A.2. ANSWERS TO EXERCISES
A.2
Answers to Exercises
Section 1.1
Try This
5.
i) Note, cos
�
cos
1. a) −0.263158, b) −0.251256,
c) −0.250125
Among the three slopes the best
approximation of the slope of the
tangent line is −0.250125
2.
the total area of the eight rectangles is a better approximation
Check-It Out
2.
1
3
True or False:
�
�
π
[
π
1 −2
2k+1
1 −2
2k
�
+2
]+2
�
Choose δ = �/2.
7.
L = 1, choose δ = 0.03.
�
�
Choose δ = Min 3� , 1 .
T,T,F,T,F,
T ,T,T,T,T
Exercises Section 1.1
1. 3
√
481
,
60
√
109
30
b)
c)
The arc length is the sum of parts
a), b), and c), which is about
1.13021.
√
17
,
16
√
√
5
41
65
b) 16
, c) 16
d) 16
The arc length is the sum of parts
a), b), c), and d), which is about
1.47428.
5. a)
19
,
25
b) same as part a)
6. a)
29
,
6
b)
2. 1
3. 4
4. no limit
lux
b) E is very large
p(4.05)−p(4)
a)
≈ $5.95/hr
0.05
p(4+h)−p(4)
b)
= 6 − h.
4
45.
If h ≈ 0, marginal profit is
$6/hr.
p(1.01)−p(1)
≈ 0.151 m/hr.
0.01
p(1+h)−p(1)
b)
= 2−49h
.
h
10
643
,
140
c) part b)
5.
5. 4
9. 2
13. 5
2. 0.25
3. 5
4. 0.125
6. −0.625
7. 9
8. 0.25
10. 2
11. 4
14. 3
b) 16.016 ft/sec
Section 1.2
Try This
i) 4,
ii) 4
3.
i) does not exist,
4.
Note, f (x) = − x1 decreases
without bound as x approaches 0
and x �= 0.
Then f (x) does not approach a
single number as x approaches 0.
−1
Thus, lim
fails to exist.
x→0 |x|
ii) 1
15. 7
17. undefined
19. 1
12. −2
0.02
3
29. L = 9, δ =
31. L = 5, δ =
1
18
1
10
ε
2
30. L = 2, δ =
32. L = 1, δ = ε
8. 1/2, 2
10. 2, 1
1
2
3.
True or False:
T,F,F,F,F,
T,T,T,T,T
0.002
7
28. L = 9, δ =
1
8
1
8
Exercises Section 1.3
1. 4
5.
2. 5 3. 2
3
8
6.
8
7
1
2
10.
4. π
7. 3
9.
34. L = −3, δ = min{ 5ε , 1}
11. 6, 10,
3
3
3 7
,
7 4
35. L =
13. 3, 1
14. 4, 4
δ = min{6ε, 1}
36. L = 1, δ = min{ε, 1}
37. L = 2, δ = min{(2 +
√
3)ε, 1}
38. L = 4, δ = min{ 3ε , 1}
√
39.
145 − 12 ≈ 0.04159 in.
�
40. 2 36 + π2 − 12 ≈ 0.1056 in.
2
π d4 = 36π ± 2
41. 4.85in. ≤ h ≤ 5.15in.
15. −4
19. 4
23.
1
4
27. 12
31. − 19
35.
1
6
39. 1
8
3
8.
√
33. L = 10, δ = min{ 8ε , 1}
1
,
3
1 1
,
4 6
5. −10
4. 7
√
26. δ = 0.005(2 2 + 3)
27. L = 2, δ =
6.
2. −1
1. 9
16. 0
22. undefined
24. δ =
√
3
, 33
3
− 14 , −1
Check It Out
18. undefined
23. δ = .0025
√
9. 1/3, 0
20. 1
25. δ =
10. a) 12.6 ft/sec, b) 12.06 ft/sec
c) 6(2 + h) ft/sec
7.
Exercises Section 1.2
1. 0.25
4. 3/2, −1/2
3. 3, does not exist
8. a) −79, 207.9 gal/hr
b) −79, 920.1 gal/hr
9. a) 16.16 ft/sec
1. 2, 2, −4
2. −2, −3/2, 3
21. no limit
2.
10
9
44. a)
Section 1.3
Try It
ε
3
F,F,T,F,F,
F,T,T,T,F
7. a) 25, b) Approximate amount of
wetlands lost to erosion
32
43. a) 0.0821(2)(293) = 48.1106 atm
b) Let V1 = 48.1106
≈ 0.828.
58.1106
Then 0.828 < V < 1.172 liter.
If h ≈ 0, instantaneous
velocity is 0.2 m/sec.
True or False:
2. a) 0.513167, b) 0.501256,
c) 0.500125
1.
T must satisfy
|T − 70| < A or 63.1 < T < 76.9
Check-It Out
1. a) 4.1, b) 4.01, c) 4.001
4. a)
42. a) L = 2(1 + 2α) ≈ 2.0000288 m
0.0001
b) Let A = 2(7.2×10
−6 ) .
46. a)
5. no limit 6. hint δ =
5
,
12
= 1 and
= −1 for all
6.
8.
3. a)
�
1
integers k.
Moreover, 2k
− 2
1
and 2k+1 − 2 approach −2 as k
increases
�
�without bound. Then
π
cos x+2
does not approach a
single number as� x appoaches
−2.
�
π
Thus, lim cos
does not
x→−2
x+2
exist.
�
�
π
=0
ii) lim cos
x→0
x+2
51
,
128
1. 22
419
12.
1 1 1
, , ,3
6 6 2
16. 3 17. 27
20. −6 21. 7
24. 6 25. 1
28. 75 29.
18. 48
22. −7
1
2
26.
3
4
30.
1
32. −4 33. − 25
36.
1
14
37.
40. −1 41.
1
3
√
3
6
38.
8
3
34. −1
1
12
√
42. 2 5
420
43.
4
5
8
44. − 17
45. 3
47. 1
48. 1 49. 4
51. 0
52. No limit
46.
50.
APPENDIX A. APPENDIX
1
4
Exercises Section 1.4
1
2
1
3
53. −1
54.
55. −3
55. 5
Section 1.5
Try This
1. Yes, f (c) = L
2. No
3. Yes, f (c) = 4
4. Yes, f (c) = 10
5. No, unless f (1) = 2
6. No, unless
lim f (x) = 4
x→3−1
1.
3.
5.
6.
−∞, ∞ 2. −∞, ∞
�
δ = 4 1/M
4. δ = (1/M )2
a) x = 2 and x = −3, b) x = nπ
x=2
7. −∞, ∞
59. a) 1, b) no limit
60. 4
7. No, unless g(0) = 0
Check-It Out
61. 0
64. 5
8. No, p(2) may or may not be defined
1. −∞, ∞, undefined
2. Line x = 5
62. 0 63. 4
9. 2, 2, 2
Section 1.4
Try This
10. 1, 1, 1
11. 0, 1, undefined
True or False
12. 3, 2, undefined
1. Nonremovable discontinuity
at x = −1
13. 0
15. Undefined
2. Removable discontinuity at x = 0.
If h(0) = 1, then h
becomes continuous at x = 0.
3. g is continuous on (−∞, ∞)
4. y = csc x is continuous in
D = {x : x �= nπ, n an integer }.
At each point not belonging to D,
y = csc x has a nonremovable
discontinuity.
17. 1
21. 0
18. −1
16. Undefined
19. −1
20. 2
23. 0
24. 0
25. 4
27. 1
28. −1
26. 5
30. −7
29. 1
31. x = −1, 2, nonrem.
32. x = ±2, nonrem.
33. x = 3, nonremov;
x = −2, remov., f (−2) =
2
27
35. x = 3, nonrem.
0, −1, 1
7.
[−1/2, 1/2], (−1, 1)
√
8. Note, 1 + x2 is defined and
2 + cos x �= 0 for all x. Since a
quotient of continuous
√ functions is
continuous, g(x) =
1+x2
2+cos x
is
continuous everywhere.
9. Continuous for all x in (−∞, ∞)
4+x−x3
10. Since g(x) =
is continuous,
g(1) = 4 and g(2) = −26, the
Intermediate Value Theorem
assures g(x0 ) = 0 for some number
x0 between 1 and 2.
36. x = 4, nonrem.
37. x = 6, nonrem.
38. x = 2, remov. , f (2) = 1
39. x =
π(2k+1)
,
4
nonrem.
40. x = π(2k + 1),
nonrem.
43. ∅
44. ∅
45. x = −2, nonrem.
47. x = 1, nonrem.
49. ∅
46. ∅
48. ∅
50. x = 3, nonrem.
51. x = 0, remov, f (0) = 1
52. x = 0, remov, f (0) = π
Check-It Out
53. (−∞, 1) ∪ (1, ∞)
1. Removable discontinuity at x = −1.
Redefine f (−1) = 12 in order to be
continuous at x = −1
54. (−∞, 0) ∪ (0, ∞)
2. 2, −2
58. (−∞, 0) ∪ (0, 2) ∪ (2, ∞)
3. (−∞, ∞)
4. 2
5. If f (x) = x3 − x − 106 , then
f (0) < 0 and f (107 ) > 0.
Intermediate Value Theorem assures f (x0 ) = 0 for some xo in
[0, 107 ]
True or False
T,F, F,T,F
T,F,F,T,F
55. (−∞, 1) ∪ (1, ∞)
56. (−∞, ∞)
59. [−4, ∞)
57. [−2, 2]
60. (−∞, 1]
61. (−∞, ∞)
63. x =
1
2
62. (−∞, ∞)
+ k, integer k
64. x = kπ, integer k
65. x = 2k, integer k
66. x = k, integer k �= 0
67. a = 1
68. a = 1
69. a = 0, b = −4
70. a =
√
3−1
,
2
b=
√
1− 3
2
F,T,F,T,F
T,T,T,F,F
Exercises Section 1.5
22. Undefined
34. x = 3, remov, f (3) =
5. Undefined, 0
6.
14. Undefined
4
5
1. ∞, −∞ undefined
2. −∞, ∞ undefined
3. −∞, ∞ undefined
4. ∞, ∞, ∞
5. −∞, −∞
6. ∞, −∞
7. −∞, ∞
8. ∞, −∞
9. Lines x = ±2
10. Lines x = −2, 3
11. Lines x = −4, 0, 2
12. Lines x = 0, ±3
13. Lines x = −3, ±2
14. Lines x = −3, ±1, 2
15. Lines x = k, integer k
, integer k
16. Lines x = π4 + kπ
2
17. Lines x = 2π
+ 2kπ,
3
x = 4π
+
2kπ,
integer k
3
18. Lines x = π6 + 2kπ,
x = 5π
+ 2kπ, integer k
6
19. Lines x = π2 + kπ
20. Lines x = kπ
21. None 22. None
23. None 24. None
25. ∞ 26. −∞ 27. −∞ 28. ∞
29. Undefined 30. Undefined
31. −∞ 32. −∞ 33. ∞ 34. 0
35. ∞ 36. Undefined
37. ∞ 38. −∞ 39. 14 40. 23
3
41. 10
42. 13 43. −∞
45. ∞ 46. ∞
47. ∞, δ = 0.2
> 0;
M
If 0 < x < δ then
48. ∞, δ =
6
M
0.2
x
44. ∞
>M
>0
If 0 < 2 − x < δ then
1
,
M −1
49. ∞, δ =
M > 1;
If 0 < −1 − x < δ
x
then 1+x
>M
50. ∞, δ = M9−3 , M > 3;
If 0 < x − 3 < δ
3x
then x−3
>M
6
2−x
>M
A.2. ANSWERS TO EXERCISES
51. −∞, δ =
�
4−
1
N
, N < 0;
1
4−x2
If 0 < x − 2 < δ, then
<N
1
(π
2
59. a)
8
√
3x
− α − sin α), c) 2
47. −∞
51.
Then
4
M −1
16. −∞, −∞
30. −1
1
2
33. 2
34.
37. 0
38. 0
19. − 19
23. 0
27. 1
20.
24. ∞
28. 2
31. − 15
35. 2
5
9
36.
32. 0
1
4
39. Continuous on (−∞, ∞)
40. Continuous on (−∞, ∞)
41. Discontinuous at x = −6, −1
removable discontinuity at x = −6,
redefine f (−6) = 59
42. Nonremovable discontinuity
at x = 0
43. In proof, let δ = 3ε
Then |f (x) − 35| = 3|x − 12| < ε
for 0 < |x − 12| < δ
� ε�
44. In proof, let δ = Min 1, 14
Then |f (x) − 10| =
|x − 2||3x + 5| < ε
for 0 < |x − 2| < δ
55. Find the δ’s for 2 + x2
and 2 + x4 .
�
Then choose �
the
� of the δ s,
√ smaller
√
4
i.e., δ = Min
ε, ε .
√
56. a = 3 9 − 3
57. [0.84375, 0.8125]
58. If f (x) ≥ x for all x then
f (1) ≥ 1. Thus, f (1) = 1.
If f (x) ≤ x for all x then
f (0) ≤ 0. Thus, f (0) = 0.
Else, f (x1 ) > 0 and f (x2 ) < 0
for some 0 ≤ x1 , x2 ≤ 1.
Thus, by Intermediate Value
Theorem, f (x) = x has a solution.
Then A1 =
12
.
tan(α/2)
144 cos3 (α/2)
.
sin(α/2)
2
b) The area of a sector is θr2 .
�
�
1
Then A2 = 144 tan(α/2)
− π−α
.
2
c)
A1
A2
→ 3 as α →
16.
17.
18.
19.
20.
C
B
B
A
D
π
2
60. a) v(2) = −64 ft/sec
b) 3 sec at the ground for s(3) = 0
61. a) R(100) = $150,
M R(100) = 0.5 dollars/brush
b) R(125) = $156.25,
M R(125) = 0 dollars/brush
2. 6, −12
3. −9, −1
f (0 + ∆x) − f (0)
4. lim
=
∆x→0
∆x
√
3
∆x
lim
=
∆x→0 ∆x
1
lim �
=∞
∆x→0 3
(∆x)2
7.
>M
59. a) Note BC =
D
A
B
C
C
5. 4, 2x
whenever 0 < 9 − x < δ.
14. ∞, −∞
29. 1
− x.
Note if 8 < x < 10 then
1
√x
> M if √9−x
> M
.
8
9−x
12. Does not exist
26. 0
> x − 4 > 0.
√x
9−x
11.
12.
13.
14.
15.
6. A
7. B
8. D
9. C
10. C
1. −3
> M whenever
0 < 1 − x < δ.
� � � �
8 2
54. Let δ = Min 1, M
.
Thus,
A
B
C
A
A
Section 2.1
Try This
4
.
M −1
12
< N if − 12
>1
x−1
N
12
Thus, x−1 < N whenever
11. Does not exist
25. 0
x
x−4
Note
10. 5
1
12
50. ∞
�
�
53. Let N < 0, δ = Min 1, − 12
.
N
8. 237, 241, 241, 121, 121, 123
Approx. limits: 241,121,
and does not exist
22.
49. ∞
Let δ = √4 .
M
Then x162 > M
52. Let M > 1 and δ =
7. 0.0303, 0.0030, 0.0003,
0.9994, 0.9940, 0.9412
Approx. limits: 0,1,
and does not exist
21. − 12
48. −∞
<ε
whenever 0 < |x| < δ
6. 1.7948,1.81188,1.81361
1.81399,1.81572,1.83318,
Approx. limit is 1.814
18. −5
1.
2.
3.
4.
5.
|x−3|
√
x+1+2
for 0 < |x − 3| < δ
5. 1.05573,1.00505,1.0005,
0.9995,0.995049,0.954451
Approx. limit is 1
17. −5
<ε
for 0 < |x − 9| < δ
Then |f (x) − 2| =
4. a) 0.249224, b) 0.249992,
1
c) √4+h+2
15. −∞, ∞
Chapter 1 Multiple Choice Test
46. In proof, let
� √
�
δ = Min 1, ( 3 + 2)ε
3. a) 2.1, b) 2.01, c) 2 + h
13. ∞, ∞
|x−9|
√
x+3
<N
Chapter 1 Review Exercises
√
√
1. 3 + 2 + 3 2. 513
25
9. 1
62. ∞
45. In proof, let
� √
�
δ = Min 1, ( 8 + 3)ε
Then |f (x) − (−1)| =
3
8
52. −∞, δ = − N
3 , N < 0;
If 0 < −x < δ, then
421
g � (x)
6. m =
1
,
4
y=
1
x
4
+2
2/x2 ;
=
Since the tangent lines
to the graph of g(x) = −2/x have
positive slopes and g � (x) is the
slope of the tangent line at x, we
obtain g � (x) > 0. Consequently,
the points (x, g � (x)) on the graph
of y = g � (x) lie above the x-axis.
8. If f (x) = 2|x − 1|, then
f (x) − f (1)
lim
=
x−1
x→1+
2|x − 1|
lim
= 2 and
x→1+ x − 1
f (x) − f (1)
lim
=
x−1
x→1−
2|x − 1|
lim
= −2.
x→1− x − 1
Since the one-sided limits are not
equal, f � (x) does not exist.
If g(x) = [[x + 2]], then
g(x) − g(0)
lim
=
x−0
x→0+
[[x + 2]] − 2
lim
=
x
x→0+
2−2
lim
= 0 and
x
x→0+
g(x) − g(0)
lim
=
x−0
x→0−
1−2
−1
lim
= lim
, which
x
x→0−
x→0− x
�
does not exist. Thus, g (0) does
not exist.
f (x) − f (−1)
9. f � (−1) = lim
x→−1
x+1
√
3
x+1
= lim
x→−1 x + 1
422
APPENDIX A. APPENDIX
1
= ∞.
(x + 1)2
�
Then f (−1) does not exist.
�
= lim
x→−1 3
43.
f (x) − f (0)
=0
x
f (x) − f (0)
lim
=0
x
x→0−
55. a)
lim
x→0+
y
y
Check-It Out
1. 6
2. 10x
x
True or False:
3
T,T,T,T,T,
T,F,T,F,F
2
x
1
b)
Exercises Section 2.1
1. 10
5. 0
9.
3
7
2.
6. 0
2
3
3. −8
7.
17. 0
21. 1
8.
1
− 14
14.
18. −2
22. 5
25. 6x
4. 6
1
4
11. −10
10. 2
13. −3
1
4
12. 9
15.
19. 3
23. 3
29.
1
√1
30. √x−1
2 x+4
√
√
3x
32. 2 x2x
2x
6x2 34. 4x3
35. 6x − 1
37. − x42
39.
41.
1
12
16.
√2
x
28.
20. −2
2
3
−2
(t+1)2
3
40.
56. a)
2
y
x
1
36. −2x + 2
38.
x
1
y
26. 2x − 6
−1
(t+4)2
33.
1
4
24.
27.
31.
y
f (x) − f (0)
44. lim
=0
x
x→0+
f (x) − f (0)
lim
=1
x
x→0−
45.
−8
(x−4)2
lim
x→−2
x
f (x) − f (−2)
=∞
x+2
y
√1
2 x+9
3
f (x) − f (1)
=1
x−1
f (x) − f (1)
lim
= −1
x−1
x→1−
lim
2
x→1+
�3
x
�1
y
b)
y
�2
46. lim
x→1
f (x) − f (1)
= −∞
x−1
y
2
1
x
x
3
2
42.
f (x) − f (1)
=2
x−1
f (x) − f (1)
lim
= −2
x−1
x→1−
lim
2
x→1+
3
x
57. a) D, E; b) less; c) see graph below
y
�2
C
D
y
47. x = 3
49. x = −2
51. x = 1
2
1
x
B
48. x = ±2
50. x = 3
x
52. x = ±3
A
53. a) f (1) = 3, f (3) = −2;
b) 5;
c) The line through
(1, 3) and (3, −2)
54. a)
f (1)−f (0)
;
1−0
b)
E
f (2)−f (3)
2−3
59. Yes
61. a =
60. No
3
,
2
b = − 12 , c = 0
A.2. ANSWERS TO EXERCISES
f (x) − f (0)
=
x
1 − cos x
1
lim
=
x→0
x2
2
�
�2
sin x
1
1
lim
=
x→0
x
1 + cos x
2
35. y = 8x − 12
62. f � (0) = lim
x→0
37. y =
39.
40.
41.
Section 2.2
Try This
5
x
2
+ 10
44. ave. rate =
2
√1 ,
x3
1
12
6. f � (x) = 42x5 − 32,
−5
g � (x) = − x288 + 16x
3
7. y � = −π sin x,
y � = 2 cos(x) − π,
√
y � = 12 (cos x − 3 sin x)
√
√
2(4+π)
8. y = − 2x +
4
9. −94.4 ft/sec, −95.84 ft/sec
10. −64 ft/sec
51. −64 ft/sec
3. 48x3
5. −33.6 ft/sec
53. 64 ft/sec
4. − sin x
55. (±2, 4)
6. −32 ft/sec
3. 3x2 + 2x
6. −π sin t
17. 32
25.
+
8
15
x
8
x3
18. 7
2
− 27
√
− 22
27. −2
29.
8.
1
√
t
22.
2
√
−
x
+
16.
1
19. − 16
1
− 64
√
− 2x12
20.
23. −2
−
30. 0
33. y = 6x − 9
3π
8x4
11
4
24. 2
26. −10 2
28. −1
dy
dx
2.
dy
dx
= 36x2 + 14x + 1
dy
dx
32. − 17
4
= x cos x + sin x,
34. y = −4x + 5
11
= − (3x−4)
2
−nx
x2n
8.
=
1. x4 cos x + 4x3 sin x
2.
x(3x+2)
(3x+1)2
3. − 2x54
5
dy
dx
n
= − xn+1
2(tan(x)+1)
tan(x)−1
4. −32 ft/sec2
True or False
F,T,F,T,F
T,T,T,F,T
Exercises Section 2.3
1. y � = 3 − 4x
2. y � = 8x3 − 24x2 − 1
3. f � (x) = 3x2 − a2 − 1
4. f � (x) = 4x3 − 2(a2 + 1)x + a
5. y � = x sec2 x + tan x
�
�
6. x x sec2 x + 2 tan x
7. x sec x (2 + x tan x)
8. p� (x) = cos2 x − sin2 x
9. f � (x) = 4(2x − 5)
10. y � = 24x2 + 24x + 6
1
11. f � (x) = −
(x − 1)2
12
12. f � (x) =
(3 − 2x)2
x(2a3 − x3 )
(x3 + a3 )2
−4c2 x
(x2 − c2 )2
x sin x + cos x
15. −
x2
sin x − x cos x
16.
sin2 x
8x
2x3 + 3x2 + 1
17.
18.
2
2
(x + 5)
(x + 1)2
14. g � (x) =
25.
=
20
3(10+x)2
6. y � = 32 x7 , y � = − 2x13 , y � = 2
� 1 �
n
n−1
d
= x ·0−1·nx
7. dx
=
xn
x2n
dy
dx
2
2
21.
= cos2 x − sin2 x = cos 2x
�
�
3. h� (x) = 40 x3 cos x + 3x2 sin x ,
h� (x) = 40x3 cos x + 120x2 sin x
4.
= csc x(csc x − cot x)
10
√
3
27x2 x2
19. 1
n−1
31. 0
66. f (x) = x3 + 2x
dy
dx
√4
x
15
x4
√ �
2
2
Try This
1.
1−cos x
sin2 x
13. g � (x) =
Section 2.3
− cos t
1
2
10. √
− √
3
5
3 x2
5 x3
1
√
− 1√
12. x − 32
3
2t t
3 t2
√
3
6 x − 2√
+ 2x1√x
x
14. −15 x −
21.
65. f (x) = x2
4. 16x + 10
√
15.
56. (5, 1/5)
64. y = − x2 + 2
63. (2, 1)
− x12
− x22
54. 19.6 m/sec
60. a = 1, b = −3, c = 2,
and y = −x + 1
�
61. y = − x2 + 32 62. 12 ,
2. −15x4 + π
7. 1 + 3 sin x
13.
52. −90.2 m/sec
59. a = 3, b = −2, c = 1
Exercises Section 2.2
11.
√
58. y = 2x, y = 2x + 4/3
T,F, F,F,T,
F,F,T,T,T
9.
=−
√
57. y = 7x ± 4
True or False
=
Check-It Out
49. 2.5 sec, v(2.5) = −56 ft/sec
√
50. −128 2 ft/sec
Check-It Out
5. 4 cos x
= 30
ft , s� (0.49) = 14.32 ft ,
46. 14.16 sec
sec
ft
|s� (0.49)| = 14.32 sec
m
47. −33.751 sec
m,
�
s (5.99) ≈ −33.70 sec
m
�
|s (5.99)| ≈ 33.70 sec
m,
48. ≈ 7.50 sec
m,
�
s (0.25) = 7.55 sec
m
�
|s (0.25)| = 7.55 sec
√6
x3
1. 6x2 − 5
11.
√
6−6 2
, f � (π/4)
π
√
24 2−24
,
π
dy
dx
10. Average acceleration and acceleration are both −32 ft/sec per sec
−1
ft , s� (1) = −12 ft ,
45. −12.32 sec
sec
ft
�
|s (1)| = 12 sec
4. y = 12x − 16
2. 1
9.
f � (π/6) = 2 3
p� (t) = 1
5. f � (x) = 12x3 , g � (x) = − x124 ,
1. 0
x
− 16
42. ave. rate ≈ 47.12, f � (4) = 47
2. f � (x) = 5x4 , g � (x) = − x23 ,
y� =
38. y =
√
y = 3x + 1 − 63 π
√
√
y = − 2 3 3 x + 4 3π−6
9
ave. rate = 30.3, f � (5)
43. ave. rate =
3. 6, −2,
36. y = −2x + 9
√
1. p� (x) = 0, a� (t) = 0
y� = −
423
20. − 3x12 −
3x
2
27 −4
31. −7
35. −1
37. −2
12x2
5
22.
2 cos x
5
−
28.
1
3
43. y =
23. − 3x43
26.
− 12
32. 0
33.
π
36. −1 −
2
38. −2
4
− 25
x
+
4
x
3
2π
9
√
−
44. y = 2 3x −
24.
6
x5
2
sec x
2
29. −17
39. y = − 21 x + 1
41. y =
4
3x3
49
− 900
30. −3
34. − π42
40. y = x
12
25
+
2
√
42. y = − 43 x +
√
3
3
3π
3
+2
7
3
424
√
√
45. y = − 2x +
2π
4
+
√
− 3
8π
3
46. y = −4x +
√
F,F,F,T,F,
T,T,T,T,T
48.
49. b − 3a
50. 2(3a + b)
52. 9x2 + 22x + 5
53. a) a(t) = 6t − 12; b) t = 2 sec
54. a) ≈ −0.51 in./sec2
b) −0.5 in./sec2
55. 121 ft
57. 6
59.
56. 60 ft
(−1)n
60.
61. −5 cos x
n!
xn+1
f (x) = x3 , g(x) = 5x − 2,
√
b) f (x) = x, g(x) = 1 − x2 ,
c) f (x) = tan x, g(x) = 2πx
2 sec 2x tan 2x,
15(5x − 2)2 , c)
2
√3x
2
1+x3
2(sin2 x − cos2 x),
b) 12 sin x(1 − cos x)2
− (x212x
, b)
−1)3
−
4x
(1+x2 )3/2
6. Vertical tangent lines at (3, 0)
and (−2, 0),
�
f � (x) = 0 at the (1/2, − 3 25/4)
√7
7x+4
3(
16. −
4x
5/3
3(x2 +4)
cos2 x+2 sin x(1+sin x)
√
2 cos2 x 1+sin x
(sin x+1)3/2
2 cos2 x
−1
(1+x)2
�
1+x
1−x
10. a) −x sin(2x2 ),
b) 2 sec(2x + 1) tan(2x + 1),
2 √
3x
− csc√
3
3 x2
11. f � (x) = 12 sin 6x
Section 2.5
�
3
Try This
dy
3 dx
, b)
1. a)
1)2
29.
+ sin 3x)
5. a)
(x+1)2
2.
−8 sec2 (cos(2x)) sin 2x
4. y = x
−6y
,
(1+3y 2 )2
x
2
�
b)
−1
y3
3
2
dy
b) 8y dx
+ 2x
�
�
dy
11
c) 12(x + y)
1 + dx
3.
−6y
(3y 2 +1)3
True or False
5x+2
√
3(x+1)2/3 2x−1
�
�
�
31. − tan cos x2 sec2 cos
�
�
�
�
32. 12 cos x4 sec2 sin x4
y2
1−2xy
b)
dy
1. a) 2 dx
+3
−3 cot2 (sin x) csc2 (sin x) cos x
30.
1
,
3y 2 −1
2
dy
1+y 3 dx
y
√
3
Check It Out
(x+1)2
�
�
�
�
2x
2x
cot x+1
csc x+1
3(x−4)x
2(x−3)3/2
dy
6y 2 dx
,
dy
x dx
+ y, d)
c)
1
(cos(2x) − 2x) 24. 12 (x
2
�
�
�
�
x
x
sec x+1
tan x+1
26. −
x4 −x3 +x−1
x3 A(x)
d� =
3. − 21
28.
4. 2x cos(x2 )
x2
(x3 +1)2/3
22. 6 cot 3x csc2 3x
Check-It Out
3. 24x(x2 − 1)3
2
1−2w
A� (w) = 1 + √
1−w2
√
2
55. d = x − 18x + 121
d� = √ 2 x−9
x −18x+121
�
�
�2
56. d = (x − 1)2 + x1 − 1
2. a)
27.
2. f (x) = 10 csc x, g(x) = 12x
2
2−4w
A� (w) = √
1−w2
√
54. A(w) = w(1 + 1 − w2 ),
21. 10 sec2 x tan x
25.
48. 98
52. f (1) = −1,
(f −1 )� (−1) = f �1(1) = − 18
√
53. A(w) = 2w 1 − w2 ,
2x
� 3�
x
csc2
3
2
47.
50. g �� (x) = 3x2 since g � (x) = x3 by
Exercise 49
20. −8 cos(4x) sin(4x)
23.
46. 6
)
19. 6 cos(3x) sin(3x)
1. 24(4x +
45. 12
7/3
2x2 +x−7
8. f � (x) =
c)
(9−x)2
4(4x+1)
−6x2
√
51. f (1) = 2, g � (2) = 1/8
15. −
17. 6x2 sec
40. y = −7x + 15
csc(x2 )
14.
�
2
5
2
41. y = 3 3x + 1 − 63 π
√
√
42. y = −6 3x + 2(1 + 3π)
√
√
43. y = 6 3x + 3 − 3π
√
√
44. y = −12 3x + 6 + 2 3π
c) y = u2 , u = 2 sin x + 1
�
�3
11. 15(3x + 1)4
12. 8x x2 − 3
2
−
49. Apply chain rule to (f ◦ g)(x) = x
18.
9. h� (x) =
√
9. a) y = u4 , u = 3x + 1
b) y = tan u, u = x2
c) y = 3 cos2 u, u = 5x
√
10. a) y = u, u = 2x + 1
x
b) y = sin u, u = x+1
13.
3
x
2
tan(x3
7. y � = 3(3x − 1)3 (x + 4)2 (7x + 15)
or f � (x) =
39. y =
8. (f ◦ g)(x) =
2
(f ◦ g)� (x) =
2
−x csc(x ) cot(x2 )
1. (f ◦ g)(x) = 3(5x − 7)4 ,
(f ◦ g)� (x) = 60(5x − 7)3
5. a)
for integer n
38. vertical none; horizontal at
(nπ, 4(−1)n ) for integer n
7. (f ◦ g)(x) =
− 10)
(f ◦ g)� (x) =
3x2 sec( x3 − 10)
Try This
4. a)
37. vertical (±nπ, 0); horizontal at
�
�
�
�
(4n+1)π
(4n+3)π
, 1 and
, −1
2
2
6. (f ◦ g)(x) = −3 cos(8x)
(f ◦ g)� (x) = 24 sin(8x)
Section 2.4
b)
1. (f ◦ g)(x) = 2(4x − 5)3
(f ◦ g)� (x) = 24(4x − 5)2
5. (f ◦ g)(x) = 2 sin(12x)
(f ◦ g)� (x) = 24 cos(12x)
64. 3.276t2 + 6.3t + 5.2
3. a)
Exercises Section 2.4
3
62. −2 cos x
63. R(x) = x − 0.002x2
R� (x) = 1 − 0.004x
2. a)
35. vertical (−5, 0), (4, 0);
� �
�
horizontal − 21 , −3 5 81
4
�
�
√
3
36. vertical (±2, 0); 0, 6 2
2. (f ◦ g)(x) = 12x2 + 1
(f ◦ g)� (x) = 24x
√
3. (f ◦ g)(x) = 1 − 2x
−1
�
(f ◦ g) (x) = √1−2x
√
4. (f ◦ g)(x) = 3 9 − x
−1
(f ◦ g)� (x) = √
3
58. 7! or 5040
− x245
33. vertical (2, 0); horizontal (0, −8)
�
�
34. vertical none; horizontal 0, 14
True or False
3a−b
9
47. 3a + b
51. 2ax − 3x2
2
APPENDIX A. APPENDIX
sin
�x�
2
F,F,T,T,F,
T,T,T,T,T
A.2. ANSWERS TO EXERCISES
425
Exercises Section 2.5
Section 2.6
Investigation Projects
1. − x
y
Try This
1. P � (2) = −2000 bacteria/minute,
3.
2.
x
y
−2x−4y
3y 2 +4x
5. −
�
y
x
3
2
6.
9y+2x
7. − 9(2y+x)
9. −1
�
3. −80 in.2 /min
y
x
5. 1 ft/sec
−y−2y
xy
√
x(1+4y xy)
1. 6
10. −1
3.
11.
x
cos(2y)+y cos(y 2 )
12.
−x
sin(2y)−y sin(y 2 )
13.
1
2 cos(2y)−1
decreasing
4.
6. 5/π
1
ft/hr
4π
3
ft /hr
3. a) C � (2000) = $2.40/item
2. 6π
− 23
4. 4
in.2 /min
b) C(2001) − C(2000) = $2.3998
c) C � (x) is the approximate cost
True or False
of the (x + 1)st item
F,T,T,F,T,
T,F,T,F,T
y+sin(x−y)
14. − x−sin(x−y)
4. a) R(x) = 3x2 + 200x − 0.02x3
b) R� (100) = $200/item
16.
1
y3
Exercises Section 2.6
x4
12y 3
18.
6y 3 −8x2
9y 5
1. 12
19.
20.
6x(2y+1) −18x
(2y+1)3
21. − 67
25. 2
22. 0
4
−3
√
x
5
31. y = − 5x
+
4
33. y = 2x − 1
28. y =
√4
5
9
4
−2x+14
5
√
32. y =
−5x
√
3
4x
27
+
+ 10
19
27
34. y = 3x − 2
36. y = −x + 3
√
√
37. (2, ±2 2) 38. (1, 2)
√
√
43. Horizontal at (0, 0) and (3 3 2, 3 3 4)
√
√
3
3
Vertical at (0, 0) and (3 4, 3 2)
y
35. y =
�3 2 , 3 4 �
3
3
3 4
7. 4.5
3
3
3 4
in.3 /hr
19. 72π ft3 /hr
21.
23.
115
mph 24. 10 7 mph
13
33
√
units/sec 26. −4 units/sec
5
√
25
ft/sec 28. 2 2 miles/min.
13
dS
2
= dV
dt
dt r
dr
= k where V = 34 πr3 ,
dt
dV
= k(4πr2 ),
dt
10
ft/sec 32. 36
ft/sec
13
13
π
yd/sec
5
−2400 lb/ft3 per min
25.
27.
29.
33.
ft/min
22.
√3
2π
36. 2.5 ft/sec =
2
�2
�1
�1
�2
2
x
y
=
1. a) max 2, min 1; b) no max , min 1;
c) no max , no min;
d) max 2, no min;
2. Max g(1) = 8, no min
3. a) f � (2) = 0, b) f � (0) is undefined
4. a) x = 2, −4, b) x = 0, c) None
5. Max = 18, min = −10
√
√
6. Max = 4 − 3 4, min = −2 3/9
7. Max f (π/6) = 1.5,
use 82 + y 2 = x2
37. decreasing at
81
39.
1
=
min f (0) = f (π/2) = f (π) = 1
dy
,
dt
√
3−216
2
Check-It Out
cm2 /sec
38. 2.5 ft/min
1
y
15
−x +
4
�
√
1� �
4 17 + 285 x−
8
�
��
√
17 15 + 285
�√
�
1�
−4
285 − 17 x+
8
�√
��
17
285 − 15
Try This
m/hr
√
35. 4 in.
Vertical at (2, 1), (−2, −1)
y
=
Section 3.1
20. 22π ft3 /hr
20
9π
34.
44. Horizontal at (1, 2) and (−1, −2)
y
18. −2.5 radians/hr, use
cot α = −m where α
is the angle in the problem
31.
x
14. 120 in.2 /hr
3
ft/min
16. −270
100
√
2 3
− 9 radians/min,
dy
= − √2 ft/min, and
dt
3
√
r = 5 5/2
7. The tangent lines are
√
area increases at 2 3 sq. ft/min
3
3
6. (h, k) = (−4, 3.5),
12. 8π in.2 /min
30.
�3 4 , 3 2 �
8.
3
16
10. 2 ft/hr.
3x − 2
30. y =
5. b) 8 cm/min
4. 40
11. 32π ft3 /hr
17.
−
6. 21
c) C(101) − C(100) = $196.98
9. 16 in.2 /min
15.
1
8
24.
26. − 12
29. y =
1
36
3.
13. 216 in.3 /min
23. 1
27. y = −x + 1
2. 6
5. 11
2(y 2 −1)2 −8x2 y
(y 2 −1)3
2
b) N (31) − N (30) = 2699 people
c) Yes, if t = 10 and t = 50 days
15. − y43
17.
2. a) N � (30) = 2700 people/day
Check-It Out
2√
8.
2. −2.5
1. 1.75
y−2x+5
2y−x
4.
1
30π
cm3 /min 40. −4 cm2 /min
1. 0, 1
2. π/3, 5π/3
3. Max f (2) = 8, min f (1) = −3
Chapter 2 Multiple Choice Test
1.
2.
3.
4.
5.
A
D
C
C
C
6. B
7. B
8. C
9. B
10. B
11.
12.
13.
14.
15.
A
D
D
C
C
16.
17.
18.
19.
20.
C
B
C
B
B
True or False
T,F,F,F,T,
T,F,T,T,T
426
Exercises Section 3.1
1. max 5 at x = 3, min 1 x = 1
2. max 5 at x = 3, no min
3. no max, min 1 at x = 0
4. max 4 at x = 3, no min
5. f � (2) = f � (−2) = 0
6. f � (1) = f � (2) = 0
7. f � (0) undefined,
vertical tangent line
APPENDIX A. APPENDIX
37. Let x and y be the dimensions of
the rectangle. Assume x is the radius of the cylinder and y is height.
Since V (x) = πx2 y = πx2 (3 − x)
and V � = π3x(2−x), the maximum
is V (2) = 4π.
38. The minimum of
�
�
f (x) = 2 1 + x2 + 1 + (2 − x)2
is f (0.461736) ≈ 4.03764
8. f � (0) undefined,
vertical tangent line
Section 3.2
Try This
9. x = ±4
1. a) c = 1, b) c = 3π/4
11. x = 0,
2
5
10. x = −6, 4
12. x = 0,
13. x = −2, 0
15. x =
17. x =
18. x =
12
7
14. x = −2, 4
π 5π
,
16. x = π3 , 5π
6 6
3
π 3π 7π 11π
,
,
,
2 2
6
6
2π 4π
, 3
3
19. Max f (4) = 0, min f (2) = −8
20. Max f (−2.5) = 12.5, min f (0) = 0
21. Max f (−3) = 59, min f (3) = −49
22. Max f (1) = f (−0.5) = 3,
min f (−1) = f (0.5) = 1
25. Max g(2) = 9, min g(1) = 0
26. Max g(−1) = 0,
min g( 18 ) = − 2187
≈ −1.1
2048
27. Max f (0) = 0,
28. Max f ( 23 ) =
2
√
9
3
,
√
min f (−1) = − 2
29. Max f (0) = f (2π) = 1,
min f (π) = −1
30. Max f (− 14 ) = 2, min f ( 14 ) = 0
31. Max f (1) = f (−1) = 0,
min f (0) = −1
32. Max f (2) =
1
,
4
min f (0) = 0
√
33. Max k( π3 ) = 3 3 − π,
min k(π) = −3π
34. Max M ( π6 ) = 1.5, min M (0) = 1
35. Let x ∈ [30, 50] be the diagonal.
The cost function is
�
C(x) = 12x + 4[40 − x2 − 900].
Minimum
45
C( √
)
2
≈ $499.41
36. Let x be the distance between P
and the shorter antenna.
The
√
x2 + 502 +
length
is L(x) =
�
(100 − x)2 + 752 .
√
The minimum is L(40) = 25 41
≈ 160 ft. Thus, P must be 40 ft
from the shorter antenna.
22.
p(12)−p(10)
2
= f � (c) by the Mean
Value Theorem. Then f (6) ≤ 14.
= p� (c) by the Mean
Value Theorem. Then p(12) ≥ 8.
Section 3.3
Try This
1. a) increasing on (−∞, 3),
decreasing on (3, ∞);
b) increasing on (−∞, 0)
and (4, ∞), decreasing on (0, 4)
√
2+ 3
,
2
√
2+ 3
=− 2
3. Rel. max k(5π/3) =
rel. min k(4π/3)
4. Rel. max f (0) = 6.25,
rel. min f (±1) = 6
5. 8 yards
3. c = 1/4
4. Let s(t) be the distance driven in t
hours. Suppose s(t) is differentiable. By the Mean Value Theorem, there exists c in (0, 4) where
s(4) − s(0)
s (c) =
=
4−0
�
300 − 0
= 75 mph.
4
Thus, the instantaneous velocity is
75 mph at some time c.
Check It Out
√
min f ( 23 ) = − 4 9 6 ≈ −1.1
f (6)−f (1)
5
2. Rel. max g(1) = 1/20,
rel. min g(0) = 0
2. Since f (0) = 4 and f (1) = −6,
f has a zero in (0, 1). Since
f � (x) = −9 − 3x2 has no zero,
f has exactly one zero by Rolle’s
Theorem. Then f has only one
x-intercept.
23. Max g(−5) = 110, min g(1) = 2
24. Max g(−1) = 2,
min g( 13 ) = 22
≈ 0.8
27
21.
1. c = 1
2. c = 1
Check It Out
1. inc (−2, 2), dec (−∞, −2) and (2, ∞)
2. max f (−2) = 27, min f (1) = 0
√
√
3. max f ( π4 ) = 2, min f ( 5π
)=− 2
4
4.
1
4
True or False
T,F,T,T,F,
F,F,T,T,T
Exercises Section 3.3
3. By the Intermediate Value Theorem,
f has a zero for f (1) > 0 and
f (−1) < 0. If f (a) = f (b) = 0
and a �= b, then f � (c) = 0 for
some c by Rolle’s theorem. But
f � (x) = − sin x + 2 has no zero.
Then f has exactly one zero.
True or False
1. inc (−∞, 1), dec (1, ∞)
2. inc (−∞, −1), dec (−1, ∞)
3. inc (2, ∞), dec (−∞, 2)
4. inc (3, ∞), dec (−∞, 3)
5. inc (−∞, − 31 ) and ( 12 , ∞)
dec (− 13 , 12 )
T,F,F,F,T,
T,T,F,F,T
6. inc (−∞, − 12 ) and
Exercises Section 3.2
7. inc (− 21 , 13 ),
1. c = 3
3. c =
5. c =
( 34 , ∞), dec (− 21 , 34 )
2. c =
2
3
π
3
4. c =
7. c = 0, ±
9. c =
2
3
11. c =
6. c =
�
π
2
dec (−∞, − 12 ) and ( 13 , ∞)
1
6
12. c =
13. c = 1 +
√
15. c = 1
16. c =
2
dec (−∞, 13 ) and ( 32 , ∞)
9. inc (−∞, ∞)
8. c = 2
10. c = − 53
√
0, ± 22
8. inc ( 13 , 32 ),
√
6− 3
3
π
4
14. c =
1
4
11. dec (−∞,
√
2(3− 3)
3
√
10 − 2
3
)
2
10. dec (−∞, ∞)
and ( 32 , ∞)
12. inc (−∞, −4) and (−4, ∞)
13. inc (−∞, −3) and (−3, 0)
dec (0, 3) and (3, ∞)
14. inc (0, 2) and (2, ∞)
dec (−∞, −2) and (−2, 0)
A.2. ANSWERS TO EXERCISES
π
) and (π, 3π
),
2
2
π
3π
dec ( 2 , π) and ( 2 , 2π)
inc ( π2 , π) and ( 3π
, 2π)
2
dec (0, π2 ) and (π, 3π
)
2
3π
7π
inc (0, 4 ) and ( 4 , 2π)
dec ( 3π
, 7π
)
4
4
π
inc (0, 4 ) and ( 5π
, 2π)
4
dec ( π4 , 5π
)
4
inc (0, π2 ), ( 7π
, 3π
), ( 11π
, 2π),
6
2
6
π 7π
11π
dec ( 2 , 6 ) and ( 3π
,
)
2
6
7π 11π
inc ( 6 , 6 ),
dec (0, 7π
) and ( 11π
, 2π)
6
6
15. inc (0,
16.
17.
18.
19.
20.
21. min f (2) = −16, no max
22. min g(−1) = −12, no max
23. rel max h(−3) = 42,
rel min h(1) = 10
24. rel max y(−1) = 0,
rel min h( 53 ) = − 256
27
25. rel max y(− 59 ) =
256
,
81
427
43. max v( π6 ) = 2, min v( 7π
) = −2
6
√
44. max A( 2π
) = 3 + π3
3
√
2π
min A(− 3 ) = − 3 − π
45.
) = f ( 11π
) = 54 ,
rel max f ( 7π
6
6
π
rel min f ( 2 ) = −1, f ( 3π
)=
2
f ( π3 )
f ( 5π
)
3
46. rel max
=
rel min f (π) = −1
47. − 14
=
1
5
,
4
48. 2
√
50. 16 ft, 4 A ft
�
51. a) inc (− 52 , 0) and ( 52 , ∞);
�
�
dec on (−∞, − 52 ) and (0, 52 )
�
b) (± 52 , 32 )
49. 9
2
ft2 , p16
ft2
�
52. a) inc ( 12 , ∞), dec (0, 12 )
�
b) ( 12 , 12 )
53. x = 2, y = 3
54. x = 2, y = 4
55. a) Possibly p(x) = (x − 2)2 + 1
y
43
,
16
27. rel max
=
rel min f (3) = −52
and f (0) = 2
√
64 10
,
25
33. rel max
=
min f (0) = f (2) = 0
b) Possibly p(x) = (x − 1)3 − 3
y
�3
5 3
56. a) Possibly v(t) = − 16
t +
15
t
4
y
�2
x
2
�5
b) Possibly v(t) =
y
−t4
+
2t2
1
�1
1
x
�1
63. y =
38.
rel max f (−2) = − 32
,
3
32
rel min f (2) = 3
√
39. rel max f (2) = 3 4,
rel min f (0) = f (3) = 0
40. rel min f ( 17 ) = −
216
√
343 7
2. (0, 2)
True or False
F,T,F,T,T,
F,T,F,T,T
64. y =
1
(x2
4
1. concave up (−∞, ∞)
2. concave down (−∞, ∞)
4. concave up (π, 2π),
concave down (0, π),
inflection pt (π, 0)
5. concave up (−∞, 1) and (3, ∞);
concave down (1, 3);
inflection pts (1, 14) and (3, 30)
5
35. max f (4) = 6, no rel min
26
37. rel max f (− 27
) = 55
,
27
rel min f (−1) = 2
1. concave up (−∞, −2) and (0, ∞)
concave down (−2, 0)
3. concave up (1, ∞),
concave down (−∞, 1),
inflection pt (1, 2)
x
1
34. rel max f (0) = 0,
min f (±3) = −81
36. min f (0) = 1
Check It Out
x
2
32. rel max f (−1) = −2,
rel min f (1) = 2
f ( 25 )
7. rel max f (π) = 3, rel min f ( π3 ) = − 32
1
28. rel max f (0) = − 14
29. min f (0) = −9
√
√
3
30. rel max f ( 3 2) = 62
√
√
3
31. rel min k(− 3 4) = − 124
6. Since s� (t) = 4t(4 − t2 ) and
s�� (t) = 16 − 12t2 , the rel. max.
value is s(±2) = 21 and the rel.
min. value is s(0) = 5.
Exercises Section 3.4
26. rel max y( 32 ) = 20,
f ( 12 )
10(x−1)
5. Since y �� =
, the graph of y is
9x4/3
concave down on (−∞, 1), concave
upward on (1, ∞), and the point of
inflection is (1, −5).
3. rel max f (0) = 0, rel min f (±2) = −4
rel min y( 13 ) = 0
rel min y(− 32 ) = −34
4. Concave up on (0, 1),
concave down (−∞, 0) and (1, ∞),
pts of inflection (0, −2) and (1, −1)
− 3x + 6)
3
−4x +30x2 −48x+103
27
Section 3.4
Try This
1. Concave up (−∞, −1) and (0, ∞),
concave downward on (−1, 0)
41. max g( π2 ) = g( 3π
)=5
2
min g(π) = 4
2. Concave up (−∞, 1) and (1, ∞),
concave down on (−1, 1)
42. max g(0) = −2
min g(± π2 ) = −3
3. Concave up on (−∞, 1),
concave down on (1, ∞)
6. concave up (0, ∞);
concave down (−∞, 0);
inflection pt (0, 2)
7. concave up (−∞, −2) and (0, ∞);
concave down (−2,
√ 0); inflection pt
(0, 0) and (−2, 6 3 2)
8. concave up (−∞, 0) and (1, ∞);
concave down (0, 1);
inflection pt (0, 0) and (1, 3)
9. concave up (8, ∞);
concave down (−∞, 8);
inflection point (8, −48)
10. concave up (2, ∞);
concave down (−∞, 2);
√
inflection point (2, 24 3 2)
11. concave up ( π4 , 3π
); concave down
4
(0, π4 ) and ( 3π
,
π);
inflection pts
4
3
( π4 , 32 ) and ( 3π
,
)
4 2
12. concave up ( π4 , 3π
); concave down
4
,
π);
inflection pts
(0, π4 ) and ( 3π
4
3π 2+3π
( π4 , 2+π
)
and
(
,
)
4
4
4
428
13. concave up ( 3π
, π);
4
concave down (0,
3π
);
4
inflection pts ( 3π
, 0)
4
( π4 , 5π
); concave down
4
( 5π
,
2π);
inflection pts
4
14. concave up
(0, π4 ) and
, 0)
( π4 , 0), ( 5π
4
15. concave up (−∞, −2) and (2, ∞);
concave down (−2, 2); inflection
points (±2, 4)
16. concave up (−∞, −5) and (5, ∞);
concave down (−5, 5);
inflection points (±5, 21 )
17. concave up (−4, 0) and (4, ∞);
concave down (−∞, −4) and (0, 4);
inflection points (−4, −2), (0, 0),
(4, 2)
18. concave up (−7, 0) and (7, ∞);
concave down (−∞, −7) and (0, 7);
inflection points (−7, −1), (0, 0),
(7, 1)
19. concave up (−∞, 0); concave down
(0, ∞); inflection pts (0, 0)
20. concave down (−1, ∞);
no inflection pt
APPENDIX A. APPENDIX
√
√
36. rel max g(−3 3) = 9 2 3 ,
√
√
rel min g(3 3) = − 9 2 3
2. a) 3, b) 3
38. max M (±8) = 16,
rel min g(0) = 0
3. a) 0, b) 0
4. a) 0, b) 0
39. f (x) = (x − 1)2 + 2
5. a) 2, b) −2
40. f (x) = −(x − 3)2 − 1
41. f (x) =
1
(x3
6
√
16 4 3
3
15. ∞
19. 2
2
3
24.
26. 0
29. 0
30. ∞
27. 0
35.
38. 0
1
2
18. ∞
22. 2
32. ∞
36. 3
8
3
39.
40.
2
3
42. y = ±1
44. y = 2
45. Possibly f (x) =
x
f (x)
1
2
28. 3
31. ∞
34. 0
43. y = 0
14.
21. −4
25. 0
37. −2
3
2
17. 0
1
3
4
3
20.
41. y = −3
3x2
x2 +1
y
3
�1
10
0.198
100
0.02
1000
0.0002
1
x
46.
y
b) limit is 0
x
f (x)
−10
−0.198
−100
−0.02
2. If x > M =
29. rel max g( π4 ) = 3,
7. N (24) ≈ 44, 083,
N (t) → 50, 000 as t → ∞
35. rel max f (0) = 1, rel min f (4) = 9
23.
13.
16. 0
54. For instance, f , g, f � , g � , f �� , and
g �� are all positive functions
26. rel max f (−2) = 27,
rel min f ( 13 ) = 386
27
√
27. rel max C(±3 2) = 6724;
rel min C(0) = 6400
√
28. rel max A(±6 2) = 9409;
rel min A(0) = 4225
34. rel max Q(−5) = 1,
rel min Q(5) = −1
12. 1
53. Show d2 y/dx2 > 0. The converse
is false, try f (x) = x1 for x > 0
1. a) limit is 0
33. rel max y(4) = 1,
rel min y(−4) = −1
10. 1
11. 1
33. 0
23. Rel. max. value: f (0) = 5
rel. min. value: f (2) = 1
√
31. rel max L(− π3 ) = − 3 + 4π
,
3
√
π
4π
rel min L( 3 ) = 3 − 3
√
32. rel max R( 2π
) = − 3 + 8π
,
3
3
√
rel min R( π3 ) = 3 + 4π
3
9. 2
51. False; f (x) = x4
√
52. False; f (x) = − x
Try This
30.
8. a) 0, b) 0
√ √
334
2
50. (2, 1/8), (−2, −1/8)
Section 3.5
rel min g( 3π
) = −3
4
2π
rel max f ( 3 ) = 5,
rel min f ( π3 ) = −3
7. a) 0, b) 0
15
x
2
43. (1, 1), (−1, −1) 44.
√
45.
5 46. ( π2 , 1)
√
√
4
47. 8 3 3
48. 2 2
49.
6. a) 1, b) 1
− 3x2 + 12)
42. f (x) = − 52 x3 +
22. concave up (0, 1); concave down
(−1, 0); inflection pt (0, 0)
25. rel max g(−2) = 32,
rel min g( 34 ) = 364
27
1. a) −2, b) −2
37. max K(1) = 3, no rel min
21. concave up (−∞, 1); no inflection pt
24. Rel. max. value: f (1) = 5
rel. min. value: f (0) = 4
Exercises Section 3.5
1
,
ε
then
1
x
3
−1000
−0.0002
2
�1
1
3. a) 0, b) 0, c) undefined
47. Possibly f (x) = arctan x
y
Π
4. a) − 12 , b) 0
2
5. a) 1, b) − 13
x
6. a) 2, b) −2
�
1. a) 0, b) − 12 , c) −∞, d) −2
3. −10400
True or False
F,T,T,F,T,
T,T,F,T,F
2
y
4
2
9. a) ∞, b) −∞
Check It Out
Π
48.
8. a) −∞, b) ∞
2. y = −1
x
<ε
�1
1
x
49. P (t) → 14 as t → ∞.
About 25% of adults will have
sinus problems towards the end
of spring
50.
1
2
51. a) C(x) = 40, 000 + 5x
b) As x → ∞,
C(x)
x
→ $5/item
52. a) A(11) ≈ 17.96 PSI; b) 25 PSI
A.2. ANSWERS TO EXERCISES
Section 3.6
Check It Out
Try This
1.
F � (w)
5. Intercept (0, 0), max f (0) = 0,
asymptotes x = ±1, y = 1
range (−∞, 0] ∪ (1, ∞)
1. D = {x : x �= ±1}
5w3 (w
− 4),
= 20w2 (w − 3),
rel max F (0) = 0,
rel min F (4) = −256,
point of inflection (3, −162)
=
F �� (w)
y
3
4
asymp. lines x = ±1
symmetric about the origin
�256
R = (−∞, −4] ∪ [0, ∞)
inc (−∞, −2) and (0, ∞)
dec (−2, −1) and (−1, 0)
3. critical num t =
x−1
,
2x3/2
2. R� (x) =
R�� (x) =
3−x
,
4x5/2
relative minimum R(1) = 2,
√
point of inflection (3, 4/ 3)
y
π
) and
4
π 5π
(4, 4 )
inc (0,
dec
π 5π
,
4 4
( 5π
, 2π)
4
4
3
2
F,T,F,F,F
6t
,
(t2 −1)2
3. g � (t) =
T,F,F,T,F
x
3
g �� (t) = −
6(3t2 +1)
(t2 −1)3
Exercises Section 3.6
1
2
3
�4
6. Intercept (0, 0),
inflection pt (0, 0),
asymptotes x = ±2, y = 0,
range (−∞, ∞)
10
5
�10
�5
5
10
�5
7. Intercept (0, 0),
√
√
rel max f (3 3) = −9 3,
√
√
rel min f (−3 3) = 9 3,
inflection pt (0, 0),
asymptotes x = ±3
and y = 2x, range (−∞, ∞)
30
20
relative minimum g(0) = 4
√
10
1. Max f (± 3) = 11,
y
�1
�10
True or False
1
�2
�2
conc down (−∞, −1)
�162
2
�3
2. D = (−∞, ∞)
conc up (−1, ∞)
x
4
429
�10
�5
5
10
�10
3
2
2
3
x
lim f (x) = −∞,
√
2(1− 3 t)
,
t2/3
√
3
2(
t−2)
H �� (t) =
,
3t5/3
x→∞
4. H � (t) =
y
3
2
x
8
lim f (x) = −∞,
√
range (−∞, 3]
√
√
2. Rel max f (− 3) = − 3 2 3 ,
√
√
rel.min f ( 3) = 3 2 3 ,
√
√
inc on (−∞, − 3) and ( 3, ∞),
√
dec√on (− 3, −1), (−1, 1), and
(1, 3), lim f (x) = ∞,
x→−∞
rel max H(1) = 3,
point of inflections (0, 0), (8, 0)
1
rel.min f (0) = 2,
√
√
inc on (−∞, − 3) and (0, 3),
√
√
dec on (− 3, 0) and ( 3, ∞),
x→∞
5.
N � (x)
=
lim f (x) = −∞,
2 sin x+1
− (2+sin
,
x)2
N �� (x) =
x→−∞
2 cos x(sin x−1)
,
(2+sin x)3
√
rel max T (11π/6 + 2kπ) = √3/3,
rel min T (7π/6 + 2kπ) = − 3/3,
points of inflection (π/2 + kπ, 0)
y
7Π
3Π
11 Π
2
6
2
6
x
3
3
t(t−6)
,
(t−3)2
6. g � (t) =
18
g �� (t) = − (t−3)
3
rel max g(0) = 0,
rel min g(6) = 12
5
�4
�2
inc on (−∞, 0) and (2, ∞),
lim f (x) = −∞,
x→−∞
range (−∞, ∞)
√
4. Max f (π/4) = 2,
dec on
�10
√
9. Min f (4) = 2 2,
x
√
( 5π
, 2π),
4
lim f (x) do not exist,
√ √
range [− 2, 2]
x→−∞
√
6
),
√
asymp x = 2, range [2 2, ∞)
3
2
4
8
10. Intercept (0, 0),
max f (1) =
√
2
,
2
√
2
,
2
√
inflection pts (0, 0), ( 4 5,
√
√
4
and (− 4 5, − √ 5 )
6
asymp y = 0, range [−
lim f (x) and
6
4
inflection pt (8,
1
x→∞
12
4
min f (−1) = −
min f (5π/4) = − 2,
π
) and
4
π 5π
( 4 , 4 ),
2
�5
√
rel.min f (2) = −2 3 4,
inc on (0,
y
10
2
x→∞
Π
�30
8. Intercept (0, 0),
rel max f (0) = 0,
√
rel min f (± 2) = 4,
asymptotes x = ±1,
range (−∞, 0] ∪ [4, ∞)
4
3. Rel max f (0) = 0,
dec on (0, 2), lim f (x) = ∞,
3
3
�
range (−∞, ∞)
�20
4
� 5
�1
1
4
5
√
√
4
√ 5 ),
6
√
2
, 22 ]
2
430
17. Max ( π3 ,
11. Intercepts (0, 0), (8, 0),
p� (x) = − 43 x−2
2/3 ,
max p(2) = 6 2,
inflection pts (0, 0)
√
and (−4, −12 3 4),
√
range (−∞, 6 3 2]
6
4Π
2
4π
),
3
−
Max S( π4 ) = S( 3π
)=
4
3
3 �
4Π
Π
�
2
Π
3
Min S(0) = S(π) = 0,
Π
Π
3
2
1
3
3
√
π 2π
18. Rel. max (− 12
, 3 − 3),
√
π
rel. min ( 12
, 3 − 2π
),
3
8
inflection pt (0, 0),
2Π
12. Intercepts (0, 0), (36, 0),
3
4(x−9)
,
27x2/3
4(x+18)
−
,
81x5/3
F � (x) = −
�
Π
�
4
�
Π
Π
3Π
4
2
4
30. y � = −3(sin 3t + sin 4t) =
2 sin( 7x
) cos( x2 ),
2
3
Π
Π
Π
12
12
4
3 �
critical nos x = 0,
2π 4π 6π
, 7 , 7
7
2Π
3
7
4
√
3
max f (9) = 3 9,
inflection pts (0, 0) and
√
√
(−18, −6 3 18), range (−∞, 3 3 9]
19. y =
x
x−3
21. y =
1
x
x
x2 −1
20. y =
+x+1
1
3
22. y =
1
x+1
+
2Π
4Π
6Π
7
7
7
Π
x
x
|x|
23. Possibly f (x) = x2
3
3 9
�18
9
32. Inflection pt (− k3 ,
3
�6 18
1
1
√
13. Intercepts (0, 0), (±2 6, 0),
rel max f (0) = 0,
√
rel min f (±2 3) = −6,
inflection pts (−2, − 10
)
3
(2, − 10
),
3
24. Possibly f (x) = x3
If k > 0, local max f (− 2k
)=
3
local min f (0) = 0;
4k3
,
27
If k < 0, local min f (− 2k
)=
3
local max f (0) = 0
4k3
,
27
2
33. G(t) =
1
range [−6, ∞)
2k3
)
27
If k = 0, no local extrema.
3
and
√
2
,
3
1
�
�12 22�3
F �� (x) =
29. S � = (cos 3θ + cos θ)/2
3),
rel min S( π2 ) =
�
3
2
�4
√
inflection pt (0, 0)
√
3
3
4π
−
3
√
π
(− 3 , 3
min
x
− 49 x+4
,
x5/3
p�� (x) =
APPENDIX A. APPENDIX
1
t3 −3t2 −9t+19
8
3
3
6
�2
�2
3
2
2
�
25. Possibly f (x) = x4
3
10
2
�6
14.
2
√
�1
1
Intercepts (0, 0), (± 330 , 0),
√
√
rel max f (−2 2) = 6415 2 ,
√
√
rel min f (2 2) = − 6415 2 ,
inflection pts (0, 0), (2, − 56
),
15
(−2, 56
),
range
(−∞,
∞)
15
64
1
3
Section 3.7
26. Possibly f (x) = x3 − 3x2
2
�1
1
Try This
√
1. 25 ft by 25 ft 2. (3.5, 3.5)
√
√
3. Length 4 2, width 2 2
3
�3
2
2
�
56
15
27. s0 = 25 ft, max ht s(3) = 169 ft
√
( π3 , 4 3 3 ),
15. Max
min
inflection pt (0, 0)
√
( 5π
, − 4 3 3 ),
3
Π
5Π
Π
3
3
2Π
rel max F ( π3 ) = − sin(
3
16. Max ( π6 ,
5
√
3
),
3
√
min ( 5π
, − 5 3 3 ),
6
inflection pt ( π2 , 0), ( 3π
, 0),
2
5
rel min
F ( 2π
)
3
=
√
3
2
√
sin( 23
−
−
π
),
6
π
),
6
�
5
3
3
Π
5Π
3Π
2
6
2
√
3
π2
π
cm
2. p(4) = 16, minimum value of p
1
True or False
3
6
10
1. x = 200 ft, y = 600 ft
3
Π
=
Check It Out
Min F (π) = −1
4
10
√
3π
[sin(θ + sin 2θ) + 1]
Max F (0) = 1,
3
4. x = 6 feet from the 3-ft post
5. r = h =
28. Use F � = − [2 cos(2θ) + 1] ×
4
�
3
1
15
�2
1
�1
2Π
�1
Π
2Π
3
3
Π
F, T,F,F,F,
T,F,F,T,F
A.2. ANSWERS TO EXERCISES
Exercises Section 3.7
Subtract the two equations above
to obtain
√
r = 12 − 2 6h − k.
1. D = 1.25 when x = 1
Rewrite the second equation as
√
(2 6−h)2 +(k−3)2 = ((r+2)+1)2 .
2. D = 1 when x = 0
3. A = 25 when w = 5
√
4. A = 9/ 2 when b = 3
Then substitute the first equation
to obtain
√
(2 6 − h)2 + (k − 3)2 =
�
( h2 + (k − 2)2 + 1)2 .
5. S = π + 2 when r = 1
6. V = 16π when r = 2
7. a) ( 17
,
2
√
34
),
2
b) (0, 0)
√
8. a) (2, 2 2), b) (0, 0)
9.
1
2
√
3
3
10.
√
√
11. x = 100 6 ft, y = 50 6 ft
12. r =
14.
20
4+π
√
4 √
3
9+4 3
13. 2 in.
or
√
48 3−64
11
15. (0, 0), (8, 0), (0, 4); not the minimum perimeter
√
√
3
3
√
2
2
16. 3 3 17. r = 15π
, h = 2 15π
π
π
18. radius 100
m, length of one
π
side of rectangle is 100 m
18
miles from the point on
7
the highway closest to Town A
√
20. BP = 3 miles
19.
21.
x2
8
square feet.
22. base b, perimeter p, two sides
of the same length p−b
2
23. x = 4 workers
24. 220 students
25. 165 ft by 300 ft
Solving for k, we find
√
√
192 6 − 336h + 23 6h2
√
k=
.
24( 6 − h)
If h = 0 then k = 8 and r = 4.
Thus, the radius of the third circle
is r = 4 if its center lies in the
y-axis.
We skip the details in evaluating
the limit of the points of contact of
the first and third circles. As the
radius of the third circle increases
to ∞, the limit is
�
�
√
8 6 96
−
,
.
25 25
Try This
2. x4 ≈ 0.739085, |x4 − x3 | ≈ 0.00003
�
�
� f (x)f �� (x) �
3. Let g(x) = � (f � (x))2 �.
The graph of g is dashed below:
28.
√
m2 +1
√
2π (3− 6)
3
29. θ =
31. c =
32.
4
27
π
3
√
8
7
radians ≈ 33◦
30.
32πr 3
81
27 when x = 3
33. 2 radians
34. C = 54
f �x� � x2 � 60
0.2
|mp−q+b|
35. K = −3
36. max A + B, min A − B
37. max A + B, min − B
2
+8A2
8A
38. Draw a right triangle through the
centers of the first and second
circles. By the Pythagorean
Theorem, √
the center of the second
circle is (2 6, 3).
Let (h, k) be the center, and let r
be the radius of the third circle.
Draw a right triangle that joins the
centers of the first and third circles.
h2 + (k − 2)2 = (r + 2)2 .
Similarly, we obtain
√
(2 6 − h)2 + (k − 3)2 = (r + 3)2 .
x
g�x�
Then g(x) < 0.2 for x in (7, 8).
By Theorem 3.11, we can apply
Newton’s Method√with x1 = 8.
Thus, lim xn = 60.
n→∞
�
�
� f (x)f �� (x) �
4. Let g(x) = � (f � (x))2 �.
The graph of g is dashed below:
y
0.2
f �x� � sin�x� � x � 2
g�x�
1
5
4
x
3
4
2. x3 =
17
12
True or False
T,T,F,T,F,T
Exercises Section 3.8
1. x2 = 3.00, x3 = 2.800
2. x2 = −2.750, x3 ≈ −2.342
3. x2 = 1.500, x3 ≈ 1.296
4. x2 = 1.500, x3 ≈ 1.153
5. x2 = 1.000, x3 = −1.000
6. x2 ≈ 2.759, x3 ≈ 2.755
7. x2 ≈ 3.134, x3 ≈ 3.142
8. x2 ≈ 0.167, x3 = 0.164
9. x1 = −1, xn = (−1)n
10. x1 = 1, xn = (−1)n 2n−1
11. x1 = 1, xn = (−1)n+1
12. x2 is undefined for f � (−3) = 0
13. x1 = 5; as n → ∞ then
xn = 2x1 − x21 → −∞
Then g(x) < 0.2 for x in (1,
By Theorem 3.11, we can apply
Newton’s Method with x1 = 1.
Thus, lim xn = c where f (c) = 0.
n→∞
x2 ≈ 1.10292
16. x5 ≈ 3.9330
17. x3 ≈ 3.1416
18. x4 ≈ 3.1416
19. Let f (x) = x2 − 10 and
�
�
� f (x)f �� (x) �
g(x) = � (f � (x))2 �.
From the graphs, we see g(x) ≤ 0.3
when 3 ≤ x ≤ 3.5.
By Theorem 3.11, xn converges
to a zero of f (x) if x1 = 3.
Then x4 ≈ 3.162.
y
f �x�
0.3
g�x�
3
x
3.5
20. The graphs of f (x) = x3 + 5x + 10
�
�
� f (x)f �� (x) �
and g(x) = � (f � (x))2 � are shown.
Note g(x) ≤ 0.6 for x in [−2, −1].
By Theorem 3.11, xn converges to
a zero of f (x) if x1 = −1. Then
x4 ≈ −1.423.
y
5
).
4
Also,
1. x2 =
15. x4 ≈ 1.7100
1. x4 ≈ 1.25992
y
Check It Out
14. x1 = 2.5,� as n →
� ∞
2�
�
|xn | = � 1−3x
�→∞
2
Section 3.8
26. y = 20 ft, x = 7.5 ft
27.
431
f �x�
0.6
g�x�
�2
�1
x
432
APPENDIX A. APPENDIX
21. Let f (x) = cos(πx2 ) and
�
�
� f (x)f �� (x) �
g(x) = � (f � (x))2 �. From the
graphs, if 0.35 ≤ x ≤ 0.45 then
g(x) ≤ 0.7. By Theorem 3.11,
xn converges to a zero of f (x) if
x1 = 0.35. Then x4 ≈ 0.399.
y
0.7
f �x�
g�x�
0.35
0.45
x
x2
22. Let f (x) =
sin(x) − 2x and
�
�
� f (x)f �� (x) �
g(x) = � (f � (x))2 �.
In [6, 7], we find g(x) ≤ 0.9.
By Theorem 3.11, xn converges
to a zero of f (x) if x1 = 7.
Thus, x4 ≈ 6.591.
y
0.9
g�x�
14. �y = csc
Check-It Out
f �x�
1. L(x) = x4 + 1;
√
4.1 ≈ L(4.1) = 2.025
7
x
25. Let f (x) = ax2 + bx + c, a > 0,
f (x)f �� (x)
and g(x) = (f � (x))2 .
Then g(x) → 12 as x → ∞.
2a(b2 −4ac)
Note, g � (x) = (2ax+b)2 > 0
b
if x �= − 2a
. Thus, g(x) is increasb
b
ing on (− 2a
, ∞) and (−∞, − 2a
).
1
In particular, |g(x)| < 2 if
b
x �= − 2a
.
Hence, Newton’s
method converges to a zero of f (x)
b
if x �= − 2a
.
−0.12
15. �y ≈ −0.238; dy = −0.24
2. Let y = g(x) = tan x and
dx = 0.005. Then
dy = g � ( π4 )dx = 0.01
16. �y ≈ −0.0175; dy ≈ −0.0177
17. 5.1
19.
3. Let f (x) = x3 be the volume
of a cube with side x, and
let �x be the error in the
measurement of the side
1
where |�x| ≤ 16
.
The propagated error is
|f (12 + �x) − f (12)| ≈
1
f � (12) · 16
≈ 13 cu. in.
√
3
2
18. 1.92
−
π
360
√
3π
180
20. 2 −
21. 32.8
≈ 0.857
≈ 1.97
22. 9.11
23. 3 sq. in.
24. 0.785 sq. in.
25. 50 cubic inch
27. 349 ft
26. 3.75 sq. in.
28. 0.013 inch
29. 0.08 inch
30.
1
radian
40
True or False
≈ 1.4◦
Chapter 3 Multiple Choice Test
T,T,F,F,T
F,T,T,F,T
1.
2.
3.
4.
5.
1. L(x) = 12x − 6;
(2.1)3 ≈ L(2.1) = 9.2
2. L(x) =
− x4
1
(1.9)2
+
7.
8.
9.
+
√2
3
11.
12.
13.
14.
15.
A
D
C
C
C
16.
17.
18.
19.
20.
B
C
A
B
B
Section 4.1
Try This
1. a) πx + C
π
;
9
π
L( 6
−
sec( π6 + 0.2) ≈
+ 0.2)
√2 + 2 ≈ 1.288
15
3
x
L(x) = − 12
+ 2;
√
3
7 ≈ L(1) = 23
≈ 1.917
12
x
1
L(x) = − 64 + 2 ;
√2 ≈ L(1) = 31 ≈ 0.484
64
17
√
√
3+ 3π
3
L(x) = − 2 x +
;
6
π
cos(61◦ ) ≈ L( π3 + 180
)=
1
π√
−
≈
0.48
2
120 3
x
√
√ ;
+ 4−π
2
4 2
π
π
◦
sin(44 ) ≈ L( 4 − 180
)
180−π
√ ≈ 0.69
180 2
b) y = x3 + C
2. a) 3x5 + x2 + C b) 2 sin x + C
�
−2
3. a) x−3 dx = x−2 + C =
5. L(x) = 2x + 1 − π2 ;
tan( π4 + 0.1) ≈
L( π4 + 0.1) = 1.2
2x
3
6. C
7. B
8. B
9. C
10. C
3. 2000 books
≈ L(1.9) = 0.275
x
4. L(x) = 48
+ 83 ;
√
3
70 ≈ L(70) ≈ 4.125
6. L(x) =
A
A
C
C
B
Investigation Projects
3
;
4
x
3. L(x) = 34
+ 17
;
2
√
285 ≈ L(285) ≈ 16.88
Section 3.9
Try This
1. Linearization L(x) = x;
π
π
π
tan 24
≈ L( 24
) = 24
2. dy = −0.05,
�y = √1 − 1 ≈ −0.0465
1.1
√
3. Let√f (x) = x. Then
15 ≈ f (16) + f � (16)�x =
31
= 3.875 where �x = −1.
8
� π �
4. sin 29◦ ≈ sin π6 + cos π6 − 180
=
√ �
�
1
3
π
+
−
≈
0.4849
2
2
180
5. If x is the height, then
f (x) = 16x is the volume.
The propagated error is
|�f | = 16 |�x| ≈ 1.6
since |�x| ≤ 0.1.
The
relative error
� percentage
�
�
�
is � f�f
·
100%
or
�
(6)
�
�
� 16�x �
�
�
� 16(6) � · 100% ≈ 1.7%.
− 2 ≈ −0.11;
45
√
− π453 ≈
dy =
Exercises Section 3.9
6
� 8π �
=
10. L(x) =
=
11. �y = 1.01; dy = 1
�
10
12. �y =
− 12 ≈ −0.006;
41
1
dy = − 160
≈ −0.006
� 23π �
13. �y = cot 90 − 1 ≈ −0.034;
π
dy = − 90
≈ −0.035
− 2x12 + C
�
b) x1/3 dx =
3 4/3
x
4
x4/3
4/3
+C =
+C
�
c) 4x−1/2 dx =
√
8 x+C
4. 2x4 − 8x2 + K
4x1/2
1/2
5. 4x +
+C =
x3
3
+K
6. sin t + C
7. y = 2t + sec t + 1
8. 5 sec
Check-It Out
b) 5x2 − x + C,
3
c) −2 cos t + K, d) − 3 + C
x
1. a) 4x + C,
2. y = x2 + x + 3
True or False
F,T,T,T,F,
T,F,F,T,F
A.2. ANSWERS TO EXERCISES
Exercises Section 4.1
Check-It Out
1. 10x + C
1. 2 +
2. πx + C
3. 20t2 + C
4. 6t2 + C
5. 5 sin t + C
6. 3 tan t + C
7. sec t + C
9.
− w13
+C
11.
1
− 16w
2
13.
3
m10/3
10
3.
+
8. − 13 cos t + C
−1
10. 3w
3 +C
C
12. 17 w7 + C
+D
√
14. 23 (m3/2 − m) + D
15.
5 9/5
m
9
16.
72 11/6
m
+K
11
1 3
1 2
s − 2 s − 2s
3
17.
+K
19.
− 25 s5/2
20.
+
3s2
C
− 8s3/2 + 8s + C
21. −3 cos θ − tanθ + C
1 3
x
3
39.
4(n+1)(2n+1)
3n2
43.
47.
51.
T,T,F,F,F,
F,T,T,T,T
1. 14
2.
5. 68
6.
29
20
i−1
3
10.
i+1
n
i=3
i=1
�� �
�
n
�
3i 2
3i
3
11.
−
·
n
n
n
i=1
�
sin
13. 670
24. 2 − csc α + G
kπ
n
�
n+1
,
n
18.
27. − cot β + P
3n+1
,
2n
19.
n−1
,
n
29. y = 2x + x2
20.
14n2 +9n+1
,
2n2
21.
(n+1)(2n+1)
,
6n2
28. − csc β + cot β + P
30. y =
1 4
x
4
√
− 13 x3 +
√
1−6 2
12
31. y = 2 t
6 11/6
32. y = − 11
t
+ 37 t7/3 +
9
77
34. y = − cot x +
√
3
3
35. a) v(t) = −32t + 16
b) s(t) = −16t2 + 16t + 96
c) s(0.5) = 100, highest point
36. a) s(6) = 102 ft, highest point
√
√
b) v(6 + 586
) = −8 586
4
≈ −194 ft/sec
37. 1 +
√
22
4
39. y =
x2
2
≈ 2.2 sec
−
x3
6
+x+
15
2
4.
5.
S=
6. 8
(n+1)(2n+1)
,
6n2
7. 16/3
2
s=
(n−1)(2n−1)
6n2
a+b
2
5.
2, 1.5
1. 182
4. 4, 10
6. 0, −4
(b−a)−(b3 −a3 )
3
2. 3
3. 9.5
T,F,T,T,T,
F,F,T,F,F
limit
1. 12
2. 12
5. 10
6. 0
7.
1
3
9.
limit 4
19. 25
1
4
23.
32.
29.
9 7
,
30. 446
, 374
2 2
27
27
√
√
(1+ 3)π (3+ 3)π
,
12
36
√
√
(9+ 3)π (7+ 3)π
,
12
12
31.
√
√
10 + 11),
√
√
10 + 11)
33.
35.
43.
14n2 +3n+1
,
6n2
2
−3n+1
= 14n 6n
2
36. S(n) =
4(n+1)
n2
, s(n) =
b) 0 c) 8,
b) −10
20. 8
d) 5
21. 1
56.
4(n−1)
n2
d) 8
22. 6
5
24. 6
25. 18
26. 4
2
112
16
28. 3
29. 16
30. 6
3
�5
(2x
−
5)dx
−2
�4
2
0 3x(4 − x) dx
�3√
�
2
5 + x dx
30. 25 x22 dx
0
8
36. 20
37. 32
38. 2
π
4
+
3
2
π
4
55. S(n) =
2
16. −4
c) 20,
44.
5
2
π
2
45. 1 +
46.
�1 2
47. 0 x dx = 31
� √
48. 01 x dx = 23
√
49. 12
50.
2−1
4
s(n) = 6 − n
�
�
�
�
1
1
35. S(n) = 1 − 16 2 − n
1− n
�
��
�
1
1
s(n) = 1 − 16 2 + n
1+ n
37. S(n) =
9
2
π+2
40. π + 4
4
41. π4
42. 32
4
,
n
2
π
4
39.
�
�
1
33. S(n) = 9 1 + n
,
�
�
1
s(n) = 9 1 − n
4. 3
10. 1 11. −4 12. −12
18. a) 31,
1
(3 + 2 3 +
8
√
1
(3 + 2 2 +
8
√
8. 2 −
17. a) −4,
limit 24
limit
25π
4
9π
2
3. −3
13. −6 14. −4 15.
28.
s(n)
√
π(1+2 3)
6
True or False
27.
34. S(n) = 6 +
2. 8
3. 151
,3
50
√
√
√
√
2+ 2+ 3<A<3+ 2+ 3
1.
3.
36 31
,
26. 533
, 319
25 25
840 420
√
√
(9+ 3)π (7+ 3)π
,
12
12
27.
2.
Check-It Out
limit 7
(n+2)(n+1)(n−1)
,
4n3
32.
Try This
1
7. 2(b3 − a3 ),
limit 1
24.
5
3
Section 4.2
1.
Exercises Section 4.3
3(8n2 +9n+3)
,
n2
31.
40.
i
n
3
2
limit
23.
25.
50.
limit 1
8n +3n+1
,
2n2
38. 144 ft
25
6
1+
11
8
π
n
22.
33. y = cos x
8.
�
16. 14,760
17.
26. sin α − tan α + G
4. 3 + 6a
14. 1225
15. 2640
25. − csc x + C
7. 60
n
�
k=1
+ tan x − x + C
46.
5
6
5
6
35
2
Try This
49
36
3.
20
�
n
�
42.
Section 4.3
21
2
12.
22. θ − sec θ + C
23.
11
6
2.
True or False
9.
(7n−1)(25n+7)
,
4n2
(7n+1)(25n−7)
s(n) =
4n2
1
14
40.
41. 12
3
3
7
44.
16
45.
39
2
20
16
48. 3
49. 52
2
52. 56
3
38. S(n) =
Exercises Section 4.2
+m+K
1 2
s − 25 s5/2 + K
2
1 3
s − 12 s2 + s +
12
18.
5
n
433
(2
√
5π
,
6
3−1)π
24
s(n) =
π
3
434
APPENDIX A. APPENDIX
Section 4.4
Try This
64.
66.
1. 130 and 1
3. a/2
2. 2.5
4. 1/4
6.
3, 2, 1.5
�x
2
1−( 0 sin(t2 )dt)
�
�
2 2
1+( 0x sin(t2 )dt)
1.
3.
5.
Check-It Out
10
3
2
π
2.
√
3.
7.
2
1
(x2
5
14
9
8.
1.
2. 15
5. − 23
6. − 74
3
√
10. −2 3
9. 3
1
6
3. 9
7
2
8. 5
19. 2
23. 16
24.
25. 12
26.
32
3
31
12
18.
35. 10
36.
8
3
4
3
38.
40.
1
1 + x2
31.
4
3
5.
28.
32.
4
3
1
(2b + dm + cm)
2
c 2
(c + cd + d2 )
3
3
41. π2
42.
2
√
44.
3
4
1
3
51.
3
2
50.
52. ±
53. ±1, 0
55.
54.
f (q(x))q � (x)
56. p�
�� x
1
√
0
13.
1
4
−
sin(x3 )
f (p(x))p� (x)
17.
K(b − a)3
6
1
=
=
.
12n2
12(62 )
72
1
.
72
1 − w(w + 2) + C
T, T, T, T, F, F, T, T, T, T
2.
1
6(9−x3 )2
4. − sin
�1�
t
+C
+C
True or False
1 5
(1+ w
)
+C
5
�
�4
1 − w12
+C
21.
1
(1 + tan α)4 +
4
− 13 csc(3α) + C
+C
23.
2
(2y
5
24.
2
(1 + 3y)3/2
3
1
26. 255
10
32
20.
1
3
6. a) 2.97842, b) 0.26247
7. a) 2.62331, b) 2.60391
8. a) 0.35233, b) 0.35698
2
2 cos α+1
9. a) 8.15121, b) 8.14594
+C
+C
C
10. a) 3.10675, b) 3.09894
11. ET ≈ 0.00116, (b) ES ≈ 0.00002
using |f �� (x)| ≤ 29 and
|f (4) (x)| ≤
64
5
27. 2 28.
√
2
6
2. a) 1.11667, b) 1.10000
5. a) 3.06198, b) 3.14876
− 2(1 + 3y)1/2 + C
31.
1. a) 1.68333, b) 1.62222
4. a) 3.06198, b) 3.11013
− 1)5/2 + 23 (2y − 1)3/2 + C
29. 1 30.
Exercises Section 4.6
3. a) 0.765496, b) 0.77753
(y 5 −5y 3 +4)9
+C
5
−2
1 √
√
+C
18. 10
( x − 1)5
x+1
19.
25.
sin(πx)
x
1 √
( 1 + √4 + √2 + √4 +
3
2
5
10
17
√2 + √4 + √1 ) ≈ 1.76327
26
37
50
+ 4)3/2 (3w − 8) + C
+C
−2
1+sin θ
22.
2.
0.088333
�
�
f (1)+2 f (1.2)+f (1.4)+f (1.6)+
�
�
f (1.8) + f (2) ≈ −0.418021
1
10
An upper bound is
16. −
58. f (x) = xn + C
� �
�
59. Hint: t12 = 1 if 2/3 ≤ t ≤ 1.
�
�
62. 01 x2 dx = 13
63. 01 x4 dx = 15
57. 1
1
2
1.
4. Since
|f �� (x)| = | − 4x2 cos x2 − 2 sin x2 |
we find |f �� (x)| ≤ 6 = K on [0, 1].
Then
� 2
�6
(6−x3 )5
1
x + 4 +C
8. − 15 +C
12
�√ �
1
sin(3t) + C
10. −2 cos
t +C
3
12.
Section 4.6
Try This
3.
15. 2(x4 + x2 + 1)4 + C
3
−
�
f (t)dt · f (x)
3
44. 2
where f (x) =
tan(4θ) + C
� �
14. −2 cot θ2 − θ + C
3
3
√
7.
2√
11. −
x2 + 1
2x sin(x6 )
2
(w
15
2
3
5. Since
|f �� (x)| = | − 4x2 sin x2 + 2 cos x2 |
we obtain |f �� (x)| ≤ 6 on [0, 1].
�
Choose n ≥ 105 /2 or n ≥ 224.
(2t2 +3)4
16
6. −
9.
45. − tan x
√
47. 4 1 − tan3 x sec2 x
sin x cos x
48.
1 + sin5 x
49.
46
15
3
42. 13
43.
2
1
A
46.
W
2
1
− 1b
48. 12 B
a
|ET | ≤
√
3. 2 sin t + C
1 + x2
46. −
x
65x3
10. 0,
Exercises Section 4.5
2
3
1.
4 + π2
2
16
27. 20
3
30. 4
34.
1
3
F,T,F,T,T,
T,T,F,F,T
20. ] 1
√
√
22. 3 2 − 2 3
21. 1
9.
True or False
√
π−2
4
17. 7
π+2
14.
4
15. 78
16. 43
43.
4. 2
12. 4 − 2 2
13.
39.
7.
cos(x3 −3x2 )+C
√
6. 2 tan t + C
1
(5 + 3x2 )π+1 + C
π+1
−1 3
2.
(x + 3x) + C
3
�
�
1 √
3.
2 2−1 +C
4. 0
3
Exercises Section 4.4
1. 9
1
3
4.
28
15
F,T,F,T,T,
F,T,T,F,T
37.
41.
2. − cos x2 + C
+C
+ 1)5 + C
Check-It Out
33.
37. 0 38. 0 39. 100 40. 2
47.
True or False
29.
· sin(x2 )
45.
2
(x2 + 4)3/2
3
−1
+C
3(3x2 +1)
1
3
34. 1
√
√
35. 3 + 2 3
36. 2 2/3
Try This
, f (x) + x sin2 x
√
− sin(x) cos x
x4 +1
8.
11.
33.
x
7.
1.
�
x
65. − cos
x2
sin x dx = 0
0
Section 4.5
−10.875 meters/sec
5.
� 2π
32.
1
2
56
81
12. ET ≈ 0.16667, (b) ES ≈ 0.00417
using |f �� (x)| ≤ 4 and
|f (4) (x)| ≤ 6
13. ET ≈ 0.01563, (b) ES ≈ 0.00011
using |f �� (x)| ≤ 3 and
|f (4) (x)| ≤ 5
A.2. ANSWERS TO EXERCISES
31. min f (e−1 ) = −e−1 ,
14. ET ≈ 0.00926, (b) ES ≈ 0.00069
using |f �� (x)| ≤ 29 and
Check-It Out
15. a) n ≥ 366, b) n ≥ 26
4. y = 5x − 2 ln |x| − 5
80
81
|f (4) (x)| ≤
1. − tan x
435
2. 1
3. y =
16. a) n ≥ 130, b) n ≥ 11
5. 3e3 − 2e2 − e3 + e2
18. a) n ≥ 255, b) n ≥ 9
True or False
x
2e4
+
3
2
concave up (0, ∞)
y
2
17. a) n ≥ 238, b) n ≥ 19
19. 14 both by TR and SR
F,T,T,T,T,
20. 132
21. TR 27 12 ft, SR 26 23 ft
�
22. 010 60D(x)dx ≈ 2910 ft3 by TR,
2900 ft3 by SR
� 30
23. 0 f (t)π( 12 )2 dt ≈ 138 ft3 by TR,
137 ft3 by SR
24.
� 12
0
w(x)dx ≈ 126 ft2 by TR,
126 23 ft2 by SR
Chapter 4 Multiple Choice Test
1. B
2. C
3. B
4. A
5. C
21. B
6. A
7. A
8. B
9. B
10.B
22.D
11.
12.
13.
14.
15.
B
A
B
B
B
16.
17.
18.
19.
20.
D
B
B
B
C
Investigation Projects
1.
π
4
2. 8x4 f (x4 ) + 2
�1
� x4
0
f (s)ds
(16x3 − 4x)dx = 2
4. sin2 1
�x√
5. If f (x) = 2 t3 + 1 dt, then the
limit is f � (2) = 3
� 1 dx
7
6. 0 (1+x)4 = 24
� 2x
7. If f (x) = x t cos t dt, then the limit
is f � (x) = 3x cos x
3.
0
8. a) 0
1
b) does not exist
T,T,T,T,T
� �1
Exercises Section 5.1
1. a) ln x + 2 ln y + ln z
b) 2 ln x + ln y − ln z
1.
2.
3.
ln 3 + ln(x + 1), ln x − ln 4, ln
20.086, 0.247, 1.386, −0.693
sec x csc x,
3
,
3x−1
4.
3
e2
6.
y� =
2
x
7.
y� =
�
2
x2/3
√ sin x ·
x+1
8.
9.
10.
y�
−1
3.
3
x
2
2x+1
+ 2 cot x −
=
2−2x
2x−3−x2
y = 4 ln |x| −
1
2
+
2
3x
9−x
4.
7.
x
x2 +1
9.
2−ln
√x
2x x
11.
t
t2 +1
8.
2 cos(2x)
x
14.
15.
1
x2 −1
16.
1
x2 +1
−
y = ln | ln x| + 1
� �1�2
√
4+x2 +x
4+x2 +x
√
28.
�
y
1−ln
√t
2t3/2 ln t
(ln x)2 +2
x
�
1
x((ln x)2 +1)
22. y =
3
x
2
30.
�
x2 + 1 � x
+
2
x2 − 1 x +1
x
x2 −1
(x − 2)(x + 3)
×
(x + 2)(x − 3)
�
1
1
1
+ x+3
− x+2
−
x−2
√
3x − 2
x(x + 1)2
�
3/2
+C
x(x − 1)
√
x+1
�
x
33. max f (e) = e, concave down (0, ∞)
4+x2
−
3
2
24. y = − 32 x + 4
�
√
25. x2 x2 − 1 x2 + x2x−1
26.
1
sin(2t)
t
17. − csc x 18. sec t 19.
29.
1 2
x
2
1
t
− 2 sin(2x) ln(x2 )
1
x ln x
tan x
1
2(x+1)
−
1
sin x(1+ln | sec x|)
cos2 x
10.
13.
20.
1
, ∞)
e3/2
y
cos(ln x)
x
6.
12. 2(ln t) cos(2t) +
27.
�
concave up (
1
)
e3/2
3(ln x)2
x
5. − tan x
x + 2x ln x
5. y = x
−
√
concave down (0,
x +1
23. y = 8x − 5
Try This
1
32. min f (e−1/2 ) = − 2e
c) ln x + ln y − 13 ln z
��
�
x2 (x+1)
2. a) ln 3 x2 +3
�
�
�
2 −4
b) ln √ 83
c) ln x
x2 +1
21. y = x + 1
Section 5.1
x
1
�
x
34. min f (e−4 ) = −e−4 ,
concave up (0, ∞)
1
x−3
�
3
1
2
− −
2(3x − 2)
x
x+1
�
�
1
y
�
3
1
1
−
−
x
2(x − 1)
2(x + 1)
(2x − 1)(x + 2)(x − 3) ×
�
�
1
2
1
1
+ x+2
+ x−3
2 2x−1
0.5
�
� �4
x
436
35. min f (1) = −1,
conc. down
APPENDIX A. APPENDIX
� e2
47.
ln x
dx
x
1
(0, e−1 ),
conc. up (e−1 , ∞),
4. 2x −
=2
y
ln |3x + 2| + C s
5. y = − ln2x + 3
6. y = − ln | ln x| + e
0.5
y
4
3
True or False
� �1
x
1
�2
1
�e
2
e−1 x dx
48.
�1
Exercises Section 5.2
1. π ln |t| + C
=4
y
36. min f (1) = 1, concave up (0, ∞),
� �1
1
x
1
39. y = 3 + ln x
1 2
(t
2
40. y =
1
(ln t)2
2
− (ln t)2 + 1)
�
x
49. If (a, ln a) is the point of tangency,
then the y-intercept of the tangent
line is (0, ln(a) − 1). Also, (0, ln a)
is the y intercept of the horizontal line through the point of tangency. The distance between the
y-intercepts is 1 unit. See figure
below.
y
42. y = x ln x − e
43. y = x + ln x − 1
44. y = 4 ln x +
45.
�2
1
1
x
ln(x)dx = ln(4) − 1
a
ln�a��1
5. 23 ln 52
6. ln 1331
1
7. 16 (8t + 3 − 3 ln |8t + 3|) +
8. t2 + 31 ln |3t + 2| − 49 + C
9. π1 ln | sec(πx)| + C
1
10. 2π
ln | sin(2πθ)| + C
3
11. ln 2
12. 12 ln 53
13.
ln 3
4
15.
1
3
=
ln |2 + 3 sec t| + C
16. − ln(1 + cos2 t) + C
17.
1
2
ln | tan 2x| + C
18. ln | tan x| + C
19.
1
4
20. (ln 2)2
√
25. ln | cos θ + θ sin θ| + C
x
26. − ln |α cos α − sin α| + C
30. 5 ln 52
√
ln(2 + 3)
32. ln
29. ln 2
y(3x−2)
x(2y 2 +1)
51. a) ∞
31.
b) 0
1
2
34. y = t − ln |t + 1| + 1
35. y =
1
ln(x)dx =
x
2
1
(1
2
− ln 2)
y
1
2
1
x
Try This
1. a) 5 ln |x + 1| + C;
b) 2 ln |x3 + 1| + C
2. 4 ln 17
3. 4 + ln 16
4. y = 3 −
3
ln x
7.
ln 2
4
8.
4
3
33. y = −3 ln |2 − x|
Section 5.2
1/2
ln 3
2π
28. ln | sin x + 1| + C
dy
dx
50.
46. −
14.
C
27. ln | sin x + cos x| + C
y
�1
2
ln |x| + C
5
1
ln |4x + 1|
4
21. 2( 2 − 1)
22. 8
√
23. 2 ln |1 + x| + C
√
√
24. x − 2 x + 2 ln |1 + x| + C
�a, ln�a��
ln�a�
−4
2.
3. ln |x + 5| + C 4.
4
y
41. y =
x
T,F,T,F,F,
F,T,F,T,T
6(
√
3−
π
√
2)
Check-It Out
1. ln(t2 + 9) + C
1
2
ln | sec 2x|
36. y = ln |1 − sin t|
37. s(t) = −10 ln |t| + 18
38. s(t) = − 91 ln |3t + 1| + 13 t
−1
2
40. √
1 + x2
1 − 4x2
1
41. √
x 4 + ln x
√
√
9+(ln x)2
42.
9 + x2 −
x
39.
43.
ln(3)
3
3
π
44.
√
√3+1
3−1
2.
1
4
ln |1 + 4 sec x| + C
45.
3.
1
4
ln 2
47. (1, 0)
ln
ln 2
2
46.
2 ln(2)
π
48. − 41 < m < 0
+C
A.2. ANSWERS TO EXERCISES
Section 5.3
437
20. If R(t1 ) = R(t2 ) then
4t1
4t2
= 4−3t
.
4−3t1
2
Cross-multiply:
16t1 − 12t1 t2 = 16t2 − 12t1 t2
9. yes
y
Try This
1. yes, yes 2. yes, no, yes
3. p−1 (x) = x+1
with domain
3
and range (−∞, ∞);
f −1 (x) =
1
x
1
1
− 1 with domain
x
21.
{x : x �= 0} and range
{y : y �= −1}
4. p(1) = 2, p−1 (2) = 1,
(p−1 )� (2) = −1/6
10. yes
y
Check-It Out
22.
1
,
1−x
D = {x : x �= 1},
1. f −1 (x) =
1
R = {y : y �= 0}
1
−4
2. Since
=
< 0,
(x − 4)2
g is decreasing and one-to-one.
g � (x)
3. p−1 (3) = 0, (p−1 )� (3) =
4.
D
R
12.
F, F, T, T, T,
F, T, T, T, F
Exercises Section 5.3
5. yes
6. yes
3. no
g −1 (t)
=
t+4
1−t
D = {t|t �= 1}
R = {y|y �= −1}
√
13. M −1 (x) = − 1 − x
D = {x|x ≤ 1}
R = {y|y ≤ 0}
True or False
2. yes
5x
7−2x
= {x|x �= 72 }
= {y|y �= − 52 }
11. f −1 (x) =
1
2
2
5
1. yes
x
4. no
14. N −1 (x) = 2 − x2
D = {x|x ≤ −1}
R = {y|y ≤ 1}
15. C −1 (t) = t2 + 3
D = {t|t ≥ 0}
R = {y|y ≥ 3}
√
16. R−1 (t) = t − 2
D = {t|t ≥ 3}
R = {y|y ≥ 1}
7. yes
y
1
x
1
17. If g(x1 ) = g(x2 ) then
πx1 − 3 = πx2 − 3.
Thus, x1 = x2 .
Hence g is one-to-one.
18. If f (x1 ) = f (x2 ) then
3
x + ln 2 = 37 x + ln 2.
7 1
8. yes
y
Thus, 37 x1 = 37 x2 and x1 = x2 .
Hence f is one-to-one.
19. If s(t1 ) = s(t2 ) then
t1
= 2tt2+3 .
2t +3
1
1
1
x
2
Cross-multiply:
2t1 t2 + 3t1 = 2t1 t2 + 3t2
Thus, t1 = t2 .
Hence s is one-to-one.
23.
25.
27.
29.
30.
31.
Thus, t1 = t2 .
Hence R is one-to-one.
If C(w1 ) = C(w2 ) then
w1 − 1
w2 − 1
=
.
w1 + 1
w2 + 1
Cross-multiply:
w1 w2 + w1 − w2 − 1 =
w 1 w2 − w 1 + w 2 − 1
Thus, 2w1 = 2w2 or w1 = w2 .
Hence C is one-to-one.
If f (t1 ) = f (t2 ) then
1 − 2t1
1 − 2t2
=
.
1 + 2t1
1 + 2t2
Cross-multiply:
1 + 2t2 − 2t1 − 4t1 t2 =
1 − 2t2 + 2t1 − 4t1 t2
Thus, 4t2 = 4t1 or t1 = t2
Hence f is one-to-one.
1
24. −1
11
16
1
−
26.
3
4
1
4
−
28.
2
3
y is increasing for
dy
4x(x + 1)
=
>0
dx
(2x + 1)2
when x > 0. Then y has an
inverse function.
−9
Differentiate (F ◦ F )(x) = x to
obtain f (F (x))f (x) = 1.
Integrate to show F (x) = x.
32. 32
34. f (x) = x if 0 ≤ x < 1, and
f (x) = 3 − x if 1 ≤ x < 2
Section 5.4
Try This
1. t =
3.
dy
dx
A−100
2. w = e4 − 10
5
x
cos
x
=e
(cos x − x sin x),
dy
dx
= esin x x cos x + esin x
�
�
4. Rel. min. point 1, e−1/2 ,
�
�
relative max. point −1, −e−1/2
5. T (5) ≈ 87◦ F ; since T � (5) ≈ −3.3◦ F ,
the soup cools at the rate of 3.3◦ F
per minute.
6. 2ex/2 + C,
7.
1
(1
2
1 x3
e
3
− e−1 )
+C
8. y = 1 + t + e2t
Check It Out
1
ln(y − 1)
2
�
f (x) = e2x + e−2x
1. x =
2.
438
3.
1
(e
3
− 1)
APPENDIX A. APPENDIX
1
(e
2
41.
4. y = − sin t +
e−3t
True or False:
46. y =
47.
�
y+1
3
�
5. h = e6P +4
6. w = e(5−10V )/2
t (2 − t)
2
7. y � = 2xex +2 8. f � (t) =
et
1 − 2t
�
9. g (t) =
e2t
�
10. y = 10ex sin x (x cos x + sin x)
11. R� (x) = e3x (5 − 3x)
1 − w ln w
wew
�
cos
2x
13. y = −2e
sin 2x
12. P � (w) =
14.
y�
= −1
4e2x
(ex − 1)(1 + ex )(1 + e2x )
�
�
16. f � (x) = 20 e20x − e−20x
15. L� (x) =
17.
r� (x)
24e4x
= √
1 + e4x
18. A� (t) = esin t (cos t cos 3t − 3 sin 3t)
1 2x
e +C
20. −ecos x + C
2
2
1
21. − e−t + C
2
e2t
22. 2et +
+t+C
2
√
� t
�
2 1 − et e − 1
23.
+C
3
3
24. x2 − x + C
2
19.
25. 2e
√
w
√
�
1
2
+
�
1−
1
e16
1
4
�
48.
49.
50.
51.
52.
Check It Out
1. 2(e
e−x
2
1+e−2t
2
)
≈ 167.6
26. ln e2w + e−2w + C
e−1
27.
28. 3(e − 1)
2
�
√
1+e
29. e 3 − e
30. ln 3
2
31. y = 3x + 1
32. y = −2x + 2
1
33. y = x + 2
34. y = 8e3 x − 3e3
2
√
√
3e
π√
x+ e−
3e
35. y =
2
12
36. y = −1
37. e − e−1
38. 3(e − 1)
39.
40. 10 ln(5) − 8
2
(2t−4) ln 3
2. a) 3π 3x ln π, b)
ln[(e2t + 1)(e−2t + 1)] − ln 2
3. a)
π 5s
,
5 ln π
3
4 ln 2
b)
True or False:
F,T,F,F,T,
T,T,F,T,F
Exercises Section 5.5
1. −3
2. −1
log(7)−1
3
≈ −0.1
56. max (1, 1e ), inflec. pt (2,
6.
log3 (8/5)
2
≈ 0.2
57.
7.
ln 5
365 ln(1+0.12/365)
8.
32
3
9.
√
55. min. (0, 1)
2
)
e2
1
max (1, √e ), min (−1, − √1e ),
√ √3
inflec. pts (0, 0), ( 3, 3/2
),
e
√
√
3
and (− 3, − 3/2 )
e
58. max ( π4 ,
e−π/4
√
),
2
min ( 5π
,−e
4
−5π/4
√
59.
60.
63.
1
1
4.
2
y�2 x
1
1
64. x = µ; x = µ ± σ
12.
1
M � (x)
�
sin x sin x
2
,
ln 10
x
√
3− 3
ln 3
y�log3 x
1
3
x
≈ 0.1
√
12 t√ln 12
,
2 t
1
√
= 2 ln(10)(x+
x)
p� (t) =
6.
x
2
y
0.3 ln 200
≈ 2.29 mm
ln 2
1
3
(log2 (10) + 1) ≈ 1.4, 25
3
�
2
tan
x
y = sec (x)2
ln 2,
5. x
≈ 6.5
y
Try This
3.
−1
11.
Section 5.5
2.
≈ 13.4
101 − 1 ≈ 9.0
3
inflec. pt (e3/2 , 3/2
)
2e
�
�
√
2
A+ A +4
61. a = ln
2
1
3
4.
≈ 10.7
10. e( ln 2 − ln 3 )
),
2
inflec. pt ( π2 , e−π/2 ),
and ( 3π
, −e−3π/2 )
2
1
min (e−1/2 , − 2e
),
−3/2
inflec. pt (e
, − 2e33 )
min. pt. (e, e−1 ),
1
2
3.
5.
53.
1. r = 4 ln 1.02 ≈ 0.079
+C
19e6 −1
2e6
1
2
dy
1
=
dx
1 + 2e2xy (x + 2)
dy
6x + yexy
=
dx
4y − xexy
dy
5 + e2y
=
dx
4 − 2xye2y
dy
ex + y
=
dx
3 − x + e−y
dy
2xyey
=
dx
1 − x2 yey
dy
2x − yexy
=
dx
xexy − 2y
y = 4x − 4
54. y = x
Exercises Section 5.4
ln y
2. x = ln
�
�
5
3. t = 10 ln A+4
�
�
20
4. t = 5 ln 30−A
ex
45. y = ln
F,T,T,T,T,
F,T,F,T,T
1
2
42.
43. y = − cos t + 14 e−2t −
+1
44. y =
1. x =
− e−1 )
+ cos(x) ln x
13.
y
�
y�log ��x�
1
7. a) $1,051,161.90 b) $1,051,267.50
8. $12,460.77
9. P (15) ≈ 27, 167
P (t) → 40, 000 as t → ∞
�10
1
x
2
3
A.2. ANSWERS TO EXERCISES
14.
28.
y
10
x
y�10 �3
3
1
15.
x
2
29.
1
2(x−1) ln a
31.
cot(x)
ln 10
33.
2x cot(x)−1
x ln 4
1
1
y�log2�x�
x
42. y =
θ tan(θ)−1
θ ln 8
34.
y�4 x �1
4
1
2
x
4.
�2
y
10t
ln 10
√
47.
63x
3 ln 6
+C
48.
23x
3 ln 2
49.
ln(1+52x )
2 ln 5
3sin t
ln 3
53.
−2
54.
61.
ln 2
+C
+C
√
8cos t +1
ln 8
50.
+C
ln(9+32x )
2 ln 3
+C
5
−2 ln 5
52.
Exercises Section 5.6
+C
5. y =
+C
y��1�3�
1
19. 2x ln 2
21.
√
3 t
x�1
�
�
x
63. xx (ln(x) + 1)
64. x1/x 1−ln
x2
�
�
2
65. xln x lnxx
�
�
66. (2x+4)3x+6 3x+6
+ 3 log(2x + 4)
x+2
1+ln(log2 x)
x ln 3
�2
2
x
68. (log x)10 ×
�
10x (ln 10) ln(log x) +
69.
≈ 258 years
17. a) 13.2 mg
19. H = 280 (2/7)t/20 + 70
H(60) ≈ 76.5◦ F
20. H = 325 − 265 (175/265)t/2.5
H(3.4287) ≈ 175◦ F
21. b) $182.21
25. a) v(t) =
74. a) 200
t
26.
ln(s) +
t+2
(t2 +t) ln 4
�
1
s ln 10
b) C =
ln 3
100
(ln 4)2 2
x
2
− x ln 2 + 1
√
76. a = 1, b = ln 2 = −c
75.
77. Hint: odd function
78. Hint: f (±1) ≤ a, f � (0) = 0.
b) 50 years
18. 15,683 years
72. $818.84; $97.77 per year at the end
of 6 years
73. P (0) = 1000
P � (0) = 200 units/minute
3
1 − x3
16. about 14.4 hours
20. 5 · 3t ln 3
23. 3t 103 ln(3) ln(10)
�
�x � 12 �
24. ln1212 ln ln(12)
27. 10
100 ln 0.9
ln 0.96
23. P (t) =
2
�
s
1+ln x
1+ln y
�
15.
71. $182.20; $109.33 per year at the end
of 10 years
ln 3
√
2 t
1−ln x
x2
70.
6. y =
8. y = ln(2 − e−t )
x
4
9. y = −2et /4
10. y =
1+x
1
11. y = 2 sec t
12. y =
1 − ln(sin x)
�
13. y = 1 + 2 sin2 t
2
14. y =
1 + cos t
�2
22. 3x 24x (ln 16 + x ln 9)
25.
y(x ln y−y)
x(ln x−x)
10x
x(log x) ln 10
2
1 − 2x
7. y = t
x
3
4. −4◦ F/min
F, F, T, T, T,
T, F, F, F, T
� tan t
�3/2
2
3
+1
+C
3 ln 3
18
4
3
− ln 2
56. ln 4
ln 3
15
58. π −2
ln 4
ln 10
e−1
60. ln19
2
√
4−2 2
62. log4 (1 + 4x )
ln 2
67. (log2 x)log3 x ·
b) y 2 = x2 + 3
True or False:
52x
2 ln 5
(2−x)2
/ ln 2 ,
3. −58 mg/hr
+C
−
x
2. 36,744,344
+C
t+1
2
51.
1. a) y = Ce2
1
ln 2
44.
ln(1/2) ≈
2
−114.7 units/year
6. 48 minutes after it was taken from the
oven
20
ln 10
46.
59.
x
4
y�log1�2�x�4�
ln 10
+1−
+C
x2
x−
� 1 �1/5.3
Check It Out
+C
57.
1
2x
ln 2
x
ln 4
+2−
�
1
1000
5.3
5. t ≈ 3.83 hr, or 3 hr and 50 min
1
ln 5
2
ln 3
+2−
5
2 ln 5
55.
y
x
25 ln 5
2
x
ln 27
45.
y
18.
� √
3
2
1. y = Cex −x
2. y = 4 t + 1
�
�2
17,639
3. 15871 15,871
≈ 19, 604
2 tan(2w)
− ln 3
32.
�
41. y = 2 1 +
43.
�4
1
x ln(10) ln(x)
30.
35. y = 10x ln 4 + 2 − ln 4
40. y =
1
Section 5.6
Try This
(x2 +1)2
39. y =
2
17.
2x (x2 ln(2)−2x+ln(2))
36. y = 28 ln(7)x + 14(1 − ln 7)
√
37. y = 8 3 ln(2)x + 4 − 8π√ln 2
3
√
√
√
30 ln 10
ln 10
38. y =
x + 10 − 30π
2
12
y
16.
439
1
(t
256
22. b) $64.82/year
+ 6400)2
24. 71.7%
b) a(1)
20
ft/sec
1+6t
= − 120
≈ 2.4
49
ft2 /sec
26. Yes since s(3) = 99 where
s(t) = −11t2 + 66t
27. P � (2) = 101 people/year
since P (t) = (t + 200)2 /4
√
28. a) 6 miles since P (h) = 30( 3 0.5)h
b) −1.7 inches of mercury per mile
√
29. 4π(0.15)2 e ≈ 0.47 cm2 /sec
440
APPENDIX A. APPENDIX
Section 5.7
Try This
10. y =
2. y = 1 −
11. y =
2
e−x
3. Amount of sugar in tank is
t3 −150t2 +4375t+31250
6250
y(t) =
29. t ≈ 7.6 sec,
13. y =
4. y(4) ≈ 143.8 mg, since
�
�
60
y(t) = ln
4 + 2−t/24 (ln 2 − 4)
2
5. I(t) = 6 − 6e−5t amperes, where t
is in seconds
6. I(t) = 6 − 6et/2 amperes, where t
is in seconds
7.
y
2
since
2t3 + 3t2 − 6
6(1 + t)
x2 + x + 4
14. y = √
2x + 1
15. y = xe−2x (1 + ln x)
e−3x + 1
3x
1
x
17. A, y = − − + Ce2x
4
2
16. y =
18. B, y 2 = 2(x + C)
�
�
−1
19. C, y = ln
x+C
2
x
�2
2
x
9. (0.1, 0.2), (0.2, 0.42), (0.3, 0.662)
�2
Check It Out
2
22. y = e1−x − 1 + x
5
y
2
1. y = e1−cos x − 1
2. y(10) ≈ 2.6 lb of salt
where y = 3 − 3e−0.2t
�
�
6
3. Q(t) = 25
1 − e−5t
4. ye (1) = 3, ye (2) = 7
�2
11
100
34. Q(t) =
5 −25t/3
e
7
1
2
x
True or False:
�2
Exercises Section 5.7
1
1
24.
= +2
y
x
1. y = 3 − 3e−x
3. y = (x +
2. y =
1 2x
(e − 1)
2
2
1)e−x /2
2
4. y = (3x + 2)e−x
�
�
�
�
1
5. y = 5 t −
6. y = 4 t2 − t
t
7. y =
1−
cos t2
t
10π + sin t5
8. y =
5t4
9. y = sin t + 2 cos t
7t
3
−1
�
Section 5.8
Try This
π π 4
, ,
3 4 5
3. 0
23. y = x
e
≈ 866 mg
�
�
since y(t) = 1200
1 − 2−t/24
ln 2
39. y(t) = $1, 000, 0000
since dy
= 0.05y − 5(104 ),
dt
y(0) = 106
40. y(4) = $22, 662.80
since dy
= 0.2y − 4(104 ),
dt
y(0) = 105 , and
�
�
y(t) = − 100,000
−4 + e3t/10
3
41. y = ex − x − 1
1.
F, F, T, T, T,
T, T, F, T, T
1−e
�
600
ln 2
38. y(24) =
0.5
8. C, A, B
−100t
3
33. Q(t) =
max Q(t) ≈ .086 coulombs
when t = 37 ln 25
18
�
�
35. I(t) = 52 1 − e−10t
�
�
36. I(t) = 94 e−3t e8t/3 − 1 , max is
√
I(t) = 2/ 4 3 ≈ 1.5 amperes
if t = 34 ln 3
�
�
15
37. y(t) = ln
32 − 32−t/12
2
21. y = 2ex−0.5 − x − 1
y
2
�2
=5
30. y(50) = 22.5 kg of sugar,
400t + t2
where y =
since
4 (200 + t)
dy
y
= 0.5 −
, y(0) = 0
dt
200 + t
31. 37.5 sec, where
2500 + 75t − t2
y=
since
250
dy
2y
= 0.5 −
, y(0) = 10
dt
100 − t
32. 1.5 lb
�
�
20. D, not separable or linear
�2
2(625+100t+t2 )
5(50+t)
dy
y
= 0.8 − 50+t
, y(0)
dt
where y =
x4
+ 3x
3
�
��
t3 �
12. y =
1 − π ln cos t2
2π
C
x
1. y = x +
1 + sin2 x
2 sin x
25. ye (1) = 1.5, ye (2) = 2.75,
ye (3) = 5.125
√
26. ye (1) = 2, ye (2) = 2 + 2/2,
√
ye (3) = 3 + 2/2,
√
ye (4) = 3 + 2
�
�
27. y = 40 1 − e−t/20
dy
5y
since
=2−
,
dt
100
y(0) = 0
�
�
28. y = 2.5 5 − 3e−2t/25
dy
2y
since
=1−
, y(0) = 5
dt
25
5.
4.
5
,
1+25x2
5
2
√
2. − π3 ,
√
7
4
3
e6x −1
6. −1 ≤ x ≤ 1, 0 < x < 1
7. x = 2, θ = tan−1 2 − tan−1 0.5 ≈
0.64 radians ≈ 37◦
Check It Out
√
10
7
√
= 33
1. a)
2. x
3. a)
4.
√
3
3
2
b) − π6
2x
x4 +1
b)
+
π
6
True or False:
F, T, T, T, F,
F T, T, T, T
2 tan−1 (t)
t2 +1
A.2. ANSWERS TO EXERCISES
Exercises Section 5.8
√
√
1. 2 1313
2. −2 2
√x
1−x2
√
3 3−4
10
3.
4. √ x 2
1+x
√
169 3−240
5.
6.
69
√
7. 14
8. 1 − 23
9. − π6
10. 53
√
√
3
√
11.
12. 43
1− 3
13.
4
5
40. θ = x
θ� (
�
4x
12
√
√3
x2 −9
�
1−
√
39
18
13) =
if θ = π3
25−x2
24
�2
for x =
�−1
√
13
46. No, if x < 0
2
2. a)
π
4
π
6
b)
√
+ ln 2
2
√
3π
6
True or False:
47. a)
5
F,F,F,T,T,
T,T,T,T,F
1
−1
| tan y|
�
Check It Out
� �
1. a) sin−1 x5 + C
� �
b) 12 tan−1 2t + C
�
�
c) √1 sec−1 √s + C
3.
1 − 25x2
−ecot x
24. y � =
1 + x2
x
π−2
25. y = +
2
4
√
π
26. y = −x + 3 +
6
√
π
27. y = x − 3 +
3
√
√
3
28. y =
x+π− 3
2
1
29. y � = �
2 1 − (x + y)2 − 1
�
1 − x2 y 2 + y
30. y � = �
1 − x2 y 2 − x
�
�
1
31. y � = ey ex − 1+x
2
�
�−1
2
sec
√ y
32. y � = y
−x
2
tan y−1
√
33. 10 7 ≈ 26.5 feet
√
34.
10/2 ≈ 1.6 yards
35. y = tan 2 ≈ 1.107, when x = 1
�
�√
√
2+2 5
36. rel max pt ≈
, −0.60578 ,
2
� √
�
√
5
rel min pt ≈ − 2+2
,
3.74737
2
52π
3
39. θ � (t) =
y
−3x2
16. y � = √
1 − x6
3
�
17. y =
9 + x2
2
18. y � = √
x(1 + x)
2
19. y � = √
x x4 − 1
−1
20. y � = √
1 − x2
−1
√
21. y � =
|x| 4x2 − 1
−1
√
22. y � =
|x| (arccsc x) x2 − 1
√
x − 1 − x2 sin−1 x
√
23. y � =
x2 1 − x2
37.
39
38. − 610
≈ −0.0639 radian/sec
14. − 13
15. y � = √
441
≈ 54 miles/min
1
y�sin�sin�1�x��
�1
x
�1
Exercises Section 5.9
� �
1. sin−1 x2 + C
�
�
2. sin−1 √s + C
3
�
�
1
t
3. √ tan−1 √
+C
5
5
y
2
�1
Рy�sin �sin �x��
�Π
2
7.
3Π
2
Π
2
x
�
�
tan−1 (cot x) = −1
domain is
{x : x �= nπ, n is an integer}
y
y�tan�1�cot x�
Π
x
�Π
2
Section 5.9
Try This
� �
� �
1. sin−1 x4 + C, 15 arctan 5t + C,
� �
1
sec−1 w
+C
7
7
�
�
π
2. π
3. 2 − sin−1 e−3 ≈ 1.5
�
�
4. 13 tan−1 x3 + ln(9 + x2 ) + C
�
�
5. 15 tan−1 x−3
+C
6. π/3
5
� 2x �
1
−1
7. 2 tan
e
+C
�
�
1
4x + C
ln
1
+
e
4
8.
1
,
3
−(ln 2)2 /4
1
2
sec−1 (e2x ) + C
12.
13.
14.
π
15
π
3
π
6
π
4
15. ln
�3�
2
tan−1
16.
�
√
2
�
17.
+C
�√
�
18. − √1 tan−1
2 cos θ + C
2
�√
�
−1
19. √
arctan
2 cot x + C
1
6
Π
2
�Π
arcsin(2t + 1) + C
10. sec−1 (2ex ) + C
11.
d
dx
1
2
8. sin−1 (tan x) + C
9.
�Π
2
48.
2
6. arcsec (4|x|) + C
2
�3 Π
2
1
3
tan−1 (3t) + C
�
�
|x|
5. √1 sec−1 √ + C
4.
b)
3 sin θ
2
2
� sec2 θdθ
20. Rewrite integral:
=
1+2 tan2 θ
√
√
2
−1
tan ( 2 tan θ) + C
2
�
�
21. 13 tan−1 t+1
+C
3
�
�
s−3
1
−1
22. 4 tan
+C
4
� x+1 �
−1
23. sin
+C
2
�
�
24. sin−1 w−3
+C
2
� x+3 �
25. arcsin 4 + C
�
�
26. 13 arctan x−2
+C
3
29.
π
20
π
3
31.
1
2
27.
�
π
2
π
48
28.
30.
�
sec−1
|x+3|
2
�2
�
+C
sec−1 x + C
�
�2
33. − 41 cos−1 (2x) + C
�
�
34. 21 ln tan−1 (2s) + C
32.
442
APPENDIX A. APPENDIX
35. y = 1 − sin−1 x
� �
36. y = 8 tan−1 2t − 2π
37. y =
38. y =
2 tan−1
39. y = t2 e− tan
40. y =
π
6
41.
1.
−1
2esec t
�
x+
−1
x cos−1 x−
x
42. π8
45. a = tan(1)
1−x2
π
6
43.
5
4
√
2. − 43
44.
π
12
4. 0
ln 3
2
6.
2.
sinh
√ x,
2 x
2
41.
9. y � = 1 − 2xsech2 x2
42.
10. y � =
43. sinh x +
y�
=
csch 2 (1/x)
x2
sech (x2 )−
44.
14.
4. ln(5/4)
√
√
√
√
5. (ln(2+ 3), 4 3), (ln(2− 3), −4 3)
7. a =
�
8. 15 cosh−1 121
− cosh−1
81
�
�
9. 12 cosh−1 sin32s + C
10.
Since tanh−1 x =
1
2
100
81
ln
�
24.
a) − 19 sech
+C
� �9
b) cosh−1 9t + C
True or False:
F, T, T, T, T,
T, T, F, T,T
y�
= 6x cosh
2
(x2
y�
27. y � = √
2x
−
51.
1
4
cosh 4x + C
8
25
48. tanh 1
4+coth 4−coth 8
4
3
5
50.
52.
1
2
cosh2 2 − cosh 2 +
3
3
53. ln(cosh 1)
�
54. 2 tan−1 e −
π
4
�
1
ln 3
2
56. rel max (0, −1),
rel min (1, − sinh 1)
55. x = −
57. 1 − e−1
� �
58. 8 tan−1 e2 − 2π
a) f −1 (x) = sinh2
− 2)
�x�
2� �
x
2
b) g −1 (x) = 2 tanh−1
68. y = ±1
Chapter 5 Multiple Choice Test
1.
2.
3.
4.
5.
1
30. y � = √
2 x2 − x
2csch−1 x
√
|x| 1 + x2
y � = −2 sec 2x
33. y � = 2 sec 2x
34. y � = sinh−1 x
35. y � = √−1
x
sinh3 x + C
cosh3 4x −
49. ln 2 −
29. y � = | sec x|
32.
1
12
1
3
6
4x2 + 1
2
28. y � =
4 − x2
31. y � = −
cosh3 (x − 1) + C
59.
2) sinh(x2
= cosh x − sinh x
�
�
26. f � (t) = e2t 2 coth t − csch2 t
25.
a) 2 cosh 3x sinh 3x
� �
−1 x
cosh x
x
23. y � = 2 coth 2x
1.
2.
− sinh x)
16. y � = sech x
�
+ sinh(x) ln x
�
�
x
= xsinh x sinh
+ cosh(x) ln x
x
�
x
= √
sinh x2 + 1
2
x +1
1
= cosh (ln x)
x
�
��
�
2
2
2
= −2xex sech ex
tanh ex
�
sinh2 (2x) + C
45. −2 coth + C
√
46.
2 sinh x + C
√
47. 2 2 sinh x2 + C
22. y � = 0
Check It Out
16x +1
10x
1−25x4
19. y �
21. y �
1+x
,
1−x
b) − [2sech 2x coth 3x] esech
c) √ 4 2
18. y �
20. y �
1+r
we find d = ln 1−r
=
�
�
1+r
1
2 2 ln 1−r = 2 tanh−1 r
d)
=
e−x (cosh x
17. y � = xcosh x
3. 0
2e
e2 +1
y�
1
4
1
3
x
2
√
√
−csch ( 3 x) coth( 3 x)
√
3 2
3 x
15. y � = sech t
sinh x
√
cosh x
20
9
38.
8. y � = 10 cosh 5x
13. y � = tanh x
−2xcsch2 (x2 ),
1
4
1
3
cosh(3x) + C
√
40. 2 sinh x + C
12. y � =
1. ln 19 ≈ 2.94, ln 199 ≈ 5.29,
ln 1999 ≈ 7.60;
d → ∞ as x → 1−
37.
39.
2x2 sech (x2 ) tanh(x2 )
Section 5.10
Try This
3
x −1
7. y � = 3 sinh(3x + 1)
11.
46. b = 3
12
13
3.
5. ln( 2 − 1)
π
2
t
√
36. y � = √−4x
8
Exercises Section 5.10
1−x2
C
A
B
C
B
6. A
7. A
8. C
9. B
10.B
11.
12.
13.
14.
15.
B
C
A
C
A
Investigation Projects
2. hint: x ≈ 4.965
3.
a) T ≈ 5, 796 K
b) T ≈ 11, 146 K
16.
17.
18.
19.
20.
A
D
A
B
D