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 π(π₯) = π/2, given π(π₯) = β7 sin π₯ + 7 cos π₯, π₯ β€ π/2 and π(π₯) = 6 cos 2π₯ β 1, π₯ > π/2.

04:35

### Video Transcript

Discuss the continuity of the function π at π₯ is equal to π by two given π of π₯ is equal to negative seven sin of π₯ plus seven cos of π₯ if π₯ is less than or equal to π by two and π of π₯ is equal to six cos of two π₯ minus one if π₯ is greater than π by two.

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

We need to check these three conditions for our function π when π₯ is equal to π by two. So letβs set our value of π equal to π by two. This then gives us the following. We can now check each of these conditions separately. Letβs start by checking that π by two is in the domain of our function π. To do this, we know that π is a piecewise-defined function. So we need to see which subdomain π by two lies in. And we can see that π by two lies in the first subdomain of this subfunction. So π by two is in the domain of π.

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

So weβve shown the first condition of continuity is true. Now letβs move on to the second condition. We want to show the limit as π₯ approaches π by two of π of π₯ exists. And since π by two lies at the endpoint of the subdomains of our piecewise-defined function, π of π₯ will have a different definition to the left and to the right of π₯ is equal to π 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 by two. We need to see if these are equal.

So letβs start with the limit as π₯ approaches π by two from the left of π of π₯. Our values of π₯ are going to be less than π by two. And we can see when π₯ is less than π by two, our function π of π₯ is equal to its first subfunction. And since the functions are equal when π₯ is less than π by two, their limits as π₯ approaches π by two from the left will also be equal. So we need to determine the limit as π₯ approaches π 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 by two into this function. However, weβve already done this above. Itβs equal to negative seven sin of π by two plus seven cos of π by two, which weβve already shown is negative seven.

Letβs now check that the limit as π₯ approaches π by two from the right is equal to negative seven. This time, our values of π₯ will be greater than π by two. And we can see that this means our function π of π₯ will be equal to its second subfunction. Once again, this tells us their limits as π₯ approaches π 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 by two, we get six cos of two times π by two minus one. And two times π by two is equal to π, and the cos of π is negative one. So we get negative six minus one, which is equal to negative seven.

Therefore, weβve shown the limit as π₯ approaches π by two from the left of π of π₯ and the limit as π₯ approaches π by two from the right of π of π₯ are both equal. And since both of these are negative seven, we can conclude the limit as π₯ approaches π of π₯ at π by two exists.

And for our third condition, weβve already shown that π evaluated at π by two is negative seven. And since the limit as π₯ approaches π by two from the left of π of π₯ and the limit as π₯ approaches π by two from the right of π of π₯ are both negative seven, we can conclude the limit as π₯ approaches π by two of π 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 by two. And therefore, by checking all three continuity conditions, we were able to show the function π of π₯ is equal to negative seven sin of π₯ plus seven cos of π₯ when π₯ is less than or equal to π by two and π of π₯ is equal to six times the cos of two π₯ minus one when π₯ is greater than π by two is continuous when π₯ is equal to π by two.

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