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
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?
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.