## Calculus (3rd Edition)

The maximum is $f(4/9) \approx0.1481$; the minimum is $f(2) \approx-0.828$
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 $f(2) \approx-0.828$.