Video: Determining the Domain of Rational Functions

Identify the domain of 𝑛(π‘₯) = (9π‘₯ + 8)/(3π‘₯ + 2).

02:06

Video Transcript

Identify the domain of 𝑛 of π‘₯ is equal to nine π‘₯ plus eight all divided by three π‘₯ plus two.

In this question, we’re given a function 𝑛 of π‘₯, which we can see is the quotient of two linear functions. In other words, this is a rational function. And we’re asked to identify the domain of this function, and there’s a few different ways of doing this. Let’s start by recalling exactly what we mean by the domain of a function.

We recall that when we say the domain of a function, what we mean is the set of inputs for that function. So in this question, when we’re asked to find the domain of our function 𝑛 of π‘₯, we want to find all of the possible inputs for this function. Now we could do this formally by looking for all of the values where our function 𝑛 of π‘₯ is not defined and the values of π‘₯ where it is defined. However, we’ve already done this in general for all rational functions, so we can actually make this a lot simpler.

We’ve actually shown that for a rational function to not be defined, the denominator has to be equal to zero. In other words, the domain of any rational function is all real values except the zeros of the denominator. Therefore, instead of finding the domain of our function 𝑛 of π‘₯ by using the definition of a domain, all we need to do is solve the denominator is equal to zero. Setting the denominator equal to zero, we get three π‘₯ plus two is equal to zero. And of course, we can solve this equation for π‘₯. We subtract two from both sides of the equation, giving us three π‘₯ minus two, and then we divide through by three. This gives us that π‘₯ is equal to negative two over three.

Therefore, our function 𝑛 of π‘₯ is not defined when π‘₯ is equal to negative two over three since we’re dividing by zero. And for all other input values of π‘₯, we’ll be taking the quotient of two real numbers where the denominator is not equal to zero. So our function is defined for these values of π‘₯. So we’ve shown the domain of our function 𝑛 of π‘₯ is all real numbers except for negative two over three. And of course, we can represent this in set notation as the set of all real numbers excluding negative two over three.

And it’s important we do this because, remember, the domain of a function is a set. Therefore, we were able to find the domain of the function 𝑛 of π‘₯ is equal to nine π‘₯ plus eight all divided by three π‘₯ plus two by finding the zeros of the denominator. We were able to show the domain of this function is the set of all real numbers excluding negative two over three.

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