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Unformatted text preview: Math Review 2  Solutions due: Jan, 21/22 The math review is collected on Tuesday / Monday, before the sections starts. Note that the math review will be collected with certainty. Some students asked me for books that would help them with calculus. I think, Barrons How to prepare for the AP Calculus 2008 is a good source. It provides lots of sample questions for training and you can get it for about $12. However, the book contains much more than we need. The relevant chapters are 1,3ah, and 4 ai, L. You dont need to worry about trigonometric functions. 1. Concavity For each of the following functions f ( x ), check whether the function is concave, strictly concave, convex or strictly convex. Let x > 0: Solution : To check for concavity/convexity, we need to find and evaluate the second deriva tives. (a) f ( x ) = 4 x f = 4 , f 00 = 0 Concave and convex (weakly). (b) f ( x ) = 3 x f = 1 3 x 2 3 , f 00 = 2 9 x 5 3 < Strictly concave. (c) f ( x ) = ( x 2 + 1) 2 x 2 f = 2( x 2 + 1)(2 x ) 2 x = 4 x 3 + 2 x f 00 = 12 x 2 + 2 > Strictly convex. (d) f ( x ) = ln x f = x 1 , f 00 = x 2 < Strictly concave. (e) f ( x ) = ax 2 ( a 6 = 0) f = 2 ax, f 00 = 2 a Strictly convex if a > 0, stirctly concave if a < 0. 1 2. Finding exteremum points Find the extreme points (local and global max/mins) and the points of inflection for the following functions. Let x ( , ). Recall your knowledge from your calculus class: (a) f ( x ) = x + 1 2 x 2 f = 1 + x f 00 = 1 x = 1 is the only critical point, it is a local and global minimum. There are no points of inflection. (b) f ( x ) = x 3 5 x 2 + 6 f = 3 x 2 10 x f 00 = 6 x 10 x = 0 and x = 10 3 are the two critical points. x = 5 3 is an inflection point. By inspection of the second derivative at the critical points, x = 0 is a local maximum and x = 10 3 is a local minimum. (c) f ( x ) = x 2 + 2 x = x 2 + 2 x 1 f = 2 x 2 x 2 f 00 = 2 + 4 x 3 Note first that this funciton is undefined at x = 0. To find cricial points, set the first derivative equal to zero: 2 x 2 x 2 = 0 2 x 3 2 = 0 x = 1 By inspection of the second derivative at the critical point, this is a local minimum. To find points of inflection, set the second derivative equal to zero: 2 + 4 x 3 = 0 2 x 3 + 4 = 0 x = ( 2) 1 3 . (d) f ( x ) = x 2 1 30 x 3 f = 2 x 1 10 x 2 f 00 = 2 1 5 x The critical points are x = 0 and x = 20. By inspection of the second derivative, x = 0 is a local minimum and x = 20 is a local maximum. 2 3. Convex Combinations: Solution : Point C is a convex combination of points A and B if there exists some (0 , 1) such that C = A +(1 ) B . Specifically x C 1 = x A 1 +(1 ) x B 1 and x C 2 = x A 2 +(1 ) x B 2 ....
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This note was uploaded on 04/09/2008 for the course ECON 401 taught by Professor Kuhn during the Winter '08 term at University of Michigan.
 Winter '08
 KUHN

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