TITLE: If $f:\mathbb{R}\to\mathbb{R}$ is infinitely differentiable, then for all $x\in[0,1]$, $f^{(m)}(x)\neq 0$ for some $m$?
QUESTION [6 upvotes]: This is question #2 from the Fall 2003 qualifying exam at my university.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be an infinitely differentiable function such that for all $x\in[0,1]$, there exists $m>0$ such that $f^{(m)}(x)\neq 0$. Prove that in fact there exists $M$ such that for all $x\in[0,1]$ there exists $0<m\leq M$ such that $f^{(m)}(x)\neq 0$.

I had this idea; for each $x_i\in[0,1]$, there exists $m_i$ such that $f^{(m_i)}(x_i)\neq 0$. Since $f$ is infinitely differentiable, $f^{(m_i)}$ is continuous, so there exists $\epsilon_i$ such that $f^{(m_i)}\neq 0$ on $(x_i-\epsilon_i,x_i+\epsilon_i)$. Then
$$
[0,1]\subseteq\bigcup_{i\in I}(x_i-\epsilon_i,x_i+\epsilon_i)$$
is an open covering of $[0,1]$, so by compactness, 
$$
[0,1]\subseteq\bigcup_{i=1}^n(x_i-\epsilon_i,x_i+\epsilon_i)$$
after relabelling. Set $M=\max\{m_1,\dots,m_n\}$. For arbitrary $x\in [0,1]$, $x\in (x_j-\epsilon_j,x_j+\epsilon_j)$ for some $j$. Then $f^{(m_j)}(x)\neq 0$ on $(x_j-\epsilon_j,x_j+\epsilon_j)$, and $m_j\leq M$. 
Is this solution correct or can it be made better?

REPLY [3 votes]: The solution is correct, as far as I can tell. This appears to be the expected approach, and I don't believe that the answer can be made better.