final - Granillo Yvette Final 1 Due 1:00 pm Inst Edward...

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Granillo, Yvette – Final 1 – Due: Dec 14 2005, 1:00 pm – Inst: Edward Odell 1 This print-out should have 25 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 Below are the graphs of functions f and g . 4 8 - 4 4 8 - 4 - 8 f : g : Use these graphs to determine lim x → - 2 { f ( x ) + g ( x ) } . 1. limit = 3 2. limit = 0 3. limit does not exist correct 4. limit = 4 5. limit = 7 Explanation: From the graph it is clear that lim x → - 2 { f ( x ) + g ( x ) } does not exist . keywords: Stewart5e, limit of a sum 002 (part 1 of 1) 10 points Determine lim x 0 f ( x ) when f ( x ) = x - 1 x 2 ( x + 4) . 1. lim x 0 f ( x ) = 0 2. lim x 0 f ( x ) = 1 3. lim x 0 f ( x ) = 4. lim x 0 f ( x ) = - 1 4 5. lim x 0 f ( x ) = -∞ correct Explanation: Now lim x 0 x - 1 = - 1 . On the other hand, x 2 ( x + 4) > 0 for all small x , both positive and negative, while lim x 0 x 2 ( x + 4) = 0 . Thus lim x 0 f ( x ) = -∞ . keywords: Stewart5e, evaluating limit, nu- meric 003 (part 1 of 1) 10 points Determine lim x 0 e 8 x - 8 x - 1 4 x 2 · . 1. limit = 17 2 2. limit = 8 correct 3. limit doesn’t exist
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Granillo, Yvette – Final 1 – Due: Dec 14 2005, 1:00 pm – Inst: Edward Odell 2 4. limit = 15 2 5. limit = 9 Explanation: The limit in question is of the form: lim x 0 f ( x ) g ( x ) where f and g are differentiable functions such that lim x 0 f ( x ) = 0 , lim x 0 g ( x ) = 0 . Thus L’Hospital’s rule can be applied, so lim x 0 f ( x ) g ( x ) = lim x 0 f 0 ( x ) g 0 ( x ) . But lim x 0 f 0 ( x ) = 0 , lim x 0 g 0 ( x ) = 0 , in which case L’Hospital’s rule has to be ap- plied again. Consequently, lim x 0 f ( x ) g ( x ) = lim x 0 f 00 ( x ) g 00 ( x ) . Since f 00 ( x ) = 64 e 8 x , g 00 ( x ) = 8 , it now follows that lim x 0 e 8 x - 8 x - 1 4 x 2 = 8 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Determine lim x 0 2 x tan - 1 (5 x ) . 1. limit = 2 2. limit does not exist 3. limit = 1 5 4. limit = 0 5. limit = 5 2 6. limit = 2 5 correct Explanation: Since the limit has the form lim x 0 2 x tan - 1 (5 x ) = 0 0 , we use L’Hospital’s Rule with f ( x ) = 2 x, g ( x ) = tan - 1 (5 x ) . For then lim x 0 f ( x ) g ( x ) = lim x 0 f 0 ( x ) g 0 ( x ) = lim x 0 2(1 + (5 x ) 2 ) 5 . Consequently, limit = 2 5 . keywords: Stewart5e, 005 (part 1 of 1) 10 points Find all values of x at which the function f defined by f ( x ) = x 2 - 2 x - 15 x 2 - 7 x + 10 , x 6 = 5, 8 3 , x = 5, is continuous, expressing your answer in in- terval notation. 1. ( -∞ , - 2) ( - 2 , )
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Granillo, Yvette – Final 1 – Due: Dec 14 2005, 1:00 pm – Inst: Edward Odell 3 2. ( -∞ , - 2) ( - 2 , 5) (5 , ) 3. ( -∞ , 2) (2 , 5) (5 , ) 4. ( -∞ , 2) (2 , ) correct 5. ( -∞ , 5) (5 , ) Explanation: After factorization f becomes f ( x ) = ( x - 5)( x + 3) ( x - 2)( x - 5) = x + 3 x - 2 for values of x 6 = 5. But lim x 5 f ( x ) = lim x 5 x + 3 x - 2 = 8 3 = f (5) , so f is defined and continuous at x = 5. Since elsewhere f is a rational function, the values of x at which it will fail to be continuous will be at zeros of the denominator as f will not be defined at those points. Hence f fails to be continuous only at x = 2; in other words, it is continuous on ( -∞ , 2) (2 , ).
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