How do I find the value of $f(0)$? Let $f$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in R$, then $f(0)$ is $?$
So this is what I've tried so far,
$f(f(0))=1$ so $f(f(0))=4f(0)+1$ (*)
Also, $f(f(1))=5$,this implies $f(5)=16f(0)+5$ Now how should I find the value of $f(5)$ in terms of $f(1)$ so that I can solve (*)? Any help is appreciated .
The answer given is $1/2$
 A: $f(x) = ax + b$ $(*)$
$f(f(x)) = a^2x + ab + b = 4x + 1$
Comparing coefficients,
$b(a+1) = 1$
$a^2 = 4$
$\implies a = 2, b = \frac{1}{3}$
($a$ is not $-2$ by the fact that $f(0) > 0$, $a = -2$ would result in $b = -\frac{1}{3} \implies f(0) = -\frac{1}{3}$)
Therefore $f(x) = 2x + \frac{1}{3}$ and $f(0) = \frac{1}{3}$
(($*$) comes from the fact that if a polynomial $f(x)$ is degree $n$, then the polynomial $f(f(x))$ is of degree $n^2$)
A: Let $f(x) = a_kx^k + a_{k-1}x^{k-1} + ..... + a_0$.
Then $f(f(x) = a_k(a_kx^k + a_{k-1}x^{k-1} + ..... + a_0)^k + a_{k-1}(a_kx^k + a_{k-1}x^{k-1} + ..... + a_0)^{k-1} + ..... + a_0$
If we expand that out we will get a very large polynomial of degree $k^2$.
But $f(f(x)) = 4x + 1$ which is of degree $1$.
So $k^2 = 1$.  And as $k \ge 0$ we have $k = 1$.
So let $f(x) =ax+b$.  And $f(f(x)) = a(ax+b) + b = a^2 +(ab + b) = 4x + 1$.
So we have $a^2 =4$ and $ab+b = b(a+1) = 1$.
So $a = \pm 2$.  $a+1=-1,3$ and $b = \frac 1{a+1}= -1, \frac 13$.
As $f(0) = a\cdot 0 + b = b > 0$ and $-1 < 0; \frac 13 > 0$ we must have $b > 0$ so $b = \frac 13$ and $f(x) = 2x + \frac 13$ and $f(0) = 2\cdot 0 + \frac 13 = \frac 13$.
[Note  $f(f(x)) = 2(2x + \frac 13) + \frac 13 = 4x + \frac 23 + \frac 13 = 4x + 1$.
