Look at the graph.
Doesn't the derivative exist everywhere?
I have this problem I haven been trying to solve for a while:
"Check if the following function is continuous and/or differentiable :"
/ (x^2-1) /2 , |x|=< 1
f(x) = \ |x| -1 , |x| > 1
So I managed to prove it is continuous for all x by checking the limits as x -> 1 from both directions = 0
and the limit as x -> 0 from both directions = -1/2 (is that necessary?)
from that point it's continuous for all x as a polynomial in either branch.
is that correct so far?
now the problem starts with the derivative check...
I get that the f'(x) = x , |x| < 1
or f'(x) = x/|x| , |x| > 1
so does that alone means the function isn't differentiable in x = 0 ?
Thank you for your help!
it is clearer (it is also correct?) to write it like this:
|x| -1 , |x| >1 <=> / x-1 , x > 1
...........................\-x-1 , x < -1
Thanks a lot!