Question

Given the function `f(x)=5-4/x`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,4] and find the c in the conclusion?

Answer 1

The answer below is for the mean value theorem for derivatives.

Explanation:

You determine whether it satisfies the hypotheses by determining whether `f(x) = 5-4/x` is continuous on the interval `[1,4]` and differentiable on the interval `(1,4)`. (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(4)-f(1))/(4-1)` on the interval `(1,4)`.

Answers

`f` is continuous on its domain, which includes `[1,4]`.

So `f` is continuous on `[1,4]`.

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

So `f` is differentiable on `(1,4)`.

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

To find `c` solve the equation `f'(x) = (f(4)-f(1))/(4-1)`. Discard any solutions outside `(1,4)`.

You should get `c = 2`. The solution `x=-2` is not in the interval `(1,4)`.)

Answer 2

The answer below is for the Mean Value Theorem for integrals for `f(x) = 5 - 1/x` on `[1,4]`.

Explanation:

`f(x) = 5-1/x` is continuous on the interval `[1,4]` because the only place it is not continuous is `0` which is not in `[1,4]`.

Therefore, there is a `c` in `[1,4]` with

`f(c) = "average value of " f " on " [1,4] = 1/(4-1)int_1^4 f(x) dx`

Now,

`{:int_1^4 (5-1/x) dx = 5x-ln(x)|:}_1^4 = 5-1/3ln4`

To find `c`, solve `f(x) = 5-1/3ln4` in `[1,4]`

`5-1/x = 5-1/3ln4` if and only if

`1/x = (ln4)/3` if and only if

`x = 3/ln4`

So we want `c = 3/ln4`