Answer
The maximum is $f(4/9) \approx0.1481$; the minimum is $f(2) \approx-0.828 $
Work Step by Step
Given $$f(x)=x-x^{3 / 2}, \quad[0,2]$$
Then
$$ f'(x)= 1-\frac{3}{2}x^{1/ 2}$$
We find the critical points
\begin{align*}
f'(x)&=0\\
1-\frac{3}{2}x^{1/ 2}&=0
\end{align*}
Then $f(x)$ has a critical point at $ x=4/9$.
We test the values at the critical point and end points:
\begin{align*}
f(0)&=0 \\
f(4/9)&\approx0.1481 \\
f(2)&\approx-0.828
\end{align*}
Then the maximum is $f(4/9) \approx0.1481$ and the minimum is $f(2) \approx-0.828$.
