Continuity and Differentiability of Piecewise Defined Functions

[ryu1]

Homework Statement

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

The Attempt at a Solution

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!

 

[LCKurtz]

Homework Statement

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

The Attempt at a Solution

So I managed to prove it is continuous for all x by checking the limits as x -> 1 from both directions = 0

You also need to check at x = -1

and the limit as x -> 0 from both directions = -1/2 (is that necessary?)

No. It is a polynomial there.

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 ?

There is no problem at x=0. The problem is at x = 1 and -1 where the two functions piece together. You need to check the function values and slopes there.

 

[ryu1]

THANKS a lot you helped me solved this at last!