Answer
Inflection points $(-\frac{\pi}{2},-\frac{\pi}{2}),(0,0),(\frac{\pi}{2},\frac{\pi}{2})$
$y(-\frac{2\pi}{3})=-\frac{2\pi}{3}+\frac{\sqrt 3}{2}$ and $y(\frac{\pi}{3})=\frac{\pi}{3}+\frac{\sqrt 3}{2}$ are local maxima,
$y(-\frac{\pi}{3})=-\frac{\pi}{3}-\frac{\sqrt 3}{2}$ and $y(\frac{2\pi}{3})=\frac{2\pi}{3}-\frac{\sqrt 3}{2}$ are local minima.
Concave up on $(-\frac{\pi}{2},0), (\frac{\pi}{2},\frac{2\pi}{3})$ and concave down on $(-\frac{2\pi}{3},-\frac{\pi}{2}),(0,\frac{\pi}{2})$
Work Step by Step
Step 1. Given the function $y=x+sin2x, -\frac{2\pi}{3}\leq x \leq\frac{2\pi}{3}$, we have $y'=1+2cos2x$ and $y''=-4sin2x$
Step 2. The inflection points can be found when $y''=0$ or it does not exist. We have $sin2x=0, x=0,\pm\frac{\pi}{2}$ which gives coordinates $(-\frac{\pi}{2},-\frac{\pi}{2}),(0,0),(\frac{\pi}{2},\frac{\pi}{2})$
Step 3. The extrema happen when $y'=0$, undefined, or at endpoints; thus we have $cos2x=-1/2$ and the critical points within the domain are $x=-\frac{2\pi}{3}, -\frac{\pi}{3},\frac{\pi}{3}, \frac{2\pi}{3}$.
Step 4. We have $y(-\frac{2\pi}{3})=-\frac{2\pi}{3}+\frac{\sqrt 3}{2}$, $y(-\frac{\pi}{3})=-\frac{\pi}{3}-\frac{\sqrt 3}{2}$, $y(\frac{\pi}{3})=\frac{\pi}{3}+\frac{\sqrt 3}{2}$, $y(\frac{2\pi}{3})=\frac{2\pi}{3}-\frac{\sqrt 3}{2}$.
Step 5. Test signs of $y'$ across critical points $(-\frac{2\pi}{3})..(-)..(-\frac{\pi}{3})..(+)..(\frac{\pi}{3})..(-)..(\frac{2\pi}{3})$; thus we have $y(-\frac{2\pi}{3})=-\frac{2\pi}{3}+\frac{\sqrt 3}{2}$ and $y(\frac{\pi}{3})=\frac{\pi}{3}+\frac{\sqrt 3}{2}$ as local maxima, $y(-\frac{\pi}{3})=-\frac{\pi}{3}-\frac{\sqrt 3}{2}$ and $y(\frac{2\pi}{3})=\frac{2\pi}{3}-\frac{\sqrt 3}{2}$ as local minima.
Step 6. We can identify $y(\frac{\pi}{3})=\frac{\pi}{3}+\frac{\sqrt 3}{2}$ as an absolute maximum and $y(-\frac{\pi}{3})=-\frac{\pi}{3}-\frac{\sqrt 3}{2}$ 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 $(-\frac{2\pi}{3})..(-)..(-\frac{\pi}{2})..(+)..(0)..(-)..(\frac{\pi}{2})..(+)..(\frac{2\pi}{3})$,
Step 8. The function is concave up on $(-\frac{\pi}{2},0), (\frac{\pi}{2},\frac{2\pi}{3})$ and concave down on $(-\frac{2\pi}{3},-\frac{\pi}{2}),(0,\frac{\pi}{2})$
