Question

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

Answer 1

$f\left(x\right)=\frac{x}{x^2+1}$ for all $x>0$

$f\left(x\right)\in (0,1)$ for $x=0$

$f\left(x\right)=0$ for $x<0$

$f\left(x\right)\in (-1,0)$

Range is $(-1,1)$ so not onto since domain is $(R,-R)$

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

We get $f^{\prime }\left(x\right)=\frac{1-x^2}{{(x^2+1)}^2}$

$f^{\prime }\left(x\right)=0$ at $x=1$

So, slope changes after $x=1$

So, there exists $x_1$ and $x_2$ for which $f\left(x_1\right)=f\left(x_2\right)$

Hence $f\left(x\right)$ is not one-one.