Video: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Point

Discuss the continuity of the function 𝑓, given 𝑓(π‘₯) = 3 + sin π‘₯, if 0 ≀ π‘₯ < πœ‹/2 and 𝑓(π‘₯) = 4 + (π‘₯ βˆ’ (πœ‹/2))⁸, if π‘₯ β‰₯ πœ‹/2.

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Video Transcript

Discuss the continuity of the function 𝑓, given that 𝑓 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œ‹ by two and 𝑓 of π‘₯ is equal to four plus π‘₯ minus πœ‹ by two to the eighth power if π‘₯ is greater than or equal to πœ‹ by two.

The question wants us to discuss the continuity of a function 𝑓. And we can see that this function is defined piecewise. To discuss the continuity of our function 𝑓, let’s recall what it means for our function to be continuous at a point π‘₯ is equal to π‘Ž. We say 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, both 𝑓 evaluated at π‘Ž and this limit must exist.

Because we’re 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 𝑓 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.

Let’s start by checking the domain of our function 𝑓 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œ‹ by two, our function 𝑓 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 it’s defined for all values of π‘₯ greater than or equal to zero and less than πœ‹ by two. So our function 𝑓 of π‘₯ is defined in this interval.

Next, we can do the same when π‘₯ is greater than or equal to πœ‹ by two. We get 𝑓 of π‘₯ is equal to four plus π‘₯ minus πœ‹ 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 it’s defined for all values of π‘₯ greater than or equal to πœ‹ by two. So we’ve shown our function 𝑓 of π‘₯ is defined on both of these intervals. So the domain of our function 𝑓 of π‘₯ is both of these intervals combined. That’s π‘₯ greater than or equal to zero. We now want to check the continuity of our function 𝑓 of π‘₯ on each of these intervals.

Let’s start by checking the continuity for π‘₯ greater than or equal to zero and π‘₯ less than πœ‹ by two. On this interval, our function 𝑓 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 they’re 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 it’s continuous. And this means our function 𝑓 of π‘₯ is continuous over this interval. However, we must be careful to remove any endpoints where our function 𝑓 of π‘₯ changes definition.

We can now do the same for values of π‘₯ greater than or equal to πœ‹ by two. We can see, for these values of π‘₯, 𝑓 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 𝑓 of π‘₯ is continuous when π‘₯ is strictly greater than πœ‹ by two. Since our function 𝑓 of π‘₯ changes definition when π‘₯ is less than πœ‹ by two and when π‘₯ is greater than πœ‹ by two.

So we now have our third and final step. We need to check that the endpoints of our function 𝑓 of π‘₯ match. And we can see the only endpoint where our function changes definition is at π‘₯ is equal to πœ‹ by two. Let’s start with the endpoint of three plus the sin of π‘₯ when π‘₯ is less than πœ‹ 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œ‹ 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œ‹ by two, our endpoint would just be where π‘₯ is equal to πœ‹ by two. So we just directly substitute π‘₯ is equal to πœ‹ by two into four plus π‘₯ minus πœ‹ by two to the eighth power. This gives us four plus πœ‹ by two minus πœ‹ 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. They’re both equal to four. So this tells us that 𝑓 is continuous at π‘₯ is equal to πœ‹ by two. So when we check the entire domain of our function 𝑓 of π‘₯, we actually show that 𝑓 of π‘₯ was continuous on its entire domain.

Therefore, we’ve shown that the function 𝑓 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œ‹ by two. And 𝑓 of π‘₯ is equal to four plus π‘₯ minus πœ‹ by two to the eighth power if π‘₯ is greater than or equal to πœ‹ by two. Is continuous for all values of π‘₯ greater than or equal to zero.

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