Answer
We have:
Inflection points $(-\frac{\pi}{2},\frac{\sqrt 2\pi}{2}),(\frac{\pi}{2},-\frac{\sqrt 2\pi}{2})$
$y(-\pi)=\sqrt 2\pi-2$, $y(-\frac{\pi}{4})=\sqrt 2+\frac{\sqrt 2\pi}{4}$, $y(\frac{3\pi}{2})=-\frac{3\sqrt 2\pi}{2}$ as local maxima,
$y(-\frac{3\pi}{4})=\frac{3\sqrt 2\pi}{4}-\sqrt 2$, $y(\frac{5\pi}{4})=-\sqrt 2-\frac{5\sqrt 2\pi}{4}$ as local minima.
$y(-\frac{\pi}{4})=\sqrt 2+\frac{\sqrt 2\pi}{4}$ as an absolute maximum
$y(\frac{5\pi}{4})=-\sqrt 2-\frac{5\sqrt 2\pi}{4}$ as an absolute minimum in the interval.
Concave up on $(-\pi, -\frac{\pi}{2}), (\frac{\pi}{2},\frac{3\pi}{2})$ and concave down on $(-\frac{\pi}{2},\frac{\pi}{2})$.
Work Step by Step
Step 1. Given the function $y=2cos(x)-\sqrt 2x, -\pi\leq x \leq\frac{3\pi}{2}$, we have $y'=-2sin(x)-\sqrt 2$ and $y''=-2cos(x)$
Step 2. The inflection points can be found when $y''=0$ (excluding end points) or it does not exist. We have $cos(x)=0, x=\pm\frac{\pi}{2}$ which gives coordinates $(-\frac{\pi}{2},\frac{\sqrt 2\pi}{2}),(\frac{\pi}{2},-\frac{\sqrt 2\pi}{2})$
Step 3. The extrema happen when $y'=0$, undefined, or at endpoints; thus we have $sin(x)=-\sqrt 2/2$ and the critical points within the domain are $x=-\pi, -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{5\pi}{4}, \frac{3\pi}{2}$.
Step 4. We have $y(-\pi)=\sqrt 2\pi-2$, $y(-\frac{3\pi}{4})=\frac{3\sqrt 2\pi}{4}-\sqrt 2$, $y(-\frac{\pi}{4})=\sqrt 2+\frac{\sqrt 2\pi}{4}$, $y(\frac{5\pi}{4})=-\sqrt 2-\frac{5\sqrt 2\pi}{4}$, $y(\frac{3\pi}{2})=-\frac{3\sqrt 2\pi}{2}$.
Step 5. Test signs of $y'$ across critical points $(-\pi)..(-)..(-\frac{3\pi}{4})..(+)..(-\frac{\pi}{4})..(-)..(\frac{5\pi}{4})..(+)..(\frac{3\pi}{2})$; thus we have $y(-\pi)=\sqrt 2\pi-2$, $y(-\frac{\pi}{4})=\sqrt 2+\frac{\sqrt 2\pi}{4}$, $y(\frac{3\pi}{2})=-\frac{3\sqrt 2\pi}{2}$ as local maxima, $y(-\frac{3\pi}{4})=\frac{3\sqrt 2\pi}{4}-\sqrt 2$, $y(\frac{5\pi}{4})=-\sqrt 2-\frac{5\sqrt 2\pi}{4}$ as local minima.
Step 6. We can identify $y(-\frac{\pi}{4})=\sqrt 2+\frac{\sqrt 2\pi}{4}$ as an absolute maximum and $y(\frac{5\pi}{4})=-\sqrt 2-\frac{5\sqrt 2\pi}{4}$ as an absolute minimum in the interval.
Step 7. 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 $(-\pi)..(+)..(-\frac{\pi}{2})..(-)..(\frac{\pi}{2})..(+)..(\frac{3\pi}{2})$,
Step 8. The function is concave up on $(-\pi, -\frac{\pi}{2}), (\frac{\pi}{2},\frac{3\pi}{2})$ and concave down on $(-\frac{\pi}{2},\frac{\pi}{2})$
