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$f$ is differentiable on $\mathbb{R}$ such that $f$ and $f'$ has no common $0$ in $[0,1]$, we need to show $f$ can not have infinitely many $0$ in $[0,1]$.could any one give me hint?

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HINT: Let $Z=\{x\in[0,1]:f(x)=0\}$. If $Z$ is infinite, it has a cluster point $p$. Show that $f(p)=f\,'(p)=0$.

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okay as every bounded infinite subset of $\mathbb{R}$ has a limit pt. in it, so $f(p)=0$ by continuity, now $f'(p)=\lim_{h\rightarrow 0}\frac{f(p+h)-f(p)}{h}=0$..am i going in right path? –  Bunuelian Trick Jan 28 '13 at 5:43
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@Panu: Yes, you are. Note that there are arbitrarily small $|h|>0$ such that $f(p+h)=0$. –  Brian M. Scott Jan 28 '13 at 5:45
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and hence that is a contradiction and we are done! thank you Brian –  Bunuelian Trick Jan 28 '13 at 5:45
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@Panu: Exactly. (Actually, this is technically a proof of the contrapositive: if $Z$ is infinite, then $f$ and $f\,'$ have a $0$ in commmon. But it can also be phrased and thought of as a proof by contradiction.) –  Brian M. Scott Jan 28 '13 at 5:46

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