# Construct suitable cutoff function

Let $$\bar x \in \mathbb R$$. Is there a cut-off function such that $$\phi_\epsilon \in C^\infty(\mathbb R)$$, $$0 \le \phi \le 1$$, and $$\phi_\epsilon(x) = \begin{cases} 1 &\text{ if } |x - \bar x| \ge \epsilon\\\ 0 &\text{ if }|x-\bar x|\le \epsilon/2 \end{cases}$$ and $$\phi' \le c_\epsilon \phi$$?

Indeed, let $$f:=\phi=\phi_\epsilon$$. Let $$a:=\sup\{x\colon f(x)=0\}$$. Then $$a$$ is real, $$f(a)=0$$ and $$f>0$$ on the interval $$(a,\infty)$$. Without loss of generality, $$a=0$$, so that $$f(0)=0$$ and $$f>0$$ on the interval $$(0,\infty)$$.
Suppose now that for some real $$c>0$$ we have $$f'\le cf$$. Then $$(\ln f)'\le c$$ on $$(0,\infty)$$, whence $$\ln f(1)-\ln f(0+)\le c$$, that is, $$0=f(0+)\ge f(1)e^{-c}>0$$, a contradiction. $$\quad\Box$$
Under the described hypotheses, the function $$f: x \in \mathbf{R} \mapsto e^{-cx} \phi(x)$$ would be decreasing, because $$f'(x) = e^{-cx} (-c \phi(x) + \phi'(x)) \leq 0$$ for all $$x \in \mathbf{R}$$. As $$f(0) = 0$$ and $$f \geq 0$$, one would have $$f(x) = 0$$ for all $$x \geq 0$$. For this reason a cutoff function cannot have $$\phi' \leq c \phi$$.
Here for simplicity, we let $$\overline{x} = 0$$, $$\epsilon = 1$$, and wrote $$c = c_{\epsilon}$$.