MATH SOLVE

10 months ago

Q:
# This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.Consider the equation below.f(x) = 2 sin(x) + 2 cos(x), 0 ≤ x ≤ 2πExercise (a)Find the interval on which f is increasing. Find the interval on which f is decreasing.Exercise (b)Find the local minimum and maximum values of f.Exercise (c)Find the inflection points. Find the interval on which f is concave up. Find the interval on which f is concave down.

Accepted Solution

A:

Answer:Step-by-step explanation:Given that there is a function of x,[tex]f(x) = 2sin x + 2cos x,0\leq x\leq 2\pi[/tex]Let us find first and second derivative for f(x)[tex]f'(x) = 2cosx -2sinx\\f"(x) = -2sinx-2cosx[/tex]When f'(x) =0 we have tanx = 1 and hence a) f'(x) >0 for I and III quadrant Hence increasing in [tex](0, \pi/2) U(\pi,3\pi/2)\\[/tex]and decreasing in [tex](\pi/2, \pi)U(3\pi/2,2\pi)[/tex][tex]x=\frac{\pi}{4}, \frac{3\pi}{4}[/tex][tex]f"(\pi/4) <0 and f"(3\pi/4)>0[/tex]Hence f has a maxima at x = pi/4 and minima at x = 3pi/4b) Maximum value = [tex]2sin \pi/4+2cos \pi/4 =2\sqrt{2}[/tex]Minimum value = [tex]2sin 3\pi/4+2cos 3\pi/4 =-2\sqrt{2}[/tex]c)f"(x) =0 gives tanx =-1[tex]x= 3\pi/4, 7\pi/4[/tex]are points of inflection.concave up in (3pi/4,7pi/4)and concave down in (0,3pi/4)U(7pi/4,2pi)