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.

  • $\begingroup$ What is the question? Why don't you need differentiability at the endpoints? $\endgroup$ – gt6989b Dec 18 '13 at 17:35
  • $\begingroup$ The point is that differentiability on the closed interval is not required. $\endgroup$ – André Nicolas Dec 18 '13 at 17:35
  • $\begingroup$ What is the reason for that? Why cannot I say closed interval simply for both continuity and differentiability? $\endgroup$ – Anirban Ghosh Dec 18 '13 at 17:37
  • 1
    $\begingroup$ You can, but then you get a less general theorem. $\endgroup$ – André Nicolas Dec 18 '13 at 17:38

The only difference is that the usual way applies to more functions. There are functions which are continuous on a closed interval and differentiable on the open interval, but not differentiable on the closed interval, and we'd like to apply the mean value theorem to them as well. Since the theorem is true when we only assume differentiability on the open interval, that's what we use.

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 $(a,b)$.


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