The domain of the rational function f(x) = (x-1)/x2-9 is determined by looking at what will cause the denominator to go to zero and excluding the number(s) that do. You can't divide by zero.
If you factor the denominator
x2-9 = (x + 3)(x-3) you see that it will go to zero when x = +3, -3, so these need to be excluded from the domain. Therefore the domain is the set of x's divided into three parts:
<-------------- -3 ----------------- +3 ---------------->
(-infinity, -3) (-3, 3) (3, +infinity)
Note that in interval notation parentheses ( ) means up to but not including the endpoints while brackets [ ] means including the endpoints.
For the domain of f(x) = sqrt (4-5x) you need to look at any x's that will make what is under the square root negative and exclude those because the square root of a negative number is imaginary.
Set 4 - 5x < = 0 and solve for x, giving you x < = 4/5 , which in interval notation would be (-infinity, 4/5]
