Answer
$f(0)=0$ is a local minimum and $f(1)=3/2$ is a local maximum; there are no absolute extrema.
inflection point is at $x=-1/2$.
See graph.
Work Step by Step
Step 1. Given the function $y=x^{2/3}(\frac{5}{2}-x)$, we have $y'=\frac{2}{3}x^{-1/3}(\frac{5}{2}-x)-x^{2/3}=\frac{5-2x}{3\sqrt[3] x}-\frac{x}{\sqrt[3] x}=\frac{5(1-x)}{3\sqrt[3] x}=\frac{5}{3}(1-x)x^{-1/3} $ and $y''=-\frac{5}{3}x^{-1/3}-\frac{5}{9}(1-x)x^{-4/3}=-\frac{5}{3\sqrt[3] x}-\frac{5(1-x)}{9\sqrt[3] {x^4}}=-\frac{15x+5-5x}{9\sqrt[3] {x^4}}=-\frac{5(2x+1)}{9\sqrt[3] {x^4}} $
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0,1$ as critical points with $f(0)=0, f(1)=3/2$
Step 3. Examine the sign change of $y'$ across the critical points: $..(-)..(0)..(+)..(1)..(-)$; we can identify $f(0)=0$ as a local minimum and $f(1)=3/2$ as a local maximum. The function does not have absolute extrema as the end behaviors are $x\to\pm\infty, y\to\mp\infty$.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $x=-1/2, 0$ as possible inflection points.
Step 5. To identify the intervals on which the functions are concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $..(+)..(-1/2)..(-)..(0)..(-)..$,
Step 6. The function is concave up on $(-\infty,-1/2)$ and concave down on $(-1/2,0), (0,\infty)$; this also means that $x=0$ is not an inflection point (no concavity change) and the inflection point is at $x=-1/2$.
Step 7. See graph.
