Question Video: Simplifying and Determining the Domain of a Product of Two Rational Functions Mathematics

Simplify the function 𝑛(π‘₯) = ((π‘₯Β² + 16π‘₯ + 64)/(π‘₯Β² + 8π‘₯)) Γ— ((7π‘₯ βˆ’ 56)/(64 βˆ’ π‘₯Β²)), and determine its domain.

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Video Transcript

Simplify the function 𝑛 of π‘₯ equals π‘₯ squared plus 16π‘₯ plus 64 over π‘₯ squared plus eight π‘₯ times seven π‘₯ minus 56 over 64 minus π‘₯ squared, and determine its domain.

Let’s begin by inspecting the function 𝑛 of π‘₯ in a little more detail. It’s the product of a pair of rational functions, a rational function of course being the quotient of two polynomials. Before we simplify it, it always makes sense to first find the domain. So what do we mean by the domain of a function? The domain is the set of possible inputs to that function, the set of π‘₯-values. And when we’re dealing with the combination of a pair of functions, in this case a product, the domain of the product is the intersection of the domains of the respective functions. So let’s begin by identifying the domains of each rational function. Let’s define the first function π‘₯ squared plus 16π‘₯ plus 64 over π‘₯ squared plus eight π‘₯ to be 𝑓 of π‘₯ and the second function to be 𝑔 of π‘₯.

We can find their domains by considering what we know about the domain of a rational function. It’s the quotient of a pair of polynomials. And so its domain is actually the set of real numbers, but we exclude any values of π‘₯ that make the denominator equal to zero. Essentially, we don’t want to be putting ourselves into a position where we’re dividing by zero. So let’s consider 𝑓 of π‘₯, the first rational function. We know its domain is the set of real numbers excluding values of π‘₯ that make the denominator π‘₯ squared plus eight π‘₯ equal to zero.

To find such values, somewhat counterintuitively we’re going to set it equal to zero and solve for π‘₯. And on the left-hand side, we notice we can factor by taking out a common factor of π‘₯. When we do, we get π‘₯ times π‘₯ plus eight equals zero. And for the product of this pair of expressions to be zero, either one or other of the expressions must itself be equal to zero. That is, π‘₯ equals zero or π‘₯ plus eight equals zero. And if we solve our second equation, we find π‘₯ is equal to negative eight. So these are the values of π‘₯ that we exclude from the domain of 𝑓 of π‘₯. So its domain is the set of real numbers minus the set containing the elements negative eight and zero.

Let’s now repeat this with the function 𝑔 of π‘₯. Its domain is going to be the set of real numbers not including any values of π‘₯ that make the denominator 64 minus π‘₯ squared equal to zero. So once again, let’s set this equal to zero and solve. We notice that 64 is a square number. It’s eight squared. So this expression in fact is the difference of two squares. When we factor then, we get eight minus π‘₯ times eight plus π‘₯. Then, we set each factor equal to zero and solve for π‘₯, so eight minus π‘₯ equals zero or eight plus π‘₯ equals zero. This time, the two values of π‘₯ that we need to exclude from the domain of the function are π‘₯ equals eight and π‘₯ equals negative eight. The domain is the set of real numbers minus the set containing negative eight and eight.

We need to find the intersection, the overlap between these two domains, to find the domain of 𝑛 of π‘₯. We know they share the set of real numbers. But we need to exclude all three values of π‘₯ that we identified make the denominator equal to zero. So it’s the set of real numbers minus the set containing negative eight, zero, and eight. Now that we have the domain of 𝑛 of π‘₯, let’s clear a little bit of space and simplify. We’ll keep on screen the factored expressions for our denominator. And this is because when we simplify a function, as requested of us in this question, we look to begin by factoring wherever possible.

Let’s now factor the numerator of 𝑓 of π‘₯. That’s π‘₯ squared plus 16π‘₯ plus 64. We want to find a pair of numbers whose product is 64 and whose sum is 16. Well, that’s eight and eight. So when we factor this expression, we get π‘₯ plus eight times π‘₯ plus eight. Similarly, the expression seven π‘₯ minus 56, the numerator of 𝑔 of π‘₯, factors to seven π‘₯ minus eight. Let’s now replace each of these factored expressions in our function 𝑛 of π‘₯. So 𝑛 of π‘₯ is π‘₯ plus eight times π‘₯ plus eight over π‘₯ times π‘₯ plus eight times seven times π‘₯ minus eight over eight minus π‘₯ times eight plus π‘₯.

In order to simplify, we start to cancel any common factors. So in this first fraction, we can divide through by π‘₯ plus eight. Remember, we’re allowed to do this because we said that π‘₯ equals negative eight is not in the domain of the function. This means we never end up with a situation where we’re dividing zero by zero, which is undefined. Similarly, we can also cross cancel another factor of π‘₯ plus eight. But what about π‘₯ minus eight and eight minus π‘₯? Let’s think about eight minus π‘₯ as being negative one times π‘₯ minus eight. Then, we can divide through by a factor of π‘₯ minus eight. And that leaves us with one over π‘₯ times seven over negative one, which is simply negative seven over π‘₯.

So 𝑛 of π‘₯ is negative seven over π‘₯. And its domain is the set of real numbers minus the set containing negative eight, zero, and eight.

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