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Unformatted text preview: mao (tm23477) – HW13 – gualdani – (57180) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine which, if any, of f ( x ) = 256(16 − 3 x ) , g ( x ) = parenleftbigg 1 16 parenrightbigg 3 x − 2 , h ( x ) = 256(4 − 6 x ) , define the same function. 1. f, g, and h correct 2. only g, h 3. only g, f 4. only f, h 5. none of f, g, or h Explanation: By the laws of Exponents f ( x ) = 256(16 − 3 x ) = 16 2 (16 − 3 x ) = 16 2 − 3 x , while g ( x ) = parenleftbigg 1 16 parenrightbigg 3 x − 2 = 16 − (3 x − 2) = 16 2 − 3 x and h ( x ) = 256(4 − 6 x ) = 16 2 ( 4 2 ) − 3 x = 16 2 (16 − 3 x ) = 16 2 − 3 x Consequently, all of f, g, and h define the same function. 002 10.0 points If a, b are the solutions of the exponential equation 3 x 2 = 9 5 x − 8 , calculate the value of ab . 1. ab = 8 2. ab = 10 3. ab = 16 correct 4. ab = 16 5. ab = 12 Explanation: By properties of exponents, 9 5 x − 8 = 3 10 x − 16 . Thus the equation can be rewritten as 3 x 2 = 3 10 x − 16 , which after taking logs to the base 3 becomes x 2 = 10 x 16 . This equation factors as ( x 2)( x 8) = 0 , and so its solutions are 2 , 8. Hence ab = 16. 003 10.0 points Which function has mao (tm23477) – HW13 – gualdani – (57180) 2 2 4 2 4 2 4 2 4 as its graph? 1. f ( x ) = 2 − x − 1 2 2. f ( x ) = 2 x − 1 3 3. f ( x ) = 2 2 − x − 1 4. f ( x ) = 2 3 − x 5. f ( x ) = 3 − x 2 6. f ( x ) = 3 x 3 correct 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 . 004 10.0 points Determine if lim x →−∞ parenleftbigg 3 e 3 x 5 e − 3 x 5 e 3 x e − 3 x parenrightbigg exists, and if it does, find its value. 1. limit does not exist 2. limit = 3 3. limit = 5 correct 4. limit = 1 5. limit = 0 6. limit = 3 5 Explanation: Since lim x →−∞ e − ax = ∞ , lim x →−∞ e ax = 0 when a > 0, evaluating the limit directly gives lim x →−∞ parenleftbigg 3 e − 3 x 5 e 3 x 5 e 3 x e − 3 x parenrightbigg = ∞ ∞ , which doesn’t make any sense. (And we can’t just cancel the ∞ ’s because infinities don’t work like that.) So we try to get rid of terms that go to ∞ and leave terms that go to zero. To achieve this, multiply top and bottom by e 3 x . Then e 3 x e 3 x parenleftbigg 3 e 3 x 5 e − 3 x 5 e 3 x e − 3 x parenrightbigg = 3 e 6 x 5 5 e 6 x 1 , and so lim x →−∞ 3 e − 3 x 5 e 3 x 5 e 3 x e − 3 x = lim x →−∞ 3 e 6 x 5 5 e 6 x 1 . mao (tm23477) – HW13 – gualdani – (57180) 3 But, as we noted, lim x →−∞ e 6 x = 0 , so by properties of limits, lim x →−∞...
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This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas.
 Fall '08
 schultz

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