Answer
See explanations.
Work Step by Step
Step 1. Given the first derivative $y'=(x-2)^{-1/3}$, we have $y''=-\frac{1}{3}(x-2)^{-4/3}$
Step 2. One possible critical point is at $x=-2$. Determine the increasing ($y'\gt0$) and decreasing ($y'\lt0$) regions by signs of $y'$: $..(-)..(2)..(+)..$; thus the function decreases on $(-\infty,2)$ and increases on $(2,\infty)$. A local and absolute minimum can be found as $f(2)=0$. There is no local or absolute maximum.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $x=-2$: $..(-)..(2)..(-)..$. Thus, the function is concave down on $(-\infty,2)$ and $(2,\infty)$, and $x=2$ is not an inflection point.
Step 4. Sketch the function based on the above results as shown.
