Question Video: Discussing the Continuity of a Piecewise-Defined Functions Involving Trigonometric Ratios at a Point | Nagwa Question Video: Discussing the Continuity of a Piecewise-Defined Functions Involving Trigonometric Ratios at a Point | Nagwa

Question Video: Discussing the Continuity of a Piecewise-Defined Functions Involving Trigonometric Ratios at a Point Mathematics • Second Year of Secondary School

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Discuss the continuity of the function 𝑓(π‘₯) = πœ‹/2, given 𝑓(π‘₯) = βˆ’7 sin π‘₯ + 7 cos π‘₯, π‘₯ ≀ πœ‹/2 and 𝑓(π‘₯) = 6 cos 2π‘₯ βˆ’ 1, π‘₯ > πœ‹/2.

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Video Transcript

Discuss the continuity of the function 𝑓 at π‘₯ is equal to πœ‹ by two given 𝑓 of π‘₯ is equal to negative seven sin of π‘₯ plus seven cos of π‘₯ if π‘₯ is less than or equal to πœ‹ by two and 𝑓 of π‘₯ is equal to six cos of two π‘₯ minus one if π‘₯ is greater than πœ‹ by two.

In this question, we’re given a piecewise function 𝑓 of π‘₯ and we need to determine the continuity of this function 𝑓 at π‘₯ is equal to πœ‹ by two. So, to answer this question, let’s start by recalling what it means for a function 𝑓 to be continuous at a point. And we say that a function 𝑓 is continuous at a point when π‘₯ is equal to π‘Ž if the following three conditions hold. First, 𝑓 evaluated at π‘Ž needs to be defined, and this is the same as saying that π‘Ž is in the domain of our function 𝑓. Second, we need the limit as π‘₯ approaches π‘Ž of our function 𝑓 of π‘₯ to exist. And it’s worth noting we can check this by considering the left and right limits. And third, we need the limit as π‘₯ approaches π‘Ž of our function 𝑓 of π‘₯ to be equal to 𝑓 evaluated at π‘Ž.

We need to check these three conditions for our function 𝑓 when π‘₯ is equal to πœ‹ by two. So let’s set our value of π‘Ž equal to πœ‹ by two. This then gives us the following. We can now check each of these conditions separately. Let’s start by checking that πœ‹ by two is in the domain of our function 𝑓. To do this, we know that 𝑓 is a piecewise-defined function. So we need to see which subdomain πœ‹ by two lies in. And we can see that πœ‹ by two lies in the first subdomain of this subfunction. So πœ‹ by two is in the domain of 𝑓.

So our first condition is true. However, it’s a good idea to evaluate 𝑓 at πœ‹ by two since we’ll need it later anyway. To do this, we substitute π‘₯ is equal to πœ‹ by two into the first subfunction. We get negative seven times the sin of πœ‹ by two plus seven multiplied by the cos of πœ‹ by two. And we know the sin of πœ‹ by two is one and the cos of πœ‹ by two is zero. So this simplifies to give us negative seven.

So we’ve shown the first condition of continuity is true. Now let’s move on to the second condition. We want to show the limit as π‘₯ approaches πœ‹ by two of 𝑓 of π‘₯ exists. And since πœ‹ by two lies at the endpoint of the subdomains of our piecewise-defined function, 𝑓 of π‘₯ will have a different definition to the left and to the right of π‘₯ is equal to πœ‹ by two. This should indicate to us that we need to look at the left and right limits of our function at π‘₯ is equal to πœ‹ by two. We need to see if these are equal.

So let’s start with the limit as π‘₯ approaches πœ‹ by two from the left of 𝑓 of π‘₯. Our values of π‘₯ are going to be less than πœ‹ by two. And we can see when π‘₯ is less than πœ‹ by two, our function 𝑓 of π‘₯ is equal to its first subfunction. And since the functions are equal when π‘₯ is less than πœ‹ by two, their limits as π‘₯ approaches πœ‹ by two from the left will also be equal. So we need to determine the limit as π‘₯ approaches πœ‹ by two from the left of negative seven sin of π‘₯ plus seven cos of π‘₯. But this is the sum of trigonometric functions. So we can just do this by using direct substitution. We would substitute π‘₯ is equal to πœ‹ by two into this function. However, we’ve already done this above. It’s equal to negative seven sin of πœ‹ by two plus seven cos of πœ‹ by two, which we’ve already shown is negative seven.

Let’s now check that the limit as π‘₯ approaches πœ‹ by two from the right is equal to negative seven. This time, our values of π‘₯ will be greater than πœ‹ by two. And we can see that this means our function 𝑓 of π‘₯ will be equal to its second subfunction. Once again, this tells us their limits as π‘₯ approaches πœ‹ by two from the right will be equal. And once again, we can evaluate this by using direct substitution since this is just a trigonometric function. Substituting π‘₯ is equal to πœ‹ by two, we get six cos of two times πœ‹ by two minus one. And two times πœ‹ by two is equal to πœ‹, and the cos of πœ‹ is negative one. So we get negative six minus one, which is equal to negative seven.

Therefore, we’ve shown the limit as π‘₯ approaches πœ‹ by two from the left of 𝑓 of π‘₯ and the limit as π‘₯ approaches πœ‹ by two from the right of 𝑓 of π‘₯ are both equal. And since both of these are negative seven, we can conclude the limit as π‘₯ approaches 𝑓 of π‘₯ at πœ‹ by two exists.

And for our third condition, we’ve already shown that 𝑓 evaluated at πœ‹ by two is negative seven. And since the limit as π‘₯ approaches πœ‹ by two from the left of 𝑓 of π‘₯ and the limit as π‘₯ approaches πœ‹ by two from the right of 𝑓 of π‘₯ are both negative seven, we can conclude the limit as π‘₯ approaches πœ‹ by two of 𝑓 of π‘₯ is negative seven. So both of these are negative seven. So our third condition of continuity is also true.

Hence, the function must be continuous at π‘₯ is equal to πœ‹ by two. And therefore, by checking all three continuity conditions, we were able to show the function 𝑓 of π‘₯ is equal to negative seven sin of π‘₯ plus seven cos of π‘₯ when π‘₯ is less than or equal to πœ‹ by two and 𝑓 of π‘₯ is equal to six times the cos of two π‘₯ minus one when π‘₯ is greater than πœ‹ by two is continuous when π‘₯ is equal to πœ‹ by two.

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