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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?

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    $\begingroup$ Where did you get that rule from? What is its precise formulation? $\endgroup$ – Hagen von Eitzen Sep 24 '16 at 18:05
  • $\begingroup$ Where are these infinite points of intersection? You can see that the graphs never touch. $\endgroup$ – mathematician Sep 24 '16 at 18:05
  • $\begingroup$ @mathematician: $f(x)=-1/x$ is its own inverse, just as $g(x)=1/x$ is. $\endgroup$ – Ted Shifrin Sep 24 '16 at 18:06
  • $\begingroup$ @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). $\endgroup$ – MrKagouris Sep 24 '16 at 18:35
  • $\begingroup$ What do you mean by "inverse"? Inverse with respect to function multiplication, or inverse with respect to function composition? $\endgroup$ – Alex M. Sep 24 '16 at 20:16
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The rule is correct. The function $f(x)=-1/x$ is neither ascending nor descending on its entire domain.

Comment: At first I thought the domain of the function had to be an interval (i.e., no gaps) for the rule to apply, but this is not necessary.

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  • $\begingroup$ 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. $\endgroup$ – Alex M. Sep 24 '16 at 20:46

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