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Show that the function $f : (-\infty , 0) \rightarrow (-1, 0)$ defined by $f(x) \frac{x}{1+\left | x \right |} , x \in (-\infty , 0)$ is one-one and onto.

Answers (1)

Given:

$f(x) \frac{x}{1+\left | x \right |} , x \in (-\infty , 0)$

$f : (-\infty , 0) \rightarrow (-1, 0)$

To prove: f(x) is one-one and onto.

For one-one

$\\ $ Let $ x_{1}, x_{2} \in(-\infty, 0)$ and

$f\left(x_{1}\right)=f\left(x_{2}\right)$

$\frac{{x}_{1}}{1+\left|{x}_{1}\right|} =\frac{{x}_{2}}{1+\left|x_{2}\right|}$

$\\ \frac{x_{1}}{1-{x}_{1}} =\frac{{x}_{2}}{1-{x}_{2}}$

$\\ {x}_{1}-{x}_{1} {x}_{2}={x}_{2}-{x}_{1}{x}_{2}$

$\\ {x}_{1} ={x}_{2}$

$\Rightarrow $f(x) is one-one$

For onto

$y \in(-1,0)$

$y=\frac{x}{1+|x|}$$

$y=\frac{x}{1-x}$

$y-yx =x$

$y=(1+y)x$

$x=\frac{y}{1+y}$

For each $y \in(-1,0)$ there exist $x \in(-\infty,0)$

$f(x)=f(\frac{y}{1+y})$

$\\f(x)=\frac{\frac{y}{1+y}}{1+\left|\frac{y}{1+y}\right|}$

$\\\\ f(x)=\frac{\frac{y}{1+y}}{1-\frac{y}{1+y}}$

$\\\\ f(x)=y$

$\Rightarrow $f(x) is onto$

Hence proved

Posted by

Safeer PP

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