### Video Transcript

Find the absolute maximum and
minimum values of the function 𝑓 of 𝑥 equals 𝑥 cubed plus three 𝑥 squared if 𝑥
is less than or equal to zero, 𝑥 squared minus six 𝑥 if 𝑥 is greater than zero in
the closed interval negative 5 to 13.

To find the absolute maximum and
minimum values of a function, we need to consider a number of things. But firstly, we’ll look for
critical points of the function. These are points where our first
derivative is equal to zero. They can indicate relative maxima
and relative minima. If we’re looking for absolute
extremer, we also need to consider the endpoints of our function. So, what we’ll do is begin by
differentiating our function with respect to 𝑥.

Now, of course, our function is
piecewise. So, we’re going to do this for both
parts. Let’s begin by differentiating 𝑥
cubed plus three 𝑥 squared with respect to 𝑥. The derivative of 𝑥 cubed is three
𝑥 squared. And then, when we differentiate
three 𝑥 squared, we multiply the entire term by the exponent and reduce the
exponent by one. So, we get two times three 𝑥,
which is six 𝑥. And so, when 𝑥 is less than or
equal to zero, the derivative is three 𝑥 squared plus six 𝑥.

We’ll now differentiate 𝑥 squared
minus six 𝑥 with respect to 𝑥. The derivative of 𝑥 squared is two
𝑥 and the derivative of negative six 𝑥 is negative six. And so, we found the first
derivative. We’re now going to set each part of
this function equal to zero and solve for 𝑥. Let’s set three 𝑥 squared plus six
𝑥 equal to zero. We’re going to factor that
expression on the left-hand side. But when we do, we see that three
𝑥 times 𝑥 plus two is equal to zero. Now, of course, for this statement
to be true, either three 𝑥 is equal to zero, which means 𝑥 must itself be equal to
zero or 𝑥 plus two must be equal to zero, which means 𝑥 itself must be negative
two.

Let’s set the other part of our
derivative equal to zero. That’s two 𝑥 minus six equals
zero. This time, we solve by adding six
to both sides. And then, we divide through by
two. And we find that 𝑥 equals three is
the location of the third critical point. To establish the absolute minimum
and maximum values of our function. We’re going to work out the value
of the function at each of the three critical points and at the two endpoints. So, we have critical points 𝑥
equals zero, negative two, and three and the endpoints of our intervals are 𝑥
equals negative five and 𝑥 equals 13.

We will need to be a little bit
careful which part of the function we use. Remember, it’s a piecewise
function. So, it has a different function
depending on the value of 𝑥. When 𝑥 is equal to zero, the
function we use is 𝑥 cubed plus three 𝑥 squared. 𝑓 of zero is, therefore, zero
cubed plus three times zero squared, which is zero. Well, similarly, when 𝑥 is less
than zero, we use that same function. Here, 𝑥 is equal to negative
two. So, we get negative two cubed plus
three times negative two squared, which is equal to four. At our third critical point, 𝑥 is
equal to three. That’s greater than zero. So, we use the second part of our
piecewise function. We get three squared minus six
times three, which is negative nine.

We go back to the first function,
where 𝑥 is equal to negative five. And we substitute that into 𝑥
cubed plus three 𝑥 squared and we find that 𝑓 of negative five is negative 50. Our final value to check is 𝑥
equals 13. Again, that’s greater than
zero. So, we substitute that into 𝑥
squared minus six 𝑥. And that gives us 91.

We said that the absolute maximum
and minimum values of the function will either occur at the critical points or at
the endpoints. Well, here, we see that the
absolute minimum value is negative 50. That occurs when 𝑥 is equal to
negative five, whereas the absolute maximum is 91.

The absolute minimum value is
negative 50 and the absolute maximum value of our function is 91.