Question

Find the absolute maximum of (x^2-1)^3 in the interval [-1,2]

Answer 1

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.