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Inverse Functions

  1. Feb 12, 2009 #1
    what is the inverse of f(x)=-2/3(x+5)^2 - 5/3

    so far i got x=-2/3(y+5)^2 -5/3
    bring 5/3 over to x side
    x+5/3=-2/3(y+5)^2

    although i know what to do, can someone just explain how they will go on from here
     
  2. jcsd
  3. Feb 12, 2009 #2
    What you want to be doing, I think, is to substitute f(x)=y. Then, you need to switch x and y in the equation and rearrange it to get y=something. Did I explain myself clearly?
     
  4. Feb 12, 2009 #3
    it is stated that i have begun doing that, so now i know i will have to divide -2/3 by x+5/3, then i will sqareroot the left side, as the right side is being squared, and the finally bring over the 5 which will reslut in -5

    overall equation = (sqareroot)(x+5/3) - 5 =y
    ........................................ -2/3

    (x+5/3 is being divided by -2/3)
     
  5. Feb 12, 2009 #4
    i believe i have to get ride of the -2/3 how will i do so?
    ( i am trying to graph it so i should be able to apply the transformations)
     
  6. Feb 12, 2009 #5
    I can only help you find the inverse in terms of y=something man. Anything other than that and I'm sorry -- have to wait for someone else to help you... :(
     
  7. Feb 12, 2009 #6

    Mark44

    Staff: Mentor

    So far, so good. Now multiply both sides by -3/2, which gives you
    -3/2 * (x + 5/3) = (y + 5)^2

    If you want, you can switch sides to get

    (y + 5)^2 = -3/2 * (x + 5/3)

    Take the square root of both sides (but don't forget that you'll have +/-).
    Then subtract 5 and you'll have y = f^(-1)(x)
     
  8. Feb 13, 2009 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Strictly speaking, that doesn't have an inverse because of the square. As Mark44 said, when you take the square root, you will have a "[itex]\pm[/itex]" so you do not have a well defined function.
     
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