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M408K Ex. 3 Solutions

# M408K Ex. 3 Solutions - Mahon Kevin Exam 3 Due Dec 1 2005...

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Mahon, Kevin – Exam 3 – Due: Dec 1 2005, 11:00 pm – Inst: Edward Odell 1 This print-out should have 18 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 f ( x ) when f 0 ( x ) = 8 cos x - 9 sin x and f (0) = 5. 1. f ( x ) = - 8 cos x + 9 sin x + 13 2. f ( x ) = 8 cos x + 9 sin x + 13 3. f ( x ) = 8 cos x + 9 sin x - 3 4. f ( x ) = 8 sin x + 9 cos x - 3 5. f ( x ) = 8 sin x + 9 cos x - 4 correct 6. f ( x ) = - 8 sin x + 9 cos x - 4 Explanation: Since d dx sin x = cos x, d dx cos x = - sin x , we see immediately that f ( x ) = 8 sin x + 9 cos x + C with C an arbitrary constant. The condition f (0) = 5 determines C : f (0) = 5 = 9 + C = 5 since sin 0 = 0 , cos 0 = 1 . Consequently, f ( x ) = 8 sin x + 9 cos x - 4 . keywords: Stewart5e, 002 (part 1 of 1) 10 points If the graph of f is the bold horizontal line shown in black in x y which one of the following contains only graphs of anti-derivatives of f ? 1. x y 2. x y 3. x y 4. x y

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Mahon, Kevin – Exam 3 – Due: Dec 1 2005, 11:00 pm – Inst: Edward Odell 2 5. x y correct Explanation: If f ( x ) = m is a constant function, then its most general anti-derivative F is of the form F ( x ) = mx + C with C an arbitrary constant. Thus the graph of F is a straight line having slope m , and so the family of anti-derivatives of f ( x ) = m consists of all parallel lines, each having slope m . Now for the given f ( x ) = m , clearly m > 0. Consequently, only x y consists entirely of graphs of anti-derivatives of f . keywords: Stewart5e, 003 (part 1 of 1) 10 points Find the value of f ( - 1) when f 00 ( t ) = 4(3 t - 2) and f 0 (1) = 4 , f (1) = 1 . 1. f ( - 1) = - 19 2. f ( - 1) = - 18 3. f ( - 1) = - 16 4. f ( - 1) = - 17 5. f ( - 1) = - 15 correct Explanation: The most general anti-derivative of f 00 has the form f 0 ( t ) = 6 t 2 - 8 t + C where C is an arbitrary constant. But if f 0 (1) = 4, then f 0 (1) = 6 - 8 + C = 4 , i.e., C = 6 . From this it follows that f 0 ( t ) = 6 t 2 - 8 t + 6 . The most general anti-derivative of f is thus f ( t ) = 2 t 3 - 4 t 2 + 6 t + D , where D is an arbitrary constant. But if f (1) = 1, then f (1) = 2 - 4 + 6 + D = 1 , i.e., D = - 3 . Thus f ( t ) = 2 t 3 - 4 t 2 + 6 t - 3 , and so f ( - 1) = - 15 . keywords: Stewart5e, 004 (part 1 of 1) 10 points A trailer rental agency rents 16 trailers per day at a rate of \$30 per day. It discovers that for each \$5 increase in rate, one fewer trailer is rented. Determine the the maximum income, I max , the rental agency can obtain. 1. I max = \$594 2. none of these 3. I max = \$616
Mahon, Kevin – Exam 3 – Due: Dec 1 2005, 11:00 pm – Inst: Edward Odell 3 4. I max = \$605 correct 5. I max = \$627 Explanation: Let r be the daily rate of renting a trailer.

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M408K Ex. 3 Solutions - Mahon Kevin Exam 3 Due Dec 1 2005...

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