Video: Simplifying and Determining the Domain of a Difference of Two Rational Functions

Simplify the function 𝑛(π‘₯) = ((π‘₯ βˆ’ 7)/(π‘₯Β² βˆ’ 3π‘₯ βˆ’ 28)) βˆ’ ((π‘₯ βˆ’ 7)/(7 βˆ’ π‘₯)), and determine its domain.

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Video Transcript

Simplify the function 𝑛 of π‘₯ equals the fraction π‘₯ minus seven over π‘₯ squared minus three π‘₯ minus 28 minus the fraction π‘₯ minus seven divided by seven minus π‘₯, and determine its domain.

After we copy down the function, the first thing that we’re going to wanna do is see if we can factor this denominator, π‘₯ squared minus three π‘₯ minus 28. Let’s do that! Can we factor π‘₯ squared minus three π‘₯ minus 28? Since the leading coefficient is one, we know that it’ll be π‘₯ as the first term of both factors. Now I need the two factors of 28 that multiply together to be negative 28, and those same factors need to add together to equal negative three.

First step, one and 28. Then we have two and 14. Three isn’t a factor. From there, we’ll take four and seven. I recognize that the difference between seven and four is three. If I make the seven negative and the four positive and add them together, I’ll get negative three. And if we multiply them together, four and negative seven, we’ll get negative 28. These will be the second terms in our two factors here.

We take those two factors and we put them in the denominator to simplify the function. From here, let’s stop and talk about determining the domain. The domain is the set of all π‘₯-values for which output a real 𝑦-value. When we’re working with fractions that have variables in the denominator as functions, we need to be careful to remember that our denominator cannot be equal to zero.

This means that π‘₯ plus four cannot be equal to zero; π‘₯ minus seven cannot be equal to zero; and seven minus π‘₯ cannot be equal to zero. What we want to do to determine the domain is to find the places that would make π‘₯ plus four equals zero and find the places that π‘₯ minus seven would be equal to zero. We need to solve for π‘₯ here. To do that I can subtract four from both sides of the equation, and then I determine that π‘₯ cannot be equal to negative four.

To solve for π‘₯ minus seven, we add seven to both sides, and we say that π‘₯ cannot be equal to seven. Let’s go ahead and check one last place, seven minus π‘₯. We can start by subtracting seven from both sides. We see that negative π‘₯ cannot equal negative seven, and then we multiply both sides of the equation by negative one, and we see that π‘₯ cannot equal seven. What we can say about our domain is that it’s true for all real numbers with the exception of negative four and seven.

So we write all reals minus the set of negative four and seven. Okay, let’s get back to simplifying this function. What we can see in one of our terms is that we have π‘₯ minus seven in the numerator and π‘₯ minus seven in the denominator. These values will cancel each other out; π‘₯ minus seven over π‘₯ minus seven equals one. This will leave us with one over π‘₯ plus four. Now I look at this second term, and I wonder what to do. π‘₯ minus seven over seven minus π‘₯. Can we simplify this?

With the help of this subtraction symbol, we can simplify π‘₯ minus seven over seven minus π‘₯. Here’s what we’ll do. Instead of saying minus here, we’ll say plus negative one times π‘₯ minus seven. What you should recognize here is that we have not changed the value of this problem at all. We’ve simply change the format. Instead of using subtraction, we’ll use addition, and we’ll multiply by negative one.

I’ll multiply π‘₯ by negative one, which gives me negative π‘₯. Then I’ll multiply negative one by negative seven, which gives m positive seven. Bring down the denominator. If we look closely at negative π‘₯ plus seven, we see that we can actually change the order here. Instead of saying negative π‘₯ plus seven, we can say seven minus π‘₯ over seven minus π‘₯. Okay, bring down the one over π‘₯ plus four.

And if we clean this problem up and just copy down seven minus π‘₯ over seven minus π‘₯, do you see that that equals one? Now we’re trying to add one over π‘₯ plus four plus one. But to add a fraction, we need to have a common denominator. So I’m going to write one as π‘₯ plus four over π‘₯ plus four. It’s still equal to one, but now we can add these two pieces together. When we’re adding fractions with a common denominator, we add the numerator, one plus π‘₯ plus four, and the denominator will stay the same, π‘₯ plus four.

We can add the one and the four together. One plus four equals five, and our denominator stays the same, π‘₯ plus four. The function 𝑛 of π‘₯ could be simplified to π‘₯ plus five over π‘₯ plus four, and the domain here is all real numbers minus the set of π‘₯ equal to negative four and seven.

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