Inverse Functions and their intersection points 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!
 A: 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)
A: 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:

A: 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. 
A: 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 $g$.
A: 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.
A: 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 $f^{-1}(x)$:

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