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Thread: Proof that f is differentiable at 0

  1. #1
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    Proof that f is differentiable at 0

    So the question states " If f satisfies |f(x)|</= |x|^(7/2) for all x, prove that f is differentiable at 0".
    We just learned about Mean Value Theorem, Intermediate Value Theorem, Triangle inequality, Rolle's Theorem, and Extreme Value Theorem.
    I don't even know where to start with this question. Any help would be greatly appreciated.
    ((I apologize in advance for the format that this is typed in, I am not that great at typing math stuff. (It is supposed to be less than or equal to; and it is the absolute value of x to the power of (7/2))).
    Thank you for your time!
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  2. #2
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    Re: Proof that f is differentiable at 0

    ??? The statement "If f satisfies |f(x)|\le |x|^{7/2} for all x then f is differentiable at x= 0" cannot be proven because it is obviously untrue. f(x)= |x| satisfies |f(x)|\le |x|^{7/2} for all x, but is not differentiable at x= 0.
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  3. #3
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    Re: Proof that f is differentiable at 0

    Halls - for any positive x with x <1, $|x|\leq |x^{7/2}|$ is false!

    Hints:
    1. First show f(0)=0.
    2. Now show the absolute value of the difference quotient satisfies:
    $$0\leq\left|{f(0+h)-f(0)\over h}\right|\leq|h^{5/2}|$$
    3. Conclude by the "squeeze" theorem that in fact f'(0)=0.
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