Question

How do you find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function `f(x)= 6 cos x` on the interval`[-pi/2, pi/2]`?

Answer 1

Solve `f'(x) = (f(pi/2) - f(-pi/2))/((pi/2) - (-pi/2))` for values in `(-pi/2, pi/2)`

Explanation:

I would begin by checking that the hypotheses are true.

This function is continuous on `[-pi/2, pi/2]` and differentiable on `(-pi/2, pi/2)`, so the hypotheses of the Mean Value Theorem are true.

The conclusion is that there is a number `c` in `(-pi/2, pi/2)` with

`f'(c) = (f(pi/2) - f(-pi/2))/((pi/2) - (-pi/2))` .

We have been asked to find the number(s) whose existence the theorem asserts.

For `f(x) = 6cosx`, we have `f(pi/2) = f(-pi/2) = 0`, so we need to solve

`f'(c) = 0`

Furthermore `f'(x) = -6sinx` which is `0` at integer multiples of `pi`

The only integer multiple of `pi` in `(-pi/2, pi/2)` is `0 pi = 0`

`c = 0`