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Hint: We need to find the domain and range of the given function f(x) so the domain refers to the set of possible values of x for which the function f(x) will be defined and the range refers to the possible range of values that the function f(x) can attain for those values of x which are in the domain of f(x).

Complete step-by-step answer:

Given real function is

$f\left( x \right) = \sqrt {x - 1} $

As we know if the value of the square root is negative then the function is not defined.

So, the value of the square root is always greater than or equal to zero.

$ \Rightarrow \sqrt {x - 1} \geqslant 0$………….. (1)

Now squaring on both sides we have,

$\left( {x - 1} \right) \geqslant 0$

$ \Rightarrow x \geqslant 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 = $\left[ {1,{\text{ }}\infty } \right)$ (domain is greater than or equal to 1 so at 1 there is a closed interval and at infinity there is an open interval).

Now, from equation (1)

$\sqrt {x - 1} \geqslant 0$

But, $f\left( x \right) = \sqrt {x - 1} $

$ \Rightarrow f\left( x \right) \geqslant 0$

Therefore the domain of f is the set of all real numbers greater than or equal to 0.

I.e. the domain of f = $\left[ {0,{\text{ }}\infty } \right)$ (domain is greater than or equal to 0 so at 0 there is a closed interval and at infinity there is an open interval).

So, this is the required answer.

Note: Whenever we face such type of problems the key concept is to be clear about the dentitions of domain and range as in this question since we know that $\sqrt x $ will only be defined when x is greater than or equal to 0, for negative values square root is not denied extending this concept we can easily find the range of the given function. This understanding of basic definitions will help you get the right answer.

Complete step-by-step answer:

Given real function is

$f\left( x \right) = \sqrt {x - 1} $

As we know if the value of the square root is negative then the function is not defined.

So, the value of the square root is always greater than or equal to zero.

$ \Rightarrow \sqrt {x - 1} \geqslant 0$………….. (1)

Now squaring on both sides we have,

$\left( {x - 1} \right) \geqslant 0$

$ \Rightarrow x \geqslant 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 = $\left[ {1,{\text{ }}\infty } \right)$ (domain is greater than or equal to 1 so at 1 there is a closed interval and at infinity there is an open interval).

Now, from equation (1)

$\sqrt {x - 1} \geqslant 0$

But, $f\left( x \right) = \sqrt {x - 1} $

$ \Rightarrow f\left( x \right) \geqslant 0$

Therefore the domain of f is the set of all real numbers greater than or equal to 0.

I.e. the domain of f = $\left[ {0,{\text{ }}\infty } \right)$ (domain is greater than or equal to 0 so at 0 there is a closed interval and at infinity there is an open interval).

So, this is the required answer.

Note: Whenever we face such type of problems the key concept is to be clear about the dentitions of domain and range as in this question since we know that $\sqrt x $ will only be defined when x is greater than or equal to 0, for negative values square root is not denied extending this concept we can easily find the range of the given function. This understanding of basic definitions will help you get the right answer.

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