Calculus I - Brian Veitchbrianveitch.com/calculus/lectures/calculus_1.pdfBrian E. Veitch 1 Review,...

Calculus I

Brian E. Veitch

August 4, 2013

Contents

1 Review, Limits, and Continuity 4

1.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Types of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 Horizontal Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2.2 Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Piecewise Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.4 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5 Composition of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

1.6 Tangents and Velocity Problems . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.7 Limit of a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.8 Limit Laws and Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.8.1 The Limit Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

1.8.2 Limits by Cancellation . . . . . . . . . . . . . . . . . . . . . . . . . . 51

1.8.3 Limits by Simplifying . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

1.8.4 Limits using the conjugate . . . . . . . . . . . . . . . . . . . . . . . . 56

1.8.5 Limits using common denominators . . . . . . . . . . . . . . . . . . . 58

1.8.6 Squeeze Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

1.9 The Precise Definition of a Limit . . . . . . . . . . . . . . . . . . . . . . . . 64

1.10 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

1

CONTENTS Brian E. Veitch

1.10.1 Intermediate Value Theorem . . . . . . . . . . . . . . . . . . . . . . . 77

2 Derivatives 81

2.1 Derivatives and Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . 81

2.2 Derivative as a Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

2.2.1 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

2.3 Differentiation Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

2.3.1 Power Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

2.3.2 Sum / Difference Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 110

2.3.3 Product Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

2.3.4 Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

2.3.5 Higher Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

2.4 Derivatives of Trig Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

2.4.1 Derivatives of Trig Functions . . . . . . . . . . . . . . . . . . . . . . . 127

2.4.2 Product Rule for Three Functions . . . . . . . . . . . . . . . . . . . . 129

2.5 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

2.5.1 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

2.5.2 Chain Rule with a composition of three functions . . . . . . . . . . . 136

2.6 Implicit Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

2.7 Rates of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

2.8 Related Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

2.9 Linear Approximations and Differentials . . . . . . . . . . . . . . . . . . . . 178

2.9.1 Linear Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 178

2.9.2 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

3 Application of Differentiation 189

3.1 Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . 189

3.1.1 The Extreme Value Theorem . . . . . . . . . . . . . . . . . . . . . . 190

3.1.2 Fermat’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

3.1.3 Steps to find the absolute max or absolute min . . . . . . . . . . . . 195

3.1.4 Steps to find the local maximums and minimums . . . . . . . . . . . 197

2

CONTENTS Brian E. Veitch

3.2 Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

3.3 Using Derivatives to analyze a Function . . . . . . . . . . . . . . . . . . . . . 210

3.3.1 First Derivative Test . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

3.3.2 Concavity and Inflection Points . . . . . . . . . . . . . . . . . . . . . 220

3.4 Limits at Infinity - Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . 234

3.4.1 Examples of Curve Sketching with Asymptotes . . . . . . . . . . . . 243

3.5 Summary of Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . 246

3.6 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

3.7 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270

3.7.1 Newton’s Method Formula . . . . . . . . . . . . . . . . . . . . . . . . 272

3.8 Anti-Derivatives (Integration) . . . . . . . . . . . . . . . . . . . . . . . . . . 278

3.8.1 Power Rule for Anti-Derivatives with Trig . . . . . . . . . . . . . . . 280

4 Integration 287

4.1 Areas and Distances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

4.1.1 The Distance Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 303

4.2 Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

4.2.1 Midpoint Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319

4.2.2 Properties of the Definite Integral . . . . . . . . . . . . . . . . . . . . 322

4.3 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . 329

4.3.1 The Fundamental Theorem of Calculus, Part 1: . . . . . . . . . . . . 337

4.3.2 Fundamental Theorem of Calculus Part 2: . . . . . . . . . . . . . . . 341

4.4 Indefinite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348

4.4.1 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351

4.5 Integration by Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357

4.5.1 Integration by Substitution Rule . . . . . . . . . . . . . . . . . . . . 358

4.5.2 Definite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363

4.5.3 u-Substitution for Definite Integrals . . . . . . . . . . . . . . . . . . . 364

3

Brian E. Veitch

1 Review, Limits, and Continuity

1.1 Functions

Definition 1.1 (Function). - a function is a rule that associates to each x-value one y-value.

Common functions you’ll see are

1. y =√x− 1

2. y = x2

3. y =1

x+ 4

Since we’re on the topic of functions, here’s an example of one that isn’t a function.

y2 = x

Why?

Remember, every function associates an x-value to ONE y-value. Another way to think

about it is

If two different y-values can come from the same x-value, it’s not a function.

x y

4 −2

4 2

4

1.1 Functions Brian E. Veitch

See what happened? One x-value, two y-values. That’s bad. So this isn’t a function.

Let’s move on!

Domain Issues

Let’s stop for a moment and look at the domain. In this course, domain problems don’t

come up too much. We focus a lot on functions that ’work.’ But the first week or so, we’ll

be looking at piece-wise functions and these functions have domain issues all over the place.

Definition 1.2 (Domain). This is the collection of all legal inputs. If you can plug in a

number and get a ’number’ out, then you’re in the domain.

I always think of the domain as stuff that doesn’t break the function.

So what are some common domain issues?

1. Square rooting a negative number: y =√x− 1

Setting x − 1 ≥ 0 gives us a domain of x ≥ 1. Plug any number less than 1 and you

square root a negative.

2. Dividing by 0

Examples:

(a) y =1

x

(b) y =x

x2 − 1

5


(c) y =x√

x− 2− 4

Of course, dividing by 0 is the worst one. But you’ll see that dividing by 0 shows up

quite a bit in this course.

Solution to 2c

You got two things to worry about.

1. When does the denominator equal 0

2. When are we square rooting a negative

Ok, let’s get started.

1. Set√x− 2− 4 = 0.

√x− 2− 4 = 0

→√x− 2 = 4

→ x− 2 = 16

→ x = 18

2. Set√x− 2 ≥ 0

x− 2 ≥ 0

x ≥ 2

Easy, right?

6


So the overall domain would look something like this:

Domain: [2, 18) ∪ (18,∞)

7

1.2 Types of Functions Brian E. Veitch

1.2 Types of Functions

Let’s run through all the different types of functions you’ll see. You should have most of

these memorized. You should also know the basic properties of each of the functions.

1. Linear: f(x) = mx+ b

Graph 1.2

Slope: m =y2 − y1x2 − x1

orf(b)− f(a)

b− a

Get used to the second version.

2. Quadratic: f(x) = ax2 + bx+ c

Graph 1.3

8


Warning: Know how to factor quadratics. It comes up.

3. General Polynomial: f(x) = anxn + an−1x

n−1 + an−2xn−2 + ...+ a1x+ a0

4. Power Functions: f(x) = xn

Even Power Odd Power

5. Root Functions: f(x) =√x, 3√x, 4√x, ..., n

√x

Or their equivalent fractional exponents: x1/2, x1/3, x1/4, ..., x1/n

9


6. Rational Functions: f(x) =P (x)

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

Note: These do have domain issues when Q(x) = 0. Watch out for those. Rational

functions tend to look like this.

Graph 1-6a Graph 1-6b

Rational Functions tend to have horizontal and vertical asymptotes. Our calculus way

of defining these will be many pages later. From algebra you had shortcuts to deter-

mine these.

1.2.1 Horizontal Asymptotes

(a) If the degree of P (x) (top polynomial) is bigger than the degree of Q(x) (bot-

tom polynomial), then there is no horizontal asymptote. These would be oblique

asymptotes, which sadly are skipped over in algebra classes.

Graph 1-6a: f(x) =x2

2x+ 2is example of a rational function with an oblique

asymptote. Notice how the graph will go to ±∞ when x gets really large.

(b) If the degree of P (x) (top polynomial) is the same as the degree of Q(x) (bottom

polynomial), then the horizontal asymptote is the ratio of the leading coefficients.

10


Example 1.1. f(x) =4x3 − 15x+ 7

1− 3x3The horizontal asymptote will be y =

4

3

(c) If the degree of P (x) is smaller than the degree of Q(x), then the horizontal

asymptote is y = 0. This one is easy!

1.2.2 Vertical Asymptotes

(a) Make sure f(x) =P (x)

Q(x)is simplified.

(b) Find what x-values make Q(x) = 0. These are potential vertical asymptotes.

Suppose Q(a) = 0.

(c) If those x-values make P (x) 6= 0, then x = a is a vertical asymptote.

Example 1.2. f(x) =x2 − 4

x2 − 1

Since x2 − 1 = 0 when x = 1,−1, x = 1 and x = −1 are potential vertical asymptote.

Since x2 − 4 6= 0 when x = 1 or x = −1, we know x = 1 and x = −1 are vertical

asymptotes.

Here’s the graph

11


7. Trig Functions

Graph 1-8 f(x) = sin(x) f(x) = cos(x) f(x) tan(x)

12

1.3 Piecewise Functions Brian E. Veitch

1.3 Piecewise Functions

Ok, let’s just take a look at a piece-wise graph.

Graph 1.1

Now a piece-wise function is just one large function f(x) made up of smaller functions

on different parts of the domain.

Example 1.3.

f(x) =

1x2, x < 0

1x, 0 < x < 1

2x− 4, [1, 3) ∪ (3,∞)

There are a couple of ways of graphing a piece-wise function. If it’s your first time or you

haven’t done it in a while, just graph all the functions and then erase the part that doesn’t

count (i.e. when it’s not in its part of the domain).

Example 1.4. Graph f(x) =

(x+ 2)2 − 1, x < −1

2, x = −1

−2x+ 3, −1 < x < 1

√x, x ≥ 1

13


We need to take this one piece at a time. Get it... one ’piece.’ You know... because it’s

a piece-wise function. Ok moving on!

1. Let’s start with (x+ 2)2 − 1. We haven’t covered transformations yet, but you proba-

bly remember a little bit about transforming f(x) = x2. Let’s start with graphing the

whole thing.

Now erase what we don’t need.

It’s now restricted to x < −1.

Note the open dot at (−1, 0). It’s open because the interval does not include x = −1.

2. Next up...2?. What this really means is when x = −1, the y-value is 2. It’s just a

closed point on the graph.

14


All we did here is add the point (−1, 2).

3. On to −2x+ 3. Let’s add that to the graph.


The graph only shows −2x+ 3 when −1 < x < −1.

4. Finally we add√x.

15



We are finally done. Notice how that open dot at (1, 1) is now a closed dot.

16

1.4 Transformations Brian E. Veitch

1.4 Transformations

When we talk about transformations in this class we are referring to shifting, stretching,

compressing, and rotating basic functions. What is a basic function? Here’s a list.

1. y = x2 or y = x3

2. y =√x or y = 3

√x

3. y =1

x

4. y = |x|

5. y = cos(x) or y = sin(x)

Let’s go through the list of our transformations.

For c > 0. Let f(x) be a basic function.

Function Effect Example

y = f(x) + c Vertical Shift up c units y = x2 + 1

y = f(x)− c Vertical Shift down c units y =√x− 4

y = f(x− c) Horizontal Shift right c units y = |x− 4|

y = f(x+ c) Horizontal Shift left c units y =1

x+ 3

Now...on to some examples

17


Example 1.5.

1. Sketch y = (x+ 2)2

This is a horizontal shift left 2 units.

y = x2 y = (x+ 2)2

2. Sketch y =√x− 1 + 2

This has two transformations. Start with the one closest to the x. So it goes...

(a) Shift right 1 unit

(b) Shift up 2 units

y =√x y =

√x− 1 y =

√x− 1 + 2

18


So those were the shifts. Let’s move on to the stretches and compressions

For c > 1. Let f(x) be a basic function.


y = cf(x) Stretch vertically by a factor of c y = 3|x|

y =1

cf(x) Compress vertically by a factor of c y =

1

4x2

y = f(cx) Compress Horizontally y = cos(4x)

y = f

(1

cx

)Stretch horizontally y = sin

(x2

)

Example 1.6.

1. Sketch y = 3|x|

This is a vertical stretch by a factor of 3.

y = |x| y = 3|x|

The way this works is take any point on the graph and multiply the y-value by a factor

of 3. Note how the point (1,1) is now at (1,3).

2. Sketch y = sin(x

2

)

19


I think it’s easiest to see why this is a horizontal stretch when you use sin(x). Recall

that sin(x) has a period of 2π. It means it will complete a full period between 0 to

2π. The left graph shows 2 full periods. When you multiple the ’inside’ by 1/2, it

means you’ll complete 1/2 as many periods, i.e., the period length should be ”larger”

or ”stretched.” So in this case, the new graph should have only 1 period (half as many

as the first graph). Let’s take a look.

y = sin(x) y = sin(x

2

)

Try to sketch:

1. y =1

3

√x

2. y = (2x)2

Our last transformations are reflections.


y = −f(x) Flip over y-axis y = −x2

y = f(−x) Flip over x-axis y =√−x

20


Example 1.7.

1. Sketch y =√−x

The basic function here is y =√x. The (-) sign inside of the basic function, so it’s

a reflection over the x-axis. Another way to look at it is follow some points on the

original function and the transformed function.

y =√x y =

√−x

Example 1.8. For the final sketch, let’s add a bunch of transformations together. Sketch

y = −2|x− 1|+ 3

To complete this graph, let’s map out the steps. Always start with the transformations

closest to x.

1. Shift 1 unit to the right

2. Vertical Stretch by a factor of 2

3. Reflect over y-axis

4. Shift up 3 units

21


Let’s get started. I will plot the point (1,1). We will follow the point as it gets trans-

formed.

1. Shift 1 unit to the right.

2. Vertical Stretch by a factor of 2.

22


3. Reflect over the y-axis.

4. Shift up 3 units

Alright! We are now done with transformations.

23

1.5 Composition of Functions Brian E. Veitch

1.5 Composition of Functions

Definition 1.3 (Composition of Functions). For functions f(x) and g(x), the composite

function f ◦ g is

f ◦ g = f (g(x))

People struggle with composition for many reasons. Let’s go through a series of examples

that will build up to the definition we just stated for the composition of functions.

Function Value

f(x) 2x2 − cos(x) + 4x− 3

f(2) 2(2)2 − cos(2) + 4(2)− 3

f(−6) 2(−6)2 − cos(−6) + 4(−6)− 3

f(a) 2(a)2 − cos(a) + 4(a)− 3

f(2x+ 1) 2(2x+ 1)2 − cos(2x+ 1) + 4(2x+ 1)− 3

f(g(x)) 2(g(x))2 − cos(g(x)) + 4(g(x))− 3

All of these are composition of functions. The last two are more obvious. A composition

of functions is just evaluating a function, like f(2). The difference is now we evaluate a

function with another function. So f(2x + 1) is a composition of functions. Remember, all

you’re doing is plugging one function into another, just like how you would evaluate any

function.

Example 1.9.

1. Let f(x) = 3x2 − 4x+ 1 and g(x) = 3x− 5. Find

(a) f ◦ g:

First, always rewrite f ◦ g as f (g(x)).

24

1.5 Composition of Functions Brian E. Veitch

So, f ◦ g = f (g(x)) = f (3x− 5) = 3 (3x− 5)2 − 4 (3x− 5) + 1.

(b) (f ◦ g) (2). This notation is really just asking for f (g(2)).

f (g(2)) = f(1) = 3(1)2 − 4(1) + 1 = 0

2. Let f(x) =1

xand g(x) = x+ 1. Find the domain of

(a) f ◦ g:

Let’s just take a look at what f (g(x)) is.

f (g(x)) = f(x+ 1) =1

x+ 1

There are two ways of doing this. One way is just to find f ◦ g (do NOT simplify

it) and simply find it’s domain. From above you can see x 6= −1. Another way

is to do it in steps. First, we look at the domain of g(x). Since g(x) = x+ 1, we

have no domain issues. Ok, so we can plug anything into g(x), but what about

f(x). Notice that we can’t plug 0 into f(x). And what do I plug into g(x) that

will give me g(x) = 0?

(b) g ◦ f :

g (f(x)) = g

(1

x

)=

1

x+ 1

We can never have a denominator equal 0, so x 6= 0. So the domain is all reals

except x = 0.

25

1.6 Tangents and Velocity Problems Brian E. Veitch

1.6 Tangents and Velocity Problems

Tangent Line

Suppose you have a graph of a function. And suppose you want to find a line that touches

the graph at a certain point so that the slope of the line is the same as the slope of the graph

at that point. It would look something like this:

So how can we do this? It comes down to some algebra and the slope formula of a line.

We’ll come back to this idea in a bit.

Velocity Problems

We know how to find average velocity. Change in distance over change in time. In cal-

culus, we want to go a bit deeper into the idea of velocity. Suppose you’re driving on the

highway and you look at your speedometer. It says you’re going 55 mph. This is the speed

you are traveling at that very moment. But how do you calculate it? Since at that very

moment the change in time is 0, how do we actually calculate velocity at any given moment

(called instantaneous velocity)?

The Answer

26


To find the slope of a tangent (and the slope of a graph at a given point P ), we find the

slopes of many secant lines. What’s a secant line you ask? I’ll show you.

A secant line is simply a line connecting two points on a graph. So how does that help

us find the tangent line? Here’s the idea:

1. Make a secant line with the point P and some other point on the graph (call it Q1)

and find the slope.

2. Pick a point closer to P . Call this Q2. Find the slope of this line.

3. Pick a point even closer to P . Call it Q3. Find the slope.

4. Continue this until your second point is extremely close to P .

5. The slope of these secant lines should be very close to the slope of the graph at P .

Can you see when the point Q gets closer to P , the secant lines are getting closer to

looking like the tangent line? Since you know how to find the slope of a line, you then

27


should be able to estimate the slope of the tangent line by using all the secant lines.

Instantaneous Velocity

The idea behind finding instantaneous velocity is very similar. Change in distance over

change in time. Sound familiar? Because that’s the slope formula (change in y over change

in x). Let’s take a look at an example.

Example 1.10.

Suppose you want to estimate the instantaneous velocity of a car after t = 1 seconds.

For the sake of simplicity, the distance traveled is modeled after the following function

s(t) = 10t− 1.86t2

You start by finding the average velocity during the following time intervals.

1. t = 1 to t = 2 seconds

Average Velocity =s(2)− s(1)

2− 1=

12.56− 8.14

1= 4.42 m/s

2. t = 1 to t = 1.5 seconds

Average Velocity =s(1.5)− s(1)

1.5− 1=

10.815− 8.14

.5= 5.35 m/s

3. t = 1 to t = 1.1 seconds


1.1− 1=

8.7494− 8.14

.1= 6.094 m/s

28


4. t = 1 to t = 1.001 seconds


1.001− 1= 6.27814 m/s

5. t = 1 to t = 1.0001 seconds


1.0001− 1= 6.279814 m/s

It looks like the velocity at t = 1 seconds is really close to 6.279814 m/s. So what’s the

real instantaneous velocity at t = 1 seconds? It’s 6.28 m/s. Pretty close?

29

1.7 Limit of a Function Brian E. Veitch

1.7 Limit of a Function

We will discuss the following in this section:

1. Limit Notation

2. Finding a limit numerically

3. Right and Left Hand Limits

4. Infinite Limits

Consider the following graph

Notation: limx→a

f(x), pronounced ”limit of f(x) as x approaches a.”

• limx→a

f(x) = L

means as x gets really close to a (but not equal to a), the corresponding y-values get

really close to L.

30


• Think of the limit as the expected y-value.

Example 1.11. Find limx→2

x2 − x+ 2

Solution: You’ll see that the first step in evaluating limits is simplify plug in what x is

approaching. In this case, it’s x = 2. Don’t think too hard about these limits. It’s asking,

”What would you expect the y-value to be x is really close to 2.” I’m hoping your response

to this is, ”Uh, can’t we just plug in x = 2.” This should make sense. To find the expected

y-value, why wouldn’t you just plug in x = 2?” If you’re taking the limit of a nice function,

plug in the x-value, and get a nice y-value, then that’s the limit.

limx→2

x2 − x+ 2 = 22 − 2 + 2 = 4

But let’s pretend like I can’t do this. I’d like to stay with this ”really close to” concept.

What else could I do? Well...if you want to know what the y value would be when x ap-

proaches 2, let’s plug in x values close to 2. Not too hard, right?

Numerically:

31


x y x y

1 2 3 8

1.5 2.75 2.5 5.75

1.9 3.71 2.1 4.31

1.99 3.9701 2.01 4.0301

1.999 3.997001 2.001 4.003001

↓ ↓ ↓ ↓

2 4 2 4

Left-Hand Limit Right-Hand Limit

I know I’m skipping ahead a bit, but I think this is as good a place as any to introduce the

idea of left and right hand limits. Since a limit is the expected y-value when x approaches a,

a left hand limit is the expected y-value when x approaches a from the left (or less than a).

A right hand limit is the expected y-value when x approaches a from the right (or greater

than a).

So you can see why the left column in the above table is a left hand limit. All the x-values

are less than 2. The right column of the table is the right hand limit because all the x-values

are greater than 2.

Notation

• Left Hand Limit: limx→a−

f(x) Note the ’exponent’ on a

• Right Hand Limit: limx→a+

f(x) Note the ’exponent’ on a

The exponent on a is the only difference between the one-sided limits. Be careful! It’s

easy to miss them or confuse them.

32


Ok, so why don’t we just plug in x = a and get the y-value all the time? It seems like

it’s the way to go. Let’s take a look at the next example.


sin(x)

x

Notice how we can’t just plug in x = 0. If we did plug in x = 0, we’d getsin(0)

0=

0

0.

What is0

0? Usually

0

0is telling us something weird is happening with our function. Since

we can’t just evaluate the function, let’s plug some numbers in for x close to 0.

x y x y

.1 .998334 -.1 .998334

.01 .99998 -.01 .99998

.001 .9999999 -.001 .9999999

↓ ↓ ↓ ↓

0 1 0 1

As x→ 0, we see the y-values are approaching 1. Therefore,

limx→0

sin(x)

x= 1

Remember this!

Maybe your next question to me is, ”Ok, so if I can’t plug in x = a (because I’d divide

by 0 or something bad like that), I could just plug in x-values close to a. Right?

Wrong. Take a look at our next example.

33


Example 1.13. limx→0

√x2 + 9− 3

x2

• We can’t just plug in x = 0 because we’d divide by 0 and that’s bad.

• So let’s try plugging in x-values close to 0

x y x y

.1 .1666 -.1 .1666

.01 .1667 -.01 .1667

.001 .166667 -.001 .16667

↓ ↓ ↓ ↓

.1666... .1666...

• So based on the table, it seems like limx→0

√x2 + 9− 3

x2= .1666...

• Now let’s take a graph of this function.

It seems like this graph confirms limx→0

√x2 + 9− 3

x2= .1666...

34


• Let’s dig deeper. Let’s zoom in.

We are now at a window of [-.3, .3] x [-.03, .2]. So far so good.

• Zoom in!!

We are at a window of [-.00000001, 0.00000001] x [-.1, .3]. This looks bad. Apparently

if we choose x-values extremely close to 0, we start getting some weird y-values. So

does that mean the limit isn’t .16666...

No. The answer is

limx→0

√x2 + 9− 3

x2= .1666...

35


The point I wanted to make is if you plug in numbers close to x = a, it’s possible you

can get something weird (wrong).

Are you sufficiently frustrated yet? I hope so! But that’s the fun of it. We’re seeing that

just simply plugging in x-values close to a may lead to a problem. We need other ways to

calculate limits. I’m jumping ahead as that is in the next section. But at this point just

know we have methods that take care of these pesky problems.

Before moving on to those wonderful limit methods, let’s look at some limits visually.

Example 1.14. Consider the following graph.

Find the following limits:

1. limx→−1−

f(x)

Just follow the graph and approach x = −1 from the left. You’d expect the y-value to

be 1. The open dot doesn’t matter. I still expected it to be y = 1.

limx→−1−

f(x) = 1

36


2. limx→−1+

f(x)

Follow the graph and approach x = −1 from the right. You’d expect the y-value to

be -2. The closed dot doesn’t matter. I expected the y value to be -2 from the right.

Just because it is actually -2 had nothing to do with the limit.

limx→−1+

f(x) = −2

3. limx→−1

f(x)

This is not a one-sided limit. This is a general limit. A general limit exists when it’s

left and right hand limits exist and they equal equal each other. Since

limx→−1−

f(x) = 1 and limx→−1+

f(x) = −2, the general limit

limx→−1

f(x) does not exist

4. limx→1

f(x)

Note, this is NOT a one-sided limit. This is a general limit. A general limit only exists

if it’s left and right hand limits equal each other. So follow the graph to x = 1 from

both sides. Do they approach the same y-value? Yes they do.

limx→1

f(x) = 2

37


5. limx→2

f(x)

This is another general limit. Let’s take a look at the left and right hand limits. If you

follow the graph to x = 2, you see an open dot there. Again, it doesn’t matter if it’s

open or closed. The question is, following the graph, where do I expect the y value to

be when I get close to x = 2. The answer is

limx→2

f(x) = 1

6. limx→−2−

f(x)

Think of ±∞ as a location or place. We are concerned about the behavior of a function.

If there’s an asymptote, we don’t want to say a limit does not exist. You can see that

the function is doing something as it approaches x = −2. It’s heading up to ∞.

limx→−2−

f(x) =∞

7. limx→−2+

f(x)

You can see from the graph there is a vertical asymptote at x = −2. When following

graph to x = −2 from the right, you see the graph heads down to −∞.

limx→−2+

f(x) = −∞

38


8. limx→−2

f(x)

This a general limit. So ask yourself the following question, ”Does the left hand limit

equal the right hand limit?”

limx→2−

f(x) = limx→−2+

f(x)?

From (6) and (7),

limx→−2−

f(x) = ∞ and limx→−2+

f(x) = −∞. So the general limit does not exist since the

left and right hand limits are not the same.

limx→−2

f(x) does not exist

Example 1.15. Consider the following piece-wise function.

f(x) =

x2, x < −1

−x, −1 ≤ x < 2

√x− 2 x ≥ 2

Find the following limits.

1. limx→−1+

f(x) =

This is a right hand limit. We don’t have a graph to look at, but we do have a

function. Since it’s a right hand limit, it means we’re using x-values slightly larger

than -1. According to the function, which ’piece’ do we use for x-values slightly larger

than -1?

limx→−1+

f(x) = limx→−1+

−x = −(−1) = 1

39


2. limx→−1−

f(x)

This is a left hand limit. It means we’re using x-values slightly less than -1. According

to the function, f(x) = x2 when x < −1.

limx→−1−

f(x) = limx→−1−

x2 = (−1)2 = 1

3. limx→−1

f(x)

This is the general limit. Do the left and right hand limits exist and equal each other?

From (1) and (2), we see they do. Therefore,

limx→−1

f(x) = 1

4. limx→2

f(x)

This is a general limit. We haven’t computed the left and right hand limits yet. Let’s

do that now.

Left-Hand Limit

limx→2−

f(x) = limx→2−−x

= −2

40


Right-Hand Limit

limx→2+

f(x) = limx→2+

√x− 2 = 0

Since limx→2−

f(x) 6= limx→2+

f(x), the general limit

limx→2

f(x) does not exist

Example 1.16. Find limx→1−

5 + x

x− 1and lim

x→1+

5 + x

x− 1

Notice that when you plug in x = 1, you divide by 0. You’ll learn in a little bit that when

you plug in an x-value and you get0

0, it means you have more work to do. These methods

we’ll learn a bit later. If you plug in an x-value and get a non-zero over 0, like5

0or−3

0, it

means the limit is −∞, ∞, or it does not exist.

1. limx→1−

5 + x

x− 1

Plug in x = 1. You get6

0. So what’s the limit? Is it −∞, ∞, or DNE?

It won’t be DNE. When you have a one-sided limit and something like6

0, it’s either

−∞ or ∞. We just have to figure out which one. The easiest way is to check the sign

of the top and bottom. Choose an x-value close to 1 but less than 1.

limx→1−

5 + x

x− 1→ +

−

41


The ratio will be (−) negative. Therefore, limx→1−

5 + x

x− 1= −∞

2. limx→1+

5 + x

x− 1

We approach this the same way as (1). After plugging in x = 1, we get6

0. So it’s

either −∞ or ∞. Let’s look at the sign of the top and bottom.

limx→1+

5 + x

x− 1→ +

+

The ratio will be (+) positive. Therefore, limx→1+

5 + x

x− 1=∞

3. limx→1

5 + x

x− 1

This is the general limit. Since the limx→1−

5 + x

x− 16= lim

x→1+

5 + x

x− 1

limx→1

5 + x

x− 1does not exist

Example 1.17. Find limx→−5

x− 1

(5 + x)2

You almost always start with just plugging in x = −5. If fact, that’s exactly what we do

here.

−5− 1

5 + (−5)2=−6

0

42


This means the one-sided limits could be −∞ or ∞. Let’s find out.

Left-Hand Limit - Check the sign of the top and bottom.

limx→−5−

x− 1

(5 + x)2→ −

+

Therefore,

limx→−5−

x− 1

(5 + x)2= −∞

Right-Hand Limit - Check the sign of the top and bottom.

limx→−5+

x− 1

(5 + x)2→ −

+

Therefore,

limx→−5+

x− 1

(5 + x)2= −∞

Since the left and right hand limits equal each other, the answer is

limx→−5

x− 1

(5 + x)2= −∞

43



sin(πx

)I think looking at the graph could shed a little light.

You can’t quite tell what’s happening near the origin. Let’s zoom in a bit.

Ok, that didn’t seem to help. Let’s zoom in maybe a bit more. Zoom!

44


Well that’s odd. The window is set now at [-.01, .01] x [-2, 2]. It actually seems like it’s

getting more chaotic as I zoom it. In fact, that’s exactly what it’s doing. The limit does not

exist because it will never approach a point. Let’s try to approach this a different way.

1. Let’s look at justπ

x.

As x→ 0+,π

x→∞.

2. What happens to sin function when the ’inside’ goes to ∞? It oscillates between -1

and 1 forever.

3. That’s exactly what is happening to limx→0

sin(πx

). As x → 0, the sin function keeps

oscillating between -1 and 1 never approaching a y-value no matter how close we get

to x = 0.

4. Therefore, limx→0

sin(πx

)does not exist.

Example 1.19. Sketch a graph satisfying all the following conditions.

45


1. limx→0

f(x) = 1

2. limx→3−

f(x) = −2

3. limx→3+

f(x) = 2

4. limx→−2

f(x) = −∞

5. f(0) = −1

6. f(3) = 1

Let’s start with (5) and (6) since these are just points on a graph.

The following should be new to you. Since a limit is an expected y-value, we treat the

limits as coordinates. BUT! we treat them as open dots since limits don’t actually tell us

the true y value, just what we would expect.

46


1. limx→0

f(x) = 1

This is a general limit. It means we should expect a y value of 1 when we approach

x = 0 from the left and right. Notice I put an open dot at (0,1) and drew small lines

going left and right from that point.

2. limx→3−

f(x) = −2

This is a left hand limit. We place an open dot at (3,-2) and draw a line approaching

this point from the left.

3. limx→3+

f(x) = 2

47


This is a right hand limit. We place an open dot at (3,2) and draw a line approaching

this point from the right.

4. limx→−2

f(x) = −∞

This is a general limit. Also, this is not a point on the graph but an asymptote at

x = −2. Since it’s a general limit, it means both the left and right hand limits are

−∞. It would look something like this.

5. Now, we just connect all the dots to finish the graph.

48


Wasn’t that fun!

49

1.8 Limit Laws and Methods Brian E. Veitch

1.8 Limit Laws and Methods

Theorem 1.1. If f(x) is a rational function (polynomials are rational BTW) and a is in

the domain of f , then

limx→a

f(x) = f(a)

So what does this mean?

It basically means, if you can plug in x = a and get a ’legal’ number, then that’s the

limit value.


x

x+ 2

Since f(x) is a rational function, limx→3

x

x+ 2=

3

3 + 2=

3

5

Example 1.21. limx→5

3

√4x+ 44

6x− 29

This one isn’t a rational function, but the idea is still the same. If it’s a nice, non-piece-

wise function and a is in the domain, then just plug in x = 5. If you get a nice value out,

then that’s the limit.

limx→5

3

√4x+ 44

6x− 29= 3

√4(5) + 44

6(5)− 29

=3√

64

= 4

50


1.8.1 The Limit Laws

Suppose limx→a

f(x) = L and limx→a

g(x) = M

1. limx→a

f ± g = L±M

The limit of a sum or difference of functions is the sum or difference of their limit values.

2. limx→a

c · f(x) = c · L = c · limx→a

f(x)

The constant can be pulled out of a limit.

3. limx→a

f(x) · g(x) = limx→a

f(x) · limx→a

g(x) = L ·M

Please note that you can only break a limit of a product into a product of limits pro-

vided the individual limits both exist.

4. limx→a

f(x)

g(x)=

limx→a f(x)

limx→a g(x)=

L

M, provided lim

x→ag(x) 6= 0.

5. limx→a

c = c

This should make sense. We are taking a limit of a constant. No matter what x is, the

function value is always c.

1.8.2 Limits by Cancellation

Example 1.22. Consider the limit limx→1

x2 − 6x+ 5

x− 1

51


If you get0

0, it means the limit may still exist.

What do you do? Factor, simplify, and try plugging in x = 1 again.

1. Factor:

limx→1

x2 − 6x+ 5

x− 1= lim

x→1

(x− 1)(x− 5)

x− 1

2. Simplify:

= limx→1

x− 5

3. Try again:

= 1− 5

= −4

Example 1.23. Find limx→−3

2x2 + 5x− 3

x2 − 9

1. Plug in x = −3

limx→−3

2x2 + 5x− 3

x2 − 9=

0

0

2. Since we get0

0, we factor f(x)

52


limx→−3

2x2 + 5x− 3

x2 − 9= lim

x→−3

(2x− 1)(x+ 3

(x− 3)(x+ 3)

3. Simplify

limx→−3

(2x− 1)(x+ 3

(x− 3)(x+ 3)= lim

x→−3

2x− 1

x− 3

4. Try again. Plug in x = −3

limx→−3

2x− 3

x− 3=

2(−3)− 1

−3− 3=

7

6

5. Final Answer:

limx→−3

2x2 + 5x− 3

x2 − 9=

7

6


x2 − 1

x2 − 2x+ 1

1. Plug in x = 1

53


limx→1

x2 − 1

x2 − 2x+ 1=

0

0

2. Since we get0

0, we factor f(x)

limx→1

x2 − 1

x2 − 2x+ 1= lim

x→1

(x− 1)(x+ 1)

(x− 1)(x− 1)

3. Simplify

limx→1

(x− 1)(x+ 1)

(x− 1)(x− 1)= lim

x→1

x+ 1

x− 1

4. Try again. Plug in x = 1

limx→1

x+ 1

x− 1=

1 + 1

1− 1→ 2

0

5. We get anon-zero

0. This means the limit is −∞,∞, or DNE. Since it’s a general limit,

let’s find the left and right hand limits.

54


6. Left Hand Limit

limx→1−

x+ 1

x− 1= −∞

7. Right Hand Limit

limx→1+

x+ 1

x− 1=∞

8. Since

limx→1+

x+ 1

x− 16= lim

x→1−

x+ 1

x− 1

limx→1

x2 − 1

x2 − 2x+ 1does not exist

1.8.3 Limits by Simplifying

Example 1.25. Evaluate limh→0

(3 + h)2 − 9

h

1. Plug in h = 0.

55


limh→0

(3 + h)2 − 9

h=

0

0

Recall,0

0means something weird is happening. There isn’t anything to factor, but we

can foil and simplify.

2. Foil the top and simplify

limh→0

(3 + h)2 − 9

h= lim

h→0

(9 + 6h+ h2)− 9

h

= limh→0

h(6 + h)

h

= limh→0

6 + h

3. Try again. Plug in h = 0

limh→0

6 + h = 6 + 0 = 6

Therefore,

limh→0

(3 + h)2 − 9

h= 6

1.8.4 Limits using the conjugate


√4 + x− 2

x

56


1. Plug in x = 0.

limx→0

√4 + x− 2

x=

0

0

Since we get0

0, we must try to simplify this somehow.

2. Multiply by the conjugate

limx→0

√4 + x− 2

x= lim

x→0

√4 + x− 2

x·√

4 + x+ 2√4 + x+ 2

= limx→0

(4 + x) + 2√

4 + x− 2√

4 + x− 4

x(√

4 + x+ 2)

3. Simplify

= limx→0

(4 + x) + 2√

4 + x− 2√

4 + x− 4

x(√

4 + x+ 2)

= limx→0

4 + x− 4

x(√

4 + x+ 2)

= limx→0

x

x(√

4 + x+ 2)

= limx→0

1√4 + x+ 2

4. Try again. Plug in x = 0.

limx→0

1√4 + x+ 2

=1√

4 + 2=

1

4

57


Therefore, limx→0

√4 + x− 2

x=

1

4

1.8.5 Limits using common denominators

Example 1.27. Find limt→0

(1

t− 1

t2 + t

)

1. As always, try plugging in t = 0.

limt→0

(1

t− 1

t2 + t

)=

1

0− 1

0

These are not numbers you can subtract. We are dividing by zero, so we should try to

simplify this. Let’s simplify by making this into one fraction.

2. Make into one fraction by using a common denominator

limt→0

(1

t− 1

t2 + t

)= lim

t→0

(1

t− 1

t(t+ 1)

)= lim

t→0

(1

t· t+ 1

t+ 1− 1

t(t+ 1)

)= lim

t→0

(t+ 1

t(t+ 1)− 1

t(t+ 1)

)= lim

t→0

(t

t(t+ 1)

)Simplify = lim

t→0

1

t+ 1

Plug in t = 0 =1

0 + 1

= 1

58


Therefore, limt→0

(1

t− 1

t2 + t

)= 1

Example 1.28. Find limx→a

5x

x+ 3− 5a

a+ 3x− a

1. As usual, plug in x = a.

limx→a

5x

x+ 3− 5a

a+ 3x− a

=

5a

a+ 3− 5a

a+ 3a− a

=0

0

2. Since we get0

0, let’s simplify by multiplying the top and bottom by a common denom-

inator.

limx→a

5x

x+ 3− 5a

a+ 3x− a

= limx→a

5x

x+ 3− 5a

a+ 3x− a

· (a+ 3)(x+ 3)

(a+ 3)(x+ 3)

= limx→a

5x(a+ 3)− 5a(x+ 3)

(x− a)(a+ 3)(x+ 3)

= limx→a

5ax+ 15x− 5ax− 15a

(x− a)(a+ 3)(x+ 3)

= limx→a

15x− 15a

(x− a)(a+ 3)(x+ 3)

3. At this point I hope you see that when you have0

0, the goal is to find a way to divide

out (cancel) the denominator that’s giving us 0. Look back at all the other problems

we’ve done and you’ll see that’s exactly what we did to evaluate the limits. This prob-

lem is no different. We need a way to divide out or cancel the denominator x− a.

59


If our goal is to cancel the x − a on the bottom, does the numerator have a factor of

x− a? It sure does.

limx→a

15x− 15a

(x− a)(a+ 3)(x+ 3)= lim

x→a

15(x− a)

(x− a)(a+ 3)(x+ 3)

= limx→a

15

(a+ 3)(x+ 3)

4. Try again. Plug in x = a.

limx→a

15

(a+ 3)(x+ 3)=

15

(a+ 3)2

If you want a really hard one, evaluate the following:

limx→0

(1

x√

1 + x− 1

x

)

This one uses common denominators, conjugates, and simplifying. Show that

limx→0

(1

x√

1 + x− 1

x

)=−1

2

Theorem 1.2. If f(x) ≤ g(x) when x is near a then

limx→a

f(x) ≤ limx→a

g(x)

60


1.8.6 Squeeze Theorem

If f(x) ≤ h(x) ≤ g(x) when x is near a and limx→a

f(x) = limx→a

g(x) = L, then

limx→a

h(x) = L

A picture would be good here to show you how the squeeze theorem works.

If h(x) must remain between f(x) and g(x) when x approaches a, and f(x) and g(x) are

expected to be the same value as x approaches a, then it forces h(x) to approach the same

limit value. In other words, where else can h(x) go if it must stay between f and g near x = a?

Based on the graph, you can probably guess why we call it the Squeeze Theorem.

Example 1.29. Show limx→0

x2 sin

(1

x

)= 0

Let’s take a look at the graph.

61


Based on the graph, we would probably guess limx→0

x2 sin

(1

x

)= 0.

We need to find two functions, one bigger than x2 sin(1/x) and one less than x2 sin(1/x).

We can’t just plug in x = 0 because we’ll divide by 0. Since sin(1/x) is what’s giving us

trouble, let’s try to get rid of it. Let’s bound sin(1/x).

We know from earlier that

−1 ≤ sin(u) ≤ 1

which means

−1 ≤ sin(1/x) ≤ 1

Next, we multiply all sides by x2.

−x2 ≤ x2 sin(1/x) ≤ x2

So how does this help us?

1. Notice the middle function is our function from the limit. Think of this as our h(x).

62


2. Notice that limx→0−x2 = lim

x→0x2 = 0. Think of these as our f(x) and g(x), where f(x) ≤

g(x) near x = 0. Let’s graph all three functions.

Can you see how −x2 and x2 trap or ’squeeze’ x2 sin(1/x)?

3. The inequality shows us that f(x) ≤ h(x) ≤ g(x) near x = 0 and limx→0−x2 = lim

x→0x2 = 0.

Using the Squeeze Theorem, we can also conclude limx→0

x2 sin(1/x) = 0

63

1.9 The Precise Definition of a Limit Brian E. Veitch

1.9 The Precise Definition of a Limit

This section introduces us to a very formal way of defining a limit. It’s really hit or miss

on whether students catch on. If you are a math major, you will encounter this topic again.

There are two types of problems in this section. For now, I’m only going to cover one of

them. You can read the textbook on the 2nd type.

First, we need to introduce some new notation.

• ε - epsilon

• δ - delta

• |f(x)− L| ⇒ represents the distance between f(x) and L.

• |x− a| ⇒ represents the distance between x and a.

Definition 1.4 (Precise Defn of a Limit). Let f be a function on some open interval that

contains a. We say the limit of f(x) as x→ a is L

limx→a

f(x) = L

if for every ε > 0, there is a δ > 0 such that if |x− a| < δ, then |f(x)− L| < ε

64


Let’s break down the definition and apply it to the graph.

1. ε and δ represent distances. These numbers are very small positive numbers.

2. |x− a| < δ means x gets really close to a (close enough so the distance between x and

a is less than δ).

3. What our definition really states is,

If x is really close to a, inside a tiny interval around a, then the y-values associated

with those x-values must be within ε from L. That’s the meaning of |f(x)− L| < ε.

If you are claiming that the limit is L, then I can pick any distance from L and you

should be able to show me that there are x-values whose y-values are that close to L.

I’m finding it a bit difficult to explain this in typed notes. My explanation will be

better in class since I can point to things and answer questions. Let’s just move on to

an example.

NOTE TO ME: Show them a graph where the limit exists and a graph where a limit

does not exist.

Example 1.30. Consider limx→2

x2 = 4

Let ε = 0.1

So if I give you ε (the distance away from the limit value) which creates an interval 3.9

and 4.1 around L = 4, you must have me an interval around x = 2 that gets me y-values

65


between 3.9 and 4.1. Based on the graph,you can see I found the x-values that give me a

y-value of 3.9 and 4.1. This means if I choose any x-value between 1.9748 and 2.0248, I will

get a y-value between 3.9 and 4.1.

δ is how close the x-values need to be from 2 to guarantee y-values between 3.9 and 4.1.

Let δ1 = |1.9784 − 2| = 0.0252 and δ2 = |2.0248 − 2| = 0.0248. The formal δ is the

minimum of {δ1, δ2}.

So δ = 0.0248.

Summary: If I’m ε = .1 away from L = 4, then I only need to be δ = 0.0248 away from

x = 2.

Every ε can have a different δ. As ε gets smaller, so does δ. You can imagine that by just

making the lines at 3.9 and 4.1 get closer together. But this is the point. Given any ε, if

you can find me a δ that can guarantee me y-values close to L, then L is truly the limit value.

I understand that I’m not explaining this that great in my notes. If you have any ques-

tions, please come see me. It so much easier to explain this in person.

66


Example: Consider limx→3

√x+ 1.

Given ε = 0.04, find δ. What does this mean?

We are allowed to be at a distance of ε = 0.04 away from the limit value, which is 2.

The question is, ”How far can x be from 3 so that the y-values are less than 0.04 away from

L = 2.” Let’s take a look at the graph.

So what’s x1 and x2?

You can use your calculator or solve the following equations:

√x1 + 1 = 1.96

and

√x2 + 1 = 2.04

Solving these two equations, we have x1 = 2.8416 and x2 = 3.1616.

Now δ1 represents the distance from x1 to 3 and δ2 represents the distance from x2 to 3.

67


δ1 = |2.8416− 3| = 0.1584

δ2 = |3.1616− 3| = 0.1616

And δ = min{δ1, δ2}

So,

δ = 0.1584

68

1.10 Continuity Brian E. Veitch

1.10 Continuity

Definition 1.5. A function is continuous at x = a if

1. f(a) exists

2. limx→a

f(x) exists

3. limx→a

f(x) = f(a)

If any of these conditions fail, f is discontinuous.

Note: From algebra you probably said a function is not continuous if when tracing the

graph with your pencil, you have to lift your pencil off the paper. Even though this isn’t a

formal way of defining continuity, it’s a good way of looking at it.

Example 1.31. Where is f discontinuous? Continuous?

Let’s start with finding all the places f is discontinuous. If f is discontinuous, it must

fail at least one condition.

69


1. Discontinuous at x = −2.

(a) f(−2) exists? Yes, and f(−2) = 4. Check!

(b) Does limx→−2

f(x) exist? Yes, and limx→−2

f(x) = 2. Check!

(c) Does limx→−2

f(x) = f(−2)? No!

It fails condition (3). Therefore, f is not continuous at x = −2.

2. Discontinuous at x = 1.

(a) f(1) exists? No. There is no y-value for x = 1.

It fails condition (1). Therefore, f is not continuous at x = 1.

3. Discontinuous at x = 3.

(a) f(3) exists? Yes, and f(3) = −1. Check!

(b) Does limx→1

f(x) exist? No. Let’s look at the left and right hand limits.

i. limx→3−

f(x) = −1.

ii. limx→3+

f(x) = 4.

70


Since limx→3−

f(x) 6= limx→3+

f(x), the general limit limx→3

f(x) does not exist.

It fails condition (2). Therefore, f is not continuous at x = 3.

There are no other problem points on the graph. Use your algebra definition of continuity

and trace the graph. You only have to lift your pencil off the paper at x = −2, x = 1, and

x = 3.

This means f in continuous everywhere except x = −2, x = 1, and x = 3. We can write

it in interval notation: (−∞,−2) ∪ (−2, 1) ∪ (1, 3) ∪ (3,∞).

So what type of functions do we typically have continuity issues?

1. Piece-wise functions

2. Rational functions

Example 1.32. Let f(x) =

5− 4x, x ≤ 1

√x+ 1, x > 1

I hope at this point we know that polynomials, radicals, and rational functions are con-

tinuous everywhere on their domain.

1. 5− 4x is continuous everywhere, especially when x < 1

2.√x+1 is continuous everywhere on its domain (x ≥ 0). So it’s continuous when x > 1.

71


The only problem we may have is when we jump from 5− 4x to√x+ 1 at x = 1. Let’s

go through the conditions.

1. Does f(1) exist? Yes and f(1) = 5− 4(1) = 1.

2. Does limx→1

f(x) exist?

Here we have to be careful. A limit exists when its left and right hand limits exist

and equal each other. We use a different function for the left hand limit than the right

hand limit.

(a) Left Hand Limit:

limx→1−

f(x) = limx→1−

5− 4x = 5− 4(1) = 1

(b) Right Hand Limit:

limx→1+

f(x) = limx→1+

√x+ 1 =

√1 + 1 = 2

Since limx→1

f(x) does not exist, f(x) is not continuous at x = 1.

Example 1.33. Let f(x) =x2 − x− 2

x− 2. Where is f continuous?

It’s easier to look for when f is discontinuous. You should be able to see right away

that x 6= 2. Also, note that f(x) is a rational function. Rational functions are continuous

72


everywhere on their domain. Since the only x-value we aren’t allowed to plug in is x = 2,

we conclude f(x) is continuous everywhere except x = 2.

(−∞, 2) ∪ (2,∞)

Example 1.34. Consider the piece-wise function f(x) =

x2 − x− 2

x− 2, x 6= 2

3, x = 2

Where is f continuous?

Note that f(x) is the rational functionx2 − x− 2

x− 2everywhere except x = 2. So just from

that, we know f(x) is continuous everywhere except possibly at x = 2. Let’s go through the

continuity conditions.

1. Does f(2) exist? Yes and f(2) = 3.

2. Does limx→2

f(x) exist?

This is a problem from a couple sections ago. When trying to evaluate a general limit,

just plug in x = 2 and see what happens.

22 − 2− 2

2− 2=

0

0

Ok, so we get0

0. That means we should try one of our limit techniques. In this case

it’s limits with cancellation (factoring)

73


limx→2

x2 − x− 2

x− 2= lim

x→2

(x− 2)(x+ 1)

x− 2

= limx→2

x+ 1

= 3

So limx→2

f(x) exists.

3. Does limx→2

f(x) = f(2)?

Yes. Since f(x) satisfies all three conditions at x = 2, we conclude f(x) is continuous

at x = 2.

Therefore, f(x) =

x2 − x− 2

x− 2, x 6= 2

3, x = 2

is continuous everywhere.

Example 1.35. Is f(x) =

√

4 + x− 2

x, x 6= 0

4, x = 0

continuous at x = 0?

1. Does f(0) exist? Yes, f(0) = 4.

2. Does limx→0

f(x) exist?

74


limx→0

√4 + x− 2

x= lim

x→0

√4 + x− 2

x·√

4 + x+ 2√4 + x+ 2

= limx→0

(4 + x)− 4

x(√

4 + x+ 2)

= limx→0

x

x(√

4 + x+ 2)

= limx→0

1√4 + x+ 2

=1√

4 + 0 + 2

=1

4

So limx→0

f(x) exists.

3. Does limx→0

f(x) = f(0)? No.

f(x) violates condition (3), so f(x) is not continuous at x = 0.

Example 1.36. Let f(x) =sin(x)

2 cos(x)− 1. Where is f(x) continuous?

You should already know that sin(x) and cos(x) are continuous everywhere. The only

thing we’re worried about is when the denominator is zero. So let’s solve the following

equation:

2 cos(x)− 1 = 0

Let’s get started.

2 cos(x)− 1 = 0

2 cos(x) = 1

cos(x) = 1/2

75


So when does cos(x) = 1/2?

x =π

3

and

x =5π

3

But this is only between [0, 2π]. Every full revolution around the unit circle starting with

x =π

3or x =

5π

3is also a solution to the equation.

Therefore, f(x) =sin(x)

2 cos(x)− 1is continuous everywhere except when

x =π

3± 2nπ, n = 0, 1, 2, ...

x =5π

3± 2nπ, n = 0, 1, 2, ...


(x− a)2, x < 0

cos(x), 0 ≤ x < π/2

What value of a would make f(x) continuous at x = 0?

In order to be continuous at x = 0, f(x) needs to satisfy the three conditions of continuity.

1. Does f(0) exist?

Yes, f(0) = cos(0) = 1

76


2. Does limx→0

f(x) exist?

(a) Left Hand Limit

limx→0−

f(x) = limx→0−

(x− a)2 = (0− a)2 = a2

(b) Right Hand Limit

limx→0+

f(x) = limx→0+

cos(x) = cos(0) = 1

So if f(x) is continuous, the left and right hand limits must equal. This means a2 = 1

Solving this, we get a = −1 or a = 1.

1.10.1 Intermediate Value Theorem

Theorem 1.3 (Intermediate Value Theorem). Suppose f is continuous on [a, b] and let N

(a y-value) be any real number between f(a) and f(b), where f(a) 6= f(b), there exists c in

(a, b) such that

f(c) = N

Let’s take a look at a graph.

77


In this example, there are actually three c values that satisfy f(ci) = N . The point is if

you have a continuous function then you must hit every y-value between f(a) and f(b). This

means for every y-value between f(a) and f(b) there must be at least one x-value between

a and b that gets you the y-value.

So why do we care about this theorem? It’s extremely useful when you’re trying to find

roots of a function. Just let N = 0. Take a look at the following graph.

Since f is continuous, f(a) > 0 and f(b) < 0, there must be some c, (x-value) in (a, b)

such that f(c) = 0.

Another way of wording it is,

78


If f is continuous and f(a) and f(b) have opposite signs, then a root must exist in the

interval (a, b).

Concerns about trying to use IVT

What happens if f(a) and f(b) have the same sign, does that mean there won’t be a root

in the interval (a, b). To answer this, look at these two graphs.

In both of these examples, f(a) and f(b) are positive. In the first graph, however, we do

see roots. In the second graph, we don’t have any roots. So what can I take away from this?

If you don’t satisfy the conditions to a theorem (in this case f(a) and f(b) have opposite

signs), you cannot say anything about the conclusion to the theorem. You cannot conclude

there are no roots or that there are roots. Theorems work only when you satisfy their

conditions. Don’t satisfy the conditions, you know absolutely knowing about the conclusion.

Example 1.38. Show there is a solution to x = sin(x).

1. If we want to use the Intermediate Value Theorem, we need a function.

Let f(x) = x− sin(x)

79


2. Choose an interval. Sometimes the interval is given. If it’s not given, try something

simple like (0, 1) or (−2, 2).

3. I’ll try the interval (−1, 1).

• f(−1) = −1− sin(−1) = −0.15853

• f(1) = 1− sin(1) = 0.15853

4. We should also note that f(x) = x− sin(x) is continuous on the interval (−1, 1).

5. So have we satisfied the requirements we need?

(a) f(x) is continuous on the interval? Check!

(b) f(a) and f(b) have opposite signs? Check!

Therefore, somewhere in the interval (−1, 1) the function f(x) must cross the x-axis.

Conclusion: A root must exist in the interval (−1, 1).

80

Brian E. Veitch

2 Derivatives

2.1 Derivatives and Rates of Change

Suppose you have a graph of a function. And suppose you want to find a line that touches

the graph at a certain point so that the slope of the line is the same as the slope of the graph

at that point. It would look something like this:

To find the slope of a tangent (and the slope of a graph at a given point P ), we find the

slopes of many secant lines. What’s a secant line you ask? I’ll show you.

A secant line is simply a line connecting two points on a graph. So how does that help

us find the tangent line? Here’s the idea:

1. Make a secant line with the point P and some other point on the graph (call it Q1)

and find the slope.

2. Pick a point closer to P . Call this Q2. Find the slope of this line.

81

2.1 Derivatives and Rates of Change Brian E. Veitch

3. Pick a point even closer to P . Call it Q3. Find the slope.

4. Continue this until your second point is extremely close to P .

5. The slope of these secant lines should be very close to the slope of the graph at P .

Can you see when the point Q gets closer to P , the secant lines are getting closer to

looking like the tangent line? Since you know how to find the slope of a line, you then

should be able to estimate the slope of the tangent line by using all the secant lines.

Slope of a Secant Line for points (a, f(a) and (x, f(x)).

Slope: m =f(x)− f(a)

x− a

And you saw from above as the point (x, f(x)) moves closer to (a, f(a)), the secant lines

get closer to looking like the tangent line. We spent the last 20 or so pages going over the

82


concept of a limit. The slope of a tangent line is a limit of the slope of the secant lines.

Slope of Tangent Line

mtangent = limx→a

f(x)− f(a)

x− a

Definition 2.1 (Tangent Line). The tangent line to the curve f(x) at point (a, f(a)) is the

line through (a, f(a)) with slope

m = limx→a

f(x)− f(a)

x− a

Example 2.1. Find an equation of the tangent line to y = x2 at point (2,4)

1. Find the slope of the secant line joining (1, 1) to (2, 4).

83


Slope: m =f(2)− f(1)

2− 1=

4− 1

2− 1= 3

2. Find the slope of the secant line joining (1.5, 2.25) to (2, 4).

Slope: m =f(2)− f(1.5)

2− 1.5=

4− 2.25

2− 1.5= 3.5


84


Slope: m =f(2)− f(1.9)

2− 1.9=

4− 3.61

2− 1.9= 3.9


Slope: m =f(2)− f(1.99)

2− 1.99=

4− 3.9601

2− 1.99= 3.99

5. You see as our second point gets closer to (2, 4), the slope of the secant lines appear

to approach m = 4. Let’s find the slope of the tangent line using our limit notation.

Slope of the Tangent Line:

85


limx→2

f(x)− f(a)

x− a= lim

x→2

x2 − 22

x− 2

= limx→2

x2 − 4

x− 2

= limx→2

(x− 2)(x+ 2)

x− 2

= limx→2

x+ 2

= 4

6. Find the equation of the tangent line at (2, 4).

We need two things to find the equation of a line.

(a) The Slope

m = 4

(b) A point

(2, 4)

(c) Now we use the point-slope formula

y − y1 = m(x− x1)

y − 4 = 4(x− 2)

y = 4x− 4

Example 2.2. Find the slope of the tangent line on the curve y = x2 at the point (0,0),

(1,1), (-3, 9), (4,16).

86


Whoa! Do I have to use the limit every time to find the slope of the tangent line? The

answer is no. You really only have to do it once.

Let’s word this question a different way. Since I want the slope at a variety of points,

let’s find the slope of the tangent line at a generic point (a, f(a)).

m = limx→a

f(x)− f(a)

x− a

= limx→a

x2 − a2

x− a

= limx→a

(x− a)(x+ 1)

x− a= lim

x→ax+ a

= 2a

If you want the slope of the tangent line at the point (0,0), let a = 0.

m = 2(0) = 0

If you want the slope of the tangent line at the point (-3,9), let a = −3.

m = 2(−3) = −6

What’s the equation of the tangent line through the point (−3, 9).

y − y1 = m(x− x1)

y − 9 = −6(x− (−3))

87


y = −6x− 9

Another Version of the Tangent Line Slope

Instead of using points (a, f(a)) and (x, f(x)), we use (a, f(a)) and (a+ h, f(a+ h)).

• We let x = a+ h.

• x→ a is the same thing as h→ 0

• Instead of f(x), we use f(a+ h).

• The slope formula is now either one

m = limx→a

f(x)− f(a)

x− a

⇒ m = limh→0

f(a+ h)− f(a)

h

88


Example 2.3. Using this new form, find the slope of the tangent line on the curve y = x2

at the point (-3,9). BTW, we know it should be m = −6.

Let a = −3.

m = limh→0

f(a+ h)− f(a)

h

= limh→0

f(−3 + h)− f(−3)

h

= limh→0

(−3 + h)2 − 9

h

= limh→0

(9− 6h+ h2)− 9

h

= limh→0

−6h+ h2

h

= limh→0

h(−6 + h)

h

= limh→0−6 + h

= −6

Wahoo! The slope matches. Math is awesome! ... too much?

Example 2.4. Find an equation of the tangent line on y =3

xat the point (3,1). Use both

limit versions of the slope formula.

1. Find the slope using limx→a

f(x)− f(a)

x− a

89


m = limx→3

f(x)− f(3)

x− 3

= limx→3

3x− 1

x− 3

ClearDenominators = limx→3

3x− 1

x− 3· xx

= limx→3

3− xx(x− 3)

= limx→3

−(x− 3)

x(x− 3)

= limx→3

−1

x

= −1

3

2. Find the slope using limh→0

f(a+ h)− f(a)

h

m = limh→0

f(3 + h)− f(3)

h

= limh→0

33+h− 1

h

ClearDenominators = limh→0

33+h− 1

h· 3 + h

3 + h

= limh→0

3− (3 + h)

h(3 + h)

= limh→0

−hh(3 + h)

= limh→0

−1

3 + h

= −1

3

3. We confirmed through both limit formulas that the slope is m = −1

3. To find the

equation of the tangent line, we use the point-slope formula.

y − y1 = m(x− x1)

90


y − 1 = −1

3(x− 3)

y = −1

3x+ 2

There is another interpretation of the slopes of the secant lines and the tangent line.

f(b)− f(a)

b− ais the change in f divided by the change in x. It represents the average rate of change of

f as x goes from a to b.

So what else can we call the slope of the tangent line at x = a? It represents the instan-

taneous rate of change at x = a.

Example 2.5. Let f(x) =3

x.

1. Find the average rate of change of f(x) =3

xover the interval [1, 4].

f(3)− f(1)

3− 1=

33− 3

1

2= −1

You an see the slope here

91


So what does an average rate of change of −1 mean? It means if f continued to change

at the same rate (in this case, -1), then for every unit x increases, the function would

decrease by 1 unit.

2. Find the instantaneous rate of change at x = 3.

We already did this. It’s m = −1

3.

This means if f continued at the same rate, then for every 3 units that x increases,

the function decreases by 1 unit.

But you see by looking at the graph, at different points of the graph, you’ll have a

different slope. So the slope is actually a function of x.

What does this mean? It means if you’re given a function f , there is another function

(denoted f ′) then tells us the slope.

Definition 2.2. The derivative of a function f at x = a, denoted f ′(a) is

f ′(a) = limh→0

f(a+ h)− f(a)

h

f ′ is pronounced f prime.

92


We refer to this formula as the limit definition of a derivative. Luckily, we will have

shortcuts to finding these derivatives without limits.

We already used derivatives. Recall when f(x) = x2. From a couple of examples ago, we

found that the slope at the point (a, f(a)) is 2a. Using our current terminology, if f(x) = x2,

the derivative at x = a is f ′(a) = 2a. This derivative function will tell us the slope at any

point on f(x).

Example 2.6. Find f ′(a) when f(x) = 3− 2x+ 3x2.

f ′(a) = limh→0

f(a+ h)− f(a)

h

= limh→0

[3− 2(a+ h) + 3(a+ h)2]− [3− 2a+ 3a2]

h

= limh→0

3− 2a− 2h+ 3(a2 + 2ah+ h2)− 3 + 2a− 3a2

h

= limh→0

3− 2a− 2h+ 3a2 + 6ah+ 3h2 − 3 + 2a− 3a2

h

= limh→0

−2h+ 6ah+ 3h2

h

= limh→0

h(−2 + 6a+ 3h)

h

= limh→0−2 + 6a+ 3h

= −2 + 6a

If f(x) = 3−2x+3x2, we can find the slope at any point (a, f(a)) by using f ′(a) = −2+6a.

1. Find the slope at (0,3).

f ′(0) = −2 + 6(0) = −2

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2. Find the slope at (-2, 19).

f ′(−2) = −2 + 6(−2) = −14

Example 2.7. Find the tangent line on y =x+ 4

x− 2at (4,4).

First, let’s find the slope. We’ll use the following limit formula,

f ′(a) = limh→0

f(a+ h)− f(a)

h

Before starting, note that we already know a. If you’re trying to find the slope at only

one point, you might as well use it.

m = limh→0

(4+h)+4(4+h)−2 −

4+44−2

h

= limh→0

h+8h+2− 4

h

Clear the Denominators → limh→0

h+8h+2− 4

h· h+ 2

h+ 2

= limh→0

h+ 8− 4(h+ 2)

h(h+ 2)

= limh→0

h+ 8− 4h− 8

h(h+ 2)

= limh→0

−3h

h(h+ 2)

= limh→0

−3

h+ 2

= −3

2

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Now, use the point-slope formula to find the equation of the tangent line.

y − y1 = m(x− x1)

y − 4 = −3

2(x− 4)

y = −3

2x+ 10

Velocity

If s is a position function, then the average velocity over t = a to t = a+ h is

vavg =s(a+ h)− s(a)

h

As h→ 0; i.e., limh→0

s(a+ h)− s(a)

h, this is called instantaneous velocity at t = a.

Think of s(t) as the position of an object moving in one dimension (moving left or right

on the x-axis or straight or down).

Example 2.8. Suppose a ball is dropped from an observation deck, 450 m above ground.

1. What is the velocity after 5 seconds? Use s(t) = 4.9t2.

2. Find v(a) - the velocity function

95


v(a) = limh→0

s(a+ h)− s(a)

h

= limh→0

4.9(a+ h)2 − 4.9a2

h

= limh→0

4.9(a2 + 2ah+ h2)− 4.9a2

h

= limh→0

9.8ah+ 4.9h2

h

= limh→0

h(9.8a+ 4.9h)

h

= limh→0

9.8a+ 4.9h

= 9.8a

So the velocity after 5 seconds is

v(5) = 9.8(5) = 49 m/s

3. Find the velocity when it hits the ground.

(a) How long will it take to hit the ground?

Since we are 450 m above ground, solves s(t) = 4.9t2 = 450.

4.9t2 = 450

4.9t2 − 450 = 0

t2 = 450/4.9

t = 9.58 seconds

96


(b) The velocity at 9.58 seconds is

v(9.58) = 9.8(9.58) = 93.88 m/s

97

2.2 Derivative as a Function Brian E. Veitch

2.2 Derivative as a Function

Recall that we defined the derivative as

f ′(a) = limh→0

f(a+ h)− f(a)

h

But since a is really just an arbitrary number that represents an x-value, why don’t we

just use x instead. We already discussed that the derivative is a function of x in the previous

section.

Given f(x), the derivative function is defined to be

f ′(x) = limh→0

f(x+ h)− f(x)

h

and f ′(x) is interpreted as the slope any a point on f(x).

Since the derivative is a function, it also has a graph. Let’s see if we can graph one. Take

a look at the graph of f(x) below.

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When trying to graph the derivative, say this to yourself over and over

The slope on f is the y-value on f ′

Start by looking for easy slopes. The easiest slope to identify is when the slope is 0. That

means the graph goes horizontal (i.e,. the tangent line to the graph is horizontal). You can

see this happens twice. One at x = −0.5 and the other at x = 0.5. That means the y-values

on the derivative should be 0.

Let’s try a couple other places. When x = 0, it looks like the graph has a slope of

m = −1. I drew the tangent line there to get an estimate of the slope. So let’s plot the

point (0,-1) on f ′(x).

99


You can see we now have three points on f ′(x). Let’s try to find two more to make sure

we have a good sketch for f ′(x).

The following graph shows tangent lines at x = −1.3 and x = 1. I’d estimate the slope

at x = −1.3 to be about 4 and the slope at x = 1 to be about 2. Keep in mind, we are

estimating these slopes. If you want the exact slope, we’ll need to find the derivative through

the limit definition.

Adding the points (-1.3, 4) and (1, 2) to f ′(x) we have the following graph,

By connecting the dots, here is our graph of the derivative f ′(x),

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So how good was our graph? Well the original graph was f(x) = x3 − x + 1. Using the

limit definition to find the derivative, which I leave as an exercise to you, we have

f ′(x) = 3x2 − 1

Here’s what we have when we graph these two functions on the same graph,

Wow! We were pretty spot on. Now if you’d like to practice doing this, go to this site

http://webspace.ship.edu/msrenault/GeoGebraCalculus/derivative_try_to_graph.html

101


Let’s do a couple more derivative problems before moving on.

Example 2.9. If f(x) =√x+ 1, find f ′(x) and its domain.

f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

√x+ h+ 1−

√x+ 1

h

= limh→0

√x+ h+ 1−

√x+ 1

h·√x+ h+ 1 +

√x+ 1√

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

2√x+ 1

So f ′(x) =1√x+ 1

, and its domain x > −1.

Let me note something here. The domain of f(x) =√x+ 1 is x ≥ −1, but the domain

of f ′(x) is x > −1. So what happened at x = −1? Let’s take a look at the graph.

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If you following the function moving right to left, you’ll notice the slope getting very

steep. Take a look at these tangent lines near x = −1.

You see how the tangent lines are getting very steep as you get closer to x = −1. In fact,

as you get closer and closer to x = −1, those tangent lines are heading towards a vertical

line. And what’s the slope of a vertical line? Anyone? Anyone? It’s undefined. From algebra

we always said the slope of a vertical line is not defined. That is why the derivative is not

defined there. It’s because that tangent line is a vertical line! And yes... I was yelling.

You might be asking yourself, ”what other weird things can happen that makes a deriva-

tive not exist?” There are three main types that cause a function to not be differentiable at

a point.

1. Vertical Tangent Lines

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2. Corners

Remember that a derivative was defined to be limit of slopes. In this example, when

you approach x = 1 from the left, what would you expect the slope to be? It appears

to be m = −1. What about when you approach x = 1 from the right? It appears to

be m = 1

As you get closer to x = 1, the left and right hand slopes will not match. This is why

a derivative cannot exist. The slopes must match from the left and right.

Let’s take a look at another example.


2x, x < 0

x− x2, x ≥ 0

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When you look at the graph you can probably guess where we have a problem with

differentiability.

It appears we may have a corner at the point (0,1). We shouldn’t be surprised. This

is a piece-wise function and they almost always have problems.

Let’s find f ′(x).

f ′(x) =

2, x < 0

1− 2x, x > 0

So if we are approaching x = 0 from the left, we would expect the slope to be m = 2.

If we approach from the right, we expect the slope to be m = 1−2(0) = 1. Since these

are not the same slopes, we have a corner at x = 0. Here’s the graph of f ′(x).

105


The solid lines are f ′(x) whereas the dotted lines are f(x). There’s a more formal way

of showing f(x) is not differentiable at x = 0. But I hope the graphs help demonstrate

what’s going on with the slopes.

At this point, I want to introduce other notations for derivatives. Instead of f ′(x), you

may also see,

1. y′

This is used when you define a function like y = x2 − 3x+ 1

2.dy

dx

Again, this is used when you define a function as y = f(x).

(a) dy means differentiate the function y

(b) dx means differentiate with respect to x.

3.df

dx

We use this notation when you define the function using f(x). For example, finddf

dx

when f(x) =1

x.

4.d

dx

This is used when you have a function but didn’t use y or f(x). For example,

d

dx

[1

x

]

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This is the same thing as saying find f ′(x) when f(x) =1

x.

5.d

dx[f(x)]

2.2.1 Higher Derivatives

If f is differentiable, then f ′ is a function.

Since f ′ may be a function, it could have its own derivative.

1. f ′(x) = first derivative

2. (f ′(x))′ = f ′′(x) = second derivative

3. (f ′′(x))′ = f ′′′(x) = third derivative

4. f 4(x) = fourth derivative

5. f 10(x) = tenth derivative

After the third derivative we stop using ′ (prime) notation. Writing f ′′′′′′(x) seems a bit

ridiculous.

We’ll do some examples of these after we get our differentiation formulas.

107

2.3 Differentiation Formulas Brian E. Veitch

2.3 Differentiation Formulas

In this section we introduce shortcuts to finding derivatives. Up to this point, we had to find

a derivative using the limit definition. We will derive the shortcuts using the limit definition.

Once we’ve done that, we will use the shortcuts from then on.

Recall:

f ′(x) = limh→0

f(x+ h)− f(x)

h

Take a look at the graph of f(x) = 3 or y = 3. This is called constant function.

Without using the derivative, you should be able to see the slope of a constant function,

i.e., a horizontal line, is 0. Let’s go ahead and prove that.

Theorem 2.1. If f(x) = c, then f ′(x) = 0.

We will use the limit definition to derive the conclusion.

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f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

c− ch

= limh→0

0

h

= limh→0

0

= 0

Let’s move on to power functions. Recall that a power function is f(x) = xn. Let’s take

a look at the following table. You can verify these derivatives on your own.

f(x) = x2 → f ′(x) = 2x

f(x) = x3 → f ′(x) = 3x2

f(x) = x4 → f ′(x) = 4x3

f(x) = x100 → f ′(x) = 100x99

Do you see a pattern for the derivative of a power function?

2.3.1 Power Rule

d

dx[xn] = nxn−1

To find the derivative of a power function, bring the exponent down in front and subtract

the exponent by 1. Let’s prove this.

Proof: Let f(x) = xn. We are going to use the other version of the limit definition.

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f ′(a) = limx→a

f(x)− f(a)

x− a

= limx→a

xn − an

x− a

= limx→a

(x− a)(xn−1 + xn−2a+ xn−3a2 + ...+ xan−2 + an−1)

x− a= lim

x→axn−1 + xn−2a+ xn−3a2 + ...+ xan−2 + an−1

= an−1 + an−2a+ an−3a2 + ...+ aan−2 + an−1︸︷︷︸n terms

= an−1 + an−1 + an−1 + ...+ an−1 + an−1

= nan−1

Remember that a is just an arbitrary letter to represent an x-value. So if f ′(a) = nan−1,

then we really just showed that f ′(x) = nxn−1.

The following rule allows us to differentiate any combination of power functions.

2.3.2 Sum / Difference Rule

If f and g are both differentiable, then

d

dx[f ± g] =

d

dxf ± d

dxg = f ′(x)± g′(x)

In other words, you can differentiate each term one at a time.

Example 2.11. Findd

dx[x8 + 12x5 − 4x4 + 10x3 + 5]

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= 8x7 + 5 · 12x4 − 4 · 4x4 + 3 · 10x3 + 0

= 8x7 + 60x4 − 16x3 + 30x2 + 0

Example 2.12. Find all points on the curve y = x4 − 8x2 + 4, where the tangent line is

horizontal.

Before doing any calculus work, let’s take a look at the graph of y = x4 − 8x2 + 4.

When it asks you to find all places where the tangent line is horizontal, it’s really asking

you where is f ′(x) = 0. From the graph, it appears this happens at x = −2, 0, 2. Let’s verify

that now.

1. Find f ′(x)

f ′(x) = 4x3 − 2 · 8x+ 0

= 4x3 − 16x

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2. To find where you have a slope of 0, set f ′(x) = 0

4x3 − 16x = 0

4x(x2 − 4) = 0

4x(x− 2)(x+ 2) = 0

We have a slope of 0 at x = −2, 0, 2.

The power rule applies to all numbers for n except n = 0.

d

dx[xn] = nxn−1 for all n 6= 0

Example 2.13. Differentiate f(x) =√x

1. Rewrite f(x) as xn

f(x) =√x = x1/2

2. Now user the power rule to find f ′(x)

f ′(x) =1

2x−1/2

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3. You may rewrite this in radical form or without negative exponents. Sometimes you

may have to.

f ′(x) =1

2x1/2or f ′(x) =

1

2√x

Example 2.14. Differentiate f(x) =5

8x7


f(x) =5

8x−7

2. User the power rule to find f ′(x)

f ′(x) = −7 · 5

8x−8

Remember to subtract 1 from the exponent → −7− 1 = −8

3. Clean up the derivative

f ′(x) = − 35

8x8or f ′(x) = −35

8x−8

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Example 2.15. Differentiate f(x) = − 3

10 5√x


f(x) = − 3

10x−1/5

2. User the power rule to find f ′(x)

f ′(x) =1

5· 10

3x−6/5

Note: −1

5− 1 = −6

5

3. Clean up the derivative

f ′(x) =2

3x−6/5

Sometimes you come across derivatives that appear complicated at first, but after sim-

plifying turn out to be fairly nice. Consider the following.

Example 2.16. Let f(x) =x3 − 2 3

√x

x2. Find f ′(x).

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This is a good practice problem for simplifying. If the denominator has one term, dis-

tribute that as a denominator to all the terms in the numerator.

f(x) =x3 − 2 3

√x

x2=x3

x2− 2 3√x

x2

Now simplify each fraction. You probably should rewrite all the terms so they are in the

xn form.

f(x) =x3

x2− 2x1/3

x2

f(x) = x− 2x−5/3

Now differentiate using the power rule.

f ′(x) = 1 +5

3· 2x−8/3

And clean up

f ′(x) = 1 +10

3x−8/3

So much easier...right?

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2.3.3 Product Rule

Now we want to differentiate functions that are defined as a product. For example, f(x) =

(x2 + 5x + 2)(x9 − 3x8). Do you see how f(x) is a product of two functions (x2 + 5x + 2)

and (x9 − 3x8)?

Differentiate using the Product Rule

d

dx[f(x) · g(x)] = f(x) · g′(x) + f ′(x) · g(x)

Example 2.17. Findd

dx[(x2 + 5x+ 2)(x9 − 3x8)]

y′ = (x2 + 5x+ 2) · ddx

[x9 − 3x8

]+

d

dx

[x2 + 5x+ 2

]· (x9 − 3x8)

= (x2 + 5x+ 2) ·(9x8 − 24x7

)+ (2x+ 5) · (x9 − 3x8)

You can foil and attempt to simplify. For now, I’ll leave the answer like this.

One of the biggest mistakes students have with the product rule is the following.

Product Rule - the wrong way

d

dx

[(x2 + 5x+ 2)(x9 − 3x8)

]

An incorrect way of using the product rule is to differentiate each factor and then just

multiply them together. For example,

116


y′ 6= d

dx(x2 + 5x+ 2) · d

dx(x9 − 3x8)

= (2x+ 5)(9x8 − 24x7)

Please do not do this. It’s a good way of getting 0 points.

Example 2.18. Find f ′(x) when f(x) =

(1

x2+ 4x3 − x5/3

)(3x−1/2 − 5

√x2)

1. Before attempting the product rule, rewrite all terms into the xn form.

f(x) = (x−2 + 4x3 − x5/3)(3x−1/2 − x2/5)

If u = x−2 + 4x3 − x5/3 and v = 3x−1/2 − x2/5, then

f ′(x) = uv′ + u′v

2. Now use the product rule

f ′(x) = (x−2 + 4x3 − x5/3) · ddx

[3x−1/2 − x2/5

]+

d

dx

[x−2 + 4x3 − x−5/3

]· (3x−1/2 − x2/5)

= (x−2 + 4x3 − x5/3)(−3

2x−3/2 − 2

5x−3/5

)+

(−2x−3 + 12x2 +

5

3x−8/3

)(3x−1/2 − x2/5)

3. You can clean this up a bit but the goal of this section is to show you how to properly

use the product rule. Simplifying some of these will come later.

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2.3.4 Quotient Rule

If you haven’t guessed already, the quotient rule allows us to differentiate functions that look

like quotients. For example, we can differentiate f(x) =3x3 + x

x2 + 10

Differentiating using the Quotient Rule

d

dx

[f(x)

g(x)

]=g(x) · f ′(x)− f(x) · g′(x)

[g(x)]2

One way students remember this formula is by remembering this,

d

dx

[hi

lo

]=lo-D-hi− hi-D-lo

lo2

If you say it enough times, it should stick. After 10 years, I still say this when using the

quotient rule (out loud mostly).

Example 2.19. Findd

dx

[3x3 + x

x2 + 10

]

1. Make sure all terms are in the xn form. It appears they are, so let’s move on.

2. Use the quotient rule.

Let v = 3x3 + x and u = x2 + 10. Then

f ′(x) =uv′ − vu′

u2

118


f ′(x) =(x2 + 10) · d

dx[3x3 + x]− (3x3 + x) · d

dx[x2 + 10]

[x2 + 10]2

=(x2 + 10) · (9x2 + 1)− (3x3 + x)(2x)

(x2 + 10)2

3. You may want practice distributing and simplifying the numerator. At some point we

will have to set the numerator equal to 0.

Example 2.20. Differentiate y =5x2 − 4

√x

x2 − 15x2

1. Change all terms into the xn form.

y =5x2 − x1/4

x2 − 15x−2

2. Use the Quotient Rule

y′ =(x2 − 15x−2) · d

dx

[5x2 − x1/4

]− (5x2 − x1/4) · d

dx[x2 − 15x−2]

(x2 − 15x−2)2

=(x2 − 15x−2) · (10x− 1

4x−3/4)− (5x2 − x1/4) · (2x+ 30x−3)

(x2 − 15x−2)2

Example 2.21. Find the equation of the tangent line to y = (1 + 2x)2 at (1,9).

119


1. To find the slope of the tangent line, we need to find y′.

2. We have two options to find y′.

(a) Foil and use the power rule → y = (1 + 2x)2 = (1 + 2x)(1 + 2x) = 1 + 4x+ 4x2

y′ = 4 + 8x

(b) Differentiate using the product rule → y = (1 + 2x)(1 + 2x)

y′ = (1 + 2x) · ddx

(1 + 2x) + (1 + 2x) · ddx

(1 + 2x)

y′ = (1 + 2x)(2) + (1 + 2x)(2)

y′ = 8x+ 4

Either way, we get y′ = 8x+ 4

3. To find the slope at (1,9), we find f ′(1)

f ′(1) = 8(1) + 4 = 12

4. Use the point-slope formula to find the equation of the tangent line

y − y1 = m(x− x1)

y − 9 = 12(x− 1)

y = 12x− 3

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Note, when differentiating, don’t use more than one rule. It will take some practice

but eventually you’ll know which rule to use.

2.3.5 Higher Derivatives

I won’t spend too much time on this section. We will do lots of higher derivatives in the

coming sections.

Recall

1. f ′(x) = first derivative

2. (f ′(x))′ = f ′′(x) = second derivative

3. (f ′′(x))′ = f ′′′(x) = third derivative

4. f 4(x) = fourth derivative

5. f 10(x) = tenth derivative

Example 2.22. Let f(x) = x3 − 4x2 + 3x1/2 − 1

x+ 4. Find f ′ and f ′′.

1. As always, rewrite so each term is of the form xn.

f(x) = x3 − 4x2 + 3x1/2 − x−1 + 4

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2. Find f ′

f ′(x) = 3x2 − 8x+1

2x−1/2 + x−2 + 0

3. Find f ′′

f ′′(x) = 6x− 8− 1

4x−3/2 − 2x−3

122

2.4 Derivatives of Trig Functions Brian E. Veitch

2.4 Derivatives of Trig Functions

Before we go ahead and derive the derivative for f(x) = sin(x), let’s look at its graph and

try to graph the derivative first.

f(x) = sin(x)

Window [−2π, 2π], unit - π/2

1. Remember that the slope on f(x) is the y-value on f ′(x).

2. Identify the easy slopes first. I see a slope of 0 at x = π/2 and x = 3π/2.

f(x) = sin(x) f ′(x) = ?

123


3. Estimate some other slopes. For example, at x = 0, x = −2π, and x = 2π, I’d estimate

the slope to be about m = 1.

f(x) = sin(x) f ′(x) = ?

4. Estimate some other slopes. For example, at x = −pi and x = π, I’d estimate the

slope to be about m = −1.

f(x) = sin(x) f ′(x) = ?

5. Ok, that’s probably good. Try to connect the dots using a smooth curve.

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f ′(x) =?? f ′(x) = cos(x)

It appears that if f(x) = sin(x), then f ′(x) = cos(x). Let’s see if that’s true.

Let f(x) = sin(x). Using the limit definition of the derivative, we get the

125


f ′(x) = limh→0

f(x+ h)− f(x)

h

= limh→0

sin(x+ h)− sin(x)

h

Use Trig Identity → sin(x+ h) = sin(x) cos(h) + sin(h) cos(x)

= limh→0

sin(x) cos(h) + sin(h) cos(x)− sin(x)

h

= Rearrange the numerator

= limh→0

sin(x) cos(h)− sin(x) + sin(h) cos(x)

h

= limh→0

sin(x) cos(h)− sin(x)

h+

sin(h) cos(x)

h

= limh→0

sin(x)cos(h)− 1

h+ lim

h→0cos(x)

sin(h)

h

= sin(x) limh→0

cos(h)− 1

h+ cos(x) lim

h→0

sin(h)

h

= sin(x) · 0 + cos(x) · 1

= cos(x)

Note: limh→0

sin(h)

h= 1 and lim

h→0

cos(h)− 1

h. We already knew the first limit. The second

limit can be found in most calculus texts.

I’m pretty sure we are officially done using the limit definition of the derivative. Yay!!

126


Let’s go through the derivatives of the six trig functions.

2.4.1 Derivatives of Trig Functions

1.d

dx[sin(x)] = cos(x)

2.d

dx[cos(x)] = − sin(x)

3.d

dx[tan(x)] = sec2(x)

4.d

dx[sec(x)] = sec(x) tan(x)

5.d

dx[cot(x)] = − csc2(x)

6.d

dx[csc(x)] = − csc(x) cot(x)

Example 2.23. Findd

dx[5x3 cos(x)]

1. First, identify which rule you should use. I see a power function 5x3 and a trig function

cos(x) and they’re being multiplied.

2. Use the product rule.

f ′(x) = 5x3 · ddx

[cos(x)] + cos(x) · ddx

[5x3]

= 5x3 · (− sin(x)) + cos(x) · 15x2

= −5x3 sin(x) + 15x2 cos(x)

127


3. You probably should start simplifying your derivatives. In this case, I’ll factor out the

GCF.

f ′(x) = −5x2(x sin(x)− 3 cos(x))

Example 2.24. If f(x) =

√x− sec(x)

tan(x) + 1400

1. First, make sure all non-trig terms are in the proper xn form.

f(x) =x1/2 − sec(x)

tan(x) + 1400

2. Use the Quotient Rule

f ′(x) =(tan(x) + 1400) · d

dx

[x1/2 − sec(x)

]− (x1/2 − sec(x)) · d

dx[tan(x) + 1400]

(tan(x) + 1400)2

=

(tan(x) + 1400) ·(

1

2x−1/2 − sec(x) tan(x)

)− (x1/2 − sec(x)) · (sec2(x))

(tan(x) + 1400)2

Example 2.25. Findd

dx[4x3 sin(x) cot(x)]

1. Notice that f(x) is a product of three functions. We don’t have a rule for three

functions, only two. This means you need to rewrite this as two functions. I’d do

something like this,

128


f(x) =(4x3 sin(x)

)· cot(x)

Now it’s a product of two functions, 4x3 sin(x) and cot(x).

2. Start the product rule. Since this is a complicated problem, show all steps.

f ′(x) =(4x3 sin(x)

)· ddx

[cot(x)] + cot(x) · ddx

[4x3 sin(x)

]

=(4x3 sin(x)

)·[− csc2(x)

]+ cot(x) · [PRODUCT RULE AGAIN]

= −(4x3 sin(x)

)· csc2(x) + cot(x) ·

[4x3 · d

dx[sin(x)] + sin(x) · d

dx

[4x3]]

= −(4x3 sin(x)

)· csc2(x) + cot(x) ·

(4x3 · cos(x) + 12x2 · sin(x)

)

If you want, you can distribute cot(x) in the second half. It is neither required. The goal

of this section is to get you used to the differentiation formulas. The derivatives are messy

and complicated. Some really aren’t meant to be simplified. Just be careful and take your

time. Show as much work as you can. You’re less likely to miss something if you’re writing

everything out. You’ll have plenty of practice simplifying derivatives later. Trust me...

2.4.2 Product Rule for Three Functions

d

dx[f(x) · g(x) · h(x)] = f ′(x)g(x)h(x) + f(x)g′(x)h(x) + f(x)g(x)h′(x)

129

2.5 The Chain Rule Brian E. Veitch

2.5 The Chain Rule

This is our last differentiation rule for this course. It’s also one of the most used. The best

way to memorize this (along with the other rules) is just by practicing until you can do it

without thinking about it.

Ok, so what’s the chain rule? It’s the rule that allows us to differentiate a composition

of functions. One way to think about a composition of functions to think of it having an

’outside’ function and ’inside’ function. For example,

1. y = (3x8 +√x)

25

(a) The outside function: f(x) = x25

(b) The inside function: g(x) = 3x8 +√x

To check this, what is f ◦ g?

f ◦ g = f (g(x))

= f(3x8 +

√x)

=(3x8 +

√x)25

2. y =√x3 + 1

(a) The outside function: f(x) =√x

130


(b) The inside function: g(x) = x3 + 1

(c) f (g(x)) =√x3 + 1

3. y = cos4(x)

(a) Remember that cos4(x) is a special notation. It represents (cos(x))4

(b) The outside function: f(x) = x4

(c) The inside function: g(x) = cos(x)

(d) f (g(x)) = (cos(x))4

4. y = sec2(√

(x))

(a) A composition of functions doesn’t have to be just a composition of two functions.

In this example, it’s a composition of three functions.

(b) Rewrite y as y = (sec(√x))

2

(c) The outside function: f(x) = x2

(d) 1st inside function: g(x) = sec(x)

(e) 2nd inside function: h(x) =√x

131


(f) The composition is y = f (g (h(x)))

We went through all those examples because it’s important you know how to identify the

composition. What is the outside function? What is the inside function? Is there another

inside function?

2.5.1 Chain Rule

If g is differentiable at x and f is differentiable at g(x), then the composite function f (g(x))

is differentiable and

d

dx[f (g(x))] = f ′ (g(x)) · g′(x)

Steps:

1. Differentiate the outside function f

2. Leave the inside function alone g

3. Multiply by the derivative of the inside function g′

Example 2.26. Find y′ when y = (x3 + 5x)5

1. Identify the outside function f and the inside function g.

132


(a) Outside function: f(x) = x5, with f ′(x) = 5x4

(b) Inside function: g(x) = x3 + 5x, with g′(x) = 3x2 + 5

2. Write out the chain rule formula.

y′ = f ′ (g(x)) · g′(x)

3. Fill in the parts (You won’t be doing it like this once you get better at it)

y′ = f ′ (g(x)) · g′(x)

= 5 (g(x))4 · g′(x)

= 5(x3 + 5x

)4 · (3x2 + 5)

Example 2.27. If y = cos2(x), find y′.

1. Rewrite y as y = (cos(x))2

2. Identify the outside and inside functions.

(a) The outside function: f(x) = x2, with f ′(x) = 2x

(b) The inside function: g(x) = cos(x), with g′(x) = − sin(x)

133


3. Use the chain rule

y′ = 2 (cos(x)) · − sin(x)

y′ = −2 cos(x) sin(x)

Example 2.28. If y = sec(5x), find y′.

You may be surprised that many students get this wrong. It’s because of the derivative

of sec(x). Let’s get started.

1. Identify the outside and inside functions

(a) The outside function: f(x) = sec(x), with f ′(x) = sec(x) tan(x).

(b) The inside function: g(x) = 5x, with g′(x) = 5.

2. Use the Chain Rule

y′ = f ′ (g(x)) · g′(x)

= sec (g(x)) tan (g(x)) · g′(x)

= sec(5x) tan(5x) · 5

= 5 sec(5x) tan(5x)

Can you find y′′? It’s not that easy. Notice that it is now a product of functions. This

means you’ll have to do the product rule and the chain rule in the same problem.

134


Example 2.29. Findd

dx[cos(x−5 + sin(x))]

1. Identify the outside and inside functions

(a) The outside function: f(x) = cos(x), with f ′(x) = − sin(x)

(b) The inside function: g(x) = x−5 + sin(x), with g′(x) = −5x−6 + cos(x)

2. When I do the chain rule, I say the following in the head,

(a) Differentiate the outside function and leave the inside alone

(b) Multiply by the derivative of the inside


y′ = − sin(x−5 + sin(x)

)·(−5x−6 + cos(x)

)

So far we’ve differentiated a composition of two functions. Recall that a composition of

functions can have any number of functions. Here’s how the chain rule looks when you have

a composition of three functions.

135


2.5.2 Chain Rule with a composition of three functions

d

dx[f (g (h(x)))] = f ′ (g (h(x))) · g′(h(x)) · h(x)

Example 2.30. Find y′ when y = cos5(√x).

1. Anytime you’re asked to differentiate a trig function with an exponent like this, rewrite

it.

y =(cos(√x))5

2. Identify the inside and outside functions. I don’t know if you can tell yet, but there’s

more than one inside function.

(a) The outside function: f(x) = x5, with f ′(x) = 5x4

(b) The inside function: g(x) = cos(√x). Note that g′(x) requires the chain rule.


y′ = 5(cos(√x)4 · d

dx

[cos(√x)]

= 5(cos(√x)4 · − sin(

√x) · d

dx

[√x]

= 5(cos(√x)4 · − sin(

√x) · −1

2x−1/2

=5

2x−1/2 · (cos(

√x))4 · sin(

√x)

136


Now it’s time for some more complicated derivatives. These involve using the chain,

product, and quotient rule at the same time.

Example 2.31. Find y′ when y′ =(cos(x) + x3)4

tan(x2).

1. Notice that when we differentiate the numerator (cos(x) + x3) and the denominator

tan(x2), we will use the chain rule. But the overall function is a quotient. So which

one goes first?

2. Since the main function is a quotient, we use the quotient rule.

3. Let’s get started with the quotient rule.

y′ =tan(x2) · d

dx[(cos(x) + x3)4]− (cos(x) + x3)4 · d

dx[tan(x2)]

(tan(x2))2

4. I STRONGLY recommend showing all your work. Notice that all I have to do now

is fill in thosed

dxwith the correct derivatives. Breaking it up like this is very useful.

d

dx

[(cos(x) + x3)4

]= 4

(cos(x) + x3

)3 · ddx

[cos(x) + x3

]= 4

(cos(x) + x3

)3 · (− sin(x) + 3x2)

d

dx

[tan(x2)

]= sec2(x2) · d

dx

[x2]

= sec2(x2) · 2x

137


5. Now just replace thed

dx’s and we’re done.

y′ =tan(x2) · d

dx[(cos(x) + x3)4]− (cos(x) + x3)4 · d

dx[tan(x2)]

(tan(x2))2

=tan(x2) ·

[4 (cos(x) + x3)

3 · (− sin(x) + 3x2)]− (cos(x) + x3)4 · [sec2(x2) · 2x]

(tan(x2))2

Example 2.32. Find y′ when y = (x3 − 2x11)4 · (1− 15x2)100.

1. The overall function is a product of two functions (x3− 2x11)4 and (1− 15x2)100. We’ll

start with the product rule.

2. Product Rule

y′ = (x3 − 2x11)4 · ddx

[(1− 15x2)100

]+ (1− 15x2)100 · d

dx

[(x3 − 2x11)4

]

3. We have two chain rules to work on now.

(a)d

dx[(1− 15x2)100]

d

dx

[(1− 15x2)100

]= 100(1− 15x2)99 · d

dx

[1− 15x2

]= 100(1− 15x2)99 · (−30x)

138


(b)d

dx[(x3 − 2x11)4]

d

dx

[(x3 − 2x11)4

]= 4(x3 − 2x11)3 · d

dx

[x3 − 2x11

]= 4(x3 − 2x11)3 · (3x2 − 22x10)

4. Now put it all together

y′ = (x3 − 2x11)4 · ddx

[(1− 15x2)100

]+ (1− 15x2)100 · d

dx

[(x3 − 2x11)4

]= (x3 − 2x11)4 ·

[100(1− 15x2)99 · (−30x)

]+ (1− 15x2)100 ·

[4(x3 − 2x11)3 · (3x2 − 22x10)

]

You have some common factors you can pull out. I’ll leave that to you as an exercise.

At some point in the near future, you will have to know how to factor this.

Example 2.33. Find y′ when y =

√x2 + 1

x3 − 2 cos(x)

1. Make sure everything is written in the correct form.

y =

(x2 + 1

x3 − 2 cos(x)

)1/2

2. This is a chain rule first, then a quotient rule.

3. Start the chain rule

139


y′ =1

2

(x2 + 1

x3 − 2 cos(x)

)−1/2· ddx

[x2 + 1

x3 − 2 cos(x)

]

4. So now we just have to findd

dx

[x2 + 1

x3 − 2 cos(x)

]

d

dx

[x2 + 1

x3 − 2 cos(x)

]=

(x3 − 2 cos(x)) · (2x)− (x2 + 1) · (3x2 + 2 sin(x))

(x3 − 2 cos(x))2

5. Put it all together

y′ =1

2

(x2 + 1

x3 − 2 cos(x)

)−1/2· ddx

[x2 + 1

x3 − 2 cos(x)

]=

1

2

(x2 + 1

x3 − 2 cos(x)

)−1/2· ddx

[(x3 − 2 cos(x)) · (2x)− (x2 + 1) · (3x2 + 2 sin(x))

(x3 − 2 cos(x))2

]

Again, I’m not focused with simplifying at the moment. You will have the rest of the

semester simplifying these things. Plus, I’m also giving examples that are nasty to

simplify.

Example 2.34. If h(x) = g(f(x)), m(x) = g(x3), f(2) = 8, g′(8) = −2, and f ′(2) = 3, find

h′(2).

1. Find h′(x) using the chain rule

h′(x) = g′(f(x)) · f ′(x)

140


2. Plug in x = 2

h′(2) = g′(f(2)) · f ′(2)

= g′(8) · 3

= −2 · 3

= −6

Example 2.35. Using the previous example’s information, find m′(2).

1. Find m′(x)

m′(x) = g′(x3) · ddx

[x3]

= g′(x3) · 3x2

2. Plug in x = 2

m′(2) = g′(23) · 3(2)2

= g′(8) · 12

= −2 · 12

= −24

As I stated before, practice these rules. In your textbook there are plenty of problems to

practice.

141

2.6 Implicit Differentiation Brian E. Veitch

2.6 Implicit Differentiation

The number one thing to remember about implicit functions is that y is a function of x.

Sometimes we know what y is and sometimes we don’t. For example, if y = x2 +x, we know

y′ = 2x+ 1. But what happens when we don’t know what y is? Then we just use y′ as the

derivative. That’s it!

1. Explicit functions

Explicit functions are functions where we know exactly what y is in terms of x.

(a) y = x2 + 5x+ 1 describes y as an explicit function of x

(b) y =√x2 − 1 describes y as an explicit function of x

(c) y =1

x

2. Implicit Functions

Implicit functions do not tell us what y is in terms of x. But that’s ok. Anytime we

have to differentiate y when we don’t know what it is, just write y′. Here are some

examples of implicit functions.

(a) x2 + y2 = 1

(b) 20x− y2 = 2xy

142


(c) 6x− 12y = 3

(d) xy = 1

One thing to note is implicit functions can sometimes be written in an explicit form. For

example,

1. xy = 1 can be written as y =1

x

2. x2 + y2 = 1 can be written as y = ±√

1− x2

So if you’re asked to differentiate an implicit function, you may be able to rewrite the

function. However, this is not always the case. For example,

20x− y2 = 2xy

cannot be written as an explicit function of x. But that’s what this section is all about.

How do we differentiate functions defined implicitly?

Before we go ahead and differentiate complicated implicit functions, let’s start small.

1.d

dx[y] = y′

2.d

dx[y2] = 2y · y′

Note, that this is actually a chain rule. Maybe it would help if you look at y2 as [y]2,

where

143


(a) The outside function is x2, withd

dx[x2] = 2x

(b) The inside function is y, withd

dx[y] = y′

3.d

dx[y3] = 3y2 · y′

4.d

dx[10y5] = 50y4 · y′

5.d

dx[sin(y)] = cos(y) · y′

6.d

dx

[y2/3

]=

2

3y−1/3 · y′

7.d

dx[6xy]

Note, this is actually a product rule of two functions 6x and y. Use the product rule

as you normally would. When it’s time to differentiate y use y′.

d

dx[6xy] = 6x · d

dx[y] + y · d

dx[6x]

d

dx[6xy] = 6x · y′ + 6y

8.d

dx[5x4y8]

Again, this is actually a product rule of two functions 5x4 and y8.

144


d

dx

[5x4y8

]= 5x4 · d

dx

[y8]

+ y8 · ddx

[5x4]

= 5x4 ·[8y7 · y′

]+ y8 ·

[20x3

]

= 40x4y7 + 20x3y8

9.d

dx[tan(x3y2)]

This is a chain rule first, with a product rule inside.

d

dx

[tan(x3y2)

]= sec(x3y2) · d

dx

[x3y2

]

= sec(x3y2) ·[x3 · d

dx[y2] + y2 · d

dx[x3]

]

= sec(x3y2) ·[x3 · (2y · y′) + y2 · (3x2)

]

= sec(x3y2) ·[6x3yy′ + 3x2y2

]

Just to show you that implicit and explicit functions are really doing the same thing,

let’s differentiate two functions which represent the same half circle. Then find the slope at

the point (3,4). Here are the two functions,

y =√

25− x2 and x2 + y2 = 25

145


Here’s a picture of what’s being asked.

Find the slope of the tangent line shown.

1.d

dx

[y =√

25− x2]

This is a chain rule problem. You should rewrite y as y = (25− x2)1/2

y′ =1

2(25− x2)−1/2 · d

dx

[25− x2

]=

1

2(25− x2)−1/2 · [−2x]

= −1x(25− x2)−1/2

Plug in x = 3 for the slope at the point (3, 4).

y′(3) = −1(3)(25− (3)2)−1/2 = −3(16)−1/2 =−3

161/2= −3

4

146


2.d

dx[x2 + y2 = 25]

2x+ 2y · y′ = 0

Solve for y′

2y · y′ = −2x

y′ = −2x

2y

y′ = −xy

Plug in x = 3 and y = 4.

y′ = −3

4

So you see that if you have the same function, one defined explicitly and the other im-

plicitly, you still get the same answer.

Example 2.36. Find y′ when x2y3 = 2x+ 1.

1. Differentiate both sides

x2 · ddx

[y3]

+ y3 · ddx

[x2]

= 2

x2 ·[3y2 · y′

]+ y3 · [2x] = 2

147


2. Solve for y′

3x2y2 · y′ + 2xy3 = 2

3x2y2 · y′ = 2− 2xy3

y′ =2− 2xy3

3x2y2

Example 2.37. Finddy

dxwhen y cos(x) = 1 + sin(xy)

1. Differentiate both sides

y · ddx

[cos(x)] + cos(x) · ddx

[y] = cos(xy) · ddx

[xy]

−y · sin(x) + cos(x) · y′ = cos(xy) · [xy′ + 1y]

−y sin(x) + cos(x) · y′ = x cos(xy) · y′ + y cos(xy)

2. Solve for y′ by bringing all terms with y′ to the left side.

cos(x) · y′ − x cos(xy) · y′ = y sin(x) + y cos(xy)

3. Factor out y′ and solve

148


y′ [cos(x)− x cos(xy)] = y sin(x) + y cos(xy)

y′ =y sin(x) + y cos(xy)

cos(x)− x cos(xy)

Example 2.38. Find the slope of the tangent line to the curve x2 − 2xy − y2 = 1 at (5,2).

1. Differentiate x2 − 2xy − y2 = 1

2x− [2x · y′ + 2y]− 2y · y′ = 0

2x− 2x · y′ − 2y − 2y · y′ = 0

2. Solve for y′

−2x · y′ − 2y · y′ = 2y − 2x

y′ · (−2x− 2y) = 2y − 2x

y′ =2y − 2x

−2x− 2y

3. To find the slope at (5,2), plug in x = 5 and y = 2.

y′(5, 2) =2(2)− 2(5)

−2(5)− 2(2)

=−6

−14

=3

7

149

2.7 Rates of Change Brian E. Veitch

2.7 Rates of Change

In this section we’ll go over velocity and acceleration. We will also see how the second

derivative can be useful to describe the behavior of velocity. For example, acceleration can

tell us if the velocity is increasing or decreasing over time.

Definition 2.3 (Instantaneous Rate of Change). If s(t) is the position function that moves

in a straight line (left or right), then limδt→0

δs

δtrepresents the average velocity over δt (some

change in time).

1. v(t) =ds

dtis the instantaneous velocity and

v(t) = s′(t)

Velocity is the derivative to the position function. This should make sense as a deriva-

tive is a rate of change and the change in position over change in time is velocity.

2. a(t) =dv

dt= v′(t) is acceleration.

Recall from previous math or physics courses that the change in velocity over change in

time is acceleration. Acceleration tells us if an object is increasing its speed, decreasing

its speed, or staying constant.

For example, a car traveling at exactly 50 mph (never changing its speed) has an ac-

celeration of 0.

The next example is very long. Are you ready? Here we go!

150


Example 2.39. The position of a particle is given by s(t) = t3 − 6t2 + 9t

1. Find the velocity at time t.

When you’re asked to find something at time t, it’s just asking for that function. In

this case, just find v(t). Recall that v(t) is s′(t).

v(t) = s′(t) = 3t2 − 12t+ 9

2. What is the velocity after 2 seconds?

We have a velocity function. Just plug in t = 2 seconds into v(t).

v(2) = 3(2)2 − 12(2) + 9 = −3 m/s

What does the negative sign mean?

A negative sign on velocity indicates its movement (direction) is the opposite direc-

tion of what we consider forward. In this case, the negative sign means the particle is

moving to the left (backwards).

3. What is the velocity after 4 seconds?

151


v(4) = 3(4)2 − 12(4) + 9 = 9 m/s

Consider these two velocities for a moment. At 2 seconds the particle is moving to

the left at a speed of 3 m/s. At 4 seconds it’s moving to the right at 9 m/s. That

means at some point between 2 and 4 seconds, the particle had to stop and turn around.

4. When is the particle at rest?

As I mentioned in the last part, somewhere between 2 and 4 seconds, the particle had

to stop and turn around. We want to know when the particle stops. Common sense

says a particle is at rest when its velocity is 0 (i.e., not moving). So let’s set v(t) = 0.

3t2 − 12t+ 9 = 0

3(t2 − 4t+ 3) = 0

3(t− 3)(t− 1) = 0

Solving for t, we get t = 3 and t = 1 seconds.

The particle is at rest after 1 seconds and 3 seconds. This makes sense that the particle

stops at t = 3 seconds since we knew the particle had to stop somewhere between 2

and 4 seconds.

5. When is the particle moving forward?

152


A particle moves forward when its velocity is positive (+), and moving backwards when

its velocity is negative (-). We are going to use a number line to figure this out. Here’s

what we’re going to do. Plot on a number line when the velocity is 0 (when the particle

is at rest).

When you evaluate v(t), it will tell you its velocity. We are concerned if that velocity

is positive (+) or negative (-). So here’s what we’re going to do,

(a) Pick a point in each region.

(b) Evaluate those points with v(t) and determine if the velocity is (+) or (-).

v(.5) = (+)

v(2) = (−)

v(4) = (+)

153


(c) Mark this on the number line

(d) So what does this tell us?

The particle is moving forward between 0 and 1 seconds and after 3 seconds.

The particle is moving backwards between 1 and 3 seconds.

6. Draw a diagram to represent the motion of the particle.

Note, the particle doesn’t actually move up. I only draw it like that so we don’t overlap

the path of the particle.

7. Find the total distance traveled in the first 5 seconds.

154


This is a little trickier than just evaluating s(5) = 20m. What does s(5) represent?

It represents the distance away from the starting point (displacement). It doesn’t

represent an accumulation of the distance traveled over 5 seconds. The diagram is

useful because it helps visually where the particle moved and how far it moved.

(a) At t = 1 the particle is 4m away from the starting point. So the accumulated

distance traveled so far is 4m.

(b) At t = 3 seconds it’s back at the starting point. We know this because s(3) = 0.

This means it’s traveled another 4m, for a total of 8m.

**Just for the sake of argument, suppose s(3) = −1 instead of s(3) = 0. This

means at t = 3 seconds, the particle is 1m behind the starting point. So between

t = 1 and t = 3 seconds, the particle actually moved 5m. But, s(3) = 0 so it only

traveled 4m back.

(c) At t = 5 the particle is 20m away from the starting point. Since at t = 3 seconds

it was at the starting point, it means between 3 and 5 seconds the particle traveled

another 20m.

**Again for the sake or argument, suppose s(3) = −1. That would mean between

3 and 5 seconds the particle traveled 20 + 1 = 21m since it was starting 1m

behind the starting point.

(d) Anyhow, the accumulated distance traveled is 4 + 4 + 20 = 28m.

8. Find the acceleration at time t.

155


Recall that a(t) = v′(t).

a(t) = v′(t) = 6t− 12

9. Find the acceleration at t = 1.5 seconds, 2 seconds, 4 seconds.

(a) t = 1.5 seconds

a(1.5) = 6(1.5)− 12 = −3m

s2

(b) t = 2 seconds

a(2) = 6(2)− 12 = 0m

s2

(c) t = 4 seconds

a(4) = 6(4)− 12 = 12m

s2

10. When is the speeding up? When is it slowing down?

Recall that acceleration tells us the rate of change in velocity. The following are the

rules to determine if the particle is speeding up or slowing down.

156


(a) Speeding up: Both v(t) and a(t) must have the same sign.

v(t) > 0 and a(t) > 0

v(t) < 0 and a(t) < 0

(b) Slowing Down: v(t) and a(t) must have opposite signs.

v(t) > 0 and a(t) < 0

v(t) < 0 and a(t) > 0

(c) The way I think of it is v(t) and a(t) have to be working together to speed up

(which is why they need to have the same sign). If they have opposite signs, it

means acceleration is working against velocity (which is why it slows the particle

down).

(d) So...let’s find out when a(t) is positive (+) and negative (-). We do this the same

way we did it for v(t). Start with a number line and plot the points when a(t) = 0.

First, we need to solve a(t) = 6t− 12 = 0.

6t− 12 = 0

t = 2

157


Now plot t = 2 on the number line.

(e) Pick a point in each region and determine if acceleration is positive (+) or nega-

tive (-).

a(1.5) = −3(−)

a(4) = 12(+)

(f) Mark the regions with (+) or (-)

(g) Next, place the v(t) number line above the a(t) number line. Use this to see how

the signs of v(t) and a(t) compare in different regions.

158


i. Between 0 and 1 seconds, a(t) and v(t) have opposite signs. So it’s slowing

down.

ii. Between 1 and 2 seconds, a(t) and v(t) have the same sign. So it’s speeding

up.

iii. Between 2 and 3 seconds, a(t) and v(t) have opposite signs. So it’s slowing

down.

iv. After 3 seconds, a(t) and v(t) have the same sign. So it’s speeding up.

Yay! We are finally done with that problem. I hope you now have a decent understand-

ing of the relationship between the position function s(t), the velocity function v(t), and the

acceleration function a(t).

Example 2.40. If a ball is thrown vertically upward from the surface of the moon with a

velocity 10 m/s, its height after t seconds is h(t) = 10t− 0.83t3 (position function).

1. What is the max height reached?

Take a look at the figure below.

159


The max height is reached when the ball stops moving up and begins to move down.

In order to do that, it must stop moving. That’s exactly when v(t) = 0.

(a) Find v(t)

v(t) = h′(t) = 10− 1.66t

(b) Set v(t) = 0

10− 1.66t = 0

−1.66t = −10

t =10

1.66

t = 6.02 seconds

(c) Let’s use the number line to confirm the ball is moving up from 0 to 6.02 seconds

and then down after 6.02 seconds. Pick a point from each region.

160


v(1) = 8.34 (+)

v(7) = −1.62 (−)

This confirms the ball is moving up from 0 to 6.02 seconds and down after 6.02

seconds.

(d) Plug t = 6.02 seconds into h(t)

h(6) = 10(6.02)− .83(6.02)2 = 30.12 m

The ball reaches a max height of 30.12 meters at 6.02 seconds.

2. What is the velocity when the ball is 25 meters above the ground?

(a) Find out when the ball is 25 m above the ground. Set h(t) = 25.

(b) Solve h(t) = 25

10t− .83t2 = 25

−.83t2 + 10t− 25 = 0

Solving this quadratic, we get t = 3.54 seconds and t = 8.51 seconds.

161


(c) Find the velocity at t = 3.54 and t = 8.51

v(3.54) = 10− 1.66(3.54) = 4.1236 m/s

v(8.51) = 10− 1.66(8.51) = −4.1266 m/s

3. What’s a(t)?

a(t) = v′(t) = −1.66

So acceleration is constant. That makes sense because it’s acceleration due to gravity

(which is constant).

Also, notice that it’s negative. Between 0 and 6.02 seconds, v(t) > 0. But a(t) < 0.

Since they have opposite signs, the ball is slowing down. After 6.02 seconds, v(t) < 0.

Now that v(t) and a(t) have the same sign, the ball is speeding up. That matches what

we know happens on Earth. When we throw a ball up in the air, it slows down until

it reaches its peak. After it reaches its peak, it starts to fall slowly. Over time it picks

up speed.

162

2.8 Related Rates Brian E. Veitch

2.8 Related Rates

The related rates section is a word problem section using implicit functions. Most of the

functions in this section are functions of time t. It’s probably best demonstrated through

example.

Example 2.41. A stone is dropped into a pool of water. A circular ripple spreads out in

the pool. The radius of the ripple increases at a rate of 5 ft / second. How fast is the area

of the circle increasing? After 4 seconds?

There are two rates involved here. One is the rate the radius increases and the other is

the rate the area increases. One we know and the other we don’t. Our goal is to find some

sort of equation or formula that relates all the information we’re given.

Procedure

1. Write down all the information we have and what we need to find.

Known:dr

dt= 5 ft/s

Unknown:dA

dt=?

2. Find a relationship between the variables.

Since we’re dealing with circles, area, and a radius, let’s use

A = πr2

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3. Draw a ”snapshot” at some typical instant t to get an idea of what it looks like.

You can draw the picture first or after you identify some of the variables needed in the

problem. It’s up to you.

4. Differentiate the relationship with respect to t.

This is the tricky part. You have two variables, A and r. These variables are actually

functions of time t. It might be a better idea to write

A(t) = πr(t)2

dA

dt= π ·

(2r · dr

dt

)= 2πr

dr

dt

But we knowdr

dt.

dA

dt= 2πr · (5)

164


dA

dt= 10πr ft2/sec

We are done with the first part. Notice that the rate at which the area increases is a

function of the radius (which is a function of time). So if we want to know how fast

it’s increasing at a specific time, we need a time.

Note: You are not allowed to plug information about specific instances until after you

differentiate. For example, we are asked to find how fast the area is increasing at t = 4

seconds. We only use this information after we differentiated and founddA

dt.

To find how fast the area is increasing after 4 seconds, we need to know the radius

after 4 seconds. Since the radius increases at a rate of 5 ft/sec, the radius should be

20 feet.

dA

dt= 10π(20) = 200π ft2/sec

Example 2.42. Air is being pumped into a spherical balloon so that its volume increases at a

rate of 100 cm3/s. How fast in the radius of the balloon increasing when its diameter is 50cm.

As I mentioned in the previous example, any information about a specific point in time is

not used until after you differentiate. So we do not use the information about the diameter

being 50cm until later.

1. We need a formula that relates all the information together. We’re dealing with a

sphere and its volume. So let’s use

165


V =4

3πr3

2. What is known and unknown?

Known:dV

dt= 100

Unknown:dr

dt=?

3. Differentiate V =4

3πr3 with respect to t.

dV

dt=

4

3π

(3r2

dr

dt

)= 4πr2 · dr

dt

4. Solve fordr

dt

dr

dt=

dVdt

4πr2

5. We want how fast the radius increases when the diameter is 50cm. We don’t have a

diameter in my function, but we do have the radius. If the diameter is 50 cm, the

radius is 25. We also knowdV

dt= 100.

166


dr

dt=

100

4π(25)2= 0.0127 cm/s

Also note that as time goes on,dr

dtis getting smaller. That means the radius keeps

getting bigger, but much more slowly.

Example 2.43. Car A is traveling west at 50 mph and Car B is traveling north at 60 mph.

Both are heading for the same intersection. At what rate are the cars approaching each

other when Car A is .3 miles and Car B is .4 miles from the intersection?

1. Draw a picture:

Let z represent the distance between Cars A and B. The rate at which the cars ap-

proach each other isdz

dt.

2. What do we know?

Known:dx

dt= −50

Known:dy

dt= −60

167


Unknown:dz

dt=?

Anyone notice thatdx

dtand

dy

dtare negative? Even though Car B is traveling up to-

wards the intersection, velocity measures the rate of change in position. The cars are

moving towards the intersection. This means their position from the intersection is

decreasing, hence the negative velocity.

3. What relationship brings all of our variables together.

Do you see from the picture we formed a right triangle. So the relationship we are

going to use is

x2 + y2 = z2

4. Differentiate and solve fordz

dt

2x · dxdt

+ 2y · dydt

= 2z · dzdt

x · dxdt

+ y · dydt

= z · dzdt

Solving fordz

dt, we get

dz

dt=x · dx

dt+ y · dy

dt

z

168


5. To finddz

dtat the specific moment the question asked we need the following information,

• x

• y

• z

• dx

dt

• dy

dt

We know everything but z. So let’s use that right triangle and solve for z.

x2 + y2 = z2

(.3)2 + (.4)2 = z2

.25 = z2

Solving for z, we get z = .5.

169


6. Plug everything in to finddz

dt

dz

dt=

(.3)(−50) + (.4)(−60)

.5= −78 mph

Example 2.44. A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder

slides away at 1 ft/sec, how fast is the top sliding down the wall when the bottom is 6 ft

from the wall?

1. Draw a picture

2. What do we know?

Known:dx

dt= 4

Unknown:dy

dt=?

170


3. The relationship

x2 + y2 = 102

4. Differentiate with respect to t

2x · dxdt

+ 2y · dydt

= 0

Solving fordy

dtwe get

dy

dt= −x

y· dxdt

5. We need to find y

Using the Pythagorean Theorem, y = 8.

171


6. Plug everything in to finddy

dt

dy

dt= −x

y· dxdt

= −6

8· 1 = −3

4ft/sec

That’s interesting. The rate at which the ladder is sliding away from the wall is not

necessarily the rate at which the ladder slides down.

Also, what do you think happens when the top of the ladder gets close to the ground

(i.e., when y gets close to 0)?

Example 2.45. A balloon is rising straight up from a level field and is being tracked by a

camera 500 ft from the point of liftoff. At the instant when the camera’s angle of elevation

is π/4, that angle is increasing at a rate of 0.14 radians/min. How fast is the balloon rising

at that distance?

1. Draw a picture

172


2. What do we know?

Known:dθ

dt= 0.14 when θ = π/4

Unknown:dy

dt=??

3. Relationship between the variables θ and y

tan(θ) =y

500

4. Differentiate with respect to t

sec2(θ) · dθdt

=1

500· dydt

5. Solve fordy

dt

dy

dt= 500 sec2(θ) · dθ

dt

6. Plug in θ = π/4 anddθ

dt= 0.14, and we get

dy

dt= 500 sec2(π/4) · 0.14 = 140 ft/min

173


Example 2.46. A water tank has the shape of an inverted circular cone with a base radius

of 2 meter and a height of 4m. If water is being pumped into the tank at a rate of 2 m3/min,

find the rate at which the water is rising when the water is 3 m deep.

1. Let’s take a look at a picture shall we?

2. What do we know?

Known:dV

dt= 2

Unknown:dh

dt=??

3. The relationship (formula) is the volume of a circular cone

V =1

3πr2 · h

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4. It may not be obvious yet but you don’t want to differentiate this yet. You have three

functions in there, V (t), r(t), and h(t). After you differentiate, you’ll havedV

dt,dr

dt,

anddh

dtfloating around somewhere. We know

dV

dtand we’re trying to solve for

dh

dt,

but we know nothing aboutdr

dt. This means we need to somehow get rid of r(t). If

you’re still unclear, let’s go ahead and differentiate what we have. You’ll see what I

mean.

5. Differentiate V =1

3πr2︸︷︷︸ · h︸︷︷︸. BTW, it’s a product of two functions, we we’ll use the

product rule.

dV

dt=

1

3πr2 · dh

dt+ h ·

[2

3πr · dr

dt

]

Do you see what I mean aboutdr

dt? We have no information on it. So again, we need

to get rid of r(t).

6. If you look at the picture again,

you’ll notice we actually have two similar triangles.

175


Using this, we can writer

h=

2

4. Solving for r and we get

r =h

2

7. Rewrite our volume formula.

V =1

3πr2 · h =

1

3π

(1

2h

)2

· h =1

12πh3

8. Now we differentiate,

dV

dt=

1

12π · 3h2 · dh

dt=π

4h2 · dh

dt

9. Now plug indV

dt= 2, h = 3, and solve for

dh

dt.

176


2 =π

4· 32 · dh

dt

dh

dt=

8

9π= .28 m/min

When the height of the water is 3 m high, the rate at which the height is rising is 0.28

m/min.

177

2.9 Linear Approximations and Differentials Brian E. Veitch

2.9 Linear Approximations and Differentials

2.9.1 Linear Approximation

Consider the following graph,

Recall that this is the tangent line at x = a. We had the following definition,

f ′(a) = limx→a

f(x)− f(a)

x− a

So for x close to a, we have the following

f ′(a) ≈ f(x)− f(a)

x− a

After rearranging the terms, we get an estimate for f(x) when x is near a.

178


f(x) = f(a) + f ′(a)(x− a)

This is called the linearization of f(x) near x = a or linear approximation of f(x) near

x = a. You may not recognize it, but this is the equation of the tangent line at x = a. It’s

just written with different notation.

So how can this be useful? Suppose you wanted to find the value f(b) where b is really

close to a. Instead of using the function f(x) to evaluate it, we can just the tangent line.

From the graph, you can see that the tangent line and the function f(x) look very similar if

you focus only on the area near x = a. Here’s that graph focused near x = a.

You can see the true value of f(b) and the estimate you get from the tangent line are

pretty close. Now as you move away from x = a, the tangent line and the function deviate

quite a bit. So a linear approximation is only useful when evaluating near x = a.

Example 2.47. Find the linearization of f(x) =√x+ 3 at a = 1 and use it to approximate

the numbers√

3.98 and√

4.05.

179


Recall the linearization of f(x) near x = a is f(x) ≈ L(x) = f(a)+f ′(a)(x−a). So what

do we need?

• a

• f(a)

• f ′(x)

• f ′(a)

We know a = 1, and f(a) = f(1) =√

1 + 3 = 2, and

f ′(x) =1

2√x+ 3

so,

f ′(1) =1

2√

1 + 3=

1

4

Putting all this together, we get

f(x) ≈ L(x) = 2 +1

4(x− 1)

180


So how do we estimate√

4.05? If you plug in a = 1, we get√

4. So what do you have to

plug in for x to get√

4.05? You need to plug in x = 1.05, right?

√4.05 = f(1.05) ≈ L(1.05) = 2 +

1

4(1.05− 1)

√4.05 ≈ 2.0125

Now, let’s do the same thing and estimate√

3.98. Again, we need to figure out what is

x. What do we need to plug in to√x+ 3 to get

√3.98? I’m hoping you say x = 0.98. If

you did, that’s great.

√3.98 = f(3.98) ≈ L(3.98) = 2 +

1

4(.98− 1) = 1.995

So how close are our estimates? Let’s find their relative error.

Relative Error:

∣∣∣∣TRUE − APPROX

APPROX

∣∣∣∣

So for√

3.98, the relative error is

RE =

∣∣∣∣∣√

3.98− 1.995√3.98

∣∣∣∣∣ = 0.000003141

181


And for√

4.05, the relative error is

RE =

∣∣∣∣∣√

4.05− 2.0125√4.05

∣∣∣∣∣ = 0.00001929

These aren’t bad estimates. Now what if I try to estimate√

10 using the current lin-

earization formula?

√10 = f(7) ≈ L(7) = 2 +

1

4(7− 1) = 3.5

The relative error is

RE =

∣∣∣∣∣√

10− 3.5√10

∣∣∣∣∣ = .106797 or 10.6797%

Obviously, this estimate isn’t as good as the previous two. However, it’s still not bad. If

you want to experiment more, try estimating√

100 or something higher.

Example 2.48. Using linearization, estimate sin(π/180).

1. First, the whole point of learning about linearization is to estimate something compli-

cated with something easy. In the last example, we used√

4 to estimate√

4.05.

182


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

3. We need a value for a. π/180 is close to 0, so let’s use a = 0. As a bonus (more of a

requirement really), we know sin(0).

4. Recall that L(x) = f(a) + f ′(a)(x− a). We need to find f ′(a).

f ′(x) = cos(x)

so,

f ′(0) = cos(0) = 1

5. Let’s put this all together to find L(x).

sin(x) ≈ L(x) = sin(0) + cos(0)(x− 0) = x

That’s interesting. When x is near 0, we just showed sin(x) ≈ x.

Our objective was to estimate sin(π/180). Based on what we just saw,

sin(π/180) ≈ π/180

183


I’d like to verify sin(x) ≈ x by looking at the graph.

The graph on the right is zoomed in near x = 0 to show you that the function f(x) = x

is a good approximation for f(x) = sin(x).

184


2.9.2 Differentials

Differentials can be used to do exactly what we just did with linearization. Differentials help

us estimate the change in function values. Let’s look at some new notation.

• ∆x - is the true change in x

• dx - is our independent variable that represents the chagen in x. We let dx = δx.

• ∆y - is the true change in y

• dy - is the estimated change in y

Without looking at the graph yet, does it make sense that the change in y depends on

the change in x? This is why dx is an independent variable and dy is the dependent variable.

Remember that ∆y is the true change in y. Based on the graph, we see that ∆y =

f(x + ∆x) − f(x). This involves us knowing the exact value of f(x + ∆x). Back in the

185


linearization part, knowing f(x+ ∆x) is like knowing√

4.05. Instead I want to estimate ∆y

with dy. Let’s start with something we know,

dy

dx= f ′(x)

This was another way of notating the derivative. Now if we treat dx as an independent

variable, we can rearrange this as

dy = f ′(x) · dx

But this should make sense.dy

dxis the derivative at x. It represents the slope (i.e, the

change in y for every unit change in x). Recall that dx is the change in x.

Let’s look at a simple example. Suppose I know at x = 1, the y-value is 5. How can I

use the differential formula to estimate f(1.7)?

If the slope is 3 and the change in x, dx, is 0.7, then the change in y from x = 1 to

x = 1.7 is 3 · 0.7 = 2.1.

186


dy = f ′(1) · dx = 3 · 0.7 = 2.1

So if you’re starting at a y-value of 5 and move up 2.1 units, then the new y-value is 7.1.

Example 2.49. Let’s go back to a previous problem and estimate√

4.05 using the function

f(x) =√x+ 3.

1. We need a starting point. We will use a = 1 since f(1) =√

4 and that’s close to what

we want.

2. Next, we need to find dx. What do we have to plug into f(x) to get√

4.05? We need

to plug in x = 1.05. However, dx is just the change in x from your starting point a.

So dx = 0.05.

3. Now we need the slope, f ′(1), at x = 1.

f ′(x) =1

2√x+ 3

So,

f ′(1) =1

2√

1 + 3=

1

4

187


4. So the estimated change in y is

dy = f ′(1) · (0.05) =1

4· 0.05 = .0125

Horray! We have the estimated change in y. But what is this really? dy tells us how

much y changes from the original y value. For us, the original y value is√

4 = 2. So

√4.05 ≈

√4 + dy = 2 + 0.0125 = 2.0125

Guess what? That’s exactly what we got using linearization!

188

Brian E. Veitch

3 Application of Differentiation

3.1 Maximum and Minimum Values

Consider the following graph,

Definition 3.1 (Absolute Extrema). A function f has an absolute max at x = a, if f(a) ≥

f(x) for all x in the domain. In other words, f(a) has the largest y-value. f(a) is the absolute

max value.

A function f has an absolute min at x = a, if f(a) ≤ f(x) for all x in the domain. f(a)

is the absolute min value.

There can be at most one absolute max and at most one absolute min. From the

graph you can see there can be many local max and local mins. Think of a local max as a

peak. There can be many peaks on a graph. But there can be only be one LARGEST peak.

Also note that an absolute max or absolute min can be an endpoint. We typically do not

call endpoints local max or local mins even if they are the absolute max or absolute min.

All of these max and mins (local or absolute) are called Extreme Values.

Some functions can have an absolute max and no absolute min (or vice versa). There

are functions that may not have any absolute max or absolute mins. The following are three

189

3.1 Maximum and Minimum Values Brian E. Veitch

functions that fit these scenarios.

Consider this graph again,

Why are we guaranteed an absolute max and absolute min in this graph? I haven’t

mentioned the theorem yet, but this graph satisfies certain requirements that guarantee an

absolute max AND an absolute min.

3.1.1 The Extreme Value Theorem

Theorem 3.1 (Extreme Value Theorem). If f is continuous on a closed interval [a,b],

then f has an absolute max and absolute min somewhere in [a,b].

190


So that’s it! As long as your function is continuous on a closed interval, it’s guaranteed

to have an absolute max and absolute min. Here are a couple of examples that satisfy the

theorem’s conditions.

So why are the two conditions in the theorem necessary? Take a look at the following

two graphs.

The first graph is on a closed interval [a,b] but is not continuous. You can see where the

absolute max should have been. The discontinuity there messed it up. The second graph is

continuous but not on a closed interval. We don’t have an absolute min because it would

have been at the endpoint b. Since its an open dot, we are never actually allowed to equal

that y-value.

I want to make something very clear about theorems. If you satisfy the conditions of

the theorem, you are guaranteed the conclusion. If f is continuous on a closed interval,

191


we’re guaranteed extreme values. Notice in the second graph, it does have an absolute max.

The first graph does have an absolute min. Just because you don’t satisfy the conditions,

doesn’t mean you can’t have extreme values. Basically, if you don’t satisfy the conditions of

a theorem, then anything goes. Maybe you have extreme values, maybe you don’t. Satisfy

the conditions, and you’re guaranteed extreme values.

Example 3.1. Let f(x) = 3x4 − 16x3 + 18x2 on the interval [−1, 4]. Find the local max,

local min, absolute max, and absolute min (if they exist). Another way of wording this

question is, ”Find all extrema.”

Since we don’t have a way to find local maximums and minimums yet, we’ll just take a

look at its graph.

1. Let’s start with the absolute maximum.

An absolute maximum is the largest (highest) y-value. Based on the graph, the abso-

lute maximum occurs at the point (-1,37).

(a) What is the absolute max? 37

192


(b) Where is the absolute max? x = −1

2. The absolute minimum occurs at the point (3,-27). Therefore, the absolute minimum

is -27.

3. Local max?

A local max is any point that is a peak. The graph must go down on both sides of the

point. There is only one local max and occurs at the point (1,5).

4. Local min?

A local min is the bottom of a dip. The graph must go up on both sides of the point.

There are two local mins. One is at (0,0) and the other is at (3,-27).

5. I mentioned this already, but I don’t allow endpoints to be local max or mins.

3.1.2 Fermat’s Theorem

Theorem 3.2 (Fermat’s Theorem). If f has a local minimum or local maximum at x = c,

and if f ′(c) exists, then f ′(c) = 0.

What does this theorem really tell us? If f ′(c) = 0, it’s a POTENTIAL local max or

local min. Take a look at the graph below.

193


It follows the theorem. We can see that it has a local max, and the derivative exists

there. The conclusion from Fermat’s Theorem guarantees us that the slope (f ′(c) = 0).

As I stated already, if f ′(c) = 0, it does not mean it’s a local max or min, only that it

could be. Consider the function f(x) = x3.

Differentiate f(x) = x3 and we get f ′(x) = 3x2. Plug in x = 0 and we get f ′(0) = 3(0)2 =

0. This shows we have a slope of 0 at x = 0; however, it is not a local max or local min. It

is neither a peak nor the bottom of a dip.

If f ′(c) = 0, it does not mean there is a local max or min.

194


If we want to find possible local max and local min locations, we find all places

where f ′(x) = 0 or where f ′(x) does not exist. They may not be local max or mins.

But if we have them, they must be there.

3.1.3 Steps to find the absolute max or absolute min

1. Find all x-values where f ′(x) = 0 or f ′(x) does not exist.

These x-values are called critical values.

2. Find the y-value for each of the critical points in the interval. If no interval is given,

assume (−∞,∞).

3. Find the y-value for the endpoints. If the interval is open, we still find the y-value. If

we find that the absolute max occurs at an endpoint that is open, then we conclude

the function has no absolute max.

4. The largest y-value is the absolute max.

5. The smallest y-value is the absolute min.

Example 3.2. Find the absolute max and min (if they exist) of f(x) = x3− 3x2 + 1 on the

interval[−1

2, 4].

Note, we have a continuous function on a closed interval. The Extreme Value Theorem

concludes we must have an absolute max and absolute min.

195


1. Find the critical values.

Set f ′(x) = 0 and solve

f ′(x) = 3x2 − 6x

3x2 − 6x = 0

3x(x− 2) = 0

Our critical values are x = 0 and x = 2. Since both are in the interval [−1/2, 4], we

use them both. Note, if any one of them wasn’t in the interval, we don’t use it.

2. Find the y-values for the critical values.

x = 0 :→ f(0) = 03 − 3(0)2 + 1 = 1

x = 2 :→ f(2) = 23 − 3(2)2 + 1 = −3

3. Find the y-values at the endpoints.

x = −1

2:→ f(−1/2) =

1

8

x = 4 :→ f(4) = 17

4. Find the absolute max and absolute min.

(a) Absolute max: (4,17)

(b) Absolute min: (2,−3)

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3.1.4 Steps to find the local maximums and minimums

This material shows up again in section 3, but we might as well cover it now. I’ve already

stated that to find a local max or min, we first find all x-values where f ′(x) = 0 or f ′(x)

does not exist. Let’s continue with the previous example.

1. Find the critical values.

We already solved f ′(x) = 0 and found critical values at x = 0 and x = 2. Since f(x)

a polynomial, the derivative will exist everywhere. So we don’t have to worry about

f ′(x) not existing.

2. Plot the critical values on a number line.

3. Pick a number from each region and plug it into f ′(x). We just need to know if f ′(x)

is positive (+) or negative (-).

Remember that f ′(x) tells us the slope of the graph. If the slope is positive, the func-

tion is going up. If the slope is negative, the function is going down.

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f ′(−1) = 3(−1)(−1− 2) = 9 (+)

f ′(1) = 3(1)(1− 2) = −3 (−)

f ′(3) = 3(3)(3− 2) = 9 (+)

This means the graph has to have the following form. It has to be increasing to x = 0

and then start decreasing. That makes a peak (local max) at x = 0.

It decreases to x = 2 and then begins to increase. That makes a dip (local min) at

x = 2.

4. Find the local max.

Now that we know the local max occurs at x = 0, we need to find its y-value. DO

NOT PLUG x = 0 into f ′(x). To find the y-value, we need to plug it in f(x).

f(0) = 1

Local Max: (0,1)

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5. Find the local min.

Now that we know the local min occurs at x = 2, we need to find its y-value. To find

the y-value, we need to plug it in f(x).

f(2) = −3

Local Max: (2,-3)

Example 3.3. Let’s jump into a harder example. It’s harder because solving f ′(x) = 0 is

difficult. Anyhow, let’s get started. Find the local extrema for f(x) = x4/5(x− 4)2.

1. Find f ′(x). Use the product rule.

f ′(x) = x4/5 · ddx

[(x− 2)2

]+ (x− 4)2 · d

dx

[x4/5

]= x4/5 · [2(x− 4) · 1] + (x− 4)2 · 4

5x−1/5

Since we have to set f ′(x) = 0, we need to simplify f ′(x).

x4/5 · [2(x− 4) · 1] + (x− 4)2 · 4

5x−1/5 = 0

2x−1/5(x− 4)

[x+

2

5(x− 4)

]& = 0

2x−1/5(x− 4)

[7

5x− 8

5

]

There are three factors here.

(a) x−1/5: This means f ′(x) does not exist at x = 0 since 0−1/5 does not exist.

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(b) (x− 4): Setting this equal to 0, we get x = 4.

(c)

[7

5x− 8

5

]: Setting this equal to 0, and we get x =

8

7or x ≈ 1.14286

2. Plot the critical values on the number line.

3. Pick a number from each region and plug it in f ′(x).

x = −1;→ f ′(−1) = (−1)−1/5(2.8(−1)2 − 14.4(−1) + 12.8) = −30

x = 1;→ f ′(1) = (1)−1/5(2.8(1)2 − 14.4(1) + 12.8) = 1.2

x = 3;→ f ′(3) = (3)−1/5(2.8(3)2 − 14.4(3) + 12.8) = −4.174256

x = 5;→ f ′(5) = (5)−1/5(2.8(5)2 − 14.4(5) + 12.8) = 7.82762

Our number line now looks like this,

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4. Find the local maximums.

There is a local max at x = 1.14286. Plug x = 1.14286 into f(x) to find the y-value.

Local Max: (1.14286, 9.083588)

5. Find the local minimums.

There are local minimums at x = 0 and x = 4.

Warning: The critical value x = 0 came from when f ′(0) did not exist. That means

it’s possible f(0) may not exist. If f(0) does not exist, then its not a local min. If it

does exist, it is a local min. Fortunately for us, f(0) does exist.

Local Mins: (0, 0) and (4,0)

Let’s go ahead and look at the graph of f(x) = x4/5(x− 4)2 to verify our findings.

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Sweet! Also, do you see how the graph makes a corner at (0,0). That’s why f ′(0) did

not exist. Hopefully you can recall the different scenarios why a derivative may not exist.

202

3.2 Mean Value Theorem Brian E. Veitch

3.2 Mean Value Theorem

Let’s begin the section with an easier extension of the Mean Value Theorem.

Theorem 3.3 (Rolle’s Theorem). 1. f is continuous on the interval [a, b].

2. f is differentiable on the interval (a, b).

3. f(a) = f(b)

If the above conditions are satisfied, then there exists a c in (a, b) such that f ′(c) = 0 (f

must have a slope of 0).

It’s easy to see that this is true by looking at some examples.

Example 3.4.

Notice that all conditions to Rolle’s Theorem are satisfied. We can clearly see an x-value

in (a, b) where f ′(x) = 0.

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This example is a bit trivial since the entire line has a slope of 0. But, it satisfied the

conditions so there must be a place where f ′(x) = 0. It just so happens we have a slope of

0 everywhere.

There really isn’t a limit to how many places we could have a slope of 0. The theorem

doesn’t guarantee the number of places. It only guarantees the existence of at least one.

So what are some of the uses for Rolle’s Theorem? First let’s recap what’s really hap-

pening. If f(a) = f(b), then we must have a slope of 0 somewhere between a and b. Another

way of saying this is f(x) must turn around somewhere between a and b.

Example 3.5. Show that x3 + x− 1 = 0 has exactly one real root (solution).

Solution:

Well to verify this, we first need to show it has at least one root. A while ago, we dis-

cussed the Intermediate Value Theorem. One of the reasons we used this theorem was to

204


show when a function has a root.

By the way, f(x) = x3 + x − 1 satisfies the conditions of the theorem (it’s continuous).

Now let x = 0 and x = 2. If you’re not given an interval, make one up.

x = 0 : f(0) = 03 + 0− 1 = −1 (negative value)

x = 2 : f(2) = 23 + 2− 1 = 9 (positive value)

Since f(0) and f(2) have opposite signs, IVT says we must have a root between 0 and 2.

Now to have a second root, our function must turn around. Assume f(x) has two roots.

In order to have two roots (i.e., reach the x-axis again), it must turn around.

A differentiable function can only turn around when the slope is 0. So let’s check out the

derivative.

f ′(x) = 3x2 + 1 = 0

If we try to set this function equal to 0

3x2 + 1 = 0

we see it has no solution. So what does that mean for us?

If f ′(x) 6= 0, then the function has no way of turning around. If the function cannot turn

around, it has no way of getting a second root.

We showed that f(x) has at least one root by using the Intermediate Value Theorem.

Now we showed using Rolle’s Theorem that f(x) can’t turn around, thus eliminating the

possibility for a second root. Therefore, x3 + x− 1 has only one root.

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Theorem 3.4 (The Mean Value Theorem). .

1. f is continuous on [a, b].

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

Then there exists a c in (a, b) such that f ′(c) =f(b)− f(a)

b− a.

The picture below shows why this makes sense. Note thatf(b)− f(a)

b− ais the slope of the

line connecting the endpoints.

This theorem states that there must be another place in (a, b) that has a tangent line

with the same slope as the secant line connection (a, f(a)) to (b, f(b)).

Example 3.6. Let f(x) = x3 + 3x2 on the interval [−5, 1]. Since f is a polynomial, it is

continuous and differentiable on the domain.

f(b)− f(a)

b− a=f(1)− f(−5)

1− (−5)= 9

The Mean Value Theorem tells me there exists a number c somewhere between -5 and 1

where f ′(c) = 9. Let’s go ahead and find out.

f ′(c) = 3c2 + 6c = 9

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Solving 3c2 + 6c = 9, we get

3c2 + 6c− 9 = 0

3(c2 + 2c− 3) = 0

(c+ 3)(c− 1) = 0

The solutions are c = −3 and c = 1. Unfortunately, c = 1 is not between -5 and 1, so we

don’t count it. So according to the mean value theorem, the only value in (−5, 1) that gives

us a slope of 9 is at x = −3.

Let’s take a look at the graph and verify this.

Example 3.7. Show x3 − 15x+ c = 0 has at most one root in [−2, 2].

1. Note that it doesn’t ask us if there are any roots. It just wants us to show there can’t

be two or more.

2. Let’s assume our equation does have two roots in [−2, 2]. That means there are two

points, let’s call them x1 and x2, such that f(x1) = f(x2) = 0.

207


3. Rolle’s Theorem says there must exist a c in (x1, x2) such that f ′(c) = 0. Let’s find

f ′(x).

f ′(x) = 3x2 − 15

4. Solving for x, we have x = ±√

5. Both of these x values are NOT in [−2, 2].

5. We conclude that it must be impossible for our equation to have two roots in [−2, 2].

And with that, we are done.

6. Just to be clear, all we’ve done is shown the equation cannot have two or more roots.

We don’t know (nor care) if it has zero or one root.

Here are some other theorems that will help us in the next few sections.

Theorem 3.5. If f ′(x) = 0 for all x in (a, b), then f is constant.

Example 3.8. Consider the following function y = f(x) = 3 ( a constant function).

This theorem should make sense. If the function wasn’t constant (a horizontal line), how

could f ′(x) = 0?

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Let’s move onto another theorem. This theorem will show its face again when we get

into integration. Don’t worry, that’s still a ways away.

Theorem 3.6. If f ′(x) = g′(x) for all x in (a, b), then f(x) − g(x) is a constant. Another

way of concluding the theorem is,

f(x) = g(x) + C

Again, let’s take a look at another graph to check this out. Note that if two functions

have the exact same derivative (i.e., they have the exact same slopes), they have to look the

same. They just might be off by a constant.

Note, the difference between f(x) and g(x) is the same.

209

3.3 Using Derivatives to analyze a Function Brian E. Veitch

3.3 Using Derivatives to analyze a Function

What does f ′ say about f? It tells us about the slope of f . But even more, it tells us when

f(x) is increasing or decreasing. This is extremely useful when trying to figure out what the

graph looks like.

Let’s go over this with a bit more formal language.

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

2. If f ′(x) < 0 on an interval I, then f is deceasing.

I’d like to go over the proof for increasing. The proof for the decreasing part is similar.

Proof. Suppose that f ′(x) > 0 for all x in the interval I. The Mean Value Theorem tells us

that for any two points in I (let’s call them x1 and x2) where x1 < x2, there exists a c in I,

such that

210


f(x2)− f(x1) = f ′(c)(x2 − x1)

Since c is between x1 and x2, then f ′(c) > 0. Plus, (x2 − x1) > 0. Therefore,

f(x2)− f(x1) > 0

By definition, this means f(x) is increasing.

3.3.1 First Derivative Test

Recall that when f ′(c) = 0 or f ′(c) does not exist, then x = c is called a critical value.

Suppose c is a critical value of a continuous function. This means f ′(c) = 0.

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

maximum at x = c.

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

minimum at x = c.

3. If f ′(x) does not change signs, f has neither a max nor min at x = c

Example 3.9. Find where f(x) = 3x4 − 4x3 − 12x2 + 5 is increasing / decreasing.

1. Find all critical values.

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(a) Find f ′(x)

f ′(x) = 12x3 − 12x2 − 24x

(b) Find where f ′(x) does not exist.

Since, f ′(x) exists everywhere, we do not get any critical values from this.

(c) Set f ′(x) = 0 for other critical values.

12x3 − 12x2 − 24x = 0

12x(x2 − x− 2) = 0

12x(x− 2)(x+ 1) = 0

We have critical values at x = 0,−1, 2.

2. Use a number line to determine increasing and decreasing intervals.

(a) Draw a number line with all the critical values plotted.

(b) Pick a point in each region. Plug that number into f ′(x) to determine if it’s

positive (+) or negative (−).

I’ll choose x = −2,−1/2, 1, 3

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f ′(−2) < 0

f ′(−1/2) > 0

f ′(1) < 0

f ′(3) > 0


This shows f(x) is

(a) decreasing on the intervals (−∞,−1) and (0, 2)

(b) increasing on the intervals (−1, 0) and (2,∞).

Based on the First Derivative Test, f(x) has a maximum x = 0. The point is (0, 5).

When we ask for the local maximum, it’s better to list it as a point.

There are two local minimums. One is at x = −1. The other is at x = 2. As points,

the local minimums are (−1, 0) and (2,−27).

Remember, you need to plug x = −1 and x = 2 into the original f(x) to get the y-value.

Looking at the graph, it appears our analysis is correct.

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Example 3.10. Find all local max/mins of f(x) = cos2(x) + sin(x) on the interval [0, π].

1. Find all critical values.

(a) Find f ′(x)

f ′(x) = −2 cos(x) · sin(x) + cos(x)

(b) Find where f ′(x) does not exist.

Since, f ′(x) exists everywhere, we do not get any critical values from this.

(c) Set f ′(x) = 0 for other critical values.

−2 cos(x) · sin(x) + cos(x) = 0

−2 cos(x) (sin(x)− 1/2) = 0

−2 cos(x) = 0 or sin(x) = 1/2

−2 cos(x) = 0 gives us x = π/2.

215


sin(x) = 1/2 gives us x = π/6 and x = 5π/6.

2. Use a number line to determine increasing and decreasing intervals. e

(a) Draw a number line with all the critical values plotted.

(b) Pick a point in each region. Plug that number into f ′(x) to determine if it’s

positive (+) or negative (−).

I’ll choose x = π/12, π/3, 2π/3, π

f ′(π/12) > 0

f ′(π/3) < 0

f ′(2π/3) > 0

f ′(11π/12) < 0


This shows f(x) is

216


(a) decreasing on the intervals (π/6, π/2) and (5π/6, π)

(b) increasing on the intervals (0, π/6) and (π/2, 5π/6).

Based on the First Derivative Test, f(x) has a maximum x = π/6 and x = 5π/6.

The points are (π/6, 1.25) and (5π/6, 1.25). When we ask for the local maximum, it’s

better to list it as a point.

The local minimum is at x = π/2. As a point, the local minimum is (π/2, 1).

Looking at the graph of f(x) = cos2(x) + sin(x), we see that our analysis is correct.

Example 3.11. Sketch a graph satisfying the following conditions.

1. f(0) = 1

2. f(2) = 3

3. f(5) = −1

217


4. f ′(x) > 0 on (0, 2) and (5,∞)

5. f ′(x) < 0 on (−∞, 0) and (2, 5)

Ok, so the first three conditions are just points. Just like when we sketched graphs with

limits, we’ll follow the same procedure here. Let’s plot those three points.

The fourth condition: f ′(x) > 0 means the function is increasing. So let’s place an in-

creasing line in those regions.

The fifth condition: f ′(x) < 0 means the function is decreasing. So let’s place a decreas-

ing line in those regions.

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You can connect all these together with a smooth curve.

Now how did I know how to connect the pionts. Originally, I connected them with a

straight line. Then I connected them with a curved line. There are two ways of connecting

two points with a curved line. We get this information from the second derivative.

Recall, a derivative will determine when the original function is increasing or decreasing.

So a second derivative will determine if the first derivative (slopes) are getting increasing

or decreasing.

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3.3.2 Concavity and Inflection Points

Let’s take a look at a region where f(x) is increasing. There are two ways of connecting

these points.

Notice that in the first graph, the slopes are getting smaller (decreasing). This means

f ′′(x) < 0. We call this concave down

In the second graph, the slopes are increasing. This means f ′′(x) > 0. We call this

concave up.

Now let’s look at two decreasing lines and how concavity works.

In the first graph, the slopes are decreasing (remember, −9 < −1). So larger negative

numbers are actually ”smaller.” Since f ′′(x) < 0, we call this concave down.

220


In the second graph, the slopes are increasing. Note, a slope of -1 is actually larger than

-9, since you had to increase from -9 to -1 on the number line. Since f ′′(x) > 0, we call this

concave up.

I tend to remember concave up or down based on parabolas. From algebra, we learned

that a parabola can open ”up” or ”down.” Concavity works exactly like that.

Take a look at the two concave down graphs. If you extend their lines, they look like a

parabola opening down (i.e., concave down).

You can see it’s very similar for concave up. Concave up graphs look very similar to

parabolas opening up.

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Definition 3.2 (Inflection Point). A point on f(x) is called an inflection point if f(x) is

continuous there, and it changes concavity.

Now it’s time to sketch some graphs. We’ll start with some easy ones and work our way

to the more challenging ones.

Example 3.12. Let f(x) = 2x3 − 3x2 − 12x.

• Find where f is increasing / decreasing.

• Find all local maximums and minimums

• Find intervals of concavity and inflection points

• Use all the above information to sketch the graph

1. First, we need to find the derivative

f ′(x) = 6x2 − 6x− 12

Next, figure out when f ′(x) = 0 or when f ′(x) does not exist.

Since f ′(x) exists everywhere, we’ll just go ahead and solve

6x2 − 6x− 12 = 0

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6(x2 − x− 2) = 0

6(x− 2)(x+ 1) = 0

Solving this, we get critical values at x = −1 and x = 2.

2. Next, we set up the number line with our critical values.

3. Pick a number from each region, determine if f ′(x) is positive or negative. I’ll let you

plug in any number from each region. If you do it correctly, you get the following

So f(x) is

(a) Increasing on the intervals (−∞,−1) and (2,∞)

(b) Decreasing on the interval (−1, 2)

4. We see from the number line we have a local maximum at x = −1 and a local minimum

at x = 2. Write these as points (since we will have to plot them soon).

Local Max: (−1, 7)

Local Min: (2,−20)

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5. Find the intervals of concavity. To do this, we need to find f ′′(x).

f ′′(x) = 12x− 6

We need to find the ”critical values” to f ′′(x).

12x− 6 = 0

Solving this, we get x = 1/2.

6. We use the number line, like we did with the first derivative, to find when f ′′(x) > 0

or when f ′′(x) < 0.

So f(x) is concave down on the interval (−∞, 1/2)

and f(x) is concave up on the interval (1/2,∞)

7. To find the inflection point, we need to satisfy two conditions.

(a) Does concavity change?

Yes, concavity changes at x = 1/2.

(b) Does the point exist at x = 1/2?

Yes, plug x = 1/2 into f(x) and we get the point (1/2,−6.5).

224


8. Now it’s time to put all of this together to sketch the graph.

(a) Plot all important points (local max/min, inflection points, intercepts, etc.)

Note, I also added a curve at the local max and min. I do this because it reminds

me that they are local max / mins.

(b) Lastly, I connect all the points using the appropriate concavity.

Example 3.13. Let f(x) = x4 − 4x3.



225





f ′(x) = 4x3 − 12x2

Next, figure out when f ′(x) = 0 or when f ′(x) does not exist.

Since f ′(x) exists everywhere, we’ll just go ahead and solve

4x3 − 12x2 = 0

4x2(x− 3) = 0

Solving this, we get critical values at x = 0 and x = 3.




So f(x) is

(a) Increasing on the intervals (−∞, 0) and (0, 3). In this case, you can also write

(−∞, 3).

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(b) Decreasing on the interval (3,∞)

4. We see from the number line we have a local maximum at x = −1 and a local minimum

at x = 2. Write these as points (since we will have to plot them soon).

Local Max: None

Local Min: (3,−27)

5. Find the intervals of concavity. To do this, we need to find f ′′(x).

f ′′(x) = 12x2 − 24x = 12x(x− 2)


12x(x− 2) = 0

Solving this, we get x = 0 and x = 2. These are our possible points of inflection.



So f(x) is concave down on the interval (0, 2)

and f(x) is concave up on the interval (−∞, 0) and (2,∞)

227




Yes, concavity changes at x = 0 and x = 2.

(b) Does the point exist at x = 0 or x = 2?

Yes, we get the following points: (0, 0) and (2,−16)

8. Now it’s time to put all of this together to sketch the graph.



228


Example 3.14. Sketch f(x) = x1/3(x+ 4).






f ′(x) = x1/3 · (1) + (x+ 4) · 1

3x−2/3

Next, figure out when f ′(x) = 0 or when f ′(x) does not exist. Before doing that, we

need to simplify this. This means ALGEBRA. If you cannot do this now, make sure

you can by exam time!

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f ′(x) = x1/3 · (1) + (x+ 4) · 1

3x−2/3

= x1/3 +1

3x−2/3(x+ 4)

= x−2/3(x+

1

3(x+ 4)

)= x−2/3

(4

3x+

4

3

)=

4

3x−2/3(x+ 1)

=4(x+ 1)

3x2/3

f ′(x) does not exist at x = 0, since it makes the denominator equal to 0.

Next we set f ′(x) = 0. We do that by setting the numerator equal to 0.

4(x+ 1)

3x2/3= 0

4(x+ 1) = 0

Solving this, we get critical values at x = −1.




230


So f(x) is

(a) Increasing on the intervals (−1, 0) and (0,∞). In this case, you can also write

(−1,∞). You need to make sure x = 0 is in the domain (which it is).

(b) Decreasing on the interval (−∞,−1)

4. We see from the number line we have a local minimum at x = −1 and no local

maximum.

Local Max: None

Local Min: (−1,−3)

5. Find the intervals of concavity. To do this, we need to find f ′′(x). Let’s use f ′(x) =4

3x−2/3(x+ 1).

f ′′(x) =4

3x−2/3 · (1) + (x+ 1) · −8

9x−5/3

f ′′(x) =4

9x−5/3(3x− 2(x+ 1))

f ′′(x) =4

9x−5/3(x− 2)

f ′′(x) =4(x− 2)

9x5/3


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4(x− 2)

9x5/3= 0

Solving this, we get x = 2.

Our other critical value is x = 0 because that’s when f ′′(x) does not exist.



So f(x) is concave down on the interval (0, 2)

and f(x) is concave up on the interval (−∞, 0) and (2,∞)



Yes, concavity changes at x = 0 and x = 2.

(b) Does the point exist at x = 0 or x = 2?

Yes, we get the following points: (0, 0) and (2, 7.56)

8. We should probably find some other points too. We should try to find x and y inter-

cepts. Typically we do this at the beginning of the problem.

232


• x-intercepts: (0, 0) and (−4, 0)

• y-intercept: (0, 0)

Now it’s time to put all of this together to sketch the graph.



233

3.4 Limits at Infinity - Asymptotes Brian E. Veitch

3.4 Limits at Infinity - Asymptotes

Definition 3.3. If f is a function defined on some interval (a,∞), then limx→∞

f(x) = L means

that values of f(x) are very close to L (keep getting closer to L) as x→∞.

The line y = L is called a horizontal asymptote of f(x) if either

limx→∞

f(x) = L

or

limx→−∞

f(x) = L

Here’s an example of a function with ONE asymptote

Notice that both graphs have a horizontal asymptote at y = 1. The first graph has the

horizontal asymptote as x → −∞ and x → ∞. The second graph only has the horizontal

asymptote as x→ −∞. So what does this mean?

When you determine an asymptote, don’t always assume it’s on both sides of the graph.

Here’s an example of a graph that has two horizontal asymptotes.

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Do you see how this graph has two horizontal asymptotes? One is y = −1 as x → −∞

and the other is y = 1 as x→∞.

At this point, you probably know about vertical asymptotes, but I want to go over

them briefly.

If limx→a

f(x) = ±∞, then x = a is a vertical asymptote.

One way to find them is to set the denominator equal to 0. If x = a makes the denomi-

nator 0 but NOT the numerator, then x = a is a vertical asymptote.

Warning! If x = a makes the numerator and denominator both 0, it means it may or

may not be a vertical asymptote. You have to follow through with the limit to confirm.

Example 3.15.

1. f(x) =1

x− 1has a vertical asymptote at x = 1

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2. f(x) =x

x2 − 4has vertical asymptotes at x = −2 and x = 2

3. f(x) =x− 2

x2 − 4has a vertical asymptote at x = −2 but not at x = 2. Why is that?

Even though the denominator is 0 when you plug in x = 2, look at the limit

limx→2

x− 2

x2 − 4= lim

x→2

x− 2

(x− 2)(x+ 2)

= limx→2

1

x+ 2

=1

4

6= ∞

236


You can see that we do NOT have a vertical asymptote at x = 2.

The shortcut methods to finding vertical and horizontal asymptotes can be found in

’Types of Functions’

Before moving on to sketching graphs with asymptotes, I want to do some examples of

finding them without the shortcuts. They involve a bit of algebra.

Example 3.16.

1. Find the horizontal asymptote(s) of f(x) =3x2 − x− 2

5x2 + 4x+ 1

We need to find two different limits.

(a) As x→∞

limx→∞

3x2 − x− 2

5x2 + 4x+ 1= lim

x→∞

3− 1x− 2

x2

5 + 4x

+ 1x2

=3− 0− 0

5 + 0 + 0

=3

5

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(b) As x→ −∞

limx→−∞

3x2 − x− 2

5x2 + 4x+ 1= lim

x→−∞

3− 1x− 2

x2

5 + 4x

+ 1x2

=3− 0− 0

5 + 0 + 0

=3

5

So there is only one horizontal asymptote, y =3

5

2. Find the horizontal and vertical asymptotes of f(x) =

√2x2 + 1

3x− 5.

(a) Let’s start with the vertical asymptote. What makes the denominator 3x−5 = 0?

x =5

3

Since x =5

3does not make the numerator 0, we know x =

5

3is a vertical asymp-

tote.

238


(b) Now on to the horizontal asymptotes. This one is going to be tricky.

limx→∞

√2x2 + 1

3x− 5= lim

x→∞

√2x2

(1 +

1

2x2

)3x− 5

= limx→∞

√2|x|

√1 +

1

2x2

3x− 5

Recall: |x| = x when x > 0

= limx→∞

√2x

√1 +

1

2x2

3x− 5

= limx→∞

√2

√1 +

1

2x2

3− 5

x

=

√2

3

239


limx→−∞

√2x2 + 1

3x− 5= lim

x→−∞

√2x2

(1 +

1

2x2

)3x− 5

= limx→−∞

√2|x|

√1 +

1

2x2

3x− 5

Recall: |x| = −x when x < 0

= limx→−∞

√2(−x)

√1 +

1

2x2

3x− 5

= limx→−∞

−√

2

√1 +

1

2x2

3− 5

x

=−√

2

3

So we have two horizontal asymptotes. Now this seemed a bit algebraic intensive.

Any way to make it easier? The answer is yes, but make sure you can do the

above method.

Shortcut

limx→∞

√2x2 + 1

3x− 5= lim

x→∞

√2x2

3x= lim

x→∞

√2x

3x=

√2

3

240


limx→−∞

√2x2 + 1

3x− 5= lim

x→−∞

√2x2

3x= lim

x→−∞

√2(−x)

3x=−√

2

3

Here’s the graph of f(x) =

√2x2 + 1

3x− 5

3. Evaluate the limit limx→∞

√x2 + 6x− x.

If you try to ”evaluate” this, you get ∞−∞. This does not necessarily mean 0. Let’s

go ahead and find out.

241


limx→∞

√x2 + 5x− x = lim

x→∞

(√x2 + 5x− x

)·√x2 + 5x+ x√x2 + 5x+ x

= limx→∞

x2 + 6x− x2√x2 + 5x+ x

= limx→∞

6x√x2 + 5x+ x

= limx→∞

6x√x2(1 + 5

x) + x

= limx→∞

6x

x(√

1 + 5x

+ 1)

= limx→∞

6√1 + 5

x+ 1

=6

2

= 3

4. Evaluate limx→∞

x− x2

The above method doesn’t really work. Instead, note that we turn this into a product

of two functions by factoring out x.

limx→∞

x(1− x) =∞ · −∞ = −∞

Keep in mind that my notation isn’t formal. But it should help you understand what

just happened. It’s a product of two functions. So I get a very large number times

a very large negative number. Common sense says this should equal an even larger

negative number.

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3.4.1 Examples of Curve Sketching with Asymptotes

Refer to the section Summary of Curve Sketching for an outline and guide to help you

with curve sketching.

Sketch y =2x2

x2 − 1

1. Domain: x 6= ±1

2. Intercepts:

x− intercept: (0, 0)

y − intercept: (0, 0)

3. Asymptotes:

(a) Vertical Asymptote: x = −1 and x = 1

(b) Horizontal Asymptote: y = 2 because limx→±∞

2x2

x2 − 1= 2

4. Find y′

y′ =(x2 − 1) · (4x)− 2x2 · (2x)

(x2 − 1)2

=−4x

(x2 − 1)2

We have critical values at x = 0, x = −1 and x = 1.

5. Use the number line to determine when y is increasing or decreasing.

243


Increasing: (−∞,−1) and (−1, 0). You can’t write (−∞, 0) because x = −1 is not in

the domain.

Decreasing: (0, 1) and (1,∞). You can’t write (0,∞) because x = 1 is not in the

domain.

We have a local max at x = 0. As a point, it’s (0, 0).

6. Find y′′

Rewrite f ′(x) as f ′(x) = −4x(x2 − 1)−2 so you can use the product rule.

f ′′(x) = −4x · −2(x2 − 1)−3 · 2x+ (x2 − 1)−2 · (−4)

= −4(x2 − 1)−3[−4x2 + (x2 − 1)

]= −4(x2 − 1)−3

[−3x2 − 1

]=

4(3x2 + 1)

(x2 − 1)3

We have critical values at x = −1 and x = 1. Notice that we get no critical values

from the numerator since 3x2 + 1 6= 0.

7. Use the number line to determine when y is concave up or down.

Concave Up: (−∞,−1) and (1,∞)

Concave Down: (−1, 1)

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We have a change of concavity at x = −1 and x = 1, but these are asymptotes. So

they cannot be points of inflection.

8. Now, let’s piece everything together to sketch the graph. First, let’s start with our inter-

cepts, important points like local extreme and points of inflection, and the asymptotes.

9. Now finish the graph using your knowledge of when the function is increasing / de-

creasing and its concavity.

245

3.5 Summary of Curve Sketching Brian E. Veitch

3.5 Summary of Curve Sketching

Follow these steps to sketch the curve.

1. Domain of f(x)

2. x and y intercepts

(a) x-intercepts occur when f(x) = 0

(b) y-intercept occurs when x = 0

3. Symmetry: Is it even or odd or neither. This usually isn’t of help.

If f(−x) = −f(x), then f(x) is symmetric about the origin.

If f(−x) = f(x), then f(x) is symmetric about the y-axis.

4. Find any vertical or horizontal asymptotes.

(a) Vertical Asymptote: Find all x-values where limx→a

f(x) = ±∞. Usually when the

denominator is 0 and the numerator is not 0

(b) Horizontal Aymptotes: Find limx→∞

f(x) and limx→−∞

f(x).

5. Find f ′(x)

(a) Find the critical values, all x-values where f ′(x) = 0 or when f ′(x) does not exist.

(b) Find increasing / decreasing intervals using numberline

(c) Find local maximums / minimums (if any exist). Remember to write them as

points.

i. Local Max at x = c: f ′(x) changes from (+) to (-) at x = c.

ii. Local Min at x = c: f ′(x) changes from (-) to (+) at x = c.

(d) Plot them

6. Find f ′′(x)

(a) Find all x-values where f ′′(x) = 0 or when f ′′(x) does not exist.

246


(b) Find intervals of concavity using the number line

(c) Find points of inflection

i. Must be a place where concavity changes

ii. The point must exist (i.e, can’t be an asymptote, discontinuity)

(d) Plot them

7. Sketch

247


Example 3.17. Sketch y =x√x2 + 1

It’s probably best to rewrite f(x) as f(x) =x

(x2 + 1)1/2

1. Domain: There are no domain issues.

2. Intercepts:

y − intercept: (0, 0)

x− intercept: (0, 0)

3. Symmetry:

f(−x) =(−x)√

(−x)2 + 1=

−x√x2 + 1

= −f(x)

This is an odd function. This means once we finish sketching, if the graph is not

symmetric about the origin we did something wrong.

4. Asymptotes:

(a) Horizontal Asymptote:

limx→∞

x√x2 + 1

= limx→∞

x√x2

= limx→∞

x

x

= 1

limx→−∞

x√x2 + 1

= limx→−∞

x√x2

= limx→∞

x

−x= −1

Recall:√x2 =

x, x > 0

−x, x < 0

248


(b) Vertical Asymptote:

There are none. Nothing makes√x2 + 1 = 0.

5. Find y′. Let’s use the Quotient Rule.

y′ =(x2 + 1)1/2 · (1)− x · 1

2(x2 + 1)−1/2 · 2x

[(x2 + 1)1/2]2

=(x2 + 1)1/2 − x2(x2 + 1)−1/2

x2 + 1

=(x2 + 1)−1/2 [(x2 + 1)− x2]

x2 + 1

=(x2 + 1)−1/2

x2 + 1

=1

(x2 + 1)3/2

Critical Values: There are no critical values.

Since there are no critical values. Check any point in the domain, the graph is increasing

the entire time.

Increasing: (−∞,∞)

6. Find y′′: Rewrite y′ = (x2 + 1)−3/2

y′′ = −3

2(x2 + 1)−5/2 · (2x)

=−3x

(x2 + 1)5/2

Critical Values: x = 0

249


Concave Up: (−∞, 0)

Concave Down: (0,∞)

Point of Inflection: (0, 0)

7. Sketch the graph

Example 3.18. Sketch y = 2√x− x

1. Domain: x ≥ 0

2. Intercepts:

(a) y-intercept: (0, 0)

(b) x-intercepts: Set 2√x− x = 0

250


2√x− x = 0

2√x = x

(2√x)2 = (x)2

4x = x2

0 = x2 − 4x

0 = x(x− 4)

which gives us x-intercepts (0, 0) and (4, 0).

3. There is no symmetry.

4. Asymptotes:

(a) There are no vertical asymptotes.

(b) There are no horizontal asymptotes.

limx→∞

2√x− x = −∞

We already did an example of this type of limit. Use that technique to show it’s

−∞.

5. Find y′

y′ = 2 · 1

2x−1/2 − 1

= x−1/2 − 1

=1√x− 1

We have a critical value at x = 0 because y′ does not exist when x = 0. Now we need

to solve y′ = 0

251


1√x− 1 = 0

1√x

= 1

1 =√x

1 = x

The other critical value is at x = 1.

6. Use the number line to determine where y is increasing or decreasing.

Note, we did not have to pick a number in the region less than 0 since that region is

not in the domain.

Increasing: (0, 1)

Decreasing: (1,∞)

Local Maximum: (1, 1)

Local Minimum: None

7. Find y′′. First, rewrite y′ as y′ = x−1/2 − 1.

y′′ = −1

2x−3/2

y′′ = − 1

2x3/2

There is a critical value at x = 0.

8. Use the number line to determine concavity.

252


Concave Down: (0,∞)

Concave Up: Never

No points of inflection

9. Sketch the graph

Example 3.19. Sketch y = 1 +1

x+

1

x2

1. Domain: x 6= 0

2. Intercepts:

(a) y-intercept: None, because x 6= 0

(b) x-intercept(s): Let’s set y = 0

1 +1

x+

1

x2= 0

Multiply through by x2

x2 + x+ 1 = 0

There are no solutions to this equation. This means y does not have any x-

intercepts.

253


3. There is no symmetry.

4. Asymptotes:

(a) Vertical Asymptote: x = 0

(b) Horizontal Asymptote:

limx→∞

1 +1

x+

1

x2= 1

limx→−∞

1 +1

x+

1

x2= 1

So there is one horizontal asymptote at y = 1.

5. Find y′. Rewrite y as y = 1 + x−1 + x−2.

y′ = −x−2 − 2x−3

To find critical values, we need to find out when y′ = 0 or y′ does not exist.

We see y′ does not exist when x = 0. Next, let’s set y′ = 0

−x−2 − 2x−3 = 0

−x−3 (x+ 2) = 0

So we have critical values at x = 0 and x = −2.

6. Use the number line to determine when y is increasing / decreasing.

254


Increasing: (−2, 0)

Decreasing: (−∞,−2) and (0,∞)

Local Minimum: (−2, 34)

There is no local maximum. The number line indicates there would be one at x = 0.

But recall that x = 0 is not in the domain.

7. Find y′′

y′′ = 2x−3 + 6x−4

Critical values occur when y′′ = 0 and y′′ does not exist. You can see y′′ does not exist

when x = 0.

2x−3 + 6x−4 = 0

2x−4 (x+ 3) = 0

So we have critical values at x = 0 and x = −3

8. Use the number line to determine concavity.

Concave Down: (−∞,−3)

Concave Up: (−3, 0) and (0,∞). You cannot write (−3,∞) because x 6= 0.

We have one point of inflection at (−3, 79).

9. Time to sketch!

255


256

3.6 Optimization Brian E. Veitch

3.6 Optimization

Problems in this section are worded very similarly. The objective to these problems is to

maximize or minimize a certain quantity. For example,

• How should I create a cylinder from a sheet of metal to maximize the volume?

• How much should I sell this product so I minimize cost?

• Where should an access road connect to the highway to minimize the time it takes

from point A to B.

Steps

1. Understand the problem. Can you explain the problem to someone else? If you can’t,

you don’t understand the problem.

(a) What’s known?

(b) What’s unknown?

(c) What is your objective function?

(d) Are you minimizing or maximizing?

2. Draw a picture / diagram / chart / etc. Identify what’s known and unknown on your

picture.

3. Assign a letter to represent the quantity (objective function) to be maximized or min-

imized. Common letters are

• A = Area

• V = Volume

• P = Perimeter

• T = Time

• P = Profit

257


• Q = Quantity

4. Express your objective function (A, V, T, ...) in terms of the variables, unknowns,

knowns, etc.

5. If your objective function is in terms of more than one variable, write all of the variables

in terms of one variable.

6. Find the absolute max or min (on the respective domain).

Example 3.20. Find two positive integers whose product is 100 and whose sum is a mini-

mum.

1. Understand the problem? Give yourself some examples so you get an idea of what’s

going on.

10 · 10 = 100and the sum is 20

100 · 1 = 100and the sum is 101

20 · 5 = 100and the sum is 25

25 · 4 = 100and the sum is 29

Based on this, it appears 10 and 10 do the trick. But these are just some of the

examples. I need to use the optimization method from above to prove it’s 10 and 10.

2. Let x = one integer and y = the other integer.

Known: x · y = 100

Objective: Minimize the sum S = x+ y

3. Since the objective function is in terms of two variables, I need to write one variable

in terms of the other. That’s where x · y = 100 comes in.

y =100

x

258


The new objective function is

Minimize S = x+100

x

4. To find the minimum of S = x+100

x, we need to find S ′

S ′ = 1− 100

x2

Now, we find the critical values. One critical value is x = 0, but we don’t count it

because we know our integers have to be positive.

1− 100

x2= 0

1 =100

x2

x2 = 100

x = ±10

We only take x = 10 as a critical value because x = −10 is not positive.

5. Just to make sure we have our minimum at x = 10, let’s use the number line to check.

So we have a local minimum at x = 10. We confirmed our initial guess that the

numbers are 10 and 10. But actually, at this point, we’ve only confirmed one integer

is 10. To find the other integer y, use

259


y =100

x

y =100

10= 10

Therefore, two numbers that multiply 100 and whose sum is at a minimum is 10 and

10.

Example 3.21. Suppose I want to enclose off an area for a garden. It will be up against my

house, so I don’t have to fence the side that’s against the house. I have 24 feet of fencing.

How should I construct this rectangular garden so I get the largest area?

1. Let’s start with a picture. This should help you understand the problem.

Known: 2W + L = 24

Objective: Maximize A = L ·W

2. Notice that there are two variables L and W . We will use 2W + L = 24 to write L in

terms of W .

2W + L = 24

L = 24− 2W

260


Our new objective is

Maximize A = (24− 2W ) ·W = 24W − 2W 2

3. To maximize A = 24W − 2W 2, we need to find A′

A′ = 24− 4W

To find the critical values, set A′ = 0

24− 4W = 0

gives us a critical value at W = 6. Verify by using the number line and we’ll find that

W = 6 is a local maximum.

If W = 6, we use L = 24− 2W to find L.

L = 24− 2(6) = 12

To get the largest area, I should fence off a rectangular region with the dimensions

W = 6 and L = 12

Example 3.22. Your task is to build an access road joining a small village to a highway

that enables drivers to reach a large city in the shortest time. A picture is given below. How

should this be done is the speed limit on the access road is 35 mph and the speed limit on

261


the highway is 55 mph. The shortest possible distance from the village to the highway is 15

miles and the city is 45 miles down the highway.

1. Let x = the distance down the highway where the access road connects to the highway.

As a function of x, the length of the access road is

√152 + x2

The distance remaining to the city is 45− x.

The total distance traveled is

√152 + x2 + (45− x)

2. But we are concerned with minimizing the time it takes to get from the village to the

city. So we need a formula for time traveled. The formula for time is

t =distance

speed=d

s

262


Since the speed is different for the two roads, we need to split it up into two times, one

for the access road, and one for the highway.

3. Time traveled on the access road is

√152 + x2

35.

4. Time traveled on the highway is45− x

55

5. Our objective is to

Minimize T (x) =

√152 + x2

35+

45− x55

over the interval 0 ≤ x ≤ 50. Keep in mind, it’s very possible that to minimize time

the access road can connect to the highway along the shortest possible distance (x = 0)

or the access road can go all the way to the city (x = 50).

6. Let’s go ahead and find T ′(x). Let’s rewrite T (x) so it’s easier to differentiate,

T (x) =1

35

(152 + x2

)1/2+

1

55(45− x)

T ′(x) =1

35· 1

2(152 + x2)−1/2 · (2x) +

1

55· (−1)

=x

30(152 + x2)1/2− 1

55

7. To find the critical values, set T ′(x) = 0.

263


x

30(152 + x2)1/2− 1

55= 0

x

30(152 + x2)1/2=

1

55

cross multiply

55x = 30(152 + x2)1/2

11x = 6(152 + x2)1/2

square both sides

121x2 = 36(152 + x2)

121x2 = 8100 + 36x2

85x2 = 8100

x2 =8100

85

x = 9.76

8. Since we’re looking for the ABSOLUTE minimum, we check the time traveled at the

endpoints, x = 0 and x = 50, and the critical value at x = 9.76.

T (0) =1

35

(152 + 02

)1/2+

1

55(45− 0) = 1.25 hours

T (9.76) =1

35

(152 + 9.762

)1/2+

1

55(45− 9.76) = 1.15 hours

T (45) =1

35

(152 + 452

)1/2+

1

55(45− 45) = 1.36 hours

The time traveled is minimized if the access road is connected to the highway 9.76

miles from P .

264


Example 3.23. Suppose you have a cylinder with a volume of 355 cm3 or 21.66 in3. What

should be the radius and height of the cylinder so that it uses the least amount of material.

1. Before we begin, using the least amount of material means minimizing surface area of

the material used.

2. Let’s take a look at some examples.

3. So we need two formulas. One is the formula for the volume of a cylinder. The second

is the formula for the surface area of a cylinder.

Volume: V = πr2h

Surface Area: S = 2πrh+ 2πr2

265


4. Our objective is to minimize S = 2πrh+ 2πr2

5. The objective function is in terms of two variables, r and h. We need to rewrite one

in terms of the other. Since r shows up more often in the formula, let’s rewrite h in

terms of r. We will use the volume formula to do it.

V = 21.66 = πr2h

h = 21.66/(πr2)

6. So our new objective is to minimize

S = 2πr ·(

21.66

πr2

)+ 2πr2 =

43.32

r+ 2πr2

7. Let’s find S ′ and set it equal to 0.

S ′ = −43.32

r2+ 4πr

Let’s set S ′ = 0

−43.32

r2+ 4πr = 0

4πr =43.32

r2

4πr3 = 43.32

r3 = 3.45

r = 1.51

266


8. We minimize the material used if the cylinder has a radius of 1.511 inches. You can

verify it’s a local minimum by using the number line. Now we find h

h =21.66

π(1.51)2

h = 3.02 inches

By the way, 21.66 in3 is the volume of a soda can. The dimensions of a soda can differ

slightly from the dimensions we found that minimizes the material used. But there

are a couple of things to note. Soda cans are not perfectly cylindrical. We found the

radius should be 1.51 in, but a soda can is actually about 1.3 in. It’s possible a soda

can has a smaller radius so it’s easier to hold.

Example 3.24. During the summer Samantha makes and sells necklaces. Last summer she

sold each one for $10 and averaged 20 sales per day. When she increased the price by $1,

the sales decreased by two per day.

1. Find the demand function.

2. If material costs $6, what should the selling price by to maximize profit?

These are good problems because in real life trying to maximize profit consists of some

trial and error. After a couple of different scenarios, you try to find which one makes you

the most money.

Note:

1. Profit = Revenue - Cost

2. Revenue = x · p(x), where x = the number sold and p(x) is the price.

3. The demand function is p(x). The price of a necklace depends on the number of

necklaces sold per day.

267


So how do we find the demand function, p(x). Notice that we have two pieces of data.

We know p(x) must go through the points (20, 10) and (18, 11). So we can estimate p(x)

by finding the equation of a line that passes through these two points.

Slope: m =11− 10

18− 20= −1

2

Using the point-slope formula

y − y1 = m(x− x1)

y − 10 = −1

2(x− 20)

y = −1

2x+ 20

Therefore, the demand function is p(x) = −1

2+ 20. What does this mean? It means for

every extra 2 necklaces Samantha sells, the price must go down by $1.

Let’s move to maximizing profit.

P (x) = R(x)− C(x)

P (x) = x · p(x)− C(x)

P (x) = x ·(−1

2x+ 20

)− (6x)

P (x) = −1

2x2 + 20x− 6x

P (x) = −1

2x2 + 14x

Note, C(x) = 6x because each necklace costs $6 to make.

268


To maximize P (x), we need to find P ′(x).

P ′(x) = −x+ 14

Setting P ′(x) = 0, we find x = 14. You can verify this is the maximum by using the

number line.

So to maximize profit, Samantha needs to sell 14 necklaces. The price per necklace is

P (14) = −1

2(14) + 20 = 13

Conclusion: Samantha needs to sell 14 necklaces at $13 to maximize profit.

269

3.7 Newton’s Method Brian E. Veitch

3.7 Newton’s Method

Finding interest on a loan payment is by no means an easy task. Often, the variable we

need to solve for is hiding in the exponent, and if its combined with a sum or difference of

multiple terms, no easy way to solve (even taking logs wouldn’t help). How do we solve

73x(1 + 2x)39 + (1 + 2x)39 − 1,

if it doesn’t factor (and it doesn’t)? Since f is a polynomial (after we substitute

u = 1 + 2x), it is still a 39th degree polynomial and there is no formula to solve greater

than a 4th degree, and even that is awful. Instead, we can do better, and with less pain, by

approximating – for instance, this is what your calculator would do, basically the graph and

zoom in method after it plotted a series of points. But how does it do this?

The most commonly used method to find these approximations is called Newton’s

Method, after the cookie guy. No, the apple guy. Yeah, him. Newton’s method involves

trying to find the root r of a function, and we demonstrate with a picture below:

We make a guess to start, and we call that guess x1. Then, we consider the tangent line

L to y = f(x) at this point, (x1, f(x1)). Now, we need to find the x-intercept of this tangent

270


line L, the point where L crosses the x-axis. We call this point x2 – this is how Newton’s

method works; the tangent line gives an x-intercept closer to the root than our initial guess

was. Since the tangent line is close to the curve, the x-intercept of the tangent line should

be close to the x-intercept of the curve, r. Even better, since L is just a straight line, we

can easily find it’s x-intercept.

To do this, recall that the slope of the tangent line is the derivative of the function, so

the slope of L is f ′(x1). Point-slope form gives the equation

y − f(x1) = f ′(x1)(x− x1)

Since we know that the x-intercept of L is x2, we can set the point (x, y) = (x2, 0):

0− f(x1) = f ′(x1)(x2 − x1)

For anything but a horizontal tangent line, we have f ′(x1) 6= 0, we can solve for x2:

x2 = x1 −f(x1)

f ′(x1).

Then, we can use x2 as an approximation to r, repeating this procedure:

Using the tangent line at (x2, f(x2)) gives a third approximation

271


x3 = x2 −f(x2)

f ′(x2)

And we can keep repeating this process over and over to get a sequence of approxima-

tions, x1, x2, x3, .... In general, we let the nth approximation be xn and if f ′(xn) 6= 0, then

the next approximation is given by

3.7.1 Newton’s Method Formula

xn+1 = xn −f(xn)

f ′(xn)

As we let n→∞, then xn → r. However, it is possible that the limit does not head to r

– such things happen if f ′(x1) ≈ 0, and x2 is much further away from r. In order to fix this

issue, we can simply choose a different, and better, approximation for x1.

Example 3.25. Let x1 = 3. Find the third approximation to the equation x3−3x2 + 6 = 0.

We apply Newton’s Method with

f(x) = x3 − 3x2 + 6

and

272


f ′(x) = 3x2 − 6x

We have

xn+1 = xn −f(xn)

f ′(xn)

x2 = x1 −f(x1)

f ′(x1)

= 3− f(3)

f ′(3)

= 3− 27− 27 + 6

27− 18

= 3− 6

9=

7

3

Then, we let n = 2 and we get

x3 = x2 −f(x2)

f ′(x2)

=7

3− f(7/3)

f ′(7/3)

=7

3− (7/3)3 − 3(7/3)2 + 6

3(7/3)2 − 6(7/3)

=7

3− 64/27

7/3=

83

63

While this is the most direct use of Newton’s Method, another quite valuable one is to

know how far we have to go until we are a certain level of accuracy away from the actual

root. If we want to know when we have accuracy to, say, 6 decimal places, we wait until xn

and xn+1 agree to 6 decimal places. For instance...

Example 3.26. Use Newton’s Method to find 3√

2 to 6 decimal places.

Finding the decimal value to 3√

2 is the same as finding the root to

273


x3 − 2 = 0.

So we let f(x) = x3 − 2 and f ′(x) = 3x2, choosing x1 = 1 as our guess, we have

xn+1 = xn −f(xn)

f ′(xn)

xn+1 = xn −x3n − 2

3x2n

x2 = 1− 13 − 2

3(1)2= 1− −1

3=

4

3

x3 =4

3− (4/3)3 − 2

3(4/3)2=

91

72

x4 = 1.259933

x5 = 1.25992105

x6 = 1.2599210499

Thus, we have 3√

2 ≈ 1.259921.

Now, perhaps needless to say, we did use a calculator for that one. As we probably

will for most of our Newton’s method problems. Just is the issue with approximating, very

difficult to do by hand. Instead, we end this section with two applications.

Example 3.27. Use Newton’s Method to find the absolute maximum of f(x) = x cos(x) on

[0, π].

274


We first get f ′(x) = cos(x)−x sin(x). Since f ′(x) exists for all x, we can find the absolute

maximum of f by finding the zeros of f ′. Thus, here, we need

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

We start by choosing x = .9:

xn+1 = xn −f ′(xn)

f ′′(xn)

xn+1 = xn −cos(x)− x sin(x)

−2 sin(x)− x cos(x)

x2 = .9− cos(x)− x sin(x)

−2 sin(x)− x cos(x)= .8607807

x3 = .86033365

x4 = .86033359

Then, since x3 ≈ x4 ≈ .86033359, we can evaluate f(.86033359) = .561096. Further,

since we have a closed interval, we check f(0) = 0 and f(π) = π(−1) = −π, we have a

maximum value at (.86033359, .561096).

Example 3.28. Try to find a root of f(x) = x3 − 3x+ 8 = 0 starting with x1 = 1.

We have f ′(x) = 3x2 − 3, and we have

xn+1 = xn −f ′(xn)

f ′′(xn)= xn −

x3 − 3x+ 8

3x2 − 3.

275


Then

x2 = 1− 1− 3 + 8

3− 3= 1− 6

0,

which is quite bad. Yeah, never use a starting approximation which gives a horizontal

tangent line. You get a divide by 0 problem, and have to start over.

There are other situations where Newton’s Method fails. The previous example failed

because you made an initial guess that gave us a horizontal tangent line. That means the

tangent line will never go back to the x-axis, which means we’ll never get our next approxi-

mation.

Take a look at the graph. I want to make sure you understand what’s happening here.

Another problem is Newton’s Method can get stuck in a cycle that does not get you a

better approximation. Consider the following equation, x1/3 = 0. Below is a graph of a

few tangent lines. New approximations come from a thicker tangent line. Notice how each

tangent line is going away from the root.

276


The last problem I want to mention is when Newton’s Method works but finds the a

different root.

277

3.8 Anti-Derivatives (Integration) Brian E. Veitch

3.8 Anti-Derivatives (Integration)

We have seen that if we have a function which gives us the position of an object, we can

find the velocity by taking the derivative of the function. However, can we turn this around?

Can we start with velocity and get an equation for the position? In this case, we want to

find a function whose derivative is the equation for velocity, v(t). If there is a function F

such that F ′(x) = v(x), then we have a definition:

Definition 3.4. A function F is called an integral of f on an interval I if F ′(x) = f(x).

Another term for an integral is an anti-derivative, which the text uses. Technically, the

integral has a geometric definition which we cover in Chapter 4, but we will be able to use

the terms interchangeably.

Let us examine f(x) = x4. We want to find F such that F ′(x) = x4. We do know the

power rule, and since g is a polynomial, this seems like a good place to start. We need a

number to be killed off by the exponent and the exponent should be one bigger. A function

that works is F (x) = 15x5, such that

d

dx

[1

5x5]

=1

5· 5x5−1 = x4.

However, we also have F (x) = 15x5 + 70, and as such

d

dx

[1

5x5 + 70

]=

1

5· 5x5−1 + 0 = x4.

278


These are both integrals of f . In general, we could have any function F (x) = 15x5 + c,

where c is any constant. This is a general integral of f . The question is, are there other

general ones?

To answer this, think back to the Mean Value Theorem – a result from this is that is any

two functions have the same derivative, then they must differ by a constant. Thus, if f and

F are any two anti-derivatives of g, then

F ′(x) = g(x) = f ′(x),

and thus since they have the same derivative, F and f can differ only by a constant.

Thus, there is only one general integral.

Theorem 3.7. If F is an integral (anti-derivative) of f over an interval I, then the most

general integral of f on I is

F (x) + c,

where F ′(x) = f(x) and c is any arbitrary constant.

The only difference between different values of c is that the graph of F shifts up or down,

but the general slope and shape of the graph is identical.

Example 3.29. Find the most general integral for each of the following functions:

f(x) = sin(x).

If F (x) = − cos(x), then F ′(x) = −(− sin(x)) = sin(x). The most general integral func-

tion is F (x) = − cos(x) + C.

279


h(x) =1

x2.

If F (x) = −x−1, then F ′(x) = −(−1)x−2 = 1x2

. The most general integral function is

F (x) =−1

x+ C

.

3.8.1 Power Rule for Anti-Derivatives with Trig

If f(x) = xn, then the general anti-derivative of f is

F (x) =xn+1

n+ 1+ C

Every single one of our differentiation formulas gives rise to an anti-differentiation for-

mula, just when read backwards. Below we list some of the more common formulas for

anti-derivatives, including a constant and a sum rule! (But no product or quotient...no no

no):

Function Antiderivative

cf(x) cF (x)

f(x) + g(x) F (x) +G(x)

xn, n 6= −11

n+ 1xn+1

cos(x) sin(x)

sin(x) − cos(x)

sec2(x) tan(x)

sec(x) tan(x) sec(x)

csc2(x) − cot(x)

csc(x) cot(x) − csc(x)

280


Example 3.30. Find a general function f such that

f ′(x) = 3 cos(x) +3x4 −

√x

x2.

We break the above function up so we do not have a complex fraction:

f ′(x) = 3 cos(x) +3x4

x2−√x

x2= 3 cos(x) + 3x2 − x−3/2.

Then, we can find an anti-derivative above by using reverse-derivative rules in the table

above:

f(x) =3 sin(x) + 3 · 1

2 + 1x2+1 − 1

−3/2 + 1x−3/2+1 + c

=3 sin(x) + x3 + 2x−1/2 + c

In Calculus, we have this situation an awful lot, where we want to find a function given

some information about its derivative. The general solution usually involves some arbitrary

constant, but we may be given some additional conditions to find the exact value of this

constant.

Example 3.31. Find f if

f ′(x) = 20 sin(x) + 5x3,

and f(0) = 3.

The general antiderivative of

f ′(x) = 20 sin(x) + 5x3

is

f(x) = −20 cos(x) +5

4x4 + C.

281


In order to find the exact value of c, we use the fact that f(0) = 3. We let x = 0 and

solve for c:

3 =− 20 · cos(0) +5

4(0)4 + c

3 =− 20 + c

c =23

Example 3.32. Find f if

f ′′(x) = 5x2 + 2x− 1,

f ′(0) = 2 and f(1) = 3.


f ′′(x) = 5x2 + 2x− 1,

is

f ′(x) =5

3x3 + x2 − x+ C1.

In order to find the exact value of C1, we use the fact that f ′(0) = 2. We let x = 0 and

solve for C1:

2 =5

3(0)3 + 02 − 0 + C1

C1 =2

Then, we have

f ′(x) =5

3x3 + x2 − x+ 2,

and we have general integral

f(x) =5

12x4 +

1

3x3 − 1

2x2 + 2x+ C2.

282


In order to find the exact value of c2, we use the fact that f(1) = 3. We let x = 1 and

solve for C2:

3 =5

12(1) +

1

3(1)− 1

2(1) + 2(1) + C2

3 =5

12+

4

12− 6

12+

24

12+ C2

3 =27

12+ C2

C2 =9

4− 3 =

−3

4

Then, we have

f(x) =5

12x4 +

1

3x3 − 1

2x2 + 2x− 3

4.

As we started this section, we talked about finding the position of an object if we start

with the velocity. For that matter, we could start with the acceleration and go up from there.

Example 3.33. Suppose a velociraptor accelerates according to the function

a(t) = 12t− 8.

If the initial position is s(0) = 10 and the initial velocity if v(0) = 1, find the position

function of the raptor s(t).


a(t) = 12t− 8

is

v(t) =12

2t2 − 8t = 6t2 − 8t+ C1.

283


In order to find the exact value of C1, we use the fact that v(0) = 1. We let t = 0 and

solve for C1:

1 =6(0)2 − 8(0) + C1

C1 =1

Then, we have

v(t) = 6t2 − 8t+ 1,

and we have general integral

s(t) =6

3t3 − 8

2t2 + t+ c2 = 2t3 − 4t2 + t+ c2.

In order to find the exact value of c2, we use the fact that s(0) = 10. We let x = 0 and

solve for c2, which gives

c2 = 10.

Thus,

s(t) = 2t3 − 4t2 + t+ 10.

Example 3.34. We throw a watermelon up in the air at an initial speed of 80ft/sec off the

top of a 9 story building which is 160 feet off the ground. If we assume that the watermelon

accelerates at −32 ft/sec2, when does the watermelon reach its maximum height, and when

does it hit the ground?

In order to find the maximum height, we need the velocity function v(t) and to find when

the ball hits the ground, we need the position function s(t). We have

s′′(t) = a(t) = −32.

284


Taking an antiderivative, we have

s′(t) = v(t) = −32t+ C,

and we can find c by using the initial speed, v(0) = 80. Thus,

s′(t) = v(t) = −32t+ 80.

Then, we can anti-differentiate again and get

s(t) = −16t2 + 80t+ C1,

and to find C1, we use the fact that s(0) = 160, so

s(t) = −16t2 + 80t+ 160.

Now, we can find the maximum height using the velocity function – the maximum height

is reached when v(t) = 0:

v(t) =0

−32t+ 80 =0

t =80

32=

5

2

Thus, the maximum height is reached after 5/2 seconds. In order to find when the ball

hits the ground, we solve s(t) = 0:

s(t) =0

−16t2 + 80t+ 160 =0

t2 − 5t− 10 =0

t =5±

√25− 4(1)(−10)

2(1)=

5±√

65

2

285


Since 5−√

65 < 0 and negative time makes no sense, we discard this solution. Thus, the

watermelon hits the ground at

t =5 +√

65

2≈ 6.53 seconds.

Example 3.35. Find f when f ′′(x) = x−√x

1. Rewrite f ′′ as f ′′(x) = x− x1/2.

2. Find f ′(x)

f ′(x) =1

2x2 − 2

3x3/2 + C1

3. Since we’re not given any information about f ′(x), we move on to finding f(x).

f(x) =1

2· 1

3x3 − 2

3x5/2 · 2

5+ C1x+ C2

f(x) =1

6x3 − 4

15x5/2 + C1x+ C2

286

Brian E. Veitch

4 Integration

4.1 Areas and Distances

We begin Chapter 4 by trying to solve a simple question: given a function f(x) over an

interval (a, b), what is the area of the region under f between a and b?

If we had an easy region under a curve, that of a simple polygon, this question is easy to

answer – a rectangle has area base time height, a triangle is 12bh, and many polygons can be

broken into a sum of triangles. But in our picture above, we have a curved top side, which

is not a polygon.

When we were looking for the slope of a tangent line, we took the limit of slopes of secant

lines – a limit of an approximation. We do the same here, approximating the area under the

curve using the easiest figure we have, the rectangle.

Example 4.1. Estimate the area under the curve y = x2 on (0, 2) with 4 rectangles.

Begin by drawing a picture of the function:

287

4.1 Areas and Distances Brian E. Veitch

Since we can put the curve entirely inside a rectangle of width 2 and height 4, of course

the area under the curve would be smaller than 2 · 4 = 8, but that’s so crap, we can do

better. Chop the region under the curve S into four strips, S1, ..., S4 by drawing three evenly

spaced vertical lines x = 12, x = 1 and x = 3

2.

Then, we can make an approximation of this area of these four strips by creating rectan-

gles. One way to do this is to extend the upper-right corner of each strip until we reach the

next vertical line, a rectangle whose height is the same as the right edge. Since the function

is f(x) = x2, the height of these rectangles will just be the value of f(x) when evaluated at

the right endpoint of the intervals [0, 1/2], [1/2, 1], [1, 3/2], [3/2, 2].

Each rectangle has a width of 1/2, and if we let Ai be the area of the i-th rectangle, we get

288


A1 = f

(1

2

)·(

1

2

)=

(1

2

)2

·(

1

2

)=

1

8

A2 = f (1) ·(

1

2

)= (1)2 ·

(1

2

)=

1

2

A3 = f

(3

2

)·(

1

2

)=

(3

2

)2

·(

1

2

)=

9

8

A4 = f (2) ·(

1

2

)= (2)2 ·

(1

2

)= 2

R4 =1

8+

1

2+

9

8+ 2 =

15

4.

Now because we counted alot of area above the curve in this approximation, we know

that

A < R4 =15

4.

However, this is not the only approximation we could do. Instead of extending the upper-

right corner, we could extend the upper-left corner and create rectangles that way. If the

heights of the rectangles are equal to the function evaluated at the left endpoint, we would

have the following picture:

289


These rectangles still have width 1/2, but if we let the area of all four be L4, then we

again add base times height 4 times:

L4 =

(1

2

)(0)2 +

(1

2

)·(

1

2

)2

+

(1

2

)· (1)2 +

(1

2

)·(

3

2

)2

=7

4

Just as the area of A was smaller then R4, we know that the area of A is larger than L4, so

L4 =7

4< A.

Combining these, we have upper and lower bounds on the area of S:

7

4< S <

15

4.

We could repeat this procedure with more and more rectangles, and the smaller the width

of the rectangle it makes sense that the approximation will get better.

Example: Let’s go ahead and try to estimate the area under y = x2 on [0,2], but with

8 rectangles. We’ll see if we get a better estimate.

290


1. Using the right hand endpoints

Each rectangle has a width of 1/4, and if we let Ai be the area of the i-th rectangle, we get

A1 = f

(1

4

)·(

1

4

)=

(1

4

)2

·(

1

4

)= 0.015625

A2 = f

(2

4

)·(

1

4

)=

(2

4

)2

·(

1

4

)= 0.0625

A3 = f

(3

4

)·(

1

4

)=

(3

4

)2

·(

1

4

)= 0.140625

A4 = f

(4

4

)·(

1

4

)=

(4

4

)2

·(

1

4

)= 0.25

A5 = f

(5

4

)·(

1

4

)=

(5

4

)2

·(

1

4

)= 0.390625

A6 = f

(6

4

)·(

1

4

)=

(6

4

)2

·(

1

4

)= 0.5625

A7 = f

(7

4

)·(

1

4

)=

(7

4

)2

·(

1

4

)= 0.765625

A1 = f

(8

4

)·(

1

4

)=

(8

4

)2

·(

1

4

)= 1

291


R8 = 0.015625 + 0.0625 + 0.140625 + 0.25 + 0.390625 + 0.5625 + 0.765625 + 1 = 3.1875.

One thing to note is7

4< 3.1875 <

15

4. This means we do have a better estimate. We

can keep doing this to get better estimates. Here’s what it would look like by taking more

rectangles.

1. 12 rectangles

Area ≈ 3.0093

2. 20 rectangles

292


Area ≈ 2.87

3. 50 rectangles

Area ≈ 2.747

293


You can see that we are getting better estimates. However, it appears that I’d have to

take a lot more rectangles to get really close to 2.666. If I used 100 rectangles, the approxi-

mation is 2.7068.

Example 4.2. For the region in the previous problem, take an indefinite number of right-

edge rectangles and find

limn→∞

Rn.

Okay, this is harder to draw. We have the same curve, but we are taking basically

infinitely many rectangles to approximate the area. Let’s start with just some number n

rectangles. Since we are chopping the interval [0, 2] into n even pieces, each rectangle will

have the same width. The formula for the width of the rectangles is

∆x =b− an

which in our case is,2− 0

n=

2

n

Thus, the right endpoints will be

x1 = a+ 1∆x → 0 + 1 · 2

n=

2

n

x2 = a+ 2∆x → 0 + 2 · 2

n=

4

n

x3 = a+ 3∆x → 0 + 3 · 2

n=

6

n

xi = a+ i∆x → 0 + i · 2

n=

2i

n

xn = a+ n∆x → 0 + n · 2

n=

2n

n= 2

294


The area of the rectangles are

1st rectangle:A1 = f(x1) ·∆x



and so on...

Area of all the rectangles are

A ≈ f(x1) ·∆x+ f(x2) ·∆x+ f(x3) ·∆x+ ...+ f(xn) ·∆x

So the rectangles are

A1 = f

(2

n

)· 2

n=

(2

n

)2

· 2

n

A2 = f

(2

n

)· 2

n=

(4

n

)2

· 2

n

A3 = f

(6

n

)· 2

n=

(6

n

)2

· 2

n

Ai = f

(2i

n

)· 2

n=

(2i

n

)2

· 2

n

...

An = f

(2n

n

)· 2

n=

(2n

n

)2

· 2

n

Let’s write it as one long sum and simplify

295


Rn =

(2

n

)2

· 2

n+

(4

n

)2

· 2

n+

(6

n

)2

· 2

n+ . . .+

(2n

n

)2

· 2

n

=2

n· 1

n2

(22 + 42 + 62 + . . .+ (2n)2

)

factor out 22 = 4 from every term

=8

n3

(12 + 22 + 32 + . . .+ n2

)

In order to finish, we need to use a commonly known identity – the sum of the squares

of the first n integers:

12 + 22 + 32 + 42 + . . .+ n2 =n(n+ 1)(2n+ 1)

6.

Thus, we can substitute this into our equation for Rn and we have

Rn =8

n3· n(n+ 1)(n+ 2)

6=

4n(n+ 1)(2n+ 1)

3n3.

296


Thus, to find the exact value, we need to take a limit as n→∞:

A = limn→∞

Rn = limn→∞

4n(n+ 1)(2n+ 1)

3n3

=4

3limn→∞

(n+ 1)(2n+ 1)

n2

=4

3limn→∞

2n2 + 3n+ 1

n2

=4

3limn→∞

2 + 3/n+ 1/n2

1

=8

3

≈ 2.667

Thus, it seems that the exact area under the curve f(x) = x2 over the interval (0, 2) is8

3.

This would work for the left endpoint rectangles too, Ln, and this is exactly how we define

the area under a curve:

A = limn→∞

Ln = limn→∞

Rn.

Example 4.3. Estimate the area under the graph of y =1

1 + x2from -2 to 2 using 6

rectangles.

1. Width: ∆ =b− an

=2− (−2)

6=

2

3

2. Find x1, x2, x3, x4, x5, x6

297


x1 = a+ 1∆x = −2 + 1 · 2

3= −4

3

x2 = a+ 2∆x = −2 + 2 · 2

3= −2

3

x3 = a+ 3∆x = −2 + 3 · 2

3= 0

x4 = a+ 4∆x = −2 + 4 · 2

3=

2

3

x5 = a+ 5∆x = −2 + 5 · 2

3=

4

3

x6 = a+ 6∆x = −2 + 6 · 2

3= 2

3. Now just plug everything in

A = f(x1) ·∆x+ f(x2) ·∆x+ f(x3) ·∆x+ f(x4) ·∆x+ f(x5) ·∆x+ f(x6) ·∆x

=1

1 + (−4/3)2· 2

3+

1

1 + (−2/3)2· 2

3+

1

1 + (0)2· 2

3+

1

1 + (2/3)2· 2

3+

1

1 + (4/3)2· 2

3

+1

1 + (2)2· 2

3

=2

3[0.36 + 0.6923 + 1 + 0.69203 + 0.36 + 0.2]

= 2.203

The exact value for the area is approximately 2.2143.

298


We now consider a more general problem, so suppose we have the following general

function f(x) over the interval (a, b), and we divide it into n rectangles:

Again, all of these intervals have equal width, and since the width of the entire interval

is b− a, the width of each is

∆x =b− an

.

These n strips divide the interval [a, b] into n subintervals: [x0, x1], [x1, x2] . . . [xn−1, xn].

as per our diagram, we have x0 = a and xn = b. We have the following right endpoints for

these intervals:

x1 = a+ ∆x

x2 = a+ 2∆x

x3 = a+ 3∆x

... =...

If we wish to consider approximating the ith strip of the area, we know it has width ∆x

and height f(xi), where xi if the value of f at the right endpoint. Then, the area of the ith

rectangle is

Ai = ∆xf(xi),

299


and we can get Rn by summing together these rectangles:

Rn = f(x1)∆x+ f(x2)∆x+ . . .+ f(xn)∆x.

The more and more rectangles we take, as we let n→∞, the better the approximation

we will get. And this gives us our definition:

Definition 4.1. The area of the region A that lies underneath the graph of a positive

continuous function f is the limit of the sum of approximating rectangles:

A = limn→∞

Rn = limn→∞

(f(x1)∆x+ f(x2)∆x+ . . .+ f(xn)∆x) .

Nice, as long as the function f is continuous, the above limit will always exist. Also, we

get out the exact same value if we use the left endpoints instead:

A = limn→∞

Ln = limn→∞

(xf(x0)∆ + f(x1)∆x+ . . .+ f(xn−1∆x)) .

As a matter of fact, there’s nothing special at all about choosing left or right endpoints,

all that matters is that we choose some points, as long as it is in the correct interval. So if

x∗i is in [xi−1, xi], we’re good. We call the points x∗1, x∗2, ..., x

∗n the sample points, and we get

a very general expression for the area under the curve:

A = limn→∞

(xf(x∗1)∆ + f(x∗2)∆x+ . . .+ f(x∗n)∆x) .

In order to write this a little cleaner, we use what is known as sigma notation for a summa-

tion:

300


n∑i=1

f(x∗i )∆x = f(x∗1)∆x+ f(x∗2)∆x+ . . .+ f(x∗n)∆x.

Thus, we can write

A = limn→∞

n∑i=1

f(xi)∆x

A = limn→∞

n∑i=1

f(xi−1∆x)

A = limn→∞

n∑i=1

f(x∗i )∆x

We will also need these later.

n∑i=1

i =n(n+ 1)

2.

n∑i=1

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

6.

n∑i=1

i3 =

[n(n+ 1)

2

]2.

Example 4.4. Let A be the area under the curve of f(x) = sin(x) over [0, π]. Find the

expression for A as a limit, without evaluating that limit.

Since a = 0 and b = π, the width of the interval is π. The width of each subinterval is

∆x =π − 0

n=π

nand xi = a+ i ·∆x

Thus, we have x1 = π/n, x2 = 2π/n, x3 = 3π/n, ..., xi = iπ/n. The sum of the areas of the

301


approximating rectangles is

Rn =n∑i=1

f(xi)∆x

=∆x ·n∑i=1

sin(xi)

=π

n

(sin(πn

)+ sin

(2π

n

)+ . . .+ sin

(nπn

))By definition, the area under the curve is

A = limn→∞

Rn = limn→∞

π

n

(sin(πn

)+ sin

(2π

n

)+ . . .+ sin

(nπn

)).

With sigma notation, we have

A = limn→∞

π

n

n∑i=1

sin

(iπ

n

).

This is NOT an easy sum or limit to try to do by hand, so we won’t. Instead, what we

can do is approximate this a little - let us do an approximation with n = 4 rectangles, each

with width π/4. The four intervals are then

[0,π

4

],[π

4,π

2

],

[π

2,3π

4

],[π

4, π].

We can choose any points we like to try to approximate, but since we only know the

exact values of sin with denominators of 2, 3, 4 or 6, we use a right endpoint approximation:

R4 =4∑i=1

f(xi)δx

=4∑i=1

sin(xi)π

4

=π

4

(sin(π

4

)+ sin

(π2

)+ sin

(3π

4

)+ sin(π)

)=π

4

(√2

2+ 1 +

√2

2+ 0

)=π

4

(√2 + 1

)302


But, this may not be so good, so let us consider a left endpoint approximation, and we will

average the two:

L4 =4∑i=1

f(xi−1)δx

=3∑i=0

sin(xi)π

4

=π

4

(sin(0) + sin

(π4

)+ sin

(π2

)+ sin

(3π

4

))=π

4

(0 +

√2

2+ 1 +

√2

2

)=π

4

(√2 + 1

)

Oh, that’s weird! They’re the same! This does NOT mean that this is the exact answer.

Much more likely is that the two approximations have the same error.

4.1.1 The Distance Problem

Suppose that we know the velocity of an object at all times, meaning that we know the value

of v(t) for all possible values of t in the domain of the function. If we have constant velocity,

then the distance is very easy to compute, we simply take distance equals velocity times time.

d = vt

But, besides that, it isn’t so simple.

Example 4.5. Suppose the mileage on a car doesn’t work and we want to estimate the

distance we drive over a minute-long time interval. We get readings off the speedometer

303


every 10 seconds and record them as:

Time (sec) 0 10 20 30 40 50 60

Velocity (ft/sec) 22 34 46 42 38 28 34

(Note that if the units are not consistent, you must change them so that they are). In

order to find how far we have traveled, we can assume that the velocity does not change over

a 10 second period, and then jumps at the 10 second mark. For example, we can assume

that for the first 10 seconds, we travel at 22 ft/sec. Thus, we travel

10 · 22 = 220

feet in the first 10 seconds. Then, for the next 10 seconds, we assume we travel at a

constant speed of 34 ft/sec, so we have another

10 · 34 = 340

feet in the second 10 seconds. If we add up all the left endpoints (t = 0, 10, 20, 30, 40, 50)

we get

D = 10(22) + 10(34) + 10(46) + 10(42) + 10(38) + 10(28) = 2100.

we could just as well add up all the velocities on the right endpoints, and get an approx-

imation that way:

D = 10(34) + 10(46) + 10(42) + 10(38) + 10(28) + 10(34) = 2220.

This should remind you of the sums we took in the area under the curve example. If we

shorten our intervals, and take readings every 5, 2 or 1 second, we would get a more accurate

304


distance. Simply, what we are doing is taking the area underneath the velocity curve with

rectangles equally spaced apart.

1. Using the Left Hand Endpoints

2. Using the Right Hand Endpoints

305


In general, suppose that we have an object moving with velocity v = f(t) over the interval

[a, b] and we need to assume, for now, that f(t) ≥ 0 for all t ∈ [a, b]. We take readings for

velocity at times a = t0, t1, t2, ..., tn = b. If we let all of the times be equally spaced, then

the time between any two consecutive readings is

∆t =b− an

.

Since the velocity during the first time interval is approximately f(t0), the distance

traveled is f(t0) ·∆t – this is true for any of the time intervals, and thus, the total distance

traveled is approximately

f(t0) ·∆t+ f(t1) ·∆t+ f(t2) ·∆t+ . . .+ f(tn−1) ·∆t.

Alternately, if we consider right endpoints, then the total distance traveled is

f(t1) ·∆t+ f(t2) ·∆t+ f(t3) ·∆t+ . . .+ f(tn) ·∆t.

And since the more often we measure velocity, the more accurate our distance becomes,

the distance is the limit of the expression above:

d = limn→∞

n∑i=1

f(ti−1)∆t = limn→∞

n∑i=1

f(ti)∆t,

the exact same setup as the area under a curve problem.

306

4.2 Definite Integral Brian E. Veitch

4.2 Definite Integral

When we compute the area under a curve, we obtained a limit of the form

limn→∞

n∑x=1

f(x∗i )∆x.

This same limit shows up when we consider finding the distance given the velocity, and

it turns our that this limit shows up in plenty of other situations. As it will appear for the

next, ohhh, forever, we give it a special definition:

Definition 4.2. If f is a function defined over the interval [a, b], we divide [a, b] into n subin-

tervals of equal width ∆x = b−an

. We let x0 = a, xa, x2, ..., xn−1, xn = b be the endpoints

of these subintervals and we let x∗1, x∗2, ..., x

∗n be any sample points in the subintervals, so

x∗i ∈ [xi−1, xi].

Then, the definite integral of f from a to b is

∫ b

a

f(x)dx = limn→∞

n∑i=1

f(x∗i )∆x,

∆x =b− an

xi = a+ i∆x for right-hand endpoints

provided that this limit exists and gives the same value for all possible choices of sample

points. If this limit does exist, we say that f is integrable over [a, b].

For notation,

∫ b

a

f(x)dx,

307


we say that∫

is the integral sign, a is the lower limit and b is the upper limit of integration.

The dx indicates that the independent variable is x – all other variables may be treated as

constants. Also, the integral∫ baf(x)dx is a number – it doesn not depend on x, and there is

nothing special about x.

For the function f , we have

∫ b

a

f(x)dx =

∫ b

a

f(t)dt =

∫ b

a

f(y)dy,

for any variable as a placeholder. Lastly, we say that the sum

n∑i=1

f(x∗i )∆x

is called a Riemann sum. If our function of interest, f(x), is always non-negative, then

we can treat the Riemann sum as a sum of areas of rectangles. Since the definite integral is

the limit of a Riemann sum, a definite integral is the limit of the sum of area of rectangles,

and thus the area under the curve f(x) from x = a to x = b.

If we consider a function f that does take on negative values, such as a function like the

one below,

308


we cannot just take the heights of all the approximating rectangles – some of these heights

would be negative, and that wouldn’t make a whole lot of sense to have negative area. The

way we can get around this is to consider the net area – we find the area of the curve above

the x-axis. Then, separately, find the area of the curve under the x-axis. We then add these

two areas together. So if A1 is the area above the x-axis but under f(x) computed by base

times height, ∆x · f(x∗i ) and A2 is the area under the x-axis but above f(x), also computed

by base times height, ∆x · f(x∗i ), we have

Area of shaded region = A1 − A2,

since A2 is negative.

Example 4.6. Evaluate

∫ 5

0

(3− x) dx and the area of the shaded region.

1. Let’s take a look at the graph

309


Do you see how some of the shaded region is above the x-axis and some of it is below.

When we evaluate

∫ 5

0

(3− x) dx, it would consider the area below negative. We can

do it two different ways at this point. Let’s do it without calculus.

2. Do you see how the two shaded regions are triangles? Guess what? We know the

formula for the area of a triangle, A =1

2bh.

Area of top triangle: A =1

2(3)(3) =

9

2

Area of bottom triangle: A =1

2(2)(2) = 2

So our integral is

∫ 5

0

(3− x) dx =9

2− 2 =

5

2= 2.5

3. Now if you want area of the shaded region (pretending like the area below the x-axis

is also positive), then what we really want to evaluate is

∫ 5

0

|3− x| dx

310


So the area of the shaded region is

∫ 5

0

|3− x| dx =9

2+ 2 =

13

2= 6.5

Theorem 4.1. If f is continuous on [a, b], or if f has only a finite number of jump discon-

tinuities, then f is integrable on [a, b] – meaning that the definite integral∫ baf(x)dx exists.

This theorem is NOT easy to prove, by any stretch of the word easy. We do not do it

here, but it does tell us something. It says that there are functions which are not integrable.

In order to simplify the calculations, we can choose specific sample points, and the right

most endpoints are as good as any other. So, we have a half definition, half theorem:

Defi-thereom: If f is integrable on [a, b], then∫ b

a

f(x)dx = limn→∞

n∑i=1

f(xi)∆x

where

∆x =b− an

311


and xi = a+ i∆x.

In order to evaluate these integrals, we have to be able to work with some very commonly

seen sums – the following three equations will be insanely valuable in doing this:

n∑i=1

1 =n

n∑i=1

i =n(n = 1)

2

n∑i=1

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

6

n∑i=1

i3 =

(n(n+ 1)

2

)2

The remaining rules that will help evaluate sums are very similar to rules that let us evaluate

limits:

n∑i=1

c =nc

n∑i=1

cai =cn∑i=1

ai

n∑i=1

(ai + bi) =n∑i=1

ai +n∑i=1

bi

n∑i=1

(ai − bi) =n∑i=1

ai −n∑i=1

bi

Example 4.7. Let’s find the Riemann sum for f(x) = 3−x by taking right-hand endpoints

over the interval [0,5]. By the way, we know the answer should be5

2. I’ll take you through

the procedure. Do this whenever you’re asked to find evaluate the Riemann Sum.

1. Find ∆x:

∆x =b− an

=5− 0

n=

5

n

312


2. Find xi:

xi = a+ i∆x = 0 + i · 5

n=

5i

n

3. Find f(xi):

f(xi) = 3− xi

= 3−(

5i

n

)

4. Find Ai:

Ai = f(xi) ·∆x

=

(3− 5i

n

)· 5

n

=15

n− 25i

n2

5. Find∑n

i=1Ai:

n∑i=1

Ai =n∑i=1

15

n− 25i

n2

=15

n

n∑i=1

1− 25

n2

n∑i=1

i

=15

n· n− 25

n2·(n(n+ 1)

2

)= 15− 25n(n+ 1)

2n2

6. Our final step is take the limit as n→∞.

True Area = limn→∞

15− 25n(n+ 1)

2n2= 15− 25

2= 2.5

which is exactly what we got from the before.

313


Example 4.8. Find the Riemann sum for f(x) = 2x3 − 4x by taking sample points to be

right endpoints, where a = 0, b = 2 and n = 4.

With n = 4, the interval width is

2− 0

4=

1

2,

and the right endpoints are x1 = .5, x2 = 1, x3 = 1.5 and x4 = 2. The Riemann sum is

R4 =4∑i=1

f(xi)∆x

=∆x (f(0.5) + f(1) + f(1.5) + f(2))

=1

2

((1

4− 2

)+ (2− 4) +

(27

4− 6

)+ (16− 8)

)=

5

2

Note that since this function does dip negative, this value is NOT the approximation for

the area under the curve. However, it does represent the difference in positive and negative

areas of the approximating rectangles of the curve. But, this is just an approximation. Now,

let’s evaluate ∫ 2

0

2x3 − 4xdx.

With n subintervals, we have

∆x =b− an

=2

n.

and we have x0 = 0, x1 = 2/n, x2 = 4/n, and in general, xi = 2i/n. We use right endpoints

314


and our theorem to let us evaluate∫ 2

0

2x3 − 4xdx = limn→∞

n∑i=1

f(xi)∆x

= limn→∞

n∑i=1

f

(2i

n

)· 2

n

= limn→∞

2

n·

n∑i=1

(2

(2i

n

)3

− 4

(2i

n

))

= limn→∞

2

n·

n∑i=1

(16i3

n3− 8i

n

)

= limn→∞

2

n·

(16

n3

n∑i=1

i3 − 8

n

n∑i=1

i

)

= limn→∞

2

n·

(16

n3·(n(n+ 1)

2

)2

− 8

n·(n(n+ 1)

2

))

= limn→∞

2

n·(

4(n+ 1)2

n− 4(n+ 1)

)=8 lim

n→∞

(n+ 1)2

n2− n+ 1

n

=8 · limn→∞

1− 2/n− 1/n2

1− 1 + 1/n

1

=8 · (1− 1)

=0

Clearly, this integral cannot be interpreted as the area under a curve, since this curve clearly

does not have 0 area. However, it can be interpreted as a difference between the positive

and negative area. If we graph the function, we should see that the positive and negative

areas are identical:

315


This is NOT how we will be computing these soon. But just like as with derivatives, we

have to go through it the long way first before we get to the shortcuts.

Example 4.9. Set up an expression for

∫ 7

3

x7 dx as a limit of sums. Do not evaluate.

We let f(x) = x7, a = 3 and b = 7. Thus,

∆x =7− 3

n=

4

n.

We have x1 = 3 + 1 · 4

n, x2 = 3 + 2 · 4

n, ..., and we have a generic term

xi = 3 +4i

n.

By our theorem, we have ∫ 7

3

x7dx = limn→∞

n∑i=1

f(xi)∆x

= limn→∞

n∑i=1

f

(3 +

4i

n

)· 4

n

= limn→∞

4

n

n∑i=1

(3 +

4i

n

)7

This is not very easy. Soon enough we’ll get to the shortcuts that allow us to evaluate

this quickly. By the way,

∫ 7

3

x7 dx = 719780

316



∫ 2

0

√4− x2dx

by interpreting the integral as an area.

Since f(x) =√

4− x2 ≥ 0 for x in [0, 2], we can interpret the integral as the area under

y =√

4− x2 from 0 to 2. We can rewrite this as y2 = 4−x2, which gives x2 + y2 = 4, which

shows that the graph of f is part of a circle with radius√

4 = 2. Since our part of the circle

is stuck in Quadrant I, we have a quarter circle. Thus, we take a quarter of the area:

A =1

4

(π · r2

)=

1

4

(π · 22

)= π.

Example 4.11. Evaluate ∫ 3

0

3x− 5 dx.

317


We are benefited by graphing this first:

In order to find the integral, note that we get positive values for the triangle above the

x-axis and negative values from the triangle below the x-axis. We find the areas of these two

triangles, and subtract. First, find the x-intercept of 3x − 5 = 0, which gives x = 53. The

upper triangle then has base equal to 3− 53

= 43, with a height of f(3) = 3(3)− 5 = 4. Thus,

the upper triangle has area

AT =1

2· 4

3· 4 =

8

3.

Similarly, the lower triangle, which has base 53

and height f(0) = −5 has area

AB =1

2· 5

3· | − 5| = 25

6.

Thus, the integral is

318


∫ 3

0

3x− 5dx = AT − AB =8

3− 25

6=−9

6=−3

2.

4.2.1 Midpoint Rule

These are all good integrals, but for some we still need to approximate, anything with a

curved side, as a matter of fact, which is not part of a circle. We so far have chosen just left

and right endpoints, and we saw in Section 4.1 how those gave us rather bad approximations.

Better, perhaps, to use the midpoint of the intervals!

Definition 4.3 (The Midpoint Rule:).∫ b

a

f(x)dx = limn→∞

n∑i=1

f(x̄i)∆x = ∆x (f(x̄1) + f(x̄2) + . . .+ f(x̄n)) ,

where

∆x =b− an

and

x̄i =1

2(xi−1 + xi),

which is the midpoint of the interval [xi−1, xi].

Example 4.12. Use the midpoint rule with n = 4 to approximate∫ 1

0

1

2xdx.

We start with ∆x = 1−04

= 14. The righthand endpoints of the 4 intervals are x1 =

1/4, x2 = 1/2, x3 = 3/4, x4 = 1. To find the midpoints, add consecutive pairs and divide by

2:

319


x̄1 =0 + 1/4

2=

1

8

x̄2 =1/4 + 1/2

2=

3

8

x̄3 =1/2 + 3/4

2=

5

8

x̄4 =3/4 + 1

2=

7

8

Thus, the midpoint rule gives

∫ 1

0

1

2xdx =∆x (f(1/8) + f(3/8) + f(5/8) + f(7/8))

=1

4

(1

2 · 1/8+

1

2 · 3/8+

1

2 · 5/8+

1

2 · 7/8

)=

1

4

(1

1/4+

1

3/4+

1

5/4+

1

7/4

)=

1

4

(4 +

4

3+

4

5+

4

7

)=

176

105

≈1.68

Now, since the function is always positive, we can think of this as the area under the

curve. However, we still have no idea how good of an approximation this is. In this case,

it’s horrible since the area is ∞.

Example 4.13. Let’s evaluate

∫ 2

0

x2 dx using the midpoint and 6 rectangles.

Recall we did this example using left and right hand endpoints. It took us 100 rectangles

to get an estimate of 2.747. We proved the exact area is8

3≈ 2.67.

320


1. ∆x =2− 0

6=

1

3

2. Find xi

x1 = 0 + 1 · 1

3=

1

3

x2 = 0 + 2 · 1

3=

2

3

x3 = 0 + 3 · 1

3=

3

3

x4 = 0 + 4 · 1

3=

4

3

x5 = 0 + 5 · 1

3=

5

3

x6 = 0 + 6 · 1

3=

6

3

so the midpoints are

x̄1 =1

6

x̄2 =3

6

x̄3 =5

6

x̄4 =7

6

x̄5 =9

6

x̄6 =11

6

321


3. The approximate area is

A ≈ 1

3[f(x1) + f(x2) + f(x3) + f(x4) + f(x5) + f(x6)]

A ≈ 1

3

[(1

6

)2

+

(3

6

)2

+

(5

6

)2

+

(7

6

)2

+

(9

6

)2

+

(11

6

)2]

A ≈ 2.648

So only using 6 rectangles we got much closer to the true area than we did with 100

rectangles! I believe you need somewhere around 190 righthand rectangles for the same ap-

proximation as 6 midpoint rectangles!

4.2.2 Properties of the Definite Integral

When we defined the integral ∫ b

a

f(x)dx,

we have so far defined this when a < b. However, if b < a, the Riemann Sum definition still

works, but ∆x changes from b−an

to a−bn

= −(b−a)n

. Thus,

∫ b

a

f(x)dx = −∫ a

b

f(x)dx.

Further, if a = b, then ∆x = b−bn

= 0, and as such∫ b

b

f(x) dx = 0.

322


Further, to help evaluate a definite integral, we have the following four properties, as

long as f and g are continuous functions:

1.

∫ b

a

c dx = c(b− a), for any constant c

2.

∫ b

a

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

∫ b

a

f(x) dx+

∫ b

a

g(x) dx

3.

∫ b

a

cf(x) dx = c

∫ b

a

f(x) dx for any constant c

4.

∫ b

a

(f(x)− g(x)) dx =

∫ b

a

f(x) dx−∫ b

a

g(x) dx

Proving these is a quite simple task. Property 1 follows from the fact that we are integrating

a constant height, and our figure would just be a rectangle – area of that is height c times

width b− a. Properties 2, 3 and 4 are all proven similarly, and we prove property 3 here:∫ b

a

cf(x)dx = limn→∞

n∑i=1

(cf(xi)) ∆x

= limn→∞

c ·n∑i=1

f(xi)∆x

=c · limn→∞

n∑i=1

f(xi)∆x

=c ·∫ b

a

f(x)dx

323


Thus, a constant, but ONLY a constant can be pulled out in front of an integral sign.

Example 4.14. Use the above properties to find∫ 7

2

3− 6x2dx.

Using the difference property, we have

∫ 7

2

3− 6x2 dx =

∫ 7

2

3 dx−∫ 7

2

6x2 dx

By property 1, we have ∫ 7

2

3 dx = 3(7− 2) = 15.

Also, ∫ 7

2

6x2 dx = 6

∫ 7

2

x2 dx,

but to evaluate∫ 7

2x2dx, we need to treat this as a Riemann Sum, with ∆x =

7− 2

n=

5

n

and xi = 2 +5i

n

324


∫ 7

2

x2dx = limn→∞

n∑i=1

(2 +

5i

n

)2

· 5

n

= limn→∞

5

n

n∑i=1

4 +20i

n+

25i2

n2

= limn→∞

5

n

[n∑i=1

4 +n∑i=1

20i

n+

n∑i=1

25i2

n2

]

= limn→∞

5

n

(4n+

20

n· n(n+ 1)

2+

25

n2· n(n+ 1)(2n+ 1)

6

)= lim

n→∞

5

n

(4n+ 10(n+ 1) +

25(n+ 1)(2n+ 1)

6n

)=5 · lim

n→∞4 + 10 · n+ 1

n+ 25 · 2n2 + 3n+ 1

6n2

=5 ·(

4 + 10 +25

3

)=

335

3

Thus, we have ∫ 7

2

3− 6x2dx = 15− 6 ·(

335

3

)= 15− 670 = −655.

A fifth property involves combining the bounds on two integrals over the same function:

∫ c

a

f(x)dx+

∫ b

c

f(x)dx =

∫ b

a

f(x)dx,

which can easily be seen by the picture below:

325


After all, if we want to find the area under f(x) from a to b, we can split it somewhere

in the middle and add those two areas.

Example 4.15. Suppose that∫ 10

2

f(x) dx = 13 and

∫ 7

2

f(x) dx = −3.

Find ∫ 10

7

f(x)ldx.

The answer here lies in a simple application from the following equation:

∫ 10

2

f(x) dx =

∫ 7

2

f(x) dx+

∫ 10

7

f(x) dx

13 = −3 +

∫ 10

7

f(x) dx

Therefore,

∫ 10

7

f(x) dx = 16

326


This property is true regardless if a < c < b or not. It doesn’t matter at all. Any other

configuration would simply be a rearrangement of the variables. However, the following three

properties are only true for a < b:

6) If f(x) ≥ 0 for a ≤ x ≤ b, then∫ baf(x) dx ≥ 0

7) If f(x) ≥ g(x) for a ≤ x ≤ b, then∫ baf(x) dx ≥

∫ bag(x) dx

8) If m ≤ f(x) ≤M for a ≤ x ≤ b, then m(b− a) ≤∫ baf(x) dx ≤M(b− a).

Property 6 implies a positive function gives positive areas. Property 7 says that a bigger

(higher) function will have larger area. Duh. Property 8 states that is a function is trapped

between two horizontal lines, and as such, the area under the curve is trapped between the

areas of two rectangles. None of these really need to be proven, but we we can demonstrate

pictorially:

Example 4.16. Estimate the value of ∫ 2

0

e−x2

dx.

Take a look at the graph,

327


From the graph, you can tell that it is a DECREASING function. You can also find this

out by the first derivative test. Unfortunately, we do not learn the derivative of exponential

functions until next semester.

You can see from the graph, our function has the smallest y value at x = 2, which is e−4.

It has its highest y-value at x = 0, which is e0 = 1. Therefore,

e−4 < e−x2

< e0 = 1.

Thus, by property 8, we have

e−4(2− 0) ≤∫ 2

0

e−x2

dx ≤ 1(2− 0)

which gives

2e−4 ≤∫ 2

0

e−x2

dx ≤ 2.

328

4.3 The Fundamental Theorem of Calculus Brian E. Veitch

4.3 The Fundamental Theorem of Calculus

As you may have noticed, differentiation and integration are quite closely related. We finally

get to that connection, the connection at the heart of calculus now – differentiation arose

from the tangent line problem discussed way back when. Integration arose from the area

problem previously discussed. The main theorem of this section will allow us to find areas

under the curve without having to compute the limits of sums, which will be nice.

We first consider functions of the form

g(x) =

∫ x

a

f(t)dt,

where f(t) is a continuous function on the interval [a, b] and we also let x vary between a

and b – as x is a bound on the integration, this makes sense. Note that g depends only on

x, not on t. If we fix x as a number, then∫ xaf(t)dt is a number as well.

If we suppose that f(x) ≥ 0, since a ≤ x, g(x) can be thought of as the area under the

curve f(x) over the interval [a, x], as the “area so far.”

Example 4.17. One of the things students will struggle with is how x as a bound works

and how it affects g(x). Let’s take a look at the following graph.

329


Let g(x) =

∫ x

0

f(x) dx. Find g(0), g(2), g(3), g(4), g(5), g(6), g(7).

1. g(0): This represents the area from 0 to 0 -

∫ 0

0

f(x) dx = 0. Therefore,

g(0) = 0

2. g(2): This represents the area from 0 to 2 -

∫ 2

0

f(x) dx

We’ll assume that f(x) = x2 on the interval [0, 2]. And we know from previous sections

that

∫ 2

0

x2 dx =8

3. Therefore,

330


g(2) =8

3

3. g(3): This represents the following region:

The region from 2 to 3 is a rectangle that has an area of 1 · 4 = 4. It’s a good idea to

think of g(x) as an accumulation of area (negative or positive) as x increases. So g(3)

equals all the area up to x = 3, which is

g(3) =8

3+ 4 =

20

3≈ 6.67


331


The region from 3 to 4 is a triangle with base = 1 and height = 4. Its area is1

2·1·4 = 2.

Therefore,

g(4) =8

3+ 4 + 2 =

26

3≈ 8.67

Notice that when we previously moved over a unit, we added an area of 4. This time

we moved over a unit and added an area of 2. Even though the area function g(x)

increased, it didn’t increase as much. In other words, (and this is very important),

g(x) is increasing but concave down.


The region from 4 to 5 is a quarter of a circle with radius 1. So the area is1

4π ≈ 0.785.

But since it’s below the x-axis, it’s negative. Therefore,

g(5) =8

3+ 4 + 2− 0.785 = 7.88

332


Notice that g(x) decreased. This should make sense since the area that we’re accumu-

lating now is negative.


The region from 5 to 6 is another quarter of a circle with radius 1. So the area is1

4π ≈ 0.785. But since it’s below the x-axis, it’s negative. Therefore,

g(6) =8

3+ 4 + 2− 0.785− 0.785 = 7.095


333


The region from 6 to 7 is a triangle with area1

2.Therefore,

g(7) =8

3+ 4 + 2− 0.785− 0.785 +

1

2= 7.595

If you were to track g(x) on a its own graph, it would look something like this

So where did g(x) reach its maximum? Does it have a local minimum? From the graph

it appears to have a maximum at x = 4. Again, if you think of g(x) as the accumulation of

area, g(x) begins accumulating negative area after x = 4. Therefore g(x) begins decreasing.

So what did we call a point where g(x) is increasing and then begins to decrease? Oh yeah...

a local maximum.

It also appears to have a local minimum at x = 6. What’s so special about x = 6? Notice

that this is when g(x) starts to accumulate positive area again. Therefore, g(x) changed from

decreasing to increasing and thus a local minimum.

It’s actually no coincidence that the area function g(x) reaches a local maximum / min-

imum when the original function f(x) intersects the x-axis. In fact, g(x) will have critical

values every time f(x) = 0. That’s odd. We know from previous chapters that g(x) has

334


critical values when g′(x) = 0. Could g′(x) = f(x)?

Before answering this EXTREMELY IMPORTANT question, let’s try one more ex-

ample.

Example 4.18. Suppose that f(x) = x. Then, find

g(x) =

∫ x

0

f(t) dt =

∫ x

0

t dt.

In order to find this, we still have to treat this as a limit of a Riemann Sum: we have

∆x =x− 0

n=x

nand xi = 0 + i

x

n=xi

n.

g(x) = limn→∞

n∑i=1

f(xi) ·∆x

= limn→∞

n∑i=1

ix

n· xn

= limn→∞

x2

n2

n∑i=1

i

=x2 · limn→∞

1

n2

(n(n+ 1)

2

)=x2 · lim

n→∞

n+ 1

2n

=x2

2

Thus,

g(x) =x2

2.

Note something important here:

d

dxg(x) =

d

dx

x2

2=

2x

2= x = f(x).

335


In other words, g′ = f , and if we define g to be the integral of f , then the derivative of g

is f , so we view the anti-derivative AS the integral, and the interchangeability of the words

make sense here. We now look to see why this may generally be true.

Begin by considering any continuous function f such that f(x) ≥ 0. Then, define

g(x) =

∫ x

a

f(t)dt,

which can be interpreted as the area under the graph of f from a to x. In order to compute

g′(x) from the definition of a derivative, we first consider for h > 0, the numerator of a

difference quotient, g(x+ h)− g(x), which is obtained by subtracting areas:

for h quite small, we have approximately a rectangle, which has height f(x) and width

h:

g(x+ h)− g(x) ≈ hf(x),

so we haveg(x+ h)− g(x)

h≈ f(x).

Thus, if we consider the limit of the above as h→ 0, we have

g′(x) = limh→0

g(x+ h)− g(x)

h= f(x).

And thus, we have our theorem:

336


4.3.1 The Fundamental Theorem of Calculus, Part 1:

If f is continuous on [a, b], then the function g defined by

g(x) =

∫ x

a

f(t)dt

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

Proof. If we let x and x+ h in the interval (a, b), then

g(x+ h)− g(x) =

∫ x+h

a

f(t)dt−∫ x

a

f(t)dt

=

(∫ x

a

f(t)dt+

∫ x+h

x

f(t)dt

)−∫ x

a

f(t)dt

=

∫ x+h

x

f(t)dt

and thus, for h 6= 0, we have

g(x+ h)− g(x)

h=

1

h

∫ x+h

x

f(t)dt.

In order to make life a little simpler, we consider only h > 0 now. Since f is continuous on

[x, x + h], we apply the Extreme Value Theorem and have that there are numbers u and v

in [x, x+ h] such that f(u) = m and f(v) = M , where m and M are the absolute minimum

and maximum values, respectively, of f on [x, x+ h]. Then, we have

mh ≤∫ x+h

x

f(t)dt ≤Mh,

which is to say

f(u)h ≤∫ x+h

x

f(t)dt ≤ f(v)h.

Now, since h > 0, we can divide by h without reversing the inequalities:

f(u) ≤ 1

h

∫ x+h

x

f(t)dt ≤ f(v).

We can replace the middle of this inequality by the interior of a difference quotient, from an

earlier equation:

f(u) ≤ g(x+ h)− g(x)

h≤ f(v).

337


(We can obtain a similar inequality if h < 0 as well.) From here, we take a limit as h → 0.

Since the interval [x, x + h] is squeezed down to a single point, we have u → x and v → x.

This gives

limh→0

f(u) = limu→x

f(u) = f(x),

and

limh→0

f(v) = limv→x

f(v) = f(x).

Thus, by the Squeeze Theorem, we have

g′ = limh→0

g(x+ h)− g(x)

h= f(x).

Lastly, if x = a or x = b, then we can view the above equation as a one-sided limit, as we

would let h→ 0±, to show that g is continuous, albeit not differentiable, on (a, b).

We can rewrite part 1 of the FTC using Leibniz notation as well, and we have

d

dx

∫ x

a

f(t)dt = f(x),

for a continuous f – meaning that the derivative of an integral (with variable bounds) is

the original function.

You just saw a formal proof of the FTC, I. Proofs can be very tedious and difficult to

follow. It’s ok if you don’t follow the formal one. The explanation that lead up to stating the

FTC, I is sufficient. You will see the formal proof again in an upper level advanced calculus

course. Have fun with that math majors.

Example 4.19. Findd

dx

∫ x

1

√t2 − t+ 1 cos(t) dt.

338


Since f(t) =√t2 − t+ 1 cos(t), which is continuous for t ≥ 1, Part 1 of the FTC gives

d

dx

∫ x

1

√t2 − t+ 1 cos(t) dt =

√x2 − x+ 1 cos(x).

Example 4.20. Findd

dx

∫ 4

x

e−t2

dt

Note this isn’t in the correct form. The bound x must be the upper bound. We use one

of our previous integral properties to rearrange the bounds.

d

dx

(−∫ x

4

e−t2

dt

)= − d

dx

(∫ x

4

e−t2

dt

)= −e−x2

Example 4.21. Consider the function, used often in actuarial work and statistics, given by

S(x) =

∫ x

0

sin

(πt2

2

)dt.

By Part 1 of the FTC, we can differentiate it, so that

S ′(x) = sin

(πx2

2

).

S(x) is a funny function to graph – I recommend you do it sometime, if you have access to

Maple or some other non-calculator software

339


Example 4.22. Find

d

dx

∫ x3

2

sin(2t)dt.

This is a bit more complicated, and requires more then just Part 1 of the FTC – it

requires the Chain rule as well. Begin by letting u = x3, and we know that dudx

= 3x2. Then,

d

dx

∫ x3

2

sin(2t) dt =d

dx

∫ u

2

sin(2t) dt

=d

du

(∫ u

2

sin(2t) dt

)du

dx

= sin(2u)du

dx

= sin(2x3) · 3x2

If you’re not a fan of this way, we can approach this a little differently. In order to do this

same problem, what was our biggest hurdle? It was the bound x3. So instead, let’s define

g(x) =

∫ x

2

sin(2t) dt

That way, we know g′(x) =d

dx

∫ x

2

sin(2t) dt = sin(2x).

So now consider g(x3). Note that g(x3) =

∫ x3

2

sin(2t) dt, which is what we’re trying to

work with.

What’sd

dx[g(x3)]?

d

dx

[g(x3)

]= g′(x3) · 3x2 from the chain rule

340


Since g′(x) = sin(2x), we know g′(x3) = sin(2(x3)). Therefore,

d

dxg(x3) =

d

dx

∫ x3

2

sin(2t) dt = sin(2x3) · 3x2

Either method works. Choose one that you’re comfortable with and go with it. Both are

a bit weird at first, so try to work with the one that feels more natural to you.

We now come to the second part, the better part, of the FTC, which lets us avoid calcu-

lating integrals as limits of Riemann Sums, which is a long and painful process.

4.3.2 Fundamental Theorem of Calculus 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 – where F is a function such that F ′ = f .

Proof. We start by defining

g(x) =

∫ x

a

f(t)dt,

and form Part 1 of the FTC, we know that g′(x) = f(x). Thus, g is an anti-derivative of

f . If F is any other anti-derivative of f , then we had a result that said that g and F differ

only by a constant, thus

F (x) = g(x) + c,

341


for some constant c and for x ∈ (a, b) (which is where g is differentiable). This holds for

any x on a < x < b. However, both F and g are continuous on [a, b], and we take limits of

both sides of the above equation (as x → a+ and x → b−), we also see that this is true at

x = a and x = b, so the equation holds for x ∈ [a, b].

If we now let x = a in the original formula for g(x), we get

g(a) =

∫ a

a

f(t)dt = 0

by one of our properties of integration from earlier (same bounds). Then, we can apply the

equation F (x) = g(x) + c with x = a and x = b and we get

F (b)− F (a) = (g(b) + c)− (g(a) + c)

=

∫ b

a

f(t)dt+ c− 0− c

=

∫ b

a

f(t)dt

Thus, by Part 2 of the FTC, if we know an anti-derivative of f(x), we can easily eval-

uate∫ baf(x)dx simply by evaluating that anti-derivative at the bounds of integration and

subtracting. It may seem a little strange that all that rectangle summing and limiting can

be taken care of simply by evaluating a different function and two points, but such is the

power of the integral.

To understand better why it works, suppose that v(t) is the velocity of an object at

time t, and s(t) is the position of the object at time t, then v(t) = s′(t). Thus, s is an

anti-derivative of v, and as such, we can evaluate a certain pesky integral rather easily:∫ b

a

v(t)dt = s(b)− s(a),

by Part 2 of the FTC.

342



3

x7 dx.

Remember this problem? We know that f(x) = x7 is continuous everywhere, so it is

definitely continuous on [2, 7]. So what’s an anti-derivative of x7. Why not use the easiest

one, F (x) =1

8x8. Thus.

∫ 7

3

x7 dx = F (7)− F (3) =1

8(7)8 − 1

8(3)8 = 719780.

Note that we could have used ANY anti-derivative, so we just went with the easiest. The

same answer would have been obtained with F (x) =1

8x8 + c for any c, as we would’ve had

∫ 7

3

x7dx = F (7)− F (3) =

(1

8(7)8 + c

)−(

1

8(3)8 + c

)=

1

8(7)8 − 1

8(3)8.

As a shortcut notation in the evaluation process for integration, we say that

F (x)|ba = F (b)− F (a).

We can write the equation in Part 2 of the FTC as

∫ b

a

f(x) dx = F (x)|ba ,

where F ′ = f .

Example 4.24. Find the area under y = x3 − 3 from x = 0 to x = 2.

343


An antiderivative of y = x3 − 3 we can use is

F (x) =1

4x4 − 3x.

We find the area using Part 2 of the FTC:

A =

∫ 2

0

x3 − 3x dx =1

4x4 − 3x

∣∣∣∣20

=

(1

424 − 3(2)

)−(

1

404 − 3(0)

)= −2

Example 4.25. Find the area under y = sin(x) from x = 0 to x = c, where 0 ≤ c ≤ π.

We know that the derivative of cos(x) = − sin(x), so to get out just sin(x), we let

F (x) = − cos(x). Then

d

dx− cos(x) = − (− sin(x)) = sin(x) = f(x).

Then ∫ c

0

sin(x)dx = − cos(x)|c0 = − cos(c) + cos(0) = 1− cos(c).

Thus, really, the integral of the sine function just gives back the cosine function at c,

with a bit of arithmetic involved.

Further, by letting c = π, we see that∫ π

0

sin(x)dx = 1− cos(π) = 2.

Example 4.26. Find

∫ 9

0

√t(1 + t) dt

344


First, we don’t have an anti-derivative for a product of two functions. And you can’t

just anti-differentiate each and multiply them together. That’s an easy way to earn 0 points.

Integration is a bit more strict than differentiation. At this point, we only know the power

rule,

∫xn dx =

xn+1

n+ 1+ c

So we need to multiply out√t(1 + t)

∫ 9

0

√t(1 + t) dt =

∫ 9

0

√t+ t3/2 dt

=

∫ 9

0

t1/2 + t3/2 dt

=t3/2

3/2+t5/2

5/2

∣∣∣∣90

=2

3t3/2 +

2

5t5/2∣∣∣∣90

=

(2

3(9)3/2 +

2

5(9)5/2

)−(

2

3(0)3/2 +

2

5(0)5/2

)= 115.2

Example 4.27. Find ∫ 3

−1

1

x4dx.

Well since f(x) = x−4, we have the anti-derivative

F (x) =x−3

−3=−1

3x3.

Then, FTC Part 2 gives ∫ 3

−1

1

x4dx =

−1

3 · 27− −1

−3=−28

81.

345


Wait a minute, that makes no sense! We have a function f(x) ≥ 0 for all values of x!

WTF? Well, remember the conditions of FTC Part 2 – we need the function to be continuous

over [a, b], and boy, oh boy do we have a discontinuity at x = 0, which is smack in the middle

of this interval. And it’s a hell of a discontinuity – it’s an infinite discontinuity, one we cannot

resolve in any way, shape or form. We cannot integrate the above function:∫ 3

−1

1

x4dx

does not exist. Too bad, so sad.

Just to be clear, the anti-derivative to f(x) =1

x4is − 1

3x3. But we just can’t evaluate it

using the current endpoints.

Since the FTC has two parts, we look at them together to end this section. Recall that

the FTC states that if f is continuous on [a, b],

1. If g(x) =

∫ x

a

f(t) dt, then g′(x) = f(x).

2.

∫ b

a

f(x) dx = F (b)− F (a), for any anti-derivative F (x) of f(x).

Part 1 can be written asd

dx

∫ x

a

f(t)dt = f(x),

which says that if we integrate and then differentiate f , we get back f . Also, since F ′(x) =

f(x) from Part 2, we write part 2 as∫ b

a

F ′(x)dx = F (b)− F (a).

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Here, if we take a function, differentiate it, then integrate, we get the original function

F back, but in the form of a difference between two numbers. Thus, by combining both

parts of the FTC, we see that differentiation and integration are inverse functions of each

other. And, armed with this little tidbit of knowledge, we will dive into integration tricks,

shortcuts and the realization that life isn’t as easy as it was with differentiation.

347

4.4 Indefinite Integrals Brian E. Veitch

4.4 Indefinite Integrals

Section 4.3 gave us our main tool for evaluating integrals, as long as we can find a function

whose derivative gives us the integrand. Here, we run through a collection of those functions

and state the previous Part 2 of the FTC so that we can apply it to more application-type

problems.

We have the connection between integration and differentiation – they are inverse oper-

ations – but we still have clunky notation. In order to fix this, we define the

Definition 4.4 (Indefinite integral).∫f(x)dx = F (x)

such that F ′(x) = f(x).

For example, sinced

dx

[x9]

= 9x8,

then ∫9x8 dx = x9 + C,

where C is some constant, as we discussed in earlier sections. Also, since

d

dx[sin(x)] = cos(x),

then ∫cos(x) dx = sin(x) + C.

Because of these +C’s, we don’t get out just one single function, but a whole family of

functions – one function for each possible value of C. Now, there is something to be careful

about. When we talked about definite integrals,∫ baf(x)dx, the result was always a number.

348


And it is, for definite integrals. For indefinite integrals, since there are no bounds of inte-

gration, the result is always a function of the integrating variable.

We now give a table of integration formulas, which really is the same as the table of

anti-differentiation formulas we saw in Section 4.9. We can verify any of the formulas by

differentiating the right side to obtain the function inside the integral on the left side.

Integral∫cf(x) dx = c

∫f(x)dx∫

f(x)± g(x) dx =∫f(x) dx±

∫g(x) dx∫

xn dx =xn+1

n+ 1+ c, n 6= −1∫

sin(x) dx = − cos(x) + c∫cos(x) dx = sin(x) + c∫

k dx = kx+ c∫sec2(x) dx = tan(x) + c∫

sec(x) tan(x) dx = sec(x) + c∫csc2(x) dx = − cot(x) + c∫

csc(x) cot(x) dx = − csc(c) + c

From these formulas, it should be clear that the most general form of an integral has

an arbitrary constant added to the end of it. This will need to be done for ALL indefinite

integrals you encounter, with no exceptions, period.

Example 4.28. Find the general, indefinite integral∫15x4 + 8x3 − cos(x)dx.

349


We can break this into a sum/difference of three integrals, and evaluate them each ac-

cording to the rules in the previous table:∫15x4 + 8x3 − cos(x) dx =15

∫x4 dx+ 8

∫x3 dx−

∫cos(x) dx

=15

5x5 +

8

4x4 − (sin(x)) + c

=3x5 + 2x4 − sin(x) + c

If we wish to check our answer, we can differentiate it. Remember, that since c is a constant,d

dx[c] = 0:

d

dx

[3x5 + 2x4 − sin(x) + c

]= 3 · 5x4 + 2 · 4x3 − cos(x) + 0 = 15x4 + 8x3 − cos(x).

Example 4.29. Evaluate ∫sin(x)

cos2(x)dx.

So far, there is no rule for integrating a quotient of two functions, so we cannot integrate

as is. Instead, use some trig identities to simplify and try to get one of the known forms in

the table: ∫sin(x)

cos2(x)dx =

∫sin(x)

cos(x)· 1

cos(x)dx

=

∫tan(x) sec(x)dx

= sec(x) + c

Example 4.30. We can also use our handy table for definite integrals: evaluate∫ 2π

π

3x4 + cos(x) dx

350


We have, from the formulas in our table,∫ 2π

π

3x4 + cos(x) dx =3

5x5 + sin(x)) |2ππ

=

(3

5(2π)5 + sin(2π)

)−(

3

5(π)5 + sin(π)

)=

3π5

5(32− 1)

=91π5

5

Example 4.31. Evaluate ∫2y2 − 3

√y + 7y10

y2dy.

First, the variable of integration is y – treat is no differently than any other, it’s just a

variable. Second, we have a quotient again, and we cannot integrate that. All we can do is

some algebra first, to get rid of that quotient:∫y − 3

√y + 7y10

y2dy =

∫2y2

y2dy − 3

∫y1/2

y2dy + 7

∫y10

y2dy

=

∫2dy − 3

∫y−3/2dy + 7

∫y8dy

=2y − 3

−1/2y−1/2 +

7

9y9 + c

=2y + 6y−1/2 +7

9y9 + c

4.4.1 Applications

We know that given a function f(x), f ′(x) represents the rate of change of y = f(x) with

respect to x and that f(b) − f(a) is the total change in y when x changes from a to b –

note that this is the net change, not the distance traveled. We could increase, decrease and

increase again and though we move a lot, the net change could be quite low. The FTC is

351


very much this net change result:

Definition 4.5 (The Net Change Theorem). The integral of a rate of change is the net

change in the function: ∫ b

a

f ′(x)dx = f(b)− f(a).

There are all kinds of examples of this; we give some common ones now, for future

problem use:

• If V (t) is the volume of water at time t, then V ′(t) is the change in volume. The net

change of our container of water is∫ b

a

V ′(t)dt = V (b)− V (a),

from starting time t = a to ending time t = b.

• If the mass of a rod from one end to a point in the middle is m(x), then we have the

linear density as ρ(x) = m′(x). Thus,∫ b

a

ρ′(x)dx = m(b)−m(a)

is the mass of the rod between x = a and x = b.

• If population grows at a rate of dPdt

, then∫ b

a

dP

dtdt = P (b)− P (a)

is the net change in population from time t = a to t = b.

352


• If an object moves along a straight path by the function s(t), we know its velocity is

v(t) = s′(t), so ∫ b

a

v(t) dt = s(b)− s(a)

is the net change in distance the object traveled. If, on the other hand, we want to find

the total amount of distance the object has traveled, we have to consider two different

problems – the interval when v(t) ≥ 0 and the interval when v(t) < 0. To do this,

compute the integral ∫ b

a

|v(t)|dt.

If we think about this in terms of the area under a curve,∫ b

a

v(t)dt

is the net area, when the area above the x-axis is subtracted by the area below the

x-axis. However, ∫ b

a

|v(t)|dt

is the total area between the function and the x-axis, which always must be positive.

• The acceleration of an object at time t is given by a(t) = v′(t), and thus the net change

in velocity from t = a to t = b is∫ b

a

a(t) dt = v(b)− v(a).

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Example 4.32. An object moves along a line with acceleration given by a(t) = 2t+1. Find

the displacement of the object from t = 1 to t = 4 and then find the total distance traveled

over the same time, if the velocity at t = 0 is -12.

Since we begin with acceleration, we need to integrate to get velocity as a function, then

take a definite integral to get the displacement.

v(t) =

∫a(t) dt

=

∫2t+ 1dt

=t2 + t+ c

v(0) = −12 =(0)2 + 0 + c

c = −12

v(t) =t2 + t− 12

s(4)− s(1) =

∫ 4

1

t2 + t− 12dt

=1

3t3 +

1

2t2 − 12t

∣∣∣∣41

=

(64

3+ 8− 48

)−(

1

3+

1

2− 12

)=−15

2

is the total displacement of the object.

In order to find the distance traveled, we need the intervals where the function is positive

and where it is negative, To find these, factor v(t) = t2 + t − 12 = (t + 4)(t − 3), but since

time can only be positive, we consider the intervals (1, 3) and (3, 4), as t = 1 is the lower

bound on the integral and t = 4 is the upper bound on the integral. Since v(t) ≤ 0 for t in

(1, 3) by the test-point method, the total distance traveled is

354


∫ 4

1

|v(t)|dt =

∫ 3

1

−v(t)dt+

∫ 4

3

v(t)dt

=

∫ 3

1

−t2 − t+ 12dt+

∫ 4

3

t2 + t− 12dt

=

[−1

3t3 − 1

2t2 + 12t

]31

+

[1

3t3 +

1

2t2 − 12t

]43

=

(−9− 9

2+ 36 +

1

3+

1

2− 12

)+

(64

3+ 8− 48− 9− 9

2+ 36

)=− 9− 9

2+ 36 +

1

3+

1

2− 12 +

64

3+ 8− 48− 9− 9

2+ 36

=91

6

Example 4.33. Suppose that the linear density of a rod of length 3π/2 meters is given by

ρ(x) = | sin(x)|.

Find the total mass of the rod.

We need to integrate the linear density from one end to the other, from x = 0 to x = 3π/2:

∫ 3π/2

0

| sin(x)| dx

But we need to know when sin(x) is positive and negative on the interval [0, 3π/2]. Recall

from trig that sin(x) > 0 on the interval (0, π) and sin(x) < 0 on the interval (π, 3π/2).

355


∫ 3π/2

0

| sin(x)|dx =

∫ π

0

sin(x)dx+

∫ 3π/2

π

− sin(x)dx

= [− cos(x)]π0 + [cos(x)]3π/2π

= (− cos(π) + cos(0)) + (cos(3π/2)− cos(π))

=(−(−1) + 1) + (0− (−1))

=3

356

4.5 Integration by Substitution Brian E. Veitch

4.5 Integration by Substitution

Since the fundamental theorem makes it clear that we need to be able to evaluate integrals

to do anything of decency in a calculus class, we encounter a bit of a problem when we have

an integral like ∫(2x+ 1) cos(x2 + x) dx.

We cannot compute this integral, since the integrand is a product, and we have no

integration rule which tells us how to deal with a product. This is the first section we have

which covers a special technique of integration, one that helps us to reconcile integrals such

as the above one. This first technique involves the introduction of a new variable – here we

let the new variable, u, equal the value in the cosine function, x2 + x. Then, the differential

of u is du = (2x+ 1) dx, so doing a substitution of u = x2 + x and dx =du

2x+ 1, we have

∫(2x+ 1) cos(x2 + x) dx =

∫(2x+ 1) cos(u)

du

2x+ 1

=

∫cos(u) du

= sin(u) + C = sin(x2 + x) + C.

But is this right? There is only one way to check, by differentiating:

d

dxsin(x2 + x) + C = cos(x2 + x) · d

dx

[x2 + x

]= cos(x2 + x)(2x+ 1)

So, yes this works. As a matter of fact, this will always work for every integral which has

the form

357


∫f(g(x))g′(x)dx.

After all, if F ′ = f , then ∫F ′(g(x))g′(x)dx = F (g(x)) + C,

because, by the Chain Rule, we have

d

dx(F (g(x))) = F ′(g(x)) · d

dxg(x) = F ′(g(x)) · g′(x).

So, what seems more clear now, is that we have to make a “change of variable,” a

substitution from the variable x to a variable u, where u = g(x). Then, we have

∫F ′(g(x))g′(x)dx = F (g(x)) + C = F (u) + C =

∫F ′(u)du,

by writing F ′ = f , we have∫f(g(x))g′(x)dx =

∫f(u)du.

This proves the following integration technique:

4.5.1 Integration by Substitution Rule

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.

358


Note that we had to use the Chain Rule to prove this – meaning that once we define u,

we need to use the chain rule to find du as well, which will have a dx in it. We do allow alge-

bra with these differentials in order to solve for dx, which will help in the substitution process.

Example 4.34. Find ∫5x3 sec2(x4 − 7)dx.

A common guideline for this u-substitution is to let u equal the inside of the most

complicated function. After all, functions don’t often get easier by differentiating! Since the

sec2() is more complicated and uglier than a cubic, we let u = x4 − 7. Then, du = 4x3dx.

We solve for dx:

dx =du

4x3

and we now make our substitutions to let us integrate:

∫5x3 sec2(x4 − 7)dx =

∫5x3 sec2(u) · du

4x3

=5

4

∫sec2(u)du

=5

4tan(u) + C

=5

4tan(x4 + 7

)+ C

Note that at the end of this problem, we MUST put the integral back in terms of its

original variable, x.

The main idea behind the u-substitution rule is the replace a complicated function with

something much easier to work with, which we do by replacing a function of x by a variable

u. The main difficulty here is to determine the function for which we are going to substitute.

If you can identify part of the integrand as the derivative of another part, that will be the

du piece and the other the u piece. However, if you cannot, a common theme is to let a

complicated part of the integrand be u. There is no exact method to find a u that works, its

more like finding how much powdered sugar goes in cake icing – we start simply by guessing.

359


Example 4.35. Evaluate ∫4x√

8x2 + 11dx.

Let u = 8x2 + 11, so that du = 16x dx, and we have dx =du

16x. Then by u-substitution

we have

∫4x√

8x2 + 11dx =

∫4x√udu

16x

=1

4

∫u1/2du

=1

4· 1

3/2u3/2 + C

=1

6

(8x2 + 11

)3/2+ C

If we tried the other function, and we let u = 4x, then du = 4dx, and there is nothing

we can do about the square root. Hence, we work with the more complicated function.

Example 4.36. Find ∫(x+ 3x2)dx

3√x2 + 2x3

.

We have two functions here, a numerator of x + 3x2 and a denominator of 3√x2 + 2x3.

Clearly, the denominator is more complicated, so we start there: let u = x2 + 2x3. Then

du = 2x+ 6x2, so

dx =du

2x+ 6x2,

360


and we have ∫(x+ 3x2)dx

3√x2 + 2x3

=

∫x+ 3x2

3√u

du

2x+ 6x2

=

∫(x+ 3x2) · u−1/3 · du

2(x+ 3x2)

=

∫1

2u−1/3du

=1

2

1

2/3u2/3 + C

=3

4u2/3 + C

Example 4.37. Find ∫ √3x+ 5dx.

There is really only one option here for u, the exponent: u = 3x + 5, so du = 3dx and

dx = du3

, and we have

∫ √3x+ 5dx =

∫ √udu

3=

1

3

∫ √u du =

1

3· 2

3u3/2 + C =

2

9(3x+ 5)3/2 + C.

If you get to the point where you can evaluate integrals via substitution without directly

writing down the substitution, that is fine. However, it can be dangerous to do so, in that

you can lose track of constants quite easily. Here, we will constantly, explicitly, give the

exact substitution and do the cancellation, just to make sure we all understand what’s going

on. Don’t let ego get in the way. Taking an extra minute or two to write down the explicit

substitution and do the cancellation is much better than wasting 10 minutes trying to find

where the constant error came from.

Example 4.38. Find ∫x5√

1− x3dx.

361


Now, this is interesting. Even if we let u = 1− x3, we won’t be able to clear out that x5

on the outside. But letting u = x4 won’t work either, as we won’t be able to clear out that

pesky root. We note that we can write x5 = x2 · x3, and we do see some of du there. Thus,

we let u = 1− x3 and du = −3x2dx. This gives

∫x5√

1− x3dx =

∫x3 · x2

√udu

−3x2

=−1

3

∫x3√udu

but we can write x3 = 1− u, so

=−1

3

∫(1− u) · u1/2du

=−1

3

∫u1/2 − u3/2du

=−1

3

(1

3/2u3/2 − 1

5/2u5/2

)+ C

=−1

3

(2

3u3/2 − 2

5u5/2

)+ C

=2

15u5/2 − 2

9u3/2 + C

=2

15(1− x3)5/2 − 2

9(1− x3)3/2 + C

Example 4.39. Find

∫sin(√x)√x

dx

Let u =√x and du =

1

2√xdx. This gives us dx = 2

√x du. Do not substitute both

√xs

as u. Only the one in the sine function becomes u.

362


∫sin(√x)√x

dx =

∫sin(u)√

x(2√xdu)

=

∫2 sin(u)

= − cos(u) + C

= − cos(√x) + C

Even though the substitution looked a bit tricky, it turned out ok.

Example 4.40. Find

∫csc2(x)

cot4(x)dx

We let u = cot(x) because we knowd

dxcot(x) = csc2(x) which is in the integral. You

just have to practice these in order to get better at identifying u. Then du = − csc2(x) dx,

which gives us dx =du

− csc2(x).

∫csc2(x)

cot4(x)dx =

∫csc2(x)

u4du

− csc2(x)

=

∫− 1

u4du

=

∫−u−4 du

= −u−3

−3+ C

=1

3(cot(x))−3 + C

4.5.2 Definite Integrals

We have two different methods for evaluating definite integrals. The first is quite similar to

what we have been doing with the indefinite integrals – we perform our substitution, inte-

363


grate and revert back to our original variable before we evaluate at the integration bounds.

For example,

∫ π

0

sin(3x− 1)dx =

∫ x=π

x=0

sin(u)du

3

=1

3

∫ x=π

x=0

sin(u)du

=−1

3cos(u) |x=πx=0

=−1

3cos(3x− 1) |π0

=−1

3(cos(3π − 1)− cos(1))

Which is all lovely, in that it works. However, there is a second way, which is occasionally

easier – we change the bounds on integration when we change the variable:

4.5.3 u-Substitution for Definite Integrals

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

a

f(g(x))g′(x)dx =

∫ g(b)

g(a)

f(u)du.

We give a proof, and then demonstrate on our previous function.

Proof. Let F be a function such that F ′ = f . Then, by the Chain Rule, F (g(x)) is an

anti-derivative of f(g(x))g′(x). We use Part 2 of the FTC to get∫ b

a

f(g(x))g′(x)dx = F (g(x)) |ba = F (g(b))− F (g(a)).

However, we can apply the FTC a second time, and we get∫ g(b)

g(a)

f(u)du = F (u) |g(b)g(a) = F (g(b))− F (g(a)).

Since they are identical, the result is proves.

364


Example 4.41. Using the second method for u-substitution, find∫ π

0

sin(3x− 1)dx.

We perform the same u-sub, with u = 3x− 1. Then, u(0) = −1 and u(π) = 3π− 1. This

gives ∫ π

0

sin(3x− 1)dx =

∫ u=3π−1

u=−1sin(u)

du

3

=−1

3cos(u)

∣∣∣∣3π−1−1

=−1

3(cos(3π − 1)− cos(−1))

Note that we did NOT have to return to the original variable after integrating, which makes

the problem a little easier. The method you use is entirely up to you – there will be no

assignment which forces you into one over the other. This second method is a little faster,

but also a little more apt to minor error.


0

7x

(1 + 7x2)2dx.

We let u = 1 + 7x2 so that du = 14xdx. We also have u(0) = 1 and u(3) = 1 + 63 = 64,

so we have

∫ 3

0

7x

(1 + 7x2)2dx =

∫ 64

1

7x

u2du

14x

=1

2

∫ 64

1

u−2du

=1

2·

(1

−1u−1∣∣∣∣641

)=

1

2

(−(64)−1 + (1)−1

)=

1

2

(1− 1

64

)=

63

128

365


Example 4.43. Find

∫ π/4

0

tan3(x) sec2(x) dx

Let u = tan(x). We do this becaused

dxtan(x) = sec2(x), which is in the integral. So

du = sec2(x) dx, which gives us dx =du

sec2(x).

We now change the bounds. If x = 0, then u(0) = tan(0) = 0. If x = π/4, then

u(π/4) = tan(π/4) = 1. Thus,

∫ π/4

0

tan3(x) sec2(x) dx =

∫ 1

0

u3 sec2(x)du

sec2(x)=

∫ 1

0

u3 du =u4

4

∣∣∣∣10

=1

4

We give one final theorem for this chapter, which helps simplify certain integrals if the

function satisfies a nice condition:

Theorem 4.2. Suppose that f is continuous on [−a, a].

1. If f is even (meaning that f(−x) = f(x)), then∫ a

−af(x)dx = 2

∫ a

0

f(x)dx.

2. If f is odd (meaning that f(−x) = −f(x)), then∫ a

−af(x)dx = 0.

Proof. We begin by splitting the integral into two pieces∫ a

−af(x)dx =

∫ 0

−af(x)dx+

∫ a

0

f(x)dx = −∫ −a0

f(x)dx+

∫ a

0

f(x)dx.

In the first integral, we make the substitution u = −x, so that du = −dx. The bounds

become u(0) = 0 and u(−a) = a.

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Thus, we have

∫ −a0

f(x)dx = −∫ a

0

f(−u)(−du) =

∫ a

0

f(−u)du.

Then, in total, we have

∫ a

−af(x)dx =

∫ a

0

f(−u)du+

∫ a

0

f(x)dx.

If f is even, then f(−u) = f(u), and as such

∫ a

−af(x)dx =

∫ a

0

f(−u)du+

∫ a

0

f(x)dx =

∫ a

0

f(u)du+

∫ a

0

f(u)du = 2

∫ a

0

f(u)du.

If f is odd, then f(−u) = −f(u), and as such

∫ a

−af(x)dx =

∫ a

0

f(−u)du+

∫ a

0

f(x)dx = −∫ a

0

f(u)du+

∫ a

0

f(u)du = 0.

While this won’t let us directly solve an integration too often, having a bound of 0 is

always super nice. But the real bonus comes from noticing that we have an odd function,

and if the bounds are opposites of each other, we can just say that the whole damn thing

equals 0 (as long as it’s continuous).


−3cos(x) + x6 − 10 dx.

Since f(x) = cos(x) + x6 − 10 and

f(−x) = cos(−x) + (−x)6 − 10 = f(x),

we have an even function. As such, we can rewrite

367


∫ 3

−3cos(x) + x6 − 10dx =2

∫ 3

0

cos(x) + x6 − 10dx

=2 ·[sin(x) +

1

7x7 − 10x

]30

=2 ·(

sin(3) +37

7− 30

)


∫ 5

−5

x4 tan(x)− 3x

x2 cos(x)− sin(x) tan(x)dx.

Omfg, EW! Well, the bounds are opposites, so let’s do a symmetry check:

f(−x) =(−x)4 tan(−x)− 3(−x)

(−x)2 cos(−x)− sin(−x) tan(−x)=

−x4 tan(x) + 3x

x2 cos(x)− sin(x) tan(x)= −f(x),

so we have an odd function! Thus∫ 5

−5

x4 tan(x)− 3x

x2 cos(x)− sin(x) tan(x)dx = 0,

by our theorem. Bam!

And with that, we are done! I hope you enjoyed Calculus I.

368

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