Answer
See graph and explanations.
Work Step by Step
(a) Step 1. Given $y'=4x^2-x^4$, we have $y''=8x-4x^3$
Step 2. Letting $y'=0$, we have $4x^2-x^4=0$, which gives $x=0,\pm2$ as possible critical points.
Step 3. Check the signs of $y'$ to get $..(-)..(-2)..(+)..(0)..(+)..(2)..(-)..$; thus the function decreases on $(-\infty,-2)$, $(2,\infty)$, and decreases on $(-2,2)$. This means that $x=-2$ gives a local minimum and $x=2$ gives a local maximum.
Step 4. Letting $y''=0$, we have $x=0,\pm\sqrt 2$. Check the signs of $y''$ to get $..(+)..(-\sqrt 2)..(-)..(0)..(+)..(\sqrt 2)..(-)..$, indicating $x=0,\pm\sqrt 2$ are inflection points with the function concave up for $x\lt-\sqrt 2, 0\lt x\lt\sqrt 2$, and concave down for $-\sqrt 2\lt x \lt0, x\gt\sqrt 2$.
(b) Graph the function based on the above information as shown in the figure.
