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