Discussion Forum

The function $f:R\rightarrow [-\frac{1}{2},\frac{1}{2}]$ defined as  $f(x)=\frac{x}{1+x^{2}} is$

Answer:

Explanation:

We have $f(x)=\frac{x}{1+x^{2}}$

$\therefore$   $f(\frac{1}{x})=\frac{\frac{1}{x}}{1+\frac{1}{x^{2}}}=\frac{x}{1+x^{2}}=f(x)$

$\therefore$  $f(\frac{1}{2})=f(2)$ or $f(\frac{1}{3})=f(3)$   and so on 

  So, f(x) is many -one function 

   Again , let y=f(x)

$\Rightarrow$    $y=\frac{x}{1+x^{2}}$

$\Rightarrow$     $y+x^{2}y=x$

$\Rightarrow$  $yx^{2}-x+y=0$

As,    $x\epsilon  R

$\therefore$   $(-1)^{2}-4(y)(y)\geq0$

$\Rightarrow$   $1-4y^{2}\geq 0$

$\Rightarrow$   $y\epsilon [\frac{-1}{2},\frac{1}{2}]$

$\therefore$ Range= Codomain= $[\frac{-1}{2},\frac{1}{2}]$

    So, f(x) is surjective.

   Hence , f(x) is surjective but not injective.