A problem on curve sketching

I came across an interesting problem on curve sketching a few weeks ago.

Question:


From the first two given information, we know f(x) is continuous at x=0, with value 1. From the third, we know f(x) is not differentiable at x=0, meaning we should expect a cusp at x=0. The slopes -1 and 1 should be noted as well. Using the last piece of information, horizontal asymptote: y=2. (?) f(x) also has to be an even function.

Now, sketching the curve is straightforward, but can we find functions that fit the above requirements?

We know the graph should look something like this.

First, look for functions with horizontal asymptote, examples being $\frac{1}{x^2+1}$ and arctan x. However, we can rule out $\frac{1}{x^2+1}$ because it does not fulfill the 3rd requirement as $\lim_{h \to 0} \frac{f(h)-f(0)}{h} = 0$. We now try f(x) = arctan x. To reach the desired graph, we need to transform the graph. $f(0)=1 \Rightarrow y=\arctan x +1$

Before we continue...

Prerequisite knowledge:
Range of arctan x
Since tan x would not be a function if inverted, only one cycle of tan x is used when finding its inverse, arctan x. The cycle that includes (0, 0) is chosen such that for tan x, domain: $(-\frac{\pi}{2}, \frac{\pi}{2})$ and range: (-∞, ∞) So for its inverse, domain: (-∞, ∞), range:$(-\frac{\pi}{2}, \frac{\pi}{2})$

Now, we want $\lim_{x \to \pm \infty}f(x)=2$, since the upper bound is $\frac{\pi}{2}$ $(\lim_{x \to +\infty} \arctan x=\frac{\pi}{2})$, multiply arctan x by $\frac{2}{\pi}\Rightarrow \frac{2}{\pi}\arctan (ax)+1$. Yet another requirement: f'(0)=1, $f'(x)=\frac{2}{\pi}\frac{a}{1+(ax)^2} \Rightarrow$ set $a=\frac{\pi}{2}$. Finally, $f(x)=\frac{2}{\pi}\arctan (\frac{\pi x}{2})+1$. Since it has to be an even function, we define $f(x)$ to be a piecewise function:

$$f(x)=
\begin{cases}
\frac{2}{\pi}\arctan (\frac{\pi x}{2})+1 & x \geq 0 \\
-\frac{2}{\pi}\arctan (\frac{\pi x}{2})+1 & x<0
\end{cases}$$

My question:
What are other functions that meet the 5 requirements?