Answer
See graph and explanations.
Work Step by Step
(a) Step 1. Given $y'=16-x^2$, we have $y''=-2x$
Step 2. Letting $y'=0$, we have $16-x^2=0$, which gives $x=\pm4$ as possible critical points.
Step 3. Check signs of $y'$ to get $..(-)..(-4)..(+)..(4)..(-)..$; thus the function decreases on $(-\infty,-4)$, $(4,\infty)$ and increases on $(-4,4)$ . This means that $x=-4$ gives a local minimum and $x=4$ gives a local maximum.
Step 4. Letting $y''=0$, we have $x=0$. Check the signs of $y''$ to get $..(+)..(0)..(-)..$, indicating $x=0$ is an inflection point with the function concave up for $x\lt0$ and concave down for $x\gt0$.
(b) Graph the function based on the above information as shown in the figure.
