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Math Help - Show it's twice differentiable

  1. #1
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    Show it's twice differentiable

    f(x) = 0 x=0
    e^(-1/x^2) x/=0

    how?

    to find
    f''(0) = limit( (f'(x)-f'(0))/x,x, goes to 0), but we dont know f'(0)?
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  2. #2
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    First thing first, do you know to find the derivative? To be differential at a point, it must be continuous, so we need to make sure the limit as f approaches zero is equal to zero. We also need to find the limit as f' approaches zero, and if that equals zero, then we are good to go.
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  3. #3
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    Quote Originally Posted by colby2152 View Post
    First thing first, do you know to find the derivative? To be differential at a point, it must be continuous, so we need to make sure the limit as f approaches zero is equal to zero. We also need to find the limit as f' approaches zero, and if that equals zero, then we are good to go.

    sure, i show it's continuous
    it's differentiable elsewhere. so need to show when x=0, its dirivative exists
    i show it by limit( (f(0+h)-f(0))/h),h goes to zero), this works.
    then to show it's twice diff, i used the same method, but, we dont know f'(0), do we?
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  4. #4
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    Quote Originally Posted by szpengchao View Post
    sure, i show it's continuous
    it's differentiable elsewhere. so need to show when x=0, its dirivative exists
    i show it by limit( (f(0+h)-f(0))/h),h goes to zero), this works.
    then to show it's twice diff, i used the same method, but, we dont know f'(0), do we?
    The derivative, f'(x) is equal to:

    f'(x) = lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

    To see if it is continuous at zero, take the limit as f'(x) approaches zero. It must be equal to zero.
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