# How do you find the critical points for f(x)=(x^2)(x-2)^(1/3)?

$\left(0 , 0\right)$ Maximum Point
$\left(\frac{12}{7} , - \frac{144 \sqrt[3]{98}}{343}\right)$ Minimum Point

#### Explanation:

To determine the critical points
Find first derivative $f ' \left(x\right)$ then equate to zero
that is $f ' \left(x\right) = 0$

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

$f ' \left(x\right) = 2 x {\left(x - 2\right)}^{\frac{1}{3}} + {x}^{2} \cdot \frac{1}{3} \cdot {\left(x - 2\right)}^{- \frac{2}{3}} = 0$

$2 x \sqrt[3]{x - 2} + {x}^{2} / \left(3 {\left(\sqrt[3]{x - 2}\right)}^{2}\right) = 0$

Multiply both sides by $3 {\left(\sqrt[3]{x - 2}\right)}^{2}$

$6 {x}^{2} - 12 x + {x}^{2} = 0$
$7 {x}^{2} - 12 x = 0$
solving for x values

${x}_{1} = 0$ and ${x}_{2} = \frac{12}{7}$
$y = 0$ when $x = 0$
$y = - \frac{144 \sqrt[3]{98}}{343}$ when $x = \frac{12}{7}$

The Critical Points are:

$\left(0 , 0\right)$ Maximum Point
$\left(\frac{12}{7} , - \frac{144 \sqrt[3]{98}}{343}\right)$ Minimum Point

Check the graph:

graph{y=x^2 (x-2)^(1/3) [-5, 5, -2.5, 2.5]}