Question Video: Simplifying the Sum of Two Rational Functions and Determining Its Domain | Nagwa Question Video: Simplifying the Sum of Two Rational Functions and Determining Its Domain | Nagwa

Question Video: Simplifying the Sum of Two Rational Functions and Determining Its Domain Mathematics • Third Year of Preparatory School

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Simplify the function 𝑛(π‘₯) = ((8π‘₯ + 7)/(π‘₯Β² βˆ’ 14π‘₯ + 45)) + ((3π‘₯ βˆ’ 24)/(π‘₯Β² βˆ’ 17π‘₯ + 72)), and determine its domain.

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Video Transcript

Simplify the function 𝑛 of π‘₯ equals eight π‘₯ plus seven over π‘₯ squared minus 14π‘₯ plus 45 plus three π‘₯ minus 24 over π‘₯ squared minus 17π‘₯ plus 72 and determine its domain.

The function we’ve been given, 𝑛 of π‘₯, is the sum of two rational functions. They’re each the quotient of two polynomials. In order to simplify this function, we need to combine the two quotients by finding their sum.

The two quotients though have different denominators. And we know that in order to add two quotients, we need to use a common denominator. Before we find this common denominator though, let’s begin by factoring the quadratic expression in each denominator in case either fraction can be individually simplified.

The quadratic π‘₯ squared minus 14π‘₯ plus 45 in the first denominator can be factored as π‘₯ minus nine multiplied by π‘₯ minus five, because the numbers negative nine and negative five have a sum of negative 14 and a product of positive 45.

The quadratic in the second denominator, π‘₯ squared minus 17π‘₯ plus 72, can be factored as π‘₯ minus nine multiplied by π‘₯ minus eight. The numbers negative nine and negative eight have a sum of negative 17, which is the coefficient of π‘₯, and a product of 72, which is the constant term.

So writing each denominator in its factored form, we have that 𝑛 of π‘₯ is equal to eight π‘₯ plus seven over π‘₯ minus nine multiplied by π‘₯ minus five plus three π‘₯ minus 24 over π‘₯ minus nine multiplied by π‘₯ minus eight.

We should observe though that the numerator of the second fraction can also be factored. Both terms here have a common factor of three. So this expression can be written as three multiplied by π‘₯ minus eight.

Now, we see that we have a shared factor of π‘₯ minus eight in the numerator and denominator of the second quotient. Now assuming that π‘₯ minus eight is not equal to zero β€” and we’ll discuss that in more detail later β€” we can therefore divide both the numerator and denominator of this quotient by π‘₯ minus eight in order to simplify. So 𝑛 of π‘₯ simplifies to eight π‘₯ plus seven over π‘₯ minus nine multiplied by π‘₯ minus five plus three over π‘₯ minus nine. And now, we can consider how to find a common denominator for these two fractions.

We should notice that the denominator of the second fraction, which is now just π‘₯ minus nine, is actually a factor of the denominator of the first. Let’s think about how we would tackle a problem like this in the numeric case.

Suppose we were finding the sum of one-third and one-sixth. In order to add these two fractions, we would need a common denominator, which should be the lowest common multiple of three and six. But as three is a factor of six, the lowest common multiple of these two numbers is just six itself. In the same way, as π‘₯ minus nine is a factor of π‘₯ minus nine multiplied by π‘₯ minus five, the lowest common multiple of these two expressions is simply π‘₯ minus nine multiplied by π‘₯ minus five.

So in fact, we don’t need to change the first fraction at all. For the second fraction, we need to multiply the denominator by π‘₯ minus five. But of course, if we do that to the denominator, we must do the same to the numerator. So we have eight π‘₯ plus seven over π‘₯ minus nine multiplied by π‘₯ minus five plus three multiplied by π‘₯ minus five over π‘₯ minus nine multiplied by π‘₯ minus five. As the two fractions now have the same denominator, we can add them together. We have eight π‘₯ plus seven plus three multiplied by π‘₯ minus five all over π‘₯ minus nine multiplied by π‘₯ minus five.

In order to simplify this quotient, we need to first distribute the parentheses in the numerator, which gives eight π‘₯ plus seven plus three π‘₯ minus 15. And the denominator is unchanged.

Finally, we group the like terms in the numerator. Eight π‘₯ plus three π‘₯ is 11π‘₯, and positive seven minus 15 is negative eight. So we have 11π‘₯ minus eight over π‘₯ minus nine multiplied by π‘₯ minus five. This is the fully simplified form of a function 𝑛 of π‘₯, as the numerator can’t be factored and there are no common factors other than one in the numerator and denominator.

Now, we need to consider how to find the domain of this function. We recall that the domain of a function is the set of all values on which the function acts. Or we can think of it as the set of all input values to the function. In the case of a rational function, as we have here, its domain is the set of all real numbers minus any exclusions. A value of π‘₯ must be excluded from the domain of a rational function if the function is undefined for that value of π‘₯. In this case, our function 𝑛 of π‘₯ would be undefined if either of the denominators were equal to zero because, as we know, dividing by zero is undefined.

To find any values of π‘₯ that must be excluded then, we need to find any values of π‘₯ that will make either π‘₯ squared minus 14π‘₯ plus 45 equal to zero or π‘₯ squared minus 17π‘₯ plus 72 equal to zero.

Earlier in the question, we saw how each of these quadratic expressions can be factored. So for the first quadratic, we have π‘₯ minus nine multiplied by π‘₯ minus five equals zero. And for the second, π‘₯ minus nine multiplied by π‘₯ minus eight equals zero.

Now remember that if the product of two factors is equal to zero, at least one of those factors must themselves be equal to zero. So for the first quadratic, either π‘₯ minus nine equals zero, in which case π‘₯ equals nine, or π‘₯ minus five equals zero, in which case π‘₯ is equal to five. For the second quadratic, either π‘₯ minus nine equals zero, in which case π‘₯ equals nine, or π‘₯ minus eight equals zero, in which case π‘₯ equals eight.

Now, this value in particular makes a lot of sense because remember at this stage here, we said that in order to divide both the numerator and denominator of the second quotient by π‘₯ minus eight, π‘₯ minus eight must not be equal to zero. So that’s okay because we’ve just found that the value of π‘₯ which makes π‘₯ minus eight equal to zero, that is, eight, must be excluded from the domain of this function. So we know we weren’t attempting to divide by zero at this point.

The values which must be excluded from the domain then are five, eight, and we have the value of nine twice. So the domain is the set of all real numbers minus the set containing the values five, eight, and nine.

Our final answer to the problem then is that the simplified form of the function 𝑛 of π‘₯ is 11π‘₯ minus eight over π‘₯ minus nine multiplied by π‘₯ minus five. And the domain of this function is the set of all real numbers minus the set containing the values five, eight, and nine.

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