exercises for first semester calculus v1

Exercises for First Semester Calculus v1.0

Craig Timmons

March 27, 2018

1

Copyright © Craig Timmons

Except where otherwise noted, this work is licensedunder http://creativecommons.org/licenses/by-nc-nd/3.0/

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Preface

This is a collection of exercises for a student in a first semester calculus course. Solving problems is oneof the best ways to learn calculus. A student who wants to demonstrate an understanding of calculusshould make it a goal to solve as many problems as possible. The intention of this work is to providestudents with a resource that will help them achieve this goal.

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Contents

0 Introductory Problems 6

1 Limits 12

1.1 Introduction to Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2 Limit Laws and the Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.3 Left and Right Hand Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.5 Limits that Involve Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2 Derivatives 32

2.1 Instantaneous Rate of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.2 Interpreting the Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.3 Basic Derivative Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.4 The Product Rule and the Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2.7 Logarithmic Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

2.8 Derivatives of Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.9 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

3 Derivatives and the Shapes of Graphs 60

3.1 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.2 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.3 Increasing and Decreasing Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

3.4 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

3.5 l’Hospital’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

3.6 Sketching Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3.7 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4 Antiderivatives 84

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4.1 Definite Integrals and Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.2 Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.3 The Fundamental Theorems of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

4.4 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

4.5 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

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0 Introductory Problems

1. Find the slope of the line that passes through the points (−1, 3) and (0, 5).

2. Find the equation of the line that passes through the points (1, 2) and (−3, 4).

3. Determine the x and y-intercepts of the line whose equation is y = 3x+ 5.

4. Evaluate 2−3 · (1/3)2.

5. Simplify the expressionx−3y2

xy5as much as possible. Write your solution using positive exponents

only.

6. Simplify the expression

(a4b−2

c5a

)−3as much as possible. Write your solution using positive exponents

only.

7. True or False: (a+ b)2 = a2 + b2 for every pair of real numbers a and b.

8. True or False: (a+ b)3 = a3 + b3 for every pair of real numbers a and b.

9. Evaluate log10(100).

10. Assume that x, y, and z are positive real numbers. Expand ln

(x2y3

2z

)as much as possible.

11. Assume that x, y, and z are positive real numbers. Expand log10

(x2y3

2z

)as much as possible.

12. Evaluate ln(e3).

13. True or False: ln(0) = 0.

14. True or False: ln(1) = 1.

15. Simplifyx2 + 5x+ 6

x2 − 9as much as possible.

16. Simplifyx3 − 1

x2 − xas much as possible.

17. Find all solutions to x3 + 5x2 + 6x = 0.

18. Find all solutions to x3 = 4x.

19. Find all solutions to x2 − x− 1 = 0.

20. Find all solutions tox2 + 5x+ 6

x2 − 9= 0.

21. Find all solutions tox3 − 4x

x2 + 1= 0.

22. Solve e2x+1 = e3x−4.

23. Solve 23x+1 = 25.

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24. Solve ln(2x) + ln(3) = ln(x+ 1).

25. Solve 0 = e−x3 − x2

16 e−x3

.

26. Evaluate cos(π/3) and sin(π/3).

27. Evaluate tan(π/3) and cot(π/3).

28. Evaluate cos(−20π + π/3).

29. Find all values of x in the interval [0, π] for which sinx− cosx = 0.

30. Find all values of x in the interval [2π, 3π] for which sinx+ cosx = 0.

31. Find all values of x for which tanx = 1.

32. Find the domain of the function f(x) =√x− 2.

33. Find the domain of the function f(x) = 3√x− 2.

34. Find the domain of the function f(x) =x2√x− 2

.

35. Find the domain of the function f(x) =x2

3√x− 2

.

36. Find the domain of the functions f(x) = lnx and g(x) = ex.

37. Find the domain of the function y =x2 + 5x+ 6

x2 − 9.

38. Find the domain of the function y =x2 + 1

sinx.

39. If f(x) = x2 + 2x, find f(3).

40. Suppose that (1, 5) is a point that is on the graph of y = f(x). Find f(1).

41. If f(x) = x2 and g(x) = x+ 3, find (f ◦ g)(x) and (g ◦ f)(x).

42. If f(x) = x2 + 4 and g(x) = ex + x4, find (fg)(x) and (f/g)(x).

43. Sketch the graph of y = x3.

44. Sketch the graph of y = sinx.

45. Sketch the graph of the function f(x) = lnx.

46. Sketch the graph of y = ex.

47. Use function transformation rules to sketch the graph of y = (x− 1)3 + 1.

48. Sketch the graph of the function g(x) = 2ex+1.

49. Sketch the graph of the function h(x) =√x− 4− 2.

50. Sketch the graph of y = sinx+ 4.

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51. Sketch the graph of y = |x|.

52. Let f(x) =

{2x− 1 x < 1x2 + 1 x ≥ 1.

Find f(−4), f(3), and f(1).

53. Sketch the graph of f(x) =

{x+ 1 x < 0√x x ≥ 0.

54. Solve the inequality x2 + 5x+ 6 > 0.

55. Solve the inequality (x+ 1)(x− 2)(x− 4) < 0.

Introductory Problems - Solutions

1. 2

2. y = − 12x+ 5

2

3. (0, 5) and (− 53 , 0)

4.1

72

5.1

x4y3

6.b6c15

a9

7. False

8. False

9. 2

10. 2 ln(x) + 3 ln(y)− ln(2)− ln(z)

11. 2 log10(x) + 3 log10(y)− log10(2)− log10(z)

12. 3

13. The statement ln(0) = 0 is false since ln(0) isundefined.

14. The statement ln(1) = 1 is false since ln(1) = 0.

15.x+ 2

x− 3provided x 6= −3

16.x2 + x+ 1

xprovided x 6= 1

17. 0, −2, and −3

18. 0, 2, and −2

19.1±√

5

2

20. −2

21. 0, 2, and −2

22. 5

23.log2(25)− 1

3

24.1

5

25. 4 and −4

26. cos(π/3) = 12 , sin(π/3) =

√32

27. tan(π/3) =√

3, cot(π/3) = 1√3

28.1

2

29.π

4

30.11π

4

31.π

4+ πn where n is any integer

32. [2,∞)

33. (−∞,∞)

34. (2,∞)

35. (−∞, 2) ∪ (2,∞)

36. The domain of f(x) = lnx is (0,∞). The do-main of g(x) = ex is (−∞,∞).

8

37. (−∞,−3), (−3, 3), and (3,∞)

38. All real numbers except for nπ where n is anyinteger

39. 15

40. f(1) = 5

41. (f ◦ g)(x) = (x+ 3)2, (g ◦ f)(x) = x2 + 3

42. (fg)(x) = (x2+4)(ex+x4), (f/g)(x) =x2 + 4

ex + x4

43.

−2 −1 1 2

−5

5

y

x

44.

−4 −2 2 4

−2

−1

1

2

y

x

45.

1 2 3 4 5

−2

−1

1

2

y

x

46.

−3 −2 −1 1 2 3

2

4

6

8

10

y

x

9

47.

−2 −1 1 2 3

−5

5

y

x

48.

−3 −2 −1 1 2 3

5

10

15

20

y

x

49.

4 6 8 10 12

−2

−1

1

2

y

x

50.

−4 −2 2 4

2

4

6

y

x

51.

−4 −2 2 4

2

4

y

x

52. f(−4) = −9, f(3) = 10, and f(1) = 2

10

53.

−4 −2 2 4

−4

−2

2

4

y

x

54. (−∞,−3) and (−2,∞)

55. (−∞,−1) and (2, 4)

11

1 Limits

1.1 Introduction to Limits

1. Let f(t) be the amount of oil, in gallons, that has leaked out of a tank t seconds after the tank waspunctured.

(a) Interpret the statement f(3) = 18 in practical terms.

(b) What does the quantity f(5)− f(3) represent?

(c) If f(5) = 50 and f(3) = 18, find the average rate of change of how fast the oil is leaking out of thetank from time t = 3 to time t = 5.

2. Let s(t) be the displacement, in meters, of a sprinter running the 100 meter dash. Here t is thenumber of seconds since the start of the race.

(a) What doess(2)− s(1)

2− 1represent in practical terms?

(b) If h > 0, then what doess(1 + h)− s(1)

1 + h− 1represent in practical terms?

3. Let a(t) be the amount of water, in gallons, that has flown out of a flood gate t seconds after the gatewas opened.

(a) What doesa(120)− a(60)

120− 60represent in practical terms?

(b) If h > 0, then what doesa(60 + h)− a(60)

hrepresent in practical terms?

4. Fill in the blank: If f is a function whose domain is all real numbers and h 6= 0, then

f(a+ h)− f(a)

h

is the slope of the line that passes through points

(a+ h, f(a+ h)) and .

5. Fill in the missing part of the informal definition of the limit of a function. Let f be a function definedon an open interval that contains c, except f need not be defined at c. We write

limx→c

f(x) = L

if we can make f(x) as close to L as we like by choosing x but notequal to c.

6. Let f(x) = 4x+ 1.

(a) Simplifyf(1 + h)− f(1)

has much as possible.

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(b) Estimate the limit

limh→0

f(1 + h)− f(1)

h

by computingf(1 + h)− f(1)

hfor the h values −0.1, −0.01, −0.001, 0.001, 0.01, and 0.1.

7. Let f(x) = x2.

(a) Simplifyf(1 + h)− f(1)

has much as possible.

(b) Estimate the limit

limh→0

f(1 + h)− f(1)

h

by computingf(1 + h)− f(1)

hfor the h values −0.1, −0.01, −0.001, 0.001, 0.01, and 0.1.

8. Describe two situations where limx→0

f(x) does not exist. For each situation, sketch the graph of a

function that illustrates the situation that you are describing.

9. Estimate limx→0

ex by computing ex for the x values −0.1, −0.01, −0.001, 0.001, 0.01, and 0.1.

10. Estimate limx→1

lnx by computing lnx for the x values 0.9, 0.99, 0.999, 1.001, 1.01, and 1.1.

11. Let g(x) =

{x2 x 6= 0,2 x = 0.

Determine limx→0

g(x) by sketching the graph of g, and then examining the

values of g(x) for x near 0.

12. Let

s(t) =

{2t+ 1 t < 3,−t+ 10 t > 3.

(a) Estimate limt→3

s(t) by computing s(t) for the t values 2.9, 2.99, 2.999, 3.001, 3.01, and 3.1.

(b) Does s(3) exist?

Introduction to Limits - Solutions

1. The statement f(3) = 18 means that after 3 seconds, a total of 18 gallons has leaked out of the tank.

(b) The quantity f(5)− f(3) is the amount of oil, in gallons, that has leaked out of the tank from timet = 3 to time t = 5.

(c) The average rate of change of how fast the oil is leaking out from time t = 3 to time t = 5 is

f(5)− f(3)

5− 3=

50− 18

5− 3= 16

gallons per second.

2. (a) The quotients(2)− s(1)

2− 1is the average velocity of the runner, in meters per second, from time

t = 1 to time t = 2.

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(b) The quotients(1 + h)− s(1)

1 + h− 1is the average velocity of the runner, in meters per second, from time

t = 1 to time t = 1 + h.

3. (a) The quotienta(120)− a(60)

120− 60

is the average speed that the water is flowing out of the gate, in gallons per second, during the timeperiod from the one minute mark to the two minute mark.

(b) The quotienta(60 + h)− a(60)

h

is the average speed that the water is flowing out of the gate, in gallons per second, during the timeperiod from the 60 second mark to the 60 + h second mark.

4. The quotientf(a+ h)− f(a)

his the slope of the line that passes through the points (a, f(a)) and

(a+ h, f(a+ h)).

5. Let f be a function defined on an open interval that contains c, except f need not be defined at c.We write

limx→c

f(x) = L

if we can make f(x) as close to L as we like by choosing x sufficiently close to c but not equal to c.

6. (a)f(1 + h)− f(1)

h=

(4(1 + h) + 1)− (4(1) + 1)

h=

4h

h= 4.

(b) From part (a), we see thatf(1 + h)− f(1)

h= 4 for any value of h with h 6= 0. Therefore,

limh→0

f(1 + h)− f(1)

h= 4.

7. (a)f(1 + h)− f(1)

h=

(1 + h)2 − 12

h=

1 + 2h+ h2 − 1

h=

2h+ h2

h=h(2 + h)

h= 2 + h.

(b) The table below shows the values off(1 + h)− f(1)

hfor different values of h.

h −0.1 −0.01 −0.001 0.001 0.01 0.1f(1 + h)− f(1)

h1.9 1.99 1.999 2.001 2.01 2.1

Based on the table, it makes sense to say that limh→0

f(1 + h)− f(1)

h= 2.

8. One way for limx→0

f(x) to not exist is if f has a vertical asymptote at x = 0. For example, if f(x) =1

x,

then limx→0

f(x) does not exist. The graph of f(x) =1

xis shown below.

14

−3 −2 −1 1 2 3

−2

2f(x) =

1

x

y

x

Another way that limx→0

f(x) may fail to exist is if f(x) is close to 0 when x is a small negative number,

while f(x) is close to 1 when x is a small positive number. An example of such a function f is shownbelow.

−3 −2 −1 1 2 3

−2

2

y

x

9. We will make a table of values of ex. The entries in the second row of the table are rounded to fourdecimal places.

x −0.1 −0.01 −0.001 0.001 0.01 0.1ex 0.9048 0.9901 0.999 1.001 1.01 1.1052

From these values, it makes sense to say that limx→0

ex = 1.

10. We will make a table of values of lnx. The entries in the second row of the table are rounded tofour decimal places.

x 0.9 0.99 0.999 1.001 1.01 1.1lnx −0.1054 −0.0101 −0.0010 0.0010 0.0100 0.0953

From these values, it makes sense to say that limx→1

lnx = 0.

11. The graph of g is shown below.

15

−3 −2 −1 1 2 3

2

4

6

8

10

g(x)

y

x

When x is close to 0 but not equal to 0, we see that g(x) is close to 0 so limx→0

g(x) = 0.

12. (a) The table below shows the values of s(t) for certain values of t that are close to 3.

t 2.9 2.99 2.999 3.001 3.01 3.1s(t) 6.8 6.98 6.998 6.999 6.99 6.9

Based on the table, it makes sense to say that limt→3

s(t) = 7.

(b) The number 3 is not in the domain of s(t) so s(3) does not exist.

1.2 Limit Laws and the Squeeze Theorem

1. Suppose that there are real numbers L and K such that limx→c

f(x) = L and limx→c

g(x) = K. Express

each of the following limits in terms of L and K.

(a) limx→c

(f(x) + g(x)) (b) limx→c

(f(x)− g(x)) (c) limx→c

(f(x)g(x)) (d) limx→c

5f(x)

2. Suppose that f(x) and g(x) are functions with

limx→2

f(x) = 5, limx→2

g(x) = 0, limx→4

f(x) = −2, and limx→4

g(x) = 4.

Evaluate each of the following limits if possible. If the limit in question does not exist, explain why.

(a) limx→2

8f(x) + g(x)

g(x) + 1(b) lim

x→2

8g(x) + f(x)

g(x)

(c) limx→4

(√g(x)f(x)2 + g(x)2

)(d) lim

x→4

(3√g(x)f(x) + 10

)3. Suppose f(x) is a function with lim

x→1f(x) = π. Evaluate lim

x→1

(2f(x)2 − f(x) + π

).

4. Suppose f(x) and g(x) are functions with limx→0

f(x) = 4 and limx→4

g(x) = 8. Evaluate each of the

following limits if possible. If you cannot evaluate the limit, explain why.

16

(a) limx→0

f(g(x)) (b) limx→0

g(f(x))

5. Complete the statement.

(a) If n is a positive even integer and a ≥ 0, then limx→a

n√x = .

(b) If n is a positive odd integer and a is any real number, then limx→a

n√x = .

6. Evaluate the given limit.

(a) limx→3

(x3 + 2x− 8) (b) limx→0

x2 + 5

x3 − 1(c) lim

t→−83√t (d) lim

s→−4

1

s2 + 2

7. Evaluate limx→1

f(x) where

f(x) =

{x2 − 5 x ≤ 0,8x x > 0.

8. Evaluate the given limit.

(a) limx→1

3x2 − 3

x+ 2(b) lim

x→1

x− 1

3x2 − 3(c) lim

x→4

x2 + 9x+ 20

x3 + x2 − 12x

9. Evaluate the limit limh→0

(2 + h)2 − 22

h.

10. Evaluate the limit limh→0

√2 + h−

√2

h.

11. Use the Squeeze Theorem to prove that limx→0

x2 cos2(

1

x

)= 0. Be sure to use complete sentences in

your proof.

12. Use the Squeeze Theorem to prove that limx→0

x sin4

(1

x3

)= 0. Be sure to use complete sentences in

your proof.

13. Complete the statement.

(a) If f(x) is a polynomial and c is any real number, then limx→c

f(x) = .

(b) If g(x) is a rational function and c is any real number that is in the domain of g(x), then

limx→c

g(x) = .

14. (a) Evaluate the limit limx→−3

x+ 2.

(b) Evaluate the limit limx→−3

x2 + 5x+ 6

x+ 3.

(c) True or False: The functions f(x) = x+ 2 and g(x) =x2 + 5x+ 6

x+ 3define the same function. Be sure

to give a reason that supports your answer.

17

(d) Let a be any real number. Show that the two limits limx→a

x+ 2 and limx→a

x2 + 5x+ 6

x+ 3are equal.

Limit Laws and the Squeeze Theorem - Solutions

1. (a) limx→c

(f(x) + g(x)) = L+K (b) limx→c

(f(x)− g(x)) = L−K

(c) limx→c

(f(x)g(x)) = LK (d) limx→c

5f(x) = 5L

2. (a) limx→2

8f(x) + g(x)

g(x) + 1=

8 · 5 + 0

0 + 1= 40

(b) The limit does not exist since the denominator goes to 0 but the numerator goes to 5.

(c) limx→4

(√g(x)f(x)2 + g(x)2

)=√

4(−2)2 + 42 =√

16 + 16 = 20

(d) limx→4

(3√g(x)f(x) + 10

)= 3√

4(−2) + 10 = 3√−8 + 10 = −2 + 10 = 8

3. limx→1

(2f(x)2 − f(x) + π

)= 2π2 − π + π = 2π2

4. (a) In order to compute limx→0

f(g(x)), we would need to know limx→0

g(x). Since we are not given the

value of limx→0

g(x), we cannot evaluate limx→0

f(g(x)).

(b) limx→0

g(f(x)) = limx→4

g(x) = 8

5. (a) If n is a positive even integer and a ≥ 0, then limx→a

n√x = n

√a.

(b) If n is a positive odd integer and a is any real number, then limx→a

n√x = n

√a.

6. (a) limx→3

(x3 + 2x− 8) = 33 + 2(3)− 8 = 25 (b) limx→0

x2 + 5

x3 − 1=

02 + 5

03 − 1= −5

(c) limt→−8

3√t = 3√−8 = −2 (d) lim

x→−4

1

s2 + 2=

1

(−4)2 + 2=

1

18

7. limx→1

f(x) = limx→1

8x = 8 · 1 = 8

8. (a) limx→1

3x2 − 3

x+ 2=

3(1)2 − 3

1 + 2= 0

(b) limx→1

x− 1

3x2 − 3= limx→1

x− 1

3(x− 1)(x+ 1)= limx→1

1

3(x+ 1)=

1

3(1 + 1)=

1

6

(c) limx→4

x2 + 9x+ 20

x3 + x2 − 12x=

42 + 9 · 4 + 20

43 + 42 − 12 · 4=

9

4

9. We have

limh→0

(2 + h)2 − 22

h= limh→0

4 + 4h+ h2 − 4

h= limh→0

h(4 + h)

h= limh→0

4 + h = 4 + 0 = 4.

18

10. We have

limh→0

√2 + h−

√2

h= lim

h→0

√2 + h−

√2

h·√

2 + h+√

2√2 + h+

√2

= limh→0

2 + h− 2

h(√

2 + h+√

2)

= limh→0

h

h(√

2 + h+√

2)= limh→0

1√2 + h+

√2

=1

√2 + 0 +

√2

=1

2√

2

11. For any real number x 6= 0, we have −1 ≤ cos2(1x

)≤ 1. We can multiply this inequality through

by x2 to get

−x2 ≤ x2 cos2(

1

x

)≤ x2.

An important remark is that x2 is never negative and so the direction of the inequalities stay the same.Now lim

x→0−x2 = 0 and lim

x→0x2 = 0 so by the Squeeze Theorem,

limx→0

x2 cos2(

1

x

)= 0.

12. For any real number x 6= 0, we have −1 ≤ sin4(

1x3

)≤ 1. We would like to multiply this inequality

through by x to get −x ≤ x sin4(

1x3

)≤ x but the problem is that if x is negative, then we must reverse

the inequalities. Instead, we use the fact that for any real number x, the inequalities x ≤ |x| and −|x| ≤ xare true. Therefore,

x sin4

(1

x3

)≤ |x| sin4

(1

x3

)≤ |x| · 1 = |x|.

Similarly,

−|x| = −|x| · 1 ≤ x sin4

(1

x3

).

Putting all of this together gives

−|x| ≤ x sin4

(1

x3

)≤ |x|.

Since limx→0−|x| = −|0| = 0 and lim

x→0|x| = |0| = 0, we have

limx→0

x sin4

(1

x3

)= 0

by the Squeeze Theorem.

13. (a) If f(x) is a polynomial and c is any real number, then limx→c

f(x) = f(c).

(b) If g(x) is a rational function and c is any real number that is in the domain of g(x), then we havelimx→c

g(x) = g(c).

14. (a) limx→−3

x+ 2 = −3 + 2 = −1

(b) limx→−3

x2 + 5x+ 6

x+ 3= limx→−3

(x+ 3)(x+ 2)

x+ 3= limx→−3

x+ 2 = −3 + 2 = −1

(c) False. The number −3 is in the domain of f , but −3 is not in the domain of g.

(d) For any real number a, limx→a

x+ 2 = a+ 2. Similarly, limx→a

x2 + 5x+ 6

x+ 3= limx→a

x+ 2 = a+ 2.

19

1.3 Left and Right Hand Limits

1. A statement is given. Determine if the statement is always true, or sometimes false. Here f is afunction whose domain is all real numbers.

(a) If limx→a

f(x) = L, then limx→a−

f(x) = L. (b) If limx→a+

f(x) = L, then limx→a

f(x) = L.

(c) If limx→a+


f(x) = L.

2. Suppose that p(x) is a polynomial and limx→1+

p(x) = 5. Do you have enough information to determine

limx→1−

p(x)?

3. The graph of f is shown below. Evaluate each of the given limits. If the limit does not exist, thenwrite “does not exist.”

1 2 3

1

2

3

f

y

x

(a) limx→0+

f(x) (b) limx→2−

f(x) (c) limx→2+

f(x) (d) limx→2

f(x) (e) limx→3+

f(x)

4. The graph of f is shown below. Evaluate each expression. If the expression does not exist, then write“does not exist.”

−2 −1 1 2

1

2

f

y

x

(a) limx→−1−

f(x) (b) limx→−1+

f(x) (c) limx→−1

f(x) (d) f(−1) (e) f(1) (f) limx→1

f(x)

20

5. Let f(x) =

{x− 4 x ≤ 3x2 x > 3.

Evaluate each of the following.

(a) limx→3−

f(x) (b) limx→3+

f(x) (c) limx→3

f(x) (d) f(3) (e) f(1)

6. Let f(x) =

x3 x < 1−2 x = 1x x > 1.


(a) limx→1−

f(x) (b) limx→1+

f(x) (c) limx→1

f(x) (d) f(1)

7. Let f(x) =

{|x| x 6= 02 x = 0.


(a) limx→0−

f(x) (b) limx→0+

f(x) (c) limx→0

f(x) (d) f(0) (e) f(−2)

8. Let f(x) =

{2x2 + 2a x < 1x− a2 x ≥ 1.

Find a value of a for which limx→1

f(x) exists.

9. Evaluate the given limit if possible. If the limit does not exist, write “does not exist.”

(a) limx→2

3x+ 6

|x+ 2|(b) lim

x→0

3x+ 6

|x+ 2|(c) lim

x→−2

3x+ 6

|x+ 2|

10. The Lorentz factor is defined as γ =1√

1− v2/c2where v is the relative velocity between inertial

reference frames, and c is the speed of light in a vacuum (c ≈ 3.00×108 meters per second). The Lorentzfactor appears in equations in special relativity.

(a) Find limv→0+

γ.

(b) Find limv→c−

γ.

Left and Right Hand Limits - Solutions

1. (a) This statement is always true. If limx→a


f(x) = limx→a+

f(x) = L.

(b) This statement is not always true. For example, let

f(x) =

{−1 if x < 01 if x > 0.

We have limx→0+

f(x) = 1, but limx→0

f(x) does not exist.

(c) This statement is not always true. For example, let

f(x) =

{x+ 1 if x < 0x− 1 if x > 0.

21

In this case, limx→0+

f(x) = −1 and limx→0−

f(x) = 1.

2. Yes. For any polynomial p(x), the limit limx→a

p(x) exists for any a and limx→a

p(x) = p(a). In particular,

limx→a−

p(x) = limx→a+

p(x) = p(a).

Therefore limx→1−

p(x) = limx→1+

p(x) = 5.

3. (a) 2 (b) 0 (c) 0 (d) 0

(e) The limit limx→3+

f(x) does not exist because f(x) is not defined for x > 3.

4. (a) 2 (b) 1 (c) The limit limx→−1

f(x) does not exist since limx→−1−

f(x) = 2 but limx→−1+

f(x) = 1.

(d) f(−1) = 2 (e) The value f(1) does not exist since f is not defined at x = 1. (f) 1

5. (a) −1 (b) 9 (c) The limit limx→3

f(x) does not exist since limx→3−

f(x) = −1 but limx→3+

f(x) = 9.

(d) −1 (e) −3

6. (a) 1 (b) 1 (c) We have limx→1

f(x) = 1 since by parts (a) and (b), limx→1−

f(x) = 1 and limx→1+

f(x) = 1.

(d) −2

7. (a) 0 (b) 0 (c) 0 (d) 2 (e) 2

8. The limit limx→1

f(x) exists if and only if the left hand limit limx→1−

f(x) and the right hand limit limx→1+

f(x)

both exist and are equal to each other. Now limx→1−

f(x) = limx→1+

f(x) implies that

limx→1−

(2x2 + 2a) = limx→1+

(x− a2).

Evaluating the limits on both sides of this equation gives 2(1)2 + 2a = 1− a2 so that

2 + 2a = 1− a2.

We solve this quadratic equation by first rewriting it as a2 + 2a+ 1 = 0. Next, we factor the left handside to get (a + 1)2 = 0. The only solution to (a + 1)2 = 0 is a = −1. We conclude that a = −1 is theonly value of a for which lim

x→1f(x) exists.

9. (a) limx→2

3x+ 6

|x+ 2|=

3(2) + 6

|2 + 2|= 3. (b) lim

x→0

3x+ 6

|x+ 2|=

3(0) + 6

|0 + 2|= 3.

(c) We claim that the limit limx→−2

3x+ 6

|x+ 2|does not exist. To prove this, we will show that

limx→−2−

3x+ 6

|x+ 2|6= limx→−2+

3x+ 6

|x+ 2|.

An important fact that will be used in the calculation is

|y| ={y if y ≥ 0−y if y < 0.

Now as x approaches −2 from the right, we know that x > −2 so that x+2 > 0. Therefore, |x+2| = x+2and so

limx→−2+

3x+ 6

|x+ 2|= limx→−2+

3(x+ 2)

x+ 2= limx→−2+

3

1= 3.

22

On the other hand, as x approaches −2 from the left, we know that x < −2 so that x+2 < 0. Therefore,|x+ 2| = −(x+ 2) and so

limx→−2−

3x+ 6

|x+ 2|= limx→−2−

3(x+ 2)

−(x+ 2)= limx→−2−

3

−1= −3.

This shows that limx→−2−

3x+ 6

|x+ 2|6= limx→−2+

3x+ 6

|x+ 2|and so we may say that the limit lim

x→−2

3x+ 6

|x+ 2|does not

exist.

10. (a) We have

limv→0+

γ = limv→0+

1√1− v2/c2

=1√

1− 02/c2=

1√1

= 1.

What this tells us is that when the relative velocity between the two objects is going to 0, then theLorentz factor approaches 1.

(b) Computing as in (a), we find that

limv→c−

γ = limv→c−

1√1− v2/c2

=∞.

This limit is ∞ because in the fraction1√

1− v2/c2, the numerator is always 1 but the denominator

tends to 0 as v approaches c from the left. Note that since v is approaching c from the left, the value1− v2/c2 is not negative and so the expression

√1− v2/c2 is a real number.

1.4 Continuity

1. Let f be a function defined on an open interval that contains c. State, in terms of limits, what itmeans for f to be continuous at c.

2. A statement is given. Determine if the statement is always true, or sometimes false.

(a) If f is continuous at c, then limx→c−

f(x) = limx→c

f(x).

(b) If f has the property that limx→c−

f(x) = L and limx→c+

f(x) = L, then f must be continuous at c.

3. A statement is given. Determine if the statement is always true, or sometimes false.

(a) If f and g are any functions, both of which are continuous at c, then the function f + g is continuousat c.

(b) If f and g are any functions, both of which are continuous at c, then the function fg is continuousat c.

(c) If f and g are any functions, both of which are continuous at c, then the function f/g is continuousat c.

4. Suppose that f is a continuous function whose domain is [1, 3] and f(1) = 7. Find limx→1+

f(x).

23

5. The graph of a function f , whose domain is [0, 3],is shown on the right. Determine all values of c forwhich f is continuous at c.

1 2 3

1

2

3

f

y

x

6. The graph of f is shown below. Find the interval(s) on which f is continuous.

−2 −1 1 2

1

2

f

y

x

7. Let f(x) =

{2x+ 1 if x ≥ 1,x2 + 2 if x < 1.

(a) Is f continuous at 0? (b) Is f continuous at 1?

8. Let f(x) =

{sinx if x < 0,−x+ 1 if x ≥ 0.

Explain, in terms of limits, why f is not continuous at 0.

9. Find all values of a for which the given function is continuous on (−∞,∞).

f(x) =

{2x+ a if x ≥ 1,x2 if x < 1.

10. Find the interval(s) on which the given function is continuous.

(a) f(x) = x2ex (b) g(x) = x−2ex (c) h(x) =√x cos(2x) (d) r(x) = x4 sin(3x) cos(x3)

11. Find the interval(s) on which the given function is continuous.

(a) f(x) =cosx

x2 + 1(b) g(x) =

lnx

x2 + 1(c) h(x) =

cosx

x2 − 1(d) r(x) =

lnx

x2 − 1

12. Let s(t) be the distance traveled by a race car, in hundreds of miles, t hours since the start of a race.Two possible graphs of s(t) are shown below. Which is the more likely graph of s(t) and why? Be sureto give a reason that supports your answer.

24

1 2 3

1

2

3

s(t)

t1 2 3

1

2

3

s(t)

t

13. Complete the statement of the Intermediate Value Theorem. Suppose that f is a continuous functionon [a, b]. If f(a) < f(b) and y is any real number with f(a) < y < f(b), then

there is a c with .

14. Let T (t) be the temperature, measured in degrees Fahrenheit, in Sacramento t hours after 6:00 A.M.If T (0) = 65◦ and T (10) = 80◦, must there exist a time between 6:00 A.M. and 4:00 P.M. such that thetemperature in Sacramento is exactly 75.2398◦? Be sure to justify your answer.

15. Suppose that the temperature of an object during an experiment is given by the function

f(t) =100

3πt| sin t+ 2|

where t is in minutes and f(t) is in degrees Fahrenheit. Show that there is a time between 0 and 3π forwhich the temperature of the object is 150◦.

Continuity - Solutions

1. The function f is continuous at c if limx→c

f(x) = f(c).

2. (a) This statement is true. Since f is continuous at c, we know that limx→c

f(x) = f(c). In particular,

limx→c

f(x) exists so the left hand limit limx→c−

f(x) must be equal to the right hand limit limx→c+

f(x).

(b) This statement is sometimes false. For example, consider

f(x) =

{−1 if x 6= 01 if x = 0.

Now limx→0−

f(x) = −1 and limx→0+

f(x) = −1. Since the left and right hand limits at 0 are both −1, we know

that limx→0

f(x) = −1. However, f is not continuous at 0 since f(0) = 1 is not equal to limx→0

f(x) = −1.

3. (a) True (b) True

(c) False. Consider f(x) = 1 and g(x) = x2. Both of these functions are continuous at 0 but (f/g)(x) =1/x2 is not continuous at 0.

25

4. Since f is continuous on [1, 3], we know that limx→1+

f(x) = f(1) = 7.

5. The function f is continuous on the intervals [0, 2) and (2, 3].

6. The function f is continuous on the intervals [−2,−1) and (−1, 2].

7. (a) To determine if f is continuous at 0, we will determine whether or not f(0) is equal to limx→0

f(x).

Since limx→0

f(x) = 02 + 2 = 2 and f(0) = 02 + 2, we see that f is continuous at 0.

(b) To determine if f is continuous at 1, we will determine whether or not f(1) is equal to limx→1

f(x).

First we find limx→1

f(x). We have

limx→1−

f(x) = limx→1−

x2 + 2 = 3 and limx→1+

f(x) = limx→1+

2x+ 1 = 3.

Therefore, limx→1

f(x) = 3. Since f(1) = 3, we get limx→1

f(x) = f(1) so f is continuous at 1.

8. We have f(0) = −0 + 1 = 1 and limx→0−

f(x) = limx→0−

sinx = 0. Thus, f(0) is not equal to limx→0

f(x) so

f is not continuous at 0.

9. We must choose a so that limx→1−

f(x) = limx→1+

f(x). In other words, we need

limx→1−

x2 = limx→1+

2x+ a

which can be rewritten as 12 = 2(1) + a. Solving this equation for a gives a = −1. With this choice ofa, lim

x→1f(x) = f(1) so f is continuous at 1. The functions 2x − 1 and x2 are continuous so that when

a = −1, f is continuous on (−∞,∞).

10. (a) The function f(x) = x2ex is continuous on (−∞,∞).

(b) The function g(x) = x−2ex is continuous on (−∞, 0) and (0,∞). The reason that g is not continuous

at 0 is because g(x) = x−2ex =ex

x2and so g(0) is not defined.

(c) The function h(x) =√x cos(2x) is continuous on [0,∞).

(d) The function r(x) = x4 sin(3x) cos(x3) is continuous on (−∞,∞).

11. (a) The function f(x) =cosx

x2 + 1is continuous on (−∞,∞). Note that x2 + 1 > 0 for all x so that

the denominator of f is never 0.

(b) The function g(x) =lnx

x2 + 1is continuous on (0,∞). Here we are using the fact that the domain of

lnx is (0,∞).

(c) The function h(x) =cosx

x2 − 1is continuous on (−∞,−1), (−1, 1), and (1,∞). The reason that 1 and

−1 must be excluded is that x2 − 1 is 0 when x = 1 or x = −1.

(d) The function r(x) =lnx

x2 − 1is continuous on (0, 1) and (1,∞).

12. The graph on the left would be the more likely graph of s(t). The reason for this is that the graphon the right is not continuous at t = 1. What does this say about the car? Notice that for the graph

26

on the right, we have f(1) ≈ 0.75 and limt→1+

s(t) ≈ 1.5. This tells that at the 1 second mark, the car had

gone about 75 miles, but then instantaneously jumps to having gone about 150 miles the moment justpast the 1 second mark. Unless the race car has the power to teleport, the graph on the left is morelikely the graph of the distance traveled by the race car.

13. Suppose that f is a continuous function on [a, b]. If f(a) < f(b) and y is any real number withf(a) < y < f(b), then there is a c with a < c < b and f(c) = y.

14. Yes. The function T (t) is continuous and T (0) = 65 while T (10) = 80. By the Intermediate ValueTheorem, there is a t0 with 0 < t0 < 10 such that T (t0) = 75.2398.

15. The function f(t) is continuous on [0, 3π]. We have

f(0) =100

3π(0)| sin(0) + 2| = 0

and

f(3π) =100

3π(3π)| sin(3π) + 2| = 100|0 + 2| = 200.

Since 0 < 150 < 200, the Intermediate Value Theorem tells us that there is a time c between 0 and 3πfor which f(c) = 150.

1.5 Limits that Involve Infinity

1. Complete the statement - We write limx→c

f(x) =∞

if the values of f(x)

as x approaches c.

2. Complete the statement - We write limx→∞

f(x) = L

if the values of f(x)

as x approaches ∞.

3. Suppose that f is a function whose domain is all real numbers. If limx→∞

f(x) = 3, does f have a

horizontal asymptote?

4. Suppose that y = 2 is a horizontal asymptote for the function f(x). Can you say anything about thelimits lim

x→∞f(x) and lim

x→−∞f(x)?

5. Suppose that f is a function that is continuous on (−∞, 4) and (4,∞), and that f has a verticalasymptote at x = 4. Does this imply that lim

x→4f(x) must either be ∞ or −∞? Be sure to justify your

answer with a correct explanation.

6. Suppose that f is a function whose domain is all real numbers except 1 and 2, limx→1

f(x) = ∞, and

limx→2

f(x) = 3.

(a) Does f have any vertical asymptotes?

27

(b) Do you have enough information to determine whether or not f has a horizontal asymptote?

7. Determine each limit. It may help to use a graphical approach.

(a) limx→0+

lnx (b) limx→0+

1

x(c) lim

x→π2

−tanx

8. The graph of a function f is shown below. Evaluate each expression. If the expression does not exist,then write “does not exist.”

1 2 3

1

2

3

f

y

x

(a) limx→1.5−

f(x)

(b) limx→1.5+

f(x)

(c) f(1.5)

9. Determine the horizontal and vertical asymptotes of the given function.

(a) f(x) =2x2 + 7x

x2 − 1(b) g(x) =

2x3 + 2x2 + 2x

3x3 − 3x(c) h(x) =

x

x2 + 5x(d) s(t) =

t2 + 8t

t+ 1

10. Evaluate each of the following limits. How are your solutions related to your solutions from theprevious exercise?

(a) limx→∞

2x2 + 7x

x2 − 1(b) lim

x→−∞

2x3 + 2x2 + 2x

3x3 − 3x

(c) limx→−∞

x

x2 + 5x(d) lim

t→∞

t2 + 8t

t+ 1

11. The graph of f(x) =1

x− 1is shown below. Evaluate each of the given limits.

0.5 1 1.5 2 2.5

−6

−4

−2

2

4

6

y

x

(a) limx→1−

f(x)

(b) limx→1+

f(x)

(c) limx→1

f(x)

(d) limx→∞

f(x)

28


(x− 1)2is shown below. Evaluate each of the limits.

−1 1 2 3

2

4

6

8

10

y

x

(a) limx→1−

f(x) (b) limx→1+

f(x) (c) limx→1

f(x) (d) limx→∞

f(x)


ex + 2is shown below. Evaluate each of the limits.

−10 −5 5 10

0.2

0.4

0.6

y

x

(a) limx→∞

f(x) (b) limx→−∞

f(x) (c) limx→0

f(x)

14. A model for population growth is given by

p(t) =100, 000

100 + 900e−2t

where t is in years and p(t) is the size of the population at time t. The function p(t) assumes that theinitial population is 100 and the growth rate of the population is 2. Evaluate

limt→∞

100, 000

100 + 900e−2t.

29

This limit exists and is the carrying capacity which is the largest population that the environment cansupport.

Limits that Involve Infinity - Solutions

1. We write limx→c

f(x) =∞ if the values of f(x) become arbitrarily large as x approaches c.

2. We write limx→∞

f(x) = L if the values of f(x) approach L as x approaches ∞.

3. Yes. The line y = 3 is a horizontal asymptote of f .

4. We know that at least one of limx→∞

f(x) or limx→−∞

f(x) must be 2. We do not have enough information

to say more than this. For example, f(x) = 2x2+5x+1x2−11x has y = 2 as a horizontal asymptote and both of

the infinite limits limx→∞

f(x) and limx→−∞

f(x) equal 2. The function g(x) = ex + 2 also has y = 2 as a

horizontal asymptote but limx→∞

g(x) =∞ and limx→−∞

g(x) = 2.

5. We cannot say whether or not limx→4

f(x) is ∞ or −∞. In fact, this limit may not exist. For example,

if f(x) =1

x− 4, then lim

x→4−f(x) = −∞ but lim

x→4+f(x) =∞.

6. (a) Yes. The function f has a vertical asymptote at x = 1.

(b) We do not have enough information to determine if f has a horizontal asymptote.

7. (a) −∞ (b) ∞ (c) ∞

8. (a) ∞ (b) ∞ (c) does not exist

9. (a) The horizontal asymptote is y = 2. Now we find any vertical asymptotes. We begin by simplifyingf(x) as much as possible:

2x2 + 7x

x2 − 1=

x(2x+ 7)

(x− 1)(x+ 1).

The vertical asymptotes are x = 1 and x = −1.

(b) The horizontal asymptote is y =2

3. Now

2x3 + 2x2 + 2x

3x3 − 3x=

2x(x2 + x+ 1)

3x(x2 − 1)=

2(x2 + x+ 1)

3(x− 1)(x+ 1).

From this we see that the vertical asymptotes are x = 1 and x = −1.

(c) The horizontal asymptote is y = 0. Now

x

x2 + 5x=

x

x(x+ 5)=

1

x+ 5.

The vertical asymptote of h(x) is x = −5.

(d) The function s(t) does not have a horizontal asymptote (the degree of the numerator is larger thanthe degree of the denominator). As for any vertical asymptotes, since

t2 + 8t

t+ 1=t(t+ 8)

t+ 1,

30

the line t = −1 is the only vertical asymptote of s(t).

10. (a) We have

limx→∞

2x2 + 7x

x2 − 1= limx→∞

2x2/x2 + 7x/x2

x2/x2 − 1/x2= limx→∞

2 + 7/x

1− 1/x2=

2 + 0

1− 0= 2.

From the previous problem, we know that f(x) =2x2 + 7x

x2 − 1has a horizontal asymptote at y = 2.

(b)

limx→∞

2x3 + 2x2 + 2x

3x3 − 3x= limx→∞

2x3/x3 + 2x2/x3 + 2x/x3

3x3/x3 − 3x/x3= limx→∞

2 + 2/x+ 2/x2

3− 3/x2=

2 + 0 + 0

3− 0=

2

3.

From the previous problem, we know that g(x) =2x3 + 2x2 + 2x

3x3 − 3xhas a horizontal asymptote at y =

2

3.

(c)

limx→∞

x

x2 + 5x= limx→∞

x/x2

x2/x2 + 5x/x2= limx→∞

1/x

1 + 5/x=

0

1 + 0= 0.

From the previous problem, we know that h(x) =x

x2 + 5xhas a horizontal asymptote at y = 0.

(d)

limt→∞

t2 + 8t

t+ 1= limt→∞

t2/t2 + 8t/t2

t/t2 + 1/t2= limt→∞

1 + 8/t

1/t+ 1/t2=∞.

From the previous problem, we know that s(t) =t2 + 8t

t+ 1does not have a horizontal asymptote.

11. (a) −∞ (b) ∞ (c) does not exist (d) 0

12. (a) ∞ (b) ∞ (c) ∞ (d) 0

13. (a) 0 (b)1

2(c)

1

3

14. We have

limt→∞

100, 000

100 + 900e−2t=

100, 000

100 + 900(0)=

100, 000

100= 1, 000.

The carrying capacity is 1,000.

31

2 Derivatives

2.1 Instantaneous Rate of Change

1. Let f be a function defined on an open interval I that contains c.

(a) Assuming that f is differentiable at c, state the definition of f ′(c).

(b) Assuming that f is differentiable on the interval I, state the definition of f ′(x).

(c) Is f ′(c) a number or a function?

(d) Is f ′(x) a number or a function?

2. Complete the statement. Let f be a function defined on an open interval I that contains c and assumethat f is differentiable at c. If l(x) is the tangent line to the graph of f at c, then

l(x) = .

3. Let f(x) = x2.

(a) Evaluatef(2 + h)− f(2)

hfor the h values 0.1, 0.01, 0.001, and 0.0001.

(b) Based on your calculations from part (a), make a guess at the value of limh→0

f(2 + h)− f(2)

h.

(c) Compute f ′(2) using the definition of the derivative.

4. If f(x) = 3x+ 5, find f ′(x) using the definition of the derivative.

5. If s(t) =1

t+ 1, find

ds

dtusing the definition of the derivative.

6. If s(t) =√t+ 1 + 8, find s′(t) using the definition of the derivative.

7. Oil leaks out of an engine. The amount of oil A(t) that has leaked out up to time t is shown in thetable below.

Time (seconds) 0.25 0.5 0.75 1 1.25 1.5 1.75 2Amount leaked (liters) 0.072 0.536 0.762 0.931 1.075 1.140 1.242 1.338

(a) Estimate how fast the oil is leaking out of the engine at time t = 0.325 by computing the slope ofthe line through the points (0.25, 0.072) and (0.5, 0.536).

(b) Estimate how fast the oil is leaking out of the engine at time t = 1.875.

(c) Is the oil leaking out faster at time t = 0.325 or time t = 1.875?

(d) Suppose that the function A(t) is a differentiable function for t > 00. What expression gives theinstantaneous rate of change in the oil leaking out of the engine at time t = 0.325?

32

8. The graph of f is shown to the right. Sketch thegraph of f ′.

−2 −1 1 2

1

2

f

y

x

9. The graph of f is shown to the right. Sketch thegraph of f ′.

−10 −5 5 10

−10

−5

5

10

f

y

x

10. The graph of g is shown below. Find the intervals on which

(a) g′(x) > 0 (b) g(x) > 0 (c) g′(x) < 0 (d) g(x) < 0

−4 −2 2 4

−5

5

10y

x

11. The displacement of an object is given by the function

f(x) =

{x2 if 0 ≤ x ≤ 1,x if x > 1.

Here f(x) is in meters and x is the number of seconds that the object has been in motion.

(a) Find f(1).

(b) Show that f is continuous at 1.

33

(c) Find the slope of the line that passes through the points (1, f(1)) and (1.01, f(1.01)).

(d) Find the slope of the line that passes through the points (1, f(1)) and (0.99, f(0.99)).

(e) Show that f ′(1) does not exist.

(f) Sketch the graph of f from x = 0 to x = 2.

12. Suppose that f is continuous on an open interval that contains 2. If f(2) = 3 and f ′(2) = 7, findthe equation of the tangent line to f at x = 2.

13. Let f be a continuous function that is differentiable at 3. If y = 2x+5 is the equation of the tangentline to f at 3, find f ′(3) and f(3).

14. Give an example of a function f that is continuous on (−∞,∞) but is not differentiable at x = 0.You may represent your function by drawing the graph of the function.

Instantaneous Rate of Change - Solutions

1. (a) For a number c in the interval I, we define

f ′(c) = limh→0

f(c+ h)− f(c)

h.

(b) For any x in I, we define the function f ′(x) as follows:

f ′(x) = limh→0

f(x+ h)− f(x)

h.

(c) If f is differentiable at c, then f ′(c) is a number. It is the slope of the graph of f at x = c.

(d) If f is differentiable on I, then f ′(x) is a function. The function f ′(x) is the derivative of f .

2. Let f be a function defined on an open interval I that contains c and assume that f is differentiableat c. If l(x) is the tangent line to the graph of f at c, then

l(x) = f ′(c)(x− c) + f(c).

3. (a) For any h > 0,f(2 + h)− f(2)

h=

(2 + h)2 − 22

h. We now use this formula to give a table of

values off(2 + h)− f(2)

h.

h 0.1 0.01 0.001 0.0001f(2+h)−f(2)

h 4.1 4.01 4.001 4.0001

(b) Based on our work in (a), it makes sense to say that limh→0

f(2 + h)− f(2)

h= 4.

(c) Using the definition of the derivative, we find that

f ′(2) = limh→0

(2 + h)2 − 22

h= limh→0

4 + 4h+ h2 − 4

h= limh→0

h(4 + h)

h= limh→0

4 + h = 4 + 0 = 4.

34

4. Using the definition of the derivative, we have

f ′(x) = limh→0

3(x+ h) + 5− (3x+ 5)

h= limh→0

3x+ 3h+ 5− 3x− 5

h= limh→0

3h

h= limh→0

3 = 3.

5. Using the definition of the derivative,

ds

dt= lim

h→0

1t+h+1 −

1t+1

h= limh→0

t+1−t−h−1(t+1)(t+h+1)

h= limh→0

−h(t+1)(t+h+1)

h

= limh→0

−1

(t+ 1)(t+ h+ 1)=

−1

(t+ 1)(t+ 0 + 1)= − 1

(t+ 1)2.

6. From the definition of the derivative, we have

s′(t) = limh→0

√t+ h+ 1 + 8− (

√t+ 1 + 8)

h= limh→0

(√t+ h+ 1−

√t+ 1)(

√t+ h+ 1 +

√t+ 1)

h(√t+ h+ 1 +

√t+ 1)

= limh→0

t+ h+ 1− (t+ 1)

h(√t+ h+ 1 +

√t+ 1)

= limh→0

1√t+ h+ 1 +

√t+ 1

=1√

t+ 0 + 1 +√t+ 1

=1

2√t+ 1

.

7. (a) 1.856 liters per second

(b) Following (a), we compute the slope of the secant line through the points (1.75, 1.242) and (2, 1.338)to obtain the estimate 0.384 liters per second.

(c) It appears that the oil is leaking out faster at time t = 0.325.

(d) A′(0.325)

8.

−2 −1 1 2

1

2

f ′

y

x

9.

−10 −5 5 10

−10

−5

5

10

f ′

y

x

10. (a) The interval on which g′(x) > 0 is (0,∞). The reason for this is that if g′(x) > 0, then the slopeof g at x is positive. Looking at the graph of g, we see that the slopes of g are positive when x > 0.

(b) The intervals on which g(x) > 0 are (−∞,−2) and (2,∞). This is because g has positive y-valueson (−∞,−2) and (2,∞).

(c) Since the slopes of g are negative on the interval (−∞, 0), we have that g′(x) < 0 on (−∞, 0).

35

(d) (−2, 2)

11. (a) f(1) = 12 = 1

(b) We must show that limx→1

f(x) = f(1). Now

limx→1−

f(x) = limx→1−

x2 = 12 = 1

andlimx→1+

f(x) = limx→1+

x = 1.

This shows that limx→1

f(x) = 1 and by part (a), we know that f(1) = 1 so limx→1

f(x) = f(1).

(c) The slope of the line through the points (1, f(1)) and (1.01, f(1.01)) is

f(1.01)− f(1)

1.01− 1=

1.01− 12

0.01= 1.

(d) The slope of the line through the points (1, f(1)) and (0.99, f(0.99)) is

f(0.99)− f(1)

0.99− 1=

(0.99)2 − 12

−0.01= 1.99.

(e) To show that f ′(1) does not exist, we will prove that limh→0

f(1 + h)− f(1)

hdoes not exist. First, we

have

limh→0+

f(1 + h)− f(1)

h= limh→0+

1 + h− 12

h= limh→0+

h

h= 1.

On the other hand,

limh→0−

f(1 + h)− f(1)

h= limh→0−

(1 + h)2 − 12

h= limh→0−

2h+ h2

h= limh→0−

2 + h = 2 + 0 = 2.

The above two calculations show that limh→0

f(1 + h)− f(1)

hdoes not exist.

(f)

0.5 1 1.5 2

0.5

1

1.5

2

y

x

36

12. The equation of the tangent line to f at x = 2 is l(x) = 7(x− 2) + 3.

13. We have f ′(3) = 2 and f(3) = 2(3) + 5 = 11.

14. A function that is continuous on (−∞,∞) but that is not differentiable at 0 is f(x) = |x|. From thegraph of f(x) = |x|, which is shown below, we see that there is a corner at x = 0.

−2 −1 1 2

1

2

One can show using the definition of the derivative that f ′(0) does not exist. Indeed,

limh→0+

f(0 + h)− f(0)

h= limh→0+

|0 + h| − |0|h

= limh→0+

h

h= 1.

However,

limh→0−

f(0 + h)− f(0)

h= limh→0−

|0 + h| − |0|h

= limh→0−

−hh

= −1.

Since the left and right hand limits do not match, the limit limh→0

f(0 + h)− f(0)

hdoes not exist so that

f ′(0) does not exist.

Remark: In our calculations, we used the fact that |h| = h if h > 0, and |h| = −h if h < 0.

2.2 Interpreting the Derivative

1. The temperature, in degrees Fahrenheit, of a substance t minutes after the substance was placedinside an oven is given by the function T (t).

(a) Explain, in practical terms, what the statement T (4) = 150 means.

(b) Explain, in practical terms, what the statement T ′(4) = 15 means.

(c) Given T (4) = 150 and T ′(4) = 15, find the equation of the tangent line to T (t) at t = 4. Use yourequation to approximate the temperature of the substance at time t = 4.1.

(d) Suppose that the temperature of the substance before it is placed in the oven is 72◦ and the tem-perature inside the oven is 425◦. Do you expect T ′(1) to be positive or negative? Why?

2. Let d(t) be the distance traveled by a runner, in miles, t hours after the start of a marathon. Amarathon is 26.21875 miles.

37

(a) What are the units of d′(t)?

(b) What are the units of d′′(t)?

(c) If the runner finishes the race in 3.5 hours, find d(3.5).

(d) Assuming the runner is trying to finish the race as fast as possible, what do you expect the sign ofd′(t) to be for 0 < t < 3.5?

(e) The runner runs the first 1.5 hours of the race above their average training speed. As the racecontinues, the runner begins to get tired and runs below their average training speed for the final 2hours of the race. What can you say about the quantities d′(1) and d′(3)?

3. A water storage tank is in the shape of a cylinder that is 50 feet high and has a radius of 10 feet.The hydrostatic pressure P (d), in pounds per square inch, on the wall of a tank depends on the depthd which is measured in feet. Assume that the tank is filled with water. What do you expect the sign ofP ′(d) to be? Why?

4. Let V (t) be the value of a share of stock, in dollars, at time t where t is the number of days since thestock became available for purchase.

(a) What are the units of V (t)?

(b) What are the units of V ′(t)?

(c) Suppose V (100) = 200 and V ′(100) = −150. Would you purchase the stock at time t = 100 for $125per share? Why or why not?

5. Let D(t) be the displacement of a particle measured in centimeters from the initial position of theparticle, and t is the number of seconds since the particle was sent into motion.

(a) What are the units of D′(t)?

(b) What are the units of D′′(t)?

(c) Suppose D(2) = 4 and at time t = 2, the particle is in the process of returning to its initial position.What is the sign of D′(2)?

(d) Interpret the statement D′(3) = 0 in practical terms.

Interpreting the Derivative - Solutions

1. (a) Four minutes after the substance was placed in the oven, its temperature is 150 degrees.

(b) Four minutes after the substance was placed in the oven, its temperature is increasing at a rate of15 degrees per minute.

(c) The equation of the tangent line to T (t) at t = 4 is

l(t) = 15(t− 4) + 150.

An approximation to the temperature of the substance at time t = 4.1 is

T (4.1) ≈ l(4.1) = 15(4.1− 4) + 150 = 151.5.

38

(d) We expect T ′(1) to be positive since it seems likely that the temperature of the object is increasingafter it has been in the oven for one minute. This is because the original temperature of the object ismuch lower than the temperature of the oven. If the original temperature of the object was much hotter,say 1000◦, it might be cooling off after it has been in the oven for one minute since the temperature ofthe oven is only 425◦.

2. (a) The units of d′(t) are miles per hour.

(b) The units of d′′(t) are miles per hour2.

(c) d(3.5) = 26.21875

(d) Positive. If the runner continues to run the entire time, then the rate of change in the distance therunner has traveled will always be positive.

(e) The value d′(1) should be bigger than the value d′(3) since the runner is running faster at the 1 hourmark than they are at the 3 hour mark.

3. Since P ′(d) is the rate of change in pressure with respect to depth, we expect P ′(d) to be positive.The pressure on the wall increases as we go deeper into the tank.

4. (a) The units of V (t) are dollars.

(b) The units of V ′(t) are dollars per day.

(c) While the value of the stock at time t = 100 is V (100) = 200$, I would not purchase the stock for125$. The reason for this is that V ′(100) = −150 which means that at time t = 100, the value of thestock is decreasing by 150$ dollars per day.

5. (a) The units of D′(t) are centimeters per second.

(b) The units of D′′(t) are centimeters per second2.

(c) The sign of D′(2) would be negative. This is because the distance from the particle to its startingposition is decreasing at time t = 2 as the particle is returning to the origin.

(d) When t = 3, the rate of change in the displacement of the particle is 0 centimeters per second.

2.3 Basic Derivative Rules

1. Complete the statement of each derivative formula.

(a) If c is a constant, thend

dxc = .

(b) If n is any nonzero real number, thend

dxxn = .

(c)d

dxex =

(d)d

dxlnx =

2. Find the derivative of the given function.

39

(a) y = x3 + e2 + x (b) f(x) = 4x8 − 1

2x+ 1 (c) g(t) =

1

3t2 +

1

3t3 (d) r(x) = 3

√x+ 10x2/3

3. Finddy

dx.

(a) y = 3ex + cosx− lnx (b) y = 12 sinx+ 8− ex (c) y = x3 + 3x− 32

4. Find f ′(3) if f(x) = 4 lnx+ 9.

5. Find the slope of the tangent line to the graph of y = 3ex + cosx at x = 0.

6. Find y′ given y. Hint: Before taking the derivative of y, try to rewrite y in a different form.

(a) y = (x+ 3)2 (b) y =x2 + 8x− 1

x(c) y = cos2 x+ sin2 x+ ln(x2)

7. Find all values of x for which the tangent line to the graph of y = 3x2 + 2x− 1 is horizontal.

8. Find all values of x for which the tangent line to the graph of y = 3x2 + 2x− 1 is parallel to the liney = x.

9. Find the equation of the tangent line to y = cosx+√

2 at x =π

4.

10. Find the equation of the tangent line to f(x) =√x at x = 4.

11. If f(x) = 8x2 + sinx− ex, find f ′′(x).

12. Is there a value c such that the tangent line to the curve y = lnx at x = c has a negative slope?

13. The distance traveled by a race car during the first 6 minutes of a race is given by

d(t) = 272.143t2 + 72.5357t

where d(t) is in miles and t is in hours. How fast, in miles per hour, is the car going when exactly 4minutes has passed since the race began?

14. Give an example of a function f(x) such that f ′(x) = 3 for all x.

15. Suppose a(t) is the amount of water, in gallons, that has leaked from a pipe t seconds after a crackin the pipe ruptured. Find the units of a′(t) and interpret that statement that a′(10) = 3 in practicalterms.

16. The displacement of a particle t seconds after it was put into motion is given by the functions(t) = 1

100 t3 + 1

2 t2 + t where s(t) is in centimeters. Find the acceleration of the object at time t = 3.

Basic Derivative Rules - Solutions

1. (a) If c is a constant, thend

dxc = 0.

(b) If n is any nonzero real number, thend

dxxn = nxn−1.

40

(c)d

dxex = ex

(d)d

dxlnx =

1

x

2. (a) y′ = 3x2 + 1 (b) f ′(x) = 32x7 − 1

2(c) g′(t) =

2

3t+ t2 (d) r′(x) =

3

2x−1/2 +

20

3x−1/3

3. (a)dy

dx= 3ex − sinx− 1

x(b)

dy

dx= 12 cosx− ex (c)

dy

dx= 3x2 + 3

4. Since f ′(x) =4

x, we have f ′(3) =

4

3.

5. First note that y′ = 3ex − sinx. The slope of the tangent line to the graph of y at x = 0 is

y′(0) = 3e0 − sin 0 = 3.

6. (a) We can rewrite y as y = x2 + 6x+ 9 so y′ = 2x+ 6.

(b) We can rewrite y as

y =x2 + 8x− 1

x=x2

x+

8x

x− 1

x= x+ 8− 1

x= x+ 8− x−1.

Therefore, y′ = 1 + x−2 or y′ = 1 +1

x2.

(c) We can rewrite y as y = 1 + 2 lnx so that y′ =2

x.

7. The slope of the tangent line to y = 3x2 + 2x− 1 at x = c is y′(c) = 6c+ 2. A line is horizontal if ithas slope 0. This means that we need to solve 6c+ 2 = 0 since we want the slope of the tangent line atx = c to be 0. The only solution to 6c+ 2 = 0 is c = − 1

3 . The only value of x for which the tangent lineto the graph of y = 3x2 + 2x− 1 is horizontal is x = − 1

3 .

8. The line y = x has slope 1. Since y′ = 6x+ 2, we need to find all c for which 6c+ 2 = 1. Solving thisequation for c gives c = − 1

6 . The only value of x for which the tangent line to the graph of y = 3x2+2x−1is parallel to y = x is x = − 1

6 .

9. First we find y′ which is y′ = − sinx. We have y(π/4) = cos(π/4) +√

2 =√22 +

√2 = 3

√2

2 . Also,

y′(π/4) = − sin(π/4) = −√22 . The equation of the tangent line to y at x = π

4 is

l(x) = −√

2

2

(x− π

4

)+

3√

2

2.

10. By the Power Rule, f ′(x) =1

2x−1/2 so f ′(x) =

1

2√x

. Since f(4) = 2 and f ′(4) =1

2√

4=

1

4, the

equation of the tangent line to f(x) =√x at x = 4 is

l(x) =1

4(x− 4) + 2.

11. First we find f ′(x) which is f ′(x) = 16x+ cosx− ex. From this, we see that f ′′(x) = 16− sinx− ex.

12. Note that y′ =1

x. The domain of lnx is (0,∞) and for any number c in this interval, y′(c) =

1

c> 0.

Thus, the slope of y = lnx at any x value in the domain of lnx is positive. There can be no value of c

41

such the tangent line to y = lnx at x = c has a negative slope. Another way to see that this is true isto consider the graph of y = lnx and observe that the graph is always rising. More precisely, y = lnx isstrictly increasing on its domain.

13. We need to compute d′(4/60). Since d′(t) = 544.286t + 72.5357, we have d′(4/60) ≈ 108.821. Thecar is going about 109 miles per hour when exactly 4 minutes has passed since the beginning of the race.

14. The function f(x) = 3x has f ′(x) = 3 for all x. Similarly, if f(x) = 3x + 22, then f ′(x) = 3 for allx. The condition f ′(x) = 3 for all values of x means that the slope of the tangent line to the graph of fat any point must be 3. In general, if f(x) is any line with slope 3, then f ′(x) = 3.

15. The units of a′(t) are gallons per second. The statement a′(10) = 3 means that at the 10 secondmark, the water is leaking out of the pipe at a rate of 3 gallons per second.

16. We have s′(t) = 3100 t

2 + t and s′′(t) = 350 t + 1. The acceleration of the object at time t = 3 is

s′′(3) = 950 + 1 = 59

50 centimeters per second2.

2.4 The Product Rule and the Quotient Rule

1. Complete the statement of the Product Rule. If f and g are functions that are differentiable on I,then the function f(x)g(x) is differentiable on I and

d

dx(f(x)g(x)) = .

2. Complete the statement of the Quotient Rule. If f and g are functions that are differentiable on I

and g(x) 6= 0 for all x in I, then the function f(x)g(x) is differentiable on I and

d

dx

(f(x)

g(x)

)= .


(a) y = 3x4ex (b) F (x) = x1/2 sinx (c) h(x) =3x4

ex + 1(d) g(x) =

x3/2 − 1

−2 sinx


(a) y = (x3 + x+ 1)(x2 − 2x+ 8) (b) h(s) =s3 + 4s− 1

s2 − 8(c) F (r) =

2 ln r + 8

r9


(a) G(t) =t2et

sin t(b) F (v) =

ev

v2 sin v(c) y = u2eu cosu

6. (a) Compute the derivative of y = (x2 − 1)(x3 + 2) using the Product Rule.

42

(b) Compute the derivative of y = (x2 − 1)(x3 + 2) by first expanding (x2 − 1)(x3 + 2) and then usingthe Power Rule.

(c) Verify that the answers in (a) and (b) are the same.

7. Compute the derivative of y =x4 + 5x2 − 8

x2in two ways.

8. Find the derivative of the given function using the Product Rule.

(a) y = e2x (b) y = sin2 x

9. Find the equation of the tangent line to f(x) = x3ex at x = 0.

10. Find the equation of the tangent line to y =lnx

8x+ 1at x = 1.

11. If F (x) =f(x)g(x)

h(x) + f(x), find F ′(x). Your answer may involve the functions f(x), g(x), h(x), f ′(x),

g′(x), and h′(x).

12. Assume that n is a positive integer. Use the Quotient Rule and the Power Rule for positive exponentsto prove the formula

d

dxx−n = −nx−(n+1).

13. Suppose that f(x) = g(x)h(x), g′(3) = g(3) = 4, h(3) = 9, and f ′(3) = −16. Find h′(3).

14. If f(x) =x2

h(x), h′(−2) = 3, and h(−2) = 1, find f ′(−2).

15. Suppose that f(x) = g(x)h(x). If g′(x) = h(x) and g(x) = h′(x), show that f ′(x) ≥ 0 for all x.

The Product and the Quotient Rule - Solutions

1. If f and g are functions that are differentiable on I, then the function f(x)g(x) is differentiable on Iand

d

dx(f(x)g(x)) = f ′(x)g(x) + f(x)g′(x).

2. If f and g are functions that are differentiable on I and g(x) 6= 0 for all x in I, then the function f(x)g(x)

is differentiable on I andd

dx

(f(x)

g(x)

)=g(x)f ′(x)− f(x)g′(x)

g(x)2.

3. (a) y′ = 12x3ex + 3x4ex (b) F ′(x) = 12x−1/2 sinx+ x1/2 cosx

(c) h′(x) =(ex + 1)(12x3)− 3x4ex

(ex + 1)2(d) g′(x) =

(−2 sinx)( 32x

1/2)− (x3/2 − 1)(−2 cosx)

4 sin2 x

4. (a) y′ = (3x2+1)(x2−2x+8)+(x3+x+1)(2x−2) (b) h′(s) =(s2 − 8)(3s2 + 4)− (s3 + 4s− 1)(2s)

(s2 − 8)2

43

(c) F ′(r) =r9( 2

r )− (2 ln r + 8)(9r8)

r18

5. (a) G′(t) =(sin t)(2tet + t2et)− t2et(cos t)

sin2 t(b) F ′(v) =

(v2 sin v)ev − ev(2v sin v + v2 cos v)

v4 sin2 v

(c) y′ = 2u(eu cosu) + u2(eu cosu− eu sinu)

6. (a) y′ = (2x)(x3 + 2) + (x2 − 1)(3x2)

(b) We have y = (x2 − 1)(x3 + 2) = x5 − x3 + 2x2 − 2 so y′ = 5x4 − 3x2 + 4x.

(c) From (a), we have y′ = (2x)(x3 + 2) + (x2 − 1)(3x2) = 2x4 + 4x+ 3x4 − 3x2 = 5x4 − 3x2 + 4x whichagrees with our calculation from (b).

7. Computing y′ using the quotient rule gives

y′ =x2(4x3 + 10x)− (x4 + 5x2 − 8)(2x)

x4.

We could also rewrite y as y = x2 + 5− 8

x2and then use the Power Rule to obtain y′ = 2x+

16

x3.

8. (a) Since y = e2x = exex , we find that y′ = exex + exex = 2e2x

(b) We have y = sin2 x = sinx sinx so by the Product Rule, y′ = cosx sinx+ sinx cosx = 2 sinx cosx.

9. By the Product Rule, f ′(x) = 3x2ex + x3ex so f ′(0) = 0 and f(0) = 0. The equation of the tangentline to f(x) at x = 0 is l(x) = 0.

10. Using the Quotient Rule, y′ =(8x+ 1)(1/x)− (lnx)(8)

(8x+ 1)2. Putting 1 in for x gives

y′ =9− 8 ln(1)

92=

1

9.

Since y(0) = 0, the equation of the tangent line to y at x = 1 is l(x) =1

9(x− 1).

11. F ′(x) =(h(x) + f(x))(f ′(x)g(x) + f(x)g′(x))− f(x)g(x)(h′(x) + f ′(x))

(h(x) + f(x))2

12. Using the Quotient Rule and the Power Rule (for positive exponents), we have

d

dxx−n =

d

dx

(1

xn

)=xn · 0− 1nxn−1

x2n= − n

xn+1= −nx−(n+1).

13. By the Product Rule, f ′(x) = g′(x)h(x) + g(x)h′(x) so f ′(3) = g′(3)h(3) + g(3)h′(3). Substitutingthe given values into this equation gives −16 = 4 · 9 + 4h′(3). Solving for h′(3) gives h′(3) = −13.

14. By the Quotient Rule, f ′(x) =h(x)2x− x2h′(x)

(h(x))2. When x = −2, we get

f ′(−2) =h(−2)2(−2)− (−2)2h′(−2)

(h(−2))2=−4− 4(3)

12= −16.

15. By the Product Rule, f ′(x) = g′(x)h(x) + g(x)h′(x). From the assumptions g′(x) = h(x) andg(x) = h′(x), we can rewrite f ′(x) as f ′(x) = h(x)h(x) + h′(x)h′(x) = (h(x))2 + (h′(x))2. Since(h(x))2 + (h′(x))2 ≥ 0 for all x, we have that f ′(x) ≥ 0 for all x.

44

2.5 The Chain Rule

1. Complete the statement of the Chain Rule. If f and g are functions where g is differentiable at x andf is differentiable at g(x), then the function f(g(x)) is differentiable at x and

d

dx(f(g(x))) = .

2. Suppose that g is a function that is differentiable at x. Use the Chain Rule to complete each statement.

(a)d

dxeg(x) = .

(b)d

dxsin(g(x)) = .

(c)d

dx(g(x))40 = .

3. Complete the statement. If a is a positive real number, then

d

dxax = .


(a) y = (ex + 1)8 (b) F (t) = 3 sin8 t (c) H(r) = (ln r − 1)8 (d) f(x) = (3x4 − x+ 3)8


(a) s(t) = et2+t−1 (b) r(t) = 3esin t−cos t (c) f(s) = ee

s+2 (d) r(s) = −12e−3


(a) r(x) = ln(x4 + x2 + 8) (b) F (v) = cos(3v2 + 8) (c) g(y) = sin(ey + y2)


(a) r(x) = tan(3x+ 2) (b) F (t) = sec(3t2 − et) (c) G(s) = 3 ln(es + s2)

8. Finddy

dx.

(a) y = 2x + x2 (b) y = 2x2

+ (x2)2 (c) y = 2sin x + sin2 x

9. Find f ′(x) given f(x).

(a) f(x) = (x sinx)4 (b) f(x) =

(x

sinx+ ex

)4

(c) f(x) = xe3x4+5

45

10. Let f(x) = e2x. Find f ′′′(x). Can you find a formula for f (n)(x)?

11. Suppose that n is a positive even integer with n ≥ 2 and f(x) = (g(x))n. If g(3) = −1 and g′(3) = 2,find f ′(3).

12. Let f(x) = g(h(x)) and suppose that f ′(3) 6= 0, f ′(3) = h′(3), and h(3) = 4. Find g′(4).

13. Suppose that k is a non-zero real number and f(x) = g(kx). Assuming that f is differentiable, usethe Chain Rule to show that f ′(x) = kg′(kx).

14. Find the equation of the tangent line to the graph of y = sin(3x) at x = −π3

.

15. Let f(x) = x√

1 + x2. Find the values of x for which the slope of the tangent to f at x is 0.

16. Let L and k be positive real numbers and p(t) =L

1 + e−kt. The function p(t) is an example of

a logistic function and can be used to model the growth of a population where size of the populationcannot exceed L.

(a) Show that p(t) < L for all t.

(b) Show that limt→∞

p(t) = L.

(c) Find p′(t).

Remark: In terms of population sizes, parts (a) and (b) show that the size of the population is alwaysless than L, and as time goes by, the size of the population gets closer and closer to L.

17. Let A and ω be non-zero real numbers and y(t) = A cos(ωt). The function y(t) describes the motionof an object that is moving in simple harmonic motion. The value A controls the amplitude of themotion while the value ω controls the frequency. Show that y satisfies the equation

y′′ = −ω2y

by computing y′′, and verifying that y′′ is equal to −ω2y.

The Chain Rule - Solutions

1. If f and g are functions where g is differentiable at x and f is differentiable at g(x), then the functionf(g(x)) is differentiable at x and

d

dx(f(g(x))) = f ′(g(x))g′(x).

2. (a)d

dxeg(x) = eg(x)g′(x) (b)

d

dxsin(g(x)) = cos(g(x))g′(x)

(c)d

dx(g(x))40 = 40(g(x))39g′(x)

3. If a is a positive real number, thend

dxax = ax ln(a).

4. (a) y′ = 8(ex + 1)7ex (b) F ′(t) = 24(sin t)7(cos t) (c) H ′(r) = 8(ln r − 1)7 1r

46

(d) f ′(x) = 8(3x4 − x+ 3)7(12x3 − 1)

5. (a) s′(t) = et2+t−1(2t+ 1) (b) r′(t) = 3esin t−cos t(cos t+ sin t) (c) f ′(s) = ee

s+2es (d) r′(s) = 0

6. (a) r′(x) =4x3 + 2x

x4 + x2 + 8(b) F ′(v) = − sin(3v2 + 8)(6v) (c) g′(y) = cos(ey + y2)(ey + 2y)

7. (a) r′(x) = sec2(3x+ 2)(3) (b) F ′(t) = sec(3t2 − et) tan(3t2 − et)(6t− et) (c) G′(s) =3(es + 2s)

es + s2

8. (a) y′ = 2x ln 2 + 2x (b) y′ = 2x2

(ln 2)(2x) + 4x3 (c) y′ = 2sin x(ln 2)(cosx) + 2 sinx cosx

9. (a) f ′(x) = 4(x sinx)3(sinx+ x cosx) (b) f ′(x) = 4

(x

sinx+ ex

)3(sinx+ ex − x(cosx+ ex)

(sinx+ ex)2

)(c) f ′(x) = e3x

4+5 + xe3x4+5(12x3)

10. We compute three derivatives to find f ′′′(x) = 23e2x. Indeed, f and its first three derivatives are

f(x) = e2x, f ′(x) = 2e2x, f ′′(x) = 22e2x, and f ′′′(x) = 23e2x.

In general, f (n)(x) = 2ne2x for any positive integer n.

11. By the Chain Rule, f ′(x) = n(g(x))n−1g′(x). When x = 3, we get

f ′(3) = n(g(3))n−1g′(3) = n(−1)n−1(2).

When n is even, n− 1 is odd so (−1)n−1 = −1. Therefore,

f ′(3) = n(−1)n−1(2) = n(−1)2 = −2n.

12. By the Chain Rule, f ′(x) = g′(h(x))h′(x). When x = 3, we have f ′(3) = g′(h(3))h′(3). Sincef ′(3) = h′(3) and f ′(3) is not zero, we can cancel f ′(3) from both sides of f ′(3) = g′(h(3))h′(3) to get1 = g′(h(3)). Substituting h(3) = 4, we get 1 = g′(4).

13. By the Chain Rule, f ′(x) = g′(kx)k = kg′(kx).

14. By the Chain Rule, y′ = 3 cos(3x). We have

y(−π/3) = sin(−π) = 0 and y′(−π/3) = 3 cos(−π) = −3.

The equation of the tangent line to y = sin(3x) at x = −π3 is

l(x) = −3(x−

(−π

3

))+ 0 = −3

(x+

π

3

).

15. From the Chain Rule and the Product Rule, we find that

f ′(x) =√

1 + x2 + x(1/2)(1 + x2)−1/2(2x) =√

1 + x2 + x

(2x

2√

1 + x2

)=√

1 + x2 +x2√

1 + x2.

Next we solve

0 =√

1 + x2 +x2√

1 + x2.

47

This equation is equivalent to

−√

1 + x2 =x2√

1 + x2

which has no solutions since the left hand side is always negative and the right hand side is nevernegative. The slope of the tangent line to f(x) = x

√1 + x2 is never equal to 0.

16. (a) Since e−kt > 0 for all t, we have 1 + e−kt > 1. Therefore ,

p(t) =L

1 + e−kt<L

1= L

and this holds for all t.

(b) Using the fact that limt→∞

e−kt = 0 since k > 0, we have

limt→∞

p(t) = limt→∞

L

1 + e−kt=

L

1 + 0= L.

(c) p′(t) =(1 + e−kt) · 0− L(e−kt(−k))

(1 + e−kt)2=

Lke−kt

(1 + e−kt)2

17. Using the Chain Rule, we find that y′(t) = −Aω sin(ωt) and y′′(t) = −Aω2 cos(ωt). Since

−ω2y(t) = −ω2A cos(ωt) = −Aω2 cos(ωt) = y′′(t),

we have −ω2y = y′′.

2.6 Implicit Differentiation


(a) f(x) =−8√x

(b) G(t) = 7t5/2 cos(−t) (c) s(t) = 2teet3


(a) r(x) = 3√

3x2 + 1 (b) t(x) =1

2(x2 − 1)3/2(c) f(x) = 8x7/3 + 4x3/7

3. Finddy

dxusing implicit differentiation.

(a) x+ y3 = y2 + 1 (b) sinx+ y3 = y2 + 1

4. Finddy

dxusing implicit differentiation.

(a) xy + x2 + y2 = x (b) y sinx+ x cos y = 0

48

5. Find y′ using implicit differentiation.

(a) ln(x4 + y2) = x (b) (x4 + y2)8 = 10 (c) exy = x+ y

6. Find the equation of the tangent line to the curve x4 + y2 = 2 at the point (1,−1).

7. Find the equation of the tangent line to the curve x sin y + y sinx = x

√2

2at the point (π, π/4).

8. Find all points on the unit circle x2 + y2 = 1 in which the slope of the tangent line to the circle is 1.

Implicit Differentiation - Solutions

1. (a) First we rewrite f(x) as f(x) = −8x−1/2. By the Power Rule, f ′(x) = 4x−3/2 or, equivalently,

f ′(x) =4√x3

.

(b) Applying the Product Rule and Chain Rule gives G′(t) =35

2t3/2 cos(−t) + 7t5/2 sin(−t).

(c) By the Product Rule and Chain Rule, s′(t) = 2ete−1et3

+ 2teet3

(3t2).

2. (a) First we rewrite r(x) as r(x) = (3x2 + 1)1/3. By the Power Rule and Chain Rule,

r′(x) = (1/3)(3x2 + 1)−2/3(6x) or r′(x) =2x

(3x2 + 1)2/3.

(b) First we rewrite t(x) as t(x) = 12 (x2 − 1)−3/2. Applying the Power Rule and Chain Rule,

t′(x) =

(1

2

)(−3

2

)(x2 − 1)−5/2(2x) or t′(x) =

−3x

2(x2 − 1)5/2.

(c) f ′(x) =56

3x4/3 +

12

7x−4/7 or f ′(x) =

56

3x4/3 +

12

7x4/7

3. (a) We begin by taking the derivative of both sides of x+ y3 = y2 + 1:

d

dx(x+ y3) =

d

dx(y2 + 1) =⇒ 1 + 3y2y′ = 2yy′ =⇒ 1 = (2y − 3y2)y′.

Solving this last equation for y′ gives y′ =1

2y − 3y2.

(b) We use the same strategy that was used in part (a).

d

dx(sinx+ y3) =

d

dx(y2 + 1) =⇒ cosx+ 3y2y′ = 2yy′ =⇒ cosx = (2y − 3y2)y′.

Solving for y′ gives y′ =cosx

2y − 3y2.

4. (a) We begin by taking the derivative of both sides of xy + x2 + y2 = x:

d

dx(xy + x2 + y2) =

d

dxx =⇒ y + xy′ + 2x+ 2yy′ = 1 =⇒ (x+ 2y)y′ = 1− y − 2x.

49

Here we have used the Product Rule to take the derivative of xy. Upon solving this last equation for y′

we obtain

y′ =1− y − 2x

x+ 2y.

(b) Using the Product Rule and Implicit Differentiation,

d

dx(y sinx+ x cos y) =

d

dx0 =⇒ y′ sinx+ y cosx+ cos y + x(− sin y)y′ = 0.

Solving for y′ gives y′ =−y cosx− cos y

sinx− x sin y.

5. (a) We haved

dx(ln(x4 + y2)) =

d

dxx =⇒ 1

x4 + y2(4x3 + 2yy′) = 1.

Here we have used the Chain Rule to differentiate ln(x4 + y2). The equation

1

x4 + y2(4x3 + 2yy′) = 1

can be rewritten as 4x3 + 2yy′ = x4 + y2. Solving this equation for y′ gives y′ =x4 + y2 − 4x3

2y.

(b) We haved

dx(x4 + y2)8 =

d

dx10 =⇒ 8(x4 + y2)7(4x3 + 2yy′) = 0.

Since (x4 + y2)7 is not zero (the reason that this is true is because (x4 + y2)8 = 10), we may cancel

8(x4 + y2)7 to get 4x3 + 2yy′ = 0. Solving this equation for y′ gives y′ = −4x3

2y.

(c) We haved

dxexy =

d

dx(x+ y) =⇒ exy(y + xy′) = 1 + y′.

This equation can be rewritten as exyy + exyxy′ = 1 + y′ so yexy − 1 = y′ − xexyy′. Factoring out they′ on the right leads to yexy − 1 = y′(1− xexy) so

y′ =yexy − 1

1− xexy.

6. The first step is to find the slope of curve at the point (1,−1). By Implicit Differentiation,

d

dx(x4 + y2) =

d

dx2 =⇒ 4x3 + 2yy′ = 0 =⇒ 2y = −4x3 =⇒ y = −2x3

y.

At the point (1,−1) we know that x = 1 and y = −1 so

y′ = −2(1)3

−1= 2.

The equation of the tangent line to x4 + y2 = 2 at the point (1,−1) is l(x) = 2(x− 1) + (−1).

7. Taking the derivative of both sides of x sin y + y sinx = x

√2

2gives

sin y + x(cos y)y′ + y′ sinx+ y cosx =

√2

2.

50

We solve this equation for y′ to get

y′ =

√22 − y cosx− sin y

x cos y + sinx.

To find y′ at the point (π, π/4) we substitute π for x and π/4 for y:

y′ =

√22 − (π/4) cos(π)− sin(π/4)

π cos(π/4) + sin(π)=

√22 + π

4 −√22

π(√

2/2) + 0=

π4

π√2

2

=1

2√

2.

The equation of the tangent line to the curve x sin y + y sinx = x

√2

2at the point (π, π/4) is

l(x) =1

2√

2(x− π) +

π

4.

8. Using Implicit Differentiation,

d

dx

(x2 + y2

)=

d

dx1 =⇒ 2x+ 2yy′ = 0 =⇒ 2yy′ = −2x =⇒ y′ = −x

y.

Now we set y′ = 1 since we are looking for the values of x and y for which the slope of the tangent line

to x2 + y2 = 1 at (x, y) is 1. Setting y′ = 1 gives 1 = −xy

so y = −x. We know substitute y = −x into

x2 + y2 = 1 to get

x2 + (−x)2 = 1 =⇒ 2x2 = 1 =⇒ x2 =1

2=⇒ x = ± 1√

2.

If x =1√2

, then y = − 1√2

. If x = − 1√2

, then y =1√2

. The points on the unit circle x2 + y2 = 1 where

the slope of the tangent line is 1 are

(1√2,− 1√

2

)and

(− 1√

2,

1√2

).

2.7 Logarithmic Differentiation

1. Suppose that x and y are positive real numbers. Indicate whether or not each statement is alwaystrue or sometimes false.

(a) ln(xy) = lnx+ ln y

(b) ln(x+ y) = lnx+ ln y

(c) ln

(x

y

)= lnx− ln y

(d) If n is a real number, then ln(xn) = n lnx.

2. If y = x2x, find y′ using Logarithmic Differentiation.

3. Find f ′(x) if f(x) = xsin x.

51

4. Use Logarithmic Differentiation to find y′ if y = tt cos t.

5. Let y =

√x− 1

x+ 1.

(a) Find y′ using the Chain Rule together with the Quotient Rule.

(b) Using properties of logarithms, show that ln y = 12 ln(x− 1)− 1

2 ln(x+ 1).

(c) Use Implicit Differentiation and the equation ln y = 12 ln(x − 1) − 1

2 ln(x + 1) to find y′. Be sure towrite y′ as a function of x.

6. Find y′ if y =x1/2(x+ 1)3/2

(x− 1)1/3.

7. Let a be a positive real number and let y = ax.

(a) Show that ln y = x ln a.

(b) Differentiate both sides of ln y = x ln a using Implicit Differentiation.

(c) Use your calculation from (b) to conclude that

d

dxax = ax ln a.

8. Find y′ given y.

(a) y = 3x (b) y =

(1

3

)x2+1

(c) y = x2x

Logarithmic Differentiation - Solutions

1. (a) For any positive real numbers x and y, it is always true that ln(xy) = lnx+ ln y.

(b) If x and y are positive real numbers, then it is not always true that ln(x+y) is the same as lnx+ln y.For example, ln(1 + e) ≈ 1.3133 but ln(1) + ln e = 0 + 1 = 1.

(c) The statement ln

(x

y

)= lnx− ln y is always true for any pair of positive real numbers x and y.

(d) If n is a real number and x is a positive real number, then it is always true that ln(xn) = n lnx.

2. We start by taking the natural logarithm of both sides of y = x2x to get ln y = ln(x2x). By propertiesof logarithms, we can rewrite this equation as ln y = 2x lnx. Next we differentiate both sides usingImplicit Differentiation:

d

dxln y =

d

dx(2x lnx) =⇒ 1

yy′ = 2 lnx+ 2x(1/x).

Therefore, y′ = y(2 lnx+ 2). We finish by replacing y with x2x to get y′ = x2x(2 lnx+ 2).

52

3. Taking the natural logarithm of both sides of f(x) = xsin x gives ln f(x) = ln(xsin x). Using propertiesof logarithms, we rewrite this equation as ln f(x) = sinx lnx. Differentiate both sides using ImplicitDifferentiation:

d

dxln f(x) =

d

dx(sinx lnx) =⇒ 1

f(x)f ′(x) = cosx lnx+ sinx(1/x).

Solving for f ′(x) gives f ′(x) = f(x)(cosx lnx+ 1x sinx) and since f(x) = xsin x,

f ′(x) = xsin x(

cosx lnx+1

xsinx

).

4. We follow the same strategy as in the previous two problems:

ln y = ln(tt cos t) =⇒ ln y = t cos t ln t =⇒ d

dtln y =

d

dtt cos t ln t

We use the Product Rule two times to differentiate t cos t ln t and so

1

yy′ = (cos t− t sin t) ln t+ (t cos t)

1

t.

Upon solving for y′ and replacing y with tt cos t we obtain

y′ = tt cos t(

(cos t− t sin t) ln t+ (t cos t)1

t

).

5. (a) y′ =1

2

(x− 1

x+ 1

)−1/2((x+ 1)− (x− 1)

(x+ 1)2

)=

1

2

(x− 1

x+ 1

)−1/2(2

(x+ 1)2

)

(b) We have ln y = ln

((x− 1

x+ 1

)1/2)

so that

ln y =1

2ln

(x− 1

x+ 1

).

By properties of logarithms, we can rewrite this last equation as ln y = 12 ln(x− 1)− 1

2 ln(x+ 1).

(c) By taking the derivative of both sides of ln y = 12 ln(x− 1)− 1

2 ln(x+ 1), we get

1

yy′ =

1

2(x− 1)− 1

2(x+ 1).

Multiplying both sides by y and replacing y with√

x−1x+1 gives

y′ =

√x− 1

x+ 1

(1

2(x− 1)− 1

2(x+ 1)

).

6. One can certainly use the Chain Rule, the Product Rule, and the Quotient Rule to find y′. Instead,we will use Logarithmic Differentiation. Taking the natural logarithm of both sides gives

ln y = ln

(x1/2(x+ 1)3/2

(x− 1)1/3

).

We use properties of logarithms to expand the right hand side and get

ln y =1

2lnx+

3

2ln(x+ 1)− 1

3ln(x− 1).

53

Next, differentiate both sides of this equation to obtain

1

yy′ =

1

2x+

3

2(x+ 1)− 1

3(x− 1).

Multiplying both sides by y and replacing y with y =x1/2(x+ 1)3/2

(x− 1)1/3gives

y′ =

(x1/2(x+ 1)3/2

(x− 1)1/3

)(1

2x+

3

2(x+ 1)− 1

3(x− 1)

).

7. (a) Taking the natural logarithm of both sides of y = ax gives ln y = ln(ax) which can be rewrittenas ln y = x ln a.

(b) Taking the derivative of both sides of ln y = x ln a, we get 1yy′ = ln a so y′ = y ln a.

(c) From part (b), we have y′ = y ln a = ax ln a so thatd

dxax = ax ln a.

8. (a) y′ = 3x ln 3

(b) y′ =

(1

3

)x2+1

ln

(1

3

)(2x)

(c) y′ = 2x + x2x ln 2

2.8 Derivatives of Inverse Functions

1. For −1 < x < 1, we have the equation sin(sin−1 x) = x and this will be the starting point for a proofof the formula

d

dxsin−1 x =

1√1− x2

.

(a) By differentiating both sides of sin(sin−1 x) = x, show that

d

dxsin−1 x =

1

cos(sin−1 x).

(b) Since sin(sin−1 x) = x =x

1=

oppositeadjacent

, we can label the sides of a right triangle as shown below.

��

��

��

��

��

��

�

sin−1 x

x

1

54

Find the length of the side that is adjacent to the angle sin−1 x in terms of x, and then write

cos(sin−1 x) =adjacent

hypotenuse

as an expression in x that does not involve any trigonometric functions.

(c) Combine your work from (a) and (b) to prove the formula

d

dxsin−1 x =

1√1− x2

.

2. Prove the formulad

dxcos−1 x = − 1√

1− x2. You may want to start by differentiating both sides of

cos−1 x = x which is true if −1 < x < 1.

3. Proved

dxtan−1 x =

1

1 + x2.


(a) y = secx+ tan−1(x2) (b) u(t) = 3 cos−1(4t) (c) H(r) = r2 sin−1(−2r)

5. Find the derivative of the function.

(a) q(x) = tan−1(x3 + 1) sin−1 x (b) H(y) = y−1 cos−1 y

Derivatives of Inverse Functions - Solutions

1. (a) Differentiating both sides of sin(sin−1 x) = x using the Chain Rule gives

cos(sin−1 x)d

dxsin−1 x = 1.

Divide both sides by cos(sin−1 x) to get

d

dxsin−1 x =

1

cos(sin−1 x).

(b) If y is the length that we are looking for, then by the Pythagorean Theorem,

x2 + y2 = 12

so y =√

1− x2. Here we took only the positive square root since y is a distance. The length of the sidethat is adjacent to the angle sin−1 x is

√1− x2. From this we now have

cos(sin−1 x) =adjacent

hypotenuse=

√1− x2

1=√

1− x2.

(c) By parts (a) and (b),d

dxsin−1 x =

1

cos(sin−1 x)=

1√1− x2

.

55

This formula is true for −1 < x < 1.

2. Taking the derivative of both sides of cos(cos−1 x) = x gives

− sin(cos−1 x)d

dxcos−1 x = 1.

This equation can be rewritten asd

dxcos−1 x =

1

− sin(cos−1 x). Since

cos(cos−1 x) = x =x

1=

adjacent

hypotenuse,

we can label a right triangle as shown below.

��

��

��

��

��

��

cos−1 x

√1− x2

1

x

The Pythagorean Theorem was used to find the length√

1− x2. Using this right triangle, we can rewritesin(cos−1 x) as

sin(cos−1 x) =opposite

hypotenuse=

√1− x2

1=√

1− x2.

Therefore,d

dxcos−1 x = − 1

sin(cos−1 x)= − 1√

1− x2.

3. Differentiating both sides of tan(tan−1 x) = x gives sec2(tan−1 x)d

dxtan−1 x = 1. We can rewrite

this equation asd

dxtan−1 x =

1

sec2(tan−1 x). Using the right triangle shown below, we can rewrite

sec2(tan−1 x) as

sec2(tan−1 x) =

(hypotenuse

adjacent

)2

=

(√1 + x2

1

)2

= 1 + x2.

Therefore,d

dxtan−1 x =

1

sec2(tan−1 x)=

1

1 + x2.

��

��

��

��

��

��

tan−1 x

x

√1 + x2

1

4. (a) y′ = secx tanx+2x

1 + x4(b) u′(t) = 3

(− 1√

1− 16t2

)· 4

56

(c) H ′(r) = 2r sin−1(−2r) + r2(

1√1− 4r2

)(−2)

5. (a) q′(x) =

(1

1 + (x3 + 1)2

)3x2 sin−1 x+ tan−1(x3 + 1)

1√1− x2

(b) H ′(y) = −y−2 cos−1 y + y−1

(− 1√

1− y2

)

2.9 Related Rates

1. The radius of a circle is increasing at a rate of 2 centimeters per second.

(a) Find the rate of change in the circumference of the circle.

(b) At the moment that the radius is 10 centimeters, what is the rate of change in the area of the circle?

2. (a) The volume V of a sphere of radius r is given by the formula V = 43πr

3. If the radius of thesphere is increasing at a rate of 3 inches per second, find a formula that gives the rate of change in thevolume of the sphere in terms of r.

(b) If the radius of a sphere is increasing at a rate of 3 inches per second, find how fast the volume isincreasing when the radius is 19 inches.

3. A ladder that is 12 feet long leans against a building. If the top of the ladder slides down the buildingat a rate of 2 feet per second, how fast is the bottom of the ladder sliding away from the building whenthe bottom of the ladder is 8 feet from the wall?

4. A ladder that is 12 feet long leans against a building. If the top of the ladder slides down the buildingat a rate of 2 feet per second, how fast is the bottom of the ladder sliding away from the building whenthe top of the ladder is 2 feet above the ground?

5. An airplane is flying from east to west at a speed of 500 miles per hour and at an altitude of 1.5 miles.The airplane is approaching a radar station. At the radar station, a camera is tracking the airplane asit flies. How fast is the camera rotating when the plane is .25 miles east of the station?

6. The surface area A of a sphere of radius r is A = 4πr2. Suppose that the surface area of the sphereis shrinking at a rate of 3 in2 per minute.

(a) Find the rate of change in the radius when the radius is 100 inches.

(b) Find the rate of change in the radius when the radius is 10 inches.

Related Rates - Solutions

1. (a) If C is the circumference of the circle and r is the radius, then C = 2πr. Taking the derivative ofboth sides of this equation with respect to t yields dC

dt = 2π drdt . We are given that drdt is equal to 2 cm

per second sodC

dt= 2π(2) = 4π.

57

The rate of change in the circumference is 4π cm per second.

(b) If A is the area of the circle and r is the radius, then A = πr2. Taking the derivative of both sides ofthis equation with respect to t gives dA

dt = 2πr drdt . We are given that drdt = 2 and we are at the moment

in time when r = 10. We havedA

dt= 2π(10)(2) = 40π.

The rate of change in the area when the radius is 10 cm is 40π cm2 per second.

2. (a) Taking the derivative of both sides of V = 43πr

3 with respect to t gives

dV

dt= 4πr2

dr

dt.

We are given drdt = 3 so we have dV

dt = 4πr2(3) = 12πr2 in3 per second.

(b) Using the equationdV

dt= 12πr2 from part (a), we put 19 in for r to get

dV

dt= 12π(19)2 = 4332π.

The volume of the sphere is increasing at a rate of 4332π in3 per second.

3. Let x be the height of the ladder against the wall and y be the distance between the wall and thebottom of the ladder. By the Pythagorean Theorem, x2 + y2 = 122. Taking the derivative of both sidesof this equation with respect to t gives 2xdxdt + 2y dydt = 0 which can be rewritten as xdxdt + y dydt = 0. We

are given y = 8 and dxdt = −2 (the negative sign is here because the ladder is sliding down the wall so its

height against the wall is decreasing as time goes by). We can substitute these values into xdxdt +y dydt = 0to obtain

−2x+ 8dy

dt= 0.

We have only one equation but two unknowns. We can find x using x2 + y2 = 122 and y = 8. This givesx2 + 64 = 144 so x =

√80. We took only the positive square root here since x is representing a distance.

Now we have

−2√

80 + 8dy

dt= 0.

We can isolatedy

dtto get

dy

dt=

2√

80

8=

2√

16 · 58

=8√

5

8=√

5.

The bottom of the ladder is sliding away from the wall at a rate of√

5 feet per second.

4. Let x be the height of the ladder against the wall and y be the distance between the wall and thebottom of the ladder. As in the previous problem, we have x2 + y2 = 122 and xdxdt + y dydt = 0. We

are given dxdt = −2 and we want to find dy

dt when x = 2. If x = 2, then from x2 + y2 = 122, we find

that y =√

140. As in the previous problem, we take only the positive square root since y represents adistance. Using these values in xdxdt + y dydt = 0 gives

2(−2) +√

140dy

dt= 0

so dydt = 4√

140= 2√

35. The bottom of the ladder is sliding away from the wall at a rate of 2√

35feet per

second.

5. Let x be the horizontal distance between the radar station and the plane. Let θ be the angle madeby the the line of sight from the station to the plane and the ground.

58

��

��

��

��

��

��

�

θ

1.5

xstation

plane

From the right triangle shown above, we see that tan θ =1.5

x. We want to find dθ

dt when x = 0.25. We

know that dxdt = −500. There is a negative sign here since x is decreasing as time goes by. Taking the

derivative of both sides of tan θ =1.5

xwith respect to t gives

sec2 θdθ

dt= −1.5

x2dx

dt.

When x = 0.25, the hypotenuse of the right triangle is√

1.52 + 0.252. Therefore, when x = 0.25,

sec2 θ =1

cos2 θ=

(hypotenuse

adjacent

)2

=

(√1.52 + 0.252

0.25

)2

=1.52 + .252

.252.

Using this equation together with dxdt = −500 and x = 0.25, we substitute into sec2 θ dθdt = − 1.5

x2dxdt to get

1.52 + .252

.252dθ

dt= − 1.5

.252(−500).

We solve this equation for dθdt to find that the camera is rotating at a rate of

500(1.5)

1.52 + .252≈ 324 radians

per hour when the plane is 0.25 miles east of the camera.

6. (a) From A = 4πr2, we get dAdt = 8πr drdt . We are given dA

dt = −3 and we are asked to find drdt when

r = 100. Upon substitution, we find

−3 = 8π(100)dr

dt

so drdt = − 3

800π . The rate of change in the radius of the sphere when r = 100 is−3

800π≈ −0.0012 inches

per minute.

(b) A similar calculation as in (a) shows that the rate of change in the radius of the sphere when r = 10

is−3

80π≈ −0.012 inches per minute.

59

3 Derivatives and the Shapes of Graphs

3.1 Maximum and Minimum Values

1. Complete the statement of the Extreme Value Theorem. If f is a functionon the interval [a, b], then f has a maximum value on [a, b] and a minimum value on [a, b].

2. (a) True or False: The function f(x) = ex has a maximum value on [0, 3].

(b) True or False: The function f(x) = ex has a maximum value on (−∞,∞).

(c) True or False: The function f(x) = ex has a minimum value on (−∞,∞).

3. True or False: There is only one value of x in the interval [−2, 2] for which f(x) = x2 attains itsmaximum value.

4. Find the absolute minimum and the absolute maximum of f(x) = 10x3 + 15x2 − 360x + 32 on theinterval [−2, 4].

5. Find the absolute minimum and the absolute maximum of f(x) = 2x+8

xon the interval [1, 9].

6. Find the extreme values of f(x) = lnx+1

xon the interval [e−3, e2].

7. Find the extreme values of f(x) = 2x4 − 16x2 + 24 on the interval [−1, 3].

8. A ball thrown from the top of ladder that is 5.5 feet tall. The height of the ball t seconds after it isreleased is h(t) = −2t2 + 6t+ 11

2 for 0 ≤ t ≤ 32 +√

5. Find the maximum height reached by the ball andthe time at which the ball is at its maximum height.

9. The depth, in feet, of water in a tank at time x, in hours, is f(x) =√

3 cosx+ sinx+ 100. Find themaximum depth reached by the water from time x = 0 to time x = π.

Maximum and Minimum Values - Solutions

1. If f is a continuous function on the interval [a, b], then f has a maximum value on [a, b] and a minimumvalue on [a, b].

2. (a) True. The function f(x) = ex is continuous on [0, 3]. By the Extreme Value Theorem, it has amaximum value and a minimum value on [0, 3].

(b) False. As x tends to infinity, the function f(x) = ex also tends to infinity. Therefore, there is no realnumber x0 for which f(x0) ≥ f(x) for all x in (−∞,∞).

(c) False. The function f(x) = ex does not have a minimum value on (−∞,∞).

3. False. The function f(x) = x2 attains its maximum value on [−2, 2] at x = −2 and at x = 2.

4. We first evaluate f at the endpoints of the interval:

f(−2) = 732 and f(4) = −528.

60

Next we find all of the critical points of f that are in the interval [−2, 4]. We have f ′(x) = 30x2+30x−360.Now we solve f ′(x) = 0:

0 = 30x2 + 30x− 360 = 30(x2 + x− 12) = 30(x+ 4)(x− 3).

The only critical point of f in the interval [−2, 4] is x = 3. Evaluating f at x = 3 gives f(3) = −643.The absolute minimum of f(x) on [−2, 4] is f(3) = −643 and the absolute maximum of f(x) on [−2, 4]is f(−2) = 732.

5. Evaluating f at the endpoints of the interval gives f(1) = 10 and f(9) = 1709 . Now we find the critical

points of f in the interval [1, 9].

f ′(x) = 0 ⇒ 2− 8

x2= 0 ⇒ 2 =

8

x2⇒ x2 = 4

The solutions to x2 = 4 are x = 2 and x = −2 but −2 is not in the interval [1, 9]. Since f(2) = 8, the

absolute minimum of f(x) on [1, 9] is f(2) = 8. The absolute maximum of f(x) on [1, 9] is f(9) =170

9.

6. We have f(e−3) = −3 +1

e−3= −3 + e3 and f(e2) = 2 +

1

e2. Next we find the critical points of f in

the interval [e−3, e2].

f ′(x) = 0 ⇒ 1

x− 1

x2= 0 ⇒ x− 1 = 0⇒ x = 1

To go from the equation1

x− 1

x2= 0 to x− 1 = 0, we multiplied through by x2 which can be done since

x 6= 0 on the interval [e−3, e2]. Since f(1) = 1, the absolute minimum of f(x) on [e−3, e2] is f(1) = 1.The absolute maximum of f(x) on [e−3, e2] is f(e−3) = −3 + e3.

7. We have f(−1) = 10 and f(3) = 42. The next step is to find the critical points of f in the interval[−1, 3].

f ′(x) = 0 ⇒ 8x3 − 32x = 0 ⇒ 8x(x2 − 4) = 0 ⇒ 8x(x− 2)(x+ 2) = 0

The critical points of f in [−1, 3] are x = 0 and x = 2. Evaluating f at these points gives f(0) = 24 andf(2) = −8. The absolute minimum of f(x) on [−1, 3] is f(2) = −8 and the absolute maximum of f(x)on [−1, 3] is f(3) = 42.

8. We want to find the maximum value of h(t) = −2t2 + 6t+ 112 on the interval [0, 32 +

√5]. Note that

h(0) = 112 and h( 3

2 +√

5) = 0. Solving h′(t) = 0 gives t = 32 . This value is in the interval [0, 32 +

√5]

and h(3/2) = 10. The maximum height reached by the ball is 10 feet. The maximum height is reachedwhen t = 3

2 seconds.

9. Evaluating f at the endpoints of the interval [0, π] gives f(0) =√

3 + 100 and f(π) = −√

3 + 100.Next we find the critical points of f in the interval [0, π].

f ′(x) = 0 ⇒ −√

3 sinx+ cosx = 0 ⇒ cosx =√

3 sinx

The only solution to cosx =√

3 sinx with 0 ≤ x ≤ π is x = π/6. We conclude that the only critical

point of f in the interval [0, π] is x = π/6 and f(π/6) =√

3 ·√32 + 1

2 + 100 = 102. The maximum depthof the water during the time interval [0, π] is 102 feet.

61

3.2 The Mean Value Theorem

1. Complete the statement of Rolle’s Theorem. If f is a continuous function on [a, b] that is differentiableon (a, b) and f(a) = f(b), then

there is a number c with a < c < b such that .

2. The function f(x) = |x| is continuous on [−1, 1] and f(−1) = f(1). Is there a number c with−1 < c < 1 such that f ′(c) = 0? If so, find such a c. If not, explain why this does not contradict Rolle’sTheorem.

3. Complete the statement of the Mean Value Theorem. If f is a continuous function on [a, b] that isdifferentiable on (a, b), then

there is a number c with a < c < b such that .

4. Let f(x) = sinx. Show that there is a number c with 0 < c <π

2such that f ′(c) =

2

π.

5. Let s(t) be the displacement of an object, measured in feet, at time t where t is in seconds. Ifs(1) = 0.5 and s(4) = 0.5, must there be a time t0 between 1 and 4 seconds such that the velocity of theobject is 0 feet per second at time t0?

6. The speed limit on a highway is 65 miles per hour. Let s(t) be the distance traveled on the highwayby a motorcycle. Here the units of t are hours and the units of s(t) are miles. If s(1/60) = 1 ands(1/48) = 1.3, must there exists a time when the car is speeding? You may assume that s(t) is continuouson [1/60, 1/48] and differentiable on (1/60, 1/48).

7. (a) Suppose that f is a continuous function on [a, b] and f is differentiable on (a, b). If f ′(x) = 0 forall x with a < x < b, show that f(b) = f(a).

(b) Suppose that f is a continuous function on [a, b] and f is differentiable on (a, b). If f ′(x) > 0 for allx with a < x < b, show that f(a) < f(b).

8. Suppose that f is a continuous function on [1, 2] and f is differentiable on (1, 2). If f(1) = 5 andf ′(x) ≥ 3 for all x with 1 < x < 2, show that f(2) ≥ 8.

The Mean Value Theorem - Solutions

1. If f is a continuous function on [a, b] that is differentiable on (a, b) and f(a) = f(b), then there is anumber c with a < c < b such that f ′(c) = 0.

2. There is no number c with −1 < c < 1 such that f ′(c) = 0. One way to see this is by considering thegraph of f(x) = |x|. The function f(x) = |x| is not differentiable at 0, and f ′(c) = −1 for −1 < c < 0,and f ′(c) = 1 if 0 < c < 1. The reason that this does not contradict Rolle’s Theorem is that one of theassumptions that is needed to apply Rolle’s Theorem is that f is differentiable on (a, b). Since f(x) = |x|is not differentiable on (−1, 1), Rolle’s Theorem cannot be applied to the function f(x) = |x| on theinterval [−1, 1].

3. If f is a continuous function on [a, b] that is differentiable on (a, b), then there is a number c witha < c < b such that

f ′(c) =f(b)− f(a)

b− a.

62

4. The function f(x) = sinx is continuous on [0, π/2] and is differentiable on (0, π/2). By the Mean

Value Theorem, there is a number c with 0 < c <π

2such that

f ′(c) =f(π/2)− f(0)

π/2− 0=

1− 0

π/2=

2

π.

5. Assuming that s(t) is continuous on [1, 4] and differentiable on (1, 4), then by Rolle’s Theorem, thereis a time t0 such that s′(t0) = 0. In practical terms, the velocity of the object is 0 feet per second attime t0.

If s(t) does not satisfy the assumptions of Rolle’s Theorem, then we cannot be sure that there is a timebetween 1 and 4 seconds where the velocity of the object is 0. For example, if s(t) = (1/3)|t − (5/2)|(see the graph below), then s(1) = 0.5, and s(4) = 0.5, but there is no value of t0 in [1, 4] such thats′(t0) = 0.

1 2 3 4

0.2

0.4

0.6

0.8

1

f

s(t) = (1/3)|t− (5/2)|

y

x

6. By the Mean Value Theorem, there is a number c with 1/60 < c < 1/48 such that

s′(c) =s(1/48)− s(1/60)

1/48− 1/60=

1.3− 1

1/48− 1/60= 72.

We conclude that at some time t between 1/60 and 1/48, the motorcycle was going 72 miles per hour.

7. (a) By The Mean Value Theorem, there is a number c with a < c < b such that f ′(c) =f(b)− f(a)

b− a.

By our assumption, f ′(c) = 0 so that 0 =f(b)− f(a)

b− a. This equation implies that 0 = f(b) − f(a) so

f(b) = f(a).

(b) By the Mean Value Theorem, there is a number c with a < c < b such that f ′(c) =f(b)− f(a)

b− a. By

our assumption, we have that f ′(c) > 0 so that

0 <f(b)− f(a)

b− a.

We multiply both sides of this inequality by b− a (which is not zero since b > a) to get 0 < f(b)− f(a).This last inequality is equivalent to f(a) < f(b).

63

8. By the Mean Value Theorem, there is a c with 1 < c < 2 such that f ′(c) =f(2)− f(1)

2− 1= f(2)−f(1).

By our assumption, f ′(c) ≥ 3 so that

3 ≤ f ′(c) = f(2)− f(1) = f(2)− 5.

Add 5 to both sides to get that 8 ≤ f(2).

3.3 Increasing and Decreasing Functions

1. Complete the statement. If f is a function with f ′(x) > 0 for all x in (a, b), then

f is on (a, b).

2. Sketch the graph of a continuous function f whose domain is [−2, 1] and satisfies both of the conditions

f ′(x) > 0 for −2 < x < 0 and f ′(x) < 0 for 0 < x < 1.

3. Sketch the graph of a function f that satisfies f ′(x) > 0 for all x and limx→∞

f(x) = 2.

4. Suppose that f is a continuous function on [0, 5] and that f is differentiable on (0, 5). If f(0) = 3 andf ′(x) < 0 for 0 < x < 5, what can you say about f(4)?

5. The graph of f is shown to the right. Find theinterval(s) on which f is increasing.

−2 −1 1 2 3

−1

1

2

3

f

y

x

6. The graph of f is shown to the right. Find theinterval(s) on which f ′(x) < 0.

−0.5 0.5 1 1.5

−2

−1

1

2

f

y

x

64

7. Let f(x) = 2x3 + 3x2 − 36x+ 42. Find the interval(s) on which f is increasing and the interval(s) onwhich f is decreasing.

8. Let f(t) = cos t− sin t on [0, 2π]. Find the interval(s) in [0, 2π] on which f is increasing.

9. Let g(x) = 2ex + 2e−x. Find the interval(s) on which g is increasing and the interval(s) on which gis decreasing.

10. Determine the local extrema of the function f(x) = 2x3 + 3x2 − 36x+ 42.

11. Determine the local extrema of the function g(x) = 2ex + 2e−x.

12. Find the interval(s) on which f(x) =x− 2

x− 1is increasing.

13. Let f(t) =1

t2 + 4. Determine the local extrema of f .

14. If h(t) = 3t ln t, find the interval(s) on which h is increasing and determine the local extrema of h.

15. Let r(x) = x +1

x. Find domain of r, and determine the interval(s) on which r is increasing, the

interval(s) on which r is decreasing, and the local extrema of r.

16. Sketch the graph of a function that is increasing on (−1, 0) and is decreasing on (0, 1), but does nothave a local maximum at x = 0. Your function need not be continuous at x = 0.

17. Suppose f ′(x) < 0 for x < 5 and f ′(x) > 0 for x > 5. Does it follow that f has a local minimum atx = 5?

Increasing and Decreasing Functions - Solutions

1. If f is a function with f ′(x) > 0 for all x in (a, b), then f is increasing on (a, b).

2. The condition f ′(x) > 0 for −2 < x < 0 means that f is increasing on the interval (−2, 0). Thecondition f ′(x) < 0 for 0 < x < 1 means that f is decreasing on the interval (0, 1). An example of sucha function is shown below.

−2 −1 1 2

−2

−1

1

2

f

y

x

3. The condition f ′(x) > 0 for all x means that f is always increasing. The condition limx→∞

f(x) = 2

means that the values of f get closer and closer to 2 as x tends to infinity. An example of such a functionis shown below.

65

−4 −2 2 4

−4

−2

2

4

f

y

x

4. Since f is strictly decreasing on (0, 5), we can say that f(4) < f(0). We are given that f(0) = 3 andso we may say that f(4) < 3.

5. The function f is increasing on the interval (1, 3).

6. The condition f ′(x) < 0 means that f is decreasing at x. By looking at the graph of f , we find thatf is decreasing on the interval (0, 1.5)

7. Our goal is to build a sign chart for f so that we may determine the interval(s) on which f is increasingand the interval(s) on which f is decreasing. We start by finding f ′(x) which is f ′(x) = 6x2 + 6x− 36.Next we find all of the critical points of f . Since f is a polynomial, there are no critical points c where cis in the domain of f and f ′(c) does not exist. The only types of critical points we need to look for arethose where f ′(c) = 0. Setting f ′(x) = 0 gives 6x2 + 6x− 36x = 0 so 6(x− 2)(x+ 3) = 0. The criticalpoints of f are x = 2 and x = −3. We now build our sign chart for f ′(x).

−3 2

f ′(−4) > 0 f ′(0) < 0 f ′(3) > 0

Sign chart for f ′(x).

From our sign chart, we see that f is increasing on (−∞,−3) and (2,∞), and f is decreasing on (−3, 2)

8. We will follow the same strategy as in the solution to the previous problem but we must keep in mindthat we are only working on the interval [0, 2π]. Here f ′(t) = − sin t − cos t and if − sin t − cos t = 0,then − sin t = cos t. The solutions to this equation in [0, 2π] are t = 3π

4 and t = 7π4 . We must now build

a sign chart for f ′(t).

3π4

7π4

f ′(π/4) < 0 f ′(π) > 0 f ′(2π) < 0

0 2π

Sign chart for f ′(t).

From our sign chart we see that f is increasing on (3π/4, 7π/4), and f is decreasing on (0, 3π/4) and(7π/4, 2π).

9. We have g′(x) = 2ex − 2e−x. If g′(x) = 0, then 2ex − 2e−x = 0 so that 2ex = 2e−x. Cancelling the2 and taking the natural logarithm of both sides gives x = −x so 2x = 0. The only solution to thisequation is x = 0. Next we construct a sign chart for g′(x).

66

0

g′(−1) < 0 g′(1) > 0

Sign chart for g′(x).

The function g is decreasing on (−∞, 0) and increasing on (0,∞)

10. In a previous problem, we showed that the sign chart for f ′(x) is

−3 2

f ′(−4) > 0 f ′(0) < 0 f ′(3) > 0


From this sign chart, we see that there is a local maximum at (−3, 123) and a local minimum at (2,−2).

11. In a previous problem, we showed that the sign chart for g′(x) is

0

g′(−1) < 0 g′(1) > 0

Sign chart for g′(x).

From this sign chart, we wee that there is a local minimum at (0, 4). We can actually say that (0, 4) isan absolute minimum since g is decreasing on (−∞, 0) and increasing on (0,∞).

12. By the Quotient Rule,

f ′(x) =(x− 1)(1)− (x− 2)(1)

(x− 1)2=x− 1− x+ 2

(x− 1)2=

1

(x− 1)2.

As f ′(x) > 0 for all x, other than x = 1, we have that f is increasing on (−∞, 1) and (1,∞). There is avertical asymptote at x = 1 so it would be incorrect to say that f is increasing on (−∞,∞).

13. By the Quotient Rule,

f ′(t) =(t2 + 4) · 0− 1(2t)

(t2 + 4)2=

−2t

(t2 + 4).

The denominator of f ′(t) is always positive. Focusing on the numerator, we see that f ′(t) > 0 on(−∞, 0), f ′(0) = 0, and f ′(t) < 0 on (0,∞). We conclude that f has a local maximum at (0, 1/4).

14. By the Product Rule, h′(t) = 3 ln t+ 3t(1/t) = 3 ln t+ 3. Next we solve h′(t) = 0.

h′(t) = 0 ⇒ 3 ln t+ 3 = 0 ⇒ 3 ln t = −3 ⇒ ln t = −1 ⇒ t = e−1.

Now we build the sign chart for h′(t).

e−1

h′(e−2) < 0 h′(e) > 0

0

67

Sign chart for h′(t).

From the sign chart, we see that h is decreasing on (0, 1/e), increasing on (1/e,∞). There is a localminimum at (1/e,−3/e)

15. The domain of r(x) is all real numbers except for 0. We have r′(x) = 1− 1

x2. We now solve r′(x) = 0.

r′(x) = 0 ⇒ 1− 1

x2= 0 ⇒ x2 = 1 ⇒ x = ±1

Now we build a sign chart for r′(x) keeping in mind that x = 0 is not in the domain of r(x).

0−1 1

r′(− 12 ) < 0 r′( 1

2 ) < 0r′(−2) > 0 r′(2) > 0

Sign chart for r′(x).

The function r is increasing on (−∞,−1) and (1,∞). It is decreasing on (−1, 0) and (0, 1). There is alocal maximum at (−1,−2) and a local minimum at (1, 2)

16.

−3 −2 −1 1 2 3

1

2

3

4

y

x

17. No. It could be the case that f has a vertical asymptote at x = 5. For example, the function

f(x) = − 1

(x− 5)2does not have a local minimum, but satisfies f ′(x) < 0 for x < 5 and f ′(x) > 0 for

x > 5.

3.4 Concavity

1. Complete the statement. If f is a function with f ′′(x) > 0 for all x in the interval (a, b), then

68

the graph of f is on (a, b).

2. Sketch the graph of a continuous function f that satisfies both of the conditions

f ′′(x) > 0 for −2 < x < 0 and f ′′(x) < 0 for 0 < x < 1.

3. Suppose that f is a continuous function whose domain is [0, 5] and f ′′(x) > 0 for 0 < x < 5. Does itfollow that f is increasing on some interval I contained in [0, 5]?

4. The graph of f is shown below. Determine the interval(s) on which the graph of f is concave up.

−1 1 2 3

−6

−4

−2

2

4

6

f

y

x

5. The graph of f is shown below. Determine the interval(s) on which f ′′(x) < 0.

−3 −2 −1 1 2 3

−6

−4

−2

2

4

6

f

y

x

6. Let f(x) = 2x3 + 3x2 − 12x+ 15.

(a) Find the interval(s) on which f is increasing.

(b) Find the interval(s) on which the graph of f is concave up.

7. Let s(t) = sin t+ cos t.

(a) Find the interval(s) in [0, 2π] on which the graph of s(t) is concave down.

69

(b) Find any inflection points of s(t) in the interval [0, 2π].

8. Let r(x) = ex + e−x. Find the interval(s) on which the graph of r(x) is concave up.

9. Let f(x) = − 1

x2.

(a) Find the intervals(s) on which the graph of f(x) is concave up.

(b) Find all inflection points of f .

10. Find the critical points of f(x) =64

x+ x and use the 2nd Derivative Test to classify them as local

minima or local maxima.

11. Let f(x) =√

3 sinx + cosx. Find the critical points of f in the interval [0, π] and use the SecondDerivative Test to classify each critical point in [0, π] as a local minimum or local maximum.

12. If f(x) = x lnx, find the domain of f and the interval(s) on which the graph of f is concave up.

13. Let s(u) = ueu. Find all of the inflection points of s.

Concavity - Solutions

1. If f is a function with f ′′(x) > 0 for all x in the interval (a, b), then the graph of f is concave up on(a, b).

2. The graph of f shown to the right satisfiesf ′′(x) > 0 for −2 < x < 0 and f ′′(x) < 0 for0 < x < 1. The graph of f is concave up on (−2, 0)and is concave down on (0, 2).

−2 −1 1 2

−4

−2

2

4

f

y

x

3. No. All we can say is that f is concave upon (0, 5). An example of a function that is de-creasing on [0, 5] but is concave up on (0, 5) isf(x) = (x − 5)2. The graph of this function isshown to the right.

1 2 3 4 5

5

10

15

20

25

f

y

x

70

4. The graph of f is concave up on the interval (1, 3).

5. The graph of f is concave down on the interval (−∞,−1).

6. (a) First we find the interval(s) on which f is increasing. We have f ′(x) = 6x2 +6x−12. If f ′(x) = 0,then

6x2 + 6x− 12 = 0 =⇒ 6(x2 + x− 2) = 0 =⇒ 6(x+ 2)(x− 1) = 0.

From this we see that f ′(x) = 0 when x = −2 or x = 1. Now we build our sign chart for f ′(x):

−2 1

f ′(−3) > 0 f ′(0) < 0 f ′(2) > 0


From our sign chart we see that the function f is increasing on (−∞,−2) and (1,∞).

(b) Next we determine the interval(s) on which the graph of f is concave up. Our goal is to determinewhen f ′′(x) is positive and when f ′′(x) is negative. Now f ′(x) = 6x2 + 6x− 12 and f ′′(x) = 12x+ 6. If12x + 6 > 0, then x > − 1

2 . If 12x + 6 < 0, then x < − 12 . The graph of f is concave up on (−1/2,∞).

Remark: we could have also solved this part of the problem by building a sign chart for f ′′(x) but sinceit is easier and faster to just solve the linear inequality 12x+ 6 > 0, we did not build a sign chart.

7. (a) We must build a sign chart for s′′(t). We have s′(t) = cos t− sin t and s′′(t) = − sin t− cos t. Nextwe solve the equation s′′(t) = 0:

0 = − sin t− cos t =⇒ sin t = − cos t.

The solutions to sin t = − cos t in the interval [0, 2π] are t = 3π/4 and t = 7π/4.

3π4

7π4

s′′(π/4) < 0 s′′(π) > 0 s′′(2π) < 0

0 2π

Sign chart for s′′(t).

In the interval [0, 2π], the graph of s(t) is concave down on (0, 3π/4) and (7π/4, 2π).

(b) The inflection points in the interval [0, 2π] are at t = 3π/4 and t = 7π/4.

8. We have r′(x) = ex − e−x and r′′(x) = ex + e−x. Since ex and e−x are positive for all x, the sumex + e−x is also always positive. We conclude that r′′(x) > 0 for all x so the graph of r(x) is alwaysconcave up.

9. (a) We have f ′(x) = 2x−3 and f ′′(x) = −6/x4. For any x with x 6= 0, f ′′(x) < 0 and so the graph off is concave down on (−∞, 0) and (0,∞).

(b) The graph of f has no inflection points.

10. We have f ′(x) = −64/x2 + 1. The critical points of f are the values of c in the domain of f forwhich f ′(c) is undefined or for which f ′(c) = 0. Note that f ′(0) is undefined but 0 is not in the domain

71

of f so 0 is not a critical point of f . Next we solve f ′(x) = 0:

−64

x2+ 1 = 0 =⇒ − 64

x2= −1 =⇒ 64 = x2 =⇒ ±

√64 = x =⇒ ± 8 = x.

For the Second Derivative Test, we need to compute f ′′(8) and f ′′(−8). Since f ′′(x) = 128x3 , we have

f ′′(8) > 0 and f ′′(−8) < 0. There is a local minimum at (8, 16) and there is a local maximum at(−8,−16).

11. We have f ′(x) =√

3 cosx − sinx. We need to find the critical points of f that are in the interval[0, π]. If 0 =

√3 cosx − sinx, then sinx =

√3 cosx so that tanx =

√3. There are many solutions to

this equation, but the only solution in the interval [0, π] is x = π3 . Now f ′′(x) = −

√3 sinx− cosx and

f ′′(π/3) = −√

3 sin(π/3)− cos(π/3) = −√

3

(√3

2

)− 1

2= −2.

By the Second Derivative Test, there is a local maximum at x = π3 .

12. The domain of f(x) is (0,∞). The first derivative of f is f ′(x) = lnx+ 1 and the second derivativeis f ′′(x) = 1/x. Since f ′′(x) = 1/x > 0 for all x > 0, the graph of f is concave up on (0,∞).

13. We need to construct a sign chart for s′′. Now s′(u) = eu + ueu and s′′(u) = eu + eu + ueu. We canrewrite s′′(u) as 2eu + ueu. Now we solve s′′(u) = 0:

s′′(u) = 0 =⇒ 2eu + ueu = 0 =⇒ eu(2 + u) = 0 =⇒ u = −2.

Next we build a sign chart for s′′(u):

−2

s′′(−3) < 0 s′′(−1) > 0

Sign chart for s′′(u).

The graph of s is concave up on (−2,∞). The only inflection point of s is at u = −2.

3.5 l’Hospital’s Rule

1. Write down four different indeterminate forms.

2. Evaluate each of the following limits. You should not use l’Hospital’s Rule to evaluate these limits.

(a) limx→0+

lnx (b) limx→∞

lnx (c) limx→∞

tan−1 x (d) limx→−∞

tan−1 x

3. Evaluate each of the following limits. You should not use l’Hospital’s Rule to evaluate these limits.

(a) limx→−∞

ex (b) limx→∞

ex (c) limx→−∞

e−x (d) limx→∞

e−x2

72

4. Find each of the following limits.

(a) limx→∞

tan−1 x

x(b) lim

x→∞

3x2 + 1

−5x2 + x− 2(c) lim

x→∞

3x+ 1

−5x2 + x− 2

5. Evaluate the given limits.

(a) limx→0

4 sinx− x2

3x(b) lim

x→0

4 sinx− x3x2

6. Evaluate the given limits.

(a) limx→0

tan(5x)

6x(b) lim

x→0

tan−1(5x)

6x(c) lim

x→π/2

(x− π

2

)secx

7. Determine the given limits.

(a) limx→1

ex−1 − 1

x− 1(b) lim

x→4

8√x− x2

ln(x− 3)

8. Find the given limits.

(a) limx→∞

xe−2x (b) limx→−∞

xex3

(c) limx→∞

xe2x

9. Find the given limits.

(a) limx→0+

4x lnx (b) limx→0+

x2 lnx (c) limx→0+

√x lnx

10. Let r > 0 be a positive real number. Use l’Hospital’s Rule to prove that

limx→0+

xr lnx = 0.

11. Determine limx→∞

ln(x3)

2x.

12. Find the given limit.

(a) limx→∞

(1 +

4

x

)3x

(b) limx→0+

(1

x

)x2

(c) limx→0+

(1

x2

)1/x

l’Hospital’s Rule - Solutions

1. The forms 00 , ∞∞ , 1∞, and ∞0 are four indeterminate forms. Two others are 0 · ∞ and ∞−∞.

73

2. (a) −∞ (b) ∞ (c)π

2(d) −π

2

Remark: To understand where these answers are coming from, it is best to think of the graph of y = lnxand the graph of y = tan−1 x.

3. (a) 0 (b) ∞ (c) ∞ (d) 0

Remark: To understand where these answers are coming from, it is best to think of the graph of y = ex.

4. (a) 0 (b) −3

5(c) 0

5. (a) Using l’Hospital’s Rule, we find that

limx→0

4 sinx− x2

3x

l’H= lim

x→0

4 cosx− 2x

3=

4 cos 0− 2(0)

3=

4

3.

(b) Again using l’Hospital’s Rule,

limx→0

4 sinx− x3x2

l’H= lim

x→0

4 cosx− 1

6x.

As x goes to 0, the numerator goes to 3 while the denominator goes to 0. This limit does not exist.

6. (a) limx→0

tan(5x)

6x

l’H= lim

x→0

sec2(5x)(5)

6=

5 sec2(0)

6=

5

6

(b) limx→0

tan−1(5x)

6x

l’H= lim

x→0

11+(5x)2 · 5

6=

51+0

6=

5

6

(c) limx→π/2

(x− π

2

)secx = lim

x→π/2

x− π2

cosx

l’H= lim

x→π/2

1

− sinx=

1

−1= −1

7. (a) By l’Hospital’s Rule,

limx→1

ex−1 − 1

x− 1

l’H= lim

x→1

ex−1

1=e1−1

1=e0

1=

1

1= 1.

(b) Applying l’Hospital’s Rule,

limx→4

8√x− x2

ln(x− 3)

l’H= lim

x→4

4x−1/2 − 2x

1/(x− 3)=

4(4)−1/2 − 2(4)

1/(4− 3)=

4/√

4− 8

1=

2− 8

1= −6.

8. (a) The limit limx→∞

xe−2x is of the form ∞ · 0. Using l’Hospital’s Rule,

limx→∞

xe−2x = limx→∞

x

e2xl’H= lim

x→∞

1

2e2x= 0.

(b) The limit limx→−∞

xex3

is of the form −∞ · 0. By l’Hospital’s Rule,

limx→−∞

xex3

= limx→−∞

x

e−x3

l’H= lim

x→−∞

1

−3x2e−x3 = 0.

(c) limx→∞

xe2x =∞.

74

9. (a) The limit limx→0+

4x lnx is of the form 0 · (−∞). Applying l’Hospital’s Rule,

limx→0+

4x lnx = limx→0+

4 lnx

1/x

l’H= lim

x→0+

4/x

−1/x2= limx→0+

−4x = 0.

(b) limx→0+

x2 lnx = limx→0+

lnx

x−2l’H= lim

x→0+

1/x

−2x−3= limx→0+

1

x· x

3

−2= limx→0+

x2

−2= 0

(c) limx→0+

√x lnx = lim

x→0+

lnx

x−1/2l’H= lim

x→0+

1/x

− 12x−3/2 = lim

x→0+

1

x· −2x3/2

1= limx→0+

−2x1/2 = 0

10. Let r > 0 be a positive real number. We have

limx→0+

xr lnx = limx→0+

lnx

x−rl’H= lim

x→0+

1/x

−rx−(r+1)= limx→0+

1

x· x

r+1

−r= limx→0+

xr

−r= 0.

11. First we note that ln(x3) = 3 lnx. By l’Hospital’s Rule,

limx→∞

3 lnx

2x

l’H= lim

x→∞

3/x

2= limx→∞

3

2x= 0.

12. (a) The given limit is an indeterminate of the form 1∞. We have

limx→∞

(1 +

4

x

)3x

= limx→∞

eln((1+ 4

x )3x

)= limx→∞

e3x ln(1+ 4x ).

Now we compute limx→∞

3x ln

(1 +

4

x

)using l’Hospital’s Rule:

limx→∞

3x ln

(1 +

4

x

)= limx→∞

3 ln(1 + 4

x

)1/x

l’H= lim

x→∞

31+ 4

x

(−4/x2)

−1/x2= limx→∞

12

1 + 4/x= 12.

Therefore,

limx→∞

(1 +

4

x

)3x

= limx→∞

e3x ln( x+4x ) = e12.

(b) The limit limx→0+

(1

x

)x2

is an indeterminate of the form ∞0. We have

limx→0+

(1

x

)x2

= limx→0+

eln((1/x)x2 ) = lim

x→0+ex

2 ln(1/x).

Now we compute the limit limx→0+

x2 ln(1/x) using l’Hospital’s Rule. It will be easier to rewrite ln(1/x)

as − lnx which follows from properties of logarithms. Indeed, ln(1/x) = ln 1 − lnx = 0 − lnx = − lnxor ln(1/x) = ln(x−1) = − lnx. We now have

limx→0+

x2 ln(1/x) = limx→0+

− lnx

x−2l’H= lim

x→0+− 1/x

−2/x3= limx→0+

x3

2x= limx→0+

−x2

2= 0.

Therefore,

limx→0+

(1

x

)x2

= limx→0+

ex2 ln(1/x) = e0 = 1.

(c) Since limx→0+

1

x2=∞ and lim

x→0+

1

x=∞, we have lim

x→0+

(1

x2

)1/x

=∞.

75

3.6 Sketching Graphs of Functions

1. Let f(x) =x

x2 − 1.

(a) Determine the domain of f and find all vertical asymptotes and horizontal asymptotes of f .

(b) Find f ′(x) and determine the interval(s) on which f is increasing and the interval(s) on which f isdecreasing.

(c) Determine the limits limx→−1−

f(x), limx→−1+

f(x), limx→1−

f(x), and limx→1+

f(x).

(d) Compute f(−2), f(0), and f(2).

(e) Use parts (a), (b), (c), and (d) to sketch the graph of f .

2. Let f(x) = xe−x.

(a) Find f ′(x) and determine the interval(s) on which f is increasing and the interval(s) on which f isdecreasing.

(b) Find any local extrema of f .

(c) Determine the limits limx→−∞

xe−x and limx→∞

xe−x.

(d) Find f(0).

(e) Use parts (a), (b), (c), and (d) to sketch the graph of f(x) = xe−x.

3. Let f(x) = 15x

5 − x.

(a) Find f ′(x) and the interval(s) on which f is increasing and the interval(s) on which f is decreasing.

(b) Find all local extrema of f .

(c) Find f ′′(x) and the interval(s) on which the graph of f is concave up and the interval(s) on whichthe graph of f is concave down.

(d) Find f(0).

(e) Use parts (a), (b), (c), and (d) to sketch the graph of f(x) = 15x

5 − x.

Sketching Graphs of Functions - Solutions

1. (a) The domain of f is all real numbers except 1 and −1. Since the degree of the numerator ofx

x2 − 1is less than the degree of the denominator, the line y = 0 is a horizontal asymptote. Lastly, f has verticalasymptotes at x = 1 and x = −1. This is because

f(x) =x

x2 − 1=

x

(x− 1)(x+ 1)

and so 1 and −1 are values that make the denominator 0 but the numerator is not 0 at these values.

(b) Using the Quotient Rule we find that f ′(x) =−x2 − 1

(x2 − 1)2. Since −x2 − 1 < 0 for all x and (x2 − 1)2

76

is positive whenever x 6= ±1, we see that f ′(x) < 0 for all x except for x = 1 and x = −1. This tells usthat f is decreasing on the intervals (−∞,−1), (−1, 1), and (1,∞).

(c) First rewrite f(x) as f(x) =x

(x− 1)(x+ 1).

For the limit limx→−1−

x

(x− 1)(x+ 1), we note that if x is close to −1 but less than −1, then

x

(x− 1)(x+ 1)≈ −1

(−2)(small negative number).

This is not a precise mathematical statement but it is useful because from it we deduce that

limx→−1−

f(x) = −∞.

Next we determine the limit limx→−1+

f(x). If x is close to −1 but greater than −1, then

x

(x− 1)(x+ 1)≈ −1

(−2)(small positive number).

Once again, this is not a precise mathematical statement, but from it we infer that limx→−1+

f(x) =∞.

Using similar arguments we find that limx→1−

f(x) = −∞ and limx→1+

f(x) =∞.

(d) f(−2) =−2

(−2)2 − 1= −2

3, f(0) = 0, f(2) =

2

22 − 1=

2

3

(e)

−2 −1 1 2

−4

−2

2

4

f

y

x

2. (a) Using the Product Rule, we find that f ′(x) = e−x − xe−x. If f ′(x) = 0, then 0 = e−x(1 − x).Since e−x is never 0, we must have 0 = 1− x so that x = 1. The sign chart for f ′(x) is

0

f ′(−1) > 0 f ′(1) < 0


77

The function f is increasing on (−∞, 1) and is decreasing on (1,∞).

(b) From part (a), we know that f is increasing on (−∞, 1) and decreasing on (1,∞). Therefore, f hasan absolute maximum at x = 1 and f(1) = e−1.

(c) Since limx→−∞

x = −∞ and limx→−∞

e−x =∞, we have limx→−∞

xe−x = −∞.

Using l’Hospital’s Rule, limx→∞

xe−x = limx→∞

x

exl’H= lim

x→∞

1

ex= 0.

(d) f(0) = 0

(e)

−3 −2 −1 1 2 3

−4

−2

2

4

f

y

x

3. (a) We have f ′(x) = x4 − 1 and now we solve f ′(x) = 0

x4 − 1 = 0 =⇒ (x2 − 1)(x2 + 1) = 0 =⇒ (x− 1)(x+ 1)(x2 + 1) = 0.

Since x2 + 1 is never 0, the solutions to f ′(x) = 0 are x = 1 and x = −1. The next step is to determinethe sign chart for f ′(x) which is

−1 1

f ′(−1) > 0 f ′(0) < 0 f ′(1) > 0


The function f is decreasing on (−1, 1) and increasing on (−∞,−1) and (1,∞).

(b) From part (a), we know that f has a local maximum at x = −1 and a local minimum at x = 1.Therefore, (−1, 4/5) is a local maximum and (1,−4/5) is a local minimum.

(c) We have f ′′(x) = 4x3. The only solution to f ′′(x) = 0 is x = 0. We know that f ′′(x) = 4x3 ispositive when x > 0 and is negative when x < 0. The graph of f is concave up on (0,∞) and is concavedown on (−∞, 0).

(d) f(0) = 0.

78

(e)

−3 −2 −1 1 2 3

−4

−2

2

4

f

y

x

3.7 Optimization

1. Suppose that x and y are two real numbers with x ≥ 1, y ≥ 3, and x + y = 10. Find the maximumvalue of the product xy.

2. Suppose that x and y are two real numbers with x ≥ 1, y ≥ 6, and x + y = 10. Find the maximumvalue of the product xy.

3. Find the maximum area of a rectangle that is inscribed in an isosceles triangle of height 6 ft and baselength 4 ft.

4. A box has a volume of 1000cm3 and the depth of the box is twice the width. Find the minimumsurface area of such a box.

5. A cylinder is placed inside a cone of radius 2 m and height 8 m so that the volume of the cylinder isas large as possible. Find the radius and the height of the cylinder.

6. A cylinder has a volume of V = 32π in3. Find the dimensions of the cylinder that minimizes thesurface area of the cylinder.

7. A rectangle whose base lies on the x axis is inscribed inside the region bounded by the graph ofy = −x2 + 2 and the x axis (see the figure below). Find the maximum area of such a rectangle.

79

−2 −1 1 2

1

2

3

y

x

Optimization - Solutions

1. First note that the condition y ≥ 3 implies that x ≤ 7. Indeed, if x > 7, then x+ y would be biggerthan 10 since y ≥ 3.

We want to find the maximum value of xy. Since x+ y = 10, we have y = 10− x so that

xy = x(10− x) = −x2 + 10x.

Our optimization problem is to

maximize f(x) = −x2 + 10x subject to 1 ≤ x ≤ 7.

Now f ′(x) = −2x+ 10 so that f ′(x) = 0 when x = 5. The only critical point of f that is in the interval[1, 7] is at x = 5. We then compute f(5) = 25, f(1) = 9, and f(7) = 21. The maximum value of theproduct xy subject to the given constraint is 25 and this occurs when x = 5 and y = 5.

2. First note that the condition y ≥ 6 implies that x ≤ 4. Indeed, if x > 4, then x+ y would be biggerthan 10. Since x+ y = 10, we have y = 10− x so that

xy = x(10− x) = −x2 + 10x.


maximize f(x) = −x2 + 10x subject to 1 ≤ x ≤ 4.

The interval [1, 4] does not contain any critical points of f (we saw in the previous problem that f ′(x) = 0only when x = 5). We compute f(1) = 9 and f(4) = 24. The maximum value of the product xy subjectto the given constraint is 24. This occurs when x = 4 and y = 6.

3. We first draw a picture of the rectangle inscribed in the triangle.

80

��

BBBBBBBBBBBBB

4

The key idea is to introduce coordinates so that we may set up an optimization problem which we thensolve using calculus. There is more than one way to do this. We will let the origin be the midpoint ofthe bottom of the triangle. One does not have to choose the midpoint of the bottom to be the origin.Another natural choice is to let the bottom left corner of the triangle be the origin. However, we will letthe origin be the midpoint of the bottom of the triangle and this allows us to label other points of thetriangle as shown below.

��

BBBBBBBBBBBBB

(0, 0) r r

r

(2, 0)

(0, 6)

(−2, 0) rr (x, y)

We have labeled the point where the rectangle touches the triangle on the right side with the coordinates(x, y). The equation of the line through the points (0, 6) and (2, 0) is y = −3x+ 6 and so we have y interms of x. The area of the rectangle is base times height which in this case is

(x+ x)y = 2xy = 2x(−3x+ 6) = −6x2 + 12x.


maximize f(x) = −6x2 + 12x subject to 0 ≤ x ≤ 2.

We find that f ′(x) = −12x+ 12 and so f ′(x) = 0 when x = 1. Since f(0) = f(2) = 0 and f(1) = 6, weconclude that the maximum area of a rectangle inscribed in an isosceles triangle of height 6 and baselength 4 is obtained when x = 1. The area of the rectangle is f(1) = 6 ft2.

4. Let x be the width of the box, y be its height, and 2x be its depth. The volume V of the box is widthtimes height times depth which is

V = xy(2x) = 2x2y.

We are given that V = 1000 so that 1000 = 2x2y. The area of the top of the box is 2x2 and this is alsothe area of the bottom of the box. The area of the front of the box is xy and this is also the area of theback of the box. The area of the right side of the box is 2xy and this is also the area of the left side ofthe box. Thus, the surface area of the box is

2(2x2) + 2(xy) + 2(2xy) = 4x2 + 6xy.

81

We can solve 1000 = 2x2y for y to get y =1000

2x2=

500

x2. Substituting into 4x2 + 6xy gives

4x2 + 6x

(500

x2

)= 4x2 +

3000

x.

Our goal then is to

minimize f(x) = 4x2 +3000

xsubject to x > 0.

We have

f ′(x) = 8x− 3000

x2.

Solving f ′(x) = 0 gives x =

(3000

8

)1/3

= 5 3√

3. Using the formula for f ′(x), we find that f is decreasing

on (0, 5 3√

3) and increasing on (5 3√

3,∞). Thus, the minimum value of f on (0,∞) occurs when x = 5 3√

3.The minimum surface area of such a box is

f(53√

3) = 4(53√

3) +3000

5 3√

3cm2.

5. Let r be the radius of the cylinder and h be the height. A side view of the cylinder inside the conewould look like

��

CCCCCCCCCCCCCCCC

(0, 0)r r

r

(2, 0)

(0, 8)

(−2, 0) r

r (r, h)

We have h = 8− 4r since the equation of the line through the points (0, 8) and (2, 0) is h = 8− 4r. Thevolume of the cylinder is V = πr2h = πr2(8− 4r). Our goal is to

maximize f(r) = πr2(8− 4r) subject to 0 ≤ r ≤ 2.

Now f(r) = 8πr2− 4πr3 so f ′(r) = 16πr− 12πr2. Solving f ′(r) = 0 gives r =4

3which is in the interval

[0, 2]. Now f(0) = 0, f(4/3) = 128π/27, and f(2) = 0. Thus, the maximum volume is 128π/27 m3 andthis occurs when

r =4

3m and h = 8− 4(4/3) =

8

3m.

82

6. Let h be the height of the cylinder and r be the radius of the cylinder. The volume of the cylinder

is 32π = πr2h so that h =32

r2. The surface area of the cylinder is 2πr2 + 2πrh. Using the equation

h =32

r2, we can rewrite 2πr2 + 2πrh as 2πr2 +

64π

r. Our goal is to minimize

s(r) = 2πr2 +64π

rsubject to r > 0.

We compute s′(r) = 4πr − 64π

r2. The only solution to s′(r) = 0 for r > 0 is r = 3

√16 so that the

only critical point we need to consider is when r = 3√

16. Using the formula for s′(r), we find that s(r)is decreasing on (0, 3

√16) and increasing on ( 3

√16,∞). It follows that s has an absolute minimum on

(0,∞) at r = 3√

16. The dimensions of the cylinder that minimize the surface area are r = 3√

16 in and

h =32

( 3√

16)2in.

7. Let x be the distance from (0, 0) to the bottom right corner of the rectangle and let y be the heightof the rectangle. The area of the rectangle is

2xy = 2x(−x2 + 2) = −2x3 + 4x.

We need to find the maximum value of A(x) = −2x3 + 4x subject to the condition that 0 ≤ x ≤√

2.To see that this is the correct range for x, we note that the graph of −x2 + 2 crosses the x axis at thepoints (

√2, 0) and (−

√2, 0). Now A(0) = 0 and A(

√2) = 0. Next we find the critical points of A(x)

that are in the interval [0,√

2]. We have A′(x) = −6x2 + 4 and

−6x2 + 4 = 0 =⇒ x2 =2

3=⇒ x = ±

√2

3.

The only critical point of A(x) in the interval [0,√

2] is√

23 and

A(√

2/3) = −2(√

2/3)3 + 4√

2/3 =8

3

√2

3≈ 2.177.

The maximum area of such a rectangle is8

3

√2

3.

83

4 Antiderivatives

4.1 Definite Integrals and Areas

1. Suppose that f is a continuous function on the interval [1, 4] and f(x) > 0 for 1 ≤ x ≤ 4. Explain

what

∫ 4

1

f(x)dx represents in geometric terms.

2. Suppose that the velocity of an object is v(t) = −12t+ 18 where t is in seconds and v(t) is in feet persecond.

(a) Find the initial velocity of the object.

(b) Find the velocity of the object at time t = 4.

(c) Find the displacement of the object from time t = 0 to time t = 1.

(d) Find the displacement of the object from time t = 0 to time t = 2.

(e) Find the values t0 for which the displacement of the object from time t = 0 to time t = t0 is 0.

3. Recall that the area A of a circle of radius r is given by the formula A = πr2. Use this fact to evaluate

the definite integral

∫ 1

−1

√1− x2dx.

4. Evaluate each definite integral

∫ b

a

f(x)dx by computing the area of the region above the x-axis and

below the graph of f(x) from x = a to x = b .

(a)

∫ 1

1

4dx (b)

∫ 2

1

4dx (c)

∫ 3

1

4dx (d)

∫ 3

1

4 + xdx

5. The velocity of an object thrown straight up into the air is v(t) = −32t+ 64 where v(t) is in feet persecond. Here t is the number of seconds since the object was thrown.

(a) What is the maximum velocity of the object?

(b) How long will it take for the object to return to its initial height?

6. Let v(t) be the rate, in gallons per second, that water flows out of a flood gate t seconds after thegate was opened.

(a) If v(t) = 100 for all t ≥ 0, then find the amount of water that went out of the gate from time t = 1to time t = 3.

(b) If v(t) = 100 + 10t for all t ≥ 0, then find the amount of water that went out of the gate from timet = 1 to time t = 3.

7. Suppose that f is a continuous function on [1, 3] with

∫ 2

1

f(x)dx = 8 and

∫ 3

2

f(x)dx = −2. Evaluate

the definite integral

∫ 3

1

5f(x)dx.

84

8. Assume that f is a continuous function on [−3, 2] with

∫ 2

−3f(x)dx = 4 and

∫ 0

−3f(x)dx = 12. Find

the value of the definite integral

∫ 2

0

f(x)dx.

9. Assume that f and g are continuous functions on [0, 1] with

∫ 1

0

f(x)dx = π and

∫ 1

0

g(x)dx = e. Find

the value of the definite integral

∫ 1

0

3f(x) + 2g(x)dx.

Definite Integrals and Areas - Solutions

1. The definite integral

∫ 4

1

f(x)dx represents the area of the region in the plane that is bounded by the

x-axis, the graph of f , the vertical line x = 1, and the vertical line x = 4.

2. (a) The initial velocity of the object is v(0) = 18 feet per second.

(b) The velocity of the object at time t = 4 is v(4) = −30 feet per second.

(c) The displacement of the object from time t = 0 to time t = 1 is the value of the definite integral∫ 1

0

−12t+ 18dt.

This value is the signed area of the shaded region shown below. This region is a trapezoid whose verticesare (0, 0), (0, 18), (1, 0), and (1, 6). The formula for the area of a trapezoid with base length b andheights h1 and h2 is 1

2b(h1 + h2). Using this formula, we find that the area of this trapezoid is

1

2(1)(18 + 6) = 12.

Therefore, the displacement of the object from time t = 0 to t = 1 is 12 feet.

0.5 1 1.5 2

5

10

15

(d) The displacement of the object from time t = 0 to time t = 2 is the value of the definite integral∫ 2

0

−12t + 18dt. This is the signed area of the shaded region shown below. This region consists of

85

two triangles. The area of the first triangle is 12 (1.5)(18) = 27/2. The area of the second triangle is

12 (.5)(6) = 3/2. Since the second triangle lies below the x-axis, we must count its area as negativebecause we are computing the signed area of the shaded region. Therefore, the displacement of theobject from time t = 0 to t = 2 is

27

2− 3

2= 12 feet.

0.5 1 1.5 2

−5

5

10

15

(e) We need to find the values of t0 for which

∫ t0

0

−12t+ 18dt = 0. One value of t0 for which this is true

is t0 = 0. By sketching the graph of v(t) = −12t + 18 from t = 0 to t = 3, we see that the area abovethe x-axis and below v(t) from t = 0 to t = 1.5 is the same as the area below the x-axis and above v(t)

from t = 1.5 to t = 3 (see the figure below). Thus,

∫ 3

0

−12t+ 18dt = 0.

0.5 1 1.5 2 2.5 3

−10

10

The values of t0 for which the displacement of the object from time t = 0 to time t = t0 is 0 are t0 = 0and t0 = 3.


∫ 1

−1

√1− x2dx represents the area of half of a circle of radius 1. Using the

formula A = πr2, we have

∫ 1

−1

√1− x2dx =

1

2π(1)2 =

1

2π.

86

4. (a) The definite integral

∫ 1

1

4dx represents the area between the lines y = 0 and y = 4 from x = 1 to

x = 1. The area of this region is 0 so

∫ 1

1

4dx = 0.

(b) The definite integral

∫ 2

1

4dx represents the area between the lines y = 0 and y = 4 from x = 1 to

x = 2. This region forms a rectangle with base length 1 and height 4 so that

∫ 2

1

4dx = 1 · 4 = 4.

(c) As in (b), the definite integral

∫ 3

1

4dx represents the area between the lines y = 0 and y = 4 from

x = 1 to x = 3. This region forms a rectangle with base length 2 and height 4 so

∫ 3

1

4dx = 2 · 4 = 8.

(d) The definite integral

∫ 3

1

4 + xdx represents the area of a trapezoid whose vertices are (1, 0), (1, 5),

(3, 7), and (3, 0) (see the figure below). The area of such a trapezoid is 2( 12 (5 + 7)) = 12 so that∫ 3

1

4 + xdx = 12.

1 2 3 4

2

4

6

8

5. (a) The maximum velocity is the largest value of v(t) = −32t+ 64 for t ≥ 0. The maximum value ofv(t) = −32t+ 64 for t ≥ 0 is v(0) = 64 feet per second.

(b) The object will return to its initial height when the displacement is 0. Therefore, we are looking forthe time t0 > 0 for which ∫ t0

0

−32t+ 64dt = 0.

By sketching the graph of v(t) = −32t+ 64, we see that the area above the x-axis and below v(t) fromt = 0 to t = 2 is the same as the area below the x-axis and above v(t) from t = 2 to t = 4 (see the figure

below). We conclude that

∫ 4

0

−32t + 64dt = 0 and so it will take 4 seconds for the object to return to

its initial height.

87

1 2 3 4

−60

−40

−20

20

40

60

6. (a) The amount of water that went out of the gate from time t = 1 to time t = 3 is∫ 3

1

v(t)dt =

∫ 3

1

100dt.

This signed area is the same as the area of a rectangle with height 100 and base length 2. We concludethat 200 gallons of water went out during this time period.

(b) As in (a), we need to find ∫ 3

1

v(t)dt =

∫ 3

1

100 + 10tdt.

This signed area is the same as the area of a trapezoid with base length 2 whose heights are 110 and130. The area of such a trapezoid is 2 · 12 (110 + 130) = 240. Therefore, 240 gallons of water went out ofthe gate from time t = 1 to time t = 3.

7.

∫ 3

1

5f(x)dx = 5

∫ 2

1

f(x)dx+ 5

∫ 3

2

f(x)dx = 5 · 8 + 5(−2) = 30

8.

∫ 2

0

f(x)dx =

∫ 2

−3f(x)dx−

∫ 0

−3f(x)dx = 4− 12 = −8

9.

∫ 1

0

3f(x) + 2g(x)dx = 3

∫ 1

0

f(x)dx+ 2

∫ 1

0

g(x)dx = 3π + 2e

4.2 Riemann Sums

1. Let a1, a2, a3, . . . be a sequence of real numbers and k be a constant.

(a) Write the sum

6∑i=2

ai in expanded form.

(b) Write the sum a1 + a2 + a3 + a4 + a5 + a6 + a7 using summation notation.

(c) Write the sum

6∑i=2

kai in expanded form.

88

(d) Fill in the blank. For any positive integer m,

m∑i=1

kai =m∑i=1

ai.

2. Write the sum in expanded form.

(a)

5∑i=1

(2i+ 1) (b)

6∑i=3

(i2 − 1) (c)

4∑j=0

j

j + 1(d)

5∑n=0

(1

2

)n

3. Write the sum using summation notation.

(a) 1 + 2 + 3 + 4 + · · ·+ 87 (b) 2 + 4 + 6 + 8 + · · ·+ 40 (c) 1 + 4 + 9 + 16 + 25 + · · ·+ 121

4. Write the sum using summation notation.

(a) 0 +1

4+

2

4+

3

4(b)

1

4+

2

4+

3

4+ 1

(c) 0 +1

5+

2

5+

3

5+

4

5(d) 1 +

4

3+

5

3+ 2 +

7

3+

8

3

5. Are the sums

10∑i=1

(2i− 1) and

9∑i=0

(2i+ 1) the same?

6. Write the sum

100∑j=1

(3j + 2) = 5 + 8 + 11 + · · ·+ 302

using summation notation in which the index begins at 0 instead of 1.

7. Find the approximation to

∫ 1

−1x2dx using the Right Hand Rule with 4 rectangles.


∫ 3

1

cosxdx using the Left Hand Rule with 6 rectangles.


∫ 3

1

1

xdx using the Right Hand Rule with 3 rectangles.

10. Consider the definite integral

∫ 1

0

xdx. If Rn is the approximation to this definite integral using the

Right Hand Rule with n rectangles, then

R2 = 1−02

(12 + 2

2

)and R3 = 1−0

3

(13 + 2

3 + 33

).

(a) Find a formula for R4 that is similar to the formulas for R2 and R3 given above.

(b) Find a formula for R5 that is similar to the formulas for R2 and R3 given above.

89

(c) Using summation notation, write down a formula for Rn.

11. Consider the definite integral

∫ 1

0

cosxdx. If Rn is the approximation to this definite integral using

the Right Hand Rule with n rectangles, then

R2 = 1−02

(cos( 1

2 ) + cos(22 ))

and R3 = 1−03

(cos( 1

3 ) + cos( 23 ) + cos(3

3 )).

(a) Find a formula for R4 that is similar to the formulas for R2 and R3 given above.

(b) Using summation notation, write down a formula for Rn.

12. Let f be a continuous function on [a, b]. The approximation to

∫ b

a

f(x)dx using the Right Hand

Rule with n rectangles isn∑i=1

f

(a+ i

(b− an

))· b− a

n.

(a) List the first three terms of this sum.

(b) Each term in this sum is of the form f(a+ i

(b−an

))· b−an and this term corresponds to the area of

the i-th rectangle in the approximation. What does f(a+ i

(b−an

))represent in terms of rectangles?

13. The approximation to

∫ 3

1

x2dx using the Left Hand Rule with n rectangles is

n−1∑i=0

(1 +

2i

n

)22

n.

What is the approximation to

∫ 3

1

x2dx using the Right Hand Rule with n rectangles?

14. The approximation to

∫ 2

1

xdx using the Right Hand Rule with n rectangles is

n∑i=1

(1 +

i

n

)1

n.

(a) Using the formula

n∑i=1

i =n(n+ 1)

2, evaluate lim

n→∞

n∑i=1

(1 +

i

n

)1

n.

(b) Find

∫ 2

1

xdx by sketching the graph of y = x and using the area of a trapezoid formula.

Riemann Sums - Solutions

1. (a) In expanded form, the sum

6∑i=2

ai is a2 + a3 + a4 + a5 + a6.

90

(b) In summation notation, the sum a1 + a2 + a3 + a4 + a5 + a6 + a7 is

7∑i=1

ai.

(c) ka2 + ka3 + ka4 + ka5 + ka6

(d) For any positive integer m,

m∑i=1

kai = k

m∑i=1

ai.

2. (a) Using the definition of summation notation we find that

5∑i=1

(2i+ 1) = (2(1) + 1) + (2(2) + 1) + (2(3) + 1) + (2(4) + 1) + (2(5) + 1) = 3 + 5 + 7 + 9 + 11.

(b)

6∑i=3

(i2 − 1) = (32 − 1) + (42 − 1) + (52 − 1) + (62 − 1) = 8 + 15 + 24 + 35

(c)

4∑j=0

j

j + 1= 0 +

1

2+

2

3+

3

4+

4

5

(d)

5∑n=0

(1

2

)n=

(1

2

)0

+

(1

2

)1

+

(1

2

)2

+

(1

2

)3

+

(1

2

)4

+

(1

2

)5

= 1 +1

2+

1

4+

1

8+

1

16+

1

32

3. (a) 1 + 2 + 3 + 4 + · · ·+ 87 =

87∑i=1

i (b) 2 + 4 + 6 + 8 + · · ·+ 40 =

20∑i=1

2i

(c) 1 + 4 + 9 + 16 + 25 + · · ·+ 121 =

11∑i=1

i2

4. (a) 0 +1

4+

2

4+

3

4=

3∑i=0

i

4(b)

1

4+

2

4+

3

4+ 1 =

4∑i=1

i

4

(c) 0 +1

5+

2

5+

3

5+

4

5=

4∑i=0

i

5(d) 1 +

4

3+

5

3+ 2 +

7

3+

8

3=

5∑i=0

(1 +

i

3

)5. The two sums are the same. In expanded form, the first sum is

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19.

This is the same as the expanded form of the second sum.

6.

99∑j=0

(3j + 5)

7.1− (−1)

4

((−.5)2 + 02 + (.5)2 + 12

)=

3

4

8.1− (−1)

6(cos(−1) + cos(−2/3) + cos(−1/3) + cos 0 + cos(1/3) + cos(2/3)) ≈ 1.6673

9.3− 1

3

(1

5/3+

1

7/3+

1

3

)=

286

315

91

10. (a) R4 =1− 0

4

(1

4+

2

4+

3

4+

4

4

)(b) R5 =

1− 0

5

(1

5+

2

5+

3

5+

4

5+

5

5

)

(c) Rn =1− 0

n

n∑i=1

i

n

11. (a) R4 =1− 0

4

(cos

(1

4

)+ cos

(2

4

)+ cos

(3

4

)+ cos

(4

4

))(b) Rn =

1− 0

n

n∑i=1

cos

(i

n

)12. (a) The first three terms of the sum are

f

(a+ 1 · b− a

n

)b− an

, f

(a+ 2 · b− a

n

)b− an

, and f

(a+ 3 · b− a

n

)b− an

.

(b) The term f

(a+ i · b− a

n

)represents the height of the i-th rectangle in the approximation.

13.

n∑i=1

(1 +

2i

n

)22

n

14. (a) We have

limn→∞

n∑i=1

(1 +

i

n

)1

n= lim

n→∞

n∑i=1

(1

n+

i

n2

)= limn→∞

(n∑i=1

1

n+

n∑i=1

i

n2

)

= limn→∞

(1

n

n∑i=1

1 +1

n2

n∑i=1

i

)= limn→∞

(1

nn+

1

n2n(n+ 1)

2

)= lim

n→∞

(1 +

n2 + n

2n2

)= limn→∞

(1 +

1 + 1/n

2

)= 1 +

1 + 0

2=

3

2.

(b) The integral

∫ 2

1

xdx is the signed area of trapezoid shown below. This area is1

2(1)(1 + 2) =

3

2.

0.5 1 1.5 2 2.5 3

1

2

3

92

4.3 The Fundamental Theorems of Calculus

1. Complete the statement of the Fundamental Theorem of Calculus, Part 1. Let f be a continuousfunction on [a, b] and let

g(x) =

∫ x

a

f(t)dt

for a ≤ x ≤ b. The function g(x) is continuous on [a, b], differentiable on (a, b) and

d

dx

∫ x

a

f(t)dt = .

2. Complete the statement of the Fundamental Theorem of Calculus, Part II - Let f be continuous on[a, b]. If F is any function with F ′(x) = f(x), then∫ b

a

f(x)dx = .

3. Evaluate the definite integral.

(a)

∫ π/3

0

cos tdt (b)

∫ 1

0

x3 + 3x2 + 1dx (c)

∫ 2

−1exdx

4. Find the area between the x-axis and the graph of y =1

xfrom x = 1 to x = e3.

5. Find the area between the x-axis and the graph of y = cosx from x = 0 to x = π/2.

6. Let v(t) be the velocity of water flowing out of a flood gate in gallons per second where t is the number

of seconds since the gate was opened. What is the practical interpretation of

∫ 3

1

v(t)dt?

7. Suppose that f(t) is the velocity of a race car, in miles per hour, during a race where the time t = 0corresponds to the time when the race began. Assume that the race is 270 miles, and the race car finishesthe race in exactly 3 hours.

(a) What is the practical interpretation of

∫ 2

1

f(t)dt?

(b) Find

∫ 3

0

f(t)dt.

(c) What is the practical interpretation of the statement that

∫ 1

0

f(t)dt = 95?

8. Let f(t) be the rate, in gallons per second, that water flows out of a fire hydrant t seconds after thehydrant broke. Set up an expression involving f(t) that gives the total amount of water that has flownout of the hydrant in the first 10 seconds.


(a)

∫ 7

−2πdx (b)

∫ 2

1

√xdx (c)

∫ 2

1

x2 + x

xdx

93

10. You are trapped at the bottom of a well. To get out of the well, you must evaluate the definiteintegral ∫ 2

1

xdx.

You have two choices. You may evaluate the definite integral using the Fundamental Theorem of CalculusPart 2, or you may evaluate the definite integral by computing

limn→∞

n∑i=1

(1 +

i

n

)1

n.

Escape the well by evaluating

∫ 2

1

xdx . State which method you used and why.

11. A hot cup of tea is placed in a refrigerator. The rate of change of the temperature of the tea, indegrees Fahrenheit per second, t seconds after it was placed in the refrigerator is given by the functionf(t).

(a) What is the practical interpretation of

∫ 60

0

f(t)dt?

(b) Suppose T is a very large number. What do you expect the value of

∫ T+60

T

f(t)dt to be?


(a) F (x) =

∫ x

2

et sin tdt (b) H(t) =

∫ t

−2

√x2 + x4 + 1dx (c) G(r) =

∫ r

0

ex3+1dx


(a) F (x) =

∫ x3

2

et sin tdt (b) H(t) =

∫ −2t

√x2 + x4 + 1dx (c) G(r) =

∫ 0

r

ex3+1dx

14. Find the derivative of F (x) =

∫ x2

x

et sin tdt.

The Fundamental Theorems of Calculus - Solutions

1. Let f be a continuous function on [a, b] and let g(x) =

∫ x

a

f(t)dt for a ≤ x ≤ b. The function g(x) is

continuous on [a, b], differentiable on (a, b) and

d

dx

∫ x

a

f(t)dt = f(x).

2. Let f be continuous on [a, b]. If F is any function with F ′(x) = f(x), then∫ b

a

f(x)dx = F (b)− F (a).

94

3. (a)

∫ π/3

0

cos tdt = sin t∣∣∣π/30

= sin(π/3)− sin 0 =

√3

2

(b)

∫ 1

0

x3 + 3x2 + 1dx =

(1

4x4 + x3 + x

) ∣∣∣10

=

(1

4(1)4 + 13 + 1

)−(

1

4(0)4 + 03 + 0

)=

9

4

(c)

∫ 2

−1exdx = ex

∣∣∣2−1

= e2 − e−1

4. The area between the x-axis and the graph of y =1

xfrom x = 1 to x = e3 is the value of the definite

integral ∫ e3

1

1

xdx

which is

∫ e3

1

1

xdx = (ln |x|)

∣∣∣e31

= ln(e3)− ln(1) = 3− 0 = 3.

5. The area between the x-axis and the graph of y = cosx from x = 0 to x = π/2 is∫ π/2

0

cosxdx = (sinx)∣∣∣π/20

= sin(π/2)− sin 0 = 1.


∫ 3

1

v(t)dt is the amount of water, in gallons, that has flown out of the gate from

the 1 second mark to the 3 second mark.

7. (a) The integral

∫ 2

1

f(t)dt represents the displacement of the race car between the one hour mark and

the two hour mark. Assuming that the race car never goes backwards during the race, then

∫ 2

1

f(t)dt

is the distance traveled by the car between the one hour mark and the two hour mark.

(b) 270

(c) The displacement of the race car in the first hour of the race is 95 miles.

8.

∫ 10

0

f(t)dt

9. (a)

∫ 7

−2πdx = (πx)

∣∣∣7−2

= π(7)− π(−2) = 9π.

(b)

∫ 2

1

√xdx =

∫ 2

1

x1/2dx =2

3x3/2

∣∣∣21

=2

3(2)3/2 − 2

3(1)3/2 =

4√

2

3− 2

3.

(c) To evaluate the integral, we first rewrite

∫ 2

1

x2 + x

xdx as

∫ 2

1

x2 + x

xdx =

∫ 2

1

x2

x+x

xdx =

∫ 2

1

x+ 1dx.

Using the Fundamental Theorem of Calculus Part II, we get that

∫ 2

1

x+ 1dx =

(x2

2+ x

) ∣∣∣21

=5

2.

95

10. It is much easier to use the Fundamental Theorem of Calculus Part II. Doing so, we get∫ 2

1

xdx =x2

2

∣∣∣21

=(2)2

2− (1)2

2=

3

2.

11. (a) Let F (t) be the temperature of the coffee t seconds after it was placed in the refrigerator. ThenF ′(t) = f(t) and so by the Fundamental Theorem of Calculus Part 2,∫ 60

0

f(t)dt = F (60)− F (0).

From this we see that the integral

∫ 60

0

f(t)dt represents the difference between the temperature of the

coffee at time t = 0 and at time t = 60. Informally, F (60)−F (0) tells us how much the coffee has cooledin the first minute.

(b) The integral

∫ T+60

T

f(t)dt represents the difference between the temperature of the coffee at time T

and one minute later at time T + 60. If T is large, we do not expect the coffee to cool very much during

this time interval since the coffee has been in the refrigerator for a long time. Thus,

∫ T+60

T

f(t)dt should

be a number that is very close to 0. If the temperature of the coffee has reached the temperature inside

the refrigerator, we expect

∫ T+60

T

f(t)dt to be zero.

12. (a) By the Fundamental Theorem of Calculus Part I,

F ′(x) =d

dx

∫ x

2

et sin tdt = ex sinx.

(b) H ′(t) =√t2 + t4 + 1 (c) G′(r) = er

3+1

13. (a) Using the Chain Rule together with the Fundamental Theorem of Calculus Part I, we find that

F ′(x) =d

dx

∫ x3

2

et sin tdt = ex3

sin(x3) · 3x2.

(b) Using the property

∫ b

a

f(x)dx = −∫ a

b

f(x)dx together with the Fundamental Theorem of Calculus

Part I, we find that

H ′(t) =d

dt

∫ −2t

√x2 + x4 + 1dx = − d

dt

∫ t

−2

√x2 + x4 + 1dx = −

√t2 + t4 + 1.

(c) Using the same approach that was used in (b), we find that G′(r) = −er3+1.

14. Using properties of definite integrals and the Fundamental Theorem of Calculus Part I,

F ′(x) =d

dx

∫ x2

x

et sin tdt =d

dx

(∫ 0

x

et sin tdt+

∫ x2

0

et sin tdt

)

=d

dx

(−∫ x

0

et sin tdt+

∫ x2

0

et sin tdt

)= −ex sinx+ ex

2

sin(x2)(2x).

96

The two properties that we used are

∫ b

a

f(x)dx = −∫ a

b

f(x)dx and

∫ b

a

f(x)dx =

∫ c

a

f(x)dx+

∫ b

c

f(x)dx.

We also used the Chain Rule when we took the derivative of

∫ x2

0

et sin tdt.

4.4 Indefinite Integrals

1. Suppose that F (x) is a function with F ′(x) = f(x). Evaluate

∫f(x)dx.

2. Evaluate each indefinite integral.

(a)

∫3x5 + 1dx (b)

∫2

x4− 1

4x2dx (c)

∫ex7 − 2x2 + 4x−2dx


(a)

∫3√x+

1√x− 2dx (b)

∫5

7√x3dx (c)

∫5

7√x3

+2

5√xdx


(a)

∫3 sinx+ secx tanxdx (b)

∫− cos θ + 6dθ (c)

∫1

x3+

4

3sinxdx


(a)

∫5

1 + x2dx (b)

∫1

x+

1√1− x2

dx (c)

∫7r + 4r7dr


(a)

∫5

x− 2exdx (b)

∫−8 cosx+ sin2 x+ cos2 x+

1

2secx tanxdx (c)

∫sec2 θ + eθdθ

7. Evaluate each indefinite integral. Hint: Try to use algebra to rewrite the integrand before integrating.

(a)

∫t2 + t

8t3dt (b)

∫(u− 2)2du (c)

∫4

u+ tanu cosudu

8. Find f(x) if f ′(x) = 3 cosx and f(π) = −4.

97

Indefinite Integrals - Solutions

1. If F ′(x) = f(x), then

∫f(x)dx = F (x) + C where C is any constant.

2. (a)1

2x6 + x+ C

(b)

∫2

x4− 1

4x2dx =

∫2x−4 − 1

4x2dx =

2x−3

−3− 1

4· x

3

3+ C = − 2

3x3− x3

12+ C

(c)

∫ex7 − 2x2 + 4x−2dx =

ex8

8− 2x3

3+

4x−1

−1+ C =

ex8

8− 2x3

3− 4

x+ C

3. (a) ∫3√x+

1√xdx− 2dx =

∫x1/3 + x−1/2 − 2dx =

x4/3

4/3+x1/2

1/2− 2x+ C

=3

4x4/3 + 2x1/2 − 2x+ C =

3

4

3√x4 + 2

√x− 2x+ C

(b)

∫5

7√x3dx =

∫5x3/7dx =

5x10/7

10/7+ C =

7

2x10/7 + C

(c) ∫5

7√x3

+2

5√xdx =

∫5x−3/7 +

2

5x−1/2dx =

5x4/7

4/7+

2

5· x

1/2

1/2+ C

=35

4x4/7 +

4

5x1/2 + C =

35

4

7√x4 +

4

5

√x+ C

4. (a) −3 cosx+ secx+ C (b) − sin θ + 6θ + C

(c)

∫1

x3+

4

3sinxdx =

∫x−3 +

4

3sinxdx =

x−2

−2− 4

3cosx+ C = − 1

2x2− 4

3cosx+ C

5. (a) 5 tan−1 x+ C (b) ln |x|+ sin−1 x+ C (c)7r

ln 7+

1

2r8 + C

6. (a) 5 ln |x| − 2ex + C

(b) ∫−8 cosx+ sin2 x+ cos2 x+

1

2secx tanxdx =

∫−8 cosx+ 1 +

1

2secx tanxdx

= −8 sinx+ x+1

2secx+ C

(c) tan θ + eθ + C

7. (a)

∫t2 + t

8t3dt =

∫t2

8t3+

t

8t3dt =

∫1

8t+

1

8t−2dt =

1

8ln |t| − 1

8t+ C

(b)

∫(u− 2)2du =

∫u2 − 4u+ 4du =

u3

3− 2u2 + 4u+ C

98

(c)

∫4

u+ tanu cosudu =

∫4

u+

sinu

cosu· cosudu =

∫4

u+ sinudu = 4 ln |u| − cosu+ C

8. We have

f(x) =

∫3 cosxdx = 3 sinx+ C

for some constant C. We use the condition f(π) = −4 to determine C:

−4 = f(π) = 3 sin(π) + C = 0 + C = C.

We conclude that C = −4 so that f(x) = 3 sinx− 4.

4.5 Integration by Substitution

1. Complete the statement. Let f and g be functions that are differentiable on an open interval I. If Fis a function with d

dxF (x) = f(x), then

∫f(g(x))g′(x)dx = .

2. Evaluate the indefinite integral.

(a)

∫2xex

2

dx (b)

∫2x cos(x2)dx (c)

∫x sin(x2)dx


(a)

∫ 2

0

xex2

dx (b)

∫ √π/40

x cos(x2)dx (c)

∫ √e1

x

x2 + 1dx

4. Evaluate the indefinite integral.

(a)

∫(4x3 + 2x+ 1)(x4 + x2 + x)8dx (b)

∫(x3 + 1)ex

4+4xdx (c)

∫sinx cos(cosx)dx


(a)

∫ 2

0

x

x2 + 1dx (b)

∫ π

−πsinx cosxdx (c)

∫ 1

0

x8ex9

dx

6. Evaluate

∫(4x+ 3)13dx.

7. Let k be a non-zero real number. Use the substitution technique to prove the given formula.

(a)

∫ekxdx =

1

kekx + C

99

(b)

∫sin(kx)dx = −1

kcos(kx) + C

(c)

∫cos(kx)dx =

1

ksin(kx) + C

8. Let f(t) =100

(t+ 1)2+ 50 be the rate, in gallons per second, at which water flows out of a flood gate.

Find the amount of water that has flown out of the gate from time t = 1 to time t = 2.

Integration by Substitution - Solutions

1. Let f and g be functions that are differentiable on an open interval I. If F is a function withddxF (x) = f(x), then

∫f(g(x))g′(x)dx = F (g(x)) + C.

2. (a) If u = x2, then dudx = 2x. The differential du is du = 2xdx. Therefore,∫

2xex2

dx =

∫eudu = eu + C = ex

2

+ C.

(b) If u = x2, then du = 2xdx so that∫2x cos(x2)dx =

∫cos(u)du = sin(u) + C = sin(x2) + C.

(c) If u = x2, then du = 2xdx so that 12du = xdx. Using this equation together with u = x2, we have∫

x sin(x2)dx =1

2

∫sin(u)du = −1

2cos(u) + C = −1

2cos(x2) + C.

3. (a) Let u = x2. First we will change the limits of integration. When x = 0, we have u = 02 = 0.When x = 2, we have u = 22 = 4. Also, du = 2xdx so that 1

2du = xdx. Thus,∫ 1

0

xex2

dx =1

2

∫ 4

0

eudu =1

2(eu)

∣∣∣40

=1

2(e4 − e0) =

1

2(e4 − 1).

(b) Let u = x2. First we will change the limits of integration. If x = 0, then u = 02 = 0. If x =√π/4,

then u = (π/4)2 = π/4. Also, du = 2xdx so that 12du = xdx. Thus,∫ √π/4

0

x cos(x2)dx =1

2

∫ π/4

0

cosudu =1

2(sinu)

∣∣∣π/40

=1

2(sin(π/4)− sin(0)) =

√2

4.

(c) Let u = x2 + 1. If x = 1, then u = 2. If x =√e, then u = e+ 1. Also, du = 2xdx so that 1

2du = xdx.Therefore, ∫ √e

1

x

x2 + 1dx =

1

2

∫ e+1

2

1

udu =

1

2(ln |u|)

∣∣∣e+1

2=

1

2(ln(e+ 1)− ln(2)).

100

4. (a) If u = x4 + x2 + x, then du = (4x3 + 2x+ 1)dx and∫(4x3 + 2x+ 1)(x4 + x2 + x)8dx =

∫u8du =

1

9u9 + C =

1

9(x4 + x2 + x)9 + C.

(b) If u = x4 + 4x, then du = (4x3 + 4)dx so 14du = (x3 + 1)dx. We have∫

(x3 + 1)ex4+4xdx =

1

4

∫eudu =

1

4eu + C =

1

4ex

4+4x + C.

(c) If u = cosx, then du = − sinxdx so that∫sinx cos(cosx)dx = −

∫cosudu = − sinu+ C = − sin(cosx) + C.

5. (a) Let u = x2 + 1. First we change the limits of integration. If x = 0, then u = 1. If x = 1, thenu = 5. Furthermore, du = 2xdx so 1

2du = xdx. We have∫ 2

0

x

x2 + 1dx =

1

2

∫ 5

1

1

udu =

1

2(ln |u|)

∣∣∣51

=1

2(ln(5)− ln(1)) =

1

2ln 5.

(b) Let u = sinx. If x = π, then u = 0. If x = −π, then u = 0. Furthermore, du = cosxdx so∫ π

−πsinx cosxdx =

∫ 0

0

udu = 0.

(c) Let u = x9. If x = 0, then u = 09 = 0. If x = 1, then u = 19 = 1. Also, du = 9x8dx so 19du = x8dx.

Thus, ∫ 1

0

x8ex9

dx =1

9

∫ 1

0

eudu =1

9(eu)

∣∣∣10

=1

9(e1 − e0) =

1

9(e− 1).

6. If u = 4x+ 3, then du = 4dx and 14du = dx. Therefore,∫

(4x+ 3)13dx =1

4

∫u13du =

1

4· 1

14u14 + C =

1

56(4x+ 3)14 + C.

7. (a) If u = kx, then du = kdx and so 1kdu = dx. Therefore,∫

ekxdx =

∫eudu

k=

1

k

∫eudu =

1

keu + C =

1

kekx + C.

(b) If u = kx, then du = kdx and so 1kdu = dx. Therefore,∫

sin(kx)dx =

∫sin(u)

du

k=

1

k

∫sinudu = −1

kcosu+ C = −1

kcos(kx) + C.

(c) If u = kx, then du = kdx and so 1kdu = dx. Therefore,∫

cos(kx)dx =

∫cos(u)

du

k=

1

k

∫cosudu =

1

ksinu+ C =

1

ksin(kx) + C.

101

8. To find the desired amount of water, we need to evaluate the definite integral

∫ 2

1

100

(t+ 1)2+ 50. If

u = t+ 1, then du = dt. If t = 1, then u = 2 and if t = 2, then u = 3. We have∫ 2

1

100

(t+ 1)2+50dt =

∫ 3

2

100

u2+50du =

(−100

u+ 50u

) ∣∣∣32

=

(−100

3+ 50(3)

)−(−100

2+ 50(2)

)=

200

3.

The amount of water that has flown out of the gate from time t = 1 to time t = 2 is 2003 gallons.

102

exercises for first semester calculus v1

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