Video: APCALC04AB-P1A-Q02-742101687279

Given that 𝑓(π‘₯) = {π‘₯Β² + 2 for π‘₯ ≀ 3, and 6π‘₯ βˆ’ 7 for π‘₯ > 3, which of the following is true? [A] 𝑓(π‘₯) is not continuous at π‘₯ = 3. [B] lim_(π‘₯ β†’ 3) 𝑓(π‘₯) does not exist. [C] 𝑓(π‘₯) is continuous at π‘₯ = 3 but 𝑓′(π‘₯) does not exist. [D] 𝑓′(3) = 6.

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Video Transcript

Given that 𝑓 of π‘₯ equals π‘₯ squared plus two for values of π‘₯ less than or equal to three and six π‘₯ minus seven for values of π‘₯ greater than three, which of the following is true? Is it a) 𝑓 of π‘₯ is not continuous at π‘₯ equals three? b) The limit as π‘₯ approaches three of 𝑓 of π‘₯ does not exist? Is it c) 𝑓 of π‘₯ is continuous at π‘₯ equals three but 𝑓 prime of π‘₯ does not exist? Or d) 𝑓 prime of three is equal to six?

Here, we’ve been given a piecewise function and a number of statements about continuity and differentiability. Let’s recall what we know about continuity first. We say that continuity at a point exists when the left- and right-sided limits match the function evaluated at that point. In other words, a function is continuous at π‘Ž if the limit as π‘₯ approaches π‘Ž from the left of 𝑓 of π‘₯ is equal to 𝑓 of π‘Ž, which is also equal to the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯. So let’s begin by evaluating the limit of our function as π‘₯ approaches three from the left.

We see that when π‘₯ is less than or equal to three, 𝑓 of π‘₯ is equal to π‘₯ squared plus two. So we need to evaluate the limit as π‘₯ approaches three from the left of π‘₯ squared plus two. To achieve this, we can simply use direct substitution. The limit as π‘₯ approaches three from the left is three squared plus two, which is equal to 11. In fact, the function 𝑓 of π‘₯ is equal to π‘₯ squared plus two at the point that we’re interested in, at π‘₯ equals three. So 𝑓 of three is also 11. So far, so good. Next, we need to evaluate the limit as π‘₯ approaches three from the right of our function. This time for values of π‘₯ greater than three, the function 𝑓 of π‘₯ is equal to six π‘₯ minus seven.

So we’re going to evaluate the limit as π‘₯ approaches three from the right of six π‘₯ minus seven, which we can once again evaluate using direct substitution. It’s six times three plus seven, which is equal to 11. We have satisfied the criteria that the limit as π‘₯ approaches three from the left of our function is equal to 𝑓 of three, which is equal to the limit as π‘₯ approaches three from the right of the function. So 𝑓 of π‘₯ is continuous at π‘₯ equals three. Statement a cannot therefore be true. We’ll, now consider what it means for the limit as π‘₯ approaches three of the function to exist.

If the limit as π‘₯ approaches π‘Ž from the left of the function is equal to the limit as π‘₯ approaches π‘Ž from the right of a function, then we say that the limit exists. If the converse is true, if they’re not equal, then the limit does not exist. Well, we just showed that the limit as π‘₯ approaches three from the left of the function is 11 and the limit as π‘₯ approaches from the right of the function is also 11. This means the limit of the function as π‘₯ approaches three does actually exist and statement b cannot be true. Part c says 𝑓 of π‘₯ is continuous at π‘₯ equals three, but 𝑓 prime of π‘₯ does not exist. 𝑓 prime of π‘₯ is the derivative of our function.

Now, we’ve already established that 𝑓 of π‘₯ is actually continuous at π‘₯ equals three. So how do we establish whether the derivative exists. Well firstly, the function does need to be continuous. So that satisfied that criteria. So next we recall the definition of the derivative of a function at π‘Ž. It’s the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ minus 𝑓 of π‘Ž over π‘₯ minus π‘Ž. For a function to be differentiable at π‘Ž, the limit as π‘₯ approaches π‘Ž from the left of 𝑓 of π‘₯ minus 𝑓 of π‘Ž over π‘₯ minus π‘Ž must be equal to the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯ minus 𝑓 of π‘Ž over π‘₯ minus π‘Ž.

Let’s begin by evaluating the limit as π‘₯ approaches three from the left. It’s of 𝑓 of π‘₯ minus 𝑓 of three over π‘₯ minus three. For values of π‘₯ less than three, the function is π‘₯ squared plus two. And we already calculated 𝑓 of three to be 11. So we’re looking to find the limit of π‘₯ squared plus two minus 11 over π‘₯ minus three, which simplifies to π‘₯ squared minus nine over π‘₯ minus three. We’re not yet going to use direct substitution. If we were to substitute π‘₯ equals three into this expression, we get zero over zero, which we know to be undefined. So instead we look to manipulate the expression by factoring the numerator. π‘₯ squared minus nine is the same as π‘₯ plus three times π‘₯ minus three. And then we spot that we can simplify by dividing through by π‘₯ minus three. And so we’re looking to find the limit as π‘₯ approaches three from the left of π‘₯ plus three.

Now, we use direct substitution. It’s three plus three, which is equal to six. We’re now going to evaluate. This limit as π‘₯ approaches three from the right. Once again, it’s the limit of 𝑓 of π‘₯ minus 𝑓 of three over π‘₯ minus three. Here, 𝑓 of π‘₯ is equal to six π‘₯ minus seven. We’re now interested in values of π‘₯ greater than three. 𝑓 of three is still 11. And so we get six π‘₯ minus 18 over π‘₯ minus three. And this time, when we factor our numerator, we get six times π‘₯ minus three. Once again, we divide through by π‘₯ minus three. And we see that our limit is simply equal to six. We have shown that the limit as π‘₯ approaches three from the left of 𝑓 of π‘₯ minus 𝑓 of three of π‘₯ minus three is equal to the limit as π‘₯ approaches three from the right of 𝑓 of π‘₯ minus 𝑓 of three over π‘₯ minus three.

So in fact, 𝑓 prime of π‘₯ a derivative does exist here. c cannot be true. We’re now going to check d. This asks us whether 𝑓 prime of three is equal to six. Well, we have actually just shown that that is indeed true. Remember, the definition for the derivative of the function at π‘Ž is the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ minus 𝑓 of π‘Ž all over π‘₯ minus π‘Ž. We’ve just shown that that derivative does indeed exist and that 𝑓 of π‘₯ minus 𝑓 of three over π‘₯ minus three from the left and the right side is six. So the statement that’s true is d. 𝑓 prime of three is equal to six.

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