There is a smooth function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ Suppose that we know everything about behaviour of this function $f(x)$ for $x\ge M$ where $M$ is real number. To know everything i mean we know its values , derivatives and etc
Let consider a zero of this function such it lies in set of real numbers $A$ and $sup A<M$. How it would impact on behaviour of $f$ so that we could detect that from the region of the domain that we know well ?.
I know it is quite general question, but lets consider a polynomial $f\in \mathbb{R}[x]$ and $A=[a,b]$ and $M$ a number greater than $b$. What can we say about answer of question above but for this case? What if $f$ is a power series?
Regards
EDIT: Suppose that smooth function $f$ has root exactly at $a$; $a<M$. How it will affect on behaviour of $f(x)$ for $x\ge M$?
If you leave a gap of positive width between $A$ and $M$, then we can glue two arbitrary smooth functions together so that one function appears on $A$ and the other function appears on $[M,\infty)$, so the behaviour in one region is completely uninformative about the behaviour in the other.
Let $g$ be a smooth function with domain containing $(-\infty, 1]$ and let $h$ be a smooth function with domain containing $[-1,\infty)$. Let $$ b(x) = \begin{cases}\mathrm{e}^{-\tan^2 (\pi x/2)} ,& -1 \leq x \leq 1 \\ 0 ,& \text{otherwise} \end{cases} $$ be a bump function, $\hat{b}(x) = \frac{1}{\int_{-\infty}^\infty b(x) \,\mathrm{d}x}b(x)$ be a bump function with unit area, and let $s(x) = \int_{-\infty}^x \; \hat{b}(t) \,\mathrm{d}t$ be a smooth step function of height $1$. Notice $b$, $\hat{b}$, and $s$ are smooth, $s$ is constantly $0$ on $(-\infty, -1]$, constantly $1$ on $[1,\infty)$, and increases smoothly between these values on $[-1,1]$. Here are plots of $b$ and $s$.
Since the product and sum of smooth functions is smooth, consider the combination $s(-x)g(x) + s(x)h(x)$, where the values of $h$ and $g$ are taken to be zero on $x$ outside of their domains. This is a smooth function that exactly agrees with $g$ on $(-\infty, -1]$ and exactly agrees with $h$ on $[1, \infty)$ and smoothly transitions between the two on $[-1,1]$. By linearly re-scaling $x$, we can move this interval $[-1,1]$ to land on $[\sup A,M]$, so we get $g$ on $A$ and $h$ on $[M,\infty)$. Since $g$ and $h$ are independent, neither is informative about the location of zeroes of the other.
Here's an plot with $g(x) = \frac{1}{16}(x^2+1)$ and $h(x) = \sin x$. On the left we have something that we know has no zeroes. On the right, we have something that we know has infinitely many zeroes. In between, we smoothly transition from one to the other.
Nothing we can learn from the polynomial on the region left of $-1$ will tell us about zeroes of this combined function to the right of $1$ and nothing we can learn from sine on the right of $1$ will tell us about the absence of zeroes of this combined function to the left of $-1$.