consider the following. (if an answer does not exist, enter dne.)\n$f(x)=7\\sin(x)+7\\cos(x),\\ 0\\leq…

consider the following. (if an answer does not exist, enter dne.)\n$f(x)=7\\sin(x)+7\\cos(x),\\ 0\\leq x\\leq 2\\pi$\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum values of $f$.\nlocal minimum value\nlocal maximum value

consider the following. (if an answer does not exist, enter dne.)\n$f(x)=7\\sin(x)+7\\cos(x),\\ 0\\leq x\\leq 2\\pi$\n(a) find the interval(s) on which $f$ is increasing. (enter your answer using interval notation.)\n\n(b) find the interval(s) on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(c) find the local minimum and maximum values of $f$.\nlocal minimum value\nlocal maximum value

Answer

Explanation:

Step1: Find the derivative of (f(x))

Using the sum rule and the derivatives (\frac{d}{dx}(\sin x)=\cos x) and (\frac{d}{dx}(\cos x)=-\sin x), we have (f^\prime(x)=7\cos x - 7\sin x).

Step2: Set (f^\prime(x) = 0)

[ \begin{align*} 7\cos x-7\sin x&=0\ \cos x&=\sin x\ \tan x& = 1 \end{align*} ] On the interval (0\leq x\leq2\pi), (x=\frac{\pi}{4}) and (x=\frac{5\pi}{4}).

Step3: Test intervals

  • For the interval (\left[0,\frac{\pi}{4}\right)), let (x = 0). Then (f^\prime(0)=7\cos(0)-7\sin(0)=7>0).
  • For the interval (\left(\frac{\pi}{4},\frac{5\pi}{4}\right)), let (x=\frac{\pi}{2}). Then (f^\prime\left(\frac{\pi}{2}\right)=7\cos\left(\frac{\pi}{2}\right)-7\sin\left(\frac{\pi}{2}\right)=- 7<0).
  • For the interval (\left(\frac{5\pi}{4},2\pi\right]), let (x=\frac{3\pi}{2}). Then (f^\prime\left(\frac{3\pi}{2}\right)=7\cos\left(\frac{3\pi}{2}\right)-7\sin\left(\frac{3\pi}{2}\right)=7>0).

Step4: Find increasing and decreasing intervals

  • (f(x)) is increasing on (\left[0,\frac{\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right]).
  • (f(x)) is decreasing on (\left(\frac{\pi}{4},\frac{5\pi}{4}\right)).

Step5: Find local extrema

  • Evaluate (f(x)) at critical points (x = \frac{\pi}{4}) and (x=\frac{5\pi}{4}) [ \begin{align*} f\left(\frac{\pi}{4}\right)&=7\sin\left(\frac{\pi}{4}\right)+7\cos\left(\frac{\pi}{4}\right)=7\sqrt{2}\ f\left(\frac{5\pi}{4}\right)&=7\sin\left(\frac{5\pi}{4}\right)+7\cos\left(\frac{5\pi}{4}\right)=-7\sqrt{2} \end{align*} ]

Answer:

(a) (\left[0,\frac{\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right]) (b) (\left(\frac{\pi}{4},\frac{5\pi}{4}\right)) (c) local minimum value: (-7\sqrt{2}), local maximum value: (7\sqrt{2})