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 order to simplify the function 𝑓 of π‘₯, we’ll need to factorize the numerator and denominator. In order to do this, we need to recognize that on the numerator, we have the difference of two squares. This tells us that any quadratic of the form π‘₯ squared minus π‘Ž squared is equal to π‘₯ plus π‘Ž multiplied by π‘₯ minus π‘Ž. We know that nine squared is equal to 81. This means that π‘₯ squared minus 81 is equal to π‘₯ plus nine multiplied by π‘₯ minus nine. The denominator π‘₯ cubed plus 729 is written in the form π‘₯ cubed plus π‘Ž cubed.

Once again, π‘Ž is equal to nine as nine cubed is 729. π‘₯ cubed plus π‘Ž cubed is equal to π‘₯ plus π‘Ž multiplied by π‘₯ squared minus π‘Žπ‘₯ plus π‘Ž squared. This means that π‘₯ cubed plus 729 is equal to π‘₯ plus nine multiplied by π‘₯ squared minus nine π‘₯ plus 81. We can divide the numerator and denominator by π‘₯ plus nine, so these terms cancel. The simplified version of 𝑓 of π‘₯ is therefore equal to π‘₯ minus nine over π‘₯ squared minus nine π‘₯ plus 81.

The second part of our question asks us to find the domain. The domain of a function is the set of π‘₯-values we can substitute or input into 𝑓 of π‘₯. As the denominator of a fraction cannot be equal to zero, the domain will be all real values apart from those which make the denominator zero. To calculate these values, we set π‘₯ cubed plus 729 equal to zero. We already know this is equal to π‘₯ plus nine multiplied by π‘₯ squared minus nine π‘₯ plus 81. Let’s consider the linear term π‘₯ plus nine first. When π‘₯ plus nine is equal to zero, π‘₯ is equal to negative nine. Therefore, negative nine cannot be in the domain of 𝑓 of π‘₯. The quadratic π‘₯ squared minus nine π‘₯ plus 81 cannot be factored or factorized.

We can actually go one stage further here as it actually has no real solutions. We know that any quadratic of the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐 has no real solutions when its discriminant, 𝑏 squared minus four π‘Žπ‘, is less than zero. In this case, we have π‘Ž is equal to one, the coefficient of π‘₯ squared, 𝑏 is equal to negative nine, the coefficient of π‘₯, and 𝑐 is 81. We need to calculate negative nine squared minus four multiplied by one multiplied by 81. This is equal to negative 243. As this is less than naught, the quadratic π‘₯ squared minus nine π‘₯ plus 81 equals zero has no solutions.

This means that the only value such that π‘₯ cubed plus 729 equals zero is π‘₯ is equal to negative nine. The domain of 𝑓 of π‘₯ is therefore equal to all real values apart from negative nine. We can write this in set notation as shown.

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