44
MATH 140 Homework Problems Contents 1 Limits and Continuity 2 2 Derivatives 9 The Limit Definition of the Derivative .................................. 9 Sum, Difference, Constant Multiple, and Reciprocal Rule. Derivatives of Power, Exponential, and Trigonometric Functions ....................................... 10 Product and Quotient Rule ........................................ 11 Chain Rule ................................................. 12 Derivatives of Inverse Functions. Derivatives of Logarithms and Inverse Trigonometric Functions . 14 Implicit and Logarithmic Differentiation ................................. 16 Higher Derivatives ............................................. 17 3 Applications of the Derivative 18 Curve Sketching .............................................. 18 Answers Answers Section 1 21 Answers Section 2 23 Answers Section 3 30

MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

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Page 1: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

MATH 140 Homework Problems

Contents

1 Limits and Continuity 2

2 Derivatives 9The Limit Definition of the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Sum, Difference, Constant Multiple, and Reciprocal Rule. Derivatives of Power, Exponential, and

Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Product and Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Derivatives of Inverse Functions. Derivatives of Logarithms and Inverse Trigonometric Functions . 14Implicit and Logarithmic Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Applications of the Derivative 18Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Answers

Answers Section 1 21

Answers Section 2 23

Answers Section 3 30

Page 2: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

2 MATH 140 HOMEWORK PROBLEMS

1 Limits and Continuity

For the function y = f(x) whose graph is shown find all specified quantities (if a quantity does not existwrite “DNE”).

1. (a) f(0)

(b) limx→0

f(x)

(c) f(1)

(d) limx→1−

f(x)

(e) limx→1+

f(x)

(f) limx→1

f(x)

x

y

−1 1 2 3

1

2

3

2. (a) f(0)

(b) limx→0

f(x)

(c) f(2)

(d) limx→2−

f(x)

(e) limx→2+

f(x)

(f) limx→2

f(x)

x

y

−1 1 2 3

1

2

3

3. (a) f(2)

(b) limx→2−

f(x)

(c) limx→2+

f(x)

(d) limx→2

f(x)

x

y

−1 1 2 3

−1

1

2

3

4

4. (a) f(1) (b) limx→1

f(x)

x

y

−1 1 2 3

−1

1

2

3

4

Page 3: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

1 LIMITS AND CONTINUITY 3

5. Sketch the graph of a function that has all the specified properties.

(a) limx→0

f(x) =∞

(b) f(0) = 1

(c) limx→1+

f(x) = −∞

(d) limx→1−

f(x) = 2

(e) limx→(−1)−

f(x) = −2

(f) limx→(−1)+

f(x) = 0

(g) f(−1) is undefined.

Find the limit, using the limit computation laws.

6. limx→2

(x2 + x− 2)

7. limx→−3

(x+ 2

x+ 1

)58. lim

x→−1(√x2 + 3− 3

√x− 7)

9. limx→√2(x2 +

√x4 + 21)

10. limx→5

x+ |x− 5|x2 − 1

11. limx→π

sin(x/3) cos(x/6)

1 + (x/π)2

12. limx→ln(2)

(3ex − e2x)

13. limx→e3

2 ln(x) + ln(√x)

x+ 1

14. limx→2

2x − 3x2

log2(x) + 1

15. limx→1/2

(x arcsin(x) + arctan(−2x))

Suppose that limx→2

f(x) = 3, limx→2

g(x) = 5, and limx→3

h(x) = 7. Find the limit.

16. limx→2

h(f(x))

17. limx→2

f(x)g(x)

18. limx→2

f(x)

g(x)

19. limx→1

f(x2 + 2x− 1)− xx2 + h(x+ 2)

20. limx→1

f(x2 − 1

x− 1

)h(x2 + x− 2

x2 − x

)21. lim

x→22 arctan

(√f(x)

)Find the indeterminate limit.

22. limx→1

x2 − 1

x− 1

23. limx→−2

3x+ 6

x2 + x− 2

24. limx→3

x2 + 2x− 15

x2 − 11x+ 24

25. limx→−1

√x+ 10− 3

x+ 1

26. limx→7

√43− x−

√29 + x√

2x+ 2−√

23− x

27. limx→−12

√13 + x−

√2x+ 25

x2 + 15x+ 36

28. limx→2

√x2 + x+ 10−

√2x2 + 8

x2 + x− 6

29. limx→2

12x+6 −

110

x− 2

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4 MATH 140 HOMEWORK PROBLEMS

30. limx→3

2x2+x −

13x−3

x2 + 4x− 21

31. limx→−4

1x+2 −

13x+10√

x2 − 3−√

17 + x

32. limx→0

ex+1(x3 + x2)

x4 + 3x2

33. limx→2

x sin(πx2 − 3πx+ 2π

x2 − x− 2

)34. lim

x→−3e(√4+x−1)/(x2−9)

35. limx→6

x3 − 4x2 − 11x− 6

cos(πx/2)(x2 − 36)

36. limx→2

(x4 − 5x3 + 9x2 − 8x+ 4

x4 − 2x3 + x− 2

)2

37. limx→e5

ln3(x)− 4 ln2(x)− 4 ln(x)− 5

ln2(x)− 25

38. limx→ln(3)

√ex + 1−

√2ex − 2

e2x + 8ex − 33

39. limx→ln(2)

e2x − 2ex − xex + 2x

e2x − ex − 2

40. limx→π/6

2 sin2(x) + sin(x)− 1

2 sin2(x)− 7 sin(x) + 3

41. limx→π/4

cos(2x)

cos(x)− sin(x)

42. limx→0

cos(6x)− 1

cos(3x)− 1

43. limx→√3

9 arcsin(x/2)− 3π

3 arcsin(x/2)2 + (15− π) arcsin(x/2)− 5π

Find the one-sided limit.

44. limx→1+

2x+ |x− 1| − 2

2|1− x|

45. limx→1−

2x+ |x− 1| − 2

2|1− x|

46. limx→3+

|x2 − 9|x− 3

47. limx→3−

|x2 − 9|x− 3

48. limx→(−2)+

|x2 − x− 6|−2x2 + x+ 10

49. limx→(−2)−

|x2 − x− 6|−2x2 + x+ 10

Find the indicated limits for the piecewise defined function.

50.

f(x) =

x2 + 3x− 2 x ≤ 1

(x+ 1)2 − 4

2x− 2x > 1

(a) limx→1−

f(x)

(b) limx→1+

f(x)

(c) limx→1

f(x)

(d) limx→0

f(x)

(e) limx→2

f(x)

51.

f(x) =

√8− x+ x− 2

x2 − 1x < −1

ln(x2 + 1) −1 ≤ x ≤ 2|2− x|(3− x)

x2 − 4x > 2

(a) limx→(−1)−

f(x)

(b) limx→2−

f(x)

(c) limx→2+

f(x)

(d) limx→3

f(x)

Page 5: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

1 LIMITS AND CONTINUITY 5

Use the Squeeze Theorem to find the limit.

52. limx→0

x2 cos(

1x2

)53. lim

x→0sin(x) cos(2/x3)

54. limx→0+

√xesin(1/x)

55. limx→1

cos(

(x− 1) sin2(

1x2−1

))56. Let f be a function such that x2 ≤ f(x) ≤ x2 − 8x+ 12

|x− 2|for all 0 < x < 2. Find lim

x→2−f(x). Carefully

explain your reasoning!

Find the trigonometric limit.

57. limx→0

sin(2x)

x

58. limx→π

sin(x)

x

59. limx→0

x cot(x)

60. limx→1

sin(x− 1)

3x− 3

61. limx→0

tan(x)

sin(2x)

62. limx→π

cosx− 1

x

63. limx→0

cos(x)− 1

5 sin(x)

64. limx→0

sin(3x) + tan(6x)

sin(8x)− tan(5x)

65. limx→3

sin(x2 − 9)

x− 3

66. limx→1

tan(√x− 1)

x2 − 1

67. limx→2

√2−√x

sin(x− 2)

68. limx→1

sin(√x+ 3− 2)

tan(√x2 + 5x+ 3− 3)

69. limx→0

x cot(2x) csc(3x) tan(5x)

70. limx→0

tan(5x) sin(x2) csc(x)

x3 csc(2x)

71. limx→0

cot(2x) cot(π2 − x)

72. limx→π/2

cosx

2x− π

73. limx→π/2

tan(2x)

x− π2

74. limx→0

arcsinx

x

75. limx→0

3x

arctan(2x)

Find the infinite limit.

76. limx→1+

ln(x− 1)

77. limx→2

5

(x− 2)2

78. limx→(−2)−

x+ 6x+ 8

x2 + 4x+ 4

79. limx→1+

√x2 − 1

x2 + x− 2

80. limx→0−

sin(x)

x2

81. limx→0+

cot(x) csc(x) tan(x− π/2)

Page 6: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

6 MATH 140 HOMEWORK PROBLEMS

82. limx→(π/2)+

tanx+ 2x

1 + x2

83. limx→0

ln |x|x3 arcsin(x)

84. limx→1−

x2 + x

π − 2 arcsin(x)

85. limx→(√3)−

ln(x)

3 arctan(x)− π

Find the limit as x→∞ or x→ −∞, respectively.

86. limx→∞

x+ 1

x2 + 3x+ 5

87. limx→−∞

3x3 − 5x2 + x− 3

7x3 − 2x2 + x+ 5

88. limx→−∞

(2x2 + 5x+ 1)2(5x+ 1)

(x+ 1)(2x4 + x− 3)

89. limx→∞

5x√x+ 2x1/3 + 1√4x3 − x+ 5

90. limx→∞

3x+ 1√x2 − 1

91. limx→−∞

3x+ 1√x2 − 1

92. limx→−∞

2x+ 3√x2 + 2x+ 1 +

√4x2 + 3

93. limx→∞

(x−√x2 + x+ 2

)94. lim

x→−∞

(√4x2 + 5x+ 1−

√4x2 − 4x+ 5

)95. lim

x→−∞

x2 + 5√x4 + x+ 1

96. limx→∞

x2 + 1

x2 +√x4 + 3 + x

√9x2 + 5

97. limx→−∞

x2 + 1

x2 +√x4 + 3 + x

√9x2 + 5

98. limx→−∞

2x+ 33√

27x3 − 2x2 + x+ 15

99. limx→∞

arctan(x)

ex + 2

100. limx→−∞

arctan(x)

ex + 2

101. limx→∞

3e2x + 5ex + 1 + 2e−x

9e2x − ex + e−x

102. limx→−∞

3e2x + 5ex + 1 + 2e−x

9e2x − ex + e−x

103. limx→∞

5 ln2(x)− 2 ln(x) + 1√16 ln4(x) + ln2(x) + 1

104. limx→∞

sinx

x

105. limx→−∞

ex cos(x2 + 1)

106. limx→−∞

cos(x)− 1

x

107. limx→∞

2x2 + x sin(x) + 1

4x2 + cos(x)

108. limx→−∞

sin(x)ex + 3 + e−2x

5ex + cos(x2) + 3e−2x

Carefully explain why the following limits do not exist.

109. limx→4

|x2 − 16|x− 4

110. limx→0

sin(1/x)

111. limx→0

cos(x) sin(1/x2)

112. limx→0

x cot(x) csc(x)

113. limx→∞

arccos(x)

x2

114. limx→∞

tan(x)

Page 7: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

1 LIMITS AND CONTINUITY 7

115. limx→∞

ex cos(x2 + 2x+ 5)

116. limx→∞

cos(x)

arctan(x)

117. limx→−∞

cos(x)x2 + x+ 5

3x2 + 15

Find the vertical asymptotes of the function (if any).

118. f(x) = ln(1− x2)

119. f(x) =x− 1√x2 − 1

120. f(x) = x cotx

121. f(x) =sinx

x2 + x

122. f(x) =x3 + x2 + x

sin(x/2) cos(x), −π < x < π

123. f(x) = e−1/x2

124. f(x) = e−1/(x+1)

125. f(x) = 3

(−x− 1

x4 − x2)

Find the horizontal asymptotes of the function (if any).

126. f(x) =5√x+ 3− x2

(5x)2

127. f(x) =√x2 + 5x+ 1−

√x2 + x+ 6

128. f(x) =arctan(x)x2 + 5

2x2 + 1

129. f(x) =x−2 + 3x−1

2x−2 + x−1

130. f(x) =5 ln3(x) + ln(x) + 1

10 ln3(x) + ln2(x) + 5

131. f(x) = e−1/x2

132. f(x) = cos(πe2x + ex − πe−x

3e2x + ex + 6e−x

)133. f(x) = e−x

3+50x+1 sin(x3)

134. Sketch the graph of a function y = f(x) that has all the specified properties:

(a) y = 1 is a horizontal asymptote as x→ −∞

(b) y = 0 is a horizontal asymptote as x→∞

(c) x = 1 is a vertical asymptote

(d) f(1) = 2

(e) f is discontinuous at x = 2

(f) limx→3

f(x) = 1

(g) f(3) is undefined

135. Find the value of a such that the function

f(x) =

x+ sinx

3x2 + xfor x 6= 0,

a for x = 0

is continuous at x = 0.

Page 8: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

8 MATH 140 HOMEWORK PROBLEMS

136. Find the value of a such that the function

f(x) =

{e− 1

(x−1)4 · cos(

πx−1

)for x 6= 1,

a for x = 1

is continuous at x = 1.

137. Find the values of a and b such that the function

f(x) =

sinx

x+ a for x < 0,

3 sec(x) + b for 0 ≤ x < π/3,

5ax2 + x for x ≥ π/3

is continuous at x = 0 and x = π/3.

138. Find the values of a and b such that the function

f(x) =

a · x

2 + x− 6

x2 − 6x+ 8for x < 2,

bx2 − x+ 1 for 2 ≤ x ≤ 3,√x+ 6−

√x2 − x+ 3

x2 − 9for x > 3

is continuous on the entire real number line.

139. Find the value of b such that the function

f(x) =

{logb

(x2+5x−14

x−2

)for x > 2,

2 sin(πx

)for 0 < x ≤ 2

is continuous at x = 2.

Show that the equation has a real solution.

140. cosx = x

141. 2 sinx+ 1 = x2

142. arctanx = 2− x2

143. arcsinx = 1− x

144. ex + 1 = secx

145. 5x7 − πx3 + 2 = 3x2 + ex

146. Show that every polynomial function of odd degree has at least one real root. Is this also true forpolynomial functions of even degree? If yes, show that it is true, and if not give a counterexample.

147. Let f be a continuous function such that 0 ≤ f(x) ≤ 1 for all 0 ≤ x ≤ 1. Show that there is a solution tothe equation f(x) = x in the interval [0, 1].

148. Let f be a continuous function on [0, 2], and suppose that f(0) = f(2). Show that there is a solution tothe equation f(x) = f(x− 1) in the interval [1, 2].

149. Let f be a continuous function that satisfies f(x) · f(f(x)) = 1 for all −∞ < x < ∞. Suppose thatf(10) = 9. What is f(5)?

150. Show that on every circle of latitude on Earth there are two diametrically opposite locations where thetemperature is the same.

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2 DERIVATIVES 9

2 Derivatives

The Limit Definition of the Derivative

Use the limit definition of the derivative to find f ′(a) at the given point a.

1. f(x) = 2x+ 3, a = 5

2. f(x) = x2 + 2, a = −1

3. f(x) =17π

5π3 − 1, a = e

4. f(x) =√π − 3x, a = 0

5. f(x) = sinx, a = 0

6. f(x) = tanx, a = 0

Find an equation for the tangent line to the graph of the function y = f(x) at the point where x = a. Youshould use the limit definition of the derivative when finding the slope!

7. f(x) =√

5x2 + 4x, a = 1

8. f(x) =3

2x+ 5, a = −3

9. f(x) =1

x2 + x, a = 2

10. f(x) =5√

2x2 + 7, a = 3

Use the limit definition of the derivative to finddf

dx.

11. f(x) = 1 +4

x− x2

12. f(x) = (πx− ln 2)2

13. f(x) = sin(2x)

14. f(x) = tan(πx)

Use the limit definition of the derivative to find y′.

15. y =5s

s2 + 2

16. y =u+ 1√3u− 10

17. y = 5x3 − 2x+ 1

18. y =√x+√x

19. y = 3√x

Hint : (u− a)(u2 + au+ a2) = u3 − a3

20. y = 2x

Hint : Use that limu→0

eu − 1

u= 1.

21. y = x cos(x)

Show that the function is not differentiable, yet continuous, at the given point a.

22. f(x) = 2|x|+ 1, a = 0

23. f(x) = |x− 3|+ 2x, a = 3

24. a = 0, and

f(x) =

{x sin(1/x) for x 6= 0,

0 for x = 0,

25. Show that

f(x) =

{x2 cos(1/x) for x 6= 0,

0 for x = 0

is differentiable at x = 0, and find f ′(0).

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10 MATH 140 HOMEWORK PROBLEMS

26. Consider the function y = f(x) whose graph is shown below.

x

y

−1 1 2 3 4 5 6 7 8 9

−2

−1

1

2

3

4

(a) Find all x-values where the function f is discontinuous.

(b) Find all x-values where the function f is not differentiable.

(c) Find all x-values where f ′(x) = 0.

(d) Find f ′(4).

Sum, Difference, Constant Multiple, and Reciprocal Rule. Derivatives of Power,Exponential, and Trigonometric Functions

Differentiate the function.

27. f(x) = 3x11 − 3x7 + 8x3 + 1

28. g(t) = 5√t− 3

3√t− 15t3/2

29. ϕ(s) = (2s3 + 1)2

30. T (ω) = (√ω − 0.1)(1.5ω8/7 − 5ω)2

31. f(γ) = 3ln(5)

32. G(x) =

√17

x5/3− ln(15)x−7.385 + π2x

33. h(u) =5u√u− 3u2

u33√

8u2

34. y =1

3x+ 1

35. u =17

8x3 − 5.2x− e

36. f(x) = 5ex + x3.127

37. f(x) = 3ex + (3e)x

38. f(t) = 3t+1 + tln(7)

39. y =

√7es − 1.93e2s

8e3s+2

40. H(u) = 3u√2 + 5ue − 172u + e3u−2

Find the indicated derivative.

41.d

dx

(2 sinx− 3 tanx

)

42.d

du

(eu + 5u3 + cos(u)

)43.

d

ds

(3s−1 + 5 cos s− 3.8 csc s

)

44.d

(5 sin(α)− 3

17 cos(α)

)

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2 DERIVATIVES 11

45.d

dx

((5 cotx− 2 secx

)−1)

46.d

dx

(2 sin(x) + cos(x)

)∣∣∣x=π/3

47.d

du

(5u2 + sec(u)− tan(u))

)∣∣∣u=11π/6

48.d

d`

( π

3 sin `+ 5 tan `

)∣∣∣`=28π/3

Evaluate the limit by recognizing it as a derivative of some function.

49. limx→2

x10 − 1024

x− 2

50. limx→3

2x − 8

x− 3

51. limh→0

sec(π+3h

3

)− 2

h

52. limx→3

1x3−3x+1 −

119

2x− 6

Find an equation of the tangent line to the graph of the function y = f(x) at the given point.

53. y = 5x2 − x3, (1, 4)

54. y = x+ 2 3√x, (1, 3)

55. y = 5x− 2ex, (ln(3), 5 ln(3)− 6)

56. y = 3 cos(x)− 2 sin(x), (π/2,−2)

57. Find the x-coordinates of all points on the curve y = 8x3 + 12x2 − 48x − 10 where the tangent line ishorizontal.

58. Find the points on the curve y = 2ex − 5x where the tangent line is horizontal.

59. Find all x-values where f ′(x) = 0 for f(x) = 2 cos(x)− x.

Let f and g be differentiable functions such that f(−1) = −7, f ′(−1) = 2, and g(−1) = 17, g′(−1) = −3.Find h′(−1) for the function h.

60. h(x) = 3f(x)− 2g(x).

61. h(x) = 3x2 + f(x).

62. h(x) = 5f(x)− 1

g(x)

63. h(x) = 5 sin(πx) +2

f(x)

Note: You will need to use the limit definition of

the derivative to findd

dxsin(πx).

Product and Quotient Rule

Differentiate the function.

64. f(x) = (x2 + 3)(x3 − 15)

65. F (u) = (u+ 3√u)(15 3

√u− 5)

66. h(x) =(

5x3/2 − 2

x

)(2xπ − 3xln(5))

67. Q(v) = (3v + 5)(17v2 −√

3)(5v3 − 2v2 + 5v − 1)

68. g(q) = (√

2q3 − ln(3))(πq3 + 0.1q − 3)2

69. f(y) = yey + (y2 + 1)(√y − 5y)

70. s = (5et + t)(3t2 + 1)− t2 csc(t)

71. a = (4v3 − 5 sin v)(2 cos v + v tan v)

72. f(x) = (ex sin(x)− x2 cos(x))(tan(x)− 3x cot(x))

73. F (α) = sin(α) cos(α)5α

74. u = sec2(z)(sin(z)− 5ze6z+1)

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12 MATH 140 HOMEWORK PROBLEMS

Finddy

dx.

75. y =x− 3

x+ 5

76. y =2x3 −

√6

x7 + 15

77. y =3√

2 · x3 + x

− x2

5x3 − 3.56x2 − x

78. y =x

(2x2 − 5)(3x3 − 2x+ 16)

79. y =(x3 − 2)2 − x

(x+√x)(x+ 1)2

80. y =3πx+ sin(1)

(√x+ x)2(x5/6 + ln(17))

81. y =5ex − 3x

2 · 7x − 15

82. y =x2ex − x2x

(5x2 − 7x)2

83. y =ex + 3 sin(x)

x2 + 5

84. y =x cos(x)

x2ex − 1

85. y =(3x2 +

√2) sin(x)

5 + 23x−5

86. y =5 csc(x)ex − 2x+cos(x)

13 sin(x)+x2

x3 sec(x) + 5 tan(x)

Find the x-coordinates of all points on the graph y = f(x) where the tangent line is horizontal, i.e., find allx-values where f ′(x) = 0 (if any).

87. f(x) = ex(x2 − 3x+ 1)

88. f(x) =x

x+ 2

89. f(x) =x2

x+ 1

90. f(x) = x2 − 4 sinx+ 4x cosx

91. f(x) = ex cos(x)

92. f(x) = x sinx, −π2 < x < π2

Let f and g be differentiable functions such that f(3) = −2, g(3) = 2, f ′(3) = −1 and g′(3) = −3. Find anequation of the tangent line to the curve y = h(x) at the point where x = 3.

93. h(x) = f(x)g(x)

94. h(x) =f(x)

g(x)

95. h(x) = exf(x) + x2g(x)

96. h(x) = sin(πx)(f(x) + x)

97. h(x) =xf(x) + cos(πx)

g(x)ex + x2

Chain Rule

Find y′.

98. y = (x2 + 5x+ 1)3

99. y = (2s4 − 5s3 − 2s+ 1)11

100. y = t(3t2 − 5t)3

101. y = (x2 + 5)3(7x3 − 8x2 + π)7

102. y =( u+ 5

2u3 + 7u− 1

)2103. y =

((x5 − 2x)3 + (7x− 5)5 + 1

)10104. y =

2(5 + (v + 1)3

)5

Page 13: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

2 DERIVATIVES 13

105. y =20w(

w + (w + 1)2)2

106. y =7x2 − (3x− 2)7

1 + (x2 + 5)3

107. y =√

5x7 − 3x2

108. y =√ω +√ω

109. y =1

4√

5x7 + 2x− 1

110. y =(2x1/3 − 7x8/7

)1/13111. y =

s1/9

(3s6/7 + 2s)3

112. y = 3

√5x7 + 3x+ 1√

x+ 5x2

113. y =7

√2x3 + 3

√5x−

√x2 +

√x

114. y = 8 cos θ − 2 sin 5θ

115. y = sin(x)(cos(x2) + 5x)

116. y = 5 sin3(α)− cos(3α2 + α)

117. y = x sin3(5x)

118. y = (4t5 − 1)3 sin(√t)

119. y =sin(x+

√x)

1 + tan(x2)

120. y =(sec2(ex)− tan2(ex)

)−5121. y =

sec(5ex) + x

cos(x) + csc(x2 + 1)

122. y =√

(x2 + 1) tan(5x)

123. y = cot( θ√

θ2 + 2

)tan(θ +

2

θ2

)124. y = sec(sin(csc(tan(2r))))

125. y = ex2+1

126. y = e3 sin(5x)+1

127. y = (x2 + 1)3ecsc(x)+x

128. y = 3x−1 + sin(x)51−x2

129. y = tan(x32x+1

)130. y =

√esin(x) + 3

√72x+1 + cos(x2 + 1)

131. y = e

(x+2 cos(x)

sin(2x)ex+1

)Find an equation of the tangent line to the curve at the given point.

132. y = (1 + 3x)8, (0, 1)

133. y = x2 sin(π

3x2

), at the point where x = 1.

134. y = sin(sinx), at the point where x = π.

135. y = sin(2x) + sin2(x), (0, 0)

Find the x-coordinates of all points on the curve where the tangent line is horizontal.

136. y = x√

18− x2

137. y = x2√

9− x2

138. y = xa(1− x)b, where a, b > 0,0 < x < 1

139. y = cos(x)− cos2(x)

140. y = sec3(x)− 3 tan2(x)

141. y = sin(2x)− 2 sin(x)

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14 MATH 140 HOMEWORK PROBLEMS

Let f and g be differentiable functions. Find an expression for h′(x). Simplify as much as possible.

142. h(x) = xf(x2)

143. h(x) = g(3x2 + 1)− f(1− x)

144. h(x) = sin(f(x) + x)

145. h(x) = f(3x)g(cos(x2) + 1)

146. h(x) = g(ex + 1

)√(f(x))2 + ex

147. h(x) = f( x

g(2x− 1)

)148. h(x) = ef(x)+x

2

149. h(x) = g(x) · 52f(x)

150. Let f be a differentiable function, and let g(x) = f(2√x). If f ′(6) = 12, find g′(9).

151. Let f be a differentiable function, and let g(x) = f(f(x)). Given that f(1) = 2, f(2) = 3, f(3) = 4,f ′(1) = −2, f ′(2) = 5, and f ′(3) = 2, find g′(2).

Derivatives of Inverse Functions. Derivatives of Logarithms and Inverse Trigono-metric Functions

Show that the following functions y = f(x) are one-to-one on the given interval I by finding the inversefunction f−1 explicitly.

152. f(x) = x2 − 3x+ 5, I = (−∞, 32 )

153. f(x) =2x√x2 − 1

, I = (−∞,−1)

154. f(x) =ex − e−x

2, I = (−∞,∞)

155. f(x) =2ex

ex + 1, I = (−∞,∞)

Show that the following functions y = f(x) are one-to-one on the given interval I by showing that f ′(x) 6= 0for all x in I. Do not attempt to find f−1 explicitly!

156. f(x) = 2x3 − 2x2 + 10x− 3,I = (−∞,∞)

157. f(x) = tan(x) + sec(x),I = (−π2 ,

π2 )

158. f(x) = 7x+ sin(3x), I = (−∞,∞)

159. f(x) = x2ex, I = (0,∞)

The following functions y = f(x) are one-to-one on the given interval I. Find (f−1)′(a) at the point(s) a.

160. f(x) = xex, I = (−1,∞).Find (f−1)′(0).

161. f(x) =√x+ tan(x), I = (π2 ,

3π2 ).

Find (f−1)′(√π).

162. f(x) = x3 + x2 + 2x+ 3, I = (−∞,∞).Find (f−1)′(7).

163. f(x) =√

3 sin(3x)− 3 cos(3x),I = (− π

18 ,5π18 ). Find (f−1)′(

√3) and (f−1)′(3).

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2 DERIVATIVES 15

164. The following diagram shows the graph of an invertible function y = f(x) (drawn solid), and the line thatis tangent to the graph at the point (2, 1) (drawn dashed). What is (f−1)′(1)?

x

y

1 2 3 4 5

−1

1

2

3

4

Differentiate the function.

165. f(x) = 5 ln(x) + log2(x)

166. f(x) =√x ln(x)

167. f(s) = ln(s+√s2 − 4

)168. h(u) =

ln(u)

u+ ln(3 + 2u)

169. g(x) = ln(1 + ln(ln(x))

)170. f(x) = x2 ln(x) + arcsin(x)

171. h(t) = sin(t) arccos(t)

172. m(r) =er + ln(r2 + 1)

sin(r) + 1

173. g(s) = arcsin(ln(s))

174. k(u) = arctan(u2) + u arccot(u)

175. f(x) =csc2 x

1 + arcsec(x2)

176. h(q) = arctan(√q)− ln(arcsin(q))

177. f(x) = ln(xx)

178. f(x) = ln(ex(x2 + 1)2

xsin x

)Find an equation for the tangent line to the graph of the function at point where x = a.

179. f(x) = arcsin(2x), a = 14 .

180. f(x) =1

x√

lnx, a = e.

181. f(x) =1

(1 + lnx)2, a = 1

e2 .

182. f(x) = ln(x) arctan(x), a = 1.

183. f(x) = x2 log7(x), a = 17 .

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16 MATH 140 HOMEWORK PROBLEMS

Implicit and Logarithmic Differentiation

Find y′.

184. y2 = 1 + x4

185. 2xy + 3y3 = 15− x2

186.y

1 + y2 + x4+

1

y2 + 1=

1

2

187. 5x sin(y) cos(x) + y = 1

188.x2

1 + x+ y2= 1− sin(x2y2)

189. x2 + arctan(x+ y4) = y

190. sin(x2 + 3x2y3 + 15y2) + cos(5x2 + 8xy+ 2y2) = 0

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2 DERIVATIVES 17

Finddy

dx.

191. tan(x/y

)= 2x+ y

192. x2 − 2x+ y4 = ln(x+ y3)− 1

193. 2xy = cot(x/y2

)194. arctan

(x/y

)= x2 + y3

Find an equation of the tangent line to the curve at the given point.

195. xy2 + x sin(y) = π2, (1, π)

196. x2 + 4xy + y − y2 = 4, (2, 0)

197. x2 + y2 = (x2 + y2 − x)2, (0, 1)

198. sin(πy) + x2 + 1 = exy, (0, 1)

199. x2 − 4y2 = cos(πxy),(1, 12).

200. arctan(x2y) + x =y + 2

x+ 1, (1, 0)

Differentiate the function.

201. f(x) = x2x

202. f(x) =(tanx

)x2

203. f(x) =(x2 + lnx

5x4 + 1

)sin x204. f(x) = (2x2 + 1)

ln x+1ex+cos x

205. f(x) = x+ xx

206. f(x) = x2 −(cosx

)x+ xcos x

207. f(x) =(x4 + x)5(x2 + x+ 1)10(cosx+ 1)3

(x− 1)3(tan(2x) + x)5(lnx+ 1)7

208. f(x) =(x+ 1)x(x5 − 1)3ex

(x2 + 1)2(sinx)x

Higher Derivatives

Find y′′.

209. y = 5x3 − 2x+ 1

210. y = ex2

211. y = x2 sin(x) + 5x7

212. y =x+ 1

cos(x) + x

213. y = arcsin(2x) + cos(x2)

214. y = e2x + ln(x+ 1)

215. y = arctan(x+ ln(x))

216. y = xarccos x

Differentiate the function several times and identify a pattern in the resulting formulas. Then use the patternto determine the indicated derivatives.

217. f (203)(x) for f(x) = xex

218.d105f

dx105for f(x) =

1

(1 + x)2

219. f (3204)(x) for f(x) = x sin(x)

220.d46y

dx46for y = sin(x) cos(x)

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18 MATH 140 HOMEWORK PROBLEMS

Findd2y

dx2.

221. x3 + 2y3 = 8xy

222.√x+ y = 1 + x2y2

223. 2 cos(x) sin(y) = 3x+ y

224. sin(xy2) = ln(x+ y)

3 Applications of the Derivative

Curve Sketching

The derivative of the function y = f(x) is given. Find the intervals on which the function is increasing ordecreasing, and find the x-coordinates of the points where local maxima or minima occur.

1. f ′(x) = (x+ 3)2(x+ 1)(x− 5)3

2. f ′(x) = (2 cos(x)− 1)(sin(x) + 1)3,0 ≤ x ≤ 2π.

3. f ′(x) = sin2(x)− 3 sin(x) + 2− cos2(x),0 ≤ x ≤ π.

4. f ′(x) = cos2(x)− sin2(x) + sin(2x),0 < x < π

2

The second derivative of the function y = f(x) is given. Find the intervals on which the function is concaveupward or downward, and find the x-coordinates of any inflection points.

5. f ′′(x) = (2x+ 1)(sec2(x)− 2),−π2 < x < π

2 .

6. f ′′(x) = ln3(x)− 2 ln2(x)− 3 ln(x)

7. f ′′(x) = (ln(x)− 1)(e2x+5 − 1)

8. f ′′(x) = 2 · 25x − 5x+1 − 3

9. The function y = f(x) is continuous on (−∞,∞), and the graph of its derivative is shown.

x

y

−2 −1 1 2 3 4 5 6

−1

1

2

y = f ′(x)

(a) Find the critical points of f .

(b) Find the intervals where f is increasing or decreasing.

(c) Find the x-coordinates of the points where f has any local maxima or minima.

(d) Find the intervals where f is concave upward or downward.

(e) Find the x-coordinates of any inflection points.

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3 APPLICATIONS OF THE DERIVATIVE 19

10. Sketch the graph of a function y = f(x) that satisfies all the criteria.

• f(−3) = f(0) = 0

• limx→1−

f(x) = −∞, limx→1+

f(x) =∞

• limx→∞

f(x) = 0

• f ′(−2) = f ′(0) = 0, and f ′(3) does not exist

• f ′ > 0 on (−2, 0), and on (2, 3)

• f ′ < 0 on (−∞,−2), (0, 1), (1, 2), and (3,∞)

• f ′′ > 0 on (−∞,−1), (1, 3), and (3,∞)

• f ′′ < 0 on (−1, 1)

11. Let T (t) be the temperature at time t. In (a)–(d) below, data about the derivatives of T (t) at time t = 2are given. In each case, determine whether the temperature at time t = 2 is increasing or decreasing.Moreover, determine whether it continues to increase/decrease more rapidly or more gently (over a brieftime period).

(a) T ′(2) = 5, T ′′(2) = 3.

(b) T ′(2) = 5, T ′′(2) = −3.

(c) T ′(2) = −5, T ′′(2) = 3.

(d) T ′(2) = −5, T ′′(2) = −3.

12. Suppose that f is increasing with f ′(x) > 0 on (−∞,∞).

(a) Show that g(x) = ef(x) is increasing on (−∞,∞)

(b) Is g(x) = (f(x))2 increasing? What if f(x) > 0 for all x? Explain!

For each of the following functions, do the following:

(a) Find the domain.

(b) Find the x- and y-intercepts (if any).

(c) Determine whether the function is even, odd, or neither.

(d) Find the horizontal and vertical asymptotes (if any).

(e) Find the intervals on which the function is increasing or decreasing.

(f) Find the local maxima and minima (if any).

(g) Find the intervals of concavity.

(h) Find the inflection points (if any).

(i) Sketch a qualitative graph of the function based on the above information.

13. f(x) = x3 − 6x2 + 9x

14. f(x) = x3 − 3x2 + 3x− 9

15. f(x) = x4 − 6x2

16. f(x) = x5 − 5x4

17. f(x) =x2 + 2x+ 2

x− 1

18. f(x) =2x

x2 + 2

19. f(x) =x

4x3 + 1

20. f(x) =x2 + 2x+ 1

x2 + 2x− 15

21. f(x) = x2 +1

x2

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20 MATH 140 HOMEWORK PROBLEMS

22. f(x) = 3x2/3 − 2x

23. f(x) = x(8− x)3/5

24. f(x) = (x− 3)2(x+ 1)2/3

25. f(x) =√x2 − 2x+ 2− x

26. f(x) = (x2 + 5x+ 4)4/5

27. f(x) =x+ 4√x2 + 8

28. f(x) =x− 1√x2 − 1

29. f(x) =√

3 sin(x) + cos(x)−π ≤ x ≤ π

30. f(x) = 2 cos(x) + x0 ≤ x ≤ 4π

31. f(x) = 2x− tan(x)−π2 < x < π

2

Note: You may use that the x-intercepts of f are0 and ±1.166 (approximately).

32. f(x) = 2 cos(x) + cos2(x)0 ≤ x ≤ 2π

33. f(x) = 3 sin(x)− 4 sin3(x)−π ≤ x ≤ π

34. f(x) =cos(x)

2 + sin(x)0 ≤ x ≤ 2π

35. f(x) = x− sin(x) cos(x) + 2 cos(x)0 < x < π

36. f(x) = x2 − 2 lnx

37. f(x) = ln(3− 2x− x2)

38. f(x) = ln(x4 + 3)

39. f(x) = ln(x+√x2 + 1)

40. f(x) =lnx

x2

41. f(x) =x

lnx

42. f(x) = ln(

1 +1

x

)43. f(x) = ln(1− lnx)

44. f(x) = xe−x

45. f(x) = (2x2 + x+ 1)ex

46. f(x) = e−x2

47. f(x) =e2x

1− ex

48. f(x) = sin(x)ex

−π ≤ x ≤ π

49. f(x) = e1/x

50. f(x) = x arctan(x)

51. f(x) = earctan(x)

52. f(x) = xx

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ANSWERS SECTION 1 21

Answers

Note: Many of these answers were generated by a computer. The format of computer-generated answerscan be quite different from the format that you get by applying the rules of Calculus manually (the answersare equivalent, though, just like f(x) = 2x(x2 + 3)5 and f(x) = 10x3 + 30x are two equivalent mathematicalexpressions that define the same function f). When in doubt whether a stated answer here is equivalent tothe answer that you have found, ask!

Answers Section 1

1a DNE

1b 12

1c DNE

1d 32

1e 2

1f DNE

2a 1

2b 1

2c 2

2d 1

2e DNE

2f DNE

3a 1

3b 2

3c ∞

3d DNE

4a 3

4b −∞

6 4

7 132

8 4

9 7

10 524

11 38

12 2

13 152e3+2

14 −4

15 −π616 7

17 15

18 35

19 14

20 21

21 2π3

22 2

23 −1

24 − 85

25 16

26 − 49

27 118

28 − 340

29 − 150

30 − 1720

31 −√139

32 e3

33√

3

34 e−112

Page 22: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

22 MATH 140 HOMEWORK PROBLEMS

35 − 4912

36 0

37 3110

38 − 156

39 2−ln(2)3

40 − 35

41√

2

42 4

43 9π+15

44 32

45 − 12

46 6

47 −6

48 59

49 − 59

50a 2

50b 2

50c 2

50d −2

50e 52

51a − 512

51b ln(5)

51c 14

51d 0

52 0

53 0

54 0

55 1

56 4

57 2

58 0

59 1

60 13

61 12

62 − 2π

63 0

64 3

65 6

66 14

67 − 12√2

68 314

69 56

70 10

71 12

72 − 12

73 2

74 1

75 32

76 −∞

77 ∞

78 −∞

79 ∞

80 −∞

81 −∞

82 −∞

83 −∞

84 ∞

85 −∞

86 0

Page 23: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 2 23

87 37

88 10

89 52

90 3

91 −3

92 − 23

93 − 12

94 − 94

95 1

96 15

97 −1

98 23

99 0

100 −π4

101 13

102 2

103 54

104 0

105 0

106 0

107 12

108 13

118 x = −1 and x = 1

119 x = −1

120 x = nπ, n ∈ Z, n 6= 0

121 x = −1

122 x = −π2 and x = π2

123 None

124 x = −1

125 x = −1

126 y = − 125 as x→∞, none as x→ −∞

127 y = 2 as x→∞, y = −2 as x→ −∞

128 y = π4 as x→∞, y = −π4 as x→ −∞.

129 y = 3 both as x→ ±∞

130 y = 12 as x→∞, none as x→ −∞

131 y = 1 both as x→ ±∞

132 y = 12 as x→∞, y =

√32 as x→ −∞.

133 y = 0 as x→∞, none as x→ −∞

135 a = 2

136 a = 0

137 a = − 3(π−12)5π2−9 , b = − 3π+10π2−54

5π2−9

138 a = 26405 , b = 17

81

139 b = 3

149 f(5) = 15

Answers Section 2

1 f ′(5) = 2

2 f ′(−1) = −2

3 f ′(e) = 0

4 f ′(0) = − 32√π

5 f ′(0) = 1

6 f ′(0) = 1

7 y = 73x+ 2

3

8 y = −6x− 21

9 y = − 536x+ 4

9

Page 24: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

24 MATH 140 HOMEWORK PROBLEMS

10 y = − 625x+ 43

25

11 dfdx = − 4

x2 − 2x

12 dfdx = 2π(πx− ln 2)

13 dfdx = 2 cos(2x)

14 dfdx = π sec2(πx)

15 y′ = − 5(s2−2)(s2+2)2

16 y′ = 3u−232(3u−10)3/2

17 y′ = 15x2 − 2

18 y′ = 2√x+1

4√x+√x√x

19 y′ = 1

33√x2

20 y′ = ln(2)2x

21 y′ = cos(x)− x sin(x)

25 f ′(0) = 0

26a x = 2

26b x = 2 and x = 5.

26c x = 1

26d f ′(4) = − 13

27 f ′(x) = 33x10 − 21x6 + 24x2

28 g′(t) = 52 t−1/2 + t−4/3 − 45

2 t1/2

29 ϕ′(s) = 12s4 + 6s

30 T ′(ω) = 35156 ω

2514 − 555

14 ω2314 + 125

2 ω32 − 18

35 ω97 +

4514 ω

87 − 5ω

31 f ′(γ) = 0

32 G′(x) = π2 − 5√17

3 x83

+ 7.385 ln(15)x8.385

33 h′(u) = 5

2u83− 65

12u196

34 y′ = − 3(3x+1)2

35 u′ = − 17(24x2−5.2)(8x3−5.2x−e)2

36 f ′(x) = 5ex + 3.127x2.127

37 f ′(x) = 3ex + ln(3e)(3e)x

38 f ′(t) = 3 ln(3)3t + ln(7)tln(7)−1

39 y′ = −√74 e−2s−2 + 193

800e−s−2

40 H ′(u) = 3√

2u√2−1 + 5eue−1 − 2 ln(17)172u +

3e3u−2

41 2 cosx− 3 sec2 x

42 eu + 15u2 − sin(u)

43 ln(3)3s−1 − 5 sin s+ 3.8 csc s cot s

44 517 sec2(α)− 3

17 sec(α) tan(α)

45 5 csc2(x)+2 sec(x) tan(x)(5 cot x−2 sec x)2

46 −√32 + 1

47 55π3 − 2

48 − 74π147

49 5120

50 8 ln(2)

51 2√

3

52 − 12361

53 y = 7(x− 1) + 4

54 y = 53 (x− 1) + 3

55 y = (−1)(x− ln(3)) + 5 ln(3)− 6

56 y = −3(x− π/2)− 2

57 x = −2 and x = 1

58 (ln(5/2), 5− 5 ln(5/2))

59 x = 7π6 + (2π)n, x = 11π

6 + (2π)n, n ∈ Z

60 12

61 −13

62 2887289

63 −5π − 449

64 f ′(x) = 5x4 + 9x2 − 30x

65 F ′(u) = 20u13 + 75

2u16− 15

2√u− 5

Page 25: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 2 25

66 h′(x) = − 12

(15√x+ 4

x2

)(3xln(5) − 2xπ

)+(5x

32 − 2

x

)(2πxπ−1 − 3xln(5)−1 ln (5)

)67 Q′(v) = 1530 v5 + 1615 v4 − 20

(3√

3− 17)v3 − 3

(19√

3− 374)v2 − 10

(√3 + 17

)v − 22

√3

68 g′(q) = 3100 (10πq3 + q − 30)

2√2q2 + 1

50 (30πq2 + 1)(√

2q3 − ln(3))(10πq3 + q − 30)

69 f ′(y) = −15 y2 + yey + 52 y

32 + 1

2√y + ey − 5

70 s′(t) = (5 et + 1)(3 t2 + 1

)+ (t+ 5 et)6t+ t2 csc (t) cot (t)− 2 t csc (t)

71 a′(v) = (v tan (v) + 2 cos (v))(12 v2 − 5 cos (v)) + (4 v3 − 5 sin (v))(sec2 (v)v + tan (v)− 2 sin (v))

72 f ′(x) = (tan (x)− 3x cot (x)) · (x2 sin (x)− 2x cos (x) + ex sin (x) + ex cos (x))− (x2 cos (x)− ex sin (x)) ·(3x csc2 (x) + sec2(x)− 3 cot (x))

73 F ′(α) = 5α ln (5) sin (α) cos (α)− 5α sin2 (α) + 5α cos2 (α)

74 u′(z) = −2(5ze6 z+1 − sin (z)

)· tan (z) sec2 (z)−

(5ze6 z+1 ln (5) + 5z · 6 · e6 z+1 − cos (z)

)sec2 (z)

75 8(x+5)2

76 − (8 x7−7√6x4+90)x2

(x7−15)2

77 25 (375 x2−178 x−25)x2

(125 x3−89 x2−25 x)2 −50 x

125 x3−89 x2−25 x + 213

x+3 −2

13 x

(x+3)2

78 − 2 (12 x5−19 x3+16 x2+40)

(2 x2−5)2(3 x3−2 x+16)2

79 −( 1√

x+2)((x3−2)2−x)

2 (x+1)2(x+√x)2

+ 6 (x3−2)x2−1(x+1)2(x+

√x)− 2 ((x3−2)2−x)

(x+1)3(x+√x)

80 −( 1√

x+2)(3πx+sin(1))

(x56 +ln(17))(x+

√x)3

+ 3π

(x56 +ln(17))(x+

√x)2− 5 (3πx+sin(1))

6 (x56 +ln(17))

2(x+√x)2x

16

81 2 (3x−5 ex)7x ln(7)

(2·7x−15)2 − 3x ln(3)−5 ex2·7x−15

82 − 2xx ln(2)−x2ex−2 xex+2x

(7x−5 x2)2− 2 (7x ln(7)−10 x)(x2ex−2xx)

(7x−5 x2)3

83 ex+3 cos(x)x2+5 − 2 (ex+3 sin(x))x

(x2+5)2

84 − x sin(x)x2ex−1 −

(x2ex+2 xex)x cos(x)

(x2ex−1)2 + cos(x)x2ex−1

85 − 3 (√2+3 x2)23 x−5 ln(2) sin(x)

(23 x−5+5)2+

6 x sin(x)+(√2+3 x2) cos(x)

23 x−5+5

86 −(5 ex csc(x)− 2 x+cos(x)

x2+13 sin(x))(x3 tan(x) sec(x)+3 x2 sec(x)+5 tan(x)2+5)

(x3 sec(x)+5 tan(x))2

−5 ex csc(x) cot(x)−5 ex csc(x)− sin(x)−2

x2+13 sin(x)− (2 x+cos(x))(2 x+13 cos(x))

(x2+13 sin(x))2

x3 sec(x)+5 tan(x)

Page 26: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

26 MATH 140 HOMEWORK PROBLEMS

87 x = −1 and x = 2

88 None

89 x = −2 and x = 0

90 x = 0, x = π6 + (2π)n, x = 5π

6 + (2π)n, n ∈ Z

91 x = π4 + nπ, n ∈ Z

92 x = 0

93 y = 4(x− 3) + (−4)

94 y = (−2)(x− 3) + (−1)

95 y = (−3e3 − 15)(x− 3) + (18− 2e3)

96 y = −π(x− 3)

97 y = −17e3−3(2e3+9)2 (x− 3)− 7

2e3+9

98 3(x2 + 5x+ 1)2(2x+ 5)

99 11(2s4 − 5s3 − 2s+ 1)10(8s3 − 15s2 − 2)

100 (3t2 − 5t)3 + t · 3(3t2 − 5t)2(6t− 5)

101 7(x2 + 5

)3(21x2 − 16x

)(π + 7x3 − 8x2

)6+ 6

(x2 + 5

)2(π + 7x3 − 8x2

)7x

102 − 2 (u+5)2(6u2+7)(2u3+7u−1)3 + 2 (u+5)

(2u3+7u−1)2

103 10(

35 (7x− 5)4

+ 3(5x4 − 2

)(x5 − 2x

)2)((7x− 5)

5+(x5 − 2x

)3+ 1)9

104 − 30 (v+1)2

((v+1)3+5)6

105 − 20 (3w2+3w−1)(w2+3w+1)3

1066 (x2+5)

2((3 x−2)7−7 x2)x

((x2+5)3+1)2 − 7 (3 (3 x−2)6−2 x)

(x2+5)3+1

107 35 x6−6 x2√5 x7−3 x2

1081√ω+2

4√ω+√ω

109 − 35 x6+2

4 (5 x7+2 x−1)(54 )

110 −2

(12 x(

17 )− 1

x23

)

39

(−7 x(

87 )+2 x(

13 ))( 12

13 )

111 −6

(9

s17

+7

)s(

19 )

7

(2 s+3 s(

67 ))4 + 1

9

(2 s+3 s(

67 ))3

s(89 )

112

2 (35 x6+3)5 x2+

√x−

(20 x+ 1√

x

)(5 x7+3 x+1)

(5 x2+√x)2

(5 x2+√x)(

23 )

6 (5 x7+3 x+1)(23 )

Page 27: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 2 27

113

72 x2−

4 x+ 1√x√

x2+√x−20

(5 x−√x2+√x)

23

84

((5 x−√x2+√x)( 1

3 )+2 x3

)( 67 )

114 −8 sin θ − 10 cos 5θ

115 −2x sin(x2)

sin (x) + 5x cos (x) + cos(x2)

cos (x) + 5 sin (x)

116 15 sin (α)2

cos (α) + 6α sin ((3α+ 1)x) + sin ((3α+ 1)α)

117 15x sin (5x)2

cos (5x) + sin (5x)3

118 60(4 t5 − 1

)2t4 sin

(√t)

+(4 t5−1)

3cos(√t)

2√t

119

(1√x+2)cos(x+

√x)

2 (tan(x2)+1) − 2 sec2(x2)x sin(x+√x)

(tan(x2)+1)2

120 0

121 5 ex tan(5 ex) sec(5 ex)+1cos(x)+csc(x2+1) +

(x+sec(5 ex))(2 x csc(x2+1) cot(x2+1)+sin(x))(cos(x)+csc(x2+1))2

1225 sec2(5x)(x2+1)+2 x tan(5 x)

2√

(x2+1) tan(5 x)

123

θ2(θ2+2

) 32

− 1√θ2+2

tan

(θ + 2

θ2

)csc

θ√θ2+2

2

−(tan

(θ + 2

θ2

)2+ 1

)(4θ3− 1

)cot

θ√θ2+2

124 −2 sec2(2r) cos (csc (tan (2 r))) tan (sin (csc (tan (2 r)))) sec (sin (csc (tan (2 r)))) csc (tan (2 r)) cot (tan (2 r))

125 ex2+12x

126 e3 sin(5x)+115 cos(5x)

127 −(csc (x) cot (x)− 1)(x2 + 1

)3e(x+csc(x)) + 6

(x2 + 1

)2xe(x+csc(x))

128 −2x5(−x2+1) ln (5) sin (x) + 3(x−1) ln (3) + 5(−x2+1) cos (x)

129 sec2(x3(2 x+1)

) (3(2 x+1) + 2x3(2 x+1) ln (3)

)

130

3 esin(x) cos(x)−2 (x sin(x2+1)−7(2 x+1) ln(7))

(7(2 x+1)+cos(x2+1))23

6

√(7(2 x+1)+cos(x2+1))(

13 )+esin(x)

131 e

(x+2 cos(x)

sin(2x)ex+1

)(− 2 sin(x)−1ex sin(2 x)+1 −

(x+2 cos(x))(ex sin(2 x)+2 ex cos(2 x))

(ex sin(2 x)+1)2

)132 y = 24x+ 1

133 y =(√

3− π3

)(x− 1

)+√32

134 y = −(x− π)

135 y = 2x

136 x = −3 and x = 3

137 x = −√

6, x = 0, x =√

6

Page 28: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

28 MATH 140 HOMEWORK PROBLEMS

138 x = aa+b

139 x = nπ, x = π3 + (2π)n, x = −π3 + (2π)n, n ∈ Z

140 x = nπ, x = π3 + (2π)n, x = −π3 + (2π)n, n ∈ Z

141 x = (2π)n, x = 2π3 + (2π)n, x = − 2π

3 + (2π)n,n ∈ Z

142 h′(x) = 2x2f ′(x2) + f(x2)

143 h′(x) = 6xg′(3x2 + 1) + f ′(1− x)

144 h′(x) = cos(f(x) + x)(f ′(x) + 1)

145 h′(x) = −2x sin(x2)f (3x) · g′

(cos(x2)

+ 1)

+

3 g(cos(x2)

+ 1)· f ′ (3x)

146 h′(x) =

√f (x)

2+ exexg′ (ex + 1) +

(2 f(x)f ′(x)+ex)g(ex+1)

2√f(x)2+ex

147 h′(x) = −(

2 xg′(2 x−1)g(2 x−1)2 −

1g(2 x−1)

)· f ′(

xg(2 x−1)

)148 h′(x) = (2x+ f ′ (x))e(x

2+f(x))

149 h′(x) = 2 · 5(2 f(x)) ln (5) g (x) f ′ (x) +5(2 f(x))g′ (x)

150 4

151 10

152 f−1(x) = 3−√4x−112 , −∞ < x < 11

4

153 f−1(x) = x√x2−4 , −∞ < x < −2

154 f−1(x) = ln(x+√x2 + 1

), −∞ < x <∞

155 f−1(x) = ln(

x2−x

), 0 < x < 2

160 (f−1)′(0) = 1

161 (f−1)′(√π) = 2

√π

2√π+1

162 (f−1)′(7) = 17

163 (f−1)′(√

3) = 19 and (f−1)′(3) =

√39

164 23

165 f ′(x) = 5x + 1

x ln(2)

166 f ′(x) = ln x+22√x

167 f ′(s) =1+ s√

s2−4

s+√s2−4

168 h′(u) =(u+ln(3+2u)) 1

u−ln(u)(

22u+3+1

)(u+ln(3+2u))2

169 g′(x) = 1(1+ln(ln(x))) ln(x)x

170 f ′(x) = 2x ln(x) + x+ 1√1−x2

171 h′(t) = cos(t) arccos(t)− sin(t)√1−t2

172 m′(r) =2 rr2+1

+er

sin(r)+1 −(er+ln(r2+1)) cos(r)

(sin(r)+1)2

173 g′(s) = 1

s√

1−ln2(s)

174 k′(u) = − uu2+1 + 2u

u4+1 + arccot (u)

175 f ′(x) = − 2 csc2(x) cot(x)arcsec(x2)+1 −

2x csc2(x)

(arcsec(x2)+1)2x2√x4−1

176 h′(q) = 12 (q+1)

√q −

1

arcsin(q)√

1−q2

177 f ′(x) = ln(x) + 1

178 f ′(x) = − ln (x) cos (x) + 4 xx2+1 −

sin(x)x + 1

179 y = 4√3

(x− 1

4

)+ π

6

180 y = − 32e2 (x− e) + 1

e

181 y = 2e2(x− 1

e2

)+ 1

182 y = π4 (x− 1)

183 y = − 2 ln(7)−17 ln(7)

(x− 1

7

)− 1

49

184 2 x3

y

185 − 2 (x+y)9 y2+2 x

186 − 4 x3y

(x4+y2+1)2(

2 y2

(x4+y2+1)2+ 2 y

(y2+1)2− 1x4+y2+1

)

Page 29: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 2 29

187 5 (x sin(x) sin(y)−sin(y) cos(x))5 x cos(x) cos(y)+1

188 −2 xy2 cos(x2y2)+ 2 x

y2+x+1− x2

(y2+x+1)2

2

(x2y cos(x2y2)− x2y

(y2+x+1)2

)

189 −2 x+ 1

(y4+x)2+1

4 y3

(y4+x)2+1−1

190 − 2 ((5 x+4 y) sin(5 x2+8 xy+2 y2)−(3 xy3+x) cos(3 x2y3+x2+15 y2))4 (2 x+y) sin(5 x2+8 xy+2 y2)−3 (3 x2y2+10 y) cos(3 x2y3+x2+15 y2)

191sec2(x/y)

y −2sec2(x/y)x

y2+1

192 −2 x− 1

y3+x−2

4 y3− 3 y2

y3+x

193 −2 y+

csc

(xy2

)2y2

2

x− x csc

(xy2

)2y3

194 −2 x− 1(

x2

y2+1

)y

3 y2+ x(x2

y2+1

)y2

195 y = − π2

2π−1 (x− 1) + π

196 y = − 49 (x− 2)

197 y = x+ 1

198 y = − 1πx+ 1

199 y = 4+π2(4−π) (x− 1) + 1

2

200 y = −3(x− 1)

201 f ′(x) = 2(ln(x) + 1)x2x

202 f ′(x) =(

sec2(x)x2

tan(x) + 2x ln(tan(x)))(

tanx)x2

203 f ′(x) = −

(5 x4+1)

(20 (x2+ln(x))x3

(5 x4+1)2−

2 x+ 1x

5 x4+1

)sin(x)

x2+ln(x) − ln(x2+ln(x)5 x4+1

)cos (x)

(x2+ln(x)5 x4+1

)sin(x)

204 f ′(x) = −((

(ln(x)+1)(ex−sin(x))(ex+cos(x))2

− 1(ex+cos(x))x

)ln(2x2 + 1

)− 4 (ln(x)+1)x

(2 x2+1)(ex+cos(x))

)·(2x2 + 1

)( ln(x)+1ex+cos(x) )

205 f ′(x) = 1 + (ln(x) + 1)xx

206 f ′(x) =(x sin(x)cos(x) − ln (cos (x))

)cos (x)

x −(

ln (x) sin (x)− cos(x)x

)xcos(x) + 2x

207 f ′(x) =(x4 + x)5(x2 + x+ 1)10(cosx+ 1)3

(x− 1)3(tan(2x) + x)5(lnx+ 1)7

(− 3 sin(x)

cos(x)+1 + 10 (2 x+1)x2+x+1 −

5 (2 sec(2 x)2+1)x+tan(2 x) +

5 (4 x3+1)x4+x − 3

x−1 −

7(ln(x)+1)x

)208 f ′(x) =

(x+ 1)x(x5 − 1)3ex

(x2 + 1)2(sinx)x

(15 x4

x5−1 −x cos(x)sin(x) + x

x+1 −4 xx2+1 + ln (x+ 1)−

ln (sin (x)) + 1)

209 30x

210 4x2e(x2) + 2 e(x

2)

Page 30: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

30 MATH 140 HOMEWORK PROBLEMS

211 210x5 − x2 sin (x) + 4x cos (x) + 2 sin (x)

212 2 (sin(x)−1)2(x+1)

(x+cos(x))3+ (x+1) cos(x)

(x+cos(x))2+ 2 (sin(x)−1)

(x+cos(x))2

213 −4x2 cos(x2)

+ 8 x

(−4 x2+1)32− 2 sin

(x2)

214 − 1(x+1)2

+ 4 e(2 x)

215 − 2 ( 1x+1)

2(x+ln(x))

((x+ln(x))2+1)2 − 1

((x+ln(x))2+1)x2

216(

ln(x)√−x2+1

− arccos(x)x

)2xarccos(x) −

(x ln(x)

(−x2+1)32

+ arccos(x)x2 + 2√

−x2+1x

)xarccos(x)

217 f (203)(x) = (x+ 203)ex

218 d105fdx105 = − 2·3·...·105·106

(x+1)107

219 f (3204)(x) = x sin(x)− 3204 cos(x)

220d46y

dx46= −246 sin(x) cos(x)

221 −3 (3 x2−8 y)

2y

(3 y2−4 x)2+6 x+

8 (3 x2−8 y)3 y2−4 x

2 (3 y2−4 x)

222 −8 (x+y)(

32 )y2+

(4√x+yxy2−1)

28 (x+y)

( 32 )x2+1

(4√x+yx2y−1)2

−2 (4√x+yxy2−1)

16 (x+y)( 3

2 )xy+1

4√x+yx2y−1

+1

2

(4 (x+y)(

32 )x2y−x−y

)

2232

(2 (2 sin(x) sin(y)+3) sin(x) cos(y)

2 cos(x) cos(y)−1+

(2 sin(x) sin(y)+3)2 sin(y) cos(x)

(2 cos(x) cos(y)−1)2+sin(y) cos(x)

)2 cos(x) cos(y)−1

224(

2

(2 ((xy2+y3) cos(xy2)−1)(x2y+2 xy2+y3)

2 (x2y+xy2) cos(xy2)−1 − ((xy2+y3) cos(xy2)−1)2(x3+2 x2y+xy2)

(2 (x2y+xy2) cos(xy2)−1)2

)cos(xy2)

+(x2y4 + 2xy5 + y6 − 4 ((xy2+y3) cos(xy2)−1)(x3y3+2 x2y4+xy5)

2 (x2y+xy2) cos(xy2)−1 +4 ((xy2+y3) cos(xy2)−1)

2(x4y2+2 x3y3+x2y4)

(2 (x2y+xy2) cos(xy2)−1)2

)sin(xy2)

+2 ((xy2+y3) cos(xy2)−1)2 (x2y+xy2) cos(xy2)−1 −

((xy2+y3) cos(xy2)−1)2

(2 (x2y+xy2) cos(xy2)−1)2 − 1)(

2(x3y + 2x2y2 + xy3

)cos(xy2)− x− y

)−1Answers Section 3

1 Increasing on: (−∞,−1), (5,∞)Decreasing on: (−1, 5)Local maxima: f(−1)Local minima: f(5)

2 Increasing on: [0, π/3), (5π/3, 2π]Decreasing on: (π/3, 5π/3)Local maxima: f(π/3)

Local minima: f(5π/3)

3 Increasing on: [0, π/6), (5π/6, π]Decreasing on: (π/6, 5π/6)Local maxima: f(π/6)Local minima: f(5π/6)

4 Increasing on: (0, 3π/8)

Page 31: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 3 31

Decreasing on: (3π/8, π/2)Local maxima: f(3π/8)Local minima: None

5 Concave up on: (−π/4,−1/2), (π/4, π/2)Concave down on: (−π/2,−π/4), (−1/2, π/4)Inflection point at: x = ±π/4, x = −1/2

6 Concave up on: (e−1, 1), (e3,∞)Concave down on: (0, e−1), (1, e3)Inflection point at: x = e−1, x = 1, x = e3

7 Concave up on: (e,∞)Concave down on: (0, e)Inflection point at: x = e

8 Concave up on: (log5(3),∞)Concave down on: (−∞, log5(3))Inflection point at: x = log5(3)

9a x = −2, x = 0, x = 1, x = 2, x = 4

9b Increasing on: (−∞,−2), (0, 1), (2, 4)Decreasing on: (−2, 0), (1, 2), (4,∞)

9c Local maxima at x = −2, x = 1, and x = 4, localminima at x = 0 and x = 2

9d Concave up on (−1, 1), (1, 3)Concave down on (−∞,−1), (3,∞)

9e x = −1, x = 3

10 There are infinitely many correct graphs. Here isone:

x

y

−4 −3 −2 −1 1 2 3 4 5 6

−3

−2

−1

1

2

3

4

11a Increases rapidly

11b Increases gently

11c Decreases gently

11d Decreases rapidly

13 Domain: (−∞,∞)

f(x) = (x− 3)2x

f ′(x) = 3 (x− 3)(x− 1)

f ′′(x) = 6x− 12

y-intercept: (0, 0)

x-intercept(s): (0, 0), (3, 0)

Symmetry: None

Vertical asymptotes: None

Horizontal asymptotes: None

Increasing on: (−∞, 1), (3,∞)

Decreasing on: (1, 3)

Local maxima: f(1) = 4

Local minima: f(3) = 0

Concave up on: (2,∞)

Concave down on: (−∞, 2)

Inflection points: (2, 2)

x

y

1

1

2

2

3

3

4

4

14 Domain: (−∞,∞)

f(x) = (x− 3)(x2 + 3

)f ′(x) = 3 (x− 1)

2

f ′′(x) = 6x− 6

y-intercept: (0,−9)

x-intercept(s): (3, 0)

Symmetry: None.

Vertical asymptotes: None

Horizontal asymptotes: None

Increasing on: (−∞,∞)

Decreasing on: Nowhere

Local maxima: None

Local minima: None

Concave up on: (1,∞)

Concave down on: (−∞, 1)

Inflection points: (1,−8)

Page 32: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

32 MATH 140 HOMEWORK PROBLEMS

x

y

1 2 3

−10

−8

−6

−4

−2

2

15 Domain: (−∞,∞)

f(x) =(x2 − 6

)x2

f ′(x) = 4(x2 − 3

)x

f ′′(x) = 12 (x− 1)(x+ 1)

y-intercept: (0, 0)x-intercept(s): (±

√6, 0), (0, 0)

Symmetry: EvenVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (−

√3, 0), (

√3,∞)

Decreasing on: (−∞,−√

3), (0,√

3)Local maxima: f(0) = 0Local minima: f(−

√3) = −9, f(

√3) = −9

Concave up on: (−∞,−1), (1,∞)Concave down on: (−1, 1)Inflection points: (−1,−5), (1,−5)

x

y

−3 −2 −1 1 2 3

−9

−7

−5

−3

−1

1

3

16 Domain: (−∞,∞)

f(x) = (x− 5)x4

f ′(x) = 5 (x− 4)x3

f ′′(x) = 20 (x− 3)x2

y-intercept: (0, 0)x-intercept(s): (0, 0), (5, 0)Symmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (−∞, 0), (4,∞)Decreasing on: (0, 4)Local maxima: f(0) = 0Local minima: f(4) = −256Concave up on: (3,∞)Concave down on: (−∞, 3)Inflection points: (3,−162)

x

y

−3−2−1 1 2 3 4 5

−250

−200

−150

−100

−50

50

17 Domain: (−∞, 1) ∪ (1,∞)

f(x) =x2 + 2x+ 2

x− 1

f ′(x) =x2 − 2x− 4

(x− 1)2

f ′′(x) =10

(x− 1)3

y-intercept: (0,−2)x-intercept(s): NoneSymmetry: NoneVertical asymptotes: x = 1Horizontal asymptotes: NoneIncreasing on: (−∞, 1−

√5), (1 +

√5,∞)

Decreasing on: (1−√

5, 1), (1, 1 +√

5)Local maxima: f(1−

√5) = 4− 2

√5

Local minima: f(1 +√

5) = 4 + 2√

5Concave up on: (1,∞)Concave down on: (−∞, 1)Inflection points: None

Page 33: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 3 33

x

y

−8 −6 −4 −2 2 4 6 8

−12

−9

−6

−3

3

6

9

12

15

18

18 Domain: (−∞,∞)

f(x) =2x

x2 + 2

f ′(x) = −2(x2 − 2

)(x2 + 2)

2

f ′′(x) =4(x2 − 6

)x

(x2 + 2)3

y-intercept: (0, 0)

x-intercept(s): (0, 0)

Symmetry: Odd

Vertical asymptotes: None

Horizontal asymptotes: y = 0 both as x→ ±∞Increasing on: (−

√2,√

2)

Decreasing on: (−∞,−√

2), (√

2,∞)

Local maxima: f(√

2) =√22

Local minima: f(−√

2) = −√22

Concave up on: (−√

6, 0), (√

6,∞)

Concave down on: (−∞,−√

6), (0,√

6)

Inflection points: (−√

6,−√64 ), (0, 0), (

√6,√64 )

x

y

−10−8−6−4−2 2 4 6 8 10

−1

1

19 Domain: (−∞,− 13√4

) ∪ (− 13√4,∞)

f(x) =x

4x3 + 1

f ′(x) = −(2x− 1)

(4x2 + 2x+ 1

)(4x3 + 1)

2

f ′′(x) =48(2x3 − 1

)x2

(4x3 + 1)3

y-intercept: (0, 0)x-intercept(s): (0, 0)Symmetry: NoneVertical asymptotes: x = − 1

3√4

Horizontal asymptotes: y = 0 both as x→ ±∞Increasing on: (−∞,− 1

3√4), (− 1

3√4, 12 )

Decreasing on: ( 12 ,∞)

Local maxima: f(1/2) = 1/3Local minima: NoneConcave up on: (−∞,− 1

3√4), ( 1

3√2,∞)

Concave down on: (− 13√4, 1

3√2)

Inflection points: ( 13√2, 13 3√2

)

x

y

−4 −3 −2 −1 1 2 3 4

−2

−1

1

2

20 Domain: (−∞,−5) ∪ (−5, 3) ∪ (3,∞)

f(x) =(x+ 1)

2

(x− 3)(x+ 5)

f ′(x) = − 32 (x+ 1)

(x− 3)2(x+ 5)

2

f ′′(x) =32(3x2 + 6x+ 19

)(x− 3)

3(x+ 5)

3

y-intercept: (0,− 115 )

x-intercept(s): (−1, 0)Symmetry: NoneVertical asymptotes: x = −5 and x = 3

Page 34: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

34 MATH 140 HOMEWORK PROBLEMS

Horizontal asymptotes: y = 1 both as x→ ±∞Increasing on: (−∞,−5), (−5,−1)

Decreasing on: (−1, 3), (3,∞)

Local maxima: f(−1) = 0

Local minima: None

Concave up on: (−∞,−5), (3,∞)

Concave down on: (−5, 3)

Inflection points: None

x

y

−10 −8 −6 −4 −2 2 4 6 8 10

−10

−8

−6

−4

−2

2

4

6

8

10

21 Domain: (−∞, 0) ∪ (0,∞)

f(x) =x4 + 1

x2

f ′(x) =2 (x− 1)(x+ 1)

(x2 + 1

)x3

f ′′(x) =2(x4 + 3

)x4

y-intercept: None

x-intercept(s): None

Symmetry: Even

Vertical asymptotes: x = 0

Horizontal asymptotes: None

Increasing on: (−1, 0), (1,∞)

Decreasing on: (−∞,−1), (0, 1)

Local maxima: None

Local minima: f(−1) = 2, f(1) = 2

Concave up on: (−∞, 0), (0,∞)

Concave down on: Nowhere

Inflection points: None

x

y

−3 −2 −1 1 2 3

1

3

5

7

9

11

22 Domain: (−∞,∞)

f(x) = −2x+ 3x( 23 )

f ′(x) = −2(x( 1

3 ) − 1)

x13

f ′′(x) = − 2

3x( 43 )

y-intercept: (0, 0)x-intercept(s): (0, 0), (27/8, 0)Symmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (0, 1)Decreasing on: (−∞, 0), (1,∞)Local maxima: f(1) = 1Local minima: f(0) = 0Concave up on: NowhereConcave down on: (−∞, 0), (0,∞)Inflection points: None

x

y

−2 −1 1 2 3 4 5

13579

11

23 Domain: (−∞,∞)

f(x) = (−x+ 8)(35 )x

f ′(x) = − 8 (x− 5)

5 (−x+ 8)(25 )

f ′′(x) = − 24 (x− 10)

25 (−x+ 8)(25 )(x− 8)

y-intercept: (0, 0)x-intercept(s): (0, 0), (8, 0)

Page 35: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 3 35

Symmetry: None

Vertical asymptotes: None

Horizontal asymptotes: None

Increasing on: (−∞, 5)

Decreasing on: (5,∞)

Local maxima: f(5) = 5 · 33/5Local minima: None

Concave up on: (8, 10)

Concave down on: (−∞, 8), (10,∞)

Inflection points: (8, 0), (10,−10 · 23/5)

x

y

2 4 6 8 10

−15

−12

−9

−6

−3

3

6

9

12

24 Domain: (−∞,∞)

f(x) = (x− 3)2(x+ 1)(

23 )

f ′(x) =8 (x− 3)x

3 (x+ 1)(13 )

f ′′(x) =8(5x2 − 9

)9 (x+ 1)(

43 )

y-intercept: (0, 9)

x-intercept(s): (−1, 0) and (3, 0)

Symmetry: None

Vertical asymptotes: None

Horizontal asymptotes: None

Increasing on: (−1, 0), (3,∞)

Decreasing on: (−∞,−1), (0, 3)

Local maxima: f(0) = 9

Local minima: f(−1) = 0, f(3) = 0

Concave up on: (−∞,− 3√5

5 ), ( 3√5

5 ,∞)

Concave down on: (− 3√5

5 ,−1), (−1, 3√5

5 )

Inflection points: (− 3√5

5 , ( 5−3√5

5 )2/3 725 ),

( 3√5

5 , ( 5+3√5

5 )2/3 365 )

x

y

−2 −1 1 2 3 4 5

1

3

5

7

9

11

25 Domain: (−∞,∞)

f(x) = −x+√x2 − 2x+ 2

f ′(x) =x−√x2 − 2x+ 2− 1√x2 − 2x+ 2

f ′′(x) =1

(x2 − 2x+ 2)32

y-intercept: (0,√

2)x-intercept(s): (1, 0)Symmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: y = −1 as x→∞Increasing on: NowhereDecreasing on: (−∞,∞)Local maxima: NoneLocal minima: NoneConcave up on: (−∞,∞)Concave down on: NowhereInflection points: None

x

y

−2 −1 1 2 3 4 5−1

1

3

5

26 Domain: (−∞,∞)

f(x) =(x2 + 5x+ 4

) 45

f ′(x) =4 (2x+ 5)

5 (x2 + 5x+ 4)(15 )

f ′′(x) =12(2x2 + 10x+ 5

)25 (x2 + 5x+ 4)(

65 )

Page 36: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

36 MATH 140 HOMEWORK PROBLEMS

y-intercept: (0, 44/5)

x-intercept(s): (−4, 0), (−1, 0)

Symmetry: None

Vertical asymptotes: None

Horizontal asymptotes: None

Increasing on: (−4,−5/2), (−1,∞)

Decreasing on: (−∞,−4), (−5/2,−1)

Local maxima: f(−5/2) = (9/4)4/5

Local minima: f(−4) = 0, f(−1) = 0

Concave up on: (−∞, −5−√15

2 ), (−5+√15

2 ,∞)

Concave down on: (−5−√15

2 ,−4), (−4,−1),

(−1, −5+√15

2 )

Inflection points: (−5−√15

2 , (3/2)4/5), (−5+√15

2 , (3/2)4/5)

x

y

−6 −5 −4 −3 −2 −1 1

2

4

6

27 Domain: (−∞,∞)

f(x) =x+ 4√x2 + 8

f ′(x) = − 4 (x− 2)

(x2 + 8)32

f ′′(x) =8 (x− 4)(x+ 1)

(x2 + 8)52

y-intercept: (0,√

2)

x-intercept(s): (−4, 0)

Symmetry: None

Vertical asymptotes: None

Horizontal asymptotes: y = 1 as x → ∞, y = −1 asx→ −∞Increasing on: (−∞, 2)

Decreasing on: (2,∞)

Local maxima: f(2) =√

3

Local minima: None

Concave up on: (−∞,−1), (4,∞)

Concave down on: (−1, 4)

Inflection points: (−1, 1), (4, 2√6

3 )

x

y

−20 −15 −10 −5 5 10 15 20

−1

1

2

28 Domain: (−∞,−1) ∪ (1,∞)

f(x) =x− 1√x2 − 1

f ′(x) =1

(x+ 1)√x2 − 1

f ′′(x) = − 2x− 1

(x− 1)(x+ 1)2√x2 − 1

y-intercept: Nonex-intercept(s): NoneSymmetry: NoneVertical asymptotes: x = −1Horizontal asymptotes: y = 1 as x → ∞, y = −1 asx→ −∞Increasing on: (1,∞)Decreasing on: (−∞,−1)Local maxima: NoneLocal minima: NoneConcave up on: NowhereConcave down on: (−∞,−1), (1,∞)Inflection points: None

x

y

−5 −3 −1 1 3 5

−3

−1

1

29 Domain: [−π, π]

f(x) =√

3 sin (x) + cos (x)

f ′(x) =√

3 cos (x)− sin (x)

f ′′(x) = −√

3 sin (x)− cos (x)

y-intercept: (0, 1)x-intercept(s): (−π6 , 0), ( 5π

6 , 0)Symmetry: None

Page 37: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 3 37

Vertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (− 2π

3 ,π3 )

Decreasing on: [−π,− 2π3 ), (π3 , π]

Local maxima: f(−π) = −1, f(π/3) = 2Local minima: f(−2π/3) = −2, f(π) = −1Concave up on: [−π,−π6 ), ( 5π

6 , π]Concave down on: (−π6 ,

5π6 )

Inflection points: (−π6 , 0), ( 5π6 , 0)

x

y

−π− 2π3−π3

π3

2π3

π

−2

−1

1

2

30 Domain: [0, 4π]

f(x) = x+ 2 cos (x)

f ′(x) = −2 sin (x) + 1

f ′′(x) = −2 cos (x)

y-intercept: (0, 2)x-intercept(s): NoneSymmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: [0, π/6), (5π/6, 13π/6), (17π, 6, 4π]Decreasing on: (π/6, 5π/6), (13π/6, 17π/6)Local maxima: f(π/6) = π/6 +

√3, f(13π/6) =

13π/6 +√

3, f(4π) = 4π + 2Local minima: f(0) = 2, f(5π/6) = 5π/6 −

√3,

f(17π/6) = 17π/6−√

3Concave up on: (π/2, 3π/2), (5π/2, 7π/2)Concave down on: [0, π/2), (3π/2, 5π/2), (7π/2, 4π]Inflection points: (π/2, π/2), (3π/2, 3π/2),(5π/2, 5π/2), (7π/2, 7π/2)

x

y

π3

2π3

π 4π3

5π3

2π 7π3

8π3

3π 10π3

11π3

2

4

6

8

10

12

14

31 Domain: (−π2 ,π2 )

f(x) = 2x− tan (x)

f ′(x) = −(tan (x)− 1)(tan (x) + 1)

f ′′(x) = −2 sec (x)2

tan (x)

y-intercept: (0, 0)x-intercept(s): (0, 0), (±1.166, 0)Symmetry: NoneVertical asymptotes: x = ±π2Horizontal asymptotes: NoneIncreasing on: (−π4 ,

π4 )

Decreasing on: (−π2 ,−π4 ), (π4 ,

π2 )

Local maxima: f(π/4) = π/2− 1Local minima: f(−π/4) = 1− π/2Concave up on: (−π2 , 0)Concave down on: (0, π2 )Inflection points: (0, 0)

x

y

−π2 −π4π4

π2

−6

−4

−2

2

4

6

32 Domain: [0, 2π]

f(x) = (cos (x) + 2) cos (x)

f ′(x) = −2 (cos (x) + 1) sin (x)

f ′′(x) = 2 sin (x)2 − 2 cos (x)

2 − 2 cos (x)

y-intercept: (0, 3)x-intercept(s): (π/2, 0), (3π/2, 0)Symmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (π, 2π]Decreasing on: [0, π)Local maxima: f(0) = 3, f(2π) = 3Local minima: f(π) = −1

Page 38: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

38 MATH 140 HOMEWORK PROBLEMS

Concave up on: (π/3, 5π/3)

Concave down on: [0, π/3), (5π3, 2π)Inflection points: (π/3, 5/4), (5π/3, 5/4)

x

y

π3

2π3

π 4π3

5π3

2π−1

1

2

3

33 Domain: [−π, π]

f(x) = −(

4 sin (x)2 − 3

)sin (x)

f ′(x) = −3 (2 sin (x)− 1)(2 sin (x) + 1)

· cos (x)

f ′′(x) = 3(

4 sin (x)2 − 8 cos (x)

2 − 1)

· sin (x)

y-intercept: (0, 0)

x-intercept(s): (−π, 0), (−2π/3, 0), (−π/3, 0), (0, 0),(π/3, 0), (2π/3, 0), (π, 0)

Symmetry: Odd

Vertical asymptotes: None

Horizontal asymptotes: None

Increasing on: (−5π/6,−π/2), (−π/6, π/6),(π/2, 5π/6)

Decreasing on: [−π,−5π/6), (−π/2,−π/6),(π/6, π/2), (5π/6, π]

Local maxima: f(−π) = 0, f(−π/2) = f(π/6) =f(5π/6) = 1

Local minima: f(−5π/6) = f(−π/6) = f(π/2) =−1, f(π) = 0

Concave up on: [−π,−2π/3), (−π/3, 0), (π/3, 2π/3)

Concave down on: (−2π/3,−π/3), (0, π/3), (2π/3, π]

Inflection points: (−2π/3, 0), (−π/3, 0), (0, 0),(π/3, 0), (2π/3, 0)

x

y

−π − 2π3−π3

π3

2π3

π

−1

1

Note: In fact, f(x) = sin(3x), which – if recognized atthe beginning – substantially simplifies the graphingdiscussion for this function.

34 Domain: [0, 2π]

f(x) =cos (x)

sin (x) + 2

f ′(x) = − 2 sin (x) + 1

(sin (x) + 2)2

f ′′(x) =2 (sin (x)− 1) cos (x)

(sin (x) + 2)3

y-intercept: (0, 1/2)x-intercept(s): (π/2, 0), (3π/2, 0)Symmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (7π/6, 11π/6)Decreasing on: [0, 7π/6), (11π/6, 2π]Local maxima: f(0) = 1/2, f(11π/6) =

√3/3

Local minima: f(7π/6) = −√

3/3, f(2π) = 1/2Concave up on: (π/2, 3π/2)Concave down on: [0, π/2), (3π/2, 2π]Inflection points: (π/2, 0), (3π/2, 0)

x

y

π3

2π3

π 4π3

5π3

−1

1

35 Domain: (0, π)

f(x) = − sin (x) cos (x) + x+ 2 cos (x)

f ′(x) = sin (x)2 − cos (x)

2

− 2 sin (x) + 1

f ′′(x) = 4 sin (x) cos (x)− 2 cos (x)

y-intercept: Nonex-intercept(s): NoneSymmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: NowhereDecreasing on: (0, π)Local maxima: NoneLocal minima: NoneConcave up on: (π/6, π/2), (5π/6, π)Concave down on: (0, π/6), (π/2, 5π/6)

Inflection points: (π/6, π/6 + 3√3

4 ), (π/2, π/2),

(5π/6, 5π/6− 3√3

4 )

Page 39: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 3 39

x

y

π3

2π3

π

1

2

36 Domain: (0,∞)

f(x) = x2 − 2 ln (x)

f ′(x) = 2x− 2

x

f ′′(x) =2

x2+ 2

y-intercept: Nonex-intercept(s): NoneSymmetry: NoneVertical asymptotes: x = 0Horizontal asymptotes: NoneIncreasing on: (1,∞)Decreasing on: (0, 1)Local maxima: NoneLocal minima: f(1) = 1Concave up on: (0,∞)Concave down on: NowhereInflection points: None

x

y

1 2 3

1

3

5

7

37 Domain: (−3, 1)

f(x) = ln(−x2 − 2x+ 3

)f ′(x) =

2 (x+ 1)

(x− 1)(x+ 3)

f ′′(x) = −2(x2 + 2x+ 5

)(x− 1)

2(x+ 3)

2

y-intercept: (0, ln(3))x-intercept(s): (−1−

√3, 0), (

√3− 1, 0)

Symmetry: None

Vertical asymptotes: x = −3, x = 1

Horizontal asymptotes: None

Increasing on: (−3,−1)

Decreasing on: (−1, 1)

Local maxima: f(−1) = ln(4)

Local minima: None

Concave up on: Nowhere

Concave down on: (−3, 1)

Inflection points: None

x

y

−3 −2 −1 1

−6

−5

−4

−3

−2

−1

1

2

38 Domain: (−∞,∞)

f(x) = ln(x4 + 3

)f ′(x) =

4x3

x4 + 3

f ′′(x) = −4(x2 − 3

)(x2 + 3

)x2

(x4 + 3)2

y-intercept: (0, ln(3))

x-intercept(s): None

Symmetry: Even

Vertical asymptotes: None

Horizontal asymptotes: None

Increasing on: (0,∞)

Decreasing on: (−∞, 0)

Local maxima: None

Local minima: f(0) = ln(3)

Concave up on: (−√

3,√

3)

Concave down on: (−∞,−√

3), (√

3,∞)

Inflection points: (−√

3, ln(12)), (√

3, ln(12))

Page 40: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

40 MATH 140 HOMEWORK PROBLEMS

x

y

−3 −2 −1 1 2 3

1

2

3

4

39 Domain: (−∞,∞)

f(x) = ln(x+

√x2 + 1

)f ′(x) =

1√x2 + 1

f ′′(x) = − x

(x2 + 1)32

y-intercept: (0, 0)x-intercept(s): (0, 0)Symmetry: OddVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (−∞,∞)Decreasing on: NowhereLocal maxima: NoneLocal minima: NoneConcave up on: (−∞, 0)Concave down on: (0,∞)Inflection points: (0, 0)

x

y

−3 −2 −1 1 2 3

−2

−1

1

2

40 Domain: (0,∞)

f(x) =ln (x)

x2

f ′(x) = −2 ln (x)− 1

x3

f ′′(x) =6 ln (x)− 5

x4

y-intercept: Nonex-intercept(s): (1, 0)

Symmetry: None

Vertical asymptotes: x = 0

Horizontal asymptotes: y = 0 as x→∞Increasing on: (0,

√e)

Decreasing on: (√e,∞)

Local maxima: f(√e) = 1/(2e)

Local minima: None

Concave up on: (e5/6,∞)

Concave down on: (0, e5/6)

Inflection points: (e5/6, 56e−5/3)

x

y

1 2 3

−11

−9

−7

−5

−3

−1

1

41 Domain: (0,∞)

f(x) =x

ln (x)

f ′(x) =ln (x)− 1

ln (x)2

f ′′(x) = − ln (x)− 2

x ln (x)3

y-intercept: None

x-intercept(s): None

Symmetry: None

Vertical asymptotes: x = 1

Horizontal asymptotes: None

Increasing on: (e,∞)

Decreasing on: (0, 1), (1, e)

Local maxima: None

Local minima: f(e) = e

Concave up on: (1, e2)

Concave down on: (0, 1), (e2,∞)

Inflection points: (e2, e2

2 )

Page 41: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 3 41

x

y

2 4 6 8 10 12 14

−11

−9

−7

−5

−3

−1

1

3

5

7

9

11

42 Domain: (−∞,−1) ∪ (0,∞)

f(x) = ln

(x+ 1

x

)f ′(x) = − 1

(x+ 1)x

f ′′(x) =2x+ 1

(x+ 1)2x2

y-intercept: Nonex-intercept(s): NoneSymmetry: NoneVertical asymptotes: x = −1, x = 0Horizontal asymptotes: y = 0 both as x→ ±∞Increasing on: NowhereDecreasing on: (−∞,−1), (0,∞)Local maxima: NoneLocal minima: NoneConcave up on: (0,∞)Concave down on: (−∞,−1)Inflection points: None

x

y

−3 −2 −1 1 2 3

−3

−2

−1

1

2

3

43 Domain: (0, e)

f(x) = ln (− ln (x) + 1)

f ′(x) =1

(ln (x)− 1)x

f ′′(x) = − ln (x)

(ln (x)− 1)2x2

y-intercept: None

x-intercept(s): (1, 0)

Symmetry: None

Vertical asymptotes: x = 0, x = e

Horizontal asymptotes: None

Increasing on: Nowhere

Decreasing on: (0, e)

Local maxima: None

Local minima: None

Concave up on: (0, 1)

Concave down on: (1, e)

Inflection points: (1, 0)

x

y

1 2 3

−5−4−3−2−1

12

44 Domain: (−∞,∞)

f(x) = xe(−x)

f ′(x) = −(x− 1)e(−x)

f ′′(x) = (x− 2)e(−x)

y-intercept: (0, 0)

x-intercept(s): (0, 0)

Symmetry: None

Vertical asymptotes: None

Horizontal asymptotes: y = 0 as x→∞Increasing on: (−∞, 1)

Decreasing on: (1,∞)

Local maxima: f(1) = 1/e

Local minima: None

Concave up on: (2,∞)

Concave down on: (−∞, 2)

Inflection points: (2, 2/e2)

Page 42: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

42 MATH 140 HOMEWORK PROBLEMS

x

y

−1 1 2 3 4

−3

−2

−1

1

45 Domain: (−∞,∞)

f(x) =(2x2 + x+ 1

)ex

f ′(x) = (x+ 2)(2x+ 1)ex

f ′′(x) = (x+ 1)(2x+ 7)ex

y-intercept: (0, 1)x-intercept(s): NoneSymmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: y = 0 as x→ −∞Increasing on: (−∞,−2), (−1/2,∞)Decreasing on: (−2,−1/2)Local maxima: f(−2) = 7e−2

Local minima: f(−1/2) = e−1/2

Concave up on: (−∞,−7/2), (−1,∞)Concave down on: (−7/2,−1)Inflection points: (−7/2, 22e−7/2), (−1, 2/e)

x

y

−6 −5 −4 −3 −2 −1 1

1

3

5

7

9

11

46 Domain: (−∞,∞)

f(x) = e(−x2)

f ′(x) = −2xe(−x2)

f ′′(x) = 2(2x2 − 1

)e(−x

2)

y-intercept: (0, 1)x-intercept(s): None

Symmetry: Even

Vertical asymptotes: None

Horizontal asymptotes: y = 0 both as x→ ±∞Increasing on: (−∞, 0)

Decreasing on: (0,∞)

Local maxima: f(0) = 1

Local minima: None

Concave up on: (−∞,−1/√

2), (1/√

2,∞)

Concave down on: (−1/√

2, 1/√

2)

Inflection points: (−1/√

2, e−1/2), (1/√

2, e−1/2)

x

y

−2 −1 1 2

1

47 Domain: (−∞, 0) ∪ (0,∞)

f(x) = − e(2 x)

ex − 1

f ′(x) = − (ex − 2)e(2 x)

(ex − 1)2

f ′′(x) = −(e(2 x) − 3 ex + 4

)e(2 x)

(ex − 1)3

y-intercept: None

x-intercept(s): None

Symmetry: None

Vertical asymptotes: x = 0

Horizontal asymptotes: y = 0 as x→ −∞Increasing on: (−∞, 0), (0, ln(2))

Decreasing on: (ln(2),∞)

Local maxima: f(ln(2)) = −4

Local minima: None

Concave up on: (−∞, 0)

Concave down on: (0,∞)

Inflection points: None

Page 43: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

ANSWERS SECTION 3 43

x

y

−2 −1 1 2

−9

−7

−5

−3

−1

1

3

5

7

48 Domain: [−π, π]

f(x) = ex sin (x)

f ′(x) = (sin (x) + cos (x))ex

f ′′(x) = 2 ex cos (x)

y-intercept: (0, 0)x-intercept(s): (−π, 0), (0, 0), (π, 0)Symmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (−π/4, 3π/4)Decreasing on: [−π,−π/4), (3π/4, π]

Local maxima: f(−π) = 0, f(3π/4) =√22 e

3π/4

Local minima: f(π) = 0Concave up on: (−π/2, π/2)Concave down on: [−π,−π/2), (π/2, π]Inflection points: (−π/2,−e−π/2), (π/2, eπ/2)

x

y

−π− 3π4−π2−

π4

π4

π2

3π4

π−1

1

2

3

4

5

6

7

8

49 Domain: (−∞, 0) ∪ (0,∞)

f(x) = e1x

f ′(x) = −e1x

x2

f ′′(x) =(2x+ 1)e

1x

x4

y-intercept: Nonex-intercept(s): NoneSymmetry: NoneVertical asymptotes: x = 0Horizontal asymptotes: y = 1 both as x→ ±∞Increasing on: NowhereDecreasing on: (−∞, 0), (0,∞)Local maxima: NoneLocal minima: NoneConcave up on: (−1/2, 0), (0,∞)Concave down on: (−∞,−1/2)Inflection points: (−1/2, e−2)

x

y

−5 −3 −1 1 3 5

2

4

6

8

50 Domain: (−∞,∞)

f(x) = x arctan (x)

f ′(x) =x2 arctan (x) + x+ arctan (x)

x2 + 1

f ′′(x) =2

(x2 + 1)2

y-intercept: (0, 0)x-intercept(s): (0, 0)Symmetry: EvenVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (0,∞)Decreasing on: (−∞, 0)Local maxima: None

Page 44: MATH 140 Homework Problems - Pennsylvania …4 MATH 140 HOMEWORK PROBLEMS 30.lim x!3 2 x2+x 1 3x 3 x2 + 4x 21 31.lim x! 4 1 x+2 1 p3x+10 x2 3 17 + x 32.lim x!0 ex+1(x3 + x2) x4 + 3x2

44 MATH 140 HOMEWORK PROBLEMS

Local minima: f(0) = 0Concave up on: (−∞,∞)Concave down on: NowhereInflection points: None

x

y

−3 −2 −1 1 2 3

1

2

3

4

51 Domain: (−∞,∞)

f(x) = earctan(x)

f ′(x) =earctan(x)

x2 + 1

f ′′(x) = − (2x− 1)earctan(x)

(x2 + 1)2

y-intercept: (0, 1)x-intercept(s): NoneSymmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: y = e−π/2 as x → −∞,y = eπ/2 as x→∞Increasing on: (−∞,∞)Decreasing on: NowhereLocal maxima: NoneLocal minima: NoneConcave up on: (−∞, 1/2)Concave down on: (1/2,∞)Inflection points: (1/2, earctan(1/2))

x

y

−3 −1 1 3 5 7 9

1

2

3

4

5

52 Domain: (0,∞)

f(x) = xx

f ′(x) = (ln (x) + 1)xx

f ′′(x) =(x ln (x)

2+ 2x ln (x) + x+ 1

)x(x−1)

y-intercept: Nonex-intercept(s): NoneSymmetry: NoneVertical asymptotes: NoneHorizontal asymptotes: NoneIncreasing on: (1/e,∞)Decreasing on: (0, 1/e)Local maxima: NoneLocal minima: f(1/e) = (1/e)1/e

Concave up on: (0,∞)Concave down on: NowhereInflection points: None

x

y

1 2

1

2

3

4

5