Question

Show that the function $f:R\to R$ defined by $f\left(x\right)=\frac{x}{x^2+1}$ $\forall x\in R$ is neither one-one nor onto.

Answer 1

put $x-a,x=b$

let $f\left(a\right)=f\left(b\right)$

$\frac{a}{a^2+1}=\frac{b}{b^2+1}$

$a{b}^2+a=b{a}^2+b$

$ab(b-a)=b-a$

$ab=0$

so $f\left(x\right)$ is not one one function

now $f\left(x\right)=\frac{x}{x^2+1}=y$

x value cannot be found that is $f^{-1}\left(x\right)$ does not exist so $f\left(x\right)$ is not onto function