Let f(x) = (x ² - 1)³. Find the critical points of f in the interval [-1, 2]:
f '(x) = 3 (x ² - 1)² (2x) = 6x (x ² - 1)² = 0
6x = 0 or (x ² - 1)² = 0
x = 0 or x ² = 1
x = 0 or x = 1 or x = -1
Check the value of f at each of these critical points, as well as the endpoints of the given domain:
f (-1) = 0
f (0) = -1
f (1) = 0
f (2) = 27
So max{f(x) | -1 ≤ x ≤ 2} = 27.
