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WZ
Numerade Educator

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Problem 47 Hard Difficulty

(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 $

Answer

(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 increasing
on $(\pi, 2 \pi)$ 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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Video Transcript

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