determine the location and value of the absolute extreme values of f on the given interval, if they…
determine the location and value of the absolute extreme values of f on the given interval, if they exist.\nf(x) = cos 3x on - \\frac{\\pi}{4}, \\frac{\\pi}{2} \nwhat is/are the absolute maximum/maxima of f on the given interval? select the correct choice below and, if necessary, fill in the answer boxes \na. the absolute maximum/maxima is/are at x = \n(use a comma to separate answers as needed. type an exact answer, using \\pi as needed.)\nb. there is no absolute maximum of f on the given interval.
Answer
Explanation:
Step1: Find the derivative of (f(x))
The derivative of (y = \cos(ax)) is (y^\prime=-a\sin(ax)). For (f(x)=\cos(3x)), (f^\prime(x)=- 3\sin(3x)).
Step2: Find the critical points
Set (f^\prime(x) = 0), so (-3\sin(3x)=0), which implies (\sin(3x)=0). Then (3x = k\pi), (x=\frac{k\pi}{3}), (k\in\mathbb{Z}). For the interval (\left[-\frac{\pi}{4},\frac{\pi}{2}\right]), when (k = 0), (x = 0); when (k=- 1), (x=-\frac{\pi}{3}\notin\left[-\frac{\pi}{4},\frac{\pi}{2}\right]); when (k = 1), (x=\frac{\pi}{3}\in\left[-\frac{\pi}{4},\frac{\pi}{2}\right]).
Step3: Evaluate (f(x)) at critical points and endpoints
- At (x =-\frac{\pi}{4}): (f\left(-\frac{\pi}{4}\right)=\cos\left(-\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}).
- At (x = 0): (f(0)=\cos(0) = 1).
- At (x=\frac{\pi}{3}): (f\left(\frac{\pi}{3}\right)=\cos(\pi)=-1).
- At (x=\frac{\pi}{2}): (f\left(\frac{\pi}{2}\right)=\cos\left(\frac{3\pi}{2}\right)=0).
Answer:
A. The absolute maximum/maxima is/are (1) at (x = 0).