Let f(x)= 1/√(x+|x|), then find the domain and range of f.

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{\mathrm{x}+\left|\mathrm{x}\right|}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \left|\mathrm{x}\right|=\left\{\begin{array}{ll}\mathrm{x}& \mathrm{x}<0\\ -\mathrm{x}& \mathrm{x}\ge 0\end{array}\right. \\ \mathrm{When} \mathrm{x}\le 0 \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{\mathrm{x}-\mathrm{x}}}=\frac{1}{\sqrt{0}} \mathrm{which} \mathrm{is} \mathrm{not} \mathrm{defined}. \\ \mathrm{Hence} \mathrm{x}\le 0 \mathrm{will} \mathrm{not} \mathrm{be} \mathrm{part} \mathrm{of} \mathrm{domain} \\ \mathrm{When} \mathrm{x}>0 \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{\mathrm{x}+\mathrm{x}}}=\frac{1}{\sqrt{2\mathrm{x}}} \\ \mathrm{Hence} \mathrm{x}>0, \mathrm{hence} 2\mathrm{x}>0 \mathrm{which} \mathrm{means} \mathrm{root} \mathrm{will} \mathrm{be} \mathrm{defined}. \\ \mathbf{\Rightarrow }\mathbf{Domain}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{f}\left(\mathbf{x}\right)\mathbf{ }\mathbf{is}\mathbf{ }\left(\mathbf{0}\mathbf{,}\mathbf{\infty }\right) \\ \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\sqrt{2\mathrm{x}}} \\ \mathrm{As} \mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{i}.\mathrm{e}. \frac{1}{\sqrt{2\mathrm{x}}} \mathrm{is} \mathrm{always} \mathrm{greater} \mathrm{than} 0 \\ \mathrm{Hence} \mathrm{L}.\mathrm{H}.\mathrm{S}. \mathrm{will} \mathrm{also} \mathrm{be} \mathrm{greater} \mathrm{than} 0. \\ \mathbf{\Rightarrow }\mathbf{Range}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{f}\left(\mathbf{x}\right)\mathbf{ }\mathbf{is}\mathbf{ }\left(\mathbf{0}\mathbf{,}\mathbf{\infty }\right) \end{gathered}$$

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