### Video Transcript

Discuss the continuity of the
function π, given that π of π₯ is equal to three plus the sin of π₯ if π₯ is
greater than or equal to zero and π₯ is less than π by two and π of π₯ is equal to
four plus π₯ minus π by two to the eighth power if π₯ is greater than or equal to
π by two.

The question wants us to discuss
the continuity of a function π. And we can see that this function
is defined piecewise. To discuss the continuity of our
function π, letβs recall what it means for our function to be continuous at a point
π₯ is equal to π. We say π is continuous at π₯ is
equal to π if π evaluated at π is equal to the limit as π₯ approaches π of π of
π₯. And, of course, both π evaluated
at π and this limit must exist.

Because weβre dealing with the
continuity of a piecewise-defined function, we can check this in the following three
steps. First, we want to find the domain
of our function π of π₯, since a function can only be continuous on its domain. Second, we check the continuity on
each interval of our piecewise-defined function. Finally, we just need to check that
the endpoints of these intervals, our functions, match.

Letβs start by checking the domain
of our function π of π₯. To do this, we need to check the
values of π₯ for which our piecewise-defined function is defined. We see when π₯ is greater than or
equal to zero and π₯ is less than π by two, our function π of π₯ is equal to three
plus the sin of π₯. And three plus the sin of π₯ has a
domain of the entire set of real numbers. So in particular, this means itβs
defined for all values of π₯ greater than or equal to zero and less than π by
two. So our function π of π₯ is defined
in this interval.

Next, we can do the same when π₯ is
greater than or equal to π by two. We get π of π₯ is equal to four
plus π₯ minus π over two to the eighth power. And this is a polynomial. So its domain is, again, the entire
set of real numbers. And, in particular, this means itβs
defined for all values of π₯ greater than or equal to π by two. So weβve shown our function π of
π₯ is defined on both of these intervals. So the domain of our function π of
π₯ is both of these intervals combined. Thatβs π₯ greater than or equal to
zero. We now want to check the continuity
of our function π of π₯ on each of these intervals.

Letβs start by checking the
continuity for π₯ greater than or equal to zero and π₯ less than π by two. On this interval, our function π
of π₯ is equal to three plus the sin of π₯. And we know the sin of π₯ is
continuous on the set of real numbers because all trig functions are continuous on
their domains and the sin of π₯ is defined for all real numbers. Next, constant functions are just
polynomials. So theyβre continuous on the set of
real numbers. Finally, this means three plus the
sin of π₯ is the sum of two continuous functions, which means that itβs
continuous. And this means our function π of
π₯ is continuous over this interval. However, we must be careful to
remove any endpoints where our function π of π₯ changes definition.

We can now do the same for values
of π₯ greater than or equal to π by two. We can see, for these values of π₯,
π of π₯ is a polynomial. And, of course, we know all
polynomials are continuous on the set of real numbers. And we must be careful at this
point, since this only tells us our function π of π₯ is continuous when π₯ is
strictly greater than π by two. Since our function π of π₯ changes
definition when π₯ is less than π by two and when π₯ is greater than π by two.

So we now have our third and final
step. We need to check that the endpoints
of our function π of π₯ match. And we can see the only endpoint
where our function changes definition is at π₯ is equal to π by two. Letβs start with the endpoint of
three plus the sin of π₯ when π₯ is less than π by two. This is just a constant plus a
trigonometric function. So we can do this by direct
substitution. We get three plus the sin of π by
two, which we can evaluate to just give us four.

The endpoint of our second interval
is even easier. Since π₯ must be greater than or
equal to π by two, our endpoint would just be where π₯ is equal to π by two. So we just directly substitute π₯
is equal to π by two into four plus π₯ minus π by two to the eighth power. This gives us four plus π by two
minus π by two to the eighth power, which we can evaluate to give us four. And we can see, in this case, that
these endpoints match. Theyβre both equal to four. So this tells us that π is
continuous at π₯ is equal to π by two. So when we check the entire domain
of our function π of π₯, we actually show that π of π₯ was continuous on its
entire domain.

Therefore, weβve shown that the
function π of π₯, which is equal to three plus the sin of π₯ if π₯ is greater than
or equal to zero and π₯ is less than π by two. And π of π₯ is equal to four plus
π₯ minus π by two to the eighth power if π₯ is greater than or equal to π by
two. Is continuous for all values of π₯
greater than or equal to zero.