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Unformatted text preview: Version 137 – K Exam 1 – Hamrick – (54868) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points After t seconds the displacement, s ( t ), of a particle moving rightward along the t axis is given (in feet) by s ( t ) = t 3 4 . Find its instantaneous velocity when t = 2. 1. instant. vel. = 3 ft/sec correct 2. instant. vel. = 2 ft/sec 3. instant. vel. = 5 2 ft/sec 4. instant. vel. = 3 2 ft/sec 5. instant. vel. = 1 ft/sec Explanation: The instantaneous velocity of the particle at time t = 2 is the value the average velocity approaches as the average is taken over time intervals [2 , 2 + h ] and h approaches 0. Now, over [2 , 2 + h ] the average velocity is given by dist traveled time taken = s (2 + h ) − s (2) h = (2 + h ) 3 − 2 3 4 h . But (2 + h ) 3 − 2 3 = 12 h + 6 h 2 + h 3 = h (12 + 6 h + h 2 ) , so s (2 + h ) − s (2) h = 12 + 6 h + h 2 4 . Now as h approaches 0 the right hand side approaches 3. Consequently, when t = 2 the particle has instantaneous velocity = 3 ft/sec . keywords: displacement, polynomial, instan taneous velocity, monomial 002 10.0 points At which point on the graph P Q R S T U is the slope smallest ( i.e. , most negative)? 1. R correct 2. U 3. T 4. S 5. P 6. Q Explanation: By inspection the point is R . 003 10.0 points Below is the graph of a function f . Version 137 – K Exam 1 – Hamrick – (54868) 2 2 4 6 − 2 − 4 − 6 2 4 6 8 − 2 − 4 Use the graph to determine lim x → − 3 f ( x ). 1. lim x → − 3 f ( x ) = 2 2. lim x → − 3 f ( x ) = 12 3. lim x → − 3 f ( x ) does not exist 4. lim x → − 3 f ( x ) = 8 5. lim x → − 3 f ( x ) = 6 correct Explanation: From the graph it is clear the f has both a left hand limit and a right hand limit at x = − 3; in addition, these limits coincide. Thus lim x →− 3 f ( x ) = 6. 004 10.0 points Determine lim x → x − 1 x 2 ( x + 3) . 1. limit = ∞ 2. limit = −∞ correct 3. limit = − 1 3 4. none of the other answers 5. limit = 1 6. limit = 0 Explanation: Now lim x → x − 1 = − 1 . On the other hand, x 2 ( x + 3) > 0 for all small x , both positive and negative, while lim x → x 2 ( x + 3) = 0 . Consequently, limit = −∞ . keywords: evaluate limit, rational function 005 10.0 points Determine if lim x → √ 1 + x − √ 1 − x x exists, and if it does, find its value. 1. limit = − 2 2. limit = 1 correct 3. limit = − 1 4. limit = 1 √ 2 5. limit does not exist 6. limit = − 1 √ 2 7. limit = 2 Explanation: After rationalization we see that √ 1 + x − √ 1 − x = (1 + x ) − (1 − x ) √ 1 + x + √ 1 − x = 2 x √ 1 + x + √ 1 − x . Version 137 – K Exam 1 – Hamrick – (54868) 3 Thus √ 1 + x − √ 1 − x x = 2 √ 1 + x + √ 1 − x for all x negationslash...
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This note was uploaded on 03/10/2011 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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