The function f(x) satisfies f(f(x))=6x−2005for all real numbers x. There exists an integer n such that f(n)=6n−2005. Find n.
Solution — click to reveal
Setting x=n, we get f(f(n))=6n−2005,so f(6n−2005)=6n−2005. Then f(f(6n−2005))=f(6n−2005)=6n−2005.But f(f(6n−2005))=6(6n−2005)−2005. Solving 6(6n−2005)−2005=6n−2005,we find n=401.