In this video, we're gonna go over
how to find an absolute maximum
or an absolute minimum
of a function on an interval.
Now, Fermat's theorem tells us that
the only place where you can have
maxima or minima—or the
general term is extrema—
is either at the endpoints,
or at critical points.
Critical points are where the derivative
is either 0, or doesn't exist.
And some people call critical points
critical numbers; it means the same thing.
So our recipe is pretty simple.
First thing is we have to
find the critical points.
Second thing is we have to find
the endpoints—that part's pretty easy.
Then we just compare the values,
and the biggest one wins.
And the point is there are very few
candidates for where the maximum can be.
You know, you just list all the
candidates, and compare them.
The biggest one wins if you're
looking for a maximum,
the smallest one wins if you're
looking for a minimum.
So let's work an example.
So we're gonna look for
maximum and minimum values
of the function 10x/(x^2 + 1).
So most—and we're gonna do this
on the interval from 0 to 3.
Now between 0 and 3,
this function is continuous.
Since it's continuous on a closed
interval, it has a maximum value,
it has a minimum value.
That's the extreme value theorem.
So we have to find the critical points.
Well that's where the derivative
either doesn't exist, or is 0.
Now, this is a perfectly good function.
The quotient rule will
give us a derivative.
So the derivative always exists.
So it basically boils down to finding out
where the derivative is 0.
So let's take the derivative.
The derivative of a quotient is the
bottom times the derivative of the top,
minus the top times the
derivative of the bottom,
all divided by the bottom squared.
And the derivative of 10x is 10,
the derivative of x^2 + 1 is 2x,
and you multiply it all out,
and you get (10 - 10x^2)/(x^2 + 1)^2,
or 10(1- x^2)/(x^2 + 1)^2.
So where is that 0?
Well the only place that a ratio can be 0
is if the numerator is 0.
So to—this function is 0—the derivative
is 0 exactly when 1 - x^2 is 0.
So the derivative always exists,
it equals 0 at x = +-1,
so +-1 are our critical numbers,
or our critical points.
Except they don't both count.
One of them is in the interval.
One is between 0 and 3,
but -1 isn't.
The only critical point that's actually
in our interval is x = 1.
We don't care about x = -1,
it's not in the interval.
Find the endpoints,
well that's pretty easy.
And then we just make a table.
The only candidates for maximum
and minimum are the endpoints 0 and 3
and the critical number 1.
And now you look at these and say
hmm, which is the biggest of 0, 5, and 3?
Well, 5 is the biggest...
0 is the smallest,
and we're done.
The maximum value
of the function is 5.
Where does the function hit
that maximum value?
It hits the maximum value at x = 1.
Now there's sometimes some confusion
in the terminology when people say,
"Where's the maximum?"
People might say, "Oh the maximum
is 1," or "The maximum is 5."
Be careful about this.
The maximum value of the function is 5.
It happens at x = 1.
And the minimum value is 0,
and it happens at x = 0.
Okay, one more example.
And this example involves logs.
So we want to find the maximum and
the minimum values of xln(x),
on the interval from 1 to 3.
So step one, we find the critical points.
We want to take the derivative
and set it equal to 0.
The derivative, we use the product rule.
And you get ln(x) + 1.
So where is that 0?
Well let's see.
If 0 was ln(x) + 1,
then ln(x) must be -1,
so x must be e^(-1),
in other words, 1/e.
But that isn't in the interval.
1/e is less than 1.
So in fact, there are no
critical points in the interval.
So the only candidates are our endpoints,
we compare our values,
and we say hey, at one endpoint
the value is 0,
at the other endpoint,
the value is 3ln(3).
The minimum value is 0,
and it occurs at x = 1,
the maximum value is 3ln(3),
and that occurs at x = 3.
That's it.
