Question

Find the range of $f\left(x\right)=\sqrt{x^2-9}$

Answer 1

$f\left(x\right)=\sqrt{x^2-9}$

$x^2-9\ge 0$

$\Rightarrow$ $x\le -3$ or $x\ge 3$

$Domain\left(y\right)=(-\infty ,-3]\cup [3,\infty )$

$\Rightarrow$ $Range\left(f\right)=[0,\infty )$