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

Jan 7, 2017

${x}_{c} = \sqrt[3]{2}$ is a local minimum

#### Explanation:

We can find the critical points by equating the first derivative of $g \left(x\right)$ to zero:

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

$\frac{1}{3} = \frac{2}{3 {x}^{3}}$

${x}^{3} = 2$

${x}_{c} = \sqrt[3]{2}$

We can easily see that $g ' \left(x\right) < 0$ for $x < {x}_{c}$ and viceversa $g ' \left(x\right) > 0$ for $x > {x}_{c}$ , so #x_c is a local minimum.

graph{x/3+x^(-2)/3 [-10, 10, -5, 5]}