Hi!

I must prove that the function f(x) is continuously differentiable function (just first derivate) on interval (-1,1).

for all other values.

I do that:

Is it ok?

Thanks in advance

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- January 8th 2011, 07:58 AMtom27differentiable function
Hi!

I must prove that the function f(x) is continuously differentiable function (just first derivate) on interval (-1,1).

for all other values.

I do that:

Is it ok?

Thanks in advance - January 8th 2011, 08:43 AMFernandoRevilla
If the question it is only related to then, it is right (evidently is continuous in ).

Fernando Revilla - January 8th 2011, 08:54 AMtom27
The basic question is if the f(x) is a part of space?

- January 8th 2011, 09:09 AMHallsofIvy
That's actually a different question from what you first asked. In your first post you said "continuously differentiable function (just first derivate) on interval (-1,1)". Now you are saying on all of . You had already answered the first question- f(x) is continuously differentiable on (-1, 1). Since f is defined to be a constant (0) for or , it is certainly continuously differentiable there. But now the question is about x= -1 and x= 1. What are the derivatives at those points? What are the derivatives on either side? Is the derivative continuous at x= -1 and x= 1?

- January 8th 2011, 09:11 AMFernandoRevilla
In that case, you should prove and continuous at (for the rest is almost trivial)

Fernando Revilla

P.S. I suppose means (the standard notation for the set of continuosly differentiable functions on ) .

Edited: Sorry, I didn't see**HallsofIvy**'s post - January 8th 2011, 11:14 AMtom27
How can I prove ?

I tried with limits:

and

but this certainly is not the correct way. - January 8th 2011, 04:02 PMHallsofIvy
While derivatives are not necessarily continuous, they do obey the "intermediate value property"- f'(x) takes on all values between f'(a) and f'(b) for x between a and b. In particular, that means that f'(c) exists if and only if . That is, determine the derivatives for x< -1, and x>-1 and determine if the limits at x= -1 are the same. Do the same for x= 1.

By the way, you**don't**want to prove that " ". You want to prove that and . You don't need to use the limit definition of the derivative- but, of course, h goes to 0, not 1 and -1. - January 9th 2011, 03:40 AMtom27
So and i must calculate:

but how can I do that? The exp function is not defined in those two points! - January 10th 2011, 07:35 AMcantona
Fuction with exp is not define in those two points so limit does not exist. But wikipedia says: the function is smooth so is continuosly differentiable on . But I dont know how to prove that if one limit does not exist so ?