Solutions Exam 1

Solutions Exam 1 - Version 069 – Exam 1 – gualdani...

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Unformatted text preview: Version 069 – Exam 1 – gualdani – (57180) 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 U T S R Q P is the slope greatest ( i.e. , most positive)? 1. S correct 2. U 3. P 4. T 5. Q 6. R Explanation: By inspection the point is S . 002 10.0 points If a ball is thrown into the air with a velocity of 50 ft/sec, its height in feet after t seconds is given by y = 50 t − 16 t 2 . Find the average velocity of the ball for the time period beginning at t = 1 and lasting 1 / 4 seconds. 1. average vel. = 11 ft/sec 2. average vel. = 13 ft/sec 3. average vel. = 10 ft/sec 4. average vel. = 12 ft/sec 5. average vel. = 14 ft/sec correct Explanation: If the height of the ball after t seconds is given by y = y ( t ) = 50 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 ) = 50(1 + h ) − 16(1 + h ) 2 = 50(1 + h ) − 16(1 + 2 h + h 2 ) = 34 + 18 h − 16 h 2 , while y (1) = 50 − 16 = 34 . Thus y (1 + h ) − y (1) h = 18 h − 16 h 2 h = 18 − 16 h. Consequently, when h = 1 / 4, therefore, average velocity = 14 ft/sec . 003 10.0 points Below is the graph of a function f . Version 069 – Exam 1 – gualdani – (57180) 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 = 6 2. limit does not exist correct 3. limit = 9 4. limit = 18 5. limit = 4 Explanation: From the graph it is clear the f has a left hand limit at x = 4 which is equal to 4; and a right hand limit which is equal to 0. 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. neither L 1 nor L 2 exists 2. L 1 doesn’t exist, but L 2 does correct 3. L 1 exists, but L 2 doesn’t 4. both L 1 and L 2 exist 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 → 2 braceleftBig 1 x − 2 − 2 x 2 − 2 x bracerightBig ....
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This note was uploaded on 05/12/2010 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.

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Solutions Exam 1 - Version 069 – Exam 1 – gualdani...

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