Video: Discussing the Continuity of the Composite Function of Two Piecewise-Defined Functions at a Given Point

Suppose 𝑓(π‘₯) = βˆ’6 if π‘₯ β‰  βˆ’3, 𝑓(π‘₯) = 6 if π‘₯ = βˆ’3, and 𝑔(π‘₯) = 9π‘₯ + 12 if π‘₯ β‰  βˆ’3, 𝑔(π‘₯) = 15 if π‘₯ = βˆ’3. What can be said of the continuity of 𝑔(𝑓(π‘₯)) at π‘₯ = βˆ’3? [A] The function is discontinuous at π‘₯ = βˆ’3 because lim_(π‘₯ β†’ βˆ’3) 𝑔(𝑓(π‘₯)) does not exist. [B] The function is continuous at π‘₯ = βˆ’3. [C] The function is discontinuous at π‘₯ = βˆ’3 because 𝑔(𝑓(βˆ’3)) is undefined. [D] The function is discontinuous at π‘₯ = βˆ’3 because lim_(π‘₯ β†’ βˆ’3) 𝑔(𝑓(π‘₯)) β‰  𝑔(𝑓(βˆ’3)).

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Video Transcript

Suppose 𝑓 of π‘₯ equals negative six, if π‘₯ is not equal to negative three and six, if π‘₯ equals negative three. And 𝑔 of π‘₯ equals nine π‘₯ plus 12, if π‘₯ is not equal to negative three and 15, if π‘₯ equals negative three. What can be said of the continuity of 𝑔 of 𝑓 of π‘₯ at π‘₯ equals negative three? We have four options here. Option A: The function is discontinuous at π‘₯ equals negative three because the limit of 𝑔 of 𝑓 of π‘₯, as π‘₯ approaches negative three, does not exist. Option B: The function is continuous at π‘₯ equals negative three. Option C: The function is discontinuous at π‘₯ equals negative three because 𝑔 of 𝑓 of negative three is undefined. And finally option D: The function is discontinuous at π‘₯ equals negative three because the limit of 𝑔 of 𝑓 of π‘₯, as π‘₯ approaches negative three, is not equal to 𝑔 of 𝑓 of negative three.

To summarise, we’re looking at the continuity of a composite function, 𝑔 of 𝑓 of π‘₯ at π‘₯ equals negative three. And there are two possibilities, either the function is continuous at π‘₯ equals negative three or it isn’t. But from the options we see that if the function is discontinuous, we have to explain why. It’s for one of three reasons.

So now let’s clear the options from our screen, so we have space to actually answer the question. We’re going to need the definition of continuity at a point. A function 𝑓 is said to be continuous at a number 𝑐, if the limit of 𝑓 of π‘₯, as π‘₯ approaches 𝑐, is just 𝑓 of 𝑐. The function in our problem is 𝑔 of 𝑓, or 𝑔 composed with 𝑓. And we’re trying to see if this is continuous at π‘₯ equals negative three. So 𝑐 is negative three. So asking if 𝑔 of 𝑓 of π‘₯ is continuous at π‘₯ equals negative three is the same as asking if the limit of 𝑔 of 𝑓 of π‘₯, as π‘₯ approaches negative three, is equal to 𝑔 of 𝑓 of negative three.

We note that there are essentially three different reasons why this equation could fail to hold. And hence, three reasons why our function could not be continuous at π‘₯ equals negative three. Firstly, it could happen that this limit doesn’t exist. Secondly, the value of 𝑔 of 𝑓 of negative three could be undefined. It could be that negative three is not in the domain of our function. Or thirdly, these two things could both have real values. But they could have different real values and so not be equal. And if you think back to the options of the question, these three cases were given as separate options. Of course, another possibility is that both the limit of 𝑔 of 𝑓 of π‘₯, as π‘₯ approaches negative three, and 𝑔 of 𝑓 of negative three are undefined.

Let’s start by trying to evaluate this limit. How do we do this? Well, this is the limit as π‘₯ approaches negative three. And so importantly, π‘₯ is never equal to negative three at any point. It’s just getting closer and closer to that value. And we know what 𝑓 of π‘₯ is when π‘₯ is not equal to negative three. We’re told that it is negative six. So this limit is the limit of 𝑔 of negative six, as π‘₯ approaches negative three. We now need to ask what is 𝑔 of negative six? We have the definition of 𝑔 of π‘₯. And negative six is not equal to negative three. So we can see how to evaluate 𝑔 of negative six. It’s nine times negative six plus 12. And evaluating this, we get negative 42. The limit of a constant function like negative 42 is just that constant. So our limit is negative 42. Let’s clear away our working and find 𝑔 of 𝑓 of negative three. We found the left-hand side of the equation that we need to be true for 𝑔 of 𝑓 of π‘₯ to be continuous at π‘₯ equals negative three. This limit exists and its value is negative 42.

Now we move on to finding the right-hand side of the equation. We want to find 𝑔 of 𝑓 of negative three. So our first step is to find 𝑓 of negative three. And we can see from the definition of 𝑓 in the question that 𝑓 of π‘₯ is six, if π‘₯ is equal to negative three. In other words, that 𝑓 of negative three is six. And so 𝑔 of 𝑓 of negative three is 𝑔 of six. Now what is 𝑔 of six? We look at the definition of 𝑔 of π‘₯. And we see that 𝑔 of π‘₯ is nine π‘₯ plus 12, if π‘₯ is not equal to negative three. And six is not equal to negative three. So 𝑔 of six is nine times six plus 12.

Now we’re ready to answer our question. Is the limit of 𝑔 of 𝑓 of π‘₯, as π‘₯ approaches negative three, equal to 𝑔 of 𝑓 of negative three? In other words, is the function 𝑔 of 𝑓 continuous at π‘₯ equals negative three? Well, no. The left-hand side is negative 42. And the right-hand side is 66. What are our conclusions? The function is discontinuous at π‘₯ equals negative three. And why? Because although the limit of 𝑔 of 𝑓 of π‘₯, as π‘₯ approaches negative three, exists and although 𝑔 of 𝑓 of negative three is defined, the two values are not equal. And we need these two quantities not only to exist or be defined, but also to be equal to satisfy the definition of continuity. This was option D in our list of options.

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