 Hi everyone, we're going to take a look at a problem from the 1989 AP Calculus A-B exam. This is the problem as I like to say started it all in terms of asking students to sketch a possible curve of an original function from that of the derivative or even seek information about an original function from the graph of the derivative. So what you see here is exactly how it appeared on the AP exam in 1989. We were given a figure that is the graph of F prime, the derivative of a function F. The domain of F is the set of all real numbers and the graph we're looking at specifically is from negative 10 to positive 10. You'll notice a couple details here. They make it very, very prominent that you are looking at the graph of the derivative, not that of the original, even to this day when problems similar to this show up on the AP exam you will see something similar as the note you see here. The first thing we were asked to do is to determine the values of X for which the graph of the original function F is going to have horizontal tangents. Now as we write up this answer there definitely are certain things that we look for when we grade the AP exam. So that's what we'll be focusing on here. We want to know where the original function F has a horizontal tangent. So let's go back to the graph. Remember once again you are looking at the graph of the derivative. Horizontal tangents on an original function occur where the slope of the tangent line is zero. Now given that you're looking at the graph of the derivative, sync on the original function where is it that the original function would have a slope of zero. It would have to be at the X intercepts of the derivative graph because the X intercepts of the derivative graph is where F prime equals zero which means the slope would be zero on the original function curve F. So you can see from this graph we have four locations at which that occurs. We just simply need to write it up correctly. Therefore we can say that F will have horizontal tangents where F prime equals zero. Therefore that would be at X equals negative seven, negative one, four, and eight. Notice a few details about this. Be very specific about which graph will have the horizontal tangents. Don't simply say it. I like to say it is a two letter word, we don't use it. Specifically say it is the function F that will have the horizontal tangents. And notice how you state why it is you know that or where that occurs on the derivative graph. Essentially what you're doing is creating a link between what the derivative graph is telling you and what you therefore know about the original graph. That is the important piece to create that link, the connection between the two. And please do remember that as we go through and answer these questions we're going to see. The second question that was asked of students that year was to determine the values of X in the interval from negative ten to ten where the original function F is going to have a relative maximum. Think about what we've talked about so far in terms of how it is a relative maximum comes to be. If we have a relative maximum the slope is changing from positive on the left to negative on the right. As we talked about the derivative, the slope aligned tangent to the curve is positive to the left and the derivative is negative to the right. So we want to go back to the graph and see if we can determine where that happens on the derivative graph. Can you locate any X values at which we know the derivative is changing from positive on the left to negative on the right which will therefore create the relative maximum on the original function F. Hopefully you've identified two places at X equals negative one. If you take a look at the graph it's changing from positive the derivatives are positive to the left negative to the right and also at positive eight. So we have two X values on the original function F at which we're going to have a relative maximum. Let's talk about how we will write that up. Again we're going to be very specific in terms of what function is having a relative maximum. So F will have a relative maximum. Remember we talked about the need to create a link between what it is we're seeing on the derivative graph and what that tells us about the original function. F is going to have a relative maximum where F prime changes from positive to negative. Remember we want to create the link. So if F prime is changing from positive to negative that means the original function F is changing from increasing to decreasing and that's what creates the relative maximum. And as we said that occurs at X equals negative one and at eight. Remember to be specific in terms of specifying are you talking about F versus F prime and to make that link between what you're seeing on the derivative graph and what that tells us about the original. The last thing for which students were asked were the values of X for which the graph of the original function F is concave down. One thing you need to recognize about concavity and this is one thing a lot of students forget. Concavity occurs over an interval. You are not going to have specific isolated X values. You are going to have an interval for the answer even though they don't remind you of that. That is one mistake we see a lot of students make on the AP exam is they itemize X values. No, concavity is over an interval. So that would be the appropriate way to answer a question like this. This is where you need to recall what you've discovered in this lesson about how you know from a first derivative graph where the original is going to be concave down. Hopefully what you recall is that an original function F will be concave down where the graph of the derivative is decreasing. That's one of the things that hopefully you've come to know throughout the activities we've done this lesson. Let's go back to the graph and we need to identify the intervals on which the derivative graph is decreasing. There are two of them. Looks like it goes from negative 3 here down to about 2 and then from 6. And remember how they define this function was only negative 10 to 10. So we do need to have an end point for that second interval. Remember also that concavity intervals are always open intervals. It would be incorrect to state a concavity interval as a closed interval. They must be open intervals. So we have 2 in this case. We have the interval from negative 3 to 2 and from 6 to 10. As I mentioned, I did add a fourth question. So we can take a look at how we could come up with a possible graph of the original function F from the graph of its derivative. And this will hopefully give you an opportunity to really tie together everything we've been talking about. Questions like this still show up on the AP exam to this day. To get started on sketching a possible graph of the original, let's start with some key points that we know are going to be on the graph. We know that the x-intercepts of the derivative graph become our max min points on the original. Let's put those in place first and we'll work left to right. Our first x-intercept is negative 7 and notice what's happening. We have the graph there changing from negative to positive. So think about what that means about the original function. If the derivative is going negative to positive, that means the original is going decreasing to increasing. It must be a relative minimum. So what I'm going to do on my graph over here at x equals negative 7, I'm going to put a minimum point. You can put it as high or low as you want. Notice the directions. Sketch a possible graph. So the y-values, you can put them as high or low as you want. And I'm going to make myself a little note here that this is going to be a minimum point. Our next x-intercept is over at negative 1. There we have the derivative changing from positive to negative. Remember, we talked about that tells us on the original function, it's going increasing to decreasing, thereby creating a relative maximum. So over here at x equals negative 1, I'm going to put a point way up here to be my maximum. Our next x-intercept is over at 4. Here we have the derivative graph changing once again from negative to positive. So that's going to be another minimum. Maybe I'll put it here. And our last one over at 8, we have the derivative changing from positive to negative. That is going to be another maximum point. So now I have in place all of my maximum points of the graph. Second thing to put in place are inflection points. The inflection points on the original are going to come from the maximum points of the derivative. Once again, working left to right, we have a maximum here. It looks like negative 3. So I know I'm going to have an inflection point. Now, as far as the placement of your inflection points, remember, we want a nice smooth continuous curve by the time we're done this. So what I want you to do on your graph, take a look at where you put your minimum and maximum at negative 7 and negative 1. We know the x-coordinate for our inflection point is going to be negative 3. The placement of that should be about halfway in between vertically between your minimum and maximum. So if you take a look at mine, mine maybe would go right around here. So you take a look at your own graph because this will depend upon how high or low you put your maximum in points. That's going to dictate where you place your inflection point. You want it about halfway in between the vertical distance between your minimum and maximum in this case. We have another minimum on the derivative graph. Here it looks like 2. So that tells us we're going to have an inflection point at 2 on our original. I had this maximum point at negative 1. We had a minimum point at 4. We're going to have an inflection point at 2. So maybe halfway would be right around there. And we have one last one over at 6. Is going to be another inflection point on our original curve. And that needs to be located between the minimum at x equals 4 and the maximum at x equals 8. What you have now are some guidelines to use as you go to do your graph. Once again, we're going to work left to right. So over on the far left, we know that the graph is underneath the x-axis, meaning that our derivative is negative. That means the original has to be decreasing. But at the same time, because the derivative itself is increasing, that tells us the original has to be concave up. So maybe it looks like that, which is decreasing and concave up. So I hit this minimum, which corresponded to right here on the graph. From that point, f prime is positive. That tells us the original function is now increasing. Because the derivative itself is increasing, it tells us the original function continues to be concave up. So from this minimum, I'm increasing concave up. I hit this inflection point. But it's going to keep being increasing, but now it changes to concave down. The more you draw these, the better you get at them. So that accounted for from this maximum right up here to about where it hit the x-intercept. From my maximum, take a look at what's happening on the derivative graph. It's now underneath the x-axis, meaning that the derivative is negative. So that tells me my original function has to be decreasing. The derivative itself is decreasing, which tells me the original has to be concave down. So after hitting this maximum, I have to be decreasing concave down. I hit this inflection point. From there, it has to keep decreasing, but now be concave up. I think I want to move my minimum to there. I hit my minimum. Take a look at your derivative graph. It's now above the x-axis, meaning that our derivative is positive. So we have to be increasing. At the same time, the derivative graph itself is increasing, which tells us the original is concave up. So after I hit this minimum, I have to be concave up and increasing until I hit this inflection point. I keep increasing, but now I am concave down. As I mentioned, this still shows up very often on the AP exam. What we look for when we're grading them, we look for the shape. We look for correct x values for your max and min points and inflection points. On the original graph, this is why I encourage you to put these points in place first so that they can serve as a guide for you. And then, really, the shape of the rest of it. It doesn't have to be perfect. As you can see, this one's not. But as long as it conveys the right idea, that it conveys the correct characteristics of the original function F, that's the important part.