Answer
The maximum is $f(3)=21$, the minimum is $f(-1)=-11$
Work Step by Step
Given $$f(x)=x(10-x), \quad[-1,3]$$
Since
$$ f'(x)= 10-2x$$
Then $f(x)$ has a critical point at $x= 5$. However, we eliminate it because $5\notin [-1,3] $. Check the endpoints:
\begin{align*}
f(-1)&= -11\\
f(3)&= 21
\end{align*}
Then the maximum is $f(3)=21$ and the minimum is $f(-1)=-11$.
