Answer
See explanations. $h(0)=3$ absolute maximum, $h(2\pi)=-3$ absolute minimum.
Work Step by Step
Step 1. Given the function $h(\theta)=3cos\frac{\theta}{2}, 0\leq \theta\leq2\pi$, we have $h'(\theta)=-\frac{3}{2}sin\frac{\theta}{2}$.
Step 2. The extrema happen when $h'(\theta)=0$, undefined, or at endpoints; thus we have $sin\frac{\theta}{2}=0$ and the critical points within the domain are $x=0, 2\pi$.
Step 3. We have $h(0)=3, h(2\pi)=-3$. Test signs of $h'(\theta)$ across critical points $(0)..(-)..(2\pi)$; thus we have $ h(0)=3$ as a local maximum, $h(2\pi)=-3$ as a local minimum.
Step 4. We can identify $ h(0)=3$ as an absolute maximum and $h(2\pi)-3$ as an absolute minimum in the interval.
