Answer
$g(x)$ has the absolute maximum value of $2$ at $x=0$ and the absolute minimum value of $0$ at $x=-2$ in the defined domain.
Work Step by Step
$$g(x)=\sqrt{4-x^2}=(4-x^2)^{1/2}\hspace{1cm}-2\le x\le1$$
1) Find the derivative $g'(x)$: $$g'(x)=\frac{1}{2}(4-x^2)^{-1/2}(4-x^2)'$$ $$g'(x)=\frac{1}{2}(4-x^2)^{-1/2}(-2x)$$ $$g'(x)=-\frac{2x}{2\sqrt{4-x^2}}=-\frac{x}{\sqrt{4-x^2}}$$
- At $x=0$, we have $g'(x)=0$
- Also, $g'(x)$ is undefined when $\sqrt{4-x^2}=0$ or $x=\pm2$.
All these values of $x=0$ and $x=\pm2$ are critical points of $g(x)$. However, we would leave out $x=2$ here since this point does not lie in the defined interval.
2) Evaluate $g'(x)$ at the critical points and endpoints:
- For $x=-2$: $$g(-2)=\sqrt{4-(-2)^2}=\sqrt{4-4}=0$$
- For $x=0$: $$g(0)=\sqrt{4-0^2}=\sqrt{4}=2$$
- For $x=1$: $$g(8)=\sqrt{4-1^2}=\sqrt{3}$$
This means $g(x)$ has the absolute maximum value of $2$ at $x=0$ and the absolute minimum value of $0$ at $x=-2$ in the defined domain.