Answer
$f(0)=0$ is a local minimum and $f(1)=3$ is a local maximum. There are no absolute extrema.
The graph is concave down on $(-\infty,0)$ and on $(0,\infty)$; no inflection point.
See graph.
Work Step by Step
Step 1. Given the function $y=-2x+5x^{2/5}$, we have $y'=-2+2x^{-3/5}$ and $y''=-\frac{6}{5}x^{-8/5}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0,1$ as critical points with $f(0)=0, f(1)=3$
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 $f(1)=3$ 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 $x=0$ and a possible inflection point is $(0,0)$
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 $..(-)..(0)..(-)..$,
Step 6. The function is concave down on $(-\infty,0)$ and on $(0,\infty)$, this also means that $(0,0)$ is not an inflection point (no concavity change).
Step 7. See graph.
