# a differential funtion

• Aug 11th 2009, 08:41 PM
Kat-M
a differential funtion
Assume $\displaystyle f:R \rightarrow R$ has a finite derivative everywhere on $\displaystyle (a,b)$ except possibly at a point $\displaystyle c \in (a,b)$. If we have $\displaystyle lim_{x \rightarrow c} f'(x)=A$, then prove $\displaystyle f'(c)$ exists and is equal to $\displaystyle A$.
• Aug 12th 2009, 12:14 AM
flyingsquirrel
Hello,

Hint : use the mean value theorem.
• Aug 12th 2009, 06:05 AM
Kat-M
mean 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, 10:29 AM
alunw
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, 10:33 AM
Failure
Quote:

Originally Posted by Kat-M
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.

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, 10:35 AM
Opalg
Quote:

Originally Posted by Kat-M
Assume $\displaystyle f:R \rightarrow R$ has a finite derivative everywhere on $\displaystyle (a,b)$ except possibly at a point $\displaystyle c \in (a,b)$. If we have $\displaystyle lim_{x \rightarrow c} f'(x)=A$, then prove $\displaystyle f'(c)$ exists and is equal to $\displaystyle A$.

The question does not actually say that $\displaystyle f$ is continuous at $\displaystyle c$. 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 $\displaystyle \lim_{x\to c}f'(x) = A$. So given $\displaystyle \varepsilon>0$ there exists $\displaystyle \delta>0$ such that $\displaystyle |f'(\xi)-A|<\varepsilon$ whenever $\displaystyle |\xi-c|<\delta$.

Now suppose that $\displaystyle |x-c|<\delta$ and $\displaystyle |y-c|<\delta$. By the mean value theorem (as hinted by flyingsquirrel (Hi)) $\displaystyle \frac{f(x) - f(y)}{x-y} = f'(\xi)$ for some $\displaystyle \xi$ between $\displaystyle x$ and $\displaystyle y$. But if $\displaystyle x$ and $\displaystyle y$ are both within distance $\displaystyle \delta$ from $\displaystyle c$ then so is $\displaystyle \xi$. It follows that $\displaystyle \left|\frac{f(x) - f(y)}{x-y} - A\right|<\varepsilon$.

Next, let $\displaystyle y\to c$ in that last inequality. By continuity of $\displaystyle f$ at $\displaystyle c$, this gives $\displaystyle \left|\frac{f(x) - f(c)}{x-c} - A\right|\leqslant\varepsilon$. Since this holds whenever $\displaystyle |x-c|<\delta$, and $\displaystyle \varepsilon$ is an arbitrary positive number, it follows that $\displaystyle f$ is differentiable at $\displaystyle c$, with derivative $\displaystyle A$.
• Aug 12th 2009, 11:12 AM
Failure
Quote:

Originally Posted by Opalg

Now suppose that $\displaystyle |x-c|<\delta$ and $\displaystyle |y-c|<\delta$. By the mean value theorem (as hinted by flyingsquirrel (Hi)) $\displaystyle \frac{f(x) - f(y)}{x-y} = f'(\xi)$ for some $\displaystyle \xi$ between $\displaystyle x$ and $\displaystyle y$.

How can you apply the mean value theorem here if you don't know (yet) that f is differentiable in the entire intervall between x and y?
• Aug 12th 2009, 11:58 AM
Opalg
Quote:

Originally Posted by Failure
How can you apply the mean value theorem here if you don't know (yet) that f is differentiable in the entire interval between x and y?

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.