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Unformatted text preview: Mahon, Kevin – Homework 13 – Due: Nov 24 2005, 3:00 am – Inst: Edward Odell 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Determine which, if any, of the following f ( x ) = 4 x + 8 , g ( x ) = 2 2 x +4 , h ( x ) = 16 (4 x ) , define the same function. 1. only g, f 2. only f, h 3. f, g, and h 4. only g, h correct 5. none of f, g, or h Explanation: By the Laws of Exponents, f ( x ) = 4 x + 8 = (2 2 ) x + 8 = 2 2 x + 8 , while g ( x ) = 2 2 x +4 and h ( x ) = 16 (4 x ) = (2 4 )(2 2 ) x = (2 4 )(2 2 x ) = 2 2 x +4 . Thus g and h define the same function. On the other hand, f (0) = 9 , g (0) = 16 = h (0) , so neither g nor h can define the same function as f . Consequently only g, h define the same function. keywords: Stewart5e, 002 (part 1 of 1) 10 points If x 1 , x 2 are the solutions of the equation 6 3 x 2 = 1 36 5 x +2 , compute the value of  x 1 + x 2  . 1.  x 1 + x 2  = 10 3 2.  x 1 + x 2  = 2 √ 13 3 3.  x 1 + x 2  = 5 3 4.  x 1 + x 2  = 10 3 correct 5.  x 1 + x 2  = 2 √ 13 3 Explanation: Since 1 36 5 x +2 = 6 10 x 4 , the equation can be written as 6 3 x 2 = 6 10 x 4 , which in turn can be rewritten as 3 x 2 = 10 x 4 by taking logs to the base 6 of both sides. By the quadratic formula, therefore, x 1 , x 2 = 5 ± √ 13 3 . Thus  x 1 + x 2  = 10 3 . keywords: Stewart5e, 003 (part 1 of 1) 10 points Mahon, Kevin – Homework 13 – Due: Nov 24 2005, 3:00 am – Inst: Edward Odell 2 Which function has 2 4 2 4 2 4 2 4 as its graph? 1. f ( x ) = 2 x 1 3 2. f ( x ) = 2 x 1 2 3. f ( x ) = 3 x 3 correct 4. f ( x ) = 3 x 2 5. f ( x ) = 2 2 x 1 6. f ( x ) = 2 3 x Explanation: The given graph has the property that lim x →∞ f ( x ) = 3 . But lim x →∞ 2 x = 0 = lim x →∞ 3 x , while lim x →∞ 2 x = 0 = lim x →∞ 3 x , so f ( x ) must be one of 3 x 3 , 2 x 1 3 . On the other hand, the yintercept of the given graph is at y = 2. Consequently, the graph is that of f ( x ) = 3 x 3 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Find the value of lim x →∞ µ e 2 x + 2 e 2 x 2 e 2 x 5 e 2 x ¶ . 1. limit = 2 2. limit = 1 2 3. limit = 2 4. limit = 1 7 5. limit = 1 2 correct 6. limit = 1 7 Explanation: From the graph of e x it follows that lim x →∞ e kx = 0 , lim x →∞ e kx = 0 for each k > 0. Now e 2 x + 2 e 2 x 2 e 2 x 5 e 2 x = 1 + 2 e 4 x 2 5 e 4 x . But by properties of limits lim x →∞ 1 + 2 e 4 x 2 5 e 4 x = 1 2 Consequently, limit = 1 2 . keywords: Stewart5e, Mahon, Kevin – Homework 13 – Due: Nov 24 2005, 3:00 am – Inst: Edward Odell 3 005 (part 1 of 1) 10 points Find the value of f (0) when f ( x ) = 1 3 e 2 x + 1 4 e x ....
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This homework help was uploaded on 04/15/2008 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz
 Differential Calculus

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