John Postovit

**University of North Dakota**

M.Ed.,Stanford University

From over 16 years of teaching experience, he has philosophy that it takes humor, patience and understanding when teaching tough subjects.

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Welcome back. This time we are going to be covering some of the type of logic problems that they often cover on the AP test. You might be asked to do a problem where you?ll need to look at a chart, and refer some details about a graph such as slopes, concavity, that sort of thing. Or you may be given a chart and asked to actually produce a graph on the free response section. Either way, you really need to be prepared.

Well, before we start, I?ve got a joke for you. This is pretty bad. So, you know of course who the largest knight at King Arthur?s round table was. Yeah, his name was Sir Camfrence. He got that way from too much pie. Once you have recovered from laughing, then turn the video back on and we?ll start.

It?s pretty darn likely that the multiple choice section of the test is going to ask you to figure out some detail, some information about a function, without even knowing what the function is. All they're going to do is give you a data chart and you?ll have to use your knowledge of things like, the mean value theorem, slopes, concavity and so forth, to figure out what is true about the function.

So here?s a little example. We?ve got a function, we know it?s continuous and differentiable. That?s an important prelude detail. Tells us that there aren?t any gaps in there, no asymptotes. And there are no sharp points in there because it?s differentiable. So it can?t be an absolute value function with a sharp little v. It?s on the interval 1 to 5. It includes that. And what we have to do, is figure out which one of these statements is true.

We are going to make a quick graph. It doesn?t have to have a lot of detail on it. It?s just enough to put everything in context. So when x is 1, f(x) -4.

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I?ll just put that down here, good enough. When x is 2, f(x) is 1. When x is 3, f(x) is 5. X is 4, f(x) is also 4. And when x is 5, f(x) is -2. I?m going to figure out what is true about this. Could (1, -4) be the lowest value on this? Let me redo this one, I placed this a little too low. it should probably be about there, (5, -2). You know it sure looks like the lowest runner, right.

Everything in here is set up as a trap. It looks reasonable. All of them look reasonable but only one of them is true. Looks like the lowest spot, right. But you don?t really know that it makes a nice smooth curve across there. It might do something like this. It might start here and go downwards for a while and then back up. Who knows? So we have no idea. This one can?t be true.

Double derivative, concavity. So the double derivative is less than 0 anywhere between there. Okay, that means it?s concave down everywhere. Look, we already know that one is false. Here?s a little section that it?s concave. That is possible because anything can happen between those points except a sharp v or an asymptote. First derivative, slope. Slope negative everywhere, definitely not. There do have to be some sections of negative slope, because it goes downwards from here to here and here to here, but not everywhere. To go from here up to here there has to be a section of positive slope. That one is no good. So we are left with this one.

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Now, this is something that you?ve used probably about a third of the way into your school year. This is a brief statement of the mean value theorem. And what it says is that, somehow, somewhere, there has to be a number, an x number, that you can put into the derivative formula, so that you get a slope of ½. Well, here is why that has to be true. The mean value theorem says that if you take the end points of an interval and find the slope between them. Bear in mind this isn?t the scale. But if you find the slope between them, there has to be some other place in that interval, if it?s continuous and differentiable where the slope is the same.

Well, if you look at this endpoints, going from -4 to -2, the rise is 2. And going from 1 to 5, the run is 4. 2/4 that?s the slope. 2/4 is ½. So since the slope between the endpoints is ½, there has to be some other spot somewhere. Who knows where? Maybe it?s right here where the slope is the same.

One more practice problem. This is another type of multiple choice question. You could see function f is differentiable and strictly decreasing. That means that across the whole thing, there?s no gaps, no asymptotes and it?s always going downwards on that interval. So which of these could be the valley of the slope at 2? Let?s make a graph again. Doesn?t have to be good. Just enough to put things in context. (1, 30) up. This time I?m going to put the numbers in, because we will be using them for more this time. (2, 25) up. Okay, there is (2, 25).

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(3, 23) about there and (4, 22). So it looks to me like the slope is decreasing. But it is always decreasing in that interval. So the question is, which of these could be the value of f'? Because it?s strictly decreasing, there can?t be any places where it bounces back up. It always has to keep going downwards.

Between here and here, if I were to connect with a straight line I?d get a slope. We don?t really know it?s a straight line of course. But the straight line will give us the model for what we are going to do. Between there and there, there has to be a different slope. Somewhere between here and here, the slope is starting out pretty negative and getting less and less negative.

The slope right at this spot has to be somewhere between these two lines and these two lines. So it has to be continuous. All I need to do for this one, is find the two slopes. So for this one, the run is 1, and the rise is -5. You know 25 minus 30. So the slope on that part is going to be -5. And now this section here, the rise is -2 between 25 and 23. And the run is 1. So the slope on that section is going to be -2. When you see these problems on the test, it?s always going to be easy numbers. Because they tend to throw the easy in on the part where you can?t use your calculator.

Well, so between here and here is -5. Between here and here is -2.

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Somewhere in there the slope gets less and less negative. The only number that?s reasonable in there is here. You don?t really know that that?s the value, but it has to be somewhere between -5 and -2.

Another thing you might need to do on the free response section, is to produce a complete graph or an estimate of what the graph looks like, based on some information that you are given. The information you are given is always going to be something about actual locations on the graph, slopes, concavities. Let?s look.

So the goal here is to produce a reasonable graph of f(x) given the information in the table. And there?s a lot of things that can happen. You can have undifferentiable locations. You can have asymptotes, you can have minimums, maxes, everything, inflection points.

You have various intervals, in each one of those intervals you are given some information. You are told the slope and you are told the second derivative. So the second derivative tells you about concavity. First derivative tells you about slope. And the zeros tell you a lot too. For example, see all this zeros in the slope column, first derivative, well if the first derivative is zero there has to be a spot where the graph levelled off. A critical point.

Second derivative, wherever the second derivative is zero, it has to be an inflection point. Which is one of those spots where the graphs changes from going more and more positive slope, to less and less positive slope. So it?s where it stops increasing, or conversely where the slope stops decreasing.

First we want to lock down the places that we are actually given.

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We are given this spot, this spot, this spot. I?m going to put a little scale on the graph. It looks like I have to go between -2 and +3. And one, two, three, we don?t need a lot of detail here. You should label the axis and you should label at least one of the hatch marks, so that you have a scale to the graph. You don?t need to label them all.

And going up and down, we have to go as far as -5 and really not that high. 1, 2, 3, 4, -5, we?ll go a few up this way. Label the scale there 2. Label the axis, let?s go. When x is -2, f(x) has to be -3. (-2, -3) right there. And before we go on, let?s right down what?s true about this spot. We know that it?s a levelling off and we also know that it?s an inflection point. I?m going to label this IP.

Now off to the side here, I?m just going to show you what could happen to that spot. We are not filling it in yet because we don?t know much about the graph. But a place that?s a levelling of and an inflection point, can?t go like this. That?s a levelling off that?s a minimum. It can?t go like this, that?s a levelling off, that?s a maximum. This is the spot where the slope is level. It has to look something like this. The slope gets less and less, it levels off. But then it doesn?t dip back down, it starts going back up again. Or it could do that. It?s going to be one of those two at that location.

When x is 0 f(x) is 0. Let?s lock that one down. And this spot is a place with positive slope. It?s also an inflection point. Let?s label that inflection point.

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And next, when x is 1, we don?t have the location, we can?t feel it in. We?ll just know something is happening there. It looks like it?s a critical point. When x is 3, f(x) is 0, put that one down. And that one is an inflection point as well. Now we can start working out some more details about the graph. Mins and maximums.

Remember mins and maximums according to the first derivative test, happen when the slope on either side is different. Or when the concavity is positive or negative. For this one here, this critical point has power to a slope on both sides. That means that it?s neither a minimum nor a maximum. Positive slope on both slopes, there we go. We are in business. We know that going through that spot it has to be going upwards. But it levels off momentarily. I won?t fill in any more graph yet. Now let?s go to this spot right here.

At this spot the slope is positive, and it?s an inflection point. Now this one is a little different. It?s neither of these. These are the most confusing ones. It?s a place where the concavity has to momentarily change. But the slope all the way through there, is still positive. It?s going to look like this. Very weird.

So you can have an inflection point where the slope is momentarily not changing. Going through here, the slope is getting larger and larger and larger. And all through here, the concavity is positive. And then right here the concavity changes to negative. But all the way through, the slope is positive.

Next spot, let?s see. We know that at 1 something happens.

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I have to put that on the graph now. Even though I don?t really know exactly where it happens. I do know that the slope is momentarily 0 and concavity is down. That tells you that it has to be. This is kind of intuitive, but the second derivative test already tells you that it has to be a maximum. It?s got to be a maximum when the slope is momentarily 0 and the concavity is down.

Another test for this is to look at the slopes on either side. It?s going to do something like this. It?s going to reach that spot and turn back downwards. And look on the left side of it. On the left side of it between 0 and 1, the slope is positive. And then suddenly, between 1 and 3 the slope changes to negative. So it dips back down. That?s another sign that it?s a maximum. Label that as a maximum. It's a local maximum because we don?t know if it?s a global maximum.

All right let?s see, next spot right here at 3. The slope is downward throughout all of that. And the concavity is momentarily 0. It?s ended at one of these. The slope is negative all the way through there. The concavity is momentarily zero and then the concavity turns positive. Take a look at the interval between 1 and 3. The concavity right in here is negative, it?s downwards. And then between 3 and infinity, it?s concave up, so it has to be a bow shaped upwards.

Now we have one more thing to look at. As x approaches infinity, f(x) gets closer and closer to -5 and never quite gets there. Then it comes in from above. This is the sign of a horizontal asymptote. At -5.

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Draw that in and then show that it is levelling off towards that asymptote. I think we are just about done, but we need to check a couple of more things. You really should look on the intervals to make sure it?s doing what it?s supposed to be doing. Between negative infinity and -2, the slope is always positive and the concavity is always negative. So that tells me that it has to go something like this.

Let?s see if that fits. The slope is positive between here and here, and it's bow shaped downwards so the concavity is negativity. Next interval between -2 and 0. It should have positive slope and be concave upwards. See going upwards, in concave upwards, we?re good. Between 0 and 1, first derivative is positive, so the slope is positive. And it?s going to be bow shaped downwards, concave down. It?s moving upwards so it?s got that positive slope and some bow shaped downwards. So it's got the negative concavity.

Next section between 1 and 3. It has to have both negative slope and negative concavity. Moving downwards. Bow shaped downwards. And the last one, always moving downwards, negative slope, but bow shaped upwards. We are done with this one, let?s have a look at another.

Here?s another one, we are being asked to produce a reasonable graph f(x) given the information in the table. Well, you notice that it doesn?t say much about it. It doesn?t say that it?s continuous everywhere or differential anywhere. There?s some weird things that can happen. This is a little more advanced problem. It?s got some odd things you have to be aware to look for.

Let?s start like we did last time. We need to put a little scale on the graph. I know I have to go between -3 and +5 on the horizontal axis. So let?s put that in there 1, 2, 3, 4, 5. Label your axis.

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Label at least one point to give it scale. I have to go between -4. And +2 I had to bear a minimum on the vertical axis 1, 2, 3, 4. 1, 2. Of course I doesn?t stop there but we just need enough to have a few points locked down.

Now let?s have a look at the definite points. They definitely go -3,-4. Put that in. We?ll get back to that in a second. There?s weird stuff happening there. Derivatives that don?t exist. (0, 2) another definite point. And that?s it, that's all for definite points this time.

Now here?s where the really strange stuff is happening. Look at the interval between negative infinity and -3. The slope is negative all the way through there. It?s tipping downwards and this is strange. Look the double derivative is zero. The concavity is zero. For the concavity to be zero, it can?t be curving like this. It can?t be curving like this. No it?s not curving. You didn?t see that in the last one?

In the Last example we did, any place the second derivative was zero was a definite location. It wasn't an inflection point. This isn?t an inflection point, it?s a straight line. We don?t know what the slope is. They didn?t tell us. So we?ll just draw in something that?s reasonable. That?s got a negative slope, that?s all it takes.

Now at -3, the first derivative doesn?t exist. The only places that the derivative doesn?t exist are going to be asymptotes or V?s. So it?s continuous through there. But this is a spot that?s an infinitely sharp point. It?s a place where for example, you have absolute values.

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And in the interval to the right of it, we know that it is going upwards all the time. But it?s concave downwards. And we have to go all the way to zero that way. So it?s moving upwards but it?s concave downwards. We get something like that.

Now (0, 2). The slope is positive at that spot. So in other words, it?s not a minimum, it?s not a maximum. It?s not a levelling off because it?s not a critical point. Critical points happen when the first derivative is zero. But the second derivative is zero. So it is an inflection point. It is a place where the concavity changes. That?s an inflection point. So at that spot it has to be going from concave one way, to concave the other. It has to start tipping back up again. I don?t know how fast but that?s enough for now.

Now, next bit of weirdness. You didn?t see this in the last one. 5 doesn?t work in the formula at all. With this one over here, -3 did work in the formula. It really produced us a result. That?s our clue that it was a sharp point. It?s got an actual location but the derivative doesn?t exist. This one right here, it doesn?t have any actual location. This is the sign of a vertical asymptote at 5. So, sketch that in before we do any more graph. A vertical asymptote.

Now look to either side of it. This says that as you are coming towards 5, from the numbers that are smaller than 5, the function is going upwards forever. And the first derivative, the slope is positive, second derivative is positive. So it?s going upwards. The slope is positive, concave up. It?s going to have to do this.

[0:22:00]

Going upwards forever, concave up. Slope positive, there we go. Now the other side. As you are approaching +5 from the positive side, here?s what is happening. These can be confusing because you are not used to drawing graphs, from the right hand side. You usually draw from the left hand side. But as you are approaching that, you are going to get to be more and more negative. The slope is positive. Let?s just try to ignore that for now because that can really can confuse you. And it?s concave down.

We just know that as we are going towards 5, it?s getting more and more negative. So it?s going to have to do something like this. It?s going more and more negative as it gets closer. Now let?s go back and just double check that slope.

See, it makes more sense once it?s drawn. Positive slope oh yeah. As you are going from left to right, it?s going upwards. That?s a positive slope. Let?s double check the concavity. Concave down, so that means it has to be bow shaped downwards. And as you go towards infinity, the slope always stays positive. Concavity all stays negative. It?s going to go something like that.

This time we covered problems which ask you to figure out information about a function or a graph from a data chart. Now, those first two problems we did involved inferring details along the style of the multiple choice questions. Those are particularly hard problems, because they test you out on theory and there?s a lot of variations possible. You need to be aware that those particular types of problems, are ones where on the real AP test, the rate at which students get them right is relatively low. So you want to keep trying those.

Try the ones in the bonus materials. Try the ones on the official AP website. Try them out. Well, to do this, you need to know what strictly increasing is, strictly decreasing, Mins, maximums, inflection points etcetera.

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Well again thanks for watching. And with any luck your calculus knowledge is continuously increasing. And if you are really lucky, your calculus knowledge is concave upwards.