# Question Video: Finding the Type of Discontinuity in a Piecewise-Defined Function Mathematics • Higher Education

Consider the function π(π₯) = (6π₯Β² β 8π₯ + 2)/(3π₯ β 1) when π₯ β  1/3 and π(π₯) = β4/3 when π₯ = 1/3. Find the type of discontinuity that the function π has at π₯ = 1/3, if it has any.

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### Video Transcript

Consider the function π 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 π of π₯ is equal to negative four over three when π₯ is equal to one-third. Find the type of discontinuity that the function π has at π₯ is equal to one-third, if it has any.

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

If all three of these conditions are true, we say that π is continuous at π₯ is equal to one-third. If one of these conditions is not true, then we say π 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. Letβs start by checking each of these conditions in turn.

First, letβs check that π evaluated at one-third is defined. We can do this directly from the piecewise definition of our function π of π₯. We can see when π₯ is equal to one-third, π of π₯ is equal to negative four over three. So weβve shown that π evaluated at one-third is defined. Itβs equal to negative four over three. Next, we need to check the limit as π₯ approaches one-third of π of π₯ exists. So we want to evaluate the limit as π₯ approaches one-third of π 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 π of π₯, when π₯ is not equal to one-third, our function π 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 weβ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 weβ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 letβs take out this factor of three π₯ minus one from the polynomial in our numerator. Thereβ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 weβll write our second linear factor as ππ₯ plus π. 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. So equating the coefficients of π₯ squared, we get three π is equal to six. Dividing through by three gives us that π is equal to two.

We can do something similar to find the value of π. Weβ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 π. So equating our constants gave us negative π is equal to two. Multiplying through by negative one gives us that π is equal the negative two. So weβ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 thatβs our second continuity condition true. Weβve shown the limit as π₯ approaches one-third of π of π₯ exists. In fact, itβs equal to negative four divided by three.

Letβs check our final continuity condition. We need the limit as π₯ approaches one-third of π of π₯ is equal to π evaluated at one-third. Weβve actually already evaluated both of these expressions. We showed the limit as π₯ approaches one-third of π 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 π evaluated at one-third was equal to negative four over three.

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