practice1

# practice1 - Granillo Yvette – Review 1 – Due Dec 9 2005...

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Unformatted text preview: Granillo, Yvette – Review 1 – Due: Dec 9 2005, 6:00 pm – Inst: Edward Odell 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Review one covers the material on test 1, namely it covers sections 2.1-2.3,2.5,2.6,3.1- 3.4. While you can submit answers and get a score as usual this counts 0 001 (part 1 of 1) 10 points After t seconds the displacement, s ( t ), of a particle moving rightwards along the x-axis is given (in feet) by s ( t ) = 5 t 2- 8 t + 6 . Determine the average velocity of the particle over the time interval [1 , 3]. 1. average vel. = 11 ft/sec 2. average vel. = 9 ft/sec 3. average vel. = 12 ft/sec correct 4. average vel. = 8 ft/sec 5. average vel. = 10 ft/sec Explanation: The average velocity over a time interval [ a, b ] is given by dist travelled time taken = s ( b )- s ( a ) b- a . For the time interval [1 , 3], therefore, ave. vel. = s (3)- s (1) 3- 1 ft/sec . Now s (3) = 5 × 9- 8 × 3 + 6 = 27 feet , while s (1) = 5- 8 + 6 = 3 feet . Consequently, avg. vel. = 27- 3 2 = 12 ft/sec . keywords: Stewart5e, 002 (part 1 of 1) 10 points Below is the graph of a function y = f ( x ) 2 4 6- 2- 4- 6 2 4 6 8- 2- 4 Use the graph to determine the right hand limit lim x → 1+ f ( x ) . 1. lim x → 1+ f ( x ) = 0 correct 2. lim x → 1+ f ( x ) =- 4 3. lim x → 1+ f ( x ) = 6 4. lim x → 1+ f ( x ) = 3 5. the limit does not exist Explanation: From the graph lim x → 1+ f ( x ) = 0. keywords: Stewart5e, Granillo, Yvette – Review 1 – Due: Dec 9 2005, 6:00 pm – Inst: Edward Odell 2 003 (part 1 of 1) 10 points A function f is defined piecewise for all x 6 = 0 by f ( x ) = 4 + x, x <- 2 , 5 2 x, < | x | ≤ 2 , 5 + x- 1 2 x 2 , x > 2 . By first drawing the graph of f , determine all the values of a at which lim x → a f ( x ) exists, expressing your answer in interval no- tation. 1. (-∞ , 0) ∪ (0 , ∞ ) 2. (-∞ , 2) ∪ (2 , ∞ ) 3. (-∞ ,- 2) ∪ (- 2 , 0) ∪ (0 , ∞ ) 4. (-∞ ,- 2) ∪ (- 2 , ∞ ) correct 5. (-∞ , 0) ∪ (0 , 2) ∪ (2 , ∞ ) 6. (-∞ ,- 2) ∪ (- 2 , 2) ∪ (2 , ∞ ) 7. (-∞ ,- 2) ∪ (- 2 , 0) ∪ (0 , 2) ∪ (2 , ∞ ) Explanation: The graph of f is 2 4- 2- 4 2 4- 2- 4 and inspection shows that lim x → a f ( x ) will exist only for a in (-∞ ,- 2) ∪ (- 2 , ∞ ) . keywords: Stewart5e, 004 (part 1 of 1) 10 points Evaluate lim x → x 6 + 5 x 3 4 x 8 + 7 x 11 . 1. limit =-∞ 2. limit = 5 3. limit = 0 4. none of these correct 5. limit = + ∞ Explanation: Since x 6 + 5 x 3 4 x 8 + 7 x 11 = x 3 + 5 x 5 (4 + 7 x 3 ) , we see that none of 5 , + ∞ , ,-∞ can be the limit because x 3 + 5 x 5 (4 + 7 x 3 )-→ + ∞ as x → 0+, while x 3 + 5 x 5 (4 + 7 x 3 )-→ -∞ as x →- . Consequently, none of these is the correct answer. Granillo, Yvette – Review 1 – Due: Dec 9 2005, 6:00 pm – Inst: Edward Odell 3 keywords: Stewart5e, limit, infinite limit, ra-...
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practice1 - Granillo Yvette – Review 1 – Due Dec 9 2005...

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