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

Nov 5, 2015

$6 x \left({x}^{2} - 1\right)$

#### 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

$\frac{d}{\mathrm{dx}} {\left(f \left(x\right)\right)}^{n} = n {\left(f \left(x\right)\right)}^{n - 1} \cdot f ' \left(x\right)$.

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

$\frac{d}{\mathrm{dx}} {\left({x}^{2} - 1\right)}^{3} = 3 {\left({x}^{2} - 1\right)}^{2} \cdot 2 x$