Show that the function $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ defined by $f(x) = \frac{x}{1+|x|}$ for all $x \in R$ is a one-one and onto function.

(A) Given $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ defined by $f(x) = \frac{x}{1+|x|}$.
For one-one:
Suppose $f(x) = f(y)$ for $x, y \in R$.
If $x$ and $y$ have opposite signs,say $x > 0$ and $y < 0$,then $f(x) = \frac{x}{1+x} > 0$ and $f(y) = \frac{y}{1-y} < 0$. Thus $f(x) \neq f(y)$.
If $x, y \geq 0$,then $\frac{x}{1+x} = \frac{y}{1+y} \Rightarrow x + xy = y + xy \Rightarrow x = y$.
If $x, y < 0$,then $\frac{x}{1-x} = \frac{y}{1-y} \Rightarrow x - xy = y - xy \Rightarrow x = y$.
Thus,$f$ is one-one.
For onto:
Let $y \in (-1, 1)$. We want to find $x \in R$ such that $f(x) = y$.
If $y \geq 0$,let $x = \frac{y}{1-y}$. Since $0 \leq y < 1$,$x \geq 0$. Then $f(x) = \frac{\frac{y}{1-y}}{1 + \frac{y}{1-y}} = \frac{y}{1-y+y} = y$.
If $y < 0$,let $x = \frac{y}{1+y}$. Since $-1 < y < 0$,$x < 0$. Then $f(x) = \frac{\frac{y}{1+y}}{1 - \frac{y}{1+y}} = \frac{y}{1+y-y} = y$.
Since for every $y \in (-1, 1)$ there exists an $x \in R$ such that $f(x) = y$,$f$ is onto.
Therefore,$f$ is one-one and onto.