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(A) The given real function is $f(x) = \sqrt{x-1}$.For the function to be defined in the set of real numbers,the expression inside the square root mus

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Domain: $[1, \infty)$,Range: $[0, \infty)$

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(A) The given real function is $f(x) = \sqrt{x-1}$.For the function to be defined in the set of real numbers,the expression inside the square root must be non-negative:$x - 1 \geq 0 \Rightarrow x \geq 1$.Therefore,the domain of $f$ is the set of all real numbers greater than or equal to $1$,i.e.,$[1, \infty)$.Since $\sqrt{x-1} \geq 0$ for all $x \geq 1$,the output values of the function are always non-negative.As $x$ takes all values in $[1, \infty)$,$\sqrt{x-1}$ takes all values in $[0, \infty)$.Thus,the range of $f$ is $[0, \infty)$.

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

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