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Proving Differentiability of a Piece-wise Function

  1. Nov 15, 2011 #1
    1. Suppose f(x)=0 if x is irrational, and f(x)=x if x is rational. Is f differentiable at x=0?



    2. the derivative= lim[h->0] [f(a+h)-f(a)]/h



    3. I don't really know how to start, but I do know that between any two real numbers, there exists a rational and irrational number. So I'm guessing that has something to do with solving for the answer. f'(0)= lim[h->0] f(h)/h
     
  2. jcsd
  3. Nov 15, 2011 #2

    Dick

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    Ok, so there are rational numbers as close as you want to 0. And irrational numbers. Pick h to be rational. What's f(h)/h? Same question for irrational.
     
  4. Nov 15, 2011 #3
    For rational,
    f'(0)= lim[h->0] f(h)/h= h/h=1

    For irrational,
    f'(0)= lim[h->0] f(h)/h= 0/h=0

    Therefore, the two limits do not equal each other, meaning that f'(0) does not exist?
    Was it that easy?
     
  5. Nov 15, 2011 #4

    Dick

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    Yes. If you are clear why you can pick h values arbitrarily close to zero that are both rational and irrational. And I think you are.
     
  6. Nov 15, 2011 #5
    So its like doing the left-hand and right-hand limits, except we are using the piece-wise function to our advantage knowing that x can be rational and irrational, and since there are an infinite number of rational and irrational numbers approaching zero, those are our 'left' and 'right'.

    Thanks a lot.
     
  7. Nov 15, 2011 #6

    Dick

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    Exactly. Very welcome.
     
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