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Homework 5 - Garcia Ilse Homework 5 Due 3:00 am Inst Fonken...

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Garcia, Ilse – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Fonken 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Find the slope of the secant line passing through the points ( - 1 , f ( - 1)) , ( - 1 + h, f ( - 1 + h )) when f ( x ) = 2 x 2 - 3 x - 4 . 1. slope = 2 h + 7 2. slope = 2 h - 1 3. slope = 4 h + 7 4. slope = 4 h - 1 5. slope = 2 h - 7 correct 6. slope = 4 h + 1 Explanation: Since the secant line passes through the points ( - 1 , f ( - 1)) , ( - 1 + h, f ( - 1 + h )) , its slope is given by f ( - 1 + h ) - f ( - 1) h = { 2( - 1 + h ) 2 - 3( - 1 + h ) - 4 } - 1 h = 2 h 2 - 7 h h = 2 h - 7 . keywords: slope, secant line 002 (part 1 of 1) 10 points If P ( a, f ( a )) is the point on the graph of f ( x ) = 2 x 2 + 3 x + 6 at which the tangent line is parallel to the line y = 6 x + 4 , determine a . 1. a = 1 2. a = 1 4 3. a = 3 4 correct 4. a = 5 4 5. a = 1 2 Explanation: The slope of the tangent line at the point P ( a, f ( a )) on the graph of f is the value f 0 ( a ) = lim h 0 f ( a + h ) - f ( a ) h of the derivative of f at x = a . To compute the value of f 0 ( a ), note that f ( a + h ) = 2( a + h ) 2 + 3( a + h ) + 6 = 2 a 2 + h (4 a + 3) + 2 h 2 + 3 a + 6 , while f ( a ) = 2 a 2 + 3 a + 6 . Thus f ( a + h ) - f ( a ) = h { (4 a + 3) + 2 h } , in which case f 0 ( a ) = lim h 0 { (4 a + 3) + 2 h } = 4 a + 3 . If the tangent line at P is parallel to the line y = 6 x + 4 ,
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Garcia, Ilse – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Fonken 2 then they have the same slopes, so f 0 ( a ) = 4 a + 3 = 6 . Consequently, a = 3 4 . keywords: tangent line, parallel, slope, deriva- tive 003 (part 1 of 1) 10 points Find the x -intercept of the tangent line at the point P ( - 2 , f ( - 2)) on the graph of f when f is defined by f ( x ) = 2 x 2 - 3 x + 2 . 1. x -intercept = 6 11 2. x -intercept = 11 6 3. x -intercept = - 6 11 correct 4. x -intercept = - 11 6 5. x -intercept = 6 6. x -intercept = - 6 Explanation: The slope, m , of the tangent line at the point P ( - 2 , f ( - 2)) on the graph of f is the value of the derivative f 0 ( x ) = 4 x - 3 at x = - 2, i.e. , m = - 11. On the other hand, f ( - 2) = 16. Thus by the point-slope formula an equation for the tangent line at P ( - 2 , f ( - 2)) is y - 16 = - 11( x + 2) , i . e ., y = - 11 x - 6 . Consequently, x -intercept = - 6 11 . keywords: tangent line, x-intercept, slope 004 (part 1 of 1) 10 points Find an equation for the tangent line to the graph of g at the point P (1 , g (1)) when g ( x ) = 5 - x 3 . 1. y + 3 x + 7 = 0 2. y = 7 x - 3 3. y + 7 x + 3 = 0 4. y = 3 x + 7 5. y + 3 x = 7 correct Explanation: If x = 1, then g (1) = 4. Thus the Newto- nian different quotient for g ( x ) = 5 - x 3 at the point (1 , 4) becomes g (1 + h ) - g (1) h = h 5 - (1 + h ) 3 i - 4 h = 5 - h 3 - 3 h 2 - 3 h - 5 h . Thus g 0 (1) = lim h 0 ( - h 2 - 3 h - 3 ) = - 3 . Consequently, by the point-slope formula, an equation for the tangent line to the graph of g at P is y - 4 = - 3( x - 1)
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Garcia, Ilse – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Fonken 3 which after simplification becomes y + 3 x = 7 . keywords: tangent line, slope, equation 005 (part 1 of 3) 10 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?
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