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College Algebra: Find Functions to Form Composite

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About this Lesson

  • Type: Video Tutorial
  • Length: 6:33
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 70 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
Trigonometry: Full Course (152 lessons, $148.50)
College Algebra: Relations and Functions (57 lessons, $74.25)
Trigonometry: Algebra Prerequisites (60 lessons, $69.30)
College Algebra: Composite Functions (5 lessons, $7.92)

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 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

Thinkwell
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Founded in 1997, Thinkwell has succeeded in creating "next-generation" 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 technology-based 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 to get a better sense of this composition stuff, which is genuinely new and a little tricky until you're really used to it, because the idea of taking a function and putting it into another function. It's time for that ever-popular, much-beloved, early afternoon game show--that's right folks, welcome to "Composition Jeopardy," with your host, of course, Ed Burger.
Hey, so it's time for the composition jeopardy part of our show. So here's the deal, of course, I'm going to give you the answer and it's your responsibility to tell me what the function g is and what the function f is so that the answer we give you is going to be f composed with g.
So remember, we want to give g and f so that the composition ends up being the answer we show you. And please, I remind you, please, please, please, make sure you pose your answers in terms of a question.
Okay, I think we're ready to begin with our first composition jeopardy. For $500 the quantity (6x - 2)^2.
Okay, well let's see how you did. We're having fun here at the studio and we hope you're having fun playing at home. So what would I do? I'm looking for sort of an inside and outside. I'm looking for sort of a big blob that I could call that function g, because the g function is really the thing I do first here, and if you notice really, there's a function here, namely just something squared. So, in fact, that's going to be the outside function. That's the f. So it seems to me--here's my thinking--let's let f(x) = x^2. That's the outside function. And the inside function, that's the thing I have to do first, well that should be the stuff that's on the inside. So g(x) = 6x - 2.
Let's check our answer and let's see if it's really the case that f composed with g(x) equals this answer. What do I do? Well, I first take g(x), so that's this, and I plug that in to f. So when I take this and plug that in to here, what do I see? I see in placing x I put in g(x). That's 6x - 2. So I see (6x - 2)^2. In just a teeny bit more detail, what I would see is the following. I would see this equals f of, and then g is 6x - 2. And what do I do? Wherever I see an x in the f, I just replace it by this, and that would actually equal exactly (6x - 2)^2, and that's exactly what was posed to us.
So you may be saying, "Aha! The answer is f(x) = x^2 and g(x) = (6x - 2), but actually that would be wrong. You have to put your answer in terms of the question, so the correct answer is, what is f(x) = x^2 and g(x) = (6x - 2)?
Okay, well, it's time for final jeopardy. Here we go, the big question for $1,000--the answer is 2(4x - 1)^3 - (4x - 1) + 3. Good luck.
Okay, let's see you how you did. Again, we're trying to identify an inside and an outside, and the inside function is going to be the g since that comes first in here, and then the outside function is going to be the f. Well, you can see the inside part because I can sort of cover up that little piece right there. You see it appearing in both. So it looks like g(x) might be 4x - 1. And if I blob that out, what do I see? I see two blob cubed minus blob + 3. So that's a good candidate for f. f(x) = 2x^3 - x + 3.
And so my answer would be what is g(x) = 4x -1 and f(x) = 2x^3 - x + 3? And it turns out that is the correct question, and we can check that in the following way. Let's just figure out the composition.
If we figure out the composition, what would we see? What we'd see is f composed with g(x) equals f(g(x)). See, I first do the g and then I do the f. and then g is 4x - 1, so I see f(4x - 1). So what do I do now? Wherever I see an x in the f function--so here and here--I have to actually insert right there the quantity for x - 1, wherever I see that x. But when you do that, look what you see. What you see is exactly this: 2, 2 times (4x - 1)^3 - (4x - 1) +3. So, in fact, this is the correct answer, and playing here in the studio, I would have won $1,500, which the producers now owe me.
Let's see how much money you can accrue playing "Composition Jeopardy." Good luck.
Relations and Functions
Composite Functions
Finding Functions That Form a Given Composite Page [1 of 1]

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