What are the absolute extrema of f(x)=(x^2 - 1)^3 in[-oo,oo]?

1 Answer
Nov 5, 2015

6x(x^2-1)

Explanation:

To compute critical points of a function, we need to compute the first derivative, and then to find its zeroes.

As a general rule, we have that

d/dx (f(x))^n = n (f(x))^{n-1} * f'(x).

Of course, n=3, and f(x)=x^2-1, which means that f'(x)=2x. Plugging all these things into the general formula, we have

d/dx (x^2-1)^3 = 3(x^2-1)^2 * 2x