2.3. Find all functions $f:[0, \infty) \longrightarrow[0, \infty)$ such that $f(f(x))+f(x)=12 x$, for all $x \geq 0$.
2.3 Fix any $x \geq 0$. Let $f^{[0]}(x)=x$ and $f^{[1]}(x)=f(x)$. For $n \geq 1$, let $f^{[n]}(x)=f\left(f^{[n-1]}(x)\right)$. Then the above functional equation gives
$$
f^{[n+2]}(x)+f^{[n+1]}(x)=12 f^{[n]}(x) .
$$

Solving this difference equation, we have
$$
f^{[n]}(x)=C_{1} 3^{n}+C_{2}(-4)^{n} .
$$

Using the initial conditions $f^{[0]}(x)=x$ and $f^{[1]}(x)=f(x)$, we have $C_{1}=(f(x)+4 x) / 7$ and $C_{2}=(3 x-f(x)) / 7$. Therefore,
$$
f^{[n]}(x)=\frac{1}{7}(f(x)+4 x) 3^{n}+\frac{1}{7}(3 x-f(x))(-4)^{n} .
$$

Since $f(x) \geq 0, f^{[n]}(x) \geq 0$ for all $n \geq 0$. By taking $n$ even, we have $\frac{1}{7}(f(x)+4 x) 3^{n}+$ $\frac{1}{7}(3 x-f(x)) 4^{n} \geq 0$. From this, $3 x-f(x) \geq 0$. By taking $n$ odd, we have $\frac{1}{7}(f(x)+4 x) 3^{n}-$ $\frac{1}{7}(3 x-f(x)) 4^{n} \geq 0$. From this, $3 x-f(x) \leq 0$. Consequently, $f(x)=3 x$. One can easily verify that $f(x)=3 x$ satisfies the given functional equation.
Alternatively, suppose for some $a, f(a)=3 a+c$ where $c \in \mathbb{R}$. Then $f^{[n]}(a)=3^{n} a+k_{n} c$. One can obtain a recurrence relation in $k_{n}$ and use it to prove that $c=0$. Finally check that $f(x)=3 x$ satisfies the given condition.