# exam1 - Granillo, Yvette Exam 1 Due: Sep 29 2005, 11:00 pm...

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Granillo, Yvette – Exam 1 – Due: Sep 29 2005, 11:00 pm – Inst: Edward Odell 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 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 ) = 7 t 2 - 7 t + 4 . Determine the average velocity oF the particle over the time interval [1 , 2]. 1. average vel. = 15 Ft/sec 2. average vel. = 16 Ft/sec 3. average vel. = 17 Ft/sec 4. average vel. = 18 Ft/sec 5. average vel. = 14 Ft/sec correct Explanation: The average velocity over a time interval [ a, b ] is given by dist travelled time taken = s ( b ) - s ( a ) b - a . ±or the time interval [1 , 2], thereFore, average vel. = s (2) - s (1) 2 - 1 Ft/sec . Now s (2) = 7 × 4 - 7 × 2 + 4 = 18 Feet , while s (1) = 7 - 7 + 4 = 4 Feet . Consequently, average vel. = 18 - 4 = 14 Ft/sec . keywords: Stewart5e, 002 (part 1 oF 1) 10 points Below is the graph oF a Function f . 2 4 6 - 2 - 4 - 6 2 4 6 8 - 2 - 4 Use the graph to determine the leFt hand limit lim x 2 - f ( x ) . 1. the limit does not exist 2. lim x 2 - f ( x ) = 5 2 3. lim x 2 - f ( x ) = 6 correct 4. lim x 2 - f ( x ) = - 1 5. lim x 2 - f ( x ) = - 4 Explanation: As the graph shows, lim x 2 - f ( x ) = 6. keywords: Stewart5e, 003 (part 1 oF 1) 10 points

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2 When f is the function deFned by f ( x ) = 3 x - 2 , x < 1 , 5 x - 3 , x 1 , determine if the limit lim x 1 - f ( x ) exists, and if it does, Fnd its value. 1. limit does not exist 2. limit = 2 3. limit = 3 4. limit = - 1 5. limit = 1 correct 6. limit = 0 Explanation: The left hand limit lim x 1 - f ( x ) depends only on the values of f to the left of 1. Thus lim x 1 - f ( x ) = lim x 1 - 3 x - 2 . Consequently, limit = 3 × 1 - 2 = 1 . keywords: Stewart5e, 004 (part 1 of 1) 10 points Determine the limit lim x 6 3 ( x - 6) 2 . 1. limit = - 1 2 2. limit = 1 2 3. limit = -∞ 4. limit = correct 5. none of these Explanation: Since ( x - 6) 2 0 for all x , we see that lim x 6 3 ( x - 6) 2 = . keywords: Stewart5e, 005 (part 1 of 1) 10 points ±ind the value of lim x 2 1 x - 1 2 ·‡ 5 x - 2 · . 1. limit = - 5 4 correct 2. limit = 5 2 3. limit = - 5 2 4. limit does not exist 5. limit = 5 4 Explanation: After the Frst term in the product is brought to a common denominator, the given expression becomes 5(2 - x ) 2 x ( x - 2) = - 5 2 x so long as x 6 = 2. Thus lim x 2 1 x - 1 2 ·‡ 5 x - 2 · = - lim x 2 5 2 x . Consequently,
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## This test prep was uploaded on 04/16/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas at Austin.

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exam1 - Granillo, Yvette Exam 1 Due: Sep 29 2005, 11:00 pm...

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