Answer
See graph and explanations.
Work Step by Step
Step 1. Given the first derivative $y'=(x+1)^{-2/3}=\frac{1}{\sqrt[3] {(x+1)^2}}$, we have $y''=-\frac{2}{3}(x+1)^{-5/3}$
Step 2. The possible critical point is at $x=-1$. Determine the increasing $y'\gt0$ and decreasing $y'\lt0$ regions by using the signs of $y'$: $..(+)..(-1)..(+)..$; thus the function increases on the entire domain and there will be no extrema.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $x=-1$: $..(+)..(-1)..(-)..$. Thus, the function is concave up on $(-\infty,-1)$ and concave down on $(-1,\infty)$ with $x=-1$ as the inflection point.
Step 4. Sketch the function based on the above results as shown.
