Video: AP Calculus AB Exam 1 β€’ Section II β€’ Part B β€’ Question 3

A function 𝑓 is continuous and twice differentiable for all values of π‘₯. The given figure shows the graph of 𝑓′, the derivative of the function 𝑓 on the closed interval [βˆ’3, 2]. The graph has horizontal tangents at π‘₯ = βˆ’1.5 and π‘₯ = one. The areas of the regions A, B, and C are 3, 1, and 12, respectively, and 𝑓(βˆ’1) = βˆ’3. i. Find all π‘₯-values on [βˆ’3, 2] such that the function 𝑓 has a local maximum. Justify your answer. ii. Find all π‘₯-values on [βˆ’3, 2] such that the function 𝑓 has a point of inflection. Justify your answer. iii. Evaluate the lim_(π‘₯ β†’ 2) (4 𝑓(π‘₯))/(π‘₯ βˆ’ 2). Explain your reasoning. iv. Evaluate the lim_(π‘₯ β†’ 2) (15 + 𝑓(π‘₯))/(π‘₯ βˆ’ 2). Explain your reasoning.

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Video Transcript

A function 𝑓 is continuous and twice differentiable for all values of π‘₯. The given figure shows the graph of 𝑓 prime, the derivative of the function 𝑓 on the closed interval between negative three and two. The graph has horizontal tangents at π‘₯ is equal to negative 1.5 and π‘₯ is equal to one. The areas of the regions A, B, and C are three, one, and 12, respectively. And 𝑓 of negative one is equal to negative three. Part one, find all π‘₯-values on the closed interval between negative three and two such that the function 𝑓 has a local maximum. Justify your answer. Part two, find all π‘₯-values on the closed interval between negative three and two such that the function 𝑓 has a point of inflection. Justify your answer. Part three, evaluate the limit as π‘₯ tends to negative two of four 𝑓 of π‘₯ over π‘₯ minus two. Explain your reasoning. Part four, evaluate the limit as π‘₯ tends to two of 15 plus 𝑓 of π‘₯ over π‘₯ minus two. Explain your reasoning.

In order to give us some more space to answer the question, I’m just going to clear up parts two to four. And we’ll focus on part one for now. For part one, we’re required to find all π‘₯-values on the closed interval between negative three and two such that 𝑓 has a local maximum. Now if 𝑓 has a local maximum, that means there must be a critical point. And a critical point will occur when the differential 𝑓 is equal to zero. Or we could say that 𝑓 prime of π‘₯ is equal to zero.

Now, not all critical points are local maximums. In order to find out what our critical point is, we can look at the second derivative of 𝑓 with respect to π‘₯ or 𝑓 double prime of π‘₯. Now if 𝑓 has a local maximum, this tells us that, at π‘₯, our function 𝑓 will be concave down, as we can see in this little sketch. And if 𝑓 is concave down, that means that the second derivative of 𝑓 with respect to π‘₯ must be less than zero. Now, 𝑓 double prime of π‘₯ is the derivative of 𝑓 prime of π‘₯ with respect to π‘₯. So 𝑓 double prime of π‘₯ tells us about the slope of 𝑓 prime of π‘₯. Therefore, if 𝑓 double prime of π‘₯ is negative, 𝑓 prime of π‘₯ has a negative slope and is therefore decreasing.

Combining these two pieces of information, we know that 𝑓 will have a local maximum when 𝑓 prime is equal to zero and decreasing. From our graph, we can see that 𝑓 prime is equal to zero when π‘₯ is equal to negative two, π‘₯ is equal to negative one, and π‘₯ is equal to two. And the only one of these three points where it is decreasing is at π‘₯ is equal to negative one. Therefore, our solution to the first part is that the only π‘₯-value for which 𝑓 has a local maximum on the interval between negative three and two is when π‘₯ is equal to negative one.

Part two, find all π‘₯-values on the closed interval between negative three and two such that the function 𝑓 has a point of inflection. Justify your answer.

Now, point of inflection occurs when the concavity of 𝑓 changes from concave up to concave down or concave down to concave up. And we know that the concavity of 𝑓 is represented by the second derivative of 𝑓 with respect to π‘₯ or 𝑓 double prime of π‘₯. We know that when 𝑓 double prime of π‘₯ is less than zero, 𝑓 is concave down. And when 𝑓 is greater than zero, 𝑓 is concave up. So what we’re looking for is a sign change of 𝑓 double prime of π‘₯. Now our graph is of 𝑓 prime of π‘₯, not 𝑓 double prime of π‘₯. So let’s try to work out what this means for 𝑓 prime of π‘₯.

Well, 𝑓 double prime of π‘₯ tells us about the slope of 𝑓 prime of π‘₯. So when there’s a sign change to 𝑓 double prime of π‘₯, this means there’ll be a sign change in the slope of 𝑓 prime of π‘₯. So we can say that 𝑓 prime of π‘₯ changes from increasing to decreasing or decreasing to increasing. When a function changes from increasing to decreasing, we have a local maximum. And when it changes from decreasing to increasing, we have a local minimum. So we’re looking for local maximum and local minimum on 𝑓 prime of π‘₯. From our graph, we can see that there are two.

Now in order to find the values of these points, we need to refer back to the question. Since we’ve been told that the graph has horizontal tangent at π‘₯ is equal to negative 1.5 and π‘₯ is equal to one. And these horizontal tangents will occur at the critical points of our graph and the critical points of the local minimum and the local maximum on the graph. Therefore, we can deduce that the point of inflection of 𝑓 will be at π‘₯ is equal to negative 1.5. And π‘₯ is equal to one.

Part three, evaluate the limit as π‘₯ tends to negative two of four 𝑓 of π‘₯ over π‘₯ minus two. Explain your reasoning.

Let’s start by trying to find this limit using direct substitution. We obtain four timesed by 𝑓 of negative two over negative two minus two, which can be simplified to give negative 𝑓 of negative two. Now, we do not know the value of 𝑓 of negative two. We can use the fundamental theorem of calculus in order to help us find this value. The fundamental theorem of calculus tells us that the integral from π‘Ž to 𝑏 of 𝑓 prime of π‘₯ with respect to π‘₯ is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž. And the reason why we’re going to use the fundamental theorem of calculus here is because we’ve been given the graph of 𝑓 prime of π‘₯. And we’re trying to find 𝑓 of negative two. And if we choose our bounds right, we’ll be able to find 𝑓 of negative two using this method. We have also been given the value of 𝑓 of negative one in the question. So the two bounds we will use here will be negative one and negative two.

Substituting in π‘Ž is equal to negative two and 𝑏 is equal to negative one, we obtain that the integral from negative two to negative one of 𝑓 prime of π‘₯ with respect to π‘₯ is equal to 𝑓 of negative one minus 𝑓 of negative two. And we can substitute in this value of 𝑓 of negative one since it’s equal to negative three, to give us that the integral is equal to negative three minus 𝑓 of negative two. Now, let’s think about what this integral is represented by on our graph. It’s the area between the curve and the π‘₯-axis between the π‘₯-values of negative one and negative two. So that’s this region here, which is also labelled as region B.

Now, if you look back at the question, we’ve in fact been given the area of region B. And it’s equal to one. Therefore, we know that the value of this integral must be one. Therefore, we can say one is equal to negative three minus 𝑓 of negative two. Rearranging this, we obtain that 𝑓 of negative two is equal to negative four. Now, we simply substitute this value of 𝑓 of negative two back in, in order to find the value of the limit. In doing so, we evaluate the limit to be equal to four.

Part four, evaluate the limit as π‘₯ tends to two of 15 plus 𝑓 of π‘₯ over π‘₯ minus two. Explain your reasoning.

Now, let’s try to solve this limit, again using direct substitution. We obtain that it’s equal to 15 plus 𝑓 of two over two minus two. Now, the denominator of this fraction is equal to zero. Therefore, we know that this fraction will be undefined. However, if the numerator is also equal to zero, then we’ll be able to use L’HΓ΄pital’s rule. So let’s check what 𝑓 of two is equal to. Similarly to the last part, we’ll use the fundamental theorem of calculus. However, this time, we’re trying to find the value of 𝑓 of two, not 𝑓 of negative two. So we know that one of our bounds will be two. And since we’ve been given 𝑓 of negative one in the question, we can again use negative one as the other bound.

Letting π‘Ž be equal to negative one and 𝑏 be equal to two, we find that the integral from negative one to two of 𝑓 prime of π‘₯ with respect to π‘₯ is equal to 𝑓 of two minus 𝑓 of negative one. We know that 𝑓 of negative one is negative three. So we can substitute this in here. The right-hand side of our equation becomes 𝑓 of two plus three. Now, the integral from negative one to two of 𝑓 prime of π‘₯ with respect to π‘₯ is represented by the area between the curve of 𝑓 prime of π‘₯ and the π‘₯-axis between π‘₯ is equal to negative one and π‘₯ is equal to two. So that’s this area here, represented by the region C. We’ve been given in the question that the area of region C is 12. Now we must be careful here since this region is below the π‘₯-axis. Therefore, the value of its integral will be negative. And so we can say that the integral from negative one to two of 𝑓 prime of π‘₯ with respect to π‘₯ must be equal to negative 12.

We obtain that negative 12 is equal to 𝑓 of two plus three, which can be rearranged to give 𝑓 of two is equal to negative 15. Now, we can substitute this value of 𝑓 of two back in to our evaluation of the limit. We obtain that our limit is equal to 15 minus 15 over two minus two, which is equal to zero over zero. And as mentioned earlier, this enables us to use L’HΓ΄pital’s rule on this limit.

L’HΓ΄pital’s rule tells us that the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ over β„Ž of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑔 prime of π‘₯ over β„Ž prime of π‘₯. In our case, we have that 𝑔 of π‘₯ is equal to 15 plus 𝑓 of π‘₯. And β„Ž of π‘₯ is equal to π‘₯ minus two. Differentiating 𝑔 of π‘₯ with respect to π‘₯, we find that 𝑔 prime of π‘₯ is equal to 𝑓 prime of π‘₯. And differentiating β„Ž of π‘₯ with respect to π‘₯, we get that β„Ž prime of π‘₯ is equal to one.

Now, we can substitute 𝑔 prime and β„Ž prime into L’HΓ΄pital’s rule. We obtain that the limit as π‘₯ approaches two of 15 plus 𝑓 of π‘₯ over π‘₯ minus two is equal to the limit as π‘₯ approaches two of 𝑓 prime of π‘₯ over one. And here, we can use direct substitution, giving us that our limit is equal to 𝑓 prime of two. In order to find the value of 𝑓 prime of two, we can simply read it off our graph. We see that when π‘₯ is equal to two, 𝑓 prime of π‘₯ is equal to zero. Therefore, the value of this limit and the answer to part four must be zero.

We’ve now completed all four parts of the question. Let’s quickly go over our solutions. In part one, we found that 𝑓 has a local maximum at π‘₯ is equal to negative one. In part two, we found that 𝑓 has an inflection point at π‘₯ is equal to negative 1.5 and π‘₯ is equal to one. In part three, we found that the limit as π‘₯ tend to negative two of four 𝑓 of π‘₯ over π‘₯ minus two is equal to four. And in part four, we found that the limit as π‘₯ tends to two of 15 plus 𝑓 of π‘₯ over π‘₯ minus two is equal to zero.

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