Video: Simplifying and Determining the Domain of Rational Functions

Simplify the function 𝑛(π‘₯) = (π‘₯ + 1)/(π‘₯Β² + 3π‘₯ + 2) and find its domain.

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Video Transcript

Simplify the function 𝑛 of π‘₯ equals π‘₯ plus one divided by π‘₯ squared plus three π‘₯ plus two and find its domain.

In order to simplify this function, we need to factorise the denominator of the fraction. As there is no common factor other than one on the denominator, we will factorise the quadratic into two parentheses or brackets. π‘₯ multiplied by π‘₯ is π‘₯ squared. Therefore, the first term in the parentheses will be π‘₯.

We now need to find a pair of numbers that multiply to give us two and add to give us three. In this case, the only option is one and two, as one multiplied by two is equal to two and one add two is equal to three. This means that our two parentheses are π‘₯ plus one and π‘₯ plus two.

At this stage, we can see that we have π‘₯ plus one on the numerator and the denominator. This means we can divide by π‘₯ plus one. Dividing the numerator by π‘₯ plus one gives us one. And dividing the denominator by π‘₯ plus one gives us π‘₯ plus two. Therefore, the simplified version of the function 𝑛 of π‘₯ is one divided by π‘₯ plus two.

Whilst it initially appears that all real values can be contained within the domain, on closer inspection there are some values of π‘₯ that would make the denominator equal to zero. These would give undefined values. Setting the denominator equal to zero gives us two equations, π‘₯ plus one equal zero and π‘₯ plus two equal zero.

Solving these equations gives us π‘₯ equals negative one or π‘₯ equals negative two. Substituting either these two values into the function 𝑛 of π‘₯ would give an undefined output. This means that π‘₯ equals negative one and π‘₯ equals negative two cannot be contained within the domain. Therefore, the domain of 𝑛 of π‘₯ is all real values with the exception of negative one and negative two.

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