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, 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 for 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 2x 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 4x squared--
cubed, sorry.
Let's make sure.
So I should have a 4x here and then an x squared times 4x,
which is 4x 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 squared's, 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 2x 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 max's or min's 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.