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Find inverse function

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data

    Graph f(x) = sqrt(x^2 - 2x), and find an interval on which it is one-to-one. Find the inverse of the function restricted to that interval.

    2. Relevant equations

    3. The attempt at a solution

    What I can't do is really finding the inverse function. It seems very simple, but somehow I got stuck in the process.

    swap x and y in the original function
    y = sqrt(x^2-2x)
    x = sqrt(y^2-2y)
    and solve for y

    so i did
    x^2 = y^2-2y, and i tried to factor out y
    x^2 = y(y-2)
    x^2/y-2 = y
    now i am really stuck. how can i pull that y out?

    Thank you for any kind of help!!!

    ---- edited

    I was thinking about this formula: d/dy f-1(x) = 1/ f ' (y)
    i guess i can then integrate the d/dy f-1(x) and get f-1(x)?

    so i started working again
    f ' = (1/2 (x^2-2x) ^-1/2) * 2x-2
    so 1/f ' =[ 2 (x^2-2x)^1/2 ]/ 2x-2
    which is d/dy f-1

    but the integration doesn't work!
  2. jcsd
  3. Apr 7, 2010 #2


    Staff: Mentor

    Did you find the interval on which f is 1-to-1?
    No, that's not what you need to do. Instead, complete the square on the right side.
    This problem doesn't require calculus to solve.
  4. Apr 7, 2010 #3
    Hi. Mark. Yes, I did find the interval.
    And thank you for your help. I got the answer with completing the square.
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