Question Video: Determining the Common Domain of Three Rational Functions | Nagwa Question Video: Determining the Common Domain of Three Rational Functions | Nagwa

# Question Video: Determining the Common Domain of Three Rational Functions Mathematics • Third Year of Preparatory School

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Find the common domain between the functions πβ(π₯) = β9/(π₯ + 9), πβ(π₯) = 8/(π₯ + 3), and πβ(π₯) = 7π₯/(π₯Β³ β 4π₯).

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### 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 letβs 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₯, weβll 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.

Letβs 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, weβll 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 itβs 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 weβre 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 weβre given is the set of real numbers minus the set containing negative nine, negative three, negative two, zero, and two.

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