Let f be the function with f(0) = 1/ (pi)^2, f(2) = 1/(pi)^2, and the derivative given by f'(x) = (x+1)cos ((pi)(x)). How many values of x in the open interval (0, 2) satisfy the conclusion of the Mean Value Theorem for the function f on the closed interval [0, 2] ?

Answer 1

The function is the same at x = 0 and x = 2
so the mean slope is zero.
Therefore what it is asking is:
How many times is f'(x) = (x+1)cos (pi x) = 0
between pi x = 0 and pi x = 2 pi
well (x+1) is always a positive number in that interval
so
How many times is cos theta = 0 between theta = 0 and theta = 2 pi ????

Answer 2

It is just pretending to be difficult.

Answer 3

So would the answer be pi/2 and 3pi/2?

Answer 4

Off of what Damon said, you need to find how many x’s satisfy cos(pi x)=0. Not just when cosx=0. Since cosx=0 when x=pi/2 and 3pi/2, 3pi/2 is out of the interval. In order for cos(pi x)=0, x=1/2. There is ONE value of x that satisfies the MVT.

Answer 5

^ whoever answered before me was incorrect, There are 2 values of X. 3pi/2 is not off the table because the function is cos(pi * x), not just cos(x) like you implied by taking 3pi/2 off. so the x values that satisfy the MVT are 1/2 and 3/2 which is in (0,2).

Answer 6

You are correct, I apologize for my mistake earlier. The function is cos(pi * x), so we need to find where cos(pi * x) = 0 between x = 0 and x = 2.

cos(pi * x) = 0 when pi * x = pi / 2 or 3pi / 2, since cos(pi / 2) = 0 and cos(3pi / 2) = 0.

So we have pi * x = pi / 2 or pi * x = 3pi / 2, which gives x = 1/2 or x = 3/2. Both of these values are in the open interval (0,2), so there are two values of x that satisfy the conclusion of the Mean Value Theorem for the function f on the closed interval [0,2].