## Calculus (3rd Edition)

The maximum is $f(-1)=3$; the minimum is $f(1) = -1$
Given $$f(x)=x^{2 / 3}-2 x^{1 / 3}, \quad[-1,3]$$ Since $$f'(x)= \frac{2}{3x^{\frac{1}{3}}}-\frac{2}{3x^{\frac{2}{3}}}$$ We find the critical points \begin{align*} f'(x)&=0\\ \frac{2x^{\frac{1}{3}}-2}{3x^{\frac{2}{3}}}&=0\\ 2x^{\frac{1}{3}}-2&=0 \end{align*} Then $f(x)$ has critical points at $x= 1,\ x=0$. We check the values at the critical points and end points: \begin{align*} f(-1)&= 3 \\ f(0)&=0 \\ f(1)&= -1 \\ f(3)&\approx-0.804 \end{align*} Then the maximum is $f(-1)=3$ and the minimum is $f(1) = -1$