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]?

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)#