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Question

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


Solution

$$f(x) = \dfrac {x}{{1 - x}} - 1 < x < 0\,and\,f\left( x \right) = \dfrac {x}{{1 + x}},0 \le x < 1$$
$$now(i)f(x){\rm{ = }}\dfrac {x}{{1 - x}}{\rm{ - 1 < }}x < 0$$
Let $$f({x_1}) = f({x_2})$$
$$ \Rightarrow \dfrac {{{x_1}}}{{1 - {x_1}}}{\rm{ = }}\dfrac {{{x_2}}}{{1 - {x_2}}}$$
$$ \Rightarrow {x_1} - {x_1}{x_2}{\rm{ = }}x_2 - {x_1}{x_2}$$
$$ \Rightarrow {x_1}={x_2}$$
$$ \Rightarrow $$f is one-one
Let  $$y{\rm{ = }}\dfrac {x}{{1 - x}}$$
$$\Rightarrow y - xy = x$$
$$ \Rightarrow {\rm{ }}y = x + xy$$
$$ \Rightarrow y = x(1 + y)$$
$$ \Rightarrow {\rm{ }}x = \dfrac {y}{{1 + y}}$$
$$ \Rightarrow {\rm{ }}\exists {\rm{   }}x = \dfrac {y}{{1 + y}}$$ for all valves of y$$ \ne {\rm{ }} - 1$$
s.t $$f(x) = y$$
$$\Rightarrow {\rm{ }}f(x)$$ is onto
(ii)$$f(x) = \dfrac {x}{{1 + x}}{\rm{0}} \le x \le 1$$
Let $$f({x_1}) = {\rm{ }}f({x_2})$$
$$ \Rightarrow \dfrac {{{x_1}}}{{1 + {x_1}}}{\rm{ }} = {\rm{ }}\dfrac {{{x_2}}}{{1 + {x_2}}}$$
$$ \Rightarrow {x_1} + {x_1}{x_2}{\rm{ }} = {\rm{ }}{x_2} + {x_1}{x_2}$$
$$ \Rightarrow {x_1} = {x_2}$$
$$ \Rightarrow $$ f is one -one
Let $$y = \dfrac {x}{{1 + x}}$$
$$ \Rightarrow y + xy = x$$
$$ \Rightarrow y = x - xy$$
$$ \Rightarrow y = x(1 - y)$$
$$ \Rightarrow x = \dfrac {y}{{1 - y}}$$
$$ \Rightarrow $$ for all values of $$y \ne 1{\rm{ }}\exists {\rm{ }}x = \dfrac {y}{{1 - y}}$$ s.t $$f(x) = y$$
$$ \Rightarrow $$ f is onto
hence f(x) is one-one and onto proved

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