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Assume that $f$ is a bounded and differentiable function in $(0,1)$. If $f({1\over 2})=0$, prove that the equation, $$2f(x)+xf'(x)=0,$$ has at least one root in $(0,{{1}\over{2}})$.

I tried to do it using Rolle's Theorem. Because the left side of the equation looks like the derivative of some function. And if I find a function $F$ s.t. $F'(x)=2f(x)+xf(x)$ and $F(0)=F({1\over 2})$, then I can use Rolle's Theorem to get the conclusion. I've found $F$, which is $$F(x)=\int_{0}^{x}f(t)\,dt+xf(x),$$ s.t. $F'(x)=2f(x)+xf(x)$, but the only problem is that I can't ask for $F(0)=F({1\over 2})$.

Can somebody give me a hint about this problem?

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Hint: What's the derivative of $F(x)=x^2f(x)$?

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  • $\begingroup$ I see! Why didn't I realize that before!! Thank you so much! $\endgroup$ – henryforever14 May 28 '12 at 20:21
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    $\begingroup$ Well done, @henryforever14! $\endgroup$ – Jyrki Lahtonen May 28 '12 at 20:22
  • $\begingroup$ I dont understand what will be the next step! please tell me! $\endgroup$ – Marso Feb 15 '13 at 17:24
  • $\begingroup$ @CityOfGod: Apply Rolle's theorem to the $F(x)$ in my answer. $\endgroup$ – Jyrki Lahtonen Feb 15 '13 at 21:53

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