Reparametrize a function as a polynomial 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 $g$).

Such functions $g$ and $p$ won't always exist. Remember that $p$ is a polynomial. That implies that $p$ can attain any sufficiently large real value $x=p(r)$ at at most two real points $r=r_1,r_2$ (only once, if $p$ has odd degree). Here the meaning of sufficiently large will depend on $p$, but that's irrelevant.
But $f$ may attain all the values at least three times. Consider a function like
$$
f(x)=\sin x+\frac{x}{100}.
$$
The part $x/100$ won't increase noticeably for several periods of sine.
Because $g$ is increasing, $f(g(r))$ will share this property with $f$.
