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# (a) Find the intervals of increase or decrease.(b) Find the local maximum and minimum values.(c) Find the intervals of concavity and the inflection points.(d) Use the information from parts $(a) - (c)$ to sketch the graph. Check your work with a graphing device if you have one.$f(\theta) = 2 \cos \theta + \cos^2 \theta$, $0 \leqslant \theta \leqslant 2\pi$

## (a) $f(\theta)=2 \cos \theta+\cos ^{2} \theta, 0 \leq \theta \leq 2 \pi \Rightarrow f^{\prime}(\theta)=-2 \sin \theta+2 \cos \theta(-\sin \theta)=-2 \sin \theta(1+\cos \theta)$$f^{\prime}(\theta)=0 \Leftrightarrow \theta=0, \pi,$ and $2 \pi . f^{\prime}(\theta)>0 \Leftrightarrow \pi<\theta<2 \pi$ and $f^{\prime}(\theta)<0 \Leftrightarrow 0<\theta<\pi .$ So $f$ is increasingon $(\pi, 2 \pi)$ and $f$ is decreasing on $(0, \pi)$(b) $f(\pi)=-1$ is a local minimum value.(c) $f^{\prime}(\theta)=-2 \sin \theta(1+\cos \theta) \Rightarrow$\begin{aligned}f^{\prime \prime}(\theta) &=-2 \sin \theta(-\sin \theta)+(1+\cos \theta)(-2 \cos \theta)=2 \sin ^{2} \theta-2 \cos \theta-2 \cos ^{2} \theta \\&=2\left(1-\cos ^{2} \theta\right)-2 \cos \theta-2 \cos ^{2} \theta=-4 \cos ^{2} \theta-2 \cos \theta+2 \\&=-2\left(2 \cos ^{2} \theta+\cos \theta-1\right)=-2(2 \cos \theta-1)(\cos \theta+1) \\\text { since }-2(\cos \theta+1)<0 &[\text { for } \theta \neq \pi], f^{\prime \prime}(\theta)>0 \Rightarrow 2 \cos \theta-1<0 \Rightarrow \cos \theta<\frac{1}{2} \Rightarrow \frac{\pi}{3}<\theta<\frac{5 \pi}{3} \text { and } \theta\end{aligned}$f^{\prime \prime}(\theta)<0 \Rightarrow \cos \theta>\frac{1}{2} \Rightarrow 0<\theta<\frac{\pi}{3}$ or $\frac{5 \pi}{3}<\theta<2 \pi .$ So $f$ is $\mathrm{CU}$ on $\left(\frac{\pi}{3}, \frac{5 \pi}{3}\right)$ and $f$ is $\mathrm{CD}$ on $\left(0, \frac{\pi}{3}\right)$ and$\left(\frac{5 \pi}{3}, 2 \pi\right) .$ There are points of inflection at $\left(\frac{\pi}{3}, f\left(\frac{\pi}{3}\right)\right)=\left(\frac{\pi}{3}, \frac{5}{4}\right)$ and $\left(\frac{5 \pi}{3}, f\left(\frac{5 \pi}{3}\right)\right)=\left(\frac{5 \pi}{3}, \frac{5}{4}\right)$

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