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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?

Myshkin
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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 $f(p)=f\,'(p)=0$.

Brian M. Scott
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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? – Myshkin Jan 28 '13 at 05:43
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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 05:45
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    and hence that is a contradiction and we are done! thank you Brian – Myshkin Jan 28 '13 at 05:45
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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 05:46