Question

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

Answer 1

`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`