Answer
a. $g(0)=0$ local minimum, $g(1)=1/3$ local maximum.
b. $g(0)=0$ absolute minimum, no absolute maximum.
c. See graph.
Work Step by Step
a. Step 1. Given the function $g(x)=\frac{x^2}{4-x^2}, -2\lt x\leq 1$, we have $g'(x)=\frac{2x(4-x^2)-x^2(-2x)}{(4-x^2)^2}=\frac{8x-x^2}{(4-x^2)^2}$.
Step 2. The extrema happen when $g'(x)=0$, undefined, or at endpoints; thus the critical points are $x=0, 1$. We excluded some points due to the domain of $x$.
Step 3. We have $g(0)=0, g(1)=1/3$. Test signs of $g'(x)$ across critical points $ (-)..(0)..(+)..(1)$; thus we have $g(0)=0$ as a local minimum, $g(1)=1/3$ as a local maximum.
b. We can identify that $g(0)=0$ is an absolute minimum and there is no absolute maximum because $x\to-2, g(x)\to \infty$
c. See graph.
