Respuesta :
The mean value theorem states that, given a function [tex] f(x) [/tex] continuous in some interval [tex] [a,b] [/tex] and differentiable in the same open interval [tex] (a,b) [/tex], 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, the function [tex] f(x) = x+1 [/tex] is a polynomial, and as such it is continuous and differentiable infinite times on the whole real number set. So, in particular, it is continuous in [tex] [a,b] = [0,63] [/tex] and differentiable in the same open interval [tex] (a,b) = (0,63) [/tex].
Moreover, the derivative of the function is
[tex] f'(x) = 1 [/tex]
So, the statement of the mean value theorem becomes the following: there exists a point [tex] c \in (0,63) [/tex] such that
[tex] f'(c) = \cfrac{f(63)-f(0)}{63-0} [/tex]
But we know that [tex] f'(c) [/tex] is constantly 1, so we have
[tex] 1 = \cfrac{64-1}{63-0} = \cfrac{63}{63} [/tex]
which is an identity, i.e. an equation which is always verified.
So, every point in [tex] (0,63) [/tex] satisfies the mean value theorem.
Why? Well, the theorem states the existance of a point such that the tangent to that point (which is [/tex] f'(c) [/tex]) has the same slope as the secant connecting [tex] f(b) [/tex] and [tex] f(a) [/tex].
But in this case, [tex] f(x) [/tex] is a line, which means that the said secant and the function itself are actually the same thing.
