Question 400822
If your rational function is {{{sqrt((x-3)/(x^2-8x+12)) = sqrt((x-3)/((x-2)(x-6)))}}}, then the domain is obtained as follows:
From the expression {{{(x-3)/(x^2-8x+12) = (x-3)/((x-2)(x-6))}}}, the critical numbers are 2, 3, and 6.  These critical numbers partition the real number line into the intervals ({{{-infinity}}}, 2), (2, 3), (3, 6), (6, {{{infinity}}}).
At ({{{-infinity}}}, 2), {{{(x-3)/((x-2)(x-6)) < 0}}} by using the test point x = 0.
At (2, 3), {{{(x-3)/((x-2)(x-6)) > 0}}} by using the test point x = 2.5.
At (3, 6), {{{(x-3)/((x-2)(x-6)) < 0}}} by using the test point x = 4.
At (6, {{{infinity}}}), {{{(x-3)/((x-2)(x-6)) > 0}}} by using the test point x = 7.
The critical numbers 2 and 6 are not included in the domain, as these will make the denominator equal to 0.  We want those intervals that will make {{{(x-3)/((x-2)(x-6)) > 0}}}, because the whole expression is under the square root symbol. The critical number x = 3 is included in the domain.

Hence the domain of the rational function is (2,3]U(6, {{{infinity}}}).