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Inverse functions

  1. Sep 13, 2007 #1
    If h(x)=(3x-5)/(7-2x)

    Find an expression for h^-1(x)





    Here's my attempt!

    y=(3x-5)/(7-2x)

    (swap x for y): x=(3y-5)/(7-2y)

    I've tried rearranging to find y in terms of x but I can't see how to do it!

    x(-2y)=(3y-5)/7

    -2y=(3y-5)/7

    -2y/3y=-5/x
     
  2. jcsd
  3. Sep 13, 2007 #2

    Kurdt

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    You don't swap them in that manner. You try and rearrange so you get x as a function of y. More an issue of semantics I believe.
     
  4. Sep 13, 2007 #3

    EnumaElish

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    Let's say your function was y = f(x) = 1/(1+x) over x < -1 or -1 < x. Then 1/y = 1+x so x = 1/y - 1. That is x = f-1(y) = 1/y - 1 over y < 0 or 0 < y.
     
  5. Sep 13, 2007 #4
    Did you try to multiply both sides by the denominator, distribute the y, then group your terms with x, then factor out the x?
     
  6. Sep 13, 2007 #5
    Oh i think I've got it:

    y=(3x-5)/(7-2y)
    x=(3y-5)/(7-2y)
    7x-2xy=3y-5
    -2xy-3y=-5-7x
    y(-2x-3)=-5-7x
    y=(-5-7x)/(-2x-3)=h^-1(x)
     
  7. Sep 13, 2007 #6
    Thanks for your help! xxx
     
  8. Sep 13, 2007 #7

    EnumaElish

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    You should start with y = (3x-5)/(7-2x) [not y=(3x-5)/(7-2y)] then apply the rules of elementary algebra until you have x = g(y).

    I'll give you the first step: (7-2x)y = (3x-5)
     
    Last edited: Sep 13, 2007
  9. Sep 13, 2007 #8
    Some books teach the variable switch first… but normally you show it in one step, not 2.

    p.s. you probably want to factor out those negatives just to make it look pretty.
     
  10. Sep 13, 2007 #9

    EnumaElish

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    This is correct, except I would've switched the variables at the end.
     
  11. Sep 13, 2007 #10

    Kurdt

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    Thats interesting, I've never seen the variables being swapped first so that threw me a bit. Is that just to maintain the notation that we usually have y = f(x).
     
  12. Sep 13, 2007 #11
    no clue, i always thought it was silly myself. But they do.
     
  13. Sep 13, 2007 #12

    Dick

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    Yes, it is. If the initial function is given as y=f(x) the inverse function should also probably be stated as y=f^(-1)(x) so x is the independent variable in both. That's all. You can swap at the end if you like.
     
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