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Unformatted text preview: mao (tm23477) – HW4 – gualdani – (57180) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the xintercept of the tangent line at the point P ( − 2 , 5) on the graph of f when lim h → f ( − 2 + h ) − 5 h = 2 . 1. xintercept = 9 2 2. xintercept = 1 2 3. xintercept = 9 4. xintercept = − 9 5. xintercept = − 9 2 correct 6. xintercept = 1 Explanation: Since lim h → f ( − 2 + h ) − f ( − 2) h is the slope of the tangent line at P ( − 2 , 5) , an equation for this tangent line is y − 5 = 2( x + 2) , i.e. , y = 2 x + 9 . Consequently, the xintercept = − 9 2 . 002 10.0 points Consider the slope of the given curve at the five points shown. B D E A C List the five slopes in decreasing order. 1. C, A, B, E, D 2. D, A, B, C, E 3. D, B, E, C, A 4. C, B, D, E, A 5. A, C, D, B, E 6. A, C, E, B, D correct Explanation: The order will be the one from most positive slope to most negative slope. Inspection of the graph shows that this is A, C, E, B, D . 003 10.0 points Find an equation for the tangent line at the point P (2 , f (2)) on the graph of f when f is defined by f ( x ) = 3 x 2 + x + 3 . 1. y + 13 x − 9 = 0 2. y − 13 x + 15 = 0 3. y − 13 x − 15 = 0 4. y + 13 x + 9 = 0 mao (tm23477) – HW4 – gualdani – (57180) 2 5. y − 13 x + 9 = 0 correct 6. y + 13 x + 15 = 0 Explanation: The slope of the tangent line at P (2 , f (2)) is the value of the limit lim h → f (2 + h ) − f (2) h . Now f (2 + h ) = 3(2 + h ) 2 + (2 + h ) + 3 = 3 h 2 + 13 h + 17 , while f (2) = 17. Thus f (2 + h ) − f (2) h = 3 h 2 + 13 h h = 3 h + 13 . As h approaches 0, therefore, f (2 + h ) − f (2) h → 13 . Consequently, by the pointslope formula, an equation for the tangent line at P (2 , f (2)) is y − 17 = 13( x − 2) , i.e. , y − 13 x + 9 = 0 . 004 (part 1 of 3) 10.0 points A Calculus student leaves the RLM build ing and walks in a straight line to the PCL Li brary. His distance (in multiples of 40 yards) from RLM after t minutes is given by the graph 2 4 6 8 10 2 4 6 8 10 t distance i) What is his speed after 3 minutes, and in what direction is he heading at that time? 1. away from RLM at 80 yds/min correct 2. towards RLM at 40 yds/min 3. away from RLM at 40 yds/min 4. away from RLM at 60 yds/min 5. towards RLM at 80 yds/min Explanation: The graph is linear and has positive slope on [2 , 4], so the speed of the student at time t = 3 coincides with the slope of the line on [2 , 4]. Hence speed = 40 · 9 − 5 4 − 2 = 80 yds/min . As the distance from RLM is increasing on [2 , 4] the student is thus moving away from the RLM....
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This note was uploaded on 05/12/2010 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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