a.2. answers to exercises 419 · a.2. answers to exercises 419 a.2 answers to exercises section 1.1...
TRANSCRIPT
A.2. ANSWERS TO EXERCISES 419
A.2 Answers to Exercises
Section 1.1
Try This
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
2.51128 , the total area of the eight rect-
angles is a better approximation
Check-It Out
1. 22 2.13
True or False:
T,T,F,T,F,
T ,T,T,T,T
Exercises Section 1.1
1. a) 4.1, b) 4.01, c) 4.001
2. a) 0.513167, b) 0.501256,
c) 0.500125
3. a)512 , b)
√48160 , c)
√10930
The arc length is the sum of parts
a), b), and c), which is about
1.13021.
4. a)
√1716 , b)
516 , c)
√4116 d)
√65
16
The arc length is the sum of parts
a), b), c), and d), which is about
1.47428.
5. a)1925 , b) same as part a)
6. a)296 , b)
643140 , c) part b)
7. a) 25, b) Approximate amount of
wetlands lost to erosion
8. a) −79, 207.9 gal/hr
b) −79, 920.1 gal/hr
9. a) 16.16 ft/sec b) 16.016 ft/sec
10. a) 12.6 ft/sec, b) 12.06 ft/sec
c) 6(2 + h) ft/sec
Section 1.2
Try This
1. 32 2. i) 4, ii) 4
3. i) does not exist, ii) 1
4. Note, f(x) = − 1x decreases
without bound as x approaches 0
and x �= 0.
Then f(x) does not approach a
single number as x approaches 0.
Thus, limx→0
−1
|x|fails to exist.
5. i) Note, cos
�π
[12k−2]+2
�= 1 and
cos
�π�
12k+1−2
�+2
�= −1 for all
integers k. Moreover,12k − 2
and1
2k+1 − 2 approach −2 as kincreases without bound. Then
cos
�π
x+2
�does not approach a
single number as x appoaches −2.
Thus, limx→−2
cos
�π
x+ 2
�does not
exist.
ii) limx→0
cos
�π
x+ 2
�= 0
6. Choose δ = �/2.
7. L = 1, choose δ = 0.03.
8. Choose δ = Min� �
3 , 1�.
Check-It Out
1. 3 2. 1 3. 4 4. no limit
5. no limit 6. hint δ =ε3
True or False:
F,F,T,F,F,
F,T,T,T,F
Exercises Section 1.2
1. 0.25 2. 0.25 3. 5 4. 0.125
5. 4 6. −0.625 7. 9 8. 0.25
9. 2 10. 2 11. 4 12. −2
13. 5 14. 3 15. 7 16. 0
17. undefined 18. undefined
19. 1 20. 1
21. no limit 22. undefined
23. δ = .0025 24. δ =0.002
7
25. δ =0.023 26. δ = 0.005(2
√2 + 3)
27. L = 2, δ =118 28. L = 9, δ =
18
29. L = 9, δ =110 30. L = 2, δ =
18
31. L = 5, δ =ε2 32. L = 1, δ = ε
33. L = 10, δ = min{ ε8 , 1}
34. L = −3, δ = min{ ε5 , 1}
35. L =13 , 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{ ε3 , 1}
39.√145− 12 ≈ 0.04159 in.
40. 2
�36 +
2π − 12 ≈ 0.1056 in.
π d2
4 = 36π ± 2
41. 4.85in. ≤ h ≤ 5.15in.
42. a) L = 2(1 + 2α) ≈ 2.0000288 m
b) Let A =0.0001
2(7.2×10−6).
T must satisfy
|T − 70| < A or 63.1 < T < 76.9
43. a) 0.0821(2)(293) = 48.1106 atm
b) Let V1 =48.110658.1106 ≈ 0.828.
Then 0.828 < V < 1.172 liter.
44. a)109 lux b) E is very large
45. a)p(4.05)−p(4)
0.05 ≈ $5.95/hr
b)p(4+h)−p(4)
4 = 6− h.If h ≈ 0, marginal profit is
$6/hr.
46. a)p(1.01)−p(1)
0.01 ≈ 0.151 m/hr.
b)p(1+h)−p(1)
h =2−49h
10 .
If h ≈ 0, instantaneous
velocity is 0.2 m/sec.
Section 1.3
Try It
1. 2, 2, −4
2. −2, −3/2, 3
3. 3, does not exist 4. 3/2, −1/2
5.
√3
3 ,
√3
3 6.14 ,
16
7. − 14 , −1 8. 1/2, 2
9. 1/3, 0 10. 2, 1
Check It Out
1. 9 2. −1 3.12
4. 7 5. −10
True or False:
T,F,F,F,F,
T,T,T,T,T
Exercises Section 1.3
1. 4 2. 5 3. 2 4. π
5.38 6.
87 7. 3 8.
83
9.12 10.
√3
3
11. 6, 10, 37 ,
74 12.
16 ,
16 ,
12 , 3
13. 3, 1 14. 4, 4
15. −4 16. 3 17. 27 18. 48
19. 4 20. −6 21. 7 22. −7
23.14 24. 6 25. 1 26.
12
27. 12 28. 75 29.34 30.
83
31. − 19 32. −4 33. − 1
25 34. −1
35.16 36.
114 37.
13 38.
112
39. 1 40. −1 41.
√36 42. 2
√5
420 APPENDIX A. APPENDIX
43.45 44. − 8
17 45. 3 46.14
47. 1 48. 1 49. 4 50.12
51. 0 52. No limit
53. −1 54.13
55. −3 55. 5
59. a) 1, b) no limit 60. 4
61. 0 62. 0 63. 4 64. 5
Section 1.4
Try This
1. Nonremovable discontinuity
at x = −1
2. Removable discontinuity at x = 0.
If h(0) = 1, then h
becomes continuous at x = 0.
3. g is continuous on (−∞,∞)
4. y = cscx is continuous in
D = {x : x �= nπ, n an integer }.At each point not belonging to D,
y = cscx has a nonremovable
discontinuity.
5. Undefined, 0
6. 0, −1, 1
7. [−1/2, 1/2], (−1, 1)
8. Note,√1 + x2 is defined and
2 + cosx �= 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 (−∞,∞)
10. Since g(x) = 4+x−x3is continuous,
g(1) = 4 and g(2) = −26, the
Intermediate Value Theorem
assures g(x0) = 0 for some number
x0 between 1 and 2.
Check-It Out
1. Removable discontinuity at x = −1.
Redefine f(−1) =12 in order to be
continuous at x = −1
2. 2, −2 3. (−∞,∞) 4. 2
5. If f(x) = x3 − x− 106, then
f(0) < 0 and f(107) > 0.
Intermediate Value Theorem as-
sures f(x0) = 0 for some xo in
[0, 107]
True or False
T,F, F,T,F
T,F,F,T,F
Exercises Section 1.4
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 limx→3−1
f(x) = 4
7. No, unless g(0) = 0
8. No, p(2) may or may not be defined
9. 2, 2, 2 10. 1, 1, 1
11. 0, 1, undefined
12. 3, 2, undefined
13. 0 14. Undefined
15. Undefined 16. Undefined
17. 1 18. −1 19. −1 20. 2
21. 0 22. Undefined
23. 0 24. 0 25. 4 26. 5
27. 1 28. −1 29. 1 30. −7
31. x = −1, 2, nonrem.
32. x = ±2, nonrem.
33. x = 3, nonremov;
x = −2, remov., f(−2) =45
34. x = 3, remov, f(3) = 227
35. x = 3, nonrem.
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. 46. ∅47. x = 1, nonrem. 48. ∅49. ∅ 50. x = 3, nonrem.
51. x = 0, remov, f(0) = 1
52. x = 0, remov, f(0) = π
53. (−∞, 1) ∪ (1,∞)
54. (−∞, 0) ∪ (0,∞)
55. (−∞, 1) ∪ (1,∞)
56. (−∞,∞) 57. [−2, 2]
58. (−∞, 0) ∪ (0, 2) ∪ (2,∞)
59. [−4,∞) 60. (−∞, 1]
61. (−∞,∞) 62. (−∞,∞)
63. x =12 + 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 =
√3−12 , b = 1−
√3
2
Section 1.5
Try This
1. −∞, ∞ 2. −∞, ∞3. δ =
4�
1/M 4. δ = (1/M)2
5. a) x = 2 and x = −3, b) x = nπ
6. x = 2 7. −∞, ∞
Check-It Out
1. −∞, ∞, undefined
2. Line x = 5
True or False
F,T,F,T,F
T,T,T,F,F
Exercises Section 1.5
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
16. Lines x =π4 +
kπ2 , integer k
17. Lines x =2π3 + 2kπ,
x =4π3 + 2kπ, integer k
18. Lines x =π6 + 2kπ,
x =5π6 + 2kπ, integer k
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
41.310 42.
13 43. −∞ 44. ∞
45. ∞ 46. ∞47. ∞, δ =
0.2M > 0;
If 0 < x < δ then0.2x > M
48. ∞, δ =6M > 0
If 0 < 2− x < δ then6
2−x > M
49. ∞, δ =1
M−1 , M > 1;
If 0 < −1− x < δ
thenx
1+x > M
50. ∞, δ =9
M−3 , M > 3;
If 0 < x− 3 < δthen
3xx−3 > M
A.2. ANSWERS TO EXERCISES 421
51. −∞, δ =
�4− 1
N , N < 0;
If 0 < x− 2 < δ, then 14−x2 < N
52. −∞, δ = − 83
N3 , N < 0;
If 0 < −x < δ, then 83√x
< N
59. a)12 (π − α− sinα), c) 2
Chapter 1 Review Exercises
1. 3 +√2 +
√3 2.
51325
3. a) 2.1, b) 2.01, c) 2 + h
4. a) 0.249224, b) 0.249992,
c)1√
4+h+2
5. 1.05573,1.00505,1.0005,
0.9995,0.995049,0.954451
Approx. limit is 1
6. 1.7948,1.81188,1.81361
1.81399,1.81572,1.83318,
Approx. limit is 1.814
7. 0.0303, 0.0030, 0.0003,
0.9994, 0.9940, 0.9412
Approx. limits: 0,1,
and does not exist
8. 237, 241, 241, 121, 121, 123
Approx. limits: 241,121,
and does not exist
9. 1 10. 5
11. Does not exist
12. Does not exist
13. ∞, ∞ 14. ∞, −∞
15. −∞, ∞ 16. −∞, −∞
17. −5 18. −5 19. − 19 20.
59
21. − 12 22.
112 23. 0 24. ∞
25. 0 26. 0 27. 1 28. 2
29. 1 30. −1 31. − 15 32. 0
33. 2 34.12 35. 2 36.
14
37. 0 38. 0
39. Continuous on (−∞,∞)
40. Continuous on (−∞,∞)
41. Discontinuous at x = −6,−1
removable discontinuity at x = −6,
redefine f(−6) =95
42. Nonremovable discontinuity
at x = 0
43. In proof, let δ =ε3
Then |f(x)− 35| = 3|x− 12| < ε
for 0 < |x− 12| < δ
44. In proof, let δ = Min�1, ε
14
�
Then |f(x)− 10| =|x− 2||3x+ 5| < ε
for 0 < |x− 2| < δ
45. In proof, let
δ = Min
�1, (
√8 + 3)ε
�
Then |f(x)− (−1)| =|x−9|√x+3
< ε
for 0 < |x− 9| < δ
46. In proof, let
δ = Min
�1, (
√3 + 2)ε
�
Then |f(x)− 2| = |x−3|√x+1+2
< ε
for 0 < |x− 3| < δ
47. −∞ 48. −∞ 49. ∞ 50. ∞51. Let δ =
4√M
.
Then16x2 > M
whenever 0 < |x| < δ
52. Let M > 1 and δ =4
M−1 .
Thenx
x−4 > M whenever
4M−1 > x− 4 > 0.
53. Let N < 0, δ = Min�1,− 12
N
�.
Note12
x−1 < N if − 12N > 1− x.
Thus,12
x−1 < N whenever
0 < 1− x < δ.
54. Let δ = Min
�1,
� 8M
�2�.
Note if 8 < x < 10 then
x√9−x
> M if1√9−x
> M8 .
Thus,x√9−x
> M
whenever 0 < 9− x < δ.
55. Find the δ’s for 2 + x2
and 2 + x4.
Then choose the smaller of the δ�s,i.e., δ = Min
�√ε, 4
√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.
59. a) Note BC =12
tan(α/2) .
Then A1 =144 cos3(α/2)
sin(α/2) .
b) The area of a sector isθr2
2 .
ThenA2 = 144
�1
tan(α/2) − π−α2
�.
c)A1A2
→ 3 as α → π2
60. a) v(2) = −64 ft/sec
b) 3 sec at the ground for s(3) = 0
61. a) R(100) = $150,
MR(100) = 0.5 dollars/brush
b) R(125) = $156.25,
MR(125) = 0 dollars/brush
62. ∞
Chapter 1 Multiple Choice Test
1. A 6. A2. B 7. B3. C 8. D4. A 9. C5. A 10. C
11. D 16. C12. A 17. B13. B 18. B14. C 19. A15. C 20. D
Section 2.1
Try This
1. −3 2. 6, −12 3. −9, −1
4. lim∆x→0
f(0 +∆x)− f(0)
∆x=
lim∆x→0
3√∆x
∆x=
lim∆x→0
1
3�
(∆x)2= ∞
5. 4, 2x 6. m =14 , y =
14x+ 2
7. g�(x) = 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
limx→1+
f(x)− f(1)
x− 1=
limx→1+
2|x− 1|x− 1
= 2 and
limx→1−
f(x)− f(1)
x− 1=
limx→1−
2|x− 1|x− 1
= −2.
Since the one-sided limits are not
equal, f �(x) does not exist.
If g(x) = [[x+ 2]], then
limx→0+
g(x)− g(0)
x− 0=
limx→0+
[[x+ 2]]− 2
x=
limx→0+
2− 2
x= 0 and
limx→0−
g(x)− g(0)
x− 0=
limx→0−
1− 2
x= lim
x→0−
−1
x, which
does not exist. Thus, g�(0) does
not exist.
9. f �(−1) = lim
x→−1
f(x)− f(−1)
x+ 1
= limx→−1
3√x+ 1
x+ 1
422 APPENDIX A. APPENDIX
= limx→−1
1
3�
(x+ 1)2= ∞.
Then f �(−1) does not exist.
Check-It Out
1. 6 2. 10x
True or False:
T,T,T,T,T,
T,F,T,F,F
Exercises Section 2.1
1. 10 2.37 3. −8 4. 6
5. 0 6. 0 7.14 8.
14
9.23 10. 2 11. −10 12. 9
13. −3 14. − 114 15.
14 16.
112
17. 0 18. −2 19. 3 20. −2
21. 1 22. 5 23. 3 24.23
25. 6x 26. 2x− 6
27.−1
(t+4)228.
−2(t+1)2
29.1
2√x+4
30.1√x−1
31.
√3x
2x 32.2√2x
x
33. 6x2 34. 4x3
35. 6x− 1 36. −2x+ 2
37. − 4x2 38.
−8(x−4)2
39.2√x
40.1
2√x+9
41. limx→1+
f(x)− f(1)
x− 1= 1
limx→1−
f(x)− f(1)
x− 1= −1
1x
2
y
42. limx→1+
f(x)− f(1)
x− 1= 2
limx→1−
f(x)− f(1)
x− 1= −2
1x
2
y
43. limx→0+
f(x)− f(0)
x= 0
limx→0−
f(x)− f(0)
x= 0
1x
23
y
44. limx→0+
f(x)− f(0)
x= 0
limx→0−
f(x)− f(0)
x= 1
1x
23
y
45. limx→−2
f(x)− f(−2)
x+ 2= ∞
�1�3x
�2
23
y
46. limx→1
f(x)− f(1)
x− 1= −∞
2 3x
�2
23
y
47. x = 3 48. x = ±2
49. x = −2 50. x = 3
51. x = 1 52. x = ±3
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)f(2)−f(3)
2−3
55. a)
x
y
b)
1x
y
56. a)
x
y
b)
x
y
57. a) D, E; b) less; c) see graph below
B
A
C
D
E
x
y
59. Yes 60. No
61. a =32 , b = − 1
2 , c = 0
A.2. ANSWERS TO EXERCISES 423
62. f �(0) = lim
x→0
f(x)− f(0)
x=
limx→0
1− cosx
x2=
1
2
limx→0
�sinx
x
�21
1 + cosx=
1
2
Section 2.2
Try This
1. p�(x) = 0, a�(t) = 0
2. f �(x) = 5x4
, g�(x) = − 2x3 ,
y� = − 12√x3
, p�(t) = 1
3. 6,−2, 112 4. y = 12x− 16
5. f �(x) = 12x3
, g�(x) = − 12x4 ,
y� = 6√x3
6. f �(x) = 42x5 − 32,
g�(x) = − 28x8 +
16x3 − 5
7. y� = −π sinx,
y� = 2 cos(x)− π,
y� = 12 (cosx−
√3 sinx)
8. y = −√2x+
√2(4+π)
4
9. −94.4 ft/sec, −95.84 ft/sec
10. −64 ft/sec
Check-It Out
1. 0 2. 1 3. 48x3 4. − sinx
5. −33.6 ft/sec 6. −32 ft/sec
True or False
T,F, F,F,T,
F,F,T,T,T
Exercises Section 2.2
1. 6x2 − 5 2. −15x4+ π
3. 3x2+ 2x 4. 16x+ 10
5. 4 cosx 6. −π sin t
7. 1 + 3 sinx 8.1√t− cos t
9. − 1x2 10.
1
33√x2
− 2
55√x3
11.1
33√t2
− 12t
√t
12. x− 32
13. 6√x− 3
2√x+
12x
√x
14. −15√x− 2
x√
x+
4√x
15. − 2x2 +
8x3 − 15
x4 16. − 12x2 − 3π
8x4
17. 32 18. 7 19. − 116 20.
114
21. − 227 22. − 1
64 23. −2 24. 2
25. −√22 26. −10
√2
27. −2 28. −1
29.815 30. 0 31. 0 32. − 17
4
33. y = 6x− 9 34. y = −4x+ 5
35. y = 8x− 12 36. y = −2x+ 9
37. y =52x+ 10 38. y = − x
16 − 1
39. y =√3x+ 1−
√36 π
40. y = − 2√3
3 x+4√3π−69
41. ave. rate = 30.3, f �(5) = 30
42. ave. rate ≈ 47.12, f �(4) = 47
43. ave. rate =6−6
√2
π , f �(π/4) = −
√2
2
44. ave. rate =24
√2−24π ,
f �(π/6) = 2
√3
45. −12.32 ftsec
, s�(1) = −12ftsec
,
|s�(1)| = 12ftsec
46. 14.16 ftsec
, s�(0.49) = 14.32 ftsec
,
|s�(0.49)| = 14.32 ftsec
47. −33.751 msec
s�(5.99) ≈ −33.70 msec
,
|s�(5.99)| ≈ 33.70 msec
48. ≈ 7.50 msec
,
s�(0.25) = 7.55 msec
,
|s�(0.25)| = 7.55 msec
49. 2.5 sec, v(2.5) = −56 ft/sec
50. −128√2 ft/sec
51. −64 ft/sec 52. −90.2 m/sec
53. 64 ft/sec 54. 19.6 m/sec
55. (±2, 4) 56. (5, 1/5)
57. y = 7x± 4
58. y = 2x, y = 2x+ 4/3
59. a = 3, b = −2, c = 1
60. a = 1, b = −3, c = 2,
and y = −x+ 1
61. y = −x2 +
32 62.
�12 ,
√2
2
�
63. (2, 1) 64. y = −x2 + 2
65. f(x) = x2 66. f(x) = x3+ 2x
Section 2.3
Try This
1.dydx = 36x2
+ 14x+ 1
2.dydx = x cosx+ sinx,
dydx = cos
2 x− sin2 x = cos 2x
3. h�(x) = 40
�x3
cosx+ 3x2sinx
�,
h�(x) = 40x3
cosx+ 120x2sinx
4.dydx = − 11
(3x−4)25
dydx =
203(10+x)2
6. y� = 32x
7, y� = − 1
2x3 , y�= 2
7.ddx
� 1xn
�=
xn·0−1·nxn−1
x2n =
−nxn−1
x2n = − nxn+1
8.dydx =
2(tan(x)+1)tan(x)−1
9.dydx =
1−cos xsin2 x
= cscx(cscx− cotx)
10. Average acceleration and accelera-
tion are both −32 ft/sec per sec
11.10
27x2 3√x2
Check-It Out
1. x4cosx+ 4x3
sinx
2.x(3x+2)(3x+1)2
3. − 52x4 4. −32 ft/sec
2
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+ tanx
6. x�x sec
2 x+ 2 tanx�
7. x secx (2 + x tanx)
8. p�(x) = cos2 x− sin
2 x
9. f �(x) = 4(2x− 5)
10. y� = 24x2+ 24x+ 6
11. f �(x) = −
1
(x− 1)2
12. f �(x) =
12
(3− 2x)2
13. g�(x) =x(2a3 − x3
)
(x3 + a3)2
14. g�(x) =−4c2x
(x2 − c2)2
15. −x sinx+ cosx
x2
16.sinx− x cosx
sin2 x
17.8x
(x2 + 5)218.
2x3+ 3x2
+ 1
(x+ 1)2
19. 1 20. − 13x2 − 4
3x3
21.3x2 22.
12x2
5 23. − 43x3 24.
6x5
25.2 cos x
5 − 13 26.
sec2x2
27 −4 28. − 12 29. −17 30. −3
31. −7 32. 0 33. − 49900 34. − 4
π2
35. −1 36. −1−π
2
37. −2 38. −2
39. y = − 12x+ 1 40. y = x
41. y = − 425x+
1225 42. y = − 4
3x+73
43. y =43x− 2π
9 +
√33
44. y = 2√3x− 2
√3π3 + 2
424 APPENDIX A. APPENDIX
45. y = −√2x+
√2π4 +
√2
46. y = −4x+8π3 −
√3
47. 3a+ b 48.3a−b
9
49. b− 3a 50. 2(3a+ b)
51. 2ax− 3x2 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 56. 60 ft
57. 6 58. 7! or 5040
59. − 24x5 60. (−1)
n n!xn+1
61. −5 cosx 62. −2 cosx
63. R(x) = x− 0.002x2
R�(x) = 1− 0.004x
64. 3.276t2 + 6.3t+ 5.2
Section 2.4
Try This
1. (f ◦ g)(x) = 3(5x− 7)4,
(f ◦ g)�(x) = 60(5x− 7)3
2. a) f(x) = x3, g(x) = 5x− 2,
b) f(x) =√x, g(x) = 1− x2
,
c) f(x) = tanx, g(x) = 2πx
3. a) 2 sec 2x tan 2x,
b) 15(5x− 2)2, c)
3x2
2√
1+x3
4. a) 2(sin2 x− cos
2 x),
b) 12 sinx(1− cosx)2
5. a) − 12x(x2−1)3
, b) − 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. y� = 3(3x− 1)3(x+ 4)
2(7x+ 15)
8. f �(x) = cos2 x+2 sin x(1+sin x)
2 cos2 x√1+sin x
or f �(x) = (sin x+1)3/2
2 cos2 x
9. h�(x) = −1
(1+x)2
�1+x1−x
10. a) −x sin(2x2),
b) 2 sec(2x+ 1) tan(2x+ 1),
c) − csc2 3√x
33√x2
11. f �(x) = 12 sin 6x
Check-It Out
1. 24(4x+ 1)2
2. f(x) = 10 cscx, g(x) = 12x
3. 24x(x2 − 1)3 4. 2x cos(x2
)
True or False
F,F,F,T,F,
T,T,T,T,T
Exercises Section 2.4
1. (f ◦ g)(x) = 2(4x− 5)3
(f ◦ g)�(x) = 24(4x− 5)2
2. (f ◦ g)(x) = 12x2+ 1
(f ◦ g)�(x) = 24x
3. (f ◦ g)(x) =√1− 2x
(f ◦ g)�(x) = −1√1−2x
4. (f ◦ g)(x) = 3√9− x
(f ◦ g)�(x) = −1
3 3√
(9−x)2
5. (f ◦ g)(x) = 2 sin(12x)(f ◦ g)�(x) = 24 cos(12x)
6. (f ◦ g)(x) = −3 cos(8x)(f ◦ g)�(x) = 24 sin(8x)
7. (f ◦ g)(x) = tan(x3 − 10)
(f ◦ g)�(x) =3x2
sec( x3 − 10)
8. (f ◦ g)(x) = csc(x2)2
(f ◦ g)�(x) =−x csc(x2
) cot(x2)
9. a) y = u4, u = 3x+ 1
b) y = tanu, u = x2
c) y = 3 cos2 u, u = 5x
10. a) y =√u, u = 2x+ 1
b) y = sinu, u =x
x+1
c) y = u2, u = 2 sinx+ 1
11. 15(3x+ 1)4 12. 8x
�x2 − 3
�3
13.7
2√7x+4
14.x2
(x3+1)2/3
15. − 4(4x+1)
3(2x2+x−7)7/3
16. − 4x
3(x2+4)5/3
17. 6x2sec
2�2x3
�
18. −6x2csc
2�x3
�
19. 6 cos(3x) sin(3x)
20. −8 cos(4x) sin(4x)
21. 10 sec2 x tanx
22. 6 cot 3x csc23x
23.12 (cos(2x)− 2x) 24.
12 (x+ sin 3x)
25.sec
�x
x+1
�tan
�x
x+1
�
(x+1)2
26. −cot
�2x
x+1
�csc
�2x
x+1
�
(x+1)2
27. −3 cot2(sinx) csc2(sinx) cosx
28. −8 sec2(cos(2x)) sin 2x
29.3(x−4)x
2(x−3)3/230.
5x+23(x+1)2/3
√2x−1
31. − tan�cos
x2
�sec
2�cos
x2
�sin
�x2
�
32.12 cos
�x4
�sec
2�sin
x4
�
33. vertical (2, 0); horizontal (0,−8)
34. vertical none; horizontal�0, 1
4
�
35. vertical (−5, 0), (4, 0);
horizontal
�− 1
2 ,−35�
814
�
36. vertical (±2, 0);�0, 6 3
√2
�
37. vertical (±nπ, 0); horizontal at�(4n+1)π
2 , 1�and
�(4n+3)π
2 ,−1
�
for integer n
38. vertical none; horizontal at
(nπ, 4(−1)n) for integer n
39. y =32x− 5
2 40. y = −7x+ 15
41. y = 3√3x+ 1−
√36 π
42. y = −6√3x+ 2(1 +
√3π)
43. y = 6√3x+ 3−
√3π
44. y = −12√3x+ 6 + 2
√3π
45. 12 46. 6 47.32 48. 98
49. Apply chain rule to (f ◦ g)(x) = x
50. g��(x) = 3x2since g�(x) = x3
by
Exercise 49
51. f(1) = 2, g�(2) = 1/8
52. f(1) = −1,
(f−1)�(−1) =
1f �(1) = − 1
8
53. A(w) = 2w√1− w2,
A�(w) =
2−4w2√1−w2
54. A(w) = w(1 +√1− w2),
A�(w) = 1 +
1−2w2√1−w2
55. d =√x2 − 18x+ 121
d� = x−9√x2−18x+121
56. d =
�(x− 1)2 +
� 1x − 1
�2
d� = x4−x3+x−1x3A(x)
Section 2.5
Try This
1. a) 3dydx , b) 6y2 dy
dx ,
c) x dydx + y, d)
y2
3√
1+y3
dydx
2. a)1
3y2−1, b)
y2
1−2xy
3. − 12 4. y = x
5. a)−6y
(1+3y2)2, b)
−1y3
Check It Out
1. a) 2dydx + 3 b) 8y dy
dx + 2x
c) 12(x+ y)11�1 +
dydx
�
2.32 3.
−6y(3y2+1)3
True or False
F,F,T,T,F,
T,T,T,T,T
A.2. ANSWERS TO EXERCISES 425
Exercises Section 2.5
1. −xy 2.
xy
3.−2x−4y3y2+4x
4.y−2x+52y−x
5. −�
yx 6.
32
�yx
7. − 9y+2x9(2y+x) 8.
−y−2y2√xyx(1+4y
√xy)
9. −1 10. −1
11.x
cos(2y)+y cos(y2)
12.−x
sin(2y)−y sin(y2)
13.1
2 cos(2y)−1 14. − y+sin(x−y)x−sin(x−y)
15. − 4y3 16.
1y3
17.x4
12y3 18.6y3−8x2
9y5
19.2(y2−1)2−8x2y
(y2−1)3
20.6x(2y+1)2−18x4
(2y+1)3
21. − 67 22. 0 23. 1 24.
18
25. 2 26. − 12
27. y = −x+ 1 28. y =√3x− 2
29. y =−3√5x− 4√
530. y =
−5x√3
+ 10
31. y = − 5x4 +
94 32. y =
4x27 +
1927
33. y = 2x− 1 34. y = 3x− 2
35. y =−2x+14
5 36. y = −x+ 3
37. (2,±2√2) 38. (1,
√2)
43. Horizontal at (0, 0) and (33√2, 3 3
√4)
Vertical at (0, 0) and (33√4, 3 3
√2)
�3 23
, 3 43 ��3 4
3, 3 2
3 �3 43
x
3 43
y
44. Horizontal at (1, 2) and (−1,−2)
Vertical at (2, 1), (−2,−1)
1�2 �1 2x
�1
2
�2
1
y
Section 2.6
Try This
1. 1.75 2. −2.5
3. −80 in.2/min 4.
14π ft/hr
5. 1 ft/sec 6. 5/π ft3/hr
Check-It Out
1. 6 2. 6π
3. − 23 4. 4 in.
2/min
True or False
F,T,T,F,T,
T,F,T,F,T
Exercises Section 2.6
1. 12 2. 6 3.136 4. 40
5. 11 6. 21 7. 4.5 8.316
9. 16 in.2/min 10. 2 ft/hr.
11. 32π ft3/hr 12. 8π in.
2/min
13. 216 in.3/min 14. 120 in.
2/hr
15.3
100 ft/min 16. −270 in.3/hr
17. − 2√3
9 radians/min,
dydt = − 2√
3ft/min, and
area increases at 2√3 sq. ft/min
18. −2.5 radians/hr, use
cotα = −m where α
is the angle in the problem
19. 72π ft3/hr 20. 22π ft
3/hr
21.209π ft/min 22.
3√2π
m/hr
23.11513 mph 24. 10
√7 mph
25.33√5units/sec 26. −4 units/sec
27.2513 ft/sec 28. 2
√2 miles/min.
29.dSdt =
dVdt
2r
30.drdt = k where V =
43πr
3,
dVdt = k(4πr2),
31.1013 ft/sec 32.
3613 ft/sec
33.π5 yd/sec
34. −2400 lb/ft3per min
35. 4 in.
36. 2.5 ft/sec =dydt ,
use 82+ y2 = x2
37. decreasing at
81√3−2162 cm
2/sec
38. 2.5 ft/min
39.1
30π cm3/min 40. −4 cm
2/min
Chapter 2 Multiple Choice Test
1. A 6. B 11. A 16. C2. D 7. B 12. D 17. B3. C 8. C 13. D 18. C4. C 9. B 14. C 19. B5. C 10. B 15. C 20. B
Investigation Projects
1. P �(2) = −2000 bacteria/minute,
decreasing
2. a) N �(30) = 2700 people/day
b) N(31)−N(30) = 2699 people
c) Yes, if t = 10 and t = 50 days
3. a) C�(2000) = $2.40/item
b) C(2001)− C(2000) = $2.3998
c) C�(x) is the approximate cost
of the (x+ 1)st item
4. a) R(x) = 3x2+ 200x− 0.02x3
b) R�(100) = $200/item
c) C(101)− C(100) = $196.98
5. b) 8 cm/min
6. (h, k) = (−4, 3.5), r = 5√5/2
7. The tangent lines are
y = −x+15
4
y =1
8
�4
�17 +
√285
�x−
17
�15 +
√285
��
y =1
8
�−4
�√285− 17
�x+
17
�√285− 15
��
Section 3.1
Try This
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,
min f(0) = f(π/2) = f(π) = 1
Check-It Out
1. 0, 1 2. π/3, 5π/3
3. Max f(2) = 8, min f(1) = −3
True or False
T,F,F,F,T,
T,F,T,T,T
426 APPENDIX A. APPENDIX
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
8. f �(0) undefined,
vertical tangent line
9. x = ±4 10. x = −6, 4
11. x = 0, 25 12. x = 0, 12
7
13. x = −2, 0 14. x = −2, 4
15. x =π6 ,
5π6 16. x =
π3 ,
5π3
17. x =π2 ,
3π2 , 7π
6 , 11π6
18. x =2π3 , 4π
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
23. Max g(−5) = 110, min g(1) = 2
24. Max g(−1) = 2,
min g( 13 ) =2227 ≈ 0.8
25. Max g(2) = 9, min g(1) = 0
26. Max g(−1) = 0,
min g( 18 ) = − 21872048 ≈ −1.1
27. Max f(0) = 0,
min f( 23 ) = − 4√6
9 ≈ −1.1
28. Max f( 23 ) =2√
39 ,
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) = 14 , 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 C(45√2) ≈ $499.41
36. Let x be the distance between Pand the shorter antenna. The
length is L(x) =√x2 + 502 +�
(100− x)2 + 752.
The minimum is L(40) = 25√41
≈ 160 ft. Thus, P must be 40 ft
from the shorter antenna.
37. Let x and y be the dimensions of
the rectangle. Assume x is the ra-
dius of the cylinder and y is height.
Since V (x) = πx2y = 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
Section 3.2
Try This
1. a) c = 1, b) c = 3π/4
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.
3. c = 1/4
4. Let s(t) be the distance driven in thours. Suppose s(t) is differen-
tiable. By the Mean Value The-
orem, there exists c in (0, 4) where
s�(c) =s(4)− s(0)
4− 0=
300− 0
4= 75 mph.
Thus, the instantaneous velocity is
75 mph at some time c.
Check It Out
1. c = 1 2. c = 1
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) = − sinx + 2 has no zero.
Then f has exactly one zero.
True or False
T,F,F,F,T,
T,T,F,F,T
Exercises Section 3.2
1. c = 3 2. c =16
3. c =23 4. c =
6−√
33
5. c =π3 6. c =
π4
7. c = 0,±�
π2 8. c = 2
9. c =23 10. c = − 5
3
11. c = 0,±√2
2 12. c =2(3−
√3)
3
13. c = 1 +√2 14. c =
√10− 2
15. c = 1 16. c =14
21.f(6)−f(1)
5 = f �(c) by the Mean
Value Theorem. Then f(6) ≤ 14.
22.p(12)−p(10)
2 = 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. Rel. max g(1) = 1/20,
rel. min g(0) = 0
3. Rel. max k(5π/3) = 2+√3
2 ,
rel. min k(4π/3) = − 2+√3
2
4. Rel. max f(0) = 6.25,
rel. min f(±1) = 6
5. 8 yards
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( 5π4 ) = −
√2
4.14
True or False
T,F,T,T,F,
F,F,T,T,T
Exercises Section 3.3
1. inc (−∞, 1), dec (1,∞)
2. inc (−∞,−1), dec (−1,∞)
3. inc (2,∞), dec (−∞, 2)
4. inc (3,∞), dec (−∞, 3)
5. inc (−∞,− 13 ) and (
12 ,∞)
dec (− 13 ,
12 )
6. inc (−∞,− 12 ) and
(34 ,∞), dec (− 1
2 ,34 )
7. inc (− 12 ,
13 ),
dec (−∞,− 12 ) and (
13 ,∞)
8. inc (13 ,
32 ),
dec (−∞, 13 ) and (
32 ,∞)
9. inc (−∞,∞) 10. dec (−∞,∞)
11. dec (−∞, 32 ) 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 427
15. inc (0, π2 ) and (π, 3π
2 ),
dec (π2 ,π) and (
3π2 , 2π)
16. inc (π2 ,π) and (
3π2 , 2π)
dec (0, π2 ) and (π, 3π
2 )
17. inc (0, 3π4 ) and (
7π4 , 2π)
dec (3π4 , 7π
4 )
18. inc (0, π4 ) and (
5π4 , 2π)
dec (π4 ,
5π4 )
19. inc (0, π2 ), (
7π6 , 3π
2 ), (11π6 , 2π),
dec (π2 ,
7π6 ) and (
3π2 , 11π
6 )
20. inc (7π6 , 11π
6 ),
dec (0, 7π6 ) and (
11π6 , 2π)
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 ) = − 25627
25. rel max y(− 59 ) =
25681 ,
rel min y( 13 ) = 0
26. rel max y( 32 ) = 20,
rel min y(− 32 ) = −34
27. rel max f( 12 ) =4316 ,
rel min f(3) = −52
and f(0) = 2
28. rel max f(0) = − 14
29. min f(0) = −9
30. rel max f( 3√2) =
3√26
31. rel min k(− 3√4) =
− 3√412
32. rel max f(−1) = −2,
rel min f(1) = 2
33. rel max f( 25 ) =64
√10
25 ,
min f(0) = f(2) = 0
34. rel max f(0) = 0,
min f(±3) = −81
35. max f(4) = 6, no rel min
36. min f(0) = 1
37. rel max f(− 2627 ) =
5527 ,
rel min f(−1) = 2
38. rel max f(−2) = − 323 ,
rel min f(2) = 323
39. rel max f(2) = 3√4,
rel min f(0) = f(3) = 0
40. rel min f( 17 ) = − 216343
√7
41. max g(π2 ) = g( 3π2 ) = 5
min g(π) = 4
42. max g(0) = −2
min g(±π2 ) = −3
43. max v(π6 ) = 2, min v( 7π6 ) = −2
44. max A(2π3 ) =
√3 +
π3
min A(− 2π3 ) = −
√3− π
45. rel max f( 7π6 ) = f( 11π6 ) =54 ,
rel min f(π2 ) = −1, f( 3π2 ) = 1
46. rel max f(π3 ) = f( 5π3 ) =54 ,
rel min f(π) = −1
47. − 14 48. 2
49. 9 ft2,
p2
16 ft2 50. 16 ft, 4
√A ft
51. a) inc (−�
52 , 0) and (
�52 ,∞);
dec on (−∞,−�
52 ) and (0,
�52 )
b) (±�
52 ,
32 )
52. a) inc (12 ,∞), dec (0, 1
2 )
b) (12 ,
�12 )
53. x = 2, y = 3 54. x = 2, y = 4
55. a) Possibly p(x) = (x− 2)2+ 1
2x
1
y
b) Possibly p(x) = (x− 1)3 − 3
1x
�3
y
56. a) Possibly v(t) = − 516 t
3+
154 t
�2 2x
�5
5
y
b) Possibly v(t) = −t4 + 2t2
�1 1x
�1
1
y
63. y =14 (x
2 − 3x+ 6)
64. y =−4x3+30x2−48x+103
27
Section 3.4
Try This
1. Concave up (−∞,−1) and (0,∞),
concave downward on (−1, 0)
2. Concave up (−∞, 1) and (1,∞),
concave down on (−1, 1)
3. Concave up on (−∞, 1),
concave down on (1,∞)
4. Concave up on (0, 1),
concave down (−∞, 0) and (1,∞),
pts of inflection (0,−2) and (1,−1)
5. Since y�� = 10(x−1)
9x4/3 , the graph of y is
concave down on (−∞, 1), concaveupward on (1,∞), and the point of
inflection is (1,−5).
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.
7. rel max f(π) = 3, rel min f(π3 ) = − 32
Check It Out
1. concave up (−∞,−2) and (0,∞)
concave down (−2, 0)
2. (0, 2)
3. rel max f(0) = 0, rel min f(±2) = −4
True or False
F,T,F,T,T,
F,T,F,T,T
Exercises Section 3.4
1. concave up (−∞,∞)
2. concave down (−∞,∞)
3. concave up (1,∞),
concave down (−∞, 1),
inflection pt (1, 2)
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)
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 ,
3π4 ); concave down
(0, π4 ) and (
3π4 ,π); inflection pts
(π4 ,
32 ) and (
3π4 , 3
2 )
12. concave up (π4 ,
3π4 ); concave down
(0, π4 ) and (
3π4 ,π); inflection pts
(π4 ,
2+π4 ) and (
3π4 , 2+3π
4 )
428 APPENDIX A. APPENDIX
13. concave up (3π4 ,π);
concave down (0, 3π4 );
inflection pts (3π4 , 0)
14. concave up (π4 ,
5π4 ); concave down
(0, π4 ) and (
5π4 , 2π); inflection pts
(π4 , 0), (
5π4 , 0)
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, 12 )
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
21. concave up (−∞, 1); no inflection pt
22. concave up (0, 1); concave down
(−1, 0); inflection pt (0, 0)
23. Rel. max. value: f(0) = 5
rel. min. value: f(2) = 1
24. Rel. max. value: f(1) = 5
rel. min. value: f(0) = 4
25. rel max g(−2) = 32,
rel min g( 43 ) =36427
26. rel max f(−2) = 27,
rel min f( 13 ) =38627
27. rel max C(±3√2) = 6724;
rel min C(0) = 6400
28. rel max A(±6√2) = 9409;
rel min A(0) = 4225
29. rel max g(π4 ) = 3,
rel min g( 3π4 ) = −3
30. rel max f( 2π3 ) = 5,
rel min f(π3 ) = −3
31. rel max L(−π3 ) = −
√3 +
4π3 ,
rel min L(π3 ) =√3− 4π
3
32. rel max R(2π3 ) = −
√3 +
8π3 ,
rel min R(π3 ) =
√3 +
4π3
33. rel max y(4) = 1,
rel min y(−4) = −1
34. rel max Q(−5) = 1,
rel min Q(5) = −1
35. rel max f(0) = 1, rel min f(4) = 9
36. rel max g(−3√3) =
9√3
2 ,
rel min g(3√3) = − 9
√3
2
37. max K(1) = 3, no rel min
38. max M(±8) = 16,
rel min g(0) = 0
39. f(x) = (x− 1)2+ 2
40. f(x) = −(x− 3)2 − 1
41. f(x) = 16 (x
3 − 3x2+ 12)
42. f(x) = − 52x
3+
152 x
43. (1, 1), (−1,−1) 44.
√3 3√42
45.√5 46. (
π2 , 1)
47.8 4√33 48. 2
√2
49.16 4√3
3 50. (2, 1/8), (−2,−1/8)
51. False; f(x) = x4
52. False; f(x) = −√x
53. Show d2y/dx2 > 0. The converse
is false, try f(x) = 1x for x > 0
54. For instance, f , g, f �, g�, f ��
, and
g�� are all positive functions
Section 3.5
Try This
1. a) limit is 0
x 10 100 1000
f(x) 0.198 0.02 0.0002
b) limit is 0
x −10 −100 −1000
f(x) −0.198 −0.02 −0.0002
2. If x > M =1ε , then
1x < ε
3. a) 0, b) 0, c) undefined
4. a) − 12 , b) 0
5. a) 1, b) − 13
6. a) 2, b) −2
7. N(24) ≈ 44, 083,
N(t) → 50, 000 as t → ∞
8. a) −∞, b) ∞
9. a) ∞, b) −∞
Check It Out
1. a) 0, b) − 12 , c) −∞, d) −2
2. y = −1 3. −10400
True or False
F,T,T,F,T,
T,T,F,T,F
Exercises Section 3.5
1. a) −2, b) −2
2. a) 3, b) 3
3. a) 0, b) 0
4. a) 0, b) 0
5. a) 2, b) −2
6. a) 1, b) 1
7. a) 0, b) 0
8. a) 0, b) 0
9. 2 10. 1
11. 1 12. 1 13.32 14.
12
15. ∞ 16. 0 17. 0 18. ∞19. 2 20.
13 21. −4 22. 2
23.23 24.
43
25. 0 26. 0 27. 0 28. 3
29. 0 30. ∞ 31. ∞ 32. ∞33. 0 34. 0 35.
12 36. 3
37. −2 38. 0 39.83 40.
23
41. y = −3 42. y = ±1
43. y = 0 44. y = 2
45. Possibly f(x) = 3x2
x2+1
�1 1x
3
y
46.
�1 1x
2
3
y
47. Possibly f(x) = arctanx
x
Π
2
�Π2
y
48.
�1 1x
2
4
y
49. P (t) → 14 as t → ∞.
About 25% of adults will have
sinus problems towards the end
of spring
50.12
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 429
Section 3.6
Try This
1. F �(w) = 5w3
(w − 4),
F ��(w) = 20w2
(w − 3),
rel max F (0) = 0,
rel min F (4) = −256,
point of inflection (3,−162)
3 4x
�256
�162
y
2. R�(x) = x−1
2x3/2 , R��(x) = 3−x
4x5/2 ,
relative minimum R(1) = 2,
point of inflection (3, 4/√3)
1 3x
2
4
3
y
3. g�(t) = 6t(t2−1)2
, g��(t) = − 6(3t2+1)(t2−1)3
relative minimum g(0) = 4
2 3x
23
y
4. H�(t) = 2(1− 3√t)
t2/3,
H��(t) = 2( 3√t−2)
3t5/3,
rel max H(1) = 3,
point of inflections (0, 0), (8, 0)
1 8x
23
y
5. N �(x) = − 2 sin x+1
(2+sin x)2,
N ��(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)
7 Π6
11 Π6
Π
23 Π2
x
33
�33
y
6. g�(t) = t(t−6)(t−3)2
, g��(t) = − 18(t−3)3
rel max g(0) = 0,
rel min g(6) = 12
6x
12
y
Check It Out
1. D = {x : x �= ±1}asymp. lines x = ±1
symmetric about the origin
2. D = (−∞,∞)
R = (−∞,−4] ∪ [0,∞)
inc (−∞,−2) and (0,∞)
dec (−2,−1) and (−1, 0)
conc up (−1,∞)
conc down (−∞,−1)
3. critical num t = π4 ,
5π4
inc (0, π4 ) and (
5π4 , 2π)
dec (π4 ,
5π4 )
True or False
F,T,F,F,F
T,F,F,T,F
Exercises Section 3.6
1. Max f(±√3) = 11,
rel.min f(0) = 2,
inc on (−∞,−√3) and (0,
√3),
dec on (−√3, 0) and (
√3,∞),
limx→∞
f(x) = −∞,
limx→−∞
f(x) = −∞,
range (−∞,√3]
2. Rel max f(−√3) = − 3
√3
2 ,
rel.min f(√3) =
3√3
2 ,
inc on (−∞,−√3) and (
√3,∞),
dec on (−√3,−1), (−1, 1), and
(1,√3), lim
x→∞f(x) = ∞,
limx→−∞
f(x) = −∞,
range (−∞,∞)
3. Rel max f(0) = 0,
rel.min f(2) = −23√4,
inc on (−∞, 0) and (2,∞),
dec on (0, 2), limx→∞
f(x) = ∞,
limx→−∞
f(x) = −∞,
range (−∞,∞)
4. Max f(π/4) =√2,
min f(5π/4) = −√2,
inc on (0, π4 ) and (
5π4 , 2π),
dec on (π4 ,
5π4 ),
limx→∞
f(x) and
limx→−∞
f(x) do not exist,
range [−√2,
√2]
5. Intercept (0, 0), max f(0) = 0,
asymptotes x = ±1, y = 1
range (−∞, 0] ∪ (1,∞)
�3 �2 �1 1 2 3
�4
�2
2
4
6. Intercept (0, 0),
inflection pt (0, 0),
asymptotes x = ±2, y = 0,
range (−∞,∞)
�10 �5 5 10
�10
�5
5
10
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 (−∞,∞)
�10 �5 5 10
�30
�20
�10
10
20
30
8. Intercept (0, 0),
rel max f(0) = 0,
rel min f(±√2) = 4,
asymptotes x = ±1,
range (−∞, 0] ∪ [4,∞)
�4 �2 2 4
�10
�5
5
10
9. Min f(4) = 2√2,
inflection pt (8, 4√6
3 ),
asymp x = 2, range [2√2,∞)
4 8
2 2
4
10. Intercept (0, 0),
max f(1) =√2
2 ,
min f(−1) = −√2
2 ,
inflection pts (0, 0), ( 4√5,
4√5√6),
and (− 4√5,−
4√5√6)
asymp y = 0, range [−√2
2 ,√
22 ]
�1 1 54� 54
1
430 APPENDIX A. APPENDIX
11. Intercepts (0, 0), (8, 0),
p�(x) = − 43
x−2x2/3 ,
p��(x) = − 49
x+4x5/3 ,
max p(2) = 63√2,
inflection pts (0, 0)
and (−4,−123√4),
range (−∞, 6 3√2]
2�4 8
�12 22�3
6 23
12. Intercepts (0, 0), (36, 0),
F �(x) = − 4(x−9)
27x2/3 ,
F ��(x) = − 4(x+18)
81x5/3 ,
max f(9) = 33√9,
inflection pts (0, 0) and
(−18,−63√18), range (−∞, 3 3
√9]
9�18
3 93
�6 183
13. Intercepts (0, 0), (±2√6, 0),
rel max f(0) = 0,
rel min f(±2√3) = −6,
inflection pts (−2,− 103 )
and (2,− 103 ), range [−6,∞)
�2 3 2 32�2
�6
6
� 103
14. Intercepts (0, 0), (± 2√30
3 , 0),
rel max f(−2√2) =
64√2
15 ,
rel min f(2√2) = − 64
√2
15 ,
inflection pts (0, 0), (2,− 5615 ),
(−2, 5615 ), range (−∞,∞)
�2 2 2
� 5615
64 215
15. Max (π3 ,
4√3
3 ), min (5π3 ,− 4
√3
3 ),
inflection pt (0, 0)
Π
35 Π3
2 ΠΠ
4
3
� 4
3
16. Max (π6 ,
5√3
3 ), min (5π6 ,− 5
√3
3 ),
inflection pt (π2 , 0), (
3π2 , 0),
Π
65 Π6
2 Π3 Π2
Π
2
5 33
�5 33
17. Max (π3 ,
4π3 −
√3),
min (−π3 ,
√3− 4π
3 ),
inflection pt (0, 0)
Π
3�Π3
Π
2�Π2
4 Π3� 3
3 � 4 Π3
18. Rel. max (− π12 ,
2π3 −
√3),
rel. min (π12 ,
√3− 2π
3 ),
inflection pt (0, 0),
� Π12
Π
12Π
4�Π4
3 � 2 Π3
2 Π3� 3
19. y =x
x−3 20. y =x3
x2−1
21. y =1x + x+ 1 22. y =
1x+1 +
x|x|x
23. Possibly f(x) = x2
1
1
3
24. Possibly f(x) = x3
1 3
1
2
25. Possibly f(x) = x4
1
1
2
26. Possibly f(x) = x3 − 3x2
1 3�1
�3
27. s0 = 25 ft, max ht s(3) = 169 ft
28. Use F �= − [2 cos(2θ) + 1]×
[sin(θ + sin 2θ) + 1]
Max F (0) = 1,
Min F (π) = −1
rel max F (π3 ) = − sin(
√3
2 − π6 ),
rel min F (2π3 ) = sin(
√32 − π
6 ),
ΠΠ
32 Π3
1
�1
29. S�= (cos 3θ + cos θ)/2
Max S(π4 ) = S( 3π4 ) =
√23 ,
Min S(0) = S(π) = 0,
rel min S(π2 ) =13
Π
4Π
23 Π4
1
30. y� = −3(sin 3t+ sin 4t) =
2 sin(7x2 ) cos(
x2 ),
critical nos x = 0, 2π7 , 4π
7 , 6π7
Π2 Π7
4 Π7
6 Π7
1
74
32. Inflection pt (− k3 ,
2k3
27 )
If k = 0, no local extrema.
If k > 0, local max f(− 2k3 ) =
4k3
27 ,
local min f(0) = 0;
If k < 0, local min f(− 2k3 ) =
4k3
27 ,
local max f(0) = 0
33. G(t) = t3−3t2−9t+198
�1 1 3
3
1
�1
Section 3.7
Try This
1. 25 ft by 25 ft 2. (3.5,√3.5)
3. Length 4√2, width 2
√2
4. x = 6 feet from the 3-ft post
5. r = h =103√π
=10
3√π2
π cm
Check It Out
1. x = 200 ft, y = 600 ft
2. p(4) = 16, minimum value of p
True or False
F, T,F,F,F,
T,F,F,T,F
A.2. ANSWERS TO EXERCISES 431
Exercises Section 3.7
1. D = 1.25 when x = 1
2. D = 1 when x = 0
3. A = 25 when w = 5
4. A = 9/√2 when b = 3
5. S = π + 2 when r = 1
6. V = 16π when r = 2
7. a) (172 ,
√342 ), b) (0, 0)
8. a) (2, 2√2), b) (0, 0)
9.12 10.
√33
11. x = 100√6 ft, y = 50
√6 ft
12. r =20
4+π 13. 2 in.
14.4√3
9+4√3or
48√3−6411
15. (0, 0), (8, 0), (0, 4); not the mini-
mum perimeter
16. 3√3 17. r =
3√15π2
π , h =2
3√15π2
π
18. radius100π m, length of one
side of rectangle is 100 m
19.187 miles from the point on
the highway closest to Town A
20. BP =√3 miles
21.x2
8 square feet.
22. base b, perimeter p, two sides
of the same lengthp−b2
23. x = 4 workers 24. 220 students
25. 165 ft by 300 ft
26. y = 20 ft, x = 7.5 ft
27.|mp−q+b|√
m2+1
28.2π(3−
√6)
3 radians ≈ 33◦
29. θ =π3 30.
32πr3
81
31. c =√27 when x = 3
32.427 33. 2 radians
34. C = 54 35. K = −3
36. max A+B, min A−B
37. max A+B, min −B2+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 rbe 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.
Subtract the two equations above
to obtain
r = 12− 2
√6h− k.
Rewrite the second equation as
(2
√6−h)2+(k−3)
2= ((r+2)+1)
2.
Then substitute the first equation
to obtain
(2
√6− h)2 + (k − 3)
2=
(
�h2 + (k − 2)2 + 1)
2.
Solving for k, we find
k =192
√6− 336h+ 23
√6h2
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
25,96
25
�.
Section 3.8
Try This
1. x4 ≈ 1.25992
2. x4 ≈ 0.739085, |x4 − x3| ≈ 0.00003
3. Let g(x) =��� f(x)f
��(x)(f �(x))2
���.The graph of g is dashed below:
f �x� � x2 � 60
g�x� 87x
0.2
y
Then g(x) < 0.2 for x in (7, 8).
By Theorem 3.11, we can apply
Newton’s Method with x1 = 8.
Thus, limn→∞
xn =
√60.
4. Let g(x) =��� f(x)f
��(x)(f �(x))2
���.The graph of g is dashed below:
f �x� � sin�x� � x � 2g�x�
541
x
0.2
y
Then g(x) < 0.2 for x in (1, 54 ).
By Theorem 3.11, we can apply
Newton’s Method with x1 = 1.
Thus, limn→∞
xn = c where f(c) = 0.
Also, x2 ≈ 1.10292
Check It Out
1. x2 =34 2. x3 =
1712
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)n2n−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 → −∞
14. x1 = 2.5, as n → ∞|xn| =
��� 1−3x2
2
��� → ∞
15. x4 ≈ 1.7100
16. x5 ≈ 3.9330
17. x3 ≈ 3.1416
18. x4 ≈ 3.1416
19. Let f(x) = x2 − 10 and
g(x) =��� f(x)f
��(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.
f �x� g�x�3 3.5
x
0.3
y
20. The graphs of f(x) = x3+ 5x+ 10
and g(x) =��� f(x)f
��(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.
f �x�g�x�
�2 �1x
0.6
y
432 APPENDIX A. APPENDIX
21. Let f(x) = cos(πx2) and
g(x) =��� f(x)f
��(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.
f �x�g�x�
0.35 0.45x
0.7y
22. Let f(x) = x2sin(x)− 2x and
g(x) =��� f(x)f
��(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.
f �x�g�x�6 7
x
0.9y
25. Let f(x) = ax2+ bx+ c, a > 0,
and g(x) = f(x)f ��(x)(f �(x))2 .
Then g(x) → 12 as x → ∞.
Note, g�(x) = 2a(b2−4ac)(2ax+b)2
> 0
if x �= − b2a . Thus, g(x) is increas-
ing on (− b2a ,∞) and (−∞,− b
2a ).
In particular, |g(x)| < 12 if
x �= − b2a . Hence, Newton’s
method converges to a zero of f(x)if x �= − b
2a .
Section 3.9
Try This
1. Linearization L(x) = x;tan
π24 ≈ L( π
24 ) =π24
2. dy = −0.05,�y =
1√1.1
− 1 ≈ −0.0465
3. Let f(x) =√x. Then√
15 ≈ f(16) + f �(16)�x =
318 = 3.875 where �x = −1.
4. sin 29◦ ≈ sin
π6 + cos
π6
�− π
180
�=
12 +
√3
2
�− π
180
�≈ 0.4849
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 percentage relative error
is
��� �ff(6)
��� · 100% or
����16�x
16(6)
���� · 100% ≈ 1.7%.
Check-It Out
1. L(x) = x4 + 1;
√4.1 ≈ L(4.1) = 2.025
2. Let y = g(x) = tanx and
dx = 0.005. Then
dy = g�(π4 )dx = 0.01
3. Let f(x) = x3be the volume
of a cube with side x, and
let �x be the error in the
measurement of the side
where |�x| ≤ 116 .
The propagated error is
|f(12 +�x)− f(12)| ≈f �(12) · 1
16 ≈ 13 cu. in.
True or False
T,T,F,F,T
F,T,T,F,T
Exercises Section 3.9
1. L(x) = 12x− 6;
(2.1)3 ≈ L(2.1) = 9.2
2. L(x) = −x4 +
34 ;
1(1.9)2
≈ L(1.9) = 0.275
3. L(x) = x34 +
172 ;
√285 ≈ L(285) ≈ 16.88
4. L(x) = x48 +
83 ;
3√70 ≈ L(70) ≈ 4.125
5. L(x) = 2x+ 1− π2 ;
tan(π4 + 0.1) ≈
L(π4 + 0.1) = 1.2
6. L(x) = 2x3 +
2√3− π
9 ;
sec(π6 + 0.2) ≈ L(π6 + 0.2) =
2√3+
215 ≈ 1.288
7. L(x) = − x12 + 2;
3√7 ≈ L(1) = 23
12 ≈ 1.917
8. L(x) = − x64 +
12 ;
2√17
≈ L(1) = 3164 ≈ 0.484
9. L(x) = −√3
2 x+3+
√3π
6 ;
cos(61◦) ≈ L(π3 +
π180 ) =
12 − π
120√
3≈ 0.48
10. L(x) = x√2+
4−π4√2;
sin(44◦) ≈ L(π4 − π
180 ) =
180−π180
√2≈ 0.69
11. �y = 1.01; dy = 1
12. �y =
�1041 − 1
2 ≈ −0.006;
dy = − 1160 ≈ −0.006
13. �y = cot� 23π
90
�− 1 ≈ −0.034;
dy = − π90 ≈ −0.035
14. �y = csc� 8π45
�− 2 ≈ −0.11;
dy = −π√3
45 ≈ −0.12
15. �y ≈ −0.238; dy = −0.24
16. �y ≈ −0.0175; dy ≈ −0.0177
17. 5.1 18. 1.92
19.
√3
2 − π360 ≈ 0.857
20. 2−√3π
180 ≈ 1.97
21. 32.8 22. 9.11
23. 3 sq. in. 24. 0.785 sq. in.
25. 50 cubic inch 26. 3.75 sq. in.
27. 349 ft 28. 0.013 inch
29. 0.08 inch 30.140 radian ≈ 1.4◦
Chapter 3 Multiple Choice Test
1. A 6. C 11. A 16. B2. A 7. B 12. D 17. C3. C 8. B 13. C 18. A4. C 9. C 14. C 19. B5. B 10. C 15. C 20. B
Investigation Projects
3. 2000 books
Section 4.1
Try This
1. a) πx+ C b) y = x3+ C
2. a) 3x5+ x2
+ C b) 2 sinx+ C
3. a)�x−3dx =
x−2
−2 + C =
− 12x2 + C
b)�x1/3dx =
x4/3
4/3 + C =
34x
4/3+ C
c)�4x−1/2dx =
4x1/2
1/2 + C =
8√x+ C
4. 2x4 − 8x2+K 5. 4x+
x3
3 +K
6. sin t+ C
7. y = 2t+ sec t + 1 8. 5 sec
Check-It Out
1. a) 4x+ C, b) 5x2 − x+ C,
c) −2 cos t+K, d) −3
x3+ C
2. y = x2+ x+ 3
True or False
F,T,T,T,F,
T,F,F,T,F
A.2. ANSWERS TO EXERCISES 433
Exercises Section 4.1
1. 10x+ C 2. πx+ C
3. 20t2 + C 4. 6t2 + C
5. 5 sin t+ C 6. 3 tan t+ C
7. sec t+ C 8. − 13 cos t+ C
9. − 1w3 + C 10.
−13w3 + C
11. − 116w2 + C 12.
17w
7+ C
13.310m
10/3+D
14.23 (m
3/2 −√m) +D
15.59m
9/5+m+K
16.7211m
11/6+K
17.13 s
3 − 12 s
2 − 2s+K
18.12 s
2 − 25 s
5/2+K
19.112 s
3 − 12 s
2+ s+ C
20. − 25 s
5/2+ 3s2 − 8s3/2 + 8s+ C
21. −3 cos θ − tanθ + C
22. θ − sec θ + C
23.13x
3+ tanx− x+ C
24. 2− cscα+G
25. − cscx+ C
26. sinα− tanα+G
27. − cotβ + P
28. − cscβ + cotβ + P
29. y = 2x+ x2
30. y =14x
4 − 13x
3+
1−6√2
12
31. y = 2√t
32. y = − 611 t
11/6+
37 t
7/3+
977
33. y = cosx
34. y = − cotx+
√33
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 +
√5864 ) = −8
√586
≈ −194 ft/sec
37. 1 +
√224 ≈ 2.2 sec 38. 144 ft
39. y =x2
2 − x3
6 + x+53 40.
256
Section 4.2
Try This
1.152 2. 8 3.
15150 , 3
4. 2 +√2 +
√3 < A < 3 +
√2 +
√3
5. S =(n+1)(2n+1)
6n2 , s =(n−1)(2n−1)
6n2
6. 8 7. 16/3
Check-It Out
1. 2 +5n 2.
116
3.4(n+1)(2n+1)
3n2
True or False
T,T,F,F,F,
F,T,T,T,T
Exercises Section 4.2
1. 14 2.212 3.
4936 4. 3 + 6a
5. 68 6.2920 7. 60 8.
118
9.
20�
i=3
i− 1
i+ 110.
n�
i=1
3
n
�1 +
i
n
11.
n�
i=1
��3i
n
�2
−3i
n
�·3
n
12.
n�
k=1
sin
�kπ
n
�π
n
13. 670 14. 1225
15. 2640 16. 14,760
17.n+1n , limit 1
18.3n+12n , limit
32
19.n−1n , limit 1
20.14n2+9n+1
2n2 , limit 7
21.(n+1)(2n+1)
6n2 , limit13
22.8n2+3n+1
2n2 , limit 4
23.3(8n2+9n+3)
n2 , limit 24
24.(n+2)(n+1)(n−1)
4n3 , limit14
25.3625 ,
3125 26.
533840 ,
319420
27.(9+
√3)π
12 ,(7+
√3)π
12
28.18 (3 + 2
√3 +
√10 +
√11),
18 (3 + 2
√2 +
√10 +
√11)
29.92 ,
72 30.
44627 ,
37427
31.(1+
√3)π
12 ,(3+
√3)π
36
32.(9+
√3)π
12 ,(7+
√3)π
12
33. S(n) = 9�1 +
1n
�,
s(n) = 9�1− 1
n
�
34. S(n) = 6 +4n , s(n) = 6− 4
n
35. S(n) = 1− 16
�2− 1
n
� �1− 1
n
�
s(n) = 1− 16
�2 +
1n
� �1 +
1n
�
36. S(n) = 14n2+3n+16n2 ,
s(n) = 14n2−3n+16n2
37. S(n) = 4(n+1)2
n2 , s(n) = 4(n−1)2
n2
38. S(n) = (7n−1)(25n+7)4n2 ,
s(n) = (7n+1)(25n−7)4n2
39.13 40.
143 41. 12 42.
56
43.72 44. 16 45. 39 46.
56
47. 16 48.203 49.
52 50.
352
51.23 52.
56
Section 4.3
Try This
1. 1 2.π(1+2
√3)
6
3.a+b2 4. 4, 10
5. 2, 1.5 6. 0,−4
7. 2(b3 − a3), (b−a)−(b3−a3)3
Check-It Out
1. 182 2. 3 3. 9.5
True or False
T,F,T,T,T,
F,F,T,F,F
Exercises Section 4.3
1. 12 2. 12 3. −3 4. 3
5. 10 6. 0
7.25π4 8. 2− π
4
9.9π2 10. 1 11. −4 12. −12
13. −6 14. −4 15.92 16. −4
17. a) −4, b) 0 c) 8, d) 5
18. a) 31, b) −10 c) 20, d) 8
19. 25 20. 8 21. 1 22. 6
23.52 24. 6 25. 18 26. 4
27.1123 28.
163 29. 16 30. 6
31.� 5−2(2x− 5)dx
32.� 40 3x(4− x)2dx
33.� 30
√5 + x2dx 30.
� 52
2x2 dx
35. 8 36. 20 37.32 38. 2
39.π + 2
440. π + 4
41.π4 42.
32
43.π4 +
32 44.
52
45. 1 +π4 46.
π2
47.� 10 x2dx =
13
48.� 10
√x dx =
23
49.12 50.
√2− 1
55. S(n) = 5π6 , s(n) = π
3
56.(2
√3−1)π24
434 APPENDIX A. APPENDIX
Section 4.4
Try This
1. 130 and 1 2. 2.5
3. a/2 4. 1/4
5. −10.875 meters/sec
6. 3, 2, 1.5
7.x
x4+1, f(x) + x sin
2 x
8. − sin(x)√cosx
Check-It Out
1.103 2.
2π 3.
√2
True or False
F,T,F,T,T,
F,T,T,F,T
Exercises Section 4.4
1. 9 2. 15 3. 9 4. 2
5. − 23 6. − 74
3 7.72 8. 5
9. 3 10. −2√3
11.16 12. 4− 2
√2
13.π + 2
414.
π − 2
4
15.78 16.
43 17. 7 18.
23
19. 2 20. ] 1
21. 1 22. 3√2− 2
√3
23. 16 24.4 + π2
2
25. 12 26.163 27. 20 28.
34
29.323 30. 4 31.
43 32.
13
33.3112 34.
43
35. 10 36.12 (2b+ dm+ cm)
37.83 38.
c3 (c
2+ cd+ d2)
39.43 40.
32 41.
2π 42. 0
43.1
1 + x244.
√x2 + 1
45. − tanx 46. −1 + x2
x
47. 4√1− tan3 x sec
2 x
48.sinx cosx
1 + sin5 x
49. 65x3 50. 2x sin(x6)− sin(x3
)
51.32 52. ±
√3
3
53. ±1, 0 54.√3
55. f(q(x))q�(x)− f(p(x))p�(x)
56. p��� x
1 f(t)dt�· f(x)
57. 1 58. f(x) = xn+ C
59. Hint:
�1t2
�= 1 if
�2/3 ≤ t ≤ 1.
62.� 10 x2dx =
13 63.
� 10 x4dx =
15
64.� 2π0 sinx dx = 0 65. − cos x
x2
66.1−(
� x0 sin(t2)dt)
2
�1+(
� x0 sin(t2)dt)
2�2 · sin(x2
)
Section 4.5
Try This
1.23 (x
2+ 4)
3/2+ C 2. − cosx2
+ C
3.−1
3(3x2+1)+C 4.
13 cos(x3−3x2
)+C
5.15 (x
2+ 1)
5+ C 6. 2
√tan t+ C
7.149 8.
2815 9.
13 10. 0, 46
15
Check-It Out
1.1
π + 1(5 + 3x2
)π+1
+ C
2.−1
3(x3
+ 3x) + C
3.1
3
�2
√2− 1
�+ C 4. 0
True or False
F,T,F,T,T,
T,T,F,F,T
Exercises Section 4.5
1.(2t2+3)4
16 + C 2.1
6(9−x3)2+ C
3. 2√sin t+ C 4. − sin
� 1t
�+ C
5.215 (w + 4)
3/2(3w − 8) + C
6. −2
3
√1− w(w + 2) + C
7.112
�x2
+ 4�6
+C 8. − (6−x3)5
15 +C
9.13 sin(3t)+C 10. −2 cos
�√t�+C
11. − (1+ 1w )
5
5 + C
12.12
�1− 1
w2
�4+ C
13.14 tan(4θ) + C
14. −2 cot
�θ2
�− θ + C
15. 2(x4+ x2
+ 1)4+ C
16. − (y5−5y3+4)9
5 + C
17.−2√x+1
+C 18.110 (
√x− 1)
5+C
19.−2
1+sin θ + C 20.2
2 cosα+1 + C
21.14 (1 + tanα)4 + C
22. − 13 csc(3α) + C
23.25 (2y − 1)
5/2+
23 (2y − 1)
3/2+ C
24.23 (1 + 3y)3/2 − 2(1 + 3y)1/2 + C
25.110 26.
25532 27. 2 28.
645
29. 1 30.13 31.
√26 32.
12
33.13 34. 1
35. 3 + 2√3 36. 2
√2/3
37. 0 38. 0 39. 100 40. 2
41.32 42.
13 43.
23 44. 2
45.12A 46. W
47.1a − 1
b 48.12B
Section 4.6
Try This
1. 0.088333
2.110
�f(1)+2
�f(1.2)+f(1.4)+f(1.6)+
f(1.8)
�+ f(2)
�≈ −0.418021
where f(x) = sin(πx)x
3.13 (
1√2+
4√5+
2√10
+4√17
+
2√26
+4√37
+1√50
) ≈ 1.76327
4. Since
|f ��(x)| = |− 4x2
cosx2 − 2 sinx2|we find |f ��
(x)| ≤ 6 = K on [0, 1].
Then
|ET | ≤K(b− a)3
12n2=
6
12(62)=
1
72.
An upper bound is172 .
5. Since
|f ��(x)| = |− 4x2
sinx2+ 2 cosx2|
we obtain |f ��(x)| ≤ 6 on [0, 1].
Choose n ≥�
105/2 or n ≥ 224.
True or False
T, T, T, T, F, F, T, T, T, T
Exercises Section 4.6
1. a) 1.68333, b) 1.62222
2. a) 1.11667, b) 1.10000
3. a) 0.765496, b) 0.77753
4. a) 3.06198, b) 3.11013
5. a) 3.06198, b) 3.14876
6. a) 2.97842, b) 0.26247
7. a) 2.62331, b) 2.60391
8. a) 0.35233, b) 0.35698
9. a) 8.15121, b) 8.14594
10. a) 3.10675, b) 3.09894
11. ET ≈ 0.00116, (b) ES ≈ 0.00002
using |f ��(x)| ≤ 2
9 and
|f (4)(x)| ≤ 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 435
14. ET ≈ 0.00926, (b) ES ≈ 0.00069
using |f ��(x)| ≤ 2
9 and
|f (4)(x)| ≤ 80
81
15. a) n ≥ 366, b) n ≥ 26
16. a) n ≥ 130, b) n ≥ 11
17. a) n ≥ 238, b) n ≥ 19
18. a) n ≥ 255, b) n ≥ 9
19. 14 both by TR and SR
20. 132
21. TR 2712 ft, SR 26
23 ft
22.� 100 60D(x)dx ≈ 2910 ft
3by TR,
2900 ft3by SR
23.� 300 f(t)π( 12 )
2dt ≈ 138 ft3by TR,
137 ft3by SR
24.� 120 w(x)dx ≈ 126 ft
2by TR,
12623 ft
2by SR
Chapter 4 Multiple Choice Test
1. B 6. A 11. B 16. D2. C 7. A 12. A 17. B3. B 8. B 13. B 18. B4. A 9. B 14. B 19. B5. C 10.B 15. B 20. C21. B 22.D
Investigation Projects
1.π4 2. 8x4f(x4
) + 2� x4
0 f(s)ds
3.� 10 (16x3 − 4x)dx = 2 4.
sin 12
5. If f(x) =� x2
√t3 + 1 dt, then the
limit is f �(2) = 3
6.� 10
dx(1+x)4
=724
7. If f(x) =� 2xx t cos t dt, then the limit
is f �(x) = 3x cosx
8. a) 0 b) does not exist
Section 5.1
Try This
1. ln 3+ln(x+1), lnx− ln 4, ln√9− x
2. 20.086, 0.247, 1.386, −0.693
3. secx cscx, 33x−1 , x+ 2x lnx
4.3e2
− 1 5. y = x
6. y� = 2x − 2
2x+1 +12 tanx
7. y� = x2/3 sin2 x√x+1
·�
23x + 2 cotx− 1
2(x+1)
�
8. y� = 2−2x2x−3−x2
9. y = 4 ln |x|− 12x
2+ C
10. y = ln | lnx|+ 1
Check-It Out
1. − tanx 2. 1 3. y =x
2e4+
32
4. y = 5x− 2 ln |x|− 5
5. 3e3 − 2e2 − e3 + e2
True or False
F,T,T,T,T,
T,T,T,T,T
Exercises Section 5.1
1. a) lnx+ 2 ln y + ln z
b) 2 lnx+ ln y − ln z
c) lnx+ ln y − 13 ln z
2. a) ln
�3�
x2(x+1)x2+3
�
b) ln
�8√
x3+1
�c) ln
�x2−4x2+1
3.3x 4.
3(ln x)2
x
5. − tanx 6.cos(ln x)
x
7.x
x2+18.
tt2+1
− 1t
9.2−ln x2x
√x
10.sin x(1+ln | sec x|)
cos2 x
11.2 cos(2x)
x − 2 sin(2x) ln(x2)
12. 2(ln t) cos(2t) + sin(2t)t
13.1
x ln x 14.
√4+x2+x
4+x2+x√
4+x2
15.1
x2−116.
1−ln t2t3/2
√ln t
17. − cscx 18. sec t 19.(ln x)2+2
x
20.1
x2+1− 1
x((ln x)2+1)
21. y = x+ 1 22. y =32x− 3
2
23. y = 8x− 5 24. y = − 32x+ 4
25. x2√x2 − 1
�2x +
xx2−1
�
26.
�x2
+ 1
x2 − 1
�x
x2+1+
xx2−1
�
27.(x− 2)(x+ 3)
(x+ 2)(x− 3)×
�1
x−2 +1
x+3 − 1x+2 − 1
x−3
�
28.
√3x− 2
x(x+ 1)2
�3
2(3x− 2)−
1
x−
2
x+ 1
�
29.x(x − 1)3/2
√x + 1
�1
x−
3
2(x − 1)−
1
2(x + 1)
�
30.�
(2x− 1)(x+ 2)(x− 3) ×12
�2
2x−1 +1
x+2 +1
x−3
�
31. min f(e−1) = −e−1
,
concave up (0,∞)
��1 1x
1
2
y
32. min f(e−1/2) = − 1
2e
concave down (0, 1e3/2
)
concave up (1
e3/2,∞)
��1�2 1x
1
y
33. max f(e) = e, concave down (0,∞)
�1x
�
y
34. min f(e−4) = −e−4
,
concave up (0,∞)
��4x
0.5
y
436 APPENDIX A. APPENDIX
35. min f(1) = −1,
conc. down (0, e−1),
conc. up (e−1,∞),
��1 1x
�1
y
36. min f(1) = 1, concave up (0,∞),
1x
1
y
39. y = 3 + lnx 40. y =12 (ln t)2
41. y =12 (t
2 − (ln t)2 + 1)
42. y = x lnx− e
43. y = x+ lnx− 1
44. y = 4 lnx+1x − 4
45.� 21 ln(x)dx = ln(4)− 1
1 2x
y
46. −� 11/2 ln(x)dx =
12 (1− ln 2)
112
x
y
47.� e2
1ln xx dx = 2
1 �2x
0.5
y
48.� ee−1
2xdx = 4
���1x
4
y
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 horizon-
tal line through the point of tan-
gency. The distance between the
y-intercepts is 1 unit. See figure
below.
�a, ln�a��a x
ln�a�ln�a��1
y
50.dydx =
y(3x−2)x(2y2+1)
51. a) ∞ b) 0
Section 5.2
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− 3ln x
7.ln 24 8.
6(√3−
√2)
π
Check-It Out
1. ln(t2 + 9) + C
2.14 ln |1 + 4 secx|+ C
3.14 ln 2
4. 2x− 43 ln |3x+ 2|+ C s
5. y = − 2ln x + 3
6. y = − ln | lnx|+ e
True or False
T,F,T,F,F,
F,T,F,T,T
Exercises Section 5.2
1. π ln |t|+ C 2.25 ln |x|+ C
3. ln |x+ 5|+ C 4.14 ln |4x+ 1|+ C
5.23 ln
52 6. ln 1331
7.116 (8t+ 3− 3 ln |8t+ 3|) + C
8. t2 +13 ln |3t+ 2|− 4
9 + C
9.1π ln | sec(πx)|+ C
10.12π ln | sin(2πθ)|+ C
11. ln32 12.
12 ln
53
13.ln 3
414.
ln 3
2π
15.13 ln |2 + 3 sec t|+ C
16. − ln(1 + cos2 t) + C
17.12 ln | tan 2x|+ C
18. ln | tanx|+ C
19.14 20. (ln 2)
2
21. 2(√2− 1) 22. 8
23. 2 ln |1 +√x|+ C
24. x− 2√x+ 2 ln |1 +
√x|+ C
25. ln | cos θ + θ sin θ|+ C
26. − ln |α cosα− sinα|+ C
27. ln | sinx+ cosx|+ C
28. ln | sinx+ 1|+ C
29. ln 2 30. 5 ln52
31.12 ln(2 +
√3) 32. ln
43
33. y = −3 ln |2− x|34. y = t− ln |t+ 1|+ 1
35. y =12 ln | sec 2x|
36. y = ln |1− sin t|37. s(t) = −10 ln |t|+ 18
38. s(t) = − 19 ln |3t+ 1|+ 1
3 t
39.−1
1 + x240.
2√1− 4x2
41.1
x√4 + lnx
42.√9 + x2 −
√9+(ln x)2
x
43.ln(3)3 44.
ln 22
45.3π ln
√3+1√3−1
46.2 ln(2)
π
47. (1, 0) 48. − 14 < m < 0
A.2. ANSWERS TO EXERCISES 437
Section 5.3
Try This
1. yes, yes 2. yes, no, yes
3. p−1(x) = x+1
3 with domain
and range (−∞,∞);
f−1(x) = 1
x − 1 with domain
{x : x �= 0} and range
{y : y �= −1}
4. p(1) = 2, p−1(2) = 1,
(p−1)�(2) = −1/6
Check-It Out
1. f−1(x) =
1
1− x,
D = {x : x �= 1},R = {y : y �= 0}
2. Since g�(x) =−4
(x− 4)2< 0,
g is decreasing and one-to-one.
3. p−1(3) = 0, (p−1
)�(3) =
1
2
4.2
5
True or False
F, F, T, T, T,
F, T, T, T, F
Exercises Section 5.3
1. yes 2. yes 3. no 4. no
5. yes 6. yes
7. yes
1x
1
y
8. yes
1x
1
y
9. yes
1x
1
y
10. yes
1x
1
y
11. f−1(x) = 5x
7−2x
D = {x|x �= 72}
R = {y|y �= − 52}
12. 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}
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}
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
37x1 + ln 2 =
37x+ ln 2.
Thus,37x1 =
37x2 and x1 = x2.
Hence f is one-to-one.
19. If s(t1) = s(t2) then
t12t1+3 =
t22t2+3 .
Cross-multiply:
2t1t2 + 3t1 = 2t1t2 + 3t2
Thus, t1 = t2.
Hence s is one-to-one.
20. If R(t1) = R(t2) then
4t14−3t1
=4t2
4−3t2.
Cross-multiply:
16t1 − 12t1t2 = 16t2 − 12t1t2
Thus, t1 = t2.Hence R is one-to-one.
21. If C(w1) = C(w2) then
w1 − 1
w1 + 1=
w2 − 1
w2 + 1.
Cross-multiply:
w1w2 + w1 − w2 − 1 =
w1w2 − w1 + w2 − 1
Thus, 2w1 = 2w2 or w1 = w2.
Hence C is one-to-one.
22. If f(t1) = f(t2) then
1− 2t11 + 2t1
=1− 2t21 + 2t2
.
Cross-multiply:
1 + 2t2 − 2t1 − 4t1t2 =
1− 2t2 + 2t1 − 4t1t2
Thus, 4t2 = 4t1 or t1 = t2Hence f is one-to-one.
23.1
1124. −1
25. −16
326.
1
4
27. −1
228.
4
3
29. y is increasing for
dy
dx=
4x(x+ 1)
(2x+ 1)2> 0
when x > 0. Then y has an
inverse function.
30. −9
31. Differentiate (F ◦ F )(x) = x to
obtain f(F (x))f(x) = 1.
Integrate to show F (x) = x.
32.23
34. f(x) = x if 0 ≤ x < 1, and
f(x) = 3− x if 1 ≤ x < 2
Section 5.4
Try This
1. t = A−1005 2. w = e4 − 10
3.dydx = ex cos x
(cosx− x sinx),dydx = esin xx cosx+ 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◦Fper minute.
6. 2ex/2 + C,13 e
x3+ C
7.12 (1− e−1
) 8. y = 1 + t+ e2t
Check It Out
1. x =12 ln(y − 1)
2. f �(x) = e2x + e−2x
438 APPENDIX A. APPENDIX
3.13 (e− 1)
4. y = − sin t+ e−3t+ 1
True or False:
F,T,T,T,T,
F,T,F,T,T
Exercises Section 5.4
1. x =12 ln y 2. x = ln
�y+13
�
3. t = 10 ln
�5
A+4
�
4. t = 5 ln
�20
30−A
�
5. h = e6P+4 6. w = e(5−10V )/2
7. y� = 2xex2+2 8. f �
(t) =t (2− t)
et
9. g�(t) =1− 2t
e2t
10. y� = 10ex sin x(x cosx+ sinx)
11. R�(x) = e3x (5− 3x)
12. P �(w) =
1− w lnw
wew
13. y� = −2ecos 2x sin 2x
14. y� = −1
15. L�(x) =
4e2x
(ex − 1)(1 + ex)(1 + e2x)
16. f �(x) = 20
�e20x − e−20x
�
17. r�(x) =24e4x
√1 + e4x
18. A�(t) = esin t
(cos t cos 3t− 3 sin 3t)
19.1
2e2x + C 20. −ecos x + C
21. −1
2e−t2
+ C
22. 2et +e2t
2+ t+ C
23.2√1− et
�et − 1
�
3+ C
24.3
2x2 − x+ C
25. 2e√w+ C
26. ln√e2w + e−2w + C
27.e− 1
228. 3(e− 1)
29. e√3 − e 30. ln
3
�1 + e
2
31. y = 3x+ 1 32. y = −2x+ 2
33. y =1
2x+ 2 34. y = 8e3x− 3e3
35. y =
√3e
2x+
√e−
π
12
√3e
36. y = −1
37. e− e−1 38. 3(e− 1)
39.19e6−1
2e640. 10 ln(5)− 8
41.12 (e− e−1
) 42.12
�1− 1
e16
�
43. y = − cos t+ 14 e
−2t − 14
44. y = ex + e−x
45. y = ln
�2
1+e−2t
46. y =12 ln[(e2t + 1)(e−2t
+ 1)]− ln 2
47.dy
dx=
1
1 + 2e2xy(x+ 2)
48.dy
dx=
6x+ yexy
4y − xexy
49.dy
dx=
5 + e2y
4− 2xye2y
50.dy
dx=
ex + y
3− x+ e−y
51.dy
dx=
2xyey
1− x2yey
52.dy
dx=
2x− yexy
xexy − 2y53. y = 4x− 4 54. y = x
55. min. (0, 1)
56. max (1, 1e ), inflec. pt (2, 2
e2)
57. max (1, 1√e), min (−1,− 1√
e),
inflec. pts (0, 0), (√3,
√3
e3/2),
and (−√3,−
√3
e3/2)
58. max (π4 ,
e−π/4√2
),
min (5π4 ,− e−5π/4
√2
),
inflec. pt (π2 , e
−π/2),
and (3π2 ,−e−3π/2
)
59. min (e−1/2,− 12e ),
inflec. pt (e−3/2,− 32e3
)
60. min. pt. (e, e−1),
inflec. pt (e3/2, 32e3/2
)
61. a = ln
�A+
√A2+42
�
63.13 64. x = µ; x = µ± σ
Section 5.5
Try This
1. r = 4 ln 1.02 ≈ 0.079
2.0.3 ln 200
ln 2 ≈ 2.29 mm
3.13 (log2(10) + 1) ≈ 1.4, 3
25 ≈ 0.1
4. y� = sec2(x)2tan x
ln 2,
p�(t) = 12√
t ln 122√t
,
M �(x) = 1
2 ln(10)(x+√x)
5. xsin x� sin x
x + cos(x) lnx�
6.2
ln 10 ,3−
√3
ln 3
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 → ∞
Check It Out
1. 2(e2) ≈ 167.6
2. a) 3π3xlnπ, b) 2
(2t−4) ln 3
3. a)π5s
5 lnπ , b)3
4 ln 2
True or False:
F,T,F,F,T,
T,T,F,T,F
Exercises Section 5.5
1. −3 2. −1 3.1
24.
2
3
5.log(7)−1
3 ≈ −0.1
6.log3(8/5)
2 ≈ 0.2
7.ln 5
365 ln(1+0.12/365) ≈ 13.4
8.323 ≈ 10.7
9.√101− 1 ≈ 9.0
10. e(1
ln 2− 1ln 3 )
−1≈ 6.5
11.
y�2 x
1 2x
1
2
y
12.
y�log3x
1 3x
1
y
13.
y�log ��x�
�10 1x
1
y
A.2. ANSWERS TO EXERCISES 439
14.
y�10 x�3
1 2x
10
3
y
15.
y�log2�x�1x
1
2
y
16.
y�4 x �1
1 2x
1
4
y
17.
y�log1�2�x�4��4 4x
1
�2
y
18.
y��1�3�x�1�21 2
x
3
�2
y
19. 2xln 2 20. 5 · 3t ln 3
21.3
√tln 3
2√t
22. 3x2
24x
(ln 16 + x ln 9)
23. 3t10
3tln(3) ln(10)
24.� 12ln 12
�xln
�12
ln(12)
�
25.1−ln x
x2 26.t+2
(t2+t) ln 4
27. 10s�ln(s) + 1
s ln 10
�
28.2x(x2 ln(2)−2x+ln(2))
(x2+1)2
29.1
2(x−1) ln a 30.1
x ln(10) ln(x)
31.cot(x)ln 10 32. − 2 tan(2w)
ln 3
33.2x cot(x)−1
x ln 4 34.θ tan(θ)−1
θ ln 8
35. y = 10x ln 4 + 2− ln 4
36. y = 28 ln(7)x+ 14(1− ln 7)
37. y = 8√3 ln(2)x+ 4− 8π ln 2√
3
38. y =
√30 ln 10
2 x+√10−
√30π ln 10
12
39. y =x
25 ln 5 + 2− 1ln 5
40. y =2
ln 27x+ 2− 2ln 3
41. y = 2�1 +
1ln 10
�x− 20
ln 10
42. y =x
ln 4 + 1− 1ln 2
43.2x
ln 2 + C 44.10t
ln 10 + C
45.5x
2
2 ln 5 + C 46.2√
t+1
ln 2 + C
47.63x
3 ln 6 + C 48.23x
3 ln 2 − 52x
2 ln 5 + C
49.ln(1+52x)
2 ln 5 +C 50.ln(9+32x)
2 ln 3 +C
51.3sin t
ln 3 + C 52.5(2−x)2
−2 ln 5 + C
53.−2
√8cos t+1ln 8 + C
54.2
3 ln 3
�3tan t
+ 1�3/2
+ C
55.18ln 3 − 4
ln 2 56.3
ln 4
57.15ln 4 58.
−2π ln 10
59.e−12 60.
1ln 9
61.4−2
√2
ln 2 62. log4(1 + 4x)
63. xx(ln(x)+ 1) 64. x1/x
�1−ln x
x2
�
65. xln x�
ln x2
x
�
66. (2x+4)3x+6
�3x+6x+2 + 3 log(2x+ 4)
�
67. (log2 x)log3 x · 1+ln(log2 x)
x ln 3
68. (log x)10x
�10
x(ln 10) ln(log x) + 10x
x(log x) ln 10
�
69.y(x ln y−y)x(ln x−x) 70.
1+ln x1+ln y
71. $182.20; $109.33 per year at the end
of 10 years
72. $818.84; $97.77 per year at the end
of 6 years
73. P (0) = 1000
P �(0) = 200 units/minute
74. a) 200 b) C =ln 3100
75.(ln 4)2
2 x2 − x ln 2 + 1
76. a = 1, b = ln√2 = −c
77. Hint: odd function
78. Hint: f(±1) ≤ a, f �(0) = 0.
Section 5.6
Try This
1. y = Cex3−x2
2. y =
�4√t+ 1
3. 15871
�17,63915,871
�2≈ 19, 604
4.10005.3
� 12
�1/5.3ln(1/2) ≈
−114.7 units/year
5. t ≈ 3.83 hr, or 3 hr and 50 min
6. 48 minutes after it was taken from the
oven
Check It Out
1. a) y = Ce2x/ ln 2
, b) y2 = x2+ 3
2. 36,744,344
3. −58 mg/hr 4. −4◦F/min
True or False:
F, F, T, T, T,
T, F, F, F, T
Exercises Section 5.6
5. y =2
1− 2x6. y =
3
1− x3
7. y = t 8. y = ln(2− e−t)
9. y = −2et4/4 10. y =
x
1 + x
11. y = 2 sec t 12. y =1
1− ln(sinx)
13. y =
�1 + 2 sin
2 t
14. y =2
1 + cos t
15.100 ln 0.9ln 0.96 ≈ 258 years
16. about 14.4 hours
17. a) 13.2 mg b) 50 years
18. 15,683 years
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 22. b) $64.82/year
23. P (t) = 1256 (t+ 6400)
2
24. 71.7%
25. a) v(t) = 201+6t ft/sec
b) a(1) = − 12049 ≈ 2.4 ft
2/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 cm
2/sec
440 APPENDIX A. APPENDIX
Section 5.7
Try This
1. y = x+Cx
2. y = 1− e−x2
3. Amount of sugar in tank is
y(t) = t3−150t2+4375t+312506250
4. y(4) ≈ 143.8 mg, since
y(t) = 60ln 2
�4 + 2
−t/24(ln 2− 4)
�
5. I(t) = 6− 6e−5tamperes, where t
is in seconds
6. I(t) = 6 − 6et/2 amperes, where tis in seconds
7.
2�2x
2
�2
y
8. C,A,B
9. (0.1, 0.2), (0.2, 0.42), (0.3, 0.662)
Check It Out
1. y = e1−cos x − 1
2. y(10) ≈ 2.6 lb of salt
where y = 3− 3e−0.2t
3. Q(t) = 625
�1− e−5t
�
4. ye(1) = 3, ye(2) = 7
True or False:
F, F, T, T, T,
T, T, F, T, T
Exercises Section 5.7
1. y = 3− 3e−x 2. y =1
2(e2x − 1)
3. y = (x+ 1)e−x2/2
4. y = (3x+ 2)e−x2
5. y = 5
�t−
1
t
�6. y = 4
�t2 − t
�
7. y =1− cos t2
t
8. y =10π + sin t5
5t4
9. y = sin t+ 2 cos t
10. y =1 + sin
2 x
2 sinx
11. y =x4
+ 3x
3
12. y =t3
2π
�1− π ln
�cos t2
��
13. y =2t3 + 3t2 − 6
6(1 + t)
14. y =x2
+ x+ 4√2x+ 1
15. y = xe−2x(1 + lnx)
16. y =e−3x
+ 1
3x
17. A, y = −1
4−
x
2+ Ce2x
18. B, y2 = 2(x+ C)
19. C, y = ln
�−1
x+ C
�
20. D, not separable or linear
21. y = 2ex−0.5 − x− 1
�2 2x
�2
2
0.5
y
22. y =2
5e1−x − 1 + x
�2 21x
�2
2y
23. y = x
24.1
y=
1
x+ 2
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
�
sincedy
dt= 2−
5y
100,
y(0) = 0
28. y = 2.5�5− 3e−2t/25
�
sincedy
dt= 1−
2y
25, y(0) = 5
29. t ≈ 7.6 sec,
where y =2(625+100t+t2)
5(50+t)
sincedydt = 0.8− y
50+t , y(0) = 5
30. y(50) = 22.5 kg of sugar,
where y =400t+ t2
4 (200 + t)since
dy
dt= 0.5−
y
200 + t, y(0) = 0
31. 37.5 sec, where
y =2500 + 75t− t2
250since
dy
dt= 0.5−
2y
100− t, y(0) = 10
32. 1.5 lb
33. Q(t) = 11100
�1− e
−100t3
�
34. Q(t) = 57 e
−25t/3�e
7t3 − 1
�
max Q(t) ≈ .086 coulombs
when t = 37 ln
2518
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
37. y(t) = 15ln 2
�32− 32
−t/12�
38. y(24) = 600ln 2 ≈ 866 mg
since y(t) = 1200ln 2
�1− 2
−t/24�
39. y(t) = $1, 000, 0000
sincedydt = 0.05y − 5(10
4),
y(0) = 106
40. y(4) = $22, 662.80
sincedydt = 0.2y − 4(10
4),
y(0) = 105, and
y(t) = − 100,0003
�−4 + e3t/10
�
41. y = ex − x− 1
Section 5.8
Try This
1.π3 ,
π4 ,
45 2. −π
3 ,
√74
3. 0 4.52
5.5
1+25x2 ,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
1. a)2√10
7 b) −π6
2. x =
√3
3
3. a)2x
x4+1b)
2 tan−1(t)t2+1
4.
√3
3 +π6
True or False:
F, T, T, T, F,
F T, T, T, T
A.2. ANSWERS TO EXERCISES 441
Exercises Section 5.8
1.2√13
13 2. −2√2
3.x√
1−x24.
x√1+x2
5.3√3−410 6.
169√3−24069
7.14 8. 1−
√32
9. −π6 10.
53
11.
√3
1−√3
12.
√34
13.45 14. − 1
3
15. y� =5
√1− 25x2
16. y� =−3x2
√1− x6
17. y� =3
9 + x2
18. y� =2
√x(1 + x)
19. y� =2
x√x4 − 1
20. y� =−1
√1− x2
21. y� =−1
|x|√4x2 − 1
22. y� =−1
|x| (arccsc x)√x2 − 1
23. y� =x−
√1− x2 sin−1 x
x2√1− x2
24. y� =−ecot
−1 x
1 + x2
25. y =x
2+
π − 2
4
26. y = −x+√3 +
π
6
27. y = x−√3 +
π
3
28. y =
√3
2x+ π −
√3
29. y� =1
2
�1− (x+ y)2 − 1
30. y� =
�1− x2y2 + y
�1− x2y2 − x
31. y� = ey�ex − 1
1+x2
�
32. y� = y
�sec2 y
| tan y|√
tan2 y−1− x
�−1
33. 10√7 ≈ 26.5 feet
34.√10/2 ≈ 1.6 yards
35. y = tan 2 ≈ 1.107, when x = 1
36. rel max pt≈�√
2+2√5
2 ,−0.60578
�,
rel min pt≈�
−√
2+2√5
2 , 3.74737
�
37.52π3 ≈ 54 miles/min
38. − 39610 ≈ −0.0639 radian/sec
39. θ�(t) = 3
4x√
x2−9
40. θ = x
�12
�1−
�25−x2
24
�2�−1
θ�(√13) =
√39
18 for x =√13
if θ =π3
46. No, if x < 0
47. a)
y�sin�sin�1�x���1 1x
�1
1
y
b)
y�sin�1�sin �x��Π2
�Π2
3 Π2
�3 Π2
x
Π2
�Π2
y
48.ddx
�tan
−1(cotx)
�= −1
domain is
{x : x �= nπ, n is an integer}
y�tan�1�cot x�Π�Π x
Π2
�Π2
y
Section 5.9
Try This
1. sin−1
�x4
�+ C,
15 arctan
� t5
�+ C,
17 sec
−1�w
7
�+ C
2. π 3.π2 − sin
−1�e−3
�≈ 1.5
4.13 tan
−1�x3
�+ ln(9 + x2
) + C
5.15 tan
−1�x−3
5
�+ C 6. π/3
7.12 tan
−1�e2x
�+ C
14 ln
�1 + e4x
�+ C
8.13 , −(ln 2)
2/4
Check It Out
1. a) sin−1
�x5
�+ C
b)12 tan
−1� t2
�+ C
c)1√2sec
−1�
s√2
�+ C
2. a)π6 b)
√3π6
3.π4 + ln
√2
True or False:
F,F,F,T,T,
T,T,T,T,F
Exercises Section 5.9
1. sin−1
�x2
�+ C
2. sin−1
�s√3
�+ C
3.1√5tan
−1�
t√5
�+ C
4.13 tan
−1(3t) + C
5.1√2sec
−1�
|x|√2
�+ C
6. arcsec (4|x|) + C
7.12 arcsin(2t+ 1) + C
8. sin−1
(tanx) + C
9.12 sec
−1(e2x) + C
10. sec−1
(2ex) + C
11.π15
12.π3
13.π6
14.π4
15. ln� 32
�16.
√2
17.16 tan
−1�
3 sin θ2
�+ C
18. − 1√2tan
−1�√
2 cos θ�+ C
19.−1√2arctan
�√2 cotx
�+ C
20. Rewrite integral:� sec2 θdθ
1+2 tan2 θ=
√2
2 tan−1
(√2 tan θ) + C
21.13 tan
−1� t+1
3
�+ C
22.14 tan
−1� s−3
4
�+ C
23. sin−1
�x+12
�+ C
24. sin−1
�w−32
�+ C
25. arcsin�x+3
4
�+ C
26.13 arctan
�x−23
�+ C
27.π20 28.
π2
29.π3 30.
π48
31.12 sec
−1�
|x+3|2
�+ C
32.�sec
−1 x�2
+ C
33. − 14
�cos
−1(2x)
�2+ C
34.12 ln
�tan
−1(2s)
�+ C
442 APPENDIX A. APPENDIX
35. y = 1− sin−1 x
36. y = 8 tan−1
� t2
�− 2π
37. y = 2esec−1 t
38. y =
�2 tan−1 x+
π2
39. y = t2e− tan−1 t
40. y =x cos−1 x−
√1−x2
x
41.π6 42.
π8 43.
π6 44.
π12
45. a = tan(1) 46. b = 3
Section 5.10
Try This
1. ln 19 ≈ 2.94, ln 199 ≈ 5.29,
ln 1999 ≈ 7.60;
d → ∞ as x → 1−
2.sinh x2√x, −2xcsch2(x2
),sinh x
2√cosh x
3. 0 4. ln(5/4)
5. (ln(2+√3), 4
√3), (ln(2−
√3),−4
√3)
7. a =2e
e2+1
8.15
�cosh
−1 12181 − cosh
−1 10081
�
9.12 cosh
−1� sin 2s
3
�+ C
10. Since tanh−1 x =
12 ln
1+x1−x ,
we find d = ln1+r1−r =
2
�12 ln
1+r1−r
�= 2 tanh
−1 r
Check It Out
1.
a) 2 cosh 3x sinh 3x
b) − [2sech 2x coth 3x] esech 2x
c)4√
16x2+1
d)10x
1−25x4
2.
a) − 19 sech
−1�x9
�+ C
b) cosh−1
� t9
�+ C
True or False:
F, T, T, T, T,
T, T, F, T,T
Exercises Section 5.10
1.54 2. − 4
3 3.1213 4. 0
5. ln(√2− 1) 6.
ln 32
7. y� = 3 sinh(3x+ 1)
8. y� = 10 cosh 5x
9. y� = 1− 2xsech2x2
10. y� = csch2(1/x)x2
11. y� = sech (x2)−
2x2sech (x2
) tanh(x2)
12. y� = −csch ( 3√x) coth( 3√x)
33√x2
13. y� = tanhx
14. y� = e−x(coshx− sinhx)
15. y� = sech t 16. y� = sech x
17. y� = xcosh x�
cosh xx + sinh(x) lnx
�
18. y� = xsinh x�
sinh xx + cosh(x) lnx
�
19. y� =x
√x2 + 1
sinh
�x2 + 1
20. y� =1
xcosh (lnx)
21. y� = −2xex2�sech ex
2��
tanh ex2�
22. y� = 0
23. y� = 2 coth 2x
24. y� = 6x cosh2(x2 − 2) sinh(x2 − 2)
25. y� = coshx− sinhx
26. f �(t) = e2t
�2 coth t− csch
2t�
27. y� =6
√4x2 + 1
28. y� =2
4− x2
29. y� = | secx|
30. y� =1
2√x2 − x
31. y� = −2csch
−1x
|x|√1 + x2
32. y� = −2 sec 2x
33. y� = 2 sec 2x
34. y� = sinh−1 x
35. y� = −1
x√
1−x2
36. y� = −4x3√x8−1
37.14 38.
209
39.13 cosh(3x) + C
40. 2 sinh√x+ C
41.14 sinh
2(2x) + C
42.13 cosh
3(x− 1) + C
43. sinhx+13 sinh
3 x+ C
44.112 cosh
34x− 1
4 cosh 4x+ C
45. −2 cothx2 + C
46.√2 sinhx+ C
47. 2√2 sinh
x2 + C 48. tanh 1
49. ln 2− 35 50.
4+coth 4−coth 84
51.825 52.
1
3cosh
22− cosh 2 +
2
3
53. ln(cosh 1)
54. 2�tan
−1 e− π4
�
55. x = −1
2ln 3
56. rel max (0,−1),
rel min (1,− sinh 1)
57. 1− e−1
58. 8 tan−1
�e2
�− 2π
59.
a) f−1(x) = sinh
2�x
2
�
b) g−1(x) = 2 tanh
−1�x
2
�
68. y = ±1
Chapter 5 Multiple Choice Test
1. C 6. A 11. B 16. A2. A 7. A 12. C 17. D3. B 8. C 13. A 18. A4. C 9. B 14. C 19. B5. B 10.B 15. A 20. D
Investigation Projects
2. hint: x ≈ 4.965
3.
a) T ≈ 5, 796 K
b) T ≈ 11, 146 K