Solve for $f(x)$ if $f(f(x))=6x-f(x)$ If $f: [0,\infty) \rightarrow [0,\infty)$ and


*

*$f(f(x))=6x-f(x)$

*$f(x)>0$ $  \forall x \in (0,\infty) $
Find f(x)
 A: For any $x$ define a sequence, 
$$a_0=x,a_1=f(x),a_2=f(a_1) \cdots, a_{n+1}=f(a_n) $$
Putting, $a_{n}$ in equation gives, 
$$ a_{n+2}=6a_n-a_{n+1}$$ 
This is a sequence, using characteristic equation method to solve this sequence.
Since $f(x)>0$ gives, $$ a_{n} = 2^nx$$
This give $a_1=2x$ 
Checking this in functional equation, we see that it satisfies. 
And so $f(x)=2x$ is only function which satisfies the FE $\Box$
A: This approach is fatally flawed.
Here's an alternate solution via linear algebra.  Let $V=\mathbb{R}\to\mathbb{R}$ denote the vector space of real functions on the real line and define $T\in V\to V$ to be the linear transformation $$(Tg)(x)=g(f(x))$$  Finally, let $p\in V$ denote the identity function, i.e. $p(x)=x$.  Then your functional equation states $$T^2p+Tp=6p$$  Rearranging, $$0=(T^2+T-6)p=(T+3)(T-2)p$$  From the general theory of linear transformations, we know that $p$ is a sum of eigenfunctions: $p(x)=q(x)+r(x)$, where \begin{gather*}
0=(T+3)q(x)=q(f(x))+3q(x) \\
0=(T-2)r(x)=r(f(x))-2r(x)
\end{gather*}
If we are very lucky, $p$ is in fact an eigenfunction itself: $$0=(T+3)p=f(x)+3x$$ or $$0=(T-2)p=f(x)-2x$$  We can check that each of these are solutions to conditions (1-2) and only one solves condition (3).  But I do not see how to rule out other solutions arising from linear combinations; for examples where this occurs, see the comments.
