Answer
a. $f(0)=0$ maximum, $f(2)=-4$ local minimum, and $f(3)=0$ maximum.
b. $f(0)=f(3)=0$ absolute maximum, no absolute minimum.
c. See graph.
Work Step by Step
a. Step 1. Given the function $f(t)=t^3-3t^2, -\infty\lt t\leq3$, we have $f'(t)=3t^2-6t=3t(t-2)$.
Step 2. The local extrema happen when $f'(x)=0$, undefined or at endpoints; thus the critical points are $x=0, 2$, and an endpoint is at $x=3$.
Step 3. We have $f(0)=0, f(2)=-4, f(3)=0$. Test signs of $f'(x)$ across critical points $(+)..(0).. (-)..(2)..(+)..(3)$; thus we have $f(0)$ as a local maximum, $f(2)$ as a local minimum, and $f(3)$ as a local maximum.
b. We can identify that at $x=0,3, f(0)=f(3)=0$ is an absolute maximum and there is no absolute minimum because $t\to -\infty, f(t)\to -\infty$
c. See graph.
