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. 
 
 
Introduction 
Author’s 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 wouldn’t be conceited to say that math was 
a subject that came easier to me than it did to 
others. Compared to my classmates, I was 
always 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 it’s just the way my brain works. 
And yet, despite the fact that it’s always been 
easier for me, I’ve 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 you’re reading 
through an example in your textbook, hoping 
with desperation that it will show you how to 
do the problem you’re stuck on? You hang in 
there for the first few steps, and you’re like, 
“Okay awesome! I’m getting this!” And then by 
about the fourth step, you start to lose track of 
their logic and you can’t for the life of you 
figure out how they got from Step 3 to Step 4? 
It’s the worst feeling. This is the point where 
most people give up completely and just resign 
themselves to failing the final exam. 
I’ve 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 didn’t 
make sense, and they’d end up feeling 
overwhelmed and defeated before they’d ever 
really gotten started. 
I wasn’t a math major in college, but I spent a 
lot of time tutoring calculus students, and I’ve 
come to the conclusion that for most people, 
the way we teach math is fundamentally 
wrong. 
First, there’s a pretty good chance that you 
won’t 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 most 
benefit 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, there’s a lot to be said for going more in 
depth with the material, and I’d love to help 
you do that if that’s your goal. For most people 
though, a basic understanding is sufficient. 
My greatest hope for this e-book and for 
integralCALC.com is that they’ll help you in 
whatever capacity you need them. If you’re 
shooting for a C+, let’s get you a C+. I don’t 
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 you’re shooting for, I hope that this e-
book will help you get closer to it, and better 
yet, save you some stress along the way. 
Remember, if there’s anything I can 
ever do for you, please contact me 
at integralCALC.com. 
 
 
 
Words of Wisdom 
There are two pieces of advice I’d like to give 
you before we get started. 
1. Stay Positive 
More than anything, you have to stay positive. 
Don’t defeat yourself before you even get 
started. You’re smarter than you think, and 
calculus is easier than you think it is. Don’t 
panic. 
Half of the people I’ve tutored over the years 
needed a personal calculus cheerleader more 
than they needed a tutor. They’d gingerly 
proceed through a new problem… “Is this 
right? Then if I… am I still doing it right?” 
They’d doubt themselves at every step. And I 
would just stand behind them and say “Yeah, 
it’s right, you’re doing great, you’ve got it, 
you’re right,” until they’d 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. It’s understandable, but 
the more you can fight the fear that starts to 
creep in, the better off you’ll be. So take a 
deep breath. It’s going to be okay.  
2. Use Your Calculator (Or Don’t) 
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 won’t 
know if what your calculator tells you is 
correct. Nor will you be able to show any work 
if you’re required to do so on an exam, which 
could cost you big points. 
Learning the calculus itself means you’ll be 
able to show your work when you need to, and 
you’ll actually understand what you’re doing. 
Once you solve a problem, you should know 
how to punch in the equation so that you can 
look at the graph or solution to verify that the 
answer you got is the same one your calculator 
gives back to you. 
What You Won’t Find 
I’m not here to replace your textbook. Because 
this is a quick-reference guide, you won’t find 
chapter introductions full of calculus history 
you don’t care about. 
I’m also not here to replace your professor, 
nor do I expect that you’re particularly excited 
about learning calculus. If you are excited 
about calculus, that’s awesome! So am I. But if 
you’re not, this is the place to be, because, at 
least in this e-book, you won’t 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 I’ve 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 that 
information, and then do poorly on the test 
because you didn’t get all the instructions. 
What You Will Find 
This e-book should give you the most crucial 
pieces of information you’ll need for a real 
understanding of how to solve most of the 
problems you’ll encounter. I don’t want to be 
your 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.
 
(Clickable) Table of Contents 
I. Foundations of Calculus 
A. Functions 
1. Vertical Line Test 
2. Horizontal Line Test 
3. 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 Don’t 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 Discontinuity 
c. 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 Don’t 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 Applications 
1. Speed/Velocity/Acceleration 
2. L’Hopital’s Rule 
3. Mean Value Theorem 
4. Rolle’s 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
 
Foundations of Calculus 
Functions 
Vertical Line Test 
Most of the equations you’ll encounter in 
calculus are functions. Since not all equations 
are functions, it’s 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 for 
any input value of . You know that an 
equation is not a function if can be two 
different 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 the 
graph 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 a 
function can only have one unique output 
value for for any input value of , the graph 
fails the Vertical Line Test and is therefore not 
a function. We’ve now proven with both the 
graph 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 they’re very similar. You’ll recall 
that any function passing the Vertical Line Test 
can only have one unique output of for any 
single input of . 
 
This graph passes the Horizontal Line 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. 
 
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 Line 
Test because any horizontal line between 
 and would intersect it more 
than once. 
Domain and Range 
Think of the domain of a function as 
everything you can plug in for without 
causing your function to be undefined. Things 
to look out for are values that would cause a 
fraction’s denominator to equal and values 
that 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 every 
number in the domain for . 
 
Independent and Dependent Variables 
Your independent variable is , and your 
dependent variable is . You always plug in a 
value for first, and your function returns to 
you a value for based on the value you gave 
it for . Remember, if your equation is a 
function, there is only one possible output of 
for any input of . 
Linear Functions 
You’ll 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 point 
2. Another point 
3. 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 is 
the -intercept, or the point at which the 
graph crosses the -axis and where . The 
slope, represented by , is calculated using 
two points on the line, and , 
and the equation you use to calculate is 
 
 
 
To find the slope, subtract the -coordinate in 
the first point from the -coordinate in the 
second point in the numerator, then subtract 
the -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 the 
denominator of the fraction to equal . 
Because setting equal to is the only 
way to make the function undefined, 
the domain of the function is all . 
 
 
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 the 
function. To use this form, find the same 
way
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 term 
and a constant, the coefficients , and can 
be 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 Quadratic 
Formula, 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, 
you’ll end up with – for the first term in 
the numerator of the Quadratic Formula, 
which would make that term positive. 
You should also remember that in order for 
this formula to work, must be 
greater than or equal to , because you can’t 
take 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 out 
the denominator of the fraction. 
Subtract from both sides. 
Divide by to solve for . 
For the final answer, plug and back 
into the formula, leaving and as 
variables. 
 
 
 
Since you always get a fraction with a plus or 
minus sign in the numerator, the 
Quadratic Formula produces two solutions, 
which you then use to factor your polynomial. 
If your solutions are and , your factors will 
always be and . 
Completing the Square 
This method is another option you can use to 
find the solutions of a quadratic function, if 
you can’t easily factor it. Since it’s very much a 
step-by-step process, the easiest way to 
explain this method is to use an example, so 
let’s do it. 
Let’s say we have the function 
 
The first thing we want to do is set the 
function equal to . 
 
Next, we’ll take one half of the coefficient on 
the term and square it. 
 
 
 
 
We then add and subtract our result back into 
the function, so that we don’t 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. Don’t 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 to 
solve it for . 
 
This is the same process you’ll follow each 
time you use this method to solve for the roots 
of 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 you’ll 
use semi-frequently in calculus. It’s just like the 
long division you learned in fifth grade, except 
that instead of just numbers, this time you’ll 
be dividing polynomials. 
 
 
 
 
 
Example 
Use long division to convert 
First, we should keep the following in 
mind: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Example 
Factor 
Plug for , for and for into 
the Quadratic Formula. 
Simplify to find your solutions. 
Use the solutions to factor your 
function. 
 
 
Limits 
What is a Limit? 
The limit of a function is the value the function 
approaches at a given value of , regardless of 
whether the function actually reaches that 
value. 
For an easy example, consider the function 
 
When , . Therefore, is the 
limit of the function at , because is the 
value that the function approaches as the 
value of gets closer and closer to . 
I know it’s strange to talk about the value that 
a function “approaches.” Think about it this 
way: If you set in the function 
above, then . Similarly, if you 
set , then . 
You can begin to see that as you get closer to 
 , whether you’re 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 actual 
value of the function at that point; the point is 
defined. However, finding limits gets a little 
trickier when we start dealing with points of 
the graph that are undefined. 
In the next section, we’ll 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, you’re dealing with a 
general limit. 
The general limit exists at the point if 
1. The left-hand limit exists at , 
2. The right-hand limit exists at , 
and 
3. The left- and right-hand limits are 
equal. 
These are the three conditions that must be 
met 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- and 
right-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 the 
dividend). Since the answer is , we 
put that on top of our long division 
problem, and multiply it by the divisor, 
 to get , which we then 
subtract from the dividend. We bring 
down from the dividend and repeat 
the same steps until we have nothing 
left to carry down from the dividend. 
Our original problem reduces to: 
 
 
 
 
The negative
sign after the indicates that 
we’re 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 
we’re 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 limits 
both approach . On the other hand, the 
general limit does not exist at because 
the 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 Don’t 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 . The 
graph 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 limit 
does not exist at that point. The left-hand limit 
is , because that is the value that the graph 
approaches 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 graph 
approaches 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-hand 
limit is . However, the general limit does 
not 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 you 
just plug into this function, you get , which 
is the limit of the function. Below is another 
example, where you can simply plug in to 
the function to solve for the limit. 
 
Factor It 
When you can’t just plug in the value you’re 
evaluating, your next approach should be 
factoring. 
 
 
 
 
 
 
 
 
 
 
 
 
Example 
Plug in for and simplify. 
 
 
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. Here’s 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. We 
also can’t 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, that’s 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 
that’s like multiplying by , which is useful to 
us but still doesn’t 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 you’re 
evaluating and solve for the limit that way. 
Trigonometric Limits 
Trigonometric limit problems revolve around 
three formulas: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Simplify and cancel the . 
Since we’re evaluating at , plug that in 
for and solve. 
 
 
 
 
 
 
 
 
 
 
 
 
 
Example 
Multiply the numerator and 
denominator by the conjugate. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Example 
Just plugging in would give us a nasty 
 result. Therefore, we’ll try 
factoring instead. 
Cancelling from the top and 
bottom of the fraction leaves us with 
something that is much easier to 
evaluate: 
Now the problem is simple enough that 
we can just plug in the value we’re 
approaching. 
 
 
 
 
 
 
 
 
 
When solving trigonometric limit problems, 
our goal is to reduce our problem to a simple 
combination involving nothing but these 
formulas and simple constants. Here’s an 
example. 
 
 
Infinite Limits 
Infinite limits exist when we can plug in a 
number for that causes the denominator of a 
rational 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 an 
infinite limit at that point. 
Now that we’ve established that this is a 
rational function in lowest terms and that a 
vertical asymptote exists, all that’s 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, the 
limit is . 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
We can see that the result will be very large 
and positive, so we know that the limit of this 
function at is . 
Continuity 
I would give you the definition of continuity, 
but I think it’s 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. Let’s 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, we’re actually going to 
borrow 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 three 
fundamental formulas. 
 
Common Discontinuities 
Jump Discontinuity 
You’ll 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 
that’s often used to illustrate piecewise-
defined functions is the cost of postage at the 
post office. Here’s 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 less 
than 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, you’d 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 a 
point discontinuity is easy to pick out because 
it looks totally normal everywhere, except for 
a hole at a single point. 
Infinite/Essential Discontinuity 
You’ll 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. The 
vertical asymptote in the second graph below 
is not a point of discontinuity, because it 
doesn’t break up any part of the graph. 
 
A vertical asymptote at that makes the 
graph discontinuous 
 
A vertical asymptote at that does not 
make the graph discontinuous 
Removable Discontinuity 
Discontinuity is removable if you can easily 
plug in the holes in its graph by redefining the 
function. When you can’t easily plug in the 
holes because the gaps are bigger than a single 
point, you’re dealing with nonremovable 
discontinuity. Point discontinuity is removable, 
because you can easily patch the hole. 
 
Let’s 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 Theorem 
Similarly 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 
doesn’t tell us anything that we don’t 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 we’ll call (orange). 
must be between and and must be 
between and . Looking at the graph, 
isn’t that obvious? Values may or may not exist 
below and above depending on the 
graph, but must exist. 
 
The Derivative 
The derivative of a function is written as 
 , and read as “ prime of .” By 
definition, the derivative is the slope of the 
original function. Let’s 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 we’re going to 
learn later. Unbelievable, I know. 
But since our goal is just to get you a good 
grade, and not to make a big scene, we’ll learn 
how to find derivatives the long way first, then 
we’ll learn the shortcuts and things will end up 
better in the end. I promise. For now, the long 
way… 
Secant and Tangent Lines 
A 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, 
 
sometimes hitting it at multiple points, and 
looks generally meaner, like this: 
 
A secant line 
 
It’s 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 Derivative 
If we start with a point, on a graph, 
and move a certain distance, , to the right of 
that 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 to 
the 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 distance 
between 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, we’ve 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 Quotient 
To find the derivative of a function using the 
difference quotient, follow these steps: 
1. Plug in for every in your 
original 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 it from the numerator and 
denominator. 
7. Plug in the number your function is 
approaching and simplify. 
 
Example 
Find the derivative of 
 at 
 
 
When Derivatives Don’t Exist 
Before we jump into finding derivatives
with 
the shortcuts, let’s talk about instances when 
the derivative doesn’t exist. When the 
derivative doesn’t 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. You’re most likely to find sharp points in 
your function if it contains absolute values or if 
it’s a piecewise-defined function. 
 
A cusp in the graph of 
Vertical Tangent Lines 
Since 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, we’ve gotten to the point where things 
start to get easier. We’ve moved past the 
Difference Quotient, which was cumbersome 
and tedious and generally not fun. You’re 
about to learn several new derivative tricks 
that will make this whole process a whole lot 
easier. Aren’t you excited?! 
The Derivative of a Constant 
The derivative of a constant (a term with no 
variable attached to it) is always . Remember 
that 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 that 
function, and the slope of a horizontal line is , 
the derivative of any constant must be . 
The Power Rule 
The Power Rule is the tool you’ll use most 
frequently when finding derivatives. The rule 
says that for any term of the form , the 
derivative of the term is 
 
To use the Power Rule, multiply the variable’s 
exponent, , by its coefficient, , then subtract 
 from the exponent. If there is no coefficient 
(the coefficient is ), then the exponent will 
become the new coefficient. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
After replacing with in , 
plug it your answer for . Then 
plug in as-is for . Put in the 
denominator. 
Expand all terms. 
Collect similar terms together then factor 
 out of the numerator and cancel it from 
the fraction. 
For , plug in the number you’re 
approaching, in this case . Then simplify 
and solve. 
 
 
 
The Product Rule 
If a function contains two variable expressions 
that are multiplied together, you cannot 
simply 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 
Find the derivative of 
Based on the Quotient Rule formula, 
we know that is the numerator 
and therefore and that 
is the denominator and therefore that 
 . 
 is , and is . Plugging 
all of these components into the 
Quotient Rule gives 
Simplifying the result gives us our final 
answer: 
 
 
 
Example 
Find the derivative of 
The two functions in this problem are 
 and . It doesn’t matter which 
one you choose for and . 
Let’s assign to and to . 
The derivative of is . 
The derivative of is . 
According to the Product Rule, 
Simplifying the result gives us our final 
answer: 
 
 
 
Example 
Find the derivative of 
Applying Power Rule gives the 
following: 
Simplify to solve for the derivative. 
 
 
 
 
 
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 hardest 
concepts for calculus students to understand. 
It’s also one of the most important, and it’s 
used all the time, so make sure you don’t leave 
this section without a solid understanding. If 
you go through the example and you’re still 
having trouble, please e-mail me for help at 
integralCALC@gmail.com. 
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 “outside” 
function, leaving the “inside” function 
completely alone. 
2. Multiply what you got in Step by the 
derivative of the “inside” function. 
 
 
Common Operations 
Equation of the Tangent Line 
You’ll see it written different ways, but the 
most understandable tangent line formula I’ve 
found is 
 
 
 
 
 
function, , untouched. 
Taking the derivative of using the 
Power Rule gives 
Plugging back in for gives us 
Step of Chain Rule tells us to take our 
result from Step and multiply it by 
the derivative of the “inside” function. 
Our “inside” function is , and its 
derivative is Multiplying the result 
from Step by the derivative of our 
inside function, , gives: 
Simplifying the result gives us our final 
answer: 
 
Example 
Find the derivative of 
In this example, the “outside” function 
is . is representing , but 
we leave that part alone for now 
because Step of Chain Rule tells us to 
take the derivative of the outside 
function while leaving the inside 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Example 
Find the derivative of 
Applying the Reciprocal Rule gives 
 
When a problem asks you to find the equation 
of the tangent line, you’ll 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, you’ll need to plug that point into the 
original function, then substitute your answer 
for . Next you’ll take the derivative of the 
function, 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 you 
can’t solve the function for , implicit 
differentiation is the only way to take the 
derivative. 
On the left sides of these derivatives, instead 
of seeing or , you’ll find instead. 
In this notation, the numerator tells you what 
function you’re deriving, and the denominator 
tells you what variable is being derived. 
is literally read “the derivative of with 
respect 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 when 
we take the derivative of y. To use implicit 
differentiation, follow these steps: 
1. Differentiate both sides with respect to 
 . 
2. Whenever you encounter , treat it as 
a variable just like , and multiply that 
term by . 
3. Move all terms involving to the 
left side and everything else to the 
right. 
4. Factor out on the left and divide 
both sides by the other left-side factor 
so that is the only thing 
remaining on the left. 
 
 
Example 
Find the derivative of 
Our first step is to differentiate both 
sides with respect to . Treat as a 
variable just like , but whenever you 
take the derivative of a term that 
includes , multiply by . You’ll 
need to use the Product Rule for the 
right side, treating as one function 
and as another. 
 
 
 
Finally, insert both and into 
the tangent line formula, along with 
for , since this is the point at which we 
were asked to evaluate. 
You can either leave the equation in 
this 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 original 
function. 
Next, take the derivative and plug in 
 . 
 
 
 
Equation of the Tangent Line 
You may be asked to find the tangent line 
equation of an implicitly-defined function. Just 
for fun, let’s pretend you’re asked to find the 
equation of the tangent line of the function in 
the previous example at the point . You’d 
pick 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 all 
the relevant variables. 
2. Differentiate the entire equation with 
respect to (time), before plugging in 
any of the values you know. 
3. Plug in all the values you know, leaving 
only the one you’re 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 into 
the spherical balloon at a rate of 400 
cubic centimeters per second. 
In this example, we’re 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 you’ve found the slope of the 
tangent line at the point , plug the 
point and the slope into Point-Slope 
Form: 
You could leave the equation as it is 
above, or simplify it as follows: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Move all terms that include to 
the left side, and everything else to the 
right side. 
Factor out on the left, then 
divide both sides by . 
Dividing the right side by 3 to simplify 
gives us our final answer: 
 
 
Common Applications 
Speed/Velocity/Acceleration 
A common application of derivatives is the 
relationship between speed, velocity and 
acceleration. In these problems, you’re usually 
given a position equation in the form “ ” or 
“ ”, which tells you the object’s 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 accounts 
for the direction of movement, so it can be 
negative. It’s like speed, but in a particular 
direction. Speed, on the other hand, can never 
be negative because it doesn’t 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 acceleration 
as 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 the 
particle 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 we’re 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. 
 
 
L’Hopital’s Rule 
L’Hopital’s Rule is used to get you out of sticky 
situations with indeterminate limit forms, such 
as or . If you plug in the 
number you’re approaching to the function for 
which you’re trying to find the limit and your 
result is
one of the indeterminate forms above, 
you should try applying L’Hopital’s 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 you’re 
approaching. If you still get an indeterminate 
form, continue using L’Hopital’s 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 
L’Hopital’s Rule, and another and another, 
until we were able to plug in the number we’re 
approaching to get an answer that was not 
indeterminate. 
Mean Value Theorem 
This theorem guarantees that, at some point 
on 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. 
If describes the distance we’ve 
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 
L’Hopital’s Rule. 
The derivative of our numerator is . 
The derivative of our denominator is 
 . To use L’Hopital’s Rule, we 
take those derivatives and plug them in 
for the original numerator and 
denominator. 
If we now try plugging in the number 
we’re 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. 
 
 
 
Rolle’s Theorem 
Rolle’s Theorem is a specific instance of the 
Mean Value Theorem. Like the Mean Value 
Theorem, Rolle’s 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, at 
some point in the interval must be equal to . 
 
Rolle’s 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 it’s done. You 
certainly won’t 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 increasing 
and 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 as 
approaches positive and negative 
infinity. 
8. Draw the graph with the information 
we just gathered. 
Critical Points 
Critical points occur at -values where the 
function’s derivative is either equal to zero or 
undefined. 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. 
 
 
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 function’s 
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 is 
decreasing 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, we’ll 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 it’s decreasing. 
The intervals that we will test are: 
 , 
 and 
 . 
 
To test , we’ll plug 
into the derivative, since is a value 
in that range. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Using Power Rule to take the derivative 
gives 
Moving the back into the 
denominator by changing the sign on 
its exponent gives 
Now set the derivative equal to and 
solve for . 
 
 
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. We’ll learn about concavity in 
the next section. For now, let’s 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 
function’s second derivative, . The
function is concave up everywhere the second 
derivative is positive, and concave down 
everywhere the second derivative is negative. 
The following graph illustrates examples of 
concavity. 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) 
We’ll 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 , we’ll plug into 
the derivative. 
To test , we’ll 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 . 
 
 
-----------------------|--------------------|----------------- 
 
 
 
 - and -Intercepts 
To find the points where the graph intersects 
the and axes, we can plug into the 
original 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 local 
minimum exists at . 
 
If you’re 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. We’ll 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 can’t 
have a result in the denominator. 
Let’s plug in for to try for -
intercepts. 
Multiply every term by to eliminate 
the 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 
 , let’s go ahead and plot that on 
our wiggle graph now. 
 
--------------------------|----------------------- 
 
As you might have guessed, we’ll be 
testing values is the following intervals: 
 and 
 
To test , we’ll plug into 
the second derivative. 
To test , we’ll plug into 
the second derivative. 
Now we can plot the results on our 
wiggle graph 
 
--------------------------|----------------------- 
 
 
We determine that is concave 
down on the interval and 
concave up on . 
 
 
 
 
 
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 and 
a local minimum at . 
You really don’t 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 left 
of 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 of 
the critical point and negative on the 
right side of it, the graph has a local 
maximum at that point. 
As a side note, if it’s 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 original 
function. 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 at 
those critical points. If the first returns a -
value of and the second returns a -value 
of , then the first point is your global 
maximum and the second point is your local 
maximum. 
If you’re asked to determine where the 
function has its maximum/minimum, your 
answer will be in the form [value]. But if 
you’re asked for the value at the 
maximum/minimum, you’ll 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 can’t draw a 
conclusion from the Second Derivative Test, 
and you have to resort to the First Derivative 
Test to solve the problem. Let’s try it. 
 
Good news! These are the same results we got 
from the First Derivative Test! So why did we 
do 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 a 
local maximum at that point. 
Since the second derivative is positive 
at , we conclude that there is a 
local minimum at that point. 
 
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 we’ve been working 
with in this section: 
 
 
 
 
You should see immediately that we have a 
vertical asymptote at because plugging 
in for makes the denominator of the 
fraction , and therefore undefined. 
Horizontal Asymptotes 
Vertical and horizontal asymptotes are similar
in that they can only exist when the function is 
a rational function. 
When we’re looking for horizontal asymptotes, 
we only care about the first term in the 
numerator and denominator. Both of those 
terms will have what’s called a degree, which 
is the exponent on the variable. If our function 
is the following: 
 
 
 
 
then the degree of the numerator is and the 
degree of the denominator is . 
Here’s how we test for horizontal asymptotes. 
1. If the degree of the numerator is less 
than the degree of the denominator, 
then the -axis is a horizontal 
asymptote. 
2. If the degree of the numerator is equal 
to 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 we’ve been working with 
throughout this section, we’ll 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 our 
denominator is . That means that our 
numerator is one degree higher than our 
denominator, which means that this function 
does not have a horizontal asymptote. 
Slant Asymptotes 
Slant asymptotes are a special case. They exist 
when the degree of the numerator is greater 
than the degree of the denominator. Let’s take 
the example we’ve been using throughout this 
section. 
 
 
 
 
First, we’ll 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 won’t 
always happen, our function just happened to 
be the composition of the quotient and 
remainder. 
 
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 we’ve finished gathering all of the 
information we can about our graph, we can 
start sketching it. This will be something you’ll 
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 won’t cross those lines, and therefore 
they act as good guidelines. So let’s 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 can’t cross either of 
the 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 didn’t have the slant asymptote, you’d 
want to graph - and -intercepts, critical and 
inflection 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 shouldn’t be. 
Optimization only requires a few simple steps, 
all of which you already know how to do. 
To solve an optimization problem, you’ll need 
to: 
1. Write an equation in one variable that 
represents what you’re trying to 
maximize. 
2. Take the derivative, find critical points 
and draw your wiggle graph. 
3. Verify that your solutions are correct 
based on the real-life situation. 
Let’s do one of the most common examples. 
 
Example: The Open-Top Box 
I don’t know why this is such a popular 
optimization example, but I swear it’s in 
every calculus book ever written. 
Say you’re given a x piece of paper. 
You’re told to cut out squares from each 
corner with side-length , as follows, 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
If we plug into our length, 
 , we get a positive number. 
However, if we plug into our 
width, , we get a negative 
number, so cannot be a 
solution to our optimization problem. 
Now we only have to test to 
make sure it’s a local max. If it is a local 
max., then is the value of that 
maximizes 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 of 
the 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: 
 
Essential Formulas 
Foundations of Calculus 
Laws of Exponents 
 
 
 
 
 
 
 
 
 
 
 
Linear Functions 
Slope-Intercept Form 
 , where is the slope and is 
the -intercept 
Point-Slope Form 
 , where is the slope 
Slope 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