Convincing others that the method of finding the inverse function is valid 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.
 A: 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 $g(y) = g(f(x)) = x$.
A: $g(x)$ 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 $x=f(y)$.
A: 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... 
EDIT 1:
... 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.
