## Thomas' Calculus 13th Edition

The domain is $(-\infty, -5)$ and $(-5,-3]$ and $[3,5)$ and $(5, \infty)$
Since one cannot divide by 0, the graph is not defined for $4- \sqrt{x^2-9} = 0$ $\sqrt{x^2-9} = 4$ ${\sqrt{x^2-9}}^2 = 4^2$ $x^2-9 = 16$ $x^2 = 25$ $x = \sqrt{25}$ and $x = -\sqrt{25}$ $x = 5$ and $x = -5$ Moreover, the solution of the root needs to be a real number, so the graph is not defined for $x^2 - 9 < 0$ $x^2 < 9$ $-3 < x < 3$ So the domain is $(-\infty, -5)$ and $(-5,-3]$ and $[3,5)$ and $(5, \infty)$