Answer
The maximum is $f(1/2) =(1/4)^{1/3}$ and the minimum is $f( 2 ) =f(-1)=-2^{1/3}$
Work Step by Step
Given $$y=\left(t-t^{2}\right)^{1 / 3}, \quad[-1,2]$$
Since
\begin{align*}
\frac{dy}{dt}&=\frac{1-2t}{3\left(t-t^2\right)^{\frac{2}{3}}}
\end{align*}
Find the critical points
\begin{align*}
f'(t)&=0\\
\frac{1-2t}{3\left(t-t^2\right)^{\frac{2}{3}}}&=0\\
1-2t&=0
\end{align*}
We get the critical point $t=1/2$ because $1/2\in [-1,2]$. Then, we note that the derivative does not exist at $t=0,\ \ t= 1$
\begin{align*}
f(-1)&=-2^{1/3}\\
f(1/2 )&=(1/4)^{1/3}\\
f(2)&=-2^{1/3}\\
f(0)&=0\\
f(1)&=0
\end{align*}
Then the maximum is $f(1/2) =(1/4)^{1/3}$ and the minimum is $f( 2 ) =f(-1)=-2^{1/3}$
