Answer
$y(2)=2$ is a local maximum and $y(-2)=-2$ is a local minimum;
absolute minimum is at $y(-2)=-2$ and the absolute maximum is at $y(2)=2$;
concave up on $(-2\sqrt 3,0), (2\sqrt 3,\infty)$,
concave down on $(-\infty,-2\sqrt 3),(0,2\sqrt 3)$;
$x=0,\pm2\sqrt 3$ are inflection points.
See graph.
Work Step by Step
Step 1. Given the function $y=\frac{8x}{x^2+4}$, we have $y'=\frac{8(x^2+4)-8x(2x)}{(x^2+4)^2}=\frac{8(4-x^2)}{(x^2+4)^2}$ and $y''=\frac{8(-2x)(x^2+4)^2-16(4-x^2)(x^2+4)(2x)}{(x^2+4)^4}=\frac{-16x(x^2+4+8-2x^2)}{(x^2+4)^3}=\frac{16x(x^2-12)}{(x^2+4)^3}$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints. We have $x=\pm2$ as critical points with $y(-2)=-2, y(2)=2$
Step 3. Examine the sign change of $y'$ across the critical points: $..(-)..(-2)..(+)..(2)..(-)..$; we can identify $ y(2)=2$ as a local maximum and $y(-2)=-2$ as a local minimum. The absolute minimum is at $y(-2)=-2$ and the absolute maximum is at $y(2)=2$.
Step 4. The inflection points can be found when $y''=0$ or when it does not exist. We have $x=0,\pm2\sqrt 3$ 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 $..(-)..(-2\sqrt 3)..(+)..(0)..(-)..(2\sqrt 3)..(+)..$,
Step 6. The function is concave up on $(-2\sqrt 3,0), (2\sqrt 3,\infty)$ and concave down on $(-\infty,-2\sqrt 3),(0,2\sqrt 3)$, this means that $x=0,\pm2\sqrt 3$ are inflection points.
Step 7. See graph.
