Video: Redefining a Given Function to Make It Continuous

Given 𝑓(π‘₯) = (π‘₯Β² + π‘₯ βˆ’ 2)/(π‘₯ βˆ’ 1), if possible or necessary, define 𝑓(1) so that 𝑓 is continuous at π‘₯ = 1.

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Video Transcript

Given 𝑓 of π‘₯ equals π‘₯ squared plus π‘₯ minus two all over π‘₯ minus one, if possible or necessary, define 𝑓 of one so that 𝑓 is continuous at π‘₯ equals one.

So we have a function 𝑓, which is a rational function. And we’d like 𝑓 to be continuous at the point π‘₯ equals one. And we’re told that we should define 𝑓 of one to make this so, but only if it is possible or necessary. What does that mean? Well, if 𝑓 is already continuous at the point π‘₯ equals one, then it isn’t necessary to define 𝑓 of one to make this so. It’s already been done for us. On the other hand, 𝑓 could be discontinuous at the point π‘₯ equals one. But in the way that it’s not possible to things just by defining 𝑓 of one. There could be some bigger issue stopping 𝑓 from being continuous at this point.

Let’s first check whether it is necessary to define 𝑓 of one to make 𝑓 continuous at π‘₯ equals one. Is 𝑓 already continuous at π‘₯ equals one? Well, we have a checklist which allows us to find out whether a function is continuous at a certain point. 𝑓 must be defined at that point. So in our case, we need to check that 𝑓 of one is defined. And the limit of 𝑓 of π‘₯ as π‘₯ approaches that point, in our case one, must exist. And finally, these two values must be equal.

Let’s start at the top of the list. Is 𝑓 of one defined? Well, we’ll use the definition of 𝑓 of π‘₯ that we’re given in the question. Substituting one for π‘₯ gives one squared plus one minus two all over one minus one. In the numerator, one squared plus one is two. And subtracting the two gives us zero. And in the denominator, one minus one is zero. So we get the indeterminate form zero over zero. Zero over zero and hence 𝑓 of one are not defined. And so our function 𝑓 is not already continuous at π‘₯ equals one.

This isn’t necessarily a massive issue. After all, our task is to define 𝑓 of one. If this is the only thing stopping 𝑓 from being continuous at π‘₯ equals one, then we can just define 𝑓 of one. And 𝑓 will be continuous at this point, as required. We need to check then that there aren’t any other issues. We need the limit of 𝑓 of π‘₯ as π‘₯ approaches one to exist. And does it? We use the definition of 𝑓 of π‘₯ from the question. And of course, we know that direct substitution is going to give us an indeterminate form. So there must be some other way to evaluate this limit.

Well, the factor theorem tells us that as both the numerator and the denominator are zero when π‘₯ is one, both numerator and denominator must have a factor of π‘₯ minus one. And indeed, we can factor the numerator to π‘₯ plus two times π‘₯ minus one. Doing so allows us to cancel the common factor of π‘₯ minus one in the numerator and denominator. To get the limit of π‘₯ plus two as π‘₯ approaches one. This is a limit that can be evaluated using direct substitution. Substituting one for π‘₯, we get one plus two, which is three. So yes, the limit of 𝑓 of π‘₯ as π‘₯ approaches one does exist. And it equals three.

Now the last thing on our checklist is that the limit of 𝑓 of π‘₯ as π‘₯ approaches one must be 𝑓 of one. We found that the left-hand side, the limit of 𝑓 of π‘₯ as π‘₯ approaches one, is three. But the right-hand side 𝑓 of one is undefined. But our task is to define 𝑓 of one. If we define 𝑓 of one to be three, that’s the value of the limit of 𝑓 of π‘₯ as π‘₯ approaches one. Then our third item on the checklist will be satisfied. And of course, defining 𝑓 of one to be three also sorts out the first item on our checklist. 𝑓 of one is now defined. So defining 𝑓 of one to be three makes 𝑓 continuous at π‘₯ equals one. Because 𝑓 of one is now defined, the limit of 𝑓 of π‘₯ as π‘₯ approaches one, as we saw, is three. And now that we’ve defined 𝑓 of one to also be three, these two values are equal.

It might be helpful to look at the graph of 𝑓 of π‘₯ to see what we’ve done. As we saw with π‘₯ not equal to one, 𝑓 of π‘₯ is just π‘₯ plus two. And so the graph of 𝑓 of π‘₯ is just the straight line graph of 𝑦 equals π‘₯ plus two, with this hole here when π‘₯ is one. This hole in the graph comes because 𝑓 of one is not defined. And as a result, 𝑓 is not continuous at π‘₯ equals one. There is then this link between the technical definition of continuity and our intuitive understanding of what continuity means. There should be no gaps. We plug this gap and make the function continuous by defining 𝑓 of one to be three. Now, there’s no hole in the graph. And 𝑓 is continuous at π‘₯ equals one.

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