CompanionLectureNotesforMSU’sMTH132 · CHAPTER1. FUNCTIONSANDLIMITS 1.5TheLimitofaFunction...

Companion Lecture Notes for MSU’s MTH 132

Compiled by: Ryan Maccombs

Contents

Contents i

1 Functions and Limits 1Forest for the Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 The Tangent and Velocity Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.5 The Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.6 Calculating Limits Using the Limit Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.7 The Precise Definition of a Limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.8 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2 Derivatives 23Forest for the Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.1 Derivatives and Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.2 The Derivative as a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.3 Differentiation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.4 Derivatives of Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.7 Rates of Change in the Natural and Social Sciences . . . . . . . . . . . . . . . . . . . . . . 472.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.8 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.9 Linear Approximations and Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3 Applications of Differentiation 65Forest for the Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663.1 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 673.2 The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 713.3 Derivatives and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753.4 Limits at Infinity; Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . 803.5 Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 823.7 Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 873.8 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 923.9 Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

CONTENTS

4 Integrals 101Forest for the Trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1024.1 Areas and Distances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103Appendix E - Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1064.2 The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1094.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1124.4 Indefinite Integrals and Net Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1164.5 The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

5 Applications of Integration 1235.1 Area Between Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1245.5 Average Value of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

ii

Chapter 1

Functions and Limits

CHAPTER 1. FUNCTIONS AND LIMITS

Forest for the TreesThe main topics for MTH 132 are slope at a point of various functions and area under a curve (by this wemean area between a function and the x-axis).

A problem we can solve:Find the slope of f(x) =

x

2+1 at the point x = 1.

x

y

Another problem we can solve:Find the area under f(x) =

x

2+ 1 between

x = 0 and x = 2.

x

y

So for linear functions we know how to solve the key questions of calculus. However as we soon as we moveto the next easiest function, quadratics, we start to have issues.

A problem we can’t solve:

Find the slope of f(x) =x2

2+ 1 at the point

x = 1.

x

y

Another problem we can’t solve:

Find the area under f(x) =x2

2+ 1 between

x = 0 and x = 2.

x

y

In order to solve these problems we need to develop a theory of limits. In 1.4 we will get a taste for usinglimits to compute slope at a point. We will spend the rest of Chapter 1 further developing the theory of limits.

2

1.4. THE TANGENT AND VELOCITY PROBLEMS

1.4 The Tangent and Velocity Problems

Section Objective(s):

• Compute slopes of secant lines and use them to estimate the slope of a tangent line.

• Understand instantaneous velocity and how tangent lines relate.

Secant Line Tangent Line

Definition(s) 1.4.1. The average rate of change of y = f(x) between P (x1, f(x1)) and Q(x2, f(x2))

is given by

A.R.o.C. =f(x2)− f(x1)

x2 − x1

Graphically this looks like:

PICTURE

Definition(s) 1.4.2. If t has units of time and f(t) describes the movement of an object

and has units of distance then the average velocity of the object between

t1 and t2 is given by

vave =f(t2)− f(t1)

t2 − t1

3


Remark 1.4.3. Typically the slope of the tangent line (or instantaneous velocity when applicable) is well

approximated by the slope of a secant line over a small interval (the smaller the

better).

Pictures:

Example 1.4.4. If a rock is thrown upward on the planet Mars with a velocity of 10m/s, its height in meterst seconds later is well approximated by y = 10t− 2t2.

(a) Find the average velocity over the time interval [1, 2].

(b) Find the average velocity over the time interval [0, 1].

(c) Using part (a) and (b) approximate the instantaneous velocity at t = 1.

4

1.4. THE TANGENT AND VELOCITY PROBLEMS

Example 1.4.5. Suppose f(x) = 2 sin πx+ 3 cosπx.

(a) Find the slope of the secant line joining P (1, f(1)) to Q(12, f(1

2)).

(b) Find the slope of the secant line joining P (1, f(1)) to R(34, f(3

4)).

(c) Find the slope of the secant line joining P (1, f(1)) to S(56, f(5

6)).

(d) Approximate the slope of the tangent line at P .

(e) Is there a way to get a better approximation? (explain)

5


1.5 The Limit of a Function


• Have an intuitive idea of the definition of a limit.

• Find the limits (two-sided, left, and right) of the piecewise defined function given algebraically orgraphically.

• Calculate infinite limits and detect vertical asymptotes.

Definition(s) 1.5.1. Suppose f(x) is defined when x is near the

number a . Then we write:

limx→a

f(x) = L

if we can make the values of f(x) arbitrarily close

to L by taking x to be sufficiently close to a (on either side of

a) but not equal to a.

Definition(s) 1.5.2 (Left-hand limit). We write

limx→a−

f(x) = L

if we can make the values of f(x) arbitrarily close to L by taking x

to be sufficiently close to a and x less than a.

Definition(s) 1.5.3 (Right-hand limit). We write

limx→a+

f(x) = L

if we can make the values of f(x) arbitrarily close to L by taking x

to be sufficiently close to a and x greater than a.

6

1.5. THE LIMIT OF A FUNCTION

Theorem 1.5.4.

limx→a

f(x) = L if and only if(

limx→a−

f(x) = L and limx→a+

f(x) = L

)

Definition(s) 1.5.5. The line x = a is called a vertical asymptote of the curve

y = f(x) if at least one of the following statements is true:

limx→a+

f(x) =∞ limx→a−

f(x) =∞ limx→a

f(x) =∞

limx→a+

f(x) = −∞ limx→a−

f(x) = −∞ limx→a

f(x) = −∞

That is f can be made arbitrarily large (or small with−∞ ) by taking x sufficiently close

to a, but not equal to a.

Note: Vertical asymptotes most often occur at x = a when f(x) =g(x)

h(x)is fully simplified and we have

h(a) = 0 and g(a) 6= 0 .

7


Example 1.5.6. Use the given graph of f to state the valueof each quantity, if it exists. If it does not exist, explain why.

x

y

(a) limx→2−

f(x)

(b) limx→2+

f(x)

(c) limx→2

f(x)

(d) limx→4

f(x)

(e) f(2)

(f) f(4)

Example 1.5.7. Draw an example of a graph of a function f(x) that satisfies:

limx→4−

f(x) = 1, limx→4+

f(x) =∞, and f(4) = 2.

8

1.5. THE LIMIT OF A FUNCTION

Example 1.5.8. Evaluate the following limits for the function f(x) =

3x− 2 if x < 1

0 if x = 1

3− x if x > 1

(a) limx→1−

f(x)

(b) limx→1+

f(x)

(c) limx→1

f(x)

(d) limx→2

f(x)

Example 1.5.9. Find the vertical asymptotes of the function y =x2 + 2x+ 1

x+ x2

9


1.6 Calculating Limits Using the Limit Laws


• Understand the eight limit laws and use these to calculate more complicated limits.

• Recall old algebraic techniques of factoring and multiplying by a conjugate.

• Utilize the Squeeze Theorem to determine more limits.

Theorem 1.6.1 (Limit Laws). Suppose that c is a constant and the limits

limx→a

f(x) and limx→a

g(x) exist. Then:

1. limx→a

[f(x) + g(x)] = limx→a

f(x) + limx→a

g(x)

2. limx→a

[cf(x)] = c limx→a

f(x)

3. limx→a

[f(x)g(x)] = limx→a

f(x) limx→a

g(x)

4. limx→a

[f(x)

g(x)

]=

limx→a

f(x)

limx→a

g(x)provided lim

x→ag(x) 6= 0

5. limx→a

c = c

6. limx→a

x = a

7. limx→a

xn = an when appropriate.

8. limx→a

[f(x)n] =[limx→a

f(x)]n

when appropriate.

Example 1.6.2. Evaluate the limit limx→0

2x2 − x3 + x

10

1.6. CALCULATING LIMITS USING THE LIMIT LAWS

Theorem 1.6.3 (Direct Substitution Property). If f is a polynomial or a rational function and a is in the

domain of f , then

limx→a

f(x) = f(a)

Example 1.6.4. Evaluate the limit, if it exists limx→1

x2 + 2x− 3

x2 − 1

Example 1.6.5. Evaluate the limit, if it exists limh→0

√9 + h− 3

h

Theorem 1.6.6 (Cancellation). If f(x) = g(x) when x 6= a, then

limx→a

f(x) = limx→a

g(x)

provided the limit exists.

11


Theorem 1.6.7. If f(x) ≤ g(x) when x is near a (except possibly at a) and the limits of f and g both

exist as x approaches a, then

limx→a

f(x) ≤ limx→a

g(x)

Theorem 1.6.8 (Squeeze Theorem). If f(x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a) and

limx→a

f(x) = limx→a

h(x) = L

then

limx→a

g(x) exists, and limx→a

g(x) = L

Example 1.6.9. Use the Squeeze Theorem to find limx→0

[x2 cos

(2

x

)]

12

1.7. THE PRECISE DEFINITION OF A LIMIT

1.7 The Precise Definition of a Limit


• Review absolute value inequalities and properties of absolute values.

• Explain the precise definition of a limit.

• Practice using the precise definition of the limit to formally calculate two-sided limits

Theorem 1.7.1 (Old algebra).

|AB| = |A||B|

Theorem 1.7.2 (Old algebra). If k > 0 then

|x− a| < k if and only if −k < x− a < k

Which has the equivalent geometric statement:

x is less than distance k away from a .

Example 1.7.3. Write a distance statement for the inequality |4x− 5| < 6

Definition(s) 1.7.4. Let f be a function defined on some open interval that contains the number a, except

possibly at a itself. Then we say that the limit of f(x) as x approaches a is L, and we write

limx→a

f(x) = L

if for every number ε > 0 there is a number δ > 0 such that

if 0 < |x− a| < δ then |f(x)− L| < ε.

13


Intuitively that means: You can always find a small

enough interval for x that contains and is

centered at a to keep f(x) within distance ε close to

L.

x

y f(x)

L

L+ ε

L− ε

a

Example 1.7.5. Prove that limx→1

[3x− 1] = 2 using the formal definition of a limit.

Remark 1.7.6 (Strategy for solving problems). Find a δ based on ε solving the absolute value inequality.

Check your work (NOT OPTIONAL).

14

1.7. THE PRECISE DEFINITION OF A LIMIT

Remark 1.7.7 (Check your work for linear functions). If f(x) = mx+ b then δ = ε/|m| is a good

choice.

Remark 1.7.8 (δ for quadratic functions). First check if you have a perfect square , other-

wise start with δ ≤ 1 .

Example 1.7.9. Given ε = 0.1, find the largest value of δ in the proof of limx→5

(x2 − 10x+ 41) = 16.

Example 1.7.10. Prove that limx→2

x2 = 4 using the formal definition of a limit.

15


Example 1.7.11. Setup the absolute value inequalities in the proof that limx→π/2

sinx = 1.

16

1.8. CONTINUITY

1.8 Continuity


• Detect when a function is continuous and when it is discontinuous.

• Use continuity to quickly evaluate limits.

• Explain and classify the different types of discontinuities.

• Understand the statement of the Intermediate Value Theorem

• Apply the Intermediate Value Theorem to mathematically prove two functions intersect on a setinterval.

Definition(s) 1.8.1. A function f is continuous at a number a if and only

if

limx→a

f(x) = f(a)

Note additional subtleties : This requires

1. f(a) to be defined

2. limx→a

f(x) to exist.

Definition(s) 1.8.2. A function f is continuous on an interval if it is

continuous at each point in that interval.

Theorem 1.8.3. Any rational function is continuous whenever it is defined ;

that is, it is continuous on its domain.

Note: Polynomials are special case of this (and are always continuous).

Theorem 1.8.4. Trigonometric functions and root functions are also continuous everywhere in their

domain.

17


Example 1.8.5. Evaluate limx→0

x2 + 3x− 4

x2 − 1

Example 1.8.6. Consider the graph of f(x) to the right on the

domain [−2, 5]. Determine where the function is continuous.

Theorem 1.8.7. If f and g are continuous at a and c is constant, then the following are also continuous

at a.

1. f + g 2. f − g 3. cf

4. fg 5.f

g, if g(a) 6= 0

Theorem 1.8.8. If g is continuous at a and f is continuous at g(a) , then the composite function

f ◦ g given by (f ◦ g)(x) = f(g(x)) is continuous at a .

18

1.8. CONTINUITY

Definition(s) 1.8.9. If f is defined near a (except perhaps at a), we say that f is

discontinuous at a (or f has a discontinuity at a) if f is not

continuous at a.

Definition(s) 1.8.10.

1. A removable discontinuity is a discontinuity that can be

‘removed’ by redefining f at just a single number .

2. An infinite discontinuity another name for a vertical asymptote.

3. A jump discontinuity is where the function is discontinuous because it

‘jumps’ from one value to another .

Pictures

Example 1.8.11. f(x) =x3 − 1

x− 1has a removable discontinuity at x = 1. Find a function g that agrees with

f for x 6= 1 and is continuous at 1.

19


Theorem 1.8.12 (Intermediate Value Theorem (IVT)). Suppose that f is continuous on the closed

interval [a, b] and let N be any

number between f(a) and f(b), where f(a) 6= f(b). Then there exists a number c ∈ (a, b) such that

f(c) = N .

Picture:

Remark 1.8.13. The intermediate value theorem states that a continuous function takes on every

intermediate value between the function values f(a) and f(b).

Example 1.8.14. Use the Intermediate Value Theorem to show that there is a root of the function:

f(x) = x4 + x− 3 on the interval (1, 2).

20

1.8. CONTINUITY

Example 1.8.15. Explain why f(x) =

−2

3− xif x 6= 3

3 if x = 3

is discontinuous at x = 3. Write the largest

interval on which f(x) is continuous.

Example 1.8.16. Show that |x| is continuous everywhere.

21


Example 1.8.17. Explain why the function f(x) =

√1 +

1

xis continuous at every number in its domain.

State the domain of f .

Example 1.8.18. Prove that the equation cosx = x3 has at least one solution. What interval is it in?

22

Chapter 2

Derivatives

CHAPTER 2. DERIVATIVES

Forest for the Trees

Now that we have a good understanding about the subtitles of limits we can now formally compute slope at a

point or even figure out the slope at every point on the graph. In sections 1 and 2 we will do just that.

Starting in section 3 we learn tricks for calculating the derivative (slope at a point) for complicated functions.

For these sections (3, 4, and 5) we can pretend we are Batman adding new tools and weapons to our utility belt.

In section 6, 7, and 8 we unleash our new found tools to help us defeat some real world problems and help us

get ready for chapter 3, Applications of Differentiation.

2.1 2.2

2.3

2.4

2.5

2.6 2.8

2.7

2.9

2.4

>

>

>

>

>

>

>

>

>

>

>>

>>

Ch 1>

Ch 1>

Chapter 2 Map

24

2.1. DERIVATIVES AND RATES OF CHANGE

2.1 Derivatives and Rates of Change


• Use limits to calculate slopes of tangent lines or instantaneous velocity

• Calculate the derivative of a function at a specific point.

• Determine tangent line equations

Definition(s) 2.1.1. The derivative of a function f at a number c, denoted by f ′(c), is the

number

f ′(c) = limh→0

f(c+ h)− f(c)

h

if the limit exists. An equivalent formulation is

f ′(c) = limx→c

f(x)− f(c)

x− c

Remark 2.1.2. The tangent line to the graph of y = f(x) at the point (c, f(c)) is the line through

(c, f(c)) whose slope is equal to f ′(c) .

Remark 2.1.3. If f(t) measures distance of a moving object, and t is time, then the velocity

(or instantaneous velocity ) of the moving object, denoted v(t), is the limit of

the average velocities (as defined in Section 1.4).

v(t) = lims→t

f(s)− f(t)

s− t

25


Example 2.1.4. Explain why f(x) = |x| is not differentiable at x = 0.

Example 2.1.5. Are the following functions differentiable at x = 2? Why or why not?

(a) f(x) =

x if x ≤ 2

3 if x > 2

26

2.1. DERIVATIVES AND RATES OF CHANGE

(b) f(x) =

x2 if x < 2

4x− 4 if x ≥ 2

(c) f(x) =

x2 if x < 2

4 if x ≥ 2

27


Example 2.1.6. Assuming a function f(x) is differentiable at x = c, come up with a general equation for the

tangent line to f at c.

28

2.2. THE DERIVATIVE AS A FUNCTION

2.2 The Derivative as a Function


• Given a function sketch the graph of its derivative.

• Calculate the formula for the derivative given a function.

• Investigate how a function can fail to be differentiable.

• Associate higher derivatives with acceleration and jerk.

Definition(s) 2.2.1. The derivative of a function f , denoted by f ′ , is the

function

f ′(x) = limh→0

f(x+ h)− f(x)

h

Another common notation is to writedf

dxor

d

dxf(x) instead of f ′(x). The derivative at x = c in this

notation is writtendf

dx

∣∣∣∣x=c

.

Definition(s) 2.2.2. A function f is differentiable at c if f ′(c) exists. It is

differentiable on the interval (a, b) if it is differentiable at every

number in (a, b).

Theorem 2.2.3. If f is differentiable at c, then f is continuous at c.

How Can a Function Fail to be Differentiable (at a point c)?

• f is discontinuous at c

PICTURE

29


• limh→0−

f(x+ h)− f(x)

h6= lim

h→0+

f(x+ h)− f(x)

h

PICTURES

• limx→c|f ′(x)| =∞

PICTURES

Definition(s) 2.2.4. The second derivative of f is the derivative of f ′(x), denoted

by f ′′(x) ord2f

x2.

In general, the nth derivative, denoted by f (n)(x) ordnf

dxn, is the derivative of f (n−1)(x).

Remark 2.2.5. If f(t)measures position/distance of a moving object, then the acceleration

of the object, a(t), is the second derivative of f , and the first derivative of the velocity,

v(t).

That is:

a(t) = v′(t) = f ′′(t)

30

2.2. THE DERIVATIVE AS A FUNCTION

Example 2.2.6. Compute the derivative using the limit definition for the following functions:

(a) f(x) = 3x2 + x− 8

(b) f(x) =5

x− 3

(c) f(x) =1

x2

31


Example 2.2.7. Is the function f(x) =

2x+ 1 if x < 0

x2 + 1 if x ≥ 0

differentiable at x = 0? Why or why not?

Example 2.2.8. The graph of a function f(x) is shown on the left. Use it to sketch a graph of f ′(x).

x

y

x

y

32

2.3. DIFFERENTIATION FORMULAS

2.3 Differentiation Formulas


• Learn when to use each of the differentiation formulas

• Understand how select differentiation formulas can be proven.

Theorem 2.3.1 (Constant, Linear, and Power Rules). Take c and n to be any constants

d

dx(c) = 0

d

dx(x) = 1

d

dx(xn) = nxn−1

Pictures for constant and linear

Theorem 2.3.2 (Constant Multiple, Sum, and Difference Rules). Take c to be a constant and f, g both

differentiable functions, then:

d

dx(cf(x)) = cf ′(x)

d

dx(f(x) + g(x)) = f ′(x) + g′(x)

d

dx(f(x)− g(x)) = f ′(x)− g′(x)

33


Theorem 2.3.3 (Product, Quotient). Take f, g both differentiable functions, then:

d

dx(f(x)g(x)) = f ′(x)g(x) + f(x)g′(x)

d

dx

(f(x)

g(x)

)=f ′(x)g(x)− f(x)g′(x)

[g(x)]2

Remark 2.3.4. The most powerful theorem we currently have isd

dx(xn) = nxn−1 (the power rule...

pun intended). You will see that it gets used in nearly every problem. The proof of it is very complicated

and is typically broken down into steps

1. n is a positive integer (proof is in this section, need binomial theorem)

2. n is a negative integer (proof in this section, need quotient rule)

3. n is a rational number (need implicit differentiation, section 2.6)

4. n is any real number (need knowledge of logarithmic differentiation, MTH133)

Example 2.3.5. Differentiate x2 + 2x− 3√

x

34

2.3. DIFFERENTIATION FORMULAS

Example 2.3.6. Find the equation of the tangent line to the curve y =2x

x+ 1a the point (1, 1).

Example 2.3.7. For what values of x does f(x) = x3 + 3x2 + x+ 3 have a horizontal tangent line?

35


Example 2.3.8. Find the velocity and acceleration of the position function s(t) =3

2− x

Example 2.3.9. Find the equations of the tangent lines to the curve y =x− 1

x+ 1that are parallel to the line

x− 2y = 2

Example 2.3.10. Let f(x) =

x2 if x ≤ 2

mx+ b if x > 2

. Find the values ofm and b that make f differentiable

everywhere.

36

2.4. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS

2.4 Derivatives of Trigonometric Functions


• Prove some important trig limits

• Use these trig limits to prove trig derivatives

Theorem 2.4.1.

limx→0

sinx

x= 1 lim

x→0

cosx− 1

x= 0

Idea of Proof that limx→0

sinx

x= 1

Only doing proof if x > 0

37


More generally,

Theorem 2.4.2. If limx→a

f(x) = 0 then

limx→a

sin(f(x))

f(x)= 1

From these we get many new types of limit problems. Lets try a few before moving on to derivatives.

Example 2.4.3. Find the limit limx→0

sin 5x

x


sin 2x

x2 + x

Example 2.4.5. Find the limit limt→0

tan 3t

sin t

38



cosx− 1

sinx


sin(x2)

x


sin(x− 1)

x2 + x− 2

Theorem 2.4.9.

(a)d

dx(sin(x)) = cos(x)

(b)d

dx(cos(x)) = − sin(x)

39


Before trying a formal proof. Let’s try to sketch the derivative of sin(x) and cos(x) to verify that this make

intuitive sense.

2π −3π2−π −π

2π2

π 3π2

2π

f(x) = sin x

2π −3π2−π −π

2π2

π 3π2

2π

f ′(x) =?

2π −3π2−π −π

2π2

π 3π2

2π

g(x) = cos x

2π −3π2−π −π

2π2

π 3π2

2π

g′(x) =?

Proof of (a).

40


Example 2.4.10. Derive formulas for the derivatives of the following functions.

(a) Put to memory: d

dx(tan(x))

(b) Put to memory: d

dx(sec(x))

(c) Extra: d

dx(csc(x))

(d) Extra: d

dx(cot(x))

41


2.5 The Chain Rule


• Identify when a function can be decomposed into a composition of two functions.

• Understand the statement of the chain rule and how to apply it.

Theorem 2.5.1. If g is differentiable at x and f is differentiable at g(x), then the composition F =

f ◦ g defined by F (x) = f(g(x)) is differentiable at x and F ′ is given by

F ′(x) = f ′(g(x)) · g′(x)

In Leibniz notation, the same thing can be written as

dF

dx=dF

dg· dgdx

Remark 2.5.2. Although the Leibniz notation should not literally be thought of as a fraction, it still gives

a good way to remember the chain rule: it looks like the dg’s just cancel.

Remark 2.5.3. If y = f(x) is differentiable, then by combining the chain rule and the power rule, we

can differentiate powers of f :d

dx(yn) = n · yn−1 · dy

dx

Or, written in terms of f(x), this would be

d

dx(f(x)n) = n · [f(x)]n−1 · f ′(x)

42

2.5. THE CHAIN RULE

Example 2.5.4. Calculate the derivatives of the following functions by first writing them as a composition of

simpler functions, and then applying the chain rule:

(a) f(x) =(7x3 + 2x2 − x+ 3

)5

(b) g(x) = sin(3x2 + 5x+ 8

)

(c) h(x) = x tan(2√x)

+ 7

43


Example 2.5.5. Prove the quotient rule by writingf(x)

g(x)as f(x) · 1

g(x)and using the product rule and chain

rule.

Example 2.5.6. Recall that f is an even function if f(−x) = f(x) for all x, and that f is an odd function

if f(−x) = −f(x) for all x. Use the chain rule to prove that the derivative of an even function is an odd

function.

44

2.5. THE CHAIN RULE

Example 2.5.7. A table of values for f, g, f ′, and g′ is given:

x f(x) g(x) f ′(x) g′(x)

1 3 2 4 6

2 1 8 5 7

3 7 2 7 9

(a) If h(x) = f(g(x)), find h′(1).

(b) If H(x) = g(f(x)), find H ′(1).

Example 2.5.8. The graphs of two functions, f and g, are pictured below.

0 1 2 3 4 5 6 7 8 9 10 11

1

2

3

4

5

6

7

8

9

10

11

g(x)

f(x)

Let h(x) = f(g(x)), and p(x) = g(f(x)). Compute the following derivatives, or explain why they do not

exist:

(a) h′(2)

(b) h′(7)

(c) p′(1)

45


Example 2.5.9. Calculate the derivatives of the following functions:

(a) y = sin (x cos(x))

(b) y =

√x+√x

(c) F (x) =(4x− x2

)100

46

2.7. RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES

2.7 Rates of Change in the Natural and Social Sciences


• Use differentiation to solve real world problems related to physics.

Definition(s) 2.7.1. The instantaneous rate of change of y = f(x)

with respect to x is the slope of the tangent line (a.k.a. derivative). Using Leibniz notation, we

write:dy

dx= lim

∆x→0

∆y

∆x

As was brought up in Section 2.1, the units for dy/dx are the units for y divided by the

units for x .

Remark 2.7.2 (From Section 2.2). If s = f(t) is the position function of an object that is moving in a

straight line, then v(t) = s′(t) represents the velocity at time t. Also, a(t) = v′(t) = s′′(t) is

the acceleration of the object at time t.

Remark 2.7.3. Some common phrases and their mathematical interpretations

• When is the object at rest (or stand still)?

• When is the object moving forward/backward?

• When is the speeding up/down?

• Find the total distance traveled

• (Gravity Problems) When does an object achieve its maximum height?

47


Example 2.7.4. A particle moves according to the law of motion s(t) = cos(πt/4) with 0 ≤ t ≤ 10, where

t is measured in seconds and s in feet.

(a) Find the velocity at time t.

(b) What is the velocity after 3 seconds?

(c) When is the particle at rest?

(d) When is the particle moving in the positive direction?

(e) Find the total distance traveled in the first 8 seconds?

(f) When is the particle speeding up? When is it slowing down?

48

2.7. RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES

Example 2.7.5. The height (in meters) of a projectile shot vertically upward from a point 15 m above ground

level with an initial velocity of 10 m/s is h = 15 + 10t− 5t2 after t seconds.

(a) Find the velocity after 2 seconds.

(b) When does the projectile reach its maximum height?

(c) What is the maximum height?

(d) When does it hit the ground?

(e) With what velocity does it hit the ground?

49


Example 2.7.6 (FS14 E1). The figure below shows the velocity v(t) of a particle moving on a horizontal

coordinate line, for t in a closed interval [0, 10].

t

1 2 3 4 5 6 7 8 9 10

v(t)

Fill in the following blanks. No partial credit available. No work needed. Use interval notation where

appropriate.

(a) The particle is moving forward during:

(b) The particle’s speed is increasing during:

(c) The particle has positive acceleration during:

(d) The particle has zero acceleration during:

(e) The particle achieves its greatest speed at:

(f) The particle stands still for more than an instant during:

50

2.6. IMPLICIT DIFFERENTIATION

2.6 Implicit Differentiation


• Recognize the need for implicit differentiation.

• Find the slopes of various curves by applying implicit differentiation.

Definition(s) 2.6.1. Implicit Differentiation is a method of differentiating both sides of an equation

with respect to x and then solving the resulting equation for y′.

Remark 2.6.2. Be careful that you are applying power, product, quotient, and chain rules correctly.

Remark 2.6.3. Khan Academy has 7 videos and practice problems all about implicit differentiation.

Feel free to check them out at:

https://www.khanacademy.org/math/differential-calculus/taking-derivatives

Example 2.6.4. Find dy/dx of x3 + y3 = 1 using implicit differentiation.

51

https://www.khanacademy.org/math/differential-calculus/taking-derivatives


Example 2.6.5. Use implicit differentiation to find an equation of the tangent line to the curve

y sin 2x = x cos 2y at the point (π/2, π/4).

Example 2.6.6. Find dy/dx by implicit differentiation of tan(x− y) =y

1 + x2.

52

2.6. IMPLICIT DIFFERENTIATION

Example 2.6.7.

(a) Find y′ if x3 + y3 = 6xy.

(b) Find and equation of the tangent line at the point (3, 3).

(c) At what point in the first quadrant is the tangent line horizontal?

53


Remark 2.6.8. Let’s practice implicit differentiation with 3 variables: x, y, and t. Where x = x(t) and

y = y(t), that is x and y depend on t. This will lead us nicely into related rates problems.

Example 2.6.9. Suppose you have the following information:

169 = x(t)2 + y(t)2 y′(t) = −3 x(a) = 5 x(t) ≥ 0 y(t) ≥ 0

Find x′(a).

54

2.8. RELATED RATES

2.8 Related Rates


• Translate sentences into mathematical equations.

• Apply implicit differentiation and the chain rule to solve many types of related rates problems.

Remark 2.8.1 (General Technique for solving Related Rate Problems).

(i) Relating given quantities with an equation .

(ii) Differentiating the equation (using implicit differentiation and the chain rule).

(iii) Solving for some rate of change (derivative).

Remark 2.8.2. Geometric/Trig formulas you are expected to know

• Circles

C = πd d = 2r A = πr2

• Triangles

A =1

2bh c2 = a2 + b2 (for right ∆s)

sin θ =O

H(for right ∆s) cos θ =

A

H(for right ∆s) tan θ =

O

A=

sin θ

cos θ(for right ∆s)

• Rectangles (Squares)

P = 2l + 2w A = lw

• Spheres

V =4

3πr3 SA = 4πr2

• Cones

V =h

3(area of base)

• Cylinders

V = h(area of base)

55


Example 2.8.3. A 13 foot ladder is leaning against a wall, and is sliding down, with the top of the ladder

moving downwards along the wall at 3 ft/sec. How fast is the bottom of the ladder moving away from the

wall when the bottom is 5 ft from the base of the wall?

Example 2.8.4. A cone-shaped tank is filling with water at a constant rate

of 9 ft3/min. The tank is 10 ft tall, and has a base radius of 5 ft. How fast is

the water level rising when the water is 6 ft deep?10

5

56

2.8. RELATED RATES

Example 2.8.5. A sphere is growing, its volume increasing at a constant rate of 10 in3 per second. Let r(t),

V (t), and S(t), be the radius, volume, and surface area of the sphere at time t. If r(1) = 2, then compute:

(a) r′(1)

(b) S ′(1)

Example 2.8.6. A circle is growing, its radius (in inches) given by r(t) =√t for t in seconds. How fast is

the area growing at time t = 4?

57


Example 2.8.7. A balloon is rising vertically above a field, and a person 500

feet away from the spot on the ground underneath the balloon is watching it,

measuring the angle of inclination, θ. When the angle is π/4 radians, the angle

is increasing at 110

radians per minute. At that moment, how fast is the balloon

rising? 500 ftperson

balloon

θ

Example 2.8.8. Let f(x) = x−x2 = x(1−x), and let θ be the angle between the positive x-axis and the line

joining the point (x, f(x)) with the origin. At what rate is θ changing, with respect to x, when x = 1− 1√3?

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

(x, f(x))

θ

58

2.8. RELATED RATES

Example 2.8.9. A person is lifting a weight with a pulley. The pulley is 25 feet

off the ground, the rope is 30 feet long, and the person is holding the end of

the rope 5 feet off the ground. If the person is walking backwards (away from

the pulley) at a constant rate of 5 ft/sec, how fast is the weight rising when the

person is 10 feet from the spot on the ground directly under the pulley?

pulley

Example 2.8.10. A spotlight on the ground shines on a wall 12 meters away. If a man 2 meters tall walks

from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the

building decreasing when he is 4 m from the building?

59


2.9 Linear Approximations and Differentials


• Utilize the tangent line or differentials to estimate how a function is changing around a specificpoint.

Remark 2.9.1. Linear approximation and tangent line approximation are two names for using the

equation of a tangent line to approximate a function


L(x) = f(x) + f ′(a)(x− a)

is called the linearization of f at a.

Note: compare this to the equation of a tangent line through (a, f(a)):

y − f(a) = f ′(a)(x− a)

Remark 2.9.3. An equivalent notion to linearization is differentials. Consider the definitions:

dy = f ′(x)dx

Where dy and dx are considered variables in their own right.


∆x = x− a ∆y = f(x+ ∆x)− f(x)

Remark 2.9.5.

dy ≈ ∆y L(x) ≈ f(x)

60

2.9. LINEAR APPROXIMATIONS AND DIFFERENTIALS

Example 2.9.6.

(a) Find the linearization L(x) of the function f(x) =√x at a = 9

(b) Use the linearization to approximate√

10

Example 2.9.7.

(a) Find the differential dy of y = cos(πx)

(b) Evaluate dy for x = 1/3 and dx = −0.02

61


Example 2.9.8.

(a) Find the linearization L(x) of the function f(x) = sinx at a = π/4

(b) Use the linearization to approximate sin(11π/40)

Example 2.9.9.

(a) Find the differential dy of y =√x2 + 8

(b) Evaluate dy for x = 1 and dx = 0.02

62

2.9. LINEAR APPROXIMATIONS AND DIFFERENTIALS

Example 2.9.10. Use linear approximation to estimate 3√

1001

Example 2.9.11. Use linear approximation to estimate1

4.002

63

Chapter 3

Applications of Differentiation

CHAPTER 3. APPLICATIONS OF DIFFERENTIATION


In Chapter 3 we apply our new found differentiation skills on various applications. The two main topics are

curve sketching (3.5) and optimization problems (3.7). Once we have mastered these sections we will be able

to solve problems such as:

Example: Give a detailed sketch of: y =x√x2 + 1

Example: A rain gutter is to be constructed from a metal sheet of width 30cm by bending up one-third of the

sheet on each side through an angle θ. How should θ be chosen so that the gutter will carry the maximum

amount of water?

Some more minor topics for this chapter will include: The Mean Values Theorem, Newton’s Method, and

Anti-Derivatives.

3.1

3.9

3.4

3.8

3.2

3.3

3.5

3.7>

>

>

>

>

Ch 2>

Ch 2>

Ch 2>

Ch 2>

Chapter 3 Map

66

3.1. MAXIMUM AND MINIMUM VALUES

3.1 Maximum and Minimum Values


• Explain the Extreme Value Theorem.

• Use the Closed Interval Method to identify absolute maxima and minima.

Definition(s) 3.1.1. Let c be a number in the domain D of a function f . Then f(c) is the

• absolute (global) maximum value of f on D if f(c) ≥ f(x) for all x

in D.

• absolute (global) minimum value of f onD if f(c) ≤ f(x) for all x in

D.

• local maximum value of f on D if f(c) ≥ f(x) for all x near c.

• local maximum value of f on D if f(c) ≤ f(x) for all x near c.

Maximums and minimums are often referred to as extreme values.

Pictures:

Remark 3.1.2. The book uses “near c” to mean technically that the statement is true in some open interval

containing c. Sometimes these definitions can make you crazy.

67


Theorem 3.1.3 (Extreme Value Theorem (EVT)). If f is continuous on a closed interval [a, b],

then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some

numbers c and d in [a, b].

Definition(s) 3.1.4. A critical number of a function f is a number c in the domain of

f such that either f ′(c) = 0 or f ′(c) does not exist.

Remark 3.1.5. If f has a local maximum or minimum at c, then c is a critical number of f .

Theorem 3.1.6. To find the absolute maximum and minimum values of a continuous function f on a

closed interval [a, b]:

1. Find the values of f at the critical numbers of f in (a, b).

2. Find the values of f at the endpoints (a and b).

3. The largest of the values from above is the absolute maximum value; the smallest is the

absolute minimum value.

Remark 3.1.7. In Section 3.3 we will find a way to classify local maximums and minimums on any domain!

(not just closed intervals)

Example 3.1.8. Find the absolute maximum an minimum values of f(x) = x3 − 12x+ 1

on the interval [1, 4]

68

3.1. MAXIMUM AND MINIMUM VALUES

Example 3.1.9. Sketch a graph of a function f that is continuous on [1, 5] and has all of the following

properties

• An absolute minimum at 2

• An absolute maximum at 3

• A local minimum at 4

Example 3.1.10. Sketch a graph of a function f who has domain [−2, 4] that has an absolute maximum but

no local maximum.

69


Example 3.1.11. Find the critical numbers for the function f(x) =x− 1

x2 + 4

Example 3.1.12. Find the absolute maximum an minimum values of f(t) = 2 cos t+ sin(2t) on the interval

[0, π/2]

70

3.2. THE MEAN VALUE THEOREM

3.2 The Mean Value Theorem


• State the Mean Value Theorem and draw pictures to help us understand its meaning.

• Identify points on the correct interval that satisfy the Mean Value Theorem.

Theorem 3.2.1 (Rolle’s Theorem). Let f(x) be a function which satisfies the following three properties:

(1) f(x) is continuous on the interval [a, b]

(2) f(x) is differentiable on (a, b)

(3) f(a) = f(b)

Then there is a number c in (a, b) such that f ′(c) = 0 .

Remark 3.2.2. The conclusion of Rolle’s Theorem says that if the function values agree at the endpoints,

then there is a place in between where the tangent line is horizontal.

71


Theorem 3.2.3 (Mean Value Theorem (MVT)). Let f(x) be a function which satisfies the following

two properties:

(1) f(x) is continuous on the interval [a, b]

(2) f(x) is differentiable on (a, b)

Then there is a number c in (a, b) such that

f ′(c) =f(b)− f(a)

b− a

Remark 3.2.4. The conclusion of The Mean Value Theorem says that there is a place in the interval where

the tangent line is parallel to the secant line between the endpoints.

Remark 3.2.5. Notice that Rolle’s Theorem and The Mean Value Theorem tell you that “there exists” a

number c with certain properties, but neither theorem tells you what that the value of c is, or how to find it.

Picture:

Corollary 3.2.6. If f ′(x) = 0 for all x in an interval (a, b), then f(x) must be

constant on (a, b).

Corollary 3.2.7. If f ′(x) = g′(x) for all x in an interval (a, b), then f(x) = g(x) + c

for some constant c.

72

3.2. THE MEAN VALUE THEOREM

Example 3.2.8. For each of the following functions (on the given domain), tell whether the hypotheses of

the Mean Value Theorem are satisfied. If so, try to find the value of c guaranteed by the theorem.

(a) f(x) = |x| on the domain [−3, 3]

(b) f(x) = x2 + 3x+ 5 on the domain [0, 1]

(c) f(x) = 1xon the domain [1, 3]

73


Example 3.2.9. Prove that the equation x3 + x+ 1 = 0 has exactly one solution.

74

3.3. DERIVATIVES AND GRAPHS

3.3 Derivatives and Graphs


• Utilize the derivative to determine when a function is increasing or decreasing.

• Use the second derivative to determine when a function is concave up or down.

Theorem 3.3.1.

(a) If f ′(x) > 0 on (a, b), then f(x) is increasing on (a, b).

(b) If f ′(x) < 0 on (a, b), then f(x) is decreasing on (a, b).

Theorem 3.3.2 (First Derivative Test). Suppose that f(x) is a function and that c is a

critical point (or “critical number ”) of f(x).

(a) If f ′(x) changes from positive to negative at x = c, then f(x) has a local

maximum at x = c.

(b) If f ′(x) changes from negative to positive at x = c, then f(x) has a local

minimum at x = c.

(c) If f ′(x) does not change sign at x = c, then f(x) has neither a local maximum

nor a local minimum at x = c.

Pictures:

75


Example 3.3.3. For the following functions, find the intervals on which it is increasing and decreasing, and

find where the local maximum and local minimum values occur.

(a) f(x) = 2x3 + 3x2 − 36x on the domain (−∞,∞)

(b) f(x) =x

x2 + 1on the domain (−∞,∞)

(c) f(x) = cos2(x)− 2 sin(x) on the domain [0, 2π]

76


Definition(s) 3.3.4. If the graph of f lies above all of its tangents on an interval I , then it is called

concave up on I . If the graph of f lies below all of its tangents on I , it is called

concave down on I .

Picture:

Theorem 3.3.5 (Concavity Test).

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

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

Definition(s) 3.3.6. A point P on a curve y = f(x) is called an inflection point if f is

continuous there and the curve changes from concave upward to concave downward or from concave

downward to concave upward at P .

Theorem 3.3.7 (Second Derivative Test). Suppose f ′′ is continuous near c.

(a) If f ′(c) = 0 and f ′′(c) > 0 , then f has a local minimum at c.

(b) If f ′(c) = 0 and f ′′(c) < 0 , then f has a local maximum at c.

77


Example 3.3.8. Suppose that the graph to the right is of is of f .

For which values of x does f have an inflection point?x

y

1 2 3 4 5 6 7

Example 3.3.9. Suppose that the graph to the right is of is of f ′.


y

1 2 3 4 5 6 7

Example 3.3.10. Suppose that the graph to the right is of is of f ′′.


y

1 2 3 4 5 6 7

Example 3.3.11. Find where f(x) = x√

6− x is concave up and where it is concave down. Where are the

inflection points?

78


Example 3.3.12. Find where f(x) = x− 4√x is concave up and where it is concave down. Where are the

inflection points?

Example 3.3.13. Use the second derivative test to classify the critical points for f(x) =x2

x− 1

79


3.4 Limits at Infinity; Horizontal Asymptotes


• Recognize horizontal asymptotes of a function given graphically.

• Investigate horizontal asymptotes of a function given algebraically by using limits at infinity.

Definition(s) 3.4.1. Let f be a function defined on some interval (a,∞). Then limx→∞

f(x) = L

means that the values of f(x) can be made arbitrarily close to L by taking x

sufficiently large .

Definition(s) 3.4.2. Let f be a function defined on some interval (−∞, a). Then limx→−∞

f(x) = L

means that the values of f(x) can be made arbitrarily close to L by taking x

sufficiently negative .

Definition(s) 3.4.3. The line y = L is called a horizontal asymptote of the

curve y = f(x) if either

limx→−∞

f(x) = L or limx→∞

f(x) = L

Theorem 3.4.4. If r > 0 is a rational number, then

limx→∞

1

xr= 0

If r > 0 is a rational number such that xr is defined for all x, then

limx→−∞

1

xr= 0

80

3.4. LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES

Example 3.4.5. Find the limits or show that they do not exist

(a) limx→∞

2x− 1

5x+ 3

(b) limx→∞

x sin1

x

(c) limx→∞

√x− x2

2x+ x2

(d) limx→−∞

x2

√3x4 + 1

81


3.5 Curve Sketching


• Summarize all of our current algebra and calculus knowledge to sketch accurate graphs of functions.

Remark 3.5.1 (General Guidelines for Curve Sketching).

When trying to sketch the graph of a given function f(x), ask yourself the following:

1. What is the domain of f?

2. What are the x and y intercepts of the graph?

3. Does the graph have any symmetry (is f an odd function? an even function?

neither?)

a) Even if f(−x) = f(x) for all x

b) Odd if f(−x) = −f(x) for all x

4. Does the graph have any asymptotes (horizontal, vertical, slant)?

a) Has horizontal asymptote y = c if limx→∞

f(x) = c or if limx→−∞

f(x) = c.

b) Has vertical asymptote x = c if either of limx→c+

f(x) or limx→c−

f(x) are equal to either∞ or −∞.

c) Has slant asymptote y = mx+ b if

limx→∞

(f(x)− (mx+ b)) = 0 or if limx→−∞

(f(x)− (mx+ b)) = 0

5. Where is f(x) increasing and decreasing?

a) See Theorem 3.3.1

6. Where are the local maximum and minimum points?

a) Use First Derivative Test or Second Derivative Test

82

3.5. CURVE SKETCHING

7. Where is the graph concave up and where is it concave down?

a) See Theorem 3.3.5

8. Where are the inflection points?

a) Find where f ′′(x) changes sign

Example 3.5.2. Sketch graphs of the following:

(a) y = x(x− 4)3

83


(b) y =x3

x2 + 1

84

3.5. CURVE SKETCHING

(c) y =x√x2 + 1

85


(d) y = sin2(x) on the interval [0, 2π]

86

3.7. OPTIMIZATION PROBLEMS

3.7 Optimization Problems


• Analyze real world problems and transform statements into mathematical equations.

• Apply our maxima/minima knowledge to help solve optimization problems.

Remark 3.7.1. Related rates problems are to implicit differentiation as optimiza-

tion problems are to minima and maxima problems.

Example 3.7.2. Find the maximum area of a rectangle inscribed in an equilateral triangle of side length 6

and one side of the rectangle lies along the base of the triangle.

87


Theorem 3.7.3 (Steps in Solving Optimization Problems).

1. Understand the problem.

• Read the problem through in its entirety

• Determine what is given and what is unknown

2. Draw a Diagram

• This is useful in most problems

3. Introduce Notation

• Assign symbols to what needs maximized or minimized.

• Select symbols for other quantities and label the diagram when appropriate.

4. Find an equation that relates the quantities with what needs to be maximized/minimized

5. Use restricting equation(s) to reduce down to one variable (when applicable)

6. Use the methods in 3.1/3.3 to find an absolute maximum/minimum. In particular, if the domain is

closed then the Closed Interval Method in Section 3.1 can be used.

Example 3.7.4. Find a point on the line y = 2x+ 3 that is closest to the origin.

88


Example 3.7.5. A fish tank with a square base is to be made of glass sides, plastic on the base, and an open

top. The fish tank needs to hold 5 cubic feet of water. Glass costs $3 per square foot and plastic cost $2 per

square foot. What is the cheapest the tank can cost?

Example 3.7.6. A right circular cylinder is inscribed in a right circular cone of radius 2 and height 3. Find

the largest possible volume of such a cylinder.

89


Example 3.7.7. A right circular cylinder is inscribed in a sphere of radius 2. Find the largest possible volume

of such a cylinder.

Example 3.7.8. The sum of two positive numbers is 16. What is the smallest possible value of the sum of

their squares?

90


Example 3.7.9. A rain gutter is to be constructed from a metal sheet of width 30cm by bending up one-third

of the sheet on each side through an angle θ. How should θ be chosen so that the gutter will carry the

maximum amount of water?

91


3.8 Newton’s Method


• Recognize how and why Newton’s Method finds intersections between functions.

• Analyze how Newton’s Method may fail.

Remark 3.8.1. Newton’s method is an extremely powerful technique used in finding roots of equations - in

general the convergence is quadratic: as the method converges on the root, the difference between the root

and the approximation is squared (the number of accurate digits roughly doubles) at each step. However,

there are some difficulties with the method which we will discuss below. Newton’s method can be altered to

help approximate solutions to many equations.

Pictures of the idea of Newton’s Method

Theorem 3.8.2 (Newton’s Method). If x1 is the initial guess of some root of f(x) then

xn+1 = xn −f(xn)

f ′(xn)

92

3.8. NEWTON’S METHOD

Remark 3.8.3. Newton’s method may fail to converge to an answer or may find the wrong root. See the

pictures below for how this can happen. When this occurs it can usually be fixed by selecting an alternative

initial guess (x1).Horizontal tangent Never converges Converges to wrong root

Example 3.8.4. Approximate the root of f(x) = x3 + x+ 4 by taking the initial guess x1 = −1.

Compute x2 and x3. Don’t simplify x3.

93


Example 3.8.5. Use Newton’s method to approximate the positive root of x2 = sinx. Take x1 = π/3.

Calculate x2.

Example 3.8.6. Use Newton’s method to approximate 3√

29. Calculate up to x3. Do not simplify.

94

3.8. NEWTON’S METHOD

Example 3.8.7. Approximate√

5 using

(a) Linearization with a = 4

(b) Newton’s method with x1 = 2. Compute x2 and x3.

95


3.9 Antiderivatives


• Compute general antiderivatives for many types of functions.

• Solve initial value problems for particular antiderivative functions.

• Use antiderivatives to calculate velocity or position from acceleration.

Definition(s) 3.9.1. A function F is called an antiderivative of f on an interval I if

F ′(x) = f(x) for all x in I .

Theorem 3.9.2. If F is an antiderivative of f on an interval I , then the most

general antiderivative of f on I is

F (x) + C

where C is an arbitrary constant.

Example 3.9.3. Find the most general antiderivative of the functions below.

(a) f(x) = π +1

x2

(b) f(x) =√x(6 + 7x)

96

3.9. ANTIDERIVATIVES

Definition(s) 3.9.4. A differential equation is an equation involving the deriva-

tives of an unknown function.

Definition(s) 3.9.5. An initial value problem is a differential equation for y =

f(x) along with an initial condition , such as f(c) = a for some constants c and a. The

solution to the initial value problem is a solution to the differential equation that also satisfies the initial

condition.

Remark 3.9.6. In an initial value problem, if the unknown function f(t) represents position as a function of

time, then the initial condition f(t0) = a means that at time t = t0, the object is at position a. If the unknown

function represents velocity, then the same initial condition means the velocity is a at time t = t0.

Example 3.9.7. Below is a graph of f ′. Sketch the graph of f if f is continuous and f(0) = −2

x

x

−1

1

2

1 2 3

97


Example 3.9.8. A particle is moving with velocity v(t) = sin t − cos t, and has initial position s(0) = 0.

Find the position function of the particle.

Example 3.9.9. Solve the initial value problem

f ′(x) = 1 + 3√x, f(4) = 25

98

3.9. ANTIDERIVATIVES

Example 3.9.10. Solve the initial value problem

f ′′(x) = 2 + cos(x), f(0) = −1, f(π/2) = 0

99

Chapter 4

Integrals

CHAPTER 4. INTEGRALS


In Chapters 4 and 5 we will bring up the second main topic: area ‘under’ a function. We will see that

surprisingly this is related to derivatives. We will learn how to find the area ‘under’ more complicated

functions and the area trapped between two functions. Additionally we will learn how to apply this to real

world problems in physics. Once we have mastered these sections we will be able to solve problems such as:

Example: An object moves along a line with velocity v(t) = t(9− t) inches per second. At time t = 0, the

object is 2 inches to the right of the origin. How far did the object move during those 15 seconds?

Example: Find the area of the region bounded by the graphs of y = 5x− x2 and y = x.

Example: Find the area between the curve x cos(x2) and the x-axis between the vertical lines x = 0 and

x =√π

4.1 Appendix E4.2 4.3 4.4 4.5

5.1

5.5

Ch 3> > > > >

>

>

>

>

Chapter 4 and 5 Map

102

4.1. AREAS AND DISTANCES

4.1 Areas and Distances


• Estimate the area under a curve using rectangles with heights given by left endpoints or rightendpoints.

• Solve for over/under estimates for the area under a curve using rectangles.

The goal of this section is to understand how to approximate areas with rectangles. Suppose that we want

to approximate the area between the graph of a continuous function f(x) and the x-axis between x = a

and x = b (suppose for now that f is positive). Let’s sub-divide the interval [a, b] into n sub-intervals

[x0, x1] through [xn−1, xn] of equal width ∆x. If we pick a point x∗i in each interval [xi, xi+1], then we

can estimate the area under the graph by the sum of the areas of the rectangles with width ∆x and height

f(x∗i ):

Area ≈ f(x∗0)∆x+ f(x∗1)∆x+ · · ·+ f(x∗n−1)∆x


• An upper sum is when the x∗i are all chosen to be the global maximum on [xi, xi+1]

for each i.

• A lower sum is when the x∗i are all chosen to be the global minimum on [xi, xi+1] for

each i.

• A left-hand sum is when x∗i is the left endpoint of [xi, xi+1] for each i.

• A right-hand sum is when x∗i is the right endpoint of [xi, xi+1] for each i.

Remark 4.1.2. To get better approximations of area, use more rectangles .

103


Remark 4.1.3. If the velocity of an object is constant, then we have

distance = velocity × time

We can think of this product as being the area of a rectangle of width “time” and height “velocity”. Thought

of in this way, the distance traveled by an object from time t = a to t = b is given by the area under the graph

of the velocity from t = a to t = b.

Example 4.1.4. Approximate the area under the graph of f(x) =√x from x = 0 to x = 4 using 4 rectangles

of equal width, using a:

(a) left-hand sum

(b) right-hand sum

104


Example 4.1.5. The speed of a runner (in ft/s) is given in the table below at different times (in seconds).

t 0 0.5 1.0 1.5 2.0 2.5 3.0

v(t) 0 6.2 10.8 14.9 18.1 19.4 20.2

(a) Give an upper estimate of the distance traveled by the runner.

(b) Give a lower estimate of the distance traveled by the runner.

Example 4.1.6. Below is the graph of the velocity of a car (in ft/s) as it is coming to a stop.

Estimate the distance the car travels as it comes

to a stop using the various types of sums.

t

v(t)

10203040

1 2 3 4 5 6

105


Appendix E - Sigma Notation


• Express sums using sigma notation.

• Memorize a few common finite sums.

• Understand basic properties of finite sums and use them to compute more complicated finite sums.

Definition(s) 4.1.7. If am, am+1, . . . , an−1, an are real numbers andm and n are integers such thatm ≤ n,

then n∑i=m

ai = am + am+1 + · · ·+ an−1 + an

Definition(s) 4.1.8. This way of short-handing sums of many numbers is called

sigma notation (uses the Greek letter Σ “Sigma”). The letter i above is called

the index of summation and it takes on consecutive integer values starting with

m and ending with n.

Theorem 4.1.9. If c is any constant then:

(a)n∑

i=m

cai = cn∑

i=m

ai

(b)n∑

i=m

(ai + bi) =n∑

i=m

ai +n∑

i=m

bi

(c)n∑

i=m

(ai − bi) =n∑

i=m

ai −n∑

i=m

bi

Theorem 4.1.10. Let c be a constant and n a positive integer. Then

(a)n∑i=1

1 = n

(b)n∑i=1

i =n(n+ 1)

2

(c)n∑i=1

i2 =n(n+ 1)(2n+ 1)

6

106


Example 4.1.11. Evaluate the following sums

(a)4∑

k=0

2k − 1

2k + 1

(b)4∑i=0

(2− 3i)

(c)38∑i=1

(3i − 3i−1)

107


Example 4.1.12. Write the sum:√

3 +√

4 + · · ·+√

25 in sigma notation

Example 4.1.13. Write the sum:√

3−√

5 +√

7−√

9 + · · ·+√

27 in sigma notation

108

4.2. THE DEFINITE INTEGRAL

4.2 The Definite Integral


• Use the limit of finite sums to calculate the definite integral of a function.

• Identify how the definite integral relates with area under the curve.

• Understand intuitively the properties of definite integrals.

a b a b a b

Theorem 4.2.1 (Definite Integral). If f is integrable on [a, b], then

∫ b

a

f(x) dx = limn→∞

n∑i=1

f(xi)∆x

where

∆x =b− an

and xi = a+ i∆x

Remark 4.2.2. The definite integral,∫ b

a

f(x) dx gives the net area between the curve f and

the x− axis.

109


Example 4.2.3. Use the definition of the definite

integral to verify that∫ 3

1

(x+ 1

2

)dx = 3

1 3

Example 4.2.4. Evaluate the definite integral∫ 7

1

f(x) dx.

Where f(x) is given by the function to the right.

110

4.2. THE DEFINITE INTEGRAL

Example 4.2.5. Evaluate the definite integral∫ 4

−1

(x2 − x+ 1

)dx

111


Theorem 4.2.6 (Properties of Definite Integrals).

1.∫ a

a

f(x) dx = 0

2.∫ b

a

f(x) dx = −∫ a

b

f(x) dx

3.∫ b

a

[cf(x) + dg(x)]dx = c

∫ b

a

f(x) dx+ d

∫ b

a

g(x) dx

4.∫ b

a

f(x) dx+

∫ c

b

f(x) dx =

∫ c

a

f(x) dx

5. If f(x) ≥ g(x) for a ≤ x ≤ b, then∫ b

a

f(x) dx ≥∫ b

a

g(x) dx.

6.∫ b

a

c dx = c(b− a)

Example 4.2.7. Suppose

∫ 3

1

f(x) dx = 3

∫ 1

5

f(x) dx = 10

Find∫ 5

3

[2f(x) + 1]dx

112

4.3. THE FUNDAMENTAL THEOREM OF CALCULUS

4.3 The Fundamental Theorem of Calculus


• Relate slopes and areas through the two parts of the Fundamental Theorem of Calculus.

• Develop the idea of a proof for the two parts of the Fundamental Theorem of Calculus.

• Use the chain rule and properties of definite integrals so solve more complicated problems.

• Use the antiderivative to calculate definite integrals.

Theorem 4.3.1 (FTC, Part 1). If f is continuous on [a, b], then the function g defined by

g(x) =

∫ x

a

f(t) dt, a ≤ x ≤ b

is continuous on [a, b] and differentiable on (a, b), and g′(x) = f(x)

Remark 4.3.2. Here is an idea of the proof:

113


Theorem 4.3.3 (FTC, Part 2). If f is 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 such that F ′ = f .

Remark 4.3.4. Here is an idea of the proof:

Remark 4.3.5. The two parts of the FTC together state that differentiation and integration are inverse

processes.

Remark 4.3.6. From our perspective FTC, Part 2 is the most important because it allows us to calculate

definite integrals without having to take limits of Riemann sums!

114

4.3. THE FUNDAMENTAL THEOREM OF CALCULUS

Example 4.3.7. Evaluate the following integrals

(a)∫ 9

1

√x dx

(b)∫ 5

−1

(u+ 1)(u− 2) du

(c)∫ 2

1

x4 + 1

x2dx

(d)∫ 2

−1

|x| dx

115


Example 4.3.8. Find the derivatives of the following functions:

(a) f(x) =

∫ x

1

t2 dt

(b) g(x) =

∫ 2

x

√t2 + 3 dt

(c) H(x) =

∫ x2

tanx

1√2 + t4

dt

Example 4.3.9. Identify what is wrong with the evaluation:

∫ 1

−1

1

x2dx =

[−x−1

]1−1

=[−(1)−1 + (−1)−1

]= [−1− 1] = −2

116

4.4. INDEFINITE INTEGRALS AND NET CHANGE

4.4 Indefinite Integrals and Net Change


• Use the Net Change Theorem to continue calculating definite integrals.

• Given a velocity function and a set time interval find the total distance traveled.

Definition(s) 4.4.1. An indefinite integral of f , written∫f(x) dx, is an

antiderivative of f . In other words,

F (x) =

∫f(x) dx means F ′(x) = f(x)

Theorem 4.4.2. (Linearity Properties of Indefinite Intagrals)

For functions f(x) and g(x), and any constant k ∈ R,

∫[f(x) + g(x)] dx =

∫f(x)dx+

∫g(x)dx

and∫kf(x) dx = k

∫f(x)dx

Theorem 4.4.3. (Net Change Theorem)

The net change of a quantity F (x) from x = a to x = b is the integral of its derivative from a to b.

That is, if F ′(x) = f(x), then

F (b)− F (a) =

∫ b

a

F ′(x) dx =

∫ b

a

f(x) dx

In other words,

F (b) = F (a) +

∫ b

a

F ′(x) dx = F (a) +

∫ b

a

f(x) dx

117


Remark 4.4.4. If s(t) represents the position of a moving object, and v(t) the velocity, then the Net Change

Theorem says we can compute the net displacement of the object from time t = a to time t = b by integrating

the velocity:

s(b)− s(a) =

∫ b

a

s′(t)dt =

∫ b

a

v(t)dt

Example 4.4.5. Find the most general function F (x) with the property that:

(a) F ′(x) =7

x7− 5

x4+ 18

(b) F ′(x) =1√x

+ cos(x)

(c) F ′(x) = sec(x) (tan(x) + sec(x))

118

4.4. INDEFINITE INTEGRALS AND NET CHANGE

Example 4.4.6. An object moves along a line with velocity v(t) = t(9− t) inches per second. At time t = 0,

the object is 2 inches to the right of the origin.

(a) What is the position of the object at time t = 15 seconds?

(b) How far did the object move during those 15 seconds?

Example 4.4.7. Find the net change of the function f(x) from x = 0 to x = 1, if f ′(x) = x2 + x+ 1.

119


4.5 The Substitution Rule


• Develop a substitution rule to find antiderivatives of more complicated functions.

• Use properties of definite integrals and symmetric functions to create a few more theorems.

Theorem 4.5.1. If u = g(x) is a differentiable function whose range is an interval I and f is continuous

on I , then ∫f(g(x))g′(x)dx =

∫f(u)du

Remark 4.5.2. The Substitution Rule says essentially that is is permissible to operate with dx and du afterthe integral signs as if they were differentials.

Theorem 4.5.3. If g′(x) is continuous on [a, b] and f is continuous on the range of u = g(x), then

∫ b

a

f(g(x))g′(x)dx =

∫ g(b)

g(a)

f(u)du

Remark 4.5.4. This change in the limits of integration is annoying and somewhat unnecessary right now (solong as you change your variable back) however in Calc II once you start doing trig substitutions it becomesextremely useful. WeBWorK will force you to practice changing the limits of integration so we will practicehere too.

Definition(s) 4.5.5. Recall again that f is called even if f(−x) = f(x)

and called odd if f(−x) = −f(x) .

Theorem 4.5.6. Suppose f is continuous on [−a, a] then:

(a) If f is even, then∫ a

−af(x) dx = 2

∫ a

0

f(x) dx

(b) If f is odd, then∫ a

−af(x) dx = 0

120

4.5. THE SUBSTITUTION RULE

Example 4.5.7. Evaluate the following indefinite integrals:

(a)∫x2√x3 + 1 dx

(b)∫

dt

(1− 3t)5

(c)∫

sec3 x tanx dx

(d)∫

sin t

cos2(cos(t))dt

121


Example 4.5.8. Evaluate the following definite integrals:

(a)∫ 3

0

x√1 + 5x

dx

(b)∫ π/4

−π/4(1 + x+ x2 tanx) dx

122

4.5. THE SUBSTITUTION RULE

Example 4.5.9. Evaluate the following definite integrals by changing the limits of integration appropriately:

(a)∫ 1

0

3√

1 + 7x dx

(b)∫ √π

0

x cos(x2) dx

123

Chapter 5

Applications of Integration

CHAPTER 5. APPLICATIONS OF INTEGRATION

5.1 Area Between Curves


• Express the area bounded by two curves as a definite integral and evaluate.

• Identify when it is advantageous to integrate with respect to y instead of x.

Theorem 5.1.1. If f(x) and g(x) are continuous functions on [a, b] where f(x) ≥ g(x) for all x in [a, b],

then the area of the region in between the graphs of y = f(x) and y = g(x) between x = a and x = b is

given by

Area =

∫ b

a

(f(x)− g(x)) dx

Remark 5.1.2. This theorem only applies if f(x) ≥ g(x) on [a, b]. In the more general case (where the

graphs cross), we can use the following theorem.

Theorem 5.1.3. If f(x) and g(x) are continuous functions on [a, b], then the area of the region in between

the graphs of y = f(x) and y = g(x) between x = a and x = b is given by

Area =

∫ b

a

|f(x)− g(x)| dx

Remark 5.1.4. If it is more convenient, you can think of x as a function of y, and integrate with respect to

y. For example, to find the area between the graphs of x = f(y) and x = g(y) between y = a and y = b,

compute

Area =

∫ b

a

|f(y)− g(y)| dy

126

5.1. AREA BETWEEN CURVES

Example 5.1.5. Find the area between the graphs of y = −x and y = cos(x) between x = 0 and x = π2.

Example 5.1.6. Find the area enclosed by the curves y = sec2(x) and y = 8 cos(x) on the interval[−π

3, π

3

].

Example 5.1.7. Find the area of the region bounded by the graphs of y = 5x− x2 and y = x.

127


Example 5.1.8. Find a positive value of c so that the area between y = x2 + 1 and y = x− c

from x = 0 to x = 1 is equal to 1.

Example 5.1.9. Find the area enclosed by the line y = x− 1 and the parabola y2 = 2x+ 6.

128

5.5. AVERAGE VALUE OF A FUNCTION

5.5 Average Value of a Function


• Calculate the average value of a function over an interval.

• Create a Mean Value Theorem for Integrals.

Definition(s) 5.5.1. The average value of a function f(x) on the interval [a, b] is defined

to be:

fave =1

b− a

∫ b

a

f(x)dx

Theorem 5.5.2. (Mean Value Theorem for Integrals)

If f is continuous on [a, b], then there exists a number c in [a, b] such that

f(c) = fave =1

b− a

∫ b

a

f(x)dx

In other words, there exists a number c in [a, b] such that

∫ b

a

f(x)dx = f(c) · (b− a)

Example 5.5.3. Find the average value of the following functions on the given interval:

(a) f(x) =√x on [0, 4]

(b) f(x) = sin(x) on [0, π]

129


Example 5.5.4. Find the average value of the given function on the given interval. Also, find the value of c

guaranteed by the Mean Value Theorem for Integrals.

(a) f(x) = 3√x on [0, 8]

(b) f(x) = sec2(x) on[−π

4, π

4

]

130

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