Let $L>0$. Is it possible to find $0<\varepsilon <L$ and a cut off function $\psi: [0,L]\rightarrow \mathbb{R}$, such that
i) $\psi \in C^3([0,L])$,
ii) $\psi(x)=0$ for all $x \in [0,\varepsilon]$,
iii) $\psi(x)=1$ for all $x \in [L-\varepsilon, L]$,
and $$|\psi'(x)||\psi(x)| < \frac{1}{6}, \quad \forall x \in [0,L]?$$
Remark: If $L \leq 3$, such function doesn't exist. (In this case is easy to obtain a contradiction)
Start with the density function $f(t) = 140 t^3 (1-t)^3$, $t\in [0,1]$. Then the CDF $F(x) = \int_0^x f(t) \, \mathrm{d} t = 35 x^4 - 84 x^5 + 70 x^6 - 20 x^7$, satisfies: \begin{align} 1) & \quad F(0) = 0,\, F(1) = 1\\ 2) & \quad F^{(k)}(0) = F^{(k)}(1) = 0, \quad k\in{1,2,3}\\ 3) & \quad F(x), F^\prime(x) \ge 0 \mbox{ for } x\in[0,1] \\ 4) & \quad \max_{x\in[0,1]}{F(x) F^\prime(x)} \le 2 \end{align}
Now you can define (taking $\varepsilon < \frac12 L$): $$ \psi(x)=\begin{cases} 0 & x<\varepsilon;\\ F\left(\frac{x-\varepsilon}{L-2\varepsilon}\right) & \varepsilon\le x\le L-\varepsilon;\\ 1 & x>L-\varepsilon. \end{cases} $$ By the scaling of the function $\psi(x)$ (and its derivative), we have $\max_{x\in[0,L]} \psi(x) \psi^\prime(x) \le \frac{2}{L-2\varepsilon}$. So this will be small if $L$ is sufficiently large.