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HW04-solutions

# HW04-solutions - le(chl528 HW04 seckin(57195 This print-out...

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le (chl528) – HW04 – seckin – (57195) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the x -intercept of the tangent line at the point P ( 4 , 3) on the graph of f when lim h 0 f ( 4 + h ) 3 h = 5 . 1. x -intercept = 23 5 2. x -intercept = 23 3. x -intercept = 23 5 correct 4. x -intercept = 23 5. x -intercept = 17 5 6. x -intercept = 17 Explanation: Since lim h 0 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 x -intercept = 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

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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 0 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 point-slope 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 -1 0 1 2 3 4 5 6 7 8 9 10 11 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 . As the distance from RLM is increasing on [2 , 4] the student is thus moving away from the RLM. 005 (part 2 of 3) 10.0 points ii) What is his speed after 9 minutes, and in what direction is he heading at that time? 1. towards RLM at 40 yds/min 2. away from RLM at 20 yds/min .
le (chl528) – HW04 – seckin – (57195) 3 3. away from RLM at 5 yds/min .

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