Video: Discussing the Continuity of a Piecewise-Defined Functions Involving Trigonometric Ratios on an Interval

Discuss the continuity of the function 𝑓 on ℝ, given 𝑓(π‘₯) = βˆ’7 sin 5π‘₯, if βˆ’πœ‹/7 ≀ π‘₯ < πœ‹/4 and 𝑓(π‘₯) = βˆ’ 5 cos 3π‘₯, if πœ‹/4 ≀ π‘₯ ≀ 2πœ‹.

05:04

Video Transcript

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

The question wants us to discuss the continuity of the function 𝑓 on the set of real numbers. In other words, we need to find the real values of π‘₯ where 𝑓 is continuous. And we recall we say a function 𝑓 is continuous at π‘₯ is equal to π‘Ž if 𝑓 evaluated at π‘Ž is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯. And of course, we also need both of these to exist. In this case, we’re 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 it’s 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 let’s start by finding the domain of our function 𝑓 of π‘₯. First, we can see when π‘₯ is greater than or equal to negative πœ‹ by seven and π‘₯ is less than πœ‹ by four, our function 𝑓 of π‘₯ is equal to negative seven times the sin of five π‘₯. So on this interval, our function 𝑓 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 𝑓 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œ‹ by four and π‘₯ is less than or equal to two πœ‹, we see that our function 𝑓 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, it’s defined for all values of π‘₯ on this interval. And since we know 𝑓 of π‘₯ is equal to negative five times the cos of three π‘₯ on this interval, this also tells us that 𝑓 of π‘₯ is defined for all values of π‘₯ on this interval. So our function 𝑓 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 𝑓 of π‘₯. This gives us the closed interval from negative πœ‹ by seven to two πœ‹.

We now want to check the continuity of our function 𝑓 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 𝑓 of π‘₯ is continuous on both of the intervals. However, we need to be careful with what exactly we’ve shown.

Remember, when our piecewise-defined function changes from one interval to another, we need to check that the endpoints match. In 𝑓 of π‘₯, this happens when π‘₯ is equal to πœ‹ by four. So we can’t yet conclude anything about the continuity of our function 𝑓 of π‘₯ when π‘₯ is equal to πœ‹ by four.

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

Let’s start with our first interval. Since π‘₯ is strictly less than πœ‹ by four, we want the limit as π‘₯ approaches πœ‹ 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œ‹ 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œ‹ by four, we can just directly substitute π‘₯ is equal to πœ‹ by four into our function. This gives us negative five times the cos of three πœ‹ 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 𝑓 of π‘₯ is not continuous when π‘₯ is equal to πœ‹ by four. In actuality, since 𝑓 of π‘₯ is continuous when π‘₯ is less than πœ‹ by four and 𝑓 of π‘₯ is continuous when π‘₯ is greater than πœ‹ by four and both of these endpoints are defined. We actually know that this is a jump discontinuity.

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

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.