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Math Help - maximum value of func.

  1. #1
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    maximum value of func.

     f(x)=2(a-x)(\sqrt{x^2+b^2} + x)

    for all real x,
    show the maximum value of f(x) is ( a^2+b^2)
    and  x=\frac{a^2-b^2}{2a} when f(x) is maximum
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  2. #2
    Senior Member
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    Hi stud_02!

    Quote Originally Posted by stud_02 View Post
     f(x)=2(a-x)(\sqrt{x^2+b^2} + x)

    for all real x,
    show the maximum value of f(x) is ( a^2+b^2)
    and  x=\frac{a^2-b^2}{2a} when f(x) is maximum
    For all real x? Your Solution seems to be wrong,
    if a=0 it does not make any sense

    And by the way, if b = 1 and a = 0
    f(x) = -2x(\sqrt{x^2+1}+x) has its maximum in \approx (-0.34 ; -1.66)
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  3. #3
    Super Member
    Joined
    Jan 2009
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    Quote Originally Posted by stud_02 View Post
     f(x)=2(a-x)(\sqrt{x^2+b^2} + x)

    for all real x,
    show the maximum value of f(x) is ( a^2+b^2)
    and  x=\frac{a^2-b^2}{2a} when f(x) is maximum
    Substitute
     t = \sqrt{x^2+b^2} + x
    \implies
     t^2 - 2tx + x^2 = x^2 + b^2
    \implies

     x = \frac{t^2-b^2}{2t}

    Go back to the function ,

    we will have
     f(t) = -t^2 + 2at + b^2

     = -(t-a)^2 + b^2 + a^2

    so it has max. value  a^2 + b^2 when  t = a
     x = \frac{a^2-b^2}{2a}
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