Infinitely many times differentiable function, whose every derivative approaches zero at x boundaries

Question

Prove that there exist such nondecreasing, infinitely many times differentiable function $f: \Bbb{R}\to[0,1]$ that $f(x)=0$ when $x\le0 $ and $f(x)=1$ when $x\ge1$.

I've come up with necessary condition that derivative of any order has to approach $0$ when $x \to0^+$ and $x \to1^-$, but I can't make any usage of it. I've already tried to fit there some exponential function, but I failed since it would require the argument to be streched, and that didn't go well with initial condition.

Any help and hints will be appreciated.

Answer 1

Define $\varphi\colon\mathbb{R}\longrightarrow\mathbb R$ by$$\varphi(x)=\begin{cases}e^{-1/x}&\text{ if }t\geqslant0\\0&\text{ otherwise.}\end{cases}$$Now, let $\displaystyle f(t)=\frac{\varphi(t)}{\varphi(t)+\varphi(1-t)}$.