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Differentiability of weird function

  1. Oct 12, 2005 #1
    Let f(x) = x if x rational and f(x) = 0 if x is irrational

    Let g(x) = x^2 if x rational and g(x) = 0 if x is irrational.

    Both functions are continuous at 0 and discontinuous at each x != 0.

    How do I show that f is not differentiable at 0?
    How should I show that g is differentiable at 0? Give g'(0) as well.

    Thanks...
     
  2. jcsd
  3. Oct 12, 2005 #2

    HallsofIvy

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    Use the definition: [tex]f'(0)= lim_{h->0}\frac{f(h)}{h}[/tex].
    but [tex]\frac{f(h)}{h}[/tex] is 1 if x is rational, 0 if x is irrational. What does that tell you about the limit?

    [tex]\frac{g(h)}{h}[/tex] is h is x is rational, 0 if x is irrational.
     
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