Video: APCALC02AB-P1A-Q24-727123078563

Given that 𝑓(π‘₯) = 3π‘₯ βˆ’ 4 π‘₯ > 2, and 𝑓(π‘₯) = 3π‘₯Β² βˆ’ 5π‘₯ π‘₯ ≀ 2, which of the following statements is true? [A] lim_(π‘₯β†’2) 𝑓(π‘₯) does not exist. [B] 𝑓(π‘₯) is continuous at π‘₯ = 2. [C] 𝑓(π‘₯) is differentiable at π‘₯ = 2 . [D] 𝑓(π‘₯) is not continuous at π‘₯ = 2.

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Video Transcript

Given that 𝑓 of π‘₯ is equal to three π‘₯ minus four when π‘₯ is greater than two and three π‘₯ squared minus five π‘₯ when π‘₯ is less than or equal to two, which of the following statements is true? a) the limit as π‘₯ tends to two of 𝑓 of π‘₯ does not exist. b) 𝑓 of π‘₯ is continuous at π‘₯ equals two. c) 𝑓 of π‘₯ is differentiable at π‘₯ equals two. And d) 𝑓 of π‘₯ is not continuous at π‘₯ equals two.

Let’s work through each of these statements in turn. We can see that 𝑓 of π‘₯ is a piecewise function. It’s made up of two different functions. For part a, we’re being asked to work out whether the limit, as π‘₯ tends to two, of 𝑓 of π‘₯ actually exists. Let’s evaluate then each part of our function as π‘₯ tends to two.

We can begin by evaluating three π‘₯ minus four as π‘₯ tends to two. As π‘₯ tends to two, our limit tends to three times two minus four, which is equal to two. So it appears that the limit, as π‘₯ tends to two, of the first part of our piecewise function does indeed exist. So let’s evaluate the limit, as π‘₯ tends to two, of three π‘₯ squared minus five π‘₯. We get three times two squared minus five times two, which is also two. So we can see that the limit as π‘₯ tends to two exists. So statement a cannot be true. And we’re going to use the information we just found to evaluate whether part b is true.

This says that 𝑓 of π‘₯ is continuous at π‘₯ equals two. Firstly, we can see that there are no gaps on the domain of our function. It’s equal to three π‘₯ minus four, when π‘₯ is greater than two, and three squared minus five π‘₯, when π‘₯ is less than or equal to two. So that two is included. We can also see that as we move along our function from the right-hand side, our function itself tends to two. And as we move along it from the left-hand side, our function also tends to two. And this tells us that there will be a point where the graph of three π‘₯ minus four and three π‘₯ squared minus five π‘₯ meet. And so, 𝑓 of π‘₯ must be continuous at π‘₯ equals two. And that, in turn, ensures that part d is incorrect.

Now, let’s sketch the graph of our function out. This will help us to see what this looks like. But it will also allow us to check whether part c is indeed incorrect. First, we have the graph of 𝑦 equals three π‘₯ minus four over the domain π‘₯ is greater than two. We then have the curve of 𝑦 equals three π‘₯ squared minus five π‘₯ over the domain π‘₯ is less than or equal to two.

We can actually see that our function 𝑓 of π‘₯ cannot be differentiable at π‘₯ equals two. And that’s because we have this sharp turn. There’s no unique tangent line that we could draw to evaluate the rate of change of our function at this point. And we’ve seen that c is indeed false. So the statement that’s true is b, 𝑓 of π‘₯ is continuous at π‘₯ equals two.

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