Answer
$ h(0)=0$ minimum, $h(\pi)=5$ maximum.
Work Step by Step
Step 1. Given the function $h(\theta)=5sin\frac{\theta}{2}, 0\leq \theta\leq\pi$, we have $h'(\theta)=\frac{5}{2}cos\frac{\theta}{2}$.
Step 2. The extrema happen when $h'(\theta)=0$, undefined, or at endpoints; thus we have $cos\frac{\theta}{2}=0$ and the critical points within the domain are $\theta=0, \pi$.
Step 3. We have $h(0)=0, h(\pi)=5$. Test signs of $h'(\theta)$ across critical points $(0)..(+)..(\pi)$; thus we have $ h(0)=0$ as a local minimum, $h(\pi)=5$ as a local maximum.
Step 4. We can identify $ h(0)=0$ as an absolute minimum and $h(\pi)=5$ as an absolute maximum in the interval.
