Answer
$y(0)=0$ and $y(\frac{3\pi}{4})=-\frac{1}{2}$ as local minima,
$y(\frac{\pi}{4})=\frac{1}{2}$ and $y(\pi)=0$ as local maxima,
$y(\frac{\pi}{4})=\frac{1}{2}$ as an absolute maximum,
$y(\frac{3\pi}{4})=-\frac{1}{2}$ as an absolute minimum in the interval.
inflection point: $(\frac{\pi}{2},0)$.
concave down on $(0, \frac{\pi}{2})$ and concave up on $(\frac{\pi}{2}, 0)$
See graph.
Work Step by Step
Step 1. Given the function $y=sin(x)cos(x), 0\leq x \leq \pi$, we have $y'=cos^2x-sin^2x=cos2x$ and $y''=-2sin2x$
Step 2. The extrema happen when $y'=0$, undefined, or at endpoints; thus we have $cos2x=0, x=\pi/4, 3\pi/4$ and the critical points within the domain are $x=0, \frac{\pi}{4}, \frac{3\pi}{4}, \pi$.
Step 3. We have $y(0)=y(\pi)=0$, $y(\frac{\pi}{4})=\frac{1}{2}$, $y(\frac{3\pi}{4})=-\frac{1}{2}$.
Step 4. Test signs of $y'$ across critical points $(0)..(+)..(\frac{\pi}{4})..(-)..(\frac{3\pi}{4})..(+)..(\pi)$; thus we have $y(0)=0$ and $y(\frac{3\pi}{4})=-\frac{1}{2}$ as local minima, and $y(\frac{\pi}{4})=\frac{1}{2}$ and $y(\pi)=0$ as local maxima,
Step 5. We can identify $y(\frac{\pi}{4})=\frac{1}{2}$ as an absolute maximum and $y(\frac{3\pi}{4})=-\frac{1}{2}$ as an absolute minimum in the interval.
Step 6. The inflection points can be found when $y''=0$ or it does not exist. We have $sin2x=0, x=\frac{\pi}{2}$ which gives coordinates $(\frac{\pi}{2},0)$
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 $(0)..(-)..(\frac{\pi}{2})..(+)..(\pi)$,
Step 8. The function is concave down on $(0, \frac{\pi}{2})$ and concave up on $(\frac{\pi}{2}, 0)$
Step 9. See graph.
