It is a well-known fact that the graphs of $y=f(x)$ and $y=f^{-1}(x)$ are symmetrical about the line $y=x$. Thus we can say the solution of $f(x)=x$ must satisfy $f(x)=f^{-1}(x)$ but converse is not true. But I have seen that many authors find the solution of $f(x)=x$ when we are rquired to solve $f(x)=f^{-1}(x)$. But this method does not give us a guarantee to give complete answer.
As an example, suppose we want to solve $(1-x)^3=1-x^{1/3}$. We can easily observe that the equation is written in the form $f(x)=f^{-1}(x)$ where $f(x)=(1-x)^3$. Let's see the solution of this equation by drawing their graphs by using a graphing calculator.
We can see that the equation $f(x)=f^{-1}(x)$ has 5 solutions but the equation $f(x)=x$ has only one solution.
The 'extra' solutions (except from $f(x)=x$) exist because there are some points on the graph of $y=f(x)$ which are image pairs (means, they themselves are symmetrical) about the line $y=x$. In the given example, we notice that the points (C, F) and (D, E) are image pairs.
My question is that if we are given to solve $f(x)=f^{-1}(x)$ then
- Do we have some easy method to solve it without actually solving $f(x)=f^{-1}(x)$?
- If we don't such an easy method, then can we have some easy method to figure out whether the graph has some image pairs or not? If we could figure out that the graph does not have any image pair then we can safely solve $f(x)=x$.
HINT:
It is known that an inverse function of $y=f(x)$ is $x=f(y)$ or vice-versa obtained by swapping $(x,y)$.
One possibility of generating a self inverse function is by rotation of graph of its y- or x- even symmetric function. What I mean is
Rotate by angle $\pi/4$ for graphs symmetric on x-axis. For that plug in:
$$x\to (x+y),~y\to (x-y)$$
to get a self-invertible function.
$\{$ This is same as rotating by $-\pi/4$ for even function graphs symmetric on y-axis $\}$
The $\sqrt 2 $ scaling is still be effected.