Show that the function $f:R \to \{ x \in R: - 1 < x < 1\}$ defined by f(x) $= \cfrac{x}{{1 + |x|}},x \in R$is one-one and onto function.

We have : f (x)$= \cfrac{x}{{1 + |x|}} = \left\{ {\begin{array}{llllllllllllllllllll}{\cfrac{x}{{1 + x}},}&{if}&{x \ge 0}\\{\cfrac{x}{{1 - x}},}&{if}&{x < 0}\end{array}} \right.$

Here, Domain of f $=$ R

To prove : f is one$-$one
Let x, y $\in$ Domain of f$=$R, such that x $\ne$ y
Here, four cases arise.

Case I : When $x \ge 0,y \ge 0$

If $x \ne y \Rightarrow 1 + x \ne 1 + y$
$\Rightarrow$ $\cfrac{1}{{1 + x}} \ne \cfrac{1}{{1 + y}} \Rightarrow \cfrac{1}{{1 + x}} \ne \cfrac{{ - 1}}{{1 + y}}$

$\Rightarrow$ $1 - \cfrac{1}{{1 + x}} \ne 1 - \cfrac{1}{{1 + y}} \Rightarrow \cfrac{x}{{1 + x}} \ne \cfrac{y}{{1 + y}}$

$\Rightarrow$ $f(x) \ne f(y).$

Case II : When x $\ge$ 0 and y $<$ 0
Then, f(x) $=$ $\cfrac{x}{{1 + x}}$ $\ge$ 0 and $f(y) = \cfrac{y}{{1 - y}} < 0$

$\Rightarrow$ $f(x) \ne f(y)$.

Case III : When $x < 0$ and y $\ge$ 0

Then, $f(x) < 0$ and f (y)$\ge$ 0 [As in Case II]

$\Rightarrow$ $f(x) \ne f(y)$

Case IV : When x $\le$0 and y $\le$ 0

If $x \ne y \Rightarrow - x \ne - y$
$\Rightarrow$ $1 - x \ne 1 - y \Rightarrow \cfrac{1}{{1 - x}} \ne \cfrac{1}{{1 - y}}$

$\Rightarrow$ $\cfrac{1}{{1 - x}} - 1 \ne \cfrac{1}{{1 - y}} - 1 \Rightarrow \cfrac{x}{{1 - x}} \ne \cfrac{y}{{1 - y}}$
$\Rightarrow$ $f(x) \ne f(y).$

Thus, in each case, $x \ne y \Rightarrow f(x) \ne f(y)$.

Hence, f is one$-$one.

To prove : f is onto

Let y $\in$ R, where y is arbitrary.

Then,$y = f(x) = \cfrac{x}{{1 + |x|}} = \left\{ {\begin{array}{llllllllllllllllllll}{\cfrac{x}{{1 + x}} < 1,}&{if}&{x \ge 0}\\{\cfrac{x}{{1 - x}} > - 1,}&{if}&{x \le 0}\end{array}} \right.$

Case I When y$=$ $\cfrac{x}{{1 + x}},$ where y $\ge$ 0

y + xy $=$ x or y $=$ x (1$-$y) or x $=$ $\cfrac{y}{{1 - y}} \ge$0

Case II When y $=$ $\cfrac{x}{{1 + x}}$, where $y < 0$

y $-$xy $=$ x or y $=$ x + xy or x $=$ $\cfrac{y}{{1 + y}} < 0$

Thus, when y $\ge$ 0, there is $\cfrac{y}{{1 - y}} \in$ Domain of f $=$ R such that

$f\left( {\cfrac{y}{{1 - y}}} \right) = \cfrac{{\cfrac{y}{{1 - y}}}}{{1 + \cfrac{y}{{1 - y}}}} = \cfrac{y}{{1 - y + y}} = \cfrac{y}{1} = y$

and when $y < 0,$ there is $\cfrac{y}{{1 + y}} \in$ Domain of f$=$R such that
$f\left( {\cfrac{y}{{1 + y}}} \right) = \cfrac{{\cfrac{y}{{1 + y}}}}{{1 - \cfrac{y}{{1 + y}}}} = \cfrac{y}{{1 + y - y}}$

$= \cfrac{y}{1} = y$

Hence, f is onto.