Answer
See explanations.
Work Step by Step
Step 1. Given the first derivative $y'=x^{-4/5}(x+1)=x^{1/5}+x^{-4/5}$, we have $y''=\frac{1}{5}x^{-4/5}-\frac{4}{5}x^{-9/5}=\frac{1}{5}x^{-9/5}(x-4)$
Step 2. Possible critical points are at $x=-1,0$. Determine the increasing ($y'\gt0$) and decreasing ($y'\lt0$) regions by signs of $y'$: $..(-)..(-1)..(+)..(0)..(+)..$; thus the function decreases on $(-\infty,-1)$ and increases on $(-1,0), (0,\infty)$. A local and absolute minimum can be found at $x=-1$. There is no local or absolute maximum.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $x=0,4$: $..(+)..(0)..(-)..(4)..(+)..$. Thus, the function is concave up on $(-\infty,0)$,$(4,\infty)$ and concave down on $(0,4)$. Thus $x=0$ and $x=4$ are inflection points.
Step 4. Sketch the function based on the above results as shown.
