Video: Finding the Type of Discontinuity in a Piecewise-Defined Function

Consider the function 𝑓(π‘₯) = (6π‘₯Β² βˆ’ 8π‘₯ + 2)/(3π‘₯ βˆ’ 1) when π‘₯ β‰  1/3 and 𝑓(π‘₯) = βˆ’4/3 when π‘₯ = 1/3. Find the type of discontinuity that the function 𝑓 has at π‘₯ = 1/3, if it has any.

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Video Transcript

Consider the function 𝑓 of π‘₯ is equal to six π‘₯ squared minus eight π‘₯ plus two divided by three π‘₯ minus one when π‘₯ is not equal to one-third, and 𝑓 of π‘₯ is equal to negative four over three when π‘₯ is equal to one-third. Find the type of discontinuity that the function 𝑓 has at π‘₯ is equal to one-third, if it has any.

The question gives us a piecewise-defined function 𝑓 of π‘₯. It wants us to check if there is a discontinuity and what type of discontinuity the function 𝑓 has when π‘₯ is equal to one-third. To answer this question, we first need to recall what it means for a function 𝑓 to be continuous at π‘₯ is equal to one-third. We say a function 𝑓 is continuous when π‘₯ is equal to one-third if it satisfies the following three conditions. First, 𝑓 evaluated at one-third must be defined. Second, the limit as π‘₯ approaches one-third of 𝑓 of π‘₯ must exist. Third, the limit as π‘₯ approaches one-third of 𝑓 of π‘₯ must be equal to 𝑓 evaluated at one-third.

If all three of these conditions are true, we say that 𝑓 is continuous at π‘₯ is equal to one-third. If one of these conditions is not true, then we say 𝑓 is discontinuous at π‘₯ is equal to one-third. And depending on which of these conditions failed and how they failed, we have different names for these types of discontinuities. Let’s start by checking each of these conditions in turn.

First, let’s check that 𝑓 evaluated at one-third is defined. We can do this directly from the piecewise definition of our function 𝑓 of π‘₯. We can see when π‘₯ is equal to one-third, 𝑓 of π‘₯ is equal to negative four over three. So we’ve shown that 𝑓 evaluated at one-third is defined. It’s equal to negative four over three. Next, we need to check the limit as π‘₯ approaches one-third of 𝑓 of π‘₯ exists. So we want to evaluate the limit as π‘₯ approaches one-third of 𝑓 of π‘₯. Since π‘₯ is approaching one-third, π‘₯ is getting closer and closer to one-third. π‘₯ is never equal to one-third.

And we can see from our piecewise definition of the function 𝑓 of π‘₯, when π‘₯ is not equal to one-third, our function 𝑓 of π‘₯ is exactly equal to six π‘₯ squared minus eight π‘₯ plus two over three π‘₯ minus one. And if these functions are exactly the same when π‘₯ is not equal to one-third, their limits as π‘₯ approaches one-third will be equal.

Now we see we’re trying to evaluate the limit of a rational function. We can attempt to do this by direct substitution. Substituting π‘₯ is equal to one-third, we get six times one-third squared minus eight times one-third plus two divided by three times one-third minus one. And evaluating this expression gives us the indeterminate form zero divided by zero. So substituting π‘₯ is equal to one-third gave us an indeterminate form. This means we’re going to need to use the factor theorem. The factor theorem tells us that three π‘₯ minus one must divide the polynomial in our numerator.

So let’s take out this factor of three π‘₯ minus one from the polynomial in our numerator. There’s a few different ways of doing this. For example, we could use polynomial division. However, we know a linear function multiplied by a linear function will give us a quadratic. So we’ll write our second linear factor as π‘Žπ‘₯ plus 𝑏. All we need to do now is equate the coefficient. The coefficient of π‘₯ squared in our quadratic is six. And when we multiply our linear factors, the coefficient of π‘₯ squared will be three π‘Ž. So equating the coefficients of π‘₯ squared, we get three π‘Ž is equal to six. Dividing through by three gives us that π‘Ž is equal to two.

We can do something similar to find the value of 𝑏. We’ll equate the constants. The constant in our quadratic is two. And when we multiply our two linear factors, we see the constant will be negative 𝑏. So equating our constants gave us negative 𝑏 is equal to two. Multiplying through by negative one gives us that 𝑏 is equal the negative two. So we’ve now rewritten our limit as the limit as π‘₯ approaches one-third of three π‘₯ minus one times two π‘₯ minus two all divided by three π‘₯ minus one.

We can then cancel the shared factor of three π‘₯ minus one in our numerator and our denominator. This gives us the limit as π‘₯ approaches one-third of two π‘₯ minus two. And we see this is the limit of a linear function. We can do this by direct substitution. Substituting π‘₯ is equal to one-third, we get two times one-third minus two, which we can calculate, gives us negative four divided by three. So that’s our second continuity condition true. We’ve shown the limit as π‘₯ approaches one-third of 𝑓 of π‘₯ exists. In fact, it’s equal to negative four divided by three.

Let’s check our final continuity condition. We need the limit as π‘₯ approaches one-third of 𝑓 of π‘₯ is equal to 𝑓 evaluated at one-third. We’ve actually already evaluated both of these expressions. We showed the limit as π‘₯ approaches one-third of 𝑓 of π‘₯ is equal to negative four over three when we were checking our second continuity condition. And when we were checking our first continuity condition, we showed that 𝑓 evaluated at one-third was equal to negative four over three.

So, in fact, both of these terms are equal. They’re both equal to negative four over three. So all three of our continuity conditions are true. So, in fact, 𝑓 is continuous when π‘₯ is equal to one-third. And, of course, this also means the function 𝑓 does not have a discontinuity at π‘₯ is equal to one-third.

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