 In this video, I want to talk about solving rational inequalities. What happens if you have a rational function involved in your inequalities? How do you go about solving it? Turns out the strategy we're going to use is the strategy that we've been using throughout this entire series. If ever you have to solve the inequality f of x is, say, greater than zero, well, the idea is f of x just represents your y-coordinate, y is greater than zero, and so if y is greater than zero, you're looking for those points which are above the x-axis. That is, we're going to solve this using the graph of the function, and what's the function in play here? It's the left-hand side. When you're solving these inequalities, you want to compare the right-hand side to zero, whether that's greater than zero, greater than equal to zero, less than zero, doesn't matter. Compare the left hand side to zero on the right-hand side, and so then the left-hand side becomes the function in play here. So we have x plus 3 times 2 minus x over x minus 1 squared. We want to graph this function. So let's think about graphing this thing. So what do we know about this function? Well, we know that if you look at the numerator, it is a rational function. The numerator gives us x-intercepts. So we have an x-intercept at negative 3, and we have an x-intercept at positive 2. When we put those things together, x minus 3 and x plus 2, notice that both of these have odd powers. Therefore, we're going to cross the x-axis when we get to negative 3 and positive 2. We also have a vertical asymptote sitting here. It comes from the x minus 1 in the denominator. So that sits between negative 3 and 1. So I'm just going to put a dash strut line right here. x equals 1 here. So in the past, as we've been working with equation, in a clause, we often solve the equation. Basically, solving the equation means we're looking for the x-intercepts. When it comes to a rational function, we have to look for where does the function go to zero, thus the x-intercepts. But we have to also look for where is the function undefined, because it turns out that a function can switch at signs in one of two places. A function switches at signs when we're at an x-intercept, because you can cross the x-axis to go from one side to the other. But you can also take the back door through infinity. And so you can switch from negative infinity to positive infinity. So it's important we pay attention to all the places where the function is equal to zero, but we have to also look at all the places where the function is undefined. These are the points which are going to be critical to solving these inequalities here. All right. So when it comes to x-1, though, the multiplicity there is even. So this is an even multiplicity. This is going to tell us we're going to touch infinity, not cross it. So there's not going to be a sign change there. The last thing to focus here on would be, I'd say, the in behavior. If we let x approach plus or minus infinity, then our function will be approximately the same thing as just the leading terms. We get x from the x plus 3. We get a negative x from them, 2 minus x. And then we're going to get an x squared from the denominator. This simplifies just to be negative 1. And this gives us a horizontal asymptote. So we get something below the x-axis right here. Now, if I was looking for the best graph in the world, I would probably ask myself, does this function cross its horizontal asymptote? It turns out for inequalities, because my horizontal asymptote is not the x-axis, I really don't care. Because if you're a little bit above negative 1 or a little bit below negative 1, you are still negative 1. So that's a little bit of details I don't need. So what we do know is that as you go to the far left, you're going to have to get close to negative 1, which is negative. So our graph's going to have to kind of do something like the following, right? We connect these dots. We're going to have to get a picture like this. So as we go to the far left, we have to be close to our asymptote negative 1. We have to cross the x-intercept negative 3 because it has odd multiplicity. And then we have to go off towards infinity. We can't turn around and go down because that would require an x-intercept we don't have. So we have to go off towards infinity. So as x approaches 1 from the left, that means that our y will approach positive infinity. That sign's important. But then, since the x-intercept has even multiplicity, we touch infinity. So we're going to come back from the other side, go through our x-intercept, and then approach the horizontal asymptote right there. As we're close to negative 1. Now, it turns out that if I were to look into a little bit more detail, it turns out my graph actually does cross its horizontal asymptote and it comes back around. But that little nuance, that little subtle down the graph, doesn't make any difference on the answer to this inequality. So if I skip that step, I would be perfectly fine. So it's probably not worth our time doing such a thing. We now have a picture of our graph. And what are we looking for? We're looking for where are we above? Where are we above the x-axis? Okay? Well, you'll notice that the function's above the x-axis when we're to the left of negative 3 before we hit the vertical asymptote because it went off to positive infinity. We're also above the x-axis when we're between the vertical asymptote and the x-intercept there. And so when we put this information together, we're going to be above the x-axis from negative 3 to 1, union 1 to 2, right? So notice that the inequality is greater than 0. It's not equal to 0. Greater than or equal to 0 is not allowed here. So we can't allow our x-intercepts inside of the solution set because the x-intercepts give us a point where the y-coordinate is 0. We need to be greater than 0. 0 is not greater than 0. So if you have x-intercepts, you do not include them if you're doing greater than 0 or less than 0. You need greater than or equal to or less than or equal to 0 to include an x-intercept. On the other hand, what about the vertical asymptote? Vertical asymptotes will never be part of the solution set because our function f is not defined at 1 and 1, you know, undefined is not greater than 0. So we can't include it inside of the solution set. The vertical asymptotes will never be part of the solution set. Let's look at another example here. This time we have x minus 1 over x squared minus 4 is less than or equal to 0. So some things I can see already. If we're looking for less than or equal to 0, then I want to find the points which are going to be below or on the x-axis when I graph this thing. But what are we graphing? We want to graph the function f of x equals x minus 1 over x squared minus 4. As we learned about when we graph these things, we want the numerator and denominator to be factored. So we're going to get in the denominator, it's a difference of squares. You get x minus 2 and x plus 2. So this informs us about the picture and play right here. So things we need to be looking out for. There's an x-intercept at x equals 1. So let's just put that right here. x equals 1. This is going to be a odd multiplicity. It's a crossing. And then we're going to have vertical asymptotes at x equals 2 and x equals negative 2. So add those to the graph here. And this vertical asymptote right here at x equals 2. And we have this other one over here at x equals negative 2. Your picture does not have to be perfect. We just need to have an idea about whether we're above or below the x-axis. That's really what we're looking for here. Both of these multiplicities are likewise odd. So we're going to cross infinity at these points. I would look for some type of in behavior. So notice that as x approaches plus or minus infinity, our function will be approximately the same thing as just the leading terms. So you have x on top, you have x times x on the bottom. So just looking at these portions right here. So this gives us a 1 over x. So our function is going to look like 1 over x. And that's going to be a positive 1 over x. 1 over x, of course, it approaches on the right-hand side from above and it approaches from below on the left-hand side. So that's the in behavior we're expecting right here. There's going to be a horizontal asymptote on the x-axis. And if you want to, we can put some dashed lines here. Now if you're wondering, does my function cross the horizontal asymptote? It turns out I already have the information for that. The answer is yes. The horizontal asymptote is the x-axis and I know that I have an x-intercept. So that's the only place we're going to cross that. That's okay. So going from here, what are our possibilities? Well, like I said, we know that on the right-hand side, we're going to have to be approaching infinity. We have to approach infinity because that's how 1 over x acts. It'll approach as x goes to infinity. We'll approach y equals 0 from above. Well, that means we have to connect together. We have to go towards our vertical asymptote. We're going to have to get a picture that looks something like this. Just using the in behavior to get started. Then because we cross infinity, we're going to flip to the other side. Our spaceship wraps around the screen. And so we get something like this. We're going to then cross our x-intercept. So we crossed infinity. We're going to cross our x-intercept. We have to then go towards infinity. So there's going to be some type of inflection going on there. But we're going to then go off towards infinity again. Then cross to the other side. We then finish up our picture like so. And that gives us the in behavior we wanted. So I didn't even use a y-intercept or anything like that. Just using the behavior in your intercepts, asymptotes, and the end, we were able to determine the graph of this function here. So where is it? Where is it below the x-axis? It's going to be below the x-axis in this sector. It's going to be below the x-axis in this sector right here. And so what does that tell us about the solution? Well, so for the first sector right here, we're going from negative infinity up to that asymptote, which is that negative two. So from anything from negative infinity to negative two was part of the solution. Then we have a big jump because there's a positive region. Then we're going to go from one to two. I'm going to go from one to two in this region right here. Now you'll notice I put a bracket next to the one, and I put a parenthesis next to the two. Why the difference there? Well, if you look at negative two and positive two, these are vertical asymptotes. These are locations outside the domain of our function. Does not exist is not great. Does not exist is not less than or equal to zero because this is not a number. It's not comparable to zero. So the vertical asymptotes will never be part of the solution set. On the other hand, x equals zero is an acceptable wide coordinate. If you plug in x, sorry, if x equals one, you get y equals zero. If you plug in x equals one here, you're going to end up with zero, which is less than equal to zero. Equality is okay here. So the x-intercepts will be allowed. Those points on the x-axis will be allowed. So we put a bracket on the x-intercepts, but we will always put parenthesis by vertical asymptotes, just like we always put it by infinity or negative infinity, because those are not points inside of the graph. All right. So the next one right here, solve the inequality for x plus five over x plus two is greater than or equal to three. This one's going to be a little bit more challenging compared to the previous two because we're not greater than or equal to zero. We need to be greater than or equal to zero. So there's a couple of ways we could do this. We could take four x plus five over x plus two. We could just attract three from both sides like so. And then we could try to just add that together the fractions. And that's the approach I really would recommend here. You want to add together the fractions. So we're going to get four x plus five over x plus two. Then we're going to get minus three times x plus two and x plus two. This is really what you want to do. You want to add together the fractions. Don't try to multiply things too much. Otherwise, you might get a little bit confused with the graph here. So we're going to get x plus five, excuse me, four x plus five in the numerator. Distribute the three right here. We end up with four x plus five. We're going to get minus three x minus six over x plus two. This should be greater than or equal to zero. In combining terms, you're going to get an x minus one and then the denominator would be x plus two. And this is greater than or equal to zero. And then from here, we're in a position where we can start graphing this thing. So consider the function f of x equals x minus one over x plus two. What do we see about this function? It has an x intercept at one. It has a vertical asymptote at negative two. So we put that on the screen. x equals negative two. And then we also think about it in behavior as x approaches plus or minus infinity. Our function f of x will be approximately the same thing as the ratio of its leading terms, x over x. This gives us x over x, which gives us one. So our horizontal asymptote would look like y equals one. So we get something like this, y equals one. So starting at the x intercept, as x goes to the right, we have to get closer to that horizontal asymptote. So we're going to have to see something like this. At x equals one, of course, we have an odd multiplicity because the power is odd there. So it's going to cross the x-axis and go towards its vertical asymptote. At the vertical asymptote, we also have odd multiplicity. So we're going to cross infinity and then approach our horizontal asymptote. So our picture's going to look something like this. And so then, where are we greater than or equal to zero? We're looking for things which are above or on the x-axis. That's what we need right here. And so then looking for that, we're above the x-axis when we are to the right of one. And we're also above the x-axis when we're to the left of the vertical asymptote. We do allow the x-intercept because that's greater than or equal to zero, the y-coordinate is. So then we record our answer as negative infinity up to negative two. Union one, oops, forgot the number there, one to infinity. And so that gives us the solution right here. So we can solve, and this is, I honestly think the best way to solve these rational inequalities, solve them by graphing these things. Now, at the very beginning of this thing, I suggested that we move the three over and add the fractions together. If you wanted to, you could try to solve the equation associated here, right? So that's something I've always advertised that if you try to solve an inequality, you want to solve the equation. So you have 4x plus five plus five over x plus two, this equals three. We could try to solve the equation by clearing the denominators. You could also could think of as like cross multiply in this situation. So you get 4x plus five equals three times x plus two, right? So this would give us 4x plus five equals 3x plus six. This looks a little bit familiar when you combine like terms, you're going to end up with x equals one. So this does find the x intercept of the graph, right? You have to also look at the domain that x can't equal negative two. So what you could do from this information right here is you could proceed to just build a sign chart of some kind. This is always an alternative, right? Where you have a sign chart at, you look at a negative two and you have one. So we could do this without the graph entirely, but then we have to pick some test points. So we could be like, okay, I'm going to take x equals zero, I'm going to take x equals two, I'm going to take x equals negative three, plug them into the rational function and see what happens. The original rational expression, right? If you plug in zero into the original expression, you get five halves is greater than or equal to two, a greater than or equal to three. Well, that's not true. If you stick in two, you'll end up with, let's see, eight plus five is 13 over four. Is that greater than or equal to three? That one is true. So you'd grab that sector. And then for the other one, you plug in like negative three, you're going to get negative 12 plus five, which is a negative seven over negative one. So that's a positive seven. Is that greater than or equal to three? And the answer is yes. So you could solve this using a sign chart if you prefer. To solve for the sign chart, you have to figure out what makes the denominator go to zero. So you can see that from the original expression. Solve the equation to find the x intercepts and go from there. Personally, I like the graphing approach. I think we can get it all at once. And it's less error prone to as, as compared to this sign chart approach. But I do want to present as an alternative, if that's something that you would rather do to solve these inequalities.