Find the domain and the range of the real function f defined by $f\left(x\right)=\sqrt{\left(x-1\right)}$.

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$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}-1} \\ \mathrm{For} \mathrm{f}\left(\mathrm{x}\right) \mathrm{to} \mathrm{be} \mathrm{defined}, \mathrm{quantity} \mathrm{inside} \mathrm{root} \mathrm{must} \mathrm{be} \mathrm{positive} \\ \mathrm{x}-1\ge 0 \\ \Rightarrow \mathrm{x}\ge 1 \\ \mathrm{Hence} \mathrm{domain} \mathrm{is} \mathrm{x}\in [1,\infty ) \\ \mathrm{Range} \mathrm{of} \mathrm{square} \mathrm{root} \mathrm{is} \mathrm{non}-\mathrm{negative}, \mathrm{hence} \mathrm{range} \mathrm{is} [0,\infty ) \end{gathered}$$

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