mathematics103–appliedcalculusi solutionsmanualuregina.ca/~pso748/math103_sol.pdf · 2020. 4....

Mathematics 103 – Applied Calculus I

Solutions Manual

Department of Mathematics and Statistics

University of Regina

By Paul Arnaud Songhafouo Tsopméné

Contents

Introduction 5

1 Practice Problems 7

1.1 PB1: Factorization and Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 PB2: The graph of a function, Lines and linear functions, and Limits . . . . . . . . . . . . . . 8

1.3 PB3: Limits (continued) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4 PB4: One-sided limits, Continuity, and the Derivative . . . . . . . . . . . . . . . . . . . . . . 12

1.5 PB5: Techniques of Differentiation, Product and Quotient Rules . . . . . . . . . . . . . . . . 13

1.6 PB6: Techniques of Differentiation, Product and Quotient Rules (continued) and the ChainRule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.7 PB7: Implicit Differentiation and Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.8 PB8: Increasing and Decreasing Functions, Local Extrema, Concavity, and Points of Infection 17

1.9 PB9: Curve Sketching and Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . . 18

1.10 PB10: Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.11 PB11: Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.12 PB12: The Definite Integral and the Fundamental Theorem of Calculus . . . . . . . . . . . . 24

2 Solutions to Practice Problems 27

2.1 Solution to PB1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.2 Solution to PB2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.3 Solution to PB3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.4 Solution to PB4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.5 Solution to PB5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2.6 Solution to PB6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

2.7 Solution to PB7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

2.8 Solution to PB8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

2.9 Solution to PB9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3

4 CONTENTS

2.10 Solution to PB10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

2.11 Solution to PB11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

2.12 Solution to PB12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Introduction

This short introduction is just to say a few words about the present manual that I wrote while teachingApplied Calculus I (for Business, Economics, and the Social and Life Sciences) at the University of Regina(U of R) in the fall of 2019.

What is in this manual?

• A wide variety of practice problems organized in such a way that students can learn gradually.Of course, these cover all the material of the Applied Calculus I course delivered at the University ofRegina.

• Detailed solutions to practice problems. These are well explained step by step so that students caneasily understand what is going on.

• A review of the theory. The theory related to every concept is briefly reviewed along the way (andat the right moment), the idea being not only to show to students how to apply it but also to facilitatethe understanding of solutions.

5

6 CONTENTS

Chapter 1

Practice Problems

1.1 PB1: Factorization and Functions

Sections covered: Section 1.1: Functions

Review: Factoring

Factor each expression.

1. x2 − 5x

2. −6x2 − 9x+ 4x+ 6

3. x2 − 25

4. −9x2 + 49

5. x2 + x− 2

6. 2x2 − 7x+ 3

Section 1.1: Functions

1.1.1. Evaluating a function

Compute the indicated values of the given function.

1. f(x) = x3 + 2x− 7, f(0), f(−1), f(2)

2. f(x) = xx2+1 , f(2), f(0), f(−1)

3. f(t) =√t2 + 2t+ 3, f(2), f(0), f(−4)

4. f(x) ={−2x+ 5 if x ≤ 2

x2 + 1 if x > 2,f(0), f(2), f(3)

7

8 CHAPTER 1. PRACTICE PROBLEMS

1.1.2. Domain of a function

Find the domain of each of the following functions.

1. f(x) = 2x

2. f(x) = x3

x2−x−2

3. f(x) = x5 − x2 + 7

4. f(x) =√x− 3

5. f(x) =√−2x+ 8

6. f(x) =√x2 − 4

7. f(x) = x√25−x2

1.1.3. Composition of functions

Find the composite function f(g(x)).

1. f(x) = x− 2 and g(x) = x2

2. f(x) = x2 − x+ 3 and g(x) = 2x+ 1

3. f(x) =√x2 + 1 and g(x) =

√x

4. f(x) = x2

x+2 and g(x) = 2x+ 7

1.1.4. Finding the difference quotient

Find and simplify the difference quotient of the given function.

1. f(x) = x+ 1

2. f(x) = 2x− 5

3. f(x) = 3x2 + 1

4. f(x) = 1x+2

1.2 PB2: The graph of a function, Lines and linear functions, andLimits

Sections covered:

• Section 1.2: The graph of a function.

• Section 1.3: Lines and linear functions.

• Section 1.5: Limits

1.2. PB2: THE GRAPH OF A FUNCTION, LINES AND LINEAR FUNCTIONS, AND LIMITS 9

Section 1.2: The graph of a function

1.2.1. Intercepts

Find the intercepts of the graph of each of the following functions.

1. f(x) = 2x− 6

2. f(x) = x2 + x− 2

3. f(x) = 4x2 − 4x+ 1

4. f(x) = x2 + x+ 3

5. f(x) = 3x−15x−8

6. f(x) = −x5 + 4x3

1.2.2. Parabola

Sketch the graph of f(x).

1. f(x) = −x2 + x+ 2

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

1.2.3. Finding points of intersection

Find points of intersection of f(x) and g(x), and sketch the graphs.

1. f(x) = 3x− 3 and g(x) = −x+ 4

2. f(x) = −x2 − 2x and g(x) = −x− 2

3. f(x) = 2x2 − x− 1 and g(x) = 2x+ 1

Section 1.3: Lines and linear functions

1.3.1. The slope of a line

Find the slope of the line passing through the points.

1. (1, 2) and (3, 4)

2. (2,−5) and (0, 1)

3. (− 12 ,

13 ) and ( 34 ,−1)


1.3.2. Finding the equation of a line

Find the equation of the line passing through the points.

1. (−1, 2) and (0, 5)

2. (7,−1) and (9, 12)

Section 1.5: Limits

1.5.1. Examining the concept of limit

Use a table to estimate the following limits.

1. limx→2

x2−4x−2

2. limx→1

√x−1x−1

1.3 PB3: Limits (continued)

Sections covered: Section 1.5: Limits

Section 1.5: Limits (continued)

1.5.2. Computing limits using a variety of techniques

Evaluate the following limits.

1. limx→7

3x

2. limx→4

x3 + x

3. limx→2

x5 − 4x3 − 7x+ 4

4. limx→−2

− 2x3 + 5x− 9

5. limx→1

9x2−4x+1

6. limx→2

x4−3x2+1x+8

7. limx→1

√−9x3 + 4x2 + 6

8. limx→−3

3√x2−1(−x3+8x−1)4

9. limx→1

9x2−4x+1

1.3. PB3: LIMITS (CONTINUED) 11

10. limx→3

x2−2x−3x2−9

11. limx→16

16−x√x−4

12. limx→0

1−√1−x2

x

13. limx→−5

x2+3x−10x2+6x+5

14. limx→4

√x−2

−3x+7

15. limx→2

√2x+5−

√x+7

x−2

16. limx→−2

−5x2−10x6+3x

17. limx→0

3√−x2 + 4x+ 2

18. limx→7

1x−

17

x−7

19. limx→3

x−3√x2−5x+10−2

1.5.3. Limits involving infinity

Find the following limits.

1. limx→∞

− 2x10

2. limx→−∞

− 5x10

3. limx→−∞

5x9

4. limx→∞

3x2

5. limx→∞

3√x

6. limx→∞

− 4x3 + x− 1

7. limx→−∞

1− 8x3 + 2x2

8. limx→∞

2x− x3 + 7x5 − 1

9. limx→−∞

13x4

10. limx→∞

1√x

11. limx→∞

−x2+13x2+x+1

12. limx→−∞

7x2

x2−4

13. limx→∞

3x2−2x−5x3+x2−x

14. limx→∞

3x2+1(x−1)2

15. limx→∞

√x+x2

2x−x2


1.4 PB4: One-sided limits, Continuity, and the Derivative

Sections covered:

• Section 1.6: One-sided limits and Continuity.

• Section 2.1: The Derivative

Section 1.6: One-sided limits and continuity

1.6.1. Evaluating one-sided limits

Evaluate the following limits (if they exist).

1. limx→2

f(x), where f(x) ={ √

−2x+ 4 if x ≤ 2

x− 2 if x > 2

2. limx→5

f(x), where f(x) ={x2 + 1 if x < 5

x+ 20 if x ≥ 5

3. limx→0

f(x), where f(x) =

{ √x−1x−1 if x > 0

x2 − x+ 12 if x ≤ 0

4. (a) limx→−3

f(x), (b) limx→0

f(x) , where f(x) =

2− x2 if x < −34x+ 3 if −3 < x < 0x2+2x−3x−1 if x ≥ 0

1.6.2. Continuity

1. Let f(x) = x2−16x−4 . Is f continuous at 4?

2. Let f(x) ={x2 + 2x if x ≤ −2√x+ 3 if x > −2. Is f continuous at −2?

3. Let f(x) =

x3 − x+ 1 if x < 0

0 if x = 0

1− x2 if x > 0.

Is f continuous at 0?

4. Let f(x) =

x2+4x−5

3−3x if x < 1

−2 if x = 1

2x− 4 if x > 1.


5. Find c such that the function f(x) ={cx2 − 1 if x ≤ 3

x+ c if x > 3is continuous at 3.

6. Find c such that the function f(x) =

x3 − x− 2 if x < 1

c2 − 3c if x = 1

−2x if x > 1

is continuous at 1.

1.5. PB5: TECHNIQUES OF DIFFERENTIATION, PRODUCT AND QUOTIENT RULES 13

Section 2.1: The derivative

2.1.1. Finding a derivative (using the limit definition)

Find the derivative of each of the following functions from the limit definition of the derivative.

1. Let f(x) = −3x+ 4. Find f ′(x).

2. Let f(x) = −5x2 + 7. Find f ′(x).

3. Let f(x) = 12x−1 . Find f

′(x).

4. Let f(x) = 1x2 . Find f ′(x).

5. Let f(x) =√3x+ 2. Find f ′(x).

2.1.2. Tangent line and rate of change

Let f(x) = −x2 + 3x and g(x) =√x− 1.

1. Find the equation of the tangent line to the graph of f at the given point.

(a) (0, 0)

(b) (1, 2)

2. Find the rate of change of g(x) at the given point.

(a) x = 2

(b) x = 5

1.5 PB5: Techniques of Differentiation, Product and Quotient Rules

Section covered:

• Section 2.2: Techniques of Differentiation

• Section 2.3: Product and Quotient Rules; Higher-Order Derivatives

Sections 2.2 & 2.3: Techniques of Differentiation, Product and Quotient Rules

Find the derivative of each of the following functions.

1. f(x) = 2018

2. f(x) = 34

3. f(x) = π2

4. f(x) = x6


5. f(x) = 3x−2

6. f(x) = x2

3

7. f(x) = 12x

43

8. f(x) =√x3

9. f(x) = 2x3

10. f(x) = 13√x

11. f(x) = 3√x5

12. f(x) =8√x

2

13. f(x) = π2

x

14. f(x) = x+ 1

15. f(x) = x2 − 2x5

16. f(x) = x−√x

2

17. f(x) = −3x4 − 2x3 + x2 − 1

18. f(x) = 2x3−3x2

4

19. f(x) = x53 − x 2

3

20. f(x) = x2 − 1x

21. f(x) = 1.4x5 − 2.5x2 + 3.8

22. f(x) = x√x

23. f(x) =√x+xx2

24. f(x) = (3x+ 4)(x− 5)

25. f(x) = (5x2 − 2)(x3 + 3x)

26. f(x) = (x3 + 1)(2x2 − 4x− 1)

27. f(x) = x2+4x+3√x

28. f(x) = ( 1x2 − 3

x4 )(x+ 5x3)

29. f(x) = 5x+15x−1

30. f(x) = 1+2x3−4x

31. f(x) = x2+1x3−1

32. f(x) = x3+3xx2−4x+3

33. f(x) = 1x3+2x2−1

34. f(x) =√x

3+x

35. f(x) = 2x5+x4−6xx

36. f(x) = xx+ 1

x

1.6. PB6: TECHNIQUES OF DIFFERENTIATION, PRODUCT AND QUOTIENT RULES (CONTINUED) AND THE CHAIN RULE15

1.6 PB6: Techniques of Differentiation, Product and Quotient Rules(continued) and the Chain Rule

Sections covered:

• Section 2.3 (continued): Product and Quotient Rules; Higher-Order Derivatives.

• Section 2.4: The Chain Rule.

Section 2.3: Product and Quotient Rules; Higher-Order Derivatives (Continued)

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

(a) y = 2xx+1 , P (1, 1).

(b) y = 2x3 − x2 + 2, P (1, 3).

2. Find the points on the curve y = x2−2x+1x−3 where the tangent line is horizontal.

3. Find the second derivative of each of the following functions.

(a) f(x) = 2x5 − 3x+ 1.

(b) g(x) = 1x2 .

4. Find the fifth derivative of f(x) = 3x4 − x3 + 7x2 − 8x+ 10.

Section 2.4: The Chain Rule


1. f(x) = (2x)5

2. f(x) = (−3x+ 4)8

3. f(x) = 3(5x− 1)7

4. f(x) = (x2 − 1)32

5. f(x) = (−x3 + 2x+ 1)15

6. f(x) = −2(5x6 − 2x)4

7. f(x) =√x2 − x

8. f(x) = 13√x2−1

9. f(x) = (−5x+ 4)(x3 + 1)6

10. f(x) = x√2− x2

11. f(x) = (x2 + 1)3(x2 + 2)6

12. f(x) = (x+1)5

x5+1


13. f(x) = x2√x3+1

14. f(x) =√

xx2+3

15. f(x) =(x4+1x2+1

)5

1.7 PB7: Implicit Differentiation and Related Rates

sections covered: Section 2.6: Implicit Differentiation and Related Rates

Section 2.6: Implicit Differentiation and Related Rates

2.6.1. Implicit Differentiation

1. Find dydx for each of the following functions.

(a) x2 + y2 = 25

(b) x3 + y3 = 1

(c) 2x2 − y2 = x

(d) x4 + 3y3 = 5y

(e)√x− y = y2 + 3

(f) xy = 5

(g) 3x2 + 2xy + y2 = 2

(h) −5x2 + xy − y3 = 1

(i) x3y2 + y4 = x

(j) x+yx−y = 1

2. Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

(a) x2y3 − xy = −y + 8 at (1, 2)

(b) xy − 3 =√y − 2y at (2, 1)

2.6.2. Related Rates

1. If V is the volume of a cube with edge length x and the cube expands as time passes, find dVdt in terms

of dxdt .

2. A large ship in the middle of the ocean is leaking oil. The oil is forming in a circle on the ocean aroundthe ship. When the radius of this circle is 50m, the radius is growing at a rate of 1 metre per hour. Atthis point, how fast is the area of the circle growing?

3. The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when thediameter is 80 mm?

4. A tiny spherical balloon is inserted into a clogged artery and is inflated at the rate of 0.002πmm3/min.How fast is the radius of the balloon growing when the radius is R = 0.005 mm?

1.8. PB8: INCREASING AND DECREASING FUNCTIONS, LOCAL EXTREMA, CONCAVITY, AND POINTS OF INFECTION17

5. A cylindrical tank with radius 5 m is being filled with water at a rate of 3m3/min. How fast is theheight of the water increasing?

6. A plane flying horizontally at an altitude of 3 mi and a speed of 500 mi/h passes directly over a radarstation. Find the rate at which the distance from the plane to the station is increasing when it is 5 miaway from the station.

1.8 PB8: Increasing and Decreasing Functions, Local Extrema,Concavity, and Points of Infection

Sections covered:

• Section 3.1: Increasing and Decreasing Functions, Local Extrema.

• Section 3.2: Concavity and Points of Infection.

Sections 3.1 & 3.2: Increasing and Decreasing Functions, Local Extrema, Concavity, and Pointsof Inflection

3.1.1. Critical Numbers

Find the critical numbers of each of the following functions

1. f(x) = −2x2 + 8x− 3

2. f(x) = 3√x

3. f(x) = x3 + x2 + 5

4. f(x) = x3 + 6x2 − 15x

5. f(x) =√1− x2

6. f(x) = 2x+1x−3

3.1.2. Absolute Maximum and Minimum Values

Find the absolute maximum and absolute minimum values of f on the given interval. Also state the locationsof these absolute extrema.

1. f(x) = 12 + 4x− x2 on [0, 5]

2. f(x) = −x2 + 2x+ 3 on [0, 2]

3. f(x) = x3 − 3x+ 5 on [0, 3]

4. f(x) = −x3 + 3x2 + 1 on [−1, 2]

5. f(x) = 2x3 − 3x2 − 12x+ 1 on [−2, 1]


6. f(x) = x2−4x2+4 on [−4, 4]

7. f(x) = (x2 − 4)3 on [−2, 3]

8. f(x) = xx2−x+1 on [0, 3].

3.1 & 3.2: Increasing/Decreasing, Local Extrema, Concavity, and Points of Inflection

For each of the following functions, find where it is increasing and where it is decreasing. Also find the localmaximum and minimum values. Also find the intervals of concavity and the inflection points.

1. f(x) = −x2 + 4x− 3

2. f(x) = x3 − 3x2 − 9x+ 4

3. f(x) = −x4 + 4x3 + 1

4. f(x) = xx2+1

1.9 PB9: Curve Sketching and Optimization Problems

Sections covered:

• Section 3.3: Curve Sketching.

• Sections 3.4 & 3.5: Optimization Problems.

Section 3.3: Curve sketching

1. Consider the following functions.

(a) f(x) = −x3 + 3x2 − 1

(b) f(x) = x4 − 4x3 + 21

In each case, find (i) the domain; (ii) the y- intercept; (iii) the intervals of increase and decrease; (iv)the local maximum and minimum values; (v) the intervals of concavity; (vi) the inflection points. (vii)Use the above information to sketch the graph of f(x).

2. Consider the function f(x) = 3x−6x+2 . (a) Give the x-intercept and the y-intercept. (b) What is the

vertical asymptote? (c) What is the horizontal asymptote? (d) Using the first derivative, show thatf(x) is an increasing function. (e) Using the above information, sketch f(x).

3. Consider the function f(x) = xx2+1 . Note that f ′(x) = 1−x2

(1+x2)2 and f ′′(x) = 2x(x2−3)(x2+1)3 . Determine

the following: (a) domain; (b) x- and y-intercepts; (c) the horizontal and vertical asymptotes; (d)the intervals of increase and decrease, local maximums and minimums; (e) intervals of concavity andinflection points. (f) Sketch the graph of f(x).

1.10. PB10: EXPONENTIAL AND LOGARITHMIC FUNCTIONS 19

Sections 3.4 & 3.5: Optimization problems

1. A compagny has determined that its total revenue (in dollars) for a product can be modelled byR(x) = −x3 + 450x2 + 52500x, where x is the number of units produced and sold. What productionlevel x will yield a maximum revenue?

2. A farmer has 2000 ft of fencing and wants to fence off a rectangular field that borders a straight river.He needs no fence along the river. What are the dimensions of the field that maximize the area?

3. A farmer wants to fence in a rectangular field beside a river. No fencing is required along the riverand the farmer’s neighbour will pay half of the cost of one of the sides perpendicular to the river. Iffencing costs 20 per linear meter and the field must have an area of 600 m2, what are the dimensionsof the field that will minimize the cost to the farmer?

4. The volume V of a cylinder of height h and radius r is V = πr2h, whereas the area of the cylinder’ssurface, including top and bottom, is A = 2πr2 + 2πrh. Of all cylinders of volume V = 1, determinethe height and radius of the cylinder that has minimal surface area.

5. Your ice cream cones are sold to unicorns for 50 cents each. At this price 1000 unicorns are willingto buy an ice cream cone. For every 5 cents decrease in price, there are 100 more unicorns willing tobuy your ice cream cones. What selling price will produce the maximum revenue and what will themaximum revenue be?

6. If 1200 cm2 of material is available to make a box with a square base and an open top, find the largestpossible volume of the box.

7. I am building a closed box with a square base and a fancy top. The material for the bottom and forthe sides costs 5 cents/cm2, while the material for the top costs 7 cents/cm2. The length (and thewidth, since it is square) of the base of the box is x, and the height of the box is h.

(a) Find a formula for the cost of this box in terms of x and h.

(b) Find the formula for the volume of the box in terms of x and h.

(c) If the volume must be 150cm3, express h in terms of x.

(d) Assume the volume of the box must be 150cm3. Find the value of x that should be used for thecheapest box.

1.10 PB10: Exponential and Logarithmic Functions

Sections covered:

• Section 4.1: Exponential Functions.

• Section 4.2: Logarithmic Functions.

• Section 4.3: Differentiation of Exponential and Logarithmic Functions.

Section 4.1: Exponential Functions

1. Solve the following equations.

(a) e2x = ex


(b) ex2

= e4

(c) ex = 1

(d) e4x2 − e36 = 0

(e) ex2

ex = e6

(f) ex2

e3x−2 = 1

(g) (ex)x2

= ex

(h) (2x− 1)ex = 0

2. Find the following limits.

(a) limx→∞

e−2x

(b) limx→∞

e4x

(c) limx→−∞

e7x

(d) limx→−∞

8e−3x

(e) limx→∞

2xex

(f) limx→−∞

9xe0.5x

(g) limt→∞

(t+ 1)e−0.75t

(h) limt→∞

5 + 6(2− 3t)e−10t

3. The concentration of a certain drug in an organ t minutes after an injection is given by C(t) =

0.05− 0.04(1− e−0.03t) grams per cubic centimeter (g/cm3).

(a) What is the initial concentration of the drug (when t = 0)?

(b) What is the concentration 10 minutes after an injection? After 1 hour?

(c) What is the average rate of change of concentration during the first hour?

(d) What happens to the concentration of the drug in the long run (as t→∞)?

Section 4.2: Logarithmic Functions

1. Find

(a) ln(e3)

(b) ln 3√e

(c) ln( 1√e)

(d) eln 5

2. Simplify the following expressions.

(a) lnx2019

(b) ln(x2y−7)

(c) ln( y8

x3 )

(d) ln(x8e−x3

)

(e) ln 3√x2 − 3x

1.10. PB10: EXPONENTIAL AND LOGARITHMIC FUNCTIONS 21

(f) ln(

(3x−1)4x2

(2x+1)5

)3. Solve the following equations.

(a) ex = 3

(b) lnx = 6

(c) ln( x20 ) = −1

(d) e4x = 3

(e) 3 lnx = 1

(f) ln(2x− 3) = 1

(g) x ln(x+ 2) = 0


(a) ln(2x+ 1) = ln(x+ 5)

(b) ln(x2 + x) = ln(2x+ 6)

(c) ln(2x+ 1)− ln(x− 1) = 1

(d) 2e−3x + 5 = 19

5. An investment firm estimates that the value of its portfolio after t years is A million dollars, whereA(t) = 300 ln(t+ 3).

(a) What is the value of the account when t = 0?

(b) How long does it take for the account to double its initial value?

Section 4.3: Differentiation of Exponential and Logarithmic Functions

1. Find the derivative of the following functions.

(a) f(x) = 3ex

(b) f(x) = 1 +√2x− ex

(c) f(x) = x3ex

(d) f(x) = x2

ex+3


(a) f(x) = e3x

(b) f(x) = −e−x

(c) f(x) = ex2+5x+1

(d) f(x) = e2x + e−2x

(e) f(x) = xex3

(f) f(x) = e−2x

x2+5

3. For the function f(x) = xe3x, find the intervals of increase and decrease. Also find the local maximumand minimum values. Also find the intervals of concavity and the inflection points.



(a) f(x) = 3 lnx

(b) f(x) = x2 lnx

(c) f(x) = ln xx3


(a) f(x) = ln(7x)

(b) f(x) = ln(x2 + 3)

(c) f(x) = 5 ln(1 + ex)

(d) f(x) = x ln(x2 + 1)

(e) f(x) = ln(2x+1)3x+1

(f) f(x) = ln(2x)e−3x

6. Find the absolute maximum and minimum of f(x) = ln(x+1)x+1 on the interval [0, 2].

7. Use logarithmic differentiation to find the derivative of f(x) = (2x+ 1)2(x− 5x2)12 .

1.11 PB11: Indefinite Integrals

Sections covered:

• Section 5.1: Indefinite Integration.

• Section 5.2: Integration by Substitution.

Section 5.1: Indefinite Integration

Evaluate the following indefinite integrals.

1.∫5dx

2.∫x8dx

3.∫

1x3 dx

4.∫

3√xdx

5.∫

16√xdx

6.∫

x3√xdx

7.∫

4√x5dx

8.∫ 3√

x2

x4 dx

9.∫e3xdx

10.∫

1e4x dx

11.∫4xdx

1.11. PB11: INDEFINITE INTEGRALS 23

12.∫ √

3xdx

13.∫(−3x2 + 5x− 1)dx

14.∫(x2 +

√x− 1)dx

15.∫

5xdx

16.∫(e−2x + 3− 2

x )dx

17.∫(3− 1

ex )dx

18.∫ex(1− e−2x)dx

19.∫(ln(ex

3

)− eln x2

)dx

20.∫

4x−3√xdx

21.∫

2−√2t+

3√t2√

tdt

22.∫2x(3− x−3)dx

23.∫(t+ 4)(2t+ 1)dt

Section 5.2: Integration by Substitution.


1.∫(1− 2x)9dx

2.∫2x(x2 + 3)15dx

3.∫x2(x3 − 7)6dx

4.∫x√1− x2dx

5.∫e1−xdx

6.∫x3ex

4

dx

7.∫

xx2+3dx

8.∫

x−1e−x2+2x

dx

9.∫

ex

1+ex dx

10.∫e2x−e−2x

e2x+e−2x dx

11.∫ ln(3x)

x dx

12.∫ −3x4

(x5+1)7 dx

13.∫ −4x+4

3√

(x2−2x+1)2dx

14.∫x√x+ 7dx

15.∫

x√1+2x

dx


1.12 PB12: The Definite Integral and the Fundamental Theoremof Calculus

Sections covered:

• Section 5.3: The Definite Integral and the Fundamental Theorem of Calculus.

• Section 5.4: Applying Definite Integration: Area Between two Curves.

Section 5.3: The Definite Integral and the Fundamental Theorem of Calculus

5.3.1. The Fundamental Theorem of Calculus

Evaluate the following definite integrals

1.∫ 1

−3 5dx

2.∫ 4

13xdx

3.∫ 9

41√xdx

4.∫ 2

1(−x2 + 3x− 1)dx

5.∫ 2

1

(1x2 − 4

x3

)dx

6.∫ 3

−2(x2 − 3)dx

7.∫ 4

14+6x√xdx

8.∫ 8

12+t3√t2dt

9.∫ 2

0(2x− 3)(4x2 + 1)dx

10.∫ 4

0(4− x)

√xdx

11.∫ 1

0e−2xdx

12.∫ ln 2

05e3xdx

13.∫ 2

1(x2 − 3

x )dx

14.∫ e1x2−xx2 dx

5.3.2. Substitution in a Definite Integral

1.12. PB12: THE DEFINITE INTEGRAL AND THE FUNDAMENTAL THEOREM OF CALCULUS 25

Evaluate the following definite integrals.

1.∫ 2

02x(x2 − 2)3dx

2.∫ 1

0(4t− 1)50dt

3.∫ 1

03√26x+ 1dx

4.∫ 1

0xex

2

dx

5.∫ 1

213

1x2 e

1x dx

6.∫ 2

1x2

(x3+1)2 dx

7.∫ 8

01√

1+2xdx

8.∫ 2

0t2√1 + t3dt

9.∫ 2

1x+1√

x2+2x−1dx

10.∫ −1−2

xx2+1dx

11.∫ ln 4

0ex

5−ex dx

12.∫ 4

0x√

1+2xdx

5.3.3. Area Under a Curve

Find the area of the region that lies under the given curve y = f(x) over the indicated interval a ≤ x ≤ b.

1. y = x3, 1 ≤ x ≤ 2

2. y =√8x+ 9, 0 ≤ x ≤ 2

3. y = e4x, 0 ≤ x ≤ ln 3

4. y = x2ex3

, 0 ≤ x ≤ 1

5. y = 45−2x , −2 ≤ x ≤ 1

Section 5.4: Applying Definite Integration: Area Between two Curves

1. Find the area bounded above by y = 2x+ 5 and below by y = x3 on [0, 2].

2. Find the area of the region enclosed by the curves y = x2 and y = x+ 2.

3. Find the area of the region enclosed by the curves y = x2 − 8 and y = −x2 + 10. Include a sketch ofthe relevant region as part of your solution.

Chapter 2

Solutions to Practice Problems

2.1 Solution to PB1

Review: Factoring

Factor each expression.

1. x2 − 5x

Solution. One can see that x is a common factor. Taking that out, we get x2 − 5x = x(x− 5).

2. −6x2 − 9x+ 4x+ 6

Solution. One can factor this by grouping: −6x2 − 9x + 4x + 6 = −6x2 − 9x︸︷︷︸+4x+ 6︸︷︷︸ = −3x(2x +

3) + 2(2x+ 3) = (2x+ 3)(−3x+ 2).

3. x2 − 25

Solution. First we recall the difference of squares which states that for any numbers a and b, one hasa2 − b2 = (a− b)(a+ b). Applying this, we get x2 − 25 = x2 − 52 = (x− 5)(x+ 5).

4. −9x2 + 49

Solution. We have −9x2 + 49 = 49− 9x2 = 72 − (3x)2 = (7− 3x)(7 + 3x).Warning! A common mistake is to write −9x2+49 = (−3x− 7)(−3x+7). This is wrong because theexpansion of the right hand side gives 9x2 − 49, and this latter is not equal to the original expression.

5. x2 + x− 2

Solution. There are several common methods to factor polynomials of the form ax2 + bx + c, witha 6= 0 including the quadratic formula, the ac-method, the trial-and-error method, and so on. Let usfocus on the ac-method whose procedure is as follows.

step 1. Find two integers whose product is ac and whose sum is b. Here a = 1, b = 1, and c = −2. Theproduct ac is equal to −2. One can see that the numbers we are looking for are −1 and 2 since−1× 2 = −2 = ac and −1 + 2 = 1 = b.

step 2. Rewrite the middle term bx as the sum of the terms whose coefficients are the integers found instep 1. Here the middle term is x, and this can be rewritten as x = (−1)x+ 2x = −x+ 2x.

step 3 Factor by grouping. We have x2 +x− 2 = x2 − x︸︷︷︸+2x− 2︸︷︷︸ = x(x− 1)+2(x− 1) = (x− 1)(x+2).

27

28 CHAPTER 2. SOLUTIONS TO PRACTICE PROBLEMS

6. 2x2 − 7x+ 3

Solution. First, one has a = 2, b = −7, and c = 3. We are looking for two integers whose productis ac = 6 and whose sum is b = −7. After trying a number of combinations, we get that the relevantnumbers are −1 and −6. So 2x2−7x+3 = 2x2 − x︸︷︷︸− 6x+ 3︸︷︷︸ = x(2x−1)−3(2x−1) = (2x−1)(x−3).

Section 1.1: Functions

1.1.1. Evaluating a function

Compute the indicated values of the given function.

1. f(x) = x3 + 2x− 7, f(0), f(−1), f(2)Solution. We have f(0) = 03+2(0)−7 = 0+0−7 = −7, f(−1) = (−1)3+2(−1)−7 = −1−2−7 = −10,and f(2) = 8 + 4− 7 = 5.

2. f(x) = xx2+1 , f(2), f(0), f(−1)

Solution. We have f(2) = 222+1 = 2

4+1 = 25 , f(0) =

002+1 = 0

1 = 0, and f(−1) = −1(−1)2+1 = −1

1+1 = −12 .

3. f(t) =√t2 + 2t+ 3, f(2), f(0), f(−4)

Solution. We have f(2) =√4 + 4 + 3 =

√11, f(0) =

√0 + 0 + 3 =

√3, and f(−4) =

√16− 8 + 3 =√

11.

4. f(x) ={−2x+ 5 if x ≤ 2

x2 + 1 if x > 2,f(0), f(2), f(3)

Solution. This is the so-called piecewise function. Here there are two pieces: −2x + 5 and x2 + 1.The first piece corresponds to all x less than or equal to 2, and the second to all x > 2. To findf(0), we need to use the first piece because 0 ≤ 2. So f(0) = −2(0) + 5 = 0 + 5 = 5. Likewisef(2) = −2(2) + 5 = −4 + 5 = 1. For f(3), we need to use the second piece since 3 > 2. Sof(3) = (3)2 + 1 = 9 + 1 = 10.

1.1.2. Domain of a function

Find the domain of each of the following functions.

1. f(x) = 2x

Solution. Recall that the domain of f(x) is the set of all real numbers x for which f(x) is defined.Because division by 0 is not possible, the domain of f(x) = 2

x is the set of all real numbers x except0. One can rewrite this as D = R\{0}, where “R” represents as usual the set of all real numbers, and“\” means except. Another way of writing this domain is: D = (−∞, 0) ∪ (0,∞), where “∪” is themathematical symbol for the union.

2. f(x) = x3

x2−x−2Solution. Because division by 0 is not possible, the domain of f is the set of all numbers x such thatx2 − x − 2 6= 0. Factoring x2 − x − 2, we get (x + 1)(x − 2), which is equal to zero when x = −1 orx = 2. So D = R\{−1, 2}.

3. f(x) = x5 − x2 + 7

Solution. Because it is possible to evaluate f(x) = x5 − x2 + 7 at any number, the domain of f is R,that is, D = R. Actually, the domain of every polynomial is R.

2.1. SOLUTION TO PB1 29

4. f(x) =√x− 3

Solution. Because the square root is defined for any non negative number, the domain of f is the setof all x such that x− 3 ≥ 0; that is, x ≥ 3. Another way of rewriting this is: D = [3,∞).

5. f(x) =√−2x+ 8

Solution. The function f(x) is defined if and only −2x + 8 ≥ 0. To solve an inequality of the formg(x) ≥ 0 (or g(x) > 0 or g(x) ≤ 0 or g(x) < 0), one can proceed as follows.

step 1. Solve the corresponding equation, that is, the equation g(x) = 0. Here g(x) = −2x + 8, and theequation −2x + 8 = 0 is equivalent to −2x = −8 or x = −8

−2 = 4. So the equation g(x) = 0 hasone solution: x = 4.

step 2. Find the sign of g(x), that is, the intervals where g(x) is positive and the intervals where g(x) isnegative. (It is possible that the function is positive everywhere or negative everywhere.) Howto get the sign? First put on the x-axis the solution(s) found in the previous step. Here wehave to put 4 on the x-axis. We now have two intervals: (−∞, 4) and (4,∞). To get the signin each of these, pick any number, say a, and evaluate the function g(x) at a. If g(a) > 0, thefunction is positive, if g(a) < 0, then it is negative. For example, let’s pick 0 in the interval(−∞, 4). Evaluating g at 0, we get g(0) = −2(0) + 8 = 8. Because g(0) > 0, the function g(x)is positive in the interval (−∞, 4). To get the sign in the other interval, let’s pick for example5. Then g(5) = −2(5) + 8 = −10 + 8 = −2. So g(x) is negative in the interval (4,∞). All thisis summarized in Figure 2.1. From that figure, we deduce that the domain of g is the interval

Figure 2.1:

D = (−∞, 4]. (Notice that 4 is included because f(4) =√−2(4) + 8 =

√0 = 0 is defined.)

6. f(x) =√x2 − 4

Solution. The function f is defined if and only if x2 − 4 ≥ 0. As before, to solve this inequality weneed to find the sign of g(x) = x2 − 4, which is given in Figure 2.2. From that figure, one can deducethat the domain of f is (−∞,−2] ∪ [2,∞).

Figure 2.2:

7. f(x) = x√25−x2

Solution. Here the function is defined if and only if 25 − x2 > 0. (Note that the denominator hasto be different from 0. Because of that the inequality here is “>”, and not “≥” as before.) The signof g(x) = 25 − x2 is given in Figure 2.3. From this, it follows that the domain of f is the interval


D = (−5, 5). (One can notice that −5 and 5 are both excluded because these make the denominatorequal to zero.)

Figure 2.3:

1.1.3. Composition of functions

Find the composite function f(g(x)).

1. f(x) = x− 2 and g(x) = x2

Solution. We have f(g(x)) = f(x2) = x2 − 2.

2. f(x) = x2 − x+ 3 and g(x) = 2x+ 1

Solution. We have f(g(x)) = f(2x+ 1) = (2x+ 1)2 − (2x+ 1) + 3 = (2x+ 1)(2x+ 1)− 2x− 1 + 3 =

4x2 + 2x+ 2x+ 1− 2x+ 2 = 4x2 + 2x+ 3.

3. f(x) =√x2 + 1 and g(x) =

√x

Solution. We have f(g(x)) = f(√x) =

√(√x)2 + 1 =

√x+ 1.

4. f(x) = x2

x+2 and g(x) = 2x+ 7

Solution. We have f(g(x)) = f(2x+ 7) = (2x+7)2

(2x+7)+2 = (2x+7)2

2x+9 .

1.1.4. Finding the difference quotient

Find and simplify the difference quotient of the given function.

1. f(x) = x+ 1

Solution. Recall that the difference quotient of a function f is given by the formula f(x+h)−f(x)h .

Here f(x) = x + 1. So f(x + h) = (x + h) + 1 = x + h + 1. Therefore the difference quotient of f is:f(x+h)−f(x)

h = x+h+1−(x+1)h = x+h+1−x−1

h = hh = 1 if h 6= 0. (Note that it is important to write the

condition h 6= 0 when simplifying the numerator and the denominator by h.)

2. f(x) = 2x− 5

Solution. We have f(x+h)−f(x)h = 2(x+h)−5−(2x−5)

h = 2x+2h−5−2x+5h = 2h

h = 2 if h 6= 0.

3. f(x) = 3x2 + 1

Solution. We have

f(x+ h)− f(x)h

=3(x+ h)2 + 1− (3x2 + 1)

h=

3(x2 + 2xh+ h2) + 1− 3x2 − 1

h=

3x2 + 6xh+ 3h2 + 1− 3x2 − 1

h=

6xh+ 3h2

h=h(6x+ 3h)

h= 6x+ 3h

if h 6= 0.


4. f(x) = 1x+2

Solution. We have

f(x+ h)− f(x)h

=1

x+h+2 −1

x+2

h=

x+2−(x+h+2)(x+h+2)(x+2)

h=

x+ 2− x− h− 2

h(x+ h+ 2)(x+ 2)=

−hh(x+ h+ 2)(x+ 2)

=−1

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

if h 6= 0.

2.2 Solution to PB2

Section 1.2: The graph of a function

1.2.1. Intercepts

Find the intercepts of the graph of each of the following functions.

1. f(x) = 2x− 6

Solution. Recall: Given a function f , an intercept of f is a point of intersection between the graphof f and the axis. A point where the graph crosses the x-axis is called x-intercept. A point wherethe graph crosses the y-axis is called y-intercept. The term “intercepts” refers to both x-intercepts andy-intercept.

• To find the x-intercepts of f , set f(x) = 0 and solve the equation for x.Note: Finding x-intercepts may be difficult.

• The y-intercept of f is f(0).

For the function f(x) = 2x − 6, the y-intercept is f(0) = 2(0) − 6 = −6. Regarding the x-intercepts,we need to solve the equation 2x−6 = 0. This is equivalent to 2x = 6 or x = 6

2 = 3. So the x-interceptis 3 and the y-intercept is −6.

2. f(x) = x2 + x− 2

Solution.

• x-intercepts: Need to solve the equation x2 + x− 2 = 0. This is a quadratic equation, and one ofthe best ways to solve it is to use the quadratic formula stated as follows. A quadratic equationax2 + bx+ c = 0 has solutions of the form

x =−b±

√b2 − 4ac

2a.

If b2 − 4ac is negative, there are no real solutions.For the equation x2 + x − 2 = 0, a = 1, b = 1, and c = −2. Applying the quadratic formula, weget

x =−1±

√12 − 4(1)(−2)2(1)

=−1±

√1 + 8

2=−1±

√9

2=−1± 3

2.

So x = −1+32 = 1 or x = −1−3

2 = −2, and these are the x-intercepts.

• y-intercept: The y-intercept is f(0) = 0 + 0− 2 = −2.


3. f(x) = 4x2 − 4x+ 1

Solution.

• x-intercepts: Need to solve the equation 4x2 − 4x + 1 = 0. We have a = 4, b = −4, and c = 1.Applying the quadratic formula, we get

x =−(−4)±

√(−4)2 − 4(4)(1)

2(4)=

4±√16− 16

8=

4±√0

8=

4± 0

8.

So x = 4+08 = 1

2 or x = 4−08 = 1

2 . Thus there is only one x-intercept, namely 12 .

• The y-intercept is f(0) = 4(0)− 4(0) + 1 = 0− 0 + 1 = 1.

4. f(x) = x2 + x+ 3

Solution.

• x-intercepts: Need to solve the equation x2 + x + 3 = 0. We have a = 1, b = 1, and c = 3.Applying the quadratic formula, we have

x =−1±

√12 − 4(1)(3)

2(1)=−1±

√1− 12

2=−1±

√−11

2.

Because√−11 is undefined, it follows that the equation x2 + x + 3 = 0 has no solutions at all.

Therefore there is no x-intercepts. Graphically, this means that the graph of f does not cross thex-axis.

• The y-intercept is f(0) = 02 + 0 + 3 = 3.

5. f(x) = 3x−15x−8

Solution.

• To get the x-intercepts, we set 3x−15x−8 = 0 and find that 3x− 1 = 0 or 3x = 1 or x = 1

3 .

• The y-intercept is f(0) = 3(0)−15(0)−8 = −1

−8 = 18 .

6. f(x) = −x5 + 4x3

Solution.

• To find the x-intercepts, we set −x5 + 4x = 0 and solve for x. Factoring the left hand side, weget −x3(x2 − 4) = 0 or −x3(x− 2)(x+ 2) = 0. This implies that x = 0 or x = 2 or x = −2.• The y-intercept is f(0) = −05 + 4(0) = 0.

1.2.2. Parabola

Sketch the graph of f(x).

1. f(x) = −x2 + x+ 2

Solution. To sketch the parabola f(x) = ax2 + bx+ c (a 6= 0), we can proceed as follows.

step 1. Find the sign of a. If a > 0, then the parabola opens up. If a < 0, it opens down. For the functionf(x) = −x2 + x+ 2, a = −1 is negative. So the parabola opens down.

step 2. Find the location of the vertex, which is given by the formula (−b2a , f(−b2a )). Here a = −1, b = 1,

c = 2, −b2a = −12(−1) =

12 , and

f(1

2) = −1

4+

1

2+ 2 = −1

4+

2

4+

8

4=−1 + 2 + 8

4=

9

4.

So the vertex is the point ( 12 ,94 ).


step 3. Find any two additional points of the parabola (often these are intercepts). For f(x) = −x2+x+2,the y-intercept is f(0) = 2. To get the x-intercepts, we set −x2 + x+2 = 0 and find that x = −1or x = 2.

step 4. Sketch the graph. This is given in Figure 2.4.

Figure 2.4:

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

Solution. This parabola opens up because a = 1 > 0. For the vertex, we have −b2a = −12(1) = −1

2 ,and f(−12 ) = 1

4 −12 + 1 = 1

4 −24 + 4

4 = 34 . So the vertex is (−12 ,

34 ). The y-intercept is f(0) = 1, and

to find the x-intercepts we set x2 + x + 1 = 0 and find that this equation has no solutions bceauseb2 − 4ac = 12 − 4(1)(1) = 1− 4 = −3 < 0. This implies that there is no x-intercepts. We then need tofind another point of the parabola. To this end, pick randomly a number, for example 1, and evaluatethe function at that number: f(1) = 1 + 1 + 1 = 3. So (1, 3) belongs to the parabola. The graph off(x) = x2 + x+ 1 is given in Figure 2.5.

Figure 2.5:


1.2.3. Finding points of intersection

Find points of intersection of f(x) and g(x), and sketch the graphs .

1. f(x) = 3x− 3 and g(x) = −x+ 4

Solution.

• To find the points of intersection between f(x) and g(x), we need to solve the equation f(x) = g(x)

for x, and then find the corresponding y coordinates. Setting f(x) = g(x), we get that 3x− 3 =

−x + 4, which is equivalent to 3x + x = 4 + 3 or 4x = 7 or x = 74 . To get the corresponding y

coordinate we substitute into f(x) or g(x) (choose any of those functions): f( 74 ) = 3( 74 ) − 3 =214 − 3 = 21

4 −124 = 9

4 . So f and g have one point of intersection, namely ( 74 ,94 ).

• The functions f(x) and g(x) are both of the form mx + b, and we know that such expression isthat of a straight line or just a line. To sketch the graph of a line, we need to find two points. Forf(x) = 3x − 3, if x = 0, f(0) = −3. If x = 1, f(1) = 3 − 3 = 0. So the points (0,−3) and (1, 0)

belong to the line f(x) = 3x − 3. The graph of f is then obtained by joigning these two points.Likewise, the graph of g is obtained for example by joingning the points (0, 4) and (2, 2). Thosegraphs are given in Figure 2.6.

Figure 2.6:

2. f(x) = −x2 − 2x and g(x) = −x− 2

Solution.

• To get the points of intersection, we solve −x2 − 2x = −x − 2 for x. This latter equationis equivalent to the quadratic equation −x2 − x + 2 = 0 whose solutions are 1 and −2. Thecorresponding y coordinates are g(1) = −1− 2 = −3 and g(−2) = −(−2)− 2 = 2− 2 = 0. So thepoints of intersection are (1,−3) and (−2, 0).

• The graphs are given in Figure 2.7.


Figure 2.7:

3. f(x) = 2x2 − x− 1 and g(x) = 2x+ 1

Solution.

• The equation 2x2−x−1 = 2x+1 is equivalent to 2x2−3x−2 = 0, and the solutions to this latterequation are − 1

2 and 2. The corresponding y coordinates are g(− 12 ) = 2(− 1

2 ) + 1 = −1 + 1 = 0,and g(2) = 2(2) + 1 = 4 + 1 = 5. So the points of intersection are (− 1

2 , 0) and (2, 5).

• The graphs are given in Figure 2.8.

Figure 2.8:


Section 1.3: Lines and linear functions

1.3.1. The slope of a line

Find the slope of the line passing through the points.

1. (1, 2) and (3, 4)

Solution. Recall: The slope of the line through (x1, y1) and (x2, y2) is given by the formula

m =y2 − y1x2 − x1

.

Applying this formula, we get that the slope of the line (1, 2) and (3, 4) is m = 4−23−1 = 2

2 = 1.

2. (2,−5) and (0, 1)

Solution. The slope is m = 1−(−5)0−2 = 1+5

−2 = 6−2 = −3.

3. (− 12 ,

13 ) and ( 34 ,−1)

Solution. The slope is

m =−1− 1

334 − (− 1

2 )=−1− 1

334 + 1

2

=−33 −

13

34 + 2

4

=−4354

=−43× 4

5=−1615

.

1.3.2. Finding the equation of a line

Find the equation of the line passing through the points.

1. (−1, 2) and (0, 5)

Solution. Recall: The point-slope form of the equation of the line through (x1, y1) with slope m is:y − y1 = m(x− x1).

• First we need to find the slope, which is m = 5−20−(−1) =

5−20+1 = 3

1 = 3.

• Then the equation of the line through (−1, 2) with slope m = 3 is: y − 2 = 3(x − (−1)) ory − 2 = 3(x+ 1). (You could go further if you want.)Note: If one uses the point (0, 5) in the place of (−1, 2), we will end up with the same equation.So the point you choose to use doesn’t matter.

2. (7,−1) and (9, 12)

Solution. First, the slope is m = 12−(−1)9−7 = 12+1

9−7 = 132 . Then the required equation is y − 12 =

132 (x− 9).

Section 1.5: Limits

1.5.1. Examining the concept of limit


Use a table to estimate the following limits.

1. limx→2

x2−4x−2

Solution. Let f(x) = x2−4x−2 , and consider the following table. (The idea is to choose numbers closer

and closer to 2, and see what happens to f(x).)

x 1.9 1.99 1.999 2 2.001 2.01 2.1

f(x) 3.9 3.99 3.999 Undefined 4.001 4.01 4.1

The numbers on the bottom line of this table suggest that f(x) approaches 4 as x gets closer and closerto 2. So lim

x→2

x2−4x−2 = 4.

2. limx→1

√x−1x−1

Solution. Let f(x) =√x−1x−1 , and consider the following table.

x 0.9 0.99 0.999 1 1.001 1.01 1.1

f(x) 0.5131 0.5012 0.5001 Undefined 0.4998 0.4987 0.4880

This table suggests that limx→1

√x−1x−1 = 0.5 = 1

2 .

2.3 Solution to PB3

Section 1.5: Limits (continued)

1.5.2. Computing limits using a variety of techniques

Evaluate the following limits.

1. limx→7

3x

Solution. limx→7

3x = 3(7) = 21.

2. limx→4

x3 + x

Solution. limx→4

x3 + x = 43 + 4 = 64 + 4 = 68.

3. limx→2

x5 − 4x3 − 7x+ 4

Solution. limx→2

x5 − 4x3 − 7x+ 4 = 25 − 4(23)− 7(2) + 4 = 32− 4(8)− 14 + 4 = 32− 32− 10 = −10.

4. limx→−2

− 2x3 + 5x− 9

Solution. limx→−2

− 2x3 + 5x− 9 = −2(−2)3 + 5(−2)− 9 = −2(−8)− 10− 9 = 16− 19 = −3.

5. limx→1

9x2−4x+1

Solution. limx→1

9x2−4x+1 = 9(1)2−4

1+1 = 52 .


6. limx→2

x4−3x2+1x+8

Solution. limx→2

x4−3x2+1x+8 = 16−3(4)+1

2+8 = 16−12+110 = 5

10 = 12 .

7. limx→1

√−9x3 + 4x2 + 6

Solution. limx→1

√−9x3 + 4x2 + 6 =

√−9 + 4 + 6 =

√1 = 1.

8. limx→−3

3√x2−1(−x3+8x−1)4


3√x2−1(−x3+8x−1)4 =

3√9−1

(−(−27)+8(−3)−1)4 =3√8

(27−24−1)4 = 224 = 2

16 = 18 .

9. limx→1

9x2−4x+1 .

Solution. We have limx→1

9x2−4x+1 = 9(1)2−4

1+1 = 9−41+1 = 5

2 .

10. limx→3

x2−2x−3x2−9 .

Solution. After substituting, we get 00 , which is a problem. One way of getting rid of that is to

factor the numerator and the denominator, and then simplify. In class I recalled the general methodof factoring. Here I am going to give you another method (in fact it is a TRICK), which is easier.To factor a polynomial of degree two p(x) = ax2 + bx + c knowing a root r (recall that a root is anumber that makes p(x) equal to zero), we can proceed as follows. First of all, since r is a root, wehave p(x) = (x − r)(αx + β), where α and β are to be found. The unknown α is the solution to theequation α×1 = a; so α = a. The unknown β is the solution to the equation β× (−r) = c; so β = − c

r .Thus p(x) = (x− r)(ax− c

r ).

• Factoring the numerator p(x) = x2 − 2x − 3. Here a = 1, b = −2, and c = −3. Since p(3) = 0,we have r = 3. So p(x) = (x − 3)(αx + β) with α = a = 1 and β = − c

r = −−33 = 1. Thusp(x) = (x− 3)(x+ 1).

• Factoring the denominator x2 − 9. This is the difference of two squares as x2 − 9 = (x)2 − (3)2.By the remarkable identity A2 −B2 = (A−B)(A+B), we have x2 − 9 = (x− 3)(x+ 3).

• Simplifying. Using the above factorizations, we get x2−2x−3x2−9 = (x−3)(x+1)

(x−3)(x+3) =x+1x+3 for x 6= 3.

Now the limit is:

limx→3

x2 − 2x− 3

x2 − 9= limx→3

x+ 1

x+ 3=

3 + 1

3 + 3=

4

6=

2

3.

11. limx→16

16−x√x−4

Solution. After substituting, we get 00 , which is a problem. Since there is a square root, one way to

get rid of the problem is to rationalize the numerator (because the square root appears there). Recallsome rationalization formulas:

√A+B

C=

(√A+B)(

√A−B)

C(√A−B)

=A−B2

C(√A−B)

,

√A−BC

=(√A−B)(

√A+B)

C(√A+B)

=A−B2

C(√A+B)

,

A√B + C

=A(√B − C)

(√B + C)(

√B − C)

=A(√B − C)

B − C2,


A√B − C

=A(√B + C)

(√B − C)(

√B + C)

=A(√B + C)

B − C2.

Also recall that (A−B)(A+B) = A2 −B2. We come back to the limit. We have

limx→16

16− x√x− 4

= limx→16

(16− x)(√x+ 4)

(√x− 4)(

√x+ 4)

=

limx→16

(16− x)(√x+ 4)

(√x)2 − 42

= limx→16

(16− x)(√x+ 4)

x− 16=

limx→16

−(x− 16)(√x+ 4)

(x− 16)= limx→16

−(√x+ 4)

1= −(

√16 + 4) = −(4 + 4) = −8.

12. limx→0

1−√1−x2

x .

Solution. We have

limx→0

1−√1− x2x

= limx→0

(1−√1− x2

) (1 +√1− x2

)x(1 +√1− x2

) = limx→0

12 −(√

1− x2)2

x(1 +√1− x2

) =

limx→0

1− (1− x2)x(1 +√1− x2

) = limx→0

x2

x(1 +√1− x2

) = limx→0

x

1 +√1− x2

=0

1 +√1− 02

=0

2= 0.

13. limx→−5

x2+3x−10x2+6x+5 .


x2+3x−10x2+6x+5 = lim

x→−5(x+5)(x−2)(x+5)(x+1) = lim

x→−5x−2x+1 = −5−2

−5+1 = −7−4 = 7

4 .

14. limx→4

√x−2

−3x+7

Solution. We have limx→4

√x−2

−3x+7 =√4−2

−3(4)+7 = 2−2−12+7 = 0

−5 = 0.

15. limx→2

√2x+5−

√x+7

x−2 .

Solution. Rationalizing, we get

limx→2

√2x+ 5−

√x+ 7

x− 2= limx→2

(√2x+ 5−

√x+ 7)(

√2x+ 5 +

√x+ 7)

(x− 2)(√2x+ 5 +

√x+ 7)

=

limx→2

(√2x+ 5)2 − (

√x+ 7)2

(x− 2)(√2x+ 5 +

√x+ 7)

= limx→2

(2x+ 5)− (x+ 7)

(x− 2)(√2x+ 5 +

√x+ 7)

=

limx→2

x− 2

(x− 2)(√2x+ 5 +

√x+ 7)

= limx→2

1√2x+ 5 +

√x+ 7

=

1√2(2) + 5 +

√2 + 7

=1√

9 +√9=

1

3 + 3=

1

6.

16. limx→−2

−5x2−10x6+3x .

Solution. We have limx→−2

−5x2−10x6+3x = lim

x→−2(x+2)(−5x)

3(x+2) = limx→−2

−5x3 = −5(−2)

3 = 103 .

17. limx→0

3√−x2 + 4x+ 2.

Solution. limx→0

3√−x2 + 4x+ 2 = lim

x→0

3√−(0)2 + 4(0) + 2 = 3

√0 + 0 + 2 = 3

√2.


18. limx→7

1x−

17

x−7 .

Solution. We have

limx→7

1x −

17

x− 7= limx→7

7−x7x

x− 7= limx→7

7− x7x(x− 7)

= limx→7

−(x− 7)

7x(x− 7)= limx→7

−17x

=−17(7)

= − 1

49.

19. limx→3

x−3√x2−5x+10−2 .

Solution. After substituting, we get 00 . To get rid of this problem, we rationalize the denominator.

limx→3

x− 3√x2 − 5x+ 10− 2

= limx→3

(x− 3)(√x2 − 5x+ 10 + 2)

(√x2 − 5x+ 10− 2)(

√x2 − 5x+ 10 + 2)

=

limx→3

(x− 3)(√x2 − 5x+ 10 + 2)

(x2 − 5x+ 10)− 4= limx→3

(x− 3)(√x2 − 5x+ 10 + 2)

x2 − 5x+ 6=

limx→3

(x− 3)(√x2 − 5x+ 10 + 2)

(x− 3)(x− 2)= limx→3

√x2 − 5x+ 10 + 2

x− 2=

√(3)2 − 5(3) + 10 + 2

3− 1=

√9− 15 + 10 + 2

1=√4 + 2 = 2 + 2 = 4.

1.5.3. Limits involving infinity

Find the following limits.

1. limx→∞

− 2x10

Solution. First we recall the following basic limits.

• If r > 0 is a positive real number, then limx→∞

xr =∞.

• Let n be a positive integer. Then

limx→−∞

xn =

{∞ if n is even−∞ if n is odd.

Also recall the following properties. If k > 0, then k ×∞ = ∞, and k × (−∞) = −∞. If k < 0, thenk ×∞ = −∞, and k × (−∞) =∞.

We come back to the question. We have limx→∞

− 2x10 = −2 limx→∞

x10 = −2(∞) = −∞.

2. limx→−∞

− 5x10

Solution. We have limx→−∞

− 5x10 = −5 limx→−∞

x10 = −5(∞) = −∞.

3. limx→−∞

5x9


5x9 = 5 limx→−∞

x9 = 5(−∞) = −∞.

4. limx→∞

3x2

Solution. We have limx→∞

3x2 =∞.


5. limx→∞

3√x


3√x = lim

x→∞x

13 =∞.

6. limx→∞

− 4x3 + x− 1

Solution. Limits of polynomials will follow the behavior of the term with respect to the highest power.So lim

x→∞− 4x3 + x− 1 = lim

x→∞(−4x3) = −∞.

7. limx→−∞

1− 8x3 + 2x2


1− 8x3 + 2x2 = limx→−∞

(−8x3) = −8(−∞) =∞.

8. limx→∞

2x− x3 + 7x5 − 1


2x− x3 + 7x5 − 1 = limx→∞

(7x5) =∞.

9. limx→−∞

13x4

Solution. First recall the following basic limits at infinity:

• If r is a positive real number, then limx→∞

1xr = 0.

• If n is a positive integer, then limx→−∞

1xn = 0.

Now we have

limx→−∞

1

3x4=

1

3

(lim

x→−∞

1

x4

)=

1

3(0) = 0.

10. limx→∞

1√x


1√x= limx→∞

1

x12= 0. (Here r = 1

2 .)

11. limx→∞

−x2+13x2+x+1

Solution. One way to find the limit of a rational function 1 at infinity is to proceed as follows:

step 1 Divide every term by the largest power of x in the denominator. Here f(x) = −x2+13x2+x+1 , and the

largest power of x in the denominator is x2. The terms of the function are: −x2, 1, 3x2, x, and1. Dividing each term by x2, we get

limx→∞

−x2 + 1

3x2 + x+ 1= limx→∞

−x2

x2 + 1x2

3x2

x2 + xx2 + 1

x2

.

step 2 Simplify each term whenever it is possible. For example, the term −x2

x2 becomes −1, the term 1x2

remains the same, etc. So we get

limx→∞

−x2 + 1

3x2 + x+ 1= limx→∞

−1 + 1x2

3 + 1x + 1

x2

.

step 3 Use the basic limits above to get the final answer.

limx→∞

−x2 + 1

3x2 + x+ 1=−1 + 0

3 + 0 + 0=−13.

1A rational function is a function of the form P (x)Q(x)

, where P (x) and Q(x) are both polynomials.


12. limx→−∞

7x2

x2−4

Solution. We have

limx→−∞

7x2

x2 − 4= limx→−∞

7x2

x2

x2

x2 − 4x2

= limx→−∞

7

1− 4x2

=7

1− 0= 7.

13. limx→∞

3x2−2x−5x3+x2−x

Solution. We have

limx→∞

3x2 − 2x

−5x3 + x2 − x= limx→∞

3x2

x3 − 2xx3

− 5x3

x3 + x2

x3 − xx3

=

limx→∞

3x −

2x2

−5 + 1x −

1x2

=0− 0

−5 + 0 + 0=

0

−5= 0.

14. limx→∞

3x2+1(x−1)2

Solution. First, we need to distribute the denominator. Using the formula (a− b)2 = a2 − 2ab+ b2,we get (x− 1)2 = x2 − 2x+ 1. The limit becomes

limx→∞

3x2 + 1

(x− 1)2= limx→∞

3x2 + 1

x2 − 2x+ 1= limx→∞

3x2

x2 + 1x2

x2

x2 − 2xx2 + 1

x2

=

limx→∞

3 + 1x2

1− 2x + 1

x2

=3 + 0

1− 0 + 0=

3

1= 3.

15. limx→∞

√x+x2

2x−x2

Solution. We have

limx→∞

√x+ x2

2x− x2= limx→∞

√xx2 + x2

x2

2xx2 − x2

x2

= limx→∞

x12

x2 + 12x − 1

= limx→∞

1

x32+ 1

2x − 1

=0 + 1

0− 1= −1.

2.4 Solution to PB4

Section 1.6: One-sided limits and continuity

1.6.1. Evaluating one-sided limits

Evaluate the following limits (if they exist).

1. limx→2

f(x), where f(x) ={ √

−2x+ 4 if x ≤ 2

x− 2 if x > 2.

Solution. We first need to find one-sided limits. Recall that “x → c−” means that x approaches cfrom the left (and therefore x < c). And “x→ c+” means that x approaches c from the right (thereforeand x > c.) So as x → 2−, we have x < 2, and therefore f(x) =

√−2x+ 4. As x → 2+, we have

x > 2, and therefore f(x) = x− 2. Thus,

limx→2−

f(x) = limx→2−

√−2x+ 4 =

√−2(2) + 4 =

√0 = 0,


andlimx→2+

f(x) = limx→2+

(x− 2) = 2− 2 = 0.

Since the limit from the left is equal to the limit from the right, it follows that the limit of f(x) as xapproaches 2 exists and is equal to lim

x→2f(x) = 0.

2. limx→5

f(x), where f(x) ={x2 + 1 if x < 5

x+ 20 if x ≥ 5.

Solution. The one-sided limits are:

limx→5−

f(x) = limx→5−

(x2 + 1) = (5)2 + 1 = 25 + 1 = 26

andlimx→5+

f(x) = limx→5+

(x+ 20) = 5 + 20 = 25.

Because these are not equal, it follows that limx→5

f(x) does not exist.

3. limx→0

f(x), where f(x) =

{ √x−1x−1 if x > 0

x2 − x+ 12 if x ≤ 0.

Solution. The one-sided limits are:

limx→0−

f(x) = limx→0−

(x2 − x+1

2) = 02 − 0 +

1

2=

1

2

and

limx→0+

f(x) = limx→0+

√x− 1

x− 1= limx→0+

(√x− 1)(

√x+ 1)

(x− 1)(√x+ 1)

= limx→0+

x− 1

(x− 1)(√x+ 1)

=

= limx→0+

1√x+ 1

=1√0 + 1

= 1.

Since the limit from the left is not equal to the limit from the right, it follows that limx→0

f(x) does notexist.

4. (a) limx→−3

f(x), (b) limx→0

f(x) , where f(x) =

2− x2 if x < −34x+ 3 if −3 < x < 0x2+2x−3x−1 if x ≥ 0

Solution.

(a) limx→−3

f(x). The one-sided limits are

limx→−3−

f(x) = limx→−3−

(2− x2) = 2− (−3)2 = 2− 9 = −7,

andlim

x→−3+f(x) = lim

x→−3+(4x+ 3) = 4(−3) + 3 = −12 + 3 = −9.

These are not the same, so the limit of f(x) when x approaches −3 does not exist.

(b) limx→0

f(x). The one-sided limits are

limx→0−

f(x) = limx→0−

(4x+ 3) = 4(0) + 3 = 0 + 3 = 3,

and

limx→0+

f(x) = limx→0+

x2 + 2x− 3

x− 1= limx→0+

(x− 1)(x+ 3)

x− 1= limx→0+

x+ 3

1= 0 + 3 = 3.

These are the same, so the limit limx→0

f(x) exists and is equal to 3.


1.6.2. Continuity

1. Let f(x) = x2−16x−4 . Is f continuous at 4?

Solution. Recall that a function f is continuous at a point c if limx→c

f(x) = f(c). Notice that thisdefinition implicitly requires three things if f is continuous at c:

(1) f(c) is defined (that is, c is in the domain of f)

(2) limx→c

f(x) exists

(3) limx→c

f(x) = f(c).

The function f(x) = x2−16x−4 is not continuous at 4 since it is not defined at 4 (substituting x by 4, we

get 00 , which is undefined).

2. Let f(x) ={x2 + 2x if x ≤ −2√x+ 3 if x > −2. Is f continuous at −2?

Solution. Condition (1) is satisfied since f is defined at −2 (actually, f(−2) = (−2)2 + 2(−2) =

4− 4 = 0). For condition (2), we need to find one-sided limits and see if they are equal.

limx→−2−

f(x) = limx→−2−

(x2 + 2x) = (−2)2 + 2(−2) = 4− 4 = 0,

andlim

x→−2+f(x) = lim

x→−2+

√x+ 3 =

√−2 + 3 =

√1 = 1.

Since these limits are not equal, it follows that limx→−2

f(x) does not exist, which implies that (2) is not

satisfied. So f is not continuous at −2.

3. Let f(x) =

x3 − x+ 1 if x < 0

0 if x = 0

1− x2 if x > 0.


Solution. Clearly, condition (1) is satisfied as f is defined at 0 (f(0) = 0). For (2), we have

limx→0−

f(x) = limx→0−

(x3 − x+ 1) = 03 − 0 + 1 = 1,

andlimx→0+

f(x) = limx→0+

(1− x2) = 1− 02 = 1.

Because these are the same, the limit limx→0

f(x) exists and therefore, (2) is satisfied. However, condition(3) does not hold as the limit lim

x→0f(x) = 1 is not equal to f(0) = 0. Thus, f is not continuous at 0.

4. Let f(x) =

x2+4x−5

3−3x if x < 1

−2 if x = 1

2x− 4 if x > 1.


Solution. Clearly f is defined at 1 since f(1) = −2. For (2), we have

limx→1−

f(x) = limx→1−

x2 + 4x− 5

3− 3x= limx→1−

(x− 1)(x+ 5)

−3(x− 1)= limx→1−

x+ 5

−3=

1 + 5

−3= −2,

andlimx→1+

f(x) = limx→1+

(2x− 4) = 2(1)− 4 = −2.

Since the one-sided limits are the same, it follows that the limit of f(x) as x approaches 1 exists, andis equal to −2. Since this latter limit is equal to f(1), we can conclude that f is continuous at 1.


5. Find c such that the function f(x) ={cx2 − 1 if x ≤ 3

x+ c if x > 3is continuous at 3.

Solution. If we want f to be continuous at 3, then the three conditions above have to be satisfied.For condition (1), we have f(3) = c(3)2 − 1 = 9c − 1. For (2), we have lim

x→3−f(x) = lim

x→3−(cx2 − 1) =

c(3)2 − 1 = 9c− 1, and limx→3+

f(x) = limx→3+

(x+ c) = 3 + c. The limit exists if and only if the left-sided

limit is equal to the right-sided limit, that is, if and only if 9c − 1 = 3 + c. This latter equation isequivalent to 8c = 4, so that c = 4

8 = 12 . Condition (3) is clearly satisfied when c = 1

2 . So for c = 12 ,

the function f is continuous at 3.

6. Find c such that the function f(x) =

x3 − x− 2 if x < 1

c2 − 3c if x = 1

−2x if x > 1

is continuous at 1.

Solution. We proceed in the same way as before. First we have f(1) = c2 − 3c. Next the one-sidedlimits are: lim

x→1−f(x) = lim

x→1−(x3 − x− 2) = (1)3 − 1− 2 = 1− 3 = −2, and lim

x→1+f(x) = lim

x→1+(−2x) =

−2(1) = −2. So limx→1

f(x) = −2. Condition (3) holds if and only if c2−3c = −2, that is, c2−3c+2 = 0,or (c− 1)(c− 2) = 0. Solving this latter equation for c, we get c = 1 or c = 2.

Section 2.1: The derivative

2.1.1. Finding a derivative (using the limit definition)

Find the derivative of each of the following functions from the limit definition of the derivative.

1. Let f(x) = −3x+ 4. Find f ′(x).

Solution. Recall that the derivative of a function f at any point x is given by

f ′(x) = limh→0

f(x+ h)− f(x)h

.

How to get f (x + h)? To get f(x+ h) we substitute x by x+ h. This meansthat anytime you see “x”, replace it by “x+ h”. For example,

• if f(x) = x2, then f(x+ h) = (x+ h)2.

• If f(x) = x2 − 3x, then f(x+ h) = (x+ h)2 − 3(x+ h).

• Another example: if f(x) = −x3+2x−52x4−x , then f(x+ h) = −(x+h)3+2(x+h)−5

2(x+h)4−(x+h) . And so on...

We come back to the question. We have

f ′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

[−3(x+ h) + 4]− [−3x+ 4]

h=

limh→0

(−3x− 3h+ 4)− (−3x+ 4)

h=

limh→0

−3x− 3h+ 4 + 3x− 4

h= limh→0

−3hh

= limh→0

(−3) = −3.


2. Let f(x) = −5x2 + 7. Find f ′(x).

Solution. We have

f ′(x) = limh→0

[−5(x+ h)2 + 7]− [−5x2 + 7]

h= limh→0

[−5(x2 + 2xh+ h2) + 7]− [−5x2 + 7]

h=

limh→0

−5x2 − 10xh− 5h2 + 7 + 5x2 − 7

h= limh→0

−10xh− 5h2

h=

limh→0

h(−10x− 5h)

h= limh→0

(−10x− 5h) = −10x− 5(0) = −10x− 0 = −10x.

3. Let f(x) = 12x−1 . Find f

′(x).

Solution. We have

f ′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

12(x+h)−1 −

12x−1

h=

limh→0

12x+2h−1 −

12x−1

h= limh→0

2x−1−(2x+2h−1)(2x+2h−1)(2x−1)

h=

limh→0

2x−1−2x−2h+1(2x+2h−1)(2x−1)

h= limh→0

−2h(2x+2h−1)(2x−1)

h= limh→0

−2hh(2x+ 2h− 1)(2x− 1)

=

limh→0

−2(2x+ 2h− 1)(2x− 1)

=−2

(2x+ 2(0)− 1)(2x− 1)=

−2(2x− 1)(2x− 1)

=−2

(2x− 1)2.

4. Let f(x) = 1x2 . Find f ′(x).

Solution. We have

f ′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

1(x+h)2 −

1x2

h= limh→0

x2−(x+h)2x2(x+h)2

h= limh→0

x2 − (x+ h)2

hx2(x+ h)2=

limh→0

x2 − (x2 + 2xh+ h2)

hx2(x+ h)2= limh→0

x2 − x2 − 2xh− h2

hx2(x+ h)2= limh→0

−2xh− h2

hx2(x+ h)2=

limh→0

h(−2x− h)hx2(x+ h)2

= limh→0

−2x− hx2(x+ h)2

=−2x− 0

x2(x+ 0)2=−2xx2(x)2

=−2xx4

=−2x3.

5. Let f(x) =√3x+ 2. Find f ′(x).

Solution. We have

f ′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

√3(x+ h) + 2−

√3x+ 2

h=

limh→0

√3x+ 3h+ 2−

√3x+ 2

h=

limh→0

(√3x+ 3h+ 2−

√3x+ 2)(

√3x+ 3h+ 2 +

√3x+ 2)

h(√3x+ 3h+ 2 +

√3x+ 2)

=

limh→0

3x+ 3h+ 2− (3x+ 2)

h(√3x+ 3h+ 2 +

√3x+ 2)

= limh→0

3x+ 3h+ 2− 3x− 2

h(√3x+ 3h+ 2 +

√3x+ 2)

=

limh→0

3h

h(√3x+ 3h+ 2 +

√3x+ 2)

= limh→0

3√3x+ 3h+ 2 +

√3x+ 2

=

3√3x+ 3(0) + 2 +

√3x+ 2

=3√

3x+ 2 +√3x+ 2

=3

2√3x+ 2

.


2.1.2. Tangent line and rate of change

Let f(x) = −x2 + 3x and g(x) =√x− 1.

1. Find the equation of the tangent line to the graph of f at the given point.

(a) (0, 0)

Solution. We first need find the slope at (0, 0). Recall that the slope m of the tangent line at apoint (c, f(c)) is given by the derivative at c, that is,

SLOPE: m = f ′(c).

To find f ′(c), we first need to find the derivative f ′(x), and then evaluate this at c.

• By the definition, we have

f ′(x) = limh→0

f(x+ h)− f(x)h

= limh→0

−(x+ h)2 + 3(x+ h)− (−x2 + 3x)

h=

limh→0

−(x2 + 2xh+ h2) + 3(x+ h)− (−x2 + 3x)

h=

limh→0

−x2 − 2xh− h2 + 3x+ 3h+ x2 − 3x

h=

limh→0

−2xh− h2 + 3h

h= limh→0

h(−2x− h+ 3)

h= limh→0

(−2x− h+ 3) =

−2x− 0 + 3 = −2x+ 3

• The slope of the tangent line at (0, 0) is: m = f ′(0) = −2(0) + 3 = 3.• Equation of the tangent line at (0, 0): recall that the point-slope form of the equation of a

line through (x1, y1) with slope m is given by y − y1 = m(x− x1). Here (x1, y1) = (0, 0) andm = 3, so the equation we are looking for is: y − 0 = 3(x− 0) or simply y = 3x.

(b) (1, 2)

Solution.

• Derivative: from the previous part, we have that f ′(x) = −2x+ 3.• Slope: here c = 1, so the slope is m = f ′(1) = −2(1) + 3 = −2 + 3 = 1.• Equation of the tangent line at (1, 2): y − 2 = 1(x− 1) or y = x+ 1.

2. Find the rate of change of g(x) at the given point.

(a) x = 2.Solution. First, recall that the rate of change of a function g(x) at x = c is nothing but thederivative of g at c, that is, g′(c). So to find the rate of change of a function at some point, wefirst need to find the derivative.

• By the definition, the derivative of g(x) =√x− 1 is

g′(x) = limh→0

g(x+ h)− g(x)h

= limh→0

√x+ h− 1−

√x− 1

h=

limh→0

(√x+ h− 1−

√x− 1)(

√x+ h− 1 +

√x− 1)

h(√x+ h− 1 +

√x− 1)

=

limh→0

x+ h− 1− (x− 1)

h(√x+ h− 1 +

√x− 1)

= limh→0

x+ h− 1− x+ 1

h(√x+ h− 1 +

√x− 1)

=


limh→0

h

h(√x+ h− 1 +

√x− 1)

= limh→0

1√x+ h− 1 +

√x− 1

=

1√x+ 0− 1 +

√x− 1

=1√

x− 1 +√x− 1

=1

2√x− 1

.

• The rate of change of g at x = 2 is:

g′(2) =1

2√2− 1

=1

2√1=

1

2.

(b) x = 5. The rate of change of g at x = 5 is:

g′(5) =1

2√5− 1

=1

2√4=

1

2(2)=

1

4.

2.5 Solution to PB5

Sections 2.2 & 2.3: Techniques of Differentiation, Product and Quotient Rules


1. f(x) = 2018

Solution. First recall that the derivative of the constant function is 0. That is, if c is a constant, then

d

dx[c] = 0.

Since f(x) = 2018 is constant, it follows that f ′(x) = 0.

2. f(x) = 34

Solution. f ′(x) = 0 since 34 is a constant (34 is a constant because it does not depend on x).

3. f(x) = π2

Solution. Since π = 3.14159 · · · is a constant, it follows that π2 = π×π is also a constant. Therefore,we have f ′(x) = 0 (and NOT 2π !).

4. f(x) = x6

Solution. First recall the following formula

The Power Rule:d

dx[xn] = nxn−1 for any real number n

Here n = 6. Applying the power rule, we get f ′(x) = 6x6−1 = 6x5.

5. f(x) = 3x−2.

Solution. First recall the following property. Let c be a constant, and let u be a differentiable function.

The Constant Multiple Rule:d

dx[cu(x)] = c

d

dx[u(x)].

Here c = 3 and u(x) = x−2. By applying the constant multiple rule, we get f ′(x) = 3 ddx [x

−2]. By thepower rule, we have d

dx [x−2] = −2x−2−1 = −2x−3. So f ′(x) = 3(−2x−3) = −6x−3.


6. f(x) = x2

3

Solution. First we can rewrite f(x) as f(x) = 13x

2. Now we have

f ′(x) =d

dx

[1

3x2]=

1

3

d

dx

[x2]=

1

3(2x2−1) =

1

3(2x1) =

2

3x.

7. f(x) = 12x

43

Solution. Recall the addition and subtraction of fractions.

a

b+c

d=ad+ bc

bd,

a

b− c

d=ad− bcbd

,a

b− 1 =

a− bb

.

We havef ′(x) =

d

dx

[1

2x

43

]=

1

2

d

dx

[x

43

]=

1

2

(4

3x

43−1)

=4

6x

13 =

2

3x

13

8. f(x) =√x3

Solution. The idea is to first rewrite f(x) on the form xn, and then apply the power rule. To do that,recall the following identity:

√xm = x

m2 . (2.5.1)

Applying that identity, we get f(x) = x32 , so that

f ′(x) =d

dx

[x

32

]=

3

2x

32−1 =

3

2x

12 =

3

2

√x.

9. f(x) = 2x3

Solution. First recall the identity:

1

xm= x−m. (2.5.2)

Using that identity, we can rewrite f(x) as f(x) = 2 1x3 = 2x−3. So

f ′(x) =d

dx

[2x−3

]= 2

d

dx[x−3] = 2(−3)x−3−1 = −6x−4 =

−6x4.

10. f(x) = 13√x


k√x = x

1k . (2.5.3)

Using that identity, we can rewrite f(x) as f(x) = 1

x13. Using the identity (2.5.2), we have 1

x13= x−

13 .

Sof ′(x) =

d

dx

[x−

13

]= −1

3x−

13−1 = −1

3x−

43 = −1

3

1

x43

=−13x

43

.

11. f(x) = 3√x5


k√xm = x

mk . (2.5.4)

Using that identity, we can rewrite f(x) as f(x) = x53 . So

f ′(x) =d

dx

[x

53

]=

5

3x

53−1 =

5

3x

23 .


12. f(x) =8√x

2

Solution. First, by using the identity (2.5.3) above, we have f(x) = x18

2 = 12x

18 . So

f ′(x) =d

dx

[1

2x

18

]=

1

2

d

dx

[x

18

]=

1

2

(1

8x

18−1)

=1

16x−

78 =

1

16

1

x78

=1

16x78

.

13. f(x) = π2

x

Solution. We have

f ′(x) =d

dx

[π2

x

]= π2 d

dx

[1

x

]= π2 d

dx

[x−1

]= π2

((−1)x−1−1

)=

π2(−x−2) = π2

(− 1

x2

)= −π

2

x2.

14. f(x) = x+ 1

Solution. First recall the following formula.

The Sum Ruled

dx[u(x) + v(x)] =

d

dx[u(x)] +

d

dx[v(x)] .

Here u(x) = x and v(x) = 1. Applying the sum rule, we get

f ′(x) =d

dx[x+ 1] =

d

dx[x] +

d

dx[1] = 1 + 0 = 1.

15. f(x) = x2 − 2x5


The Difference Ruled

dx[u(x)− v(x)] = d

dx[u(x)]− d

dx[v(x)] .

Here u(x) = x2 and v(x) = 2x5 . Applying the difference rule, we get

f ′(x) =d

dx

[x2 − 2x

5

]=

d

dx

[x2]− d

dx

[2x

5

]= 2x− 2

5

d

dx[x] = 2x− 2

5(1) = 2x− 2

5.

16. f(x) = x−√x

2

Solution. First we can rewrite f(x) as f(x) = 12 (x−

√x).

f ′(x) =d

dx

[1

2

(x−√x)]

=1

2

d

dx

[x−√x]=

1

2

(d

dx[x]− d

dx

[√x])

=

1

2

(1− d

dx

[x

12

])=

1

2

(1− 1

2x−

12

)=

1

2

(1− 1

2

1

x12

)=

1

2

(1− 1

2√x

).

17. f(x) = −3x4 − 2x3 + x2 − 1

Solution. We have

f ′(x) =d

dx

[−3x4 − 2x3 + x2 − 1

]=

d

dx

[−3x4

]− d

dx

[2x3]+

d

dx

[x2]− d

dx[1]

= −3(4x3)− 2(3x2) + 2x− 0 = −12x3 − 6x2 + 2x.


18. f(x) = 2x3−3x2

4

Solution. We have

f ′(x) =d

dx

[1

4

(2x3 − 3x2

)]=

1

4

d

dx

[2x3 − 3x2

]=

1

4

(2(3x2)− 3(2x)

)=

1

4(6x2 − 6x).

19. f(x) = x53 − x 2

3

Solution. We have

f ′(x) =d

dx

[x

53 − x 2

3

]=

5

3x

53−1 − 2

3x

23−1 =

5

3x

23 − 2

3x−

13 .

20. f(x) = x2 − 1x

Solution. We have

f ′(x) =d

dx

[x2 − 1

x

]=

d

dx[x2]− d

dx

[1

x

]= 2x− (− 1

x2) = 2x+

1

x2.

For the derivative ddx

[1x

], see question 13.

21. f(x) = 1.4x5 − 2.5x2 + 3.8

Solution. We havef ′(x) = 1.4(5x4)− 2.5(2x) + 0 = 7x4 − 5x.

22. f(x) = x√x

Solution. First recall the identity

xk

xm= xk−m. (2.5.5)

We havef ′(x) =

d

dx

[x

x12

]=

d

dx

[x1−

12

]=

d

dx

[x

12

]=

1

2x−

12 =

1

2√x.

23. f(x) =√x+xx2

Solution. We have

f ′(x) =d

dx

[√x+ x

x2

]=

d

dx

[√x

x2+

x

x2

]=

d

dx

[x

12

x2+

x

x2

]=

d

dx

[x

12−2 + x1−2

]=

d

dx

[x−

32 + x−1

]=

d

dx

[x−

32

]+

d

dx

[x−1

]=

−3

2x−

32−1 + (−1)x−1−1 = −3

2x−

52 − x−2.

24. f(x) = (3x+ 4)(x− 5)

Solution. First we recall the following formula.

The Product Rule:d

dx[u(x)v(x)] = u′(x)v(x) + u(x)v′(x) =

(d

dx[u(x)]

)v(x) + u(x)

(d

dx[v(x)]

).

Here u(x) = 3x+ 4 and v(x) = x− 5. Applying the product rule, we have

f ′(x) =

(d

dx[3x+ 4]

)(x− 5) + (3x+ 4)

(d

dx[x− 5]

)=

(3 + 0)(x− 5) + (3x+ 4)(1− 0) = 3x− 15 + 3x+ 4 = 6x− 11.


25. f(x) = (5x2 − 2)(x3 + 3x)

Solution. Applying the product rule, we have

f ′(x) =

(d

dx

[5x2 − 2

])(x3 + 3x) + (5x2 − 2)

(d

dx

[x3 + 3x

])=

10x(x3 + 3x) + (5x2 − 2)(3x2 + 3) = 10x4 + 30x2 + 15x4 + 15x2 − 6x2 − 6 = 25x4 + 39x2 − 6.

26. f(x) = (x3 + 1)(2x2 − 4x− 1)

Solution. Applying the product rule, we have

f ′(x) = 3x2(2x2 − 4x− 1) + (x3 + 1)(4x− 4) = (6x4 − 12x3 − 3x2) + (4x4 − 4x3 + 4x− 4) =

10x4 − 16x3 − 3x2 + 4x− 4.

27. f(x) = x2+4x+3√x

Solution. There are many ways of calculating f ′(x): we can use the quotient rule or separate firstand use the power rule or use the product rule. We will use the separation method because it is easier.We have

f(x) =x2 + 4x+ 3√

x=

x2√x+

4x√x+

3√x=x2

x12

+ 4x

x12

+3

x12

= x32 + 4x

12 + 3x−

12 .

Now, by applying the sum rule and the power rule, the derivative is:

f ′(x) =3

2x

12 + 4

(1

2x−

12

)+ 3

(−1

2x−

32

)=

3

2x

12 + 2x−

12 − 3

2x−

32 .

28. f(x) = ( 1x2 − 3

x4 )(x+ 5x3)

Solution. Applying the product rule, we get

f ′(x) =

(d

dx

[1

x2− 3

x4

])(x+ 5x3) + (

1

x2− 3

x4)

(d

dx

[x+ 5x3

])=

(d

dx

[x−2 − 3x−4

])(x+ 5x3) + (x−2 − 3x−4)

(d

dx

[x+ 5x3

])=(

−2x−3 − 3(−4x−5

))(x+ 5x3) + (x−2 − 3x−4)(1 + 5(3x2)) =

(−2x−3 + 12x−5)(x+ 5x3) + (x−2 − 3x−4)(1 + 15x2) =

−2x−2 − 10x0 + 12x−4 + 60x−2 + x−2 + 15x0 − 3x−4 − 45x−2 =

14x−2 + 5x0 + 9x−4 =14

x2+ 5 +

9

x4.

Note that x0 = 1.

29. f(x) = 5x+15x−1


The Quotient Rule:d

dx

[u(x)

v(x)

]=u′(x)v(x)− u(x)v′(x)

(v(x))2.

Here u(x) = 5x + 1 and v(x) = 5x − 1. So u′(x) = 5 and v′(x) = 5. Applying the quotient rule, wehave

f ′(x) =5(5x− 1)− (5x+ 1)(5)

(5x− 1)2=

25x− 5− (25x+ 5)

(5x− 1)2=

25x− 5− 25x− 5

(5x− 1)2=

−10(5x− 1)2

.


30. f(x) = 1+2x3−4x

Solution. Applying the quotient rule, we have

f ′(x) =(1 + 2x)′(3− 4x)− (1 + 2x)(3− 4x)′

(3− 4x)2=

2(3− 4x)− (1 + 2x)(−4)(3− 4x)2

=

6− 8x− (−4− 8x)

(3− 4x)2=

6− 8x+ 4 + 8x

(3− 4x)2=

10

(3− 4x)2.

31. f(x) = x2+1x3−1


f ′(x) =(x2 + 1)′(x3 − 1)− (x2 + 1)(x3 − 1)′

(x3 − 1)2=

2x(x3 − 1)− (x2 + 1)(3x2)

(x3 − 1)2=

2x4 − 2x− (3x4 + 3x2)

(x3 − 1)2=

2x4 − 2x− 3x4 − 3x2

(x3 − 1)2=−x4 − 2x− 3x2

(x3 − 1)2.

32. f(x) = x3+3xx2−4x+3


f ′(x) =(x3 + 3x)′(x2 − 4x+ 3)− (x3 + 3x)(x2 − 4x+ 3)′

(x2 − 4x+ 3)2=

(3x2 + 3)(x2 − 4x+ 3)− (x3 + 3x)(2x− 4)

(x2 − 4x+ 3)2=

(3x4 − 12x3 + 9x2 + 3x2 − 12x+ 9)− (2x4 − 4x3 + 6x2 − 12x)

(x2 − 4x+ 3)2=

3x4 − 12x3 + 9x2 + 3x2 − 12x+ 9− 2x4 + 4x3 − 6x2 + 12x

(x2 − 4x+ 3)2=x4 − 8x3 + 6x2 + 9

(x2 − 4x+ 3)2.

33. f(x) = 1x3+2x2−1


f ′(x) =(1)′(x3 + 2x2 − 1)− (1)(x3 + 2x2 − 1)′

(x3 + 2x2 − 1)2=

0(x3 + 2x2 − 1)− (3x2 + 4x)

(x3 + 2x2 − 1)2=

0− 3x2 − 4x

(x3 + 2x2 − 1)2=

−3x2 − 4x

(x3 + 2x2 − 1)2.

34. f(x) =√x

3+x


f ′(x) =(√x)′(3 + x)−

√x(3 + x)′

(3 + x)2=

12√x(3 + x)−

√x(1)

(3 + x)2=

3+x2√x−√x

(3 + x)2=

(3+x)−2x2√x

(3 + x)2=

3 + x− 2x

2√x(3 + x)2

=3− x

2√x(3 + x)2

.


35. f(x) = 2x5+x4−6xx

Solution. Here it is easier to first separate, and then use the sum and power rules (instead of usingthe quotient rule). We have

f ′(x) =d

dx

[2x5 + x4 − 6x

x

]=

d

dx

[2x5

x+x4

x− 6x

x

]=

d

dx

[2x4 + x3 − 6

]=

2(4x3) + 3x2 − 0 = 8x3 + 3x2.

36. f(x) = xx+ 1

x

Solution. First we have f(x) = xx+ 1

x

= xx2+1x

= x xx2+1 = x2

x2+1 . Now we have the following derivative

(by applying the quotient rule):

f ′(x) =d

dx

[x2

x2 + 1

]=

(x2)′(x2 + 1)− x2(x2 + 1)′

(x2 + 1)2=

2x(x2 + 1)− x2(2x)(x2 + 1)2

=

2x3 + 2x− 2x3

(x2 + 1)2=

2x

(x2 + 1)2.

2.6 Solution to PB6

Section 2.3: Product and Quotient Rules; Higher-Order Derivatives (Continued)

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

(a) y = 2xx+1 , P (1, 1).

Solution. Let f(x) = 2xx+1 . We need two things: the point and the slope. We already have the

point, which is P (1, 1). For the slope, we need to find the derivative f ′(x).

• Applying the quotient rule, we get

f ′(x) =2(x+ 1)− 2x(1)

(x+ 1)2=

2x+ 2− 2x

(x+ 1)2=

2

(x+ 1)2.

• Slope at P (1, 1): m = f ′(1) = 2(1+1)2 = 2

4 = 12 .

• Equation of the tangent line: y − y1 = m(x − x1), where x1 = 1, y1 = 1, and m = 12 . So we

have y − 1 = 12 (x− 1).

(b) y = 2x3 − x2 + 2, P (1, 3).Solution. Let f(x) = 2x3 − x2 + 2. As before, there are three steps:

• Derivative: f ′(x) = 2(3x2)− 2x+ 0 = 6x2 − 2x+ 0 = 6x2 − 2x.• Slope at (1, 3): m = f ′(1) = 6(1)2 − 2(1) = 6− 2 = 4.• Equation of the tangent line at (1, 3): y − 3 = 4(x− 1).

2. Find the points on the curve y = x2−2x+1x−3 where the tangent line is horizontal.

Solution. Let f(x) = x2−2x+1x−3 . Horizontal tangents occur where the derivative is zero. Applying the

quotient rule, we get

f ′(x) =(2x− 2)(x− 3)− (x2 − 2x+ 1)(1)

(x− 3)2=

2x2 − 6x− 2x+ 6− x2 + 2x− 1

(x− 3)2=


x2 − 6x+ 5

(x− 3)2=

(x− 1)(x− 5)

(x− 3)2.

We are looking for points that make the derivative f ′(x) equal to 0. That is, we are looking for pointsx such that (x−1)(x−5)

(x−3)2 = 0. This latter equation is equivalent to (x− 1)(x− 5) = 0. So x = 1 or x = 5.(Note that x = 3 is not a solution since f ′(x) is undefined when x = 3).

• For x = 1, f(x) = f(1) = 12−2(1)+11−3 = 0

−2 = 0. This gives the point P (1, 0).

• For x = 5, f(5) = 52−2(5)+15−3 = 16

2 = 8. This gives the point Q(5, 8).

Conclusion: The tangent line is horizontal at the points P (1, 0) and Q(5, 8).

3. Find the second derivative of each of the following functions.

(a) f(x) = 2x5 − 3x+ 1.

Solution. Recall: The second derivative of a function y = f(x), denoted f ′′(x) or d2ydx2 , is the

derivative of its derivative. That is,

f ′′(x) =d

dx[f ′(x)].

For the function f(x) = 2x5− 3x+1, the derivative is f ′(x) = 2(5x4)− 3+ 0 = 10x4− 3, and thesecond derivative is given by

f ′′(x) =d

dx[f ′(x)] =

d

dx[10x4 − 3] = 10(4x3)− 0 = 40x3.

(b) g(x) = 1x2 .

Solution. First we can rewrite the function as g(x) = x−2. Using the Power Rule, we getg′(x) = −2x−3. We now find the second derivative:

g′′(x) =d

dx[g′(x)] =

d

dx[−2x−3] = −2 d

dx[x−3] = −2(−3)x−4 = 6x−4 =

6

x4.

4. Find the fifth derivative of each of f(x) = 3x4 − x3 + 7x2 − 8x+ 10.

Solution. Recall: Let n be a positive integer. The nth derivative of a function y = f(x), denotedf (n)(x) or dny

dxn , is obtained from f(x) by differentiating successively n times.To get the fifth derivative of f(x) = 3x4 − x3 + 7x2 − 8x+ 10, we need to differentiate 5 times.

(1) f ′(x) = 3(4x3)− 3x2 + 7(2x)− 8 + 0 = 12x3 − 3x2 + 14x− 8

(2) f ′′(x) = ddx [f

′(x)] = ddx [12x

3 − 3x2 + 14x− 8] = 12(3x2)− 3(2x) + 14− 0 = 36x2 − 6x+ 14

(3) f (3)(x) = ddx [f

′′(x)] = ddx [36x

2 − 6x+ 14] = 36(2x)− 6 + 0 = 72x

(4) f (4)(x) = ddx [f

(3)(x)] = ddx [72x] = 72

(5) f (5)(x) = ddx [f

(4)(x)] = ddx [72] = 0.

So the fifth derivative of f(x) = 3x4 − x3 + 7x2 − 8x+ 10 is zero.

Section 2.4: The Chain Rule



1. f(x) = (2x)5

Solution. First recall the General Power Rule, which is a combination of the Power Rule with theChain Rule:

d

dx[(g(x))n] = n(g(x))n−1

d

dx[g(x)].

Here g(x) = 2x and n = 5. So

f ′(x) = 5(2x)5−1d

dx[2x] = 5(2x)4(2) = (5)(2)(2x)4 = 10(2x)4.

2. f(x) = (−3x+ 4)8

Solution. We have

f ′(x) = 8(−3x+ 4)7d

dx[−3x+ 4] = 8(−3x+ 4)7(−3) = 8(−3)(−3x+ 4)7 = −24(−3x+ 4)7.

3. f(x) = 3(5x− 1)7

Solution. We havef ′(x) =

d

dx

[3(5x− 1)7

]= 3

d

dx

[(5x− 1)7

]=

3(7)(5x− 1)6d

dx[5x− 1] = 21(5x− 1)6(5) = 105(5x− 1)6.

4. f(x) = (x2 − 1)32

Solution. We have

f ′(x) =3

2(x2 − 1)

32−1

d

dx[x2 − 1] =

3

2(x2 − 1)

12 (2x) =

3

2(2x)(x2 − 1)

12 = 3x(x2 − 1)

12 .

5. f(x) = (−x3 + 2x+ 1)15

Solution. We havef ′(x) = 15(−x3 + 2x+ 1)14

d

dx[−x3 + 2x+ 1] =

15(−x3 + 2x+ 1)14(−3x2 + 2) = 15(−3x2 + 2)(−x3 + 2x+ 1)14.

6. f(x) = −2(5x6 − 2x)4

Solution. We have

f ′(x) =d

dx

[−2(5x6 − 2x)4

]= −2 d

dx

[(5x6 − 2x)4

]= −2(4)(5x6 − 2x)3

d

dx[5x6 − 2x] =

−8(5x6 − 2x)3(30x5 − 2) = −8(30x5 − 2)(5x6 − 2x)3.

7. f(x) =√x2 − x

Solution. First we have f(x) =√x2 − x = (x2 − x) 1

2 . Now we have

f ′(x) =1

2(x2 − x) 1

2−1d

dx[x2 − x] = 1

2(x2 − x)− 1

2 (2x− 1) =

1

2(2x− 1)

1

(x2 − x) 12

=12 (2x− 1)√x2 − x

=2x− 1

2√x2 − x

.


8. f(x) = 13√x2−1

Solution. First we have f(x) = 1

(x2−1)13= (x2 − 1)−

13 . Now we have

f ′(x) = −1

3(x2 − 1)−

13−1

d

dx[x2 − 1] = −1

3(x2 − 1)−

43 (2x) = −2

3x(x2 − 1)−

43 .

9. f(x) = (−5x+ 4)(x3 + 1)6

Solution. If we look at f(x), we can see many rules including the General Power Rule (G.P.R), theProduct Rule, etc. The G.P.R applies only to the second term, while the Product Rule applies to thewhole function. So we are going to first use the Product Rule.

f ′(x) =d

dx[−5x+ 4] (x3 + 1)6 + (−5x+ 4)

d

dx

[(x3 + 1)6

]=

−5(x3 + 1)6 + (−5x+ 4)6(x3 + 1)5(3x2) =

−5(x3 + 1)6 + 18x2(−5x+ 4)(x3 + 1)5 = (x3 + 1)5(−5(x3 + 1) + 18x2(−5x+ 4)

)=

(x3 + 1)5(−5x3 − 5− 90x3 + 72x2) = (x3 + 1)5(−95x3 + 72x2 − 5).

10. f(x) = x√2− x2

Solution. First we have f(x) = x(2− x2) 12 . As before, we first use the Product Rule.

f ′(x) =d

dx[x](2− x2) 1

2 + xd

dx

[(2− x2) 1

2

]= (1)(2− x2) 1

2 + x(1

2)(2− x2)− 1

2 (−2x) =

(2− x2) 12 − x2 1

(2− x2) 12

=√

2− x2 − x2√2− x2

=(2− x2)− x2√

2− x2=

2− 2x2√2− x2

=

2(1− x2)√2− x2

=2(1− x)(1 + x)√

2− x2.

11. f(x) = (x2 + 1)3(x2 + 2)6

Solution. We have

f ′(x) =d

dx

[(x2 + 1)3

](x2 + 2)6 + (x2 + 1)3

d

dx

[(x2 + 2)6

]=

3(x2 + 1)(2x)(x2 + 2)6 + (x2 + 1)36(x2 + 2)5(2x) = 6x(x2 + 1)2(x2 + 2)6 + 12x(x2 + 1)3(x2 + 2)5 =

6x(x2 + 1)2(x2 + 2)5(x2 + 2 + 2(x2 + 1)

)= 6x(x2 + 1)2(x2 + 2)5(x2 + 2 + 2x2 + 2) =

6x(x2 + 1)2(x2 + 2)5(3x2 + 4).

12. f(x) = (x+1)5

x5+1

Solution. Using the Quotient Rule, we get

f ′(x) =ddx

[(x+ 1)5

](x5 + 1)− (x+ 1)5 d

dx [x5 + 1]

(x5 + 1)2=

5(x+ 1)4(1)(x5 + 1)− (x+ 1)5(5x4)

(x5 + 1)2=

(x+ 1)4(5(x5 + 1)− (x+ 1)(5x4)

)(x5 + 1)2

=(x+ 1)4(5x5 + 5− 5x5 − 5x4)

(x5 + 1)2=

(x+ 1)4(5− 5x4)

(x5 + 1)2.


13. f(x) = x2√x3+1

Solution. First we have f(x) = x2

(x3+1)12. Now, by using the Quotient Rule, we get

f ′(x) =2x(x3 + 1)

12 − x2 1

2 (x3 + 1)−

12 (3x2)

(√x3 + 1)2

=

2x√x3 + 1− 3

2x4 1√

x3+1

x3 + 1=

2x(x3 + 1)− 32x

4

(x3 + 1)√x3 + 1

=

2x4 + 2x− 32x

4

(x3 + 1)√x3 + 1

=12x

4 + 2x

(x3 + 1)√x3 + 1

=12x(x

3 + 4)

(x3 + 1)√x3 + 1

=x(x3 + 4)

2(x3 + 1)√x3 + 1

.

14. f(x) =√

xx2+3

Solution. First we have f(x) =(

xx2+3

) 12

. Now we have

f ′(x) =1

2

(x

x2 + 3

)− 12 d

dx

[x

x2 + 3

]=

1

2

(x

x2 + 3

)− 12(1(x2 + 3)− x(2x)

(x2 + 3)2

)=

1

2

(x

x2 + 3

)− 12 (−x2 + 3)

(x2 + 3)2=−x2 + 3

2(x2 + 3)2

(x

x2 + 3

)− 12

.

15. f(x) =(x4+1x2+1

)5Solution. We have

f ′(x) = 5

(x4 + 1

x2 + 1

)4d

dx

[x4 + 1

x2 + 1

]= 5

(x4 + 1

x2 + 1

)4(4x3(x2 + 1)− (x4 + 1)2x

(x2 + 1)2

)=

5

(x4 + 1

x2 + 1

)4(4x5 + 4x3 − 2x5 − 2x

(x2 + 1)2

)= 5

(x4 + 1

x2 + 1

)4(2x5 + 4x3 − 2x

(x2 + 1)2

)

5

(2x(x4 + 2x2 − 1)

(x2 + 1)2

)(x4 + 1

x2 + 1

)4

=10x(x4 + 2x− 1)

(x2 + 1)2

(x4 + 1

x2 + 1

)4

.

2.7 Solution to PB7

Section 2.6: Implicit Differentiation and Related Rates

2.6.1. Implicit Differentiation

1. Find dydx for each of the following functions.

(a) x2 + y2 = 25

Solution. Taking the derivative of both sides with respect to x, we get ddx [x

2 + y2] = ddx [25],

which is equivalent to ddx [x

2] + ddx [y

2] = 0 or 2x+ 2y dydx = 0. Solving now this latter equation fordydx , we get 2y dydx = −2x, so that dy

dx = − 2x2y = −xy . Thus

dydx = −xy .


(b) x3 + y3 = 1

Solution. We haved

dx[x3 + y3] =

d

dx[1] =⇒ d

dx[x3] +

d

dx[y3] = 0 =⇒

3x2 + 3y2dy

dx= 0 =⇒ 3y2

dy

dx= −3x2 =⇒ dy

dx=−3x2

3y2=⇒ dy

dx=−x2

y2.

(c) 2x2 − y2 = x

Solution. We haved

dx[2x2 − y2] = d

dx[x] =⇒ d

dx[2x2]− d

dx[y2] = 1 =⇒ 4x− 2y

dy

dx= 1 =⇒

−2y dydx

= 1− 4x =⇒ dy

dx=

1− 4x

−2y.

(d) x4 + 3y3 = 5y

Solution. We haved

dx[x4 + 3y3] =

d

dx[5y] =⇒ d

dx[x4] + 3

d

dx[y3] = 5

dy

dx=⇒

4x3 + 9y2dy

dx= 5

dy

dx=⇒ 9y2

dy

dx− 5

dy

dx= −4x3 =⇒

(9y2 − 5)dy

dx= −4x3 =⇒ dy

dx=−4x3

9y2 − 5.

(e)√x− y = y2 + 3

Solution. We haved

dx[√x− y] = d

dx[y2 + 3] =⇒ d

dx[√x]− d

dx[y] =

d

dx[y2] +

d

dx[3] =⇒

1

2√x− dy

dx= 2y

dy

dx+ 0 =⇒

−dydx− 2y

dy

dx= − 1

2√x

=⇒ (−1− 2y)dy

dx= − 1

2√x

=⇒

dy

dx=− 1

2√x

−1− 2y=

−12√x(−1− 2y)

=1

2√x(1 + 2y)

.

(f) xy = 5

Solution. Taking the derivative of both sides, and applying the Product to the lefthand side, weget

d

dx[xy] =

d

dx[5] =⇒ d

dx[x]y + x

d

dx[y] = 0 =⇒ (1)y + x

dy

dx= 0 =⇒

xdy

dx= −y =⇒ dy

dx=−yx

(g) 3x2 + 2xy + y2 = 2

Solution. We haved

dx[3x2 + 2xy + y2] =

d

dx[2] =⇒ 3

d

dx[x2] + 2

d

dx[xy] +

d

dx[y2] = 0 =⇒

6x+ 2

(y + x

dy

dx

)+ 2y

dy

dx= 0 =⇒ 6x+ 2y + 2x

dy

dx+ 2y

dy

dx= 0 =⇒

(2x+ 2y)dy

dx= −6x− 2y =⇒ dy

dx=−6x− 2y

2x+ 2y=−3x− yx+ y

.


(h) −5x2 + xy − y3 = 1

Solution. We have

d

dx[−5x2 + xy − y3] = d

dx[1] =⇒ −5 d

dx[x2] +

d

dx[xy]− d

dx[y3] = 0 =⇒

−10x+

(y + x

dy

dx

)− 3y2

dy

dx= 0 =⇒ (x− 3y2)

dy

dx= 10x− y =⇒ dy

dx=

10x− yx− 3y2

.

(i) x3y2 + y4 = x

Solution. We have

d

dx[x3y2 + y4] =

d

dx[x] =⇒ d

dx[x3y2] +

d

dx[y4] = 1 =⇒

d

dx[x3]y2 + x3

d

dx[y2] + 4y3

dy

dx= 1 =⇒ 3x2y2 + x3(2y)

dy

dx+ 4y3

dy

dx= 1 =⇒

(2x3y + 4y3)dy

dx= 1− 3x2y2 =⇒ dy

dx=

1− 3x2y2

2x3y + 4y3.

(j) x+yx−y = 1

Solution. We have

d

dx

[x+ y

x− y

]=

d

dx[1] =⇒

ddx [x+ y](x− y)− (x+ y) ddx [x− y]

(x− y)2= 0 =⇒

d

dx[x+ y](x− y)− (x+ y)

d

dx[x− y] = 0 =⇒

(1 +dy

dx)(x− y)− (x+ y)(1− dy

dx) = 0 =⇒

x− y + xdy

dx− y dy

dx−(x+ y − xdy

dx− y dy

dx

)= 0 =⇒

x− y + xdy

dx− y dy

dx− x− y + x

dy

dx+ y

dy

dx= 0 =⇒

−2y + 2xdy

dx= 0 =⇒ 2x

dy

dx= 2y =⇒ dy

dx=

2y

2x=y

x.

From the first line to the second we used the fact that if a fraction AB = 0, then A = 0.

2. Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

(a) x2y3 − xy = −y + 8 at (1, 2)Solution.

• First, we need to find the derivative dydx .

d

dx[x2y3 − xy] = d

dx[−y + 8] =⇒ d

dx[x2y3]− d

dx[xy] =

d

dx[−y] + d

dx[8] =⇒

d

dx[x2]y3 + x2

d

dx[y3]−

(d

dx[x]y + x

d

dx[y]

)= −dy

dx+ 0 =⇒

2xy3 + x2(3y2dy

dx)−

(y + x

dy

dx

)= −dy

dx=⇒

2xy3 + 3x2y2dy

dx− y − xdy

dx= −dy

dx=⇒


3x2y2dy

dx− xdy

dx+dy

dx= −2xy3 + y =⇒

(3x2y2 − x+ 1)dy

dx= −2xy3 + y =⇒

dy

dx=−2xy3 + y

3x2y2 − x+ 1

• Substituting x by 1 and y by 2 into the dydx , we get the slope

m =−2(1)(2)3 + 2

3(1)2(2)2 − 1 + 1=−16 + 2

12− 1 + 1=−1412

=−76.

• The equation of the tangent line at (1, 2) is given by y − 2 = − 76 (x− 1).

(b) xy − 3 =√y − 2y at (2, 1)

Solution. As before, we need to find the derivative first.

d

dx[xy − 3] =

d

dx[√y − 2y] =⇒ d

dx[xy]− d

dx[3] =

d

dx[√y]− d

dx[2y] =⇒

d

dx[x]y + x

d

dx[y]− 0 =

d

dx[y

12 ]− 2

d

dx[y] =⇒

(1)y + xdy

dx=

1

2y−

12dy

dx− 2

dy

dx=⇒

xdy

dx− 1

2y−

12dy

dx+ 2

dy

dx= −y =⇒

dy

dx

(x− 1

2y−

12 + 2

)= −y =⇒

dy

dx=

−yx+ 2− 1

2y− 1

2

=−y

x+ 2− 12

1

y12

=−y

x+ 2− 12√y

.

The slope at (2, 1) is

m =−1

2 + 2− 12√1

=−1

4− 12

=−172

= −127=−27.

The equation of the tangent line at (2, 1) is: y − 1 = − 27 (x− 2).

2.6.2. Related Rates

1. If V is the volume of a cube with edge length x and the cube expands as time passes, find dVdt in terms

of dxdt .

Solution. Recall: if y is a quantity that depends on time t, then by the General Power Rule, we have

d

dt[yn] = nyn−1

dy

dt.

We know that the volume of a cube with edge x is given by the formula V = x3. Differentiating bothsides with respect to t, we get

dV

dt=

d

dt[x3] = 3x2

dx

dt.


2. A large ship in the middle of the ocean is leaking oil. The oil is forming in a circle on the ocean aroundthe ship. When the radius of this circle is 50m, the radius is growing at a rate of 1 metre per hour. Atthis point, how fast is the area of the circle growing?

Solution. How to solve a related rates problem?

1. Read the problem carefully.

2. Draw a diagram if possible.

3. Introduce notation. Assign symbols to all quantities that are functions of time t.

4. Express the given information and the required rate in terms of derivatives.

5. Write an equation that relates the various quantities of the problem.

6. Use the Chain Rule to differentiate both sides of the equation with respect to t.

7. Substitute the given information into the resulting equation and solve for the unknown rate.

Here the quantities that are functions of time are the radius and the area of the circle. Let r denotethe radius, and let A be the area.

Given:dr

dt= 1 m/h Unknown:

dA

dtwhen r = 50 m.

The equation that relates A and r is: A = πr2 (this is nothing but the formula for the area of a circleof radius r). Differentiating both sides, we get

dA

dt=

d

dt[πr2] = π

d

dt[r2] = π2r

dr

dt= 2πr

dr

dt.

Substituting, we get dAdt = 2π(50)(1) = 100π m2/h.

3. The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when thediameter is 80 mm?

Solution. Here the quantities that are functions of time are the volume and the radius. Let V denotethe volume, and let r denote the radius.

Given:dr

dt= 4 mm/s Unknown:

dV

dtwhen r = 40 mm.

Equation that relates V and r: we know that the volume of a sphere of radius r is given by the formula:V = 4

3πr3. Differentiating both sides with respect to time t, we get

dV

dt=

d

dt

[4

3πr3]=

4

3πd

dt[r3] =

4

3π(3r2)

dr

dt= 4πr2

dr

dt.

Substituting, we get

dV

dt= 4π(40)24 = 16π40× 40 = 16π × 1600 = 256π × 100 = 25600π mm3/s.

4. A tiny spherical balloon is inserted into a clogged artery and is inflated at the rate of 0.002πmm3/min.How fast is the radius of the balloon growing when the radius is r = 0.005 mm?

Solution. Let V be the volume of the balloon.

Given:dV

dt= 0.002π mm3/min Unknown:

dr

dtwhen r = 0.005 m.


We have V = 43πr

3. Differentiating both sides, we get

dV

dt=

d

dt[4

3πr3] =

4

3πd

dt[r3] =

4

3π3r2

dr

dt= 4πr2

dr

dt.

So dVdt = 4πr2 drdt . Solving this latter equation for dr

dt , we get drdt =

14πr2

dVdt . Substituting, we get

dr

dt=

1

4π(0.005)2(0.002π) =

1

4(0.005)2(0.002) = 20 mm/min.

5. A cylindrical tank with radius 5 m is being filled with water at a rate of 3m3/min. How fast is theheight of the water increasing?

Solution. Here the quantities that depend on time are the volume V , and the height h. (Note thatthe radius is constant here: r = 5 m.)

Given:dV

dt= 3 m3/min Unknown:

dh

dt.

We know that the volume of a cylinder with radius r and height h is given by the formula: V = πr2h.Differentiating both sides with respect to time t, we get

dV

dt=

d

dt[πr2h] = πr2

dh

dt.

So dVdt = πr2 dhdt . Solving this latter equation for dh

dt , we get dhdt = 1

πr2dVdt . Now, by substituting, we get

dhdt = 1

π(5)2 (3) =3

25π m/min.

6. A plane flying horizontally at an altitude of 3 mi and a speed of 500 mi/h passes directly over a radarstation. Find the rate at which the distance from the plane to the station is increasing when it is 5 miaway from the station.

Solution. Consider Figure 2.9.

Figure 2.9:

Given:dx

dt= 500 mi/h Unknown:

dz

dtwhen z = 5 mi.

Equation relating x and z: x2 + 9 = z2. Differentiating both sides, we get 2xdxdt + 0 = 2z dzdt , so thatdzdt =

2x2z

dxdt = x

zdxdt . Using the Pythagorean theorem, one can see that x = 4 when z = 5. Thus

dz

dt=x

z

dx

dt=

4

5(500) = 400 mi/h.

2.8 Solution to PB8

Sections 3.1 & 3.2: Increasing and Decreasing Functions, Local Extrema, Concavity, and Pointsof Inflection


3.1.1. Critical Numbers

Find the critical numbers of each of the following functions

1. f(x) = −2x2 + 8x− 3

Solution. Recall that a number c in the domain of a function f is called a critical number if f ′(c) = 0

or f ′(c) is undefined. So to find critical numbers,

• we first find the derivative f ′(x), and then we solve the equation f ′(x) = 0, and

• we find all numbers c such that f ′(c) is undefined and c lies in the domain of f .

For f(x) = −2x2 + 8x − 3, we have f ′(x) = −4x + 8. Solving the equation −4x + 8 = 0, we get−4x = −8 or x = −8

−4 = 2. So 2 is the only number that makes the derivative equal to 0. Since f ′(x)is defined everywhere (polynomial), it follows that there is no c such that f ′(c) is undefined. Thus fhas only one critical number: 2.

2. f(x) = 3√x

Solution. First we have f(x) = x13 . Now we have

f ′(x) =1

3x−23 =

1

3

1

x23

=1

3x23

.

If 1

3x23= 0, then 1 = 0 2, which is impossible. So there is no number that makes the derivative equal

to 0. However we can observe that f ′(x) is undefined when x = 0. Since 0 is in the domain of f , itfollows that it is a critical number of f .

3. f(x) = x3 + x2 + 5

Solution. We have f ′(x) = 3x2 + 2x = x(3x+ 2) [Note: In order to find critical numbers, it is usefulto factor the derivative whenever it is possible.]. f ′(x) = 0 is equivalent to x = 0 or 3x+2 = 0, so thatx = 0 or x = −2

3 , which are the critical numbers.

4. f(x) = x3 + 6x2 − 15x

Solution. We have

f ′(x) = 3x2 + 12x− 15 = 3(x2 + 4x− 5) = 3(x− 1)(x+ 5).

So f ′(x) = 0 is equivalent to x − 1 = 0 or x + 5 = 0, so that x = 1 or x = −5, which are the criticalnumbers of f .

5. f(x) =√1− x2

Solution. Firs we have f(x) = (1− x2) 12 . Now we have

f ′(x) =1

2(1− x2)− 1

2d

dx[1− x2] = 1

2(1− x2)− 1

2 (−2x) = −x 1

(1− x2) 12

=−x√1− x2

.

f ′(x) = 0 is equivalent to −x = 0 or x = 0. So 0 is a critical number. Furthermore, we can observe thatthe derivative is undefined when its denominator is equal to zero: 1−x2 = 0, that is, (1−x)(1+x) = 0

or x = 1 or x = −1. Since −1 and 1 belong to the domain of f , it follows that they are also criticalnumbers of f . Hence the critical numbers of f are: 0,−1, and 1.

2If a fraction AB

= 0, then A = 0


6. f(x) = 2x+1x−3


f ′(x) =2(x− 3)− (2x+ 1)(1)

(x− 3)2=

2x− 6− 2x− 1

(x− 3)2=

−7(x− 3)2

.

f ′(x) = 0 is equivalent to −7 = 0. This is impossible; so there is no number that makes the derivativeequal to 0. However, one can see that f ′(x) is undefined when x = 3. Since 3 does not lie in thedomain of f (3 also makes the denominator of f(x) equal to 0), it follows 3 is not a critical number.Thus f has no critical number.

3.1.2. Absolute Maximum and Minimum Values

Find the absolute maximum and absolute minimum values of f on the given interval. Also state the locationsof these absolute extrema.

1. f(x) = 12 + 4x− x2 on [0, 5]

Solution. To find the absolute maximum and absolute minimum values of f on a closed interval, wecan proceed as follows (this method is called the Closed Interval Method).

Step 1. Find the critical numbers of f on the open interval. First we need to find the derivative: f ′(x) =4− 2x = 2(2−x). The equation f ′(x) = 0 is equivalent to 2(2−x) = 0 or x = 2. So 2 is a criticalnumber of f . In fact it is the only one. Now we have to check whether that critical number liesin the open interval (0, 5). Clearly 2 belongs to (0, 5).

Step 2. Find the values of f at the critical numbers that lie in the open interval. We have f(2) =

12 + 4(2)− (2)2 = 12 + 8− 4 = 16.

Step 3. Find the values of f at the endpoints of the interval. (The endpoints of the interval here are 0

and 5.) We have f(0) = 12 and f(5) = 12 + 20− 25 = 7.

Step 4. The absolute maximum (abbreviated abs max) value of f is the largest value from steps 2 and3. So abs max = 16 located at 2 (it is located at 2 because f(2) = 16). The absolute minimum(abbreviated abs min) is the smallest value from steps 2 and 3. So abs min = 7 located at 5.

2. f(x) = −x2 + 2x+ 3 on [0, 2]

Solution. We have f ′(x) = −2x+ 2 = −2(x− 1). f ′(x) = 0 is equivalent to x = 1. Does 1 lie in theopen interval (0, 2)? Yes. So 1 is the only critical number of f in (0, 2). Now the value of f at thatnumber is f(1) = −1 + 2 + 3 = 4, and the values of f at the endpoints of the interval are: f(0) = 3

and f(2) = −4 + 4 + 3 = 3. Thus abs max = 4 located at 1 and abs min = 3 located at 0 and 2.

3. f(x) = x3 − 3x+ 5 on [0, 3]

Solution. We have f ′(x) = 3x2 − 3 = 3(x2 − 1) = 3(x− 1)(x+1). f ′(x) = 0 is equivalent to x = 1 orx = −1. Does 1 lie in (0, 3)? Yes. Does −1 lie in (0, 3)? No. So we throw away −1. Now the value off at 1 is f(1) = 1−3+5 = 3, and its values at the endpoints are: f(0) = 5 and f(3) = 27−9+5 = 23.Thus abs max = 23 located at 3, and abs min = 3 located at 1.

4. f(x) = −x3 + 3x2 + 1 on [−1, 2]Solution. We have f ′(x) = −3x2 + 6x = −3x(x − 2). The critical numbers are then 0 and 2. Wecan see that 0 is the only one that belongs to the open interval (−1, 2). The value of f at that criticalnumber is f(0) = 1, and its values at the endpoints are: f(−1) = −(−1)3 +3(−1)2 +1 = 1+3+1 = 5

and f(2) = −8 + 12 + 1 = 5. Thus abs max = 5 located at −1 and 2, and abs min = 1 located at 0.


5. f(x) = 2x3 − 3x2 − 12x+ 1 on [−2, 1]

Solution. The derivative is f ′(x) = 6x2−6x−12 = 6(x2−x−2) = 6(x+1)(x−2). This implies that−1 and 2 are the critical numbers of f . But 2 does not lie in (−2, 1), so we throw it away. Now we havef(−1) = −2− 3 + 12 + 1 = 8, and f(−2) = −16− 12 + 24 + 1 = −3 and f(1) = 2− 3− 12 + 1 = −12.This shows that abs max = 8 located at −1 and abs min = −12 located at 1.

6. f(x) = x2−4x2+4 on [−4, 4]

Solution. Using the Quotient Rue, we have

f ′(x) =2x(x2 + 4)− (x2 − 4)(2x)

(x2 + 4)2=

2x3 + 8x− 2x3 + 8x

(x2 + 4)2=

16x

(x2 + 4)2.

f ′(x) = 0 is equivalent to 16x = 0 or x = 0. Does 0 lie in (−4, 4)? Yes. Now we have f(0) = 0−40+4 = −1

and f(−4) = 16−416+4 = 12

20 = 35 , and f(4) = 3

5 . Thus abs max = 35 located at −4 and 4, and abs min

= −1 located at 0.

7. f(x) = (x2 − 4)3 on [−2, 3]

Solution. Using the General Power Rule, we get

f ′(x) = 3(x2 − 4)2(2x) = 6x(x2 − 4)2 = 6x[(x+ 2)(x− 2)]2 = 6x(x+ 2)2(x− 2)2.

So the critical numbers are 0,−2, and 2. The values of f at the critical numbers in the open intervalare: f(0) = (−4)3 = −64, and f(2) = (4 − 4)3 = 0. The values of f at the endpoints are: f(−2) = 0

and f(3) = (9− 4)3 = 53 = 125. Thus abs max = 125 located at 3, and abs min = −64 located at 0.

8. f(x) = xx2−x+1 on [0, 3].


f ′(x) =(1)(x2 − x+ 1)− x(2x− 1)

(x2 − x+ 1)2=

−x2 + 1

(x2 − x+ 1)2=−(x2 − 1)

(x2 − x+ 1)2=−(x− 1)(x+ 1)

(x2 − x+ 1)2.

f ′(x) = 0 is equivalent to −(x− 1)(x+1) = 0, so that x = 1 or x = −1, which are the critical numbersof f . One has to throw away −1 since it is not a member of the open interval (0, 3). Now we havef(1) = 1

1−1+1 = 1, and f(0) = 00−0+1 = 0

1 = 0 and f(3) = 39−3+1 = 3

7 . Thus abs max = 1 located at 1and abs min = 0 located at 0.

3.1 & 3.2: Increasing/Decreasing, Local Extrema, Concavity, and Points of Inflection

For each of the following functions, find where it is increasing and where it is decreasing. Also find the localmaximum and minimum values. Also find the intervals of concavity and the inflection points.

1. f(x) = −x2 + 4x− 3

Solution. We need to find the first derivative for increasing/decreasing, local max and min, and weneed the second derivative for the concavity and inflection points. Recall the following.

Increasing/Decreasing Test:

(a) If f ′(x) > 0 on an interval, then f is increasing on that interval.

(b) If f ′(x) < 0 on an interval, then f is decreasing on that interval.

(c) If f ′(x) = 0 on an interval, then f is constant on that interval.


Local maximum and minimum values: Suppose that c is a critical number of a continuous functionf .

(a) If f ′ changes from positive to negative at c, then f has a local maximum at c. In that case, f(c)is a local maximum value.

(b) If f ′ changes from negative to positive at c, then f has a local minimum at c. In that case, f(c)is a local minimum value.

(c) If f ′ is positive to the left and right of c, or negative to the left and right of c, then f has no localmaximum or minimum.

Concavity Test:

(a) If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave upward on I.

(b) If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave downward on I.

So to find the intervals of concavity, we need to first find the second derivative f ′′(x), then find thepoints that make f ′′(x) equal to zero. And then deduce the sign of f ′′(x).

Inflection point: An inflection point is a point where the concavity changes. That is, a point wherethe graph changes from concave upward to concave downward or from concave downward to concaveupward.

We come back to the question.

• First derivative: we have f ′(x) = −2x+ 4.

• Critical numbers: −2x+ 4 = 0 gives −2x = −4, that is, x = −4−2 = 2.

• Second derivative: f ′′(x) = −2.

• Table: see Figure 2.10. How to fill out that table? At the top row, we have the domain of f(which is R = (−∞,∞)) and the critical numbers (here we have only one critical number = 2).At the second row we have the sign of the first derivative. To get the sign on the interval (−∞, 2),pick any number in that interval and plug-in f ′(x). If the result is positive then f ′(x) > 0. If theresult is negative then f ′(x) < 0. For example, if we pick 0, we get f ′(0) = 4 > 0, so the sign is“+” on (−∞, 2). We use the same technique to get the sign of f ′(x) on the interval (2,∞): if forexample, we pick 3, we get f ′(3) = −6 + 4 = −2 < 0. So the sign is “−” on (2,∞). Thus f isincreasing on the interval (−∞, 2) and decreasing on (2,∞). The table shows that f has a localmaximum at 2; so f(2) = −4 + 8 − 3 = 1 is a local maximum value. However, there is no localminimum.

How about the concavity? Since f ′′(x) = −2 < 0 is negative everywhere, it follows that f is concavedownward on R. So there is no inflection point.

2. f(x) = x3 − 3x2 − 9x+ 4

Solution.

• Critical numbers: the first derivative is f ′(x) = 3x2 − 6x− 9 = 3(x2 − 2x− 3) = 3(x+ 1)(x− 3).This is equal to 0 if x = −1 or x = 3. So we have two critical numbers: −1 and 3. Value of f at−1: f(−1) = −1− 3 + 9 + 4 = 9. Value of f at 3: f(3) = 27− 27− 27 + 4 = −23.

• Second derivative: we have f ′′(x) = 6x − 6 = 6(x − 1). Points that make f ′′(x) equal to zero:f ′′(x) = 0 gives 6(x− 1) = 0, that is x = 1. Value of f at that point: f(1) = 1− 3− 9 + 4 = −7.


Figure 2.10:

• Table: see Figure 2.11. The sign of f ′(x) is obtained by using the same technique as before. Forthe sign of f ′′(x) we still use the same method. For example, to get the sign of f ′′(x) on theinterval (−∞, 1), we pick for example 0, and plug-in f ′′(x): f ′′(0) = −6, which is negative. Forthe sign on the interval (1,∞), we have f ′′(2) = 12 − 6 = 6 > 0. That table shows that f isincreasing on the intervals (−∞,−1) and (3,∞), and decreasing on (−1, 3). The function f hasone local maximum at −1, so the local maximum value is f(−1) = 9. f has one local minimumat 3, so the local minimum value is f(3) = −23. Again from the same table, we can see that fis concave downward on the interval (−∞, 1) and concave upward on (1,+∞). So the concavitychanges at 1, which means that the point (1, f(1)) = (1,−7) is an inflection point.

Figure 2.11:

3. f(x) = −x4 + 4x3 + 1

Solution.

• Critical numbers: The first derivative is f ′(x) = −4x3 + 12x2 = 4x2(−x + 3). f ′(x) = 0 gives4x2 = 0 or −x+ 3 = 0, that is, x = 0 or x = 3, which are the critical numbers. Now the value off at 0 is: f(0) = 1. The value at 3 is: f(3) = −81 + 108 + 1 = 28.

• Second derivative: we have f ′′(x) = −12x2 + 24x = 12x(−x + 2). Points that make f ′′(x)equal to zero: f ′′(x) = 0 gives x = 0 or x = 2. Values of f at those points: f(0) = 1 andf(2) = −16 + 32 + 1 = 17.


• Table: see Figure 2.12. From that table, one can see that f is increasing on the interval (−∞, 3)and decreasing on (3,∞). The number 28 is a local maximum value. There is no local minimum.For the concavity, one can again see from the same table that f is concave downward on (−∞, 0),concave upward on (0, 2), and again concave downward on (2,∞). So the concavity changes twice:at 0 and 2. So we have two inflection points: (0, 1) and (2, 17).

Figure 2.12:

4. f(x) = xx2+1

Solution.

• Critical numbers: using the Quotient Rule, we have

f ′(x) =(1)(x2 + 1)− x(2x)

(x2 + 1)2=x2 + 1− 2x2

(x2 + 1)2=−x2 + 1

(x2 + 1)2.

f ′(x) = 0 gives −x2 + 1 = 0, that is, −(x2 − 1) = 0 or −(x− 1)(x+ 1) = 0, which gives x = −1or x = 1. Values of f at those critical numbers: f(−1) = −1

1+1 = −12 , and f(1) = 1

2 .

• Second derivative: again by using the Quotient Rule, we have

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

(x2 + 1)4=

(x2 + 1)[−2x(x2 + 1)− 4x(−x2 + 1)]

(x2 + 1)4=

(x2 + 1)(−2x3 − 2x+ 4x3 − 4x)

(x2 + 1)4=

(x2 + 1)(2x3 − 6x)

(x2 + 1)4=

(x2 + 1)2x(x2 − 3)

(x2 + 1)4.

Points that make f ′′(x) equal to zero: f ′′(x) = 0 gives (x2 +1)2x(x2 − 3) = 0, that is, x2 +1 = 0

or 2x = 0 or x2 − 3 = 0, that is, x = 0 or x = ±√3. [Note that x2 + 1 = 0 has no solution

at all as x2 + 1 is always positive, and then can not be equal to 0.] Values of f at those points:f(−√3) = −

√3

3+1 = −√34 , f(0) = 0

1 = 0, and f(√3) =

√34 .

• Table: see Figure 2.13. That table shows that f is increasing on the interval (−1, 1) and decreasingon (−∞,−1) and (1,∞). The number − 1

2 is a local minimum value, while 12 is a local maximum

value. For the concavity, f is concave upward on (−√3, 0) and (

√3,+∞), and concave downward

on (−∞,−√3) and (0,

√3). So we have three inflection points: (−

√3,−

√34 ), (0, 0) and (

√3,√34 ).


Figure 2.13:

2.9 Solution to PB9

Section 3.3: Curve sketching

1. Consider the following functions.

(a) f(x) = −x3 + 3x2 − 1

(b) f(x) = x4 − 4x3 + 21

In each case, find (i) the domain; (ii) the y- intercept; (iii) the intervals of increase and decrease; (iv)the local maximum and minimum values; (v) the intervals of concavity; (vi) the inflection points. (vii)Use the above information to sketch the graph of f(x).

Solution.

(a) f(x) = −x3 + 3x2 − 1.

(i) The function f is defined everywhere since it is a polynomial. So its domain is R = (−∞,∞).(ii) The y-intercept is f(0) = 03 + 3(0)2 − 1 = −1.(iii) For the intervals of increase and decrease, we need to find the first derivative. We have

f ′(x) = −3x2 + 6x. The equation −3x2 + 6x = 0 gives −3x(x − 2) = 0, that is, −3x = 0

or x − 2 = 0. The former equation gives x = 0 and the latter gives x = 2. So we have twocritical numbers: 0 and 2. For the sign of f ′(x), see the table from Figure 2.14. (To learnhow to make the table, we refer the reader to the Solution to Practice Problems 8). From thetable, we deduce that f is decreasing on (−∞, 0) ∪ (2,∞) and increasing on (0, 2).

(iv) From Figure 2.14, we can see that the sign of f ′(x) changes from − to + at the critical number0. So there is a local min at 0, and the local min value at that point is f(0) = −1. Againfrom the same table, we can see that the sign of f ′(x) changes from + to − at the criticalnumber 2. So there is a local max at 2, and the local max value is f(2) = −8 + 12− 1 = 3.

(v) For the concavity, we need to find the second derivative. We have f ′(x) = −3x2 + 6x andf ′′(x) = −6x+ 6. Setting −6x+ 6 = 0, we get −6x = −6, that is, x = −6

−6 = 1. The sign off ′′(x), and the concavity of f are provided by Figure 2.14.

(vi) Since the concavity changes at 1 and since 1 lies in the domain, we have an inflection point at1. The corresponding y-coordinate is f(1) = −1 + 3− 1 = 1. So the inflection point is (1, 1).


Figure 2.14:

(vii) The graph of f(x) = −x3 + 3x2 − 1 is given by Figure 2.15.

Figure 2.15: Graph of f(x) = −x3 + 3x2 − 1

(b) f(x) = x4 − 4x3 + 21

(i) As before the domain of f(x) is R.(ii) The y-intercept is f(0) = 21.(iii) For the intervals of increase and decrease, we have f ′(x) = 4x3 − 12x2. Setting f ′(x) = 0,

we get 4x2(x − 3) = 0, that is, 4x2 = 0 or x − 3 = 0. The former equation gives x = 0,and the latter gives x = 3. So there are two critical numbers: 0 and 3. The coorespondingy-coordinates are f(0) = 21 and f(3) = 81 − 108 + 21 = −6. The intervals of increase anddecrease can be found in Figure 2.16.

(iv) There is no local min and no local max at x = 0 since the sign of f ′(x) is the same from bothsides of 0 (see Figure 2.16). But f has a local min at 3 since the sign of f ′(x) changes from− to + at that point.

(v) For the concavity, we need to find the second derivative, which is f ′′(x) = 12x2 − 24x =

12x(x − 2). Setting f ′′(x) = 0, we get x = 0 or x = 2. The corresponding y-coordinates aref(0) = 21 and f(2) = 16−32+21 = 5. The intervals of concavity can be found in Figure 2.16.

(vi) From the concavity, we deduce that there are two inflection points, namely (0, 21) and (2, 5).


Figure 2.16:

Figure 2.17: Graph of f(x) = x4 − 4x3 + 21

(vii) The graph of f(x) = x4 − 4x3 + 21 is given by Figure 2.17.

2. Consider the function f(x) = 3x−6x+2 . (a) Give the x-intercept and the y-intercept. (b) What is the

vertical asymptote? (c) What is the horizontal asymptote? (d) Using the first derivative, show thatf(x) is an increasing function. (e) Using the above information, sketch f(x).

Solution. We want to sketch the graph of f(x) = 3x−6x+2 .

(a) • For the x-intercept, we need to solve the equation f(x) = 0 for x. This is equivalent to solving3x−6x+2 = 0, which implies that 3x− 6 = 0. The latter equation gives 3x = 6 or x = 6

3 = 2. Sothe x-intercept is 2. Note that if a fraction A

B = 0, then the numerator A is equal to 0 (notthe denominator!).

• The y-intercept is f(0) = 3(0)−60+2 = −6

2 = −3.(b) Vertical asymptotes. First, recall that the line x = c is a vertical asymptote (VA) of the graph

of f(x) if at least one of the one-sided limits of f(x) as x approaches c is ∞ or −∞. A quick wayto find VA is to proceed as follows.

• Set the denominator equal to zero, and solve for x.• The solutions found in the previous step are the candidates for being a VA. To know whether

a candidate, say x = c, is a VA or not, we need to substitute x by c into f(x). There are twopossibilities:


- if f(c) is of the form 00 , then the line x = c is not a VA;

- if f(c) is of the form k0 , where k 6= 0, then the line x = c is a VA.

In the present case, the denominator is x + 2, and the equation x + 2 = 0 leads to x = −2.Substituting this into f(x), we get 3(−2)−6

−2+2 = −120 , which is something of the form k

0 . So the linex = −2 is a VA.

(c) Horizontal asymptotes. First, recall that the line y = b is a horizontal asymptote (HA) of thegraph of f(x) if the limit of f(x) as x goes to ∞ or −∞ is equal to b. So to find the HA, we needto find two limits: lim

x→−∞f(x) and lim

x→∞f(x).

• We have

limx→−∞

f(x) = limx→−∞

3x− 6

x+ 2= limx→−∞

3xx −

6x

xx + 2

x

= limx→−∞

3− 6x

1 + 2x

=3− 0

1 + 0=

3

1= 3.

• Likewise, limx→∞

f(x) = 3.

So the line y = 3 is a HA.

(d) Applying the quotient rule, we get

f ′(x) =3(x+ 2)− (3x− 6)(1)

(x+ 2)2=

3x+ 6− 3x+ 6

(x+ 2)2=

12

(x+ 2)2.

Since 12 is a positive number, and since (x+2)2 is a square (which is always greater than or equalto zero), it follows that f ′(x) is positive everywhere in the domain of f(x). This implies that f(x)is an increasing function. Another way to see this is to make the diagram of the sign of f ′(x).You will see that f ′(x) > 0 for all x in the domain of f(x).

(e) The graph of f(x) is given by Figure 2.18.

Figure 2.18: Graph of f(x) = 3x−6x+2

3. Consider the function f(x) = xx2+1 . Note that f ′(x) = 1−x2

(1+x2)2 and f ′′(x) = 2x(x2−3)(x2+1)3 . Determine

the following: (a) domain; (b) x- and y-intercepts; (c) the horizontal and vertical asymptotes; (d)the intervals of increase and decrease, local maximums and minimums; (e) intervals of concavity andinflection points. (f) Sketch the graph of f(x).

Solution.


(a) For the domain, let’s try to find numbers that make the denominator equal to zero. Settingx2 + 1 = 0, we get x2 = −1, which is impossible. So there is no number that makes the bottomequal to 0. This means that f is defined everywhere, so that D = R.

(b) The y-intercept is f(0) = 01 = 0. For the x-intercept, we need to solve the equation x

x2+1 = 0.This implies that x = 0, which is the required x-intercept.

(c) For the asymptotes, we have that the line y = 0 is a horizontal asymptote since limx→∞

f(x) = 0 and

limx→−∞

f(x) = 0. There is no vertical asymptote since the equation x2 + 1 = 0 has no solutions atall.

(d)-(e) For the intervals of increase, decrease, for the local max and local min values, and for theconcavity and inflection points, see the solution to Practice Problems 8 – 3.1 & 3.2 – Question 4.All those information are summarized in Figure 2.19. The graph of f is given by Figure 2.20.

Figure 2.19:

Figure 2.20: Graph of f(x) = xx2+1

Sections 3.4 & 3.5: Optimization problems

1. A compagny has determined that its total revenue (in dollars) for a product can be modelled byR(x) = −x3 + 450x2 + 52500x, where x is the number of units produced and sold. What productionlevel x will yield a maximum revenue?

Solution. We want to find x that maximixes R(x). To do this, we need to find the local extremumsof f(x).


• Derivative of R(x): R′(x) = −3x2 + 900x+ 52500.

• Critical Numbers: consider the equation −3x2 +900x+52500 = 0. Factoring out 3, the equationbecomes 3(−x2+300x+17500) = 0. Dividing by 3, this in turn becomes −x2+300x+17500 = 0.This latter equation seems difficult to factor further, so we are going to use the quadratic formula(−b±

√b2−4ac2a ). The solutions are

−300±√

3002 − 4(−1)(17500)2(−1)

=−300±

√160000

−2=−300± 400

−2.

So x = −300+400−2 = −50 or x = −300−400

−2 = 350. From the problem x ≥ 0 since x is the numberof untits (it is not possible to have a negative number of units!). So the solution x = −50 is to berejected. Thus x = 350.

• Intervals of increase and decrease. (See Figure 2.21.)

Figure 2.21: Sign of R′(x)

• Figure 2.21 tells us that the production x = 350 will yield a maximum revenue.

2. A farmer has 2000 ft of fencing and wants to fence off a rectangular field that borders a straight river.He needs no fence along the river. What are the dimensions of the field that maximize the area?

Solution. We are going to solve this step by step.

Step 1 Draw a picture (see Figure 2.22).

Figure 2.22:

Step 2 Introduce notation. Assign a symbol to the quantity that is to be maximized or minimized. Alsoselect symbols for other unknown quantities. Here we want to maximize the area that we denoteA. The other unknown are x and y as shown Figure 2.22.

Step 3 Express the quantity you want to maximize (or minimize) in terms of the other variables fromstep 2. Here we have A = xy, which is the formula for the area of a rectangle of sides x and y.

Step 4 If there is more than one variable, in step 3, use the given information to find relationshipsamong these variables (called constraint), and try to write A as a function of one variable.


Here the constraint is that the total length of the fencing is 2000 ft, that is, 2x+ y = 2000. (Notethat from the problem no fence is needed along the river. This is why the variable y appears onlyonce in the constraint.) From the constraint we have y = 2000 − 2x. Substituting this into theexpression for A, we get A = x(2000− 2x) = 2000x− 2x2.

Step 5 Find the interval and use the methods of Section 3.1 (the Closed Interval Method) or Section 3.3(The First Derivative Test or the Second Derivative Test) to find absolute max (or absolute min).To get the interval, the following two questions might be helpful. Question 1: What is the leastvalue you could make x? Answer: 0 (this is because x can’t be negative). Question 2: What isthe maximum value you could make x? Answer: 1000 (this uses all the fence for the depth andnone for the width). So the interval is [0, 1000]. Now the problem is to find the absolute maxvalue of the function A(x) = 2000x− 2x2 on the interval [0, 1000]. Since the interval is closed, wewill use the Closed Interval Method (this was introduced in the Solution to Practice Problems 8–Section 3.1 – Question 2).

• Critical numbers: We have A′(x) = 2000− 4x. Setting this to 0, we get −4x = −2000, thatis, x = 500.• Values of A at the critical numbers: A(500) = 500(2000− 1000) = 500000.• Values of A at the endpoints: A(0) = 0 and A(1000) = 1000(2000− 2000) = 0.• The absolute max is then 500000 located at x = 500. The corresponding value for y isy = 2000− 2(500) = 1000.

Conclusion: The maximum area is 500000 ft2 when x = 500 ft and y = 1000 ft.

3. A farmer wants to fence in a rectangular field beside a river. No fencing is required along the riverand the farmer’s neighbour will pay half of the cost of one of the sides perpendicular to the river. Iffencing costs 20 per linear meter and the field must have an area of 600 m2, what are the dimensionsof the field that will minimize the cost to the farmer?

Solution. Consider the same picture as before (see Figure 2.22). Let C denote the cost, which is thequantity we want to minimize.

• Equation. We have C = 20x+ 20y + 20x2 = 20x+ 20y + 10x = 30x+ 20y. This equation has two

variables, so we have to eliminate one of them by using the constraint, which states the area mustbe equal to 600, that is, xy = 600 or y = 600

x . Substituting this into the formula for the cost, weget

C = 30x+ 20× 600

x= 30

(x+ 20× 20

x

)= 30

(x+

400

x

).

So the function we want to minimize is C(x) = 30(x+ 400x )

• Interval. For the interval, we can’t use the argument of Question 1 as there is no constraint hereabout the total length of the fencing. Since x can’t be 0 here (otherwise the area would be 0,which would contradict the fact that the area is 600), and since there is no maximum value for x,the physical domain for x is the open interval (0,∞).

• Critical numbers. The first derivative is C ′(x) = 30(1 − 400x2 ) = 30

(x2−400x2

). Setting C ′(x) = 0,

we get x2 − 400 = 0, that is, x2 = 400. This latter equation gives x = −20 or x = 20. Rejectinga negative length leaves x = 20. So the only critical number is x = 20.

• Absolute min. To find the absolute min, we can’t use the Closed Interval Method here since theinterval is open. We are going to use the Second Derivative Test, which states that if f ′(c) = 0 andf ′′(c) > 0, then f has a local min at c. (The second part states that if f ′(c) = 0 and f ′′(c) < 0,then f has a local max at c.) The second derivative is C ′′(x) = 30

(800xx4

). Since C ′(20) = 0 and

C ′′(20) > 0, the function C has a local min at 20. That local min is actually an absolute local


min since C ′′(x) > 0 on the interval (0,∞) (which means that C is concave up on (0,∞)). So theabsolute min cost occurs when x = 20. The value of y corresponding to x = 20 is y = 600

20 = 30.

Hence the dimensions of the field that will minimize the cost to the farmer are x = 20 m and y = 30

m.

4. The volume V of a cylinder of height h and radius r is V = πr2h, whereas the area of the cylinder’ssurface, including top and bottom, is A = 2πr2 + 2πrh. Of all cylinders of volume V = 1, determinethe height and radius of the cylinder that has minimal surface area.

Solution. The quantity we want to minimize is the surface area A = 2πr2 +2πrh, and the constraintstates that the volume must be equal to 1, that is, πr2h = 1, or h = 1

πr2 . Substituting this into theexpression for A, we get

A = 2πr2 + 2πr(1

πr2) = 2πr2 +

2

r.

Since the radius r can’t be negative, and since r 6= 0 (otherwise, the volume would be 0, and this wouldcontradict the fact that the volume is 1), we have r > 0. So the interval for r is (0,∞). Now, we findthe critical numbers. The first derivative is

A′(r) = 4πr − 2

r2=

4πr3 − 2

r2.

Setting this to 0, we get 4πr3 − 2 = 0, that is, r3 = 12π . Taking the cube root, we have r = 3

√12π ,

which is the only critical number. To see that A has an absolute min at that number, we can use theSecond Derivative Test by arguing in the same way as we did in Question 2. Namely, one can easilysee that the second derivative, A′′(r) = 4π + 4r

r4 is positive on (0,∞). The value of h corresponding to

r = 3

√12π is

h =1

πr2=

1

π(

3

√12π

)2 .Thus to minimize the surface area, the radius should be r = 3

√12π and the height should be h =

1

π(

3√

12π

)2 .

5. Your ice cream cones are sold to unicorns for 50 cents each. At this price 1000 unicorns are willingto buy an ice cream cone. For every 5 cents decrease in price, there are 100 more unicorns willing tobuy your ice cream cones. What selling price will produce the maximum revenue and what will themaximum revenue be?

Solution. Let x be the price of one ice cream, and let R(x) be the corresponding revenue. ThenR(x) = N(x) × x, where N(x) is the number of ice creams sold. What is N(x)? From the problem,we have the following.

• For x = 50, N(50) = 1000 = 1000.

• For x = 45, N(45) = 1000 + 100

• For x = 40, N(x) = 1000 + 200 = 1000 + 100(2)

• For x = 35, N(x) = 1000 + 300 = 1000 + 100(3)

· · ·

So N(x) = 1000+ 100M(x), where M(x) = number of 5 cents decreases. If one thinks a little bit, onewill see that M(x) = 50−x

5 . This implies that

N(x) = 1000 + 100(50− x

5) = 1000 + 20(50− x) = 1000 + 1000− 20x = −20x+ 2000.


Therefore, R(x) = (−20x+ 2000)x = −20x2 + 2000x.We now proceed to find x that maximizes R(x). The derivative is R′(x) = −40x+200, and the criticalnumber is x = 50. The diagram of the sign of R′(x), which is easy to make, tells us that the sellingprice x = 50 will produce the maximum revenue.

6. If 1200 cm2 of material is available to make a box with a square base and an open top, find the largestpossible volume of the box.

Solution. Let V denote the volume of the box, which is the quantity we want to maximize. Let xdenote the length of side of the base (which is a square), and let y denote the height of the box. Wehave V = x× x× y = x2y. (Recall that the volume of a box of width a, length b, and height c is givenby the formula V = abc.) From the problem, the constraint is that the total surface is 1200, that is,the sum of areas of all faces (except the top) is equal to 1200, or x2 + xy + xy + xy + xy = 1200, thatis, x2 + 4xy = 1200. Solving this for y, we get y = 1200−x2

4x . Substituting this into the expression forV , we get

V = x2(1200− x2

4x

)=x(1200− x2)

4=

1200x− x3

4=

1

4(1200x− x3).

The derivative is V ′ = 14 (1200− 3x2). Setting V ′ = 0, we get 1200− 3x2 = 0, that is, x2 = 1200

3 = 400,or x = ±20. Rejecting a negative length leaves x = 20 and y = 1200−x2

4x = 1200−(20)24(20) = 1200−400

80 =80080 = 10. So the largest possible volume of the box is V = x2y = (20)210 = 400× 10 = 4000 cm3.

7. I am building a closed box with a square base and a fancy top. The material for the bottom and forthe sides costs 5 cents/cm2, while the material for the top costs 7 cents/cm2. The length (and thewidth, since it is square) of the base of the box is x, and the height of the box is h.

(a) Find a formula for the cost of this box in terms of x and h.

(b) Find the formula for the volume of the box in terms of x and h.

(c) If the volume must be 150cm3, express h in terms of x.

(d) Assume the volume of the box must be 150cm3. Find the value of x that should be used for thecheapest box.

Solution.

(a) We want to find a formula for the cost C.

• The area of the bottom is x2. Since the meterial for the bottom costs 5 cents/cm2, it followsthat: cost for the bottom = 5x2.

• The area of one side is xh. So the area of the four sides is 4xh. Since the meterial for thesides costs 5 cents/cm2, it follows that: cost for the sides = 4xh(5) = 20xh.

• The area of the top is x2. Since the meterial for the top costs 7 cents/cm2, it follows that:cost for the top = 7x2.

Adding up these, we get the cost of the box C = 5x2 + 20xh+ 7x2 = 12x2 + 20xh.

(b) The volume of the box is V = x2h.

(c) If V = 150, then x2h = 150, which implies that h = 150x2 .

(d) Substituting h into the formula for C, we get C(x) = 12x2+20x( 150x2 ) = 12x2+ 3000x . The interval

for x is (0,∞), while the derivative is C ′(x) = 24x− 3000x2 = 24x3−3000

x2 . Setting this equal to 0, wetget 24x3 − 3000 = 0 or x3 = 125. This latter equation implies that x = 3

√125 = 5, which is the

only critical number. The second derivative is C ′′(x) = 24 + 6000x3 . Since C ′(5) > 0, it follows by

the second derivatve test (for absolute extrema) that the absolute minimum of C(x) is at x = 5.So the value x = 5 is the one that should be used for the cheapest cost.


2.10 Solution to PB10

Section 4.1: Exponential Functions


(a) e2x = ex

Solution. First, recall the exponential rule saying that ex = ey if and only if x = y. That is,

Rule 1: ex = ey if and only if x = y.

Using this, the equation e2x = ex becomes 2x = x, which is equivalent to 2x− x = 0 or x = 0.

(b) ex2

= e4

Solution. This is equivalent to x2 = 4 or x = ±2.

(c) ex = 1

Solution. To use the rule, we need to write the righthand side of this equation on the form ey.Since e0 = 1, the original equation becomes ex = e0, and this implies that x = 0.

(d) e4x2 − e36 = 0

Solution. This implies that e4x2

= e36, which in turn implies that 4x2 = 36. Simplifying thislatter equation by 4, we get x2 = 9. So x = ±3.

(e) ex2

ex = e6

Solution. Recall:Rule 2: exey = ex+y.

Using this rule, the equation ex2

ex = e6 becomes ex2+x = e6. Using now Rule 1, we get x2+x = 6

or x2 + x− 6 = 0. Factoring this, we get (x− 2)(x+ 3) = 0, which implies that x = 2 or x = −3.

(f) ex2

e3x−2 = 1

Solution. Recall:

Rule 3:ex

ey= ex−y.

Using this rule, the original equation becomes ex2−(3x−2) = 1 or ex

2−3x+2 = e0 or x2−3x+2 = 0.Factoring this latter equation, we get (x− 1)(x− 2) = 0, which gives x = 1 or x = 2.

(g) (ex)x2

= ex

Solution. Recall:Rule 4: (ex)y = exy.

Using this rule, the equation (ex)x2

= ex becomes ex3

= ex. Using now Rule 1, we get x3 = x orx3 − x = 0. Factoring this, we get x(x2 − 1) = 0 or x(x − 1)(x + 1) = 0. So x = 0 or x = 1 orx = −1.

(h) (2x− 1)ex = 0

Solution. This implies that 2x − 1 = 0 or ex = 0. The former equation gives x = 12 , while the

latter is impossible since the exponential of any number is always positive–for every real numberx, ex > 0.

2. Find the following limits.


(a) limx→∞

e−2x

Solution. First, recall the Basic Limits. Let k be a positive number (k > 0). Then

(BL1) limx→−∞

ekx = 0 (BL2) limx→∞

ekx =∞ (BL3) limx→∞

e−kx = 0

(BL4) limx→−∞

e−kx =∞ (BL5) limx→−∞

xekx = 0 (BL6) limx→∞

xe−kx = 0.

By (BL3), we have that limx→∞

e−2x = 0.

(b) limx→∞

e4x

Solution. Using (BL2), we get limx→∞

e4x =∞.

(c) limx→−∞

e7x

Solution. By (BL1), we have limx→−∞

e7x = 0.

(d) limx→−∞

8e−3x


8e−3x = 8 limx→−∞

e−3x = 8(∞) =∞.

(e) limx→∞

2xex


2xex = 2 limx→∞

xex = 2(∞) =∞.

(f) limx→−∞

9xe0.5x


9xe0.5x = 9 limx→−∞

xe0.5x = 9(0) = 0. Here we used (BL5) to conclude

that limx→−∞

xe0.5x = 0.

(g) limt→∞

(t+ 1)e−0.75t

Solution. Using the algebra of limits and basic limits, we have

limt→∞

(t+ 1)e−0.75t = limt→∞

te−0.75t + e−0.75t = limt→∞

te−0.75t + limt→∞

e−0.75t = 0 + 0 = 0.

(h) limt→∞

5 + 6(2− 3t)e−10t

Solution. Again, by using the algebra of limits and basic limits, we have

limt→∞

5 + 6(2− 3t)e−10t = limt→∞

5 + limt→∞

6(2− 3t)e−10t = 5 + 6 limt→∞

(2− 3t)e−10t =

5 + 6 limt→∞

(2e−10t − 3te−10t) = 5 + 6(2 limt→∞

e−10t − 3 limt→∞

te−10t) =

5 + 6(2(0)− 3(0)) = 5 + 0 = 5.

3. The concentration of a certain drug in an organ t minutes after an injection is given by C(t) =

0.05− 0.04(1− e−0.03t) grams per cubic centimeter (g/cm3).

(a) What is the initial concentration of the drug (when t = 0)?

(b) What is the concentration 10 minutes after an injection? After 1 hour?

(c) What is the average rate of change of concentration during the first hour?

(d) What happens to the concentration of the drug in the long run (as t→∞)?

Solution.

(a) The initial concentration of the drug si C(0) = 0.05−0.04(1−e−0.03(0)) = 0.05−0.04(1−e0) = 0.05.


(b) • The concentration 10 minutes after an injection is C(10) = 0.05 − 0.04(1 − e−0.03(10)) =

0.05− 0.04(1− e−0.3) = 0.0396.• The concentration after 1 hour (that is, after 60 minutes) is C(60) = 0.05 − 0.04(1 −e−0.03(60)) = 0.05− 0.04(1− e−1.8) = 0.0166.

(c) Recall that the average of change of a quantity Q(t) during a period t ∈ [t1, t2] is given by theformula

A =Q(t2)−Q(t1)

t2 − t1.

Here the period is [0, 60]. So A = C(60)−C(0)60−0 = 0.0166−0.05

60 = −0.0005.

(d) We need to find the limit of C(t) as t goes to infinity. Using the same process as before (seeQuestion 2 – (h)), we have that

limt→∞

0.05− 0.04(1− e−0.03t) = 0.05− 0.04(1− 0) = 0.05− 0.04 = 0.01.

So the concentration of the drug in the long run tends to 0.01g/cm3.

Section 4.2: Logarithmic Functions

1. Find

(a) ln(e3)

Solution. Recall:Rule 5: ln(ex) = x for any real number x.

Using this rule, we get ln(e3) = 3.

(b) ln 3√e

Solution. We have ln 3√e = ln e

13 = 1

3 .

(c) ln( 1√e)

Solution. We have ln( 1√e) = ln( 1

e12) = ln(e−

12 ) = − 1

2 .

(d) eln 5

Solution. Recall:Rule 6: eln x = x for any positive number x.

Using this, we get eln 5 = 5.

2. Simplify the following expressions.

(a) lnx2019

Solution. Recall:Rule 7: ln(xr) = r lnx for any real number r.

Using this, we get lnx2019 = 2019 lnx.

(b) ln(x2y−7)

Solution. Recall:Rule 8: ln(xy) = lnx+ ln y.

Using this, we have that ln(x2y−7) = lnx2 + ln y−7 = 2 lnx− 7 ln y.


(c) ln( y8

x3 )

Solution. Recall:Rule 9: ln(

x

y) = lnx− ln y.

Using this, we get ln( y8

x3 ) = ln y8 − lnx3 = 8 ln y − 3 lnx.

(d) ln(x8e−x3

)

Solution. Using Rule 8 (and also Rules 7 & 5), we have that ln(x8e−x3

) = lnx8 + ln e−x2

=

8 lnx− x2.

(e) ln 3√x2 − 3x

Solution. We have ln 3√x2 − 3x = ln(x2 − 3x)

13 = 1

3 ln(x2 − 3x).

(f) ln(

(3x−1)4x2

(2x+1)5

)Solution. We have

ln

((3x− 1)4x2

(2x+ 1)5

)= ln((3x− 1)4x2)− ln(2x+ 1)5 =

ln(3x− 1)4 + lnx2 − ln(2x+ 1)5 = 4 ln(3x− 1) + 2 lnx− 5 ln(2x+ 1).


(a) ex = 3

Solution. Recall:Definition: ex = y if and only if x = ln y.

Using this, the equation ex = 3 implies that x = ln 3.

(b) lnx = 6

Solution. Again, by using the definition, the equation lnx = 6 implies that x = e6.

(c) ln( x20 ) = −1Solution. This implies that x

20 = e−1. Multiplying both sides by 20, we get x = 20e−1.

(d) e4x = 3

Solution. This implies that 4x = ln 3, which in turn implies x = ln 34 .

(e) 3 lnx = 1

Solution. Isolating first lnx, we get lnx = 13 . Isolating now x, we get x = e

13 .

(f) ln(2x− 3) = 1

Solution. This implies that 2x− 3 = e1, that is, 2x = e+ 3 or x = e+32 .

(g) x ln(x+ 2) = 0

Solution. We have x = 0 or ln(x+2) = 0. The latter equation implies that x+2 = e0 (remembere0 = 1) or x = 1− 2 = −1. So the solutions to the equation x ln(x+ 2) = 0 are: 0 and −1.


(a) ln(2x+ 1) = ln(x+ 5)

Solution. Recall:Rule 10: lnx = ln y if and only if x = y.

Using this, the equation ln(2x+1) = ln(x+5) implies 2x+1 = x+5. Solving this latter equation,we get x = 4.


(b) ln(x2 + x) = ln(2x+ 6)

Solution. This implies that x2 + x = 2x + 6 or x2 − x − 6 = 0. Factoring the latter, we get(x+ 2)(x− 3) = 0. So x = −2 or x = 3.

(c) ln(2x+ 1)− ln(x− 1) = 1

Solution. First, using Rule 9, we get ln( 2x+1x−1 ) = ln e. Applying now Rule 10, we get 2x+1

x−1 = e.Multiplying both sides by x− 1, we get 2x+1 = e(x− 1) or 2x+1 = ex− e or 2x− ex = −e− 1.Solving this for x, we get x(2− e) = −e− 1 or x = −e−1

2−e .

(d) 2e−3x + 5 = 19

Solution. Subtracting 5 from both sides, we get 2e−3x = 14. Dividing by 2, we get e−3x = 7.Using now the definition, we get −3x = ln 7 or x = ln 7

−3 .

5. An investment firm estimates that the value of its portfolio after t years is A million dollars, whereA(t) = 300 ln(t+ 3).

(a) What is the value of the account when t = 0?Solution. The value of the account when t = 0 is A(0) = 300 ln 3.

(b) How long does it take for the account to double its initial value?Solution. We need to solve the equation A(t) = 2A(0) for t. This is the same as 300 ln(t +

3) = 2A(0). Dividing both sides by 300, we get ln(t + 3) = 2A(0)300 = 2(300 ln 3)

300 = 2 ln 3. Soln(t+ 3) = 2 ln 3. Using the definition, we get t+ 3 = e2 ln 3 or

t = e2 ln 3 − 3 = (eln 3)2 − 3 = 32 − 3 = 9− 3 = 6.

Conclusion: it takes 6 years for the account to double its initial value.

Section 4.3: Differentiation of Exponential and Logarithmic Functions


(a) f(x) = 3ex

Solution. Recall: the derivative of ex is ex, that is,

Derivative of ex :d

dx[ex] = ex.

Using this, we have f ′(x) = ddx [3e

x] = 3 ddx [e

x] = 3ex.

(b) f(x) = 1 +√2x− ex

Solution. We have

f ′(x) =d

dx[1] +

d

dx[√2x]− d

dx[ex] = 0 +

d

dx[(2x)

12 ]− ex =

1

2(2x)−

12 (2)− ex =

(2x)−12 − ex =

1

(2x)12

− ex =1√2x− ex =

1− ex√2x√

2x.

(c) f(x) = x3ex

Solution. Using the product rule, we get

f ′(x) =d

dx[x3]ex + x3

d

dx[ex] = 3x2ex + x3ex = (3x2 + x3)ex.


(d) f(x) = x2

ex+3

Solution. Using the quotient rule, we get

f ′(x) =ddx [x

2](ex + 3)− x2 ddx [e

x + 3]

(ex + 3)2=

2x(ex + 3)− x2(ex)(ex + 3)2

=

2xex + 6x− x2ex

(ex + 3)2=x(2ex + 6− xex)

(ex + 3)2.


(a) f(x) = e3x

Solution. Recall:

Chain Rule for eu(x) :d

dx[eu(x)] =

d

dx[u(x)]eu(x) = u′(x)eu(x).

Using this, we get f ′(x) = ddx [3x]e

3x = 3e3x.

(b) f(x) = −e−x

Solution. We have

f ′(x) =d

dx[−e−x] = − d

dx[e−x] = − d

dx[−x]e−x = −(−1)e−x = e−x.

(c) f(x) = ex2+5x+1

Solution. We have

f ′(x) =d

dx[x2 + 5x+ 1]ex

2+5x+1 = (2x+ 5)ex2+5x+1.

(d) f(x) = e2x + e−2x

Solution. We have

f ′(x) =d

dx[e2x] +

d

dx[e−2x] =

d

dx[2x]e2x +

d

dx[−2x]e−2x = 2e2x − 2e−2x.

(e) f(x) = xex3

Solution. Using the product rule (and then the chain rule), we get

f ′(x) =d

dx[x]ex

3

+ xd

dx[ex

3

] = 1ex3

+ x(3x2)ex3

= ex3

+ 3x3ex3

= (1 + 3x3)ex3

.

(f) f(x) = e−2x

x2+5

Solution. Using the quotient rule (and then the chain rule), we get

f ′(x) =ddx [e

−2x](x2 + 5)− e−2x ddx [x

2 + 5]

(x2 + 5)2=−2e−2x(x2 + 5)− e−2x(2x)

(x2 + 5)2=

[−2(x2 + 5)− 2x]e−2x

(x2 + 5)2=

(−2x2 − 10− 2x)e−2x

(x2 + 5)2.

3. For the function f(x) = xe3x, find the intervals of increase and decrease. Also find the local maxi-mum and minimum values. Also find the intervals of concavity and the inflection points. Find theasymptotes, and sketch the graph of f(x).

Solution.


• Intervals of increase and decrease. First, we need to find the critical numbers.

- Using the product rule, we get

f ′(x) =d

dx[x]e3x + x

d

dx[e3x] = 1e3x + x(3e3x) = e3x + 3xe3x = (1 + 3x)e3x.

- Setting this to 0, we have (1 + 3x)e3x = 0, which implies that 1 + 3x = 0 or e3x = 0. Theformer equation is equivalent to 3x = −1 or x = −1

3 . The latter equation is impossible. So,x = −1

3 , is the only critical number.

- From Figure 2.23, it follows that f(x) is decreasing on (−∞,− 13 ) and increasing on (− 1

3 ,∞).

Figure 2.23:

• Local max and min. Figure 2.23 tells us that the sign of the derivative changes from − to+ at − 1

3 . This implies that f(x) has a local min at x = − 13 . The corresponding y value is

f(− 13 ) = −

13e

3(− 13 ) = − 1

3e−1. Again from Figure 2.23, one can see that there is no local max.

• Concavity.

- First, we need to find the second derivative.

f ′′(x) =d

dx[(1 + 3x)]e3x + (1 + 3x)

d

dx[e3x] = 3e3x + (1 + 3x)(3e3x) =

3e3x + (3 + 9x)e3x = (3 + 3 + 9x)e3x = (9x+ 6)e3x.

- Setting this to 0, we get (9x + 6)e3x = 0, which implies that 9x + 6 = 0 or e3x = 0. Theformer equation is equivalent to 9x = −6 or x = −6

9 = −23 , while the latter is impossible.

- Figure 2.24 tells us that f(x) is concave downward on (−∞,− 23 ) and concave upward on

(− 23 ,∞).

Figure 2.24:

• Inflection points. Since the concavity changes at − 23 , there is an inflection point at − 2

3 . Thecorresponding y value is f(− 2

3 ) = −23e−2. So (− 2

3 ,−23e−2) is the only inflection point of f(x).

• Asymptotes.


- Horizontal asymptotes. We need to find two limits:

limx→−∞

f(x) = limx→−∞

xe3x = 0,

andlimx→∞

f(x) = limx→∞

xe3x =∞.

So the line y = 0 is a horizontal asymptote when x goes to −∞. There is no horizontalasymptote when x goes to ∞ because the limit lim

x→∞f(x) is not a number.

- Vertical asymptotes. Usually, to find the vertical asymptotes we set the denominator to 0

and solve the resulting equation for x. Here f(x) = xe3x, and its denominator is just 1 whichis a constant (there is no x in there). Setting 1 = 0, we get an equation that has no solutionsat all. So there is no vertical asymptotes. Another way to see this is the fact that the domainof f(x) is R (f(x) is defined everywhere).

• Graph of f(x). This is shown in Figure 2.25.

Figure 2.25:


(a) f(x) = 3 lnx

Solution. Recall: the derivative of lnx is 1x . That is,

Derivative of lnx :d

dx[lnx] =

1

x.

Using this, we have f ′(x) = ddx [3 lnx] = 3 d

dx [lnx] = 3 1x = 3

x .

(b) f(x) = x2 lnx


f ′(x) =d

dx[x2] lnx+ x2

d

dx[lnx] = 2x lnx+ x2

1

x= 2x lnx+ x

(c) f(x) = ln xx3


f ′(x) =ddx [lnx]x

3 − lnx ddx [x

3]

(x3)2=

1xx

3 − lnx(3x2)

(x3)2=

x2 − 3x2 lnx

(x3)2=x2(1− 3 lnx)

x6=

1− 3 lnx

x4.



(a) f(x) = ln(7x)

Solution. Recall:

Chain Rule for lnu(x) :d

dx[ln(u(x))] =

ddx [u(x)]

u(x)=u′(x)

u(x).

Using this we have, f ′(x) =ddx [7x]

7x = 77x = 1

x .

(b) f(x) = ln(x2 + 3)

Solution. We have

f ′(x) =ddx [x

2 + 3]

x2 + 3=

2x

x2 + 3.

(c) f(x) = 5 ln(1 + ex)

Solution. We have

f ′(x) =d

dx[5 ln(1 + ex)] = 5

d

dx[ln(1 + ex)] = 5

ddx [1 + ex]

1 + ex= 5

0 + ex

1 + ex=

5ex

1 + ex.

(d) f(x) = x ln(x2 + 1)


f ′(x) =d

dx[x] ln(x2 + 1) + x

d

dx[ln(x2 + 1)] = (1) ln(x2 + 1) + x

ddx [x

2 + 1]

x2 + 1=

ln(x2 + 1) + x2x

x2 + 1= ln(x2 + 1) +

2x2

x2 + 1.

(e) f(x) = ln(2x+1)3x+1


f ′(x) =ddx [ln(2x+ 1)](3x+ 1)− ln(2x+ 1) ddx [3x+ 1]

(3x+ 1)2=

22x+1 (3x+ 1)− 3 ln(2x+ 1)

(3x+ 1)2=

2(3x+1)−3(2x+1) ln(2x+1)(2x+1)

(3x+ 1)2=

2(3x+ 1)− 3(2x+ 1) ln(2x+ 1)

(2x+ 1)(3x+ 1)2

(f) f(x) = ln(2x)e−3x

Solution. Using the product rule, we have

f ′(x) =d

dx[ln(2x)]e−3x + ln(2x)

d

dx[e−3x] =

2

2xe−3x + ln(2x)(−3e−3x) =

1

xe−3x − 3e−3x ln(2x) = (

1

x− 3 ln(2x))e−3x =

(1− 3x ln(2x))e−3x

x.

6. Find the absolute maximum and minimum of f(x) = ln(x+1)x+1 on the interval [0, 2].

Solution.


• Critical numbers.

f ′(x) =ddx [ln(x+ 1)](x+ 1)− ln(x+ 1) ddx [x+ 1]

(x+ 1)2=

1x+1 (x+ 1)− ln(x+ 1)

(x+ 1)2=

1− ln(x+ 1)

(x+ 1)2.

Setting this to 0, we get 1−ln(x+1)(x+1)2 = 0, which implies that 1 − ln(x + 1) = 0 or ln(x + 1) = 1 or

x + 1 = e1. This latter equation is equivalent to x = e − 1 ≈ 1.7182. Does this lie in the giveninterval? Yes. So x = e− 1 is a critical number.• Value of the function at the critical number.

f(e− 1) =ln(e− 1 + 1)

e− 1 + 1=

ln e

e=

1

e≈ 0.3678.

• Values of the function at the endpoints.

f(0) =ln 1

1=

0

1= 0 and f(2) =

ln 3

3≈ 0.3662.

• The absolute minimum is 0 when x = 0, and the absolute maximum is 1e when x = e− 1.

7. Use logarithmic differentiation to find the derivative of f(x) = (2x+ 1)2(x− 5x2)12 .

Solution. To find the derivative using the method of logarithmic differentiation, one can proceed asfollows.

step 1. Take logarithms of both sides of the expression for f(x) and simplify.

For f(x) = (2x+ 1)2(x− 5x2)12 , we have

ln f(x) = ln[(2x+ 1)2(x− 5x2)

12

]= ln(2x+ 1)2 + ln(x− 5x2)

12 =

2 ln(2x+ 1) +1

2ln(x− 5x2).

So ln f(x) = 2 ln(2x+ 1) + 12 ln(x− 5x2).

step 2. Use the chain rule to differentiate both sides of the simplified equation.

We haved

dx[ln f(x)] =

d

dx

[2 ln(2x+ 1) +

1

2ln(x− 5x2)

]=

2d

dx[ln(2x+ 1)] +

1

2

d

dx

[ln(x− 5x2)

]=

2ddx [2x+ 1]

2x+ 1+

1

2

ddx [x− 5x2]

x− 5x2= 2

2

2x+ 1+

1

2

1− 10x

x− 5x2=

4

2x+ 1+

1− 10x

2x− 10x2.

On the other side, ddx [f(x)] =

f ′(x)f(x) . So

f ′(x)

f(x)=

4

2x+ 1+

1− 10x

2x− 10x2.

step 3. Multiply both sides by f(x) to isolate f ′(x).

Multiplying the latter equation by f(x), we get

f ′(x) = f(x)

(4

2x+ 1+

1− 10x

2x− 10x2

)=

(2x+ 1)2(x− 5x2)12

(4

2x+ 1+

1− 10x

2x− 10x2

).



Section 5.1: Indefinite Integration


1.∫5dx

Solution. Let’s first recall the Table of Indefinite integrals.

(1)∫kdx = kx+ C for all constant k (2)

∫xrdx = 1

r+1xr+1 + C (r 6= −1)

(3)∫

1xdx = ln |x|+ C (for all x 6= 0) (4)

∫ekxdx = 1

kekx + C (k 6= 0)

Using rule (1) with k = 5, we get∫5dx = 5x+ C.

2.∫x8dx

Solution. Using rule (2) with r = 8, we get∫x8dx = 1

8+1x8+1 + C = 1

9x9 + C.

3.∫

1x3 dx

Solution. This integral does not fit into the table, but by using basic algebra, we can make it easier.Recall the formula

1

xm= x−m.

Using that formula, we get∫

1x3 dx =

∫x−3dx. This latter integral fits into the table (in fact, it

corresponds to∫xrdx with r = −3). Therefore∫1

x3dx =

∫x−3dx =

1

−3 + 1x−3+1 + C =

1

−2x−2 + C = − 1

2x2+ C.

Note. In order to do integral, it must fit into the integration table. So the idea is to try to manipulateand make it fit somehow.

4.∫

3√xdx

Solution. First recall the formulam√x = x

1m .

Using that formula, we get∫

3√xdx =

∫x

13 dx, which fits into the table (here r = 1

3 ). So∫3√xdx =

∫x

13 dx =

113 + 1

x13+1 + C =

143

x43 + C =

3

4x

43 + C.

5.∫

16√xdx

Solution. We have∫16√xdx =

∫1

x16

dx =

∫x−

16 dx =

1−16 + 1

x−16+1 + C =

156

x56 + C =

6

5x

56 + C.

6.∫

x3√xdx

Solution. First recall the formulaxn

xm= xn−m.

We have ∫x3√xdx =

∫x

x13

dx =

∫x1−

13 dx =

∫x

23 dx =

153

x53 + C =

3

5x

53 + C.


7.∫

4√x5dx

Solution. First recall the formulam√xn = x

nm .

Now we have ∫4√x5dx =

∫x

54 dx =

194

x94 + C =

4

9x

94 + C.

8.∫ 3√

x2

x4 dx

Solution. We have ∫ 3√x2

x4dx =

∫x

23

x4dx =

∫x

23−4dx =

∫x−

103 dx =

1

− 103 + 1

x−103 +1 + C =

1

− 73

x−73 + C = −3

7× 1

x73

= − 3

7x73

+ C.

9.∫e3xdx

Solution. Using rule (4), we get∫e3xdx = 1

3e3x + C.

10.∫

1e4x dx

Solution. Recall that 1ea = e−a. Using this, we have∫

1

e4xdx =

∫e−4xdx =

1

−4e−4x + C.

11.∫4xdx

Solution. First we recall the basic properties of integrals.

(a)∫kf(x)dx = k

∫f(x)dx. In words, this property says that if we have the integral of a constant

times a function, we can pull the constant outside.

(b)∫[f(x) + g(x)]dx =

∫f(x)dx+

∫g(x)dx. The integral of a sum is the sum of integrals.

(c)∫[f(x)− g(x)]dx =

∫f(x)dx−

∫g(x)dx. The integral of a difference is the difference of integrals.

(d) WARNING! The equality∫f(x)g(x)dx =

(∫f(x)dx

) (∫g(x)dx

)is not right in general. In other

words, the integral of a product in NOT equal to the product of integrals. The same remarkapplies to the quotient.

Applying the first property, we get∫4xdx = 4

∫xdx = 4x

2

2 + C = 2x2 + C.

12.∫ √

3xdx

Solution. We have∫ √3xdx =

∫ √3√xdx =

√3

∫ √xdx =

√3

∫x

12 dx =

√3x

32

32

+ C =2

3

√3x

32 + C.

13.∫(−3x2 + 5x− 1)dx

Solution. We have∫(−3x2 + 5x− 1)dx =

∫−3x2dx+

∫5xdx−

∫1dx = −3

∫x2dx+ 5

∫xdx−

∫1dx =

−3x3

3+ 5

x2

2− x+ C = −x3 + 5

2x2 − x+ C.

Note: When dealing with the sum or difference of integrals, you can add the constant C at the end.(Don’t add one constant for each integral.) This is because a constant times a number is still a constant,and the sum of constants is still a constant.


14.∫(x2 +

√x− 1)dx

Solution. We have∫(x2 +

√x− 1)dx =

∫x2dx+

∫ √xdx−

∫1dx =

∫x2dx+

∫x

12 dx−

∫1dx =

x3

3+x

32

32

− x+ C =1

3x3 +

2

3x

32 − x+ C.

15.∫

5xdx

Solution. First, we have∫

5xdx = 5

∫1xdx. Using now rule (3) from the table above, we get

∫5xdx =

5 ln |x|+ C.

16.∫(e−2x + 3− 2

x )dx

Solution. We have∫(e−2x + 3− 2

x)dx =

∫e−2xdx+

∫3dx− 2

∫1

xdx =

1

−2e−2x + 3x− 2 ln |x|+ C.

17.∫(3− 1

ex )dx

Solution. We have∫(3− 1

ex)dx =

∫3dx−

∫1

exdx =

∫3dx−

∫e−xdx = 3x− (

1

−1)e−x + C = 3x+ e−x + C.

18.∫ex(1− e−2x)dx

By multiplying, we get∫ex(1− e−2x)dx =

∫(ex − exe−2x)dx =

∫(ex − ex−2x)dx =

∫(ex − e−x)dx = ex − (

1

−1e−x) + C = ex + e−x + C.

19.∫(ln(ex

3

)− eln x2

)dx

Solution. First recall that ln(eA) = A for any number A. Also recall that elnB = B for any positivenumber B. Using this, we get∫

(ln(ex3

)− eln x2

)dx =

∫ln(ex

3

)dx−∫eln x

2

dx =

∫x3dx−

∫x2dx =

x4

4− x3

3+ C.

20.∫

4x−3√xdx

Solution. The idea is to split the function into two fractions using the following property: a−bc = a

c−bc .

We have ∫4x− 3√

xdx =

∫ (4x√x− 3√

x

)=

∫4x√xdx−

∫3√xdx =

4

∫x

x12

dx− 3

∫1

x12

dx = 4

∫x

12 dx− 3

∫x−

12 dx =

4x

32

32

− 3x

12

12

+ C = 4(2

3)x

32 − 3(

2

1)x

12 + C =

8

3x

32 − 6x

12 + C.


21.∫

2−√2t+

3√t2√

tdt

Solution. Again by splitting this, we get∫2−√2t+

3√t2√

tdt =

∫2√tdt−

∫ √2t√tdt+

∫ 3√t2√tdt =

∫2

t12

dt−∫ √

2√t√

tdt+

∫t23

t12

dt = 2

∫t−

12 dt−

∫ √2dt+

∫t16 dt =

2t12

12

−√2t+

t76

76

= 2(2

1)t

12 −√2t+

6

7t76 + C = 4t

12 −√2t+

6

7t76 + C.

22.∫2x(3− x−3)dx

Solution. We first need to distribute before taking the integral.∫2x(3− x−3)dx =

∫(6x− 2x−2)dx = 6

∫xdx− 2

∫x−2dx =

6x2

2− 2

x−1

−1+ C = 3x2 + 2x−1 + C.

23.∫(t+ 4)(2t+ 1)dt

Solution. Again, we first need to distribute.∫(t+ 4)(2t+ 1)dt =

∫(2t2 + t+ 8t+ 4)dt =

∫(2t2 + 9t+ 4)dt = 2

∫t2dt+ 9

∫tdt+

∫4dt =

2t3

3+ 9

t2

2+ 4t+ C =

2

3t3 +

9

2t2 + 4t+ C.

Section 5.2: Integration by Substitution.


1.∫(1− 2x)9dx

Solution. We will calculate this integral step by step.

Step 1 Pick a substitution u that “simplifies” the integral. Sometimes u is the “inside” of something. Forthis integral, we choose u = 1− 2x.

Step 2 Express dx in terms of du. We have that the derivative of u with respect to x is dudx = −2, so that

dx = du−2 .

Step 3 Rewrite the original integral only in terms of u and do it.∫(1− 2x)9dx =

∫(u)9

du

−2= −1

2

∫u9du = −1

2

u10

10+ C = − 1

20u10 + C.

Note: All x’s must be eliminated before you integrate.

Step 4 Translate back to x (by substituting u).∫(1− 2x)9dx = − 1

20(1− 2x)10 + C.


2.∫2x(x2 + 3)15dx

Solution. Let u = x2 + 3. Then dudx = 2x, so dx = du

2x . Therefore∫2x(x2 + 3)15dx =

∫2xu15

du

2x=

∫u15du =

u16

16+ C =

1

16(x2 + 3)16 + C.

3.∫x2(x3 − 7)6dx

Solution. Let u = x3 − 7. Then dudx = 3x2, so dx = du

3x2 . Therefore∫x2(x3 − 7)6dx =

∫x2u6

du

3x2=

1

3

∫u6du =

1

3

u7

7+ C =

1

21(x3 − 7)7 + C.

4.∫x√1− x2dx

Solution. Let u = 1− x2. Then dudx = −2x, so dx = du

−2x . Therefore∫x√1− x2dx =

∫x√udu

−2x= −1

2

∫u

12 du = −1

2

u32

32

+ C =

−1

2(2

3)u

32 + C = −1

3u

32 + C = −1

3(1− x2) 3

2 + C.

5.∫e1−xdx

Solution. If the integral contains an exponential function, it is often useful to substitute for theexponent. In this case, we choose u = 1− x so that du = −dx or dx = −du. We then have∫

e1−xdx =

∫eu(−du) = −

∫eudu = −eu + C = −e1−x + C.

6.∫x3ex

4

dx

Solution. We choose u = x4 and obtain du = 4x3dx so that dx = du4x3 . Then the integral becomes∫

x3ex4

dx =

∫x3eu

du

4x3=

1

4

∫eudu =

1

4eu + C =

1

4ex

4

+ C.

7.∫

xx2+3dx

Let u = x2 + 3. Then du = 2xdx so that dx = du2x . Thus, the integral becomes∫

x

x2 + 3dx =

∫x

u

du

2x=

1

2

∫1

udu =

1

2ln |u|+ C =

1

2ln |x2 + 3|+ C.

8.∫

x−1e−x2+2x

dx

Solution. First, by using the property 1ea = e−a, we have

∫x−1

e−x2+2xdx =

∫(x − 1)e−(−x

2+2x)dx =∫(x− 1)ex

2−2x. Now we choose u = x2 − 2x and obtain du = (2x− 2)dx so that dx = du2x−2 . Then the

integral becomes ∫x− 1

e−x2+2xdx =

∫(x− 1)eu

du

2x− 2=

∫(x− 1)eu

du

2(x− 1)=

∫eudu

2=

1

2

∫eudu =

1

2eu + C =

1

2ex

2−2x + C.


9.∫

ex

1+ex dx

Solution. Pick u = 1 + ex. Then du = exdx so that dx = dudx . Therefore,∫

ex

1 + exdx =

∫ex

u

du

ex=

∫1

udu = ln |u|+ C = ln |1 + ex|+ C.

10.∫e2x−e−2x

e2x+e−2x dx

Solution. We choose u = e2x + e−2x and obtain du = (2e2x − 2e−2x)dx so that dx = du2(e2x−e−2x) .

Then the integral becomes∫e2x − e−2x

e2x + e−2xdx =

∫e2x − e−2x

u

du

2(e2x − e−2x)=

∫1

u

du

2=

1

2

∫1

udu =

1

2ln |u|+ C =

1

2ln |e2x + e−2x|+ C.

11.∫ ln(3x)

x dx

Solution. Let u = ln(3x). Then du = 33xdx = 1

xdx so that dx = xdu. Thus, the integral becomes∫ln(3x)

xdx =

∫u

xxdu =

∫udu =

u2

2+ C =

(lnx)2

2+ C.

12.∫ −3x4

(x5+1)7 dx

Solution. First we have∫ −3x4

(x5+1)7 dx =∫−3x4(x5 + 1)−7dx. Let u = x5 + 1. Then du

dx = 5x4, sodx = du

5x4 . Therefore∫−3x4

(x5 + 1)7dx =

∫−3x4(u)−7 du

5x4= −3

5

∫u−7du = −3

5

(u−6

−6

)+ C =

1

10u−6 + C =

1

10(x5 + 1)−6 + C =

1

10(x5 + 1)6+ C.

13.∫ −4x+4

3√

(x2−2x+1)2dx

Solution. First we have∫−4x+ 4

3√

(x2 − 2x+ 1)2dx =

∫−4x+ 4

(x2 − 2x+ 1)23

dx =

∫(−4x+ 4)(x2 − 2x+ 1)−

23 dx.

Let u = x2 − 2x+ 1. Then dudx = 2x− 2, so dx = du

2x−2 . Therefore∫−4x+ 4

3√

(x2 − 2x+ 1)2dx =

∫(−4x+ 4)(x2 − 2x+ 1)−

23 dx =

∫(−4x+ 4)u−

23

du

2x− 2=

∫−2(2x− 2)u−

23

du

2x− 2=

∫−2u− 2

3 du = −2∫u−

23 du =

−2u13

13

+ C = −2(3

1

)u

13 + C = −6u 1

3 + C = −6(x2 − 2x+ 1)13 + C.

14.∫x√x+ 7dx

Solution. First we have∫x√x+ 7dx =

∫x(x+7)

12 dx. Let u = x+7. Then du

dx = 1, so dx = du1 = du.

Therefore∫x√x+ 7dx =

∫xu

12 du. It looks like it is not possible to eliminate all x’s. Since we want


to write the original integral only in terms of u, we can use the substitution formula to express x as afunction of u. This gives x = u− 7. Substituting x = u− 7 into

∫xu

12 du, we get∫

x√x+ 7dx =

∫xu

12 du =

∫(u− 7)u

12 du =

∫(u

32 − 7u

12 )du =

∫u

32 du− 7

∫u

12 du =

u52

52

− 7u

32

32

+ C =2

5u

52 − 7

(2

3

)u

32 + C =

2

5u

52 − 14

3u

32 + C =

2

5(x+ 7)

52 − 14

3(x+ 7)

32 + C.

15.∫

x√1+2x

dx

Solution. We have∫

x√1+2x

dx =∫

x

(1+2x)12dx =

∫x(1 + 2x)−

12 dx. Let u = 1 + 2x. Then du

dx = 2, so

dx = du2 . Therefore

∫x√

1+2xdx =

∫xu−

12du2 = 1

2

∫xu−

12 du. Again it is not possible to cross out x.

From the substitution formula, u = 1 + 2x, we have 2x = u − 1, that is, x = u−12 . Substituting this

into 12

∫xu−

12 du, we get∫

x√1 + 2x

dx =1

2

∫u− 1

2u−

12 du =

1

4

∫(u− 1)u−

12 du =

1

4

∫ (u

12 − u− 1

2

)du =

1

4

(∫u

12 du−

∫u−

12 du

)=

1

4

∫u

12 du− 1

4

∫u−

12 du =

1

4

u32

32

− 1

4

u12

12

+ C =1

4(2

3)u

32 − 1

4(2

1)u

12 + C =

1

6u

32 − 1

2u

12 + C =

1

6(1 + 2x)

32 − 1

2(1 + 2x)

12 + C.


Section 5.3: The Definite Integral and the Fundamental Theorem of Calculus

5.3.1. The Fundamental Theorem of Calculus

Evaluate the following definite integrals

1.∫ 1

−3 5 dx

Solution. A definite integral is an integral of the form∫ baf(x) dx. The numbers a and b are called

bounds of the integral. We can refer to a as the bottom (or lower) bound, and to b as the top (orupper) bound. Note that the result of a definite integral is a number, while that of an indefinite integral(that is, an integral of the form

∫f(x) dx without bounds) is a function or a family of functions. The

Fundamental Theorem of Calculus (FTC), gives a nice way to find a definite integral. It states that iff is a continuous on [a, b], then ∫ b

a

f(x) dx = F (b)− F (a),

where F is any antiderivative of f , that is, a function F such that F ′ = f . We will write [F (x)]ba forF (b)− F (a). That is, [F (x)]ba = F (b)− F (a). Note that when evaluating [F (x)]ba, we first plug-in thetop bound, and then the bottom bound. (Always respect that order!)


We come back to the integral∫ 1

−3 5dx. The function is f(x) = 5. Clearly an antiderivative of f isF (x) = 5x+ C, where C is a constant. So by the FTC, we have∫ 1

−35dx = [F (x)]51 = F (1)− F (−3) =

(5(1) + C)− (5(−3) + C) = 5 + C − (−15 + C) = 5 + C + 15− C = 20.

Note. For Definite Integrals, the C’s will be always cancelled out. So we don’t need to write C. Butfor indefinite integrals, we have to write C (see Practice Problems 11). The following shows how wewill take definite integrals from now on.∫ 1

−35 dx = [5x]1−3 = 5(1)− 5(−3) = 5 + 15 = 20.

When evaluating a definite integral, be careful with the signs! Add parentheses or brackets if necessary.

2.∫ 4

13x dx

Solution. We have ∫ 4

1

3x dx = 3

∫ 4

1

x dx = 3

[x2

2

]41

=

3

(42

2− 12

2

)= 3

(16

2− 1

2

)= 3

(15

2

)=

45

2.

3.∫ 9

41√xdx


4

1√xdx =

∫ 9

4

1

x12

dx =

∫ 9

4

x−12 dx =

[x

12

12

]94

=

[2√x]94= 2√9− 2

√4 = 2(3)− 2(2) = 6− 4 = 2.

4.∫ 2

1(−x2 + 3x− 1) dx


1

(−x2 + 3x− 1) dx =

∫ 2

1

−x2 dx+

∫ 2

1

3x dx−∫ 2

1

1 dx =

−∫ 2

1

x2 dx+ 3

∫ 2

1

x dx−∫ 2

1

1 dx =

−[x3

3

]21

+ 3

[x2

2

]21

− [x]21 = −

(8

3− 1

3

)+ 3

(4

2− 1

2

)− (2− 1) =

−(7

3

)+ 3

(3

2

)− (1) = −7

3+

9

2− 1 =

−14 + 27

6− 1 =

13

6− 1 =

13− 6

6=

7

6.

5.∫ 2

1

(1x2 − 4

x3

)dx

Solution. We have∫ 2

1

(1

x2− 4

x3

)dx =

∫ 2

1

(x−2 − 4x−3) dx =

∫ 2

1

x−2 dx− 4

∫ 2

1

x−3 dx =

[x−1

−1

]21

− 4

[x−2

−2

]21

=

[− 1

x

]21

− 4

−2[x−2

]21= −

[1

x

]21

+ 2

[1

x2

]21

=

−(1

2− 1

1

)+ 2

(1

4− 1

1

)= −

(−1

2

)+ 2

(−3

4

)=

1

2− 3

2=−22

= −1.


6.∫ 3

−2(x2 − 3) dx


−2(x2 − 3) dx =

[x3

3− 3x

]3−2

=

(33

3− 3(3)

)−((−2)3

3− 3(−2)

)=

(27

3− 9

)−(−83

+ 6

)= (9− 9)−

(−8 + 18

3

)= 0−

(10

3

)= −10

3.

7.∫ 4

14+6x√xdx


1

4 + 6x√x

dx =

∫ 4

1

(4

x12

+6x

x12

)dx =

∫ 4

1

(4x−

12 + 6x

12

)dx = 4

∫ 4

1

x−12 dx+ 6

∫ 4

1

x12 dx =

4

[x

12

12

]41

+ 6

[x

32

32

]41

= 4

(2

1

)[x

12

]41+ 6

(2

3

)[x

32

]41= 8

[x

12

]41+ 4

[(x

12 )3]41=

8[√x]41+ 4

[(√x)3]4

1= 8

(√4−√1)+ 4

((√4)3−(√

1)3)

=

8(2− 1) + 4(23 − 13) = 8(1) + 4(8− 1) = 8 + 4(7) = 8 + 28 = 36.

8.∫ 8

12+t3√t2dt


1

2 + t3√t2

dt =

∫ 8

1

(2

t23

+t

t23

)dt =

∫ 8

1

(2t−

23 + t

13

)dt =

2

∫ 8

1

t−23 dt+

∫ 8

1

t13 dt = 2

[t13

13

]81

+

[t43

43

]81

=

2

(3

1

)[t13

]81+

3

4

[t43

]81= 6(8

13 − 1

13 ) +

3

4

(8

43 − 1

43

)=

6(

3√8− 1

)+

3

4

((8

13

)4− 1

)= 6(2− 1) +

3

4

((2)4 − 1

)=

6(1) +3

4(16− 1) = 6 +

3

4(15) = 6 +

45

4=

24 + 45

4=

69

4.

9.∫ 2

0(2x− 3)(4x2 + 1)dx

Solution. First, we need to distribute.∫ 2

0

(2x− 3)(4x2 + 1)dx =

∫ 2

0

(8x3 + 2x− 12x2 − 3) dx =

8

∫ 2

0

x3 dx+ 2

∫ 2

0

x dx− 12

∫ 2

0

x2 dx−∫ 2

0

3 dx =

8

[x4

4

]20

+ 2

[x2

2

]20

− 12

[x3

3

]20

− [3x]20 =

8

(16

4− 0

4

)+ 2

(4

2− 0

2

)− 12

(8

3− 0

3

)− (6− 0) =

8(4) + 2(2)− 4(8)− 6 = 32 + 4− 32− 6 = −2.


10.∫ 4

0(4− x)

√xdx

Solution. Again, by distributing, we get∫ 4

0

(4− x)√xdx =

∫ 4

0

(4√x− x

√x) dx =

∫ 4

0

(4x

12 − xx 1

2

)dx = 4

∫ 4

0

x12 dx−

∫ 4

0

x32 dx =

4

[x

32

32

]40

−

[x

52

52

]40

= 4

(2

3

)[x

32

]40− 2

5

[x

52

]40=

8

3

(4

32 − 0

)− 2

5

(4

52 − 0

)=

8

3

(4

12

)3− 2

5

(4

12

)5=

8

3(2)3 − 2

5(2)5 =

8

3(8)− 2

5(32) =

64

3− 64

5= 64

(1

3− 1

5

)= 64

(5− 3

15

)=

128

15.

11.∫ 1

0e−2xdx

Solution. Recall the rule∫ekxdx = 1

kekx. Using this, we get∫ 1

0

e−2xdx =

[1

−2e−2x

]10

= −1

2

[e−2x

]10=

−1

2(e−2(1) − e−2(0)) = −1

2(e−2 − e0) = −1

2(e−2 − 1).

12.∫ ln 2

05e3xdx

Solution. We have ∫ ln 2

0

5e3xdx = 5

∫ ln 2

0

e3xdx = 5

[1

3e3x]ln 2

0

=5

3

[e3x]ln 2

0=

5

3

(e3 ln 2 − e0

)=

5

3

((eln 2)3 − 1

)=

5

3((2)3 − 1) =

5

3(7) =

35

3.

13.∫ 2

1(x2 − 3

x )dx

Solution. Recall the rule∫

1xdx = ln |x|. We have∫ 2

1

(x2 − 3

x)dx =

∫ 2

1

x2dx− 3

∫ 2

1

1

xdx =

[x3

3

]21

− 3 [ln |x|]21 =

(23

3− 13

3

)− 3 (ln |2| − ln |1|) =

(8

3− 1

3

)− 3(ln 2− ln 1) =

7

3− 3(ln 2− 0) =

7

3− 3 ln 2.

14.∫ e1x2−xx2 dx

Solution. We have ∫ e

1

x2 − xx2

dx =

∫ e

1

(x2

x2− x

x2

)dx =

∫ e

1

(1− 1

x

)dx =

[x− ln |x|]e1 = (e− ln |e|)− (1− ln |1|) = (e− 1)− (1− 0) = e− 1− 1 = e− 2.

5.3.2. Substitution in a Definite Integral


Evaluate the following definite integrals.

1.∫ 2

02x(x2 − 2)3dx

Solution. When evaluating a definite integral by substitution, two methods are possible.Method 1. Take the indefinite integral first and then evaluate it. Let us explain that with the integralof the question. We make the substitution u = x2 − 2. Then du

dx = 2x, so dx = du2x . Therefore the

indefinite integral is ∫2x(x2 − 2)3dx =

∫2xu3

du

2x=

∫u3du =

u4

4=

(x2 − 2)4

4.

Evaluating this from 0 to 2, we get∫ 2

0

2x(x2 − 2)3dx =

[(x2 − 2)4

4

]20

=

(((2)2 − 2)4

4

)−(((0)2 − 2)4

4

)=

(2)4

4− (−2)4

4=

16

4− 16

4= 0.

Method 2 (Usually Preferable). Change the bounds during substitution. Let us explain that with thesame integral. First we make the (same) substitution u = x2 − 2. Then du

dx = 2x, so dx = du2x . Then

we change the bounds.

When x = 0, u = (0)2 − 2 = −2 and when x = 2, u = (2)2 − 2 = 2.

(The new bounds are −2 and 2). Therefore∫ 2

0

2x(x2 − 2)3dx =

∫ 2

−2u3 du =

[u4

4

]2−2

=(2)4

4− (−2)4

4=

16

4− 16

4= 0.

For the following integrals, we will use the second method.

2.∫ 1

0(4t− 1)50 dt

Solution. We make the substitution u = 4t − 1. Then dudt = 4, so dt = du

4 . Now we change thebounds.

When t = 0, u = −1 and when t = 1, u = 4(1)− 1 = 3.

So the new bounds are −1 and 3. Therefore∫ 1

0

(4t− 1)50 dt =

∫ 3

−1u50

du

4=

1

4

∫ 3

−1u50du =

1

4

[u51

51

]3−1

=

1

204

[u51]3−1 =

1

204

((351 − (−1)51

)=

1

204

(351 + 1

).

3.∫ 1

03√26x+ 1 dx

Solution. We can rewrite that integral as∫ 1

03√26x+ 1 dx =

∫ 1

0(26x+1)

13 dx. Let u = 26x+1. Then

dudx = 26, so dx = du

26 . Now we change the bounds.

When x = 0, u = 1 and when x = 1, u = 27.

The integral becomes∫ 1

0

(26x+ 1)13 dx =

∫ 27

1

u13du

26=

1

26

∫ 27

1

u13 du =

1

26

[u

43

43

]271

=


1

26

(3

4

)[u

43

]271

=3

104

(27

43 − 1

43

)=

3

104

((27

13

)4− 1

)=

3

104

((3)4 − 1

)=

3

104(81− 1) =

3

104(80) =

30

13.

4.∫ 1

0xex

2

dx

Solution. Let u = x2. Then dudx = 2x so that dx = du

2x .

When x = 0, u = 02 = 0 and When x = 1, u = 12 = 1.


0

xex2

dx =

∫ 1

0

xeudu

2x=

1

2

∫ 1

0

eudu =1

2[eu]10 =

1

2(e− e0) = 1

2(e− 1).

5.∫ 1

213

1x2 e

1x dx

Solution. We make the substitution u = 1x so that du

dx = − 1x2 or dx = −x2du.

When x =1

3, u =

113

= 3 and When x =1

2, u =

112

= 2.


13

1

x2e

1x dx =

∫ 2

3

1

x2eu(−x2du) = −

∫ 2

3

eudu = −[eu]23 = −(e2 − e3) = e3 − e2.

6.∫ 2

1x2

(x3+1)2 dx

Solution. Let u = x3 + 1. Then dudx = 3x2 so that dx = du

3x2 .

When x = 1, u = 13 + 1 = 2 and When x = 2, u = 23 + 1 = 9.


1

x2

(x3 + 1)2dx =

∫ 9

2

x2

u2du

3x2=

1

3

∫ 9

2

1

u2du =

1

3

∫ 9

2

u−2du =

1

3

[1

−1u−1

]92

= −1

3

[1

u

]92

= −1

3

(1

9− 1

2

)= −1

3(− 7

18) =

7

54.

7.∫ 8

01√

1+2xdx

Solution. First we rewrite the integral as∫ 8

01√

1+2xdx =

∫ 8

0(1 + 2x)−

12 dx. Let u = 1 + 2x. Then

dudx = 2, so dx = du

2 .When x = 0, u = 1 and when x = 8, u = 17.

Therefore ∫ 8

0

1√1 + 2x

dx =

∫ 17

1

u−12du

2=

1

2

∫ 17

1

u−12 du =

1

2

[u

12

12

]171

=

1

2

(2

1

)[u

12

]171

=[u

12

]171

= 1712 − 1

12 =√17− 1.


8.∫ 2

0t2√1 + t3 dt

Solution. First we have∫ 2

0t2√1 + t3 dt =

∫ 2

0t2(1 + t3)

12 dt. We make the substitution u = 1 + t3.

Then dudt = 3t2, so dt = du

3t2 .

When t = 0, u = 1 and when t = 2, u = 9.

Therefore ∫ 2

0

t2√1 + t3 dt =

∫ 9

1

t2u12du

3t2=

1

3

∫ 9

1

u12 du =

1

3

[u

32

32

]91

=1

3

(2

3

)[u

32

]91=

2

9

(9

32 − 1

32

)=

2

9

((9

12

)3− 1

)=

2

9

((3)3 − 1

)=

2

9(27− 1) =

2

9(26) =

52

9.

9.∫ 2

1x+1√

x2+2x−1 dx


1x+1√

x2+2x−1 dx =∫ 2

1(x+ 1)(x2 + 2x− 1)−

12 dx. Let u = x2 + 2x− 1. Then

dudx = 2x+ 2, so dx = du

2x+2 .

When x = 1, u = 2 and when x = 2, u = 4 + 4− 1 = 7.

Therefore∫ 2

1

x+ 1√x2 + 2x− 1

dx =

∫ 7

2

(x+ 1)u−12

du

2x+ 2=

∫ 7

2

(x+ 1)u−12

du

2(x+ 1)=

∫ 7

2

u−12du

2=

1

2

∫ 7

2

u−12 du =

1

2

[u

12

12

]72

=1

2

(2

1

)[u

12

]72=[u

12

]72= 7

12 − 2

12 =√7−√2.

10.∫ −1−2

xx2+1dx

Solution. Let u = x2 + 1. Then dudx = 2x so that dx = du

2x .

When x = −2, u = (−2)2 + 1 = 5 and when x = −1, u = (−1)2 + 1 = 2.

The integral becomes ∫ −1−2

x

x2 + 1dx =

∫ 2

5

x

u

du

2x=

1

2

∫ 2

5

1

udu =

1

2[ln |u|]25 =

1

2(ln |2| − ln |5|) = 1

2(ln 2− ln 5).

11.∫ ln 4

0ex

5−ex dx

Solution. Let u = 5− ex. Then dudx = −ex so that dx = du

−ex .

When x = 0, u = 5− e0 = 5− 1 = 4 and when x = ln 4, u = 5− eln 4 = 5− 4 = 1.

The integral becomes ∫ ln 4

0

ex

5− exdx =

∫ 1

4

ex

u

du

(−ex)= −

∫ 1

4

1

udu =

− [ln |u|]14 = − (ln |1| − ln |4|) = −(0− ln 4) = ln 4.


12.∫ 4

0x√

1+2xdx


0x√

1+2xdx =

∫ 4

0x(1+2x)−

12 dx. Let u = 1+2x. Then du

dx = 2, so dx = du2 .


Therefore∫ 4

0x√

1+2xdx =

∫ 9

1xu−

12du2 . Solving the substitution formula for x, we get 2x = u− 1, that

is, x = u−12 . Substituting this into the integral, we get∫ 4

0

x√1 + 2x

dx =

∫ 9

1

(u− 1

2

)u−

12du

2=

1

4

∫ 9

1

(u− 1)u−12 du =

1

4

∫ 9

1

(u

12 − u− 1

2

)du =

1

4

∫ 9

1

u12 du− 1

4

∫ 9

1

u−12 du =

1

4

[u

32

32

]91

− 1

4

[u

12

12

]91

=

1

4

(2

3

)[u

32

]91− 1

4

(2

1

)[u

12

]91=

1

6(9

32 − 1

32 )− 1

2

(9

12 − 1

12

)=

1

6(27− 1)− 1

2(3− 1) =

1

6(26)− 1

2(2) =

13

3− 1 =

10

3.

5.3.3. Area Under a Curve

Find the area of the region that lies under the given curve y = f(x) over the indicated interval a ≤ x ≤ b.

1. y = x3, 1 ≤ x ≤ 2

Solution. First, recall that if f(x) is a continuous function and f(x) ≥ 0 on the interval [a, b], theregion under the curve y = f(x) over the interval a ≤ x ≤ b has area A given by the definite integral

A =

∫ b

a

f(x)dx.

Using this, the area under the curve y = x3 over the inverval 1 ≤ x ≤ 2 is given by

A =

∫ 2

1

x3dx =

[x4

4

]21

=24

4− 14

4=

16

4− 1

4=

15

4.

2. y =√8x+ 9, 0 ≤ x ≤ 2

Solution. Te area is given by the definite integral A =∫ 2

0

√8x+ 9dx. To evaluate this, we make the

substitution u = 8x+ 9 so that dudx = 8 or dx = du

8 .


The integral becomes

A =

∫ 2

0

√8x+ 9dx =

∫ 25

9

√udu

8=

1

8

∫ 25

9

u12 du =

1

8

[132

u32

]259

=

1

8

2

3

[u

32

]259

=1

4

1

3

(25

32 − 9

32

)=

1

12((25

12 )3 − (9

12 )3) =

1

12(53 − 33) =

1

12(98) =

49

6.


3. y = e4x, 0 ≤ x ≤ ln 3

Solution. The area is given by

A =

∫ ln 3

0

e4xdx =

[1

4e4x]ln 3

0

=

1

4

[e4x]ln 3

0=

1

4(e4 ln 3 − e4(0) = 1

4((eln 3)4 − e0) = 1

4(34 − 1) =

1

4(80) = 20.

4. y = x2ex3

, 0 ≤ x ≤ 1

Solution. The area is given by A =∫ 1

0x2ex

3

dx. To evaluate this, we make the substitution u = x3

so that dudx = 3x2 or dx = du

3x2 .



A =

∫ 1

0

x2ex3

dx =

∫ 1

0

x2eudu

3x2=

1

3

∫ 1

0

eudu =1

3[eu]10 =

1

3(e1 − e0) = 1

3(e− 1).

5. y = 45−2x , −2 ≤ x ≤ 1

Solution. The area is given by A =∫ 1

−24

5−2xdx. To calculate this, we make the substitution u = 5−2xso that du

dx = −2 or dx = du−2 .

When x = −2, u = 5− 2(−2) = 5 + 4 = 9 and when x = 1, u = 5− 2 = 3.


A =

∫ 1

−2

4

5− 2xdx =

∫ 3

9

4

u

du

(−2)=

4

−2

∫ 3

9

1

udu = −2 [ln |u|]39 =

−2(ln |3| − ln |9|) = −2(ln 3− ln 9) = 2(ln 9− ln 3) =

2(ln 32 − ln 3) = 2(2 ln 3− ln 3) = 2(ln 3) = 2 ln 3.

Section 5.4: Applying Definite Integration: Area Between two Curves

1. Find the area bounded above by y = 2x+ 5 and below by y = x3 on [0, 2].

Solution. Recall the formula for the area between two curves. If f and g are two continuous functionson [a, b] such that f(x) is above g(x) on [a, b]. That is, f(x) ≥ g(x) for all x in [a, b]. Then the areabetween the graph of f and that of g is given by the formula

A =

∫ b

a

[f(x)− g(x)] dx.

Note. In order to find the area, we need to determine which function is on top. Then the area is theintegral of the function on top minus the function on bottom. Here we know from the problem thaty = 2x+ 5 is on top. Again from the problem, we know that the interval is [0, 2]. So

A =

∫ 2

0

(2x+ 5− x3)dx = 2

∫ 2

0

x dx+

∫ 2

0

5 dx−∫ 2

0

x3 dx =

2

[x2

2

]20

+ [5x]20 −[x4

4

]20

= (22 − 0) + (5(2)− 0)− (24

4− 0) = 4 + 10− 4 = 10.


2. Find the area of the region enclosed by the curves y = x2 and y = x+ 2.

Solution. Let us solve this problem step by step.

Step 1. Find the x-coordinates of the intersection of the curves (set f(x) = g(x), and solve for x). Thisgives the bounds of integration. Let f(x) = x2 and g(x) = x + 2. Setting f(x) = g(x), we getx2 = x+ 2, that is, x2 − x− 2 = 0 or (x+ 1)(x− 2) = 0. So x = −1 or x = 2, and therefore theinterval is [−1, 2].

Step 2. Which function is on Top? To answer this question, we can pick a number, say c, in the intervaland plug-in into both f(x) and g(x). If f(c) > g(c) then f is on top. If g(c) > f(c) then g is ontop. Here we can pick for example 0, which is clearly between −1 and 2. Plug-in: f(0) = 0 andg(0) = 2. Since g(0) > f(0), it follows that g is on top.

Step 3. Set-up and solve. Since g is on top, the required area is the integral of g minus f on the interval[−1, 2]. Specifically,

A =

∫ 2

−1x+ 2− x2 dx =

∫ 2

−1x dx+

∫ 2

−12 dx−

∫ 2

−1x2 dx =

[x2

2

]2−1

+ [2x]2−1 −[x3

3

]2−1

=

(22

2− (−1)2

2

)+ (2(2)− 2(−1))−

(23

3− (−1)3

3

)=(

4

2− 1

2

)+ (4 + 2)−

(8

3+

1

3

)=

3

2+ 6− 3 =

3

2+ 3 =

3 + 6

2=

9

2.

3. Find the area of the region enclosed by the curves y = x2 − 8 and y = −x2 + 10. Include a sketch ofthe relevant region as part of your solution.

Solution. Let f(x) = x2 − 8 and g(x) = −x2 + 10. Setting f(x) = g(x), we get x2 − 8 = −x2 + 10,that is, 2x2 − 18 = 0 or 2(x2 − 9) = 0, that is, 2(x+ 3)(x− 3) = 0. So x = −3 or x = 3, and thereforethe interval is [−3, 3]. To see which function is on top, pick 0, which is clearly a number between −3and 3. We have f(0) = −8 and g(0) = 10. So g is on top since g(0) > f(0). Therefore the area is

A =

∫ 3

−3

[(−x2 + 10)− (x2 − 8)

]dx =

∫ 3

−3(−2x2 + 18) dx =

[−2x

3

3+ 18x

]3−3

=

(−54

3+ 54

)−(54

3− 54

)= −108

3+ 108 = 108

(−1

3+ 1

)= 108(

2

3) = (36)(2) = 72.

For the sketch, see Figure 2.26.

Figure 2.26: Graphs of f(x) = x2 − 8 and g(x) = −x2 + 10

mathematics103–appliedcalculusi solutionsmanualuregina.ca/~pso748/math103_sol.pdf · 2020. 4....

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