Absolute value and differentiability

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Homework Help Overview

The discussion revolves around the differentiability of the function f(x) = |x|, particularly at the point x = 0. Participants are exploring the implications of the derivative's behavior on either side of this point.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the necessity of evaluating the derivative at x = 0 and the conditions under which it exists. There is a focus on the definition of the derivative and the behavior of limits from both sides.

Discussion Status

Several participants have provided guidance on how to approach the problem, emphasizing the importance of using the definition of the derivative at x = 0. There is an ongoing exploration of the continuity of the function and its relation to differentiability.

Contextual Notes

There is some confusion regarding the relationship between continuity and differentiability, as well as the specific requirements for limits to exist at x = 0. Participants are questioning assumptions about the function's behavior near this point.

Niles
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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.
 
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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&#039;(0) = \lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h}

What are the requirements for this limit to exist?
 
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?
 
Niles said:
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! :smile:

As vela :smile: 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 :wink:).
 
tiny-tim said:
just use the definition of f'(0) (choose a delta, and describe what happens :wink:).

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 :confused:
 
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. :smile:
 
Last edited:
tiny-tim said:
describe how (f(h)-f(0))/h behaves around h = 0, and show that that means that it has no limit. :smile:

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?
 
Niles said:
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? :smile:
 
No, it does not, so it's not differentiable there. Thanks!
 

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