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Unformatted text preview: Granillo, Yvette Exam 3 Due: Dec 1 2005, 11:00 pm Inst: Edward Odell 1 This printout should have 18 questions. Multiplechoice 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 ( x ) = 7 cos x 2 sin x and f (0) = 3. 1. f ( x ) = 7 cos x + 2 sin x + 10 2. f ( x ) = 7 sin x + 2 cos x 4 3. f ( x ) = 7 sin x + 2 cos x + 1 4. f ( x ) = 7 cos x + 2 sin x 4 5. f ( x ) = 7 cos x + 2 sin x + 10 6. f ( x ) = 7 sin x + 2 cos x + 1 correct Explanation: Since d dx sin x = cos x, d dx cos x = sin x, we see immediately that f ( x ) = 7 sin x + 2 cos x + C with C an arbitrary constant. The condition f (0) = 3 determines C : f (0) = 3 = 2 + C = 3 since sin 0 = 0 , cos 0 = 1 . Consequently, f ( x ) = 7 sin x + 2 cos x + 1 . 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 antiderivatives of f ? 1. x y correct 2. x y 3. x y 4. x y Granillo, Yvette Exam 3 Due: Dec 1 2005, 11:00 pm Inst: Edward Odell 2 5. x y Explanation: If f ( x ) = m is a constant function, then its most general antiderivative 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 antiderivatives of f ( x ) = m consists of all parallel lines, each having slope m . Now for the given f ( x ) = m , clearly m &lt; 0. Consequently, only x y consists entirely of graphs of antiderivatives of f . keywords: Stewart5e, 003 (part 1 of 1) 10 points Find the value of f ( 1) when f 00 ( t ) = 6( t + 1) and f (1) = 5 , f (1) = 3 . 1. f ( 1) = 10 2. f ( 1) = 11 3. f ( 1) = 12 4. f ( 1) = 9 correct 5. f ( 1) = 8 Explanation: The most general antiderivative of f 00 has the form f ( t ) = 3 t 2 + 6 t + C where C is an arbitrary constant. But if f (1) = 5, then f (1) = 3 + 6 + C = 5 , i.e., C = 4 . From this it follows that f ( t ) = 3 t 2 + 6 t 4 . The most general antiderivative of f is thus f ( t ) = t 3 + 3 t 2 4 t + D , where D is an arbitrary constant. But if f (1) = 3, then f (1) = 1 + 3 4 + D = 3 , i.e., D = 3 . Thus f ( t ) = t 3 + 3 t 2 4 t + 3 , and so f ( 1) = 9 . keywords: Stewart5e, 004 (part 1 of 1) 10 points A trailer rental agency rents 12 trailers per day at a rate of $40 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....
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This test prep was uploaded on 04/16/2008 for the course M 408k taught by Professor Schultz during the Spring '08 term at University of Texas.
 Spring '08
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