Answer
a. $f(0)=-1$ and $ f(7\pi/4)=-\sqrt 2$ local minima,
$f(3\pi/4)=\sqrt 2$ and $f(2\pi)=-1$ local maxima.
$f(3\pi/4)=\sqrt 2$ absolute maximum and $f(7\pi/4)=-\sqrt 2$ absolute minimum.
b. See graph and explanations.
Work Step by Step
a. Step 1. Given the function $f(x)=sin(x)-cos(x), 0\leq x\lt2\pi$, we have $f'(x)=cos(x)+sin(x)$.
Step 2. The extrema happen when $f'(x)=0$, undefined, or at endpoints; thus we have $cos(x)+sin(x)=0, tan(x)=-1$ and the critical points within the domain are $x=0, 3\pi/4, 7\pi/4,2 \pi$.
Step 3. We have $f(0)=-1, f(3pi/4)=\sqrt 2, f(7\pi/4)=-\sqrt 2, f(2\pi)=-1$. Test signs of $f'(x)$ across critical points $(0).. (+)..(3\pi/4)..(-)..(7\pi/4)..(+)..(2\pi)$; thus we have $f(0)=-1$ and $ f(7\pi/4)=-\sqrt 2$ as local minima, $f(3\pi/4)=\sqrt 2$ and $f(2\pi)=-1$ as local maxima.
Step 4. We can identify that $f(3\pi/4)=\sqrt 2$ is an absolute maximum and $f(7\pi/4)=-\sqrt 2$ 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 the critical points where $f'=0$.
