Answer
The absolute maximum $g(0)=2$ and the absolute minimum as $g(-2)=0$.
See graph.
Work Step by Step
Step 1. Given the function $g(x)=\sqrt {4-x^2}=(4-x^2)^{1/2}, -2\leq x\leq 1$, we have $g(-2)=0, g(1)=\sqrt 3$,
Step 2. $g'(x)=\frac{1}{2}(4-x^2)^{-1/2}(-2x)=-x(4-x^2)^{-1/2}$, let $g'(x)=0$, we get a critical point at $x=0$, $g(0)=2$
Step 3. We can identify the absolute maximum as $g(0)=2$ and the absolute minimum as $g(-2)=0$ on the given interval.
Step 4. See graph.