Answer
See graph and explanations.
Work Step by Step
(a) Step 1. Given $y'=x^4-2x^2$, we have $y''=4x^3-4x$
Step 2. Letting $y'=0$, we have $x^4-2x^2=0$, which gives $x=0,\pm\sqrt 2$ as possible critical points.
Step 3. Check the signs of $y'$ to get $..(+)..(-\sqrt 2)..(-)..(0)..(-)..(\sqrt 2)..(+)..$; thus the function increases on $(-\infty,-\sqrt 2)$, $\sqrt 2,\infty)$, and decreases on $(-\sqrt 2,\sqrt 2)$ . This means that $x=-\sqrt 2$ gives a local maximum and $x=\sqrt 2$ gives a local minimum.
Step 4. Letting $y''=0$, we have $x=0,\pm1$. Check the signs of $y''$ to get $..(-)..(-1)..(+)..(0)..(-)..(1)..(+)..$, indicating $x=0,\pm1$ are inflection points with the function concave down for $x\lt-1,0\lt x\lt1$, and concave up for $-1\lt x \lt0, x\gt1$.
(b) Graph the function based on the above information as shown in the figure.
