Answer
inflection points $(-\pi,0), (\pi,0)$
$y(\pm\frac{\pi}{2})=1$ and $y(\pm2\pi)=0$ are local maxima,
$y(0)=0$ and $y(\pm\frac{3\pi}{2})=-1$ are local minima.
Concave up on $(-2\pi,-\pi),(\pi,2\pi)$, concave down on $(-\pi,0),(0,\pi)$
Work Step by Step
Step 1. Given the function $y=sin|x|, -2\pi\leq x \leq2\pi$, we have $y'=\begin{cases}cos(x)\ 0\leq x\leq2\pi\\-cos(x)\ -2\pi\leq x\lt0\end{cases}$ and $y''=\begin{cases}-sin(x)\ 0\leq x\leq2\pi\\sin(x)\ -2\pi\leq x\lt0\end{cases}$.
Step 2. The inflection points can be found when $y''=0$ or it does not exist. We have $sin(x)=0, x=0,\pm\pi,\pm2\pi$. The inflection requires a change of concavity across the point and we should exclude endpoints where only half of the curve is available; thus we have $x=\pm\pi$ with coordinates $(-\pi,0), (\pi,0)$
Step 3. The extrema happen when $y'=0$, undefined, or at endpoints; thus we have $cos(x)=0$ and the critical points within the domain are $x=0, \pm\pi/2,\pm3\pi/2,\pm2\pi$.
Step 4. We have $y(0)=0, y(\pm\frac{\pi}{2})=1$,$y(\pm\frac{3\pi}{2})=-1,y(\pm2\pi)=0$.
Step 5. Test signs of $y'$ across critical points $(-2\pi)..(-)..(-\frac{3\pi}{2})..(+)..(-\frac{\pi}{2})..(-)..(0)..(+)..(\frac{\pi}{2})..(-)..(\frac{3\pi}{2})..(+)..(2\pi)$; thus we have $y(\pm\frac{\pi}{2})=1$ and $y(\pm2\pi)=0$ as local maxima, $y(0)=0$ and $y(\pm\frac{3\pi}{2})=-1$ as local minima.
Step 6. We can identify as $y(\pm\frac{\pi}{2})=1$ as an absolute maximum and $y(\pm\frac{3\pi}{2})=-1$ as an absolute minimum in the interval.
Step 7. To identify the intervals on which the function is concave up and concave down, we need to examine the sign of $y''$ on different intervals. We have $(-2\pi)..(+)..(-\pi)..(-)..(0)..(-)..(\pi)..(+)..(2\pi)$,
Step 8. The function is concave up on $(-2\pi,-\pi),(\pi,2\pi)$ and concave down on $(-\pi,0),(0,\pi)$
