By looking at graph of $f(x) = \frac{x}{1+x^2}$ I can clearly see that there are at least 2 points $x_1$, $x_2$ where: $$f(x_1) = f(x_2), \quad x_1 \neq x_2$$
How can I prove this function is not injective?
Problem here is that in contrast with function for example $f(x) = x^2$ you can check with negative $x$ to prove non injectivity, but in this case both $x_1$, $x_2$ are either positive or negative.
You start with $f(x)=f(y)$. Thus $$yx^2-(1+y^2)x+y=0.$$ It's a second degree polynomial in $x$. You can easily see that $$\Delta = (y^2-1)^2.$$ So the two solutions for $x$ (if $y\neq 0$, otherwise the unique solution is $x=0$) are $$\frac{1+y^2\pm (y^2-1)}{2y},$$ ie $x=y$ or $x=1/y$.
You can thus remark that the only points which have only one antecedent are $f(-1),f(0)$ and $f(1)$, ie $-1/2,0$ and $1/2$.
Let $x\ge 0;$
$f(x) =\dfrac{x}{1+x^2}$, continuous.
$f(0)=0;$ $f(1)=1/2;$
$f(10)=\dfrac{10}{1+10^2}< 1/10$;
Intermediate Value Theorem for continuous functions
Consider the intervals $[0,1]$ and $[1,10].$
1)There is a $s \in [0,1]$ s.t. $f(s) = 1/4;$
2)There is a $r \in [1,10]$ $(r \not =1) $
s.t. $f(r)=1/4,$
and we are done.
Substitute $x=\tan(y), -\pi/2<y<\pi/2$, get $f(x)=g(y)=\tan(y)/(1/\cos^2(y))=1/2(\sin(2y))$. Then $\sqrt2/4=g(\pi/8)=g(3\pi/8)$. So $f(\tan(\pi/8))=f(\tan(3\pi/8))$.
