Question

The range of the function f : R → R given by $f\left(x\right)=x+\sqrt{x^2}$ is _________.

Answer 1

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

$$\begin{gathered} f\left(x\right)=x+\sqrt{x^2} \\ =x+\left|x\right| \\ =\left\{\begin{array}{l}x+x,x\ge 0\\ x-x,x<0\end{array}\right. \\ =\left\{\begin{array}{l}2x,x\ge 0\\ 0,x<0\end{array}\right. \end{gathered}$$

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

$$\begin{gathered} y=2x\mathrm{when}x\ge 0 \\ \Rightarrow x=\frac{y}{2}\ge 0 \\ \Rightarrow y\ge 0 \\ \Rightarrow y\in [0,\infty )...\left(1\right) \\ y=0\mathrm{when}x<0...\left(2\right) \\ \mathrm{Thus},\mathrm{from}\left(1\right)\mathrm{and}\left(2\right), \\ y\in [0,\infty ) \end{gathered}$$

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