I know that if $r$ is a root of $p(x)\in \mathbb{R}[x]$ of multiplicity $k+1$, then $p^{(j)}(r)=0$, for every $j\leq k$, where $p^{(j)}$ stands for the $j$-th derivative of the function $x\mapsto p(x)$.
Conversely, if $p(r)=p^{(1)}(r)=\dots=p^{(k)}(r)=0$, then does $(x-r)^{k+1}$ divide $p$?
If yes, then
Is there a notion of multiplicity for arbitrary functions? What about some relation with zeros of its derivatives?
Thank you.