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Math Help - Maximum of function

  1. #1
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    Maximum of function

    For each integer n\geq1 find the maximum of fucntion
    f(x)=\frac{x+x^2+...+x^{2n-1}}{(1+x^n)^2} for x>0 and every x>0 to which it's taken
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  2. #2
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    Re: Maximum of function

    Quote Originally Posted by geagry View Post
    For each integer n\geq1 find the maximum of fucntion
    f(x)=\frac{x+x^2+...+x^{2n-1}}{(1+x^n)^2} for x>0 and every x>0 to which it's taken
    Dear geagry,

    f(x)=\frac{x+x^2+...+x^{2n-1}}{(1+x^n)^2}~;~n\geq 1\mbox{ and }x>0

    f'(x)=\frac{(1+x^n)^2\left[1+2x+3x^2+.....+(2n-1)x^{2n-2}\right]-2(1+x^n)nx^{n-1}\left(x+x^2+....+x^{2n-1}\right)}{(1+x^n)^4}

    f'(x)=\frac{(1+x^n)\left[1+2x+3x^2+.....+(2n-1)x^{2n-2}\right]-2nx^{n-1}\left(x+x^2+....+x^{2n-1}\right)}{(1+x^n)^3}

    Substituting x=1 you will see that, f'(1)=0

    Also observe that, f(1)=\frac{2n-1}{4}

    Therefore, x=1 is an inflection point and, f(1)=\frac{2n-1}{4}

    It could be shown that at x=1 is a maximum by considering the second derivative (the calculation of f''(1) is terrible ).

    f''(1)=-\frac{n}{24}(2n-1)(n+2)<0\mbox{ since }n\geq{1}

    Therefore, at x=1 the function has a maximum and f(1)=\frac{2n-1}{4}\mbox{ where }n=1,2,3,......
    Last edited by Sudharaka; June 24th 2011 at 07:35 AM. Reason: I found a false in my previous method.
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