Video: Inequalities Involving Rational Functions

Solve the inequality (π‘₯ βˆ’ 1)/((π‘₯ + 1)(π‘₯ βˆ’ 3)) ≀ 1/3.

11:29

Video Transcript

Solve the inequality π‘₯ minus one over π‘₯ plus one times π‘₯ minus three is less than or equal to a third.

Let’s call the algebraic fraction on the left-hand side of our inequality, π‘₯ minus one over π‘₯ plus one times π‘₯ minus three, 𝑓 of π‘₯. And let’s sketch the graph of 𝑓 of π‘₯. 𝑓 of π‘₯ is a rational function, and so there’re about five steps to sketching its graph. First, we find the π‘₯-intercept, then the vertical asymptotes, then the 𝑦-intercept, then we think about the behavior as π‘₯ tends to plus and minus infinity. And then once we have all those points, we connect the dots.

First, we find the π‘₯-intercepts. They occur when 𝑓 of π‘₯ is equal to zero; so that’s when π‘₯ minus one over π‘₯ plus one times π‘₯ minus three is equal to zero. If we multiply by the denominator of the function, we get just π‘₯ minus one is equal to zero. And so the only π‘₯-intercept occurs at π‘₯ equals one which we mark on the graph.

So that the π‘₯-intercept sorted out, we can move on now to the second step which is to find the vertical asymptotes of the graph. The vertical asymptotes of the graph will occur when the denominator of the function is zero. And we’re lucky that the denominator is factorized, and so we can just read off the solutions. These solutions are π‘₯ equals negative one and π‘₯ equals three. So we mark these asymptotes on our graph, and that’s the second step completed.

Now it’s time to find the 𝑦-intercept. We do this by finding 𝑓 of zero, by substituting zero in to the expression we have for 𝑓 of π‘₯. This gives us zero minus one over zero plus one times zero minus three which is negative one over negative three which is a third. So we mark that point, the 𝑦-intercept that is, on the 𝑦-axis.

And now we continue to the behavior as π‘₯ tends to positive or negative infinity. In place of infinity, I will choose 10 to the power of 12 which is a trillion which is the biggest number I can think of. So we substitute a trillion in to the expression we have for 𝑓 of π‘₯, and we notice that it is greater than zero. Why is this? Well, the numerator is a trillion minus one which is clearly positive. Likewise, both factors of the denominator are clearly positive, they’re both are around a trillion. As those two numbers are positive, certainly their product is positive. And so we have a positive number divided by a positive number which is of course a positive number.

The other thing to notice is that it is only just positive. You can see this by putting this expression into a calculator. Alternatively, we can do some approximation. The numerator is about 10 to the power of 12, and the denominator is about 10 to the power of 12 times 10 to the power of 12. So this is about 10 to the power of 12 over 10 to the power of 24 which is 10 to the power of negative 12. So 𝑓 of a trillion is about a trillionth.

Now remember, there’s nothing particularly special about a trillion in this context. It’s just a really really big number that we’re using instead of infinity to work out what happens as π‘₯ tends to infinity. And the conclusion should be that as π‘₯ tends to infinity, 𝑓 of π‘₯ is positive but very small and getting smaller and smaller as π‘₯ tends to infinity, as π‘₯ gets bigger and bigger.

Another way of saying this, is that there is a horizontal asymptote at 𝑦 equals zero. Whatever terminology we use, the effect is the same. As π‘₯ is very large, the graph of 𝑓 of π‘₯ will be just above the π‘₯-axis and getting closer and closer to it. We can do exactly the same thing with negative a trillion, and we’ll get that 𝑓 of negative a trillion is negative a trillionth, or there about. And so as π‘₯ tends to negative infinity, the graph of 𝑓 of π‘₯ is just below the π‘₯-axis, and as π‘₯ gets more and more negative, the graph gets closer and closer to the π‘₯-axis because 𝑓 of π‘₯ gets closer and closer to zero.

So now that we’ve got all these points, we can connect the dots. Firstly in this region, just to the left of the asymptote at π‘₯ equals negative one. We know that the only time that the graph of 𝑓 of π‘₯ can cross the π‘₯-axis is at the π‘₯-intercept at one. And so the graph from across the π‘₯-axis in the region we’re looking at. And so there’s only one possible way that we can get an asymptote at negative one, and that’s if the graph continues down. We now look at the region to the right of the vertical asymptote at π‘₯ equals three, and we can have the same realisation. The graph can’t cross the π‘₯-axis in this region, and so it must continue up to get closer and closer to that vertical asymptote, like so.

And now, there’s just the region between the two vertical asymptotes to think about. First, we think about what happens near the vertical asymptote at π‘₯ equals negative one. The root of negative one in the denominator of the function is not repeated. To the left of the vertical asymptote, the curve goes down. And so to the right of the vertical asymptote, it must come down from above. And the same is true near the asymptote at π‘₯ equals three. To the right of the asymptote, the graph comes from the top. So to the left of it, it must go down to the bottom.

Now I think we really are ready to join all the dots including the π‘₯- and 𝑦-intercepts. So that’s all five steps of sketching the graph. And now we’re ready to consider the inequality.

First, let’s draw the line 𝑦 equals a third. What does this inequality mean graphically? Well, it means finding the values of π‘₯ for which the graph of 𝑓 of π‘₯ is below this line. And it looks like there are basically three regions where this happens. One of these is to the left of the vertical asymptote at π‘₯ equals negative one. And we have to decide whether this should be π‘₯ is less than negative one, or π‘₯ is less than or equal to negative one. So really the question is: Is negative one included in the solution? Well, no. 𝑓 of negative one is undefined, which is not less than or equal to a third. It’s not even a real number, it doesn’t make sense to ask.

And so the first region is: π‘₯ is less than negative one. And we can write that in interval notation as: open parentheses, negative infinity, comma, negative one, close parentheses; where the fact that we’re using these parentheses instead of square brackets tells us that neither end point, neither negative infinity nor negative one, is included in this interval.

Now we can move on to the second interval. We are very lucky that we can immediately just read off the left-hand endpoint of the interval. Normally, we’d have to solve 𝑓 of π‘₯ equals a third to find the left-hand endpoints. But we noticed that the 𝑦-intercept is a third, and so 𝑓 of zero is a third. And so the left-hand endpoint of this interval is zero. Of course, this endpoint is included because 𝑓 of zero is a third, and so it satisfies the inequality. On the right-hand side of the interval, as with the first interval, we have an asymptote. This time the asymptote is at π‘₯ equals three. And so on the right-hand, our right-hand endpoint is three. And this is not included, again for the same reason as the endpoint of the first interval was not included. And again, we can write this in interval notation. So it’s: open square bracket, zero, comma, three, close parenthesis; where the open square brackets tells us that the zero is included, and the close parenthesis tells us that the three is not included in the interval.

And finally, the third interval. This looks to be in the form of π‘₯ is greater than or equal to some number. And that some number is the π‘₯-coordinate of the intersection of the graph and the line. Of course, this number will satisfy the inequality, and so it is included. And so we have greater than or equal to instead of just greater than. To find what this number is, we have to solve 𝑓 of π‘₯ equals a third. So we use the definition of 𝑓 of π‘₯. We can multiply both sides by three times the denominator, and we can expand and simplify. If we subtract three π‘₯ minus three from both sides, we’ll get all the terms on the right-hand side. And we can easily factor the right-hand side because π‘₯ is a factor of both terms. So this equation has two solutions: π‘₯ equals zero which we stumbled upon luckily before, and π‘₯ equals five which we had to do some work for, which must be our question mark. So our third interval is given by π‘₯ is greater than or equal to five, which in interval notation is: open square bracket, five, comma, infinity, close parenthesis.

We have three intervals on which this inequality holds. Any solution to this inequality will be in one of those three intervals. And so the solution, overall, is the union of those intervals. This is our final answer; the union of three intervals.

Let’s just recap what we’ve done. We called the left-hand side of the inequality 𝑓 of π‘₯. We sketched a graph of 𝑓 of π‘₯. This along with the line 𝑦 equals a third help us identify the regions where the inequality holds. The endpoints of the intervals are either at asymptotes of the graph or at the π‘₯-coordinates of the points where the graph crosses the line. If the endpoint is as an asymptote, then that value is not included in the interval because the value of the function at the point where the asymptote occurs is undefined. However, when the endpoint of an interval is at the π‘₯-coordinate of a point of intersection of the line and the graph, then that endpoint is included in the interval. We found three intervals where the inequality holds, and so the solution was their union.

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