Question Video: Simplifying a Rational Function and Identifying Its Domain | Nagwa Question Video: Simplifying a Rational Function and Identifying Its Domain | Nagwa

Question Video: Simplifying a Rational Function and Identifying Its Domain Mathematics • Third Year of Preparatory School

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Simplify the function 𝑛(π‘₯) = (2/(π‘₯ + 2)) Γ— ((π‘₯Β² + 6π‘₯ + 8)/2π‘₯), and determine its domain.

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Video Transcript

Simplify the function 𝑛 of π‘₯ equals two over π‘₯ plus two times π‘₯ squared plus six π‘₯ plus eight over two π‘₯, and determine its domain.

It generally makes sense to determine the domain of a rational function before simplifying it. By doing so, we can avoid losing any potential values of π‘₯ that could cause us issues later down the line. But before we can simplify 𝑛 of π‘₯, let’s multiply the two fractions together. When we multiply pairs of fractions, we multiply their numerators and separately their denominators. So, 𝑛 of π‘₯ is two times the quadratic π‘₯ squared plus six π‘₯ plus eight all over two π‘₯ times π‘₯ plus two.

So, before we simplify it, let’s determine the domain of the function. The domain is said to be the set of possible inputs to the function, in other words, the possible π‘₯-values that ensure 𝑛 of π‘₯ is well defined. And since we’re dealing with a rational function, we know that the denominator of that function cannot be equal to zero. Otherwise, it’s simply the quotient of two polynomial functions. And the domain of a polynomial function is the set of real numbers. So, the domain of 𝑛 of π‘₯ will be the set of real numbers. But we will need to exclude any values of π‘₯ that make the denominator equal to zero.

To find those values of π‘₯, we’ll set the denominator equal to zero and solve for π‘₯. In other words, two π‘₯ times π‘₯ plus two equals zero. And of course, since we are finding the product of two π‘₯ and π‘₯ plus two and getting an answer of zero, that will only be true if either of those expressions itself is equal to zero, in other words, if two π‘₯ equals zero or π‘₯ plus two equals zero. Well, if two π‘₯ equals zero, then π‘₯ itself must be zero. Similarly, if we subtract two from both sides of this equation, we get π‘₯ equals negative two. So, the domain is the set of real numbers not including the set containing the elements zero and negative two.

With that in mind, we’re now ready to simplify the expression. And when we simplify a function, we’re looking to find common factors. We begin by noticing that there is a shared factor of two on the numerator and denominator of our fraction. We therefore divide through by two, leaving us with π‘₯ squared plus six π‘₯ plus eight over π‘₯ times π‘₯ plus two. Then, at this stage, we’ve actually run out of shared factors, so we look to factor the numerator. π‘₯ squared plus six π‘₯ plus eight can in fact be written as π‘₯ plus four times π‘₯ plus two.

Then we notice that the fraction π‘₯ plus four π‘₯ plus two over π‘₯ times π‘₯ plus two has a common factor of π‘₯ plus two on the numerator and the denominator. Since π‘₯ cannot be equal to negative two, we’re therefore able to divide the numerator and denominator by π‘₯ plus two. And that leaves us simply with π‘₯ plus four over π‘₯. So, the function 𝑛 of π‘₯ simplifies to π‘₯ plus four over π‘₯. And its domain is the set of real numbers minus the set containing the elements zero and negative two.

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