f(x)= square root of log₂(x² +1).find the domain and range of functions.

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)={\log }_2\left({\mathrm{x}}^2+1\right) \\ \mathrm{For} \mathrm{f}\left(\mathrm{x}\right) \mathrm{to} \mathrm{be} \mathrm{defined}, \mathrm{argument} \mathrm{of} \log \mathrm{should} \mathrm{be} \mathrm{strictly} \mathrm{positive}. \\ \Rightarrow {\mathrm{x}}^2+1>0 \\ \mathrm{which} \mathrm{is} \mathrm{true} \mathrm{for} \mathrm{x}\in \mathrm{R}, {\mathrm{x}}^2 \mathrm{is} \mathrm{never} \mathrm{negative}. \\ \mathrm{Hence} \mathrm{domain} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} \left(-\infty ,\infty \right) \mathrm{or} \mathrm{x}\in \mathrm{R}. \\ \mathrm{Now} \mathrm{minimum} \mathrm{value} \mathrm{of} {\mathrm{x}}^2+1 \mathrm{is} 1 \left(\mathrm{when} \mathrm{x}=0\right) \mathrm{and} {\log }_2\left(1\right)=0 \mathrm{and} \mathrm{as} \mathrm{value} \\ \mathrm{of} {\mathrm{x}}^2 \mathrm{increases} {\log }_2\left({\mathrm{x}}^2+1\right) \mathrm{will} \mathrm{also} \mathrm{increase}. \\ \mathrm{Hence} \mathrm{range} \mathrm{of} \mathrm{f}\left(\mathrm{x}\right) \mathrm{is} [0,\infty ) \end{gathered}$$

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