# Query on Mean Value Theorem Criteria

In the theorem why do we consider that the function is continuous on the closed interval but differentiable on the open interval? What difference it would make if in both the cases, closed interval is considered.

Explanation with an example to clear this query would be helpful.

• What is the question? Why don't you need differentiability at the endpoints? – gt6989b Dec 18 '13 at 17:35
• The point is that differentiability on the closed interval is not required. – André Nicolas Dec 18 '13 at 17:35
• What is the reason for that? Why cannot I say closed interval simply for both continuity and differentiability? – Anirban Ghosh Dec 18 '13 at 17:37
• You can, but then you get a less general theorem. – André Nicolas Dec 18 '13 at 17:38

An example of such a function is $f(x) = x\sin(\frac{1}{x})$ on the interval $(0,1]$ and $f(0) = 0$. This is continuous on $[0,1]$ and differentiable on $(0,1)$ but not differentiable at $0$.
Another reason is that the proof of the Mean Value Theorem uses Rolle's Theorem, which does not require differentiability on a closed interval $[a,b]$ but an open interval $(a,b)$.