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Math Help - Proof involving functions

  1. #1
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    Proof involving functions

    f(x) = (ax+b)/(cx+d) , x is real, x=/= -d/c , b=/=0 , c=/=0

    Prove that if

    a + d =/= 0

    and

    (a-d)^2 +4bc = 0

    then y=f(x) and y=f^-1(x) intersect at exactly one point


    So far I've tried setting f(x) = f^(x) and writing as a quadratic in x. I then tried to find the discriminant and show that it was (a-d)^2 +4bc but I just got into a big mess.

    Is this method correct or am I missing something? Is there another approach?
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  2. #2
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    Re: Proof involving functions

    Have you found f^(-1)(x)? What is it?

    (I hope you understand that f^{-1} is NOT the reciprocal "1/f(x)" but the inverse function.)
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    Re: Proof involving functions

    f^-1(x) = (b-dx)/(cx-a)
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    Re: Proof involving functions

    Quote Originally Posted by kinhew93 View Post
    f^-1(x) = (b-dx)/(cx-a)
    Good. Now how to find where they intersect:
    y = f(x) = f^{-1}(x)

    Given that they can only intersect at one point (that's why I set them equal to each other) what do a, b, c, and d have to be?

    -Dan
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  5. #5
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    Re: Proof involving functions

    So

    (ax+b)/(cx+d) = (b-dx)/(cx-a)

    (ax+b)(cx-a) = (b-dx)(cx+d)

    then all I can think to do is write as a polynamial equal to zero, and find the discriminant. This didn't get me (a-d)^2 +4bc.


    Quote Originally Posted by topsquark View Post
    Good. Now how to find where they intersect:
    y = f(x) = f^{-1}(x)

    Given that they can only intersect at one point (that's why I set them equal to each other) what do a, b, c, and d have to be?

    -Dan
    I don't know
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