Question Video: Defining the Value of π‘₯ that Makes a Rational Function 𝑓(π‘₯) Continuous

Given 𝑓(π‘₯) = (π‘₯Β² βˆ’ 16)/(π‘₯ βˆ’ 4), if possible, define 𝑓(π‘₯) so that 𝑓 is continuous at π‘₯ = 4.

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Video Transcript

Given that 𝑓 of π‘₯ is equal to π‘₯ squared minus 16 divided by π‘₯ minus four, if possible, define 𝑓 of π‘₯ so that 𝑓 is continuous at π‘₯ is equal to four.

The question gives us a rational function 𝑓 of π‘₯. And it wants us to attempt to define 𝑓 when π‘₯ is equal to four so that our function 𝑓 will be continuous when π‘₯ is equal to four.

Let’s start by recalling what it means for a function to be continuous when π‘₯ is equal to four. We say that a function 𝑔 is continuous at π‘₯ is equal to four if the following three conditions hold.

First, 𝑔 evaluated at π‘₯ is equal to four must be defined. This is the same as saying that four is in the domain of our function 𝑔. Second, we need that the limit as π‘₯ approaches four of 𝑔 of π‘₯ must exist. And it’s worth noting this is the same as saying the limit as π‘₯ approaches four from the left of 𝑔 of π‘₯ and the limit as π‘₯ approaches four from the right of 𝑔 of π‘₯ both exist and are equal. Thirdly, we need the limit as π‘₯ approaches four of 𝑔 of π‘₯ is equal to 𝑔 evaluated at four.

So let’s go through these three conditions for our function 𝑓 of π‘₯ is equal to π‘₯ squared minus 16 divided by π‘₯ minus four. Let’s start with our first condition. We need 𝑓 evaluated at four to be defined. To check this, we know that 𝑓 of π‘₯ is equal to π‘₯ squared minus 16 over π‘₯ minus four. So we substitute π‘₯ is equal to four. This gives us four squared minus 16 divided by four minus four. And if we try to calculate this, we get zero divided by zero. So our function 𝑓 of π‘₯ is not defined when π‘₯ is equal to four.

However, we don’t need to worry too much about this because the question says we’re allowed to define 𝑓 of π‘₯ when π‘₯ is equal to four to try to make it continuous. So let’s move on to the second part of our continuity condition.

We need the limit as π‘₯ approaches four of our function 𝑓 of π‘₯ to exist. We could attempt to evaluate this limit directly, for example, by using direct substitution, since our function is a rational function. However, we already showed that 𝑓 evaluated at four gave the indeterminate form zero divided by zero.

Let’s try to evaluate this limit by using our left-hand and right-hand limit method. We’ll start by calculating the limit as π‘₯ approaches four from the left of 𝑓 of π‘₯. We see this is equal to the limit as π‘₯ approaches four from the left of π‘₯ squared minus 16 divided by π‘₯ minus four. We see that our numerator of π‘₯ squared minus 16 is a difference between squares. This means we can factor our numerator. This gives us the limit as π‘₯ approaches four from the left of π‘₯ minus four times π‘₯ plus four all divided by π‘₯ minus four.

Since we’re taking the limit as π‘₯ approaches four from the left, we must have that π‘₯ is less than four. This means that π‘₯ is never equal to four. And we can cancel the shared factor of π‘₯ minus four in our numerator and our denominator. This gives us the limit as π‘₯ approaches four from the left of π‘₯ plus four. And this is a linear function, so we can evaluate this by using direct substitution.

Substituting π‘₯ is equal to four gives us four plus four, which is eight. So we’ve shown the limit as π‘₯ approaches four from the left exists. It’s equal to eight.

We now need to show the limit as π‘₯ approaches four from the right of 𝑓 of π‘₯ exists and is also equal to eight. Let’s look at our working out for the left-hand limit. And consider what would change if instead we were calculating the limit as π‘₯ approach four from the right. We could still write 𝑓 of π‘₯ as the rational function π‘₯ squared minus 16 divided by π‘₯ minus four. We could still use a difference between squares to factor our numerator.

However, instead, we would now have that π‘₯ is approaching four from the right. So π‘₯ is greater than four. It’s still not equal to zero. So we can still justify cancelling the shared factor of π‘₯ minus four in the numerator and the denominator. And we can still evaluate this by using direct substitution.

So we see that the limit as π‘₯ approaches four from the right of 𝑓 of π‘₯ is also equal to eight. So our second condition for continuity is true. The limit as π‘₯ approaches four of 𝑓 of π‘₯ is equal to eight.

Let’s now consider our third condition for continuity. We need the limit as π‘₯ approaches four of 𝑓 of π‘₯ to be equal to 𝑓 evaluated at four. Since we’ve shown that the left-hand and right-hand limit as π‘₯ approaches four of 𝑓 of π‘₯ is equal to eight. The limit as π‘₯ approaches four of 𝑓 of π‘₯ must also be equal to eight. And if we want 𝑓 of π‘₯ to be continuous, this must be equal to 𝑓 evaluated at four.

So if we define 𝑓 evaluated at four to be eight, then we have the limit as π‘₯ approaches four of 𝑓 of π‘₯ is equal to 𝑓 evaluated at four. They’re both equal to eight. And since we’re now defining 𝑓 evaluated at four to be equal to eight, the first part of our continuity condition is also true.

Therefore, we’ve shown if we define 𝑓 of π‘₯ to be equal to eight when π‘₯ is equal to four. Then our function 𝑓 of π‘₯ is equal to π‘₯ squared minus 16 divided by π‘₯ minus four will be continuous when π‘₯ is equal to four.

To help us see what’s happening, let’s sketch a graph of our function 𝑓 of π‘₯. When we were calculating our left- and right-hand limit of 𝑓 of π‘₯, we already reasoned that 𝑓 of π‘₯ is equal to the linear function π‘₯ plus four when π‘₯ is not equal to four. We did this by factoring and then cancelling the shared factor of π‘₯ minus four in our numerator and denominator. And if our function 𝑓 of π‘₯ is equal to π‘₯ plus four everywhere except when π‘₯ is equal to four, we can use this to sketch our graph.

To sketch the curve 𝑦 is equal to 𝑓 of π‘₯, we sketch the straight line 𝑦 is equal to π‘₯ plus four except we remove the point when π‘₯ is equal to four since our function 𝑓 of π‘₯ is undefined. We represent this by a hollow circle. However, we can find the coordinates of this hollow circle since it lies on the line 𝑦 is equal to π‘₯ plus four. Its π‘₯-coordinate is four and its 𝑦-coordinate is eight.

We can then look at what happens to our outputs 𝑓 of π‘₯ as π‘₯ approaches four from the left and from the right. As π‘₯ is approaching four from the left, we see that 𝑓 of π‘₯ is getting closer and closer to eight. And as π‘₯ is approaching four from the right, we see that 𝑓 of π‘₯ is still getting closer and closer to eight. This is why defining our function 𝑓 of π‘₯ to be equal to eight when π‘₯ is equal to four will make the function continuous.

In conclusion, we’ve shown we can make our function 𝑓 of π‘₯ is equal to π‘₯ squared minus 16 divided by π‘₯ minus four continuous at the point where π‘₯ is equal to four by defining 𝑓 of four to be equal to eight.

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