This problem is really weird. I have to explain why MVT applies for f(x)=2sinx-sin2x on the closed interval 7pi,8pi and then determine all values of c in the interval (7pi,8pi) that satisfies the conclusion of the theorem. However, the condition of MVT is that the function is continous, but the domain of the function is (-pi, 3pi), technically implicating that the function in discontinous on the interval (7pi,8pi). So how would I solve this problem?

Answer 1

why do you think the domain is from -pi to 3pi ? I think it is all real x

df/dx = 2 cos x -2 cos 2x (hoping you did not mean sin^2x)

at 7 pi
2sin 7pi = 0
in fact sin any number of pis is zero
so we are looking for where the slope is 0
that is where cos x = cos2x
well, that is true when cos x = 0
which is pi/2 + 2 pi n and 3 pi/2 + 2 pi n
between
7 pi and 8 pi that is
14 pi/2 + pi/2 = 15 pi/2
14pi/2 + 3pi/2 = 17 pi/2

Answer 2

I see. I thought it was -pi to 3pi because the period was 2pi when I was inputting the domain and range.

Answer 3

So to reiterate, the values of c that satisfy the conclusion of the theorem are 15pi/2 and 17pi/2?

Answer 4

f(x)=2sinx-sin2x
f(7pi) = 0
f(8pi) = 0
f(x) is continuous.
So, it satisfies the MVT. In fact, it satisfies Rolle's Theorem.

(f(8pi)-f(7pi))/(8pi-7pi) = 0

f'(x) = 2cosx - 2cos(2x)
= 2cosx - 2(2cos^2x-1)
= -4cos^2x + 2cosx + 2
= -2(2cos^2x - cosx - 1)
= -2(2cosx+1)(cosx-1)

So, you want c such that f'(c) = 0
cosx = -1/2: x = pi/3+7pi = 22pi/3
or
cosx = 1: x = 8pi

We want x in the interval (7pi,8pi), so that means x = 22pi/3 ? 23.03

http://www.wolframalpha.com/input/?i=2sinx-sin2x,+7pi+%3C%3D+x+%3C%3D+8pi

Answer 5

cos x =cos 2x
cos 2x = cos^2x -sin^2 x
where does cos^2x -(1-cos^2x) =cos x cos x = 2 cos^2x-1
2 cos^2x -cos x - 1 = 0
(2 cos x +1)(cos x -1) = 0

cos x = 1 or cos x = -1/2

x = 0 or 2 pi n
or x = 2 pi/3 +2 pi n