# CalcHW5Answers - Villarreal Natalie Homework 5 Due 3:00 am...

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Villarreal, Natalie – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Louiza Fouli 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 + x - 3 . 1. slope = 2 h - 3 correct 2. slope = 2 h + 3 3. slope = 4 h + 5 4. slope = 4 h + 3 5. slope = 4 h - 5 6. slope = 2 h - 5 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 + ( - 1 + h ) - 3 } + 2 h = 2 h 2 - 3 h h = 2 h - 3 . 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 ) = x 2 + 6 x + 1 at which the tangent line is parallel to the line y = 5 x + 3 , determine a . 1. a = 1 2 2. a = 0 3. a = - 1 4. a = 1 5. a = - 1 2 correct 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 ) = ( a + h ) 2 + 6( a + h ) + 1 = a 2 + h (2 a + 6) + h 2 + 6 a + 1 , while f ( a ) = a 2 + 6 a + 1 . Thus f ( a + h ) - f ( a ) = h { (2 a + 6) + h } , in which case f 0 ( a ) = lim h 0 { (2 a + 6) + h } = 2 a + 6 . If the tangent line at P is parallel to the line y = 5 x + 3 ,

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Villarreal, Natalie – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Louiza Fouli 2 then they have the same slopes, so f 0 ( a ) = 2 a + 6 = 5 . Consequently, a = - 1 2 . 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 ) = 3 x 2 - 4 x + 3 . 1. x -intercept = 16 9 2. x -intercept = 9 16 3. x -intercept = - 9 4. x -intercept = - 16 9 5. x -intercept = - 9 16 correct 6. x -intercept = 9 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 ) = 6 x - 4 at x = - 2, i.e. , m = - 16. On the other hand, f ( - 2) = 23. Thus by the point-slope formula an equation for the tangent line at P ( - 2 , f ( - 2)) is y - 23 = - 16( x + 2) , i . e ., y = - 16 x - 9 . Consequently, x -intercept = - 9 16 . 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 ) = 3 - x 3 . 1. y + 3 x = 5 correct 2. y + 3 x + 5 = 0 3. y = 3 x + 5 4. y = 5 x - 3 5. y + 5 x + 3 = 0 Explanation: If x = 1, then g (1) = 2. Thus the Newto- nian different quotient for g ( x ) = 3 - x 3 at the point (1 , 2) becomes g (1 + h ) - g (1) h = h 3 - (1 + h ) 3 i - 2 h = 3 - h 3 - 3 h 2 - 3 h - 3 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 - 2 = - 3( x - 1)
Villarreal, Natalie – Homework 5 – Due: Sep 25 2007, 3:00 am – Inst: Louiza Fouli 3 which after simplification becomes y + 3 x = 5 . 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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