Answer
Absolute maximum $f(-0.5)=4$, absolute minimum as $f(2)=1/4$. See graph.
Work Step by Step
Step 1. Given the function $f(x)=-\frac{1}{x^2}, -0.5\leq x\leq 2$, we have $f(-0.5)=4, f(2)=1/4$,
Step 2. $f'(x)=-\frac{2}{x^3}\ne0$; no extrema here.
Step 3. We can identify the absolute maximum as $f(-0.5)=4$ and the absolute minimum as $f(2)=1/4$ on the given interval.
Step 4. See graph.
