Existence of $f(x)$

Question

Suppose $g(x)$ is cubic which has two local extrema.

Is there differentiable function $f(x)$ which satisfies $\forall x \in \mathbb{R}, g(f(x))=x$ exist?

I know if I make $f$ piecewise inverse of $g$, then $g(f(x))=x$ But Can I make $f$ differentiable?

Answer 1

Even if you consider the three pieces independently so that the function is truly invertible, they won't be differentiable. Because

$$(f^{-1})'=\frac1{f'}$$ and the derivative cancels at the stationary points (vertical tangents).

Answer 2

Such a function cannot exist as there is not a one-to-one mapping between each of $x$ and $g(x)$ due to the conditions placed on $g(x)$.