### Video Transcript

Find the common domain between the
functions π sub one of π₯ equals negative nine over π₯ plus nine, π sub two of π₯
equals eight over π₯ plus three, and π sub three of π₯ equals seven π₯ over π₯
cubed minus four π₯.

Remember, given any number of
functions, the common domain is the intersection of the domains of the respective
functions. In this case then, we need to find
the domain of π sub one of π₯, π sub two of π₯, and π sub three of π₯. Then, we can find their
intersection. So, next, we remind ourselves how
we find the domain of a rational function. The domain of a rational function
is the set of real numbers, but we exclude any values of π₯ that make the
denominator equal to zero. So letβs take function π sub one
of π₯. Its domain is going to be the set
of real numbers, but we need to exclude values of π₯ that make π₯ plus nine equal to
zero. To solve for π₯, weβll subtract
nine from both sides, and we find the value of π₯ that satisfies this equation is
negative nine. So the domain of π sub one of π₯
is the set of real numbers minus the set containing negative nine.

Letβs now consider π sub two of
π₯. This time, we need to exclude
values of π₯ that make π₯ plus three, the denominator of π sub two of π₯, equal to
zero. The value of π₯ that satisfies this
equation is π₯ equals negative three. And so the domain of this function
is the set of real numbers minus the set containing negative three. Finally, weβll move on to π sub
three of π₯. The denominator here is π₯ cubed
minus four π₯. So we know we need to exclude any
values of π₯ that make this equal to zero. So we have the equation π₯ cubed
minus four π₯ equals zero. And how do we solve it?

Well, we might first look to factor
the expression on the left-hand side. The first step to this is to take
out a common factor of π₯. Then, we can further factor the
expression π₯ squared minus four using the difference of two squares. So π₯ cubed minus four π₯ can be
written as π₯ times π₯ plus two times π₯ minus two.

The first solution to this equation
is when π₯ is equal to zero, so itβs π₯ equals zero. The second solution is found by
setting π₯ plus two equal to zero. And when we solve that equation, we
get π₯ equals negative two. Finally, we solve the equation π₯
minus two equals zero, and we get π₯ equals two. So, finally, we have found that the
domain of π sub three of π₯ is the set of real numbers minus the set containing
these values of π₯. The common domain then is the
intersection of these three domains. So weβre going to have to take the
set of real numbers and exclude the following values of π₯: negative nine, negative
three, negative two, zero, and two. Hence, the common domain between
the three functions weβre given is the set of real numbers minus the set containing
negative nine, negative three, negative two, zero, and two.