## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

The domain is {$x|x\gt9$}.
Requirement (1) For a square root to be defined, the expression inside the radial sign must be greater than or equal to $0$. The domain is defined by applying the requirement: $x-9\geq0$ $x-9+9\ge 0+9$ $x\geq 9$ Requirement (2) For a fraction to be defined, the denominator can not be equal to $0$. Thus, $\sqrt{x-9}\ne0$ $(\sqrt{x-9})^2\ne 0^2$ $x-9\ne0$ $x-9+9\ne0+9$ $x\ne9$ The two requirements can be applied together, to define the domain: $x\geq9$ and $x\ne9$. Therefore, the domain is {$x|x\gt9$}.