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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Continuity, by the time we?ve finished today?s episode, your knowledge of continuity is going to be seamless, no gaps. We?ve seen continuity in a number of episodes. We talked about related concepts when we looked limits. When we were doing the episode when we were doing graphs from data charts, we were using some of the concepts we?re using today. But now we?re going to use them in reverse.

We?re going to be using a graph to figure out some data about the function. Finally, we?re going to be looking at the mean value theorem again. You saw the mean value theorem before when we were dealing with some problems in AP Tricks 2. Where we were integrating operates. Well you?ll see it again, and we?ll do an actual direct problem using the mean value theorem today.

Continuity, limits and differentiability. Today we?re going to be looking at concepts of continuity, in other words, whether or not the function connects up. We?re going to be looking at differentiability, in other words, can you differentiate at that spot, and we?re also going to be looking at limits. The problem I?m going to show you right now is one where we?re given a little graph, and you have to use that graph to figure out at a certain number of locations, which of any of these things are true. Is there a limit? Is it differentiable? Does the function value exist? Let?s look.

Here?s your chart. What we?re going to be doing is going through all these locations and examine the graph to find out whether or not the limit exists. To find out whether or not the function is continuous and also to find out whether or not the function itself works at that location, and whether its derivative works at that location.

[0:02:00]

I?m going to make a little correction on here, the derivative sign is hard to see. This one is f'(c). Can you see the c? Here we go. The first location we were supposed to check is when x is -3. Now looking right above that, there is an open circle. That means that the function doesn?t go to that spot.

A limit might exist though. Limit first. If you come in from the right side, from the positive side, it looks like the limit is +5. But if you come in from the left side, what do you mean come in from the left side? You can?t. So on that particular spot, the limit doesn?t exist at all. You have to get the same thing coming in from both sides for a limit to exist. Does the function value exist? No, because it?s not a solid dot, you have to have a solid line or a solid dot for the function to actually work there. Finally, is it differentiable? No. if it?s not continuous, it?s not differentiable.

Let?s go back and fill in some of the chart. The limit did not exist there and it was not continuous and neither the function value nor the derivative existed. Going back for just a second, I?m going to add something that you?ll need to ignore later, but let?s say it went like this, and it reaches that open circle from both sides. If it reaches the open circle from both sides, then the limit actually does exist. But the function value still doesn?t because there is a gap there.

We?ll look at another one of those in a moment. The next location we?re supposed to look at is when x equals -2.

[0:04:00]

You?ll notice the dotted line there. Dotted line means you have an asymptote. If there?s an asymptote, the function doesn?t work there. So f(c) doesn?t exist and f' doesn?t exist either. If there?s any gaps or asymptotes, you can?t have a derivative. The limit; the limit is tricky. As you?re going closer and closer to the asymptote the function value is going lower, lower and lower. It keeps on going lower forever to negative infinity.

On this side it keeps going lower, and lower, and lower, to negative infinity and you think, coming in form both sides and reaching negative infinity in both cases. So my limit must be negative infinity. Well it would be expect you can?t really reach infinity. So you really have to say that the limit doesn?t exist there. One way to remember this is that infinity is unlimited, it goes on forever. Let?s go back and fill in our chart. The limit does not exist, it?s not continuous because you can?t got through there and neither the function nor the derivative existed. Next location is going to be -1.

-1; you?ll notice in this case that there is an open circle with a filled in circle directly above it. Let?s deal with the limits first. If you come in from the left side, the function result gets closer and closer to four. So from this side the limit is 4. Coming in form the right tough, it gets closer and closer to a height of 2. So you have different numbers coming in from the two sides. If the limit is coming in from two different sides isn?t the same thing, the limit doesn?t exists. One side of the limit exists, yes. But if it just says does the limit exist, then it?s assumed to be a two-sided limit. So the limit doesn?t exist at that spot.

Next step.

[0:06:00]

We have to go back and find out whether it?s continuous or not. Limit didn?t exist. Is it continuous or not? No, that?s fairly easy. It takes a jump, it?s not continuous. Does the function value exist? This is tricky. If you put -1 into this formula, whatever it happens to be, the result you?re going to get into the open circle because it can?t actually reach this open circle. The result is going to be 4. So f(c) really does exist. It?s 4. If you put -1 into the formula you get 4 for the result. Last thing; is this differentiable or not? No. if it?s not continuous it?s not differentiable. So f (c) exists, it was 4. The derivative does not exist, it is not continuous.

Let?s go to +1. We?re doing so many of these because there are a lot of different variations of what can happen and you need to be aware of the different combinations. At +1, it goes along, there?s a gap, so it?s not continuous. The limit does exist. Coming in form the left side and the right side, you get to the same height of +1. So the limit exists and it?s +1. It?s not continuous though.

Does the function value exist? No. there?s an open circle there with no closed circle above or below it to fill that in. Does the derivative exist? No, it?s not continuous, then the derivative doesn?t exist. So the limit on this one actually was 1, it was not continuous, f(c) doesn?t exist, the derivative doesn?t exist.

Next location, +2. That?s looking like the same thing we have here. You?ve got open circle, closed circle, this time there?s no limit. The function value is 3. It?s not continuous, it?s not differentiable.

[0:08:00]

At 2, the limit doesn?t not exist, it?s not continuous. It has a function value. You notice we haven?t filled in anything yet for f'(c). Derivatives are touchy. You just don?t normally run across this because usually when you?re doing derivatives, you?re given functions that are differentiable. I mean if you wanted to differentiate at x equals zero you could because it?s continuous through there. Anywhere where it?s continuous you can differentiate it.

Now if we go over to +3, at +3, there is a function value of -2, and it is continuous, it goes in through there, comes out with no gap. Since it?s continuous, yes there is a limit as well. Limits can happen anywhere. You just don?t normally bother doing them for things where you can put numbers into the formula and get an answer. You normally do it for things that are impossible like this. If you had a formula and put 1 into it, it would give you an impossible result at that spot. It would be undefined.

The one thing that doesn?t work for this is differentiation. Continuity doesn?t mean it?s differentiable. It?s not reversible. If it?s differentiable then it had to be continuous, but if it?s continuous it doesn?t have to be differentiable. This is an infinitely sharp point. If you come in from this side, the slope is something negative. If you go in form this side, you have a slope that?s positive. So the slope on either side of that point instantly jumps from one slope to another, which means it?s not differentiable.

So the limit exists, it was -2. We were continuous and f(c) did exist. That was -2, but the derivative doesn?t exist.

[0:10:00]

Last one, 4; this is one that?s going to really catch you. Look at that point and then look at that curve going into it. Notice that the curve going into it, gets steeper and steeper. It?s actually meant to be vertical at that point.

Well, does the limit exist? Yes, it's 3 coming in form either side. It approaches 3 from this side, 3 from that side. Does the function value exist? Yes it's 3. Is it continuous? Yes, but it?s not differentiable again. The reason is it's vertical.

Well what is a differentiation except for a slope? And vertical slopes are infinite. They don?t exist, so it is not differentiable at that spot. So the limit exists, it's +3, it's continuous, and f(c) exists, it?s 3 and it wasn?t differentiable at any of those spot. Differentiation is tough. It?s tough stuff to do.

A really common type of problem on the AP Test involves two part functions: where you?re supposed to find out a way to make the functions neat and to test if they?re differentiable or not. That often comes in the multiple choice section. Let?s look.

You?re given a to part function; on one side of the function it?s a parabola, it?s got x² and x. The other side of the function is some kind of line but we don?t know what the slope of the line is. The place where they meet is at 1, or the place where we will need to make them meet. What we?re going to do is find out what it takes in the place of k to make these meet up. I?m going to make a little sketch of the graph.

It looks something like this.

[0:12:00]

Parabola looks something like that. It does actually go on forever, though we?re not going to have it going forever. It looks like this is supposed to stop at x is 1. So I?m going to go back up a little bit and erase just a little bit of this. There we go. It?s supposed to stop at x is 1. And it includes 1 so that part should have the closed circle, because you can really put 1 into this part of the function.

If we?re going to make these meet up, we have to have just the right slope, so that this straight line will meet it. Well really all you?re doing is when you have the same x value of 1, you?re finding out what it takes to make them have the same y value. I?m going to take both parts of the function and set them equal to each other.

The left section of the function is ?x² minus 2x plus 4, and the right side of the function is kx plus 3. We?re going to find out what it takes to make them meet at 1. That tells me I can put 1 in for both sides, including this one right here where it can?t really go in, because it?s supposed to be greater than one. That?s okay, it will work out. I?ll have -1² minus 2 times 1 plus 4 equals k times 1 plus 3. A little bit of arithmetic on this. We would have -1, -1² which is -1 minus the 2. I?m writing this out because even though I?ve been doing this for years, I find you make few eras, if you write down even this trivial arithmetic you?ve been doing since first grade. Equals k plus 3.

[0:14:00]

These are going to add up to -3 plus 4, is +1, take 3 from both sides and it looks like k is going to be -2. That?s what it takes to make this work. This section of line has to be f(x) equals -2x plus 3. The final answer really is just k equals -2, but I did this so you can see what it looks like. We?re going to be using this in a minute anyhow.-2x plus 3.

This function right here crosses the y axis at 4, so +3 intercept has to be something like this, and it?s got a -2 slope, this is actually farther down than I drew it. To make these connect up, this line will have to go like that. What happens is you have the curve, it?s going along and then it suddenly stops being a curve and changes into a straight line. Let?s get rid of this little part now so we can see what the final result is really supposed to look like. There it is.

Now we?re going to go on and do one more part with this problem. We?re going to find out what it takes to make the function differentiable if that?s even possible. Is it differentiable at x equals 1? Let?s see what it takes. To be differentiable, remember what has to be true, it has to have to have no sharp bends. This one looks like it has a sharp bend. So it?s probably not differentiable there. But we?re going to have to test that out to prove it.

[0:16:00]

Especially if you?re working on the free response section, you?re going to have to prove it. And you prove it by checking the slopes on either side. What we need to do is find the slope on either side of this and we just found out that this function is -2x plus 3. We have to use that. We have to differentiate on both sides, even though we don?t know if it?s differentiable at that spot. It is differentiable on either side of it.

On the left side, f' is going to be -2x minus 2, the derivative of +4 is zero. Right side; we're going to do the derivative of -2x plus 3. The derivative on that side works out to be just plain old -2.

These formulas are locked down. They can?t be changed. So if it doesn?t turn out to be differentiable at that spot, we can?t make them differentiable. There?s another type of problem where you can actually make them differentiable, but not this time. Now checking if it?s differentiable at x equals 1; I can substitute 1 into this now. So f'(1) using the left side of the formula is -2 times 1 minus 2, that?s -4. So the slope coming in from the left side is -4. The slope coming in form the right side is -2.

The slope has to be the same coming in from either side, so it?s not differentiable. Remember that if a slope is different coming in form either side, you?ve got some kind of a sharp bend, then it?s not differentiable.

The next topic we?re going to take up in this mixed bag of different topics that you find in the AP tricks episodes, is mean value theorem for derivatives.

[0:18:00]

We last saw the mean value theorem in AP Tricks 2, where we were doing the mean value theorem for integrals. They?re a bit different than each other. So this mean value theorem for derivatives is one that you did quite a while ago actually. Not long after you first started doing derivatives. So you haven?t seen it for a long time, which means that it?s probably review worthy.

Now, the way I like to remember this, is really all that mean value theorem says is that the slope somewhere is equal to the slope between the end points. It?s an extension of Rolle?s theorem which says that if the end points are left and right of each other, in other words they?re the same height, the slope between them is 0. And there has to be some other place on the curve where the slope is 0. If, and this is the big IF, the function is continuous, and differentiable. We?ll get back to that in a minute.

You do this only for closed intervals; closed interval in this case goes from there to there. And if I connect those up, I can calculate where those endpoints are and I could find the slope just like that. Rise over run, Y2 minus y1 equals x, y2 minus y1, over x2 minus x1, there?s your slope. Let?s label these down here. We?ll call this point (x1, y1). We?ll call this point (x2, y2).

Now notice that there is a straight line between them, it isn?t part of the function. The other part of it says that you?ve got a slope between the endpoints right there. There has to be some place in there where the slope on the function is the same, that?s what this means. The slope at some spot and that some spot is c. That some spot is the mean value.

It looks to me like if I try to draw a parallel line, one that has the same slope, it?s going to look something like that.

[0:20:00]

And the location of that spot has to be (c, f(c)) c being the mean value. So c is the value that you would put into the function, so that the derivative at that spot would have the same slope.

This is the technical form of it, which I find a little harder to remember. You?d probably do it too. It says the same thing. F(b) is y2, so this would be f(b). And instead of calling it x2, they call this location x2, they called it b. likewise over here instead of(x1, y1). You call this location (a, f(a)).

The reason I need to bring this up, is because, you?ll sometimes see multiple choice questions where you?re asked is this true somewhere in the function? And you won?t know a lot about the function at all. You will just be asked to choose from a list of multiple choice possibilities whether or not this mean value works in there or not. And usually, you just have to look to say well is it differentiable and continuous in there? And if so, then the answer should be yes.

Look in the bonus materials. I?ve got a practice problem in there that involves using this technical definition and knowing how to recognize it.

Now let?s try a practice problem. There?s a really important technicality that you need to do. You need to know how to write if you see one of these on the free response section of the AP Test.

Find the mean value of the function of f(x) equals 1/3x³ minus 2x plus 3 on the interval from 0 to +3. You have to be given the interval, because there?s a lot of mean values, and what it is depends not only on the function, but what endpoints you choose.

[0:20:00]

I made this graph go from 0 to 3 so that it was easier to see. But if you?re putting it on you?re graphing calculator, you won?t see that unless you change the window to match. That?s a nice way to look at it. So there?s a slope between there and there. And there has to be some spot, probably about there that?s going to be c, where the slope of the tangent line equals the slope of that curve. It looks like c is maybe 1.6 or 1.7, but we?ll find out.

Here is the technicality, after first verifying that the theorem is applicable. Really all you need to do for this one is to make the statement 'it?s continuous and differentiable on (0, 3). Let?s write this down.

F(x) is continuous and differentiable on [0, 3]. Notice it's x is equated than or equal to, less than or equal to; closed interval. It?s important to have a closed interval. If you?re asked to do a mean value, on what?s the question where you?re being asked to determine whether it?s possible or not, you?re probably going to be given something that?s continuous or differentiable. Because you can?t do this if it?s not continuous and differentiable. But if you need a little bit of proof, understand this is a polynomial. There?s no variables in the denominator so it?s going to be continuous. Pretty much continuous everywhere. Differentiable, yes, continuous and differentiable, you can differentiate that thing. You don?t have to have a form of proof of why it?s continuous and differentiable you just typically need to make a statement.

[0:24:00]

What we?re going to need to do before we start is figure out where the endpoints are. We?re going to use the endpoints in order to find the slope. I?m going to write it as a separate calculation, I find it more understandable this way instead of using the formal definition of the formula. We have to use the endpoints, so I need to find out what f(0) is. When you have x equals 0, f(0) if you substitute into there, not hard to do, 1/3 times 0, minus 0, its 3. So one of our end points is (0, 3) right here.

Next one; this end point is at 3, something. Let?s find out what that something is. What we need to do of course is just find out what f(3) is. F(3) is 1/3 of 3³ minus 2 times 3 plus 3. Easy enough to do. It works out to be 6. Keep those in mind, (0, 3) and (3, 6).

What we?re going to do now is go to the next slide where I?ve got some more writing room. We?ll find the slope, then we?re going to do the derivative of this and set them equal. So the slope is y2 minus y1 over x2 minus x1; 3 over 0. I always like to do slopes this way, one point at a time, put the ys and xs above and below each other. Make fewer mistakes that way. (3, 6) put the 6 right above that 3 and we get a slope of, looks like its -3 over -3, that?s nice.

[0:26:00]

It?s a slope of +1. Easy enough.Now let?s do the derivative. F'(x) is 1/3 times the derivative of x³ which works out to be x², minus 2 times x which is just -2. Good enough. We?re done with that derivative.

Last step; what we?re going to do now is, and this can be confusing, is put cs where the xs were. So this is now going to become f'(c). Just because f'(x) is the regular function, f'(c) is a particular location; the function of a particular location, that?s going to be our answer. So that?s c² minus 2 and that function at c has to equal the slope between the endpoints.

This will be easy enough to solve. Add 2 to both sides, c² equals 3. You know what you call a naughty sailor don?t you? ?Sea? squared. C is plus or minus the square root of 3. So we actually got 2 answers, but we?re going to ignore one of the answers. See this function is a cone, it?s going to curve up and down. There is another place on the curve where it has that slope. But the only one that?s in the interval is positive square root of 3. So that is our answer; c is positive square root of 3.

Well in this episode we?ve just covered a few concepts from early in your calculus course.

[0:28:00]

Concepts involving continuity, in other words, whether there is gaps. We?ve covered concepts like differentiability, in other words is the slope the same coming in from both sides at a particular location. And we?ve covered the concepts of the mean value theorem for derivatives. In other words, finding the place on the curve where the slope is the same as the slope between the endpoints.

They?re not too hard to do. But I have found that students typically forget this by April or May when the AP test happens because you do them early in the course, and there are a lot of other things that you?ve learned in the mean time.

Be sure to check out the bonus materials. There are a few review questions there and in the mean time, I?ve got a bit of a joke for you. You?ve heard about the tragedy the other day? Yes, it happened on a math class. See there is a student who?s fooling around and the teacher confiscated his rubber band, why? Why confiscate a rubber band? Because it was a weapon of math disruption.