Differentiable Function on an interval

[Punkyc7]

Let f:[a,b]\rightarrowR be continuous on [a,b] and differentiable in (a,b). Show that if lim f'(x)=A as x goes to a then f'(a) exist and equals A.


So I was thinking this has to do either with the mean value theorem or Darboux's Theorem.

I have that
f(b)-f(a)=f'(c)(b-a) by the mean value theorem.

From here I stuck on how to get the x into the equation.
Would I say that let x=c.

Then we have
f'(x)=\frac{f(b)-f(a)}{b-a}=A


If so how would I work in the f'(a)?

 

[math man]

It really comes down to the definitions of continuity and differentiability. They are asking you to show that:
if f(x) is continuous for all x \in [a,b]
and if f(x) is differentiable for all x \in (a,b)
and if lim f'(x) as x->a exists
then f'(a) exists

If the limit exists for all x0 ∈ (a, b) then f is said to be differentiable in the interval.
It is differentiable at the end points if the appropriate one-sided limits exists.
-http://www.math.ku.edu/~lerner/m500f09/Differentiability.pdf

f(x) is differentiable in the interval (a,b) and the one-sided limit as x->a of f'(x) exists, therefore f'(a) exists. If f'(a) exists, then f'(x) is differentiable at x=a. Differentiability implies continuity, and so f'(x) would also be continuous at x=a. If f'(x) is continuous at x=a, then you can say by the definition of continuity that:
\stackrel{lim}{x→a} f'(x) = f'(a) = A

 

[HallsofIvy]

It really comes down to the definitions of continuity and differentiability. They are asking you to show that:
if f(x) is continuous for all x \in [a,b]
and if f(x) is differentiable for all x \in (a,b)
and if lim f'(x) as x->a exists
then f'(a) exists


f(x) is differentiable in the interval (a,b) and the one-sided limit as x->a of f'(x) exists, therefore f'(a) exists. If f'(a) exists, then f'(x) is differentiable at x=a. Differentiability implies continuity, and so f'(x) would also be continuous at x=a. If f'(x) is continuous at x=a, then you can say by the definition of continuity that:
\stackrel{lim}{x→a} f'(x) = f'(a) = A

"If f'(a) exists, then f'(x) is differentiable at x= a" That is, that f is twice differentiable at x= a? Why is that true?

It looks to me like the statement in the first post is not even true. Let f(x)= |x|. Then f is continuous for all x and so on [0, 1]. Further, f is differentiable in (0, 1)- its derivative is just 1 for all x in (0, 1). The limit of f'(x), as x goes to 0, is, of course, 1 but f is NOT differentable at x= 0.

 

[math man]

Sorry, you are correct HallsofIvy. I don't know how I missed that. Well, at least the first part is correct, f'(a) exists. I guess the question then is does continuity and differentiability of f imply continuity of df/dx? I know that f(x) being differentiable alone doesn't necessarily prove it, but if f(x) is also continuous then I see no reason why df/dx is not continuous. I have googled it extensively with not much luck.

 

[math man]

What about this...
f'(x) = \stackrel{lim}{s→x}\frac{f(s)-f(x)}{s-x}

\stackrel{lim}{x→a}f'(x)
= \stackrel{lim}{(x→a)} \stackrel{lim}{(s→x)}\frac{f(s)-f(x)}{s-x}
= \stackrel{lim}{s→a}\frac{f(s)-f(a)}{s-a} = f'(a)

\stackrel{lim}{x→a}f'(x) = f'(a) = A