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

Discuss the continuity of the function π, given π(π₯) = 3 + sin π₯, if 0 β€ π₯ < π/2 and π(π₯) = 4 + (π₯ β (π/2))βΈ, if π₯ β₯ π/2.

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### 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, letβs 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 weβ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 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.

Letβs 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 itβs 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 itβs defined for all values of π₯ greater than or equal to π by two. So weβve 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. Thatβs 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.

Letβs 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 theyβ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 itβs 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. Letβs 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. Theyβre 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, weβve 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.