Question Video: Determining the Domain of a Rational Function | Nagwa Question Video: Determining the Domain of a Rational Function | Nagwa

Question Video: Determining the Domain of a Rational Function Mathematics

Find the domain of the function 𝑓(π‘₯) = (π‘₯ + 4)/((π‘₯ βˆ’ 8)Β²).

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Video Transcript

Find the domain of the function 𝑓 of π‘₯ equals π‘₯ plus four over π‘₯ minus eight squared.

What is the domain of our function? It’s the set of values of π‘₯ on which 𝑓 of π‘₯ is defined. We could ask ourselves, is two in the domain? That’s the same as asking, is 𝑓 of two defined? Well, let’s find out. We use our definition of 𝑓 of π‘₯. Replacing π‘₯ by two, we get that 𝑓 of two is equal to two plus four over two minus eight squared. In the numerator, two plus four is six. And in the denominator, two minus eight is negative six. And that squared is 36. So we get six over 36 which is a sixth. So clearly, 𝑓 of two is defined. And so two must be in the domain.

Let’s try another number. Is eight in the domain? That’s the same as asking if 𝑓 of eight is defined. Substituting eight in, we get that 𝑓 of eight is eight plus four over eight minus eight squared. In the numerator, eight plus four is 12. And in the denominator, eight minus eight is zero. And zero squared is also zero. And so we get 12 over zero. 12 over zero is not defined. You can’t divide a number by zero. So 𝑓 of eight is not defined. And eight is therefore, not in the domain of our function.

Okay, so we can’t go through this process for every real number. We’re going to have to be slightly cleverer than that. There was no problem with two. Two was in the domain. What was the problem was eight? Well, the problem was that we ended up with a denominator of zero. And we can’t divide by zero. It turns out that for a rational function, that is a function which is written in the form of a fraction where both the numerator and the denominator are polynomials, this is the only thing that can go wrong. The domain is all real numbers except those values which make the denominator zero. The denominator of our function is π‘₯ minus eight squared. So our domain is the set of real numbers minus the set of values of π‘₯ which make π‘₯ minus eight squared equal to zero. Which values of π‘₯ make π‘₯ minus eight squared equal to zero? Well, only π‘₯ equals eight.

Our domain is therefore, all real numbers apart from eight, which is written in set notation like this. For any value of π‘₯ apart from eight, which makes the denominator zero, the rational function 𝑓 of π‘₯ is defined.

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