Deeper meaning and intuition behind $\frac{x}{1+x^2}$ having the same values for $x$ and for $\frac{1}{x}$ I was playing around with the function $f(x)=\frac{x}{1+x^2}$ and I noticed that $f(x)$ has the same values for $x$ and for $\frac{1}{x}$. 
Is there any intuition or deeper meaning behind this?
Thanks in advance!
 A: Not sure of what you expect but your function is $\frac1{x+1/x}$.

The field of rational functions is $\Bbb{C}(x)$ and the field of rational functions invariant under $x\to 1/x$ is $\Bbb{C}(x+1/x)$. 

The proof requires a bit of field theory : the field extension $\Bbb{C}(x)/\Bbb{C}(x+1/x)$ is algebraic of degree $2$, this is because $x$ is a root of $$(T-x)(T-1/x)=T^2-(x+1/x)T+1\in \Bbb{C}(x+1/x)[T]$$
Thus there is not field between $\Bbb{C}(x)$ and $\Bbb{C}(x+1/x)$ and the latter has to be the full field of rational functions invariant under $x\to 1/x$.
A: It is interesting to note the trigonometric significance of this. If $x=\tan\theta$, then
\begin{align*}
\dfrac x{1+x^2}&=\dfrac{\tan\theta}{1+\tan^2\theta}\\[5pt]
&=\dfrac12\sin2\theta\\
&=\dfrac12\sin\left\{2\left(\dfrac\pi2-\theta\right)\right\}\\
&=\dfrac{\cot\theta}{1+\cot^2\theta}\\
&=\left(\dfrac{\dfrac1x}{1+\left(\dfrac1{x}\right)^2}\right)
\end{align*}
A: A function is even when for all x, f(-x) = f(x).
Your function is an even function in the multiplicative group of reals
A: There is no deep meaning, you can create as many such functions as you like.
Let $f(x)$ be an even function and let $g(x):=f(\log x)$. Then $g(x)=g(x^{-1})$.
For example
$$\cosh(x)=\frac{e^x+e^{-x}}2$$ is an even function and corresponds to
$$g(x):=\frac{x+x^{-1}}2=\frac{x^2+1}{2x}$$ with the desired property.
Similarly, you can create functions such that
$$f(-x)=\dfrac1{f(x)}$$ or $$f\left(\dfrac1x\right)=-f(x)$$ or $$f\left(\dfrac1x\right)=\dfrac1{f(x)}$$$$\cdots$$

Even better, every function has an even part,
$$f_e:=\frac{f(x)+f(-x)}2,$$ and you draw
$$g(x):=f_e(\log x)=\frac{f(\log x)+f(-\log x)}2.$$
A: Consider $g(x) = x + \frac{1}{x}$. Then $g(x) = g(\frac{1}{x})$, because $x + \frac{1}{x}$ = $\frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x$.
Your function $f(x) = \frac{1}{g(x)} = \frac{x}{x} \frac{1}{x + \frac{1}{x}} = \frac{x}{x(x + \frac{1}{x})} = \frac{x}{x^2 + 1}$
So some intuitions or deeper meanings behind this are: addition commutes and a reciprocal of a reciprocal is the original value (inverse of inverse). BTW note that $x\ne0$ for this to work - which we know it is, because we have already $\frac{1}{x}$.
