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January 4th, 2017, 11:22 AM   #1
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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
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January 4th, 2017, 12:02 PM   #2
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How do you know that 3 of something produces that something when divided by 3 or that -2 is cancelled by adding 2?
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January 4th, 2017, 12:32 PM   #3
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Quote:
Originally Posted by skipjack View Post
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
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January 4th, 2017, 01:07 PM   #4
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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
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January 4th, 2017, 01:51 PM   #5
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Quote:
Originally Posted by JeffM1 View Post
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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