Question

consider the following. (if an answer does not exist, enter dne.)
$f(\theta)=2\cos(\theta)+\cos^{2}(\theta),\\ 0\leq\theta\leq2\pi$
(a) find the interval(s) of increase. (enter your answer using interval notation.)
find the interval(s) of decrease. (enter your answer using interval notation.)
(b) find the local minimum value(s). (enter your answers as a comma - separated list.)
find the local maximum value(s). (enter your answers as a comma - separated list.)
(c) find the inflection points.
smaller $x$-value $(x,y)=$(
larger $x$-value $(x,y)=$(
find the interval(s) where the function is concave up. (enter your answer using interval notation.)
find the interval(s) where the function is concave down. (enter your answer using interval notation.)

Explanation:

Step1: Find the first derivative

Use the chain rule. If \(y = 2\cos\theta+\cos^{2}\theta\), then \(y^\prime=- 2\sin\theta - 2\cos\theta\sin\theta=-2\sin\theta(1 + \cos\theta)\)
Set \(y^\prime = 0\), \(\sin\theta=0\) gives \(\theta = 0,\pi,2\pi\); \(\cos\theta=- 1\) gives \(\theta=\pi\)

Step2: Determine intervals of increase and decrease

Test intervals:

• For \(0<\theta<\pi\), let \(\theta=\frac{\pi}{2}\), \(y^\prime=-2\times1\times(1 + 0)=-2<0\), function is decreasing.
• For \(\pi<\theta<2\pi\), let \(\theta=\frac{3\pi}{2}\), \(y^\prime=-2\times(-1)\times(1+0) = 2>0\), function is increasing.

Step3: Find the second derivative

\(y^\prime=-2\sin\theta(1 + \cos\theta)=-2\sin\theta-2\sin\theta\cos\theta\)
\(y^{\prime\prime}=-2\cos\theta-2(\cos^{2}\theta-\sin^{2}\theta)=-2\cos\theta-2(2\cos^{2}\theta - 1)=-4\cos^{2}\theta-2\cos\theta + 2\)
Set \(y^{\prime\prime}=0\), let \(t = \cos\theta\), then \(4t^{2}+2t - 2=0\), \(2t^{2}+t - 1=0\), \((2t - 1)(t + 1)=0\)
\(t=\cos\theta=\frac{1}{2}\) gives \(\theta=\frac{\pi}{3},\frac{5\pi}{3}\); \(t=\cos\theta=-1\) gives \(\theta=\pi\) (but \(y^{\prime\prime}\) at \(\theta = \pi\): \(y^{\prime\prime}(\pi)=-4\times(-1)^{2}-2\times(-1)+2=-4 + 2+2=0\), check the sign change around \(\theta=\pi\)

Test intervals for concavity:

• For \(0<\theta<\frac{\pi}{3}\), let \(\theta=\frac{\pi}{6}\), \(y^{\prime\prime}(\frac{\pi}{6})=-4\times(\frac{\sqrt{3}}{2})^{2}-2\times\frac{\sqrt{3}}{2}+2=-3-\sqrt{3}+2=-1-\sqrt{3}<0\)
• For \(\frac{\pi}{3}<\theta<\pi\), let \(\theta=\frac{\pi}{2}\), \(y^{\prime\prime}(\frac{\pi}{2})=-4\times0-2\times0 + 2=2>0\)
• For \(\pi<\theta<\frac{5\pi}{3}\), let \(\theta=\frac{4\pi}{3}\), \(y^{\prime\prime}(\frac{4\pi}{3})=-4\times(-\frac{1}{2})^{2}-2\times(-\frac{1}{2})+2=-1 + 1+2=2>0\)
• For \(\frac{5\pi}{3}<\theta<2\pi\), let \(\theta=\frac{11\pi}{6}\), \(y^{\prime\prime}(\frac{11\pi}{6})=-4\times(\frac{\sqrt{3}}{2})^{2}-2\times(\frac{\sqrt{3}}{2})+2=-3-\sqrt{3}+2=-1-\sqrt{3}<0\)

Step4: Find local minima and maxima

Since the function changes from decreasing (\(0<\theta<\pi\)) to increasing (\(\pi<\theta<2\pi\)), at \(\theta=\pi\), \(y(\pi)=2\cos\pi+\cos^{2}\pi=-2 + 1=-1\) (local minimum)
At endpoints: \(y(0)=2\cos0+\cos^{2}0=2 + 1=3\), \(y(2\pi)=2\cos(2\pi)+\cos^{2}(2\pi)=2 + 1=3\) (local maxima)

Step5: Find inflection points

When \(\theta=\frac{\pi}{3}\), \(y(\frac{\pi}{3})=2\cos\frac{\pi}{3}+\cos^{2}\frac{\pi}{3}=2\times\frac{1}{2}+(\frac{1}{2})^{2}=1+\frac{1}{4}=\frac{5}{4}\)
When \(\theta=\frac{5\pi}{3}\), \(y(\frac{5\pi}{3})=2\cos\frac{5\pi}{3}+\cos^{2}\frac{5\pi}{3}=2\times\frac{1}{2}+(\frac{1}{2})^{2}=\frac{5}{4}\)

Answer:

(a)

• Interval of increase: \((\pi,2\pi)\)
• Interval of decrease: \((0,\pi)\)

(b)

• Local minimum value: \(-1\)
• Local maximum value: \(3\)

(c)

• Smaller \(x -\)value (inflection point): \((\frac{\pi}{3},\frac{5}{4})\)
• Larger \(x -\)value (inflection point): \((\frac{5\pi}{3},\frac{5}{4})\)
• Concave up: \((\frac{\pi}{3},\frac{5\pi}{3})\)
• Concave down: \((0,\frac{\pi}{3})\cup(\frac{5\pi}{3},2\pi)\)