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Math Help - Prove function is differentiable

  1. #1
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    Prove function is differentiable

    Prove that the function

    f(x) = x^2 * sin (1/x) if x!=0
    = 0 if x=0

    is differetiable in (-infinity, infinity), but has no second derivative at x=0
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jzellt View Post
    Prove that the function

    f(x) = x^2 * sin (1/x) if x!=0
    = 0 if x=0

    is differetiable in (-infinity, infinity), but has no second derivative at x=0
    This is not differentiable at x=0.

    CB
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  3. #3
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    Quote Originally Posted by jzellt View Post
    Prove that the function

    f(x) = x^2 * sin (1/x) if x!=0
    = 0 if x=0

    is differetiable in (-infinity, infinity), but has no second derivative at x=0
    Sorry, Captain Black, but this function definitely is differentiable at x= 0.

    Use the definition of the derivative: \lim_{h\to 0}\frac{f(0+h)- f(0}{h}

    As long as h is not 0, f(h)= h^2 sin(1/h) so \frac{f(h)- f(0)}{h}= \frac{h^2 sin(1/h)}{h}= h sin(1/h).

    -1\le sin(1/h)\le 1 for all non-zero h while h goes to 0 so the derivative at x= 0 is f'(0)= \lim_{h\to 0} h sin(1/h)= 0.

    Notice that for x not 0, the derivative is given by the product rule:
    f'(x)= 2x sin(1/x)+ x^2 cos(1/x)(-1/x^2)= 2x sin(1/x)- cos(1/x).

    Now, that has no limit as x goes to 0 so while f is differentiable for all x, its derivative is not continuous at x= 0. That means that f' is not differentiable at x= 0 and so f is not twice differentiable at x= 0.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by HallsofIvy View Post
    Sorry, Captain Black, but this function definitely is differentiable at x= 0.

    Oppss.. I was looking at the limit of the derivative as x \to 0
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