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