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Finding the inverse of a rational function

  1. Aug 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the inverse of a rational function: [itex]f(x)=\frac{2x}{x-2}[/itex]

    3. The attempt at a solution

    1. Verify the function is one-to-one.

    Question: is there a way to do that withought drawing the graph? Some algebraic method? I know that this means that the function crosess any given x=a line only once, so, e.g. the function with two y intercepts would not be 1-1 ???

    2. show that [itex]f(x)=f^{-1}(x)[/itex]

    Q.: How to express the given equation in the form where x is given in terms of y? It is imposible since you always get either y or x on both sides.
     
  2. jcsd
  3. Aug 20, 2012 #2

    SammyS

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    What is [itex]f^{-1}(f(x))\ ?[/itex]
     
  4. Aug 20, 2012 #3

    CAF123

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    You should be able to solve for [itex] x [/itex] in terms of [itex] y. [/itex] You will need to take out a common factor at some stage.
     
  5. Aug 20, 2012 #4

    HallsofIvy

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    Technically, if there are two y intercepts, it would not be a function. That is different from saying it is not one to one. A function is not one to one if it crosses some y line more than once.

    To determine whether this function is one to one, look at what happens if two values of x give the same y: suppose 2a/(a- 2)= 2b/(b- 2). Multply on both sided by (a- 2)(b- 2) to get 2a(b- 2)= 2b(a- 2). That is the same as 2ab- 2a= 2ba- 2b. We can subtract 2ba (= 2ab) from both sides to get -2a= -2b, and finally divide both sides by -2: a= b, showing that two different values of x cannot give the same y.

    That kind of equation is the whole point of algebra! If you have, for example, 4xy= y- x, you solve for x by adding x to both sides: xy+ x= y, factoring out x: x(y+ 1)= 4, and dividing both sides by y+ 1: x= 4/(y+1).
     
  6. Aug 20, 2012 #5
    I do it thus:

    (i) [itex]y=\frac{2x}{x-2}[/itex]

    (ii) [itex]y(x-2)=2x[/itex]

    (iii) [itex]\frac{y(x-2)}{2}=x[/itex]

    Is this the right path? Probably not ...
     
  7. Aug 20, 2012 #6

    HallsofIvy

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    From (ii) y(x- 2)= 2x, multiply out the left side: xy- 2y= 2x, then get the "x"s together on one side- xy- 2x= 2y; x(y- 2)= 2y. Can you finish that?
     
  8. Aug 21, 2012 #7
    Thanks, to you and to all :)
     
  9. Aug 21, 2012 #8

    SammyS

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    Notice that for the function given,
    [itex]f(f(x))=x\ .[/itex]​
     
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