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    Function question

    Show that f(x) = x^2 - 2x (x is (negative infinity, 1) is 1:1 and find its inverse. How should I approach this question? what should I do?
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    Quote Originally Posted by ziyadbasheer View Post
    Show that f(x) = x^2 - 2x (x is (negative infinity, 1) is 1:1 and find its inverse. How should I approach this question? what should I do?
    f(x) = x^2 - 2x

     = x^2 - 2x + (-1)^2 - (-1)^2

     = (x - 1)^2 - 1.


    So the turning point is at (1, -1).

    Since the function is cut off at x = 1, that means you don't have any repeating y values. Therefore f(x) is one-to-one.


    Now, to find the inverse, swap the x and y...

    x = y^2 - 2y

    x = y^2 - 2y + (-1)^2 - (-1)^2

    x = (y - 1)^2 - 1

    x + 1 = (y - 1)^2

    -\sqrt{x + 1} = y - 1 (since we are only accepting the lower half)

    y = 1 - \sqrt{x + 1}.


    So f^{-1}(x) = 1 - \sqrt{x + 1}.
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