Answer
a. $f(0)=\sqrt 3$ and $ f(7\pi/6)=-2$ local minima, $f(\pi/6)=2$ and $f(2\pi)=\sqrt 3$ local maxima.
$f(\pi/6)= 2$ absolute maximum, $f(7\pi/6)=-2$ absolute minimum.
b. See graph and explanations.
Work Step by Step
a. Step 1. Given the function $f(x)=\sqrt 3cos(x)+sin(x), 0\leq x\lt2\pi$, we have $f'(x)=-\sqrt 3sin(x)+cos(x)$.
Step 2. The extrema happen when $f'(x)=0$, undefined, or at endpoints; thus we have $-\sqrt 3sin(x)+cos(x)=0, tan(x)=\sqrt 3/3$ and the critical points within the domain are $x=0, \pi/6, 7\pi/6,2 \pi$.
Step 3. We have $f(0)=\sqrt 3, f(\pi/6)=\sqrt 3cos(\pi/6)+sin(\pi/6)=2, f(7\pi/6)=-2, f(2\pi)=\sqrt 3$. Test signs of $f'(x)$ across critical points $(0).. (+)..(\pi/6)..(-)..(7\pi/6)..(+)..(2\pi)$; thus we have $f(0)=\sqrt 3$ and $ f(7\pi/6)=-2$ as local minima, $f(\pi/6)=2$ and $f(2\pi)=\sqrt 3$ as local maxima.
Step 4. We can identify that $f(\pi/6)= 2$ is an absolute maximum and $f(7\pi/6)=-2$ is an absolute minimum.
b. See graph. We can see that when $f'\lt0$, $f$ decreases; while when $f'\gt0$, $f$ increases (both 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 $f'=0$.
