Range of a function, with contradictory restriction I have this function:
$$f(x)=\frac{x}{x^2+1}$$
it's clear that the domain is all real numbers, because the denominator never become zero. I've done the following procedure to found the range of $f(x)$:
$\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \operatorname{Dom} f,y=f(x)}  \right \} \\
\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},y=\dfrac{x}{x^2+1}}  \right \} \\
\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\exists x\in \mathbb{R},x=\dfrac{1\pm \sqrt{1-4y}}{2y}}  \right \} \\
\operatorname{Rec} f=\left \{{y\in \mathbb{R}:\left (1-4y^2  \right )\geq 0\, \wedge\, y\neq 0 }  \right \}\\
\operatorname{Rec} f=[-\tfrac{1}{2},\tfrac{1}{2}]-\left \{ 0 \right \}$
But, the function is zero when $x=0$ $(f(0)=0)$, and the range of the function is $[-\tfrac{1}{2}, \tfrac{1}{2}]$. So I don't know where is my error in the procedure.

$$f(x)=\frac{x}{x^2+1}$$
Since the function is an odd function $(f(x)=-f(-x))$,  then the minimum value is equal to the opposite of the maximum value. We see that $f(x) \ge 0$ when $x \ge 0$, so we need to find the maximum value of $f(x)$ when $x \ge 0$.
Note that 
$\dfrac 12 - f(x) = \dfrac 12 - \dfrac{x}{x^2+1} 
                  = \dfrac{x^2+1-2x}{2(x^2+1)}
                  = \dfrac{(x-1)^2}{2(x^2+1)}$
So, when $x \ge 0$, we see that $\dfrac 12 - f(x) \ge 0$ and, when $x=1$, we see that $\dfrac 12 - f(x) = 0$.
Since $\dfrac 12 - f(x) \ge 0 \iff f(x) \le \dfrac 12$ and $f(1)= \dfrac 12$, we see that $\displaystyle \max_{x \ge 0} f(x) = \dfrac 12$. Hence, because $f(x)$ is an odd function, $\operatorname{Range} f(x) = \left[ -\dfrac 12, \dfrac 12 \right]$
