Exam 1 - Version 078 – Exam 1 – Zheng –(57255 1 This...

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Unformatted text preview: Version 078 – Exam 1 – Zheng – (57255) 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 At which point on the graph P Q R S T U is the slope greatest ( i.e. , most positive)? 1. U 2. P correct 3. Q 4. R 5. S 6. T Explanation: By inspection the point is P . 002 10.0 points If a ball is thrown into the air with a velocity of 52 ft/sec, its height in feet after t seconds is given by y = 52 t − 16 t 2 . Find the average velocity of the ball for the time period beginning at t = 1 and lasting 1 / 8 seconds. 1. average vel. = 14 ft/sec 2. average vel. = 16 ft/sec 3. average vel. = 15 ft/sec 4. average vel. = 18 ft/sec correct 5. average vel. = 17 ft/sec Explanation: If the height of the ball after t seconds is given by y = y ( t ) = 52 t − 16 t 2 , then the average velocity of the ball over a time interval [1 , t + h ] is avg. vel. = y (1 + h ) − y (1) h . Now y (1 + h ) = 52(1 + h ) − 16(1 + h ) 2 = 52(1 + h ) − 16(1 + 2 h + h 2 ) = 36 + 20 h − 16 h 2 , while y (1) = 52 − 16 = 36 . Thus y (1 + h ) − y (1) h = 20 h − 16 h 2 h = 20 − 16 h. Consequently, when h = 1 / 8, therefore, average velocity = 18 ft/sec . 003 10.0 points Below is the graph of a function f . Version 078 – Exam 1 – Zheng – (57255) 2 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 = 9 3. limit = 6 4. limit = 5 5. limit = 7 Explanation: From the graph it is clear the f has a left hand limit at x = 4 which is equal to 6; and a right hand limit which is equal to − 2. Since the two numbers do not coincide, the limit does not exist . 004 10.0 points If f oscillates faster and faster when x ap- proaches 0 as indicated by its graph determine which, if any, of L 1 : lim x → 0+ f ( x ) , L 2 : lim x → − f ( x ) exist. 1. both L 1 and L 2 exist 2. L 1 doesn’t exist, but L 2 does correct 3. neither L 1 nor L 2 exists 4. L 1 exists, but L 2 doesn’t Explanation: For x > 0 the graph of f oscillates but the oscillations do not get smaller and smaller as x approaches 0; so L 1 does not exist. But for x < 0, the graph oscillates and the oscillations get smaller and smaller as x approaches 0; in fact, the oscillation goes to 0 as x approaches 0, so L 2 exists. Consequently, L 1 does not exist, but L 2 does . 005 10.0 points Determine lim x → 1 braceleftBig 1 x − 1 − 1 x 2 − x bracerightBig ....
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Exam 1 - Version 078 – Exam 1 – Zheng –(57255 1 This...

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