Question

Find the domain of the real function, $f\left(x\right)=\frac{1}{\sqrt{x+|x|}}$

Answer 1

We have, $f\left(x\right)=\frac{1}{\sqrt{x+|x|}}$

Now, $|x|=\{\begin{array}{c}x,\text{when}x\ge 0\\ -x,\text{when}x<0\end{array}$

$\Rightarrow x+|x|=\{\begin{array}{c}x+x,\text{when}x\ge 0\\ x-x,\text{when}x<0\end{array}$

$\Rightarrow x+|x|=\{\begin{array}{c}2x,\text{when}x\ge 0\\ 0,\text{when}x<0\end{array}$

$\Rightarrow x+|x|>0,\text{when}x>0$

$\Rightarrow f\left(x\right)=\frac{1}{\sqrt{x+|x|}}$ assumes real values only when $x+|x|>0$ and this happens only when $x>0$

$\therefore \text{dom (f)}=(0,\infty )$.