# What are the critical points of  f(x)=x^(2/3)x^2 - 16?

Dec 23, 2015

$c = 0$

#### Explanation:

Critical points ($c$) occur when $f ' \left(c\right) = 0$ or when $f ' \left(c\right)$ doesn't exist.

In order to find $f ' \left(x\right)$, first simplify $f \left(x\right)$.

$f \left(x\right) = {x}^{\frac{2}{3}} {x}^{2} - 16$

$f \left(x\right) = {x}^{\frac{2}{3}} {x}^{\frac{6}{3}} - 16$

$f \left(x\right) = {x}^{\frac{8}{3}} - 16$

Use the power rule: $\frac{d}{\mathrm{dx}} \left[{x}^{n}\right] = n {x}^{n - 1}$

$f ' \left(x\right) = \frac{8}{3} {x}^{\frac{5}{3}}$

$f ' \left(x\right) = 0$ when $x = 0$ and is never undefined.

Thus, $c = 0$.

graph{x^(2/3)x^2-16 [-32.06, 32.9, -20.27, 12.2]}