Video: APCALC02AB-P1A-Q24-727123078563

Given that π(π₯) = 3π₯ β 4 π₯ > 2, and π(π₯) = 3π₯Β² β 5π₯ π₯ β€ 2, which of the following statements is true? [A] lim_(π₯β2) π(π₯) does not exist. [B] π(π₯) is continuous at π₯ = 2. [C] π(π₯) is differentiable at π₯ = 2 . [D] π(π₯) is not continuous at π₯ = 2.

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Video Transcript

Given that π 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 π of π₯ does not exist. b) π of π₯ is continuous at π₯ equals two. c) π of π₯ is differentiable at π₯ equals two. And d) π of π₯ is not continuous at π₯ equals two.

Letβs work through each of these statements in turn. We can see that π of π₯ is a piecewise function. Itβs made up of two different functions. For part a, weβre being asked to work out whether the limit, as π₯ tends to two, of π of π₯ actually exists. Letβ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 letβ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 weβre going to use the information we just found to evaluate whether part b is true.

This says that π of π₯ is continuous at π₯ equals two. Firstly, we can see that there are no gaps on the domain of our function. Itβ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, π of π₯ must be continuous at π₯ equals two. And that, in turn, ensures that part d is incorrect.

Now, letβ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 π¦ equals three π₯ minus four over the domain π₯ is greater than two. We then have the curve of π¦ 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 π of π₯ cannot be differentiable at π₯ equals two. And thatβs because we have this sharp turn. Thereβs no unique tangent line that we could draw to evaluate the rate of change of our function at this point. And weβve seen that c is indeed false. So the statement thatβs true is b, π of π₯ is continuous at π₯ equals two.