TITLE: Prove the equation has a root.
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: Hint: What's the derivative of $F(x)=x^2f(x)$?