If the question it is only related to then, it is right (evidently is continuous in ).
Fernando Revilla
If the question it is only related to then, it is right (evidently is continuous in ).
Fernando Revilla
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?
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
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.