The method of finding the inverse of a simple function $y = f(x)$ involves the following steps:-

1) Change the subject to $x$ instead of $y$.

2) Interchange $x$ and $y$.

3) The newly formed function ($y = g(x)$, say) is then the required inverse.

We know that the method works but why does it work? My question is how to convince others that the interchange part of the above can do the magic and is logically sound? A proof would be even better.

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    $\begingroup$ I don't understand what you mean. To begin with, why is $y$ initially the "subject", rather than $x$? To me both are just (related) values. Does interchange of $x$ and $y$ in the relation $y=f(x)$ give $x=f(y)$ as would seem to be what you are saying (but which is mathematically a very different relation). If not, what would interchanging mean? For instance in the concrete case $y=x^3-x$? (That is a case where the inverse is not easy to express at all, so I have my doubts about what your method could do with it.) $\endgroup$ – Marc van Leeuwen Mar 3 '16 at 16:06
  • $\begingroup$ @MarcvanLeeuwen First of all, the functions that we are talking about are some simple functions. For example, y = 2x – 1 is one. At this stage and in that form, y is called the subject. We can re-write it as x = (y + 1)/2. In which, x is now the subject. Interchanging x and y, we will get y = (x + 1)/2. $\endgroup$ – Mick Mar 3 '16 at 17:01
  • $\begingroup$ OK, I see now. What you call "subject", I would call left hand side of the equation. Then the first point "change the subject" is really the hard part: rewrite the relation (if you can) so that $x$ now appears as the left hand side of the equation. If and when this is done, you don't really need any further steps: you new right hand side defines your inverse function, as a function of the variable $y$ (the argument in an expression defining a function does not have to be called $x$, you know, and personally I prefer to use $y$ instead when the function is obtained as an inverse). $\endgroup$ – Marc van Leeuwen Mar 4 '16 at 4:58
  • $\begingroup$ @MarcvanLeeuwen The term subject comes from "subject of a formula". $\endgroup$ – Mick Mar 4 '16 at 8:31
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    $\begingroup$ Oh, that strange terminology the invent in elementary math education... (just to be sure, I checked that the WP article on "formula" does not mention its "subject" at all). $\endgroup$ – Marc van Leeuwen Mar 4 '16 at 9:03

Suppose you have $y=f(x)$. If you change the subject to $x$ and want to get it alone, what you are doing is to find a function $g$ such that $g(f(x)) = x$. To keep the equality we then see that $g(y) = g(f(x)) = x$.

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$g(x)$ is the inverse of $f(x)$ if it satisfies $x=f(g(x))$ and $x=g(f(x))$.

On base of $x=f(g(x))$ we go hunting for $g(x)$.

First we abbreviate $g(x)$ by $y$.

Now let's find $y$ on base of the equation $x=f(y)$.

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I had a (seemingly un-mathematical) practical method.

Sketch $ y = f(x) $ on a transparent plastic sheet used for projections, the edges serving as x- and y- axes. Flip the sheet swapping x and y along with rigid curve and see. It is so convincing.. no questions will be asked...


... as the operation makes it visibly obvious, so at each point you can verify :

1) x and y are interchanged

2) slope is its inverse now, no sign change $ \dfrac{dy}{dx} \rightarrow \dfrac{dx}{dy} $

3) curvature at any point is invariant except sign change, and it can be explained by differentials

$$ \frac{d^2y/dx^2}{(1+y'^{2})^{3/2}} \rightarrow \frac{-d^2x/dy^2}{(1+x'^{2})^{3/2}} $$

4) even higher order isometric invariants are conserved

4) one more flip and you are back; any double transformation annuls, i.e., inverse of an inverse function gives the starting function.

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  • $\begingroup$ I should have your question included in my post too. Yes but why does it work. And why is that there is no question asked? $\endgroup$ – Mick Mar 3 '16 at 9:35
  • $\begingroup$ Some changes are noticeable , I edited. $\endgroup$ – Narasimham Mar 3 '16 at 12:23
  • $\begingroup$ The very last point is good. But others are advanced level topics. They cannot be used as explanations to a high school student. $\endgroup$ – Mick Mar 3 '16 at 13:48

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