Polynomials, $3^x$ and the Collatz conjecture

Question

$\DeclareMathOperator\Orb{Orb}\newcommand\abs[1]{\lvert#1\rvert}$The Collatz or the $3n+1$ conjecture is open.

1. Are there non-trivial polynomials $f(x)\in\mathbb Z[x]$ and $g(x)\in\mathbb R[x]$ having unbounded $\abs{f(n)}$, $\abs{g(n)}$ as $n$ grows in $\mathbb N$ and an integer $x_0\in\mathbb N$ satisfying $\abs{g(x)}<2\abs{f(x)}+1$ for every $x>x_0$ such that for every $m\in\mathbb N_{>x_0}$ there is a $k\in \Orb(2\abs{f(m)}+1)$ such that $k<\abs{g(m)}$?

Here $\Orb(t)$ is the set of numbers $t$ traverses through as we apply Collatz transformations.

2. If $f(x)$ is the non-polynomial $(3^x-1)/2$ is there a polynomial $g(x)$ as in 1.?

Note $1$ need not be in $\Orb(\abs{g(x)})$ for every $x\in\mathbb N_{x_0}$ for both 1. and 2..

Answer 1

Point 1. is easy to satisfy even with nontrivial polynomials, since if you know $t$ modulo $2^n$ then you know for sure what the first n transformations will be.

For example. I could take $f=2^{20}x+174762$ and $g=6x+2$. Indeed I choose $f$ such that $3(2f(n)+1)+1)$ is always divisible by $2^{20}$, meaning that the first 21 transformations applied to $2f(n)+1$ are

1. $3x+1$ because $2f(n)+1$ is odd;
2. $x/2$ repeated 20 times because $3(2f(n)+1)+1)=2^{20}(6n+1)$ is divisible by $2^{20}$.

After this we will have reached $k = 6n+1$ which is smaller than $g(n)=6n+2$.

Point 2. is likely very hard and I have no idea how to tackle that. Indeed answering 1. with polynomials $f$ and $g$ with $\deg g < \deg f$ will already be very difficult. Although when taking $f$ to be the similar looking non polynomial $f=2(4^x-1)/3$ one can take $g=2$ for silly reasons.