Question Video: Finding the Value of an Unknown In a Rational Function Given Its Domain Mathematics

Given that the domain of the function 𝑛(π‘₯) = (36/π‘₯) + (20/(π‘₯ + π‘Ž)) is ℝ βˆ’ {βˆ’2, 0}, evaluate 𝑛(3).

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Video Transcript

Given that the domain of the function 𝑛 of π‘₯ equals 36 over π‘₯ plus 20 over π‘₯ plus π‘Ž is the set of real numbers minus the set containing negative two, zero, evaluate 𝑛 of three.

𝑛 of π‘₯ is the sum of a pair of rational functions. Both 36 over π‘₯ and 20 over π‘₯ plus π‘Ž are the quotient of a pair of polynomials. We also know that we can find the domain of the sum of a pair of functions by considering the intersection of those domains. So let’s begin by looking at the domains of 36 over π‘₯ and 20 over π‘₯ plus π‘Ž to the domain that we’ve been given, the set of real numbers minus the set containing negative two, zero. This will allow us to find the value of π‘Ž, which will in turn allow us to evaluate 𝑛 of three.

Let’s begin by looking at the expression 36 over π‘₯. Remember, 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. In this case, the denominator is simply π‘₯. So we set π‘₯ equal to zero, and we see that π‘₯ equals zero is a value that we’re going to exclude from the domain of this function. The domain 𝑛 of 36 over π‘₯ is the set of real numbers minus the set containing zero. We’ll now consider the second rational function. We have 20 over π‘₯ plus π‘Ž. This time, the domain is going to be the set of real numbers excluding any values of π‘₯ that make the denominator π‘₯ plus π‘Ž equal to zero.

So let’s set π‘₯ plus π‘Ž equal to zero and solve for π‘₯. If we do, we find π‘₯ is equal to negative π‘Ž. So we can say that the domain of this second function is the set of real numbers minus the set containing negative π‘Ž. The domain of 𝑛 of π‘₯ then is the intersection of these two domains. The intersection here is the set of real numbers minus the set containing negative π‘Ž, zero. Now, if we compare this to the domain we’ve been given, we can see that negative π‘Ž must be equal to negative two. And if negative π‘Ž is equal to negative two, π‘Ž itself must be equal to two. So we can rewrite 𝑛 of π‘₯ using the value we have for π‘Ž. It’s 36 over π‘₯ plus 20 over π‘₯ plus two.

We’re now ready to evaluate 𝑛 of three, and we can do so by substituting three into this equation. When we do, we get 36 over three plus 20 over three plus two. And that becomes 12 plus four, which is of course equal to 16. So, given information about the domain of our function, 𝑛 of three must be equal to 16.

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