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So I'm beginning numerical analysis and an interesting thing was brought up in class. I know the rules for MVT are:

  1. $F$ is continuous on $[a,b]$
  2. $F$ is differentiable on $(a,b)$

So a question was brought up in class I couldn't answer. Why is it that we don't consider differentiability at the end points? The professor sort of assumed we knew the answer but I didn't. My calculus is rusty.

Is there any reason we can't know why the end-points aren't differentiable? A person in class suggested it had to do something with not knowing whats on the other side of the point, but that didn't make any sense to me. Can anyone break this down for me so I can really understand it?

Thank you!

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    $\begingroup$ We don't need differentiability in the endpoints. So we don't require it. $\endgroup$
    – Daniel Fischer
    Aug 29, 2014 at 15:50
  • $\begingroup$ We dont require it, but why? To me it seems like knowing differentiability does not provide any more or less information, so why not just make rule (2) [a,b] instead of (a,b)? $\endgroup$
    – John S.
    Aug 29, 2014 at 15:55
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    $\begingroup$ One desires to make the weakest [well, as long as they're simple enough; otherwise there's a trade-off between simplicity and weakness of conditions] requirements that guarantee the conclusion. Demanding differentiability only on the interior of the interval and not specifying anything about differentiability or not in the endpoints is a weaker requirement than saying anything about differentiability in either endpoint. It suffices for the conclusion. So one does not require anything about differentiability in the endpoints in the statement of the theorem. $\endgroup$
    – Daniel Fischer
    Aug 29, 2014 at 15:59
  • $\begingroup$ So, it's entirely possible the end points of a function that satisfies the MVT are differentiable? If that's so, would you mind giving me an example? Thanks for all your help $\endgroup$
    – John S.
    Aug 29, 2014 at 16:05
  • $\begingroup$ Yes, of course. Take $f(x)=x^2$ on $[0,1]$, e.g.. $\endgroup$
    – David Mitra
    Aug 29, 2014 at 16:10

2 Answers 2

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Technically, the limit definition of the derivative involves a two-sided limit; that is, in order for the derivative to exist at $a$ (or at $b$), we must be able to get arbitrarily close to $a$ (or to $b$) from either side. However, in the case of $a$, we can only approach $a$ from the right; similarly, in the case of $b$, we can only approach $b$ from the left. Hence, without any further information, we cannot conclude that the function is differentiable at $a$ or at $b$.

That being said, we may very well have a problem where the function is differentiable over $R$ (and thus continuous over $R$) and so, differentiable at the endpoints $a$ and $b$. But that is not necessary to know for the purpose of applying the MVT.

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My reasoning on this is continuity at the points a and b is needed, but not differentiability. Geometrically, MVT says that if $f$ has a continuous curve on the interval $[a,b]$, and the curve has a tangent on every point in it with abscissa between a and b, then, there is a point c in $(a,b)$ such that the tangent to the curve at $(c,f(c))$ is parallel to the line segment joining $(a,f(a))$ and $(b,f(b))$. This requires the continuity but not the differentiability. But had $f$ been differentiable at a,b, it wouldn't make any difference anyway. It simply isn't needed for drawing the curve.

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