# Math Help - a differential funtion

1. ## a differential funtion

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

2. Hello,

Hint : use the mean value theorem.

3. ## 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.

4. 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.

5. 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.

6. Originally Posted by Kat-M
Assume $f:R \rightarrow R$ has a finite derivative everywhere on $(a,b)$ except possibly at a point $c \in (a,b)$. If we have $lim_{x \rightarrow c} f'(x)=A$, then prove $f'(c)$ exists and is equal to $A$.
The question does not actually say that $f$ is continuous at $c$. 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 $\lim_{x\to c}f'(x) = A$. So given $\varepsilon>0$ there exists $\delta>0$ such that $|f'(\xi)-A|<\varepsilon$ whenever $|\xi-c|<\delta$.

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

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

7. Originally Posted by Opalg

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

8. 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.