How to choose smooth function $f:\mathbb R \to [0,1]$?

Question

Let $\alpha, \beta >0.$

Can we expect to choose differentiable $f:\mathbb R \to [0,1]$ such that $f(x+\alpha)=0$ if $x\leq 0,$ and $f(x-\alpha)=1$ if $x\geq \beta$? If so, can we make $f$ smooth?

Answer 1

With $u<v:$ Let $f(x)=0$ for $x\leq u.$ Let $f(x)=1$ for $x\geq v.$

Let $ n\in \Bbb Z^+.$

Let $g(y)=(y-u)^{n+1} (y-v)^{n+1}.$ Let $K=\int_u^v g(y)dy.$

For $u<x<v$ let $f(x)=\frac {1}{K}\int_u^x g(y)dy.$

The $n$th derivative of $f$ exists and is continuous at all points.

Note: $K\ne 0$ because either (i) $g(y)>0$ for all $y\in (u,v)$ or (ii) $g(y)<0$ for all $y\in (u,v).$