Assume I have a function continuous functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $\lim_{x\rightarrow\pm \infty} f(x)=\pm \infty$.
I am looking for some continuous and strictly monotonous increasing function $g:\mathbb{R}\rightarrow \mathbb{R}$ with $g(0)=0$, such that $f(g(r))=p(r)$ is a polynomial. That is $g$ can be used to reparametrize $f$ into a polynomial $p$.
Example: If I take $f(x)=|x|$, then I can choose $g(r)=x\cdot|x|$, such that $f(g(r))=|x\cdot|x||=x^2=p(r)$.
I don't have a proof, but I think such a $g$ should always exist, given the conditions on $f$. If not, please let me know, which conditions I am missing.
Is there a general method for finding such a $g$? My main problem is, that I don't know what the resulting polynomial $p$ should look like, except for the degree of the polynomial.
Example: Let $f(x)=|x-a|+b$. I know that $p$ must have the form $p(r)=\beta_2 r^2+\beta_1 r + |a| + b$ from the form of $f$. But I can neither find the coefficients $\beta$ nor the function $g$ that allows for this transformation. I tried the approach $|g(r)-a|+b=\beta_2r^2+\beta_1x+|a|+b$, but I cannot solve this because I don't know $\beta$ and $|\cdot|$ is not invertible (so I can't just solve for $g$).
You cannot do this if $f$ has zeros at $0$ and distinct points $a_n$ tending to $0$ and non-zero at other points. Since $f(x)=p(g^{-1}(x))$ in some interval around $0$ we see that $f$ cannot have at zeros accumulating at $0$. [ $g$ continuous and strictly increasing so it is has an inverse on some interval $(-r,r)$].