Question

How do you find the value of c guaranteed by the Mean Value Theorem for `f(x)=(2x)/(x^2+1)` on the interval [0,1]?

Answer 1

`c = sqrt(sqrt5 - 2)` For "how" see the explanation.

Explanation:

I would begin by checking that the hypotheses are true.

This function is continuous on `[0, 1]` [because rational functions are continuous on their domains and the domain of `f` includes the closed interval]

and differentiable on `(0, 1)`, [because `f'(x) = (2-2x^2)/(x^2+1)^2` exists for all `x` in the open interval.]

so the hypotheses of the Mean Value Theorem are true.

The conclusion of the Mean Value Theorem is that there is a number `c` in `(0, 1)` with

`f'(c) = (f(1) - f(0))/((1) - (0))` .

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

For `f(x)=(2x)/[(x^2)+1]`, we have `f(1) =1` and `f(0) = 0`, so we need to solve

`f'(c) = (1 - 0)/1 = 1`

Furthermore `f'(x) = (2-2x^2)/(x^2+1)^2`

We need to solve:

`(2-2x^2)/(x^2+1)^2 = 1` clearing denominator and expanding gets us:

`2-2x^2 = x^4 + 2x^2 +1`

So `x^4 +4x-1=0`

Complete the square or use the quadratic formula to solve for `x^2`.

`x^2 = (-4 +- sqrt 20)/2 = -2+-sqrt5`

Clearly `x^2 = -2-sqrt5` will lead to imaginary values of `x`.

`x^2 = -2+sqrt5` leads to `x = +-sqrt(sqrt5 - 2)`

Again clearly, the negative root is not in `(0,1)`, so the number we are looking for must be

`c = sqrt(sqrt5 - 2)`