Question Video: Determining If a Rational Function Is Continuous at a Certain Point Mathematics • Higher Education

Video Transcript

Given that 𝑓 of 𝑥 is equal to six 𝑥 squared minus 27𝑥 plus 27 all divided by six 𝑥 minus nine, if possible or necessary, define 𝑓 evaluated at three over two so that 𝑓 is continuous at 𝑥 is equal to three over two.

The question gives us a rational function 𝑓 of 𝑥, and it wants us to define 𝑓 at three over two so that 𝑓 is continuous when 𝑥 is equal to three over two, only if this is possible or necessary. We recall we call a function 𝑓 continuous at the point 𝑥 is equal to 𝑎 if it satisfies the following three conditions.

First, 𝑓 of 𝑥 must be defined when 𝑥 is equal to 𝑎. This is the same as saying that 𝑥 is in the domain of our function 𝑓. Second, we must have that the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 exists. This condition is the same as saying the limit as 𝑥 approaches 𝑎 from the right of 𝑓 of 𝑥 and the limit as 𝑥 approaches 𝑎 from the left of 𝑓 of 𝑥 both exist and are equal. Thirdly, we need the limit as 𝑥 approaches 𝑎 of 𝑓 of 𝑥 is equal to 𝑓 evaluated at 𝑎.

Since the question wants us to make the function 𝑓 of 𝑥 continuous, we need all three of these conditions to be true. Since we want the continuity at 𝑥 is equal to three over two, we’ll set 𝑎 in our definition of continuity to be three over two.

The first condition we need is that 𝑓 evaluated at three over two is defined. At First, we might be worried. If we substitute 𝑥 is equal to three over two into our definition of 𝑓 of 𝑥, we’ll end up with the indeterminate form of zero divided by zero. However, the question tells us if it’s possible or necessary, we can define the function 𝑓 of 𝑥 at three over two. So, this tells us we’re going to need to define our function 𝑓 at three over two. This would then make our first condition for continuity true.

Since we still need our function 𝑓 to be continuous at 𝑥 is equal to three over two, we still need both of our other continuity conditions to be true. Let’s check the limit as 𝑥 approaches three over two of 𝑓 of 𝑥 exists. To do this, we’ll check the limit as 𝑥 approaches three over two from the right and the limit as 𝑥 approaches three over two from the left of 𝑓 of 𝑥. We’ll start with the limit as 𝑥 approaches three over two from the right.

Since we’re taking the limit of a rational function, we might be tempted to try direct substitution at this point. However, we’ve already shown that 𝑓 evaluated at three over two gives us the indeterminate form zero over zero. In fact, we’ve shown three over two is the root of the polynomial in our numerator and it’s the root of the linear function in our denominator. So, by the factor theorem, 𝑥 minus three over two divides both our numerator and our denominator. So, by the factor theorem, we can take a factor of 𝑥 minus three over two outside of our numerator and our denominator.

In fact, we can simplify this slightly by multiplying through by two. This factor is the same as two 𝑥 minus three. In our numerator, the leading term is six 𝑥 squared. So, two 𝑥 multiplied by the leading factor in our second term will give us six 𝑥 squared. That means it’s three 𝑥. And the constant term in our numerator is 27. So, negative three multiplied by the constant term in our second term gives us 27. That means it’s negative nine. And we see that two 𝑥 minus three multiplied by three gives us six 𝑥 minus nine.

Since our numerator and our denominator share a factor of two 𝑥 minus three, we can cancel these shared factors out. We can also divide through by three. This gives us the limit as 𝑥 approaches three over two from the right of 𝑥 minus three. And we can do this because both functions inside of our limit agree everywhere except when 𝑥 is equal to three over two. And when we’re calculating the limit as 𝑥 approaches three over two, we don’t care what happens when 𝑥 is equal to three over two. We only worry about what happens when 𝑥 is near three over two.

Finally, we can evaluate the limit of a polynomial by using direct substitution. Substituting 𝑥 is equal to three over two gives us three over two minus three, which is equal to negative three over two. So, we’ve shown the limit as 𝑥 approaches three over two from the right of our function 𝑓 of 𝑥 is equal to negative three over two. We now need to check that the limit as 𝑥 approaches three over two from the left of 𝑓 of 𝑥 exists and is equal to negative three over two.

If we check our working for the limit as 𝑥 approaches three over two from the right, we’ll notice that at no point did we actually use the fact that 𝑥 is approaching three over two from the right. If we’d had the limit as 𝑥 approaches three over two from the left, all of our lines of working would’ve been the same. This means the limit as 𝑥 approaches three over two from the left is also equal to negative three over two. Therefore, our second condition for continuity is true.

The last thing we need for 𝑓 to be continuous at 𝑥 is equal to three over two is the limit as 𝑥 approaches three over two of 𝑓 of 𝑥 is equal to 𝑓 evaluated at three over two. When we showed the limit as 𝑥 approaches three over two from the left and the right were both equal to negative three over two, this is the same as saying the limit as 𝑥 approaches three over two of 𝑓 of 𝑥 is equal to negative three over two. And since we need 𝑓 evaluated at three over two to be equal to this, we need to define 𝑓 of three over two as negative three over two.

Therefore, if we define 𝑓 evaluated at three over two to be equal to negative three over two, we’ve shown that our third continuity condition is also true. Therefore, we’ve shown defining 𝑓 of three over two to be equal to negative three over two makes the function 𝑓 continuous when 𝑥 is equal to three over two. To help us see why, let’s sketch a graph of our function 𝑓 of 𝑥.

We showed, using the factor theorem, that our function 𝑓 of 𝑥 is equal to 𝑥 minus three everywhere except when 𝑥 is equal to three over two. This means the curve 𝑦 is equal to 𝑓 of 𝑥 is the same as the curve 𝑦 is equal to 𝑥 minus three, except we removed the point when 𝑥 is equal to three over two because our function 𝑓 is not defined here. This gives us the straight line 𝑦 is equal to 𝑥 minus three with the point 𝑥 is equal to three over two removed. We represent this by using a hollow circle.

We then see by defining 𝑓 of three over two to be equal to negative three over two, we complete the line. This gives us a function which is exactly the same as 𝑥 minus three. Since we’ve shown we can redefine the function at a single point to remove this discontinuity, we call this type of discontinuity a removable discontinuity. Therefore, we’ve shown that the function 𝑓 of 𝑥 is equal to six 𝑥 squared minus 27𝑥 plus 27 over six 𝑥 minus nine can be made continuous when 𝑥 is equal to three over two by defining 𝑓 of three over two to be equal to negative three over two.