Question

Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem `f(x)= 6 cos (x)`; [-pi/2, pi/2]?

Answer 1

Please see below.

Explanation:

There is no "using the priciple of the mean-value theorem" (whatever that means) involved.

Write down what the conclusion of the theorem says:

There is a `c` in `(-pi/2, pi/2)` such that

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

Now do the calculus, algebra, and trigonometry needed to find `c` in the interval `(-pi/2, pi/2)`.

We can see that `f'(x) = -6sinx`

and `(f(pi/2)-f(-pi/2))/(pi/2-(-pi/2)) = (6cos(pi/2)-6cos(-pi/2))/pi = 0`.

So you need to solve

`-6sinx = 0` on the interval `(-pi/2, pi/2)`.

The only solution in the interval is `x=0`, so the only value of `c` in the conclusion of the theorem is `0`.