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Simplify the function π of π₯ equals π₯ squared plus 16π₯ plus 64 over π₯ squared plus eight π₯ times seven π₯ minus 56 over 64 minus π₯ squared, and determine its domain.
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Letβs begin by inspecting the function π of π₯ in a little more detail.
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Itβs the product of a pair of rational functions, a rational function of course being the quotient of two polynomials.
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Before we simplify it, it always makes sense to first find the domain.
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So what do we mean by the domain of a function?
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The domain is the set of possible inputs to that function, the set of π₯-values.
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And when weβre dealing with the combination of a pair of functions, in this case a product, the domain of the product is the intersection of the domains of the respective functions.
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So letβs begin by identifying the domains of each rational function.
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Letβs define the first function π₯ squared plus 16π₯ plus 64 over π₯ squared plus eight π₯ to be π of π₯ and the second function to be π of π₯.
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We can find their domains by considering what we know about the domain of a rational function.
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Itβs the quotient of a pair of polynomials.
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And so its domain is actually the set of real numbers, but we exclude any values of π₯ that make the denominator equal to zero.
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Essentially, we donβt want to be putting ourselves into a position where weβre dividing by zero.
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So letβs consider π of π₯, the first rational function.
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We know its domain is the set of real numbers excluding values of π₯ that make the denominator π₯ squared plus eight π₯ equal to zero.
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To find such values, somewhat counterintuitively weβre going to set it equal to zero and solve for π₯.
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And on the left-hand side, we notice we can factor by taking out a common factor of π₯.
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When we do, we get π₯ times π₯ plus eight equals zero.
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And for the product of this pair of expressions to be zero, either one or other of the expressions must itself be equal to zero.
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That is, π₯ equals zero or π₯ plus eight equals zero.
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And if we solve our second equation, we find π₯ is equal to negative eight.
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So these are the values of π₯ that we exclude from the domain of π of π₯.
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So its domain is the set of real numbers minus the set containing the elements negative eight and zero.
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Letβs now repeat this with the function π of π₯.
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Its domain is going to be the set of real numbers not including any values of π₯ that make the denominator 64 minus π₯ squared equal to zero.
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So once again, letβs set this equal to zero and solve.
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We notice that 64 is a square number.
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Itβs eight squared.
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So this expression in fact is the difference of two squares.
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When we factor then, we get eight minus π₯ times eight plus π₯.
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Then, we set each factor equal to zero and solve for π₯, so eight minus π₯ equals zero or eight plus π₯ equals zero.
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This time, the two values of π₯ that we need to exclude from the domain of the function are π₯ equals eight and π₯ equals negative eight.
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The domain is the set of real numbers minus the set containing negative eight and eight.
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We need to find the intersection, the overlap between these two domains, to find the domain of π of π₯.
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We know they share the set of real numbers.
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But we need to exclude all three values of π₯ that we identified make the denominator equal to zero.
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So itβs the set of real numbers minus the set containing negative eight, zero, and eight.
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Now that we have the domain of π of π₯, letβs clear a little bit of space and simplify.
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Weβll keep on screen the factored expressions for our denominator.
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And this is because when we simplify a function, as requested of us in this question, we look to begin by factoring wherever possible.
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Letβs now factor the numerator of π of π₯.
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Thatβs π₯ squared plus 16π₯ plus 64.
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We want to find a pair of numbers whose product is 64 and whose sum is 16.
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Well, thatβs eight and eight.
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So when we factor this expression, we get π₯ plus eight times π₯ plus eight.
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Similarly, the expression seven π₯ minus 56, the numerator of π of π₯, factors to seven π₯ minus eight.
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Letβs now replace each of these factored expressions in our function π of π₯.
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So π of π₯ is π₯ plus eight times π₯ plus eight over π₯ times π₯ plus eight times seven times π₯ minus eight over eight minus π₯ times eight plus π₯.
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In order to simplify, we start to cancel any common factors.
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So in this first fraction, we can divide through by π₯ plus eight.
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Remember, weβre allowed to do this because we said that π₯ equals negative eight is not in the domain of the function.
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This means we never end up with a situation where weβre dividing zero by zero, which is undefined.
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Similarly, we can also cross cancel another factor of π₯ plus eight.
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But what about π₯ minus eight and eight minus π₯?
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Letβs think about eight minus π₯ as being negative one times π₯ minus eight.
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Then, we can divide through by a factor of π₯ minus eight.
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And that leaves us with one over π₯ times seven over negative one, which is simply negative seven over π₯.
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So π of π₯ is negative seven over π₯.
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And its domain is the set of real numbers minus the set containing negative eight, zero, and eight.