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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 ?

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    $\begingroup$ Differentiability on $(a,b)$ does not imply continuity in $a$ or $b$. The continuity in the endpoints must be explicitly demanded (or the differentiability also in $a$ resp. $b$, but that would be a stronger hypothesis than required). $\endgroup$ – Daniel Fischer Jun 22 '14 at 20:11
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    $\begingroup$ Note the difference in the sets between $1$ and $2$ - it would be a stronger statement to say $f$ needs to be differentiable on the closed interval $[a,b]$ $\endgroup$ – DanZimm Jun 22 '14 at 20:11
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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 $[a,b]$.

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