My Math Forum Trouble understanding inverse of fractional functions

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 January 4th, 2017, 11:22 AM #1 Newbie   Joined: Oct 2016 From: UK Posts: 7 Thanks: 0 Trouble understanding inverse of fractional functions I've developed a model for finding inverse functions, as demonstrated in the image enclosed below: https://i.stack.imgur.com/flqlO.png I know there's a method for working out an inverse function by swapping out x and ys, but the model I've developed makes more sense to me from a visual standpoint. However I can't seem to apply this model to fractional functions like f(x) = (x)/(3 + x). Could somebody use the model enclosed in the image to explain how to find the inverse of function f(x) = (x)/(3 + x)? Thanks
 January 4th, 2017, 12:02 PM #2 Global Moderator   Joined: Dec 2006 Posts: 18,594 Thanks: 1492 How do you know that 3 of something produces that something when divided by 3 or that -2 is cancelled by adding 2?
January 4th, 2017, 12:32 PM   #3
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 Originally Posted by skipjack How do you know that 3 of something produces that something when divided by 3 or that -2 is cancelled by adding 2?
The -2 is cancelled by adding the two, giving me 3x, which divided by 3 gives x i.e. domain of original function

 January 4th, 2017, 01:07 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 903 Thanks: 359 At one time, I too found it strange to solve for x and then swap variables. The process did not immediately appeal to my intuition. Nevertheless, I think you would be better off trying to develop such an intuition about why that process works rather than look for a different process. Here is mine, which may of course not work for you. I start HERE. $f^{-1}(f(x)) \equiv x.$ The line above is definitional. It is what is MEANT by an inverse function. $Let\ y = f(x).$ This just simplifies notation. $ASSUME\ \exists\ m(y)\ such\ that\ y = f(x) \implies x = m(y).$ This just assumes that you can restate y = f(x), which equates y to an expression in terms of x, so that x is equated to an expression in terms of y. Because I am trying for an intuitive explanation rather than a formal proof, I am going to ignore what it is entailed if the hypothesis is false. $\therefore m(f(x)) = m(y) = x \implies m\ is\ the\ inverse\ function\ of f.$ Now a formal proof would need to deal with the problem of whether or not an inverse exists if m(y) does not exist and would also need to show that $f^{-1}(f(x)) = x = f(f^{-1}(x)).$ But the intuition for why the process works is very obvious (to me at least) once we use a simpler notation. I have found it works for young adolescents. Thanks from Tomedb
January 4th, 2017, 01:51 PM   #5
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 Originally Posted by JeffM1 At one time, I too found it strange to solve for x and then swap variables. The process did not immediately appeal to my intuition. Nevertheless, I think you would be better off trying to develop such an intuition about why that process works rather than look for a different process. Here is mine, which may of course not work for you. I start HERE. $f^{-1}(f(x)) \equiv x.$ The line above is definitional. It is what is MEANT by an inverse function. $Let\ y = f(x).$ This just simplifies notation. $ASSUME\ \exists\ m(y)\ such\ that\ y = f(x) \implies x = m(y).$ This just assumes that you can restate y = f(x), which equates y to an expression in terms of x, so that x is equated to an expression in terms of y. Because I am trying for an intuitive explanation rather than a formal proof, I am going to ignore what it is entailed if the hypothesis is false. $\therefore m(f(x)) = m(y) = x \implies m\ is\ the\ inverse\ function\ of f.$ Now a formal proof would need to deal with the problem of whether or not an inverse exists if m(y) does not exist and would also need to show that $f^{-1}(f(x)) = x = f(f^{-1}(x)).$ But the intuition for why the process works is very obvious (to me at least) once we use a simpler notation. I have found it works for young adolescents.
Ok, I see. So it's similar to composite functions in that you take the range of the original function, make it the domain of inverse function, which gives the domain of the original function, hence x and y swap. Brilliant thank you!

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