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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₯.
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Remember, the domain of a function is the set of possible inputs to that function.
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And we see that π of π₯ is the quotient of two functions.
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Itβs π one of π₯ divided by π two of π₯.
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We can therefore write it as π₯ plus nine over π₯ minus six divided by nine π₯ plus 81 divided by π₯ minus six.
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Now, before we can identify the domain of our function, letβs look to manipulate each expression somewhat.
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Letβs begin by adding the terms in our first function.
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To do so, we think about π₯ as being equivalent to π₯ over one.
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Then, we want to create a common denominator.
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That common denominator is going to be the product of the two given denominators.
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So itβs just π₯ minus six.
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To achieve the numerator to our new expression, we multiply π₯ by π₯ minus six and then we add nine or nine times one.
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And so we can rewrite π sub one of π₯ as π₯ squared minus six π₯ plus nine over π₯ minus six.
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We might then even factor the numerator to get π₯ minus three squared.
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Letβs repeat this process to add nine π₯ to 81 over π₯ minus six.
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Once again, we treat nine π₯ as nine π₯ over one.
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And the common denominator is one times π₯ minus six, which is still π₯ minus six.
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The numerator is then nine π₯ times π₯ minus six plus 81 times one.
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And if we distribute our parentheses, the numerator becomes nine π₯ squared minus 54π₯ plus 81.
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We then notice, though, that each term on our numerator is divisible by nine.
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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.
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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.
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Letβs clear some space and recall what we know about dividing by a fraction.
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Dividing by a fraction is the same as multiplying by the reciprocal of that fraction.
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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.
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And if we were looking to simplify, we now see that there are a number of common factors that we can cancel through.
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However, weβre looking to find the domain of π of π₯.
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And we should always do that before simplifying.
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So instead, weβre going to multiply our fractions by multiplying the numerators and then multiplying the denominators.
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And we get π₯ minus three squared times π₯ minus six over nine times π₯ minus six times π₯ minus three squared.
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Now, we see that we have a rational function.
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Remember, a rational function is the quotient of a pair of polynomials.
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And if we were to distribute the parentheses on the numerator and denominator of our fraction, we would find they are both polynomials.
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So what do we know about the domain of a polynomial?
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The domain of a polynomial function is the set of real numbers.
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But we donβt want to be dividing by zero.
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So we exclude any values of π₯ that make the denominator equal to zero.
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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.
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So letβs solve for π₯ to find the values of π₯ that we have to exclude from the domain.
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We notice that nine is independent of π₯.
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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.
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Adding six to both sides of our first equation, and we get π₯ equals six.
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Taking the square root and then adding three to both sides of this equation, and we get π₯ equals three.
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This means that we must exclude the values of three and six from the domain of the function π of π₯.
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And this in turn means that we can write the domain of π of π₯ as shown.
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Itβs the set of real numbers minus the set containing the elements three and six.
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And of course, once weβve identified the domain, should we wish, we can simplify the fraction.
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The question doesnβt ask us to, but letβs remind ourselves how we do it.
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We divide both the numerator and denominator by π₯ minus six.
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Since weβve excluded π₯ equals six from the domain of our function, weβre not calculating zero divided by zero, which is undefined.
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So this step is legal.
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Then we divide through by π₯ minus three all squared.
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And we see π of π₯ actually simplifies to one-nineth.
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Notice that one-nineth is a constant; itβs independent of π₯.
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And so the domain of a function just defined as one-nineth would be the set of real numbers.
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This is why itβs important to calculate the domain before simplifying any fractions.