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Math Help - Problem involving functions and inverses

  1. #1
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    Problem involving functions and inverses

    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.



    This is what I did:

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

    (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.



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

    Hey kinhew93.

    Can you solve for the inverse function? [Hint: exchange the x's and y's and solve for y = f^(-1)[x]).
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  3. #3
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    Re: Problem involving functions and inverses

    Quote Originally Posted by chiro View Post
    Hey kinhew93.

    Can you solve for the inverse function? [Hint: exchange the x's and y's and solve for y = f^(-1)[x]).
    Yeah if I understand you correctly I've already done that, but from there I am totally stuck...
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  4. #4
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    Re: Problem involving functions and inverses

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

    acx^2 + (bc-a^2)x - ab = -cdx^2 + (bc-d^2)x + bd

    (ac+cd)x^2 + (bc - a^2 - bc + d^2)x - (ab + bd) = 0

    (ac+cd)x^2 + (d^2 - a^2)x - (ab + bd) = 0

    c(a+d)x^2 + (d-a)(d+a)x - b(a+d) = 0

    since a+d \ne 0

    cx^2 + (d-a)x - b = 0

    finish it ...
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