Rolles Theorem?

Question

My daughter got stuck on her last calculus problem for her homework. She was given an example very similar to this, but did not understand how they worked it so the example did her no help. Can someone please break this down for us because it is well over my head. Thanks in advance.

A function f(x) and interval (a,b) are given. Check if Rolle's Theorem can be applied to f on (a,b). If so, find all values c in (a,b) such that f prime(c) =0. f(x)=2-2sin8x on [pi/8,pi/2]. If Rolles Theorem does not apply, DNE for the solution. Separate multiple answers by commas.

Answer 1

In order for Rolle's Theorem to be applied, the function f(x) must be continuous and differentiable on [a,b], and f(b) must be equal to f(a).

Given that we are working with f(x)=2 - 2sin(8x) on the interval [pi/8, pi/2], lets confirm that f(pi/8) = f(pi/2).

f(pi/8) = 2 - 2sin(8·(pi/8))

f(pi/8) = 2 - 2sin(pi)

f(pi/8) = 2 - 0

f(pi/8) = 2

f(pi/2) = 2 - 2sin(8·(pi/2))

f(pi/2) = 2 - 2sin(4·pi)

f(pi/2) = 2 - 0

f(pi/2) = 2

So we confirmed that f(pi/8) = f(pi/2) = 2, which means Rolle's Theorem can be applied.

Now we will find all values c, where a ≤ c ≤ b, such that f'(c) = 0.

We find the derivative of f(x) which is -16cos(8x)

Then we set the derivative equal to 0 and solve for c.

-16cos(8c) = 0

cos(8c) = 0

8c = n·pi + pi/2, where n is an integer.

c = (n/8)·pi + pi/16

Now we are only concerned with the interval [pi/8, pi/2], So all the values of c that would make f'(c) = 0 are 3pi/16, 5pi/16, 7pi/16.