Question Video: Simplifying and Determining the Domain of Rational Functions Mathematics

Simplify the function 𝑓(π‘₯) = (7π‘₯Β² + 43π‘₯ + 6)/(7π‘₯Β² + 50π‘₯ + 7), and find its domain.

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Video Transcript

Simplify the function 𝑓 of π‘₯ is equal to seven π‘₯ squared plus 43π‘₯ plus six over seven π‘₯ squared plus 50π‘₯ plus seven and find its domain.

In order to simplify the function, we need to factorize or factor the numerator and denominator and then look for common factors. A common mistake here would be to try and cancel seven π‘₯ squared on the top and bottom of our function. We can prove that this is not true by considering a numerical example.

Imagine we had the fraction seven plus three over seven plus five. This is equal to 10 over 12, which in turn simplifies to five over six. Had we tried to cancel the sevens, we would be left with three over five, which is not the same as five over six.

Let’s now consider how we can factorize the numerator of our function. The numerator is written in the form π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. One way to factorize a quadratic of this type is to firstly try and find a pair of numbers that has a product of π‘Ž multiplied by 𝑐 and a sum of 𝑏. In this case, we need two numbers that have a product of 42, seven times six, and a sum of 43. There are four factor pairs of 42: 42 and one, 21 and two, 14 and three, and seven and six. The only pair that has a sum of 43 is 42 and one. This means that we can rewrite 43π‘₯ as 42π‘₯ plus π‘₯ or plus one π‘₯.

Our next step is to find the highest common factor of the first two terms and then repeat this with the last two terms. The highest common factor of seven π‘₯ squared and 42π‘₯ is seven π‘₯. Factorizing this out, we’re left with seven π‘₯ multiplied by π‘₯ plus six. The only common factor of one π‘₯ and six is one. So this can be rewritten as one multiplied by π‘₯ plus six.

We notice that π‘₯ plus six is common to both parts of this expression. As we are multiplying π‘₯ plus six by seven π‘₯ and then by one, this can be rewritten as seven π‘₯ plus one multiplied by π‘₯ plus six. We can factorize the denominator using the same method. Seven π‘₯ squared plus 50π‘₯ plus seven is equal to seven π‘₯ plus one multiplied by π‘₯ plus seven. This means that 𝑓 of π‘₯ is equal to seven π‘₯ plus one multiplied by π‘₯ plus six over seven π‘₯ plus one multiplied by π‘₯ plus seven. We can divide the numerator and denominator by seven π‘₯ plus one. The function 𝑓 of π‘₯ in its simplest form is π‘₯ plus six over π‘₯ plus seven.

We will now clear some space so we can find the domain of the function. We know that any fraction is undefined when its denominator is equal to zero. This means that any value that makes the denominator zero will not be in the domain. We established in the first part of the question that seven π‘₯ squared plus 50π‘₯ plus seven was equal to seven π‘₯ plus one multiplied by π‘₯ plus seven. Setting this equal to zero means that either seven π‘₯ plus one equals zero or π‘₯ plus seven equals zero.

We can subtract one from both sides of the first equation such that seven π‘₯ equals negative one. Dividing both sides by seven gives us π‘₯ is equal to negative one-seventh. For our second equation, we need to subtract seven from both sides. This gives us π‘₯ is equal to negative seven. Substituting π‘₯ equals negative one-seventh or π‘₯ equals negative seven into the denominator gives an answer of zero.

As the domain is the set of π‘₯-values that we can input into our function 𝑓, the domain is equal to all real numbers apart from negative one-seventh and negative seven. This can be written in set notation as shown.

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