# Zeros of a continuously differentiable function

Let $$f: D\rightarrow \mathbb{R}$$ be a continuously differentiable function defined on a domain $$D$$. Is it true that if $$x$$ is a non critical point of $$f$$, then there is a neighborhood of $$x$$ which contains no accumulation point of the set of zeros of $$f$$?

If $$f(x)\ne0$$, then there is a neighborhood of $$x_0$$ with no zeros of $$f$$. Without loss of generality, asume $$f\colon(-a,a)\to\Bbb R$$, $$f(0)=0$$ and $$f'(0)\ne0$$. Then $$f(x)=f'(0)\,x+h(x)\quad\text{with}\quad \lim_{x\to0}\frac{h(x)}{x}=0.$$ There exists $$\delta>0$$ such that $$|x|<\delta\implies\Bigl|\frac{h(x)}{x}\Bigr|\le\frac{|f'(0)|}{2}.$$ Then, if $$0<|x|<\delta$$, $$|f(x)|\ge|f'(0)|\,|x|-|h(x)|\ge|f'(0)|\,|x|-\frac{|f'(0)|}{2}\,|x|\ge\frac{|f'(0)|}{2}\,|x|>0.$$
• Thanks for your answer. Is the statement true when $D$ is a domain in $R^n$? – Jiu Dec 5 '18 at 15:22
• No. Consider $f(x,y)=x$. Then $(0,0)$ is not a critical point, but $f(0,y)=0$ for all $y$. – Julián Aguirre Dec 5 '18 at 15:26