Calculus 1 Survival Guide
Transcript of 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,
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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
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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.
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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
Find the derivative of
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
Find the derivative of
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,
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.
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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
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:
Simplifying the result gives us our final
answer:
Example
Find the derivative of
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
Find the derivative of
Applying the Reciprocal Rule gives
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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
Find the derivative of
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
.
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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:
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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.
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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.
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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.
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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
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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
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- 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
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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
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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
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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.