Invertible "Sigmoid + x" function I need an invertible function that represents a smooth transition between two straight, parallel line segments, like this:

Depicted is $f(x) = -0.3/(1+e^{-10*(x-p)})+0.3/2+x$ (where $p$ is the location of the lowest slope), but it does not seem to be invertible in terms of standard mathematics.
Can you recommend a function that looks similar, but is invertible?
 A: I think a relatively simple way to obtain an "invertible sigmoid" is to find a suitable cubic polyomial. The method I propose here has the drawback that it does not let you control both the tangency points at the same time.
Let's first simplify the setting. Let's consider the two points $P_1=(-2,-1)$ and $P_2=(2,1)$ and the two lines $r_1:y=x+1$ and $r_2:y=x-1$. We now wish to find a cubic polynomial that is always increasing and is tangent to the lines in $P_1$ and $P_2$. In this particular situation the points are symmetrical w.r.t the origin, so our polynomial must be of the form $y=px^3+qx$.
If we impose the aforementioned conditions, we get:
1) passage through $P_1$: $px_1^3+qx_1=y_1\Rightarrow 8p+2q=1$ (this is equivalent to the request that the cubic pass through the other point).
2) tangency in $P_1$: $3px_1^2+q=1\Rightarrow 12p+q=1$ (Again, tangency in $P_2$ is automatically satisfied)
These equations completely determine $p$ ad $q$, namely $p=1/16$ and $q=1/4$, so the polynomial we are searching for is
\begin{equation}
f(x)=\frac{1}{16}x^3+\frac{1}{4}x
\end{equation}
Take this as your "base" function: we will tweak it, adjusting it to the real case. Let the two parallel lines be $r: y=ax+b$ and $s: y=ax+c$ with $c<b$ (i.e. $r$ is the "leftmost" line), with $a>0$. Their zeroes are $-b/a$ and $-c/a$ respectively, and the distance between them is $\Delta x=(b-c)/a$.
With this in mind, let's first tweak the two lines $r_1$ and $r_2$ so that the distance between their zeroes matches $\Delta x$. Let's define $\tilde{r}_1: y=kx+1$ and $\tilde{r_2}: y=kx-1$. The zeroes are respectively $-1/k$ and $1/k$; their distance is therefore $2/k$ and so in order to match $r$ and $s$ we must set $k=2a/(b-c)$. with this in mind, we can tweak our cubic, setting
\begin{equation}
f_1(x):=f(kx)=\frac{1}{16}(kx)^3+\frac{1}{4}(kx)=\frac{kx}{4}\left(\frac{k^2}{4}x^2+1\right)
\end{equation}
Now, to adjust the slope to the required slope ($a$), we can multiply the cubic equation by an appropriate factor, which is $a/k$:
\begin{equation}
f_2(x):=\frac{a}{k}f_1(x)=\frac{ax}{4}\left(\frac{k^2}{4}x^2+1\right)
\end{equation}
Now, with a suitable translation, we can move around our function so that a tangency point match a given one:
\begin{equation}
F(x):=f_2(x-x_0)+y_0=\frac{a(x-x_0)}{4}\left(\frac{k^2}{4}(x-x_0)^2+1\right)+y_0=\\
=\frac{a(x-x_0)}{4}\left(a^2\frac{(x-x_0)^2}{(b-c)^2}+1\right)+y_0
\end{equation}
(The new tangency points are now $\tilde{P}_1=(-2/k,-a/k)$ and $\tilde{P}_2=(2/k,a/k)$, so if we want, for example, move $\tilde{P}_1$ to a point on $r$ which has coordinates $(P_x,P_y)$, we will have to set in the previous equation $x_0=P_x+2/k$ and $y_0=P_y+a/k$.
You requested that this function is invertible- well, it is thanks to Cardano's formulas. Our original cubic has an inverse which is explicitly
\begin{equation}
f^{-1}(y)=\left(8y-8\sqrt{y^2+1/27}\right)^{\frac{1}{3}}-\frac{4}{3}\left(8y-8\sqrt{y^2+1/27}\right)^{-\frac{1}{3}}
\end{equation} 
To get the inverse $F^{-1}$ of the function $F$, we just have to do the transformations backwards:
\begin{equation}
F^{-1}(y)=\frac{1}{k}f^{-1}\left(\frac{k}{a}(y-y_0)\right)+x_0
\end{equation}
It might be a lenghty procedure (even if not so complex), but at least you get all analytical results.
SUMMARY: 
Let the two lines be $r:y=ax+b$ and $s:y=ax+c$, with $c<b$. Let $P=(P_x,P_y)$ be a point on $r$. Let $x_0=P_x+(b-c)/a$ and $y_0=P_y+(b-c)/2$. Then a cubic function tangent to $r$ in $P$ which joins smoothly $s$ is given by the function
\begin{equation}
F(x)=\frac{a(x-x_0)}{4}\left(a^2\frac{(x-x_0)^2}{(b-c)^2}+1\right)+y_0
\end{equation}
Its inverse is
\begin{equation}
F^{-1}(y)=\frac{b-c}{2a}f^{-1}\left(\frac{2}{b-c}(y-y_0)\right)+x_0
\end{equation}
where $f^{-1}(y)$ is given above.
