### Video Transcript

Discuss the continuity of π on the
set of real numbers, given that π of π₯ is equal to negative seven sin of five π₯
if π₯ is greater than or equal to negative π by seven and π₯ is less than π by
four and π of π₯ is equal to negative five cos of three π₯ if π₯ is greater than or
equal to π by four and π₯ is less than or equal to two π.

The question wants us to discuss
the continuity of the function π on the set of real numbers. In other words, we need to find the
real values of π₯ where π is continuous. And we recall we say a function 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, we also need both of
these to exist. In this case, weβre trying to find
the continuity of a function which is defined piecewise. And we can find the continuity of a
piecewise-defined function using the following three steps.

First, we want to find the domain
of our piecewise-defined function because remember a function can only be continuous
at values where itβs defined. Next, we want to check the
continuity of our piecewise-defined function on each interval. Finally, since our
piecewise-defined function will be made up of several different functions, we want
to check that the endpoints of these different functions match up. Otherwise, our function would not
be continuous at these points.

So letβs start by finding the
domain of our function π of π₯. First, we can see when π₯ is
greater than or equal to negative π by seven and π₯ is less than π by four, our
function π of π₯ is equal to negative seven times the sin of five π₯. So on this interval, our function
π of π₯ is equal to negative seven times the sin of five π₯. And we know that this function is
defined for all real values of π₯.

And since negative seven times the
sin of five π₯ is defined for all values of π₯ on this interval, our function π of
π₯ must also be defined for all values of π₯ on this interval. We can do the same for our second
interval. When π₯ is greater than or equal to
π by four and π₯ is less than or equal to two π, we see that our function π of π₯
is equal to negative five times the cos of three π₯.

And again, negative five times the
cos of three π₯ is defined for all real values of π₯. So in particular, itβs defined for
all values of π₯ on this interval. And since we know π of π₯ is equal
to negative five times the cos of three π₯ on this interval, this also tells us that
π of π₯ is defined for all values of π₯ on this interval. So our function π of π₯ is defined
for all values of π₯ in both of these intervals. So we just combine both of these
intervals to give us the domain of π of π₯. This gives us the closed interval
from negative π by seven to two π.

We now want to check the continuity
of our function π of π₯ on each interval. We can actually do both intervals
at the same time in this case. We can see that both negative seven
times the sin of five π₯ and negative five times the cos of three π₯ are
trigonometric functions. And we know all trigonometric
functions are continuous on their domains.

And of course, we already explained
that both of these functions are defined for all real values of π₯. So both of these functions are
continuous for all real values of π₯. In particular, this tells us our
function π of π₯ is continuous on both of the intervals. However, we need to be careful with
what exactly weβve shown.

Remember, when our
piecewise-defined function changes from one interval to another, we need to check
that the endpoints match. In π of π₯, this happens when π₯
is equal to π by four. So we canβt yet conclude anything
about the continuity of our function π of π₯ when π₯ is equal to π by four.

So, so far, weβve only shown that
π of π₯ is continuous on the closed interval from negative π by seven to two π,
where we remove the point π by four. The last thing we need to do is
check if our endpoints match.

Letβs start with our first
interval. Since π₯ is strictly less than π
by four, we want the limit as π₯ approaches π by four from the left of negative
seven times the sin of five π₯. And since this is a trigonometric
function, we can do this by direct substitution. This gives us negative seven times
the sin of five π by four. And we can evaluate this, and we
get seven root two divided by two.

Doing the same for our second
interval is even easier. Since π₯ is greater than or equal
to π by four, we can just directly substitute π₯ is equal to π by four into our
function. This gives us negative five times
the cos of three π by four. And we can evaluate this, and we
get five root two divided by two. And these two values are not
equal. This tells us our function π of π₯
is not continuous when π₯ is equal to π by four. In actuality, since π of π₯ is
continuous when π₯ is less than π by four and π of π₯ is continuous when π₯ is
greater than π by four and both of these endpoints are defined. We actually know that this is a
jump discontinuity.

So weβve shown that the function π
of π₯, which is equal to negative seven sin of five π₯ if π₯ is greater than
negative π by seven and π₯ is less than π by four. And π of π₯ is equal to negative
five cos of three π₯ if π₯ is greater than or equal to π by four and π₯ is less
than or equal to two π. Is continuous on the closed
interval from negative π by seven to two π, where we remove the point π by
four.