# How do you find the relative extrema for f(x) =2x- 3x^(2/3) +2 on the interval [-1,3]?

May 25, 2015

Find and test the critical numbers for $f$.

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

Critical numbers for $f$ are values for $x$ that are:

1. In the domain of $f$, and

2. at which $f ' \left(x\right) = 0$ or $f ' \left(x\right)$ does not exist.

(This is a wonderful example of why you cannot ignore the second kind of critical numbers.)

For this function, we have:

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

$f ' \left(x\right) = 0$ $\textcolor{w h i t e}{\text{sssssssss}}$ $f ' \left(x\right)$ does not exist

$x = 1$ $\textcolor{w h i t e}{\text{ssssssssssssss}}$ $x = 0$

Using the first derivative we find that $f$ is

increasing on $\left(- \infty , 0\right)$ (Yes, I know the domain has been restricted, but it is not necessary.)

decreasing on (0,1) so

$f \left(0\right) = 2$ is a relative maximum.

$f$ is increasing on (1,oo), so

$f \left(1\right) = 1$ is a relative minimum.