Question Video: Simplifying and Determining the Domain of Rational Functions | Nagwa Question Video: Simplifying and Determining the Domain of Rational Functions | Nagwa

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

Simplify the function 𝑓(π‘₯) = (π‘₯Β² βˆ’ 81)/(π‘₯Β³ + 729) and find its domain.

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Video Transcript

Simplify the function 𝑓 of π‘₯ is equal to π‘₯ squared minus 81 over π‘₯ cubed plus 729 and find its domain.

In this question, we will begin by finding the domain of the function and then simplify it by factoring the numerator and denominator and canceling any shared factors. We recall that the domain of a rational function is all real values of π‘₯ except where the denominator is equal to zero. We therefore need to set the cubic polynomial π‘₯ cubed plus 729 equal to zero. In order to solve this equation, we will begin by factoring the left-hand side. 729 is a cube number. It is equal to nine cubed. This means that π‘₯ cubed plus 729 is written in the form π‘₯ cubed plus π‘Ž cubed. The sum of cubes formula states that this is equal to π‘₯ plus π‘Ž multiplied by π‘₯ squared minus π‘Žπ‘₯ plus π‘Ž squared. π‘₯ cubed plus 729 is therefore equal to π‘₯ plus nine multiplied by π‘₯ squared minus nine π‘₯ plus 81.

At this stage, we have the product of a linear and quadratic term equal to zero. Setting the linear factor equal to zero gives us π‘₯ is equal to negative nine. The quadratic equation π‘₯ squared minus nine π‘₯ plus 81 equals zero has no real solutions. This is because the discriminant 𝑏 squared minus four π‘Žπ‘ is less than zero. We therefore have only one real solution to the cubic equation π‘₯ cubed plus 729 equals zero. It is π‘₯ is equal to negative nine. As this is the only value of π‘₯ that makes the function undefined, we can conclude that the domain of 𝑓 of π‘₯ is the set of all real values minus the set containing negative nine.

Let’s now move on to simplifying the function. The numerator is written in the form π‘₯ squared minus π‘Ž squared and can therefore be factored using the difference of two squares. This is equal to π‘₯ plus π‘Ž multiplied by π‘₯ minus π‘Ž. Since the square root of 81 is nine, 𝑓 of π‘₯ is equal to π‘₯ plus nine multiplied by π‘₯ minus nine over π‘₯ plus nine multiplied by π‘₯ squared minus nine π‘₯ plus 81. Since π‘₯ cannot equal negative nine, we can cancel the shared factor of π‘₯ plus nine from the numerator and denominator. 𝑓 of π‘₯ simplifies to π‘₯ minus nine over π‘₯ squared minus nine π‘₯ plus 81. And we now have answers to the two parts of the question.

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