# Continuity and differentiation

• Apr 26th 2013, 07:04 AM
ryu1
Continuity and differentiation
Hi

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 ?

• Apr 26th 2013, 07:57 AM
Plato
Re: Continuity and differentiation
Quote:

Originally Posted by ryu1
Hi

I have this problem I haven been trying to solve for a while:

"Check if the following function is continuous and/or differentiable :"
Code:

        / (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

Look at the graph.

Doesn't the derivative exist everywhere?
• Apr 26th 2013, 08:05 AM
ryu1
Re: Continuity and differentiation
it is but the abs in the domain(is that what it's called? where you write which x's can "go in the function") confused me, how to prove it ?
• Apr 26th 2013, 08:15 AM
Plato
Re: Continuity and differentiation
Quote:

Originally Posted by ryu1
it is but the abs in the domain(is that what it's called? where you write which x's can "go in the function") confused me, how to prove it ?

$f(x) = \left\{ {\begin{array}{rl}{x-1,}&{x > 1}\\{\frac{{{x^2} - 1}}{2},}&{ - 1 \le x \le 1}\\{ - x-1,}&{x < - 1}\end{array}} \right.$

So
$f'(x) = \left\{ {\begin{array}{rl}{1,}&{x > 1}\\{x,}&{ - 1 \le x \le 1}\\{ - 1,}&{x < - 1}\end{array}} \right.$

What about that can't you understand?
• Apr 26th 2013, 09:32 AM
ryu1
Re: Continuity and differentiation
how you branched the |x| > 1 domain and function?
I kind of understand but it is slightly vague to me ...

• Apr 26th 2013, 09:40 AM
Plato
Re: Continuity and differentiation
Quote:

Originally Posted by ryu1
how you branched the |x| > 1 domain and function?
I kind of understand but it is slightly vague to me ...

It is simply a matter of definition.

$|x|$ is simply the distance $x$ is from $0$.

Thus $\text{If }|x|>1\text{ then }x>1\text{ or }x<-1~.$
• Apr 26th 2013, 09:53 AM
ryu1
Re: Continuity and differentiation
Quote:

Originally Posted by Plato
It is simply a matter of definition.

$|x|$ is simply the distance $x$ is from $0$.

Thus $\text{If }|x|>1\text{ then }x>1\text{ or }x<-1~.$

I remember that, it just that the branch there gave me a small headache because it has an abs in the domain and the function itself...
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!