One of my professors told me that all values of x that satisfy (1) $f(x)- x = 0$ are also solutions of (2) $f(x)- f^{-1}(x) = 0$ provided that the function is invertible. Thinking of graphs and noting the symmetry of the function and its inverse, this statement seems obviously true. I was wondering if the converse is also true, i.e. all solutions of (2) are identical to that of (1)? If it were, then this would help save a lot of time in solving for roots of (2).
I found a few interesting cases:
- When $f(x)$ is identically equal to $f^{-1}(x)$, e.g. $f(x)=1/x$, then the roots of (2) are all elements in the domain of the function (in this case, all non-zero numbers). Hence, this is a counter-example for the converse statement. In that case can we conclude that all solutions of (1) are identical to that of (2) iff the function is not identically equal to its inverse?
- Whenever I imagine the graph of $f^{-1}(x)$ meeting $f(x)$ for normal cases, it feels like the curves have to pass through the line $y=x$ to meet, but this isn't necessary if the function is discontinuous, so can we have different conclusions for continuous and non-continuous functions?