Question Video: Simplifying the Quotient of Two Rational Functions and Identifying the Domain of the Resultant Function | Nagwa Question Video: Simplifying the Quotient of Two Rational Functions and Identifying the Domain of the Resultant Function | Nagwa

Question Video: Simplifying the Quotient of Two Rational Functions and Identifying the Domain of the Resultant Function Mathematics • Third Year of Preparatory School

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Simplify the function ๐‘›(๐‘ฅ) = (9๐‘ฅ + 72)/(๐‘ฅ + 1) รท (9๐‘ฅ + 72)/(5๐‘ฅ + 5), and determine its domain.

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Video Transcript

Simplify the function ๐‘› of ๐‘ฅ is equal to nine ๐‘ฅ plus 72 over ๐‘ฅ plus one divided by nine ๐‘ฅ plus 72 over five ๐‘ฅ plus five, and determine its domain.

We begin by recalling how to divide two rational functions. If ๐‘” of ๐‘ฅ is equal to ๐‘ of ๐‘ฅ over ๐‘ž of ๐‘ฅ and โ„Ž of ๐‘ฅ is equal to ๐‘Ÿ of ๐‘ฅ over ๐‘  of ๐‘ฅ, then their quotient ๐‘“ of ๐‘ฅ is equal to ๐‘ of ๐‘ฅ multiplied by ๐‘  of ๐‘ฅ divided by ๐‘ž of ๐‘ฅ multiplied by ๐‘Ÿ of ๐‘ฅ. When dividing two fractions, we multiply the first fraction by the reciprocal of the second fraction.

In this question, nine ๐‘ฅ plus 72 over ๐‘ฅ plus one divided by nine ๐‘ฅ plus 72 over five ๐‘ฅ plus five is equal to nine ๐‘ฅ plus 72 over ๐‘ฅ plus one multiplied by five ๐‘ฅ plus five over nine ๐‘ฅ plus 72. We can cross cancel by dividing the numerator and denominator by nine ๐‘ฅ plus 72. Factoring five ๐‘ฅ plus five on the numerator gives us five multiplied by ๐‘ฅ plus one, and this is divided by ๐‘ฅ plus one. We can then cancel the common factor of ๐‘ฅ plus one, leaving us with five. The function ๐‘› of ๐‘ฅ in its simplest form is equal to five.

We are also asked to determine the domain of the function. Using the definition of the two rational functions above, then the domain of their quotient is equal to the set of real numbers minus ๐‘ of ๐‘ž of ๐‘ฅ minus ๐‘ of ๐‘Ÿ of ๐‘ฅ minus ๐‘ of ๐‘  of ๐‘ฅ, where ๐‘ denotes the zeros of the function. For example, ๐‘ of ๐‘ž of ๐‘ฅ is the values of ๐‘ฅ for which ๐‘ž of ๐‘ฅ equals zero. We need to subtract these values from the set of real numbers as the function will not be defined at these values.

In this question, this means we need to set ๐‘ฅ plus one, nine ๐‘ฅ plus 72, and five ๐‘ฅ plus five all equal to zero. When ๐‘ฅ plus one equals zero, ๐‘ฅ is equal to negative one. This means that ๐‘ฅ equals negative one cannot be in the domain of ๐‘› of ๐‘ฅ. To solve nine ๐‘ฅ plus 72 equals zero, we begin by dividing through by nine, giving us ๐‘ฅ plus eight equals zero. And then subtracting eight from both sides, we have ๐‘ฅ is equal to negative eight. Negative eight will therefore not be included in the domain of our function. Finally, when five ๐‘ฅ plus five equals zero, we can divide through by five such that ๐‘ฅ plus one equals zero and therefore ๐‘ฅ equals negative one. This is the same solution to our first equation.

And we can therefore conclude that the domain of our function is the set of real values minus the set containing negative one and negative eight. The function ๐‘› of ๐‘ฅ is defined for all real values of ๐‘ฅ apart from negative one and negative eight. And we have now answered both parts of the question.

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