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Given that π sub one of π₯ is equal to five π₯ minus eight over 25π₯ squared minus four divided by 25π₯ squared minus 30π₯ minus 16 over 125π₯ cubed plus eight, π sub two of π₯ is equal to 25π₯ squared minus four over 50π₯ squared minus 20π₯ plus eight, and π of π₯ is equal to π sub one of π₯ times π sub two of π₯, simplify the function π and determine its domain.
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Ultimately, weβre going to be looking to find the product of the two functions given to us in this question.
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Before we do, weβre going to need to begin by simplifying our function π sub one of π₯.
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π sub one of π₯ is the quotient of two rational functions.
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Thatβs two functions that are fractions.
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And so we recall that to divide by a fraction, we multiply by the reciprocal of that fraction.
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The reciprocal of our second fraction is one over that fraction.
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But of course, to find the reciprocal, we simply reverse the numerator and the denominator.
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And so we can rewrite π sub one of π₯ as five π₯ minus eight over 25π₯ squared minus four times 125π₯ cubed plus eight over 25π₯ squared minus 30π₯ minus 16.
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Now, whenever weβre dealing with rational functions, we should always check whether we can simplify by factoring.
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In fact, this function contains a few expressions that can be factored.
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Letβs begin with the expression 25π₯ squared minus four.
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We should see that both 25π₯ squared and four are square numbers.
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So weβre actually dealing with the difference of two squares, meaning that 25π₯ squared minus four can be written as five π₯ minus two times five π₯ plus two.
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And what about this expression 25π₯ squared minus 30π₯ minus 16?
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Well, this one could be factored as five π₯ minus eight times five π₯ plus two.
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In fact, whilst it might not look like it, there is one further factorable expression here.
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This is the sum of two cubes.
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And we can factor a sum of two cubes as shown.
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We say that π cubed plus π cubed is π plus π times π squared minus ππ plus π squared.
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We can see that if we compare our expression to the expression π cubed plus π cubed, weβll need to let π be equal to five π₯ and π be equal to two.
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And so this factors to be five π₯ plus two times 25π₯ squared minus 10π₯ plus four.
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Now, if youβre unsure where these numbers come from, 25π₯ squared is the value of π squared.
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So thatβs five π₯ squared.
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Negative 10π₯ is negative ππ.
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And π is five π₯, and π is two.
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And five π₯ times two is 10π₯.
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Then π squared is four.
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And now, we see we can rewrite π sub one of π₯ as five π₯ minus eight over five π₯ minus two times five π₯ plus two times five π₯ plus two times 25π₯ squared minus 10π₯ plus four over five π₯ minus eight times five π₯ plus two.
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Next, before we multiply, weβre going to cross cancel.
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We look for common factors in the numerator of the first fraction and the denominator of the second and the denominator of the first fraction and the numerator of the second.
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We see we have a factor of five π₯ plus eight.
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Then we could actually cancel by any five π₯ plus two on our denominator.
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Now, we multiply the numerators and separately multiply the denominators.
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And we have an expression for π sub one of π₯.
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Remember, we were told that π of π₯ is equal to π sub one of π₯ times π sub two of π₯.
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So letβs find the product of the two functions in our question.
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π of π₯ is the product of 25π₯ squared minus 10π₯ plus four over five π₯ minus two times five π₯ plus two and 25π₯ squared minus four over 50π₯ squared minus 20π₯ plus eight.
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Once again, we see if we can cross cancel.
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Well, we know that 25π₯ squared minus four is the same as five π₯ minus two over five π₯ plus two.
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And so we divide through by this common factor.
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Similarly, the denominator of our second fraction can be written as shown.
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We take out a constant factor of two.
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And this means we can then divide through by 25π₯ squared minus 10π₯ plus four.
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This leaves us with one times one on our numerator and one times two on our denominator.
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So π of π₯ is simply one-half.
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But weβre not quite finished.
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We were told to determine the domain of this function.
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The domain of a function is the set of possible inputs that will yield a real output.
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When we deal with the product or quotient of two functions, the domain is the intersection of the domains of the two functions.
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If our resultant function is also a rational function, then we need to ensure that we exclude any values of π₯ that make the function in the denominator equal to zero.
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In this case, though, our function is simply one-half, so thereβs no way of making the denominator equal to zero.
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So weβll consider the domains of our two original functions, π sub one of π₯ and π sub two of π₯.
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Letβs begin by looking at the domain of π sub one of π₯.
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There are a couple of things we need to consider when looking at the domain.
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Firstly, if we have any square roots, then we know the number inside that square root must be positive.
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We donβt actually have any square roots.
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But what we do have is fractional functions, where the denominator itself could be equal to zero.
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We donβt want to be dividing by zero, so weβre going to set the denominators equal to zero to see which values of π₯ we need to exclude in our domain.
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Weβll begin by looking at the denominator of our simplified expression.
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Thatβs five π₯ minus two times five π₯ plus two.
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We set this equal to zero and find the values of π₯ that we want to exclude.
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Now, for the expression five π₯ minus two times five π₯ plus two to be equal to zero, either one or other of these expressions must itself be equal to zero.
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So five π₯ minus two is equal to zero or five π₯ plus two is equal to zero.
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By adding two to both sides of this first equation and then dividing by five, we find π₯ is equal to two-fifths.
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Similarly, we also find π₯ is equal to negative two-fifths to be a solution to this equation.
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And so these are two of the values weβre going to exclude from our domain of π sub one of π₯.
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And similarly, since weβre excluding these from the domain of π sub one, we exclude them for the domain of π.
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But letβs go back to our unsimplified version of π sub one of π₯.
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If we look at that second fraction, we remember that we multiplied by the reciprocal of that fraction.
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When multiplying by its reciprocal, we had a denominator of 25π₯ squared minus 30π₯ minus 16.
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So weβre going to find the values of π₯ such that this expression is equal to zero.
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When we factored, we got five π₯ minus eight times five π₯ plus two.
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And so the two equations that we need to solve will be five π₯ minus eight equals zero and five π₯ plus two is equal to zero.
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But of course, this second equation is a duplicate.
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And so we solve the first equation by adding eight to both sides and dividing by five.
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And we find π₯ is equal to eight-fifths.
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So currently we have three values of π₯ that we cannot include in the domain of our function π sub one of π₯ and, therefore, in the domain of the function π.
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Of course, we said that the domain of π will be the intersection of our two domains.
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So we need to consider the domain of π sub two of π₯.
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In this case, we know it will be all real numbers except those values of π₯ such that the denominator is equal to zero.
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So we want to try and solve this equation.
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We could begin by dividing through by two to get 25π₯ squared minus 10π₯ plus four equals zero.
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The problem is this expression here, 25π₯ squared minus 10π₯ plus four, is not factorable.
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One way we can check for this is to look for the discriminant.
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The discriminant of an equation in the form ππ₯ squared plus ππ₯ plus π equals zero is π squared minus four ππ.
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So here thatβs negative 10 squared minus four times 25 times four.
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That gives us a value of negative 300.
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Now, negative 300 is less than zero.
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And this tells us there are no real solutions to our equation 25π₯ squared minus 10π₯ plus four equals zero.
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In other words, there are no values of π₯ that will make this expression equal to zero.
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And so the domain of our second function, π sub two of π₯, must be the set of real numbers.
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The intersection of our two domains and, therefore, the domain of the function π can be represented using set notation as shown.
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Itβs the set of all real numbers minus the set including the numbers negative two-fifths, two-fifths, and eight-fifths.