Answer
a. $k(0)=1$ maximum.
b. $k(0)=1$ absolute maximum, no absolute minimum.
c. See graph.
Work Step by Step
a. Step 1. Given the function $k(x))=x^3+3x^2+3x+1, -\infty\lt x\leq0$, we have $k'(x)=3x^2+6x+3=3(x+1)^2$.
Step 2. The local extrema happen when $f'(x)=0$ or undefined or at endpoints; thus the critical points are $x=-1$, and an endpoint is at $x=0$.
Step 3. We have $k(-1)=0, k(0)=1$. Test signs of $f'(x)$ across critical points $(+).. (-1)..(+)..(0)$; thus we have $k(0)$ as a maximum, and $k(-1)$ is not an extreme (k' has no sign change across the point).
b. We can identify that at $x=0, k(0)=1$ is an absolute maximum and there is no absolute minimum because $x\to-\infty, k(x)\to -\infty$
c. See graph.
