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Calculus: Functions w Asymptotes, Critical Points


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About this Lesson

  • Type: Video Tutorial
  • Length: 17:20
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 187 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

Calculus (279 lessons, $198.00)
Calculus: Curve Sketching (20 lessons, $25.74)
Calculus: Asymptotes (5 lessons, $7.92)

Taught by Professor Edward Burger, this lesson comes from a comprehensive Calculus course. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers limits, derivatives, implicit differentiation, integration or antidifferentiation, L'Hopital's Rule, functions and their inverses, improper integrals, integral calculus, differential calculus, sequences, series, differential equations, parametric equations, polar coordinates, vector calculus and a variety of other AP Calculus, College Calculus and Calculus II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, "Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas" and of the textbook "The Heart of Mathematics: An Invitation to Effective Thinking". He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The "Journal of Number Theory" and "American Mathematical Monthly". His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Curve Sketching
Functions with Asymptotes and Critical Points Page [1 of 1]
We now are ready for the very last example we'll work through together on graphing functions. So let's take it slowly, and let's really rejoice in this because this is the last time together we are going to graph a function.
The one I selected - I thought you would enjoy this vintage function -. Some vintage, huh? In fact, since this is our last one, I want to do this just really write because, you know, this is the vintage. I want this to be really pretty. Look at that! And I want to shoot the works on this...let's do it all! Do it all. In fact, I'm going to use blue on this just to really celebrate.
So where do we start? First off, I see it has a denominator, so there may be asymptotes. Let's investigate that first. So the first thing I always look at are the vertical asymptotes. So for the vertical asymptotes, what do I do? Well, step one is to see if I can factor the top and the bottom. The top can't be factored, though the bottom can be. Notice that it's the difference of two perfect squares. So I could write f(x) as . I can't cancel, though, anywhere, so all I have to do now is take a look at when the bottom equals zero. So if I set the bottom equal to zero, I see that either x = -1 or x = +1. So I see (x = -1), (x = +1). That's where the denominator is equal zero, and therefore those are my vertical asymptotes.
So that was pretty easy, I found the vertical asymptotes right away. I factored the top, factored the bottom, did any cancellation I possibly could, and then after that set the bottom equal to zero and found my vertical asymptotes.
Horizontal asymptotes.... well, for the horizontal asymptotes, it's a different issue. Remember, there I'm asking what happens as you go off to the horizon? So what I want to do is make x get really, really big and see if I'm honing in on any particular y value. So there I'm interested in the growth rate. Well, the top is growing like x, the bottom is growing like x^2, so this whole thing is growing like , which is . And what happens as x gets bigger and bigger and bigger to ? Well, that's getting smaller and smaller and smaller, and it's heading towards zero. So, in fact, since this power - the bottom power - is higher than the top power, I know automatically there will be a horizontal asymptote, and it will be located at y = 0, or the x-axis.
So that was really fast. We were able to immediately knock off the asymptotes. You see there are going to be a whole bunch. Notice there are two vertical asymptotes, by the way. We have to be careful...there are two vertical asymptotes.
Okay, now it's time to shoot the works and try the hodgepodge of increasing, decreasing, and so forth. In fact, why don't you give this a shot? Since this is the last one we'll do together, let's do it together. Right now why don't you locate all of the critical points? So take the derivative of this and do all of that stuff, and then make a little sine chart. Find out where it's increasing, where the function's decreasing, and then determine if we have any max's or min's. Try that one right now, and then we'll do it together and see how you made out.
We've picked the derivative of this, which will require us to use the quotient rule. So let's use the quotient rule here. So that's the bottom, (x^2-1), multiplied by the derivative of the top, which is pretty easy, that's just one, minus the top, which is x, multiplied by the derivative of the bottom, which is just 2x. We divide all of that by the bottom squared. We can simplify that just a little teeny bit. What do we see? Well, here I see an x^2, and then I have a -2x^2. So x^2 - 2x^2 = -x^2. Then I have that minus one - and that's all I have on the top - divided by (x^2 - 1)^2. Let me caution you about something. You might say right now, "Hey, I can cancel these two things. They look the same." They are not the same. This is (-x^2 - 1); this is (x^2 - 1). There is no negative sign there, so you can't cancel. In fact, if you really want to make this clear in your mind, you might want to factor out that negative sign, and then you can really see the difference between the top and the bottom. You cannot cancel there.
Okay, now let's find out the critical points of this function. So I take the derivative and set it equal to zero. When I set the derivative equal to zero, the only way that could happen is if the top were zero. Is the top ever zero? Well, no, because if that would equal zero, that would mean that x^2 equals negative one, and x^2 is always non-negative. So, in fact, this is never zero.
Well, what are the other places to look for critical points? We have to look where the denominator is zero. Well, the denominator is zero when x equals one or minus one. Are those critical points? Well, no, because the function's not defined there. In fact, we've already seen that those points represent vertical asymptotes. So, in fact, again, no critical points. So that's sort of interesting. There will be no max and there will be no min here.
What about increasing or decreasing? Well, let's make a little sine chart. So here's f^1(x). Do we have no points to write down? That's false. We do have points to write down. You should mark down all your vertical asymptotes. So we have one at negative one, we have one at one. Now let's run the sine chart business on the derivatives. Why the derivatives? Because I'm interested in if the function is rising or falling, so how the function is moving is given by the derivative, given by the slopes of the tangents. So I go to the derivative function, which is right here, and I plug in some points. Let's plug in negative two, which is to the left of minus one. If I put a negative two in here, what do we see? Well, here we see a four, plus one is five, and there's a minus sign in front, so the top is negative - negative five. On the bottom, I see something squared. So no matter what this is, it turns out this is going to be four minus one, which is three. Three squared is positive. I have a negative over a positive; that's a net gain of negative.
What do I have when I plug in zero? If I plug in zero, I see this negative one over minus one squared, or one. So negative one over one is negative. So I see negative here. Here, I have a vertical asymptote. I mark that with a va here just to remind me there's no extrema there, that's a vertical asymptote. That's a scary point. Then pick a point here. Let's pick let's say two. If I plug in two here, this is still negative in front. If I put a two in here, that's going to be four minus one - it's still three - squared, it's positive. I see a negative, divided by a positive...that remains negative.
So what have we discovered? We've discovered that this function is first decreasing down to this vertical asymptote, then it's decreasing down to this vertical asymptote, and then decreasing down to this vertical asymptote. It's always decreasing; it's never increasing. So this is a function that's always going downhill.... not looking good.
Okay, so that's the increasing and decreasing issues. Now we want to take a look at the curvature issues. What I want for you to do right now is to try to find all of the information you can about curvature. Remember, that requires you to analyze the second derivative, find potential points of inflection, make up a little sine chart like this, plug it in, see if you can find where the functions concave up, where the functions concave down. Try it right now on your own. We'll do it together in a second.
Let's see if we can make some progress on this problem. I'm going to need to take the derivative of the derivative. Let me slide this information off here to this side. Let me remind you that the derivative that we found is equal to (-x^2 + 1) divided by (x^2 -1)^2. So now if you take the derivative of this - the second derivative - again I have to use the quotient rule, so it's the bottom times the derivative of the top, which is -2x, minus the top. So what's the top? The top is just a negative, so that makes this a positive (x^2 + 1), times the derivative of the bottom. Now that could require me to do a little of a chain rule here, so the chain rule is going to look like what? Well, the chain rule is going to be blah^2. So 2(blah^1) - peel that away - times the derivative of that, which is 2x, all divided by the bottom squared - so (x^2 - 1), all to the fourth power. That looks pretty horrific, but let's simplify that and see if we can make any progress here. In fact, you'll notice something: There's an (x^2 - 1) right here, and there's an (x^2 - 1) right here. I can actually factor those out, so I can factor out one of these (x^2 - 1)'s with this (x^2 -1). Also notice that I can factor out this 2x with that 2x, so I can do an awful lot of factoring. You might be tempted to sort of expand everything and cancel. You can do that; but if you can factor, it's always better to factor.
So let me factor out the 2x - you've got to be really careful here now - and factor out one of these terms, (x^2 - 1), and what am I left with? Well, here I've taken one of those away, but I had two to begin with, so I have one left over. So I have (x^2 - 1). Then I pull that out, but I have a negative sign there, so let me put a negative sign in front of that. So there's the negative sign, there's the 2x, and these two together give me this thing squared. Then plus...and what do I have here? If I take this out, I take that out, it leaves me with this term, so I see (2x^2 + 1). So I simplified that actually a good deal, and you'll notice that, in fact, I can actually cancel this factor with one of those. So I can cancel this factor with one of these, which leaves me with the power of three on the bottom now.
Okay, now where am I? Well, let me now untangle this a little bit. Sometimes these problems are algebraically intensive, but, of course, they're worth doing carefully. Here you want to distribute that negative sign. Be careful, that negative sign has got to hit everybody here. I see a minus x^2, and then plus 2x^2, so that gives me a net gain of just x^2. Then I have a minus and a minus is a plus one. Then here I have a two times one, which is a plus two, so plus two plus one is plus three. So after all of this simplification, I get, and I hope that you got too, this for the second derivative. Wow, okay, but now we're in a position to do all of the work that we need to do.
First of all, when does this thing equal zero? Well, this will equal zero when the top is zero. And when is the top zero? Well, either x is zero or this is zero, but notice that (x^2 + 3) is never zero because squared is always non-negative. So, in fact, the only solution that we get that we set equal to zero, we have that x equals zero. The other place we have to look is where the second derivative is undefined. The second derivative is undefined when the bottom is zero, and that could happen at either plus or minus one. It's possibly a place of contention, but remember, we've already seen that those represent vertical asymptotes. We don't have to worry about them at points of inflection.
Okay, now let's make a little sine chart here for the second derivative, and I'm going to put in zero, but remember we also put in our asymptotes, which we saw to be minus one and one. Let me remind you, these are vertical asymptotes, so you don't want to cross them. You don't want to cross vertical asymptotes; they are very mean.
So now what am I going to do? Well, now what I'm going to do is take values and plug in and see what we get, positive or negative, and determine concave up or concave down. Pick a point to the left here, like negative two. If I plug in negative two, what do I see? Well, that is always positive no matter what because (x^2 + 3) is positive. So that's always positive. Negative two makes this negative, so the whole top here is negative. If I put a negative two in here, that makes a plus four, minus one is three; three cubed is positive. So I've got a positive on the bottom; I've got a negative on the top; the net gain is negative. So this thing is falling.
What happens between minus one and zero? We've got to pick a point between there, so you might want to pick like -. Let's try - and see what happens. Again, this is always positive. The - makes this negative here. If I put a - in here, when I square it I get + . So (+ - 1) is actually - ; and -^3 is negative. So I've got a negative and a negative on the top, and this actually becomes positive. What about at ? Well, if you try and plug it in, I get a positive here, a positive here. When I put in a here, I see a ^2 - - minus one is -3/2, again negative. So I have a negative here but a positive here, so this is making this negative. And if I put something really big in here, like let's say ten - positive, positive, and if I put 10^2 and subtract one, that's positive. So, in fact, I see it's a positive.
So what have we learned? I've learned that this function is concave down up to the asymptote, then concave up to the point of zero, and then it's concave down. So zero really is a point of inflection. Then past this asymptote for a concave up again.
Okay, now we have all the information to graph this final function. So let's take a look at how we put all of this information together. Okay, so what do we have? Well, let me draw some axes in black - like Men in Black, axes in black. Now, where will we mark? Well, let's mark our asymptotes. We have a vertical one at minus one. So at minus one we have a vertical asymptote...we can't cross that line no matter what. We have one at one; we can't cross that line no matter what. We have a horizontal asymptote actually at (y = 0), which is the x-axis. I'll just draw it a little bit up so you can see it, but you have to remember it's actually on the x-axis. Okay, so there are all of the asymptotes, which you'll notice cut our world up into a whole bunch of pieces. We can't cross these lines. So we may exist here, we may exist here, but we're not going to cross over.
Okay, let's see what we've learned. I'll bring all of this information back for us quickly so you can see. We see that, in fact, we're always decreasing. So the function is always going to be going down. So what that tells me is that I'm going to be dropping here, I'm going to be dropping here, and I'm going to be dropping here, but I have to do it in these asymptotes. So somehow here I can't go way up and drop down because I have to be asymptotic here. So I must just somehow just stick up there and then come down like that and be asymptotic here. In fact, that's consistent with what we've already discovered because I've seen that we're concave down up to negative one. So therefore we're concave down - we're decreasing - so there's only one way to go, and it must be like this. Notice that it's decreasing, it's concave down, I'm asymptotic here, and I'm asymptotic here.
Okay, what about in this region? Well, what do I know? I' m still going down; but at (x = 0), I know that, in fact, I have a point of inflection, and we've already seen that here I'm concave up and here I'm going to be concave down. So how must that go? I must be asymptotic, decreasing, and concave up. There's only one way to that. Now, I'm still going to decrease...asymptotic to here, but concave down. So therefore we must have that little nice, pretty bend in there. Now we come to this region. What do we know? We know that function is still decreasing, so we're still going to go down. We can't be going down like this because I've got to be asymptotic here, so I must be way up here. And if you back that there, we see that we are still concave up in this region. So I'm concave up, asymptotic, and decreasing. All of the conditions are fulfilled, and this actually is a very accurate sketch of the function f(x). Notice how we saw this nice little bend - the curvature - here, and we see exactly where that curvature happens. It changes right here.
So there's the graph of the function. I really invite you to try these on your own. This is the only way to really nail these things. And I know full well - I'm not an idiot - I know full well this is a multi-, multi-step process. It's easy to get tired, it's easy to get frustrated, and especially if you get conflicting information, which means you've made a mistake somewhere. Just take it really slowly. A lot of times everyone has to do these things again and again to get them right. Practice them. Make this be an idea that's your own, and you've really done something big. In fact, let me just close that by saying that we've really come a long way; and, in fact, we've completely resolved the differential side of calculus. We've looked at derivatives, we've built them up, we've seen how to find them, we've seen the power of these things, now we're ready to finally move on to that very second question of calculus that I mentioned in our very first discussion, we're now going to look at a completely different realm. We'll look at areas, look at shapes, and try to figure out how to put those things together.
Congratulations, and I'll see you at our next session. Bye.

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