Let $f\colon\mathbb R\to\mathbb R$ be a surjective function. Assume that $f'$ is also surjective. Assume also that $f$ vanishe on $s=1$ of order $m$.
My question is: Are the following functions also surjective? $$f(s^2-s+1)$$ and $$(s-1)f(s)$$
At least locally near the point $s=1$.
1) Since $s^2 - s + 1 \geq 0$ for all $s \in \mathbb{R}$, the answer to the first question is "no." That is because $f(s)$ might only be surjective because of its properties on negative values of $s$. (You can come up with a specific example).
2) The idea here is to construct a function $f(s)$ that is surjective over $s \in \mathbb{R}$, is not surjective over $s\geq 2$, but has a surjective derivative over $s\geq 2$.
Specifically, to show that $(s-1)f(s)$ is not necessarily surjective, even when $f(s)$ meets the criteria, define:
$f(s) = g(s)$ if $s \in [2, \infty)$
$f(s) = c(s)$ if $s \in [0,2]$
$f(s) = s-1$ if $s \in (-\infty, 0]$
where $g(s)$ has derivatives of all orders over $s\geq 2$, is non-negative for all $s \geq 2$, and satisfies $\sup_{s\geq 2}g'(s) = \sup_{s\geq 2}g(s) = \infty$ and $\inf_{s\geq 2} g'(s) = -\infty$. You can come up with such a function $g(s)$. Then fill in $c(s)$ appropriately over $s \in [0,2]$ to ensure the differentiability and $s=1$ conditions are met. Finally, show that $f(s)$ is surjective, $f'(s)$ is surjective, but $\inf_{s\in \mathbb{R}} (s-1)f(s) > -\infty$.