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Math Help - Function having a third derivative

  1. #1
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    Function having a third derivative

    Let f: \mathbb{R} \rightarrow \mathbb{R} have a third derivative that exists at all points. Show that there must exist at least one point \xi for which
    f(\xi)f^{'}(\xi)f^{''}(\xi)f^{'''}(\xi) \geq 0
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Markeur View Post
    Let f: \mathbb{R} \rightarrow \mathbb{R} have a third derivative that exists at all points. Show that there must exist at least one point \xi for which
    f(\xi)f^{'}(\xi)f^{''}(\xi)f^{'''}(\xi) \geq 0
    Once again, any ideas?
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  3. #3
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    Start by using Mean Value Theorem?How do I start from there?

    Or which other theorem can I start with?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Markeur View Post
    Start by using Mean Value Theorem?How do I start from there?

    Or which other theorem can I start with?
    I'd use the MVT, but how?
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  5. #5
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    Hmm

    Restrict the function to f:[a,b] \rightarrow \mathbb{R}, where f, f^{'}, f^{''} are increasing in that interval and f(x) \geq 0 for x \in [a,b]. Since f is continuous, by MVT, there exists c \in [a,b] such that \frac{f(b)-f(a)}{b-a} = f^{'}(c). Then f^{'}(c) \geq 0. Since f^{'} is continuous, by MVT, there exists d \in [c,b] such that \frac{f^{'}(b)-f^{'}(c)}{b-c} = f^{''}(d). Then f^{''}(d) \geq 0. Since f^{''} is continuous, by MVT, there exists \xi \in [d,b] such that \frac{f^{''}(b)-f^{''}(d)}{b-d} = f^{'''}(\xi). Then f^{'''}(\xi) \geq 0. Since f, f^{'}, f^{''} are increasing in that interval, therefore f^{'}(\xi) \geq f^{'}(c) and f^{''}(\xi) \geq f^{''}(d). These imply that f(\xi), f^{'}(\xi), f^{''}(\xi), f^{'''}(\xi) \geq 0. Hence the conclusion holds.

    This could be one possible case out of so many different cases. Anyway, is this particular proof correct?

    Please advise. Thank you.
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  6. #6
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    Here are some vague thoughts, which you may be able to convert into a proof.

    First, if any one of the four functions f,\ f',\ f'',\ f''' vanishes at some point \xi, then f(\xi)f'(\xi)f''(\xi)f'''(\xi) = 0 and so the result is proved. It follows from the intermediate value theorem that each of the (continuous) functions f,\ f',\ f'' always has the same sign. By Darboux's theorem the same is true for f'''.

    Replacing  f by -f if necessary, you can assume that  f is always positive

    Now suppose that f' is always negative. By considering what happens as \xi\to\infty, show that f'' must be positive. The same argument, repeated, will show that f''' is negative. Therefore the product ff'f''f''' must be positive.

    By considering what happens as \xi\to-\infty, use the same ideas to show that if f' is positive then all four functions must be positive, hence so is their product.
    Last edited by Opalg; January 21st 2011 at 12:06 PM.
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