TITLE: Will inverse functions, and functions always meet at the line $y=x$?
QUESTION [14 upvotes]: If I have a function, the inverse function, by definition will be a reflection of the original function in the line $y=x$, so if I wanted to find the point of intersection, instead of solving it with equating both of the functions to equaling each other, could I assume that the point of intersection, between the two functions, will always be at the point $y=x$?
This would enable me to solve problems much easier, instead of having to solve quartic equations, for example.

REPLY [10 votes]: You can prove that if $f$ is a strictly increasing function (of course it is 1-1 and so it has an inverse), then the points of intersection of $C_f$ (the graph of $f$) and $C_{f^{-1}}$ lie on the line $y=x$. Assume a point $A(x,y)$ that is an intersection point of $C_f$ and $C_{f^{-1}}$. Since $x,y\in\mathbb{R}$, one of the following holds: either $x=y$, or $x>y$, or $x<y$. We will just show that the last two lead to a contradiction due to the fact that $f$ is increasing. First of all since $A\in C_f, C_{f^{-1}}$, it holds $y=f(x)=f^{-1}(x)$. Now we take cases:

Let $x>y$. Then $x>y \Rightarrow x>f^{-1}(x)\Rightarrow f(x) > f(f^{-1}(x)) \Rightarrow y>x $ a contradiction.
Now let $x<y$. Similarly $x<y \Rightarrow y<x$ a contradiction.

Therefore $y=x$, which means graphically that the points of intersection lie on the line $y=x$. ($f$ increasing $\Rightarrow$ (if $(x,y)\in C_f, C_{f^{-1}}$ then $y=x$)).
However the restriction that $f$ be strictly increasing is absolutely necessary since as others have mentioned it is a quite common case that the curves intersect outside $y=x$. If your problems involve increasing functions you may simply state $f(x) = f^{-1}(x) \Leftrightarrow f(x) = x$ (The converse is trivial: if $f(x)=x$, then $x = f^{-1}(x)$, so $f(x)=f^{-1}(x)$ and it holds for any 1-1 function) and reduce your equation. This technique is useful not only for polynomials but also for functions involving say logarithms and exponentials. Consider $f(x) = \ln(x-1) + x$. Without the above theorem you get to solve a system of equations involving transcendental functions, whereas with the theorem you just prove $f$ is increasing and then you immediately get that the sole point of intersection of $f$ with its inverse is $(2,2)$. This means that although you can't use it with $y = -x^3$ of mkausp, the intersection points of $y=x^3$ are bound to be on $y=x$. Hope this helps.