CHAPTER 1
Introduction to Calculus
Review of Prerequisite Skills, pp. 2–3
27 2 5
622
5 23
4 2 (24)
b. m 5
21 2 3
5 22
420
c. m 5
120
54
420
d. m 5
21 2 0
5 24
4 2 4.41
e. m 5
22 2 (22.1)
5 24.1
21 2 1
f. m 5 7 4 3 4
4 2 4
2
2
4
5
1
1
52
2
2. a. Substitute the given slope and y-intercept into
y 5 mx 1 b.
y 5 4x 2 2
b. Substitute the given slope and y-intercept into
y 5 mx 1 b.
y 5 22x 1 5
c. The slope of the line is
12 2 6
m5
4 2 (21)
6
5
5
The equation of the line is in the form
y 2 y1 5 m(x 2 x1 ). The point is (21, 6) and
m 5 65.
The equation of the line is y 2 6 5 65 (x 1 1) or
y 5 65 (x 1 1) 1 6.
824
m5
d.
26 2 (22)
5 21
1. a. m 5
Calculus and Vectors Solutions Manual
y 2 4 5 21(x 2 (22))
y 2 4 5 2x 2 2
x1y2250
e. x 5 23
f. y 5 5
3. a. f(2) 5 26 1 5
5 21
b. f(2) 5 (8 2 2)(6 2 6)
50
c. f(2) 5 23(4) 1 2(2) 2 1
5 29
d. f(2) 5 (10 1 2)2
5 144
210
4. a. f(210) 5
100 1 4
5
52
52
23
b. f(23) 5
914
3
52
13
0
c. f(0) 5
014
50
10
d. f(10) 5
100 1 4
5
5
52
5. f(x) 5 •
"3 2 x, if x , 0
"3 1 x, if x $ 0
a. f(233) 5 6
b. f(0) 5 "3
c. f(78) 5 9
d. f(3) 5 "6
1
, if 23 , t , 0
t
6. s(t) 5 μ
5, if t 5 0
t3, if t . 0
1
a. s(22) 5 2
2
b. s(21) 5 21
1-1
c. s(0) 5 5
d. s(1) 5 1
e. s(100) 5 1003 or 106
7. a. (x 2 6)(x 1 2) 5 x 2 2 4x 2 12
b. (5 2 x)(3 1 4x) 5 15 1 17x 2 4x 2
c. x(5x 2 3) 2 2x(3x 1 2) 5 5x 2 2 3x 2 6x 2 2 4x
5 2x 2 2 7x
d. (x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)
5 x 2 1 2x 2 3 2 (2x 2 1 x 2 10)
5 2x 2 1 x 1 7
e. (a 1 2)3 5 (a 1 2)(a 1 2)(a 1 2)
5 (a 2 1 4a 1 4)(a 1 2)
5 a 3 1 6a 2 1 12a 1 8
f. (9a 2 5)3 5 (9a 2 5)(9a 2 5)(9a 2 5)
5 (81a 2 2 90a 1 25)(9a 2 5)
5 729a 3 2 1215a 2 1 675a 2 125
3
8. a. x 2 x 5 x(x 2 2 1)
5 x(x 1 1)(x 2 1)
2
b. x 1 x 2 6 5 (x 1 3)(x 2 2)
c. 2x 2 2 7x 1 6 5 (2x 2 3)(x 2 2)
d. x 3 1 2x 2 1 x 5 x(x 2 1 2x 1 1)
5 x(x 1 1)(x 1 1)
3
e. 27x 2 64 5 (3x 2 4)(9x 2 1 12x 1 16)
f. 2x 3 2 x 2 2 7x 1 6
x 5 1 is a zero, so x 2 1 is a factor. Synthetic or
long division yields
2x 3 2 x 2 2 7x 1 6 5 (x 2 1)(2x 2 1 x 2 6)
5 (x 2 1)(2x 2 3)(x 1 2)
9. a. 5xPR 0 x $ 256
b. 5xPR6
c. 5xPR 0 x 2 16
d. 5xPR 0 x 2 06
e. 2x 2 2 5x 2 3 5 (2x 1 1)(x 2 3)
1
e xPR ` x 2 2 , 3 f
2
f. 5xPR 0 x 2 25, 22, 16
10. a. h(0) 5 2, h(1) 5 22.1
22.1 2 2
average rate of change 5
120
5 20.1 m> s
b. h(1) 5 22.1, h(2) 5 32.4
32.4 2 22.1
average rate of change 5
221
5 10.3 m> s
11. a. The average rate of change during the second
hour is the difference in the volume at t 5 120 and
t 5 60 (since t is measured in minutes), divided by
the difference in time.
1-2
V(120) 2 V(60)
0 2 1200
5
120 2 60
60
5 220 L>min
b. To estimate the instantaneous rate of change in
volume after exactly 60 minutes, calculate the average
rate of change in volume from minute 59 to minute 61.
1186.56 2 1213.22
V(61) 2 V(59)
8
61 2 59
2
5 213.33 L>min
c. The instantaneous rate of change in volume is
negative for 0 # t # 120 because the volume of
water in the hot tub is always decreasing during that
time period, a negative change.
y
12. a., b.
8
4
0
–2
x
2
4
6
–4
–8
The slope of the tangent line is 28.
c. The instantaneous rate of change in f(x) when
x 5 5 is 28.
1.1 Radical Expressions:
Rationalizating Denominators, p. 9
1. a. 2"3 1 4
b. "3 2 "2
c. 2"3 1 "2
d. 3"3 2 "2
e. "2 1 "5
f. 2"5 2 2"2
2. a.
"3 1 "5
"2
"6 1 "10
5
2
b.
2"3 2 3"2
"2
2"6 2 6
5
2
5 "6 2 3
?
?
"2
"2
"2
"2
Chapter 1: Introduction to Calculus
c.
4"3 1 3"2
?
2"3
12 1 3"6
5
6
5
d.
"3
"3
f.
3"3 2 2"2
3"3 1 2"2
4 1 "6
2
3"5 2 "2
2"2
?
"2
"2
4. a.
"5 2 1
4
521
5
b.
3("5 1 "2 )
3
5 "5 1 "2
2"5
2"5 1 3"2
20 2 6"10
20 2 18
5 10 2 3"10
5
5
?
2"5 2 3"2
2"5 2 3"2
5
5
"3 2 "2
c.
?
"3 2 "2
"3 1 "2 "3 2 "2
3 1 2"6 1 2
5
322
5 5 1 2"6
d.
5
2"5 2 8
2"5 1 3
5
?
20 2 22"5 1 24
20 2 9
44 2 22"5
11
5 4 2 2"5
2"3 2 "2
5"2 1 "3
"5 1 1
"5 1 1
2( 2 1 3"2 )
27
2 1 3"2
"5 1 2
?
"5 2 2
2"5 2 1 "5 2 2
524
10 2 5"5 1 2
1
5
12 2 5 !5
2"5 2 3
2"5 2 3
?
2 2 3"2 2 1 3"2
?
2
2 1 3"2
4 2 18
5. a.
8"2
"20 2 "18
5
8"40 1 8"36
20 2 18
5
16"10 1 48
2
?
"20 1 "18
"20 1 "18
5 8"10 1 24
5
e.
3"3 2 2"2
27 2 12"6 1 8
27 2 8
35 2 12"6
5
19
4("5 1 1)
1
5
!5 1 1
c.
3"3 2 2"2
5
3"10 2 2
5
4
"5 1 "2
3
3. a.
?
"5 2 "2 "5 1 "2
b.
?
?
5"2 2 "3
5"2 2 "3
10"6 2 6 2 10 1 "6
50 2 3
11"6 2 16
5
47
5
Calculus and Vectors Solutions Manual
b.
8"2
2"5 2 3"2
5
16"10 1 48
20 2 18
5
16"10 1 48
2
?
2"5 1 3"2
2"5 1 3"2
5 8"10 1 24
c. The expressions in the two parts are equivalent.
The radicals in the denominator of part a. have been
simplified in part b.
1-3
6. a.
2"2
2"3 2 "8
4"6 1 8
628
5 22"3 2 4
?
2"3 1 "8
2"3 1 "8
5
5
b.
2"6
2"27 2 "8
?
2"27 1 "8
2"27 1 "8
4"162 1 2"48
54 2 8
36"2 1 8"3
5
46
18"2 1 4"3
5
23
2"2
c.
"16 2 "12
5
5
5
2"2
4 2 2"3
?
4 1 2"3
4 1 2"3
8"2 1 4"6
16 2 12
5 2"2 1 "6
d.
3"2 1 2"3
"12 2 "8
?
"12 1 "8
"12 1 "8
3"24 1 12 1 12 1 2"24
5
12 2 8
5
24 1 15"3
4
e.
3 !5
4 !3 1 5!2
?
4!3 2 5 !2 4 !3 1 5!2
5
12"15 1 15"10
48 2 50
52
f.
5
12"15 1 15"10
2
"18 1 "12
"18 2 "12
?
"18 1 "12
"18 1 "12
18 1 2"216 1 12
18 2 12
30 1 12"6
6
5 5 1 2"6
5
1-4
7. a.
5
"a 2 2 "a 1 2
?
a24
"a 1 2
a24
( a 2 4 )("a 2 2)
1
"a 2 2
"x 1 4 2 2 "x 1 4 1 2
b.
?
x
"x 1 4 1 2
x1424
5
x("x 1 4 1 2)
x
5
x("x 1 4 1 2)
1
5
"x 1 4 2 2
!x 1 h 2 !x !x 1 h 1 !x
c.
?
h
!x 1 h 1 !x
x1h2x
5
hA !x 1 h 1 !xB
h
5
hA !x 1 h 1 !xB
1
5
!x 1 h 1 !x
1.2 The Slope of a Tangent, pp. 18–21
28 2 7
23 2 2
53
27 2 3
b. m 5 7 2 1 2
2 2 2
2 102
5 6
1. a. m 5
2
5
3
21 2 (22.6)
c. m 5
1.5 2 6.3
1
52
3
2. a. The slope of the given line is 3, so the slope
of a line perpendicular to the given line is 2 13.
b. 13x 2 7y 2 11 5 0
27y 5 213x 2 11
13
11
y5 x1
7
7
13
The slope of the given line is 7 , so the slope of a line
perpendicular to the given line is 2 137 .
52
Chapter 1: Introduction to Calculus
2 53 2 (24)
3. a. m 5 5
3 2 (24)
4
5
7
3
17
3
2
5
7
17
–2
7
(x 2 (24))
17
17y 1 68 5 7x 1 28
7x 2 17y 2 40 5 0
y
4
0
2
4
4
6
d. The line is a vertical line because both points
have the same x-coordinate.
x55
6 x
y
2
–2
–2
–4
2
–4
4
–2
x
0
–2
y 2 (24) 5
2
y
0
x
2
4
6
–2
b. The slope and y-intercept are given.
y 5 8x 1 6
y
8
–4
(5 1 h)3 2 125
h
(5 1 h 2 5)((5 1 h)2 1 5(5 1 h) 1 25)
5
h
2
h(75 1 15h 1 h )
5
h
5 75 1 15h 1 h 2
(3 1 h)4 2 81
b.
h
((3 1 h)2 2 9)((3 1 h)2 1 9)
5
h
2
(9 1 6h 1 h 2 9)(9 1 6h 1 h 2 1 9)
5
h
5 (6 1 h)(18 1 6h 1 h 2 )
5 108 1 54h 1 12h 2 1 h 3
1
21
1212h
1
c. 1 1 h
5
52
h
h(1 1 h)
11h
2
2
3(1 1 h) 2 3
3((1 1 h) 2 1)
d.
5
h
h
3(1 1 2h 1 h 2 2 1)
5
h
4. a.
4
–4
–2
0
x
2
4
–4
–8
c. (0, 23), (5, 0)
0 2 (23)
m5
520
3
5
5
3
y 2 0 5 (x 2 5)
5
3x 2 5y 2 15 5 0
Calculus and Vectors Solutions Manual
1-5
3(2h 1 h2 )
h
5 6 1 3h
5
e.
f.
3
4 1 h
2 34
h
21
2 1 h
h
5. a.
1 12
5
h
23
5
4(4 1 h)
5
22 1 2 1 h
2 (2 1 h)
h
h
5
2h(2 1 h)
1
5
4 1 2h
h
5
5
16 1 h 2 16
h( "16 1 h 1 4)
1
"16 1 h 1 4
"h 1 5h 1 4 2 2
h 2 1 5h 1 4 2 4
5
h
h("h 2 1 5h 1 4 1 2 )
h15
5
2
"h 1 5h 1 4 1 2
"5 1 h 2 "5
51h25
5
c.
h
h ("5 1 h 1 "5 )
2
b.
5
1
"5 1 h 1 "5
6. a. P(1, 3), Q(1 1 h, f(1 1 h)), f(x) 5 3x 2
3(1 1 h)2 2 3
m5
h
5 6 1 3h
b. P(1, 3), Q(1 1 h, (1 1 h)3 1 2)
(1 1 h)3 1 2 2 3
m5
h
1 1 3h 1 3h 2 1 h 3 2 1
5
h
5 3 1 3h 1 h 2
c. P(9, 3), Q (9 1 h, "9 1 h )
5
1-6
"9 1 h 2 3 "9 1 h 1 3
?
h
"9 1 h 1 3
1
"9 1 h 1 3
Q
P
12 2 12 2 3h
4 (4 1 h)
"16 1 h 2 4
m5
7. a.
Slope of Line PQ
(2, 8)
(3, 27)
19
(2, 8)
(2.5, 15.625)
15.25
(2, 8)
(2.1, 9.261)
12.61
(2, 8)
(2.01, 8.120 601)
12.060 1
(2, 8)
(1, 1)
(2, 8)
(1.5, 3.375)
9.25
(2, 8)
(1.9, 6.859)
11.41
(2, 8)
(1.99, 7.880 599)
11.940 1
7
b. 12
c. (2, 8), ((2 1 h), (2 1 h)3 )
(2 1 h)3 2 8
m5
21h22
8 1 12h 1 6h 2 1 h 3 2 8
5
h
5 12 1 6h 1 h 2
d. m 5 lim (12 1 6h 1 h 2 )
hS0
5 12
e. They are the same.
f.
y
12
8
4
–4
–2
0
x
2
4
–4
8. a. y 5 3x 2, (22, 12)
3(22 1 h)2 2 12
m 5 lim
hS0
h
12 2 12h 1 3h 2 2 12
5 lim
hS0
h
5 lim (212 1 3h)
hS0
5 212
b. y 5 x 2 2 x at x 5 3, y 5 6.
(3 1 h)2 2 (3 1 h) 2 6
m 5 lim
hS0
h
9 1 6h 1 h 2 2 3 2 h 2 6
5 lim
hS0
h
5 lim (5 1 h)
hS0
55
Chapter 1: Introduction to Calculus
c. y 5 x 3 at x 5 22, y 5 28.
(22 1 h)3 1 8
m 5 lim
hS0
h
28 1 12h 2 6h 2 1 h 3 1 8
5 lim
hS0
h
5 lim (12 2 6h 1 h 2 )
hS0
5 12
9. a. y 5 "x 2 2; (3, 1)
"3 1 h 2 2 2 1
hS0
h
m 5 lim
5 lim £
hS0
5 lim
"1 1 h 2 1
"1 1 h 1 1
3
§
h
"1 1 h 1 1
hS0 "1
1
1h11
1
2
b. y 5 "x 2 5 at x 5 9, y 5 2
5
"9 1 h 2 5 2 2
hS0
h
m 5 lim
5 lim £
hS0
"4 1 h 2 2
"4 1 h 1 2
3
§
h
"4 1 h 1 2
5 lim
hS0 "4
1
1h12
1
5
4
c. y 5 "5x 2 1 at x 5 2, y 5 3
"10 1 5h 2 1 2 3
hS0
h
m 5 lim
5 lim £
hS0
5 lim
"9 1 5h 2 3
"9 1 5h 1 3
3
§
h
"9 1 5h 1 3
5
hS0 "9
1 5h 1 3
5
5
6
8
at (2, 4)
x
8
24
m 5 lim 2 1 h
hS0
h
24
5 lim
hS0 2 1 h
5 22
8
b. y 5
at x 5 1; y 5 2
31x
8
22
m 5 lim 4 1 h
hS0
h
10. a. y 5
Calculus and Vectors Solutions Manual
5 lim
hS0
22
41h
1
2
1
1
c. y 5
at x 5 3; y 5
x12
5
1
1
2
m 5 lim 5 1 h 5
hS0
h
21
5 lim
hS0 5(5 1 h)
1
52
10
11. a. Let y 5 f(x).
52
f(2) 5 (2)2 2 3(2) 5 4 2 6 5 22
f(2 1 h) 5 (2 1 h)2 2 3(2 1 h)
Using the limit of the difference quotient, the slope
of the tangent at x 5 2 is
f(2 1 h) 2 f(2)
m 5 lim
hS0
h
(2 1 h)2 2 3(2 1 h) 2 (22)
5 lim
hS0
h
4 1 4h 1 h 2 2 6 2 3h 1 2
5 lim
hS0
h
h2 1 h
5 lim
hS0
h
5 lim (h 1 1)
hS0
5011
51
Therefore, the slope of the tangent to
y 5 f(x) 5 x 2 2 3x at x 5 2 is 1.
4
b. f(22) 5
5 22
22
4
f(22 1 h) 5
22 1 h
Using the limit of the difference quotient, the slope
of the tangent at x 5 22 is
f(22 1 h) 2 f(22)
m 5 lim
hS0
h
5 lim
hS0
5 lim
4
2 (22)
22 1 h
h
4
12
22 1 h
h
4 2 4 1 2h 1
5 lim c
? d
hS0
22 1 h
h
2h
1
5 lim c
? d
hS0 22 1 h
h
hS0
1-7
2
hS0 22 1 h
2
5
22 1 0
5 21
5 lim
1
!0 1 9 1 3
1
5
313
1
5
6
Therefore, the slope of the tangent to
y 5 f(x) 5 !x 2 7 at x 5 16 is 16.
e. Let y 5 f(x).
5
4
Therefore, the slope of the tangent to f(x) 5 x at
x 5 22 is 21.
c. Let y 5 f(x).
f(1) 5 3(1)3 5 3
f(1 1 h) 5 3(1 1 h)3
Using the limit of the difference quotient, the slope
of the tangent at x 5 1 is
f(1 1 h) 2 f(1)
m 5 lim
hS0
h
3(1 1 h)3 2 3
5 lim
hS0
h
Using the binomial formula to expand (1 1 h)3 (or
one could simply expand using algebra), the slope m is
3(h 3 1 3h 2 1 3h 1 1) 2 (3)
5 lim
hS0
h
3
2
3h 1 9h 1 9h 1 3 2 3
5 lim
hS0
h
3h 3 1 9h 2 1 9h
5 lim
hS0
h
5 lim (3h 2 1 9h 1 9)
f(3) 5 "25 2 (3)2 5 !25 2 9 5 4
f(3 1 h) 5 "25 2 (3 1 h)2
5 "25 2 (9 1 6h 1 h 2 )
5 "25 2 9 2 6h 2 h 2
5 "16 2 6h 2 h 2
Using the limit of the difference quotient, the slope
of the tangent at x 5 3 is
f(3 1 h) 2 f(3)
m 5 lim
hS0
h
"16 2 6h 2 h 2 2 4
5 lim
hS0
h
5 lim c
hS0
hS0
5 3(0) 1 9(0) 1 9
59
Therefore, the slope of the tangent to
y 5 f(x) 5 3x 3 at x 5 1 is 9.
d. Let y 5 f(x).
f(16) 5 !16 2 7 5 !9 5 3
f(16 1 h) 5 !16 1 h 2 7 5 !h 1 9
Using the limit of the difference quotient, the slope
of the tangent at x 5 16 is
f(16 1 h) 2 f(16)
m 5 lim
hS0
h
!h 1 9 2 3
5 lim
hS0
h
!h 1 9 2 3 !h 1 9 1 3
5 lim
?
hS0
h
!h 1 9 1 3
(h 1 9) 2 9
5 lim
hS0 h( !h 1 9 1 3)
h
5 lim
hS0 h( !h 1 9 1 3)
1
5 lim
hS0 !h 1 9 1 3
1-8
5 lim
hS0
5 lim
hS0
5 lim
hS0
5
"16 2 6h 2 h 2 2 4
h
"16 2 6h 2 h 2 1 4
3
d
"16 2 6h 2 h 2 1 4
16 2 6h 2 h 2 2 16
h("16 2 6h 2 h 2 1 4)
h(26 2 h)
h("16 2 6h 2 h 2 1 4)
26 2 h
"16 2 6h 2 h 2 1 4
26 2 0
"16 2 6(0) 2 (0)2 1 4
26
5
!16 1 4
26
5
8
3
52
4
Therefore, the slope of the tangent to
y 5 f(x) 5 "25 2 x 2 at x 5 3 is 2 34.
f. Let y 5 f(x).
12
418
5
52
f(8) 5
822
6
12 1 h
4 1 (8 1 h)
5
f(8 1 h) 5
(8 1 h) 2 2
61h
Chapter 1: Introduction to Calculus
Using the limit of the difference quotient, the slope
of the tangent at x 5 8 is
f(8 1 h) 2 f(8)
m 5 lim
hS0
h
12 1 h
22
5 lim 6 1 h
hS0
h
12 1 h 2 12 2 2h 1
?
5 lim
hS0
61h
h
2h
1
?
5 lim
hS0 6 1 h
h
21
5 lim
hS0 6 1 h
21
5
610
1
52
6
Therefore, the slope of the tangent to
41x
y 5 f(x) 5 x 2 2 at x 5 8 is 2 16.
12.
y
8
4
0
–4
A
x
4
8
–4
y 5 "25 2 x 2 S Semi-circle centre (0, 0)
rad 5, y $ 0
OA is a radius.
The slope of OA is 43.
The slope of tangent is 2 34.
13. Take values of x close to the point, then
Dy
determine Dx.
14.
5 lim (3 1 h)
hS0
53
The slope of the tangent is 3.
y 2 1 5 3(x 2 3)
3x 2 y 2 8 5 0
(2 1 h)2 2 7(2 1 h) 1 12 2 2
16. m 5 lim
hS0
h
2
4 1 4h 1 h 2 14 2 7h 1 10
5 lim
hS0
h
2
23h 1 h
5 lim
hS0
h
5 lim ( 2 3 1 h)
hS0
5 23
The slope of the tangent is 23.
When x 5 2, y 5 2.
y 2 2 5 23(x 2 2)
3x 1 y 2 8 5 0
17. a. f(3) 5 9 2 12 1 1 5 22; (3, 22)
b. f(5) 5 25 2 20 1 1 5 6; (5, 6)
c. The slope of secant AB is
6 2 (22)
mAB 5
523
8
5
2
54
The equation of the secant is
y 2 y1 5 mAB (x 2 x1 )
y 1 2 5 4(x 2 3)
y 5 4x 2 14
d. Calculate the slope of the tangent.
f(x 1 h) 2 f(x)
m 5 lim
hS0
h
(x 1 h)2 2 4(x 1 h) 1 1 2 (x2 2 4x 1 1)
5 lim
hS0
h
x2 1 2xh 1 h2 2 4x 2 4h 1 1 2 x2 1 4x 2 1
5 lim
hS0
h
2xh 1 h2 2 4h
5 lim
hS0
h
5 lim (2x 1 h 2 4)
hS0
Since the tangent is horizontal, the slope is 0.
(3 1 h)2 2 3(3 1 h) 1 1 2 1
15. m 5 lim
hS0
h
9 1 6h 1 h2 2 9 2 3h
5 lim
hS0
h
3h 1 h2
5 lim
hS0
h
Calculus and Vectors Solutions Manual
5 2x 1 0 2 4
5 2x 2 4
When x 5 3, the slope is 2(3) 2 4 5 2. So the
equation of the tangent at A(3, 22) is
y 2 y1 5 m(x 2 x1 )
y 1 2 5 2(x 2 3)
y 5 2x 2 8
1-9
e. When x 5 5, the slope of the tangent is
2(5) 2 4 5 6.
So the equation of the tangent at B(5, 6) is
y 2 y1 5 m(x 2 x1 )
y 2 6 5 6(x 2 5)
y 5 6x 2 24
18. a.
P
The slope is undefined.
b.
P
The slope is 0.
c.
P
The slope is about –2.5.
d.
P
The slope is about 1.
e.
P
The slope is about 2 78.
f. There is no tangent at this point.
20
, p . 1 at (5, 10)
19. D(p) 5
"p 2 1
m 5 lim
hS0
20
2 10
!4 1 h
5 10 lim
hS0
5 10 lim
hS0
10
8
5
52
4
52
1-10
h
2 2 "4 1 h
3
2 1 "4 1 h
h"4 1 h
2 1 "4 1 h
4242h
h"4 1 h( 2 1 "4 1 h )
20. C(t) 5 100t 2 1 400t 1 5000
Slope at t 5 6
Cr(t) 5 200t 1 400
Cr(6) 5 1200 1 400 5 1600
Increasing at a rate of 1600 papers per month.
21. Point on f(x) 5 3x 2 2 4x tangent parallel to
y 5 8x. Therefore, tangent line has slope 8.
3(h 1 a)2 2 4(h 1 a) 2 3(a 2 1 4a)
m 5 lim
58
hS0
h
3h 2 1 6ah 2 4h
58
lim
hS0
h
6a 2 4 5 8
a52
The point has coordinates (2, 4).
1
4
22. y 5 x 3 2 5x 2
x
3
1
1
1
(a 1 h)2 2 a 3 5 a 2h 1 ah 2 1 h 3
3
3
3
1
lim aa 2 1 ah 1 h3 b 5 a 2
hS0
3
(a 1 h) 2 (2a)
5 lim 2
5 25
hS0
h
4
4
4a 1 4a 1 4h
2
1 52
a
a1h
a(a 1 h)
4
4
lim
5 2
hS0 a(a 1 h)
a
4
m 5 a2 2 5 1 2 5 0
a
a 4 2 5a 2 1 4 5 0
2
(a 2 4)(a 2 2 1) 5 0
a 5 62, a 5 61
Points on the graph for horizontal tangents are:
( 22, 283) , ( 21, 263) , ( 1, 2 263) , ( 2, 2 283) .
1
23. y 5 x 2 and y 5 2 x 2
2
1
x2 5 2 x2
2
1
x2 5
4
1
1
x 5 or x 5 2
2
2
The points of intersection are
P( 12, 14) , Q( 2 12, 14) .
Tangent to y 5 x2:
(a 1 h)2 2 a 2
m 5 lim
hS0
h
2ah 1 h 2
5 lim
hS0
h
5 2a.
Chapter 1: Introduction to Calculus
The slope of the tangent at a 5 12 is 1 5 mp,
at a 5 2 12 is 21 5 mq.
Tangents to y 5 12 2 x 2:
S 12 2 (a 1 h)2 T 2 S 12 2 a 2 T
m 5 lim
hS0
h
2
22ah 2 h
5 lim
hS0
h
5 22a.
The slope of the tangents at a 5 12 is 21 5 Mp;
at a 5 2 12 is 1 5 Mq
mpMp 5 21 and mqMq 5 21
Therefore, the tangents are perpendicular at the
points of intersection.
24. y 5 23x 3 2 2x, (21, 5)
23(21 1 h)3 2 2(21 1 h) 2 5
m 5 lim
hS0
h
23(21 1 3h 2 3h2 1 h3 ) 1 2 2 2h 2 5
5 lim
hS0
h
2
23(21 1 3h 2 3h 1 h3 ) 1 2 2 2h 2 5
5 lim
hS0
h
2
3 2 9h 1 9h 2 3h3 1 2 2 2h 2 5
5 lim
hS0
h
2
211h 1 9h 2 3h3
5 lim
hS0
h
5 lim (211 1 9h 2 3h2 )
hS0
5 211
The slope of the tangent is 211.
We want the line that is parallel to the tangent (i.e.
has slope 211) and passes through (2, 2). Then,
y 2 2 5 211(x 2 2)
y 5 211x 1 24
25. a. Let y 5 f(x).
f(a) 5 4a 2 1 5a 2 2
f(a 1 h) 5 4(a 1 h)2 1 5(a 1 h) 2 2
5 4(a 2 1 2ah 1 h 2 ) 1 5a 1 5h 2 2
5 4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2
Using the limit of the difference quotient, the slope
of the tangent at x 5 a is
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2
5 lim c
hS0
h
(4a 2 1 5a 2 2)
2
d
h
Calculus and Vectors Solutions Manual
5 lim c
4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2
hS0
h
24a 2 2 5a 1 2
1
d
h
8ah 1 4h 2 1 5h
5 lim
hS0
h
5 lim (8a 1 4h 1 5)
hS0
5 8a 1 4(0) 1 5
5 8a 1 5
b. To be parallel, the point on the parabola and the
line must have the same slope. So, first find the
slope of the line. The line 10x 2 2y 2 18 5 0 can
be rewritten as
22y 5 18 2 10x
18 2 10x
y5
22
y 5 29 1 5x
y 5 5x 2 9
So, the slope, m, of the line 10x 2 2y 2 18 5 0 is 5.
To be parallel, the slope at a must equal 5. From
part a., the slope of the tangent to the parabola at
x 5 a is 8a 1 5.
8a 1 5 5 5
8a 5 0
a50
Therefore, at the point (0, 22) the tangent line is
parallel to the line 10x 2 2y 2 18 5 0.
c. To be perpendicular, the point on the parabola
and the line must have slopes that are negative
reciprocals of each other. That is, their product must
equal 21. So, first find the slope of the line. The
line x 2 35y 1 7 5 0 can be rewritten as
235y 5 2x 2 7
2x 2 7
y5
235
1
7
y5 x1
35
35
So, the slope, m, of the line x 2 35y 1 7 5 0 is 351 .
To be perpendicular, the slope at a must equal
the negative reciprocal of the slope of the line
x 2 35y 1 7 5 0. That is, the slope of a must equal
235. From part a., the slope of the tangent to the
parabola at x 5 a is 8a 1 5.
8a 1 5 5 235
8a 5 240
a 5 25
Therefore, at the point (25, 73) the tangent line is
perpendicular to the line x 2 35y 1 7 5 0.
1-11
1.3 Rates of Change, pp. 29–31
1. v(t) 5 0 when t 5 0 or t 5 4.
2. a.
s(9) 2 s(2)
.
7
Slope of the secant between the
points (2, s(2)) and (9, s(9)).
s(6 1 h) 2 s(6)
.
h
hS0
b. lim
Slope of the tangent at the
point (6, s(6)).
"4 1 h 2 2
.
h
hS0
3. lim
Slope of the tangent to the
function with equation y 5 !x at the point (4, 2).
4. a. A and B
b. greater; the secant line through these two points
is steeper than the tangent line at B.
c. y
y = f(x)
B
A
C
D
E x
5. Speed is represented only by a number, not a
direction.
6. Yes, velocity needs to be described by a number
and a direction. Only the speed of the school bus
was given, not the direction, so it is not correct to
use the word “velocity.”
7. s(t) 5 320 2 5t 2, 0 # t # 8
a. Average velocity during the first second:
s(1) 2 s(0)
5 5 m>s;
1
third second:
s(3) 2 s(2)
45 2 20
5
5 25 m>s;
1
1
eighth second:
s(8) 2 s(7)
320 2 245
5
5 75 m>s.
1
1
b. Average velocity 3 # t # 8
s(8) 2 s(3)
320 2 45
275
5
5
5 55 m>s
823
5
5
c. s(t) 5 320 2 5t 2
320 2 5(2 1 h)2 2 (320 2 5(2)2 )
v(t) 5 lim
hS0
h
24h 1 h 2
5 5 lim
hS0
h
5 220
Velocity at t 5 2 is 20 m>s downward.
8. s(t) 5 8t(t 1 2), 0 # t # 5
a. i. from t 5 3 to t 5 4
s(4) 2 s(3)
Average velocity
1
1-12
5 32(6) 2 24(5)
5 24(8 2 5)
5 72 km>h
ii. from t 5 3 to t 5 3.1
s(3.1) 2 s(3)
0.1
126.48 2 120
5
0.1
5 64.8 km>h
iii. 3 # t # 3.01
s(3.01) 2 s(3)
0.01
5 64.08 km>h
b. Instantaneous velocity is approximately 64 km>h.
c. At t 5 3
s(t) 5 8t 2 1 16t
v(t) 5 16t 1 16
v(3) 5 48 1 16
5 64 km>h
9. a. N(t) 5 20t 2 t 2
N(3) 2 N(2)
1
51 2 36
5
1
5 15
15 terms are learned between t 5 2 and t 5 3.
20(2 1 h) 2 (2 1 h)2 2 36
b. lim
hS0
h
40 1 20h 2 4 2 4h 2 h2 2 36
5 lim
hS0
h
16h 2 h2
5 lim
hS0
h
5 lim (16 2 h)
hS0
5 16
At t 5 2, the student is learning at a rate of 16 terms>h.
10. a. M in mg in 1 mL of blood t hours after the
injection.
1
M(t) 5 2 t2 1 t; 0 # t # 3
3
Calculate the instantaneous rate of change when t 5 2.
2 1 (2 1 h)2 1 (2 1 h) 2 (2 43 1 2)
lim 3
hS0
h
4
4
1 2
2 2 h 2 3 h 1 2 1 h 1 43 2 2
5 lim 3 3
hS0
h
1
1 2
2 h 2 3h
5 lim 3
hS0
h
1
1
5 lim a2 2 hb
hS0
3
3
1
52
3
Chapter 1: Introduction to Calculus
Rate of change is 2 13 mg> h.
b. Amount of medicine in 1 mL of blood is being
dissipated throughout the system.
s
11. t 5
Å5
Calculate the instantaneous rate of change when
s 5 125.
lim
Ä
125 1 h
125
2Ä 5
5
h
hS0
5 lim
125 1 h
25
Ä 5
h
hS0
5 lim ≥
hS0
5 lim ≥
hS0
5 lim ≥
hS0
Ä
125 1 h
25
5
h
125 1 h
2 25
5
ha
125 1 h 2 125
5
ha
5
125 1 h
15
5
125 1 h
15
Ä 5
125 1 h
1 5b
Ä 5
125 1 h
1 5b
Ä 5
¥
¥
5a
Ä
1
¥
125 1 h
1 5b
5
1 5b
Ä 5
1
5
5(5 1 5)
1
5
50
At s 5 125, rate of change of time with respect to
height is 501 s>m.
5a
125
60
12. T(h) 5 h 1 2
Calculate the instantaneous rate of change when
h 5 3.
60
60
2 (3 1 2)
(3 1 k) 1 2
lim
kS0
k
5 lim
kS0
k
212k
5 lim
kS0 k(5 1 k)
212
5 lim
kS0 (5 1 k)
12
52
5
Temperature is decreasing at 125 °C> km.
13. h 5 25t 2 2 100t 1 100
When h 5 0, 25t 2 2 100t 1 100 5 0
t 2 2 4t 1 4 5 0
(t 2 2)2 5 0
t52
Calculate the instantaneous rate of change when t 5 2.
25(2 1 h)2 2 100(2 1 h) 1 100 2 0
lim
hS0
h
100 1 100h 1 25h2 2 200 2 100h 1 100
5 lim
hS0
h
25h2
5 lim
hS0 h
5 lim 25h
kS0
hS0
1
5 lim
hS0
?
Ä
5 lim
60
60 1 12k
2
51k
51k
50
It hit the ground in 2 s at a speed of 0 m> s.
14. Sale of x balls per week:
P(x) 5 160x 2 x 2 dollars.
a. P(40) 5 160(40) 2 (40)2
5 4800
Profit on the sale of 40 balls is $4800.
b. Calculate the instantaneous rate of change when
x 5 40.
160(40 1 h) 2 (40 1 h)2 2 4800
lim
hS0
h
6400 1 160h 2 1600 2 80h 2 h2 2 4800
5 lim
hS0
h
80h 2 h2
5 lim
hS0
h
5 lim (80 2 h)
hS0
5 80
Rate of change of profit is $80 per ball.
c.
60
2 12
51k
k
Calculus and Vectors Solutions Manual
Rate of change of profit is positive when the sales
level is less than 80.
1-13
15. a. f(x) 5 2x2 1 2x 1 3; (22, 25)
f(x) 2 f(22)
lim
xS22
x12
2x 2 1 2x 1 3 1 5
5 lim
xS22
x12
2
2 (x 2 2x 2 8)
5 lim
xS22
x12
(x 2 4)(x 1 2)
5 2 lim
xS22
x12
5 2 lim (x 2 4)
xS22
56
x
,x52
x21
x
22
x21
lim
xS2
x22
x 2 2x 1 2
5 lim
xS2 (x 2 1)(x 2 2)
2 (x 2 2)
5 lim
xS2 (x 2 1)(x 2 2)
5 21
c. f(x) 5 !x 1 1, x 5 24
f(x) 2 f(24)
5 lim
xS24
x 2 24
!x 1 1 2 5 !x 1 1 1 5
5 lim
?
xS24
x 2 24
!x 1 1 1 5
x 2 24
5 lim
xS24 (x 2 24)( !x 1 1 1 5)
1
5
10
16. S(x) 5 246 1 64x 2 8.9x 2 1 0.95x3
b. f(x) 5
For the year 2005, x 5 2005 2 1982 5 23. Hence,
the rate at which the average annual salary is changing
in 2005 is
P r(23) 5 64 2 17.8(23) 1 2.85(23)2 5
$1 162 250> years since 1982
17. s(t) 5 3t 2
a. The distance travelled from 0 s to 5 s is
s(5) 5 3(5)2 5 75 m
b. s(10) 5 3(10)2 5 300 m
The rate at which the avalanche is moving from 0 s
to 10 s is
Ds
300 2 0
5
Dt
10 2 0
5 30 m> s
c. Calculate the instantaneous rate of change when
t 5 10.
3(10 1 h)2 2 300
lim
hS0
h
300 1 60h 1 3h2 2 300
5 lim
hS0
h
60h 1 3h2
5 lim
hS0
h
5 lim (60 1 3h)
hS0
5 60
At 10 s the avalanche is moving at 60 m> s.
d. Set s(t) 5 600:
3t 2 5 600
t 2 5 200
t 5 610 !2
Since t $ 0, t 5 10 !2 8 14 s.
Calculate the instantaneous rate of change.
S(x 1 h) 2 S(x)
5 lim
hS0
h
246 1 64(x 1 h) 2 8.9(x 1 h)2 1 0.95(x 1 h)3 2 (246 2 64x 2 8.9x2 1 0.95x3 )
5 lim
hS0
h
246 2 246 1 64(x 1 h 2 x) 2 8.9(x 2 1 2xh 1 h 2 2 x 2 ) 1 0.95(x3 1 3x2h 1 3xh2 1 h3 2 x3 )
5 lim
hS0
h
64h 2 8.9(2xh 1 h 2 ) 1 0.95(3x 2h 1 3xh 2 1 h 3 )
5 lim
hS0
h
5 lim 364 2 8.9(2x 1 h) 1 0.95(3x 2 1 3xh 1 h 2 )4
hS0
5 64 2 8.9(2x 1 0) 1 0.95 33x 2 1 3x(0) 1 (0)24
5 64 2 17.8x 1 2.85x 2
1-14
Chapter 1: Introduction to Calculus
18. The coordinates of the point are aa, a b . The slope
1
1
of the tangent is 2 a 2. The equation of the tangent
1
1
1
2
is y 2 a 5 2 a 2 (x 2 a) or y 5 2 a 2 x 1 a. The
intercepts are a0, a b and (22a, 0). The tangent line
and the axes form a right triangle with legs of length
2
2
a
and 2a. The area of the triangle is 2 a a b (2a) 5 2.
1 2
19. C(x) 5 F 1 V(x)
C(x 1 h) 5 F 1 V(x 1 h)
Rate of change of cost is
C(x 1 h) 2 C(x)
lim
xSR
h
V(x 1 h) 2 V(x)
5 lim
h,
xSh
h
which is independent of F (fixed costs).
20. A(r) 5 pr 2
Rate of change of area is
A(r 1 h) 2 A(r)
lim
hS0
h
p(r 1 h)2 2 pr 2
5 lim
hS0
h
(r 1 h 2 r)(r 1 h 1 r)
5 p lim
hS0
h
5 2pr
r 5 100 m
Rate is 200p m2> m.
21. Cube of dimensions x by x by x has volume
V 5 x 3. Surface area is 6x 2.
1
Vr(x) 5 3x 2 5 surface area.
2
22. a. The surface area of a sphere is given by
A(r) 5 4pr 2.
The question asks for the instantaneous rate of
change of the surface when r 5 10. This is
A(10 1 h) 2 A(10)
lim
hS0
h
4p(10 1 h)2 2 4p(10)2
5 lim
hS0
h
4p(100 1 20h 1 h 2 ) 2 4p(100)
5 lim
hS0
h
400p 1 80ph 1 4ph 2 2 400p
5 lim
hS0
h
2
80ph 1 4ph
5 lim
hS0
h
5 lim (80p 1 4ph)
hS0
Calculus and Vectors Solutions Manual
5 80p 1 4p(0)
5 80p
Therefore, the instantaneous rate of change of
the surface area of a spherical balloon as it is
inflated when the radius reaches 10 cm is
80p cm2> unit of time.
b. The volume of a sphere is given by V(r) 5 43pr 3.
The question asks for the instantaneous rate of
change of the volume when r 5 5.
Note that the volume is deflating. So, find the rate
of the change of the volume when r 5 5 and then
make the answer negative to symbolize a deflating
spherical balloon.
V(5 1 h) 2 V(5)
lim
hS0
h
4
p(5
1 h)3 2 43 p(5)3
5 lim 3
hS0
h
Using the binomial formula to expand
(5 1 h)3 (or one could simply expand using
algebra), the limit is
4
p(h 3 1 15h 2 1 75h 1 125) 2 43 p(5)3
5 lim 3
hS0
h
4
3
2
ph
1
20ph
1
100ph
1 43 p(125)
5 lim 3
hS0
h
2 43 p(125)
h
4
3
ph
1
20ph 2 1 100ph
5 lim 3
hS0
h
4 2
5 lim a ph 1 20ph 1 100pb
hS0 3
4
5 p(0)2 1 20p(0) 1 100p
3
5 100p
Because the balloon is deflating, the instantaneous rate
of change of the volume of the spherical balloon when
the radius reaches 5 cm is 2100p cm3>unit of time.
Mid-Chapter Review pp. 32–33
1. a. Corresponding conjugate: !5 1 !2.
( !5 2 !2)( !5 1 !2)
5 ( !25 1 !10 2 !10 2 !4)
5522
53
b. Corresponding conjugate: 3!5 2 2!2.
(3 !5 1 2 !2)(3 !5 2 2 !2)
5 (9 !25 2 6!10 1 6 !10 2 4 !4)
5 9(5) 2 4(2)
5 45 2 8
5 37
1-15
c. Corresponding conjugate: 9 2 2 !5.
(9 1 2 !5)(9 2 2!5)
5 (81 2 18!5 1 18!5 2 4!25)
5 81 2 4(5)
5 81 2 20
5 61
d. Corresponding conjugate: 3 !5 1 2 !10.
(3!5 2 2!10)(3 !5 1 2 !10)
5 (9!25 1 6 !50 2 6!50 2 4 !100)
5 9(5) 2 4(10)
5 45 2 40
55
6 1 !2 !3
2. a.
?
!3
!3
6!3 1 !6
5
!9
6!3 1 !6
5
3
2!3 1 4 !3
b.
?
!3
!3
2!9 1 4 !3
5
!9
6 1 4!3
5
3
5
!7 1 4
?
c.
!7 2 4 !7 1 4
5( !7 1 4)
5
!49 1 4 !7 2 4!7 2 16
5( !7 1 4)
5
7 2 16
5( !7 1 4)
52
9
2!3
!3 1 2
d.
?
!3 2 2 !3 1 2
2 !9 1 4 !3
5
!9 1 2 !3 2 2 !3 2 4
6 1 4!3
5
324
6 1 4!3
5
21
5 22(3 1 2!3)
2 !3 2 4
5!3
?
e.
2 !3 1 4 2 !3 2 4
10 !9 2 20 !3
5
4!9 2 8 !3 1 8!3 2 16
30 2 20 !3
5
12 2 16
1-16
30 2 20 !3
24
10 !3 2 15
5
2
3 !2
2 !3 1 5
f.
?
2 !3 2 5 2 !3 1 5
3 !2(2 !3 1 5)
5
4 !9 1 10 !3 2 10 !3 2 25
3 !2(2 !3 1 5)
5
4(3) 2 25
5
3 !2(2 !3 1 5)
12 2 25
3 !2(2 !3 1 5)
5
213
3 !2(2!3 1 5)
52
13
!2 !2
3. a.
?
5
!2
!4
5
5 !2
2
5
5 !2
!3
!3
b.
?
6 1 !2 !3
!9
5
!3(6 1 !2)
3
5
!3(6 1 !2)
5
!7 2 4 !7 1 4
?
5
!7 1 4
!49 1 4 !7 2 4 !7 2 16
5
5( !7 1 4)
7 2 16
5
5( !7 1 4)
9
52
5( !7 1 4)
c.
2 !3 2 5 2 !3 1 5
?
3 !2
2 !3 1 5
4 !9 1 10 !3 2 10 !3 2 25
5
3 !2(2!3 1 5)
4(3) 2 25
5
3 !2(2!3 1 5)
13
12 2 25
52
5
3 !2(2!3 1 5)
3!2(2!3 1 5)
d.
Chapter 1: Introduction to Calculus
!3 2 !7 !3 1 !7
?
4
!3 1 !7
!9 1 !21 2 !21 2 !49
5
4( !3 1 !7)
327
5
4( !3 1 !7)
4
52
4( !3 1 !7)
1
52
( !3 1 !7)
2!3 1 !7 2 !3 2 !7
f.
?
5
2 !3 2 !7
4 !9 2 2 !21 1 2 !21 2 !49
5
5(2!3 2 !7)
4(3) 2 7
5
5(2!3 2 !7)
12 2 7
5
5(2!3 2 !7)
1
5
(2!3 2 !7)
2
4. a.
m52 ;
3
2
y 2 6 5 2 (x 2 0)
3
2
y2652 x
3
2
x1y2650
3
11 2 7
4
b.
m5
5 51
622
4
y 2 7 5 1(x 2 2)
y275x22
2x 1 y 2 5 5 0
x2y1550
m54
c.
y 2 6 5 4(x 2 2)
y 2 6 5 4x 2 8
24x 1 y 1 2 5 0
4x 2 y 2 2 5 0
1
m5
d.
5
1
y 2 (22) 5 (x 2 (21))
5
1
1
y125 x1
5
5
10
1
1
2 50
2 x1y1
5
5
5
e.
Calculus and Vectors Solutions Manual
1
9
2 x1y1 50
5
5
1
9
x2y2 50
5
5
x 2 5y 2 9 5 0
5. The slope of PQ is
f(1 1 h) 2 (21)
m 5 lim
hS0
(1 1 h) 2 1
2 (1 1 h)2 1 1
hS0
h
2 (1 1 2h 1 h 2 ) 1 1
5 lim
hS0
h
21 2 2h 2 h 2 1 1
5 lim
hS0
h
22h 2 h 2
5 lim
hS0
h
5 lim (22 2 h)
5 lim
hS0
5 22 2 (0)
5 22
So, the slope of PQ with f(x) 5 2x 2 is 22.
6. a. Unlisted y-coordinates for Q are found by
substituting the x-coordinates into the given function.
The slope of the line PQ with the given points is
given by the following: Let P 5 (x1, y1 ) and
y 2y
1
Q 5 (y1, y2 ). Then, the slope 5 m 5 2
.
x2 2 x1
P
Q
Slope of Line PQ
(21, 1)
(22, 6)
(21, 1)
(21.5, 3.25)
2 4.5
(21, 1)
(21.1, 1.41)
2 4.1
(21, 1)
(21.01, 1.040 1)
2 4.01
(21, 1)
(21.001, 1.004 001)
2 4.001
P
Q
Slope of Line PQ
(21, 1)
(0, 22)
(21, 1)
(20.5, 20.75)
2 3.5
(21, 1)
(20.9, 0.61)
2 3.9
(21, 1)
(20.99, 0.9601)
2 3.99
(21, 1)
(20.999, 0.996 001)
2 3.999
25
23
b. The slope from the right and from the left appear
to approach 24. The slope of the tangent to the
graph of f(x) at point P is about 24.
c. With the points P 5 (21, 1) and
Q 5 (21 1 h, f(21 1 h)), the slope, m, of PQ is
the following:
1-17
y2 2 y1
x2 2 x1
3(21 1 h)2 2 2(21 1 h) 2 24 2 (1)
5
(21 1 h) 2 (21)
1 2 2h 1 h 2 1 2 2 2h 2 2 2 1
5
21 1 h 1 1
h 2 2 4h
5
h
5h24
d. The slope of the tangent is lim f(x).
m5
4
x22
f(6 1 h) 2 f(6)
m 5 lim
hS0
h
c. y 5 f(x) 5
5 lim
h
hS0
5 lim
hS0
hS0
In this case, as h goes to zero, h 2 4 goes to
h 2 4 5 0 2 4 5 24. The slope of the tangent to
the graph of f(x) at the point P is 24.
e. The answers are equal.
4
4
2
61h22
622
5 lim
4
4
24
h14
h
4
21
h14
h
4 2 (h 1 4) 1
5 lim a
b
hS0
h14
h
hS0
f(23 1 h) 2 f(23)
hS0
h
3(23 1 h)2 1 3(23 1 h) 2 54 2 3(23)2 1 3(23) 2 54
5 lim
hS0
h
9 2 6h 1 h 2 2 9 1 3h 2 5 2 (9 2 9 2 5)
5 lim
hS0
h
h 2 2 3h 2 5 2 (25)
5 lim
hS0
h
2h 1
2
5 lim a
b
h 2 3h
hS0 h 1 4 h
5 lim
hS0
h
21
5 lim
5 lim (h 2 3)
hS0 h 1 4
hS0
5023
21
5
5 23
014
1
1
52
b. y 5 f(x) 5
4
x
f(5 1 h) 2 f(5)
f( 13 1 h) 2 f( 13 )
d. m 5 lim
m 5 lim
hS0
h
hS0
h
!5 1 h 1 4 2 !5 1 4
1
1
5 lim
2 1
1
hS0
h
3 1 h
3
5 lim
!9
1
h
2
!9
hS0
h
5 lim
1
1
hS0
h
( 3 ) 2 ( 3 1 h)
1 1
!9
1
h23
(
1
h)
3 3
5 lim
5 lim
hS0
h
hS0
h
!9 1 h 2 3 !9 1 h 1 3
2h 1
5 lim
?
5 lim a 1 1 b
hS0
h
!9 1 h 1 3
hS0 9 1 3 h h
9 1 h 1 3 !9 1 h 2 3 !9 1 h 2 9
21
5 lim
5 lim 1 1
hS0
h( !9 1 h 1 3)
hS0 9 1 3 h
h
21
5 lim
51 1
hS0
h(
!9
1
h 1 3)
9 1 3 (0)
1
5 29
5 lim
hS0 !9 1 h 1 3
7. a. m 5 lim
1-18
Chapter 1: Introduction to Calculus
1
!9 1 0 1 3
1
5
6
8. s(t) 5 6t(t 1 1) 5 6t 2 1 6t
s(3) 2 s(2)
a. i. average velocity 5
322
5 36(3)2 1 6(3)42 36(2)2 1 6(2)4
5 6(9) 1 18 2 (24 1 12)
5 54 1 18 2 36
5 36 km> h
s(2.1) 2 s(2)
ii. average velocity 5
2.1 2 2
36(2.1)2 1 6(2.1)4 2 36(2)2 1 6(2)4
5
0.1
326.46 1 12.64 2 324 1 124
5
0.1
39.06 2 36
5
0.1
3.06
5
0.1
5 30.6 km> h
s(2.01) 2 s(2)
iii. average velocity 5
2.01 2 2
36(2.01)2 1 6(2.01)4 2 36(2)2 1 6(2)4
5
0.01
324.2406 1 12.064 2 36(2)2 1 6(2)4
5
0.01
36.3006 2 324 1 124
5
0.01
36.3006 2 36
5
0.01
0.3006
5
0.01
5 30.06 km> h
5
b. At the time t 5 2, the velocity of the car appears
to approach 30 km> h.
f(2 1 h) 2 f(2)
c. average velocity 5
(2 1 h) 2 (2)
2
36(2 1 h) 1 6(2 1 h)4 2 36(2)2 1 6(2)4
5
h
36(4 1 4h 1 h 2 ) 1 12 1 6h4 2 324 1 124
5
h
324 1 24h 1 6h 2 1 12 1 6h4 2 36
5
h
6h 2 1 30h 1 36 2 36
5
h
Calculus and Vectors Solutions Manual
6h 2 1 30h
h
5 (6h 1 30) km> h
d. When t 5 2, the velocity is the limit as h
approaches 0.
velocity 5 lim (6h 1 30)
5
hS0
5 6(0) 1 30
5 30
Therefore, when t 5 2 the velocity is 30 km> h.
9. a. The instantaneous rate of change of f(x) with
respect to x at x 5 2 is given by
f(2 1 h) 2 f(2)
lim
hS0
h
35 2 (2 1 h)24 2 35 2 (2)24
5 lim
hS0
h
5 2 (4 1 4h 1 h 2 ) 2 1
5 lim
hS0
h
5 2 4 2 4h 2 h 2 2 1
5 lim
hS0
h
2
2h 2 4h
5 lim
hS0
h
5 lim (2h 2 4)
hS0
5 2 (0) 2 4
5 24
b. The instantaneous rate of change of f(x) with
respect to x at x 5 12 is given by
f( 1 1 h) 2 f( 12 )
lim 2
hS0
h
3
3
2 1
1
1h
2
5 lim 2
hS0
h
3
26
1
1
h
2
5 lim
hS0
h
3 2 6( 12 1 h) 1
5 lim
?
1
hS0
h
2 1 h
3 2 3 2 6h 1
?
5 lim
1
hS0
h
2 1 h
26h 1
5 lim 1
?
hS0 2 1 h
h
26
5 lim 1
hS0 2 1 h
26
51
2 1 0
5 212
1-19
10. a. The average rate of change of V(t) with
respect to t during the first 20 minutes is given by
f(20) 2 f(0)
20 2 0
350(30 2 20)24 2 350(30 2 0)24
5
20
5000 2 45 000
5
20
40 000
52
20
5 22000 L> min
b. The rate of change of V(t) with respect to t at the
time t 5 20 is given by
f(20 1 h) 2 f(20)
lim
hS0
h
350(30 2 (20 1 h))24 2 350(30 2 20)24
5 lim
hS0
h
2
350(10 2 h) 4 2 350(10)24
5 lim
hS0
h
350(100 2 20h 1 h 2 )4 2 350(100)4
5 lim
hS0
h
5000 2 1000h 1 50h 2 2 5000
5 lim
hS0
h
50h 2 2 1000h
5 lim
hS0
h
5 lim 50h 2 1000
hS0
5 50(0) 2 1000
5 21000 L> min
11. a. Let y 5 f(x).
f(4) 5 (4)2 1 (4) 2 3 5 16 1 1 5 17
f(4 1 h) 5 (4 1 h)2 1 (4 1 h) 2 3
5 16 1 8h 1 h 2 1 h 1 1
5 h 2 1 9h 1 17
Using the limit of the difference quotient, the slope
of the tangent at x 5 4 is
f(4 1 h) 2 f(4)
m 5 lim
hS0
h
2
h 1 9h 1 17 2 (17)
5 lim
hS0
h
2
h 1 9h
5 lim
hS0
h
5 lim (h 1 9)
hS0
5019
59
Therefore, the slope of the tangent to
y 5 f(x) 5 x 2 1 x 2 3 at x 5 4 is 9.
1-20
So an equation of the tangent at x 5 4 is given by
y 2 17 5 9(x 2 4)
y 2 17 5 9x 2 36
29x 1 y 2 17 1 36 5 0
29x 1 y 1 19 5 0
b.
Let y 5 f(x).
f(22) 5 2(22)2 2 7 5 2(4) 2 7 5 1
f(22 1 h) 5 2(22 1 h)2 2 7
5 2(4 2 4h 1 h 2 ) 2 7
5 8 2 8h 1 2h 2 2 7
5 2h 2 2 8h 1 1
Using the limit of the difference quotient, the slope
of the tangent at x 5 4 is
f(22 1 h) 2 f(22)
m 5 lim
hS0
h
2h 2 2 8h 1 1 2 (1)
5 lim
hS0
h
2h 2 2 8h
5 lim
hS0
h
5 lim (2h 2 8)
hS0
5 2(0) 2 8
5 28
Therefore, the slope of the tangent to
y 5 f(x) 5 2x 2 2 7 at x 5 22 is 28.
So an equation of the tangent at x 5 22
is given by
y 2 1 5 28(x 2 (22))
y 2 1 5 28x 2 16
8x 1 y 2 1 1 16 5 0
8x 1 y 1 15 5 0
c. f(21) 5 3(21)2 1 2(21) 2 5 5 3 2 2 2 5
5 24
f(21 1 h) 5 3(21 1 h)2 1 2(21 1 h) 2 5
5 3(1 2 2h 1 h 2 ) 2 2 1 2h 2 5
5 3 2 6h 1 3h 2 2 7 1 2h
5 3h 2 2 4h 2 4
Using the limit of the difference quotient, the slope
of the tangent at x 5 4 is
f(21 1 h) 2 f(21)
m 5 lim
hS0
h
2
3h 2 4h 2 4 2 (24)
5 lim
hS0
h
3h 2 2 4h
5 lim
hS0
h
5 lim (3h 2 4)
hS0
5 3(0) 2 4
5 24
Chapter 1: Introduction to Calculus
Therefore, the slope of the tangent to
y 5 f(x) 5 3x 2 1 2x 2 5 at x 5 21 is 24.
So an equation of the tangent at x 5 24 is given by
y 2 (24) 5 24(x 2 (21))
y 1 4 5 24(x 1 1)
y 1 4 5 24x 2 4
4x 1 y 1 4 1 4 5 0
4x 1 y 1 8 5 0
d. f(1) 5 5(1)2 2 8(1) 1 3 5 5 2 8 1 3 5 0
f(1 1 h) 5 5(1 1 h)2 2 8(1 1 h) 1 3
5 5(1 1 2h 1 h 2 ) 2 8 2 8h 1 3
5 5 1 10h 1 5h 2 2 5 2 8h
5 5h 2 1 2h
Using the limit of the difference quotient, the slope
of the tangent at x 5 1 is
f(1 1 h) 2 f(1)
m 5 lim
hS0
h
2
5h 1 2h 2 (0)
5 lim
hS0
h
5 lim (5h 1 2)
hS0
5 5(0) 1 2
52
Therefore, the slope of the tangent to
y 5 f(x) 5 5x 2 2 8x 1 3 at x 5 1 is 2.
So an equation of the tangent at x 5 1 is given by
y 2 0 5 2(x 2 1)
y 5 2x 2 2
22x 1 y 1 2 5 0
12. a. Using the limit of the difference quotient, the
slope of the tangent at x 5 25 is
f(25 1 h) 2 f(25)
m 5 lim
hS0
h
25 1 h
25
1
5 lim a
2
b?
hS0 25 1 h 1 3
25 1 3
h
25 1 h
5
1
5 lim a
2 b?
hS0 22 1 h
2
h
210 1 2h 2 (210 1 5h)
1
5 lim a
b?
hS0
24 1 2h
h
1
210 1 2h 1 10 2 5h
5 lim a
b?
hS0
24 1 2h
h
23h
1
5 lim a
b?
hS0 24 1 2h
h
23
5 lim a
b
hS0 24 1 2h
23
5
24 1 2(0)
3
5
4
Calculus and Vectors Solutions Manual
Therefore, the slope of the tangent to
x
f(x) 5 x 1 3 at x 5 25 is 43.
So an equation of the tangent at x 5 34 is given by
5
3
y 2 5 (x 2 (25))
2
4
5
3
15
y2 5 x1
2
4
4
3
10
15
2 x1y2
2
50
4
4
4
3
25
2 x1y2
50
4
4
23x 1 4y 2 25 5 0
b. Using the limit of the difference quotient, the
slope of the tangent at x 5 21 is
f(21 1 h) 2 f(21)
m 5 lim
hS0
h
2(21 1 h) 1 5
2(21) 1 5
1
5 lim a
2
b?
hS0 5(21 1 h) 2 1
5(21) 2 1
h
22 1 2h 1 5
22 1 5
1
5 lim a
2
b?
hS0 25 1 5h 2 1
25 2 1
h
2h 1 3
3
1
5 lim a
2
b?
hS0 5h 2 6
26
h
1
1
2h 1 3
5 lim a
1 b?
hS0 5h 2 6
2
h
4h 1 6 1 5h 2 6
1
5 lim a
b?
hS0
10h 2 12
h
9h
1
5 lim a
b?
hS0 10h 2 12
h
9
5 lim a
b
hS0 10h 2 12
9
5
10(0) 2 12
9
52
12
3
52
4
Therefore, the slope of the tangent to
2x 1 5
f(x) 5 5x 2 1 at x 5 21 is 2 34.
So an equation of the tangent at x 5 2 34 is given by
1
3
y 2 a2 b 5 2 (x 2 (21))
2
4
1
3
3
y1 52 x2
2
4
4
4y 1 2 5 23x 2 3
3x 1 4y 1 2 1 3 5 0
3x 1 4y 1 5 5 0
1-21
1.4 The Limit of a Function,
pp. 37–39
1. a.
6
27
99
b. p
2. One way to find a limit is to evaluate the function
for values of the independent variable that get
progressively closer to the given value of the
independent variable.
3. a. A right-sided limit is the value that a
function gets close to as the values of the
independent variable decrease and get close
to a given value.
b. A left-sided limit is the value that a function
gets close to as the values of the independent
variable increase and get close to a given
value.
c. A (two-sided) limit is the value that a function
gets close to as the values of the independent
variable get close to a given value, regardless
of whether the values increase or decrease
toward the given value.
4. a. 25
b. 3 1 7 5 10
c. 102 5 100
d. 4 2 3(22)2 5 28
e. 4
f. 23 5 8
5. Even though f(4) 5 21, the limit is 1, since that
is the value that the function approaches from the
left and the right of x 5 4.
6. a. 0
b. 2
c. 21
d. 2
7. a. 2
b. 1
c. does not exist
8. a. 9 2 (21)2 5 8
b.
9. 22 1 1 5 5
0 1 20
5 "4
Å015
52
c. "5 2 1 5 "4
52
1-22
y
4
2
–4
–2
x
0
2
4
10. a. Since 0 is not a value for which the function is
undefined, one may substitute 0 in for x to find that
lim1 x 4 5 lim x 4
xS0
xS0
5 (0)4
50
b. Since 2 is not a value for which the function is
undefined, one may substitute 2 in for x to find that
lim2 (x 2 2 4) 5 lim (x 2 2 4)
xS2
xS2
5 (2)2 2 4
5424
50
c. Since 3 is not a value for which the function is
undefined, one may substitute 3 in for x to find that
lim2 (x 2 2 4) 5 lim (x 2 2 4)
xS3
xS3
5 (3)2 2 4
5924
55
d. Since 1 is not a value for which the function is
undefined, one may substitute 1 in for x to find that
1
1
5 lim
lim
xS1 1 x 2 3
xS1 x 2 3
1
5
123
1
52
2
e. Since 3 is not a value for which the function is
undefined, one may substitute 3 in for x to find that
1
1
lim1
5 lim
xS3 x 1 2
xS3 x 1 2
1
5
312
1
5
5
f. If 3 is substituted in the function for x, then the
function is undefined because of division by zero.
There does not exist a way to divide out the x 2 3 in
Chapter 1: Introduction to Calculus
1
the denominator. Also, lim1 x 2 3 approaches infinity,
xS3
d.
8
6
4
2
1
while lim2 x 2 3 approaches negative infinity.
xS3
1
1
1
Therefore, since lim1 x 2 3 2 lim2 x 2 3, lim x 2 3
xS3
xS3
xS3
does not exist.
11. a.
y
8
6
4
2
–8 –6 –4 –2 0
–2
–4
–6
–8
x
2 4 6 8
–8 –6 –4 –2 0
–2
–4
–6
–8
xS21
not exist.
b.
8
6
4
2
–8 –6 –4 –2 0
–2
–4
y
x
–8 –6 –4 –2 0
–2
–4
2 4 6 8
8
6
4
2
–8 –6 –4 –2 0
–2
–4
–6
–8
–8 –6 –4 –2 0
–2
–4
y
2 4 6 8
2 4 6 8
x
2 4 6 8
x
2 4 6 8
6 y
4
2
d.
x
x
6 y
4
2
c.
xS2
is equal to 2.
c.
xS20.5
6 y
4
2
b.
xS2
xS20.5
6
4
2
lim f(x) 5 lim2 f(x). Therefore, lim f(x) exists and
xS2 1
2 4 6 8
does not exist.
12. Answers may vary. For example:
a.
y
xS21
–8 –6 –4 –2 0
–2
–4
–6
–8
x
lim f(x) 2 lim 2 f(x). Therefore, lim f(x)
xS20.5 1
lim f(x) 2 lim 2 f(x). Therefore, lim f(x) does
xS21 1
y
–8 –6 –4 –2 0
–2
–4
x
2 4 6 8
13. f(x) 5 mx 1 b
lim f(x) 5 22 m 1 b 5 22
xS1
lim f(x) 5 lim12 f(x). Therefore, lim1 f(x) exists and
xS 12 1
xS 2
xS 2
is equal to 2.
Calculus and Vectors Solutions Manual
lim f(x) 5 4
xS21
2m 1 b 5 4
2b 5 2
b 5 1, m 5 23
1-23
14. f(x) 5 ax 2 1 bx 1 c, a 2 0
f(0) 5 0
c50
lim f(x) 5 5
a1b55
xS1
lim f(x) 5 8
xS22
4a 2 2b 5 8
6a 5 18
a 5 3,
b52
Therefore, the values are a 5 3, b 5 2, and c 5 0.
15. a.
y
10
8
6
4
2
x
0
–4 –2
2 4 6 8 10 12
–2
1
(6)2
tS6
12
36
531
12
5313
56
1
lim1 p(t) 5 2 1 (6)2
tS6
18
36
521
18
5212
54
c. Since p(t) is measured in thousands, right before
the chemical spill there were 6000 fish in the lake.
Right after the chemical spill there were 4000 fish
in the lake. So, 6000 2 4000 5 2000 fish were
killed by the spill.
d. The question asks for the time, t, after the chemical
spill when there are once again 6000 fish in the lake.
Use the second equation to set up an equation that is
modelled by
1
6 5 2 1 t2
18
1 2
45 t
18
72 5 t 2
!75 5 t
(The question asks for time so the negative answer
is disregarded.)
So, at time t 5 !72 8 8.49 years the population
has recovered to the level before the spill.
b. lim2 p(t) 5 3 1
1-24
1.5 Properties of Limits, pp. 45–47
1. lim (3 1 x) and lim (x 1 3) have the same value,
xS2
xS2
but lim 3 1 x does not. Since there are no brackets
xS2
around the expression, the limit only applies to 3,
and there is no value for the last term, x.
2. Factor the numerator and denominator. Cancel
any common factors. Substitute the given value of x.
3. If the two one-sided limits have the same value,
then the value of the limit is equal to the value of
the one-sided limits. If the one-sided limits do not
have the same value, then the limit does not exist.
3(2)
4. a. 2
51
2 12
b. (21)4 1 (21)3 1 (21)2 5 1
1 2
1 2
c. c "9 1
d 5 a3 1 b
3
"9
100
5
9
d. (2p)3 1 p2 (2p) 2 5p3 5 8p3 1 2p3 2 5p3
5 5p3
e. "3 1 "1 1 0 5 "3 1 1
52
26
23 2 3
5
f.
Å 2(23) 1 4 Å 22
5 "3
(22)3
5 22
5. a.
22 2 2
2
2
5
b.
!1 1 1
!2
5 "2
6. Since substituting t 5 1 does not make the
denominator 0, direct substitution works.
12125
25
5
621
5
5 21
4 2 x2
(2 2 x)(2 1 x)
5 lim
7. a. lim
xS2 2 2 x
xS2
(2 2 x)
5 lim (2 1 x)
xS2
54
2x 2 1 5x 1 3
(x 1 1)(2x 1 3)
5 lim
b. lim
xS21
x11
xS21
x11
55
x 3 2 27
(x 2 3)(x 2 1 3x 1 9)
5 lim
c. lim
xS3 x 2 3
xS3
x23
591919
5 27
Chapter 1: Introduction to Calculus
2 1 "4 1 x
2 2 "4 1 x
3
d. lim £
§
xS0
x
2 1 "4 1 x
5 lim
xS0
52
1
4
21
2 1 "4 1 x
"x 2 2
"x 2 2
5 lim
xS4 x 2 4
xS4 ("x 2 2)("x 1 2)
1
5
4
e. lim
"7 2 x 2 "7 1 x
"7 2 x 1 "7 1 x
§
3
f. lim £
x
xS0
"7 2 x 1 "7 1 x
5 lim
xS0
52
72x272x
x("7 2 x 1 "7 1 x)
1
"7
3
"
x22
xS8 x 2 8
8. a. lim
3
x. Therefore, u 3 5 x as x S 8, u S 2.
Let u 5 "
u22
1
5 lim 2
Here, lim 3
xS2 u 2 8
xS2 u 1 2u 1 4
1
5
12
1
27 2 x
Let x 3 5 u
b. lim 13
xS27 x 2 3
x 5 u3
3
u 2 27
x S 27, u S 3.
5 lim
xS3 u 2 3
(u 2 3)(u 2 1 3u 1 9)
5 2lim
xS3
u23
5 2 (9 1 9 1 9)
5 227
1
1
x6 2 1
x 6 5 u, x 5 u 6
c. lim
x S 1, u S 1
xS1 x 2 1
u21
5 lim 6
xS1 u 2 1
(u 2 1)
5 lim
5
4
3
2
xS1 (u 2 1)(u 1 u 1 u 1 u 1 u 1 1)
1
5
6
1
1
x6 2 1
Let x 6 5 u
d. lim 13
xS1 x 2 1
u6 5 x
1
u21
x 3 5 u2
5 lim 2
xS1 u 2 1
As x S 1, u S 1
Calculus and Vectors Solutions Manual
5 lim
xS1
5
1
2
e. lim
u21
(u 2 1)(u 1 1)
"x 2 2
"x 2 8
u22
5 lim 3
xS2 u 2 8
xS4
5 lim
xS2
5
1
12
3
1
Let x 2 5 u
3
x 2 5 u3
x S 4, u S 2
u22
(u 2 2)(u 2 1 2u 1 4)
1
1
(x 1 8)3 2 2
Let (x 1 8)3 5 u
x
xS0
x 1 8 5 u3
u22
x 5 u3 2 8
lim 3
xS2 u 2 8
x S 0, u S 2
1
5
12
16 2 16
9. a.
50
64 1 64
16 2 16
b.
50
16 2 20 1 6
x2 1 x
x(x 1 1)
c. lim
5 lim
xS21 x 1 1
xS21 x 1 1
5 21
"x 1 1 2 1
"x 1 1 2 1
d. lim
5 lim
x
xS0
xS0 x 1 1 2 1
"x 1 1 2 1
5 lim
xS0 ("x 1 1 2 1)("x
1 1 1 1)
1
5
2
(x 1 h)2 2 x 2
2xh 1 h 2
5 lim
e. lim
hS0
h
hS0
h
5 2x
1
1
2
ba
2
b
f. lim a
xS1 x 2 1
x13
3x 1 5
1
3x 1 5 2 2x 2 6
5 lim a
ba
b
xS1 x 2 1
(x 1 3)(3x 1 5)
1
5 lim
xS1 (x 1 3)(3x 1 5)
1
5
4(8)
1
5
32
f. lim
1-25
0x 2 50
does not exist.
xS5 x 2 5
0x 2 50
x25
lim1
5 lim1
xS5 x 2 5
xS5 x 2 5
51
0x 2 50
x25
lim2
5 lim2 2 a
b
xS5 x 2 5
xS5
x25
5 21
y
2
10. a. lim
0
–4
xS2
(x 2 2)(x 1 1)
(x 2 2)(x 1 1)
5 lim2 2
0x 2 20
xS2
(x 2 2)
5 lim2 2 (x 1 1)
xS2
5 23
2
–2
0 x 1 2 0 5 x 1 2 if x . 22
5 2 (x 1 2) if x , 22
(x 1 2)(x 1 2)2
lim
5 lim 1 (x 1 2)2 5 0
xS22 1
x12
xS22
(x 1 2)(x 1 2)2
lim
50
xS222
2 (x 1 2)
d.
0 2x 2 5 0 (x 1 1)
does not exist.
2x 2 5
xS 2
5
0 2x 2 5 0 5 2x 2 5, x $
2
(2x 2 5)(x 1 1)
lim
5x11
2x 2 5
xS 52
5
0 2x 2 5 0 5 2 (2x 2 5), x ,
2
2 (2x 2 5)(x 1 1)
lim2
5 2 (x 1 1)
2x 2 5
xS 52
y
4
b. lim5
4
2
–4
–2
x
2
4
–2
–4
x
2
4
DT
20
20
20
20
–4
20
20
2
x 2x22
(x 2 2)(x 1 1)
5 lim
xS2
0x 2 20
xS2
0x 2 20
(x 2 2)(x 1 1)
(x 2 2)(x 1 1)
lim
5 lim1
xS2 1
0x 2 20
xS2
x22
5 lim1 x 1 1
lim
xS2
53
1-26
0
11. a.
–2
c.
y
2
1
0
4
–4
–2
–2
2
–2
8
–1
–4
x
0
x
4
y
4
–4
1
–8
lim2
T
V
240
19.1482
220
20.7908
0
22.4334
20
24.0760
40
25.7186
60
27.3612
80
29.0038
DV
1.6426
1.6426
1.6426
1.6426
1.6426
1.6426
DV is constant, therefore T and V form a linear
relationship.
DV
b. V 5
?T1K
DT
DV
1.6426
5
5 0.082 13
DT
20
Chapter 1: Introduction to Calculus
V 5 0.082 13T 1 K
T50
V 5 22.4334
Therefore, k 5 22.4334 and
V 5 0.082 13T 1 22.4334.
V 2 22.4334
c. T 5
0.082 13
d. lim T 5 2273.145
f(x)
g(x)
5 1 and lim
52
xS0 x
xS0 x
g(x)
b5032
a. lim g(x) 5 lim xa
x
xS0
xS0
50
f (x)
f(x)
1
x
5 lim g (x)
5
b. lim
xS0 g(x)
xS0 x
2
15. lim
vS0
e.
!x 1 1 2 !2x 1 1
xS0 !3x 1 4 2 !2x 1 4
V
12
16. lim
10
!x 1 1 2 !2x 1 1
!x 1 1 1 !2x 1 1
8
5 lim c
6
3
xS0
4
3
2
0
T
0
2
4
6
8
10
12
x2 2 4
xS5 f(x)
lim (x2 2 4)
xS0
!3x 1 4 1 !2x 1 4
d
!3x 1 4 1 !2x 1 4
(x 1 1 2 2x 2 1)
!3x 1 4 1 !2x 1 4
d
3
(3x 1 4 2 2x 2 4)
!x 1 1 1 !2x 1 1
212
111
5 22
x 2 1 0 x 2 1 021
17. lim
xS1
0x 2 10
x S 11 0 x 2 1 0 5 x 2 1
x2 1 x 2 2
(x 1 2)(x 2 1)
5
x21
x21
x 2 1 0 x 2 1 021
lim
53
xS1 1
0x 2 10
x S 12 0 x 2 1 0 5 2x 1 1
x2 2 x
x(x 2 1)
lim2
5 lim2
xS1 2x 1 1
xS1 2x 1 1
5 21
Therefore, this limit does not exist.
y
4
52
12. lim
5
5 lim c
!x 1 1 1 !2x 1 1
!3x 1 4 2 !2x 1 4
xS5
lim f(x)
xS5
21
5
3
57
13. lim f(x) 5 3
xS4
a. lim 3 f(x)4 3 5 33 5 27
xS4
b.
3 f(x)4 2 2 x 2
( f(x) 2 x)( f(x) 1 x)
5 lim
xS4 f(x) 1 x
xS4
f(x) 1 x
5 lim ( f(x) 2 x)
lim
xS4
5324
5 21
c. lim "3f(x) 2 2x 5 "3 3 3 2 2 3 4
2
f(x)
51
14. lim
xS0 x
0
xS4
51
a. lim f(x) 5 lim c
f(x)
3 xd 5 0
x
xS0
xS0
f(x)
x f(x)
5 lim c
d 50
b. lim
x
xS0 g(x)
xS0 g(x)
Calculus and Vectors Solutions Manual
–4
–2
x
2
4
–2
–4
1-27
1.6 Continuity, pp. 51–53
1. Anywhere that you can see breaks or jumps is a
place where the function is not continuous.
2. It means that on that domain, you can trace the
graph of the function without lifting your pencil.
3. point discontinuity
10
8
6
4
2
y
hole
–2 0
–2
x
2 4 6
5. a. The function is a polynomial, so the function
is continuous for all real numbers.
b. The function is a polynomial, so the function is
continuous for all real numbers.
c. x 2 2 5x 5 x(x 2 5)
The is continuous for all real numbers except
0 and 5.
d. The is continuous for all real numbers greater
than or equal to 22.
e. The is continuous for all real numbers.
f. The is continuous for all real numbers.
6. g(x) is a linear function (a polynomial),
and so is continuous everywhere,
including x 5 2.
7.
y
8
jump discontinuity
10
8
6
4
2
4
y
–8
x
The function is continuous everywhere.
8.
y
4
y
2
–4
–2
x
1
x
0
2
4
–2
2 3 4
vertical
asymptote
4. a. x 5 3 makes the denominator 0.
b. x 5 0 makes the denominator 0.
c. x 5 0 makes the denominator 0.
d. x 5 3 and x 5 23 make the denominator 0.
e. x 2 1 x 2 6 5 (x 1 3)(x 2 2)
x 5 23 and x 5 2 make the denominator 0.
f. The function has different one-sided limits at x 5 3.
1-28
8
–8
2 4 6
infinite discontinuity
–1 0
–2
–4
4
–4
–2 0
–2
10
8
6
4
2
–4
x
0
–4
The function is discontinuous at x 5 0.
9.
y
4
2
0
x
200
400
600
Chapter 1: Introduction to Calculus
x2 2 x 2 6
xS3
xS3
x23
(x 2 3)(x 1 2)
5 lim
xS3
x23
55
Function is discontinuous at x 5 3.
11. Discontinuous at x 5 2
y
4
10. lim f(x) 5 lim
2
–4
–2
x
0
2
4
–2
–4
12. g(x) 5 e
x 1 3, if x 2 3
2 1 !k, if x 5 3
g(x) is continuous.
2 1 "k 5 6
"k 5 4, k 5 16
13.
21, if x , 0
f(x) 5 • 0, if x 5 0
1, if x . 0
a.
y
4
2
–4
–2
x
0
2
4
–2
–4
b. i. From the graph, lim2 f(x) 5 21.
xS0
ii. From the graph, lim1 f(x) 5 1.
Thus, lim f(x) 5 4. But, f(3) 5 2. Hence f is not
xS3
continuous at x 5 2 (and also not continuous over
23 , x , 8).
15. The function is to be continuous at x 5 1 and
discontinuous at x 5 2.
f(x) 5 μ
Ax 2 B
, if x # 1
x22
3x, if 1 , x , 2
Bx 2 A, if x $ 2
2
For f(x) to be continuous at x 5 1:
A(1) 2 B
5 3(1)
122
A(1) 2 B 5 23
A5B23
For f(x) to be discontinuous at x 5 2:
B(2)2 2 A 2 3(2)
4B 2 A 2 6
If 4B 2 A . 6, then
if 4B 2 A , 6, then
4B 2 (B 2 3) . 6
4B 2 B 1 3 , 6
3B 1 3 . 6
3B 1 3 , 6
3B . 3
3B , 3
B . 1 and
B , 1 and
A . 22
A , 22
This shows that A and B can be any set of real
numbers such that
(1) A 5 B 2 3
(2) 4B 2 A 2 6 (if B . 1, then A . 22 if B , 1,
then A , 22)
A 5 1 and B 5 22 is not a solution because then
the graph would be continuous at x 5 2.
2x, if 23 # x # 22
2
16. f(x) 5 • ax 1 b, if 22 , x , 0
6, if x 5 0
at x 5 22, 4a 1 b 5 2
at x 5 0, b 5 6.
a 5 21
2x, if 23 # x # 22
f(x) 5 • 2x2 1 b, if 22 , x , 0
6, if x 5 0
if a 5 21, b 5 6. f(x) is continuous.
xS0
x0 x 2 1 0
, if x 2 1
g(x) 5 • x 2 1
iii. Since the one-sided limits differ, lim f(x) does
xS0
not exist.
c. f is not continuous since lim f(x) does not exist.
17.
14. a. From the graph, f(3) 5 2.
b. From the graph, lim2 f(x) 5 4.
lim g(x) 5 21
a. xS12
¶ lim g(x)
xS1
lim1 g(x) 5 1
c. lim2 f(x) 5 4 5 lim2 f(x)
lim g(x) does not exist.
xS0
xS3
xS3
xS3
Calculus and Vectors Solutions Manual
0, if x 5 1
xS1
xS1
1-29
b.
4
y
c. h(x) 5
2
–4
–2
0
x
2
4
–2
–4
g(x) is discontinuous at x 5 1.
Review Exercise, pp. 56–59
1. a. f(22) 5 36, f(3) 5 21
21 2 36
m5
3 2 (22)
5 23
b. f(21) 5 13, f(4) 5 48
48 2 13
m5
4 2 (21)
57
c. f(1) 5 23
5(1 1 2h 1 h 2 ) 2 (23)
m 5 lim
hS0
h
2h 1 h 2
5 lim
hS0
h
5 lim 2 1 h
m 5 lim
2
2
, Pa4, b
!x 1 5
3
2
2
23
!4 1 h 1 5
h
3 2 !9 1 h
3 1 !9 1 h
d
5 2 lim c
3
hS0
3h!9 1 h
3 1 !9 1 h
1
5 2 lim c 2
d
hS0
3 !9 1 h(3 1 !9 1 h)
2
52
9(6)
1
52
27
5
5
d. f(x) 5
, Pa4, b
x22
2
5
5
2
41h22
2
m 5 lim
hS0
h
10 2 5(2 1 h)
5 lim
hS0 h(2 1 h)(2)
25h
5 lim 2
hS0
h(2 1 h)(2)
5
52
4
4 2 x2 , if x # 1
3. f(x) 5 e
2x 1 1, if x . 1
hS0
a. Slope at P(21, 3) f(x) 5 4 2 x 2
hS0
52
y 2 (23) 5 2(x 2 1)
2x 2 y 2 5 5 0
3
2. a. f(x) 5
, P(2, 1)
x11
3
21
m531h
h
1
5 lim 2
hS0
31h
1
52
3
b. g(x) 5 "x 1 2, P(21, 1)
"21 1 h 1 2 2 1
h
!h 1 1 2 1
!h 1 1 1 1
5 lim c
d
3
x
hS0
!h 1 1 1 1
1
5 lim
hS0 !h 1 1 1 1
1
5
2
m 5 lim
hS0
1-30
4 2 (21 1 h)2 2 3
hS0
h
4 2 1 1 2h 2 h 2 2 3
5 lim
hS0
h
5 lim (2 2 h)
m 5 lim
hS0
52
Slope of the graph at P(21, 3) is 2.
b. Slope at P(2, 0.5)
f(x) 5 2x 1 1
f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5
5 2h
2h
m 5 lim
52
hS0 h
Slope of the graph at P(2, 0.5) is 2.
4. s(t) 5 25t 2 1 180
a. s(0) 5 180, s(1) 5 175, s(2) 5 160
Average velocity during the first second is
s(1) 2 s(0)
5 25
1
m> s.
Chapter 1: Introduction to Calculus
Average velocity during the second second is
s(2) 2 s(1)
5 215
1
m> s.
b. At t 5 4:
s(4 1 h) 2 s(4)
5 25(4 1 h)2 1 180 2 (25(16) 1 180)
5 280 2 40h 2 5h 2 1 180 1 80 2 180
s(4 1 h) 2 s(4)
240h 2 5h 2
5
h
h
v(4) 5 lim (240 2 5h) 5 240
Velocity is 240 m> s.
c. Time to reach ground is when s(t) 5 0.
Therefore, 25t 2 1 180 5 0
t 2 5 36
t 5 6, t . 0.
Velocity at t 5 6:
s(6 1 h) 5 25(36 1 12h 1 h 2 ) 1 180
5 260h 2 5h 2
s(6) 5 0
Therefore, v(6) 5 lim (260 2 5h) 5 260.
hS0
hS0
5. M(t) 5 t 2 mass in grams
a. Growth during 3 # t # 3.01
M(3.01) 5 (3.01)2 5 9.0601
M(3) 5 32
59
Grew 0.0601 g during this time interval.
b. Average rate of growth is
0.0601
5 6.01 g> min.
0.01
c.
s(3 1 h) 5 9 1 6h 1 h 2
s(3) 5 9
s(3 1 h) 2 s(3)
6h 1 h 2
5
h
h
Rate of growth is lim (6 1 h) 5 6 g> min.
c. Present rate of change:
Q(h) 5 104 (h 2 1 15h 1 70)
Q(0) 5 104 1 70
Q(h) 2 Q(0)
lim
5 lim 104 (h 1 15)
hS0
h
hS0
5 15 3 104 t per year.
d. Q(a 1 h)
5 104 3a 2 1 2ah 1 h 2 1 15a 1 15h 1 704
Q(a) 5 104 3a 2 1 15a 1 704
104 32ah 1 h 2 1 15h4
Q(a 1 h) 2 Q(a)
5
h
h
Q(a 1 h) 2 Q(a)
lim
5 lim 104 (2a 1 h 1 15)
hS0
h
hS0
5 (2a 1 15)104
Now,
(2a 1 15)104 5 3 3 105
2a 1 15 5 30
a 5 7.5
It will take 7.5 years to reach a rate of
3.0 3 105 t per year.
7. a. From the graph, the limit is 10.
b. 7; 0
c. p(t) is discontinuous for t 5 3 and t 5 4.
8. a. Answers will vary. lim f(x) 5 0.5, f is
xS21
discontinuous at x 5 21
2
y
1
–2
–1
x
0
1
2
–1
hS0
6. Q(t) 5 104 (t 2 1 15t 1 70) tonnes of waste,
0 # t # 10
a. At t 5 0,
Q(t) 5 70 3 104
5 700 000.
700 000 t have accumulated up to now.
b. Over the next three years, the average rate of
change:
Q(3) 5 104 (9 1 45 1 70)
5 124 3 104
Q(0) 5 70 3 104
Q(3) 2 Q(0)
54 3 104
5
3
3
5 18 3 104 t per year.
–2
b. f(x) 5 24 if x , 3; f is increasing for x . 3
lim1 f(x) 5 1
xS3
4
y
2
–4
–2
0
x
2
4
–2
–4
Calculus and Vectors Solutions Manual
1-31
9. a.
4
y
13. a.
x
2
–4
–2
0
1.9
1.99
1.999
2.001
2.01
2.1
x22
0.344 83 0.334 45 0.333 44 0.333 22 0.332 23 0.322 58
x2 2 x 2 2
1
3
x
2
4
–2
–4
x 1 1, if x , 21
b. f(x) 5 • 2x 1 1, if 21 # x , 1
x 2 2, if x . 1
Discontinuous at x 5 21 and x 5 1.
c. They do not exist.
10. The function is not continuous at x 5 24
because the function is not defined at x 5 24.
(x 5 24 makes the denominator 0.)
2x 2 2
11. f(x) 5 2
x 1x22
2(x 2 1)
5
(x 2 1)(x 1 2)
a. f is discontinuous at x 5 1 and x 5 22.
2
b. lim f(x) 5 lim
xS1
xS1 x 1 2
2
5
3
2
lim f(x): 5 lim 1
5 1`
xS22
xS22 x 1 2
2
lim
5 2`
xS222 x 1 2
lim f(x) does not exist.
xS22
1
, lim f(x) does not exist.
x 2 xS0
b. g(x) 5 x(x 2 5), lim g(x) 5 0
xS0
x 3 2 27
c. h(x) 5 2
,
x 29
37
lim h(x) 5
5 5.2857
xS4
7
lim h(x) does not exist.
12. a. f(x) 5
xS23
b.
x
0.9
0.99
0.999
1.001
1.01
1.1
x21
0.526 32 0.502 51 0.500 25 0.499 75 0.497 51 0.476 19
x2 2 1
1
2
14.
20.1
x
20.01
20.001
0.001
0.01
0.1
"x 1 3 2 "3 0.291 12 0.288 92 0.2887 0.288 65 0.288 43 0.286 31
x
!x 1 3 2 !3 !x 1 3 1 !3
d
?
x
!x 1 3 1 !3
x1323
5 lim
xS0 xA !x 1 3 1 !3B
x
5 lim
xS0 xA !x 1 3 1 !3B
1
5 lim
xS0 !x 1 3 1 !3
1
5
2 !3
lim c
xS0
This agrees well with the values in the table.
15. a. f(x) 5
"x 1 2 2 2
x22
x
2.1
2.01
2.001
2.0001
f(x)
0.248 46
0.249 84
0.249 98
0.25
x 5 2.0001
f(x) 8 0.25
1-32
Chapter 1: Introduction to Calculus
b.
5 lim
xS0
1
!5
5
1
A !x 1 5 1 !5 2 xB
(x 2 2)(x 1 2)
(x 2 2)(x 2 1 2x 1 4)
x12
5 lim 2
xS2 x 1 2x 1 4
(2) 1 2
5
(2)2 1 2(2) 1 4
4
5
12
1
5
3
4 2 !12 1 x 4 1 !12 1 x
d
?
e. lim c
xS4
x24
4 1 !12 1 x
16 2 (12 1 x)
5 lim
xS4 (x 2 4)(4 1 !12 1 x)
42x
5 lim
xS4 (x 2 4)(4 1 !12 1 x)
2 (x 2 4)
5 lim
xS4 (x 2 4)(4 1 !12 1 x)
21
5 lim
xS4 4 1 !12 1 x
21
5
4 1 !12 1 (4)
21
5
414
1
52
8
1
1
1
2 b
f. lim a
x
xS0
21x
2
1
x
5 lim c 3 2
d
xS0 x
2(2 1 x)
1
5 lim c 2
d
xS0
2(2 1 x)
1
52
4
18. a. The function is not defined for x , 3, so
there is no left-side limit.
b. Even after dividing out common factors from
numerator and denominator, there is a factor of
x 2 2 in the denominator; the graph has a vertical
asymptote at x 5 2.
25, if x , 1
c. f(x) 5 e
2, if x $ 1
lim2 f(x) 5 25 2 lim1 f(x) 5 2
d. lim
lim f(x) 5 0.25
xS2
!x 1 2 2 2
!x 1 2 1 2
d
3
xS2
x22
!x 1 2 1 2
1
5 lim
xS2 !x 1 2 1 2
1
5 5 0.25
4
(5 1 h)2 2 25
16. a. lim
hS0
h
5 lim (10 1 h)
c. lim c
hS0
5 10
Slope of the tangent to y 5 x 2 at x 5 5 is 10.
"4 1 h 2 2
"4 1 h 2 2
5 lim
h
hS0 4 1 h 2 4
1
5 lim
hS0 !4 1 h 1 2
1
5
4
Slope of the tangent to y 5 "x at x 5 4 is 14.
b. lim
hS0
1
1
24
41h
4242h
hS0
h
hS0 4(4 1 h)(h)
1
5 lim 2
hS0
4(4 1 h)
1
52
16
1
Slope of the tangent to y 5 x at (x 5 4) is 2 161 .
(x 1 4)(x 1 8)
5 lim (x 1 8)
17. a. lim
xS24
x14
xS24
5 (24) 1 8
54
2
2
(x 1 4a) 2 25a
(x 2 a)(x 1 9a)
5 lim
b. lim
x
2
a
x2a
xSa
xSa
5 10a
!x 1 5 2 !5 2 x
!x 1 5 1 !5 2 x
d
3
c. lim c
x
xS0
!x 1 5 1 !5 2 x
x15251x
5 lim
xS0 xA !x 1 5 1 !5 2 xB
c. lim
5 lim
xS2
xS1
Calculus and Vectors Solutions Manual
xS1
1-33
d. The function has a vertical asymptote at x 5 2.
0x0
e. lim
xS0 x
x S 02 0 x 0 5 2x
0x0
lim2
5 21
xS0 x
0x0
lim1
51
xS0 x
0x0
0x0
lim1
2 lim2
xS0 x
xS0 x
5x2, if x , 21
f(x) 5 e
f.
2x 1 1, if x $ 21
lim 1 f(x) 5 21
xS21
lim f(x) 5 5
xS212
lim f(x) 2 lim 2 f(x)
xS21 1
xS21
Therefore, lim f(x) does not exist.
xS21
19. a.
23(1 1 h)2 1 6(1 1 h) 1 4 2 (23 1 6 1 4)
hS0
h
23 2 6h 2 h2 1 6 1 6h 1 4 2 7
5 lim
hS0
h
2h2
5 lim
hS0 h
5 lim 2h
m 5 lim
hS0
50
When x 5 1, y 5 7.
The equation of the tangent is y 2 7 5 0(x 2 1)
y57
b.
(22 1 h)2 2 (22 1 h) 2 1 2 (4 1 2 2 1)
m 5 lim
hS0
h
4 2 4h 1 h2 1 2 2 h 2 1 2 5
5 lim
hS0
h
25h 1 h2
5 lim
hS0
h
5 lim (25 1 h)
hS0
5 25
When x 5 22, y 5 5.
The equation of the tangent is y 2 5 5 25(x 1 2)
y 5 25x 2 5
6(21 1 h)3 2 3 2 (26 2 3)
c. m 5 lim
hS0
h
6(21 1 3h 2 3h2 1 h3 ) 2 3 1 9
5 lim
hS0
h
1-34
18h 2 18h2 1 6h3
hS0
h
5 lim (18 2 18h 1 6h2 )
5 lim
hS0
5 18
When x 5 21, y 5 29.
The equation of the tangent is
y 2 (29) 5 18(x 2 (21))
y 5 18x 1 9
22(3 1 h)4 2 (2162)
d. m 5 lim
hS0
h
22(81 1 108h 1 54h2 1 12h3 1 h4 ) 1 162
5 lim
hS0
h
2216h 2 108h2 2 24h3 2 2h4
5 lim
hS0
h
5 lim ( 2 216 2 108h 2 24h2 2 2h3 )
hS0
5 2216
When x 5 3, y 5 2162.
The equation of the tangent is
y 2 (2162) 5 2216(x 2 3)
y 5 2216x 1 486
20. P(t) 5 20 1 61t 1 3t 2
a. P(8) 5 20 1 61(8) 1 3(8)2
5 700 000
b.
20 1 61(8 1 h) 1 3(8 1 h)2 2 (20 1 488 1 192)
lim
hS0
h
20 1 488 1 61h 1 3(64 1 16h 1 h2 ) 2 700
5 lim
hS0
h
20 1 488 1 61h 1 192 1 48h 1 3h2 2 700
5 lim
hS0
h
2
109h 1 3h
5 lim
hS0
h
5 lim (109 1 3h)
hS0
5 109
The population is changing at the rate of
109 000>h.
Chapter 1 Test, p. 60
1
1. lim x 2 1 does not exist since
xS1
1
1
5 1` 2 lim2
5 2 `.
xS1 x 2 1
xS1 x 2 1
2. f(x) 5 5x 2 2 8x
f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36
f(1) 5 5 2 8 5 23
36 1 3
39
Slope of secant is
52
22 2 1
3
5 213
lim1
Chapter 1: Introduction to Calculus
3. a. lim f(x) does not exist.
xS1
b. lim f(x) 5 1
xS2
c. lim2 f(x) 5 1
xS4
d. f is discontinuous at x 5 1 and x 5 2.
4. a. Average velocity from t 5 2 to t 5 5:
s(5) 2 s(2)
(40 2 25) 2 (16 2 4)
5
3
3
15 2 12
5
3
51
Average velocity from t 5 2 to t 5 5 is 1 km> h.
b. s(3 1 h) 2 s(3)
5 8(3 1 h) 2 (3 1 h)2 2 (24 2 9)
5 24 1 8h 2 9 2 6h 2 h 2 2 15
5 2h 2 h 2
2h 2 h 2
v(3) 5 lim
52
hS0
h
Velocity at t 5 3 is 2 km> h.
5. f(x) 5 "x 1 11
Average rate of change from x 5 5 to x 5 5 1 h:
f(5 1 h) 2 f(5)
h
"16 1 h 2 "16
5
h
x
6. f(x) 5 2
x 2 15
Slope of the tangent at x 5 4:
41h
f(4 1 h) 5
(4 1 h)2 2 15
41h
5
1 1 8h 1 h 2
4
f(4) 5
1
41h
f(4 1 h) 2 f(4) 5
24
1 1 8h 1 h 2
4 1 h 2 4 2 32h 2 4h 2
5
1 1 2h 1 h 2
31h 2 4h 2
52
(1 1 2h 1 h 2 )
(231 2 4h)
f(4 1 h) 2 f(4)
lim
5 lim
2
hS0
h
hS0 1 1 2h 1 h
5 231
Slope of the tangent at x 5 4 is 231.
Calculus and Vectors Solutions Manual
4x 2 2 36
2(x 2 3)(x 1 3)
5 lim
xS3 2x 2 6
xS3
(x 2 3)
5 12
2x 2 2 x 2 6
(2x 1 3)(x 2 2)
5 lim
b. lim 2
xS2 3x 2 7x 1 2
xS2 (x 2 2)(3x 2 1)
7
5
5
x25
(x 2 1) 2 4
c. lim
5 lim
xS5 !x 2 1 2 2
xS5 !x 2 1 2 2
7. a. lim
5 lim
xS5
54
A !x 2 1 2 2BA !x 2 1 1 2B
!x 2 1 2 2
x3 1 1
(x 1 1)(x 2 2 x 1 1)
5
lim
4
2
xS21 x 2 1
xS21 (x 2 1)(x 1 1)(x 1 1)
3
5
22(2)
3
52
4
1
6
(x 1 3) 2 6
e. lim a
2 2
b 5 lim
xS3 x 2 3
x 29
xS3 (x 2 3)(x 1 3)
1
5 lim
xS3 x 1 3
1
5
6
(x 1 8) 2 2
(x 1 8) 2 2
f. lim
5 lim
x
xS0
xS0 (x 1 8) 2 8
1
(x 1 8)3 2 2
5 lim
1
2
1
xS0 ((x 1 8)3 2 2)((x 1 8)3 1 2(x 1 8)3 1 4)
1
5
41414
1
5
12
ax 1 3, if x . 5
8. f(x) 5 •
8, if x 5 5
2
x 1 bx 1 a, if x , 5
f(x) is continuous.
Therefore, 5a 1 3 5 8
a51
25 1 5b 1 a 5 8
5b 5 218
18
b52
5
d. lim
1
3
1
3
1-35
CHAPTER 1
Introduction to Calculus
Review of Prerequisite Skills, pp. 2–3
27 2 5
622
5 23
4 2 (24)
b. m 5
21 2 3
5 22
420
c. m 5
120
54
420
d. m 5
21 2 0
5 24
4 2 4.41
e. m 5
22 2 (22.1)
5 24.1
21 2 1
f. m 5 7 4 3 4
4 2 4
2
2
4
5
1
1
52
2
2. a. Substitute the given slope and y-intercept into
y 5 mx 1 b.
y 5 4x 2 2
b. Substitute the given slope and y-intercept into
y 5 mx 1 b.
y 5 22x 1 5
c. The slope of the line is
12 2 6
m5
4 2 (21)
6
5
5
The equation of the line is in the form
y 2 y1 5 m(x 2 x1 ). The point is (21, 6) and
m 5 65.
The equation of the line is y 2 6 5 65 (x 1 1) or
y 5 65 (x 1 1) 1 6.
824
m5
d.
26 2 (22)
5 21
1. a. m 5
Calculus and Vectors Solutions Manual
y 2 4 5 21(x 2 (22))
y 2 4 5 2x 2 2
x1y2250
e. x 5 23
f. y 5 5
3. a. f(2) 5 26 1 5
5 21
b. f(2) 5 (8 2 2)(6 2 6)
50
c. f(2) 5 23(4) 1 2(2) 2 1
5 29
d. f(2) 5 (10 1 2)2
5 144
210
4. a. f(210) 5
100 1 4
5
52
52
23
b. f(23) 5
914
3
52
13
0
c. f(0) 5
014
50
10
d. f(10) 5
100 1 4
5
5
52
5. f(x) 5 •
"3 2 x, if x , 0
"3 1 x, if x $ 0
a. f(233) 5 6
b. f(0) 5 "3
c. f(78) 5 9
d. f(3) 5 "6
1
, if 23 , t , 0
t
6. s(t) 5 μ
5, if t 5 0
t3, if t . 0
1
a. s(22) 5 2
2
b. s(21) 5 21
1-1
c. s(0) 5 5
d. s(1) 5 1
e. s(100) 5 1003 or 106
7. a. (x 2 6)(x 1 2) 5 x 2 2 4x 2 12
b. (5 2 x)(3 1 4x) 5 15 1 17x 2 4x 2
c. x(5x 2 3) 2 2x(3x 1 2) 5 5x 2 2 3x 2 6x 2 2 4x
5 2x 2 2 7x
d. (x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)
5 x 2 1 2x 2 3 2 (2x 2 1 x 2 10)
5 2x 2 1 x 1 7
e. (a 1 2)3 5 (a 1 2)(a 1 2)(a 1 2)
5 (a 2 1 4a 1 4)(a 1 2)
5 a 3 1 6a 2 1 12a 1 8
f. (9a 2 5)3 5 (9a 2 5)(9a 2 5)(9a 2 5)
5 (81a 2 2 90a 1 25)(9a 2 5)
5 729a 3 2 1215a 2 1 675a 2 125
3
8. a. x 2 x 5 x(x 2 2 1)
5 x(x 1 1)(x 2 1)
2
b. x 1 x 2 6 5 (x 1 3)(x 2 2)
c. 2x 2 2 7x 1 6 5 (2x 2 3)(x 2 2)
d. x 3 1 2x 2 1 x 5 x(x 2 1 2x 1 1)
5 x(x 1 1)(x 1 1)
3
e. 27x 2 64 5 (3x 2 4)(9x 2 1 12x 1 16)
f. 2x 3 2 x 2 2 7x 1 6
x 5 1 is a zero, so x 2 1 is a factor. Synthetic or
long division yields
2x 3 2 x 2 2 7x 1 6 5 (x 2 1)(2x 2 1 x 2 6)
5 (x 2 1)(2x 2 3)(x 1 2)
9. a. 5xPR 0 x $ 256
b. 5xPR6
c. 5xPR 0 x 2 16
d. 5xPR 0 x 2 06
e. 2x 2 2 5x 2 3 5 (2x 1 1)(x 2 3)
1
e xPR ` x 2 2 , 3 f
2
f. 5xPR 0 x 2 25, 22, 16
10. a. h(0) 5 2, h(1) 5 22.1
22.1 2 2
average rate of change 5
120
5 20.1 m> s
b. h(1) 5 22.1, h(2) 5 32.4
32.4 2 22.1
average rate of change 5
221
5 10.3 m> s
11. a. The average rate of change during the second
hour is the difference in the volume at t 5 120 and
t 5 60 (since t is measured in minutes), divided by
the difference in time.
1-2
V(120) 2 V(60)
0 2 1200
5
120 2 60
60
5 220 L>min
b. To estimate the instantaneous rate of change in
volume after exactly 60 minutes, calculate the average
rate of change in volume from minute 59 to minute 61.
1186.56 2 1213.22
V(61) 2 V(59)
8
61 2 59
2
5 213.33 L>min
c. The instantaneous rate of change in volume is
negative for 0 # t # 120 because the volume of
water in the hot tub is always decreasing during that
time period, a negative change.
y
12. a., b.
8
4
0
–2
x
2
4
6
–4
–8
The slope of the tangent line is 28.
c. The instantaneous rate of change in f(x) when
x 5 5 is 28.
1.1 Radical Expressions:
Rationalizating Denominators, p. 9
1. a. 2"3 1 4
b. "3 2 "2
c. 2"3 1 "2
d. 3"3 2 "2
e. "2 1 "5
f. 2"5 2 2"2
2. a.
"3 1 "5
"2
"6 1 "10
5
2
b.
2"3 2 3"2
"2
2"6 2 6
5
2
5 "6 2 3
?
?
"2
"2
"2
"2
Chapter 1: Introduction to Calculus
c. s(0) 5 5
d. s(1) 5 1
e. s(100) 5 1003 or 106
7. a. (x 2 6)(x 1 2) 5 x 2 2 4x 2 12
b. (5 2 x)(3 1 4x) 5 15 1 17x 2 4x 2
c. x(5x 2 3) 2 2x(3x 1 2) 5 5x 2 2 3x 2 6x 2 2 4x
5 2x 2 2 7x
d. (x 2 1)(x 1 3) 2 (2x 1 5)(x 2 2)
5 x 2 1 2x 2 3 2 (2x 2 1 x 2 10)
5 2x 2 1 x 1 7
e. (a 1 2)3 5 (a 1 2)(a 1 2)(a 1 2)
5 (a 2 1 4a 1 4)(a 1 2)
5 a 3 1 6a 2 1 12a 1 8
f. (9a 2 5)3 5 (9a 2 5)(9a 2 5)(9a 2 5)
5 (81a 2 2 90a 1 25)(9a 2 5)
5 729a 3 2 1215a 2 1 675a 2 125
3
8. a. x 2 x 5 x(x 2 2 1)
5 x(x 1 1)(x 2 1)
2
b. x 1 x 2 6 5 (x 1 3)(x 2 2)
c. 2x 2 2 7x 1 6 5 (2x 2 3)(x 2 2)
d. x 3 1 2x 2 1 x 5 x(x 2 1 2x 1 1)
5 x(x 1 1)(x 1 1)
3
e. 27x 2 64 5 (3x 2 4)(9x 2 1 12x 1 16)
f. 2x 3 2 x 2 2 7x 1 6
x 5 1 is a zero, so x 2 1 is a factor. Synthetic or
long division yields
2x 3 2 x 2 2 7x 1 6 5 (x 2 1)(2x 2 1 x 2 6)
5 (x 2 1)(2x 2 3)(x 1 2)
9. a. 5xPR 0 x $ 256
b. 5xPR6
c. 5xPR 0 x 2 16
d. 5xPR 0 x 2 06
e. 2x 2 2 5x 2 3 5 (2x 1 1)(x 2 3)
1
e xPR ` x 2 2 , 3 f
2
f. 5xPR 0 x 2 25, 22, 16
10. a. h(0) 5 2, h(1) 5 22.1
22.1 2 2
average rate of change 5
120
5 20.1 m> s
b. h(1) 5 22.1, h(2) 5 32.4
32.4 2 22.1
average rate of change 5
221
5 10.3 m> s
11. a. The average rate of change during the second
hour is the difference in the volume at t 5 120 and
t 5 60 (since t is measured in minutes), divided by
the difference in time.
1-2
V(120) 2 V(60)
0 2 1200
5
120 2 60
60
5 220 L>min
b. To estimate the instantaneous rate of change in
volume after exactly 60 minutes, calculate the average
rate of change in volume from minute 59 to minute 61.
1186.56 2 1213.22
V(61) 2 V(59)
8
61 2 59
2
5 213.33 L>min
c. The instantaneous rate of change in volume is
negative for 0 # t # 120 because the volume of
water in the hot tub is always decreasing during that
time period, a negative change.
y
12. a., b.
8
4
0
–2
x
2
4
6
–4
–8
The slope of the tangent line is 28.
c. The instantaneous rate of change in f(x) when
x 5 5 is 28.
1.1 Radical Expressions:
Rationalizating Denominators, p. 9
1. a. 2"3 1 4
b. "3 2 "2
c. 2"3 1 "2
d. 3"3 2 "2
e. "2 1 "5
f. 2"5 2 2"2
2. a.
"3 1 "5
"2
"6 1 "10
5
2
b.
2"3 2 3"2
"2
2"6 2 6
5
2
5 "6 2 3
?
?
"2
"2
"2
"2
Chapter 1: Introduction to Calculus
c.
4"3 1 3"2
?
2"3
12 1 3"6
5
6
5
d.
"3
"3
f.
3"3 2 2"2
3"3 1 2"2
4 1 "6
2
3"5 2 "2
2"2
?
"2
"2
4. a.
"5 2 1
4
521
5
b.
3("5 1 "2 )
3
5 "5 1 "2
2"5
2"5 1 3"2
20 2 6"10
20 2 18
5 10 2 3"10
5
5
?
2"5 2 3"2
2"5 2 3"2
5
5
"3 2 "2
c.
?
"3 2 "2
"3 1 "2 "3 2 "2
3 1 2"6 1 2
5
322
5 5 1 2"6
d.
5
2"5 2 8
2"5 1 3
5
?
20 2 22"5 1 24
20 2 9
44 2 22"5
11
5 4 2 2"5
2"3 2 "2
5"2 1 "3
"5 1 1
"5 1 1
2( 2 1 3"2 )
27
2 1 3"2
"5 1 2
?
"5 2 2
2"5 2 1 "5 2 2
524
10 2 5"5 1 2
1
5
12 2 5 !5
2"5 2 3
2"5 2 3
?
2 2 3"2 2 1 3"2
?
2
2 1 3"2
4 2 18
5. a.
8"2
"20 2 "18
5
8"40 1 8"36
20 2 18
5
16"10 1 48
2
?
"20 1 "18
"20 1 "18
5 8"10 1 24
5
e.
3"3 2 2"2
27 2 12"6 1 8
27 2 8
35 2 12"6
5
19
4("5 1 1)
1
5
!5 1 1
c.
3"3 2 2"2
5
3"10 2 2
5
4
"5 1 "2
3
3. a.
?
"5 2 "2 "5 1 "2
b.
?
?
5"2 2 "3
5"2 2 "3
10"6 2 6 2 10 1 "6
50 2 3
11"6 2 16
5
47
5
Calculus and Vectors Solutions Manual
b.
8"2
2"5 2 3"2
5
16"10 1 48
20 2 18
5
16"10 1 48
2
?
2"5 1 3"2
2"5 1 3"2
5 8"10 1 24
c. The expressions in the two parts are equivalent.
The radicals in the denominator of part a. have been
simplified in part b.
1-3
6. a.
2"2
2"3 2 "8
4"6 1 8
628
5 22"3 2 4
?
2"3 1 "8
2"3 1 "8
5
5
b.
2"6
2"27 2 "8
?
2"27 1 "8
2"27 1 "8
4"162 1 2"48
54 2 8
36"2 1 8"3
5
46
18"2 1 4"3
5
23
2"2
c.
"16 2 "12
5
5
5
2"2
4 2 2"3
?
4 1 2"3
4 1 2"3
8"2 1 4"6
16 2 12
5 2"2 1 "6
d.
3"2 1 2"3
"12 2 "8
?
"12 1 "8
"12 1 "8
3"24 1 12 1 12 1 2"24
5
12 2 8
5
24 1 15"3
4
e.
3 !5
4 !3 1 5!2
?
4!3 2 5 !2 4 !3 1 5!2
5
12"15 1 15"10
48 2 50
52
f.
5
12"15 1 15"10
2
"18 1 "12
"18 2 "12
?
"18 1 "12
"18 1 "12
18 1 2"216 1 12
18 2 12
30 1 12"6
6
5 5 1 2"6
5
1-4
7. a.
5
"a 2 2 "a 1 2
?
a24
"a 1 2
a24
( a 2 4 )("a 2 2)
1
"a 2 2
"x 1 4 2 2 "x 1 4 1 2
b.
?
x
"x 1 4 1 2
x1424
5
x("x 1 4 1 2)
x
5
x("x 1 4 1 2)
1
5
"x 1 4 2 2
!x 1 h 2 !x !x 1 h 1 !x
c.
?
h
!x 1 h 1 !x
x1h2x
5
hA !x 1 h 1 !xB
h
5
hA !x 1 h 1 !xB
1
5
!x 1 h 1 !x
1.2 The Slope of a Tangent, pp. 18–21
28 2 7
23 2 2
53
27 2 3
b. m 5 7 2 1 2
2 2 2
2 102
5 6
1. a. m 5
2
5
3
21 2 (22.6)
c. m 5
1.5 2 6.3
1
52
3
2. a. The slope of the given line is 3, so the slope
of a line perpendicular to the given line is 2 13.
b. 13x 2 7y 2 11 5 0
27y 5 213x 2 11
13
11
y5 x1
7
7
13
The slope of the given line is 7 , so the slope of a line
perpendicular to the given line is 2 137 .
52
Chapter 1: Introduction to Calculus
6. a.
2"2
2"3 2 "8
4"6 1 8
628
5 22"3 2 4
?
2"3 1 "8
2"3 1 "8
5
5
b.
2"6
2"27 2 "8
?
2"27 1 "8
2"27 1 "8
4"162 1 2"48
54 2 8
36"2 1 8"3
5
46
18"2 1 4"3
5
23
2"2
c.
"16 2 "12
5
5
5
2"2
4 2 2"3
?
4 1 2"3
4 1 2"3
8"2 1 4"6
16 2 12
5 2"2 1 "6
d.
3"2 1 2"3
"12 2 "8
?
"12 1 "8
"12 1 "8
3"24 1 12 1 12 1 2"24
5
12 2 8
5
24 1 15"3
4
e.
3 !5
4 !3 1 5!2
?
4!3 2 5 !2 4 !3 1 5!2
5
12"15 1 15"10
48 2 50
52
f.
5
12"15 1 15"10
2
"18 1 "12
"18 2 "12
?
"18 1 "12
"18 1 "12
18 1 2"216 1 12
18 2 12
30 1 12"6
6
5 5 1 2"6
5
1-4
7. a.
5
"a 2 2 "a 1 2
?
a24
"a 1 2
a24
( a 2 4 )("a 2 2)
1
"a 2 2
"x 1 4 2 2 "x 1 4 1 2
b.
?
x
"x 1 4 1 2
x1424
5
x("x 1 4 1 2)
x
5
x("x 1 4 1 2)
1
5
"x 1 4 2 2
!x 1 h 2 !x !x 1 h 1 !x
c.
?
h
!x 1 h 1 !x
x1h2x
5
hA !x 1 h 1 !xB
h
5
hA !x 1 h 1 !xB
1
5
!x 1 h 1 !x
1.2 The Slope of a Tangent, pp. 18–21
28 2 7
23 2 2
53
27 2 3
b. m 5 7 2 1 2
2 2 2
2 102
5 6
1. a. m 5
2
5
3
21 2 (22.6)
c. m 5
1.5 2 6.3
1
52
3
2. a. The slope of the given line is 3, so the slope
of a line perpendicular to the given line is 2 13.
b. 13x 2 7y 2 11 5 0
27y 5 213x 2 11
13
11
y5 x1
7
7
13
The slope of the given line is 7 , so the slope of a line
perpendicular to the given line is 2 137 .
52
Chapter 1: Introduction to Calculus
2 53 2 (24)
3. a. m 5 5
3 2 (24)
4
5
7
3
17
3
2
5
7
17
–2
7
(x 2 (24))
17
17y 1 68 5 7x 1 28
7x 2 17y 2 40 5 0
y
4
0
2
4
4
6
d. The line is a vertical line because both points
have the same x-coordinate.
x55
6 x
y
2
–2
–2
–4
2
–4
4
–2
x
0
–2
y 2 (24) 5
2
y
0
x
2
4
6
–2
b. The slope and y-intercept are given.
y 5 8x 1 6
y
8
–4
(5 1 h)3 2 125
h
(5 1 h 2 5)((5 1 h)2 1 5(5 1 h) 1 25)
5
h
2
h(75 1 15h 1 h )
5
h
5 75 1 15h 1 h 2
(3 1 h)4 2 81
b.
h
((3 1 h)2 2 9)((3 1 h)2 1 9)
5
h
2
(9 1 6h 1 h 2 9)(9 1 6h 1 h 2 1 9)
5
h
5 (6 1 h)(18 1 6h 1 h 2 )
5 108 1 54h 1 12h 2 1 h 3
1
21
1212h
1
c. 1 1 h
5
52
h
h(1 1 h)
11h
2
2
3(1 1 h) 2 3
3((1 1 h) 2 1)
d.
5
h
h
3(1 1 2h 1 h 2 2 1)
5
h
4. a.
4
–4
–2
0
x
2
4
–4
–8
c. (0, 23), (5, 0)
0 2 (23)
m5
520
3
5
5
3
y 2 0 5 (x 2 5)
5
3x 2 5y 2 15 5 0
Calculus and Vectors Solutions Manual
1-5
3(2h 1 h2 )
h
5 6 1 3h
5
e.
f.
3
4 1 h
2 34
h
21
2 1 h
h
5. a.
1 12
5
h
23
5
4(4 1 h)
5
22 1 2 1 h
2 (2 1 h)
h
h
5
2h(2 1 h)
1
5
4 1 2h
h
5
5
16 1 h 2 16
h( "16 1 h 1 4)
1
"16 1 h 1 4
"h 1 5h 1 4 2 2
h 2 1 5h 1 4 2 4
5
h
h("h 2 1 5h 1 4 1 2 )
h15
5
2
"h 1 5h 1 4 1 2
"5 1 h 2 "5
51h25
5
c.
h
h ("5 1 h 1 "5 )
2
b.
5
1
"5 1 h 1 "5
6. a. P(1, 3), Q(1 1 h, f(1 1 h)), f(x) 5 3x 2
3(1 1 h)2 2 3
m5
h
5 6 1 3h
b. P(1, 3), Q(1 1 h, (1 1 h)3 1 2)
(1 1 h)3 1 2 2 3
m5
h
1 1 3h 1 3h 2 1 h 3 2 1
5
h
5 3 1 3h 1 h 2
c. P(9, 3), Q (9 1 h, "9 1 h )
5
1-6
"9 1 h 2 3 "9 1 h 1 3
?
h
"9 1 h 1 3
1
"9 1 h 1 3
Q
P
12 2 12 2 3h
4 (4 1 h)
"16 1 h 2 4
m5
7. a.
Slope of Line PQ
(2, 8)
(3, 27)
19
(2, 8)
(2.5, 15.625)
15.25
(2, 8)
(2.1, 9.261)
12.61
(2, 8)
(2.01, 8.120 601)
12.060 1
(2, 8)
(1, 1)
(2, 8)
(1.5, 3.375)
9.25
(2, 8)
(1.9, 6.859)
11.41
(2, 8)
(1.99, 7.880 599)
11.940 1
7
b. 12
c. (2, 8), ((2 1 h), (2 1 h)3 )
(2 1 h)3 2 8
m5
21h22
8 1 12h 1 6h 2 1 h 3 2 8
5
h
5 12 1 6h 1 h 2
d. m 5 lim (12 1 6h 1 h 2 )
hS0
5 12
e. They are the same.
f.
y
12
8
4
–4
–2
0
x
2
4
–4
8. a. y 5 3x 2, (22, 12)
3(22 1 h)2 2 12
m 5 lim
hS0
h
12 2 12h 1 3h 2 2 12
5 lim
hS0
h
5 lim (212 1 3h)
hS0
5 212
b. y 5 x 2 2 x at x 5 3, y 5 6.
(3 1 h)2 2 (3 1 h) 2 6
m 5 lim
hS0
h
9 1 6h 1 h 2 2 3 2 h 2 6
5 lim
hS0
h
5 lim (5 1 h)
hS0
55
Chapter 1: Introduction to Calculus
c. y 5 x 3 at x 5 22, y 5 28.
(22 1 h)3 1 8
m 5 lim
hS0
h
28 1 12h 2 6h 2 1 h 3 1 8
5 lim
hS0
h
5 lim (12 2 6h 1 h 2 )
hS0
5 12
9. a. y 5 "x 2 2; (3, 1)
"3 1 h 2 2 2 1
hS0
h
m 5 lim
5 lim £
hS0
5 lim
"1 1 h 2 1
"1 1 h 1 1
3
§
h
"1 1 h 1 1
hS0 "1
1
1h11
1
2
b. y 5 "x 2 5 at x 5 9, y 5 2
5
"9 1 h 2 5 2 2
hS0
h
m 5 lim
5 lim £
hS0
"4 1 h 2 2
"4 1 h 1 2
3
§
h
"4 1 h 1 2
5 lim
hS0 "4
1
1h12
1
5
4
c. y 5 "5x 2 1 at x 5 2, y 5 3
"10 1 5h 2 1 2 3
hS0
h
m 5 lim
5 lim £
hS0
5 lim
"9 1 5h 2 3
"9 1 5h 1 3
3
§
h
"9 1 5h 1 3
5
hS0 "9
1 5h 1 3
5
5
6
8
at (2, 4)
x
8
24
m 5 lim 2 1 h
hS0
h
24
5 lim
hS0 2 1 h
5 22
8
b. y 5
at x 5 1; y 5 2
31x
8
22
m 5 lim 4 1 h
hS0
h
10. a. y 5
Calculus and Vectors Solutions Manual
5 lim
hS0
22
41h
1
2
1
1
c. y 5
at x 5 3; y 5
x12
5
1
1
2
m 5 lim 5 1 h 5
hS0
h
21
5 lim
hS0 5(5 1 h)
1
52
10
11. a. Let y 5 f(x).
52
f(2) 5 (2)2 2 3(2) 5 4 2 6 5 22
f(2 1 h) 5 (2 1 h)2 2 3(2 1 h)
Using the limit of the difference quotient, the slope
of the tangent at x 5 2 is
f(2 1 h) 2 f(2)
m 5 lim
hS0
h
(2 1 h)2 2 3(2 1 h) 2 (22)
5 lim
hS0
h
4 1 4h 1 h 2 2 6 2 3h 1 2
5 lim
hS0
h
h2 1 h
5 lim
hS0
h
5 lim (h 1 1)
hS0
5011
51
Therefore, the slope of the tangent to
y 5 f(x) 5 x 2 2 3x at x 5 2 is 1.
4
b. f(22) 5
5 22
22
4
f(22 1 h) 5
22 1 h
Using the limit of the difference quotient, the slope
of the tangent at x 5 22 is
f(22 1 h) 2 f(22)
m 5 lim
hS0
h
5 lim
hS0
5 lim
4
2 (22)
22 1 h
h
4
12
22 1 h
h
4 2 4 1 2h 1
5 lim c
? d
hS0
22 1 h
h
2h
1
5 lim c
? d
hS0 22 1 h
h
hS0
1-7
2
hS0 22 1 h
2
5
22 1 0
5 21
5 lim
1
!0 1 9 1 3
1
5
313
1
5
6
Therefore, the slope of the tangent to
y 5 f(x) 5 !x 2 7 at x 5 16 is 16.
e. Let y 5 f(x).
5
4
Therefore, the slope of the tangent to f(x) 5 x at
x 5 22 is 21.
c. Let y 5 f(x).
f(1) 5 3(1)3 5 3
f(1 1 h) 5 3(1 1 h)3
Using the limit of the difference quotient, the slope
of the tangent at x 5 1 is
f(1 1 h) 2 f(1)
m 5 lim
hS0
h
3(1 1 h)3 2 3
5 lim
hS0
h
Using the binomial formula to expand (1 1 h)3 (or
one could simply expand using algebra), the slope m is
3(h 3 1 3h 2 1 3h 1 1) 2 (3)
5 lim
hS0
h
3
2
3h 1 9h 1 9h 1 3 2 3
5 lim
hS0
h
3h 3 1 9h 2 1 9h
5 lim
hS0
h
5 lim (3h 2 1 9h 1 9)
f(3) 5 "25 2 (3)2 5 !25 2 9 5 4
f(3 1 h) 5 "25 2 (3 1 h)2
5 "25 2 (9 1 6h 1 h 2 )
5 "25 2 9 2 6h 2 h 2
5 "16 2 6h 2 h 2
Using the limit of the difference quotient, the slope
of the tangent at x 5 3 is
f(3 1 h) 2 f(3)
m 5 lim
hS0
h
"16 2 6h 2 h 2 2 4
5 lim
hS0
h
5 lim c
hS0
hS0
5 3(0) 1 9(0) 1 9
59
Therefore, the slope of the tangent to
y 5 f(x) 5 3x 3 at x 5 1 is 9.
d. Let y 5 f(x).
f(16) 5 !16 2 7 5 !9 5 3
f(16 1 h) 5 !16 1 h 2 7 5 !h 1 9
Using the limit of the difference quotient, the slope
of the tangent at x 5 16 is
f(16 1 h) 2 f(16)
m 5 lim
hS0
h
!h 1 9 2 3
5 lim
hS0
h
!h 1 9 2 3 !h 1 9 1 3
5 lim
?
hS0
h
!h 1 9 1 3
(h 1 9) 2 9
5 lim
hS0 h( !h 1 9 1 3)
h
5 lim
hS0 h( !h 1 9 1 3)
1
5 lim
hS0 !h 1 9 1 3
1-8
5 lim
hS0
5 lim
hS0
5 lim
hS0
5
"16 2 6h 2 h 2 2 4
h
"16 2 6h 2 h 2 1 4
3
d
"16 2 6h 2 h 2 1 4
16 2 6h 2 h 2 2 16
h("16 2 6h 2 h 2 1 4)
h(26 2 h)
h("16 2 6h 2 h 2 1 4)
26 2 h
"16 2 6h 2 h 2 1 4
26 2 0
"16 2 6(0) 2 (0)2 1 4
26
5
!16 1 4
26
5
8
3
52
4
Therefore, the slope of the tangent to
y 5 f(x) 5 "25 2 x 2 at x 5 3 is 2 34.
f. Let y 5 f(x).
12
418
5
52
f(8) 5
822
6
12 1 h
4 1 (8 1 h)
5
f(8 1 h) 5
(8 1 h) 2 2
61h
Chapter 1: Introduction to Calculus
Using the limit of the difference quotient, the slope
of the tangent at x 5 8 is
f(8 1 h) 2 f(8)
m 5 lim
hS0
h
12 1 h
22
5 lim 6 1 h
hS0
h
12 1 h 2 12 2 2h 1
?
5 lim
hS0
61h
h
2h
1
?
5 lim
hS0 6 1 h
h
21
5 lim
hS0 6 1 h
21
5
610
1
52
6
Therefore, the slope of the tangent to
41x
y 5 f(x) 5 x 2 2 at x 5 8 is 2 16.
12.
y
8
4
0
–4
A
x
4
8
–4
y 5 "25 2 x 2 S Semi-circle centre (0, 0)
rad 5, y $ 0
OA is a radius.
The slope of OA is 43.
The slope of tangent is 2 34.
13. Take values of x close to the point, then
Dy
determine Dx.
14.
5 lim (3 1 h)
hS0
53
The slope of the tangent is 3.
y 2 1 5 3(x 2 3)
3x 2 y 2 8 5 0
(2 1 h)2 2 7(2 1 h) 1 12 2 2
16. m 5 lim
hS0
h
2
4 1 4h 1 h 2 14 2 7h 1 10
5 lim
hS0
h
2
23h 1 h
5 lim
hS0
h
5 lim ( 2 3 1 h)
hS0
5 23
The slope of the tangent is 23.
When x 5 2, y 5 2.
y 2 2 5 23(x 2 2)
3x 1 y 2 8 5 0
17. a. f(3) 5 9 2 12 1 1 5 22; (3, 22)
b. f(5) 5 25 2 20 1 1 5 6; (5, 6)
c. The slope of secant AB is
6 2 (22)
mAB 5
523
8
5
2
54
The equation of the secant is
y 2 y1 5 mAB (x 2 x1 )
y 1 2 5 4(x 2 3)
y 5 4x 2 14
d. Calculate the slope of the tangent.
f(x 1 h) 2 f(x)
m 5 lim
hS0
h
(x 1 h)2 2 4(x 1 h) 1 1 2 (x2 2 4x 1 1)
5 lim
hS0
h
x2 1 2xh 1 h2 2 4x 2 4h 1 1 2 x2 1 4x 2 1
5 lim
hS0
h
2xh 1 h2 2 4h
5 lim
hS0
h
5 lim (2x 1 h 2 4)
hS0
Since the tangent is horizontal, the slope is 0.
(3 1 h)2 2 3(3 1 h) 1 1 2 1
15. m 5 lim
hS0
h
9 1 6h 1 h2 2 9 2 3h
5 lim
hS0
h
3h 1 h2
5 lim
hS0
h
Calculus and Vectors Solutions Manual
5 2x 1 0 2 4
5 2x 2 4
When x 5 3, the slope is 2(3) 2 4 5 2. So the
equation of the tangent at A(3, 22) is
y 2 y1 5 m(x 2 x1 )
y 1 2 5 2(x 2 3)
y 5 2x 2 8
1-9
e. When x 5 5, the slope of the tangent is
2(5) 2 4 5 6.
So the equation of the tangent at B(5, 6) is
y 2 y1 5 m(x 2 x1 )
y 2 6 5 6(x 2 5)
y 5 6x 2 24
18. a.
P
The slope is undefined.
b.
P
The slope is 0.
c.
P
The slope is about –2.5.
d.
P
The slope is about 1.
e.
P
The slope is about 2 78.
f. There is no tangent at this point.
20
, p . 1 at (5, 10)
19. D(p) 5
"p 2 1
m 5 lim
hS0
20
2 10
!4 1 h
5 10 lim
hS0
5 10 lim
hS0
10
8
5
52
4
52
1-10
h
2 2 "4 1 h
3
2 1 "4 1 h
h"4 1 h
2 1 "4 1 h
4242h
h"4 1 h( 2 1 "4 1 h )
20. C(t) 5 100t 2 1 400t 1 5000
Slope at t 5 6
Cr(t) 5 200t 1 400
Cr(6) 5 1200 1 400 5 1600
Increasing at a rate of 1600 papers per month.
21. Point on f(x) 5 3x 2 2 4x tangent parallel to
y 5 8x. Therefore, tangent line has slope 8.
3(h 1 a)2 2 4(h 1 a) 2 3(a 2 1 4a)
m 5 lim
58
hS0
h
3h 2 1 6ah 2 4h
58
lim
hS0
h
6a 2 4 5 8
a52
The point has coordinates (2, 4).
1
4
22. y 5 x 3 2 5x 2
x
3
1
1
1
(a 1 h)2 2 a 3 5 a 2h 1 ah 2 1 h 3
3
3
3
1
lim aa 2 1 ah 1 h3 b 5 a 2
hS0
3
(a 1 h) 2 (2a)
5 lim 2
5 25
hS0
h
4
4
4a 1 4a 1 4h
2
1 52
a
a1h
a(a 1 h)
4
4
lim
5 2
hS0 a(a 1 h)
a
4
m 5 a2 2 5 1 2 5 0
a
a 4 2 5a 2 1 4 5 0
2
(a 2 4)(a 2 2 1) 5 0
a 5 62, a 5 61
Points on the graph for horizontal tangents are:
( 22, 283) , ( 21, 263) , ( 1, 2 263) , ( 2, 2 283) .
1
23. y 5 x 2 and y 5 2 x 2
2
1
x2 5 2 x2
2
1
x2 5
4
1
1
x 5 or x 5 2
2
2
The points of intersection are
P( 12, 14) , Q( 2 12, 14) .
Tangent to y 5 x2:
(a 1 h)2 2 a 2
m 5 lim
hS0
h
2ah 1 h 2
5 lim
hS0
h
5 2a.
Chapter 1: Introduction to Calculus
The slope of the tangent at a 5 12 is 1 5 mp,
at a 5 2 12 is 21 5 mq.
Tangents to y 5 12 2 x 2:
S 12 2 (a 1 h)2 T 2 S 12 2 a 2 T
m 5 lim
hS0
h
2
22ah 2 h
5 lim
hS0
h
5 22a.
The slope of the tangents at a 5 12 is 21 5 Mp;
at a 5 2 12 is 1 5 Mq
mpMp 5 21 and mqMq 5 21
Therefore, the tangents are perpendicular at the
points of intersection.
24. y 5 23x 3 2 2x, (21, 5)
23(21 1 h)3 2 2(21 1 h) 2 5
m 5 lim
hS0
h
23(21 1 3h 2 3h2 1 h3 ) 1 2 2 2h 2 5
5 lim
hS0
h
2
23(21 1 3h 2 3h 1 h3 ) 1 2 2 2h 2 5
5 lim
hS0
h
2
3 2 9h 1 9h 2 3h3 1 2 2 2h 2 5
5 lim
hS0
h
2
211h 1 9h 2 3h3
5 lim
hS0
h
5 lim (211 1 9h 2 3h2 )
hS0
5 211
The slope of the tangent is 211.
We want the line that is parallel to the tangent (i.e.
has slope 211) and passes through (2, 2). Then,
y 2 2 5 211(x 2 2)
y 5 211x 1 24
25. a. Let y 5 f(x).
f(a) 5 4a 2 1 5a 2 2
f(a 1 h) 5 4(a 1 h)2 1 5(a 1 h) 2 2
5 4(a 2 1 2ah 1 h 2 ) 1 5a 1 5h 2 2
5 4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2
Using the limit of the difference quotient, the slope
of the tangent at x 5 a is
f(a 1 h) 2 f(a)
m 5 lim
hS0
h
4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2
5 lim c
hS0
h
(4a 2 1 5a 2 2)
2
d
h
Calculus and Vectors Solutions Manual
5 lim c
4a 2 1 8ah 1 4h 2 1 5a 1 5h 2 2
hS0
h
24a 2 2 5a 1 2
1
d
h
8ah 1 4h 2 1 5h
5 lim
hS0
h
5 lim (8a 1 4h 1 5)
hS0
5 8a 1 4(0) 1 5
5 8a 1 5
b. To be parallel, the point on the parabola and the
line must have the same slope. So, first find the
slope of the line. The line 10x 2 2y 2 18 5 0 can
be rewritten as
22y 5 18 2 10x
18 2 10x
y5
22
y 5 29 1 5x
y 5 5x 2 9
So, the slope, m, of the line 10x 2 2y 2 18 5 0 is 5.
To be parallel, the slope at a must equal 5. From
part a., the slope of the tangent to the parabola at
x 5 a is 8a 1 5.
8a 1 5 5 5
8a 5 0
a50
Therefore, at the point (0, 22) the tangent line is
parallel to the line 10x 2 2y 2 18 5 0.
c. To be perpendicular, the point on the parabola
and the line must have slopes that are negative
reciprocals of each other. That is, their product must
equal 21. So, first find the slope of the line. The
line x 2 35y 1 7 5 0 can be rewritten as
235y 5 2x 2 7
2x 2 7
y5
235
1
7
y5 x1
35
35
So, the slope, m, of the line x 2 35y 1 7 5 0 is 351 .
To be perpendicular, the slope at a must equal
the negative reciprocal of the slope of the line
x 2 35y 1 7 5 0. That is, the slope of a must equal
235. From part a., the slope of the tangent to the
parabola at x 5 a is 8a 1 5.
8a 1 5 5 235
8a 5 240
a 5 25
Therefore, at the point (25, 73) the tangent line is
perpendicular to the line x 2 35y 1 7 5 0.
1-11
1.3 Rates of Change, pp. 29–31
1. v(t) 5 0 when t 5 0 or t 5 4.
2. a.
s(9) 2 s(2)
.
7
Slope of the secant between the
points (2, s(2)) and (9, s(9)).
s(6 1 h) 2 s(6)
.
h
hS0
b. lim
Slope of the tangent at the
point (6, s(6)).
"4 1 h 2 2
.
h
hS0
3. lim
Slope of the tangent to the
function with equation y 5 !x at the point (4, 2).
4. a. A and B
b. greater; the secant line through these two points
is steeper than the tangent line at B.
c. y
y = f(x)
B
A
C
D
E x
5. Speed is represented only by a number, not a
direction.
6. Yes, velocity needs to be described by a number
and a direction. Only the speed of the school bus
was given, not the direction, so it is not correct to
use the word “velocity.”
7. s(t) 5 320 2 5t 2, 0 # t # 8
a. Average velocity during the first second:
s(1) 2 s(0)
5 5 m>s;
1
third second:
s(3) 2 s(2)
45 2 20
5
5 25 m>s;
1
1
eighth second:
s(8) 2 s(7)
320 2 245
5
5 75 m>s.
1
1
b. Average velocity 3 # t # 8
s(8) 2 s(3)
320 2 45
275
5
5
5 55 m>s
823
5
5
c. s(t) 5 320 2 5t 2
320 2 5(2 1 h)2 2 (320 2 5(2)2 )
v(t) 5 lim
hS0
h
24h 1 h 2
5 5 lim
hS0
h
5 220
Velocity at t 5 2 is 20 m>s downward.
8. s(t) 5 8t(t 1 2), 0 # t # 5
a. i. from t 5 3 to t 5 4
s(4) 2 s(3)
Average velocity
1
1-12
5 32(6) 2 24(5)
5 24(8 2 5)
5 72 km>h
ii. from t 5 3 to t 5 3.1
s(3.1) 2 s(3)
0.1
126.48 2 120
5
0.1
5 64.8 km>h
iii. 3 # t # 3.01
s(3.01) 2 s(3)
0.01
5 64.08 km>h
b. Instantaneous velocity is approximately 64 km>h.
c. At t 5 3
s(t) 5 8t 2 1 16t
v(t) 5 16t 1 16
v(3) 5 48 1 16
5 64 km>h
9. a. N(t) 5 20t 2 t 2
N(3) 2 N(2)
1
51 2 36
5
1
5 15
15 terms are learned between t 5 2 and t 5 3.
20(2 1 h) 2 (2 1 h)2 2 36
b. lim
hS0
h
40 1 20h 2 4 2 4h 2 h2 2 36
5 lim
hS0
h
16h 2 h2
5 lim
hS0
h
5 lim (16 2 h)
hS0
5 16
At t 5 2, the student is learning at a rate of 16 terms>h.
10. a. M in mg in 1 mL of blood t hours after the
injection.
1
M(t) 5 2 t2 1 t; 0 # t # 3
3
Calculate the instantaneous rate of change when t 5 2.
2 1 (2 1 h)2 1 (2 1 h) 2 (2 43 1 2)
lim 3
hS0
h
4
4
1 2
2 2 h 2 3 h 1 2 1 h 1 43 2 2
5 lim 3 3
hS0
h
1
1 2
2 h 2 3h
5 lim 3
hS0
h
1
1
5 lim a2 2 hb
hS0
3
3
1
52
3
Chapter 1: Introduction to Calculus
Rate of change is 2 13 mg> h.
b. Amount of medicine in 1 mL of blood is being
dissipated throughout the system.
s
11. t 5
Å5
Calculate the instantaneous rate of change when
s 5 125.
lim
Ä
125 1 h
125
2Ä 5
5
h
hS0
5 lim
125 1 h
25
Ä 5
h
hS0
5 lim ≥
hS0
5 lim ≥
hS0
5 lim ≥
hS0
Ä
125 1 h
25
5
h
125 1 h
2 25
5
ha
125 1 h 2 125
5
ha
5
125 1 h
15
5
125 1 h
15
Ä 5
125 1 h
1 5b
Ä 5
125 1 h
1 5b
Ä 5
¥
¥
5a
Ä
1
¥
125 1 h
1 5b
5
1 5b
Ä 5
1
5
5(5 1 5)
1
5
50
At s 5 125, rate of change of time with respect to
height is 501 s>m.
5a
125
60
12. T(h) 5 h 1 2
Calculate the instantaneous rate of change when
h 5 3.
60
60
2 (3 1 2)
(3 1 k) 1 2
lim
kS0
k
5 lim
kS0
k
212k
5 lim
kS0 k(5 1 k)
212
5 lim
kS0 (5 1 k)
12
52
5
Temperature is decreasing at 125 °C> km.
13. h 5 25t 2 2 100t 1 100
When h 5 0, 25t 2 2 100t 1 100 5 0
t 2 2 4t 1 4 5 0
(t 2 2)2 5 0
t52
Calculate the instantaneous rate of change when t 5 2.
25(2 1 h)2 2 100(2 1 h) 1 100 2 0
lim
hS0
h
100 1 100h 1 25h2 2 200 2 100h 1 100
5 lim
hS0
h
25h2
5 lim
hS0 h
5 lim 25h
kS0
hS0
1
5 lim
hS0
?
Ä
5 lim
60
60 1 12k
2
51k
51k
50
It hit the ground in 2 s at a speed of 0 m> s.
14. Sale of x balls per week:
P(x) 5 160x 2 x 2 dollars.
a. P(40) 5 160(40) 2 (40)2
5 4800
Profit on the sale of 40 balls is $4800.
b. Calculate the instantaneous rate of change when
x 5 40.
160(40 1 h) 2 (40 1 h)2 2 4800
lim
hS0
h
6400 1 160h 2 1600 2 80h 2 h2 2 4800
5 lim
hS0
h
80h 2 h2
5 lim
hS0
h
5 lim (80 2 h)
hS0
5 80
Rate of change of profit is $80 per ball.
c.
60
2 12
51k
k
Calculus and Vectors Solutions Manual
Rate of change of profit is positive when the sales
level is less than 80.
1-13
15. a. f(x) 5 2x2 1 2x 1 3; (22, 25)
f(x) 2 f(22)
lim
xS22
x12
2x 2 1 2x 1 3 1 5
5 lim
xS22
x12
2
2 (x 2 2x 2 8)
5 lim
xS22
x12
(x 2 4)(x 1 2)
5 2 lim
xS22
x12
5 2 lim (x 2 4)
xS22
56
x
,x52
x21
x
22
x21
lim
xS2
x22
x 2 2x 1 2
5 lim
xS2 (x 2 1)(x 2 2)
2 (x 2 2)
5 lim
xS2 (x 2 1)(x 2 2)
5 21
c. f(x) 5 !x 1 1, x 5 24
f(x) 2 f(24)
5 lim
xS24
x 2 24
!x 1 1 2 5 !x 1 1 1 5
5 lim
?
xS24
x 2 24
!x 1 1 1 5
x 2 24
5 lim
xS24 (x 2 24)( !x 1 1 1 5)
1
5
10
16. S(x) 5 246 1 64x 2 8.9x 2 1 0.95x3
b. f(x) 5
For the year 2005, x 5 2005 2 1982 5 23. Hence,
the rate at which the average annual salary is changing
in 2005 is
P r(23) 5 64 2 17.8(23) 1 2.85(23)2 5
$1 162 250> years since 1982
17. s(t) 5 3t 2
a. The distance travelled from 0 s to 5 s is
s(5) 5 3(5)2 5 75 m
b. s(10) 5 3(10)2 5 300 m
The rate at which the avalanche is moving from 0 s
to 10 s is
Ds
300 2 0
5
Dt
10 2 0
5 30 m> s
c. Calculate the instantaneous rate of change when
t 5 10.
3(10 1 h)2 2 300
lim
hS0
h
300 1 60h 1 3h2 2 300
5 lim
hS0
h
60h 1 3h2
5 lim
hS0
h
5 lim (60 1 3h)
hS0
5 60
At 10 s the avalanche is moving at 60 m> s.
d. Set s(t) 5 600:
3t 2 5 600
t 2 5 200
t 5 610 !2
Since t $ 0, t 5 10 !2 8 14 s.
Calculate the instantaneous rate of change.
S(x 1 h) 2 S(x)
5 lim
hS0
h
246 1 64(x 1 h) 2 8.9(x 1 h)2 1 0.95(x 1 h)3 2 (246 2 64x 2 8.9x2 1 0.95x3 )
5 lim
hS0
h
246 2 246 1 64(x 1 h 2 x) 2 8.9(x 2 1 2xh 1 h 2 2 x 2 ) 1 0.95(x3 1 3x2h 1 3xh2 1 h3 2 x3 )
5 lim
hS0
h
64h 2 8.9(2xh 1 h 2 ) 1 0.95(3x 2h 1 3xh 2 1 h 3 )
5 lim
hS0
h
5 lim 364 2 8.9(2x 1 h) 1 0.95(3x 2 1 3xh 1 h 2 )4
hS0
5 64 2 8.9(2x 1 0) 1 0.95 33x 2 1 3x(0) 1 (0)24
5 64 2 17.8x 1 2.85x 2
1-14
Chapter 1: Introduction to Calculus
18. The coordinates of the point are aa, a b . The slope
1
1
of the tangent is 2 a 2. The equation of the tangent
1
1
1
2
is y 2 a 5 2 a 2 (x 2 a) or y 5 2 a 2 x 1 a. The
intercepts are a0, a b and (22a, 0). The tangent line
and the axes form a right triangle with legs of length
2
2
a
and 2a. The area of the triangle is 2 a a b (2a) 5 2.
1 2
19. C(x) 5 F 1 V(x)
C(x 1 h) 5 F 1 V(x 1 h)
Rate of change of cost is
C(x 1 h) 2 C(x)
lim
xSR
h
V(x 1 h) 2 V(x)
5 lim
h,
xSh
h
which is independent of F (fixed costs).
20. A(r) 5 pr 2
Rate of change of area is
A(r 1 h) 2 A(r)
lim
hS0
h
p(r 1 h)2 2 pr 2
5 lim
hS0
h
(r 1 h 2 r)(r 1 h 1 r)
5 p lim
hS0
h
5 2pr
r 5 100 m
Rate is 200p m2> m.
21. Cube of dimensions x by x by x has volume
V 5 x 3. Surface area is 6x 2.
1
Vr(x) 5 3x 2 5 surface area.
2
22. a. The surface area of a sphere is given by
A(r) 5 4pr 2.
The question asks for the instantaneous rate of
change of the surface when r 5 10. This is
A(10 1 h) 2 A(10)
lim
hS0
h
4p(10 1 h)2 2 4p(10)2
5 lim
hS0
h
4p(100 1 20h 1 h 2 ) 2 4p(100)
5 lim
hS0
h
400p 1 80ph 1 4ph 2 2 400p
5 lim
hS0
h
2
80ph 1 4ph
5 lim
hS0
h
5 lim (80p 1 4ph)
hS0
Calculus and Vectors Solutions Manual
5 80p 1 4p(0)
5 80p
Therefore, the instantaneous rate of change of
the surface area of a spherical balloon as it is
inflated when the radius reaches 10 cm is
80p cm2> unit of time.
b. The volume of a sphere is given by V(r) 5 43pr 3.
The question asks for the instantaneous rate of
change of the volume when r 5 5.
Note that the volume is deflating. So, find the rate
of the change of the volume when r 5 5 and then
make the answer negative to symbolize a deflating
spherical balloon.
V(5 1 h) 2 V(5)
lim
hS0
h
4
p(5
1 h)3 2 43 p(5)3
5 lim 3
hS0
h
Using the binomial formula to expand
(5 1 h)3 (or one could simply expand using
algebra), the limit is
4
p(h 3 1 15h 2 1 75h 1 125) 2 43 p(5)3
5 lim 3
hS0
h
4
3
2
ph
1
20ph
1
100ph
1 43 p(125)
5 lim 3
hS0
h
2 43 p(125)
h
4
3
ph
1
20ph 2 1 100ph
5 lim 3
hS0
h
4 2
5 lim a ph 1 20ph 1 100pb
hS0 3
4
5 p(0)2 1 20p(0) 1 100p
3
5 100p
Because the balloon is deflating, the instantaneous rate
of change of the volume of the spherical balloon when
the radius reaches 5 cm is 2100p cm3>unit of time.
Mid-Chapter Review pp. 32–33
1. a. Corresponding conjugate: !5 1 !2.
( !5 2 !2)( !5 1 !2)
5 ( !25 1 !10 2 !10 2 !4)
5522
53
b. Corresponding conjugate: 3!5 2 2!2.
(3 !5 1 2 !2)(3 !5 2 2 !2)
5 (9 !25 2 6!10 1 6 !10 2 4 !4)
5 9(5) 2 4(2)
5 45 2 8
5 37
1-15
18. The coordinates of the point are aa, a b . The slope
1
1
of the tangent is 2 a 2. The equation of the tangent
1
1
1
2
is y 2 a 5 2 a 2 (x 2 a) or y 5 2 a 2 x 1 a. The
intercepts are a0, a b and (22a, 0). The tangent line
and the axes form a right triangle with legs of length
2
2
a
and 2a. The area of the triangle is 2 a a b (2a) 5 2.
1 2
19. C(x) 5 F 1 V(x)
C(x 1 h) 5 F 1 V(x 1 h)
Rate of change of cost is
C(x 1 h) 2 C(x)
lim
xSR
h
V(x 1 h) 2 V(x)
5 lim
h,
xSh
h
which is independent of F (fixed costs).
20. A(r) 5 pr 2
Rate of change of area is
A(r 1 h) 2 A(r)
lim
hS0
h
p(r 1 h)2 2 pr 2
5 lim
hS0
h
(r 1 h 2 r)(r 1 h 1 r)
5 p lim
hS0
h
5 2pr
r 5 100 m
Rate is 200p m2> m.
21. Cube of dimensions x by x by x has volume
V 5 x 3. Surface area is 6x 2.
1
Vr(x) 5 3x 2 5 surface area.
2
22. a. The surface area of a sphere is given by
A(r) 5 4pr 2.
The question asks for the instantaneous rate of
change of the surface when r 5 10. This is
A(10 1 h) 2 A(10)
lim
hS0
h
4p(10 1 h)2 2 4p(10)2
5 lim
hS0
h
4p(100 1 20h 1 h 2 ) 2 4p(100)
5 lim
hS0
h
400p 1 80ph 1 4ph 2 2 400p
5 lim
hS0
h
2
80ph 1 4ph
5 lim
hS0
h
5 lim (80p 1 4ph)
hS0
Calculus and Vectors Solutions Manual
5 80p 1 4p(0)
5 80p
Therefore, the instantaneous rate of change of
the surface area of a spherical balloon as it is
inflated when the radius reaches 10 cm is
80p cm2> unit of time.
b. The volume of a sphere is given by V(r) 5 43pr 3.
The question asks for the instantaneous rate of
change of the volume when r 5 5.
Note that the volume is deflating. So, find the rate
of the change of the volume when r 5 5 and then
make the answer negative to symbolize a deflating
spherical balloon.
V(5 1 h) 2 V(5)
lim
hS0
h
4
p(5
1 h)3 2 43 p(5)3
5 lim 3
hS0
h
Using the binomial formula to expand
(5 1 h)3 (or one could simply expand using
algebra), the limit is
4
p(h 3 1 15h 2 1 75h 1 125) 2 43 p(5)3
5 lim 3
hS0
h
4
3
2
ph
1
20ph
1
100ph
1 43 p(125)
5 lim 3
hS0
h
2 43 p(125)
h
4
3
ph
1
20ph 2 1 100ph
5 lim 3
hS0
h
4 2
5 lim a ph 1 20ph 1 100pb
hS0 3
4
5 p(0)2 1 20p(0) 1 100p
3
5 100p
Because the balloon is deflating, the instantaneous rate
of change of the volume of the spherical balloon when
the radius reaches 5 cm is 2100p cm3>unit of time.
Mid-Chapter Review pp. 32–33
1. a. Corresponding conjugate: !5 1 !2.
( !5 2 !2)( !5 1 !2)
5 ( !25 1 !10 2 !10 2 !4)
5522
53
b. Corresponding conjugate: 3!5 2 2!2.
(3 !5 1 2 !2)(3 !5 2 2 !2)
5 (9 !25 2 6!10 1 6 !10 2 4 !4)
5 9(5) 2 4(2)
5 45 2 8
5 37
1-15
c. Corresponding conjugate: 9 2 2 !5.
(9 1 2 !5)(9 2 2!5)
5 (81 2 18!5 1 18!5 2 4!25)
5 81 2 4(5)
5 81 2 20
5 61
d. Corresponding conjugate: 3 !5 1 2 !10.
(3!5 2 2!10)(3 !5 1 2 !10)
5 (9!25 1 6 !50 2 6!50 2 4 !100)
5 9(5) 2 4(10)
5 45 2 40
55
6 1 !2 !3
2. a.
?
!3
!3
6!3 1 !6
5
!9
6!3 1 !6
5
3
2!3 1 4 !3
b.
?
!3
!3
2!9 1 4 !3
5
!9
6 1 4!3
5
3
5
!7 1 4
?
c.
!7 2 4 !7 1 4
5( !7 1 4)
5
!49 1 4 !7 2 4!7 2 16
5( !7 1 4)
5
7 2 16
5( !7 1 4)
52
9
2!3
!3 1 2
d.
?
!3 2 2 !3 1 2
2 !9 1 4 !3
5
!9 1 2 !3 2 2 !3 2 4
6 1 4!3
5
324
6 1 4!3
5
21
5 22(3 1 2!3)
2 !3 2 4
5!3
?
e.
2 !3 1 4 2 !3 2 4
10 !9 2 20 !3
5
4!9 2 8 !3 1 8!3 2 16
30 2 20 !3
5
12 2 16
1-16
30 2 20 !3
24
10 !3 2 15
5
2
3 !2
2 !3 1 5
f.
?
2 !3 2 5 2 !3 1 5
3 !2(2 !3 1 5)
5
4 !9 1 10 !3 2 10 !3 2 25
3 !2(2 !3 1 5)
5
4(3) 2 25
5
3 !2(2 !3 1 5)
12 2 25
3 !2(2 !3 1 5)
5
213
3 !2(2!3 1 5)
52
13
!2 !2
3. a.
?
5
!2
!4
5
5 !2
2
5
5 !2
!3
!3
b.
?
6 1 !2 !3
!9
5
!3(6 1 !2)
3
5
!3(6 1 !2)
5
!7 2 4 !7 1 4
?
5
!7 1 4
!49 1 4 !7 2 4 !7 2 16
5
5( !7 1 4)
7 2 16
5
5( !7 1 4)
9
52
5( !7 1 4)
c.
2 !3 2 5 2 !3 1 5
?
3 !2
2 !3 1 5
4 !9 1 10 !3 2 10 !3 2 25
5
3 !2(2!3 1 5)
4(3) 2 25
5
3 !2(2!3 1 5)
13
12 2 25
52
5
3 !2(2!3 1 5)
3!2(2!3 1 5)
d.
Chapter 1: Introduction to Calculus
!3 2 !7 !3 1 !7
?
4
!3 1 !7
!9 1 !21 2 !21 2 !49
5
4( !3 1 !7)
327
5
4( !3 1 !7)
4
52
4( !3 1 !7)
1
52
( !3 1 !7)
2!3 1 !7 2 !3 2 !7
f.
?
5
2 !3 2 !7
4 !9 2 2 !21 1 2 !21 2 !49
5
5(2!3 2 !7)
4(3) 2 7
5
5(2!3 2 !7)
12 2 7
5
5(2!3 2 !7)
1
5
(2!3 2 !7)
2
4. a.
m52 ;
3
2
y 2 6 5 2 (x 2 0)
3
2
y2652 x
3
2
x1y2650
3
11 2 7
4
b.
m5
5 51
622
4
y 2 7 5 1(x 2 2)
y275x22
2x 1 y 2 5 5 0
x2y1550
m54
c.
y 2 6 5 4(x 2 2)
y 2 6 5 4x 2 8
24x 1 y 1 2 5 0
4x 2 y 2 2 5 0
1
m5
d.
5
1
y 2 (22) 5 (x 2 (21))
5
1
1
y125 x1
5
5
10
1
1
2 50
2 x1y1
5
5
5
e.
Calculus and Vectors Solutions Manual
1
9
2 x1y1 50
5
5
1
9
x2y2 50
5
5
x 2 5y 2 9 5 0
5. The slope of PQ is
f(1 1 h) 2 (21)
m 5 lim
hS0
(1 1 h) 2 1
2 (1 1 h)2 1 1
hS0
h
2 (1 1 2h 1 h 2 ) 1 1
5 lim
hS0
h
21 2 2h 2 h 2 1 1
5 lim
hS0
h
22h 2 h 2
5 lim
hS0
h
5 lim (22 2 h)
5 lim
hS0
5 22 2 (0)
5 22
So, the slope of PQ with f(x) 5 2x 2 is 22.
6. a. Unlisted y-coordinates for Q are found by
substituting the x-coordinates into the given function.
The slope of the line PQ with the given points is
given by the following: Let P 5 (x1, y1 ) and
y 2y
1
Q 5 (y1, y2 ). Then, the slope 5 m 5 2
.
x2 2 x1
P
Q
Slope of Line PQ
(21, 1)
(22, 6)
(21, 1)
(21.5, 3.25)
2 4.5
(21, 1)
(21.1, 1.41)
2 4.1
(21, 1)
(21.01, 1.040 1)
2 4.01
(21, 1)
(21.001, 1.004 001)
2 4.001
P
Q
Slope of Line PQ
(21, 1)
(0, 22)
(21, 1)
(20.5, 20.75)
2 3.5
(21, 1)
(20.9, 0.61)
2 3.9
(21, 1)
(20.99, 0.9601)
2 3.99
(21, 1)
(20.999, 0.996 001)
2 3.999
25
23
b. The slope from the right and from the left appear
to approach 24. The slope of the tangent to the
graph of f(x) at point P is about 24.
c. With the points P 5 (21, 1) and
Q 5 (21 1 h, f(21 1 h)), the slope, m, of PQ is
the following:
1-17
y2 2 y1
x2 2 x1
3(21 1 h)2 2 2(21 1 h) 2 24 2 (1)
5
(21 1 h) 2 (21)
1 2 2h 1 h 2 1 2 2 2h 2 2 2 1
5
21 1 h 1 1
h 2 2 4h
5
h
5h24
d. The slope of the tangent is lim f(x).
m5
4
x22
f(6 1 h) 2 f(6)
m 5 lim
hS0
h
c. y 5 f(x) 5
5 lim
h
hS0
5 lim
hS0
hS0
In this case, as h goes to zero, h 2 4 goes to
h 2 4 5 0 2 4 5 24. The slope of the tangent to
the graph of f(x) at the point P is 24.
e. The answers are equal.
4
4
2
61h22
622
5 lim
4
4
24
h14
h
4
21
h14
h
4 2 (h 1 4) 1
5 lim a
b
hS0
h14
h
hS0
f(23 1 h) 2 f(23)
hS0
h
3(23 1 h)2 1 3(23 1 h) 2 54 2 3(23)2 1 3(23) 2 54
5 lim
hS0
h
9 2 6h 1 h 2 2 9 1 3h 2 5 2 (9 2 9 2 5)
5 lim
hS0
h
h 2 2 3h 2 5 2 (25)
5 lim
hS0
h
2h 1
2
5 lim a
b
h 2 3h
hS0 h 1 4 h
5 lim
hS0
h
21
5 lim
5 lim (h 2 3)
hS0 h 1 4
hS0
5023
21
5
5 23
014
1
1
52
b. y 5 f(x) 5
4
x
f(5 1 h) 2 f(5)
f( 13 1 h) 2 f( 13 )
d. m 5 lim
m 5 lim
hS0
h
hS0
h
!5 1 h 1 4 2 !5 1 4
1
1
5 lim
2 1
1
hS0
h
3 1 h
3
5 lim
!9
1
h
2
!9
hS0
h
5 lim
1
1
hS0
h
( 3 ) 2 ( 3 1 h)
1 1
!9
1
h23
(
1
h)
3 3
5 lim
5 lim
hS0
h
hS0
h
!9 1 h 2 3 !9 1 h 1 3
2h 1
5 lim
?
5 lim a 1 1 b
hS0
h
!9 1 h 1 3
hS0 9 1 3 h h
9 1 h 1 3 !9 1 h 2 3 !9 1 h 2 9
21
5 lim
5 lim 1 1
hS0
h( !9 1 h 1 3)
hS0 9 1 3 h
h
21
5 lim
51 1
hS0
h(
!9
1
h 1 3)
9 1 3 (0)
1
5 29
5 lim
hS0 !9 1 h 1 3
7. a. m 5 lim
1-18
Chapter 1: Introduction to Calculus
1
!9 1 0 1 3
1
5
6
8. s(t) 5 6t(t 1 1) 5 6t 2 1 6t
s(3) 2 s(2)
a. i. average velocity 5
322
5 36(3)2 1 6(3)42 36(2)2 1 6(2)4
5 6(9) 1 18 2 (24 1 12)
5 54 1 18 2 36
5 36 km> h
s(2.1) 2 s(2)
ii. average velocity 5
2.1 2 2
36(2.1)2 1 6(2.1)4 2 36(2)2 1 6(2)4
5
0.1
326.46 1 12.64 2 324 1 124
5
0.1
39.06 2 36
5
0.1
3.06
5
0.1
5 30.6 km> h
s(2.01) 2 s(2)
iii. average velocity 5
2.01 2 2
36(2.01)2 1 6(2.01)4 2 36(2)2 1 6(2)4
5
0.01
324.2406 1 12.064 2 36(2)2 1 6(2)4
5
0.01
36.3006 2 324 1 124
5
0.01
36.3006 2 36
5
0.01
0.3006
5
0.01
5 30.06 km> h
5
b. At the time t 5 2, the velocity of the car appears
to approach 30 km> h.
f(2 1 h) 2 f(2)
c. average velocity 5
(2 1 h) 2 (2)
2
36(2 1 h) 1 6(2 1 h)4 2 36(2)2 1 6(2)4
5
h
36(4 1 4h 1 h 2 ) 1 12 1 6h4 2 324 1 124
5
h
324 1 24h 1 6h 2 1 12 1 6h4 2 36
5
h
6h 2 1 30h 1 36 2 36
5
h
Calculus and Vectors Solutions Manual
6h 2 1 30h
h
5 (6h 1 30) km> h
d. When t 5 2, the velocity is the limit as h
approaches 0.
velocity 5 lim (6h 1 30)
5
hS0
5 6(0) 1 30
5 30
Therefore, when t 5 2 the velocity is 30 km> h.
9. a. The instantaneous rate of change of f(x) with
respect to x at x 5 2 is given by
f(2 1 h) 2 f(2)
lim
hS0
h
35 2 (2 1 h)24 2 35 2 (2)24
5 lim
hS0
h
5 2 (4 1 4h 1 h 2 ) 2 1
5 lim
hS0
h
5 2 4 2 4h 2 h 2 2 1
5 lim
hS0
h
2
2h 2 4h
5 lim
hS0
h
5 lim (2h 2 4)
hS0
5 2 (0) 2 4
5 24
b. The instantaneous rate of change of f(x) with
respect to x at x 5 12 is given by
f( 1 1 h) 2 f( 12 )
lim 2
hS0
h
3
3
2 1
1
1h
2
5 lim 2
hS0
h
3
26
1
1
h
2
5 lim
hS0
h
3 2 6( 12 1 h) 1
5 lim
?
1
hS0
h
2 1 h
3 2 3 2 6h 1
?
5 lim
1
hS0
h
2 1 h
26h 1
5 lim 1
?
hS0 2 1 h
h
26
5 lim 1
hS0 2 1 h
26
51
2 1 0
5 212
1-19
10. a. The average rate of change of V(t) with
respect to t during the first 20 minutes is given by
f(20) 2 f(0)
20 2 0
350(30 2 20)24 2 350(30 2 0)24
5
20
5000 2 45 000
5
20
40 000
52
20
5 22000 L> min
b. The rate of change of V(t) with respect to t at the
time t 5 20 is given by
f(20 1 h) 2 f(20)
lim
hS0
h
350(30 2 (20 1 h))24 2 350(30 2 20)24
5 lim
hS0
h
2
350(10 2 h) 4 2 350(10)24
5 lim
hS0
h
350(100 2 20h 1 h 2 )4 2 350(100)4
5 lim
hS0
h
5000 2 1000h 1 50h 2 2 5000
5 lim
hS0
h
50h 2 2 1000h
5 lim
hS0
h
5 lim 50h 2 1000
hS0
5 50(0) 2 1000
5 21000 L> min
11. a. Let y 5 f(x).
f(4) 5 (4)2 1 (4) 2 3 5 16 1 1 5 17
f(4 1 h) 5 (4 1 h)2 1 (4 1 h) 2 3
5 16 1 8h 1 h 2 1 h 1 1
5 h 2 1 9h 1 17
Using the limit of the difference quotient, the slope
of the tangent at x 5 4 is
f(4 1 h) 2 f(4)
m 5 lim
hS0
h
2
h 1 9h 1 17 2 (17)
5 lim
hS0
h
2
h 1 9h
5 lim
hS0
h
5 lim (h 1 9)
hS0
5019
59
Therefore, the slope of the tangent to
y 5 f(x) 5 x 2 1 x 2 3 at x 5 4 is 9.
1-20
So an equation of the tangent at x 5 4 is given by
y 2 17 5 9(x 2 4)
y 2 17 5 9x 2 36
29x 1 y 2 17 1 36 5 0
29x 1 y 1 19 5 0
b.
Let y 5 f(x).
f(22) 5 2(22)2 2 7 5 2(4) 2 7 5 1
f(22 1 h) 5 2(22 1 h)2 2 7
5 2(4 2 4h 1 h 2 ) 2 7
5 8 2 8h 1 2h 2 2 7
5 2h 2 2 8h 1 1
Using the limit of the difference quotient, the slope
of the tangent at x 5 4 is
f(22 1 h) 2 f(22)
m 5 lim
hS0
h
2h 2 2 8h 1 1 2 (1)
5 lim
hS0
h
2h 2 2 8h
5 lim
hS0
h
5 lim (2h 2 8)
hS0
5 2(0) 2 8
5 28
Therefore, the slope of the tangent to
y 5 f(x) 5 2x 2 2 7 at x 5 22 is 28.
So an equation of the tangent at x 5 22
is given by
y 2 1 5 28(x 2 (22))
y 2 1 5 28x 2 16
8x 1 y 2 1 1 16 5 0
8x 1 y 1 15 5 0
c. f(21) 5 3(21)2 1 2(21) 2 5 5 3 2 2 2 5
5 24
f(21 1 h) 5 3(21 1 h)2 1 2(21 1 h) 2 5
5 3(1 2 2h 1 h 2 ) 2 2 1 2h 2 5
5 3 2 6h 1 3h 2 2 7 1 2h
5 3h 2 2 4h 2 4
Using the limit of the difference quotient, the slope
of the tangent at x 5 4 is
f(21 1 h) 2 f(21)
m 5 lim
hS0
h
2
3h 2 4h 2 4 2 (24)
5 lim
hS0
h
3h 2 2 4h
5 lim
hS0
h
5 lim (3h 2 4)
hS0
5 3(0) 2 4
5 24
Chapter 1: Introduction to Calculus
Therefore, the slope of the tangent to
y 5 f(x) 5 3x 2 1 2x 2 5 at x 5 21 is 24.
So an equation of the tangent at x 5 24 is given by
y 2 (24) 5 24(x 2 (21))
y 1 4 5 24(x 1 1)
y 1 4 5 24x 2 4
4x 1 y 1 4 1 4 5 0
4x 1 y 1 8 5 0
d. f(1) 5 5(1)2 2 8(1) 1 3 5 5 2 8 1 3 5 0
f(1 1 h) 5 5(1 1 h)2 2 8(1 1 h) 1 3
5 5(1 1 2h 1 h 2 ) 2 8 2 8h 1 3
5 5 1 10h 1 5h 2 2 5 2 8h
5 5h 2 1 2h
Using the limit of the difference quotient, the slope
of the tangent at x 5 1 is
f(1 1 h) 2 f(1)
m 5 lim
hS0
h
2
5h 1 2h 2 (0)
5 lim
hS0
h
5 lim (5h 1 2)
hS0
5 5(0) 1 2
52
Therefore, the slope of the tangent to
y 5 f(x) 5 5x 2 2 8x 1 3 at x 5 1 is 2.
So an equation of the tangent at x 5 1 is given by
y 2 0 5 2(x 2 1)
y 5 2x 2 2
22x 1 y 1 2 5 0
12. a. Using the limit of the difference quotient, the
slope of the tangent at x 5 25 is
f(25 1 h) 2 f(25)
m 5 lim
hS0
h
25 1 h
25
1
5 lim a
2
b?
hS0 25 1 h 1 3
25 1 3
h
25 1 h
5
1
5 lim a
2 b?
hS0 22 1 h
2
h
210 1 2h 2 (210 1 5h)
1
5 lim a
b?
hS0
24 1 2h
h
1
210 1 2h 1 10 2 5h
5 lim a
b?
hS0
24 1 2h
h
23h
1
5 lim a
b?
hS0 24 1 2h
h
23
5 lim a
b
hS0 24 1 2h
23
5
24 1 2(0)
3
5
4
Calculus and Vectors Solutions Manual
Therefore, the slope of the tangent to
x
f(x) 5 x 1 3 at x 5 25 is 43.
So an equation of the tangent at x 5 34 is given by
5
3
y 2 5 (x 2 (25))
2
4
5
3
15
y2 5 x1
2
4
4
3
10
15
2 x1y2
2
50
4
4
4
3
25
2 x1y2
50
4
4
23x 1 4y 2 25 5 0
b. Using the limit of the difference quotient, the
slope of the tangent at x 5 21 is
f(21 1 h) 2 f(21)
m 5 lim
hS0
h
2(21 1 h) 1 5
2(21) 1 5
1
5 lim a
2
b?
hS0 5(21 1 h) 2 1
5(21) 2 1
h
22 1 2h 1 5
22 1 5
1
5 lim a
2
b?
hS0 25 1 5h 2 1
25 2 1
h
2h 1 3
3
1
5 lim a
2
b?
hS0 5h 2 6
26
h
1
1
2h 1 3
5 lim a
1 b?
hS0 5h 2 6
2
h
4h 1 6 1 5h 2 6
1
5 lim a
b?
hS0
10h 2 12
h
9h
1
5 lim a
b?
hS0 10h 2 12
h
9
5 lim a
b
hS0 10h 2 12
9
5
10(0) 2 12
9
52
12
3
52
4
Therefore, the slope of the tangent to
2x 1 5
f(x) 5 5x 2 1 at x 5 21 is 2 34.
So an equation of the tangent at x 5 2 34 is given by
1
3
y 2 a2 b 5 2 (x 2 (21))
2
4
1
3
3
y1 52 x2
2
4
4
4y 1 2 5 23x 2 3
3x 1 4y 1 2 1 3 5 0
3x 1 4y 1 5 5 0
1-21
1.4 The Limit of a Function,
pp. 37–39
1. a.
6
27
99
b. p
2. One way to find a limit is to evaluate the function
for values of the independent variable that get
progressively closer to the given value of the
independent variable.
3. a. A right-sided limit is the value that a
function gets close to as the values of the
independent variable decrease and get close
to a given value.
b. A left-sided limit is the value that a function
gets close to as the values of the independent
variable increase and get close to a given
value.
c. A (two-sided) limit is the value that a function
gets close to as the values of the independent
variable get close to a given value, regardless
of whether the values increase or decrease
toward the given value.
4. a. 25
b. 3 1 7 5 10
c. 102 5 100
d. 4 2 3(22)2 5 28
e. 4
f. 23 5 8
5. Even though f(4) 5 21, the limit is 1, since that
is the value that the function approaches from the
left and the right of x 5 4.
6. a. 0
b. 2
c. 21
d. 2
7. a. 2
b. 1
c. does not exist
8. a. 9 2 (21)2 5 8
b.
9. 22 1 1 5 5
0 1 20
5 "4
Å015
52
c. "5 2 1 5 "4
52
1-22
y
4
2
–4
–2
x
0
2
4
10. a. Since 0 is not a value for which the function is
undefined, one may substitute 0 in for x to find that
lim1 x 4 5 lim x 4
xS0
xS0
5 (0)4
50
b. Since 2 is not a value for which the function is
undefined, one may substitute 2 in for x to find that
lim2 (x 2 2 4) 5 lim (x 2 2 4)
xS2
xS2
5 (2)2 2 4
5424
50
c. Since 3 is not a value for which the function is
undefined, one may substitute 3 in for x to find that
lim2 (x 2 2 4) 5 lim (x 2 2 4)
xS3
xS3
5 (3)2 2 4
5924
55
d. Since 1 is not a value for which the function is
undefined, one may substitute 1 in for x to find that
1
1
5 lim
lim
xS1 1 x 2 3
xS1 x 2 3
1
5
123
1
52
2
e. Since 3 is not a value for which the function is
undefined, one may substitute 3 in for x to find that
1
1
lim1
5 lim
xS3 x 1 2
xS3 x 1 2
1
5
312
1
5
5
f. If 3 is substituted in the function for x, then the
function is undefined because of division by zero.
There does not exist a way to divide out the x 2 3 in
Chapter 1: Introduction to Calculus
1
the denominator. Also, lim1 x 2 3 approaches infinity,
xS3
d.
8
6
4
2
1
while lim2 x 2 3 approaches negative infinity.
xS3
1
1
1
Therefore, since lim1 x 2 3 2 lim2 x 2 3, lim x 2 3
xS3
xS3
xS3
does not exist.
11. a.
y
8
6
4
2
–8 –6 –4 –2 0
–2
–4
–6
–8
x
2 4 6 8
–8 –6 –4 –2 0
–2
–4
–6
–8
xS21
not exist.
b.
8
6
4
2
–8 –6 –4 –2 0
–2
–4
y
x
–8 –6 –4 –2 0
–2
–4
2 4 6 8
8
6
4
2
–8 –6 –4 –2 0
–2
–4
–6
–8
–8 –6 –4 –2 0
–2
–4
y
2 4 6 8
2 4 6 8
x
2 4 6 8
x
2 4 6 8
6 y
4
2
d.
x
x
6 y
4
2
c.
xS2
is equal to 2.
c.
xS20.5
6 y
4
2
b.
xS2
xS20.5
6
4
2
lim f(x) 5 lim2 f(x). Therefore, lim f(x) exists and
xS2 1
2 4 6 8
does not exist.
12. Answers may vary. For example:
a.
y
xS21
–8 –6 –4 –2 0
–2
–4
–6
–8
x
lim f(x) 2 lim 2 f(x). Therefore, lim f(x)
xS20.5 1
lim f(x) 2 lim 2 f(x). Therefore, lim f(x) does
xS21 1
y
–8 –6 –4 –2 0
–2
–4
x
2 4 6 8
13. f(x) 5 mx 1 b
lim f(x) 5 22 m 1 b 5 22
xS1
lim f(x) 5 lim12 f(x). Therefore, lim1 f(x) exists and
xS 12 1
xS 2
xS 2
is equal to 2.
Calculus and Vectors Solutions Manual
lim f(x) 5 4
xS21
2m 1 b 5 4
2b 5 2
b 5 1, m 5 23
1-23
14. f(x) 5 ax 2 1 bx 1 c, a 2 0
f(0) 5 0
c50
lim f(x) 5 5
a1b55
xS1
lim f(x) 5 8
xS22
4a 2 2b 5 8
6a 5 18
a 5 3,
b52
Therefore, the values are a 5 3, b 5 2, and c 5 0.
15. a.
y
10
8
6
4
2
x
0
–4 –2
2 4 6 8 10 12
–2
1
(6)2
tS6
12
36
531
12
5313
56
1
lim1 p(t) 5 2 1 (6)2
tS6
18
36
521
18
5212
54
c. Since p(t) is measured in thousands, right before
the chemical spill there were 6000 fish in the lake.
Right after the chemical spill there were 4000 fish
in the lake. So, 6000 2 4000 5 2000 fish were
killed by the spill.
d. The question asks for the time, t, after the chemical
spill when there are once again 6000 fish in the lake.
Use the second equation to set up an equation that is
modelled by
1
6 5 2 1 t2
18
1 2
45 t
18
72 5 t 2
!75 5 t
(The question asks for time so the negative answer
is disregarded.)
So, at time t 5 !72 8 8.49 years the population
has recovered to the level before the spill.
b. lim2 p(t) 5 3 1
1-24
1.5 Properties of Limits, pp. 45–47
1. lim (3 1 x) and lim (x 1 3) have the same value,
xS2
xS2
but lim 3 1 x does not. Since there are no brackets
xS2
around the expression, the limit only applies to 3,
and there is no value for the last term, x.
2. Factor the numerator and denominator. Cancel
any common factors. Substitute the given value of x.
3. If the two one-sided limits have the same value,
then the value of the limit is equal to the value of
the one-sided limits. If the one-sided limits do not
have the same value, then the limit does not exist.
3(2)
4. a. 2
51
2 12
b. (21)4 1 (21)3 1 (21)2 5 1
1 2
1 2
c. c "9 1
d 5 a3 1 b
3
"9
100
5
9
d. (2p)3 1 p2 (2p) 2 5p3 5 8p3 1 2p3 2 5p3
5 5p3
e. "3 1 "1 1 0 5 "3 1 1
52
26
23 2 3
5
f.
Å 2(23) 1 4 Å 22
5 "3
(22)3
5 22
5. a.
22 2 2
2
2
5
b.
!1 1 1
!2
5 "2
6. Since substituting t 5 1 does not make the
denominator 0, direct substitution works.
12125
25
5
621
5
5 21
4 2 x2
(2 2 x)(2 1 x)
5 lim
7. a. lim
xS2 2 2 x
xS2
(2 2 x)
5 lim (2 1 x)
xS2
54
2x 2 1 5x 1 3
(x 1 1)(2x 1 3)
5 lim
b. lim
xS21
x11
xS21
x11
55
x 3 2 27
(x 2 3)(x 2 1 3x 1 9)
5 lim
c. lim
xS3 x 2 3
xS3
x23
591919
5 27
Chapter 1: Introduction to Calculus
14. f(x) 5 ax 2 1 bx 1 c, a 2 0
f(0) 5 0
c50
lim f(x) 5 5
a1b55
xS1
lim f(x) 5 8
xS22
4a 2 2b 5 8
6a 5 18
a 5 3,
b52
Therefore, the values are a 5 3, b 5 2, and c 5 0.
15. a.
y
10
8
6
4
2
x
0
–4 –2
2 4 6 8 10 12
–2
1
(6)2
tS6
12
36
531
12
5313
56
1
lim1 p(t) 5 2 1 (6)2
tS6
18
36
521
18
5212
54
c. Since p(t) is measured in thousands, right before
the chemical spill there were 6000 fish in the lake.
Right after the chemical spill there were 4000 fish
in the lake. So, 6000 2 4000 5 2000 fish were
killed by the spill.
d. The question asks for the time, t, after the chemical
spill when there are once again 6000 fish in the lake.
Use the second equation to set up an equation that is
modelled by
1
6 5 2 1 t2
18
1 2
45 t
18
72 5 t 2
!75 5 t
(The question asks for time so the negative answer
is disregarded.)
So, at time t 5 !72 8 8.49 years the population
has recovered to the level before the spill.
b. lim2 p(t) 5 3 1
1-24
1.5 Properties of Limits, pp. 45–47
1. lim (3 1 x) and lim (x 1 3) have the same value,
xS2
xS2
but lim 3 1 x does not. Since there are no brackets
xS2
around the expression, the limit only applies to 3,
and there is no value for the last term, x.
2. Factor the numerator and denominator. Cancel
any common factors. Substitute the given value of x.
3. If the two one-sided limits have the same value,
then the value of the limit is equal to the value of
the one-sided limits. If the one-sided limits do not
have the same value, then the limit does not exist.
3(2)
4. a. 2
51
2 12
b. (21)4 1 (21)3 1 (21)2 5 1
1 2
1 2
c. c "9 1
d 5 a3 1 b
3
"9
100
5
9
d. (2p)3 1 p2 (2p) 2 5p3 5 8p3 1 2p3 2 5p3
5 5p3
e. "3 1 "1 1 0 5 "3 1 1
52
26
23 2 3
5
f.
Å 2(23) 1 4 Å 22
5 "3
(22)3
5 22
5. a.
22 2 2
2
2
5
b.
!1 1 1
!2
5 "2
6. Since substituting t 5 1 does not make the
denominator 0, direct substitution works.
12125
25
5
621
5
5 21
4 2 x2
(2 2 x)(2 1 x)
5 lim
7. a. lim
xS2 2 2 x
xS2
(2 2 x)
5 lim (2 1 x)
xS2
54
2x 2 1 5x 1 3
(x 1 1)(2x 1 3)
5 lim
b. lim
xS21
x11
xS21
x11
55
x 3 2 27
(x 2 3)(x 2 1 3x 1 9)
5 lim
c. lim
xS3 x 2 3
xS3
x23
591919
5 27
Chapter 1: Introduction to Calculus
2 1 "4 1 x
2 2 "4 1 x
3
d. lim £
§
xS0
x
2 1 "4 1 x
5 lim
xS0
52
1
4
21
2 1 "4 1 x
"x 2 2
"x 2 2
5 lim
xS4 x 2 4
xS4 ("x 2 2)("x 1 2)
1
5
4
e. lim
"7 2 x 2 "7 1 x
"7 2 x 1 "7 1 x
§
3
f. lim £
x
xS0
"7 2 x 1 "7 1 x
5 lim
xS0
52
72x272x
x("7 2 x 1 "7 1 x)
1
"7
3
"
x22
xS8 x 2 8
8. a. lim
3
x. Therefore, u 3 5 x as x S 8, u S 2.
Let u 5 "
u22
1
5 lim 2
Here, lim 3
xS2 u 2 8
xS2 u 1 2u 1 4
1
5
12
1
27 2 x
Let x 3 5 u
b. lim 13
xS27 x 2 3
x 5 u3
3
u 2 27
x S 27, u S 3.
5 lim
xS3 u 2 3
(u 2 3)(u 2 1 3u 1 9)
5 2lim
xS3
u23
5 2 (9 1 9 1 9)
5 227
1
1
x6 2 1
x 6 5 u, x 5 u 6
c. lim
x S 1, u S 1
xS1 x 2 1
u21
5 lim 6
xS1 u 2 1
(u 2 1)
5 lim
5
4
3
2
xS1 (u 2 1)(u 1 u 1 u 1 u 1 u 1 1)
1
5
6
1
1
x6 2 1
Let x 6 5 u
d. lim 13
xS1 x 2 1
u6 5 x
1
u21
x 3 5 u2
5 lim 2
xS1 u 2 1
As x S 1, u S 1
Calculus and Vectors Solutions Manual
5 lim
xS1
5
1
2
e. lim
u21
(u 2 1)(u 1 1)
"x 2 2
"x 2 8
u22
5 lim 3
xS2 u 2 8
xS4
5 lim
xS2
5
1
12
3
1
Let x 2 5 u
3
x 2 5 u3
x S 4, u S 2
u22
(u 2 2)(u 2 1 2u 1 4)
1
1
(x 1 8)3 2 2
Let (x 1 8)3 5 u
x
xS0
x 1 8 5 u3
u22
x 5 u3 2 8
lim 3
xS2 u 2 8
x S 0, u S 2
1
5
12
16 2 16
9. a.
50
64 1 64
16 2 16
b.
50
16 2 20 1 6
x2 1 x
x(x 1 1)
c. lim
5 lim
xS21 x 1 1
xS21 x 1 1
5 21
"x 1 1 2 1
"x 1 1 2 1
d. lim
5 lim
x
xS0
xS0 x 1 1 2 1
"x 1 1 2 1
5 lim
xS0 ("x 1 1 2 1)("x
1 1 1 1)
1
5
2
(x 1 h)2 2 x 2
2xh 1 h 2
5 lim
e. lim
hS0
h
hS0
h
5 2x
1
1
2
ba
2
b
f. lim a
xS1 x 2 1
x13
3x 1 5
1
3x 1 5 2 2x 2 6
5 lim a
ba
b
xS1 x 2 1
(x 1 3)(3x 1 5)
1
5 lim
xS1 (x 1 3)(3x 1 5)
1
5
4(8)
1
5
32
f. lim
1-25
0x 2 50
does not exist.
xS5 x 2 5
0x 2 50
x25
lim1
5 lim1
xS5 x 2 5
xS5 x 2 5
51
0x 2 50
x25
lim2
5 lim2 2 a
b
xS5 x 2 5
xS5
x25
5 21
y
2
10. a. lim
0
–4
xS2
(x 2 2)(x 1 1)
(x 2 2)(x 1 1)
5 lim2 2
0x 2 20
xS2
(x 2 2)
5 lim2 2 (x 1 1)
xS2
5 23
2
–2
0 x 1 2 0 5 x 1 2 if x . 22
5 2 (x 1 2) if x , 22
(x 1 2)(x 1 2)2
lim
5 lim 1 (x 1 2)2 5 0
xS22 1
x12
xS22
(x 1 2)(x 1 2)2
lim
50
xS222
2 (x 1 2)
d.
0 2x 2 5 0 (x 1 1)
does not exist.
2x 2 5
xS 2
5
0 2x 2 5 0 5 2x 2 5, x $
2
(2x 2 5)(x 1 1)
lim
5x11
2x 2 5
xS 52
5
0 2x 2 5 0 5 2 (2x 2 5), x ,
2
2 (2x 2 5)(x 1 1)
lim2
5 2 (x 1 1)
2x 2 5
xS 52
y
4
b. lim5
4
2
–4
–2
x
2
4
–2
–4
x
2
4
DT
20
20
20
20
–4
20
20
2
x 2x22
(x 2 2)(x 1 1)
5 lim
xS2
0x 2 20
xS2
0x 2 20
(x 2 2)(x 1 1)
(x 2 2)(x 1 1)
lim
5 lim1
xS2 1
0x 2 20
xS2
x22
5 lim1 x 1 1
lim
xS2
53
1-26
0
11. a.
–2
c.
y
2
1
0
4
–4
–2
–2
2
–2
8
–1
–4
x
0
x
4
y
4
–4
1
–8
lim2
T
V
240
19.1482
220
20.7908
0
22.4334
20
24.0760
40
25.7186
60
27.3612
80
29.0038
DV
1.6426
1.6426
1.6426
1.6426
1.6426
1.6426
DV is constant, therefore T and V form a linear
relationship.
DV
b. V 5
?T1K
DT
DV
1.6426
5
5 0.082 13
DT
20
Chapter 1: Introduction to Calculus
V 5 0.082 13T 1 K
T50
V 5 22.4334
Therefore, k 5 22.4334 and
V 5 0.082 13T 1 22.4334.
V 2 22.4334
c. T 5
0.082 13
d. lim T 5 2273.145
f(x)
g(x)
5 1 and lim
52
xS0 x
xS0 x
g(x)
b5032
a. lim g(x) 5 lim xa
x
xS0
xS0
50
f (x)
f(x)
1
x
5 lim g (x)
5
b. lim
xS0 g(x)
xS0 x
2
15. lim
vS0
e.
!x 1 1 2 !2x 1 1
xS0 !3x 1 4 2 !2x 1 4
V
12
16. lim
10
!x 1 1 2 !2x 1 1
!x 1 1 1 !2x 1 1
8
5 lim c
6
3
xS0
4
3
2
0
T
0
2
4
6
8
10
12
x2 2 4
xS5 f(x)
lim (x2 2 4)
xS0
!3x 1 4 1 !2x 1 4
d
!3x 1 4 1 !2x 1 4
(x 1 1 2 2x 2 1)
!3x 1 4 1 !2x 1 4
d
3
(3x 1 4 2 2x 2 4)
!x 1 1 1 !2x 1 1
212
111
5 22
x 2 1 0 x 2 1 021
17. lim
xS1
0x 2 10
x S 11 0 x 2 1 0 5 x 2 1
x2 1 x 2 2
(x 1 2)(x 2 1)
5
x21
x21
x 2 1 0 x 2 1 021
lim
53
xS1 1
0x 2 10
x S 12 0 x 2 1 0 5 2x 1 1
x2 2 x
x(x 2 1)
lim2
5 lim2
xS1 2x 1 1
xS1 2x 1 1
5 21
Therefore, this limit does not exist.
y
4
52
12. lim
5
5 lim c
!x 1 1 1 !2x 1 1
!3x 1 4 2 !2x 1 4
xS5
lim f(x)
xS5
21
5
3
57
13. lim f(x) 5 3
xS4
a. lim 3 f(x)4 3 5 33 5 27
xS4
b.
3 f(x)4 2 2 x 2
( f(x) 2 x)( f(x) 1 x)
5 lim
xS4 f(x) 1 x
xS4
f(x) 1 x
5 lim ( f(x) 2 x)
lim
xS4
5324
5 21
c. lim "3f(x) 2 2x 5 "3 3 3 2 2 3 4
2
f(x)
51
14. lim
xS0 x
0
xS4
51
a. lim f(x) 5 lim c
f(x)
3 xd 5 0
x
xS0
xS0
f(x)
x f(x)
5 lim c
d 50
b. lim
x
xS0 g(x)
xS0 g(x)
Calculus and Vectors Solutions Manual
–4
–2
x
2
4
–2
–4
1-27
1.6 Continuity, pp. 51–53
1. Anywhere that you can see breaks or jumps is a
place where the function is not continuous.
2. It means that on that domain, you can trace the
graph of the function without lifting your pencil.
3. point discontinuity
10
8
6
4
2
y
hole
–2 0
–2
x
2 4 6
5. a. The function is a polynomial, so the function
is continuous for all real numbers.
b. The function is a polynomial, so the function is
continuous for all real numbers.
c. x 2 2 5x 5 x(x 2 5)
The is continuous for all real numbers except
0 and 5.
d. The is continuous for all real numbers greater
than or equal to 22.
e. The is continuous for all real numbers.
f. The is continuous for all real numbers.
6. g(x) is a linear function (a polynomial),
and so is continuous everywhere,
including x 5 2.
7.
y
8
jump discontinuity
10
8
6
4
2
4
y
–8
x
The function is continuous everywhere.
8.
y
4
y
2
–4
–2
x
1
x
0
2
4
–2
2 3 4
vertical
asymptote
4. a. x 5 3 makes the denominator 0.
b. x 5 0 makes the denominator 0.
c. x 5 0 makes the denominator 0.
d. x 5 3 and x 5 23 make the denominator 0.
e. x 2 1 x 2 6 5 (x 1 3)(x 2 2)
x 5 23 and x 5 2 make the denominator 0.
f. The function has different one-sided limits at x 5 3.
1-28
8
–8
2 4 6
infinite discontinuity
–1 0
–2
–4
4
–4
–2 0
–2
10
8
6
4
2
–4
x
0
–4
The function is discontinuous at x 5 0.
9.
y
4
2
0
x
200
400
600
Chapter 1: Introduction to Calculus
x2 2 x 2 6
xS3
xS3
x23
(x 2 3)(x 1 2)
5 lim
xS3
x23
55
Function is discontinuous at x 5 3.
11. Discontinuous at x 5 2
y
4
10. lim f(x) 5 lim
2
–4
–2
x
0
2
4
–2
–4
12. g(x) 5 e
x 1 3, if x 2 3
2 1 !k, if x 5 3
g(x) is continuous.
2 1 "k 5 6
"k 5 4, k 5 16
13.
21, if x , 0
f(x) 5 • 0, if x 5 0
1, if x . 0
a.
y
4
2
–4
–2
x
0
2
4
–2
–4
b. i. From the graph, lim2 f(x) 5 21.
xS0
ii. From the graph, lim1 f(x) 5 1.
Thus, lim f(x) 5 4. But, f(3) 5 2. Hence f is not
xS3
continuous at x 5 2 (and also not continuous over
23 , x , 8).
15. The function is to be continuous at x 5 1 and
discontinuous at x 5 2.
f(x) 5 μ
Ax 2 B
, if x # 1
x22
3x, if 1 , x , 2
Bx 2 A, if x $ 2
2
For f(x) to be continuous at x 5 1:
A(1) 2 B
5 3(1)
122
A(1) 2 B 5 23
A5B23
For f(x) to be discontinuous at x 5 2:
B(2)2 2 A 2 3(2)
4B 2 A 2 6
If 4B 2 A . 6, then
if 4B 2 A , 6, then
4B 2 (B 2 3) . 6
4B 2 B 1 3 , 6
3B 1 3 . 6
3B 1 3 , 6
3B . 3
3B , 3
B . 1 and
B , 1 and
A . 22
A , 22
This shows that A and B can be any set of real
numbers such that
(1) A 5 B 2 3
(2) 4B 2 A 2 6 (if B . 1, then A . 22 if B , 1,
then A , 22)
A 5 1 and B 5 22 is not a solution because then
the graph would be continuous at x 5 2.
2x, if 23 # x # 22
2
16. f(x) 5 • ax 1 b, if 22 , x , 0
6, if x 5 0
at x 5 22, 4a 1 b 5 2
at x 5 0, b 5 6.
a 5 21
2x, if 23 # x # 22
f(x) 5 • 2x2 1 b, if 22 , x , 0
6, if x 5 0
if a 5 21, b 5 6. f(x) is continuous.
xS0
x0 x 2 1 0
, if x 2 1
g(x) 5 • x 2 1
iii. Since the one-sided limits differ, lim f(x) does
xS0
not exist.
c. f is not continuous since lim f(x) does not exist.
17.
14. a. From the graph, f(3) 5 2.
b. From the graph, lim2 f(x) 5 4.
lim g(x) 5 21
a. xS12
¶ lim g(x)
xS1
lim1 g(x) 5 1
c. lim2 f(x) 5 4 5 lim2 f(x)
lim g(x) does not exist.
xS0
xS3
xS3
xS3
Calculus and Vectors Solutions Manual
0, if x 5 1
xS1
xS1
1-29
b.
4
y
c. h(x) 5
2
–4
–2
0
x
2
4
–2
–4
g(x) is discontinuous at x 5 1.
Review Exercise, pp. 56–59
1. a. f(22) 5 36, f(3) 5 21
21 2 36
m5
3 2 (22)
5 23
b. f(21) 5 13, f(4) 5 48
48 2 13
m5
4 2 (21)
57
c. f(1) 5 23
5(1 1 2h 1 h 2 ) 2 (23)
m 5 lim
hS0
h
2h 1 h 2
5 lim
hS0
h
5 lim 2 1 h
m 5 lim
2
2
, Pa4, b
!x 1 5
3
2
2
23
!4 1 h 1 5
h
3 2 !9 1 h
3 1 !9 1 h
d
5 2 lim c
3
hS0
3h!9 1 h
3 1 !9 1 h
1
5 2 lim c 2
d
hS0
3 !9 1 h(3 1 !9 1 h)
2
52
9(6)
1
52
27
5
5
d. f(x) 5
, Pa4, b
x22
2
5
5
2
41h22
2
m 5 lim
hS0
h
10 2 5(2 1 h)
5 lim
hS0 h(2 1 h)(2)
25h
5 lim 2
hS0
h(2 1 h)(2)
5
52
4
4 2 x2 , if x # 1
3. f(x) 5 e
2x 1 1, if x . 1
hS0
a. Slope at P(21, 3) f(x) 5 4 2 x 2
hS0
52
y 2 (23) 5 2(x 2 1)
2x 2 y 2 5 5 0
3
2. a. f(x) 5
, P(2, 1)
x11
3
21
m531h
h
1
5 lim 2
hS0
31h
1
52
3
b. g(x) 5 "x 1 2, P(21, 1)
"21 1 h 1 2 2 1
h
!h 1 1 2 1
!h 1 1 1 1
5 lim c
d
3
x
hS0
!h 1 1 1 1
1
5 lim
hS0 !h 1 1 1 1
1
5
2
m 5 lim
hS0
1-30
4 2 (21 1 h)2 2 3
hS0
h
4 2 1 1 2h 2 h 2 2 3
5 lim
hS0
h
5 lim (2 2 h)
m 5 lim
hS0
52
Slope of the graph at P(21, 3) is 2.
b. Slope at P(2, 0.5)
f(x) 5 2x 1 1
f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5
5 2h
2h
m 5 lim
52
hS0 h
Slope of the graph at P(2, 0.5) is 2.
4. s(t) 5 25t 2 1 180
a. s(0) 5 180, s(1) 5 175, s(2) 5 160
Average velocity during the first second is
s(1) 2 s(0)
5 25
1
m> s.
Chapter 1: Introduction to Calculus
b.
4
y
c. h(x) 5
2
–4
–2
0
x
2
4
–2
–4
g(x) is discontinuous at x 5 1.
Review Exercise, pp. 56–59
1. a. f(22) 5 36, f(3) 5 21
21 2 36
m5
3 2 (22)
5 23
b. f(21) 5 13, f(4) 5 48
48 2 13
m5
4 2 (21)
57
c. f(1) 5 23
5(1 1 2h 1 h 2 ) 2 (23)
m 5 lim
hS0
h
2h 1 h 2
5 lim
hS0
h
5 lim 2 1 h
m 5 lim
2
2
, Pa4, b
!x 1 5
3
2
2
23
!4 1 h 1 5
h
3 2 !9 1 h
3 1 !9 1 h
d
5 2 lim c
3
hS0
3h!9 1 h
3 1 !9 1 h
1
5 2 lim c 2
d
hS0
3 !9 1 h(3 1 !9 1 h)
2
52
9(6)
1
52
27
5
5
d. f(x) 5
, Pa4, b
x22
2
5
5
2
41h22
2
m 5 lim
hS0
h
10 2 5(2 1 h)
5 lim
hS0 h(2 1 h)(2)
25h
5 lim 2
hS0
h(2 1 h)(2)
5
52
4
4 2 x2 , if x # 1
3. f(x) 5 e
2x 1 1, if x . 1
hS0
a. Slope at P(21, 3) f(x) 5 4 2 x 2
hS0
52
y 2 (23) 5 2(x 2 1)
2x 2 y 2 5 5 0
3
2. a. f(x) 5
, P(2, 1)
x11
3
21
m531h
h
1
5 lim 2
hS0
31h
1
52
3
b. g(x) 5 "x 1 2, P(21, 1)
"21 1 h 1 2 2 1
h
!h 1 1 2 1
!h 1 1 1 1
5 lim c
d
3
x
hS0
!h 1 1 1 1
1
5 lim
hS0 !h 1 1 1 1
1
5
2
m 5 lim
hS0
1-30
4 2 (21 1 h)2 2 3
hS0
h
4 2 1 1 2h 2 h 2 2 3
5 lim
hS0
h
5 lim (2 2 h)
m 5 lim
hS0
52
Slope of the graph at P(21, 3) is 2.
b. Slope at P(2, 0.5)
f(x) 5 2x 1 1
f(2 1 h) 2 f(2) 5 2(2 1 h) 1 1 2 5
5 2h
2h
m 5 lim
52
hS0 h
Slope of the graph at P(2, 0.5) is 2.
4. s(t) 5 25t 2 1 180
a. s(0) 5 180, s(1) 5 175, s(2) 5 160
Average velocity during the first second is
s(1) 2 s(0)
5 25
1
m> s.
Chapter 1: Introduction to Calculus
Average velocity during the second second is
s(2) 2 s(1)
5 215
1
m> s.
b. At t 5 4:
s(4 1 h) 2 s(4)
5 25(4 1 h)2 1 180 2 (25(16) 1 180)
5 280 2 40h 2 5h 2 1 180 1 80 2 180
s(4 1 h) 2 s(4)
240h 2 5h 2
5
h
h
v(4) 5 lim (240 2 5h) 5 240
Velocity is 240 m> s.
c. Time to reach ground is when s(t) 5 0.
Therefore, 25t 2 1 180 5 0
t 2 5 36
t 5 6, t . 0.
Velocity at t 5 6:
s(6 1 h) 5 25(36 1 12h 1 h 2 ) 1 180
5 260h 2 5h 2
s(6) 5 0
Therefore, v(6) 5 lim (260 2 5h) 5 260.
hS0
hS0
5. M(t) 5 t 2 mass in grams
a. Growth during 3 # t # 3.01
M(3.01) 5 (3.01)2 5 9.0601
M(3) 5 32
59
Grew 0.0601 g during this time interval.
b. Average rate of growth is
0.0601
5 6.01 g> min.
0.01
c.
s(3 1 h) 5 9 1 6h 1 h 2
s(3) 5 9
s(3 1 h) 2 s(3)
6h 1 h 2
5
h
h
Rate of growth is lim (6 1 h) 5 6 g> min.
c. Present rate of change:
Q(h) 5 104 (h 2 1 15h 1 70)
Q(0) 5 104 1 70
Q(h) 2 Q(0)
lim
5 lim 104 (h 1 15)
hS0
h
hS0
5 15 3 104 t per year.
d. Q(a 1 h)
5 104 3a 2 1 2ah 1 h 2 1 15a 1 15h 1 704
Q(a) 5 104 3a 2 1 15a 1 704
104 32ah 1 h 2 1 15h4
Q(a 1 h) 2 Q(a)
5
h
h
Q(a 1 h) 2 Q(a)
lim
5 lim 104 (2a 1 h 1 15)
hS0
h
hS0
5 (2a 1 15)104
Now,
(2a 1 15)104 5 3 3 105
2a 1 15 5 30
a 5 7.5
It will take 7.5 years to reach a rate of
3.0 3 105 t per year.
7. a. From the graph, the limit is 10.
b. 7; 0
c. p(t) is discontinuous for t 5 3 and t 5 4.
8. a. Answers will vary. lim f(x) 5 0.5, f is
xS21
discontinuous at x 5 21
2
y
1
–2
–1
x
0
1
2
–1
hS0
6. Q(t) 5 104 (t 2 1 15t 1 70) tonnes of waste,
0 # t # 10
a. At t 5 0,
Q(t) 5 70 3 104
5 700 000.
700 000 t have accumulated up to now.
b. Over the next three years, the average rate of
change:
Q(3) 5 104 (9 1 45 1 70)
5 124 3 104
Q(0) 5 70 3 104
Q(3) 2 Q(0)
54 3 104
5
3
3
5 18 3 104 t per year.
–2
b. f(x) 5 24 if x , 3; f is increasing for x . 3
lim1 f(x) 5 1
xS3
4
y
2
–4
–2
0
x
2
4
–2
–4
Calculus and Vectors Solutions Manual
1-31
9. a.
4
y
13. a.
x
2
–4
–2
0
1.9
1.99
1.999
2.001
2.01
2.1
x22
0.344 83 0.334 45 0.333 44 0.333 22 0.332 23 0.322 58
x2 2 x 2 2
1
3
x
2
4
–2
–4
x 1 1, if x , 21
b. f(x) 5 • 2x 1 1, if 21 # x , 1
x 2 2, if x . 1
Discontinuous at x 5 21 and x 5 1.
c. They do not exist.
10. The function is not continuous at x 5 24
because the function is not defined at x 5 24.
(x 5 24 makes the denominator 0.)
2x 2 2
11. f(x) 5 2
x 1x22
2(x 2 1)
5
(x 2 1)(x 1 2)
a. f is discontinuous at x 5 1 and x 5 22.
2
b. lim f(x) 5 lim
xS1
xS1 x 1 2
2
5
3
2
lim f(x): 5 lim 1
5 1`
xS22
xS22 x 1 2
2
lim
5 2`
xS222 x 1 2
lim f(x) does not exist.
xS22
1
, lim f(x) does not exist.
x 2 xS0
b. g(x) 5 x(x 2 5), lim g(x) 5 0
xS0
x 3 2 27
c. h(x) 5 2
,
x 29
37
lim h(x) 5
5 5.2857
xS4
7
lim h(x) does not exist.
12. a. f(x) 5
xS23
b.
x
0.9
0.99
0.999
1.001
1.01
1.1
x21
0.526 32 0.502 51 0.500 25 0.499 75 0.497 51 0.476 19
x2 2 1
1
2
14.
20.1
x
20.01
20.001
0.001
0.01
0.1
"x 1 3 2 "3 0.291 12 0.288 92 0.2887 0.288 65 0.288 43 0.286 31
x
!x 1 3 2 !3 !x 1 3 1 !3
d
?
x
!x 1 3 1 !3
x1323
5 lim
xS0 xA !x 1 3 1 !3B
x
5 lim
xS0 xA !x 1 3 1 !3B
1
5 lim
xS0 !x 1 3 1 !3
1
5
2 !3
lim c
xS0
This agrees well with the values in the table.
15. a. f(x) 5
"x 1 2 2 2
x22
x
2.1
2.01
2.001
2.0001
f(x)
0.248 46
0.249 84
0.249 98
0.25
x 5 2.0001
f(x) 8 0.25
1-32
Chapter 1: Introduction to Calculus
b.
5 lim
xS0
1
!5
5
1
A !x 1 5 1 !5 2 xB
(x 2 2)(x 1 2)
(x 2 2)(x 2 1 2x 1 4)
x12
5 lim 2
xS2 x 1 2x 1 4
(2) 1 2
5
(2)2 1 2(2) 1 4
4
5
12
1
5
3
4 2 !12 1 x 4 1 !12 1 x
d
?
e. lim c
xS4
x24
4 1 !12 1 x
16 2 (12 1 x)
5 lim
xS4 (x 2 4)(4 1 !12 1 x)
42x
5 lim
xS4 (x 2 4)(4 1 !12 1 x)
2 (x 2 4)
5 lim
xS4 (x 2 4)(4 1 !12 1 x)
21
5 lim
xS4 4 1 !12 1 x
21
5
4 1 !12 1 (4)
21
5
414
1
52
8
1
1
1
2 b
f. lim a
x
xS0
21x
2
1
x
5 lim c 3 2
d
xS0 x
2(2 1 x)
1
5 lim c 2
d
xS0
2(2 1 x)
1
52
4
18. a. The function is not defined for x , 3, so
there is no left-side limit.
b. Even after dividing out common factors from
numerator and denominator, there is a factor of
x 2 2 in the denominator; the graph has a vertical
asymptote at x 5 2.
25, if x , 1
c. f(x) 5 e
2, if x $ 1
lim2 f(x) 5 25 2 lim1 f(x) 5 2
d. lim
lim f(x) 5 0.25
xS2
!x 1 2 2 2
!x 1 2 1 2
d
3
xS2
x22
!x 1 2 1 2
1
5 lim
xS2 !x 1 2 1 2
1
5 5 0.25
4
(5 1 h)2 2 25
16. a. lim
hS0
h
5 lim (10 1 h)
c. lim c
hS0
5 10
Slope of the tangent to y 5 x 2 at x 5 5 is 10.
"4 1 h 2 2
"4 1 h 2 2
5 lim
h
hS0 4 1 h 2 4
1
5 lim
hS0 !4 1 h 1 2
1
5
4
Slope of the tangent to y 5 "x at x 5 4 is 14.
b. lim
hS0
1
1
24
41h
4242h
hS0
h
hS0 4(4 1 h)(h)
1
5 lim 2
hS0
4(4 1 h)
1
52
16
1
Slope of the tangent to y 5 x at (x 5 4) is 2 161 .
(x 1 4)(x 1 8)
5 lim (x 1 8)
17. a. lim
xS24
x14
xS24
5 (24) 1 8
54
2
2
(x 1 4a) 2 25a
(x 2 a)(x 1 9a)
5 lim
b. lim
x
2
a
x2a
xSa
xSa
5 10a
!x 1 5 2 !5 2 x
!x 1 5 1 !5 2 x
d
3
c. lim c
x
xS0
!x 1 5 1 !5 2 x
x15251x
5 lim
xS0 xA !x 1 5 1 !5 2 xB
c. lim
5 lim
xS2
xS1
Calculus and Vectors Solutions Manual
xS1
1-33
d. The function has a vertical asymptote at x 5 2.
0x0
e. lim
xS0 x
x S 02 0 x 0 5 2x
0x0
lim2
5 21
xS0 x
0x0
lim1
51
xS0 x
0x0
0x0
lim1
2 lim2
xS0 x
xS0 x
5x2, if x , 21
f(x) 5 e
f.
2x 1 1, if x $ 21
lim 1 f(x) 5 21
xS21
lim f(x) 5 5
xS212
lim f(x) 2 lim 2 f(x)
xS21 1
xS21
Therefore, lim f(x) does not exist.
xS21
19. a.
23(1 1 h)2 1 6(1 1 h) 1 4 2 (23 1 6 1 4)
hS0
h
23 2 6h 2 h2 1 6 1 6h 1 4 2 7
5 lim
hS0
h
2h2
5 lim
hS0 h
5 lim 2h
m 5 lim
hS0
50
When x 5 1, y 5 7.
The equation of the tangent is y 2 7 5 0(x 2 1)
y57
b.
(22 1 h)2 2 (22 1 h) 2 1 2 (4 1 2 2 1)
m 5 lim
hS0
h
4 2 4h 1 h2 1 2 2 h 2 1 2 5
5 lim
hS0
h
25h 1 h2
5 lim
hS0
h
5 lim (25 1 h)
hS0
5 25
When x 5 22, y 5 5.
The equation of the tangent is y 2 5 5 25(x 1 2)
y 5 25x 2 5
6(21 1 h)3 2 3 2 (26 2 3)
c. m 5 lim
hS0
h
6(21 1 3h 2 3h2 1 h3 ) 2 3 1 9
5 lim
hS0
h
1-34
18h 2 18h2 1 6h3
hS0
h
5 lim (18 2 18h 1 6h2 )
5 lim
hS0
5 18
When x 5 21, y 5 29.
The equation of the tangent is
y 2 (29) 5 18(x 2 (21))
y 5 18x 1 9
22(3 1 h)4 2 (2162)
d. m 5 lim
hS0
h
22(81 1 108h 1 54h2 1 12h3 1 h4 ) 1 162
5 lim
hS0
h
2216h 2 108h2 2 24h3 2 2h4
5 lim
hS0
h
5 lim ( 2 216 2 108h 2 24h2 2 2h3 )
hS0
5 2216
When x 5 3, y 5 2162.
The equation of the tangent is
y 2 (2162) 5 2216(x 2 3)
y 5 2216x 1 486
20. P(t) 5 20 1 61t 1 3t 2
a. P(8) 5 20 1 61(8) 1 3(8)2
5 700 000
b.
20 1 61(8 1 h) 1 3(8 1 h)2 2 (20 1 488 1 192)
lim
hS0
h
20 1 488 1 61h 1 3(64 1 16h 1 h2 ) 2 700
5 lim
hS0
h
20 1 488 1 61h 1 192 1 48h 1 3h2 2 700
5 lim
hS0
h
2
109h 1 3h
5 lim
hS0
h
5 lim (109 1 3h)
hS0
5 109
The population is changing at the rate of
109 000>h.
Chapter 1 Test, p. 60
1
1. lim x 2 1 does not exist since
xS1
1
1
5 1` 2 lim2
5 2 `.
xS1 x 2 1
xS1 x 2 1
2. f(x) 5 5x 2 2 8x
f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36
f(1) 5 5 2 8 5 23
36 1 3
39
Slope of secant is
52
22 2 1
3
5 213
lim1
Chapter 1: Introduction to Calculus
d. The function has a vertical asymptote at x 5 2.
0x0
e. lim
xS0 x
x S 02 0 x 0 5 2x
0x0
lim2
5 21
xS0 x
0x0
lim1
51
xS0 x
0x0
0x0
lim1
2 lim2
xS0 x
xS0 x
5x2, if x , 21
f(x) 5 e
f.
2x 1 1, if x $ 21
lim 1 f(x) 5 21
xS21
lim f(x) 5 5
xS212
lim f(x) 2 lim 2 f(x)
xS21 1
xS21
Therefore, lim f(x) does not exist.
xS21
19. a.
23(1 1 h)2 1 6(1 1 h) 1 4 2 (23 1 6 1 4)
hS0
h
23 2 6h 2 h2 1 6 1 6h 1 4 2 7
5 lim
hS0
h
2h2
5 lim
hS0 h
5 lim 2h
m 5 lim
hS0
50
When x 5 1, y 5 7.
The equation of the tangent is y 2 7 5 0(x 2 1)
y57
b.
(22 1 h)2 2 (22 1 h) 2 1 2 (4 1 2 2 1)
m 5 lim
hS0
h
4 2 4h 1 h2 1 2 2 h 2 1 2 5
5 lim
hS0
h
25h 1 h2
5 lim
hS0
h
5 lim (25 1 h)
hS0
5 25
When x 5 22, y 5 5.
The equation of the tangent is y 2 5 5 25(x 1 2)
y 5 25x 2 5
6(21 1 h)3 2 3 2 (26 2 3)
c. m 5 lim
hS0
h
6(21 1 3h 2 3h2 1 h3 ) 2 3 1 9
5 lim
hS0
h
1-34
18h 2 18h2 1 6h3
hS0
h
5 lim (18 2 18h 1 6h2 )
5 lim
hS0
5 18
When x 5 21, y 5 29.
The equation of the tangent is
y 2 (29) 5 18(x 2 (21))
y 5 18x 1 9
22(3 1 h)4 2 (2162)
d. m 5 lim
hS0
h
22(81 1 108h 1 54h2 1 12h3 1 h4 ) 1 162
5 lim
hS0
h
2216h 2 108h2 2 24h3 2 2h4
5 lim
hS0
h
5 lim ( 2 216 2 108h 2 24h2 2 2h3 )
hS0
5 2216
When x 5 3, y 5 2162.
The equation of the tangent is
y 2 (2162) 5 2216(x 2 3)
y 5 2216x 1 486
20. P(t) 5 20 1 61t 1 3t 2
a. P(8) 5 20 1 61(8) 1 3(8)2
5 700 000
b.
20 1 61(8 1 h) 1 3(8 1 h)2 2 (20 1 488 1 192)
lim
hS0
h
20 1 488 1 61h 1 3(64 1 16h 1 h2 ) 2 700
5 lim
hS0
h
20 1 488 1 61h 1 192 1 48h 1 3h2 2 700
5 lim
hS0
h
2
109h 1 3h
5 lim
hS0
h
5 lim (109 1 3h)
hS0
5 109
The population is changing at the rate of
109 000>h.
Chapter 1 Test, p. 60
1
1. lim x 2 1 does not exist since
xS1
1
1
5 1` 2 lim2
5 2 `.
xS1 x 2 1
xS1 x 2 1
2. f(x) 5 5x 2 2 8x
f(22) 5 5(4) 2 8(22) 5 20 1 16 5 36
f(1) 5 5 2 8 5 23
36 1 3
39
Slope of secant is
52
22 2 1
3
5 213
lim1
Chapter 1: Introduction to Calculus
3. a. lim f(x) does not exist.
xS1
b. lim f(x) 5 1
xS2
c. lim2 f(x) 5 1
xS4
d. f is discontinuous at x 5 1 and x 5 2.
4. a. Average velocity from t 5 2 to t 5 5:
s(5) 2 s(2)
(40 2 25) 2 (16 2 4)
5
3
3
15 2 12
5
3
51
Average velocity from t 5 2 to t 5 5 is 1 km> h.
b. s(3 1 h) 2 s(3)
5 8(3 1 h) 2 (3 1 h)2 2 (24 2 9)
5 24 1 8h 2 9 2 6h 2 h 2 2 15
5 2h 2 h 2
2h 2 h 2
v(3) 5 lim
52
hS0
h
Velocity at t 5 3 is 2 km> h.
5. f(x) 5 "x 1 11
Average rate of change from x 5 5 to x 5 5 1 h:
f(5 1 h) 2 f(5)
h
"16 1 h 2 "16
5
h
x
6. f(x) 5 2
x 2 15
Slope of the tangent at x 5 4:
41h
f(4 1 h) 5
(4 1 h)2 2 15
41h
5
1 1 8h 1 h 2
4
f(4) 5
1
41h
f(4 1 h) 2 f(4) 5
24
1 1 8h 1 h 2
4 1 h 2 4 2 32h 2 4h 2
5
1 1 2h 1 h 2
31h 2 4h 2
52
(1 1 2h 1 h 2 )
(231 2 4h)
f(4 1 h) 2 f(4)
lim
5 lim
2
hS0
h
hS0 1 1 2h 1 h
5 231
Slope of the tangent at x 5 4 is 231.
Calculus and Vectors Solutions Manual
4x 2 2 36
2(x 2 3)(x 1 3)
5 lim
xS3 2x 2 6
xS3
(x 2 3)
5 12
2x 2 2 x 2 6
(2x 1 3)(x 2 2)
5 lim
b. lim 2
xS2 3x 2 7x 1 2
xS2 (x 2 2)(3x 2 1)
7
5
5
x25
(x 2 1) 2 4
c. lim
5 lim
xS5 !x 2 1 2 2
xS5 !x 2 1 2 2
7. a. lim
5 lim
xS5
54
A !x 2 1 2 2BA !x 2 1 1 2B
!x 2 1 2 2
x3 1 1
(x 1 1)(x 2 2 x 1 1)
5
lim
4
2
xS21 x 2 1
xS21 (x 2 1)(x 1 1)(x 1 1)
3
5
22(2)
3
52
4
1
6
(x 1 3) 2 6
e. lim a
2 2
b 5 lim
xS3 x 2 3
x 29
xS3 (x 2 3)(x 1 3)
1
5 lim
xS3 x 1 3
1
5
6
(x 1 8) 2 2
(x 1 8) 2 2
f. lim
5 lim
x
xS0
xS0 (x 1 8) 2 8
1
(x 1 8)3 2 2
5 lim
1
2
1
xS0 ((x 1 8)3 2 2)((x 1 8)3 1 2(x 1 8)3 1 4)
1
5
41414
1
5
12
ax 1 3, if x . 5
8. f(x) 5 •
8, if x 5 5
2
x 1 bx 1 a, if x , 5
f(x) is continuous.
Therefore, 5a 1 3 5 8
a51
25 1 5b 1 a 5 8
5b 5 218
18
b52
5
d. lim
1
3
1
3
1-35