Answer
absolute maximum $f(-2)=12$ and minimum as $f(1)=3$. See graph.
Work Step by Step
Step 1. Given the function $f(x)=4-x^3, -2\leq x\leq 1$, we have $f(-2)=4-(-2)^3=12, f(1)=4-1=3$, also let $f'(x)=-3x^2=0$ to get $x=0, f(0)=4$, and we can identify the absolute maximum as $f(-2)=12$ and the absolute minimum as $f(1)=3$ on the given interval.
Step 2. See graph.