# Homework Help: Absolute value and differentiability

1. Jan 16, 2010

### Niles

1. The problem statement, all variables and given/known data
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.

2. Jan 16, 2010

### vela

Staff Emeritus
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

$$f'(0) = \lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h}$$

What are the requirements for this limit to exist?

3. Jan 16, 2010

### 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?

4. Jan 16, 2010

### tiny-tim

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 ).

5. Jan 16, 2010

### Niles

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 wont help me that much

6. Jan 16, 2010

### 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.

Last edited: Jan 16, 2010
7. Jan 16, 2010

### Niles

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

8. Jan 16, 2010

### tiny-tim

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

9. Jan 16, 2010

### Niles

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