jmc

Substituting $f(x)$ into the equation $f^{-1}(x)=\frac{1-2x}{2x}$, and noting that $f^{-1}(f(x))=x$ for all $x$ in the domain of $f$, we get $$x=\frac{1-2f(x)}{2f(x)}.$$ Solving for $f(x)$, we get $$f(x)=\frac{1}{2x+2}.$$ Therefore, $b=\boxed{2}$.