How to create an explicit formula of a function Let $\phi:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable and nondecreasing function such that $\phi(x)=-1$ for $x<-1$, $\phi(x)=1$ for $x>1$, and $\phi^\prime(x)>0$ for $x\in(-1,1)$. 
Question  How to find some explicit formulae for $\phi$?
Thank you for kind comments.
 A: Let $f(x) = \begin{cases} {3 \over 2} ( 1-x^2), & |x| < 1 \\
0, &\text{otherwise} \end{cases}$,
and note that $f$ is continuous and $\int_{-\infty}^\infty f(x)dx = 2$.
Let $\phi(x) = -1+\int_{-\infty}^x f(t)dt$. We see that $\phi'(x) = f(x)$.
A: You can define $\phi$ in a piecewise manner as long as you make sure to maintain differentiability, particularly at the boundary points of subdomains, i.e. 
$$
\phi(x) = 
\begin{cases}
-1, & x\in(-\infty,-1) \\
h(x),  & x \in [-1, 1] \\
1, & x \in (1, +\infty)
\end{cases}
$$
If you choose $h(x)$ s.t. it is nondecreasing, differentiable, and satisfies certain conditions, which stand  for differentiability and continuity of $\phi$ at the boundary points $a=-1$ and $b = 1$. The conditions are:


*

*$\lim\limits_{x \to a^{+}} h(x) = -1 \quad$ — right limit of $h$ equals to $-1$ at the point $a = -1$,

*$\lim\limits_{x \to b^{-}} h(x) =  1 \quad$ — left limit of $h$ equals to $1$ at the point $b = 1$,

*$ \partial_{+}h(a) := \lim\limits_{x \to a^{+}} \dfrac{h(x) - h(a)}{x-a} =  0 \quad$ — right derivative of $h$ equals to $0$ at the point $a = -1$,

*$ \partial_{-}h(b) := \lim\limits_{x \to b^{-}} \dfrac{h(b) - h(x)}{b-x} =  0 \quad$ — left derivative of $h$ equals to $0$ at the point $b = 1$.


The last two conditions are actually extras as they will give you a continuous derivative of $\phi$.
I can image that one can come with a lot of versions of function $h$. For example, you can use following function
$$
h(x) :=  \operatorname{sgn}(x)\cdot \left( 1 - \mathrm{e}^\frac{x^2}{x^2-1}  \right),
$$

where $\operatorname{sgn}(x) = x/|x|$ is the signum function.
