Answer
$f(x)$ has an absolute maximum value of $12$ at $x=-2$ and an absolute minimum value of $3$ at $x=1$ on the defined domain.
Work Step by Step
$$f(x)=4-x^3\hspace{1cm}-2\le x\le1$$
1) Find the derivative $f'(x)$: $$f'(x)=-3x^2$$
Here $f'(x)=0$ only 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 $x=-2$: $$f(-2)=4-(-2)^3=4-(-8)=12$$
- For $x=0$: $$f(0)=4-0^3=4-0=4$$
- For $x=1$: $$f(1)=4-1^3=4-1=3$$
This means the function $f(x)$ has an absolute maximum value of $12$ at $x=-2$ and an absolute minimum value of $3$ at $x=1$ on the defined domain.
