Video: Discussing the Continuity of a Piecewise Defined Function at a Point

Discuss the continuity of the function ๐‘“ at ๐‘ฅ = โˆ’2, given ๐‘“(๐‘ฅ) = (๐‘ฅยณ + 8)/(๐‘ฅยฒ โˆ’ 4) if ๐‘ฅ โ‰  โˆ’2 and ๐‘“(๐‘ฅ) = โˆ’3 if ๐‘ฅ = โˆ’2. [A] The function is discontinuous at ๐‘ฅ = โˆ’2 because ๐‘“(โˆ’2) โ‰  lim_(๐‘ฅโ†’โˆ’2) ๐‘“(๐‘ฅ). [B] The function is continuous at ๐‘ฅ = โˆ’2. [C] The function is discontinuous at ๐‘ฅ = โˆ’2 because ๐‘“(โˆ’2) is undefined. [D] The function is discontinuous at ๐‘ฅ = โˆ’2 because lim_(๐‘ฅโ†’โˆ’2) ๐‘“(๐‘ฅ) does not exist.

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Video Transcript

Discuss the continuity of the function ๐‘“ at ๐‘ฅ equals negative two, given ๐‘“ of ๐‘ฅ is equal to ๐‘ฅ cubed plus eight over ๐‘ฅ squared minus four if ๐‘ฅ is not equal to negative two and itโ€™s equal to negative three, if ๐‘ฅ is equal to negative two.

So there are four options here and Iโ€™m going to start with B which is that the function is continuous at ๐‘ฅ equals negative two. And the other three options are saying that the function is discontinuous for a variety of reasons. So A says itโ€™s discontinuous at that point because the value of the function ๐‘“ of negative two is not equal to the limit of that function as ๐‘ฅ tends to negative two. Option C says that ๐‘“ of negative two is undefined. And option D says itโ€™s discontinuous because the limit doesnโ€™t exist.

These four options give four different scenarios. And I think itโ€™s worth thinking about what the graph of ๐‘“ looks like near ๐‘ฅ equals negative two for each of them. So in option A, limit as ๐‘ฅ tends to negative two if ๐‘“ of ๐‘ฅ does exist, but it doesnโ€™t equal ๐‘“ of negative two. And we can see this scenario being illustrated in the diagram in the bottom, left-hand corner.

You can see that the limit as ๐‘ฅ tends to negative two does exist. Iโ€™ll pour it on; there it is; itโ€™s somewhere maybe around minus 1.5. But unfortunately, it doesnโ€™t match the value of the function when ๐‘ฅ is negative two, which is minus three. So thatโ€™s one thing that could go wrong, one thing that could cause the function to be discontinuous.

Option B is very similar, except the limit as ๐‘ฅ tends to negative two of ๐‘“ of ๐‘ฅ is now negative three. And so that point is kind of in the right place to fill the gap in that curve and the function is continuous.

Option C is what happens when you donโ€™t define ๐‘“ of negative two. I think itโ€™s pretty clear that we have to find ๐‘“ of negative two; in fact, itโ€™s here. So we know that ๐‘“ of ๐‘ฅ is negative three if ๐‘ฅ is negative two, but had we not done that itโ€™s possible that this thing here is not defined and so the function is not defined.

And finally, option D where the limit as ๐‘ฅ tends to negative two, ๐‘“ of ๐‘ฅ does not exist. And you can see in the diagram Iโ€™ve suggested that could be because thereโ€™s a vertical asymptote and ๐‘ฅ equals negative two. And so the function is discontinuous there even though it is defined. ๐‘“ of negative two is equal to negative three, but either side of that the function is going off to plus one negative โˆž. Of course, it doesnโ€™t have to be an asymptote. It could just be at a jump gap, discontinuity. Either way, itโ€™s an option.

Step one is to find the limit as ๐‘ฅ tends to negative two of ๐‘“ of ๐‘ฅ. Why do we do that? Well, if the limit exists, but is not equal to ๐‘“ of negative two, which we said was negative three, then we know the answer is option A. If instead the limit does exist, but is equal to negative three, then the function is continuous. And so option B is our answer. Option C Iโ€™ve said we can safely discount because ๐‘“ of negative two is defined; itโ€™s negative three. And if the limit doesnโ€™t exist, then our answer is option D.

We want to find this limit as ๐‘ฅ tends to negative two and as ๐‘ฅ tends to negative two is never actually negative two. And so the value of this limit doesnโ€™t depend on the value of the function when ๐‘ฅ is equal to negative two; it only depends on the value of the function when ๐‘ฅ is not equal to negative two. And so we can safely use this formula in place of ๐‘“ of ๐‘ฅ. So we got this rational function here: ๐‘ฅ cubed plus eight over ๐‘ฅ squared minus four.

And we know that for a rational function, itโ€™s continuous wherever itโ€™s defined. So we could just try substituting in negative two to this rational function. Hope itโ€™s defined there. And if it is, then our value will also be our limit by the definition of continuity. Okay, so we substitute negative two in. And because negative two cubed is negative eight and because negative two squared is four, we get zero over zero which is indeterminate. So thereโ€™s no luck there; we canโ€™t tell what the limit is from that. But what we can tell from the factor theorem is that both the numerator and denominator of this rational function have a factor of ๐‘ฅ plus two. So letโ€™s divide that out.

So you might be able to recognize that the denominator is the difference of two squares. And so this factor here is going to be ๐‘ฅ minus two. What about the other factor of the numerator? Well, it looks like it should be a quadratic. So we could write down the general form of a quadratic and solve the coefficients. Okay, thatโ€™s the general form of a quadratic. Let me now multiply out the brackets in the numerator.

Iโ€™ve multiplied out the brackets on the numerator. And of course, I want to end up with ๐‘ฅ cubed plus eight, which it is after all what weโ€™re trying to factorize. And now I will compare coefficients. The coefficient of ๐‘ฅ cubed on the left is ๐‘Ž and on the right itโ€™s kind of hidden; itโ€™s one. So I can say that ๐‘Ž is equal to one. The coefficient of ๐‘ฅ squared on the left is two ๐‘Ž plus ๐‘ or as we know ๐‘Ž is one; itโ€™s two plus ๐‘. And on the right, itโ€™s zero. Thereโ€™s no ๐‘ฅ squared term, which is explicitly written out. So we see that two ๐‘Ž plus ๐‘ equals zero. As we said before, ๐‘Ž is one, so two plus ๐‘ equals zero and ๐‘ is equal to negative two.

Itโ€™s a very similar story for the coefficient of ๐‘ฅ, which is two ๐‘ plus ๐‘ on the left and well not mentioned on the right zero. Solving that gives ๐‘ equal to four and then our last equation the coefficient of ๐‘ฅ to the power of zero or the constant term on the left is two ๐‘, on the right is eight, two ๐‘ is eight; that agrees with what we have, ๐‘ is four. And so weโ€™ve got consistency, which is good, and our coefficients. So letโ€™s put them in. Okay, there we go.

And now Iโ€™m going to claim that I can cancel these two factors of ๐‘ฅ plus two. Why can I do that? Well, Iโ€™m saying that neither of these is zero because ๐‘ฅ is not equal to negative two because weโ€™re taking the limit as ๐‘ฅ tends to negative two, which means explicitly avoiding values of ๐‘ฅ, which are negative two.

Okay, so weโ€™ve got the limit as ๐‘ฅ tends to negative two of a slightly simpler rational expression, which we can think of as a rational function. And weโ€™re going to think about evaluating it at negative two as we tried before. And this time, weโ€™re going to hope that we donโ€™t get zero over zero. Indeterminate thing we hope that weโ€™re going to get a proper real number because if we get a real number, when we evaluate the expression we know that it will also be the limit of that expression as ๐‘ฅ tends to negative two.

So here, Iโ€™ve just substituted in negative two. Now, letโ€™s evaluate. So we simplify the numerator to find itโ€™s 12 and the denominator to find itโ€™s negative four. And 12 divided by negative four is negative three. And because itโ€™s not indeterminate or undefined, we know that thatโ€™s our limit. Okay, great! Thatโ€™s step one completed. What do we do now?

Letโ€™s have another look at the options. Remember we already discounted option C coz we knew that ๐‘“ of negative two is defined. Now, we found that the limit of ๐‘“ of ๐‘ฅ as ๐‘ฅ turns to negative two is negative three. We know that limit of ๐‘“ of ๐‘ฅ does exist. And so this option D canโ€™t be true, so thatโ€™s also eliminated as an option.

So this leaves two options: option A and option B. And as discussed before, this hinged on whether ๐‘“ of negative two was equal to limit of ๐‘“ of ๐‘ฅ as ๐‘ฅ tends to negative two. Well of course, we can see what ๐‘“ of negative two is quite clearly. Weโ€™re told itโ€™s negative three.

And so comparing that to the limit as ๐‘ฅ tends to negative two of ๐‘“ of ๐‘ฅ, we see that theyโ€™re the same. So that eliminates option A, which says that theyโ€™re different and not equal, where in fact they are equal to negative three. So the only option that weโ€™re left with, and of course the option that makes sense, is that the function is continuous at ๐‘ฅ equals negative three.

And it makes sense of course because thatโ€™s the definition of continuity of a function at a point. For function to be continuous at a point, it has to be defined at that point. And more than that, the definition of the function at that point has to be equal to the limit of the function as ๐‘ฅ approaches that point. So our answer is B. The function is continuous at ๐‘ฅ equals negative two.

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