g(x) is in the denominator, so g(x) = x-5 cannot be equal to zero. If it were, then
g(x) = 0
x-5 = 0
x = 5
Meaning that x = 5 leads to g(x) = 0
g(x) = x-5
g(5) = 5-5
g(5) = 0
This is a problem since we cannot divide by zero. We don't run into this issue for any other x value. So we must exclude the value x = 5 from the domain of (f/g)(x). Any other x value is allowed.
Domain: The set of all real numbers x such that x cannot equal 5
In set builder notation, that is written as [tex]\{x|x \in \mathbb{R}, \ x \ne 5\}[/tex]
In interval notation, that is written as [tex](-\infty, 5) \cup (5, \infty)[/tex] which tells us to glue together the two intervals from negative infinity to 5 (exclude both endpoints) and from 5 to infinity (again exclude both endpoints). This is effectively the same as taking the interval (-infinity, infinity) and poking a hole at 5.
