consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=7\\cos ^{2}(x)-14\\sin…

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=7\\cos ^{2}(x)-14\\sin (x),\\ 0\\leq x\\leq 2\\pi$\n\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection points. (order your answers from smallest to largest $x$, then from smallest to largest $y$.)\n\n$(x,y)=(\\square)$\n\n$(x,y)=(\\square)$\n\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)

consider the equation below. (if an answer does not exist, enter dne.)\n\n$f(x)=7\\cos ^{2}(x)-14\\sin (x),\\ 0\\leq x\\leq 2\\pi$\n\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\n\nlocal minimum value\n\nlocal maximum value\n\n(c) find the inflection points. (order your answers from smallest to largest $x$, then from smallest to largest $y$.)\n\n$(x,y)=(\\square)$\n\n$(x,y)=(\\square)$\n\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)

Answer

Explanation:

Step1: Find the first derivative

Using the chain - rule and basic derivative rules. The derivative of (y = 7\cos^{2}(x)-14\sin(x)) is: [ \begin{align*} f^{\prime}(x)&=7\times2\cos(x)(-\sin(x))-14\cos(x)\ &=-14\sin(x)\cos(x)-14\cos(x)\ &=-14\cos(x)(\sin(x) + 1) \end{align*} ]

Step2: Find critical points

Set (f^{\prime}(x)=0), so (-14\cos(x)(\sin(x)+1)=0). Since (\sin(x)+1\geq0) for all (x), and (\sin(x)+1 = 0) when (x=\frac{3\pi}{2}), and (\cos(x)=0) when (x=\frac{\pi}{2},\frac{3\pi}{2}).

Step3: Determine increasing and decreasing intervals

Use test - points in the intervals (\left[0,\frac{\pi}{2}\right)), (\left(\frac{\pi}{2},\frac{3\pi}{2}\right)) and (\left(\frac{3\pi}{2},2\pi\right]).

  • For (x\in\left[0,\frac{\pi}{2}\right)), let (x = \frac{\pi}{4}), (f^{\prime}\left(\frac{\pi}{4}\right)=-14\times\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}+1\right)<0)
  • For (x\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)), let (x=\pi), (f^{\prime}(\pi)=-14\times(- 1)(0 + 1)=14>0)
  • For (x\in\left(\frac{3\pi}{2},2\pi\right]), let (x=\frac{7\pi}{4}), (f^{\prime}\left(\frac{7\pi}{4}\right)=-14\times\frac{\sqrt{2}}{2}\left(-\frac{\sqrt{2}}{2}+1\right)<0)

So (f(x)) is increasing on (\left(\frac{\pi}{2},\frac{3\pi}{2}\right)) and decreasing on (\left[0,\frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2},2\pi\right])

Step4: Find local minima and maxima

Evaluate (f(x)) at critical points (x = \frac{\pi}{2},\frac{3\pi}{2}) [ \begin{align*} f\left(\frac{\pi}{2}\right)&=7\cos^{2}\left(\frac{\pi}{2}\right)-14\sin\left(\frac{\pi}{2}\right)=-14\ f\left(\frac{3\pi}{2}\right)&=7\cos^{2}\left(\frac{3\pi}{2}\right)-14\sin\left(\frac{3\pi}{2}\right)=14 \end{align*} ] So the local minimum value is (-14) and the local maximum value is (14)

Step5: Find the second derivative

[ \begin{align*} f^{\prime}(x)&=-14\cos(x)(\sin(x)+1)\ f^{\prime\prime}(x)&=14\sin(x)(\sin(x)+1)-14\cos^{2}(x)\ &=14\sin^{2}(x)+14\sin(x)-14(1 - \sin^{2}(x))\ &=28\sin^{2}(x)+14\sin(x)-14\ &=14(2\sin^{2}(x)+\sin(x)-1)\ &=14(2\sin(x)-1)(\sin(x)+1) \end{align*} ] Set (f^{\prime\prime}(x)=0), then (2\sin(x)-1 = 0) (since (\sin(x)+1\geq0) and we ignore (\sin(x)+1 = 0) for inflection - point consideration in the non - degenerate sense). So (\sin(x)=\frac{1}{2}), (x=\frac{\pi}{6},\frac{5\pi}{6}) [ \begin{align*} f\left(\frac{\pi}{6}\right)&=7\cos^{2}\left(\frac{\pi}{6}\right)-14\sin\left(\frac{\pi}{6}\right)=7\times\frac{3}{4}-7=-\frac{7}{4}\ f\left(\frac{5\pi}{6}\right)&=7\cos^{2}\left(\frac{5\pi}{6}\right)-14\sin\left(\frac{5\pi}{6}\right)=7\times\frac{3}{4}-7=-\frac{7}{4} \end{align*} ] Use test - points in the intervals (\left[0,\frac{\pi}{6}\right)), (\left(\frac{\pi}{6},\frac{5\pi}{6}\right)) and (\left(\frac{5\pi}{6},2\pi\right])

  • For (x\in\left[0,\frac{\pi}{6}\right)), let (x = 0), (f^{\prime\prime}(0)=14(0 + 0-1)<0)
  • For (x\in\left(\frac{\pi}{6},\frac{5\pi}{6}\right)), let (x=\frac{\pi}{2}), (f^{\prime\prime}\left(\frac{\pi}{2}\right)=14(2 - 1)>0)
  • For (x\in\left(\frac{5\pi}{6},2\pi\right]), let (x=\pi), (f^{\prime\prime}(\pi)=14(0 + 0 - 1)<0)

So the inflection points are (\left(\frac{\pi}{6},-\frac{7}{4}\right)) and (\left(\frac{5\pi}{6},-\frac{7}{4}\right)), (f(x)) is concave up on (\left(\frac{\pi}{6},\frac{5\pi}{6}\right)) and concave down on (\left[0,\frac{\pi}{6}\right)\cup\left(\frac{5\pi}{6},2\pi\right])

Answer:

(a) Increasing: (\left(\frac{\pi}{2},\frac{3\pi}{2}\right)); Decreasing: (\left[0,\frac{\pi}{2}\right]\cup\left[\frac{3\pi}{2},2\pi\right]) (b) Local minimum value: (-14); Local maximum value: (14) (c) Inflection points: (\left(\frac{\pi}{6},-\frac{7}{4}\right)), (\left(\frac{5\pi}{6},-\frac{7}{4}\right)); Concave up: (\left(\frac{\pi}{6},\frac{5\pi}{6}\right)); Concave down: (\left[0,\frac{\pi}{6}\right]\cup\left[\frac{5\pi}{6},2\pi\right])