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exam 1 - Version 071 Exam 1 gualdani(56455 This print-out...

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Version 071 – Exam 1 – gualdani – (56455) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Below is the graph of a function f . -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 2 4 6 2 4 6 2 4 6 8 2 4 Use the graph to determine lim x 4 f ( x ) . 1. limit does not exist correct 2. limit = 3 3. limit = 5 4. limit = 9 5. limit = 7 Explanation: From the graph it is clear the f has a left hand limit at x = 4 which is equal to 3; and a right hand limit which is equal to 1. Since the two numbers do not coincide, the limit does not exist . 002 10.0 points When f is the function defined by f ( x ) = braceleftbigg 3 x 2 , x < 2 , 4 x 3 , x 2 , determine if lim x 2 f ( x ) exists, and if it does, find its value. 1. limit = 6 2. limit = 2 3. limit = 5 4. limit = 4 correct 5. limit = 3 6. limit does not exist Explanation: The left hand limit lim x 2 f ( x ) depends only on the values of f for x > 2. Thus lim x 2 f ( x ) = lim x 2 3 x 2 . Consequently, limit = 3 × 2 2 = 4 . 003 10.0 points Determine lim x 1 x + 3 2 x 1 . 1. limit doesn’t exist 2. limit = 2 3. limit = 1 2 4. limit = 4 5. limit = 1 4 correct
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Version 071 – Exam 1 – gualdani – (56455) 2 Explanation: After rationalizing the numerator we see that x + 3 2 = ( x + 3) 4 x + 3 + 2 = x 1 x + 3 + 2 . Thus x + 3 2 x 1 = 1 x + 3 + 2 for all x negationslash = 1. Consequently, limit = lim x 1 1 x + 3 + 2 = 1 4 . 004 10.0 points Determine lim h 0 f (1 + h ) f (1) h when f ( x ) = 2 x 2 + 5 x + 5 . 1. limit = 9 correct 2. limit = 11 3. limit = 8 4. limit does not exist 5. limit = 10 6. limit = 7 Explanation: Since f (1 + h ) f (1) = 2(1 + h ) 2 + 5(1 + h ) + 5 12 = 9 h + 2 h 2 = h (9 + 2 h ) , we see that lim h 0 f (1 + h ) f (1) h = lim h 0 h (9 + 2 h ) h . Consequently, limit = 9 . 005 10.0 points Determine if the limit lim x 0 sin 3 x 7 x exists, and if it does, find its value. 1. limit = 7 3 2. limit = 3 7 correct 3. limit doesn’t exist 4. limit = 7 5. limit = 3 Explanation: Using the known limit: lim x 0 sin ax x = a, we see that lim x 0 sin 3 x 7 x = 3 7 . 006 10.0 points After t seconds the displacement, s ( t ), of a particle moving rightwards along the x -axis is given (in feet) by s ( t ) = 4 t 2 2 t + 5 . Determine the average velocity of the particle over the time interval [1 , 2]. 1. average vel. = 13 ft/sec
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Version 071 – Exam 1 – gualdani – (56455) 3 2. average vel. = 10 ft/sec correct 3. average vel. = 9 ft/sec 4. average vel. = 11 ft/sec 5. average vel. = 12 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 .
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