Answer
The absolute maximum $g(-\sqrt 5)=0$ and the absolute minimum $g(0)=-\sqrt 5$.
See graph.
Work Step by Step
Step 1. Given the function $g(x)=-\sqrt {5-x^2}=-(5-x^2)^{1/2}, -\sqrt 5\leq x\leq 0$, we have $g(-\sqrt 5)=0, g(0)=-\sqrt 5$,
Step 2. $g'(x)=-\frac{1}{2}(5-x^2)^{-1/2}(-2x)=x(5-x^2)^{-1/2}$, let $g'(x)=0$, we get a critical point at $x=0$ and $g(0)=-\sqrt 5$
Step 3. We can identify the absolute maximum as $g(-\sqrt 5)=0$ and the absolute minimum as $g(0)=-\sqrt 5$ on the given interval.
Step 4. See graph.
