How do I create a sigmoid-esque function with the following properties? For a range of $x$ values between $A$ and $B$ I would like $f(x) \rightarrow x$.  For values less than $A$ I would like $f(x)$ to exhibit a sigmoid-esque convergence to $A'$  where $A'$  is $A - \delta$ for some small $\delta$.  Similarly, for values greater than $B$ I would like $f(x)$ to converge to $B'$ where $B'$ is $B+\delta$ for some small $\delta$.
Typical values will be $A = 0.5$, $B = 1.5$, $A' = 0.4$, $B' = 1.6$.
 A: This image from Wikipedia shows several sigmoid-type functions, each with derivative 1 at the origin:

By taking any of these (choose one, and call it $g(x)$), translating, and rescaling gives you $B + \delta g((x-B)/\delta)$. This will serve as the upper end of your function. The "$B +$" is the vertical translation, the "$- B)$" is the horizontal translation, the factor of $\delta$ on the outside rescales $g$ vertically, and the factor of $\delta$ on the inside scales $g$ horizontally (making the derivative correct, so that you don't have a singular point in your piecewise function).
See this example that I typed into wolfram alpha with $g = \tanh$, $B = 1.5$, $\delta = 0.1$: http://www.wolframalpha.com/input/?i=1.5+%2B+0.1+*+tanh%28%28x-1.5%29%2F0.1%29

You can work out the lower end of your function analogously.
$$
   f(x) = \left\{
     \begin{array}{lr}
       A + \delta\tanh((x-A)/\delta), & x \leq A \\
       x, & A < x < B \\
       B + \delta\tanh((x-B)/\delta), & x \geq B \\
     \end{array}
   \right.
$$
is one such function, but you don't need to use $\tanh$. If it's unclear how I arrived at this, it could be helpful to take each of the four transformations I performed on $\tanh$ and plotting them individually to see exactly how the function changes, and how the transformations act together to give you what you need.
