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Say you have $f(x) $ and $g(x)$ and $g(x) = f^{-1}(x) $.

I observed that these two curves need not intersect, for example with $f(x) = e^x$ and $g(x) = \ln x $ never intersecting each other.

I also observed that a function can either have one, two, or three intersections with its inverse, but I was unable to find a function which has more than 3 intersection points with its inverse.

How would I prove or disprove the hypothesis that an elementary function and its intersection can only have up to 3 intersection points? Any counterexamples are appreciated!

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    $\begingroup$ How about $f(x)=g(x)=x$? $\endgroup$ – Gerhard S. Mar 4 '18 at 17:10
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Consider $f(x) = x$.

Note that $g(x) = f^{-1}(x) = x = f(x)$.

So, there are infinitely many intersections!

Good solution. I am aware of the infinite intersections solution, but does anyone have any functions which have 4 or more intersections with their inverse (but not an infinite number of intersections)?

Consider, over any finite interval say $X$, $f(x) = x + \sin x$.

Over the interval $X$ there are finitely many intersections. The exact number depends on $X$ itself. But you can have any finite number of intersections.

At first, I was unsure of how to find the inverse of that function, so I decided to graph it to verify my claim and I'm right!

http://www.wolframalpha.com/input/?i=find+inverse+of+f(x)+%3D+x%2Bsin(x)

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    $\begingroup$ Good solution. I am aware of the infinite intersections solution, but does anyone have any functions which have 4 or more intersections with their inverse (but not an infinite number of intersections)? $\endgroup$ – user536517 Mar 4 '18 at 17:27
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I have an easy counterexample:

$$f(x)=10\cos(x)$$ and its inverse, $$g(x)=\cos^{-1}(x/10)$$

Take a look at the plot on desmos, that should convince you. In fact, you can bump up the 10 to get arbitrarily many intersections!

As a more general answer to your question, the number of intersections will be related to the number of times $f$ crosses the line $y=x$ since the inverse of $f$ is just a reflection of $f$ over that line.

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I assume that $f\colon \mathbb{R}\rightarrow \mathbb{R}$ is a bijection, such that $g$ is defined on all of $\mathbb{R}$. This can be adjusted to other situations (say $\mathbb{R} \rightarrow (0,\infty)$ in the case of the exponential function, then $g$ is only defined on $(0,\infty)$.)

Assume $f(x_0) =x_0$, then $g(x_0)=g(f(x_0))=x_0=f(x_0)$, i.e. each point $x_0$ at which the graph of $f$ intersects the diagonal of $\mathbb{R}^2$ will be an intersection point of $f$ and $g$. This make sense graphically, as taking the inverse of a function corresponds to reflecting its graph at the diagonal.

This shows that you can have arbitrarily many intersection points of $f$ and $g$.

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You can in fact have arbitrary many intersections with an inverse. Consider for instance the function $f(x) = \sin(x) + x$ which has an inverse on $\mathbb R$ (because it is strictly monotonic almost everywhere as $f'(x) = \cos(x)+1 \geq 0 \forall x$ and $f'(x) = \cos(x)+1 > 0 \forall x \in \pi(\mathbb Z +1/2)$) and has infinitely many discrete intersections with its inverse:

enter image description here

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Besides the particular examples given in the precedent answer, for any function $y=f(x)$, the graph of its inverse function $f_{(-1)}(x)$ will be symmetric with respect to the diagonal line $y=x$.

If $f(x)$ does not cross the diagonal (as $e^x$), so won't the inverse. And any zero of $f(x)-x$ will be the same for $f_{(-1)}(x)-x$ , and a point of crossing of the two functions.

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A function can have over $3$ intersection points with its inverse relation.

For example, take $f(x)=x^4-3x^2+2x-3$.

Finding points of intersection of $f(x)$ and $f^{-1}(x)$:

enter image description here

So, there are $4$ points of intersection. $4>3$, thus disproven.

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  • $\begingroup$ Why the downvote? $\endgroup$ – user535339 Mar 4 '18 at 17:45
  • $\begingroup$ idk, I didnt downvote but what qualifies this as function?? $\endgroup$ – King Tut Mar 4 '18 at 18:45
  • $\begingroup$ @KingTut I thought the question meant that $f(x)$ has to be a function, but $g(x)$ might not necessarily be a function. $\endgroup$ – user535339 Mar 4 '18 at 18:47

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