Answer
(a) $f$ is decreasing on the interval $(0, \pi)$
$f$ is increasing on the interval $(\pi,2\pi)$
(b) The local minimum is $f(\pi) = -1$
(c) The graph is concave down on these intervals: $(0,\frac{\pi}{3})\cup (\frac{5\pi}{3}, 2\pi)$
The graph is concave up on this interval: $(\frac{\pi}{3},\frac{5\pi}{3})$
The points of inflection are $(\frac{\pi}{3}, \frac{5}{4})$ and $(\frac{5\pi}{3}, \frac{5}{4})$
(d) We can see a sketch of the graph below.
Work Step by Step
(a) $f(\theta) = 2~cos~\theta + cos^2~\theta,~~~~0 \leq \theta \leq 2\theta$
We can find the points where $f'(\theta) = 0$:
$f'(\theta) = -2~sin~\theta - 2~cos~\theta~sin~\theta = 0$
$(-2~sin~\theta)(1+cos~\theta) = 0$
$\theta = 0, \pi, 2\pi$
When $0 \lt \theta \lt \pi~~$ then $f'(\theta) \lt 0$
$f$ is decreasing on the interval $(0, \pi)$
When $\pi \lt \theta \lt 2\pi$ then $f'(\theta) \gt 0$
$f$ is increasing on the interval $(\pi,2\pi)$
(b) $f(\pi) = 2~cos~\pi + cos^2~\pi = -1$
The local minimum is $f(\pi) = -1$
There is no local maximum because $\theta = 0$ and $\theta = 2\pi$ are endpoints of the graph.
(c) We can find the points where $f''(x) = 0$:
$f''(\theta) = -2~cos~\theta + 2~sin^2~\theta - 2~cos^2~\theta = 0$
$-2~cos~\theta + 2(1-cos^2~\theta) - 2~cos^2~\theta = 0$
$-2~cos~\theta + 2 - 4~cos^2~\theta = 0$
$4~cos^2~\theta+2~cos~\theta - 2 = 0$
$(2~cos~\theta-1)(2~cos~\theta+2) = 0$
$\theta = \frac{\pi}{3}, \pi, \frac{5\pi}{3}$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on these intervals: $(0,\frac{\pi}{3})\cup (\frac{5\pi}{3}, 2\pi)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(\frac{\pi}{3},\frac{5\pi}{3})$
$f(\frac{\pi}{3}) = 2~cos~\frac{\pi}{3} + cos^2~\frac{\pi}{3} = \frac{5}{4}$
$f(\frac{5\pi}{3}) = 2~cos~\frac{5\pi}{3} + cos^2~\frac{5\pi}{3} = \frac{5}{4}$
Note that $\theta = \pi$ is not a point of inflection since the graph does not change concavity at this point.
The points of inflection are $(\frac{\pi}{3}, \frac{5}{4})$ and $(\frac{5\pi}{3}, \frac{5}{4})$
(d) We can see a sketch of the graph below.