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Math Help - Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

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    Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

    Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

    Please help me solve this problem I'm desperate!!!!!!
    Last edited by mr fantastic; November 27th 2008 at 05:57 PM. Reason: Added the actual question from the title
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    Quote Originally Posted by gnarlycarly227 View Post
    Given f(x) = 1/3 √(16-x^2), find f-1(x). Then state whether f-1(x) is a function.

    Please help me solve this problem I'm desperate!!!!!!
    So, if I understand correctly. You are trying to determine whether the inverse function of f(x) is function. First, you must find the inverse function. To do this, typically one "switches" the x and y values and solves for y. So, let's do that.

    y = \frac{1}{3} \sqrt{16-x^2}
    flip x and y

    x=\frac{1}{3} \sqrt{16-y^2}
    solve for y, first multiply both side by 3

    3x=\sqrt {16-y^2}
    square both sides to solve to rid of the radical

    (3x)^2=(\sqrt {16-y^2})^2
    9x^2 =16-y^2
    now solve for y^2

    -y^2 = 9x^2 - 16
    y^2=16-9x^2
    take the root to isolate y

    y=\pm \sqrt {16-9x^2}

    since for any x, y (your inverse function) can have 2 values the inverse is not a function
    Last edited by mr fantastic; November 27th 2008 at 05:58 PM. Reason: Edited quote to match edit of original post
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