TITLE: Premises of the Mean Value Theorem
QUESTION [7 upvotes]: Why does the statement of the mean value theorem requires that:
(1)The function $f$ be continuous on the closed interval $[a,b]$ 
(2)Differentiable on the open interval $(a,b)$. 
Couldn't we just require (2) and the the first premise will be met because of the fact that differientiability implies continuity ?

REPLY [8 votes]: Consider the function $$f(x) = \left\{\begin{array}{cc} \frac 1 x &: x \ne 0 \\ 0 &: x = 0\end{array}\right.$$
Then $f$ is differentiable on the interval $(0, 1)$, but not continuous on $[0,1]$; furthermore, the conclusion of the MVT does not hold on the interval $[0,1]$, since
$$f'(c) \ne 1 = \frac{f(1) - f(0)}{1 - 0}$$
for any $c \in (0, 1)$. 

Note that differentiability on $(a, b)$ implies continuity on $(a, b)$, but not on $[a,b]$.