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About this Lesson
 Type: Video Tutorial
 Length: 12:26
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 133 MB
 Posted: 06/27/2009
This lesson is part of the following series:
College Algebra: Full Course (258 lessons, $198.00)
Algebra: Exponential and Logarithmic Functions (36 lessons, $49.50)
College Algebra: Function Inverses (4 lessons, $5.94)
Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found athttp://www.thinkwell.com/student/product/collegealgebra. The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra 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 UTAustin 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, padic 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
 Thinkwell
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11/14/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
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So, if we are given two functions, how can we determine if those two functions are actually inverses of each other? Well, there's a couple of ways of thinking about it. First, let's think about it visually, graphically, and then we'll think about it algebraically and see the differences. So, first of all, what does it mean for things to be inverses, visually? Well, let's contemplate that for a second. Suppose I had a point here, let's say (a,b). Let's suppose it's way out here. So, here's a and here's b. So, (a,b) would be right here.
Now, if I wanted to undo that, what would that mean? It would mean that if I now plug in b, what should it spit out? It should spit out a. So, if you think about it, sort of the inverse of that point would be to plug back b in for x and the output should be a. So, in fact, the inverse of that would be right up here. What do you notice? If I have (a,b), and then to undo that I have to look at (b,a), what you see is I'm just taking this point and reflecting it over to here. You can see that with a mirror. I'll try this mirror thing again. Take a look at the mirror. There's the point by the way. You can see it now, pretty clearly, but look what happens. It's just literally a reflection. It's sort of, oh it's so close, oh if this mirror was just a teeny bit bigger.
You see it? It's just a reflection. Where is there a reflection? Around what line? Well, if you look straight down here, you'll see that line looks like the y = x line. It's perfectly diagonal line. So, that's exactly what happens, believe it or not. If you look at the diagonal line, y = x, the line that cuts right through both axes, half way, what you see is, if you have a point here, then the inverse would be just the flip over that. So, if you have (a,b), you'd now go to (b,a) So, graphically two functions are going to be inverse functions if whatever the function does, if you flip that over this line, you get the other function. If that's true, then those are inverse functions.
So, for two things to be inverse functions, they have to have a symmetry over this flip. Whatever you do with the first functions, if you flip it you should get the second function. So, that's the way to test. So, let's just try an example here and see that in action. Let's suppose we look at f(x) = 2x  3. The inverse function, by the way, I now want to introduce some new notation. So, new notation, this is how you denote an inverse function. I won't write g. I want us to now say it's f backwards. So, what do you do? Well, I write the following notation, f with a little 1 right on top of the f, like a superscript, like an exponent of x.
Now, I want to call your attention to a couple of quick things. These are some classic mistakes here. Classic mistake number 1these aren't real classic mistakes, like top ten lists, but just in this realm. The first thing to remember is that the notation is not this. The notation is put the 1 right by the f. The second thing is do not, do not, please, please, please do not think that since it's a 1 exponent, I can write it as 1 over. That's not true. Even though you may think that, because it looks that way. This is just a symbol. It's just a symbol. I'm telling you, it's a notation. It doesn't mean 1 over, it means inverse. So, what it means, really, is that this is what undoes that function. What we saw, or what I told you a little while ago, is that this should equal that.
So, let's see if these are really inverses by looking at their graphs. Let's take a look here. This is a straight line that has y intercept 3 and slope 2 over 1. So, 1 over, 2 up. So, there we go. There's the first line. Now, let's take a look at this second line. This second line would look like what? Well, this is going to be x + ^3/[2]. So, the y intercept is ^3/[2]. So, I go to ^3/[2]. That's 1, so you go right to here. Then what do I do? I go slope 1 over 2. So, I go 2 over and 1 up. Should be right here, 2 over, 1 up, and so on. If we connect those things, what does that look like? That looks like this.
Now, let's put in that middle line, that y = x line. That's where the reflection is supposed to happen. Where's the y = x line and what do you notice? You notice, if you look at thisand you really have be like a visual kind of person, but look, if you take the red line and now just flip it over this orange linedo you see that we get the purple line? See, this line would get flipped to that part of the purple line and this red line would get flipped to this part of the purple line. So, these are flips of each other. Which means these are inverse functions
Now, how can you see this? That's graphically. How could you see this, now, algebraically? Wells, here's the property of inverse functions. They undo each other. That was the whole point. So, therefore, f composed with f inverse should just give back x and f inverse composed with f should give back x. So, these two things have to hold. These two things have to hold in order to be inverses. Let's try and see if these two things really do hold in this example. So, the first thing I'm going to look at is ask, "What is f composed with f inverse?"
Let me remind you of what that means. It means, you take f of f ^1(x). So, what is f of f ^1(x)? Well, what do you do? What I do is, I go to the f function and in place of x I put in this new function f ^1. So, I put this in and I insert that wherever I see an x. So, I see, 2, 2, times, times, x. But now I'm putting in that whole "garble dee gook," in place of x. So, garble dee gook and then 3. Notice that the 2's cancel and after I cancel the 2's, I'm just left with x + 33and that's just x. So, that checks, because I'm suppose to have that thing give me back x. So, that checks.
What about this way? This way has to hold as well. So, let's try the other direction. What is f ^1 composed with f(x). That's composition. That means I take f ^1 of f(x). What's that? Well, now what that means is I take the f ^1 function and in place of x, I shove in f(x). I've got to push that in. So, I take this thing and stick it in here and I hope that I'm going to get just x alone. So, here I see an x. I put all that stuff in there. So, I see 2x 3, that's that x right there, then + 3 divided by 2. Well, 3 + 3 just equals 0. So, I see 2x divided by x and look. That equals x. This checks. So, these two functions really are inverses of each other and I saw that visually, by seeing that's just the flip over the y = x line and I saw it algebraically by checking that both these conditions hold. One undoes the other and the other undoes the one. So, that takes care of that.
Let's try one other little example, really fast together. Suppose f(x) = 2 over x+ 6. Let's see if this is really the inverse function, 6x + 2 all over x. Let's see that using just algebra. So, what I have to do is first of all check that f composed with f ^1 is equal to x. So, what is that? That's f composed with f ^1. So, I take f^ 1 and I stick it in wherever I see an x in the f function. This is going to get a little bit ugly, so let's stick together.
I see 2, so I have 2 divide by x. But now, in place of x, I'm putting in this entire function. So, I put all this in, just for x. So, I see 6x + 2 all over x and then don't forget the +6, because that's just the x part. All right, let's see if this works. Well, I'm going to get a common denominator here. So, I'm going to multiply top and bottom here by x. So, what do I see? I see 2 divided by 6x + 6x is 12x. Then I have a +2 and the whole thing is divided by x. So, I have that. Then if I now invert and multiply, I see 2x divided by 12x + 2. Does that equal x? No, it equals 2x divided by 12x + 2. So, in fact, this is not the inverse function. This is not the inverse function. So, I was able to check that by working this out and seeing that I'm not getting just x alone.
Maybe one quick last example. Just one quick last example, if I have the time. You know, I probably don't even have the time, but let me just set this up. We'll see. Maybe I'll do it anyway. We'll see. You tell me. We'll have a vote. You know how they have these votes on the web, like they take a survey? Are these two things inverse functions? Well, let's look at f composed with f ^1(x). That would be f of f ^1(x). That means I take f ^1 and stick it in for x in this function. So, that would be 2 times f ^1, which is x  2 and then all + 4. You see how I just took this function and in place of x I just stuck that right in. Stuck in right in. Now, what does that equal?
Well, if I distribute the 2, I've got to distribute the 2 everywhere. I would see 2 times is just 1x and this is a 4 plus this 4 and that equals x. Well, that looks good. Are we done? No, we're not done. You have to check the other way too. So, you have to check f ^1 composed with f(x) and see if that's going to give you x. In fact, I'm going leave this as an exercise right now for you to try and I'm going to call it quits for this lecture. I want you to try this. By taking f ^1 and then plugging in for x the f function, and actually verifying that this will in fact equal x. These two functions are inverse functions.
So, inverse functions, how do they look? Graphically they're going to be sort of flips of each other over the y = x line, because I'm switching the roles of x and y. Algebraically, I have to check these two conditions and make sure in both cases I get back to x. Try some of these, carefully. See what you think about inverse functions.
Exponential and Logarithmic Functions
Function Inverses
Are Two Functions Inverses of Each Other? Page [2 of 2]
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