Respuesta :
The mean value theorem requires that a function [tex] f(x) [/tex] is continuous on the interval [tex] [a,b] [/tex] and differentiable on the same open interval [tex] (a,b) [/tex].
Since the only point of discontinuity of your function is [tex] x = 0 [/tex], where the denominator of 2/x equals zero, the function is defined, continuous and differentiable on the interval [tex] [1,2] [/tex].
Since the hypothesis of the mean value theorem are verified, we can claim that there exists a point [tex] c \in (a,b) [/tex] such that
[tex] f'(c) = \cfrac{f(b)-f(a)}{b-a} [/tex]
In your case, [tex] b = 2, a = 1 [/tex], so the expression becomes
[tex] f'(c) = \cfrac{f(2)-f(1)}{2-1} = \cfrac{3-3}{1} = 0 [/tex]
Let's compute the derivative:
[tex] f(x) = x+\cfrac{2}{x} \implies f'(x) = 1-\cfrac{2}{x^2} [/tex]
So, we're looking for a point c such that
[tex] f'(c)=0 \implies 1-\cfrac{2}{c^2} = 0 \iff 1 = \cfrac{2}{c^2} \iff c^2=2 \iff c = \pm\sqrt{2} [/tex]
Since we are looking for a point in [tex] (1,2) [/tex], we accept the solution [tex] c = \sqrt{2} [/tex]
