Suppose $f \in C^{\infty}(\mathbb{R})$ suppose also $f$ has compact support, name such support $K$, defining
$$ \begin{array}{l} x_m = \inf K \\ x_M = \sup K \\ \end{array} $$
we must have $f(x_m) = f(x_M) = 0$, what I'd like to know is if for each $n$ we have $$ f^{(n)}(x_m) = f^{(n)}(x_M) = 0 $$
I think this might be true and I'm attempting to prove it. Suppose $f^{(1)}(x_m) > 0$ then we would have a neighborhood $\mathcal{U}(x_m)$ where $f$ is strictly increasing, but $f(x_m)$ is 0 which implies that there's some point $y \in \mathcal{U(x_m)}, y < x_m$ where $f(y) < f(x_m) = 0$ but then $x_m$ wouldn't be the $\inf K$. We can do the same for $x_M$, and by induction we can prove this for each $n$, since $supp (f^{(1)}) = K$ and $f^{(1)} \in C^{\infty}(\mathbb{R}^n)$.
Is this proof correct?
This is completely trivial from the definition.
Let $V=\Bbb R\setminus K$. Then $V$ is an open set and $f=0$ in $V$. Since $V$ is open this shows that $f^{(n)}=0$ in $V$, and now since $x_m,x_M\in \overline V$ and $f^{(n)}$ is continuous it follows that $f^{(n)}(x)=0$ for $x=x_m,x_M$.