Given f(x)=1/x, and g(x)=1-x. Functions f and g can be composed with themselves and each other in many ways. How many functions can result from these compositions?

Question

I know there are 6 functions that can result from listing

1) f(x)=f(f(f(x)))=f(g(g(x)))=g(g(f(x)))= 1/x

2)f(f(x))=f(f(f(f(x))))=g(g(x))=g(g(g(g(x))))=f(g(g(f(x))))=f(f(g(g(x)))=x

3)g(x)=g(g(g(x))=f(f(g(x)))=g(f(f(x)))=1-x

4)f(g(x))=f(f(f(g(x)))=f(g(f(f(x)))=1/(1-x)

5)f(g(f(x)))=g(f(g(x)))=x/(x-1)

6)g(f(x))=f(f(g(f(x))))=(x-1)/x

Repeated compositions will eventually lead to the identity function， how to explain that in a mathematical way?

Answer 1

f(x)=1/x = f^-1(x)

f is its own inverse

f(f^-1(x))=x = f^-1(f(x))

f(f(x))=x

f(f(f(f(x))))=x with 4 or an even number of compositions

odd number of compositions = 1/x= f(f(f(x)))=f(x)

same with g(x)=1-x=g^-1(x), etc.

let h(x)= g(g^-1(x))

h(x)= an identity function