Get ready Part 6

For an arbitrary but fixed positive $t$ , we will iterate $f$ on $t$ , defining the sequence $a_0=t$ and $a_{n+1}=f(a_n)$ for $n\geq 0$ . Applying the given condition to $x=a_n$ , we find the recursion $a_{n+2}=6a_n-a_{n+1}$ . The solutions of this recursion are of the form $a_n=2^np+(-3)^n q$ . If $q$ is nonzero, then $a_n$ will be negative for some $n$ , which is verboten. Thus $q=0$ , $p=t$ , $a_n=2^nt$ and $f(t)=a_1=2t$ ; this is the unique solution.