Video: Determine the Type of Discontinuity in a Piecewise Defined Function

Consider the function 𝑓(π‘₯) = βˆ’π‘₯Β² βˆ’ 1 when π‘₯ ≀ 2, and 𝑓(π‘₯) = 3π‘₯ βˆ’ 1 when π‘₯ > 2. Find the type of discontinuity that the function 𝑓 has at π‘₯ = 2, if it has any.

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to negative π‘₯ squared minus one when π‘₯ is less than or equal to two and 𝑓 of π‘₯ is equal to three π‘₯ minus one if π‘₯ is greater than two. Find the type of discontinuity that the function 𝑓 has at π‘₯ is equal to two, if it has any.

Since the question is asking us to classify the type of discontinuity of the function at π‘₯ is equal to two, if it has any, we can start by asking what does it mean for a function 𝑓 to be continuous at the point π‘₯ is equal to two. We recall that we say the function 𝑓 is continuous at the point π‘₯ is equal to π‘Ž if the following are true.

First, we must have that the function 𝑓 evaluated at π‘Ž must be defined. It’s worth noting that this is equivalent to saying that π‘Ž is in the domain of our function 𝑓. Secondly, we must have that the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž must exist. And finally, we must have that the limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž is equal to the function 𝑓 evaluated at π‘Ž.

If all three of these requirements are met, then we can say that the function 𝑓 is continuous at the point π‘₯ is equal to π‘Ž. Similarly, if any of these three requirements fail, we can use these to classify our discontinuity. Since the question is asking us about the discontinuity of the function 𝑓 when π‘₯ is equal to two, we can replace π‘Ž in our definition of continuity as two.

Let’s now start with our first requirement that the function 𝑓 evaluated at two must be defined. We can see from the question that the function 𝑓 of π‘₯ is defined piecewise and the first piece has π‘₯ is less than or equal to two. Therefore, we have that the function 𝑓 evaluated at two is equal to two substituted into negative π‘₯ squared minus one. Giving us that 𝑓 evaluated at two is equal to negative one multiplied by two squared minus one, which we can calculate to be equal to negative five. Therefore, we can conclude that the function 𝑓 evaluated at two is in fact defined. So, we can now move on to our second requirement.

We recall that we say a limit of 𝑓 of π‘₯ as π‘₯ approaches π‘Ž exists if the left-hand and right limit of the function 𝑓 of π‘₯ as π‘₯ approaches π‘Ž both exist and if these two limits are equal. What this means is to check our second requirement, we must check that the limit of 𝑓 of π‘₯ as π‘₯ approaches two from the right and the limit of 𝑓 of π‘₯ as π‘₯ approaches two from the left both exist and we must check that both of these limits are equal.

We’ll start with the limit of 𝑓 of π‘₯ as π‘₯ approaches two from the right. Since π‘₯ is approaching two from the right, we must have that π‘₯ is greater than two. And we see from the question from our piecewise definition of the function 𝑓 of π‘₯ when π‘₯ is greater than two, we have that 𝑓 of π‘₯ is exactly equal to the function three π‘₯ minus one. Therefore, we can replace the 𝑓 of π‘₯ in our right-hand limit with the function there π‘₯ minus one.

Since three π‘₯ minus one is a linear function, we can evaluate this limit by direct substitution, giving us three multiplied by two minus one, which we can calculate to give us that the limit of 𝑓 of π‘₯ as π‘₯ approaches two from the right is equal to five. We can now do something similar with the left-hand limit. Since π‘₯ is approaching two from the left, we must have that π‘₯ is strictly less than two. We can again see from our piecewise definition of the function 𝑓 of π‘₯ when π‘₯ is less than two, our function 𝑓 of π‘₯ is exactly equal to negative π‘₯ squared minus one. So, we can replace the 𝑓 of π‘₯ in our left-hand limit with negative π‘₯ squared minus one.

Since we are now evaluating the limit of a polynomial, we can use direct substitution. Giving us negative one multiplied by two squared minus one, which we can calculate to give us the limit of 𝑓 of π‘₯ as π‘₯ approaches two from the left is equal to negative five. Therefore, what we have shown is that the left-hand and right-hand limit of our function 𝑓 of π‘₯ as π‘₯ approaches two both exist. However, these two limits are not equal. We call this a jump discontinuity.

To see this, let’s sketch a graph of our function 𝑓 of π‘₯ as defined in the question. In the sketch of this graph, we can see the first segment of our graph is the same as negative π‘₯ squared minus one and the second segment is three π‘₯ minus one. We can also see that our graph 𝑦 equals 𝑓 of π‘₯ is equal to negative five when π‘₯ is equal to two, which is represented by the solid dot.

We can also see that as π‘₯ approaches two from the left, we will get closer and closer to our output of negative five, and as π‘₯ approaches two from the right, we will get closer and closer to our output of five. This gap between our outputs of five and negative five at π‘₯ is equal to two is what we describe as our jump. Therefore, we can conclude that the function 𝑓 has a jump discontinuity at π‘₯ is equal to two.

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