In this video, I will use an example to illustrate
how to calculate the maximum and the minimum of a function using derivatives.
Here is the example.
I want to find the extrema, and perhaps also the local extrema,
of the function f defined by f(x) = x^3 – 3x^2 – 9x + 35,
on the closed interval from -4 to 4.
I am going to reason in two steps.
First: f is continuous on this interval, which includes both endpoints.
I can use the Extreme Value Theorem to conclude that
f must have a maximum and a minimum on this interval.
I know some students like to skip this first step, but it is essential.
Without it, everything else fails.
Second: now that I know there is a maximum and a minimum,
we know they may be at an ENDPOINT, or at an INTERIOR POINT.
But if they are at interior points, they must also be LOCAL EXTREMA.
Then, the Local Extreme Value Theorem tells me they must be critical points.
So, I know there are extrema, and I know where to look for them.
Here is my plan.
This could work for any continuous function on a closed and bounded interval.
Find all the endpoints and critical points.
As a reminder, critical points are interior points where the derivative
is zero or does not exist.
Compare the function at these points.
One of these points must be the maximum (the one with the largest value of f),
and one of them must be the minimum (the one with the smallest value of f).
That will be it.
This bears repeating: the method only works for a continuous function
on an interval including both endpoints.
Otherwise, the function may not even have a maximum or a minimum,
and if I use this method carelessly, I may get the wrong answer.
Okay then!
Let's perform the calculations for our example.
Here is the function again.
Finding the endpoints is easy. They are -4 and 4.
As for the critical points, I need to find out when the derivative is zero.
This is a polynomial, so the derivative always exists.
The derivative: f'(x) = 3x^2 – 6x – 9.
I can factor a 3...
...and I can further factor this as (x + 1)(x – 3).
To factor, either guess and check, or use the quadratic formula, as your prefer.
So, the critical points are x = -1 and x = 3.
That's it! Those are all the points I need.
I have four candidate points.
To figure out which one is the maximum and which one is the minimum,
we evaluate f at them.
These are the values.
The largest value of f is 40, at x = -1.
That must be the maximum.
The smallest value of f is -41, at x = -4.
That must be the minimum.
We are done!
But this tells us even more about the function.
We can guess a sketch of f just from this.
Here are the four interesting points in the graph of f.
I know that between any two of them, there are no other critical points.
So that will suggest what the graph with look like.
Something like this...
So, I am guessing that at x = -1, I also have a local maximum,
and at x = 3 I have a local minimum.
Strictly speaking, this final part was not very rigorous (yet),
but geometrically it makes perfect sense.
We will learn soon how to argue this more rigorously.