Question

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

Answer 1

The given real function is $f\left(x\right)=\sqrt{x-1}$
It can be seen that $\sqrt{x-1}$ is defined for $(x-1)\ge 0$
i.e., $f\left(x\right)=\sqrt{(x-1)}$ is defined for $x\ge 1$
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 $x\ge 1$

$\Rightarrow (x-1)\ge 0$

$\Rightarrow \sqrt{x-1}\ge 0$

$f\left(x\right)\ge 0$
Therefore the range of $f$ is the set of all real numbers greater than or equal to 0
i.e., the range of f $=[0,\infty$)