calculus 1 survival guide

8/3/2019 Calculus 1 Survival Guide

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Disclaimer

This e-book is presented solely for educational purposes. While best efforts have been used in

preparing this e-book, the author makes no representations or warranties of any kind and

assumes no liabilities of any kind with respect to the accuracy or completeness of the contents.

The author shall not be held liable or responsible to any person or entity with respect to any

loss or incidental or consequential damages caused, or alleged to have been caused, directly or

indirectly, by the information contained herein.

Every student and every course is different and the advice and strategies contained herein may

not be suitable for your situation. This e-book is intended for supplemental use only. You

should always seek help FIRST from your professor and other course material regarding any

questions you may have.


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Introduction

Authors Note

As a third grader, I learned my multiplication

tables faster than anyone in my class. I was

allowed to skip all of seventh-grade math and

go straight to eighth-grade (something I think

most people would pay a lot of money for,

considering how much math sucks for pretty

much everyone). As a junior, I finished all the

math courses my high school was offereing.

It wouldnt be conceited to say that math was

a subject that came easier to me than it did to

others. Compared to my classmates, I wasalways good at it. What can I say? I got my butt

kicked by every science class I ever took, but I

was always ahead of the curve when it came to

math. I guess its just the way my brain works.

And yet, despite the fact that its always been

easier for me, Ive struggled with all kinds of

math concepts soooooo many times, and often

remember feeling totally and completely lost

in math classes.

You know that feeling when youre reading

through an example in your textbook, hoping

with desperation that it will show you how to

do the problem youre stuck on? You hang in

there for the first few steps, and youre like,

Okay awesome! Im getting this! And then by

about the fourth step, you start to lose track of

their logic and you cant for the life of you

figure out how they got from Step 3 to Step 4?

Its the worst feeling. This is the point where

most people give up completely and just resign

themselves to failing the final exam.

Ive seen this same reaction in many of the

students I tutored in calculus while I was in

college. As hard as they tried to understand,

the professor and the textbook just didnt

make sense, and theyd end up feeling

overwhelmed and defeated before theyd ever

really gotten started.

I wasnt a math major in college, but I spent a

lot of time tutoring calculus students, and Ive

come to the conclusion that for most people,

the way we teach math is fundamentally

wrong.

First, theres a pretty good chance that you

wont ever actually use what you learn in

calculus. Algebra? Definitely. Basic geometry?

Probably. But calculus? Not so much. Second,

even if it is worthwhile to learn this stuff,

trying to teach us how to work through

problems with proofs that are supposed to

illustrate how the original formulas are

derived, just seems ridiculous.

In my experience, most students get the mostbenefit out of understanding the basic steps

involved in completing the problem, and

leaving it at that. Get in, get out, escape with

your life, and hopefully your G.P.A. still intact.

Sure, theres a lot to be said for going more in

depth with the material, and Id love to help

you do that if thats your goal. For most people

though, a basic understanding is s ufficient.

My greatest hope for this e-book and for

integralCALC.com is that theyll help you in

whatever capacity you need them. If youre

shooting for a C+, lets get you a C+. I dont

want to waste your time trying to give you

more than you need. That being said though,

most of the students I tutored who came in

shooting for a C+ came out with something

closer to a B+ or an A-. If you want an A,

attaining it is easier than you think.

No matter what your skill level, or the final

grade youre shooting for, I hope that this e-

book will help you get closer to it, and betteryet, save you some stress along the way.

Remember, if theres anything I can

ever do for you, please contact me

at integralCALC.com.


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Words of Wisdom

There are two pieces of advice Id like to give

you before we get started.

1. Stay Positive

More than anything, you have to stay positive.

Dont defeat yourself before you even get

started. Youre smarter than you think, andcalculus is easier than you think it is. Dont

panic.

Half of the people Ive tutored over the years

needed a personal calculus cheerleader more

than they needed a tutor. Theyd gingerly

proceed through a new problem Is this

right? Then if I am I still doing it right?

Theyd doubt themselves at every step. And I

would just stand behind them and say Yeah,

its right, youre doing great, youve got it,

youre right, until theyd solved the problem

without my help at all.

So many students let themselves get worked

up and freaked out the moment something

starts to get difficult. Its understandable, but

the more you can fight the fear that starts to

creep in, the better off youll be. So take a

deep breath. Its going to be okay.

2. Use Your Calculator (Or Dont)

Your calculator can be your greatest ally, but it

can also be your worst enemy. As calculators

have gotten more powerful, students have

come to rely on them more and more to solve

their problems on both homework and exams.

Instead of relying on my calculator to solve

problems outright, I like to use it as a double-

check system. If you never learn how to do the

problem without your calculator, you wont

know if what your calculator tells you is

correct. Nor will you be able to show any work

if youre required to do so on an exam, which

could cost you big points.

Learning the calculus itself means youll be

able to show your work when you need to, and

youll actually understand what youre doing.

Once you solve a problem, you should know

how to punch in the equation so that you canlook at the graph or solution to verify that the

answer you got is the same one your calculator

gives back to you.

What You Wont Find

Im not here to replace your textbook. Because

this is a quick-reference guide, you wont find

chapter introductions full of calculus historyyou dont care about.

Im also not here to replace your professor,

nor do I expect that youre particularly excited

about learning calculus. If you are excited

about calculus, thats awesome! So am I. But if

youre not, this is the place to be, because, at

least in this e-book, you wont find pointless

tangents where I geek out hard core and get

really excited about proofs, and you just get

bored and confused.

The purpose of this e-book is to serve as a

supplement to the rest of your course

material, not to completely replace your

professor or your textbook.

Even though Ive tried to cover the most

common introductory calculus topics in

enough detail that you could get by with just

this e-book, neither of us can predict whether

your professor will ask you to solve a problem

with a different method on a test, or a specific

problem not covered here. The last thing I

want is for you to think that this e-book is a

replacement for going to class, miss thatinformation, and then do poorly on the test

because you didnt get all the instructions.

What You Will Find

This e-book should give you the most crucial

pieces of information youll need for a real

understanding of how to solve most of the

problems youll encounter. I dont want to beyour textbook, which is why this e-book is only

about thirty pages long. I want this to be your

quick reference, the thing you reach for when

you need a clear understanding in only a few

minutes.

For a specific list of topics covered in this e-

book, please refer to the Table of Contents.


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(Clickable) Table of Contents

I. Foundations of Calculus

A. Functions

1. Vertical Line Test

2. Horizontal Line Test3. Domain and Range

4. Independent/Dependent Variables

5. Linear Functions

a. Slope-Intercept Form

b. Point-Slope Form

6. Quadratic Functions

a. The Quadratic Formula

b. Completing the Square

7. Rational Functions

a. Long Division

B. Limits

1. What is a Limit?

2. When Does a Limit Exist?

a. General vs. One-Sided Limits

b. Where Limits Dont Exist

3. Solving Limits Mathematically

a. Just Plug It In

b. Factor It

c. Conjugate Method

4. Trigonometric Limits

5. Infinite Limits

C. Continuity

1. Common Discontinuities

a. Jump Discontinuity

b. Point Discontinuityc. Infinite/Essential Discontinuity

2. Removable Discontinuity

3. The Intermediate Value Theorem

II. The Derivative

A. The Difference Quotient

1. Secant and Tangent Lines

2. Creating the Derivative

3. Using the Difference Quotient

B. When Derivatives Dont Exist

1. Discontinuities

2. Sharp Points

3. Vertical Tangent Lines

C. On to the Shortcuts!

1. The Derivative of a Constant

2. The Power Rule

3. The Product Rule

4. The Quotient Rule

5. The Reciprocal Rule

6. The Chain Rule

D. Common Operations

1. Equation of the Tangent Line

2. Implicit Differentiation

a. Equation of the Tangent Line

b. Related Rates

E. Common Applications1. Speed/Velocity/Acceleration

2. LHopitals Rule

3. Mean Value Theorem

4. Rolles Theorem

III. Graph Sketching

A. Critical Points

B. Increasing/Decreasing

C. Inflection Points

D. Concavity

E.- and -Intercepts

F. Local and Global Extrema

1. First Derivative Test

2. Second Derivative Test

G. Asymptotes

1. Vertical Asymptotes

2. Horizontal Asymptotes

3. Slant Asymptotes

H. Putting It All Together

IV. Optimization

V. Essential Formulas


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Foundations of Calculus

Functions

Vertical Line TestMost of the equations youll encounter in

calculus are functions. Since not all equations

are functions, its important to understand

that only functions can pass the Vertical Line

Test. In other words, in order for a graph to be

a function, no completely vertical line can

cross its graph more than once.

This graph does not pass the Vertical Line Test

because a vertical line would intersect it more

than once.

Passing the Vertical Line Test also implies that

the graph has only one output value for forany input value of . You know that anequation is not a function if can be twodifferent values at a single value.

You know that the circle below is not a

function because any vertical line you draw

between and will cross thegraph twice, which causes the graph to fail the

Vertical Line Test.

You can also test this algebraically by plugging

in a point between and for , such as .

At , can be both and . Since afunction can only have one unique output

value for for any input value of , the graphfails the Vertical Line Test and is therefore not

a function. Weve now proven with both thegraph and with algebra that this circle is not a

function.

Horizontal Line Test

The Horizontal Line Test is used much less

frequently than the vertical line test, despite

the fact that theyre very similar. Youll recall

that any function passing the Vertical Line Test

can only have one unique output of for anysingle input of.

This graph passes the Horizontal L ine Test

because a horizontal line cannot intersect it

more than once.

Contrast that with the Horizontal Line Test,

which says that no value corresponds to two

different values. If a function passes the

Example

Determine algebraically whether or not

is a function.

Plug in for and simplify.


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Horizontal Line Test, then no horizontal line

will cross the graph more than once, and the

graph is said to be one-to-one.

This graph does not pass the Horizontal LineTest because any horizontal line between and would intersect it more

than once.

Domain and RangeThink of the domain of a function as

everything you can plug in for withoutcausing your function to be undefined. Things

to look out for are values that would cause afractions denominator to equal and valuesthat would force a negative number under a

square root sign.

The range of a function is then any value that

could result for from plugging in everynumber in the domain for .

Independent and Dependent Variables

Your independent variable is , and yourdependent variable is . You always plug in avalue for first, and your function returns toyou a value for based on the value you gaveit for . Remember, if your equation is afunction, there is only one possible output offor any input of.

Linear FunctionsYoull need to know the formula for the

equation of a line like the back of your hand

(actually, better than the back of your hand,

because who really knows what the back of

their hand looks like anyway?). You have two

options about how to write the equation of a

line. Both of them require that you know at

least two of the following pieces of

information about the line:

1. A point2. Another point3. The slope, 4. The y-intercept,

If you know any two of these things, you can

plug them into either formula to find the

equation of the line.

Slope-Intercept Form

The equation of a line can be written in slope-

intercept form as

,where is the slope of the function and isthe -intercept, or the point at which thegraph crosses the -axis and where . Theslope, represented by , is calculated usingtwo points on the line, and ,and the equation you use to calculate is

To find the slope, subtract the -coordinate inthe first point from the -coordinate in thesecond point in the numerator, then subtractthe -coordinate in the first point from the -coordinate in the second point in the

denominator.

Example

Describe the domain and range of the

function

In this function, cannot be equal to ,because that value causes thedenominator of the fraction to equal .Because setting equal to is the onlyway to make the function undefined,

the domain of the function is all .


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Point-Slope Form

The equation of a line can also be written in

point-slope form as

In this form, is one point on the line, and is the other. Just as with slope-intercept form, is still the slope of thefunction. To use this form, find the sameway you did in slope-intercept form, then

simply plug in your two points to the point-

slope formula.

Quadratic Functions

Quadratic Functions are functions of the

specific form

As long as you have an term and an termand a constant, the coefficients , and canbe any number.

The Quadratic Formula

The Quadratic Formula can be used to factor

and solve for the roots of a quadratic function.

To use it, plug , and into the QuadraticFormula, here:

If any terms in your quadratic function are

negative, make sure to keep the negative sign

when plugging into the formula. For example,

If is negative in your quadratic function,youll end up with for the first term inthe numerator of the Quadratic Formula,

which would make that term positive.

You should also remember that in order for

this formula to work, must begreater than or equal to , because you canttake the square root of a negative number. If

you do end up with a negative number inside

the radical, then there are no real solutions to

your quadratic function.

Example

Find the equation of the line in point-

slope form that passes through the

points and .We start by finding the slope.

Now plug in the slope and either one of

the points into the formula.

Even though we could, simplifying any

further would take this out of point

slope form, so we leave it as is.

Example

Find the equation of the line in slope-

intercept form that passes through the

points and .We start by finding the slope.

Now we can plug in our slope and

either one of our points to the formula

and solve for .

Multiply every term by to cancel outthe denominator of the fraction.

Subtract from both sides.

Divide by

to solve for

.

For the final answer, plug and backinto the formula, leaving and asvariables.


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Since you always get a fraction with a plus or

minus sign in the numerator, theQuadratic Formula produces two solutions,

which you then use to factor your polynomial.

If your solutions are and , your factors willalways be and .Completing the Square

This method is another option you can use to

find the solutions of a quadratic function, if

you cant easily factor it. Since its very much a

step-by-step process, the easiest way to

explain this method is to use an example, so

lets do it.

Lets say we have the function

The first thing we want to do is set the

function equal to .

Next, well take one half of the coefficient on

the term and square it.

We then add andsubtract our result back into

the function, so that we dont change the

value of the function.

Now we factor the quantity in parentheses and

consolidate the constants.

Now add

to both sides to move the constant

to the right side.

Take the square root of both sides to eliminate

the exponent on the left. Dont forget to add

the positive/negative sign in front of the

square root on the right side.

Finally, add to both sides of the equation tosolve it for .

This is the same process youll follow each

time you use this method to solve for the rootsof a quadratic function.

Rational Functions

A rational function is a quotient of two

polynomials (a fraction with polynomials in

both the numerator and denominator). While

polynomials themselves are defined for all

values of

, rational functions are undefined

where the denominator of the function is

equal to .Long Division

Believe it or not, long division is a skill youll

use semi-frequently in calculus. Its just like the

long division you learned in fifth grade, except

that instead of just numbers, this time youll

be dividing polynomials.

Example

Use long division to convert

First, we should keep the following in

mind:

Example

Factor Plug for , for and for intothe Quadratic Formula.

Simplify to find your solutions.

Use the solutions to factor your

function.


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Limits

What is a Limit?The limit of a function is the value the function

approaches at a given value of, regardless ofwhether the function actually reaches thatvalue.

For an easy example, consider the function

When , . Therefore, is thelimit of the function at , because is thevalue that the function approaches as the

value of

gets closer and closer to

.

I know its strange to talk about the value that

a function approaches. Think about it this

way: If you set in the functionabove, then . Similarly, if youset , then .You can begin to see that as you get closer to

, whether youre approaching it from the side or the side, the value of gets closer and closer to .

.0000 .0000

In this simple example, the limit of the

function is clearly because that is the actualvalue of the function at that point; the point isdefined. However, finding limits gets a little

trickier when we start dealing with points of

the graph that are undefined.

In the next section, well talk about when

limits do and do not exist, and some more

creative methods for finding the limit.

When Does a Limit Exist?

General vs. One-Sided Limits

When you hear your professor talking about

limits, he or she is usually talking about the

general limit. Unless a right- or left-hand limit

is specifically specified, youre dealing with a

general limit.

The general limit exists at the point if1. The left-hand limit exists at ,2. The right-hand limit exists at ,

and

3. The left- and right-hand limits areequal.

These are the three conditions that must bemet in order for the general limit to exist. The

general limit will look something like this:

You would read this general limit formula as

The limit of of as approaches equals

.

Left- and right-hand limits may exist even

when the general limit does not. If the graph

approaches two separate values at the point

as you approach from the left- andright-hand side of the graph, then separate

left- and right-hand limits may exist.

Left-hand limits are written as

Divisor Numerator Dividend Denominator

________ | -( )

-( To start our long division problem, we

determine what we have to multiply by

(in the divisor) to get (in thedividend). Since the answer is , weput that on top of our long division

problem, and multiply it by the divisor, to get , which we thensubtract from the dividend. We bring

down from the dividend and repeatthe same steps until we have nothing

left to carry down from the dividend.

Our original problem reduces to:


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The negative sign after the indicates thatwere talking about the limit as we approach from the negative, or left-hand side of the

graph.

Right-hand limits are written as

The positive sign after the 2 indicates that

were talking about the limit as we approach 2

from the positive, or right-hand side of the

graph.

In the graph below, the general limit exists at

because the left- and right- hand limitsboth approach . On the other hand, thegeneral limit does not exist at becausethe left-hand and right-hand limits are not

equal, due to a break in the graph.

Left- and right-hand limits are equal at , but not at .

Where Limits Dont Exist

We already know that a general limit does not

exist where the left- and right-hand limits are

not equal. Limits also do not exist whenever

we encounter a vertical asymptote.

There is no limit at a vertical asymptote

because the graph of a function must

approach one fixed numerical value at the

point for the limit to exist at . Thegraph at a vertical asymptote is increasing

and/or decreasing without bound, which

means that it is approaching infinity instead of

a fixed numerical value.

In the graph below, separate right- and left-

hand limits exist at but the general limitdoes not exist at that point. The left-hand limitis , because that is the value that the graphapproaches as you trace the graph from left to

right. On the other hand, the right-hand limit is

, since that is the value that the graphapproaches as you trace the graph from right

to left.

The general limit does not exist at or at .

Where there is a vertical asymptote at ,the left-hand limit is , and the right-handlimit is . However, the general limit doesnot exist at the vertical asymptote because the

left- and right-hand limits are unequal.

Solving Limits Mathematically

Just Plug It In

Sometimes you can find the limit just by

plugging in the number that your function is

approaching. You could have done this with

our original limit example, . If youjust plug into this function, you get , whichis the limit of the function. Below is another

example, where you can simply plug in tothe function to solve for the limit.

Factor It

When you cant just plug in the value youre

evaluating, your next approach should befactoring.

Example

Plug in for and simplify.


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Conjugate Method

This method can only be used when either the

numerator or denominator contains exactly

two terms. Needless to say, its usefulness is

limited. Heres an example of a great, and

common candidate for the Conjugate Method.

In this example, the substitution method

would result in a in the denominator. Wealso cant factor and cancel anything out of the

fraction. Luckily, we have the Conjugate

Method. Notice that the numerator has exactly

two terms, and .Conjugate Method to the rescue! In order to

use it, we have to multiply by the conjugate of

whichever part of the fraction contains the

two terms. In this case, thats the numerator.

The conjugate of two terms is those same two

terms with the opposite sign in between them.

Notice that we multiply both the numerator

and denominator by the conjugate, because

thats like multiplying by, which is useful tous but still doesnt change the value of the

original function.

Remember, if none of these methods work,

you can always go back to the method we

were using originally, which is to plug in a

number very close to the value youre

evaluating and solve for the limit that way.

Trigonometric LimitsTrigonometric limit problems revolve around

three formulas:

Simplify and cancel the .

Since were evaluating at , plug that infor

and solve.

Example

Multiply the numerator and

denominator by the conjugate.

Example

Just plugging in would give us a nasty result. Therefore, well tryfactoring instead.

Cancelling from the top andbottom of the fraction leaves us with

something that is much easier to

evaluate:

Now the problem is s imple enough that

we can just plug in the value were

approaching.


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When solving trigonometric limit problems,

our goal is to reduce our problem to a simple

combination involving nothing but these

formulas and simple constants. Heres an

example.

Infinite Limits

Infinite limits exist when we can plug in a

number for that causes the denominator of arational function in lowest terms to equal .Here is an example of a rational function in

lowest terms, which means that we cannot

factor and cancel anything in the fraction.

We can see that setting

gives

in the

denominator, which means that we have a

vertical asymptote at , and therefore aninfinite limit at that point.

Now that weve established that this is a

rational function in lowest terms and that a

vertical asymptote exists, all thats left to

determine is whether the limit at approaches positive or negative infinity.

In order to do that, simply plug in a number

very close to 1. If our result is positive, the

limit will be . If the result is negative, thelimit is.

We can see that the result will be very large

and positive, so we know that the limit of this

function at is.

ContinuityI would give you the definition of continuity,

but I think its confusing. Plus, you should have

some intuition about what it means for a graph

to be continuous. Basically, a function is

continuous if there are no holes, breaks,

jumps, fractures, broken bones, etc. in its

graph.

You can also think about it this way: A function

is continuous if you can draw the entire thing

without picking up your pencil. Lets take some

time to classify the most common types of

discontinuity, or what makes a function not

continuous.

As it turns out, we can now easily

evaluate our entire problem with the

three fundamental trigonometric limit

formulas, without making the

denominator .

Example

Since we have exactly two terms in the

numerator, were actually going toborrow the Conjugate Method for the

first step of this problem.

Applying the identity to the numerator gives

Notice now that we can factor out

, which is one of our threefundamental formulas.


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Common Discontinuities

Jump Discontinuity

Youll usually encounter jump discontinuities

with piecewise-defined functions. A piece-

wahoozle whatsit? you ask? Exactly. A

piecewise-defined function is a function for

which different parts of the domain are

defined by different functions. One example

thats often used to illustrate piecewise-

defined functions is the cost of postage at the

post office. Heres how the cost of postage

might be defined as a function, as well as the

graph of this function. They tell us that the

cost per ounce of any package lighter than pound is

cents per ounce, that the cost of

every ounce from pound to anything lessthan pounds is cents per ounce, etc.

A piecewise-defined function

Every break in this graph is a point of jump

discontinuity. You can remember this by

imagining yourself walking along on top of the

first segment of the graph. In order to

continue, youd have to jump up to the second

segment.

Point Discontinuity

Point discontinuity exists when there is a hole

in the graph at one point. You usually find this

kind of discontinuity when your graph is a

fraction like this:

In this case, the point discontinuity exists at , where the denominator would equal

. This function is defined and continuous

everywhere, except at . The graph of apoint discontinuity is easy to pick out becauseit looks totally normal everywhere, except for

a hole at a single point.

Infinite/Essential Discontinuity

Youll see this kind of discontinuity called both

infinite discontinuity and essential

discontinuity. In either case, it means that the

function is discontinuous at a vertical

asymptote. Vertical asymptotes are only points

of discontinuity when the graph exists on both

sides of the asymptote.

The first graph below shows a vertical

asymptote that makes the graph

discontinuous, because the function exists on

both sides of the vertical asymptote. Thevertical asymptote in the second graph below

is not a point of discontinuity, because it

doesnt break up any part of the graph.

A vertical asymptote at that makes thegraph discontinuous

A vertical asymptote at

that does not

make the graph discontinuous

Removable DiscontinuityDiscontinuity is removable if you can easily

plug in the holes in its graph by redefining the

function. When you cant easily plug in the

holes because the gaps are bigger than a single

point, youre dealing with nonremovable

discontinuity. Point discontinuity is removable,

because you can easily patch the hole.


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Lets take the function from the Point

Discontinuity section:

If we add another piece to this function as

follows, we plug the hole and the function

becomes continuous:

Jump and infinite discontinuities are always

nonremovable, because the gaps are large.

The Intermediate Value TheoremSimilarly to the definition of continuity, the

definition of the Intermediate Value Theorem

is absolutely more harmful than helpful. So

instead, consider the following graph:

The Intermediate Value Theorem

This theorem is fairly ridiculous because it

doesnt tell us anything that we dont already

know. All it says is that, when we look at a

continuous function on a closed interval

between (blue) and (purple), there will be a point in between

them, which well call (orange). must be between and and must bebetween and. Looking at the graph,isnt that obvious? Values may or may not exist

below and above depending on thegraph, but mustexist.

The Derivative

The derivative of a function is written as, and read as prime of . Bydefinition, the derivative is the slope of the

original function. Lets find out why.

The Difference Quotient

I should warn you that this is one of those

dumb things you have to learn to do before

you learn how to do it the real way. If you can

believe this, your professor will actually have

the nerve to require you on a test to find the

derivative using this method, even though you

could just use the shortcuts that were going tolearn later. Unbelievable, I know.

But since our goal is just to get you a good

grade, and not to make a big scene, well learn

how to find derivatives the long way first, then

well learn the shortcuts and things will end up

better in the end. I promise. For now, the longway

Secant and Tangent LinesA tangent line is a line that juuussst barely

touches the edge of the graph, intersecting it

at only one specific point. Tangent lines look

very graceful and tidy, like this:

A tangent line

A secant line, on the other hand, is a line that

runs right through the middle of a graph,


16/33

sometimes hitting it at multiple points, and

looks generally meaner, like this:

A secant line

Its important to realize here that the slope of

the secant line is the average rate of change

over the interval between the points where

the secant line intersects the graph. The slope

of the tangent line instead indicates an

instantaneous rate of change, or slope, at the

single point where it intersects the graph.

Creating the DerivativeIf we start with a point, on a graph,and move a certain distance, , to the right ofthat point, we can call the new point on the

graph

.

Connecting those points together gives us a

secant line, and we can use the slope equation

to determine that the slope of the secant line

is

which, when we simplify, gives us

I bet your heart just skipped a beat out of pure

excitement. No? Strange

The point of all this nonsense is that, if I take

my second point and start moving it slowly

left, closer to the original point, the slope of

the secant line becomes closer to the slope of

the tangent line at the original point.

As the secant line moves closer and closer tothe tangent line, the points where the line

intersects the graph get closer together, which

eventually reduces to .Running through this exercise allows us to

realize that if I reduce to and the distancebetween the two secant points becomes

nothing, that the slope of the secant line is

now exactly the same as the slope of the

tangent line. In fact, weve just changed the

secant line into the tangent line entirely.

That is how we create the formula above,

which is the very definition of the derivative,

which is why the definition of the derivative is

the slope of the function at a single point.

Using the Difference QuotientTo find the derivative of a function using the

difference quotient, follow these steps:

1. Plug in for every in youroriginal function.

2. Plug your answer from Step in for in the difference quotient.3. Plug your original function in for

in the difference quotient.

4. Put in the denominator.5. Expand all terms and collect like terms.6. Factor out in the numerator, then

cancel i t from the numerator and

denominator.

7. Plug in the number your function isapproaching and simplify.

Example

Find the derivative of

at


17/33

When Derivatives Dont Exist

Before we jump into finding derivatives with

the shortcuts, lets talk about instances when

the derivative doesnt exist. When thederivative doesnt exist at a point in the graph,

we say that the original function is not

differentiable there.

Discontinuities

A derivative cannot exist at a point of

discontinuity in a function. This does not mean

that the function is not differentiable at other

points in its domain, only that the function is

not differentiable at the specific point of

discontinuity.

Sharp Points

If a graph contains a sharp point (A.K.A. a

cusp), the function is not differentiable at that

point. Youre most likely to find sharp points in

your function if it contains absolute values or if

its a piecewise-defined function.

A cusp in the graph of

Vertical Tangent LinesSince the slope of a vertical line is undefined,

and a tangent line represents the slope of the

graph, a tangent line by definition cannot be

vertical, so the derivative cannot be a perfectly

vertical line.

On to the Shortcuts!

Finally, weve gotten to the point where things

start to get easier. Weve moved past the

Difference Quotient, which was cumbersome

and tedious and generally not fun. Youre

about to learn several new derivative tricks

that will make this whole process a whole lot

easier. Arent you excited?!

The Derivative of a Constant

The derivative of a constant (a term with no

variable attached to it) is always . Rememberthat the graph of any constant is a perfectly

horizontal line. Remember also that the slope

of any horizontal line is . Because the

derivative of a function is the slope of thatfunction, and the slope of a horizontal l ine is ,the derivative of any constant must be .

The Power RuleThe Power Rule is the tool youll use most

frequently when finding derivatives. The rule

says that for any term of the form , thederivative of the term is

To use the Power Rule, multiply the variables

exponent, , by its coefficient, , then subtract from the exponent. If there is no coefficient(the coefficient is ), then the exponent willbecome the new coefficient.

After replacing with in,plug it your answer for . Thenplug in as-is for. Put in thedenominator.

Expand all terms.

Collect similar terms together then factor

out of the numerator and cancel i t fromthe fraction.

For , plug in the number youreapproaching, in this case . Then simplifyand solve.


18/33

The Product Rule

If a function contains two variable expressions

that are multiplied together, you cannotsimply take their derivatives separately and

then multiply the derivatives together. You

have to use the Product Rule. Here is the

formula:

If a function

then

To use the Product Rule, multiply the first

function by the derivative of the second

function, then add the derivative of the first

function times the second function to your

result.

The Quotient Rule

Just as you must always use the Product Rule

when two variable expressions are multiplied,

you must use the Quotient Rule whenever two

variable expressions are divided. Here is the

formula:

If a function

then

The Reciprocal Rule

The Reciprocal Rule is very similar to the

Quotient Rule, except that it can only be used

with quotients in which the numerator is

exactly . It says that if

Example


Based on the Quotient Rule formula,

we know that is the numeratorand therefore and that is the denominator and therefore that

.

is , and is . Plugging

all of these components into theQuotient Rule gives

Simplifying the result gives us our final

answer:

Example


The two functions in this problem are

and . It doesnt matter which

one you choose for and.Lets assign to and to .The derivative of is .The derivative of is .

According to the Product Rule,


answer:

Example


Applying Power Rule gives the

following:

Simplify to solve for the derivative.


19/33

then

Given

as your numerator and anything at all

as your denominator, the derivative will be the

negative derivative of the denominator divided

by the square of the denominator.

The Chain Rule

The Chain Rule is often one of the hardestconcepts for calculus students to understand.

Its also one of the most important, and its

used all the time, so make sure you dont leave

this section without a solid understanding. If

you go through the example and youre still

having trouble, please e-mail me for help at

[email protected].

You should use Chain Rule anytime your

function contains something more

complicated than a single variable. The Chain

Rule says that if your function takes the form

then

The Chain Rule tells us how to take the

derivative of something where one function is

inside another one. It seems complicated,

but applying the Chain Rule requires just two

simple steps:

1.

Take the derivative of the outsidefunction, leaving the inside function

completely alone.

2. Multiply what you got in Step by thederivative of the inside function.

Common Operations

Equation of the Tangent LineYoull see it written different ways, but the

most understandable tangent line formula Ive

found is

function,, untouched.Taking the derivative of using thePower Rule gives

Plugging back in for

gives us

Step of Chain Rule tells us to take ourresult from Step and multiply it bythe derivative of the inside function.

Our inside function is , and itsderivative is Multiplying the resultfrom Step

by the derivative of our

inside function, , gives:


answer:

Example


In this example, the outside function

is . is representing , butwe leave that part alone for now

because Step of Chain Rule tells us totake the derivative of the outside

function while leaving the inside

Example


Applying the Reciprocal Rule gives


20/33

When a problem asks you to find the equation

of the tangent line, youll always be asked to

evaluate at the point where the tangent line

intersects the graph.

In order to find the equation of the tangent

line, youll need to plug that point into the

original function, then substitute your answer

for. Next youll take the derivative of thefunction, plug the same point into the

derivative and substitute your answer for

.

Implicit Differentiation

Implicit Differentiation allows you to take the

derivative of a function that contains both and on the same side of the equation. If youcant solve the function for , implicitdifferentiation is the only way to take the

derivative.

On the left sides of these derivatives, instead

of seeing or, youll find instead.In this notation, the numerator tells you whatfunction youre deriving, and the denominator

tells you what variable is being derived. is literally read the derivative of withrespect to .

One of the most important things to

remember, and the thing that usually confuses

students the most, is that we have to treat

as a function and not just as a variable like .Therefore, we always multiply by whenwe take the derivative of y. To use implicit

differentiation, follow these s teps:

1. Differentiate both sides with respect to.

2. Whenever you encounter , treat it asa variable just like, and multiply thatterm by .

3. Move all terms involving to theleft side and everything else to the

right.

4. Factor out on the left and divideboth sides by the other left-side factor

so that is the only thingremaining on the left.

Example


Our first step is to differentiate both

sides with respect to

. Treat

as a

variable just like , but whenever youtake the derivative of a term that

includes , multiply by . Youllneed to use the Product Rule for the

right side, treating as one functionand as another.

Finally, insert bothand intothe tangent line formula, along with for , since this is the point at which wewere asked to evaluate.

You can either leave the equation inthis form, or simplify it further, as

follows:

Example

Find the equation of the tangent line at

to the graph of

First, plug in to the originalfunction.

Next, take the derivative and plug in

.


21/33

Equation of the Tangent Line

You may be asked to find the tangent line

equation of an implicitly-defined function. Just

for fun, lets pretend youre asked to find the

equation of the tangent line of the function in

the previous example at the point . Youdpick up right where you left off, and plug in

this point to the derivative of the function.

Related Rates

Related Rates are an application of implicit

differentiation, and are usually easy to spot.

They ask you to find how quickly one variable

is changing when you know how quickly

another variable is changing. To solve a related

rates problem, complete the following steps:

1. Construct an equation containing allthe relevant variables.

2. Differentiate the entire equation withrespect to (time), before plugging inany of the values you know.

3. Plug in all the values you know, leavingonly the one youre solving for.

4. Solve for your unknown variable.

Example

How fast is the radius of a balloon

increasing when the radius is 100

centimeters, if air is being pumped intothe spherical balloon at a rate of 400

cubic centimeters per second.

In this example, were asked to find the

rate of change of the radius, given the

rate of change of the volume.

The formula that relates the volume

and radius of a sphere to one another

is simply the formula for the volume of

a sphere:

Before doing anything else, we use

implicit differentiation to differentiate

both sides with respect to .

Example (continued)

Now that youve found the slope of the

tangent line at the point , plug thepoint and the s lope into Point-Slope

Form:

You could leave the equation as it is

above, or simplify it as follows:

Move all terms that include tothe left side, and everything els e to the

right side.

Factor out on the left, thendivide both sides by .

Dividing the right s ide by 3 to simplify

gives us our final answer:


22/33

Common Applications

Speed/Velocity/AccelerationA common application of derivatives is the

relationship between speed, velocity and

acceleration. In these problems, youre usually

given a position equation in the form or

, which tells you the objects distance

from some reference point. This equation also

accounts for direction, so the distance could

be negative, depending on which direction

your object moved away from the reference

point.

Average speed of the object is

Average velocity of the object is

To find velocity, take the derivative of your

original position equation. Speed is the

absolute value of velocity. Velocity accountsfor the direction of movement, so it can be

negative. Its like speed, but in a particular

direction. Speed, on the other hand, can never

be negative because it doesnt account for

direction, which is why speed is the absolute

value of velocity. To find acceleration, take the

derivative of velocity.

Example

Suppose a particle is moving along the

-axis so that its position at time is

given by the formula

Compute its velocity and accelerationas functions of. Next, decide in which

direction (left or right) the particle is

moving when and whether its

velocity and speed are increasing or

decreasing.

To find velocity, we take the derivative

of the original position equation.

To find acceleration, we take the

derivative of the velocity function.

To determine the direction of theparticle at , we plug into the

velocity function.

Because is positive, we can

Now we plug in everything that we

know. Keep in mind that is the

rate at which the volume is changing,

is the rate at which the radius is

changing, and is the length of the

radius at a specific moment.

Our problem tells us that the rate of

change of the volume is 400, and that

the length of the radius at the specific

moment were interested in is 100.

Solving for gives us

Therefore, we know that the radius of

the balloon is increasing at a rate of

centimeters per second.


23/33

LHopitals Rule

LHopitals Rule is used to get you out of sticky

situations with indeterminate limit forms, such

as or . If you plug in the

number youre approaching to the function for

which youre trying to find the limit and your

result is one of the indeterminate forms above,

you should try applying LHopitals Rule.

To use it, take the derivatives of the numerator

and denominator and replace the original

numerator and denominator with their

derivatives. Then plug in the number youre

approaching. If you still get an indeterminate

form, continue using LHopitals Rule until you

can use substitution to get a prettier answer. is our final answer. However, if plugging in

had resulted in another indeterminate form,

we could have applied another round of

LHopitals Rule, and another and another,

until we were able to plug in the number were

approaching to get an answer that was not

indeterminate.

Mean Value Theorem

This theorem guarantees that, at some pointon a closed interval, the tangent line to the

graph will be parallel to the line connecting the

endpoints of that interval. The Mean Value

Theorem is the following:

Example

Pretend that we drive from Florida to

California in exactly hours, from

time to time , and travel a

distance of miles.

Ifdescribes the distance weve

traveled at time , then the Mean Value

Theorem tells us that

Example

If we try plugging in for , we get the

indeterminate form , so we know

that this is a good candidate for

LHopitals Rule.

The derivative of our numerator is .

The derivative of our denominator is

. To use LHopitals Rule, we

take those derivatives and plug them in

for the original numerator and

denominator.

If we now try plugging in the number

were approaching, we get a clear

answer.

conclude that the particle is moving in

the positive direction (toward the

right).

To determine whether velocity is

increasing or decreasing, we plug 1 into

the acceleration function, because that

will give us the rate of change of

velocity, since acceleration is the

derivative of velocity.

Since acceleration is negative at ,

velocity must be decreasing at that

point.

Since the velocity is positive and

decreasing at , that means that

speed is also decreasing at that point.


24/33

Rolles TheoremRolles Theorem is a specific instance of the

Mean Value Theorem. Like the Mean Value

Theorem, Rolles Theorem applies to a

function on a closed interval, . If

and are both equal to , meaning that

the interval starts and ends on the -axis, then

the derivative, or slope of the function, atsome point in the interval must be equal to .

Rolles Theorem - At some point between and, the slope of the derivative must be equal to and the derivative must be parallel to the -

axis.

Graph Sketching

Graph sketching is not very hard, but there are

a lot of steps to remember. Like anything, the

best way to master it is with a lot of practice.

When it comes to sketching the graph, if

possible I absolutely recommend graphing the

function on your calculator before you get

started so that you have a visual of what your

graph should look like when its done. You

certainly wont get all the information you

need from your calculator, so unfortunately

you still have to learn the steps, but your

calculator is still a good double-check system.

Our strategy for sketching the graph will

include the following steps:

1. Find critical points.

2. Determine where is increasingand decreasing.

3. Find inflection points.4. Determine where is concave up

and concave down.

5. Find - and -intercepts.6. Plot critical points, possible inflection

points and intercepts.

7. Determine behavior asapproaches positive and negative

infinity.

8. Draw the graph with the informationwe just gathered.

Critical Points

Critical points occur at -values where the

functions derivative is either equal to zero orundefined. Critical points are the only points at

which a function can change direction, and

also the only points on the graph that can be

maxima or minima of the function.

Example

Find the critical points of

Take the derivative and simplify. You

can move the in the denominator of

the fraction into the numerator by

changing the sign on its exponent from

to .

The Mean Value Theorem therefore

implies that there was an instantaneous

velocity of exactly miles/hour at

least once during the trip.


25/33

Increasing/Decreasing

A function that is increasing (moving up as you

travel from left to right along the graph), has a

positive slope, and therefore a positive

derivative.

An increasing function

Similarly, a function that is decreasing (moving

down as you travel from left to right along the

graph), has a negative slope, and therefore a

negative derivative.

A decreasing function

Based on this information, it makes sense that

the sign (positive or negative) of a functions

derivative indicates the direction of the

original function. If the derivative is positive at

a point, the original function is increasing at

that point. Not surprisingly, if the derivative is

negative at a point, the original function isdecreasing there.

We already know that the direction of the

graph can only change at the critical points

that we found earlier. As we continue with our

example, well therefore plot those critical

points on a wiggle graph to test where the

function is increasing and decreasing.

Example (continued)

Determine where

is

increasing and decreasing

First, we create our wiggle graph and

plot our critical points, as follows:

-----------------------|--------------------|-----------------

Next, we pick values on each interval of

the wiggle graph and plug them into

the derivative. If we get a positive

result, the graph is increasing. A

negative result means its decreasing.

The intervals that we will test are:

,

and .

To test , well plug

into the derivative, s ince is a value

in that range.

Using Power Rule to take the derivative

gives

Moving the back into the

denominator by changing the s ign on

its exponent gives

Now set the derivative equal to and

solve for .

h h h di i hi h i h j b i


26/33

Inflection Points

Inflection points are just like critical points,

except that they indicate where the graph

changes concavity, instead of indicating where

the graph changes direction, which is the job

of critical points. Well learn about concavity in

the next section. For now, lets find our

inflection points.

In order to find inflection points, we first take

the second derivative, which is the derivative

of the derivative. We then set the second

derivative equal to zero and solve for .

There is no solution to this equation, but we

can see that the second derivative is undefined

at . Therefore, is the only possible

inflection point.

Concavity

Concavity is indicated by the sign of the

functions second derivative, . The

function is concave up everywhere the second

derivative is positive, and concave down

everywhere the s econd derivative is negative.

The following graph illustrates examples ofconcavity. From , the graph is

concave down. Think about the fact that a

graph that is concave down looks like a frown.

Sad, I know. The inflection point at which the

graph changes concavity is at . On the

range , the graph is concave up. It

looks like a smile. Ah much better.

is concave down on the range

and concave up on the range .We can use the same wiggle graph technique,

along with the possible inflection point we just

found, to test for concavity.

Example (continued)

Well start with the first derivative, and

then take its derivative to find the

second derivative.

Now set the second derivative equal to

zero and solve for .

To test , well plug into

the derivative.

To test , well plug into

the derivative.

Now we plot the results on our wiggle

graph, and we can see that is

Increasing on ,Decreasing on and

Increasing on .

-----------------------|--------------------|-----------------

d I t t L l d Gl b l E t


27/33

- and -InterceptsTo find the points where the graph intersects

the and axes, we can plug into theoriginal function for one variable and solve for

the other.

Local and Global Extrema

Maxima and Minima (these are the plural

versions of the singular words maximum and

minimum) can only exist at critical points, but

not every critical point is necessarily an

extrema. To know for sure, you have to test

each solution separately.

Minimums exist at as well as .Based on the -values at those points, the

global minimum exists at , and a localminimum exists at .

If youre dealing with a closed interval, for

example some function on the interval to

, then the endpoints

and

are

candidates for extrema and must also be

tested. Well use the First Derivative Test to

find extrema.

First Derivative Test

Remember the wiggle graph that we created

from our earlier test for increasing and

decreasing?

Example (continued)

To find -intercepts, plug in for .

Immediately we can recognize there

are no -intercepts because we canthave a result in the denominator.

Lets plug in for to try for -intercepts.

Multiply every term by to eliminatethe fraction.

Since there are no solutions to this

equation, we know that there are no -intercepts either for this particular

function.

Example (continued)

Since our only inflection point was at

, lets go ahead and plot that onour wiggle graph now.

--------------------------|-----------------------

As you might have guessed, well betesting values is the following intervals:

and

To test , well plug intothe second derivative.

To test , well plug intothe second derivative.

Now we can plot the results on our

wiggle graph

--------------------------|-----------------------

We determine that is concavedown on the interval andconcave up on .

hi h i t l b l i d


28/33

Based on the positive and negative signs on

the graph, you can see that the function is

increasing, then decreasing, then increasing

again, and if you can picture a function like

that in your head, then you know immediately

that we have a local maximum at anda local minimum at .

You really dont even need the silly First

Derivative Test, because it tells you in a formal

way exactly what you just figured out on your

own:

1. If the derivative is negative to the leftof the critical point and positive to the

right of it, the graph has a local

minimum at that point.

2. If the derivative is positive to the left ofthe critical point and negative on the

right side of it, the graph has a local

maximum at that point.

As a side note, if its positive on both sides or

negative on both sides, then the point is

neither a local maximum nor a local minimum,

and the test is inconclusive.

Remember, if you have more than one local

maximum or minimum, you must plug in the

value of at the critical points to your originalfunction. The values you get back will tell

you which points are global maxima and

minima, and which ones are only local. For

example, if you find that your function has two

local maxima, you can plug in the value for atthose critical points. If the first returns a -value of and the second returns a -valueof , then the first point is your global

maximum and the second point is your localmaximum.

If youre asked to determine where the

function has its maximum/minimum, your

answer will be in the form [value]. But ifyoure asked for the value at the

maximum/minimum, youll have to plug in the

-value to your original function and state the-value at that point as your answer.

Second Derivative Test

You can also test for local maxima and minima

using the Second Derivative Test if it easier for

you than the first derivative test. In order to

use this test, simply plug in your critical points

to the second derivative. If your result is

negative, that point is a local maximum. If the

result is positive, the point is a local minimum.

If the result is zero, you cant draw a

conclusion from the Second Derivative Test,

and you have to resort to the First Derivative

Test to solve the problem. Lets try it.

Good news! These are the same results we got

from the First Derivative Test! So why did wedo this? Because you may be asked on a test to

use a particular method to test the extrema, so

unfortunately, you should really know how to

use both tests.

Asymptotes

Vertical Asymptotes

Vertical asymptotes are the easiest to test for,because they only exist where the function is

Example (continued)

Our critical points are

and .

Since the second derivative is negative

at , we conclude that there is alocal maximum at that point.

Since the second derivative is positive

at , we conclude that there is alocal minimum at that point.

undefined Remember a function is undefined then the axis is a horizontal the example weve been using throughout this


29/33

undefined. Remember, a function is undefined

whenever we have a value of zero as the

denominator of a fraction, or whenever we

have a negative value inside a square root sign.

Consider the example weve been working

with in this section:

You should see immediately that we have a

vertical asymptote at because pluggingin for makes the denominator of thefraction , and therefore undefined.

Horizontal AsymptotesVertical and horizontal asymptotes are similar

in that they can only exist when the function isa rational function.

When were looking for horizontal as ymptotes,

we only care about the first term in the

numerator and denominator. Both of those

terms will have whats called a degree, which

is the exponent on the variable. If our function

is the following:

then the degree of the numerator is and thedegree of the denominator is .

Heres how we test for horizontal asymptotes.

1. If the degree of the numerator is lessthan the degree of the denominator,

then the -axis is a horizontalasymptote.

2. If the degree of the numerator is equalto the degree of the denominator, then

the coefficient of the first term in the

numerator divided by the coefficient in

the first term of the denominator is the

horizontal asymptote.3. If the degree of the numerator is

greater than the degree of the

denominator, there is no horizontal

asymptote.

Using the example weve been working with

throughout this section, well determine

whether the function has any horizontal

asymptotes. We can use long division to

convert the function into one fraction. The

following is the same function as our original

function, just consolidated into one fraction:

We can see immediately that the degree of our

numerator is , and that the degree of ourdenominator is . That means that ournumerator is one degree higher than our

denominator, which means that this function

does not have a horizontal as ymptote.

Slant Asymptotes

Slant asymptotes are a special case. They exist

when the degree of the numerator is greaterthan the degree of the denominator. Lets take

the example weve been using throughout this

section.

First, well convert this function to a rational

function by multiplying the first term by .

Now that we have a common denominator, we

can combine the fractions.

We can see that the degree of our numerator

is one greater than the degree of our

denominator, so we know that we have a slant

asymptote.

To find the equation of that asymptote line,

we divide the denominator into the numerator

using long division and we get

Right back to our original function! That wont

always happen, our function just happened to

be the composition of the quotient and

remainder.

Whenever we use long division in this way to


30/33

Whenever we use long division in this way to

find the slant asymptote, the first term is our

quotient and the second term is our

remainder. The quotient is the equation of the

line representing the slant asymptote.

Therefore, our slant asymptote is the line

.

Putting It All Together

Now that weve finished gathering all of the

information we can about our graph, we can

start sketching it. This will be something youll

just have to practice and get the hang of.

The first thing I usually do is sketch any

asymptotes, because you know that your

graph wont cross those lines, and therefore

they act as good guidelines. So lets draw in

the lines and .

The asymptotes of

Knowing that the graph is concave up in the

upper right, and concave down in the lower

left, and realizing that it cant cross either ofthe asymptotes, you should be able to make a

pretty good guess that it will look like the

following:

The graph of

In this case, picturing the graph was a little

easier because of the two asymptotes, but if

you didnt have the slant asymptote, youd

want to graph - and -intercepts, critical andinflection points, and extrema, and then

connect the points using the information you

have about increasing/decreasing and

concavity.

Optimization

Optimization is one of the most feared topics

for calculus students, but it really s houldnt be.

Optimization only requires a few simple steps,

all of which you already know how to do.

To solve an optimization problem, youll need

to:

1. Write an equation in one variable thatrepresents what youre trying to

maximize.

2. Take the derivative, find critical pointsand draw your wiggle graph.

3. Verify that your solutions are correctbased on the real-life s ituation.

Lets do one of the most common examples.

Example: The Open-Top Box

I dont know why this is such a popular

optimization example, but I swear its in

every calculus book ever written.

Say youre given a x piece of paper.Youre told to cut out squares from each

corner with side-length , as follows,


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If we plug into our length, , we get a positive number.However, if we plug into ourwidth, , we get a negativenumber, so cannot be asolution to our optimization problem.

Now we only have to test tomake sure its a local max. If it is a local

max., then is the value of thatmaximizes the volume of our box.

----------------------------|-----------------------

----------------------------|-----------------------

Since our function is increasing to the

left of the critical point and decreasing

to the right of it, is the value of that maximizes our volume.

Multiplying everything together gives

Now take the derivative with respect to

.

Find critical points by setting the

equation equal to zero and solving for .

Using the Quadratic Formula gives:

Our critical points are approximately

and .

Before we draw our wiggle graph and

start testing critical points, we should

always test our answers for plausibility.

Remember, length, width and height can

never be negative.

such that folding the sides up will create

a box with no top. Your job is to find the

value of that maximizes the volume ofthe box.

As soon as you hear volume of a box,

you should immediately write down the

formula for the volume of a box:

Based on the picture we drew of our

problem, we already know our length,

width and height, so we rewrite the

formula as follows:

E i l F l


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Essential Formulas

Foundations of Calculus

Laws of Exponents

Linear Functions

Slope-Intercept Form , where is the slope and isthe -intercept

Point-Slope Form , where is the slopeSlope of a Linear Function

Quadratic Functions

Quadratic Function Quadratic Formula

Derivatives

Definition of the Derivative

Shortcut Rules

The Power Rule The Product Rule

The Quotient Rule The Chain Rule

Logarithms & Exponentials

Trigonometric Derivatives

Common Operations Test for Global Extrema


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Common Operations

Equation of the Tangent Line Average Speed

Average Velocity

Graph Sketching

Critical PointsA point is a critical point if either or does not exist.Test for Increasing/DecreasingIf on an interval, then isincreasing on that interval

If on an interval, then isdecreasing on that interval

Test for Local Extrema

has a local maximum at a point if the

value there is greater than or equal to the

values around it

has a local minimum at a point if thevalue there is less than or equal to the values

around it

Test for Global Extrema has a global maximum at a point if thevalue there is greater than all others in the

domain of the function

has a local minimum at a point if thevalue there is less than all others in the domain

of the function

Inflection PointsIf is an inflection point, then either or is undefined.ConcavityThe graph is concave up if is increasing.The graph is concave down if isdecreasing.

calculus 1 survival guide

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