Answer
$y(0)=1$ is a local maximum, no local minima;
absolute maximum $y(0)=1$, no absolute minimum;
concave up on $(-\infty,\sqrt[4] 3),(\sqrt[4] 3,\infty)$,
concave down on $(-\sqrt[4] 3,0),(0,\sqrt[4] 3)$;
$x=\pm\sqrt[4] 3$ are inflection points.
See graph.
Work Step by Step
Step 1. Given the function $y=\frac{5}{x^4+5}$, we have $y'=\frac{-5(4x^3)}{(x^4+5)^2}=\frac{-20x^3}{(x^4+5)^2}$ and $y''=\frac{-60x^2(x^4+5)^2+40x^3(x^4+5)(4x^3)}{(x^4+5)^4}=\frac{-60x^6-300x^2+160x^6}{(x^4+5)^3}=\frac{100x^2(x^4-3)}{(x^4+5)^3}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=0$ as critical points with $y(0)=1$
Step 3. Examine the sign change of $y'$ across the critical points: $..(+)..(0)..(-)..$; we can identify $y(0)=1$ as a local maximum and there is no local minima. The absolute maximum is also at $y(0)=1$ and there is no absolute minimum.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $x=0, \pm\sqrt[4] 3$ 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 $..(+)..(-\sqrt[4] 3)..(-)..(0)..(-)..(\sqrt[4] 3)..(+)..$.
Step 6. The function is concave up on $(-\infty,\sqrt[4] 3),(\sqrt[4] 3,\infty)$ and concave down on $(-\sqrt[4] 3,0),(0,\sqrt[4] 3)$, this means that $x=0$ is not an inflection point, while $x=\pm\sqrt[4] 3$ are inflection points.
Step 7. See graph.
