Answer
a. $ f(0)=0$ and $ f(3\pi/4)=-1$ local minima, $f(\pi/4)=1$ and $f(\pi)=0$ local maxima.
$f(\pi/4)=1$ absolute maximum, $f(3\pi/4)=-1$ absolute minimum.
b. See graph and explanations.
Work Step by Step
a. Step 1. Given the function $f(x)=sin2x, 0\leq x\lt\pi$, we have $f'(x)=2cos2x$.
Step 2. The extrema happen when $f'(x)=0$, undefined and at endpoints; thus the critical points within the domain are $x=0, \pi/4, 3\pi/4, \pi$.
Step 3. We have $f(0)=0, f(pi/4)=1, f(3\pi/4)=-1, f(\pi)=0$. Test signs of $f'(x)$ across critical points $(0).. (+)..(\pi/4)..(-)..(3\pi/4)..(+)..(\pi)$
Thus we have $ f(0)=0$ and $ f(3\pi/4)=-1$ as local minima, $f(\pi/4)=1$ and $f(\pi)=0$ as local maxima.
Step 4. We can identify that $f(\pi/4)=1$ is an absolute maximum and $f(3\pi/4)=-1$ is an absolute minimum.
b. See graph. We can see that when $f'\lt0$, $f$ decreases; while when $f'\gt0$, $f$ increases (with the tangent slopes corresponding to the values of $f'$). When $f'=0$, $f$ will be at an extrema with tangent slopes of zero. We can also observe the change of signs in $f'$ on different sides of critical points where $f'=0$.
