# hw13 - Granillo Yvette – Homework 13 – Due 3:00 am –...

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Unformatted text preview: Granillo, Yvette – Homework 13 – Due: Nov 24 2005, 3:00 am – Inst: Edward Odell 1 This print-out should have 23 questions. Multiple-choice 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 ) = 16 x + 12 , g ( x ) = 4 2 x +3 , h ( x ) = 64 (16 x ) , define the same function. 1. only g, f 2. f, g, and h 3. none of f, g, or h 4. only g, h correct 5. only f, h Explanation: By the Laws of Exponents, f ( x ) = 16 x + 12 = (4 2 ) x + 12 = 4 2 x + 12 , while g ( x ) = 4 2 x +3 and h ( x ) = 64 (16 x ) = (4 3 )(4 2 ) x = (4 3 )(4 2 x ) = 4 2 x +3 . Thus g and h define the same function. On the other hand, f (0) = 13 , g (0) = 64 = 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 7 3 x 2 = 1 49 4 x +2 , compute the value of | x 1 + x 2 | . 1. | x 1 + x 2 | = 8 3 correct 2. | x 1 + x 2 | = 4 3 3. | x 1 + x 2 | = 2 √ 4 3 4. | x 1 + x 2 | =- 2 √ 4 3 5. | x 1 + x 2 | =- 8 3 Explanation: Since 1 49 4 x +2 = 7- 8 x- 4 , the equation can be written as 7 3 x 2 = 7- 8 x- 4 , which in turn can be rewritten as 3 x 2 =- 8 x- 4 by taking logs to the base 7 of both sides. By the quadratic formula, therefore, x 1 , x 2 =- 4 ± √ 4 3 . Thus | x 1 + x 2 | = 8 3 . keywords: Stewart5e, 003 (part 1 of 1) 10 points Granillo, Yvette – 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- 2- x- 1 2. f ( x ) = 3 x- 3 3. f ( x ) = 2 x- 1- 3 4. f ( x ) = 2- 3- x 5. f ( x ) = 3- x- 2 correct 6. f ( x ) = 2- x- 1- 2 Explanation: The given graph has the property that lim x →∞ f ( x ) =- 2 . 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- 2 , 2- x- 1- 2 . On the other hand, the y-intercept of the given graph is at y =- 1. Consequently, the graph is that of f ( x ) = 3- x- 2 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Find the value of lim x →∞ µ e 3 x- 4 e- 3 x 2 e 3 x + e- 3 x ¶ . 1. limit =- 2 2. limit = 2 3. limit = 1 4. limit = 1 2 correct 5. limit =- 1 6. limit =- 1 2 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 3 x- 4 e- 3 x 2 e 3 x + e- 3 x = 1- 4 e- 6 x 2 + e- 6 x . But by properties of limits lim x →∞ 1- 4 e- 6 x 2 + e- 6 x = 1 2 Consequently, limit = 1 2 . keywords: Stewart5e, Granillo, Yvette – 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 2 e- 2 x ....
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hw13 - Granillo Yvette – Homework 13 – Due 3:00 am –...

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