y
8
∫
dy
= −6 xy 2
dx
dy
= ∫ − 6 x dx
y2
−
1
3x 2 + C
1
1
⇒ C =
3 =
C
3
1
3
y =
=
2
1
x
+1
9
2
3x +
3
y =
x
−4 −3
1
dy
∫4−
=
y
2
3
4
∫ dx
ln 4 − y = − x + C1
4 − y = e − x + C1
y = 4 + Ce − x
(c) At x = 1, y =
dy
= 0.25 x( 4 − y )
dx
dy
= 0.25 x dx
4− y
34.
∫
dy
=
y − 4
37. (a) Euler’s Method gives y ≈ 3.0318 when x = 2.
(b)
y 3 − 4 y = x 2 + 12 x − 13
2
(c) At x = 2,
y 3 − 4 y = 22 + 12( 2) − 13 = 15
y 3 − 4 y − 15 = 0
4
(y
2
4
38. (a) Euler’s Method gives y ≈ 1.7270 when x = 1.5.
35. (a) Euler’s Method gives y ≈ 0.1602 when x = 1.
dy
= −6 xy
dx
dy
∫ y = ∫ − 6x
ln y = −3 x 2 + C1
y = Ce−3 x
2
y (0) = 5 ⇒ C = 5
y = 5e −3 x
− 3)( y 2 + 3 y + 5) = 0 ⇒ y = 3.
Error: 3.0318 − 3 = 0.0318
x
−2
(b)
∫ (2 x + 12) dx
y(1) = 2: 23 − 4( 2) = 1 + 12 + C ⇒ C = −13
2
y
−2
− 4) dy =
y − 4 y = x 2 + 12 x + C
8
−4
2
3
= Ce −(1 8)x
y = 4 + Ce −(1 8)x
dy
2 x + 12
=
dx
3y2 − 4
∫ (3 y
1
ln y − 4 = − x 2 + C1
8
2
3
3
=
= 0.3.
9(1) + 1 10
Error: 0.3 − 0.2622 = 0.0378
1
∫ − 0.25 x dx = − 4 ∫ x dx
y − 4 = eC1 − (1 8)x
1
= −3 x 2 + C1
y
2
(c) At x = 1, y = 5e −3(1) ≈ 0.2489.
(b)
dy
= 2 x(1 + y 2 )
dx
dy
∫ 1 + y 2 = ∫ 2 x dx
arctan y = x 2 + C
arctan (0) = 12 + C ⇒ C = −1
arctan ( y ) = x 2 − 1
y = tan ( x 2 − 1)
(c) At x = 1.5, y = tan (1.52 − 1) ≈ 3.0096.
Error: 1.7270 − 3.0096 = −1.2826
Error: 0.2489 − 0.1602 ≈ 0.0887
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
39.
dy
= ky ,
dt
y = Ce kt
Differential Equations: Separation of Variables
Half-life:
dw
∫ 1200 − w
y0
= y0e k (1599)
2
1
⎛1⎞
ln ⎜ ⎟
k =
1599 ⎝ 2 ⎠
=
∫ k dt
ln 1200 − w = − kt + C1
1200 − w = e − kt + C1 = Ce − kt
w = 1200 − Ce − kt
w(0) = 60 = 1200 − C ⇒ C = 1200 − 60 = 1140
⎡ln 1 2 1599⎤⎦t
y = Ce ⎣ ( )
w = 1200 − 1140e − kt
When t = 50, y = 0.9786C or 97.86%.
(b)
40.
dw
= k (1200 − w)
dt
45. (a)
Initial amount: y(0) = y0 = C
569
1400
dy
= ky , y = Ce kt
dt
Initial conditions: y (0) = 40, y (1) = 35
0
35 = 40e k
k = ln
7
8
10
0
40 = Ce0 = C
k = 0.8
1400
Particular solution: y = 40et ln(7 8)
When 75% has been changed:
10 = 40e
t ln(7 8)
1
= et ln(7 8)
4
ln (1 4)
t =
≈ 10.38 hours
ln (7 8)
10
0
0
k = 0.9
1400
10
0
41. (a)
dy
= k ( y − 4)
dx
(b) The direction field satisfies ( dy dx) = 0 along
y = 4; but not along y = 0. Matches (a).
dy
42. (a)
= k ( x − 4)
dx
(b) The direction field satisfies ( dy dx) = 0 along
x = 4. Matches (b).
43. (a)
dy
= ky ( y − 4)
dx
(b) The direction field satisfies ( dy dx) = 0 along
y = 0 and y = 4. Matches (c).
44. (a)
0
k =1
(c) k = 0.8:
t = 1.31 years
k = 0.9:
t = 1.16 years
k = 1.0:
t = 1.05 years
(d) Maximum weight: 1200 pounds
lim w = 1200
x→∞
46. From Exercise 39:
w = 1200 − Ce − kt , k = 1
w = 1200 − Ce − t
w(0) = w0 = 1200 − C ⇒ C = 1200 − w0
w = 1200 − (1200 − w0 )e − t
dy
= ky 2
dx
(b) The direction field satisfies ( dy dx) = 0 along
y = 0, and grows more positive as y increases.
Matches (d).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
570
Chapter 6
Differential Equations
x2 + y 2 = C
47. Given family (circles):
50. Given family (parabolas):
2 x + 2 yy′ = 0
x
y
y′ = −
y′ =
Orthogonal trajectory (lines):
dy
∫y =
y′ =
y
x
∫
y 2 = 2Cx
2 yy′ = 2C
4
ln y = ln x + ln K
∫ y dy
−6
6
y = Kx
−4
51. Given family:
6
y
= − x 2 + K1
2
2 x2 + y 2 = K
y 2 = Cx3
3Cx 2
3x 2 ⎛ y 2 ⎞
3y
=
⎜ ⎟ =
2y
2 y ⎝ x3 ⎠
2x
y′ =
48. Given family (hyperbolas): x 2 − 2 y 2 = C
y′ = −
Orthogonal trajectory (ellipses):
2 x − 4 yy′ = 0
y′ =
3y2
= − x 2 + K1
2
3 y 2 + 2x2 = K
−2 y
x
dy
2
= − ∫ dx
y
x
4
−6
ln y = −2 ln x + ln k
y = kx −2 =
2x
3y
3∫ y dy = −2 ∫ x dx
x
2y
y′ =
∫
= − ∫ 2 x dx
2 yy′ = 3Cx 2
−4
Orthogonal trajectory:
2x
y
2
4
−6
y′ = −
Orthogonal trajectory (ellipse):
dx
x
C
y2 ⎛ 1 ⎞
y
=
⎜ ⎟ =
y
2x ⎝ y ⎠
2x
k
x2
6
−4
2
52. Given family (exponential functions): y = Ce x
−3
y′ = Ce x = y
3
Orthogonal trajectory (parabolas):
−2
4
∫ y dy
49. Given family (parabolas): x 2 = Cy
−6
2 x = Cy′
y′ =
Orthogonal trajectory (ellipses):
4
−6
y′ = −
6
−4
1
y
= − ∫ dx
y2
= − x + K1
2
y 2 = −2 x + K
x
2y
2∫ y dy = − ∫ x dx
6
−4
2x
2x
2y
= 2
=
C
x y
x
y′ = −
y2 = −
x2
+ K1
2
x2 + 2 y 2 = K
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
Differential Equations: Separation of Variables
571
dN
= kN (500 − N )
dt
53.
dN
∫ N (500 − N )
=
∫ k dt
1 ⎡1
1
⎤
dN =
+
500 ∫ ⎢⎣ N
500 − N ⎥⎦
∫ k dt
ln N − ln 500 − N = 500( kt + C1 )
N
= e500 kt + C2 = Ce500 kt
500 − N
N =
500Ce500 kt
1 + Ce500 kt
When t = 0, N = 100. So, 100 =
125e500 kt
500C
.
⇒ C = 0.25. Therefore, N =
1 + 0.25e500 kt
1+C
ln (8 3)
125e 2000 k
⇒ k =
≈ 0.00049.
2000 k
1 + 0.25e
2000
500
.
=
1 + 4e −0.2452t
When t = 4, N = 200. So, 200 =
Therefore, N =
125e0.2452t
1 + 0.25e0.2452t
54. The differential equation is given by the following.
dS
= kS ( L − S )
dt
dS
∫ S(L − S )
=
∫ k dt
1
⎡ln S − ln L − S ⎤⎦ = kt + C1
L⎣
S
= Ce Lkt
L − S
S =
CLe Lkt
CL
=
Lkt
C + e − Lkt
1 + Ce
When t = 0, S = 10. So, C =
Therefore, S =
10
.
L − 10
⎡⎣10 ( L − 10)⎤⎦ L
CL
10 L
=
=
.
⎡⎣10 ( L − 10)⎤⎦ + e − Lkt
C + e − Lkt
10 + ( L − 10)e − Lkt
55. The general solution is y = 1 − Ce − kt . Because
y = 0 when t = 0, it follows that C = 1.
Because y = 0.75 when t = 1, you have
0.75 = 1 − e − k (1)
56. The general solution is y = 1 − Ce − kt . Because
y = 0 when t = 0, it follows that C = 1.
Because y = 0.9 when t = 2, you have
0.9 = 1 − e − 2 k
− 0.25 = − e − k
− 0.1 = − e − 2k
0.25 = e − k
0.1 = e − 2 k
ln 0.25 = − k
ln 0.1 = − 2k
k = ln 0.25 = ln 4 ≈ 1.386.
So, y ≈ 1 − e
−1.386t
.
Note: This can be written as y = 1 − 4 − x.
k = −
1
1
ln 0.1 = ln 10 ≈ 1.151.
2
2
So, y ≈ 1 − e −1.151t .
Note: This can be written as y = 1 − 10 − x 2.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
572
Chapter 6
Differential Equations
57. The general solution is y = −
1
.
kt + C
60. The general solution is y = 5000e − Ce
. Because
y = 500 when t = 0, it follows that 500 = 5000e − C
Because y = 45 when t = 0, it follows that
1
1
45 = − and C = − .
C
45
Therefore, y = −
− kt
1 = ln 10. So, you have
which implies that C = −ln 10
y = 5000e(−ln 10)e
it follows that
1
45
=
.
kt − (1 45)
1 − 45kt
− kt
625 = 5000e(−ln 10)e
Because y = 4 when t = 2, you have
45
41
4 =
⇒ k = −
.
1 − 45k ( 2)
360
. Because y = 625 when t = 1,
−k
ln (1 8)
ln (1 10)
e− k =
⎛ ln (1 8) ⎞
≈ 0.1019.
k = −ln ⎜
⎜ ln (1 10) ⎟⎟
⎝
⎠
45
360
So, y =
=
.
1 + ( 41 8)t
8 + 41t
(−0.1019)t
So, you have y = 5000e(−2.3026)e
45
.
y
5000
4000
0
3000
3
0
2000
58. The general solution is y = −1 ( kt + C ).
1000
Because y = 75 when t = 0, you have C = −1 75.
So, y = −
1
75
=
.
kt − (1 75)
1 − 75kt
15 = 60e −2.0149e
k = −
59. Because y = 100 when t = 0, it follows that
100 = 500e − C , which implies that C = ln 5. So,
you have y = 500e
t = 2, it follows that
150 = 500e(− ln 5)e
25
t
− kt
−3 k
−3 k
1
= e −2.0149e
4
1
ln = −2.0149e −3k
4
4
(−ln 5)e− kt
20
Because y = 15 when t = 3,
75
300
.
=
1 + 5.25t
4 + 21t
0
15
Because y = 8 when t = 0,
15
≈ 2.0149.
8 = 60e − C ⇒ C = ln
2
80
0
10
61. From Example 8, the general solution is y = 60e − Ce
Because y = 12 when t = 1, you have
75
7
12 =
.
⇒ k = −
1 − 75k
100
So, you have y =
5
1 ⎛ ln (1 4) ⎞
ln ⎜
⎟ ≈ 0.1246.
3 ⎝ −2.0149 ⎠
So, y = 60e −2.0149e
. Because y = 150 when
−0.1246t
.
When t = 10, y ≈ 34 beavers.
−2 k
ln 0.3
ln 0.2
1
ln 0.3
k = − ln
≈ 0.1452.
2
ln 0.2
e −2 k =
y
So, y is given by
y = 500e
−1.6904 e−0.1451t
.
300
15
t
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
.
Section 6.3
62. From Example 8, the general solution is y = 400e − Ce
− kt
Differential Equations: Separation of Variables
.
Because y = 30 when t = 0,
⎛ 40 ⎞
30 = 400e −C ⇒ C = ln ⎜ ⎟ ≈ 2.5903.
⎝ 3⎠
64. Following Example 9, the differential equation is
dy
= ky(1 − y )( 2 − y )
dt
and its general solution is
Because y = 90 when t = 1,
90 = 400e −2.5903e
−k
−k
9
= e −2.5903e
40
⎛ 9⎞
ln ⎜ ⎟ = −2.5903e − k
⎝ 40 ⎠
−0.5519t
(0.4)(1.6)
2
(0.6)
=
y = 0.8 when t = 5 ⇒
(0.8)(1.2)
(0.2)2
= 24 =
.
y(2 − y )
So, the particular solution is
= Ce .
2 kt
(1 − y )2
(1 2)(3 2) = C ⇒ C = 3
1
when t = 0 ⇒
2
2
(1 2)
3
when
4
(3 4)(5 4)
(1 4)
2
2
(1 − y )
2
=
16 0.5205t
.
e
9
Using a symbolic algebra utility or graphing utility, you
find that when t = 8, y ≈ 0.91, or 91%.
65. (a)
dQ
Q
= −
dt
20
dQ
1
∫ Q = ∫ − 20 dt
ln Q = −
1
t + C1
20
So, the particular solution is Q = 25e −(1 20)t .
1
ln 5 ≈ 0.2012.
8
y(2 − y )
(1 − y)
2
= 3e0.4024t .
Using a symbolic algebra utility or graphing utility, you
find that when t = 10,
(1 − y )
y(2 − y )
(b) When Q = 15, you have 15 = 25e −(1 20)t .
So, the particular solution is
y(2 − y )
16 2 k (5)
e
9
Because Q = 25 when t = 0, you have 25 = C .
= 15 = 3e 2 k (4)
⇒ 5 = e8 k
⇒ k =
16
= C
9
Q = e −(1 20)t + C1 = Ce −(1 20)t
y = 0.75 =
t = 4 ⇒
= Ce 2 kt .
27
= e10 k
2
1 ⎛ 27 ⎞
ln ⎜ ⎟ ≈ 0.2603
⇒ k =
10 ⎝ 2 ⎠
63. Following Example 9, the differential equation is
dy
= ky (1 − y )( 2 − y )
dt
y =
(1 − y )2
⇒
Finally, when t = 3, y ≈ 244 rabbits.
and its general solution is
y(2 − y )
y = 0.4 when t = 0 ⇒
⎛ ln (9 40) ⎞
k = − ln ⎜
⎟ ≈ 0.5519.
⎝ −2.5903 ⎠
So, y = 400e −2.5903e
573
3
= e −(1 20)t
5
1
⎛ 3⎞
ln ⎜ ⎟ = − t
20
⎝5⎠
⎛ 3⎞
−20 ln ⎜ ⎟ = t
⎝5⎠
t ≈ 10.217 minutes
= 3e0.4024(10)
and y ≈ 0.92, or 92%.
66. Because Q′ +
1Q
20
=
5
2
is a first-order linear differential equation with P( x) =
1
20
and R( x) =
5,
2
(1 20)dt
you have the integrating factor u (t ) = e ∫
= e(1 20)t , and the general solution is
5
Q = e −0.05t ∫ e0.05t dt = e −0.05t (50e0.05t + C ) = 50 + Ce −0.05t .
2
Because Q = 0 when t = 0, you have C = −50 and Q = 50(1 − e −0.05t ). Finally, when t = 30, you have
Q ≈ 38.843 lb gal.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
574
Chapter 6
Differential Equations
69. The general solution is y = Ce kt . Because
y = 0.60C when t = 1, you have
dy
= ky
dt
dy
∫ y = ∫ k dt
67. (a)
0.60C = Ce k ⇒ k = ln 0.60 ≈ −0.5108.
So, y = Ce −0.5108t . When y = 0.20C , you have
ln y = kt + C1
0.20C = Ce −0.5108t
y = e kt + C1 = Ce kt
ln 0.20 = −0.5108t
(b) y (0) = 20 ⇒ C = 20
16
⎛ 4⎞
y (1) = 16 = 20e ⇒ k = ln
= ln ⎜ ⎟
20
⎝5⎠
t ≈ 3.15 hours.
k
70.
y = 20et ln(4 5)
⎛ 1 dy ⎞
∫ ⎜⎝ y dt ⎟⎠ dt
1
∫ y dy
When 75% has changed:
5 = 20et ln(4 5)
1
= et ln(4 5)
4
ln (1 4)
t =
≈ 6.2 hours
ln ( 4 5)
68.
=
71.
72.
N
dP =
∫ dt
N
k
dy
= − 0.2 y
dx
y = Ce −0.2 x
y(0) = 29.92 ⇒ C = 29.92 ⇒ y = 29.92e −0.2 x
(a) 8364 feet ≈ 1.5841 miles
y (1.5841) ≈ 21.80 inches
and
(b) 23,320 feet ≈ 3.8485 miles
25
−13
k =
=
≈ −11.8331.
ln ( 2C )
ln 3
Therefore, s is given by the following.
= −
1
P = Ce kt −
1 −(25 13)ln 3
≈ 0.0605
e
2
= −
1
∫ kP +
k
12 = k ln (6C ), which implies
s = −
∫ x dx
1
ln kP + N = t + C1
k
kP + N = C2e kt
s = k ln h + C1 = k ln Ch
C =
=
y = Cx
∫ h dh
Because s = 25 when h = 2 and s = 12 when
h = 6, it follows that 25 = k ln ( 2C ) and
∫ ⎜⎝ x dt ⎟⎠ dt
ln y = ln x + C1 = ln Cx
ds
k
=
dh
h
∫ ds
⎛ 1 dx ⎞
=
y (3.8485) ≈ 13.86 inches
73.
13 ⎡ h −(25 13) ln 3 ⎤
ln e
⎥
ln 3 ⎢⎣ 2
⎦
13 ⎡ h 25
⎤
ln −
ln 3⎥
ln 3 ⎢⎣ 2 13
⎦
h
1 ⎡
⎤
13 ln − 25 ln 3⎥
⎢
ln 3 ⎣
2
⎦
= 25 −
13ln ( h 2)
, 2 ≤ h ≤ 15
ln 3
∫
dA
= rA + P
dt
dA
= dt
rA + P
dA
= ∫ dt
rA + P
1
ln ( rA + P ) = t + C1
r
ln ( rA + P ) = rt + C2
rA + P = e rt + C2
A =
A =
When t = 0: A
0
A
C3e rt − P
r
P
rt
Ce −
r
= 0
P
P
= C −
⇒ C =
r
r
P rt
= (e − 1)
r
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
P rt
(e − 1)
r
275,000 0.08(10)
A =
e
− 1 ≈ $4,212,796.94
0.06
74. A =
(
)
Differential Equations: Separation of Variables
78.
575
⎛ 1000 ⎞
dy
= 0.05 y ln ⎜
⎟
dt
⎝ y ⎠
(a)
1200
75. From Exercise 73,
P rt
(e − 1).
r
A =
0
Because A = 260,000,000 when t = 8 and
r = 0.0725, you have
Ar
P = rt
e −1
(260,000,000)(0.0725)
=
e(0.0725)(8) − 1
≈ $23,981,015.77.
76. 1,000,000 =
100
0
(b) As t → ∞, y → L = 1000.
(c) Using a computer algebra system or separation of
variables, the general solution is
y = 1000e − Ce
− kt
= 1000e − Ce
−0.05t
.
Using the initial condition y(0) = 100, you obtain
100 = 1000e − C ⇒ C = ln 10 ≈ 2.3026.
So, y = 1000e −2.3026e
125,000 0.08t
(e − 1)
0.08
(d)
−0.05t
.
1200
1.64 = e0.08t
t =
ln (1.64)
0.08
≈ 6.18 years
0
77.
⎛ 5000 ⎞
dy
= 0.02 y ln ⎜
⎟
dt
⎝ y ⎠
(a)
The graph is concave upward on (0, 16.7) and
concave downward on (16.7, ∞).
5000
0
300
0
(b) As t → ∞, y → L = 5000.
(c) Using a computer algebra system or separation of
variables, the general solution is
y = 5000e
− Ce− kt
= 5000e
− Ce−0.02 t
.
Using the initial condition y (0) = 500, you obtain
500 = 5000e − C ⇒ C = ln 10 ≈ 2.3026.
So, y = 5000e −2.3026 e
(d)
100
0
−0.02 t
79. A differential equation can be solved by separation of
variables if it can be written in the form
dy
M ( x) + N ( y )
= 0.
dx
To solve a separable equation, rewrite as,
M ( x) dx = − N ( y ) dy
and integrate both sides.
80. Two families of curves are mutually orthogonal if each
curve in the first family intersects each curve in the
second family at right angles.
81. y (1 + x ) dx + x dy = 0
x dy = − y (1 + x) dx
.
(1 + x) dx
1
dy = −
y
x
5000
Separable
0
300
0
The graph is concave upward on (0, 41.7) and
concave downward on ( 41.7, ∞).
82. y′ =
dy
= y1 2
dx
dy
= dx
y1 2
Separable
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
576
83.
Chapter 6
Differential Equations
dy
+ xy = 5
dx
88.
f (tx, ty ) = t 3 x3 + 3t 4 x 2 y 2 − 2t 2 y 2
Not separable
84.
Not homogeneous
dy
= x − xy − y + 1
dx
= x(1 − y ) + (1 − y )
89.
dy
= ( x + 1) dx
1− y
x2 + y 2
t 4 x2 y2
t 2 x2 + t 2 y2
x2 y2
= t3
x2 + y2
Homogeneous of degree 3
Separable
90.
dv
85. (a)
= k (W − v)
dt
dv
∫ W − v = ∫ k dt
−ln W − v = kt + C1
v = W − Ce
x + y2
2
tx ty
t 2 x2 + t 2 y2
t 2 xy
=
t
− kt
W = 20, v = 0 when t = 0 and v = 10
xy
f ( x, y ) =
f (tx, ty ) =
x + y
2
2
= t
xy
x + y2
2
Homogeneous of degree 1
Initial conditions:
91.
f ( x, y ) = 2 ln xy
f (tx, ty ) = 2 ln[txty]
when t = 0.5 so, C = 20, k = ln 4.
= 2 ln ⎡⎣t 2 xy⎤⎦ = 2(ln t 2 + ln xy )
Particular solution:
t
⎛
⎛1⎞ ⎞
v = 20 1 − e −(ln 4)t = 20⎜1 − ⎜ ⎟ ⎟
⎜
⎝ 4 ⎠ ⎟⎠
⎝
or
)
Not homogeneous
92.
Not homogeneous
(b) s = ∫ 20(1 − e −1.386t ) dt ≈ 20(t + 0.7215e −1.386t ) + C
Because s(0) = 0, C ≈ −14.43 and you have
93.
s ≈ 20t + 14.43(e −1.386t − 1).
For graph (d), the y-intercept is (0, 1), so C = 0.5.
x
y
tx
x
= 2 ln
ty
y
Homogeneous of degree 0
For graph (a), the y-intercept is (0, 6), so C = 3.
For graph (c), the y-intercept is (0, 2), so C = 1.
f ( x, y ) = 2 ln
f (tx, ty ) = 2 ln
86. Use the y-intercepts to match the graphs with the
appropriate value of C.
For graph (b), the y-intercept is (0, 4), so C = 2.
f ( x, y ) = tan ( x + y )
f (tx, ty ) = tan (tx + ty ) = tan ⎡⎣t ( x + y )⎤⎦
v = 20(1 − e −1.386t )
87.
x2 y 2
f ( x, y ) =
f (tx, ty ) =
= ( x + 1)(1 − y )
(
f ( x, y ) = x 3 + 3 x 2 y 2 − 2 y 2
94.
y
x
ty
y
f (tx, ty ) = tan
= tan
tx
x
f ( x, y ) = tan
Homogeneous of degree 0
f ( x, y ) = x3 − 4 xy 2 + y 3
f (tx, ty ) = t 3 x3 − 4txt 2 y 2 + t 3 y 3
= t 3 ( x 3 − 4 xy 2 + y 3 )
Homogeneous of degree 3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.3
95.
(x
+ y )dx − 2 x dy = 0, y = ux, dy = x du + u dx
(x
+ ux)dx − 2 x( x du + u dx) = 0
(1 + u )dx
Differential Equations: Separation of Variables
96.
(1 + u3 )dx − u 2 ( x du + u dx) = 0
= 2 x du
dx = xu 2 du
1
2
dx =
du
1−u
x
1
1
∫ x dx = 2 ∫ 1 − u du
ln x + ln C = − 2 ln 1 − u
ln Cx = ln 1 − u
Cx =
=
Cx =
= 0, y = ux, dy = x du + u dx
⎡ x3 + (ux)3 ⎤ dx − x(ux)2 ( x du + u dx) = 0
⎣
⎦
− 2 x du − 2u dx = 0
(1 − u )dx
( x3 + y3 )dx − xy 2 dy
577
∫
dx
=
x
ln x + C1 =
1⎛ y ⎞
u3
= ⎜ ⎟
3
3⎝ x ⎠
3
y 3 = 3 x 3 ln x + Cx 3
97. ( x − y)dx − ( x + y ) dy = 0, y = ux, dy = x du + u dx
1
⎣⎡1 − ( y x)⎤⎦
2
( x − ux)dx − ( x + ux)( x du + u dx)
(1 − u )dx − (1 + u )( x du + u dx)
x2
2
x = C ( x − y)
du
⎛ y⎞
⎜ ⎟ = 3 ln x + C
⎝ x⎠
(1 − u )2
− y)
2
3
−2
1
(x
∫u
=0
=0
(1 − 2u − u )dx = x(1 + u)du
2
2
1+ u
dx
du
= 2
x
u + 2u − 1
dx
u +1
du
−∫
= ∫ 2
x
u + 2u − 1
1
− ln x + ln C = ln u 2 + 2u − 1
2
12
C
ln
= ln u 2 + 2u − 1
x
−
C2
= u 2 + 2u − 1
x2
2
C
⎛ y⎞
⎛ y⎞
= ⎜ ⎟ + 2⎜ ⎟ − 1
x2
⎝ x⎠
⎝ x⎠
C = y 2 + 2 yx − x 2
98. ( x 2 + y 2 )dx − 2 x dy = 0, y = ux, dy = x du + u dx
(x
2
)
+ (ux) dx − 2 x(ux)( x du + u dx) = 0
2
(1 + u 2 )dx − 2u( x du + u dx)
(1 − u 2 )dx
= 0
= 2ux du
− 2u
dx
=
du
x
1 − u2
− 2u du
dx
−∫
= ∫
x
1 − u2
− ln x + ln C = ln 1 − u 2 = ln ⎡⎣u 2 − 1⎤⎦ = ln u 2 − 1
C
= ln u 2 − 1
ln
x
−
2
C
⎛ y⎞
= u2 − 1 = ⎜ ⎟ − 1
x
⎝ x⎠
2
2
Cx = y − x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
578
Chapter 6
Differential Equations
99. xydx + ( y 2 − x 2 )dy = 0, y = ux, dy = x du + u dx
2
x(ux) dx + ⎡(ux) − x 2 ⎤ ( x du + u dx) = 0
⎣
⎦
u dx + (u 2 − 1)( x du + u dx) = 0
u 3 dx = − (u 2 − 1) x du
dx
1 − u2
=
du
x
u3
∫
dx
=
x
⎛
∫ ⎜⎝ u
−3
−
1⎞
⎟ du
u⎠
1
− ln u
2u 2
1
ln C1 xu = − 2
2u
ln x + ln C1 = −
ln C1 y = −
1
2( y x)
y = Ce
− x2
2
= −
x2
2 y2
(2 y 2 )
100. ( 2 x + 3 y )dx − x dy = 0, y = ux, dy = x du + u dx
(2 x
+ 3ux )dx − x( x du + u dx) = 0
(2
+ 3u )dx − x du − u dx = 0
(2
+ 2u )dx = x du
du
2dx
=
x
1+ u
1
1
du
2 ∫ dx = ∫
x
u +1
2ln x + ln C = ln u + 1
ln x 2 C = ln u + 1
1 + u = x2 C
1+
y
= x2 C
x
y
= Cx 2 − 1
x
y = Cx3 − x
101. False.
dy
x
=
is separable, but y = 0 is not a solution.
dx
y
102. True
dy
= ( x − 2)( y + 1)
dx
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.4
103. True
fg ′ + gf ′ = f ′g ′
104.
x 2 + y 2 = 2Cy
x 2 + y 2 = 2 Kx
dy
K − x
=
dx
y
dy
x
=
dx
C − y
The Logistic Equation
(f
x
K − x
Kx − x
⋅
=
C − y
y
Cy − y 2
=
2 Kx − 2 x 2
2Cy − 2 y 2
=
x2 + y2 − 2x2
x2 + y 2 − 2 y 2
=
y − x
x2 − y 2
2
2
= −1
Product Rule
− f ′) g ′ + gf ′ = 0
g′ +
f′
g = 0
f − f′
2
2
579
Need f − f ′ = e x − 2 xe x
2
= (1 − 2 x)e x
2
≠ 0, so
1
.
2
avoid x =
2
g′
f′
2 xe x
1
=
=
=1+
x2
−1
g
f′ − f
x
2
(2 x − 1)e
ln g ( x) = x +
1
ln 2 x − 1 + C1
2
g ( x ) = Ce x 2 x − 1
12
So there exists g and interval ( a, b), as long as
1
∉ ( a, b).
2
Section 6.4 The Logistic Equation
1. y =
12
1 + e− x
5. y =
Because y(0) = 6, it matches (c) or (d).
Because (d) approaches its horizontal asymptote slower
than (c), it matches (d).
2. y =
=
8
2e −2t
⋅
−2 t
(1 + e ) (1 + e−2t )
⎛ e −2t ⎞
= 2 y⎜
−2 t ⎟
⎝1 + e ⎠
12
= 3, it matches (a).
4
⎛
⎞
8
⎟
= 2 y ⎜1 −
t
−
2
⎜
8(1 + e ) ⎟⎠
⎝
12
1
1 + e− x
2
Because y (0) =
4. y =
−2
dy
= −8(1 + e −2t ) ( −2e−2t )
dt
12
1 + 3e − x
Because y (0) =
3. y =
−1
8
= 8(1 + e −2t ) ; L = 8, k = 2, b = 1
−2 t
1+ e
y⎞
⎛
= 2 y ⎜1 − ⎟
8⎠
⎝
12
= 8, it matches (b).
⎛ 3⎞
⎜ ⎟
⎝ 2⎠
y ( 0) =
8
= 4
1 + e0
12
1 + e −2 x
Because y(0) = 6, it matches (c) or (d).
Because y approaches L = 12 faster for (c), it matches
(c).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
580
Chapter 6
Differential Equations
−1
10
= 10(1 + 3e − 4t ) ;
− 4t
1 + 3e
L = 10, k = 4, b = 3
6. y =
−2
dy
= −10(1 + 3e − 4t ) ( −12e − 4t )
dt
10
12e− 4t
=
⋅
− 4t
1 + 3e
(1 + 3e− 4t )
⎛ 3e − 4t ⎞
= 4y ⋅ ⎜
− 4t ⎟
⎝ 1 + 3e ⎠
9. P(t ) =
(a) k = 0.75
(b) L = 2100
1 + 29e −0.75t = 2
10
10
5
=
=
1 + 3e0
4
2
7. y = 12(1 + 6e
)
; L = 12, k = 1, b = 6
y′ = −12(1 + 6e − t )
−2
⎛1⎞
−0.75t = ln ⎜ ⎟ = −ln 29
⎝ 29 ⎠
ln 29
≈ 4.4897 years
t =
0.75
( − 6e − t )
(e)
dP
P ⎞
⎛
= 0.75 P⎜1 −
⎟
dt
2100 ⎠
⎝
10. P(t ) =
5000
1 + 39e −0.2t
(a) k = 0.2
(b) L = 5000
⎛ 12 ⎞⎛ 6e −t ⎞
= ⎜
− t ⎟⎜
−t ⎟
⎝ 1 + 6e ⎠⎝ 1 + 6e ⎠
(c) P(0) =
1
⎛
⎞
= y ⎜1 −
⎟
1 + 6e − t ⎠
⎝
(d)
2500 =
e −0.2t =
y⎞
⎛
= y ⎜1 − ⎟
12
⎝
⎠
5000
1 + 39e −0.2t
1
39
⎛1⎞
−0.2t = ln ⎜ ⎟ = −ln 39
⎝ 39 ⎠
ln 39
t =
≈ 18.3178 years
0.2
12
12
=
1+ 6
7
8. y = 14(1 + 5e −3t ) ; L = 14, k = 3, b = 5
−1
y′ = −14(1 + 5e −3t )
5000
= 125
1 + 39
1 + 39e −0.2t = 2
⎛
⎞
12
⎟
= y ⎜1 −
−t ⎟
⎜
12(1 + 6e ) ⎠
⎝
y ( 0) =
1
29
e −0.75t =
y⎞
⎛
= 4 y ⎜1 − ⎟
10
⎝
⎠
−1
2100
1 + 29e −0.75t
1050 =
(d)
⎛
⎞
10
⎟
= 4 y ⎜1 −
− 4t ⎟
⎜
10(1 + 3e ) ⎠
⎝
−t
2100
= 70
1 + 29
(c) P(0) =
1
⎛
⎞
= 4 y ⎜1 −
⎟
1 + 3e− 4t ⎠
⎝
y ( 0) =
2100
1 + 29e −0.75t
−2
(−15e−3t )
(e)
dP
P ⎞
⎛
= 0.2 P⎜1 −
⎟
dt
5000 ⎠
⎝
⎛ 14
⎞⎛ 5e −3t ⎞
= 3⎜
−3t ⎟⎜
−3t ⎟
⎝ 1 + 5e ⎠⎝ 1 + 5e ⎠
1
⎛
⎞
= 3 y ⎜1 −
⎟
1 + 5e −3t ⎠
⎝
⎛
⎞
14
⎟
= 3 y ⎜1 −
−3t ⎟
⎜
14(1 + 5e ) ⎠
⎝
y⎞
⎛
= 3 y ⎜1 −
⎟
14 ⎠
⎝
y ( 0) =
14
7
=
1+5
3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.4
6000
1 + 4999e −0.8t
11. P(t ) =
12. P(t ) =
(a) k = 0.2
(b) L = 6000
(b) L = 1000
6000
6
=
1 + 4999
5
(d)
3000 =
(c) P(0) =
6000
1 + 4999e −0.8t
(d)
1 + 4999e −0.8t = 2
1
4999
⎛ 1 ⎞
−0.8t = ln ⎜
⎟ = −ln 4999
⎝ 4999 ⎠
(e)
13.
1000
1000
=
1+8
9
500 =
1000
1 + 8e −0.2t
1 + 8e −0.2t = 2
e −0.8t =
t =
581
1000
1 + 8e −0.2t
(a) k = 0.8
(c) P(0) =
The Logistic Equation
e −0.2t =
⎛1⎞
−0.2t = ln ⎜ ⎟ = −ln 8
⎝8⎠
ln 8
t =
≈ 10.40 years
0.2
ln 4999
≈ 10.65 years
0.8
dP
P ⎞
⎛
= 0.8 P⎜1 −
⎟
dt
6000 ⎠
⎝
1
8
(e)
dP
P ⎞
⎛
= 0.2 P⎜1 −
⎟
dt
1000 ⎠
⎝
dP
P ⎞
⎛
= 3P⎜1 −
⎟
100 ⎠
dt
⎝
(a) k = 3
(b) L = 100
(c)
P
120
100
80
60
40
20
t
1
(d)
2
3
4
5
d 2P
P ⎞
⎛
⎛ − P′ ⎞
= 3P′⎜1 −
⎟ + 3P ⎜
⎟
dt 2
100
⎝
⎠
⎝ 100 ⎠
⎡ ⎛
P ⎞⎤ ⎛
P ⎞
P ⎞⎤
P ⎞⎛
P
P ⎞
P ⎞⎛
3P ⎡ ⎛
2P ⎞
⎛
⎛
= 3⎢3P⎜1 −
⎟⎥ ⎜1 −
⎟−
⎢3P⎜1 − 100 ⎟⎥ = 9 P⎜1 − 100 ⎟⎜1 − 100 − 100 ⎟ = 9 P⎜1 − 100 ⎟⎜1 − 100 ⎟
100
100
100
⎠⎦ ⎝
⎠
⎠⎦
⎝
⎠⎝
⎠
⎝
⎠⎝
⎠
⎣ ⎝
⎣ ⎝
L
100 ⎞
d 2P
⎛
=
= 0 for P = 50, and by the first Derivative Test, this is a maximum. ⎜ Note: P = 50 =
⎟
2
2 ⎠
dt 2
⎝
14.
dP
P ⎞
⎛
= 0.5P⎜1 −
⎟
250 ⎠
dt
⎝
(a) k = 0.5
(b) L = 250
P
(c)
(d)
300
dP
250
is a maximum for P =
= 125 (see Exercise 13).
2
dt
250
200
150
100
50
4
8
12
16
20
t
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
582
15.
Chapter 6
Differential Equations
dP
= 0.1P − 0.0004 P 2
dt
= 0.1P(1 − 0.004 P )
18.
P ⎞
⎛
= 0.1P⎜1 −
⎟
250 ⎠
⎝
L
10
=
1 + be − kt
1 + be −2.8t
10
10
3
⇒ 1+b =
⇒ b =
(0, 7): 7 =
1+b
7
7
10
⎛ 3 ⎞ −2.8t
1 + ⎜ ⎟e
⎝7⎠
10
y (5) =
≈ 10.00
⎛ 3 ⎞ −2.8(5)
1 + ⎜ ⎟e
⎝7⎠
10
y (100) =
≈ 10.00
⎛ 3⎞
1 + ⎜ ⎟e −2.8(100)
⎝7⎠
(b) L = 250
(c)
Solution: y =
P
300
240
180
120
60
t
20
40
60
80
100
250
= 125
2
(Same argument as in Exercise 13)
(d) P =
16.
19.
dy
4y
y2
4 ⎛
y ⎞
=
−
= y ⎜1 −
⎟,
dt
5
150
5 ⎝
120 ⎠
k =
(a) k = 0.4
(b) L = 1600
y (0) = 8: 8 =
P
Solution: y =
1200
800
400
4
17.
8
12
16
20
t
1600
dP
is a maximum for P =
= 800
dt
2
(see Exercise 13).
dy
y⎞
⎛
= y⎜1 −
⎟,
dt
36 ⎠
⎝
k = 1, L = 36
y =
(0, 4):
120
⇒ b = 14
1+b
120
1 + 14e−0.8t
120
y (5) =
≈ 95.51
1 + 14e −0.8(5)
120
y (100) =
≈ 120.0
1 + 14e −0.8(100)
1600
(d)
y ( 0) = 8
4
= 0.8, L = 120
5
120
L
y =
=
1 + be − kt
1 + be −0.8t
dP
P ⎞
⎛
= 0.4 P − 0.00025 P 2 = 0.4 P⎜1 −
⎟
1600
dt
⎝
⎠
(c)
y ( 0) = 7
y =
1
10
(a) k = 0.1 =
dy
y⎞
⎛
= 2.8 y⎜1 − ⎟,
dt
10 ⎠
⎝
k = 2.8, L = 10
y ( 0) = 4
36
⇒ b = 8
1+ b
36
1 + 8e − t
36
y (5) =
≈ 34.16
1 + 8e −5
36
y (100) =
≈ 36.00
1 + 8e −100
Solution: y =
dy
3y
y2
3 ⎛
y ⎞
=
−
=
y⎜1 −
⎟;
dt
20 1600
20 ⎝
240 ⎠
y(0) = 15
3
, L = 240
20
240
L
y =
=
1 + be − kt
1 + be(−3 20)t
k =
L
36
=
1 + be − kt
1 + be− t
4 =
20.
y(0) = 15: 15 =
Solution: y =
240
⇒ b = 15
1+b
240
1 + 15e(−3 20)t
240
y (5) =
≈ 29.68
1 + 15e(−3 20)(5)
240
y (100) =
≈ 240.0
1 + 15e(−3 20)(100)
21. L = 250 and y(0) = 350
Matches (c).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.4
22. L = 100 and y(0) = 100
28. No, it is not possible to determine b. However,
L = 2500 and k = 0.75. You need an initial condition
to determine b.
23. L = 250 and y(0) = 50
Matches (b).
24. L = 100 and y(0) = 50
29. Yes, the logistic differential equation is separable. See
Example 1.
Matches (a).
30. Answers will vary. Sample answer: There might be
limits on available food or space.
dy
y ⎞
⎛
= 0.2 y ⎜1 −
⎟
dt
1000 ⎠
⎝
(a)
31. (a) P =
y
1000
(0, 500)
L
, L = 200, P(0) = 25
1 + be − kt
25 =
800
600
400
39 =
(0, 105)
t
20
30
40
50
(b) k = 0.2, L = 1000
y =
y ( 0) =
1+b =
b =
y =
26.
1000
1 + be −0.2t
1000
105 =
1+b
1000
200
=
105
21
179
≈ 8.524
21
1000
1 + (179 21)e −0.2t
1000
(b) For t = 5, P ≈ 70 panthers.
(c)
0
60
0
100 =
200
1 + 7e −0.264t
1 + 7e −0.264t = 2
⎛1⎞
−0.264t = ln ⎜ ⎟
⎝7⎠
t ≈ 7.37 years
dy
y ⎞
⎛
= 0.9 y ⎜1 −
⎟
dt
200 ⎠
⎝
y
(a)
200
⇒ b = 7
1+b
200
1 + 7e − k (2)
200
1 + 7e −2 k =
39
23
e −2 k =
39
1 ⎛ 23 ⎞
1 ⎛ 39 ⎞
k = − ln ⎜ ⎟ = ln ⎜ ⎟ ≈ 0.2640
2 ⎝ 39 ⎠
2 ⎝ 23 ⎠
200
P =
1 + 7e −0.2640t
200
10
583
27. L represents the value that y approaches as t approaches
infinity. L is the carrying capacity.
Matches (d).
25.
The Logistic Equation
300
(d)
240
dP
P⎞
⎛
= kP⎜1 − ⎟
dt
L⎠
⎝
P ⎞
⎛
= 0.264 P⎜1 −
⎟, P(0) = 25
200 ⎠
⎝
180
120
60
2
4
6
8
10
Using Euler’s Method, P ≈ 65.6 when t = 5.
t
(e) P is increasing most rapidly where
P = 200 2 = 100, corresponds to t ≈ 7.37 years.
(b) k = 0.9, L = 200
y =
y ( 0) =
1+b =
b =
y =
200
1 + be −0.9t
200
240 =
1+b
200
5
=
240
6
1
−
6
200
1 − (1 6)e −0.9t
300
0
0
10
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
584
Chapter 6
Differential Equations
L
, L = 20, y (0) = 1, y ( 2) = 4
1 + be − kt
20
1=
⇒ b = 19
1+ b
20
4=
1 + 19e −2 k
y =
32. (a)
1 + 19e −2 k = 5
19e −2 k = 4
1 ⎛ 4 ⎞ 1 ⎛ 19 ⎞
k = − ln ⎜ ⎟ = ln ⎜ ⎟ ≈ 0.7791
2 ⎝ 19 ⎠ 2 ⎝ 4 ⎠
20
y =
1 + 19e −0.7791t
(b) For t = 5, y ≈ 14.43 grams.
18 =
(c)
1 + 19e −0.7791t =
19e −0.7791t =
e −0.7791t =
t =
(d)
20
1 + 19e −0.7791t
20
10
=
18
9
1
9
1
171
1
⎛ 1 ⎞
ln ⎜
−
⎟ ≈ 6.60 hours
0.7791 ⎝ 171 ⎠
33. False. If y > L, then dy dt < 0 and the population
decreases.
34. True. If 0 < y < L, then dy dt > 0 and the
population increases.
35. y =
y′ =
1
1 + be − kt
−1
(1 + be− kt )
2
(−bke− kt )
=
k
be − kt
⋅
− kt
(1 + be ) (1 + be− kt )
=
1 + be − kt − 1
k
⋅
− kt
(1 + be ) (1 + be−kt )
=
1
k
⎛
⎞
⋅ ⎜1 −
− kt ⎟
− kt
1
+
be
1
+
be
⎠
(
) ⎝
= ky (1 − y )
dy
y⎞
y⎞
1 ⎛ 19 ⎞ ⎛
⎛
= ky ⎜1 − ⎟ = ln ⎜ ⎟ y⎜1 −
⎟
dt
L⎠
2 ⎝4⎠ ⎝
20 ⎠
⎝
t
0
1
2
3
4
5
Exact
1
2.06
4.00
7.05
10.86
14.43
Euler
1
1.74
2.98
4.95
7.86
11.57
(e) The weight is increasing most rapidly when
y = L 2 = 20 2 = 10, corresponding to
t ≈ 3.78 hours.
36.
dy
y⎞
⎛
= ky ⎜1 − ⎟,
dt
L⎠
⎝
y ( 0) < L
⎡
y ⎞⎤
⎛
2
⎢ − ky⎜1 − L ⎟ ⎥
d2y
y⎞
y⎞
⎛
⎛ y′ ⎞
⎝
⎠ ⎥ = k 2 ⎛1 − y ⎞ y ⎡⎛1 − y ⎞ − y ⎤ = k 2 ⎛1 − y ⎞ y⎛1 − 2 y ⎞
2 ⎛
= ky′⎜1 − ⎟ + ky⎜ − ⎟ = k y⎜1 − ⎟ + ky ⎢
⎜
⎟ ⎜
⎟
⎜
⎟ ⎜
⎟
dt 2
L⎠
L⎠
L
L ⎠ ⎢⎣⎝
L ⎠ L ⎥⎦
L⎠ ⎝
L⎠
⎢
⎥
⎝
⎝ L⎠
⎝
⎝
⎝
⎢⎣
⎥⎦
d2y
2y
L
So,
= 0 when 1 −
= 0 ⇒ y = .
dt 2
L
2
By the First Derivative Test, this is a maximum.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.5
First-Order Linear Differential Equations
585
Section 6.5 First-Order Linear Differential Equations
7. y′ − y = 16
1. x 3 y′ + xy = e x + 1
y′ +
1
1
y = 3 (e x + 1)
x2
x
Integrating factor: e ∫
ln x
2
2 x dx
= ex
Integrating factor: e ∫
ye x
2
=
∫ 10 xe
x2
9.
2 − y′
= 5x
y
(y
2
dx = 5e x + C
y = 5 + Ce − x
Not linear, because of the xy 2 -term.
2
+ 1) cos x dx = dy
y′ = ( y + 1) cos x = y cos x + cos x
y′ − (cos x) y = cos x
2 − y′ = 5 xy
y′ + 5 xy = 2
Integrating factor: e ∫
− cos x dx
= e − sin x
y′e − sin x − (cos x)e − sin x y = (cos x)e − sin x
Linear
ye − sin x =
dy ⎛ 1 ⎞
+ ⎜ ⎟ y = 6x + 2
dx ⎝ x ⎠
Integrating factor: e
xy =
dx = −16e − x + C
8. y′ + 2 xy = 10 x
= 0
3. y′ − y sin x = xy 2
5.
−x
y = −16 + Ce x
= 0
Linear
4.
∫ 16e
ye − x =
2 xy − y′ ln x = y
(ln x) y′ + (1 − 2 x) y
(1 − 2 x) y
y′ +
= e− x
e − x y′ − e − x y = 16e − x
Linear
2.
−1 dx
= e
ln x
C
y = 2x + x +
x
2
10. ⎣⎡( y − 1) sin x⎦⎤ dx − dy = 0
y′ − (sin x) y = −sin x
− sin x dx
= ecos x
Integrating factor: e ∫
dy
2
6.
+ y = 3x − 5
dx
x
yecos x =
y =
3 2
5
C
x + x + 2
4
3
x
=
3 4
5 x3
x +
+ C
4
3
∫ − sin xe
cos x
dx = ecos x + C
y = 1 + Ce − cos x
2
2 x dx
Integrating factor: e ∫
= eln x = x 2
2
∫ x (3x − 5) dx
dx
y = −1 + Cesin x
= x
= 2 x3 + x 2 + C
x2 y =
− sin x
= −e − sin x + C
∫ (1 x) dx
∫ x(6 x + 2) dx
∫ (cos x)e
11.
(x
− 1) y′ + y = x 2 − 1
⎛ 1 ⎞
y′ + ⎜
⎟y = x + 1
⎝ x − 1⎠
⎡1 ( x −1)⎤⎦ dx
Integrating factor: e ∫ ⎣
= eln x −1 = x − 1
y ( x − 1) =
y =
∫ (x
2
− 1) dx =
1 3
x − x + C1
3
x3 − 3x + C
3( x − 1)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
586
Chapter 6
Differential Equations
12. y′ + 3 y = e3 x
16. (a)
y
4
3 dx
= e3 x
Integrating factor: e ∫
ye3 x =
y =
∫e
e dx =
3x 3x
∫e
6x
dx =
1 6x
e +C
6
x
−4
1 3x
e + Ce−3 x
6
−4
3
13. y′ − 3 x 2 y = e x
(b) y′ +
− 3x
Integrating factor: e ∫
3
ye − x =
4
∫e
x3 − x3
e
dx =
y = ( x + C )e x
2 dx
= e− x
∫ dx
3
(1 x) dx
u ( x) = e ∫
= eln x = x
= x + C
y′x + y = x sin x 2
3
∫ sec
cos x
= sec x
y = sin x + C ⋅ cos x
1
cos x 2 + C
2
1⎡ 1
1⎤
− cos x 2 − ⎥
2⎦
x ⎢⎣ 2
y =
15. (a) Answers will vary.
dx = −
1 ⎡ 1
1
⎤
− cos π + C ⎥ ⇒ C = −
⎢
2
π⎣ 2
⎦
0 =
x dx = tan x + C
2
1⎡ 1
⎤
− cos x 2 + C ⎥
x ⎢⎣ 2
⎦
y =
tan x dx
= e −ln
Integrating factor: e ∫
y sec x =
∫ x sin x
yx =
14. y′ + y tan x = sec x
2
1
1
y = sin x 2 , P( x) = , Q( x) = sin x 2
x
x
(c)
4
y
5
−4
−4
x
−4
4
17. y′ cos 2 x + y − 1 = 0
−3
(b)
4
y′ + (sec 2 x) y = sec 2 x
dy
= ex − y
dx
dy
+ y = ex
dx
sec
Integrating factor: e ∫
Integrating factor: e ∫
dx
= ex
ye tan x =
∫ sec
2
2 x dx
= e tan x
xe tan x dx = e tan x + C
y = 1 + Ce − tan x
e x y′ + e x y = e 2 x
Initial condition: y(0) = 5, C = 4
( ye x ) = ∫ e2 x dx
1
ye = e 2 x + C
2
1
1
y ( 0) = 1 ⇒ 1 =
+ C ⇒ C =
2
2
1 2x
1
x
ye = e +
2
2
1 x
1 −x
1
y = e + e = (e x + e − x )
2
2
2
Particular solution: y = 1 + 4e − tan x
x
6
(c)
18. x 3 y′ + 2 y = e1 x
2
1
⎛2⎞
y′ + ⎜ 3 ⎟ y = 3 e1 x
x
⎝x ⎠
(2 x
Integrating factor: e ∫
6
−2
)
3 dx
ye−1
x2
=
1
∫ x3 dx
y = e1
−6
2
= −
x 2 ⎛ Cx
⎜
⎝
= e
( )
− 1 x2
1
+ C1
2 x2
− 1⎞
⎟
2 x2 ⎠
2
Initial condition: y(1) = e, C = 3
2
2 ⎛ 3x − 1 ⎞
Particular solution: y = e1 x ⎜
⎟
2
⎝ 2x ⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.5
tan x dx
= eln sec x = sec x
Integrating factor: e ∫
∫ sec x(sec x + cos x) dx
= tan x + x + C
Integrating factor: e ∫
ye x
2 −x
y = sin x + x cos x + C cos x
Initial condition: y(0) = 1, 1 = C
1
∫ y dy
∫ (sec x + tan x) sec x dx
= sec x + tan x + C
C
y =1+
sec x + tan x
C
Initial condition: y (0) = 4, 4 = 1 +
,C = 3
1+ 0
Particular solution:
3
3 cos x
=1+
y =1+
sec x + tan x
1 + sin x
⎛1⎞
21. y′ + ⎜ ⎟ y = 0
⎝ x⎠
x
dy
y
= −
dx
x
∫
1
− dx
x
ln y = −ln x + ln C
ln xy = ln C
xy = C
Initial condition: y( 2) = 2, C = 4
Particular solution: xy = 4
2
=
∫ (1 − 2 x) dx
yC1 = e x − x
2
y = Ce x − x
2
Initial condition: y (1) = 2, 2 = C
Particular solution: y = 2e x − x
2
23. x dy = ( x + y + 2) dx
dy
x + y + 2
y
2
=
=
+1+
dx
x
x
x
dy
1
2
− y =1+
dx
x
x
u ( x) = e ∫
−(1 x) dx
=
Linear
1
x
2⎞1
2⎞
⎛
⎛1
y = x ∫ ⎜1 + ⎟ dx = x ∫ ⎜ + 2 ⎟ dx
x⎠x
x ⎠
⎝
⎝x
Separation of variables:
∫
2 −x
ln y + ln C1 = x − x 2
Integrating factor:
sec x dx
e∫
= eln sec x + tan x = sec x + tan x
1
dy =
y
= ex
Separation of variables:
20. y′ + y sec x = sec x
(1 x) dx
Integrating factor: e ∫
= e ln
(2 x −1) dx
= C
y = Ce x − x
Particular solution: y = sin x + ( x + 1) cos x
y (sec x + tan x ) =
587
22. y′ + ( 2 x − 1) y = 0
19. y′ + y tan x = sec x + cos x
y sec x =
First-Order Linear Differential Equations
= x
−2
⎡
⎤
= x ⎢ln x +
+ C⎥
x
⎣
⎦
= −2 + x ln x + Cx
y (1) = 10 = −2 + C ⇒ C = 12
y = −2 + x ln x + 12 x
24. 2 xy′ − y = x3 − x
dy
1
x2
1
−
−
y =
dx
2x
2
2
1
−
x
dx
1
2
(
)
= 12
u ( x) = e ∫
x
Linear
⎛ x2
⎛ x3 2
1⎞ 1
x −1 2 ⎞
− ⎟ 1 2 dx = x1 2 ∫ ⎜
−
y = x1 2 ∫ ⎜
⎟ dx
2⎠x
2 ⎠
⎝2
⎝ 2
⎡ x5 2
⎤
= x1 2 ⎢
− x1 2 + C ⎥
5
⎣
⎦
=
x3
− x +C
5
y ( 4) = 2 =
y =
x
64
17
− 4 + 2C ⇒ C = −
5
5
x3
17
− x −
5
5
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
588
25.
Chapter 6
Differential Equations
dP
= kP + N , N constant
dt
dP
= dt
kP + N
1
∫ kP + N dP = ∫ dt
1
ln ( kP + N ) = t + C1
k
ln ( kP + N ) = kt + C2
kP + N = e
P =
28. 1,000,000 =
1.64 = e0.08t
t =
29. (a)
C3e
− N
( Nekt )′
N
k
20 = 75 + Ce − k ⇒ − 55 = Ce − k
For t = 20, N = 35:
55
Ce − k
11
1 ⎛ 11 ⎞
=
⇒ e19 k =
⇒ k =
ln ⎜ ⎟
40
Ce −20 k
8
19 ⎝ 8 ⎠
≈ 0.0168
Ce − k = −55
C = −55e k ≈ −55.9296
N = 75 − 55.9296 e −0.0168t
30. (a)
C3e rt − P
r
P
A = Cert −
r
dQ
= q − kQ, q constant
dt
(b) Q′ + kQ = q
A =
Let P(t ) = k , Q(t ) = q, then the integrating factor
is u (t ) = e kt .
When t = 0: A = 0
P
P
⇒ C =
r
r
P rt
(e − 1)
r
q
⎛q
⎞
Q = e − kt ∫ qe kt dt = e − kt ⎜ e kt + C ⎟ =
+ Ce − kt
k
⎝k
⎠
When t = 0: Q = Q0
q
q
+ C ⇒ C = Q0 −
k
k
q ⎛
q ⎞ − kt
+ ⎜ Q0 − ⎟e
Q =
k
k⎠
⎝
Q0 =
P rt
(e − 1)
r
275,000 0.08(10)
A =
e
− 1 ≈ $4,212,796.94
0.06
27. (a) A =
(b) A =
(
= 75 e kt + C
35 = 75 + Ce −20 k ⇒ − 40 = Ce −20 k
rA + P = e rt + C2
(
kt
(c) For t = 1, N = 20:
1
ln ( rA + P ) = t + C1
r
ln ( rA + P ) = rt + C2
A =
∫ 75 ke
N = 75 + Ce − kt
dA
= rA + P
dt
dA
= dt
rA + P
dA
= ∫ dt
rA + P
0 = C −
= 75 ke kt
Ne kt =
N
N
P0 = C −
⇒ C = P0 +
k
k
N ⎞ kt
N
⎛
P = ⎜ P0 + ⎟e −
k⎠
k
⎝
∫
≈ 6.18 years
k dt
Integrating factor: e ∫
= e kt
N ′e kt + kNe kt = 75 ke kt
When t = 0: P = P0
26.
0.08
dN
= k (75 − N )
dt
k
P = Ce kt −
ln (1.64)
(b) N ′ + kN = 75k
kt + C2
kt
125,000 0.08t
(e − 1)
0.08
)
)
(c) lim Q =
t →∞
q
k
550,000 0.059(25)
e
− 1 ≈ $31,424,909.75
0.05
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.5
First-Order Linear Differential Equations
34. I (0) = 0, E0 = 120 volts, R = 600 ohms,
31. From Example 3,
L = 4 henrys
dv
kv
+
= g
dt
m
mg
v =
(1 − e−kt m ),
k
I =
Solution
g = −32, mg = −8, v(5) = −101, m =
implies that −101 =
( 0)
−8
1
=
g
4
(
)
v = −159.47(1 − e −0.2007t ).
lim I =
t →∞
(0.90)
As t → ∞, v → −159.47 ft/sec. The graph of v is
1
=
5
e
50
−150 t
= 0.1
−150t = ln (0.1)
40
t =
−200
=
E0
+ Ce −Rt L
R
120
1
+C ⇒ C = −
600
5
1 1 −150t
− e
5 5
1
amp
5
1
0.18 = (1 − e −150t )
5
0.9 = 1 − e −150t
shown below.
0
=
I =
−8
1 − e −5k (1 4) .
k
Using a graphing utility, k ≈ 0.050165, and
32. s(t ) =
589
∫ v(t ) dt
−0.2007 t
) dt
∫ −159.47(1 − e
= −159.47t − 794.57e −0.2007t + C
s(0) = 5000 = −794.57 + C ⇒ C = 5794.57
s(t ) = −159.47t − 794.57e −0.2007t + 5794.57
The graph of s(t ) is shown below.
ln (0.1)
−150
≈ 0.0154 sec
35. Let Q be the number of pounds of concentrate in the
solution at any time t. Because the number of gallons of
solution in the tank at any time t is v0 + ( r1 − r2 )t and
because the tank loses r2 gallons of solution per minute,
it must lose concentrate at the rate
⎡
⎤
Q
⎢
⎥ r2 .
⎢⎣ v0 + ( r1 − r2 )t ⎥⎦
The solution gains concentrate at the rate r1q1. Therefore,
the net rate of change is
⎡
⎤
dQ
Q
= q1r1 − ⎢
⎥ r2
dt
⎢⎣ v0 + ( r1 − r2 )t ⎥⎦
6000
or
0
40
−500
s(t ) = 0 when t ≈ 36.33 sec.
33. L
dI
R
E
+ RI = E0 , I ′ + I = 0
dt
L
L
Integrating factor: e ∫
( R L) dt
I e Rt L =
I =
∫
= e Rt L
dQ
r2Q
+
= q1r1.
dt
v0 + ( r1 − r2 )t
36. From Exercise 35, and using r1 = r2 = r ,
dQ
rQ
+
= q1r. lim x
δ x→0
dt
v0
E0 Rt L
E
e
dt = 0 e Rt L + C
L
R
E0
+ Ce − Rt L
R
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
590
Chapter 6
Differential Equations
37. (a) From Exercise 35,
(c) The volume of the solution is given by
v0 + ( r1 − r2 )t = 100 + (5 − 3)t = 200 ⇒ t = 50
r2 Q
dQ
+
= q1r1
dt
u0 + ( r1 − r2 )t
minutes.
r2Q
= q1r1
v0 + ( r1 − r2 )t
You have Q(0)q0 = 25, q1 = 0, u0 = 200, and
Q′ +
r1 = r2 = 10. Hence, the linear differential
Q(0) = q0 = 0, q1 = 1, v0 = 100, r1 = 5, r2 = 3
equation is
Q′ +
dQ
1
+
Q = 0.
dt
20
Integrating factor is (50 + t ) .
32
By separating variables,
∫
Q(50 + t )
dQ
1
dt
= −∫
Q
20
0 = 100 + C (50)
−1t
3
1
⎛ 3⎞
(b) 15 =
⇒
= e 20 ⇒ ln ⎜ ⎟ = − t
5
5
20
⎝ ⎠
⎛ 3⎞
⇒ t = − 20 ln ⎜ ⎟ ≈ 10.2 minutes
⎝5⎠
t →∞
(uy )′
52
(50 + t )
32
= −2(50)
52
−3 2
52
(100)
−3 2
= 200 −
40. (a) At t = 0, Q = 20 pounds.
(b) The rate of solution withdrawn is greater.
(c) At t = 25, Q = 0. It takes 25 minutes to empty the
tank.
32
⎡3 (100 + 2 t )⎦⎤ dt
Integrating factor: e ∫ ⎣
= (50 + t )
∫ 2.5(50 + t )
= Q( x )u
so u′( x) = P( x)u
3
Q = 2.5
100 + 2t
=
⇒ C = −100(50)
Answer (a)
Q(0) = q0 , q0 = 0, q1 = 0.5,v0 = 100, r1 = 5,
32
−3 2
y′u + P( x) yu = Q( x )u
r2Q
= q1r1
v0 + ( r1 − r2 )t
Q(50 + t )
+C
P( x) dx
Integrating factor: u = e ∫
= 0
t = 50 minutes.
r2 = 3, Q′ +
52
39. y′ + P( x) y = Q( x)
38. (a) The volume of the solution in the tank is given by
v0 + ( r1 − r2 )t. Therefore, 100 + (5 − 3)t = 200 or
(b) Q′ +
dt = 2(50 + t )
50
2
≈ 164.64 lb (double the answer to part (b))
Q = 200 − 2(50)
−1t
25e 20
− 1t
25e 20
32
When t = 50,
−1t
20 .
C = 25. Hence, Q = 25e
−3 2
Q = 2(50 + t ) − 2(50)
The initial condition Q(0) = 25 implies that
t →∞
= ∫ 5(50 + t )
Q(0) = 0:
−1t
Ce 20 .
(c) lim Q = lim
32
Q = 2(50 + t ) + C (50 + t )
1
ln Q = − t + ln C1
20
Q =
3Q
= 5
100 + 2t
32
dt = (50 + t )
Q = (50 + t ) + C (50 + t )
52
−3 2
Initial condition:
Q(0) = 0, 0 = 50 + C (50−3 2 ), C = −505 2
+C
41.
dy
+ P( x) y = Q( x)
dx
Standard form
P( x) dx
u ( x) = e ∫
Integrating factor
42. The term “first-order” means that the derivative in the
equation is first order.
Particular solution:
Q = (50 + t ) − 50−5 2 (50 + t )
Q(50) = 100 − 505 2 (100)
= 100 −
−3 2
−3 2
25
≈ 82.32 lb
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.5
43. y′ − 2 x = 0
∫ dy
=
First-Order Linear Differential Equations
45. y′ − 2 xy = 0
∫ 2 x dx
∫
y = x + C
2
∫ 2 x dx
y = Ce x
44. y′ − 2 y = 0
∫
dy
=
y
ln y = x 2 + C1
Matches (c).
dy
=
y
591
2
Matches (a).
∫ 2 dx
46.
ln y = 2 x + C1
y = Ce
y′ − 2 xy = x
dy
∫ 2y + 1
2x
=
∫ x dx
1
1
ln ( 2 y + 1) = x 2 + C1
2
2
Matches (d).
2 y + 1 = C2 e x
y = −
2
2
1
+ Ce x
2
Matches (b).
47. (a)
10
−4
4
−6
(b)
dy
1
− y = x2
dx
x
Integrating factor: e −1 x dx = e − ln x =
1
x
1
1
y′ − 2 y = x
x
x
⎛1 ⎞
⎜ y⎟ =
⎝x ⎠
y =
∫ x dx
=
x2
+C
2
x3
+ Cx
2
1
x3
−8
− 2C ⇒ C = −4 ⇒ y =
− 4 x = x( x 2 − 8)
2
2
2
3
8
1
x
+ 2 x = x ( x 2 + 4)
(2, 8): 8 = + 2C ⇒ C = 2 ⇒ y =
2
2
2
(−2, 4):
(c)
4 =
10
−4
4
−6
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
592
Chapter 6
48. (a)
Differential Equations
5
−1
3
−1
(b) y′ + 4 x 3 y = x 3
Integrating factor: e ∫
4
4 x3 dx
4
4
4
∫x e
y′e x + 4 x3 ye x = x3e x
ye x =
y =
= ex
3 x4
1
4
+ Ce
4
dx =
1 e x4
4
+C
− x4
(0, 72 ): 72 = 14 + C ⇒ C = 134 ⇒ y = 14 + 134 e
(0, − 12 ): − 12 = 14 + C ⇒ C = − 34 ⇒ y = 14 − 34 e
− x4
(c)
− x4
5
−1
3
−1
49. (a)
3
−2
6
−3
(b) y′ + (cot x) y = 2
cot x dx
= eln sin x = sin x
Integrating factor: e ∫
y′sin x + (cos x) y = 2 sin x
y sin x =
∫ 2 sin x dx
= −2 cos x + C
y = −2 cot x + C csc x
(1, 1): 1
= −2 cot 1 + C csc 1 ⇒ C =
1 + 2 cot 1
= sin 1 + 2 cos 1
csc 1
y = −2 cot x + (sin 1 + 2 cos 1) csc x
(3, −1):
−1 = −2 cot 3 + C csc 3 ⇒ C =
2 cot 3 − 1
= 2 cos 3 − sin 3
csc 3
y = −2 cot x + ( 2 cos 3 − sin 3) csc x
(c)
3
−2
6
−3
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.5
50. (a)
First-Order Linear Differential Equations
53.
7
( y cos x
593
− cos x) dx + dy = 0
Separation of variables:
−5
∫ cos x dx
5
−3
sin x = −ln ( y − 1) + ln C
(b) y′ + 2 xy = xy 2
ln ( y − 1) = −sin x + ln C
Bernoulli equation, n = 2 letting
1− 2
−1
z = y
= y , you obtain e
2
1 2
and ∫( −1) xe− x dx = e − x .
2
The solution is:
y −1e − x
2
−2x dx
= e
y = Ce − sin x + 1
− x2
54. y′ = 2 x 1 − y 2
Separation of variables:
2
1
1
1 + 2Ce x
=
+ Ce x =
2
2
y
2
y =
2
1 + 2Ce x
y =
55. ( 2 y − e x ) dx + x dy = 0
1
⎛ 2⎞
Linear: y′ + ⎜ ⎟ y = e x
x
⎝ x⎠
(c)
2
(2 x) dx
= eln x = x 2
Integrating factor: e ∫
yx 2 =
2
1
⇒ 1 + 2C = 2 ⇒ C =
1 + 2C
2
2
1 + ex
y =
2
56.
7
(x
∫x
21 x
x
e dx = e x ( x − 1) + C
ex
C
x − 1) + 2
2(
x
x
+ y ) dx − x dy = 0
Linear: y′ −
−5
5
51. e2 x + y dx − e x − y dy = 0
Separation of variables:
e 2 x e y dx = e x e − y dy
x
dx =
∫e
−2 y
dy
e x = − 12 e −2 y + C1
2e x + e −2 y = C
52.
dy
x −3
=
dx
y ( y + 4)
∫ (y
+ 4 y ) dy =
y
1
=
x
1
∫ x dx
x(ln x
y =
∫ ( x − 3) dx
3
y
x2
+ 2 y2 =
− 3 x + C1
3
2
2 y 3 + 12 y 2 = 3x 2 − 18 x + C
−1
=
1
x
= ln x + C
+ C)
57. 3( y − 4 x 2 ) dx = − x dy
dy
= −3 y + 12 x 2
dx
3
y′ + y = 12 x
x
x
Integrating factor: e ∫
Separation of variables:
2
1
y =1
x
ln x
−(1 x) dx
Integrating factor: e ∫
= e
−3
∫e
∫ 2 x dx
y = sin ( x 2 + C )
=
=
dy =
arcsin y = x 2 + C
( )
(0, 1): 1
1 − y2
2
2
2
1
⇒ 1 + 2C =
⇒ C = −
1 + 2C
3
6
2
6
y =
=
2
2
3 − ex
1 − ex 3
(0, 3): 3
1
∫
1 − x2
e
+ C
2
=
−1
∫ y − 1 dy
=
y′x3 +
(3 x) dx
= e3 ln x = x3
3 3
x y = 12 x( x3 ) = 12 x 4
x
12 5
yx3 = ∫ 12 x 4 dx =
x + C
5
12 2
C
y =
x + 3
5
x
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594
Chapter 6
Differential Equations
58. x dx + ( y + e y )( x 2 + 1) dy = 0
63. xy′ + y = xy 3
y′ +
Separation of variables:
∫ x2
y
∫ − ( y + e ) dy
x
dx =
+1
1
1
ln ( x 2 + 1) = − y 2 − e y + C1
2
2
ln ( x + 1) + y + 2e = C
2
2
1
y = y3
x
1
,
x
n = 3, Q = 1, P =
y −2 x −2 =
y
∫ − 2x
−2
e
∫
−2
dx
x
= e −2 ln x = x −2
dx + C = 2 x −1 + C
y −2 = 2 x + Cx 2
1
2 x + Cx 2
y2 =
59. y′ + 3 x 2 y = x 2 y 3
1
= 2 x + Cx 2
y2
or
n = 3, Q = x 2 , P = 3 x 2
y −2e ∫
(−2)3 x 2 dx
=
2 ( −2)3 x
∫ (−2) x e∫
2 dx
−2 −2 x3
= −∫ 2 x e
−2 −2 x3
3
1
= e −2 x + C
3
3
1
= + Ce2 x
3
3
1
1
+ Ce 2 x +
y2
3
y e
y e
y −2
2 −2 x3
64. y′ − y = y 3
dx
dx
y −2e 2 x =
y2 =
2 x2
=
y e
∫ 2 xe
x2
dx = e
y 2 = 1 + Ce − x
x2
= ex
+ C
2
−( 2 3) dx
= e−(2 3)x
y 2 3e −(2 3)x =
x −( 2 3) x
∫ 23 e e
dx =
∫ 23 e
(1 3) x dx
y 2 3e −(2 3)x = 2e(1 3)x + C
y 2 3 = 2e x + Ce 2 x 3
y′ − 2 y = e x y −1
n = −1, Q = e x , P = −2
−1
e∫
−(1 x) dx
= e− ln x = x −1
e∫
=
1
−1 + Ce −2 x
66. yy′ − 2 y 2 = e x
n = 2, Q = x, P = x
y x
dx = −e 2 x + C
2
⎛1⎞
61. y′ + ⎜ ⎟ y = xy 2
⎝ x⎠
−1 −1
2x
= e2 x
65. y′ − y = e x 3 y , n = 13 , Q = e x , P = −1
e∫
2 x dx
∫ (−2)e
−2( −1) dx
y −2 = −1 + Ce −2 x
60. y′ + xy = xy −1
n = −1, Q = x, P = x, e ∫
e∫
n = 3, P = −1, Q = 1,
∫ − x( x ) dx
−1
2( −2) dx
= e −4 x
y 2e −4 x =
= −x + C
∫ 2e
−4 x x
e dx = − 23 e−3 x + C
y 2 = − 23 e x + Ce4 x
1
= − x 2 + Cx
y
67. False. The equation contains
1
y =
Cx − x 2
68. True. y′ + ( x − e x ) y = 0 is linear.
y.
⎛1⎞
62. y′ + ⎜ ⎟ y = x y
⎝ x⎠
1
n = , Q = x, P = x −1
2
e(1 2)(1 x) dx = e(1 2) ln x =
y1 2 x1 2 =
=
y =
x
1 12
∫ 2 x ( x) dx
x5 2 + C
1 52
x + C1 =
5
5
( x5 2
+ C)
2
25 x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.6
Predator-Prey Differential Equations
595
Section 6.6 Predator-Prey Differential Equations
1.
dx
= ax − bxy = 0.9 x − 0.05 xy
dt
dy
= − my + nxy = − 0.6 y + 0.008 xy
dt
dx
dy
=
= 0 ⇒ 0.9 x − 0.05 xy = x(0.9 − 0.05 y ) = 0
dt
dt
− 0.6 y + 0.008 xy = y( −0.6 + 0.008 x) = 0
If x = 0, then y = 0.
If y =
0.9
90
0.6
600
=
= 18, then x =
=
= 75.
0.008
8
0.05
5
Solutions: (0, 0) and (75, 18)
2.
dx
= ax − bxy = 0.75 x − 0.006 xy
dt
dy
= − my + nxy = − 0.9 y + 0.003xy
dt
dx
dy
=
= 0 ⇒ 0.75 x − 0.006 xy = x(0.75 − 0.06 y ) = 0
dt
dt
− 0.9 y + 0.003 xy = y ( − 0.9 + 0.003 x) = 0
If x = 0, then y = 0.
If y =
0.75
750
0.9
900
=
= 300.
=
= 125, then x =
0.003
3
0.006
6
Solutions: (0, 0) and (300, 125)
3.
dx
= ax − bxy = 0.5 x − 0.01xy
dt
dy
= − my + nxy = − 0.49 y + 0.007 xy
dt
dx
dy
=
= 0 ⇒ 0.5 x − 0.01xy = x(0.5 − 0.01 y ) = 0
dt
dt
− 0.49 y + 0.007 xy = y( − 0.49 + 0.007 x) = 0
If x = 0, then y = 0.
If y =
0.5
50
0.49
490
=
= 50, then x =
=
= 70.
0.01
1
0.007
7
Solutions: (0, 0) and (70, 50)
4.
dx
= ax − bxy = 1.2 x − 0.04 xy
dt
dy
= − my + nxy = −1.2 y + 0.02 xy
dt
dx
dy
=
= 0 ⇒ 1.2 x − 0.04 xy = x(1.2 − 0.04 y ) = 0
dt
dt
−1.2 y + 0.02 xy = y( −1.2 + 0.02 x) = 0
If x = 0, then y = 0.
If y =
1.2
120
1.2
120
=
= 30, then x =
=
= 60.
0.04
4
0.02
2
Solutions: (0, 0) and (60, 30)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
596
Chapter 6
5. (a)
Differential Equations
9. Critical points are ( x, y ) = (0,0) and
y
40
( x, y )
(150, 30)
30
⎛ m a ⎞ ⎛ 0.3 0.8 ⎞
= ⎜ , ⎟ = ⎜
,
⎟ = (50, 20).
⎝ n b ⎠ ⎝ 0.006 0.04 ⎠
20
10.
150
10
x(t)
x
80
(b)
160
240
320
400
40
0
11.
0
y(t)
0
36
50
400
0
y
6. (a)
0
150
0
10
8
12.
50
6
4
(15, 3)
2
4
(b)
8
12
16
20
0
x
10
0
(55, 10)
13. Critical points are ( x, y ) = (0,0) and
( x, y )
0
20
7. (a) The initial conditions are x(0) = 40 and
= (10,000, 1250).
y(0) = 20.
14.
y
⎛m a⎞
= ⎜ , ⎟
⎝ n b⎠
0.1 ⎞
⎛ 0.4
= ⎜
,
⎟
⎝ 0.00004 0.00008 ⎠
0
(b)
150
25,000
x(t)
100
80
60
0
0
y(t)
240
40
20
(40, 20)
15.
5000
x
20
40
60
80
100
8. (a) The initial conditions are x(0) = 60 and
y(0) = 10.
0
25,000
0
y
(b)
16.
5,000
100
80
60
(60, 10)
40
0
0
(4000, 1000)
25,000
20
20
40
60
80
100
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 6.6
17. Using x(0) = 50 and y(0) = 20, you obtain the
Predator-Prey Differential Equations
597
18. Using x(0) = 10,000 and y (0) = 1250, you obtain the
constant solutions x = 50 and y = 20.
constant solutions x = 10,000 and y = 1250.
15,000
60
x
x(t)
y
y(t)
0
0
36
0
The slope field is the same, but the solution curve
reduces to a single point at (50, 20).
The slope field is the same, but the solution curve
reduces to a single point at (10,000, 1250).
2,500
50
0
19.
0
150
0
240
0
(50, 20)
0
20,000
(10,000, 1,250)
dx
= ax − bx 2 − cxy = 2 x − 3 x 2 − 2 xy
dt
dy
= my − ny 2 − pxy = 2 y − 3 y 2 − 2 xy
dt
⎛ m⎞ ⎛ 2⎞ ⎛ a ⎞ ⎛ 2 ⎞
From Example 5, you have (0, 0), ⎜ 0, ⎟ = ⎜ 0, ⎟, ⎜ , 0 ⎟ = ⎜ , 0 ⎟ and
⎝ n ⎠ ⎝ 3⎠ ⎝b ⎠ ⎝ 3 ⎠
⎛ an − mc bm − ap ⎞
⎛6 − 4 6 − 4⎞
⎛ 2 2⎞
,
,
⎜
⎟ = ⎜
⎟ = ⎜ , ⎟.
bn
cp
bn
cp
9
4
9
4
−
−
−
−
⎝
⎠
⎝5 5⎠
⎝
⎠
20.
dx
= ax − bx 2 − cxy = x − 0.5 x 2 − 0.5 xy
dt
dy
= my − ny 2 − pxy = 2.5 y − 2 y 2 − 0.5 xy
dt
⎛ m⎞ ⎛ 5⎞ ⎛a
From Example 5, you have (0, 0), ⎜ 0, ⎟ = ⎜ 0, ⎟, ⎜ ,
⎝ n ⎠ ⎝ 4⎠ ⎝b
⎞
0 ⎟ = ( 2, 0) and
⎠
⎛ an − mc bm − ap ⎞ ⎛ 2 − 5 4 5 4 − 1 2 ⎞
,
,
⎟ = (1, 1).
⎜
⎟ = ⎜
⎝ bn − cp bn − cp ⎠ ⎝ 1 − 1 4 1 − 1 4 ⎠
21.
dx
= ax − bx 2 − cxy = 0.15 x − 0.6 x 2 − 0.75 xy
dt
dy
= my − ny 2 − pxy = 0.15 y − 12 y 2 − 0.45 xy
dt
⎛ m⎞ ⎛ 1⎞ ⎛ a
From Example 5, you have (0, 0), ⎜ 0, ⎟ = ⎜ 0, ⎟, ⎜ ,
⎝ n ⎠ ⎝ 8⎠ ⎝ b
⎞ ⎛1
0⎟ = ⎜ ,
⎠ ⎝4
⎞
0 ⎟ and
⎠
⎛ an − mc bm − ap ⎞
⎛ 0.18 − 0.1125 0.09 − 0.0675 ⎞
⎛3 1⎞
,
,
⎜
⎟ = ⎜
⎟ = ⎜ , ⎟ ≈ (0.1765, 0.0588).
⎝ 0.72 − 0.3375 0.72 − 0.3375 ⎠
⎝ 17 17 ⎠
⎝ bn − cp bn − cp ⎠
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
598
22.
Chapter 6
Differential Equations
dx
= ax − bx 2 − cxy = 0.025 x − 0.1x 2 − 0.2 xy
dt
dy
= my − ny 2 − pxy = 0.3 y − 0.45 y 2 − 0.1xy
dt
⎛ m⎞ ⎛ 2⎞ ⎛ a
From Example 5, you have (0, 0), ⎜ 0, ⎟ = ⎜ 0, ⎟, ⎜ ,
⎝ n ⎠ ⎝ 3⎠ ⎝b
⎞ ⎛1
0⎟ = ⎜ ,
⎠ ⎝4
⎞
0 ⎟ and
⎠
⎛ an − mc bm − ap ⎞
⎛ 0.01125 − 0.06 0.03 − 0.0025 ⎞
⎛ 39 11 ⎞
,
,
⎜
⎟ = ⎜
⎟ = (−1.95, 1.1) = ⎜ − , ⎟.
⎝ 0.045 − 0.02 0.045 − 0.02 ⎠
⎝ 20 10 ⎠
⎝ bn − cp bn − cp ⎠
23. a = 0.8, b = 0.4, c = 0.1, m = 0.3, n = 0.6,
p = 0.1
27. Assuming the initial conditions are the critical points
( x(0), y(0))
Four critical points:
3
⎛ m ⎞ ⎛ 0.3 ⎞
⎛ 1⎞
⎜ 0, ⎟ = ⎜ 0,
⎟ = ⎜ 0, ⎟
⎝ n ⎠ ⎝ 0.6 ⎠
⎝ 2⎠
⎞ ⎛ 0.8
0⎟ = ⎜ ,
⎠ ⎝ 0.4
x
⎞
0 ⎟ = ( 2, 0)
⎠
⎛ an − mc bm − ap ⎞ ⎛ 0.45 0.04 ⎞
⎛ 45 4 ⎞
,
,
⎜
⎟ = ⎜
⎟ = ⎜ ,
⎟
⎝ 23 23 ⎠
⎝ bn − cp bn − cp ⎠ ⎝ 0.23 0.23 ⎠
24.
( 4523 , 234 )
you obtain constant solutions.
(0, 0)
⎛a
⎜ ,
⎝b
=
y
0
36
0
28. Assuming the initial conditions are the critical points
( x(0), y(0))
( 12 )
= 0,
you obtain constant solutions.
10
2
x(t)
y(t)
0
x(t)
36
0
0
Both species survive.
25. a = 0.8, b = 0.4, c = 1, m = 0.3, n = 0.6, p = 1
Four critical points:
29. Yes, they are separable. See bottom of page 437.
dx
= ax − bxy = 0
dt
dy
= − my + nxy = 0
dt
⎛ m ⎞ ⎛ 0.3 ⎞ ⎛ 1 ⎞
⎜ 0, ⎟ = ⎜ 0,
⎟ = ⎜ 0, ⎟
⎝ n ⎠ ⎝ 0.6 ⎠ ⎝ 2 ⎠
⎞ ⎛ 0.8
0⎟ = ⎜ ,
⎠ ⎝ 0.4
⎞
0 ⎟ = ( 2, 0)
⎠
to obtain critical points
(0, 0)
⎛ 0.18 − 0.68 ⎞
⎛ an − mc bm − ap ⎞
,
,
⎟
⎜
⎟ = ⎜
⎝ bn − cp bn − cp ⎠
⎝ − 0.76 − 0.76 ⎠
⎛ 9 17 ⎞
= ⎜− , ⎟
⎝ 38 19 ⎠
26.
36
0
30. Solve the equations
(0, 0)
⎛a
⎜ ,
⎝b
y(t)
⎛m a⎞
and ⎜ , ⎟.
⎝ n b⎠
The solutions will be constant for these initial conditions.
31. As in Exercise 30, using any of the four critical points as
initial conditions will yield constant solutions.
10
32.
y
100
x(t)
0
y(t)
0
80
36
One species (the trout) becomes extinct.
(50, 67)
60
40
(70, 40)
(35, 34)
20
(44, 22)
20
40
60
80
100
x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
dx
x⎞
⎛
= ax⎜1 − ⎟, which is a logistic equation.
dt
L⎠
⎝
33. (a) If y = 0, then
(b)
599
dx
x ⎞
⎛
= 0.4 x⎜1 −
⎟ − 0.01xy
dt
100
⎝
⎠
dy
= − 0.3 y + 0.005 xy
dt
(0, 0)
is a critical point. If y = 0, then x = 100 and (100, 0) is a critical point. If x, y =/ 0, then
x ⎞
⎛
0.4⎜1 −
⎟ = 0.01y
100
⎝
⎠
−0.3 + 0.05 x = 0.
So, x =
60 ⎞
0.3
⎛
= 60 and 0.4⎜1 −
⎟ = 0.01 y ⇒ y = 16.
100 ⎠
0.005
⎝
The third critical point is (60, 16).
(c)
(d)
100
(e)
80
80
x
y
0
72
0
100
0
0
0
100
0
Review Exercises for Chapter 6
1. y = x3 , y′ = 3x 2
2 xy′ + 4 y = 2 x(3 x 2 ) + 4( x3 ) = 10 x3 .
6.
dy
= 2 sin x
dx
y =
Yes, it is a solution.
y = 2 sin 2 x
2.
7.
y′ = 4 cos 2 x
y′′′ = −16 cos 2 x
y′′′ − 8 y = −16 cos 2 x − 8( 2 sin 2 x) ≠ 0
8.
Not a solution
3.
4 x3
y = ∫ ( 4 x 2 + 7) dx =
+ 7x + C
3
4.
dy
= 3x3 − 8 x
dx
y =
5.
9.
∫ (3x
3
− 8 x) dx =
3 4
x − 4x2 + C
4
10.
dy
= cos 2 x
dx
y =
∫ cos 2 x dx
=
1
sin 2 x + C
2
∫e
2− x
dx = − e 2 − x + C
dy
= 2 e3 x
dx
y =
dy
= 4x2 + 7
dx
= −2 cos x + C
dy
= e2 − x
dx
y =
y′′ = −8 sin 2 x
∫ 2 sin x dx
∫ 2e
3x
dx =
2 3x
e + C
3
dy
= 2x − y
dx
x
–4
−2
0
2
4
8
y
2
0
4
4
6
8
dy dx
–10
–4
–4
0
2
8
dy
⎛π y ⎞
= x sin ⎜ ⎟
dx
⎝ 4 ⎠
x
–4
−2
0
2
4
8
y
2
0
4
4
6
8
dy dx
–4
0
0
0
–4
0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
600
Chapter 6
Differential Equations
11. y′ = 2 x 2 − x,
(0, 2)
(−1, 1)
12. y′ = y + 4 x,
(a) and (b)
(a) and (b)
y
y
(−1, 1)
(0, 2)
2
5
x
−3
3
x
−3
3
−1
−4
13. y′ = x − y, y (0) = 4, n = 10, h = 0.05
y1 = y0 + hf ( x0 , y0 ) = 4 + (0.05)(0 − 4) = 3.8
y2 = y1 + hf ( x1 , y1 ) = 3.8 + (0.05)(0.05 − 3.8) = 3.6125, etc.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
yn
4
3.8
3.6125
3.437
3.273
3.119
2.975
2.842
2.717
2.601
2.494
14. y′ = 5 x − 2 y , y (0) = 2, n = 10, h = 0.1
y1 = y0 + hf ( x0 , y0 ) = 2 + (0.1)(5(0) − 2( 2)) = 1.6
y2 = y1 + hf ( x1 , y1 ) = 1.6 + (0.1)(5(0.1) − 2(1.6)) = 1.33, etc.
15.
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
2
1.6
1.33
1.164
1.081
1.065
1.102
1.182
1.295
1.436
1.599
dy
= 2x − 5x2
dx
y =
16.
∫
dy
2
= (3 + y )
dx
17.
2
∫ (2 x − 5 x )dx
= x2 −
−2
∫ (3 + y) dy
5 3
x + C
3
− (3 + y )
dy
= y +8
dx
dy
= ∫ dx
y +8
−1
=
∫ dx
= x + C
3+ y =
−1
x + C
y = −3 −
ln y + 8 = x + C1
y + 8 = e x + C1 = Ce x
y = −8 + Ce
x
dy
= 10
dx
18.
∫y
−1 2
dy =
1
x +C
y
∫ 10 dx
2 y1 2 = 10 x + C1
C1 ⎞
⎛
⎜C =
⎟
2⎠
⎝
y1 2 = 5 x + C
y = (5 x + C )
2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
19. ( 2 + x) y′ − xy = 0
(2
+ x)
24. y = Ce kt
dy
= xy
dx
x
1
dy =
dx
y
2+ x
(0, 5): 5
= C
(5, 16 ): 16
= 5e k (5)
1
30
1
2 ⎞
⎛
dy = ⎜1 −
⎟ dx
y
2 + x⎠
⎝
y = Ce x ( 2 + x)
−2
=
2
22.
=
= Ce 4 k =
1
4
=
∫ (50k
(Alternate form: y = −
= e3 k ⇒ k =
27.
k
(50 − t )2 + C1 )
2
dP
= kp,
dh
y =
ln
3 ⎣⎡ln (20 3) 5⎦⎤t
e
4
9 0.602t
e
20
= Ce k ⇒ C = 4e − k
= Ce 4 k
1
3
ln
1
4
= − 13 ln 4 ≈ − 0.4621
P(18,000) = 30e18,000 k = 15
ln (1 2)
18,000
=
−ln 2
18,000
P( h) = 30e −(h ln 2) 18,000
P(35,000) = 30e −(35,000 ln 2) 18,000 ≈ 7.79 inches
= e5 k
1
5
9
.
20
P(0) = 30
k =
k =
ln
P( h) = 30e kh
23. y = Ce kt
20
3
(103 )
3 3
=
2 ( 10 )
3 2k
e
2
y = 6.3496e − 0.4621t
− kt )dt
k 2
t + C
2
= 34 e k (5)
=
=
So, C = 4e − k = 4e0.4621 ≈ 6.3496.
∫ k (50 − t )dt
(5, 5): 5
≈
1
2
4k
1 = ( 4e − k )(e 4 k ) = 4e3k
k
+ C
2t 2
=C
−2 k
3 −2(1 2) ln(10 3)
e
2
(1, 4): 4 = Cek (1)
(4, 1): 1 = Cek (4)
(0, 34 ): 34
( 32 e )e
= e2k ⇒ k =
9 1 2 ln(10 3)t
e
20
3 −2 k
e
2
26. y = Ce kt
dt
y = 50kt −
(4, 5): 5
y =
dy
= k (50 − t )
dt
∫ dy
= Ce 2 k ⇒ C =
So, C =
dy
k
= 3
dt
t
y = −
(2, 32 ): 32
10
3
ln y = x + ln x + C1
−3
≈ 5e − 0.6802t
25. y = Ce kt
dy
x
= ( x + 1) y
dx
dy
x +1
∫ y = ∫ x dx
y = Cxe x
= − 15 ln 30
⎡− ln (30) 5⎦⎤ t
Ce x
(2 + x)
1
30
y = 5e ⎣
20. xy′ − ( x + 1) y = 0
∫ dy = ∫ kt
= e5 k
k = 15 ln
ln y = x − 2 ln 2 + x + C1
21.
601
( 203 )
≈
28. y = Ce kt = 15e kt
3 0.379 t
e
4
7.5 = 15e k (1599)
k =
1
1599
ln
( 12 ) ≈ −0.000433
When t = 750, y = 15e −0.000433(750) ≈ 10.84 g .
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
602
29.
Chapter 6
Differential Equations
31. S = Ce k t
P = Ce0.0185t
2C = Ce0.0185t
2 = e
5 = Ce k
ln 2 = 0.0185t
t =
S = 5 when t = 1
(a)
0.0185t
lim Ce
k t
t →∞
ln 2
≈ 37.5 years
0.0185
= C = 30
5 = 30e k
k = ln
30. A = 1000e(0.04)(8) ≈ $1377.13
1
6
≈ −1.7918
S = 30e −1.7918 t
(b) When t = 5, S ≈ 20.9646 which is 20,965 units.
(c)
30
0
40
0
32. S = 25(1 − e kt )
(
)
(a) 4 = 25 1 − e k (1) ⇒ 1 − e k =
4
25
⇒ ek =
21
25
⇒ k = ln
( 2521 ) ≈ −0.1744
S = 25(1 − e −0.1744t )
(
)
(b) 25,000 units lim S = 25
t →∞
(c) When t = 5, S ≈ 14.545 which is 14,545 units.
(d)
25
0
8
0
33.
35. y′ − 16 xy = 0
dy
5x
=
dx
y
∫ y dy
=
∫ 5 x dx
5x2
y2
=
+ C1
2
2
y 2 = 5x2 + C
dy
= 16 xy
dx
1
∫ y dy
=
∫ 16 x dx
ln y = 8 x 2 + C1
e8 x
2 +C
1
= y
3
dy
x
=
dx
2 y2
34.
∫ 2y
2
dy =
∫x
3
y = Ce8 x
dx
36. y′ − e y sin x = 0
2 y3
x4
=
+ C1
3
4
8 y 3 = 3x 2 + C
2
dy
= e y sin x
dx
∫e
−y
dy =
∫ sin x dx
−e − y = −cos x + C1
ey =
1
cos x + C
y = ln
(C
= −C1 )
1
= −ln cos x + C
cos x + C
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
Initial condition: y (0) = − 2 : 4 = sin 0 + C
37. y 3 y′ − 3 x = 0, y ( 2) = 2
y3
∫y
C = 4
dy
= 3x
dx
3
Particular solution: y 2 = sin x 2 + 4
∫ 3x dx
dy =
603
4
y
3x 2
=
+ C1
4
2
y 4 = 6x2 + C
41.
∫ y dy
Initial condition: y ( 2) = 2 : 16 = 24 + C
=
∫ − 4 x dx
2
y
= −2 x 2 + C1
2
4 x2 + y 2 = C
ellipses
C = −8
Particular solution: y = 6 x − 8
4
dy
−4 x
=
dx
y
2
y
38. yy′ − 5e 2 x = 0, y (0) = − 3
y
∫
4
dy
= 5e 2 x
dx
y dy =
2x
∫ 5e dx
x
−4
y2
5
= e 2 x + C1
2
2
y 2 = 5e 2 x + C
−4
Initial condition: y (0) = − 3 : ( − 3) = 5 + C
2
C = 4
Particular solution: y = 5e
2
4
2x
+ 4
42.
dy
= 3 − 2y
dx
dy
∫ 2y − 3
=
∫ − dx
39. y 3 ( x 4 + 1) y′ − x 3 ( y 4 + 1) = 0, y (0) = 1
1
ln 2 y − 3 = − x + C1
2
ln 2 y − 3 = −2 x + 2C1
y 3 ( x 4 + 1)
2 y − 3 = C2e −2 x
∫
dy
= x 3 ( y 4 + 1)
dx
y3
dy =
4
y +1
x3
dx
4
x +1
∫
1
1
1
ln ( y 4 + 1) = ln ( x 4 + 1) + ln C1
4
4
4
2 y = 3 + C2e −2 x
y =
3
+ Ce −2 x
2
y
ln ( y 4 + 1) = ln ⎡⎣C ( x 4 + 1)⎤⎦
y 4 + 1 = C ( x 4 + 1)
Initial condition: y (0) = 1 : 1 + 1 = C (0 + 1)
x
4
C = 2
Particular solution: y 4 + 1 = 2( x 4 + 1)
y4 = 2 x4 + 1
40. yy′ − x cos x 2 = 0, y(0) = − 2
y
∫
dy
= x cos x 2
dx
y dy =
∫ x cos x
2
dx
y2
1
= sin x 2 + C1
2
2
y 2 = sin x 2 + C
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
604
Chapter 6
Differential Equations
5250
1 + 34e −0.55t
43. P(t ) =
46.
(a) k = 0.55
L
8
=
− kt
1 + be
1 + be −1.76t
8
5
y (0) = 3: 3 =
⇒ b =
1+b
3
y =
5250
= 150
1 + 34
2625 =
(d)
5250
1 + 34e −0.55t
Solution: y =
1 + 34e −0.55t = 2
1
34
−1
⎛1⎞
ln ⎜ ⎟ ≈ 6.41 yr
t =
0.55 ⎝ 34 ⎠
e −0.55t =
y =
20,400
⇒ b = 16
1+b
20,400
y (1) = 2000 =
1 + 16e − k
46
16e − k =
5
23
40
k = −ln
= ln
≈ 0.553
40
23
20,400
y =
1 + 16e −0.553t
(a) k = 0.15
(b) L = 4800
(c) P(0) =
4800
= 320
1 + 14
2400 =
(d)
14e
−0.15t
4800
1 + 14e −0.15t
(b) y(8) ≈ 17,118 trout
=1
(c) 10,000 =
1
⎛1⎞
ln ⎜ ⎟ ≈ 17.59 yr
t = −
0.15 ⎝ 14 ⎠
dP
P ⎞
⎛
(e)
= 0.15P⎜1 −
⎟
dt
4800 ⎠
⎝
dy
y⎞
⎛
45.
= y ⎜1 −
⎟,
dt
80 ⎠
⎝
20,400
1 + be − kt
y (0) = 1200 =
4800
1 + 14e −0.15t
44. P(t ) =
8
⎛ 5 ⎞ −1.76t
1 + ⎜ ⎟e
⎝ 3⎠
47. (a) L = 20,400, y (0) = 1200, y (1) = 2000
dP
P ⎞
⎛
= 0.55P⎜1 −
⎟
dt
5250
⎝
⎠
(e)
(0, 3)
k = 1.76, L = 8
(b) L = 5250
(c) P(0) =
dy
y⎞
⎛
= 1.76 y ⎜1 − ⎟,
dt
8⎠
⎝
48.
20,400
⇒ t ≈ 4.94 yr
1 + 16e −0.553t
dy
y ⎞
⎛
= 0.553 y⎜1 −
⎟, y(0) = 1200
dt
20,400 ⎠
⎝
Use Euler’s method with h = 1.
(0, 8)
t
0
2
4
6
8
k = 1, L = 80
Exact
1200
3241
7414
12,915
17,117
L
80
=
1 + be − kt
1 + be − t
80
y(0) = 8: 8 =
⇒ b = 9
1+b
Euler
1200
2743
5853
10,869
16,170
y =
Solution: y =
80
1 + 9e − t
Euler’s method gives y(8) ≈ 16,170 trout.
49.
dS
= k(L − S )
dt
dS
∫ L − S = ∫ k dt
−ln L − S = kt + C1
L − S = e − kt − C1
S = L + Ce − kt
Because S = 0 when t = 0, you have
0 = L + C ⇒ C = − L. So, S = L(1 − e − kt ).
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
50. The general solution is S = L(1 − e − kt ).
(a) Because L = 100 and S = 25 when t = 2, you
have the following.
(b) Because L = 0.80 and P = 0.25 when n = 0,
and P = 0.60 when n = 10, you have the
following.
0.80C
5
⇒ C =
C +1
11
5 11)(0.80)
(
1
5
⇒ k = − ln
≈ 0.2359
0.60 =
8
33
(5 11) + e−8k
25 = 100(1 − e −2 k )
1
4
0.25 =
= 1 − e −2 k
e −2 k =
3
4
−2k = ln
3
4
k = − 12 ln
Therefore, P =
3
4
≈ 0.1438
− 0.1438t
).
0.8
0.6
(b) Because L = 500 and S = 50 when t = 1, you
have the following.
0.4
50 = 500(1 − e − k )
0.2
= 1 − e−k
e−k =
5
9
10
− k = ln
10
9
10
15
t
20
y
53. (a)
9
10
k = − ln
4
.
5 + 11e −0.1887 n
P
So, the particular solution is S = 100(1 − e
1
10
605
4
3
= ln
10
9
≈ 0.1054
2
1
So, the particular solution is S = 500(1 − e
−0.1054t
).
x
− 4 −3 −2 −1
1
2
3
4
51. The differential equation is given by the following.
dP
= kP( L − P)
dn
1
∫ P( L − P) dP
y′ = e x 2 − y
(b)
ye x =
1
⎡ln P − ln L − P ⎤⎦ = kn + C1
L⎣
P
= Ce Lkn
L − P
CLe Lkn
CL
P =
= − Lkn
1 + Ce Lkn
e
+ C
52. The general solution is P =
=
y =
y =
(c)
(a) Because L = 1 and P = 0.50 when n = 0, and
P = 0.85 when n = 4, you have the following.
C
⇒ C =1
C +1
1
1
3
0.85 =
⇒ k = − ln
≈ 0.4337
1 + e− 4k
4
17
e dx =
x 2 x
2 e (3 2) x
3
2 ex 2
3
2 ex 2
3
∫e
dx
= ex
(3 2) x dx
+C
+ Ce − x
2
3
+ C ⇒ C = − 53
− 53 e − x =
1 ⎡2e x 2
3⎣
− 5e − x ⎤⎦
6
−9
9
−6
0.50 =
P =
∫e
y ( 0 ) = −1 =
CL
.
C + e− Lkn
Therefore,
Integrating factor: e ∫
y′ + y = e x 2 ,
= ∫ k dn
P
1
1 + e −0.4337 n
.
1
1
2
3
4
t
−1
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
606
Chapter 6
Differential Equations
54. (a)
(b)
y
= e2 x
4
(c)
( ye2 x ) = ∫ e2 x sin x dx
2
−2
2
4
x
−6
1 2x
e ( 2 sin x − cos x ) + C
5
1
y (0) = 4 ⇒ 4 = (0 − 1) + C
5
1
21
⇒ 4 = − + C ⇒ C =
5
5
1 2x
21
2x
ye = e ( 2 sin x − cos x ) +
5
5
1
21 −2 x
y = ( 2 sin x − cos x) +
e
5
5
ye 2 x =
−2
−4
y
55. (a)
2 dx
e 2 x y′ + 2e 2 x y = e 2 x sin x
4
−4
Integrating factor: e ∫
y′ + 2 y = sin x,
(b)
3
dy
= csc x + y cot x
dx
−4
(c)
dy
− (cot x ) y = csc x
dx
x
−3
3
6
3
−4.5
Integrating factor: e ∫ − cot x dx = e −ln sin x = csc x
4.5
−3
csc x ⋅ y′ − csc x cot x ⋅ y = csc 2 x
−3
( y csc x)′
= csc 2 x
y csc x =
∫ csc
2
x dx = − cot x + C
y = − cos x + C sin x
y (1) = 1 ⇒ 1 = − cos 1 + C sin 1
1 + cos 1
≈ 1.8305
sin 1
y = − cos x + 1.8305 sin x
⇒ C =
56. (a)
(b)
y
(1, 2)
4
dy
+ (cot x) y = csc x
dx
2
x
−2
2
−2
dy
= csc x − y cot x
dx
(c)
4
−4
Integrating factor: e ∫ cot x dx = e ln sin x = sin x
4
−4
sin x y′ + cos x y = 1
( y sin x)′
=1
y sin x = x + C
y (1) = 2 ⇒ 2 sin 1 = 1 + C
⇒ C = 2 sin 1 − 1 ≈ 0.683
y = x csc x + C csc x
y = x csc x + 0.683 csc x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
57. y′ − y = 10
62.
P( x) = −1, Q ( x) = 10
+ 3) y′ + 2 y = 2( x + 3)
P( x) =
1
10e − x dx
e− x ∫
= −10 + Ce x
y =
58. e y′ + 4e y = 1
x
y′ + 4 y = e − x
=
P( x) = 4, Q ( x) = e− x
=
4 dx
u ( x) = e ∫
= e4 x
y =
1
⎛1
⎞ 1
e − x e 4 x dx = e −4 x ⎜ e3 x + C ⎟ = e − x + Ce −4 x
4x ∫
e
⎝3
⎠ 3
4 y′ = e x
59.
y′ −
y =
1
e
1
−(1 4) x
∫ 4e
x 4 −(1 4) x
e
x 2 dx
1
5x
61.
(x
e
5x
1
∫ x2 e
(x
+ 3)
2
+
2
C
(x
+ 3)
2
63. y′ + 5 y = e5 x
∫e
dx =
1 e5 x
10
+ Ce −5 x
10 x
1 e10 x
10
+C
yx − a =
3 −a
∫ bx ( x ) dx
y =
bx 4
+ Cx a
4− a
=
b
x4 − a + C
4− a
= e
dx =
1 ⎛ 1 5x
1
⎞
− e + C ⎟ = − + Ce−5 x
e ⎝ 5
5
⎠
5 x⎜
Initial condition:
1
1
, Q( x) =
x − 2
x − 2
(1 x − 2) dx
u ( x) = e ∫
= eln
1
(e5 x )(e5 x ) dx
e5 x ∫
1
= 5 x ∫ e10 x dx
e
1
⎛1
⎞
= 5 x ∫ ⎜ e10 x + C ⎟
e
10
⎝
⎠
1 5x
−5 x
e + Ce
=
10
y =
5 x
dy
1
1
+
y =
dx
x − 2
x − 2
y =
⎡ ( x + 3)4
⎤
⎢
+ C⎥
2
⎥⎦
( x + 3) ⎢⎣ 2
5 dx
= e5 x
u ( x) = e∫
− 2) y′ + y = 1
P( x) =
dx
P( x ) = 5, Q( x) = e5 x
−(5
u ( x) = e ∫
y =
2
65. y′ + 5 y = e5 x , y(0) = 3
5
1
, Q( x) = 2
x2
x
)
∫ 2( x + 3)( x + 3)
− ( a x) dx
= e− a ln x = x − a
Integrating factor: e ∫
dx
dy
5y
1
− 2 = 2
dx
x
x
P( x) = −
2
⎛a⎞
64. y′ − ⎜ ⎟ y = bx3
⎝ x⎠
= e −(1 4)x
⎛1
⎞
= e(1 4)x ⎜ x + C ⎟
⎝4
⎠
1 x4
= xe + Ce x 4
4
60.
+ 3)
1
y =
1
1
P( x) = − , Q( x) = e x 4
4
4
−(1 4) dx
(x
ye5 x =
1
1
y = ex 4
4
4
u ( x) = e ∫
1
5 dx
= e5 x
Integrating factor: e∫
+ y
y
2
, Q( x) = 2( x + 3)
x +3
2
( 2 x + 3) dx
u ( x) = e ∫
= e 2 ln( x + 3) = ( x + 3)
= e x ( −10e− x + C )
x
2
dy
2
+
y = 2( x + 3)
dx
x +3
− dx
u ( x) = e ∫
= e− x
y =
(x
607
x−2
= x − 2
1 0
29
e + Ce 0 ⇒ C =
10
10
1 5x
29 − 5 x
Particular solution: y =
e +
e
10
10
y ( 0) = 3 : 3 =
1
1
⎛ 1 ⎞
(x + C)
⎜
⎟( x − 2) dx =
x − 2∫ ⎝ x − 2⎠
x−2
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
608
Chapter 6
Differential Equations
⎛ 3⎞
66. y′ − ⎜ ⎟ y = 2 x 3 , y(1) = 1
⎝ x⎠
3
P( x ) = − , Q( x ) = 2 x 3
x
− dx
u ( x) = e ∫ x = e − 3 ln x = x − 3
3
y =
1
x −3
= x3
3 −3
∫ 2 x ( x ) dx
∫ 2 dx
= x (2 x + C )
3
= 2 x 4 + Cx 3
Initial condition: y(1) = 1 :1 = 2 + C ⇒ C = −1
Particular solution: y = 2 x 4 − x3
67. Answers will vary. Sample answer: ( x 2 + 3 y 2 ) dx − 2 xy dy = 0
Solution: Let y = vx, dy = x dv + v dx.
( x2
+ 3v 2 x 2 ) dx − 2 x(vx)( x dv + v dx) = 0
( x2
+ v 2 x 2 ) dx − 2 x3v dv = 0
(1 + v 2 ) dx
∫
= 2 xv dv
dx
=
x
2v
∫ 1 + v2
dv
ln x = ln 1 + v 2 + C1
⎛
y2 ⎞
x = C (1 + v 2 ) = C ⎜1 + 2 ⎟
x ⎠
⎝
x3 = C ( x 2 + y 2 )
y⎞
⎛
68. Answers will vary. Sample answer: y′ = y ⎜1 −
⎟
40 ⎠
⎝
Solution: k = 1, L = 40
y =
70.
L
40
=
1 + be − kt
1 + be − t
69. Answers will vary.
Sample answer: x y′ + 2 x y = 1
3
2
y′ +
2
1
y = 3
x
x
(2 x) dx
u( x) = e ∫
= x2
1 1
1
y = 2 ∫ 3 ( x 2 ) dx = 2 ⎡⎣ln x + C ⎤⎦
x x
x
71. A0 = 500,000,
For this linear differential equation, you have
P(t ) = −r and Q(t ) = − P. Therefore, the integrating
− r dt
= e − rt and the solution is
factor is u ( x) = e ∫
P
⎛P
⎞
A = e rt ∫ − Pe − rt dt = e rt ⎜ e − rt + C ⎟ =
+ Ce rt .
r
⎝r
⎠
Because A = A0 when t = 0, you have
C = A0 − ( P r ) which implies that
A =
P ⎛
P⎞
+ ⎜ A0 − ⎟e rt .
r
r⎠
⎝
r = 0.10
(a) P = 40,000
A =
dA
− rA = − P
dt
2,500,000
40,000 ⎛
40,000 ⎞ 0.10t
+ ⎜ 500,000 −
= 100,000( 4 + e 0.10t )
⎟e
0.10
0.10 ⎠
⎝
The balance continues to increase.
0
30
0
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Exercises for Chapter 6
(b) P = 50,000
609
2,500,000
50,000 ⎛
50,000 ⎞ 0.10t
+ ⎜ 500,000 −
= 500,000
⎟e
0.10
0.10 ⎠
⎝
A =
The balance remains at $500,000.
0
30
0
(c) P = 60,000
2,500,000
60,000 ⎛
60,000 ⎞ 0.10t
+ ⎜ 500,000 −
= 100,000(6 − e 0.10t )
⎟e
0.10
0.10 ⎠
⎝
A =
The balance decreases and is depleted in t = (ln 6) 0.10 ≈ 17.9 years.
0
30
0
A =
72.
200,000 ⎛
200,000 ⎞ 0.14t
+ ⎜1,000,000 −
⎟e
0.14
0.14 ⎠
⎝
50 ⎞ 0.14t ⎤
⎡ 50 ⎛
0 = 200,000 ⎢
+ ⎜5 −
⎟e ⎥
7⎠
⎝
⎣7
⎦
10
e0.14t =
3
ln (10 3)
t =
≈ 8.6 years
0.14
73. (a)
75. (a)
dx
= ax − bx 2 − cxy = 3 x − x 2 − xy
dt
dy
= my − ny 2 − pxy = 2 y − y 2 − 0.5 xy
dt
(b) x′ = y′ = 0 when ( x, y ) = (0, 0),
⎛ m⎞
= ⎜ 0, ⎟ = (0, 2),
⎝ n⎠
a
( x, y ) = ⎛⎜ , 0 ⎞⎟ = (3, 0),
⎝b ⎠
( x, y )
dx
= ax − bxy = 0.3x − 0.02 xy
dt
dy
= − my + nxy = − 0.4 y + 0.01xy
dt
( x, y )
⎛ 1 1 2⎞
= ⎜ ,
⎟ = ( 2, 1).
⎝1 2 1 2 ⎠
(b) x′ = y′ = 0 when ( x, y ) = (0, 0) and
⎛m a⎞
⎛ 0.4 0.3 ⎞
= ⎜ , ⎟ = ⎜
,
⎟ = ( 40, 14).
⎝ n b⎠
⎝ 0.01 0.02 ⎠
( x, y )
⎛ an − mc bm − ap ⎞
= ⎜
,
⎟
⎝ bn − cp bn − cp ⎠
(c)
4
x(t)
(c)
y(t)
80
x(t)
0
6
0
y(t)
0
36
76. (a)
0
74. (a)
dx
= ax − bxy = 0.4 x − 0.04 xy
dt
dy
= − my + nxy = − 0.6 y + 0.02 xy
dt
(b) x′ = y′ = 0 when ( x, y ) = (0, 0),
⎛ m⎞
⎛ 17 ⎞
= ⎜ 0, ⎟ = ⎜ 0, ⎟,
⎝ n⎠
⎝ 2⎠
a
15
( x, y ) = ⎛⎜ , 0 ⎞⎟ = ⎛⎜ , 0 ⎞⎟,
⎝b ⎠
⎝2 ⎠
( x, y )
(b) x′ = y′ = 0 when ( x, y ) = (0, 0) and
( x, y )
(c)
dx
= ax − bx 2 − cxy = 15 x − 2 x 2 − 4 xy
dt
dy
= my − ny 2 − pxy = 17 y − 2 y 2 − 4 xy
dt
⎛m a⎞
⎛ 0.6 0.4 ⎞
= ⎜ , ⎟ = ⎜
,
⎟ = (30, 10).
⎝ n b⎠
⎝ 0.02 0.04 ⎠
( x, y )
50
x(t)
⎛ −38 −26 ⎞
⎛ 19 13 ⎞
= ⎜
,
⎟ = ⎜ , ⎟.
⎝ −12 −12 ⎠
⎝6 6⎠
y(t)
0
0
⎛ an − mc bm − ap ⎞
= ⎜
,
⎟
⎝ bn − cp bn − cp ⎠
24
(c)
15
y(t)
0
x(t)
0
One species, x, becomes
extinct.
4
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
610
Chapter 6
Differential Equations
Problem Solving for Chapter 6
dy
= y1.01
dt
1. (a)
∫y
−1.01
dy =
2. (a)
∫ dt
−0.01
y
= t + C1
−0.01
1
= −0.01t + C
y 0.01
1
y 0.01 =
C − 0.01t
1
y =
100
C
0.01
t)
−
(
L
C Le − kt
⎛k⎞
= ⎜ ⎟
⋅
− kt
1 + Ce − kt
⎝ L ⎠1 + Ce
L
L
⎛k⎞
⎛
⎞
= ⎜ ⎟
⋅ ⎜L −
⎟
− kt
1 + Ce − kt ⎠
⎝ L ⎠1 + Ce
⎝
k
= k1S ( L − S ), where k1 = .
L
L = 100. Also, S = 10 when t = 0 ⇒ C = 9.
4
And, S = 20 when t = 1 ⇒ k = −ln .
9
100
100
Particular Solution: S =
=
1 + 9eln(4 9)t 1 + 9e −0.8109t
1
⇒ C =1
C100
1
So, y =
.
(1 − 0.01t )100
y (0) = 1: 1 =
For T = 100, lim y = ∞.
t →T −
(b)
∫y
−(1 + ε )
dy =
∫ k dt
y −ε
= kt + C1
−ε
(b)
y −ε = −ε kt + C
y =
y(0) = y0 =
So, y =
For t →
1
(C
1ε
− ε kt )
ε
⎛1⎞
1
1
⇒ C1 ε =
⇒ C = ⎜ ⎟
C1 ε
y0
⎝ y0 ⎠
1
1ε
⎛ 1
⎞
⎜ ε − ε kt ⎟
⎝ y0
⎠
dS
= k1S ( L − S )
dt
L
is a solution because
S =
1 + Ce − kt
−2
dS
= − L(1 + Ce − kt ) (−Cke − kt )
dt
LC ke − kt
=
2
(1 + Ce− kt )
.
1
, y → ∞.
y0ε ε k
dS
= k1S (100 − S )
dt
2
⎡ ⎛ dS ⎞
d S
dS ⎤
= k1 ⎢S ⎜ − ⎟ + (100 − S ) ⎥
2
dt
dt ⎦
⎣ ⎝ dt ⎠
dS
= k1 (100 − 2S )
dt
dS
= 0 when S = 50 or
= 0.
dt
Choosing S = 50, you have:
100
50 =
1 + 9e ln(4 9)t
2 = 1 + 9e ln(4 9)t
ln (1 9)
ln ( 4 9)
= t
t ≈ 2.7 months
(This is the point of inflection.)
(c)
125
0
10
0
(d)
S
140
120
100
80
60
40
20
t
1
2
3
4
(e) Sales will decrease toward the line S = L.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 6
3. (a) y′ = x − y, y(0) = 1, h = 0.1
( fg )′
(b)
Using the modified Euler Method, you obtain:
x
y
0
1.0
0.1
0.91
0.2
0.83805
#
#
1.0
0.73708
(b)
= f ′g ′
f ′g + fg ′ = f ′g ′
f ′( g − g ′) = − fg ′
f′
g′
=
f
g′ − g
g′
ln f = ∫
dx
′
g − g
g′
f = e
∫ g ′ − g dx
(c) If g ( x) = e x , then g ′( x) − g ( x) = e x − e x = 0
1.0
Therefore, no f can exist.
The modified Euler Method
is more accurate.
0
1.0
0.5
[
[.1
[
[.2
[
[.3
[
[.4
[
[.5
[
[.6
[
[.7
[
[.8
[
[.9
[
[1.0
611
]
]
]
.82
]
]
.758
]
]
.7122
]
]
.68098
]
]
.662882
]
]
.6565938 ]
]
.66093442 ]
]
.674840978 ]
]
.6973568802]
.9
[x
[
[0
[
[.1
[
[.2
[
[.3
[
[.4
[
[.5
[
[.6
[
[.7
[
[.8
[
[.9
[
[1.0
y
]
]
1
]
]
.9100000000]
]
.8380500000]
]
.7824352500]
]
.7416039013]
]
.7141515307]
]
.6988071353]
]
.6944204575]
]
.6999505140]
]
.7144552152]
]
.7370819698]
2
⎛1⎞
5. k = ⎜ ⎟ π
⎝ 12 ⎠
g = 32
x 2 + ( y − 6) = 36
2
Equation of tank
x 2 = 36 − ( y − 6) = 12 y − y 2
2
Area of cross section: A( h) = (12h − h 2 )π
dh
= − k 2 gh
dt
1
π 64h
= −
(12h − h 2 )π dh
dt
144
1
= − h1 2
(12h − h2 ) dh
dt
18
A( h)
∫ (18h
32
− 216h1 2 ) dh =
∫ dt
36 5 2
h − 144h3 2 = t + C
5
h3 2
(36h − 720) = t + C
5
63 2
(−504) ≈ −1481.45.
5
The tank is completely drained when
When h = 6, t = 0 and C =
h = 0 ⇒ t = 1481.45 sec ≈ 24 min, 41 sec
y
6 ft
4. ⎣⎡ f ( x ) g ( x)⎤⎦′ = f ′( x) g ′( x)
(a) Let g ( x) = x, g ′( x) = 1, then
?
⎡⎣ f ( x) x⎤⎦′ = f ′( x)
f ′( x) x + f ( x) = f ′( x)
h
x
x 2 + (y − 6)2 = 36
df
( x − 1) = − f ( x)
dx
df
dx
∫ f = ∫1 − x
ln f ( x) = −ln 1 − x
f ( x) =
1
1− x
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
612
Chapter 6
Differential Equations
dh
= − k 2 gh
dt
dh
πr2
= − k 64h
dt
−8 k
h −1 2 dh =
dt = −C dt ,
πr2
6. (a) A( h)
(b)
⇒ h ≈ 7.21 ft
C =
8k
(at t
r
πr2
2 h = −Ct + C1
2 18 = C1
t = 3600 sec ⇒ 2 h = −0.000865(3600) + 6 2
= 0, h = 18)
18 ft
h
So, 2 h = −Ct + 6 2.
At t = 30(60) = 1800, h = 12:
2 12 = −1800 C + 6 2
6 2 − 4 3
= C ≈ 0.000865
1800
So, 2 h = −0.000865t + 6 2.
h = 0 ⇒ t =
7.
6 2
0.000865
≈ 9809.1 seconds ( 2 h, 43 min, 29 sec)
dh
= − k 2 gh
dt
dh
−π
=
8 h
π 64
36
dt
−1 2
−1
∫ h dh = ∫ 288 dt
−t
2 h =
+C
288
A( h)
9.
ds
= 3.5 − 0.019s
dt
− ds
(a) ∫
= − ∫ dt
3.5 − 0.019s
1
ln 3.5 − 0.019 s = −t + C1
0.019
ln 3.5 − 0.019s = −0.019t + C2
3.5 − 0.019 s = C3e −0.019t
0.019 s = 3.5 − C3e −0.019t
h = 20: 2 20 = C = 4 5
2 h =
s = 184.21 − Ce −0.019t
−t
+ 4 5
288
(b)
400
h = 0 ⇒ t = 4 5 ( 288)
≈ 2575.95 sec ≈ 42 min, 56 sec
0
1 ⎛
ln b ⎞
8. Let u = k ⎜ t −
⎟.
2 ⎝
k ⎠
1 + tanh u = 1 +
200
0
(c) As t → ∞, Ce −0.019t → 0, and s → 184.21.
eu − e − u
2
=
eu + e − u
1 + e −2u
− k t − ln b k ))
e −2 u = e ( (
= eln be − kt = be − kt
Finally,
⎛1 ⎛
1 ⎡
ln b ⎞ ⎞⎤
L
L ⎢1 + tanh ⎜ k ⎜ t −
⎟ ⎟⎥ = [1 + tanh u]
2 ⎣
2
k ⎠ ⎠⎦
⎝2 ⎝
=
2
L⎛
⎞
⎜
⎟
2 ⎝ 1 + be − kt ⎠
=
L
.
1 + be − kt
Notice the graph of the logistics function is just a shift of
the graph of the hyperbolic tangent. (See Section 5.9.)
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Problem Solving for Chapter 6
f ( 0 − ∆ x ) − f ( 0)
10. f ′(0) = lim
∆x
∆x → 0
f ′( x) = lim
= lim
∆x
∆x → 0
f ( x + ∆ x) − f ( x)
∆x
∆x → 0
f ( ∆ x) − 1
= lim
=1
f ( x) f ( ∆ x) − f ( x)
∆x
∆x → 0
613
= f ( x) lim
∆x → 0
f ( ∆ x) − 1
∆x
= f ( x)(1) = f ( x)
Finally, solve the differential equation
dy
= f ( x) = y :
dx
dy
∫ y = ∫ dx
f ′( x) =
ln y = x + k
y = Ce x
Since y (0) = 1, C = 1 and y = f ( x) = e x .
11. (a)
dC
=
C
∫
R
∫ −V
dt
R
ln C = − t + K1
V
C = Ke − Rt V
Since C = C0 when t = 0, it follows that K = C0 and the function is C = C0e − Rt V .
(b) Finally, as t → ∞, we have
lim C = lim C0e − Rt V = 0.
t →∞
t →∞
12. From Exercises 11, you have C = C0e − Rt V .
(a) For V = 2, R = 0.5, and C0 = 0.6, you have C = 0.6e −0.25t
0.8
0
4
0
(b) For V = 2, R = 1.5, and C0 = 0.6, you have C = 0.6e −0.75t .
0.8
0
4
0
1
∫ Q − RC dC
13. (a)
−
=
1
∫V
dt
1
t
ln Q − RC =
+ K1
R
V
Q − RC = e
C =
− R ⎡⎣(t V ) + K1⎤⎦
(
)
1
1
− R⎡ t V + K ⎤
Q − e ⎣( ) 1⎦ = (Q − Ke − Rt V )
R
R
Because C = 0 when t = 0, it follows that K = Q and you have C =
Q
(1 − e− Rt V ).
R
(b) As t → ∞, the limit of C is Q R.
© 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 7
Applications of Integration
Section 7.1
Area of a Region Between Two Curves............................................619
Section 7.2
Volume: The Disk Method ................................................................634
Section 7.3
Volume: The Shell Method................................................................650
Section 7.4
Arc Length and Surfaces of Revolution ............................................662
Section 7.5
Work....................................................................................................675
Section 7.6
Moments, Centers of Mass, and Centroids .......................................681
Section 7.7
Fluid Pressure and Fluid Force ..........................................................694
Review Exercises ........................................................................................................699
Problem Solving .........................................................................................................707
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C H A P T E R 7
Applications of Integration
Section 7.1 Area of a Region Between Two Curves
1. A =
∫ 0 ⎣⎡0 − ( x
2. A =
∫ − 2 ⎡⎣(2 x + 5) − ( x
=
6
− 6 x)⎤⎦ dx = − ∫ ( x 2 − 6 x) dx
0
6
2
∫ − 2 (− x
2
3
3. A =
∫0
=
3
4. A =
2
∫0
2
9.
3 ⎡⎛
⎣
+ 2 x + 1)⎤⎦ dx
y
7
+ 4) dx
2
6
5
4
3
⎡( − x 2 + 2 x + 3) − ( x 2 − 4 x + 3)⎤ dx
⎣
⎦
2
1
x
(−2 x 2 + 6 x) dx
∫ 0 (x
1
2
⎞ x⎤
− x ⎟ − ⎥ dx
⎠ 3 ⎦⎥
x3
∫ 2 ⎢⎢⎜⎝ 3
2
−1
− x 3 ) dx
10.
4
∫ − π 4 (sec
π 4
2
5
6
7
x − cos x ) dx
y
5. A = 2 ∫ 3( x3 − x) dx = 6∫
or −6 ∫
0
0
−1
−1
1
0
( x3 − x) dx
3
( x3 − x) dx
2
3
6. A = 2 ∫ ⎡( x − 1) − ( x − 1)⎤ dx
0⎣
⎦
1
7.
⎡
x⎤
dx
2 ⎥⎦
∫ 0 ⎢⎣( x + 1) −
4
−π
4
11.
x
π
4
∫ − 2 ⎡⎣(2 − y) −
1
y 2 ⎤⎦ dy
y
y
5
4
3
3
2
2
−1
1
x
1
−1
2
3
4
5
x
1
2
3
4
5
−3
8.
2
∫ −1 ⎡⎣(2 − x ) −
1
x 2 ⎤⎦ dx
12.
y
∫ 0 (2
)
4
y − y dy
y
3
4
1
−2
3
x
−1
1
−1
2
2
1
x
1
2
3
4
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
619
620
Chapter 7
Applications of Integration
13. f ( x) = x + 1
y
g ( x) = ( x − 1)
2
x = 4 − y2
15. (a)
x = y − 2
(3, 4)
3
4 − y2 = y − 2
2
A 4
y2 + y − 6 = 0
Matches (d)
(y
(0, 1)
Intersection points: (0, 2) and ( −5, − 3)
x
14. f ( x ) = 2 −
3
2
1
A =
1x
2
g ( x) = 2 −
+ 3)( y − 2) = 0
=
x
∫ − 5 ⎡⎣( x + 2) +
0
+
61
6
32
3
=
4 − x dx +
4
∫0 2
4 − x dx
125
6
y
A ≈1
6
Matches (a)
4
(0, 2)
y
4
x
−6 −4
6
(−5, −3)
3
−6
(0, 2)
(b) A =
1
(4, 0)
2
∫ − 3 ⎡⎣(4 − y ) − ( y − 2)⎤⎦ dy
2
=
125
6
x
2
1
3
(c) The second method is simpler. Explanations will
vary.
16. (a) y = x 2 and y = 6 − x
x 2 = 6 − x ⇒ x 2 + x − 6 = 0 ⇒ ( x + 3)( x − 2) = 0
Intersection points: ( 2, 4) and ( −3, 9)
y
10
(−3, 9)
8
6
(2, 4)
4
−6 −4 −2
A =
(b) A =
x
2
−2
4
6
∫ − 3 ⎡⎣(6 − x) −
2
4
∫0 2
y dy +
x 2 ⎤⎦ dx =
125
6
∫ 4 ⎡⎣(6 − y) +
9
y dy =
+
32
3
61
6
=
125
6
(c) The first method is simpler. Explanations will vary.
y
17.
A =
6
=
4
x
−2
2
−2
1
∫ 0 (− x
1
2
2
− 1)⎤⎦ dx
− x + 3) dx
1
2
−4
∫ 0 ⎡⎣(− x + 2) − ( x
4
⎡ − x3
⎤
x2
= ⎢
−
+ 3 x⎥
3
2
⎣
⎦0
13
⎛ 1 1
⎞
= ⎜ − − + 3⎟ − 0 =
6
⎝ 3 2
⎠
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 7.1
y
18.
Area of a Region Between Two Curves
20. The points of intersection are given by:
− x 2 + 3x + 1 = − x + 1
4
(−1, 3)
− x2 + 4x = 0
(1, 1)
−4
x( 4 − x) = 0 when x = 0, 4
x
−2
4
−2
y
(1, −2)
(−1, −4)
A =
=
621
4
∫ −1 ⎡⎣(− x
1
∫ −1 (− x
1
3
+ 2) − ( x − 3)⎤⎦ dx
3
(0, 1)
x
−2
− x + 5) dx
4
6
(4, −3)
1
⎡ − x4
⎤
x2
= ⎢
−
+ 5 x⎥
2
⎣ 4
⎦ −1
A =
1
1
⎛ 1
⎞ ⎛ 1
⎞
= ⎜ − − + 5 ⎟ − ⎜ − − − 5⎟ = 10
2
2
⎝ 4
⎠ ⎝ 4
⎠
=
∫ 0 ⎡⎣(− x
4
∫ 0 (− x
4
+ 3 x + 1) − (1 − x)⎤⎦ dx
2
+ 4 x) dx
2
4
⎡ − x3
⎤
= ⎢
+ 2x2 ⎥
⎣ 3
⎦0
64
32
= −
+ 32 =
3
3
19. The points of intersection are given by:
x2 + 2 x = x + 2
x2 + x − 2 = 0
(x
+ 2)( x − 1) = 0 when x = −2, 1
21. The points of intersection are given by:
y
6
x = 2 − x and
x = 0 and 2 − x = 0
x =1
x = 0
x = 2
4
y
(1, 3)
2
(−2, 0)
3
x
−4
2
4
2
−2
A =
(1, 1)
1
∫ − 2 ⎡⎣g ( x) − f ( x)⎤⎦ dx
1
(2, 0)
(0, 0) 1
= ∫ ⎡⎣( x + 2) − ( x 2 + 2 x)⎤⎦ dx
−2
1
A =
1
⎡ − x3
⎤
x2
= ⎢
−
+ 2 x⎥
3
2
⎣
⎦ −2
9
⎛ 1 1
⎞ ⎛8
⎞
= ⎜ − − + 2⎟ − ⎜ − 2 − 4⎟ =
3
2
3
2
⎝
⎠ ⎝
⎠
2
x
3
∫ 0 ⎡⎣(2 − y) − ( y)⎤⎦ dy
1
1
= ⎡⎣2 y − y 2 ⎤⎦ = 1
0
Note that if you integrate with respect to x, you need two
integrals. Also, note that the region is a triangle.
22.
y
(1, 4)
4
3
2
(4, 161 (
1
x
1
A =
2
4
∫1
3
4
dx =
x3
4
4
∫1
4 x −3 dx
4
= ⎡⎣− 2 x −2 ⎤⎦
1
4
⎡− 2⎤
= ⎢ 2⎥
⎣ x ⎦1
2
15
= −
+ 2 =
16
8
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
622
Chapter 7
Applications of Integration
23. The points of intersection are given by:
25. The points of intersection are given by:
y2 = y + 2
1
x +3
2
1
x = x
2
x2
x =
when x = 0, 4
4
x +3 =
(y
− 2)( y + 1) = 0 when y = −1, 2
y
3
(4, 2)
2
1
y
x
6
1
(4, 5)
2
3
4
5
−1
5
(1, −1)
4
−3
(0, 3)
2
1
A =
A =
x
−2 −1
−1
1
⎡
∫ 0 ⎢⎣(
4
2
3
4
5
=
⎛1
⎞⎤
x + 3 − ⎜ x + 3⎟⎥ dx
⎝2
⎠⎦
)
∫ −1 ⎡⎣( y + 2) −
2
y 2 ⎤⎦ dy
2
26. The points of intersection are given by:
2 y − y2 = − y
24. The points of intersection are given by:
3
2
⎡
9
y2
y3 ⎤
= ⎢2 y +
−
⎥ =
2
3 ⎦ −1
2
⎣
4
⎡2
16
4
x2 ⎤
= ⎢ x3 2 − ⎥ =
−4 =
4 ⎦0
3
3
⎣3
∫ −1 ⎡⎣g ( y) − f ( y)⎤⎦ dy
y( y − 3) = 0 when y = 0, 3
x −1 = x −1
x − 1 = ( x − 1) = x 3 − 3x 2 + 3 x − 1
3
y
(−3, 3)
x3 − 3 x 2 + 2 x = 0
3
x ( x 2 − 3 x + 2) = 0
x( x − 2)( x − 1) = 0
when x = 0, 1, 2
1
(0, 0)
y
−3
x
−2
1
−1
1
(2, 1)
(1, 0)
A =
x
2
1
3
x − 1⎤⎦ dx
2
∫ 0 ⎡⎣(2 y − y ) − (− y)⎤⎦ dy
=
2
∫ 0 (3 y − y ) dy
⎡⎛ 1
1
⎞ ⎛ 3 ⎞⎤
= 2 ⎢⎜ − 1 − 0 ⎟ − ⎜ − ⎟⎥ =
2
⎠ ⎝ 4 ⎠⎦
⎣⎝ 2
3
3
= ⎡⎣ 32 y 2 −
1
⎡ x2
3
4 3⎤
= 2⎢
− x − ( x − 1) ⎥
2
4
⎣
⎦0
3
=
(0, −1)
A = 2∫ ⎡⎣( x − 1) −
0
∫ 0 ⎡⎣ f ( y) − g ( y)⎤⎦ dy
3
1 3⎤
y
3 ⎦0
=
9
2
y
27.
3
(0, 2)
(5, 2)
1
x
2
−2
3
4
(0, −1)
A =
=
5
6
(2, −1)
∫ −1 ⎡⎣ f ( y) − g ( y)⎤⎦ dy
2
∫ −1 ⎡⎣( y
2
2
+ 1) − 0⎤⎦ dy
2
⎡ y3
⎤
= ⎢
+ y⎥ = 6
3
⎣
⎦ −1
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Section 7.1
28.
y
Area of a Region Between Two Curves
10
10
⇒ x =
x
y
29. y =
4
(
3
3
,3
7
623
y
)
A =
12
2
(0, 10)
1
8
(1, 10)
10 10
∫2
y
dy
= [10 ln y]2
10
= 10(ln 10 − ln 2)
6
x
−1
1
4
3
2
(0, 2)
A =
=
∫ 0 ⎡⎣ f ( y) − g ( y)⎤⎦ dy
= 10 ln 5 ≈ 16.0944
(5, 2)
3
3⎡
y
∫ 0 ⎢⎢
⎣ 16 − y
= −
2
⎤
− 0⎥ dy
⎥⎦
1
(16 − y 2 )
2∫0
3
−1 2
x
−4 −2
= ⎡− 16 − y 2 ⎤ = 4 −
⎣
⎦0
4
6
8
30. The point of intersection is given by:
4
= 4
2 − x
(−2 y ) dy
3
2
4
− 4 = 0
2 − x
7 ≈ 1.354
when x = 1
y
A =
(1, 4)
4 ⎞
⎟ dx
2 − x⎠
1
= ⎡⎣4 x + 4 ln 2 − x ⎤⎦ 0
3
= 4 − 4 ln 2
(0, 2)
≈ 1.227
1
x
−1
31. (a)
1⎛
∫ 0 ⎜⎝ 4 −
1
3
11
(3, 9)
−6
(0, 0)
(1, 1)
12
−1
(b) The points of intersection are given by:
x3 − 3 x 2 + 3 x = x 2
x( x − 1)( x − 3) = 0
A=
=
=
when x = 0, 1, 3
∫ 0 ⎡⎣ f ( x) − g ( x)⎤⎦ dx + ∫ 1 ⎡⎣g ( x) − f ( x)⎤⎦ dx
1
3
∫ 0 ⎡⎣( x
1
1
∫0
3
− 3 x 2 + 3x) − x 2 ⎤⎦ dx +
( x3 − 4 x 2 + 3x) dx +
3
∫ 1 ⎡⎣ x
2
− ( x3 − 3x 2 + 3 x)⎤⎦ dx
1
3
∫1
3
⎡ 4
⎤
⎡ 4
⎤
37
(− x3 + 4 x 2 − 3x) dx = ⎢ x4 − 43 x3 + 32 x 2 ⎥ + ⎢ −4x + 43 x3 − 32 x 2 ⎥ = 125 + 83 = 12
⎣
⎦0 ⎣
⎦1
(c) Numerical approximation: 0.417 + 2.667 ≈ 3.083
32. (a)
(b) The points of intersection are given by:
10
(−2, 8)
(2, 8)
−4
x4 − 2x2 = 2x2
4
−2
x 2 ( x 2 − 4) = 0
when x = 0, ± 2
2
(0, 0)
2
2
⎡ 4 x3
x5 ⎤
128
A = 2 ∫ ⎡⎣2 x 2 − ( x 4 − 2 x 2 )⎤⎦ dx = 2 ∫ ( 4 x 2 − x 4 ) dx = 2 ⎢
− ⎥ =
0
0
5 ⎦0
15
⎣ 3
(c) Numerical approximation: 8.533
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
624
Chapter 7
Applications of Integration
33. (a) f ( x) = x 4 − 4 x 2 , g ( x) = x 2 − 4
2
−4
(−2, 0)
(2, 0)
(−1, −3)
4
(1, − 3)
−5
(b) The points of intersection are given by:
x4 − 4x2 = x2 − 4
x4 − 5x2 + 4 = 0
( x2
− 4)( x 2 − 1) = 0 when x = ± 2, ±1
By symmetry:
A = 2 ∫ ⎡⎣( x 4 − 4 x 2 ) − ( x 2 − 4)⎤⎦ dx + 2 ∫ ⎡⎣( x 2 − 4) − ( x 4 − 4 x 2 )⎤⎦ dx
0
1
1
= 2∫
1
0
2
( x4
− 5 x 2 + 4) dx + 2 ∫
2
1
(− x 4
+ 5 x 2 − 4) dx
1
2
⎡ x5
⎤
⎡ x5
⎤
5 x3
5 x3
= 2⎢