TITLE: differentiation
QUESTION [7 upvotes]: I don't have any idea how to start solving this problem. Any help please?
Problem: 
Suppose that a differentiable function $f:\mathbb R \to \mathbb R$ and its derivative $f'$ have no common zeros. Prove that $f$ has only finitely many zeros in $[0,1]$.

REPLY [4 votes]: A different argument.
Suppose $f$ and $f'$ have no common zeroes. If $x\in\mathbb R$ be a zero of $f$, then $f'(x)\neq0$ and $f$ is therefore injective in a small neighborhood $I=(x-\varepsilon,x+\varepsilon)$ of $x$: in particular, the only zero of $f$ in $I$ is $x$ itself. This shows that the set of zeroes of $f$ is discrete in $\mathbb R$ and, therefore, intersects every bounded closed interval in a finite set.