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cinar

  • 3 years ago

I have question..

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  1. cinar
    • 3 years ago
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    Suppose that f:R to R is differentiable at c and that f(c)=0. Show that g(x):=|f(x)| is differentiable at c if and only if f'(c)=0.

  2. TuringTest
    • 3 years ago
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    \[g'(x)=\frac{f(x)}{|f(x)|}\cdot f'(x)\]so I guess I'm stuck on showing how to deal with the intederminacy of plugging in x=c=0

  3. TuringTest
    • 3 years ago
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    I guess we must show that\[\frac{f(0)}{|f(0)|}\neq0\]

  4. TuringTest
    • 3 years ago
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    can't get l'Hospital to yield anything really...

  5. cinar
    • 3 years ago
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    \[\Large f'(x)=|\frac{f(x)-f(0)}{x-c}-f'(c)|<\epsilon\]

  6. cinar
    • 3 years ago
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    this is the epsilon-delta definition of diff. at c..

  7. cinar
    • 3 years ago
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    sorry should be this\[\Large f'(x)=|\frac{f(x)-f(c)}{x-c}-f'(c)|<\epsilon\]

  8. TuringTest
    • 3 years ago
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    yeah that looks a bit better...

  9. TuringTest
    • 3 years ago
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    oh, @mahmit2012 is here, he can likely help.

  10. mahmit2012
    • 3 years ago
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    |dw:1354859726253:dw|

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