Determining the image of a function 
Determine the image of the following function without using calculus: 
  $$f\left(x\right)=\ \frac{x}{2\left|x\right|+1}$$

My attempt:
We know $\left| x \right| < 2\left| x \right| + 1$, then, $\frac{\left| x \right|}{2\left| x \right| + 1} < 1$ which implies $\left| f\left( x \right) \right| < 1$.
So $-1 < \frac{x}{2\left| x \right| + 1} < 1$.
However, the image is $-\frac{1}{2} < f\left( x \right) < \frac{1}{2}$.
What am I doing wrong? 

As discussed in the comment, we also need to show that $(-\frac{1}{2},\frac{1}{2}) \subseteq r(f)$. For each $y \in (-\frac{1}{2},\frac{1}{2})$, let
$$g(y) = \frac{y}{1-2|y|}$$
Check that $f(g(y)) = y$, so that $(-\frac{1}{2},\frac{1}{2}) \subseteq r(f)$. It follows from Walton's hint that $r(f) \subseteq (-\frac{1}{2},\frac{1}{2})$. So $r(f) = (-\frac{1}{2},\frac{1}{2})$.
As a remark, $g \circ f= \text{Id}_\Bbb{R}$ and $f \circ g = \text{Id}_{(-\frac{1}{2},\frac{1}{2}) }$ and so that $f$ is actually also injective and $g$ is the inverse of $f$.
