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Question Video: Determining the Domain of a Quotient of Two Rational Functions Mathematics • 10th Grade

Determine the domain of the function 𝑛(π‘₯) = ((3π‘₯ βˆ’ 15)/(π‘₯ βˆ’ 6)) Γ· ((6π‘₯ βˆ’ 30)/(4π‘₯ βˆ’ 24)).

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Video Transcript

Determine the domain of the function 𝑛 of π‘₯ equals three π‘₯ minus 15 over π‘₯ minus six divided by six π‘₯ minus 30 over four π‘₯ minus 24.

Let’s begin by inspecting our function 𝑛 of π‘₯. We notice that it’s a rational function divided by a second rational function. Let’s define this pair of rational functions to be 𝑓 of π‘₯ and 𝑔 of π‘₯, respectively, where we remember that a rational function is the quotient of a pair of polynomials. Then the domain of a function is said to be the set of possible inputs to that function.

So, how do we identify the domain of the quotient of two functions? The domain of the quotient of two functions, here 𝑓 divided by 𝑔, is all values common to the domain of both 𝑓 and 𝑔, but we must exclude those that make the denominator equal to zero. So, what values are in the domain of 𝑓 and 𝑔, respectively?

Let’s begin by inspecting our first function 𝑓 of π‘₯. We said this is a rational function; it’s the quotient of a pair of polynomials. And the domain of a rational function is the set of real numbers excluding any values of π‘₯ that make the denominator equal to zero. So, to find the domain of 𝑓 of π‘₯, we need to find those values of π‘₯ that make the denominator equal to zero so we can exclude them. So, we set it equal to zero, π‘₯ minus six equals zero, and then solve for π‘₯. To solve for π‘₯, we add six to both sides. And so, the domain of 𝑓 of π‘₯ is the set of real numbers minus the set containing six.

Next, let’s find the domain of 𝑔 of π‘₯. Once again, it’s the set of real numbers, but we need to exclude any values of π‘₯ that make the denominator zero. So, four π‘₯ minus 24 is equal to zero. And once again, if we solve for π‘₯, we add 24 to both sides and divide by four and we find π‘₯ is equal to six. So, the domain of 𝑔 of π‘₯ is the set of real numbers minus the set containing six.

We’re not quite finished, though. All values in the domain of 𝑓 and 𝑔 are the set of real numbers minus the set containing six. However, we need to exclude any values that make 𝑔 of π‘₯ equal to zero. Since 𝑔 of π‘₯ is a rational function, it’s the quotient of two polynomials, the only way for it to be equal to zero over the domain of our function is if the numerator is equal to zero. So, six π‘₯ minus 30 equals zero. This time, if we solve for π‘₯, we add 30 to both sides and then divide through by six. And that gives us π‘₯ equals five.

So, we have all values in the domain of 𝑓 and 𝑔. That’s the set of real numbers minus the set containing six. And then there is a single value of π‘₯ that makes 𝑔 of π‘₯ zero. It’s five. And so we combine all these pieces of information using set notation. And the domain of the function 𝑛 of π‘₯ is the set of real numbers minus the set containing elements five and six.

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