Question

Given the function `f(x)=(-x^2+9)/(4x)`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,3] and find the c?

Answer 1

The only hypothesis required by the mean value theorem is that `f(x)` has to be continuous in the interval.

Explanation:

The only hypothesis required by the mean value theorem is that `f(x)` has to be continuous in the interval.

As:

`f(x) = (9-x^2)/(4x)`

is in fact continuous in `[1,3]` the hypothesis is satisfied and we have that the is a value `c in [1,3]` for which:

`f(c) = 1/2 int_1^3 (9-x^2)/(4x)dx`

We can calculate the definite integral as:

#int_1^3 (9-x^2)/(4x)dx = int_1^3 (9/(4x)-x/4)dx=[9/4lnx-x^2/8]_1^3= 9/4ln3-9/8+1/8=9/4ln3-1#

Then we can find `c` from:

`f(c) = (9-c^2)/(4c) = 9/4ln3-1`

Answer 2

I will assume that you are referring to the Mean Value Theorem for Derivatives.

Explanation:

You determine whether it satisfies the hypotheses by determining whether `f(x) = (-x^2+9)/(4x)` is continuous on the interval `[1,3]` and differentiable on the interval `(1,3)`. (Those are the hypotheses of the Mean Value Theorem)

You find the `c` mentioned in the conclusion of the theorem by solving `f'(x) = (f(3)-f(1))/(3-1)` on the interval `(1,3)`.

`f` is continuous on its domain, which includes `[1,3]`

`f'(x) = -(x^2+9)/(4x^2)` which exists for all `x != 0` so it exists for all `x` in `(1,3)`

Therefore this function satisfies the hypotheses of the Mean Value Theorem on this interval.

To find `c` solve the equation `f'(x) = (f(3)-f(1))/(3-1)`. Discard any solutions outside `(1,3)`.

You should get `c = sqrt3`. (The solution, `-sqrt3` is not in the interval.)