### Video Transcript

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

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

In this question, nine ๐ฅ plus 72 over ๐ฅ plus one divided by nine ๐ฅ plus 72 over five ๐ฅ plus five is equal to nine ๐ฅ plus 72 over ๐ฅ plus one multiplied by five ๐ฅ plus five over nine ๐ฅ plus 72. We can cross cancel by dividing the numerator and denominator by nine ๐ฅ plus 72. Factoring five ๐ฅ plus five on the numerator gives us five multiplied by ๐ฅ plus one, and this is divided by ๐ฅ plus one. We can then cancel the common factor of ๐ฅ plus one, leaving us with five. The function ๐ of ๐ฅ 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 of ๐ฅ minus ๐ of ๐ of ๐ฅ minus ๐ of ๐ of ๐ฅ, where ๐ denotes the zeros of the function. For example, ๐ of ๐ of ๐ฅ is the values of ๐ฅ for which ๐ of ๐ฅ 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 ๐ฅ plus one, nine ๐ฅ plus 72, and five ๐ฅ plus five all equal to zero. When ๐ฅ plus one equals zero, ๐ฅ is equal to negative one. This means that ๐ฅ equals negative one cannot be in the domain of ๐ of ๐ฅ. To solve nine ๐ฅ plus 72 equals zero, we begin by dividing through by nine, giving us ๐ฅ plus eight equals zero. And then subtracting eight from both sides, we have ๐ฅ is equal to negative eight. Negative eight will therefore not be included in the domain of our function. Finally, when five ๐ฅ plus five equals zero, we can divide through by five such that ๐ฅ plus one equals zero and therefore ๐ฅ 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 ๐ฅ is defined for all real values of ๐ฅ apart from negative one and negative eight. And we have now answered both parts of the question.