Question:
Find the domain and the range of the real function $t$ defined by $f(x)=\sqrt{(x-1)}$
Solution:
The given real function is $f(x)=\sqrt{x-1}$.
It can be seen that $\sqrt{x-1}$ is defined for $(x-1) \geq 0$.
i.e., $f(x)=\sqrt{(x-1)}$ is defined for $x \geq 1$.
Therefore, the domain of $f$ is the set of all real numbers greater than or equal to 1 i.e., the domain of $f=[1, \infty)$.
As $x \geq 1 \Rightarrow(x-1) \geq 0 \Rightarrow \sqrt{x-1} \geq 0$
Therefore, the range of $f$ is the set of all real numbers greater than or equal to 0 i.e., the range of $f=[0, \infty)$.
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