Assume has a finite derivative everywhere on except possibly at a point . If we have , then prove exists and is equal to .

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- Aug 11th 2009, 09:41 PMKat-Ma differential funtion
Assume has a finite derivative everywhere on except possibly at a point . If we have , then prove exists and is equal to .

- Aug 12th 2009, 01:14 AMflyingsquirrel
Hello,

Hint : use the mean value theorem. - Aug 12th 2009, 07:05 AMKat-Mmean value theorem
i thought about it but is it possible to even use it? f has to be ontinuous on the endpoints and has to be differentiable on (a,b). but we don know if it is continuoust at a and b or it is diffentiable at c.

- Aug 12th 2009, 11:29 AMalunw
You cannot hope to prove this unless you are given that f is continuous at c. Otherwise you can have a function that is continuous and differentiable everywhere in the interval apart from c but which takes an arbitrary value at c itself, or you could have a function that looks like tan(x) but with a small section round pi/2 removed. Then there is a huge discontinuity at c, and yet the derivative is the same both sides of c.

- Aug 12th 2009, 11:33 AMFailure
We must be allowed to assume that f is continuous on (a,b), otherwise the propostion would be clearly false (because in that case f(c) could be anything).

So, assuming f is continuous on (a,b), I suggest that you use the mean value theorem, applied to (a,c] and [c,b), respectively, to show that the*left*and the*right*derivative of f at c both exist and are both equal to A. - Aug 12th 2009, 11:35 AMOpalg
The question does not actually say that is continuous at . But it better had be, as noted by

**alunw**and**Failure**. With that assumption, the result is true, but it seems to be quite a tricky piece of analysis, requiring a very careful proof.

We are told that . So given there exists such that whenever .

Now suppose that and . By the mean value theorem (as hinted by**flyingsquirrel**(Hi)) for some between and . But if and are both within distance from then so is . It follows that .

Next, let in that last inequality. By continuity of at , this gives . Since this holds whenever , and is an arbitrary positive number, it follows that is differentiable at , with derivative . - Aug 12th 2009, 12:12 PMFailure
- Aug 12th 2009, 12:58 PMOpalg
Good point. I should have said that y must be on the same side of c as x. Then I think the proof should work. But in fact your comment #5 above gives a shorter proof: once you know that f is continuous on the closed interval [c,x], you can apply the MVT on that interval and the result comes out straight away.