Show that the function defined by <img alt="f\left ( x \right )= \frac{x}{x^{2}+1},\fora

Show that the function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $f\left ( x \right )= \frac{x}{x^{2}+1},\forall \, x\ \epsilon\, \mathbb{R}$ is
neither one-one nor onto. Also, if $g:\because \mathbb{R}\rightarrow \mathbb{R}$ is defined as $g\left ( x \right )= 2x-1$ , find $fog\left ( x \right )\cdot$

Answers (1)

given $f\left ( x \right )= \frac{x}{x^{2}+1} , \forall x\, \epsilon \, R$
For one-one $f\left ( x \right )=f\left ( y \right )$
$\frac{x}{x^{2}+1}= \frac{y}{y^{2}+1}$
$\Rightarrow xy^{2}+x= yx^{2}+y$
$\Rightarrow xy^{2}- yx^{2}= y-x$
$\Rightarrow xy\left ( y-x \right )= y-x$
$\Rightarrow xy= 1,$ $x= \frac{1}{y}$
Since $x \neq y,$ therefore $f\left ( x \right )$ is not one-one for onto $f\left ( x \right )= y$
$\Rightarrow \frac{x}{x^{2}+1}= y$
$\Rightarrow x= yx^{2}+y$ $\Rightarrow xy^{2}+y-x= 0$
$\Rightarrow x$ cannot be experemed in lines of y
$\Rightarrow \therefore f\left ( x \right )$ is not outs.
As $g\left ( x \right )= 2x-1$
$\therefore fog\left ( x \right )=f\left ( g\left ( x \right ) \right )= f\left ( 2x-1 \right )$
$= \frac{2x-1}{\left ( 2x-1 \right )^{2}+1}= \frac{2x-1}{4x^{2}-4x+2}$