Introduction to Rational Functions - Video Tutorials & Practice Problems

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Intro to Rational Functions

Hey, everyone, by now, we've worked with a bunch of different polynomial functions, things like X squared plus four, X plus one or three X plus two or other polynomials of different degrees. But if we take these polynomials and we put one on the top of a fraction and the other on the bottom of a fraction, we now have a new function called a rational function that has a polynomial of any degree in both the numerator and the denominator of a fraction. Now, you might see this written in your textbooks as P of X divided by Q of X where P FX just represents any polynomial and Q of X is any other polynomial. Now, I know that fractions might not be your favorite thing to work with and taking polynomials and putting them into fractions might seem a little bit intimidating. But we've worked with polynomial functions before and we've even already worked with rational equations. So, in working with rational functions, we're going to rely on a lot of what we already know about these functions and equations that we've already worked with. And I'm gonna walk you through everything that you need to no step by step. So let's go ahead and get started. Now in working with rational equations, we said that our denominator could not be equal to zero. So for this particular equation looking at my denominator, I would say that X cannot be equal to one and that would be referred to as my restriction because one would make my denominator zero, which is not something that I want to happen. Now, the same thing is true of rational functions because the same thing is true of all fractions. The denominator can never be zero. Now because we're working with a function now that can be evaluated for any value of X. We're going to refer to this as our domain restriction. So it's the same exact idea, our denominator cannot be zero. We're now just referring to our entire domain and defining what X can and can't be based on that restriction. So looking at this function that I have here one over X minus one, I have the same restriction. But writing it in my domain, I write my domain in set notation in such that X cannot be equal to in this case one because that's what would make my denominator zero. Now we write it like this because we determine our domain the same exact way we simply set our denominator equal to zero and solve for X to get that restriction. And then we know that our domain can be any real number besides that number that makes our denominator zero. So now that we know how to find the domain, let's go ahead and find the domain of a couple of different rational functions here. Now, looking at this first rational function, I have 3/3 X minus 12. So we're simply going to take our denominator and set it equal to zero, solving for X. So subtracting 12 from both sides, I end up with three X is equal to negative 12 and then to isolate X, I want to go ahead and divide both sides by three. Now that leaves me with X is equal to negative four. And that tells me my domain restriction. I know that X can be any real number except X cannot be equal to negative four. Because if I took negative four and I plugged it back into my denominator, it would make it zero, which is what I don't want. So my domain here is X such that X cannot be equal to negative four. Now let's look at another function here. So if F of X is equal to X plus five, divided by X squared minus 25 so again, taking our denominator setting it equal to zero by add 25 to both sides. Canceling out that 25 leaving me with X squared is equal to 25 and to isolate X, I simply take the square root of both sides. Leaving me with X is equal to plus or minus the square root of 25 which I know is just five. Now, with this answer, this tells me that my domain is actually restricted by two values. It can't be five or negative five. So my domain is X such that X cannot be equal to five or negative, we can have multiple domain restrictions depending on what our denominator is. Now that we know how to find the domain of different functions. Something else that you'll be commonly asked to do when working with rational functions is to write them in lowest terms. Now, in order to write functions in lowest terms, we're going to factor something that we've done a million times before we're going to factor both the top and the bottom of our function and then simply cancel out any and factors. So we're just writing our function in the simplest terms, simplifying it as fully as we can. So let's take another look at these functions here. So it's starting again with 3/3 X plus 12. Let's go ahead and factor the top and the bottom here. Now, I can't really factor the top because it's just the constant. So I simply left with three on the top there, but on the bottom, it looks like I can pull out a greatest common factor of three from both of these turns leaving me with three times X plus four. Now, you want to go ahead and cancel out any common factors. In this case, I have a common factor in my numerator and my denominator of three. So I can go ahead and cancel that out, leaving me with my function one over X plus four as all that's left. So my function in the lowest terms here is simply one over X plus four. Now let's look at our other function. So we have of X is equal to X plus five divided by X squared minus 25. So let's go ahead and factor. Now again, my numerator can't be factored. It's already just X plus five, but my denominator is a difference of squares. So this factors to X plus five times X minus five. Now, I want to go ahead and cancel our common factors. In this case, I have a common factor of X plus five that I can cancel from both my numerator and my denominator. So canceling that factor out leaves me with a one over X minus five and that's my function in the lowest terms. Now, something that I want to point out here is that if I were to take this function in lowest terms. And so for the denominator here by setting my or solve for the domain here setting my denominator equal to zero, I would only get one of the values that appeared in my domain restriction up here. I would say that X cannot be equal to five, but I wouldn't know about that negative five that's because you always need to find your domain before you write in lowest terms, never write in lowest terms and then find the domain because you might be leaving something out. OK? Now that we know the basics of working with rational functions, let's move on. I'll see you in the next video.

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Problem

Problem

Find the domain of the rational function. Then, write it in lowest terms. $f\left(x\right)=\frac{x^2+9}{x-3}$

A

{x∣x≠0}, $f\left(x\right)=\frac{1}{x-3}$

B

{x∣x≠3}, $f\left(x\right)=\frac{x^2+9}{x-3}$

C

{x∣x≠−3}, $f\left(x\right)=\frac{x^2+9}{x-3}$

D

{x∣x≠3}, $f\left(x\right)=x+3$

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Problem

Problem

Find the domain of the rational function. Then, write it in lowest terms. $f\left(x\right)=\frac{6x^5}{2x^2-8}$