Answer
$f(x)$ has an absolute maximum value of $3$ at $x=2$ and an absolute minimum value of $-1$ at $x=0$ on the defined domain.
Work Step by Step
$$f(x)=x^2-1\hspace{1cm}-1\le x\le2$$
1) Find the derivative $f'(x)$: $$f'(x)=2x$$
Here $f'(x)=0$ when $x=0$. So $x=0$ is the only critical point of $f(x)$.
2) Evaluate $f(x)$ at the critical points and endpoints:
- For $x=-1$: $$f(-1)=(-1)^2-1=1-1=0$$
- For $x=0$: $$f(0)=0^2-1=0-1=-1$$
- For $x=2$: $$f(2)=2^2-1=4-1=3$$
This means the function $f(x)$ has an absolute maximum value of $3$ at $x=2$ and an absolute minimum value of $-1$ at $x=0$ on the defined domain.