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Math Help - Proof involving functions

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    Proof involving functions

    f(x)= a/x + b , x is real, x does not equal 0, b or -a/b

    ff(x) = f^-1(x)

    prove that a + b^2 = 0


    Would really appreciate a hint!
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    Re: Proof involving functions

    Does anyone know how to do this?
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    Re: Proof involving functions

    Quote Originally Posted by kinhew93 View Post
    f(x)= a/x + b , x is real, x does not equal 0, b or -a/b
    ff(x) = f^-1(x)
    prove that a + b^2 = 0
    Quote Originally Posted by kinhew93 View Post
    Does anyone know how to do this?

    It is against forum rules to bump. Don't do it.

    I did not choose to help because I think the question is flawed.

    If it is f(x)=\frac{a}{x}+b then f^{-1}(x)=\frac{a}{x-b}.

    BUT WHAT IS f(f(x))~?
    Last edited by Plato; December 3rd 2012 at 03:36 PM.
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    Re: Proof involving functions

    Quote Originally Posted by kinhew93 View Post
    f(x)= a/x + b , x is real, x does not equal 0, b or -a/b

    ff(x) = f^-1(x)

    prove that a + b^2 = 0


    Would really appreciate a hint!
    I agree with Plato. I did the problem and didn't come up with a proof that a + b^2 = 0 The defining equation that you have,
    f(f(x)) = f^{-1}(x)

    gives a relationship between x, a, and b, not just a and b. Something isn't right.

    -Dan
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    Re: Proof involving functions

    Sorry I didn't realise that commenting on this would send it back up the list! - It was just that I suspected that people were also unable to see a proof and was hoping that someone would confirm this for me.

    The question is designed to require more 'creative thinking' however, so I assume that there is more to it than just setting ff(x) = f^-1(x)?

    Can anyone see another relationship in here somewhere? fff(x) = x and ffff(x) = f(x) Will they help?

    There are other questions of this vein in the book so I think that there is a proof.
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    Re: Proof involving functions

    Quote Originally Posted by kinhew93 View Post
    assume that there is more to it than just setting ff(x) = f^-1(x)?
    Can anyone see another relationship in here somewhere? fff(x) = x and ffff(x) = f(x) Will they help? There are other questions of this vein in the book so I think that there is a proof.

    Here is the point: we do not know the notation.

    fff(x) is meaningless.

    Is it f\circ f\circ f(x)=f(f(f(x)))~? or fff(x)=f^3(x)=f(x)\cdot f(x)\cdot f(x)~?.
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    Re: Proof involving functions

    Ok sorry I think that the british notation I use is slightly different:

    I would write f(x) . f(x) . f(x) = (f(x))^3

    and

    fff(x) = fofof(x) = f(f(f(x))) = f(^3)(x)

    so by ff(x) I actually mean f(f(x)) = f o f(x)
    Last edited by kinhew93; December 4th 2012 at 06:47 AM.
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  8. #8
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    Re: Proof involving functions

    Okay, so have you determined what [tex]f^{-1}(x)[/itex] is? f(f(x))= \frac{a}{\frac{a}{x}+ b}+ b= \frac{ax}{a+bx}. Set that equal to f^{-1}(x), "cross multiply" so that you get two polynomials equal to each other, for all x. What does the fact that they are "equal for all x" tell you about the coefficients?
    Thanks from kinhew93
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    Re: Proof involving functions

    Thanks a lot HallsofIvy!

    Just one question though; Why is it okay to crossmultiply if the expressions are 'exactly equal' when a and b are constants?
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