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# HW6 - Jolley Garrett Homework 6 Due Oct 2 2007 3:00 am Inst...

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Jolley, Garrett – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: R Heitmann 1 This print-out should have 20 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 Determine the derivative of f when f ( x ) = 4 5 2 / 3 . 1. f 0 ( x ) = 4 5 x - 1 / 3 2. f 0 ( x ) = 2 3 4 5 - 1 / 3 3. f 0 ( x ) = 0 correct 4. f 0 ( x ) = 8 15 x - 1 / 3 5. f 0 ( x ) does not exist Explanation: The derivative of any constant function is zero. Consequently, f 0 ( x ) = 0 . keywords: derivative, constant function 002 (part 1 of 1) 10 points Find the x -coordinates of all the points on the graph of f at which the tangent line is horizontal when f ( x ) = x 3 + x 2 - x + 2 . 1. x -coord = - 1 2. x -coord = 1 3. x -coords = - 1 3 , 1 4. x -coord = 1 3 5. x -coord = - 1 3 6. x -coords = 1 3 , - 1 correct Explanation: The tangent line will be horizontal at P ( x 0 , f ( x 0 )) when f 0 ( x 0 ) = 0 . Now f 0 ( x ) = 3 x 2 + 2 x - 1 = (3 x - 1)( x + 1) . Consequently, x 0 = 1 3 , - 1 . keywords: horizontal tangent line, extrema, polynomials, derivative 003 (part 1 of 1) 10 points Differentiate the function f ( x ) = 5 x 8 1. df dx = 8 5 x 9 2. df dx = 5 8 x 7 3. df dx = - 9 5 x 9 4. df dx = - 8 5 x 9 correct 5. df dx = - 8 5 x 7 Explanation: f ( x ) = 5 x 8 = 5 x - 8 f 0 ( x ) = - 8 5 x - 9 = - 8 5 x 9

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Jolley, Garrett – Homework 6 – Due: Oct 2 2007, 3:00 am – Inst: R Heitmann 2 keywords: derivative, rational function 004 (part 1 of 1) 10 points Find the derivative of f ( x ) = ( x 5 + 1)(1 - 3 x 3 ) . 1. f 0 ( x ) = 5 x 5 + 9 x 3 - 24 x 7 2. f 0 ( x ) = 5 x 4 - 9 x 2 - 24 x 7 correct 3. f 0 ( x ) = 5 x 5 - 9 x 3 - 24 x 7 4. f 0 ( x ) = 5 x 4 - 9 x 2 - 24 x 6 5. f 0 ( x ) = 5 x 4 + 9 x 2 - 24 x 6 Explanation: By the Product rule f 0 ( x ) = 5 x 4 (1 - 3 x 3 ) - 9 x 2 ( x 5 + 1) . Thus f 0 ( x ) = 5 x 4 - 9 x 2 - 24 x 7 . keywords: derivatives, product rule 005 (part 1 of 1) 10 points Find the derivative of f when f ( x ) = x (2 x + 5) . 1. f 0 ( x ) = 4 x - 5 x x 2. f 0 ( x ) = 4 x + 5 2 x 3. f 0 ( x ) = 6 x - 5 x x 4. f 0 ( x ) = 6 x + 5 2 x correct 5. f 0 ( x ) = 6 x - 5 2 x 6. f 0 ( x ) = 4 x + 5 x x Explanation: By the Product Rule f 0 ( x ) = 2 x + 5 2 x + 2 x . After simplification this becomes f 0 ( x ) = 2 x + 5 + 4 x 2 x = 6 x + 5 2 x . keywords: derivatives, product rule 006 (part 1 of 1) 10 points Find f 0 ( x ) when f ( x ) = 3 x - 1 4 x - 1 . 1. f 0 ( x ) = - 1 (4 x - 1) 2 2. f 0 ( x ) = 1 4 x - 1 3. f 0 ( x ) = 1 (4 x - 1) 2 correct 4. f 0 ( x ) = 12 x - 3 (4 x - 1) 2 5. f 0 ( x ) = 4 - 3 x (4 x - 1) 2 Explanation: Using the Quotient Rule for differentiation we see that f 0 ( x ) = 3(4 x - 1) - 4(3 x - 1) (4 x - 1) 2 .
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