Find the absolute maximum and absolute minimum values

Question

Find the absolute maximum and absolute minimum values of

f(x) = (x2 - 1)^3 on

the interval

-1 ≤ x ≤  2

Answer 1

Absolute minimum and maximum over an interval can occur at two kinds of places:  The end points of the interval and/or values of x where the derivative is 0.  So first let's find the derivative:

 

By the chain rule, the derivative is 3(x² - 1)² * 2x = 6x (x² - 1).

 

For this to be 0 (where a minimum or maximum would be), x can be 0, 1, or -1.  All three of those points are in the interval, and one of them is at the lower endpoint of the interval. So we have to evaluate the function at each of those values of x, and also at x = 2, to get the upper end of the interval.

 

x = -1:  f(x) = 0

x = 0:  f(x) = -1

x = 1:  f(x) = 0

x = 2:  f(x) = 27

 

So the absolute minimum in the interval is -1, and the absolute maximum is 27.