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PROFESSOR: Welcome back
to recitation.

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Today what we're going to do
is use what we know about

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first and second derivatives
and what we know about

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functions from way back in
algebra and precalculus, to

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sketch a curve.

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So I want you to sketch
the curve y equals x

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over 1 plus x squared.

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

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about these derivatives, first
and second derivatives of this

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function, and what you've talked
about in the lecture to

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get a, get a pretty good
sketch of this.

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

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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, is

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I'm going to put an axis in this
region and then I'm going

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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 going to draw myself

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a nice axis here.

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And I'll give myself even
a little bit of--

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oops, that's maybe a
little off, but--

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so we'll assume every hash
mark is one unit.

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

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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, is you want

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to make sure that you understand
where the function

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is defined.

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

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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 where would have to

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check and make sure that
the input was positive.

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In this case, because we have
a rational function, we have

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to make sure that the
denominator is

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never equal to 0.

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But if you notice, the

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denominator is 1 plus x squared.

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

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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, so I don't have to

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put any vertical asymptotes.

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Some other things we think about
before we even start

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taking derivatives, or anything
I can find out about

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this function, like
end behavior.

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When we say end behavior we
mean, what happens as x goes

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to positive infinity and as x
goes to negative infinity?

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And from what you've seen
before, as x goes to positive

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infinity, because this is a
rational function, the higher

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power 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 so as x goes

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to positive infinity this
whole expression is

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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

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bigger 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 we know that

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our function is going to be
headed to 0, so it has a

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horizontal asymptote.

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OK.

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And then another thing we would,
we should notice is the

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sign of the graph.

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Notice where the sign
will change.

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This denominator is always
positive so the sign of the

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function depends completely
on the numerator.

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And so when the numerator
is positive this

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function will be positive.

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When the numerator
is negative this

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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 out a

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little bit more.

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So obviously, first I should
take some derivatives and then

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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

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squared 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--

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in the numerator--

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over 1 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 below in a

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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, is the derivative of

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the first derivative.

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So now I'm going to take this
derivative, again using the

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quotient rule, which
I used here.

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So the derivative of the top is
minus 2x and then times 1

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plus x squared squared 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 has a little chain rule

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on it, so I'm going to get
a times 2 times 1 plus x

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squared times 2x.

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

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x plus 1.

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We'll write x squared plus
1 for 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 squared from the

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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 2x cubed.

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

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this 1 minus x squared.

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So I have a minus 4x
plus 4x squared--

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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 4x,

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which is 4x cubed.

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And that sign should
be positive.

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And then I still have to divide
by 1 plus x squared to

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the fourth.

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To make this much simpler I'm
just going to divide out one

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of the 1 plus x squared's,
simplify what's inside, and

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we'll leave it that way.

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Actually, let me move
this down so there's

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a little more room.

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

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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, we

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have a maximum or minimum
for the function.

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And we know that when the second
derivative is equal to

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0, 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 and let's find

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where the second derivative
is 0.

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So I'm going to work under
each individual

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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 the

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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--

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

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You could factor it and then
solve, or you could see right

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away this is going 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 for

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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, when is the

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second derivative equal to 0?

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So the second derivative is
equal to 0, again, when the

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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

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minus 3 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, and it's going

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to be equal to 0 at plus or
minus root 3, which is sort of

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unfortunate that we don't know
exactly where that is, but we

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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 equals 0 and the

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points x equal plus or minus
square root of 3.

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So these are our places of
interest. And so let's

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evaluate at least a couple
of these places and see

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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 I didn't say

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earlier, which is, if you know
the function is defined

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everywhere, what you might want
to do is evaluate the

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function at x 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 think about we

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know where max's or min's occur
we know a max or min

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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 and I can

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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 lose track of

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what's happening.

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That looks potentially like it
could be a maximum, given sort

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of what's happening
here, 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, or from going down

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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 we know about the

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second derivative, 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 and then it's going

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to start decreasing.

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So let's look--

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I think I might have said
something a little off there,

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so I'm going to maybe come back
and see if I have to fix

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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 and then

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we're going down.

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We're going down here and then
we have to go back up because

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the end behavior.

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So we have three inflection
points-- this is what I want

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to point out-- we 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 an inflection point,

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which represents a change
in the concavity.

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Right?

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Which represents how the
derivative is going to change

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00:11:02,450 --> 00:11:05,990
the direction, whether it's
continuing to get more

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00:11:05,990 --> 00:11:09,110
negative and then getting more
positive than it was

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00:11:09,110 --> 00:11:10,060
previously.

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00:11:10,060 --> 00:11:11,910
So yeah, that's where we're
looking at where the

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00:11:11,910 --> 00:11:13,740
derivative changes sign.

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

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

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00:11:17,850 --> 00:11:21,710
Maybe right about in this x
region we want to change

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00:11:21,710 --> 00:11:23,890
concavity, and then
this x region we

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00:11:23,890 --> 00:11:25,500
want to change concavity.

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00:11:25,500 --> 00:11:29,370
So the graph will look something
like going up, going

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00:11:29,370 --> 00:11:30,120
down, going down.

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00:11:30,120 --> 00:11:32,770
And then I've tried to represent
the change in

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00:11:32,770 --> 00:11:35,510
concavity changing that
direction there.

261
00:11:35,510 --> 00:11:38,880

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

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

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00:11:47,250 --> 00:11:49,500
And in fact, one thing I didn't
tell you about this

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00:11:49,500 --> 00:11:51,940
function-- that maybe you
picked up on already--

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00:11:51,940 --> 00:11:55,160
is that when I take the right
hand side and I rotate it

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00:11:55,160 --> 00:11:58,050
about the origin I get
the left hand side.

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

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

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

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00:12:02,510 --> 00:12:04,480
Because the numerator is
an odd function and the

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00:12:04,480 --> 00:12:06,180
denominator is an
even function.

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

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00:12:09,200 --> 00:12:13,100
So this is, I would say, a
fairly good sketch of the

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00:12:13,100 --> 00:12:15,400
curve y equals x over
1 plus x squared.

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

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00:12:17,920 --> 00:12:19,710
And that's where we'll stop.

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00:12:19,710 --> 00:12:20,007