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Math Help - Extremes of f"(x)

  1. #1
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    Extremes of f"(x)

    Assuming the second derivative f"(x)>0 of a convex function f(x) exists over an interval (a,b).


    Is it correct saying that the extremes (min, max, relative or absolute) of f"(x) always occur at the same ordinates where the extremes* of f(x) occur?


    (*) Not necessarily of the same type, min or max as in f(x): can be opposite.


    I cannot depict an example that says "NO, it's not correct".


    I would appreciate either such an example or few words that help me proving the above statement is correct.
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  2. #2
    MHF Contributor arbolis's Avatar
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    Assuming the second derivative f"(x)>0 of a convex function f(x) exists over an interval (a,b).


    Is it correct saying that the extremes (min, max, relative or absolute) of f"(x) always occur at the same ordinates where the extremes* of f(x) occur?
    The answer is no. As an example I think the following function works, f(x)=x^8+1.
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    Why?

    Quote Originally Posted by arbolis View Post
    The answer is no. As an example I think the following function works, f(x)=x^8+1.
    Isaac,
    I might miss something here but I do see your example proving that my statement is just correct.

    Plot f(x) = x^8 + 1 and f"(x) = 56*x^6.

    I see that in any interval (a,b) both f(x) and f"(x) reach max or min at the same ordinate.

    Also,
    I tried defeating my statement with exponentials, logarithms, trigonometrics functions, basic polinomials.
    It always came out clean and valid.

    I'd really need to see a function that proves the contrary.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Isaac,
    I might miss something here but I do see your example proving that my statement is just correct.

    Plot f(x) = x^8 + 1 and f"(x) = 56*x^6.

    I see that in any interval (a,b) both f(x) and f"(x) reach max or min at the same ordinate.

    Also,
    I tried defeating my statement with exponentials, logarithms, trigonometrics functions, basic polinomials.
    It always came out clean and valid.

    I'd really need to see a function that proves the contrary.
    You're right. The more I'm thinking about it, the more I believe the statement is true. I'll try to prove it but I'm not sure I'll success, so I hope someone else will do it. Anyway it's an interesting problem.
    Any news I get, I post.
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    Theorem

    I started from this theorem:

    Assume f'(x) exists over (a,b). Then, if f' is monotonic over (a,b), f is certainly convex over (a,b).
    Special case: f is certainly convex over (a,b) if f" exists and is non-negative over (a,b).
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  6. #6
    MHF Contributor arbolis's Avatar
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    What about f(x)=x^2. f''(x)=2>0, so f is convex as it must.
    Now looking at
    Is it correct saying that the extremes (min, max, relative or absolute) of f"(x) always occur at the same ordinates where the extremes* of f(x) occur?


    (*) Not necessarily of the same type, min or max as in f(x): can be opposite.
    , I'd say "no" because any point given is an extrema for f'', but not for f. I mean that for example let (a,b)=(-1,1). Any given point in (-1,1) is a maximum and a minimum for f'', according to the definition of an extrema of a function. While only x=0 is a minimum in (-1,1).
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