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Video: Solving Absolute Value Nonlinear Inequalities

Tim Burnham

We explain the process of sketching the curve of a nonlinear absolute value function and use that sketch to help us select appropriate algebraic representations of the expressions in a nonlinear inequality, then use them to solve the inequality.

13:41

Video Transcript

In this video, we’re gonna look at how to solve an absolute value nonlinear inequality. And this involves expressions which are nonlinear. We evaluate them and then we just take the positive value. So if the result was negative, we just take the positive version of that number. First, we’ll go through the process of how to sketch the curve of the nonlinear absolute function. And then we’ll look at how to analyse the graph, and to solve the algebra, and go through and solve the inequality yourself.

So let’s dive in and take a look at our example. Well we’ve been asked to find the values of 𝑥 that satisfy the inequality one over the absolute value of 𝑥 minus three is greater than two. Now that’s a strict inequality, so it’s not allowed to be equal to two. Now straight away when looking at that inequality, if you look at the left-hand side, the expression on the left, one over the absolute value of 𝑥 minus three, well we’ve got a problem there. When 𝑥 is equal to three, we’d have three minus three, which will be zero. The absolute value of zero is zero, so one over zero. Now that’s not defined. So we’re not gonna be able to use 𝑥 equals three as a solution, if that does turn out to be a solution in the end.

So I’m just gonna make a note of that now, so that I remember that later on when I’m filling in my answer. Okay then, so the questions says one over the absolute value of 𝑥 minus three. Well we’ll gradually work towards that, so we’re gonna start off by considering 𝑦 equals 𝑥 minus three. We’ll then take the absolute vadule- or the modulus of that and then we’ll look at one over there. So we’re gonna do this in different stages. We’re gonna build up our solution. So thinking about just 𝑦 equals 𝑥 minus three, this is a linear function and it means one times 𝑥 minus three. So that tells us the slope is one and the 𝑦-intercept is negative three. Now it would also cut the 𝑥-axis when the 𝑦-coordinate is zero. That’s the definition of the points on the 𝑥-axis. They’ve all got a 𝑦-coordinate of zero. So putting 𝑦 equal to zero in the equation and solving, gives us 𝑥 equals three. So that gives us all we need to know. We know where it cuts the 𝑥-axis, at 𝑥 equals three, we neno- we know where it cuts the 𝑦-axis, and we know the slope of this line. We can now do a sketch of 𝑦 equals 𝑥 minus three.

Okay. So that’s what that line looks like. Now let’s try to consider what the absolute value, or modulus, of that would look like. So this is basically just like 𝑦 equals 𝑥 minus three but wherever the 𝑦-coordinate comes out as being a negative value, we would take the positive version of that. So looking at our graph, basically everything to the right of 𝑥 equals three is a positive anyway, has a positive 𝑦-value. So in that region, the graph of 𝑦 equals 𝑥 minus three looks exactly the same as the graph of 𝑦 equals the absolute value of 𝑥 minus three. If I- even when 𝑥 equals three, both of those two things generate an- a 𝑦-coordinate of zero. So even at 𝑥 equals three, that same function can be used.

Now to the left of 𝑥 equals three, every time we’ve got a negative 𝑦-coordinate, we’re gonna reflect that in the 𝑥-axis up to the corresponding positive version of that coordinate. So negative three will become positive three up here. And other v- other negative values would become their positive equivalent up here. So the graph to the left of 𝑥 equals three is going to look like this. So let’s just think what we’ve done there. We- when we took 𝑦 equals 𝑥 minus three, we got maybe a 𝑦-coordinate down here, but what we had to do to take the absolute value of that, is to take the negative value of that negative number. So what we’ve got here is the negative of 𝑦 and of course the negative of a negative number comes out to be a positive number. So that reflects that line above the 𝑥-axis. So to the right of 𝑥 equals three, in order to work out the corresponding 𝑦-coordinates on that line up there, we’re using the equation 𝑦 is equal to 𝑥 minus three. But to the left of that, we’re using 𝑦 is equal to the negative of 𝑥 minus three, which if I multiply out those parenthesis, the negative of negative three is positive three and the negative of 𝑥 is negative 𝑥. So a-a simpler way of writing that is 𝑦 equals three minus 𝑥.

So to work out these 𝑦-coordinates from these 𝑥-coordinates, we use this equation. And to work out these 𝑦-coordinates from these 𝑥-coordinates we use this equation. So just tidying that up and zooming in a little bit so we’re ready for the next stage, this is what that diagram looks like. So this is the graph of 𝑦 equals the absolute value of 𝑥 minus three. So to the right of 𝑥 equals three, we’re using this equation to plug our 𝑥-coordinates into, to generate our 𝑦-coordinates. And to the left of 𝑥 equals three, this is the equation that we’re using to plug our 𝑥-values in, to generate the corresponding 𝑦-coordinates. That fact when 𝑥 equals three, whether you put it into the left-hand equation or the right-hand equation, you’ll still get a 𝑦-coordinate of zero. So it actually doesn’t matter which of those equations you use at 𝑥 equals three.

Then what I wanna do is sketch the curve of one over the absolute value of 𝑥 minus three. So I’m gonna take the-the 𝑦-coordinates that we got in 𝑦 equals the absolute value of 𝑥 minus three, and then I’m gonna do one over those values across the range of 𝑥-values that we’ve got. Now there’s a 𝑦-coordinate of one here. So it’s getting across here and there’s a 𝑦-coordinate of one here. And one over one is one, so those are gonna stay in the same place. Now as I increase my 𝑥-coordinates to the right of that, the corresponding 𝑦-coordinates are getting bigger and bigger and bigger. So I’m doing one over a bigger and bigger number. Now that’s gonna get closer and closer to zero. It’s always a- I mean we’re doing one over it, and one is a positive number, and the bigger numbers that we’re getting are all positive. So it’s gonna be positive divided by positive, so they’re always gonna be positive. That’s never quite gonna reach zero, certainly never gonna go negative. So what’s gonna happen here, is that these numbers are gonna get closer and closer. One over the-the orange line is gonna get closer and closer to zero. The 𝑥-axis is gonna be an asymptote there, and the curve is gonna get closer and closer to the 𝑥-axis.

And the same is true to the left of this left point here. This is just a mirror image of what we’ve seen on the right. As I move in this direction here, my 𝑦-coordinates are getting bigger and bigger and more positives. So one over them is gonna get smaller and smaller, closer to zero. So we’re gonna have that kind of asymptotic effect coming down here as well, on this side. So now we need to consider the points in between the two. Now as I move in this direction on the right-hand side towards 𝑥 equals three, the 𝑦-coordinates that I’m doing one over, on that orange line, are getting smaller and smaller and smaller. And if you do one over a smaller and smaller and smaller number, you’re gonna get a bigger number.

So the curve’s gonna look something like that. Obviously when 𝑥 equals three, the 𝑦-coordinate, the orange 𝑦-coordinate, was zero, one over zero. Well it’s not defined, but it’s kind of infinitely big. So we’re never actually going to reach that line. That’s gotta be an asymptote, 𝑥 equals three. And we’re gonna have a mirror image of that, just on the other side of it, so the same pattern as we come in from here, these 𝑦-coordinates are getting smaller and smaller and smaller. We’re doing one over a smaller and smaller number, so this part of the curve is gonna get bigger and bigger and bigger. And again, we’re gonna have this kind of asymptotic effect going up to 𝑥 equals three.

So now we know what that curve is gonna look like. Let’s try to look out what equations we’re going to be using to the right and to the left of 𝑥 equals three. So to get those green curves in this case, we’re doing one over 𝑥 minus three and to the left we were doing one over three minus 𝑥. So although the curve is actually 𝑦 equals one over the absolute value of 𝑥 minus three, we’re actually practically using two different equations to generate the 𝑦-coordinates for those two different regions of the curve. So now we want to go on and actually solve the inequality that we were given in the question. And we’ve gotta have — the 𝑦-coordinate of that graph has to be greater than two. So I’ve drawn in the line 𝑦 equals two and tidied things up a little bit.

So to put that in another way, we’re looking for the points where the 𝑦-coordinates of this are greater or above the 𝑦-coordinates for this. And because it’s a strict inequality, we don’t want them to be equal we just want them to be greater than, so we’re looking at these points here on our curve which are above that black line. Now what we’re gonna do is, we’re gonna analyse the points where they’re equal, and then find out the 𝑥-coordinates of those points. And then the 𝑥-coordinates between the two, are the 𝑥-coordinates that are generating these 𝑦-coordinates that we’re interested in, the region that we’re interested in, the solution to that inequality. Of course, the one thing we do need to be careful of, we said 𝑥 equals three was not defined. So in fact, well there’s gonna be a little gap in our set of answers in the middle here, at 𝑥 equals three.

So let’s consider the left-hand point first. This is the equation that we’re using to generate the-the 𝑦-coordinates on that green curve. And we’re saying in this particular case, the 𝑦-coordinate is equal to two. So we’re saying that two is equal to one over three minus 𝑥. So multiplying both sides by three minus 𝑥, in order to get rid of that from the denominator, we’ve got two lots of three minus 𝑥 is equal to one. And now I’m gonna divide both sides by two, and add 𝑥 to both sides, and finally subtract a half from both sides. So the 𝑥-coordinate of that left-hand point that we’re interested in there, is two and a half.

Now let’s do the same for the right-hand point there. The 𝑦-coordinate is two because it’s on that black line. And the equation that we’re using for the green line is this. So we’re saying two is equal to one over 𝑥 minus three. So again, rearranging that equation, and solving for 𝑥, gives an 𝑥-coordinate of three and a half, for that right-hand point.

Right. Let’s zoom in on this area here, so that we can see what’s going on a little bit more clearly. So the question said that one over the absolute value of 𝑥 minus three, so the 𝑦-coordinate of this, has to be strictly greater than two, so a 𝑦-coordinate of this. Now we just identified this point and this point here, where the 𝑦-coordinates of those two things are equal. So really, it’s these points up here and these points up here, for which this is greater than this. That’s the region we’re looking for. So the 𝑥-values that generate those points are not equal to two and a half, not equal to three and a half, but they’re equal to everything in between. But remember, we have to make a slight adjustment because 𝑥 equals three was not defined. So we’ll remove that from our solution set.

So looking at this region first, that’s two and a half is less than 𝑥 is less than three. So they’re strict inequalities there. And then considering this region, we’ve got three is less than 𝑥 is less than three and a half. So expressing our solution as inequalities, we’ve got a pair of inequalities, it’s a noncontinuous region, so we have to express it as two separate inequalities.

Now we might also write that as a pair of intervals. And the critical values there will be two and a half, and three, and three, and three and a half. So none of those values are actually included in the region. So we’re gonna have to put round parentheses around each of them. And it’s the union of those two intervals, so we can put the union sign in between. So that’s another way of expressing that answer.

And one more way of looking at that is, we could-we could describe this region here, this interval, as one interval. But we could just subtract the value in the middle in set notation of three. So-so our solution will be the interval from two and a half to three and a half minus the one point in the set 𝑥 equals three.

So just to summarise the process then, the first stage was to go through a series of sketches, 𝑦 equals 𝑥 minus three, 𝑦 equals the absolute value of 𝑥 minus three, and 𝑦 equals one over the absolute value of 𝑥 minus three. We’ve built up this graph of what that function looks like. We also tracked which would be the equations that we would use in each region of that graph to generate the 𝑦-coordinates. We then plotted 𝑦 equals two, because that was the right-hand side of our inequality. And then we had to analyse the regions of the graph which-which areas of the graph had the green bits higher than the black bits, and so on. And then we looked at which were the corresponding 𝑥-coordinates that generated those 𝑦-coordinates that-that met the criteria in the inequality. Lastly, we had to remember that because of the particular function that we had in-in the first place, 𝑥 equals three wasn’t defined, so we had to remove that from our solution set.

So hopefully going through that example there has helped you to understand the steps and understand some of the pitfalls that you can fall into, if you do things wrong. And overall, you’ll have a much better idea of how to solve absolute value nonlinear inequalities.