### Video Transcript

Given that π sub one of π₯ equals π₯ plus nine over π₯ minus six, π sub two of π₯ equals nine π₯ plus 81 over π₯ minus six, and π of π₯ equals π sub one of π₯ divided by π sub two of π₯, identify the domain of π of π₯.

Remember, the domain of a function is the set of possible inputs to that function. And we see that π of π₯ is the quotient of two functions. Itβs π one of π₯ divided by π two of π₯. We can therefore write it as π₯ plus nine over π₯ minus six divided by nine π₯ plus 81 divided by π₯ minus six. Now, before we can identify the domain of our function, letβs look to manipulate each expression somewhat.

Letβs begin by adding the terms in our first function. To do so, we think about π₯ as being equivalent to π₯ over one. Then, we want to create a common denominator. That common denominator is going to be the product of the two given denominators. So itβs just π₯ minus six. To achieve the numerator to our new expression, we multiply π₯ by π₯ minus six and then we add nine or nine times one. And so we can rewrite π sub one of π₯ as π₯ squared minus six π₯ plus nine over π₯ minus six. We might then even factor the numerator to get π₯ minus three squared.

Letβs repeat this process to add nine π₯ to 81 over π₯ minus six. Once again, we treat nine π₯ as nine π₯ over one. And the common denominator is one times π₯ minus six, which is still π₯ minus six. The numerator is then nine π₯ times π₯ minus six plus 81 times one. And if we distribute our parentheses, the numerator becomes nine π₯ squared minus 54π₯ plus 81. We then notice, though, that each term on our numerator is divisible by nine. So we can rewrite it as nine times π₯ squared minus six π₯ plus nine, which, a little bit like π sub one of π₯, can then in turn be written as nine times π₯ minus three squared over π₯ minus six.

So replacing π sub one of π₯ and π sub two of π₯ in our equation for π of π₯ and we see that itβs equal to π₯ minus three squared over π₯ minus six divided by nine times π₯ minus three squared over π₯ minus six.

Letβs clear some space and recall what we know about dividing by a fraction. Dividing by a fraction is the same as multiplying by the reciprocal of that fraction. So we can rewrite π of π₯ as π₯ minus three squared over π₯ minus six times π₯ minus six over nine times π₯ minus three squared. And if we were looking to simplify, we now see that there are a number of common factors that we can cancel through. However, weβre looking to find the domain of π of π₯. And we should always do that before simplifying.

So instead, weβre going to multiply our fractions by multiplying the numerators and then multiplying the denominators. And we get π₯ minus three squared times π₯ minus six over nine times π₯ minus six times π₯ minus three squared. Now, we see that we have a rational function. Remember, a rational function is the quotient of a pair of polynomials. And if we were to distribute the parentheses on the numerator and denominator of our fraction, we would find they are both polynomials.

So what do we know about the domain of a polynomial? The domain of a polynomial function is the set of real numbers. But we donβt want to be dividing by zero. So we exclude any values of π₯ that make the denominator equal to zero. So the domain of π of π₯ will be the set of real numbers minus any values of π₯ that satisfy the equation nine times π₯ minus six times π₯ minus three squared equals zero.

So letβs solve for π₯ to find the values of π₯ that we have to exclude from the domain. We notice that nine is independent of π₯. So for the expression nine times π₯ minus six times π₯ minus three squared to be equal to zero, either π₯ minus six must be equal to zero or π₯ minus three squared must be equal to zero. Adding six to both sides of our first equation, and we get π₯ equals six. Taking the square root and then adding three to both sides of this equation, and we get π₯ equals three. This means that we must exclude the values of three and six from the domain of the function π of π₯. And this in turn means that we can write the domain of π of π₯ as shown. Itβs the set of real numbers minus the set containing the elements three and six.

And of course, once weβve identified the domain, should we wish, we can simplify the fraction. The question doesnβt ask us to, but letβs remind ourselves how we do it. We divide both the numerator and denominator by π₯ minus six. Since weβve excluded π₯ equals six from the domain of our function, weβre not calculating zero divided by zero, which is undefined. So this step is legal. Then we divide through by π₯ minus three all squared. And we see π of π₯ actually simplifies to one-nineth. Notice that one-nineth is a constant; itβs independent of π₯. And so the domain of a function just defined as one-nineth would be the set of real numbers. This is why itβs important to calculate the domain before simplifying any fractions.