The function $f:R\to R$ defined by $f\left(x\right)=\frac{x}{\sqrt{1+x^2}}$ is 
 

Injective let $x_1,x_2\in R$ such that
 $\begin{array}{l}F\left(x_1\right)=F\left(x_2\right)\\ \frac{x_1}{\sqrt{1+x_1^2}}=\frac{x_2}{\sqrt{1+x_2^2}}\\ \frac{x_1^2}{1+x_1^2}=\frac{x_2^2}{1+x_2^2}\\ x_1^2=x_2^2\Rightarrow x_1=x_2\end{array}$
$\therefore f\left(x\right)$ is injective surjective $y=\frac{x}{\sqrt{1+x^2}}$
 $\begin{array}{l}{y}^2(1+x^2)=x^2\Rightarrow y^2+y^2{x}^2=x^2\\ x^2(1-y^2)=y^2\Rightarrow x=\sqrt{\frac{y^2}{1-y2}}\end{array}$
$\begin{array}{l}\frac{y^2}{1-y^2}\ge 0\\ y\in (-\infty ,-1)\cup (0,1)\end{array}$
$\therefore$ Range of $f\left(x\right)=(-\infty ,-1)\cup (0,1)$
f(x) is not surjective.