Let $f_0,f_1,\ldots$ be a sequence of functions $f_n : [0,1] \rightarrow R$ defined as follows:

$$f_0(x) =1+2x$$

$$f_{n}(x) := \left\{\frac{5+t}{2} : \text{ where t solves } f_{n-1}\left(\frac{x}{t}\right) = \frac{5+t}{2}\right \}$$

In other words, $f_n(x)$ is equal to the intercept (with respect to $y$) of the functions $f_{n-1}(x/t)$ and $\frac{5+t}{2}$.

It isn't hard to prove by induction that each $f_n$ (for $n>0$) will be a continuous increasing function such that $f_n(0)=5/2$ and $f_n(1)=3$. If we fix $0<x<1$ and consider the function $f_{n-1}(x/t)$ as a function of $t$, we see that it is decreasing as $t$ increases and $f_{n-1}(x/t)=3$ for $t=x$, and $f_{n-1}(x/t) <3$ for $y=1$. From this it follows that there is a unique intercept and that $f_n$ is well defined.

One can show that: $$f_1(x)= \frac{1}{4}(\sqrt{16x+9}+7),$$ $$f_2(x) = \{\frac{5+t}{2} : \text{ where t solves } x = (t^3+3t^2)/4 \}.$$

Is there a method for finding a good approximation (upper and lower bounds) by elementary functions of the $n$-th iterate $f_n(x)$ in a neighborhood of $x=1$?

I'm interested in both a solution to this particular problem, as well as understanding methods that work in similar situations.