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Inverse Function

  1. Apr 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the inverse function.
    [Find (f^-1)(x)]
    f(x) = (-2x^5) + 1/3

    2. Relevant equations

    Find the inverse function.
    [Find (f^-1)(x)]
    f(x)=2-2x^2

    3. The attempt at a solution

    (-2x^5) + 1/3
    (-1/2x^1/5) + 3
    Answer is different, how would I get to (-1/2x + 1/6)^1/5?
     
    Last edited: Apr 7, 2009
  2. jcsd
  3. Apr 7, 2009 #2

    tiny-tim

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    uhh? :confused: if x → y is a function, then the inverse is y → x.

    so if y = -2x5 + 1/3, what is x ?
     
  4. Apr 7, 2009 #3
    Yea, not for this, just find functions.
     
  5. Apr 7, 2009 #4
    Is the answer of the referred problem even correct?
     
  6. Apr 7, 2009 #5

    Mark44

    Staff: Mentor

    If you have y = f(x) = -2x5 + 1/3, then the inverse is x = f-1(y).
    From your equation y = -2x5 + 1/3, solve for x in terms of y. That will be your inverse, as a function of y. To write the inverse as a function of x, just switch the variables (i.e., change y to x).
     
  7. Apr 7, 2009 #6

    Mark44

    Staff: Mentor

    Not even close. What you did is not how you find the inverse function. See tiny-tim's and my posts.
     
  8. Apr 7, 2009 #7
    Inverse of x=y does not = y=x.
    -y=-x is the inverse as it is perpendicular.
    Go graph it.
     
  9. Apr 7, 2009 #8
    It's the answer in the book, the reference has less than a 1/1,000 chance of being wrong.
     
  10. Apr 7, 2009 #9

    tiny-tim

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    Like this :rolleyes:
     
  11. Apr 7, 2009 #10
    I still want to know how, though.
     
  12. Apr 7, 2009 #11
    There is no reference to what x is.
     
  13. Apr 7, 2009 #12
    I am confused with all the confusion here.


    @SETHOSCOTT
    tiny-tim and Mark44 have given the perfect cues to the answer.

    It's a simple algebraic manipulation of the equation. Interchange x and y and make y the subject of the equation to get the inverse function.

    What's left to clear up?
     
  14. Apr 7, 2009 #13

    Mark44

    Staff: Mentor

    The function y = x is its own inverse. Perpendicularity has nothing to do with inverses.

    -y = -x is in fact equivalent to y = x. The graphs of these two equations are identical.
     
  15. Apr 8, 2009 #14
    I'm not going to even say anything... nevermind...
     
  16. Apr 9, 2009 #15
    PWND
    y=-2x5 + (1/3)
    - (1/3) - (1/3)
    y - (1/3) = -2x5
    * -.5 * -.5
    (1/6) - (y/2)= x5
    [(1/6) - (y/2)]1/5 = (x5)1/5
    [(1/6) - (y/2)]1/5 = x
     
  17. Apr 9, 2009 #16
    Don't kid yourself... har-har.
     
  18. Apr 9, 2009 #17

    tiny-tim

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    That's better! :smile:

    But you don't need all those intermediate steps (some of which I don't really understand) …

    it's ok if you just say

    y = -2x5 + (1/3),

    so 2x5 = (1/3) - y

    so x = (1/6 - y/2)1/5 :wink:

    (and so the inverse is f-1(x) = … ? :smile:)
     
  19. Apr 9, 2009 #18
    Wait so you KNEW this?
     
  20. Apr 9, 2009 #19
    Well, duh, switch, ummm... x and y, but, yea.
     
    Last edited: Apr 9, 2009
  21. Apr 9, 2009 #20

    tiny-tim

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    "Epiphany"

    "duh" is irrelevant …

    if you want full marks in the exam,

    you do actually need to finish the question! :wink:
     
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