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If $f: U \to \mathbb{R}$ is differentiable in a open and bounded $U \subset \mathbb{R}^m$ and, for all $a \in \overline{U}-U$ we have $f(x) \to 0, \ x \to a ;$ Then exist a $x_0 \in U$ such that $f'(x_0)=0$.

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    So...what's the question?2012-10-18

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Since $f(x)\rightarrow 0$ as we approach the boundary, we can extend $f$ to a continuous function on $\bar{U}$ by defining $f$ to be $0$ on $\bar{U}-U$. Since $U$ is bounded, so is $\bar{U}$, so $\bar{U}$ is actually compact. A continuous function must achieve its max and min on a compact set. If $f$ is constant, we're done. Otherwise, since $f$ is constant on $\bar{U}-U$ it must have either a min or max on the interior, i.e. in $U$. Use what you know about derivatives at a max or min.