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Math Help - solution to f(x) = (f^(-1))(x) using quadratic equation

  1. #1
    Senior Member furor celtica's Avatar
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    solution to f(x) = (f^(-1))(x) using quadratic equation

    Alright so this is the last problem of an exercise where it has already been determined that f(x) = 4 - (x^2), x larger than or equal to 0, and (f^(-1))(x) = sqrt(4-x), x smaller than or equal to 4:
    Show, by reference to a sketch or otherwise, that the solution to the equation f(x) = (f^(-1))(x) can be obtained from the quadratic equation (x^2) + x - 4 = 0. Determine the solution of f(x) = (f^(-1))(x), giving your value to 2 decimal places.

    From the graph of these three curves I can indeed observe that the value of x (1.56 correct to 2 d.p.) for which (x^2) + x - 4 = 0 is the same value of x for which the curves of f(x) and its inverse interesct; however I'm not sure of how to use this observation as a coherent proof.
    So I tried to prove this algabraically in the following manner:
    f(x) = 4 - (x^2), (f^(-1))(x) = sqrt(4 - x), f(x) = (f^(-1))(x)
    => 4 - (x^2) = sqrt(4 - x)
    => (4 - (x^2))^2 = 4 - x
    => (x^4) - 8(x^2) + x + 12 = 0
    => ((x^2) + x - 4)((x^2) - x - 3) = 0
    Therefore where ((x^2) + x - 4) = 0 and x larger than or equal 0, f(x) = (f^(-1))(x).
    Solving for x I then get 1.56 correct to 2 d.p.

    However you will have observed that I conveniently left out the roots provided by ((x^2) - x - 3) as they do not satisfy the equation f(x) = (f^(-1))(x); I'm pretty sure this is due to the squaring of 4 - (x^2) in the second step of my proof, and as I was once told, squaring sides of an equation often means one ends up with unwanted roots.
    I would appreciate it if anybody could explain this to me in more detail and suggest a better way of solving this problem non-graphically. Thanks!
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: solution to f(x) = (f^(-1))(x) using quadratic equation

    You get one valid and three extraneous roots, only because the inverse of f(x) and x is given as being non-negative. You seem to have done a good job algebraically.
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