Absolute value and differentiability

[Niles]

Homework Statement

Hi all

I have f(x)=|x|. This I write as

f(x) = -x for x<0
f(x) = x for x>0
f(x) = 0 for x=0

If I want to show that f(x) is not differentiable at x=0, then is it enough to show that

f'(x) = -1 for x<0
f'(x) = 1 for x>0

and from this conclude that it is not differentiable at x=0? I am a little worried about f(x) at x=0.

 

[vela]

You're close, but those two statements about f'(x) really only tell you about the derivative at points other than x=0. You need to address f'(0) explicitly and say why it doesn't exist. By definition, you have that

[tex]f'(0) = \lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h}[/tex]

What are the requirements for this limit to exist?

 

[Niles]

That the limit is the same when approaching it from both sides?

So if I write what I wrote in my OP and add that f'(0) does not exist because it is not the same when going to 0+ and 0-, then f'(0) is not defined?

 

[tiny-tim]

That the limit is the same when approaching it from both sides?

So if I write what I wrote in my OP and add that f'(0) does not exist because it is not the same when going to 0+ and 0-, then f'(0) is not defined?

Hi Niles!

As vela has indicated, you're missing the point …

you don't need to use results related to one-sided differentials (and you'd have to state those results, and prove that they give you the answer you want, which would be very roudn-about) …

just use the definition of f'(0) (choose a delta, and describe what happens ).

 

[Niles]

just use the definition of f'(0) (choose a delta, and describe what happens ).

Uhm, I'm a little confused now. I understand that my mission is to show that f'(0) does not exist.

I can show that it isn't continuous (is that what you mean when you say choose a delta?), but differentiability => continuity, so showing that it is not continuous won't help me that much

 

[tiny-tim]

You're making this too complicated …

forget f'(x), just use (f(x+h)-f(x))/h …

[STRIKE]describe how (f(x+h)-f(x))/h behaves around x= 0, and show that that means that it has no limit.[/STRIKE]

describe how (f(h)-f(0))/h behaves around h = 0, and show that that means that it has no limit.

 

[Niles]

describe how (f(h)-f(0))/h behaves around h = 0, and show that that means that it has no limit.

For x<0 it is -1, and for x>0 it is 1, but I am not sure this is what you are referring to?

 

[tiny-tim]

For x<0 it is -1, and for x>0 it is 1, but I am not sure this is what you are referring to?

Yes. And so does it have a limit as h tends to 0?

 

[Niles]

No, it does not, so it's not differentiable there. Thanks!