Question Video: Deciding If a Piecewise Function Is Continuous at a Certain Point Mathematics • Higher Education

Is 𝑓(π‘₯) = (2π‘₯ + 4)/(π‘₯ + 2) if π‘₯ < βˆ’2, 𝑓(π‘₯) = 0 if π‘₯ = βˆ’2, 𝑓(π‘₯) = (π‘₯Β²+ 6π‘₯ + 8)/(π‘₯ + 2) if π‘₯ > βˆ’2 continuous at π‘₯ = βˆ’2?

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Video Transcript

Is the function 𝑓 of π‘₯, which is piecewise defined to be two π‘₯ plus four all over π‘₯ plus two if π‘₯ is less than negative two. Zero if π‘₯ is equal to negative two. And π‘₯ squared plus six π‘₯ plus eight all over π‘₯ plus two if π‘₯ is greater than negative two, continuous at π‘₯ equals negative two.

To find out, we go through our checklist. For the function 𝑓 to be continuous at π‘₯ equals negative two, 𝑓 of negative two must be defined. The limit of 𝑓 of π‘₯ as π‘₯ approaches negative two must exist. And these values must be equal. So we start by checking that 𝑓 of negative two is defined. Well, that’s easy to see from the question. We’re told that 𝑓 of π‘₯ is zero if π‘₯ is negative two. So 𝑓 of negative two is defined. 𝑓 of negative two is equal to zero. Now, we need to check that the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two exists. There are different rules for the values of 𝑓 of π‘₯ when π‘₯ is less than negative two and when π‘₯ is greater than negative two. Which we’ll use to check the left-hand and right-hand limits, making sure that they exist and are equal.

What’s the left-hand limit? Well, to the left of negative two, 𝑓 of π‘₯ is defined by two π‘₯ plus four all over π‘₯ plus two. So we have to find the limit of two π‘₯ plus four all over π‘₯ plus two, as π‘₯ approaches negative two from the left. If we try direct substitution, we get the indeterminate form zero over zero. So we have to find this limit some other way. We do this by factoring the numerator, getting two times π‘₯ plus two. And this factor of π‘₯ plus two cancels with the factor of π‘₯ plus two in the denominator. And so our limit is that of the constant function two whose value must be two. So the left-hand limit exists. What about the right-hand one?

To the right of π‘₯ equals negative two, 𝑓 of π‘₯ is this algebraic fraction. And we can find the value of this limit in a similar way. Factoring the numerator and canceling the common factor of π‘₯ plus two, to get the limit of π‘₯ plus four as π‘₯ approaches negative two from the right. This can be solved using direct substitution. We get negative two plus four, which is two. So this right-hand limit also exists and is equal to the left-hand limit. Both of them have the value two. Hence the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two from either direction also exists and has the value two.

All that’s left to check is this third point. For 𝑓 of π‘₯ to be continuous at π‘₯ equals negative two, we need the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two to be equal to the value of 𝑓 at negative two, 𝑓 of negative two. But we’ve just seen that the value of this limit is two, whereas 𝑓 evaluated at negative two is zero. The limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is not equal to 𝑓 of negative two. Although the first two conditions on our checklist are satisfied, the last one isn’t. And so our answer is no. The function 𝑓 is not continuous at π‘₯ equals negative two, as the limit of 𝑓 of π‘₯ as π‘₯ approaches negative two is not equal to 𝑓 of negative two.

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