# How do you find the relative extrema for f(x) =2x- 3x^(2/3) +2?

May 18, 2017

Relative maximum $f \left(0\right) = 2$ and relative minimum $f \left(1\right) = 1$

#### Explanation:

The domain of $f$ is $\mathbb{R}$

$f ' \left(x\right) = 2 - 2 {x}^{- \frac{1}{3}} = \frac{2 \left(\sqrt[3]{x} - 1\right)}{\sqrt[3]{x}}$

Critical numbers for $f$ are values of $x$ in the domain of $f$ at which $f ' \left(x\right) = 0$ or $f ' \left(x\right)$ does not exist.

$f ' \left(x\right) = 0$ at $x = 1$ and $f ' \left(x\right)$ fails to exist at $x = 0$.
Both are in the domain of $f$, so both are critical numbers for $f$.

on the interval $\left(- \infty , 0\right)$, $f ' \left(x\right)$ is positive and on $\left(0 , 1\right)$ it is negative, so $f \left(0\right) = 2$ is a relative maximum.

on $\left(1 , \infty\right)$, $f ' \left(x\right)$ is again positive, so $f \left(1\right) = 1$ is a relative minimum.

Here is the graph of $f$:
graph{2x-3x^(2/3)+2 [-5.365, 5.735, -2.02, 3.53]}