The range of the function f : R → R given by

Given: $f(x)=x+\sqrt{x^{2}}$

$f(x)=x+\sqrt{x^{2}}$

$=x+|x|$

$= \begin{cases}x+x & , x \geq 0 \\ x-x & , x<0\end{cases}$

$= \begin{cases}2 x & , x \geq 0 \\ 0 & , x<0\end{cases}$

To find the range, we find the real values of y obtained.

$y=2 x$ when $x \geq 0$

$\Rightarrow x=\frac{y}{2} \geq 0$

$\Rightarrow y \geq 0$

$\Rightarrow y \in[0, \infty)$

$y=0$ when $x<0 \quad \ldots(1)$

Thus, from (1) and (2),

$y \in[0, \infty)$

Hence, the range of the function $f: R \rightarrow R$ given by $f(x)=x+\sqrt{x^{2}}$ is $[0, \infty)$.