Khan Academy Live: AP Calculus

Hi and welcome to live tutoring for the AP Calculus exams provided by Con Academy! In case you are curious, I am not Sal Con; my name is Dave. I first took the AP Calculus test back in 2006, and before joining KH Academy, I was an AP Calculus teacher.

So, I'm really looking forward to spending about the next hour together practicing a range of multiple choice and free response questions. Know that the test is right around the corner on May 9th, and we hope that this tutoring session will be as helpful as possible. Today, we're going to go in depth on one topic: analyzing graphs of functions and their derivatives. This is a topic that's on the AB and BC tests, so it should be helpful for anybody taking an AP Calculus test this year. But the items that we'll look at all come from the AB exam.

Good news! You will not need a calculator to work out the problems today, but I do recommend that you have something to write with and a piece of paper so that you can follow along as we go. Just a plug for our website: after the session is over, don't forget that you can go to khanacademy.org, and we have over 500 videos, exercises, and articles for the AB exam, and nearly 600 for the AP Calculus BC exam, including video explanations for two full free response question sections.

Okay, so let's get started. This first section, we're actually going to take about 5 to 10 minutes to review the key concepts that we'll be going over today. So, if you haven't already, grab something to write with; it will be helpful to have these notes to look back on when we start the multiple choice questions.

Let's take a look at this first graph, and I'll zoom in a little bit so that it's hopefully easier to read. Let's review how we would interpret if this were a graph of the original function f.

This is the most straightforward because we can say that this original section right here at the beginning is going up, and so we can say it's an increasing interval. Similarly, here at the end, we have another increasing interval, right? Starting simple. In the middle, we have a decreasing interval here in the color red.

What else can we conclude when we're looking at a graph of the original function f? We can say that this point up here at the top is a relative maximum; it's the highest point in that section of the graph. It's a relative maximum, and down here, opposite, we have a relative minimum—relative minimum.

What else? Well, if we look at the curvature or the concavity, this first section of the graph right here, we can say it is concave down—concave down. And maybe it's a little cheesy, but I like to write it like this instead of the letter "N," to remind myself that that section—any part of a general parabola shape in that part—is concave down.

Up here, we have concave up—concave up. Again, I can use a curve instead of the letter "U." And last, this point in the middle where the concavity switches, it looks like it's the origin on this graph, is our point of inflection. That should be a familiar term to you by this point in the year; we'll label it as our point of inflection.

So, on this graph, it's the most straightforward. When we're looking at a graph of f, what you see is pretty much what you get; you can interpret it literally. But now we're going to take a look at a second graph, and it will look exactly the same at first—same shape. But this time, we're going to label it as f prime, the graph of the first derivative of f.

How would we interpret this graph? Well, now we would say that up here where f prime is greater than zero, that gives us information about the original function f. It tells us that the original function f was increasing during those intervals. It was increasing during those intervals.

So, even though this section, if we look at it visually, is decreasing, because this is a graph of f prime, that tells us that the original function f was decreasing in that interval—or was increasing in that interval. And down here, we have the opposite; we have f prime is less than zero, and we can say that that means that the original function f was decreasing.

Alright, what other information can we identify from this graph? We've got these interesting points where we have a transition in between a negative f prime value and a positive f prime value, or vice versa.

So, let's think about this first one right here—that’s where the original function f switched from decreasing. So I’m visualizing it down here: it was decreasing, and then we had a moment where it hits a slope value of zero. Then it switches to a positive value of f prime and starts increasing, and so that moment where we had a transition is our relative minimum—our relative minimum right there.

And there's another point on here that also appears to have that same characteristic; it’s this last one over here—another relative minimum. I’ll abbreviate relative min, but our middle point is the opposite. It actually switches from f increasing, a positive value of f prime, to decreasing, a negative value of f prime.

So that point where it switches from a positive value to a negative value is our relative maximum. It looks like we have a question coming in from YouTube, asking about the difference between relative and absolute minimums and maximums.

We'll get into that difference later as we look at the free response questions, but the short answer is that a relative maximum is the highest value within an interval of the function, but it may not be the highest value overall; there could be an endpoint that reaches a higher value.

Okay, so now we've looked at the graph of f prime, and we've also looked at the graph of the original function f. We'll finally look at a graph of the second derivative f double prime. It'll be the same shape. We'll see what information we can find from that graph.

Let's take a look—here we go. f double prime, I hope you're able to record this at home on a piece of paper so that you can look back on it. This first section up here where the second derivative is positive tells us that the original function f was concave up—concave up.

And down here where the second derivative is negative, the original function f was concave down—concave down. And just like the last graph, there are these three interesting points here where the second derivative value is neither positive nor negative; it's zero, and it has a switch, a transition in sign.

So, what happens when the second derivative switches? When the second derivative switches, those three points on here are all points of inflection—all points of inflection. So we have one, two, three points of inflection for this function.

Unlike with a maximum or a minimum, there's no different vocabulary term for when it switches from negative to positive as from positive to negative; they are all equally points of inflection.

Alright, well, before we move on to multiple choice, let's take a look—there is another question coming in from YouTube. It asks, "Can you identify a point of inflection on the graph of f prime, the first derivative graph?"

So that's a good question that comes up often, and we’ll take a look back here at this graph because we did mention points of inflection for the other two graphs, but we didn't talk about them here at all. The key thing to remember for that is that the second derivative is the same thing as the slope of the first derivative.

The second derivative is the same thing as the slope of the first derivative. We just mentioned that a point of inflection occurs when the second derivative changes sign, so we can also say that a point of inflection occurs when the slope of the first derivative changes sign.

And we're looking here at a graph of the first derivative. So if we ask ourselves, where does the slope change sign? Where does the slope change sign? And right here, in a big gray circle, the slope changes sign from a positive slope to a negative slope.

And again, down here, the slope of f prime changes sign. So those two points—I’ll abbreviate point of inflection—we can identify the point of inflection from a first derivative graph. Great!

Okay, so that wraps up the notes, the concept review section. The next thing that we're going to do is look at eight different multiple choice questions. Like I mentioned, all of these questions come from past AP Calculus AB exams, and I encourage you to kind of keep track of how you're doing at home.

You can maybe tally how many of the eight you think you would have been able to get correct on your own, and you can identify the problems that were especially difficult for you.

On the first problem, I'll walk through the answer, but on the second one, I'm going to stall a little bit, move a little bit slower, so that you have a chance to work ahead and solve it before me.

Let's take a look at the first problem. Now, this is from the 1998 exam—before most, if not all of you, were born. Let’s read the problem:

The graph of f is shown in the figure above. Which of the following could be the graph of the derivative of f?

Well, before we even look at the answer choices, I'm going to zoom in on this original graph, and let's think about what we know the derivative graph would need to look like.

So this first section here is increasing; f is increasing, and that tells me that the derivative graph will need to have a positive y-value. The graph of f prime will need to have a positive y-value in that section.

So I'm going to create this little blue box right there, telling me that it will need to have a positive y-value. And on the second half here, where we have a decreasing interval, the graph of f prime will need to have negative y-values, and then we have this moment at the top where the slope is zero for an instant. The graph of f prime will need to be zero there as well.

Okay, with that in mind, let's take a look at the five answers. So, we're looking for positive, zero, then negative. See, I want to be able to see all five at once—positive, zero, and then negative.

So it looks to me like we have some that clearly don’t work; there’s no negative section here. This one, C, gets it backwards, and D is just wrong in almost each section. Hard to know where to start with D, but we do have two contenders: A and E, because they both start out positive, have a moment where it's zero derivative value of zero, and then negative.

How are we going to choose between the two? Well, if we think about what E is saying, E is saying that the very beginning of the graph has the largest positive derivative. We would expect then that the original function would be increasing at its steepest rate right there.

If E were correct, the original graph would be increasing at its steepest rate. But in fact, at the beginning, the graph is nearly flat; it is barely increasing. So we can say that A matches that because A has a small positive derivative, and it reaches its largest value somewhere in the middle of this first interval, which is where we have our steepest section on the original graph.

That's why we can say that A, in this case, is the correct answer and not E.

Okay, so let's look at a similar question—another multiple choice question—and this time, like I mentioned, I'm going to read it a little slower. It's intentional; I'm trying to give you time to solve the problem on your own and see if it makes sense to you.

So, number two—I know it says 11; you can ignore those numbers—it’s a second multiple choice problem for us. The graph of a function f is shown above. Which of the following could be the graph of f prime, the derivative of f?

And I'm going to stall for another 10 seconds. If you don't have something to write with, at least say in your head which letter answer choice you think is correct, or you could type it in the chat box and argue with other people for a second about if it's A or B or E or C or D.

And now, let's take a look. Well, the first section is increasing—a positive derivative value. The last section is also increasing—a positive derivative value. The middle segment is decreasing, so it should be a negative y-value on the graph of f prime.

If I look to see which answer has that same setup, only answer choice B matches that one, and in this case, answer choice B is correct.

So how did you do? Were you able to get that one? No problem? If so, great job! If not, there's plenty of other practice problems like this on our website so that you can get the hang of it.

You know the AP Calculus test—one of the nice things about it is it's pretty predictable what types of questions are going to be coming up on the test, and so you will be very likely to see a question that connects the graph of the original function with the derivative. I hope that when you see it, you'll approach that question with complete confidence.

Let's take a look now at a slightly different type of multiple choice problem. The graph of a twice differentiable function f is shown in the figure above. Which of the following is true?

So instead of asking us to compare two different graphs, we're asked to order the values of f, f prime, and f double prime at the x-value of 1 from least to greatest.

Well, let's start with remembering that this is the graph of the original function, and so f of 1 is the y-value right here. So we can say that’s zero. But what about the derivative value at one?

Well, there's no way we can actually calculate the precise slope at that value because we don't have the function in front of us, but we can say with certainty that the value of f prime at one would be positive—greater than zero—because the graph has a positive slope at that value, one.

And the second derivative? We also can't calculate the value, but we can say that this graph is concave down. Remember, it's part of a general N shape; it's concave down, so it's negative at that point.

And so we have one value equal to zero, one value that's positive, and one value that's negative. If we arrange those from least to greatest, we’d end up with the second derivative at one less than the original function at one, both less than the derivative first derivative function at one, and that matches up with answer choice D.

Okay, let's look at the second question in this pair, and this is very similar to the last question, but I'm going to read it a little bit slower. Take a look—there's one key difference between this problem and the last problem, so don't approach it exactly the same way, but see if you're able to get it correct without any help from me.

The graph of a differentiable function f is shown above. If h of x equals the anti-derivative from 0 to x of f of t dt, which of the following is true?

So let me give you some time to think about how you would approach that, and in just a few seconds, I'll start working it out. You can chat in the box about which answer you think is correct and see if you're able to get this one.

Okay, so first, let's think about what's different. The difference is that this time, we’re asked to evaluate h of 6, but that’s actually the anti-derivative from 0 to 6 of the graph we’re looking at.

So, h of 6 equals the anti-derivative from 0 to 6, the definite integral from 0 to 6. And when we look at the graph and we think about the definite integral value from 0 to 6, although we can't calculate the exact value of that accumulated quantity, we know that it will be negative because all of that area is below the horizontal axis.

So we can say that it's going to be less than zero, because all the area is below the horizontal axis—the x-axis. What about the derivative h prime of 6? Well, that will be equal to f of 6 at that point right here, and we know f of 6 from looking at the graph is equal to zero.

Last, the second derivative is the same thing as the first derivative of f, f prime of 6. So here we’re thinking about the slope at this value six, and the slope there is positive.

So if we arrange those from least to greatest, it actually goes in order this time. We have h of 6 being negative, h prime of 6 being zero, h double prime of 6 being positive—answer choice A.

How did you do? Did you find that one easy, medium, difficult? I hope it wasn't too bad.

Okay, so we've got four multiple choice problems done. The next four we're going to go maybe a little bit quicker, and each one will be a little bit different than the other one. So for each one, I encourage you to work a little bit faster than me and try them on your own before hearing me say the answer.

Alright, right? So question five—zoom in, make it easier to see. The graph of the derivative f prime is shown above for -2 ≤ x ≤ 5. On what intervals is f increasing?

Go ahead, type in your answer, write it down on paper, or just say it in your head, and let's take a look. So the common mistake is to look at intervals where the graph itself is increasing, but remember that we're looking at a graph of f prime.

And as we said in the initial notes, when f prime is greater than zero, that’s when the original graph f is increasing. And so keeping that in mind, we can say that this whole interval from -2 all the way to positive 3 is the interval where f is increasing. B is the correct answer for this problem.

Now, depending on how you interpret the question, you might have thought that they would split it up into two intervals—from -2 to 1 and then from 1 to 3—because we do have that moment where the derivative equals zero, but that wasn't an answer choice. So we know that the best answer is B, from -2 to 3.

Okay, number six: the graph of the derivative of a function f is shown in the figure above. The graph has horizontal tangent lines at x = -1, x = 1, and x = 3. At which of the following values of x does f have a relative maximum?

Okay, so a relative maximum, we can say that a relative maximum occurs when the derivative function switches from positive to negative. Because that's where the original function f goes from increasing to decreasing, switches from positive to negative.

So here, we have positive; here we have a switch from positive to negative. It's kind of weird because it looked like it was more like 3.95, like a little bit before four, but the closest and best answer here is going to be four only.

Alright, four only. If it was asking about relative minimum, on the other hand, then -2 only would be the best choice for that.

So we have two more multiple choice questions, right? The graph of y = f of x is shown in the figure above. On which of the following intervals are the first derivative greater than zero and the second derivative less than zero?

It's helpful to see problems that have this other derivative notation, Lib's notation. Even though the more common one that you’ll see on the AP test will be the Leibniz notation. But if we were to read this problem again, instead of reading how I just said it, let's interpret it a little bit in an easier way so that when we look at the graph, we know what to look for.

So the last sentence in the problem says, on which of the following intervals is the original function increasing and concave down—increasing and concave down. Okay, so we can tell clearly that the graph is increasing here from A to B and from B to C, and then we can also say that the graph is concave down from B to C and from C to D.

And the only section—the only interval where it is both of those two things—increasing and concave down—is the interval from B to C. And so, because it needs to be both, we have that conjunction in there; we’re going to select two only for this problem.

Alright, now I know there are a lot of questions in the comments asking about other topics—whether it's related rates or optimization or Taylor series—and all of those are concepts that we would love to review. But given just one hour, we're going to focus on graphical derivatives, and so when the hour is over, go ahead and look up those topics on KH Academy to watch the videos and exercises that we have related to the AP material for those concepts.

Let's take a look at one last multiple choice question. The graph of the function f is shown in the figure above. For which of the following values of x is the derivative f prime of x positive and increasing—positive and increasing?

So f prime of x positive—that means that the graph that I'm looking at should have a positive slope—positive slope. So that would be here and here. So it's either going to be C or E, but f prime of x needs to be increasing. That means that the second derivative is also positive; the slope of the value of f prime is positive.

And so that would be where the graph is concave up—concave up would narrow it down to just point E. At point C, we are concave down. Answer choice E for that one is correct.

This one actually a small percentage of people got it correct on the actual AP test; it appears simple at first, but recognizing that f prime of x increasing is referring to the second derivative and being concave up was the key reason I think why a lot of people mistakenly got this question wrong.

Okay, that wraps up the multiple choice section. So there were eight problems. I hope you're able to tally up how you did overall on those eight, and I hope that you have more confidence when you see problems related to graphical derivatives on the upcoming test.

We now have time to go over two free response questions, and we'll actually be writing out full answers here so that we can see how we would have scored out of the nine points available on a free response question.

So let's take a look at the first free response. This is from the 2015 exam, two years ago. The figure above shows the graph of f prime, the derivative of a twice differentiable function f. On the interval from 3 to 4, the graph of f prime has horizontal tangents at -1, 1, and 3.

The areas of the regions bounded by the x-axis and the graph of f prime on the intervals -2 to 1 and 1 to 4 are 9 and 12, respectively. Part A: find all x-coordinates at which f has a relative maximum. Give a reason for your answer.

So, I don’t know how you all compare it to the students I used to have, but many of them, when we were doing practice problems, would say, “Oh, yeah, I don’t know how to give a reason when it's the actual test.” But in these practice problems, I'm just going to write my answer. But for today's tutoring, let's actually practice writing the reason we would provide for the test so that we can see if the language we use would meet the AP standards.

So, a relative maximum—give a reason for your answer. Well, we can remember that a relative maximum is where the derivative f prime switches from positive to negative, and that on this graph occurs at -2.

So our relative maximum would be -2. Let's practice writing in our full answer, though: f prime of x changes from positive to negative at x = -2. We can say therefore—this is an abbreviation for therefore, like in conclusion—the original graph f has a relative maximum at x = -2.

So how was this problem scored on the test? This was two points. The first point you got for getting the correct answer of -2, and the second point was for providing your reason. The common mistakes that we see from students are using vague language like "the graph changes from positive to negative" instead of identifying which graph we're referring to, f prime of x.

So we want to avoid using vague words like “the graph” or “the slope.” A second mistake was not referencing this value or this graph of f prime and instead always just referring back to the original function. So that was two points.

Let’s take a look at Part B now. Part B says, "On what open intervals contained in the interval from -3 < x < 4 is the graph of f both concave down and decreasing?"

Concave down and decreasing, okay? So decreasing is going to occur when the derivative f prime is less than zero. So I'm going to highlight in blue all the spaces where the original function was decreasing. Concave down is going to happen when the derivative function—let me switch colors here to make it a little bit easier—do gray.

When the derivative function f prime is decreasing, like a negative slope of f prime, that tells us that the second derivative will be negative, and so f is concave down. This was the key reason why people made mistakes on this problem, so I'm writing out in a little more detail.

And this graph is decreasing. Let's look in that blue section because that is the only candidate for the final answer. So we have here from -2 to -1 and we also have from 1 to 3, from 1 to 3. Those two segments, from -2 to -1 and from 1 to 3, are both increasing and concave down on the graph of f.

And so writing in our answer, we can write it just in one sentence. But we have to use as precise of language as we can. So f—we can kind of copy the language they used in the actual question—f is both concave down and decreasing.

f is both concave down and decreasing on the intervals. And we saw them right above; it was from -2 to -1 and from 1 to 3. And for our rationale, we can say because the derivative function is decreasing—that's how we knew it was concave down—and negative, so we know f was decreasing there on these intervals.

This was another two-point problem. One for getting the correct intervals, and another one for having the correct rationale. And I just want to say again that f prime decreasing was our rationale for concave down, and then we knew the original function was decreasing because f prime was negative.

So those two parts are the corresponding rationale for the two criteria we were looking at. If we think about common mistakes here again, students were able to identify the intervals, but many of them struggled to express the reason for it, and oftentimes they did not say which graph or slope or derivative they were referring to.

So get in the good habit of always writing f or f prime or f double prime instead of writing “the graph.”

Alright, let's take a look now at part C of this problem. Part C is—I'll reproduce the graph so it's a little bit easier and cleaner for us to look at it again.

So part C is a two-point problem again, and it says, "Find the x-coordinates of all points of inflection for the graph of f. Give a reason for your answer." Now, we talked actually because one of you brought up a question about how to identify the points of inflection from a graph of f prime.

So see if you remember, and if you can identify the points of inflection and a rationale for them. So in about 10 seconds, I’ll jump in, but see if you can guess how many points of inflection there are on the graph of f and the one sentence you might use to justify it.

Alright, so what do we think? How many points of inflection are there? A point of inflection is going to occur where the first derivative graph changes sign.

First derivative, and I should say the slope of the first derivative changes sign—that is a point of inflection. We have that in our notes from the beginning of the session, but we'll reproduce it here. And that happens at -1; it happens again at positive 1 and for the last time at positive 3.

So there are three points of inflection on this graph, and we can write that here: f has a point of inflection at x = -1, x = 1, and x = 3. Because f prime changes from decreasing to increasing, changes from decreasing to increasing at -1 and 3, and opposite, we have f prime changing from increasing to decreasing at x = 1.

So it is a bit of writing; it’s tempting to just abbreviate and write just a really short justification, but then we run the risk of not being specific enough and losing out on what could have been guaranteed points.

Okay, so there’s a part D to this problem, but the part D doesn't relate to the concept we're going over today.

Good news, though, is that we do have a video on our website that explains the answer to part D, so just look up the AP Calculus AB videos, and this is from the 2015 exam.

Okay, the last free response question that we'll go over today is from the 2013 exam. Let’s take a look: The figure above shows the graph of f prime, the derivative of a twice differentiable function f on the closed interval 0 ≤ x ≤ 8. The graph of f prime has horizontal tangent lines at x = 1, x = 3, and x = 5.

The areas of the regions between the graph of f prime and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies the condition f(8) = 4.

There’s a lot of information; I assume we’ll use all of it eventually, but let’s take a look at part A: find all values of x on the open interval 0 < x < 8 for which the function f has a local minimum. Justify your answer.

Well, we’ve seen several problems, both multiple choice and free response, where we’re asked to identify a local optimum from a graph of f prime. And so hopefully you’re becoming really confident on how to do problems like this.

But a local minimum, as a reminder, occurs when f prime switches from negative to positive, and that occurs on this graph at the value x = 6.

So if we write our justification, and one sentence is always nice, we can say x = 6 is the only point—it’s the only point—where f prime changes sign from negative to positive. So f, the original function f, has a local minimum.

I want to directly answer the question they asked: f has a local minimum at x = 6. This was just one point out of the full nine points, so you had to get the answer and the justification to get the one point.

Oftentimes, the first part of a question is the quickest, and so we want to get through it so we can move on to the other parts of this problem.

Okay, let's take a look at Part B then. Part B says, "Determine the absolute minimum value of f on the closed interval from 0 to 8. Justify your answer."

And one of you asked earlier about the difference between relative and absolute, and so now we have a chance to talk about it.

So first, when we think about an absolute minimum value, we want to identify the candidates. There are three suspects for the absolute minimum: it could be, based on Part A, the relative minimum that we found at six. A relative minimum value could end up being the absolute value as well, but the other candidates are the endpoints—in this case, x = 0 and x = 8.

So those three points are the candidates. We can say candidates—we want to lay out our logic when you write our answer.

And the first piece of that logic is the three x values that we're considering are x = 0, x = 6, and x = 8. Now we have to actually find the value of f at each of those three values to see which one is the minimum value.

The easiest one is f of 8, and the reason it's the easiest—and I'm going to zoom out a little bit—the reason it's the easiest is because they told us f of 8 equals 4 in the problem, so I can say f of 8 = 4.

I know that value; if four ends up being lower than the other two values, then four will be my minimum value. But let’s find f of 6. So, f of 6, if we look at this graph, what would the value be right here?

Well, we know that f of 8 was equal to 4, and now we have to kind of go backwards on this graph and undo this accumulated quantity of seven. And when we go backwards, that area is really going to be treated as negative by the properties of definite integrals.

And so we can kind of think of it as four, and we're going to be subtracting that seven. How would we write out our justification for that, though?

We'd have f of 6 equals the initial value of f of 8 plus the integral from 8 to 6 of f prime of x dx. That’s a little bit strange, though, to have the bounds of integration with the larger number on the bottom.

And so that's where I mentioned that since we’re switching the order, and we would want to think about it from 6 to 8, when we switch those bounds of integration, it becomes negative; the sign changes.

So we have f prime of x dx, and now we can write out that it would be four, and this quantity right here from 6 to 8 was 7, so we have 4 minus 7, and we have 3 then for that value.

Okay, so that is lower than f of 8, so so far, the absolute minimum value may be -3. But the last one we need to test out is f of 0.

f of 0, we can use the same process as f of 6, but this time we have to go back through four different segments. So, we have minus 7, so we're moving from right to left in this case.

Well, this would be a positive 3, a -6, and a -2, -7. Positive 3, -6, -2, and if we think of all four of those together, we have -12, 4, and -12.

We have our initial value of 4, and then we have -12, so it looks like that will be the minimum value. To write out our rationale, we can follow a similar pattern as we did for f of six.

80 f prime of x dx—if we switch it then to negative, because we’re reversing the bounds of integration, you still have, oh, there’s a typo there: f of 8—there we go and we have four then and minus 12 for a value of 8.

So we’ve written quite a bit. This problem is a lot more work. Now we have to think about what our actual answer is going to be: is our answer going to be 0? Is our answer going to be 8?

If we reread the problem, it says, "Determine the absolute minimum value of f on the interval from 0 to 8." It doesn’t say "Determine the x value where that minimum value happens." And so, in this case, we want to report out the actual minimum value.

So, I’m running out of space here, so I'm going to write it right up here in this box: my final part of the answer, the minimum value—the absolute minimum value on the interval from 0 to 8 is 8.

This was a three-point problem. The first point comes from laying out the candidates, the second point comes from finding the answer, which is 8, and the third point comes from your justification.

So the math that we wrote, where we calculated the answers in these parts in the middle, the main mistakes on this problem were accidentally writing x = 0 as the final answer, so the x value instead of the actual minimum value of 8.

So reread the problem as it was written right before you write your final answer to see: did they ask for the x value, the minimum value, or the coordinate pair where the minimum value occurs? That will make sure that you're able to format your answer correctly.

So we have two more parts, part C and part D; we’re almost done. We have a little less than 10 minutes left, so we’re almost through.

Part C—so looking here, let me see if I can erase some of this work up here to make it easier to read. Alright, same graph, I've copied it here so that's easier for us to see.

What open intervals contained in 0 < x < 8 is the graph of f both concave down and increasing—both concave down and increasing. Explain your reasoning.

This is very similar to the problem in part B on the last free response, so our confidence should be higher. Take a few seconds to think through, and if you have time, you can type in what your answer would be in the box on the screen, or you can write it down on your paper—whichever is more convenient.

Where is it concave down and increasing? Okay, so increasing is going to be any time that the derivative is positive. So I'm going to highlight in blue all of the parts where the derivative is positive, which is almost all of this graph—like everything except for that section from, what is that, four to six?

The rest of it, f prime is positive, so the original graph f was increasing. But where is the graph of f concave down? f will be concave down when f prime is decreasing; f prime is decreasing.

So that, if I show that in red, we have from 0 to 1, and we have from 3 to 4.

And it is also decreasing from 4 to 5, but that wasn't part of our original interval where the graph was—the value of f prime was positive, so that's not going to be part of our final answer.

And so those sections that are both blue and red are the final answer, and we'll write it in right below. We can do it in one sentence: we have f is concave down and increasing on— and the first interval was from 0 to 1, we can see up here, from 0 to 1 and from 3 to 4.

And for our rationale, we can say because f prime is decreasing and positive—decreasing and positive on these intervals. Great!

It's a two-point question: one for the answer, one for the justification. The main mistake, actually, on this problem was that students were able to find where the graph was increasing, and they were able to find where the graph was concave down.

But many of them wrote those two answers separately instead of finding only the parts that had both characteristics—like the intersection of those two sets.

Okay, we have the last problem. This one’s kind of a challenge problem. It doesn’t relate as directly to the rest of the material we’ve looked at today, but it’s a good challenge to end with.

Part D: the function g is defined by g of x = f of x cubed if f of 3 = -52. Find the slope of the line tangent to the graph of g at x = 3.

Well, the first challenge in this problem, like in many calculus problems, is just clearly identifying what it’s asking for. It's asking for the slope of the line, so we can guess that we’re going to be taking a derivative on this problem, and it’s asking at a specific value, x = 3.

So our first step then is to try to take the derivative of g of x. The derivative of g of x. Well, here, we have f of x cubed, so we have an outside function, the cubed, and we also have an inside function, f of x.

And so this is going to require more than just the basic power rule; this is actually a—probably can guess this is a chain rule problem. One of the most common mistakes was not recognizing that this would be a chain rule problem.

So, using the power rule, we take the outside derivative, 3, keep the inside function the same, f of x, reduce the power to two, and afterward multiply by the derivative of the inside function, f prime of x.

And we can write, as a reminder, what did we just do? That was the chain rule. And now let's put in three for all of these values of x: we’ve got f of three, f prime of three.

Okay, now we have to figure out what f of three and f prime of three are. So these two that I'm underlining red, we have to somehow figure out those values. The first one’s given to me right here: f of three equals -5, so that one’s good to go.

But what about f prime of three? Well, we haven't even looked at the graph at all yet; that graph we were given at the start of the problem, that was the graph of f prime.

And so it gives me the value at three. It tells me that f prime of three is equal to 4 right there. So that’s how the graph came in handy for this last part of the problem.

We can say that that is equal to four, and from here, it’s an algebra problem; we have to square -5. Hales, that becomes—let's see—25/4, and then we have to multiply by four.

And finishing off the algebra, we have—let's see—divide by four, multiply by four—inverse operations—we'll be left with just 3 times 25.

Or, [Music]. 75! That was a three-point problem; actually, two of the three points came from just calculating the form of the derivative.

So just by using the chain rule and only one point came from the rest of the work we did in finding the final answer. And the key mistakes on this problem were, one, not using the chain rule, as we mentioned, and the second mistake that we saw was mistakes on the algebra—so maybe making a mistake in squaring the value of -5.

Hales, remember if you type that into your calculator and you forget to put in parentheses, it may tell you -25/4. So when we’re rushing, we may make algebra mistakes that mess up our final answer.

Okay, so it’s 4:58. Thank you for sticking with me for the full hour! I want to go over some final tips that we went through, and then we’ll be done.

So the first tip is when you’re looking at problems with graphs, always ask yourself right at the start: is this a graph of f, f prime, or f double prime? Have that guide your thinking for the rest of the time.

The second tip—we mentioned this a few times—is to use precise language. Never just write "slope," "graph," or "derivative." Instead, always replace it with something like "the slope of f" or "the derivative of f prime," so that the reader knows exactly what you're talking about since they won't be there to ask you any follow-up questions.

The third one is beware of the difference between positive and increased, and we've used both of those terms frequently throughout this hour. And so it was kind of a good reminder that they mean different things depending on which graph we're looking at.

And last is to analyze the original graph before you take a look at the answer choices, and that will help you more quickly eliminate wrong answers.

Now just to make you feel a little more confident: that last problem we did was nine points on the AP test; the average score was just over 2.5 out of nine.

So even if it seemed difficult, and even if you think I would have only had three or four points on that, you would have still been above the average AP test taker.

Alright, so that's all we have for today! Thank you for joining. Don't forget to check out our website, Khan Academy, where you can do hundreds of more practice problems and watch other study videos.

Keep up the effective practice! I wish you the best of luck on May 9th, and from all of us here at Khan Academy, goodbye for now!