## College Algebra 7th Edition

$\{x|x\geq 3\}$
We know that $\displaystyle \frac{\sqrt{x-3}}{x^{2}-4x+4}$ is undefined whenever the denominator is 0, so: $x^2-4x+4=(x-2)^2=0$ $x=2$ We also know that the equation is undefined for negative roots: $x-3<0$ $x<3$ So $x$ must be greater than or equal to 3 (which also ensures that it's not 2). Thus the domain is: $\{x|x\geq 3\}$