# $\frac{1}{x}$ and its inverse breaking a rule of inverse functions?

I was recently introduced to a rule for inverse functions that specifies that an ascending function and its inverse will only intersect on the $y=x$ axis if they intersect at all, while descending functions and their inverses can intersect at infinite points.

Well, below we have the familiar $\frac{1}{x}$ graph with the $y=x$ axis:

$\frac{1}{x}$

So as we see, it has infinite intersection points with its inverse, but this is normal because it is a descending function. Now let's take a look at $\frac{-1}{x}$:

$\frac{-1}{x}$

$\frac{-1}{x}$ is ascending, but it seems to also have infinite points of intersection with its inverse.

Am I making some kind of obvious mistake or is $\frac{-1}{x}$ really breaking the aforementioned rule?

• Where did you get that rule from? What is its precise formulation? – Hagen von Eitzen Sep 24 '16 at 18:05
• Where are these infinite points of intersection? You can see that the graphs never touch. – mathematician Sep 24 '16 at 18:05
• @mathematician: $f(x)=-1/x$ is its own inverse, just as $g(x)=1/x$ is. – Ted Shifrin Sep 24 '16 at 18:06
• @mathematician 1/x and -1/x obviously never touch, but -1/x and its inverse overlap perfectly. Try symmetrically mirroring -1/x on the y=x axis (the axis used for mirroring an inverse from the original function). – MrKagouris Sep 24 '16 at 18:35
• What do you mean by "inverse"? Inverse with respect to function multiplication, or inverse with respect to function composition? – Alex M. Sep 24 '16 at 20:16

The rule is correct. The function $f(x)=-1/x$ is neither ascending nor descending on its entire domain.
• It seems that the rule might have a chance to be correct provided some information about the domain of definition be added. If you take $-\dfrac 1 x$ to be defined on $(0, \infty)$, then the rule fails. – Alex M. Sep 24 '16 at 20:46