### 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. Letβs start by checking each of
these conditions in turn.

First, letβs 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 weβve shown that π evaluated at
one-third is defined. Itβs 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 weβre 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 weβre 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 letβs take out this factor of
three π₯ minus one from the polynomial in our numerator. Thereβs 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 weβll write our second linear
factor as ππ₯ 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. Weβll 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 weβve 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 thatβs our second continuity
condition true. Weβve shown the limit as π₯
approaches one-third of π of π₯ exists. In fact, itβs equal to negative
four divided by three.

Letβs 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. Weβve 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. Theyβre 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.