$\begin{aligned} f_1 (x)&=&x\\ f_2 (x)&=&1-x\\ f_3 (x)&=&\frac{1}{x}\\ f_4 (x)&=&\frac{1}{1-x}\\ f_5 (x)&=&\frac{x}{x-1}\\ f_6 (x)&=&\frac{x-1}{x}\\ \end{aligned}$

If we know that $f_6 (f_m(x))=f_4(x)$ and $f_n (f_4(x))=f_3(x)$ , find the minimum value of $m+n$ .

The answer is 11.

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The best way to approach this is to use group theory to prove that the functions form a complete group, (closed under the composite function operation):

1) Identity element is $\ f_{1}$

2) Inverses: $\ f_{1}$ , $\ f_{2}$ , $\ f_{3}$ and $\ f_{5}$ are all their own inverses and.. $f^{-1}_{4} (x) = f_{6} (x) \ .... \ f^{-1}_{6} (x) = f_{4} (x)$ 3) The function are all associative

4) A Cayley table can be used to show that the group is closed under operation such that: $(f_{6}(x))^2 = f_{4} (x) \ ....... \ f_{5} (f_{4} (x) ) = f_{3} (x)$ $\Rightarrow\ m = 6 \ , \ n = 5 \therefore \ m+n = 11$