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Unformatted text preview: le (chl528) HW04 seckin (57195) 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 ( 4 , 3) on the graph of f when lim h f ( 4 + h ) 3 h = 5 . 1. xintercept = 23 5 2. xintercept = 23 3. xintercept = 23 5 correct 4. xintercept = 23 5. xintercept = 17 5 6. xintercept = 17 Explanation: Since lim h f ( 4 + h ) f ( 4) h is the slope of the tangent line at P ( 4 , 3) , an equation for this tangent line is y 3 = 5( x + 4) , i.e. , y = 5 x + 23 . Consequently, the xintercept = 23 5 . 002 10.0 points Consider the slope of the given curve at the five points shown. C E A B D List the five slopes in decreasing order. 1. E, B, C, D, A 2. E, C, A, D, B 3. B, D, E, C, A 4. D, C, E, A, B 5. D, B, C, A, E 6. B, D, A, C, E correct Explanation: The order will be the one from most positive slope to most negative slope. Inspection of the graph shows that this is B, D, A, C, E . 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 + 11 x + 9 = 0 correct 2. y 11 x + 9 = 0 3. y 11 x 9 = 0 4. y + 11 x + 15 = 0 le (chl528) HW04 seckin (57195) 2 5. y 11 x + 15 = 0 6. y + 11 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 11 h + 13 , while f ( 2) = 13. Thus f ( 2 + h ) f ( 2) h = 3 h 2 11 h h = 3 h 11 . As h approaches 0, therefore, f ( 2 + h ) f ( 2) h 11 . Consequently, by the pointslope formula, an equation for the tangent line at P ( 2 , f ( 2)) is y 13 = 11( x + 2) , i.e. , y + 11 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 45 yds/min 2. towards RLM at 30 yds/min 3. away from RLM at 30 yds/min 4. towards RLM at 60 yds/min 5. away from RLM at 60 yds/min correct 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 8 5 4 2 = 60 yds/min ....
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 Spring '10
 RAdin

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