Question Video: Determining the Common Domain of Three Rational Functions Mathematics

Find the common domain between the functions 𝑓₁(π‘₯) = βˆ’9/(π‘₯ + 9), 𝑓₂(π‘₯) = 8/(π‘₯ + 3), and 𝑓₃(π‘₯) = 7π‘₯/(π‘₯Β³ βˆ’ 4π‘₯).

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Video Transcript

Find the common domain between the functions 𝑓 sub one of π‘₯ equals negative nine over π‘₯ plus nine, 𝑓 sub two of π‘₯ equals eight over π‘₯ plus three, and 𝑓 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 𝑓 sub one of π‘₯, 𝑓 sub two of π‘₯, and 𝑓 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 let’s take function 𝑓 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‘₯, we’ll subtract nine from both sides, and we find the value of π‘₯ that satisfies this equation is negative nine. So the domain of 𝑓 sub one of π‘₯ is the set of real numbers minus the set containing negative nine.

Let’s now consider 𝑓 sub two of π‘₯. This time, we need to exclude values of π‘₯ that make π‘₯ plus three, the denominator of 𝑓 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, we’ll move on to 𝑓 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 it’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 𝑓 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 we’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 we’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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