## University Calculus: Early Transcendentals (3rd Edition)

$\mathrm{Domain:}\ \ (-\infty ,-5)\cup (-5,-3]\cup [3,5)\cup (5,\infty )$
$\mathrm{Remember:}\$ We can take square root only of the positive numbers, i.e $\ \ \sqrt{f(x)}\quad \Rightarrow \quad \:f(x)\ge 0$ For a rational function, to find the domain, take the denominator of it and compare it to zero. Exclude the points obtained from the domain as they will be undefined. Taking the first step, we have: $x^2-9\ge 0$ $\Rightarrow\ x\le \:-3\quad \mathrm{or}\quad \:x\ge \:3$ Now take the denominator and compare it to zero. $4-\sqrt{x^2-9}=0$ $\Rightarrow\ x^2-9=16$ $\Rightarrow\ x=5,\:\:x=-5$ $\mathrm{Domain:}\ \ (-\infty ,-5)\cup (-5,-3]\cup [3,5)\cup (5,\infty )$