Answer
See graph and explanations.
Work Step by Step
(a) Step 1. Given $y'=x^2(6-4x)=6x^2-4x^3$, we have $y''=12x-12x^2$
Step 2. Letting $y'=0$, we have $x^2(6-4x)=0$, which gives $x=0,\frac{3}{2}$ as possible critical points.
Step 3. Check the signs of $y'$ to get $..(+)..(0)..(+)..(\frac{3}{2})..(-)..$; thus the function increases on $(-\infty,0)$, $(0,\frac{3}{2})$ and decreases on $(\frac{3}{2},\infty)$ . This means that only $x=\frac{3}{2}$ gives a local maximum.
Step 4. Letting $y''=0$, we have $x=0,1$. Check the signs of $y''$ to get $..(-)..(0)..(+)..(1)..(-)..$, indicating $x=0,1$ are inflection points with the function concave down for $x\lt0, x\gt1$, and concave up for $0\lt x\lt 1$.
(b) Graph the function based on the above information as shown in the figure.
