Answer
a) Increasing on $(\pi,2\pi)$,
decreasing on $(0,\pi)$
b) Local minimum value: $f(\pi) = -1$
c) Concave upward on $\left(\frac{\pi}{3},\frac{5\pi}{3}\right)$,
concave downward on $\left(0,\frac{\pi}{3}\right)\cup\left(\frac{\pi}{3},2\pi\right)$
Points of inflection: $\left(\frac{\pi}{3},\frac{5}{4}\right)$ and $\left(\frac{5\pi}{3},\frac{5}{4}\right)$
d) See graph
Work Step by Step
a)
$f(θ)$ = $2\cos θ+\cos^{2} θ$
$f'(θ)$ = $-2\sin θ-2\cos θ\sin θ$ = $-2\sin θ(1+\cos θ)$
$f'(θ)$ = $0\Rightarrow θ$ = $0,\pi,2\pi$
$f'(θ)$ $\gt$ $0$ for $ \pi$ $\lt$ $θ$ $\lt$ $2\pi$
$f'(θ)$ $\lt$ $0$ for $0$ $\lt$ $θ$ $\lt$ $\pi$
$f$ is increasing on $(\pi,2\pi)$
$f$ is decreasing on $(0,\pi)$
b)
$f(\pi)$ = $-1$ is a local minimum value
c)
$f'(θ)$ = $-2\sin θ(1+\cos θ)$
$f''(θ)$ = $2\sin θ\sin θ-2\cos θ(1+\cos θ)$ = $-2(2\cos θ-1)(2\cos θ+1)$
$f''(θ)$ $\gt$ $0$
$2\cos θ-1$ $\lt$ $0$
$\cos θ$ $\lt$ $\frac{1}{2}$ for $\frac{\pi}{3}$ $\lt$ $θ$ $\lt$ $\frac{5\pi}{3}$
$f''(θ)$ $\lt$ $0$
$\cos θ$ $\gt$ $\frac{1}{2}$ for $0$ $\lt$ $θ$ $\lt$ $\frac{\pi}{3}$ or $\frac{5\pi}{3}$ $θ$ $\lt$ $2\pi$
$f$ is concave upward on $\left(\frac{\pi}{3},\frac{5\pi}{3}\right)$
$f$ is concave downward on $\left(0,\frac{\pi}{3}\right)\cup\left(\frac{\pi}{3},2\pi\right)$
There are points of inflection at $\left(\frac{\pi}{3},\frac{5}{4}\right)$ and $\left(\frac{5\pi}{3},\frac{5}{4}\right)$
