PROFESSOR: Welcome
back to recitation.
Today what we're
going to do is use
what we know about first
and second derivatives
and what we know about
functions from way
back in algebra and
precalculus, to sketch a curve.
So I want you to sketch
the curve y equals x over 1
plus x squared.
Doesn't have to be
perfect, but try and use
what you know about
these derivatives,
first and second derivatives
of this function,
and what you've talked
about in the lecture to get
a pretty good sketch of this.
I'll give you a little
time to work on it
and then I'll be back and
I'll work on it for you.
Welcome back.
So hopefully you feel good
about the sketch you've drawn.
But just to check everything,
we can go through it together.
And what I'm going to do,
just to keep track of things,
is I'm going to put
an axis in this region
and then I'm going to do all
my work sort of off to the side
and come back slowly.
So we'll try and keep track
of everything that way.
So before I do
anything else I'm just
going to draw myself
a nice axis here.
And I'll give myself even a
little bit of-- oops, that's
maybe a little off, but-- so
we'll assume every hash mark
is one unit.
I'll just put a 1 there so
we know every hash mark here
is going to represent one unit.
And I won't write
the rest of them.
Now one of the things
you always do first,
is you want to make
sure that you understand
where the function is defined.
So we have to check right
away, are there any values
of x for which this
function is not defined?
Well, how can that happen?
If it were a logarithm or if
it were a square root function
we would have
problems in the domain
where would have to
check and make sure
that the input was positive.
In this case, because we
have a rational function,
we have to make sure that
the denominator is never
equal to 0.
But if you notice,
the denominator
is 1 plus x squared.
Well, x squared is always
bigger than or equal to 0,
and once I add 1,
I'm in the clear.
I'm always positive
in the denominator.
So the denominator
is always positive,
so I don't have to put
any vertical asymptotes.
Some other things we
think about before we even
start taking
derivatives, or anything
I can find out about this
function, like end behavior.
When we say end
behavior we mean,
what happens as x goes
to positive infinity
and as x goes to
negative infinity?
And from what
you've seen before,
as x goes to positive
infinity, because this
is a rational function,
the higher power
is going to win out.
The higher power
always wins out.
So the higher power here
is in the denominator,
so as x goes to
positive infinity
this whole expression
is going to head to 0.
For large values of x the
x squared is significantly
bigger than the x.
And so the denominator
is significantly bigger
than the numerator.
That's how we can
think about this.
So when x goes to
plus or minus infinity
we know that our function
is going to be headed to 0,
so it has a
horizontal asymptote.
OK.
And then another thing we
would-- we should notice
is the sign of the graph.
Notice where the
sign will change.
This denominator
is always positive
so the sign of the
function depends completely
on the numerator.
And so when the
numerator is positive
this function will be positive.
When the numerator is negative
this function will be negative.
So that's a little bit that
we should keep in mind.
And now let's go to using
our derivatives to figure
out a little bit more.
So obviously, first I
should take some derivatives
and then we'll look at what
we can get out of them.
So let's let f of x equal
x over 1 plus x squared.
So then f prime of
x, what do we get?
We get 1 plus x squared minus x
times 2x over 1 plus x squared,
squared.
So I'm just going to
continue that straight below.
Let's see.
I can keep this x squared
minus 2x squared, gives me a 1
minus x squared, in
the numerator, over 1
plus x squared,
quantity squared.
OK.
I'm going to keep
that right here.
We're going to do a
little bit of calculation
below in a moment, but I'm going
to record the second derivative
just to the right.
So the second
derivative, remember,
is the derivative of
the first derivative.
So now I'm going to
take this derivative,
again using the quotient
rule, which I used here.
So the derivative of
the top is minus 2x
and then times 1 plus
x squared squared
and then I subtract
the derivative
of the bottom times the top.
So I'll keep the top
here, 1 minus x squared.
And then the derivative
of the bottom
has a little chain rule on it,
so I'm going to get a times 2
times 1 plus x squared times 2x.
And then this whole thing is
over 1 plus-- whoa-- x plus 1.
We'll write x squared
plus 1 to the fourth.
Sorry to switch the direction
or the order of those.
OK.
Now I'm going to
pull out a 1 plus x
squared from the
numerator to simplify it.
And then I'm going to
see what I have left.
Here I have a 1 plus x
squared times a negative 2x.
That's going to be negative
2x minus 2 x cubed.
Here I'm going to have-- 2
times 2 is 4x times this 1
minus x squared.
So I have a minus 4x plus
4 x squared-- cubed, sorry.
Let's make sure.
So I should have a 4x here
and then an x squared times
4x, which is 4 x cubed.
And that sign
should be positive.
And then I still
have to divide by 1
plus x squared to the fourth.
To make this much
simpler I'm just
going to divide out one
of the 1 plus x squareds,
simplify what's inside, and
we'll leave it that way.
Actually, let me move this down
so there's a little more room.
So the numerator will now
be 2 x cubed minus 6x over 1
plus x squared to the third.
So these were some
tools that we needed.
Now we're going to
try and use them.
So let's recall what we know.
We know that when the
derivative is equal to 0,
we have a maximum or
minimum for the function.
And we know that when the
second derivative is equal to 0,
we have changes in concavity.
So let's find those places.
Let's find where the
first derivative is 0
and let's find where the
second derivative is 0.
So I'm going to work under each
individual function to do that.
So where is f prime equal to 0?
Well, f prime is
only equal to 0 when
the numerator is equal to 0.
So let's solve 1 minus
x squared equals 0.
Well that's-- there's a couple
ways you can think about that.
You could factor
it and then solve,
or you could see right
away this is going
to be x is plus or minus 1.
You get the same
thing if you factor.
But we see x is equal
to plus or minus 1.
So those are our maximum
values or minimum values
for the function.
OK.
So we know that this is an
important spot for the x-value
and that's an important
spot for the x-value.
Now let's just come
over here and look at,
when is the second
derivative equal to 0?
So the second derivative
is equal to 0,
again, when the
numerator is equal to 0.
So let's look at what we get.
Well, if we factor that we
get 2x times x squared minus 3
equals 0.
So this has three places
it's going to be equal to 0.
It's going to be equal
to 0 at 0, x equals 0,
and it's going to be equal to 0
at plus or minus root 3, which
is sort of unfortunate that we
don't know exactly where that
is, but we know it's
between 1 and 2.
I think it's about 1.7
or something like this.
So we know we're
interested in the point x
equals 0 and the points x equal
plus or minus square root of 3.
So these are our
places of interest.
And so let's evaluate at
least a couple of these places
and see what's going on.
Let's go back to the
graph to do this.
Now I want to
point out something
I didn't say earlier, which
is, if you know the function is
defined everywhere,
what you might
want to do is evaluate
the function at x
equals 0 right away.
It's an easy place
to evaluate it.
It gives you sort of
a launching point.
So if I evaluate this
at x equals 0 I get 0.
So I know the point
(0, 0) is on the graph.
So I know that's one point.
And now what I'm
interested in, if you
think about-- we know
where maxes or mins occur,
we know a max or min occurs
at x equals plus or minus 1.
Or we have a hope for
a max or min there.
It's a critical point, at least.
So I can evaluate the
function-- sorry--
I can evaluate the function
at 1 and at negative 1
and I can then
plot those points.
So when x is 1, I get 1 over 1
plus 1 squared, so I get 1/2.
So with input 1
I get output 1/2.
I'm going to erase
that 1 now so we don't
lose track of what's happening.
That looks potentially
like it could
be a maximum, given sort
of what's happening here,
to the left.
So let's plug in
negative 1 for x.
I get a negative 1 over 1 plus
quantity negative 1 squared.
So I get negative 1 over
2, so I get negative 1/2.
So at x equals negative
1, I get negative 1/2.
And let's recall what we know
about the end behavior, which
we said at the beginning.
The end behavior of this is as
x goes to positive infinity,
the function's outputs go to 0.
Which tells you that, in fact,
this has to be a maximum.
There are the only two places
where the function can change
direction from going
up to going down,
or from going down to going up.
So it has to be that
this is a maximum.
It has to be that
this is a minimum.
So, and also notice
0, based on what
we know about the
second derivative,
is one of the inflection points.
So that's also representing a
place where the derivative is
changing sign.
So maybe the derivative
was increasing
and then it's going
to start decreasing.
So let's look-- I think I
might have said something
a little off there, so I'm
going to maybe come back and see
if I have to fix
anything in a moment--
but let me draw a rough
sketch of what's happening.
Very rough, very roughly
we know we're going up
and then we're going down.
We're going down
here and then we
have to go back up
because the end behavior.
So we have three
inflection points--
this is what I want
to point out-- we
have three inflection points.
We have an inflection point at
0 and at plus or minus root 3.
So we said root 3 is bigger
than 1, it's less than 2.
So I know somewhere
in here I have
an inflection point,
which represents
a change in the concavity.
Right?
Which represents
how the derivative
is going to change the
direction, whether it's
continuing to get more negative
and then getting more positive
than it was previously.
So yeah, that's where--
we're looking at where
the derivative changes sign.
As I said before.
So let me point out-- this
is a change in concavity.
Maybe right about
in this x region
we want to change concavity,
and then this x region
we want to change concavity.
So the graph will look something
like going up, going down,
going down.
And then I've tried to represent
the change in concavity
changing that direction there.
And I'm doing something
that I didn't tell you yet.
But if you notice, this looks
highly symmetric, doesn't it?
And in fact, one thing I didn't
tell you about this function--
that maybe you picked
up on already--
is that when I take
the right-hand side
and I rotate it about the
origin I get the left hand side.
Why is that?
That's because this
is an odd function.
Why is it an odd function?
Because the numerator is an odd
function and the denominator
is an even function.
And so the quotient
is an odd function.
So this is, I would say,
a fairly good sketch
of the curve y equals x
over 1 plus x squared.
So hopefully yours looked
something like this.
And that's where we'll stop.
