Answer
$y(1)=3$ is a local minimum and there is no local maximum or absolute extrema;
concave up on $(-\infty,-\sqrt[3] 2)$ and $(0,\infty)$, concave down on $(-\sqrt[3] 2,0)$;
$x=-\sqrt[3] 2$ is an inflection point.
See graph.
Work Step by Step
Step 1. Given the function $y=x^2+\frac{2}{x}, x\ne0$, we have $y'=2x-\frac{2}{x^2}$ and $y''=2+\frac{4}{x^3}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=1$ as critical points with $y(1)=3$ (we exclude $x=0$ as it is an asymptote).
Step 3. Examine the sign change of $y'$ across the critical points: $(0)..(-)..(1)..(+)..$; we can identify $y(1)=3$ as a local minimum and there is no local maximum or absolute extrema because as $x\to\pm0,y\to\pm\infty$.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $x^3=-2, x=-\sqrt[3] 2$ as possible inflection points within the domain.
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 $..(+)..(-\sqrt[3] 2)..(-)..(0)..(+)..$,
Step 6. The function is concave up on $(-\infty,-\sqrt[3] 2)$, $(0,\infty)$, concave down on $(-\sqrt[3] 2,0)$, and $x=-\sqrt[3] 2$ is an inflection point.
Step 7. See graph.
