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 $g$).
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 $(-r,r)$].
The answer is essentially

$$g(r)=f^{-1}(p(r))$$ where $p$ is a polynomial that you can choose relatively freely.

More specifically, you can decompose $f$ in monotonic branches, that are each invertible, and choose a polynomial that follows the same decomposition, with extrema every time the growth direction changes.

E.g. the function $|x|$ is decomposed in a decreasing branch $-x$ in $(-\infty,0)$, and an increasing one $x$ in $(0,\infty)$. Then the polynomial $(r+2)^2$ has a similar behavior (decreasing to zero until $r=-2$ then increasing).

So we map the first branch to $(x+2)^2$ by the inverse transform of $-x$, and the second branch by $x$ (it turns out that these functions are their own inverses):

$$g(r)=\begin{cases}r&lt;2\to-(r+2)^2,\\r&gt;2\to (r+2)^2.\end{cases}$$

Notice that the ordinates of the local extrema must match those of the given function, but the abscissas a free, provided the order is kept.



Let us play this game with

$$\begin{cases}x&lt;-1&amp;\to x+2,\\-1&lt;x&lt;1&amp;\to-x,\\x&gt;1&amp;\to x-2\end{cases}.$$

We can map it to $\dfrac{r^3-3r}2$, that also has the local extrema $(-1,1)$ and $(1,-1)$.

Now

$$g(r)=\begin{cases}r&lt;-1&amp;\to\dfrac{r^3-3r}2-2,\\-1&lt;r&lt;1&amp;\to-\dfrac{r^3-3r}2,\\r&gt;-1&amp;\to\dfrac{r^3-3r}2+2.\end{cases}$$

The function $g$: