I want to modify
$$B(x) = \left\{ \begin{array}{lr} e^{-\frac{1}{x^2}} & : x > 0\\ 0 & : x \leq 0 \end{array} \right.$$
so that the new function
$$C(x) = \left\{ \begin{array}{lr} 0 & : x \leq 0\\ 1 & : x \geq 1 \end{array} \right.$$
and also that $C'(0) > 0$ if $0 < x < 1$.
I tried $B(\frac{x}{1 - x})$, but it doesn't stay at $1$ when $x \geq 1$.
Take $C(x)=\frac{B(x)}{B(x)+B(1-x)}$.