Video: Inequalities Using Rational Functions

What are all the values of π‘₯ for which it is true that (π‘₯ + 3)/(π‘₯ βˆ’ 1) > 0?

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Video Transcript

What are all the values of π‘₯ for which it is true that π‘₯ plus three over π‘₯ minus one is greater than zero? So let’s call the left-hand side of this inequality 𝑓 of π‘₯ and sketch a graph of 𝑓 of π‘₯.

We can use five steps to sketch the graph of a rational function like 𝑓 of π‘₯. We find the π‘₯-intercepts, the vertical asymptotes, the 𝑦-intercept. We consider the behavior as π‘₯ tends to positive or negative ∞. And then once we have all those features of the graph, we can connect the dots. The first step is to find the π‘₯-intercepts, and we do that by solving 𝑓 of π‘₯ is equal to zero.

We use the expression we have for 𝑓 of π‘₯, and now we can multiply both sides by π‘₯ minus one, and finally conclude that π‘₯ must be equal to negative three. This makes sense. The π‘₯-intercepts of the graph of the function occur when the function is zero. And of course this happens for a rational function when the numerator of that function is zero, so when π‘₯ plus three is equal to zero or π‘₯ is equal to negative three. We mark the π‘₯-intercept on the graph and move on to the second step.

The graph of a rational function has a vertical asymptote and values of π‘₯ for which the function itself is undefined. And that of course happens when the denominator is zero, because we get some number divided by zero which is of course undefined. That’s the only way we can get an undefined value for a rational function.

In our case, the denominator of our rational function is π‘₯ minus one, and we set this equal to zero. And we can see that we have exactly one vertical asymptote at π‘₯ equals one. So we mark that on our graph, and we’re done with the second step and can move onto the third.

To find the 𝑦-intercept, we need to find the value of the function at π‘₯ equals zero. 𝑓 of π‘₯ is π‘₯ plus three over π‘₯ minus one. So 𝑓 of zero is zero plus three over zero minus one, which is three over negative one or just negative three. So we can mark the 𝑦-intercept on our graph and continue to the fourth step.

We have to think about the behaviour as π‘₯ tends to plus or minus ∞. We can’t really ask what 𝑓 of ∞ is, but we can pick a really big number, like a trillion, and think about what’s the value of the function should be there. So 𝑓 of a trillion which is 𝑓 of 10 to the power of 12 is 10 to the power of 12 plus three over 10 to the power of 12 minus one. The numerator and denominator are both about a trillion, and so their ratio is about one. You can confirm this using a calculator if you’d like.

There wasn’t anything particularly special about a trillion. We just picked a very big number to see what would happen as π‘₯ tends to ∞. And we’ve seen that as π‘₯ tends to ∞, 𝑓 of π‘₯ becomes roughly one. And this means that there is a horizontal asymptote of 𝑦 equals one on the right-hand side of the graph. The graph of 𝑓 of π‘₯ approaches that horizontal asymptote as π‘₯ tends to positive ∞, but we still need to find which direction it approaches it from. Does it approach it from above or below?

Looking at the fraction again, we see that the numerator is greater than the denominator by four as it happens. And so the value of the fraction is greater than one. So as π‘₯ grows very large, as π‘₯ tends to ∞; in other words, 𝑓 of π‘₯ is greater than one. And so the graph of 𝑓 of π‘₯ is above the horizontal asymptote at 𝑦 equals one.

So we can sketch in a bit of the graph at the right-hand side of the π‘₯-axis. And a similar thing happens as π‘₯ tends to negative ∞, except now we have 𝑓 of π‘₯ being less than one. And so the asymptote is approached from below. You have to be quite careful to make sure that this value is indeed less than one, because we’re dividing two negative numbers. But you can confirm it either by hand or using a calculator that the asymptote is approached from below, and we sketch this in.

An alternative method is to use polynomial long division. Using this, we can write 𝑓 of π‘₯ as one plus four over π‘₯ minus one. As π‘₯ tends to either positive ∞ or negative ∞, this term four over π‘₯ minus one gets smaller and smaller, closer and closer to zero. And so 𝑓 of π‘₯ gets closer and closer to one.

And with a bit of thinking, we can work out what the sign of this term is as well, which will tell us which direction the asymptote is being approached from. Whatever method we use, we get the same result; that as π‘₯ tends to positive ∞, the graph of 𝑓 of π‘₯ approaches the horizontal asymptote at 𝑦 equals one from above. And as π‘₯ tends to negative ∞, it approaches the same horizontal asymptote at 𝑦 equals one from below.

Now that we have all the features marked on the graph, we can connect the dots. To the left of the vertical asymptote, we have to go through the π‘₯-intercept and negative three and through the 𝑦-intercept and negative three and approach that vertical asymptote.

To the right of the vertical asymptote, β€” because we can’t cross the π‘₯-axis. We know the only π‘₯-intercept is at negative three β€” we have to go up to approach the vertical asymptote at π‘₯ equals one. So we’ve sketched the graph of 𝑓 of π‘₯ using a process which works for basically any rational function.

You might prefer to think of 𝑓 of π‘₯ as being one plus four over π‘₯ minus one, and think of its graph as being a transformation of the graph of 𝑓 of π‘₯ equals one over π‘₯, the reciprocal function. If you do this and hopefully you will get the same result, but the five step sketching process works for any rational function. And so I think it’s a good idea to use that in general or at least to be comfortable using it.

Now that we’ve sketched the graph, we can go back to think about the inequality. We want to solve 𝑓 of π‘₯ is greater than zero. What does this mean in terms of our graph? Well we wanted to find all the values of π‘₯ for which the graph of 𝑓 of π‘₯ lies above the π‘₯-axis.

Looking at the graph, we can see that there are two regions where this happens, one is to the left of negative three and one is to the right of one on the π‘₯-axis. The region on the left-hand side of the graph is either π‘₯ is less than negative three or π‘₯ is less than or equal to negative three. We have to think carefully about whether negative three should be included as one of the values of π‘₯ for which it is true that the inequality holds.

Well of course being the π‘₯-intercept, 𝑓 of negative three is equal to zero and so not greater than zero. And so this region should not contain the endpoint negative three, so the inequality should just be π‘₯ is less than negative three.

A similar thing happens to the right of the vertical asymptote; π‘₯ equals one. The region is either π‘₯ is greater than one or π‘₯ is greater than or equal to one depending on whether one should be included. 𝑓 of one is undefined and so not greater than zero, and so π‘₯ equals one should not be included in the region. And the region is just π‘₯ is greater than one. We’ve looked at the graph and we’ve seen the two regions for which the inequality holds; π‘₯ is less than negative three or π‘₯ is greater than one.

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