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PROFESSOR: Welcome
back to recitation.

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Today what we're
going to do is use

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what we know about first
and second derivatives

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and what we know about
functions from way

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back in algebra and
precalculus, to sketch a curve.

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So I want you to sketch
the curve y equals x over 1

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plus x squared.

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Doesn't have to be
perfect, but try and use

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what you know about
these derivatives,

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first and second derivatives
of this function,

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and what you've talked
about in the lecture to get

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a pretty good sketch of this.

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I'll give you a little
time to work on it

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and then I'll be back and
I'll work on it for you.

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Welcome back.

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So hopefully you feel good
about the sketch you've drawn.

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But just to check everything,
we can go through it together.

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And what I'm going to do,
just to keep track of things,

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is I'm going to put
an axis in this region

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and then I'm going to do all
my work sort of off to the side

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and come back slowly.

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So we'll try and keep track
of everything that way.

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So before I do
anything else I'm just

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going to draw myself
a nice axis here.

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And I'll give myself even a
little bit of-- oops, that's

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maybe a little off, but-- so
we'll assume every hash mark

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is one unit.

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I'll just put a 1 there so
we know every hash mark here

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is going to represent one unit.

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And I won't write
the rest of them.

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Now one of the things
you always do first,

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is you want to make
sure that you understand

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where the function is defined.

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So we have to check right
away, are there any values

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of x for which this
function is not defined?

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Well, how can that happen?

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If it were a logarithm or if
it were a square root function

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we would have
problems in the domain

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where would have to
check and make sure

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that the input was positive.

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In this case, because we
have a rational function,

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we have to make sure that
the denominator is never

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equal to 0.

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But if you notice,
the denominator

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is 1 plus x squared.

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Well, x squared is always
bigger than or equal to 0,

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and once I add 1,
I'm in the clear.

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I'm always positive
in the denominator.

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So the denominator
is always positive,

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so I don't have to put
any vertical asymptotes.

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Some other things we
think about before we even

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start taking
derivatives, or anything

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I can find out about this
function, like end behavior.

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When we say end
behavior we mean,

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what happens as x goes
to positive infinity

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and as x goes to
negative infinity?

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And from what
you've seen before,

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as x goes to positive
infinity, because this

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is a rational function,
the higher power

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is going to win out.

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The higher power
always wins out.

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So the higher power here
is in the denominator,

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so as x goes to
positive infinity

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this whole expression
is going to head to 0.

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For large values of x the
x squared is significantly

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bigger than the x.

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And so the denominator
is significantly bigger

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than the numerator.

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That's how we can
think about this.

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So when x goes to
plus or minus infinity

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we know that our function
is going to be headed to 0,

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so it has a
horizontal asymptote.

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OK.

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And then another thing we
would-- we should notice

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is the sign of the graph.

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Notice where the
sign will change.

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This denominator
is always positive

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so the sign of the
function depends completely

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on the numerator.

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And so when the
numerator is positive

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this function will be positive.

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When the numerator is negative
this function will be negative.

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So that's a little bit that
we should keep in mind.

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And now let's go to using
our derivatives to figure

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out a little bit more.

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So obviously, first I
should take some derivatives

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and then we'll look at what
we can get out of them.

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So let's let f of x equal
x over 1 plus x squared.

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So then f prime of
x, what do we get?

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We get 1 plus x squared minus x
times 2x over 1 plus x squared,

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squared.

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So I'm just going to
continue that straight below.

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Let's see.

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I can keep this x squared
minus 2x squared, gives me a 1

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minus x squared, in
the numerator, over 1

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plus x squared,
quantity squared.

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OK.

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I'm going to keep
that right here.

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We're going to do a
little bit of calculation

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below in a moment, but I'm going
to record the second derivative

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just to the right.

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So the second
derivative, remember,

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is the derivative of
the first derivative.

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So now I'm going to
take this derivative,

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again using the quotient
rule, which I used here.

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So the derivative of
the top is minus 2x

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and then times 1 plus
x squared squared

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and then I subtract
the derivative

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of the bottom times the top.

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So I'll keep the top
here, 1 minus x squared.

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And then the derivative
of the bottom

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has a little chain rule on it,
so I'm going to get a times 2

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times 1 plus x squared times 2x.

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And then this whole thing is
over 1 plus-- whoa-- x plus 1.

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We'll write x squared
plus 1 to the fourth.

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Sorry to switch the direction
or the order of those.

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OK.

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Now I'm going to
pull out a 1 plus x

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squared from the
numerator to simplify it.

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And then I'm going to
see what I have left.

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Here I have a 1 plus x
squared times a negative 2x.

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That's going to be negative
2x minus 2 x cubed.

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Here I'm going to have-- 2
times 2 is 4x times this 1

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minus x squared.

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So I have a minus 4x plus
4 x squared-- cubed, sorry.

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Let's make sure.

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So I should have a 4x here
and then an x squared times

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4x, which is 4 x cubed.

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And that sign
should be positive.

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And then I still
have to divide by 1

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plus x squared to the fourth.

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To make this much
simpler I'm just

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going to divide out one
of the 1 plus x squareds,

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simplify what's inside, and
we'll leave it that way.

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Actually, let me move this down
so there's a little more room.

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So the numerator will now
be 2 x cubed minus 6x over 1

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plus x squared to the third.

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So these were some
tools that we needed.

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Now we're going to
try and use them.

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So let's recall what we know.

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We know that when the
derivative is equal to 0,

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we have a maximum or
minimum for the function.

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And we know that when the
second derivative is equal to 0,

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we have changes in concavity.

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So let's find those places.

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Let's find where the
first derivative is 0

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and let's find where the
second derivative is 0.

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So I'm going to work under each
individual function to do that.

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So where is f prime equal to 0?

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Well, f prime is
only equal to 0 when

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the numerator is equal to 0.

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So let's solve 1 minus
x squared equals 0.

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Well that's-- there's a couple
ways you can think about that.

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You could factor
it and then solve,

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or you could see right
away this is going

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to be x is plus or minus 1.

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You get the same
thing if you factor.

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But we see x is equal
to plus or minus 1.

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So those are our maximum
values or minimum values

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for the function.

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OK.

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So we know that this is an
important spot for the x-value

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and that's an important
spot for the x-value.

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Now let's just come
over here and look at,

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when is the second
derivative equal to 0?

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So the second derivative
is equal to 0,

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again, when the
numerator is equal to 0.

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So let's look at what we get.

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Well, if we factor that we
get 2x times x squared minus 3

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equals 0.

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So this has three places
it's going to be equal to 0.

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It's going to be equal
to 0 at 0, x equals 0,

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and it's going to be equal to 0
at plus or minus root 3, which

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is sort of unfortunate that we
don't know exactly where that

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is, but we know it's
between 1 and 2.

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I think it's about 1.7
or something like this.

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So we know we're
interested in the point x

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equals 0 and the points x equal
plus or minus square root of 3.

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So these are our
places of interest.

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And so let's evaluate at
least a couple of these places

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and see what's going on.

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Let's go back to the
graph to do this.

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Now I want to
point out something

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I didn't say earlier, which
is, if you know the function is

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defined everywhere,
what you might

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want to do is evaluate
the function at x

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equals 0 right away.

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It's an easy place
to evaluate it.

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It gives you sort of
a launching point.

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So if I evaluate this
at x equals 0 I get 0.

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So I know the point
(0, 0) is on the graph.

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So I know that's one point.

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And now what I'm
interested in, if you

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think about-- we know
where maxes or mins occur,

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we know a max or min occurs
at x equals plus or minus 1.

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Or we have a hope for
a max or min there.

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It's a critical point, at least.

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So I can evaluate the
function-- sorry--

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I can evaluate the function
at 1 and at negative 1

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and I can then
plot those points.

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So when x is 1, I get 1 over 1
plus 1 squared, so I get 1/2.

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So with input 1
I get output 1/2.

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I'm going to erase
that 1 now so we don't

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lose track of what's happening.

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That looks potentially
like it could

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be a maximum, given sort
of what's happening here,

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to the left.

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So let's plug in
negative 1 for x.

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I get a negative 1 over 1 plus
quantity negative 1 squared.

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So I get negative 1 over
2, so I get negative 1/2.

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So at x equals negative
1, I get negative 1/2.

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And let's recall what we know
about the end behavior, which

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we said at the beginning.

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The end behavior of this is as
x goes to positive infinity,

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the function's outputs go to 0.

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Which tells you that, in fact,
this has to be a maximum.

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There are the only two places
where the function can change

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direction from going
up to going down,

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or from going down to going up.

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So it has to be that
this is a maximum.

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It has to be that
this is a minimum.

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So, and also notice
0, based on what

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we know about the
second derivative,

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is one of the inflection points.

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So that's also representing a
place where the derivative is

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changing sign.

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So maybe the derivative
was increasing

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and then it's going
to start decreasing.

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So let's look-- I think I
might have said something

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a little off there, so I'm
going to maybe come back and see

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if I have to fix
anything in a moment--

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but let me draw a rough
sketch of what's happening.

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Very rough, very roughly
we know we're going up

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and then we're going down.

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We're going down
here and then we

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have to go back up
because the end behavior.

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So we have three
inflection points--

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this is what I want
to point out-- we

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have three inflection points.

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We have an inflection point at
0 and at plus or minus root 3.

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So we said root 3 is bigger
than 1, it's less than 2.

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So I know somewhere
in here I have

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an inflection point,
which represents

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a change in the concavity.

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00:10:57,873 --> 00:10:58,550
Right?

246
00:10:58,550 --> 00:11:00,680
Which represents
how the derivative

247
00:11:00,680 --> 00:11:05,030
is going to change the
direction, whether it's

248
00:11:05,030 --> 00:11:08,720
continuing to get more negative
and then getting more positive

249
00:11:08,720 --> 00:11:09,884
than it was previously.

250
00:11:09,884 --> 00:11:11,800
So yeah, that's where--
we're looking at where

251
00:11:11,800 --> 00:11:13,740
the derivative changes sign.

252
00:11:13,740 --> 00:11:14,620
As I said before.

253
00:11:14,620 --> 00:11:17,850
So let me point out-- this
is a change in concavity.

254
00:11:17,850 --> 00:11:21,030
Maybe right about
in this x region

255
00:11:21,030 --> 00:11:23,800
we want to change concavity,
and then this x region

256
00:11:23,800 --> 00:11:25,500
we want to change concavity.

257
00:11:25,500 --> 00:11:29,620
So the graph will look something
like going up, going down,

258
00:11:29,620 --> 00:11:30,120
going down.

259
00:11:30,120 --> 00:11:34,180
And then I've tried to represent
the change in concavity

260
00:11:34,180 --> 00:11:35,510
changing that direction there.

261
00:11:38,880 --> 00:11:43,260
And I'm doing something
that I didn't tell you yet.

262
00:11:43,260 --> 00:11:47,250
But if you notice, this looks
highly symmetric, doesn't it?

263
00:11:47,250 --> 00:11:49,900
And in fact, one thing I didn't
tell you about this function--

264
00:11:49,900 --> 00:11:51,940
that maybe you picked
up on already--

265
00:11:51,940 --> 00:11:54,410
is that when I take
the right-hand side

266
00:11:54,410 --> 00:11:58,050
and I rotate it about the
origin I get the left hand side.

267
00:11:58,050 --> 00:11:59,040
Why is that?

268
00:11:59,040 --> 00:12:00,770
That's because this
is an odd function.

269
00:12:00,770 --> 00:12:02,510
Why is it an odd function?

270
00:12:02,510 --> 00:12:05,020
Because the numerator is an odd
function and the denominator

271
00:12:05,020 --> 00:12:06,180
is an even function.

272
00:12:06,180 --> 00:12:09,200
And so the quotient
is an odd function.

273
00:12:09,200 --> 00:12:12,840
So this is, I would say,
a fairly good sketch

274
00:12:12,840 --> 00:12:15,400
of the curve y equals x
over 1 plus x squared.

275
00:12:15,400 --> 00:12:17,920
So hopefully yours looked
something like this.

276
00:12:17,920 --> 00:12:19,507
And that's where we'll stop.