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For which values of π₯ is the function π of π₯ equals π₯ squared minus 25 over π₯ squared minus 12π₯ plus 32 not defined?
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Letβs begin by inspecting the function π of π₯.
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π of π₯ is the quotient of a pair of polynomials.
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That is, it is a polynomial function divided by a second polynomial function.
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In order to identify the values of π₯ for which the function is not defined, weβll begin by considering the domain of a rational function.
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The domain of a rational function, of course, is the set of values of π₯ for which the function is defined.
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So, if we consider the set of values of π₯ for which the function is defined, weβll be able to quickly identify the values for which it is not defined.
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The domain of a rational function is the set of real numbers, but we must exclude any values of π₯ that make the denominator of that function equal to zero.
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This means that our function will be defined over the set of real numbers excluding the set of numbers that make the expression on the denominator, π₯ squared minus 12π₯ plus 32, equal to zero.
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To find such values of π₯, we will set the denominator equal to zero and solve, that is, π₯ squared minus 12π₯ plus 32 equals zero.
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Since we have a quadratic, we can attempt to solve by first factoring this quadratic expression.
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We know that we must have an π₯ at the beginning of each expression because π₯ times π₯ gives us the π₯ squared.
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Then, we need to find a pair of numbers whose product is 32 and whose sum is negative 12.
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Well, negative four times negative eight is positive 32 as required.
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But negative four plus negative eight is indeed negative 12.
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So we rewrite our equation as shown.
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π₯ minus four times π₯ minus eight is equal to zero.
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Then, of course, for the product of these two expressions to be equal to zero, we know that either one or other of those expressions must itself be zero.
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So the solutions to our equation are given by the solutions to the equations π₯ minus four equals zero and π₯ minus eight equals zero.
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We solve our first equation by adding four to both sides, so we get π₯ is equal to four.
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And we solve our second equation by adding eight to both sides, so π₯ is equal to eight.
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Remember, if weβre thinking about the domain of π of π₯, we know itβs the set of real numbers minus the set containing the numbers that make the denominator equal to zero.
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So the domain of our function is the set of real numbers minus the set containing four and eight.
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Of course, this means our function is defined over this set.
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And therefore, it must be undefined when π₯ is equal to four or π₯ is equal to eight.
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And so we see that the function π of π₯ is undefined for the set containing four and eight.