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?


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

  • $\begingroup$ I see! Why didn't I realize that before!! Thank you so much! $\endgroup$ – henryforever14 May 28 '12 at 20:21
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.