Inverse Functions and their intersection points

Say you have f(x)f(x) and g(x)g(x) and g(x)=f1(x)g(x) = f^{-1}(x) .

I observed that these two curves need not intersect, for example with f(x)=exf(x) = e^x and g(x)=lnxg(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!

Note by Oli Hohman
3 years, 4 months ago

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Consider f(x)=xf(x) = x.

Note that g(x)=f1(x)=x=f(x)g(x) = f^{-1}(x) = x = f(x).

So, there are infinitely many intersections!

Deeparaj Bhat - 3 years, 4 months ago

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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)?

Oli Hohman - 3 years, 4 months ago

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

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

Deeparaj Bhat - 3 years, 4 months ago

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Nice solution

Ian Limarta - 3 years, 4 months ago

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

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

Oli Hohman - 3 years ago

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