Answer
a. $f(-3)$ local maximum, $f(-2)$ local minimum, $f(2)$ maximum.
b. $f(2)=16$ absolute maximum, no absolute minimum.
c. See graph.
Work Step by Step
a. Step 1. Given the function $f(t)=12t-t^3, -3\leq t\lt\infty$, we have $f'(t)=12-3t^2=3(4-t^2)=3(2+t)(2-t)$.
Step 2. The local extrema happen when $f'(x)=0$ or undefined and at endpoints; thus the critical points are $x=\pm2$, and an endpoint is at $x=-3$.
Step 3. We have $f(-3)=-9, f(-2)=-16, f(2)=16$. Test signs of $f'(x)$ across critical points $(-3).. (-)..(-2)..(+)..(2)..(-)..$, thus we have $f(-3)$ as a local maximum, $f(-2)$ as a local minimum, and $f(2)$ as a maximum.
b. We can identify that at $x=2, f(2)=16$ is an absolute maximum and there is no absolute minimum because $t\to\infty, f(t)\to -\infty$
c. See graph.
