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Video: Determining the Quotient of a Function and Its Domain

Tim Burnham

Simplify the function n(x)=7 x+63/x + 6/x+9/x + 6, and find its domain in R.

08:50

Video Transcript

Simplify the function 𝑛 of π‘₯ is equal to seven π‘₯ plus 63 over π‘₯ plus six over π‘₯ plus nine over π‘₯ plus six and find its domain in the set of real numbers.

Okay. So first of all, we’re gonna have to do a bit of algebra to simplify this function here, and then we can go on to consider the domain, the set of valid input values to that function.

So first, let’s try to simplify this denominator here. So we’re adding two terms, π‘₯ and nine over π‘₯ plus six. Now the first term is π‘₯ and the second term is a fraction. So I can to- convert this into a fraction by just putting π‘₯ over one. So I’ve now got fraction plus another fraction. And by multiplying that first term by π‘₯ plus six over π‘₯ plus six, which is just equal to one so it’s not changing the size of the number, I’m gonna create a fraction which has got the same denominator as the other fraction. So just swapping around the π‘₯ and the π‘₯ plus six on the numerator there, that first term becomes π‘₯ times π‘₯ plus six over π‘₯ plus six. And the second term is still nine over π‘₯ plus six. Now I’ve got a common denominator. I can just add the numerators together. And that gives me π‘₯ times π‘₯ plus six plus nine over π‘₯ plus six. Now I can multiply out the parentheses, π‘₯ times π‘₯ gives me π‘₯ squared and π‘₯ times positive six gives me positive six π‘₯. And that gives me π‘₯ squared plus six π‘₯ plus nine over π‘₯ plus six.

Now we can also notice that π‘₯ squared plus six π‘₯ plus nine factors down to π‘₯ plus three times π‘₯ plus three. And it’s always a good idea to factor as much as we can because then hopefully some things might cancel. So now we’ve got this new expression here, which is equivalent to that expression there. So let’s just copy that across. So I’ve converted my denominator here into an expression, where I’ve got two terms multiplied together divided by another term. And hopefully, if we do a similar process to the original numerator here, we’ll generate an expression which has got things which will cancel out with our denominator down there. So let’s go ahead and have a go at that.

So we’ll start up with seven π‘₯ plus 63 over π‘₯ plus six. I can convert the first term into a fraction. So I’ve now got a fraction plus another fraction, and I need to create a common denominator. So I can multiply the first term by π‘₯ plus six over π‘₯ plus six, which again is one. So one times seven π‘₯ over one is still just seven π‘₯ over one. And that gives me seven π‘₯ times π‘₯ plus six over π‘₯ plus six plus 63 over π‘₯ plus six. Now I’ve got a common denominator. I can just add the numerators together again. And it gives us seven π‘₯ times π‘₯ plus six plus 63 all over π‘₯ plus six. Now I could multiply seven π‘₯ by π‘₯ to give me seven π‘₯ squared and seven π‘₯ by positive six to give me plus 42π‘₯ and then have plus 63, but I’m actually not gonna do that in this occasion. Because I’ve spotted that seven is a factor of 63, so I can factor out a seven. So we’ve got seven here and we’ve got seven times nine is 63. And when we work that out, we’ve got seven times π‘₯ times π‘₯ plus six plus nine over π‘₯ plus six.

Now the reason I did all that and factored out the seven, is because the contents of these parentheses here, π‘₯ times π‘₯ plus six plus nine, that’s exactly the expression that we had before in the- when we were simplifying the denominator. So when I multiply π‘₯ times π‘₯ and π‘₯ times positive six, that gives me π‘₯ squared plus six π‘₯. So we’ve got π‘₯ squared plus six π‘₯ plus nine in the parentheses there over π‘₯ plus six, and we saw that π‘₯ squared plus six π‘₯ plus nine factored down to π‘₯ plus three times π‘₯ plus three. So we’ve got seven times π‘₯ plus three times π‘₯ plus three over π‘₯ plus six.

So when we simplified that numerator, we’ve come up with an expression here which is actually very similar to the denominator that we came up with. So we’ve managed to rewrite our original expression for the function 𝑛 of π‘₯ as this expression over here, which doesn’t look particularly as though we’ve simplified it. But remember, we’ve got this numerator divided by this denominator. And when we write that out, we’ve got the function 𝑛 of π‘₯ is equal to seven times π‘₯ plus three times π‘₯ plus three over π‘₯ plus six divided by π‘₯ plus three times π‘₯ plus three over π‘₯ plus six.

And when we divide fractions, we just simply have to flip the second fraction and turn it into a multiplication. So that becomes seven times π‘₯ plus three times π‘₯ plus three over π‘₯ plus six times π‘₯ plus six over π‘₯ plus three times π‘₯ plus three. Now that turns out to be quite handy because we’ve got lots of terms multiplied together on the top and lots of terms multiplied together on the bottom. But the convenient thing is that some of the terms on the top are the same as some of the terms on the bottom. So we’re gonna be able to cancel them out.

For example, dividing the top by π‘₯ plus six gives us one here. Dividing the bottom by π‘₯ plus six gives us one here. Dividing the top by π‘₯ plus three here gives us one. Dividing the bottom by π‘₯ plus three here gives us one. Dividing the top by π‘₯ plus three gives us one. Dividing the bottom by π‘₯ plus three gives us one. So we’ve got seven times one times one times one over one times one times one.

Hey presto! This great brick expression that we had right at the very beginning, for n of π‘₯, simplifies down just to 𝑛 of π‘₯ is equal to seven. So that’s the answer to the first part of the question. The simplified version of that function is 𝑛 of π‘₯ is equal to seven.

Okay. Now let’s consider the domain of this function. What are the valid π‘₯-values to go into that function? Well, on the face of it, you might think that it’s just any real number because the function is always just equal to seven. But wait, that’s not quite right. If we look at our original expression for the function over here, if this denominator here turns out to have a value of zero, then we’ve got something divided by zero and that will be undefined. So an π‘₯-value that generates a denominator in that function of zero would need to be excluded from the domain. And likewise, up here on the numerator, we’ve actually got this term 63 over π‘₯ plus six. If the value of π‘₯ that we put into the function generates a value of zero in this position here, we’re gonna have 63 divided by zero which again is gonna be undefined.

So we need to ask ourselves, which π‘₯-values must we exclude from the domain? Well, the value of π‘₯ that makes this bit equal to zero β€” in other words, when π‘₯ plus six is equal to zero β€” needs to be excluded. And subtracting six from both sides of that equation tells us that π‘₯ would need to be negative six for that to happen. So that’s the first value of π‘₯ that we need to exclude from the domain.

But we also said that if the denominator here turns out to be zero, then whatever value of π‘₯ that makes that equal zero also needs to be excluded. So π‘₯ plus nine over π‘₯ plus six is equal to zero. And remember that we already reexpressed that in this way, π‘₯ plus three times π‘₯ plus three over π‘₯ plus six. So whatever value of π‘₯ makes that equal to zero is a value that we need to exclude as well. Well, in order for this expression here to be equal to zero, we need to make the numerator equal to zero. And both terms multiplied together there are in fact π‘₯ plus three.

Now it doesn’t matter whether we have π‘₯ plus three being equal to zero, or π‘₯ plus three being equal to zero. So long as one of those sets of parentheses is zero, then those two things multiplied together are gonna give an answer of zero. Since they’re both the same, π‘₯ plus three, it’s basically π‘₯ plus three that has to be equal to zero. And subtracting three from each side of that equation tells us that π‘₯ would need to be equal to negative three in order to make that expression equal to zero. So we can exclude π‘₯ is equal to negative three from our domain as well.

So the domain is equal to the real numbers but excluding negative six and excluding negative three as well. And one way that we can write that out is to say that the domain is equal to the set of real numbers minus the set containing negative six and negative three. So there are our answers.