Question

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.

Answer 1

Answer:

Step-by-step explanation:

Given that there is a function of x,

$f(x) = 2sin x + 2cos x,0\leq x\leq 2\pi$

Let us find first and second derivative for f(x)

$f'(x) = 2cosx -2sinx\\f"(x) = -2sinx-2cosx$

When f'(x) =0 we have tanx = 1 and hence

a) f'(x) >0 for I and III quadrant

Hence increasing in $(0, \pi/2) U(\pi,3\pi/2)$

and decreasing in $(\pi/2, \pi)U(3\pi/2,2\pi)$

$x=\frac{\pi}{4}, \frac{3\pi}{4}$

$f"(\pi/4)$

Hence f has a maxima at x = pi/4 and minima at x = 3pi/4

b) Maximum value = $2sin \pi/4+2cos \pi/4 =2\sqrt{2}$

Minimum value = $2sin 3\pi/4+2cos 3\pi/4 =-2\sqrt{2}$

c)

f"(x) =0 gives tanx =-1

$x= 3\pi/4, 7\pi/4$

are points of inflection.

concave up in (3pi/4,7pi/4)

and concave down in (0,3pi/4)U(7pi/4,2pi)