Introduction to Rational Functions - Online Tutor, Practice Problems & Exam Prep

Intro to Rational Functions

Hey, everyone. By now, we've worked with a bunch of different polynomial functions, things like \( x^2 + 4x + 1 \) or \( 3x + 2 \) or other polynomials of different degrees. But if we take these polynomials and we put 1 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 \( \frac{p(x)}{q(x)} \), where \( p(x) \) just represents any polynomial and \( q(x) \) 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 going to walk you through everything that you need to know step by step. So let's go ahead and get started. Now when working with rational equations, we said that our denominator could not be equal to 0. So for this particular equation, looking at my denominator, I would say that \( x \) cannot be equal to 1, and that would be referred to as my restriction because 1 would make my denominator 0, 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 0. 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 0. 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, \( \frac{1}{x-1} \), I have the same restriction. But writing it in my domain, I write my domain in set notation as \( x \) such that \( x \neq 1 \) because that's what would make my denominator 0. Now we write it like this because we determine our domain the same exact way. We simply set our denominator equal to 0 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 0. 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 \( \frac{3}{3x-12} \). So we're simply going to take our denominator and set it equal to 0, solving for \( x \). So subtracting 12 from both sides, I end up with \( 3x = -12 \). And then to isolate \( x \), I want to go ahead and divide both sides by 3. Now that leaves me with \( x = -4 \), and that tells me my domain restriction. I know that \( x \) can be any real number, except \( x \) cannot be equal to -4 because if I took -4 and I plugged it back into my denominator, it would make it 0, which is what I don't want. So my domain here is \( x \) such that \( x \neq -4 \). Now let's look at another function here. So I have \( f(x) = \frac{x+5}{x^2-25} \). So again, taking our denominator, setting it equal to 0, if I add 25 to both sides, canceling out that 25, leaving me with \( x^2 = 25 \). And to isolate \( x \), I simply take the square root of both sides, leaving me with \( x = \pm 5 \), which I know is just 5. Now with this answer, this tells me that my domain is actually restricted by 2 values. It can't be 5 or -5. So my domain is \( x \) such that \( x \neq 5 \) or \( -5 \). 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 common 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 starting again with \( \frac{3}{3x+12} \), 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'm simply left with 3 on the top there. But on the bottom, it looks like I can pull out a greatest common factor of 3 from both of these terms, leaving me with \( 3(x+4) \). Now we wanna go ahead and cancel out any common factors. In this case, I have a common factor in my numerator and my denominator of 3, so I can go ahead and cancel that out, leaving me with my function \( \frac{1}{x+4} \) as all that's left. So my function in lowest terms here is simply \( \frac{1}{x+4} \). Now let's look at our other function. So we have \( f(x) = \frac{x+5}{x^2-25} \). So let's go ahead and factor. Now again, my numerator can't be factored, it's already just \( x+5 \), but my denominator is a difference of squares, so this factors to \( (x+5)(x-5) \). Now I wanna go ahead and cancel our common factors. In this case, I have a common factor of \( x+5 \) that I can cancel from both my numerator and my denominator. So canceling that factor out leaves me with \( \frac{1}{x-5} \) and that's my function in lowest terms. Now something that I want to point out here is that if I were to take this function in lowest terms and solve for the denominator here by setting my or solve for the domain here, setting my denominator equal to 0, 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 5, but I wouldn't know about that -5. 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. Okay. Now that we know the basics of working with rational functions, let's move on. I'll see you in the next video.

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}, $f\left(x\right)=x+3$

Find the domain of the rational function. Then, write it in lowest terms.

$f\left(x\right)=\frac{6x^5}{2x^2-8}$

• A. ${\left\lbrace x|x\ne2,-2\right\rbrace},f(x)=\frac{3x^5}{x^2-4}$
• B. ${\left\lbrace x|x\ne2,-2\right\rbrace},f(x)=\frac{6x^5}{2x^2-8}$
• C. ${\left\lbrace x|x\ne2\right\rbrace},f(x)=\frac{3x^5}{x^2-4}$
• D. ${\left\lbrace x|x\ne2\right\rbrace},f(x)=\frac{3x^5}{x^2-8}$