# What are the critical points of  f(x) = x^(1/3)*(x+8)?

Apr 8, 2016

The critical numbers are $0$ and $- 2$.

#### Explanation:

A critical number for $f$ is a number, c, in the domain of $f$ such that f'C9)=0 or f'(c) does not exist.

For $f \left(x\right) = {x}^{\frac{1}{3}} \left(x - 8\right)$, the domain is $\mathbb{R}$.

And differentiating, we get

$f ' \left(x\right) = \frac{1}{3} {x}^{- \frac{2}{3}} \left(x - 8\right) + {x}^{\frac{1}{3}} \left(1\right)$

$= \frac{x - 8}{3 {x}^{\frac{2}{3}}} + {x}^{\frac{1}{3}}$

$= \frac{x - 8}{3 {x}^{\frac{2}{3}}} + \frac{3 x}{3 {x}^{\frac{2}{3}}}$

 = (4x+8)/(3x^(2/3)#.

So, $f ' \left(x\right) = 0$ at $x = - 2$ and $f ' \left(x\right)$ does not exists at $x = 0$.

Both are in the domain of $f$, so both are critical numbers for $f$.