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    yeah it asks for the exact value at which f(x) = f^-1(x)

    f(x) = x^2 - 4, x is less than or equal to 0
    f^-1(x) = is the inverse function of f(x)
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    (Original post by haseem)
    yeah it asks for the exact value at which f(x) = f^-1(x)

    f(x) = x^2 - 4, x is less than or equal to 0
    f^-1(x) = is the inverse function of f(x)
    I see! You're doing things the hard way (and I don't know why I didn't spot this). It would have saved a lot of bother if you'd put this in your first post!

    The graphs of y=f(x) and y=f^{-1}(x) are reflections of each other in the line y=x. That means that they can only meet on the line y=x, and so the equation you need to solve is f(x) = x (or f^{-1}(x) = x) rather than f(x) = f^{-1}(x).
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    OMG, I feel so stupid because I wrote that f(x) = x but crossed it out for some stupid reason. But thanks nuodai!
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    I get x = 1/2 (1+sqrt(17)) as one solution, but cannot manage to get the other solution
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    (Original post by haseem)
    I get x = 1/2 (1+sqrt(17)) as one solution, but cannot manage to get the other solution
    That's wrong I'm afraid. Anyway, there is only one solution because you're given that x \le 0. This also means that your inverse should actually be f^{-1}(x) = -\sqrt{x+4}, because this must be negative. (It turns out not to make a difference here when calculating the solution, although it does for checking it's correct.)

    So proceed as you did before, but take the negative root. Then check that it works.
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    How would you get x = 1/2 (-1-sqrt(13))
 
 
 
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