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Math Help - differentiability

  1. #1
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    differentiability

    Suppose g:\mathbb{R}\rightarrow{\mathbb{R}} is a twice differentiable function with g(0)=g'(0)=0 and g''(0)=14. Let f:\mathbb{R}\rightarrow{\mathbb{R}} be defined by
    f(x)=\frac{g(x)}{x} if x\neq{0}, 0 if x=0
    Prove that f is differentiable at x=0 and find f'(0).


    Can we use L' Hopital's Rule to solve this question?
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  2. #2
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    Re: differentiability

    Use L' Hopital's Rule to find the limit of f(x) at x=0 first.

    f(x) = g(x)/x, since g is a differentiable function,
    therefore f(x) must be differentiable somewhere on R.
    assume f'(x) exists at point x=k, then

    f'(k) = [k g'(k) - g(k) ]/k^2. (by quotient rule)
    take above expression become a limit which k approach 0 and apply L' Hopital's Rule to show that f is differentiable at x=0, and hence find f'(0), the answer should be 7.
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  3. #3
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    Re: differentiability

    Quote Originally Posted by piscoau View Post
    Use L' Hopital's Rule to find the limit of f(x) at x=0 first.

    f(x) = g(x)/x, since g is a differentiable function,
    therefore f(x) must be differentiable somewhere on R.
    assume f'(x) exists at point x=k, then

    f'(k) = [k g'(k) - g(k) ]/k^2. (by quotient rule)
    take above expression become a limit which k approach 0 and apply L' Hopital's Rule to show that f is differentiable at x=0, and hence find f'(0), the answer should be 7.
    \lim_{k\to0}f'(k)=\lim_{k\to0}\frac{kg'(k)-g(k)}{k^2}=\lim_{k\to0}\frac{kg''(k)+g'(k)-g'(k)}{2k}=\lim_{k\to0}\frac{g''(k)}{2}=7

    Is this correct?
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  4. #4
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    Re: differentiability

    YES
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