HW6-solutions - wei (jw35975) HW6 milburn (54685) This...

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wei (jw35975) – HW6 – milburn – (54685) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points ±ind the y -intercept oF the tangent line at the point P ( 2 , f ( 2)) on the graph oF f ( x ) = 4 x 2 3 x + 4 . 1. y -intercept = 20 2. y -intercept = 20 3. y -intercept = 19 4. y -intercept = 12 correct 5. y -intercept = 19 6. y -intercept = 12 Explanation: To use the point-slope Formula to determine an equation For the tangent line we need to know frst the point P ( 2 , f ( 2)) and the slope m = lim h 0 f ( 2 + h ) f ( 2) h oF the tangent line at P . Now f ( 2) = 26, while f ( 2 + h ) = 4 h 2 19 h + 26 . Thus m = lim h 0 4 h 2 19 h h = 19 . By the point-slope Formula, thereFore, the tan- gent line at P has equation y 26 = 19( x + 2) , which aFter simplifcation becomes y = 19 x 12 . Consequently, its y -intercept = 12 . 002 10.0 points ±ind the slope oF the secant line passing through ( 3 , f ( 3)) , (1 , f (1)) when f ( x ) = x 2 + 2 x + 2 . 1. slope = 4 correct 2. slope = 15 4 3. slope = 17 4 4. slope = 9 2 5. slope = 7 2 Explanation: Since the points ( 3 , f ( 3)) , (1 , f (1)) lie on the secant line, the slope oF that line is given by rise run = f (1) f ( 3) 1 ( 3) . Thus slope = 4 . 003 10.0 points A Calculus student leaves the RLM build- ing and walks in a straight line to the PCL Library. His distance From RLM aFter t min- utes is given by the graph
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wei (jw35975) – HW6 – milburn – (54685) 2 2 4 6 8 10 12 min yards 80 160 240 320 400 What is his speed after 3 minutes, and in what direction is he heading at that time? 1. towards RLM at 20 yds/min 2. away from RLM at 30 yds/min 3. towards RLM at 40 yds/min 4. away from RLM at 20 yds/min 5. towards RLM at 30 yds/min 6. away from RLM at 40 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 = 240 160 4 2 = 40 yds/min . As the distance from RLM is increasing on [2 , 4] the student is thus moving away from the RLM. 004 10.0 points If f is a diFerentiable function, then f ( a ) is given by which of the following? I. lim h 0 f ( a + h ) f ( a ) h II. lim x a f ( x ) f ( a ) x a III. lim x a f ( x + h ) f ( x ) h 1. I and II only correct 2. I and III only 3. I only 4. I, II, and III 5. II only Explanation: Both of f ( a ) = lim h 0 f ( a + h ) f ( a ) h and f ( a ) = lim x a f ( x ) f ( a ) x a are valid de±nitions of f
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This note was uploaded on 10/25/2011 for the course MATHEMATIC 408K taught by Professor Milburn during the Fall '11 term at University of Texas at Austin.

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HW6-solutions - wei (jw35975) HW6 milburn (54685) This...

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