Question

Determine where the global extrema of  f(x)=3x^2/3−2x

on the interval [−1,1] occur.

Answer 1

$$y=3x^{2/3}-2x, \qquad [-1,1]\\ \begin{aligned} y'(x) = &2x^{-1/3}-2\\ =& \frac{2}{\sqrt[3]{x}}-2 \end{aligned}$$

Since $y'(x)$ exists for all $x={0}$ , the critical numbers of occur when $y'(x)=0$ and $x=0$

$$\frac{2}{\sqrt[3]{x}}-2=0 \rightarrow \frac{2}{\sqrt[3]{x}}=2\rightarrow \sqrt[3]{x}=1 \rightarrow x=1\\ y(1)=1\\ y(0)=0$$

The values at the endpoints of the interval are:

$$y(-1)=5$$

Comparing these three numbers, we see that the absolute maximum value is $y(-1)=5$ and the absolute minimum value is $y(0)=0$