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Math Help - Looking for a function f(x,a) such that f(a,a) = inf(f(x,a))

  1. #1
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    Unhappy Looking for a function f(x,a) such that f(a,a) = inf(f(x,a))

    Hi!

    I've been at this all day and can't seem to find it...

    I've got a function f(x) = 1/x + 1/(1-x)
    Its minimum is x = 0.5 and f(x) = 4

    The f(x) value doesn't really matter, but the function shape does, as it reaches infinity around 0 and 1. Hence, x is in ]0;1[

    However, i'm trying to change the function so as to accept an "a" parameter, "a" being in ]0;1[ as well, such that the minimum isn't in x=0.5 but in x=a.
    i.e.
    inf(f(x,a)) = f(a,a)

    I've been going for something like f(x,a) = a/x + 1/(1-x) but here "a" goes from 0+ to infinity...

    Any ideas?
    Thanks in advance!
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  2. #2
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    If I interpret your problem correctly you have
    f'=\ln x - \ln (1-x) = 0 for x=0.5

    and you want
    f'=\ln x - \ln (1-x) = 0 for x=a

    so choose:
    g'=\ln x - \ln (1-x) -f'(a)

    or
    g=\frac 1x + \frac 1{1-x}-(\ln a - \ln (1-a))*x
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