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Homework Help: Absolute value and differentiability

  1. Jan 16, 2010 #1
    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. jcsd
  3. Jan 16, 2010 #2

    vela

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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

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

    What are the requirements for this limit to exist?
     
  4. Jan 16, 2010 #3
    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?
     
  5. Jan 16, 2010 #4

    tiny-tim

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    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:).
     
  6. Jan 16, 2010 #5
    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 :confused:
     
  7. Jan 16, 2010 #6

    tiny-tim

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    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: Jan 16, 2010
  8. Jan 16, 2010 #7
    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?
     
  9. Jan 16, 2010 #8

    tiny-tim

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    Yes. And so does it have a limit as h tends to 0? :smile:
     
  10. Jan 16, 2010 #9
    No, it does not, so it's not differentiable there. Thanks!
     
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