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Unformatted text preview: Version 111 EXAM 1 Radin (57410) 1 This printout should have 17 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points If the graph of f is which one of the following contains only graphs of antiderivatives of f ? 1. 2. 3. 4. correct 5. 6. Explanation: If F 1 and F 2 are antiderivatives of f then F 1 ( x ) F 2 ( x ) = constant independently of x ; this means that for any two antiderivatives of f the graph of one is just a vertical translation of the graph of the other. But no horizontal translation of the graph of an antiderivative of f will be the graph of an antiderivative of f , nor can Version 111 EXAM 1 Radin (57410) 2 a horizontal and vertical translation be the graph of an antiderivative. This rules out two sets of graphs. Now in each of the the remaining four fig ures the dotted and dashed graphs consist of vertical translations of the graph whose line style is a continuous line. To decide which of these figures consists of antiderivatives of f , therefore, we have to look more carefully at the actual graphs. But calculus ensures that (i) an antiderivative of f will have a local extremum at the xintercepts of f . This eliminates two more figures since they contains graphs whose local extrema occur at points other than the xintercepts of f . (ii) An antiderivative of f is increasing on interval where the graph of f lies above the xaxis, and decreasing where the graph of f lies below the xaxis. Consequently, of the two remaining figures only consists entirely of graphs of antiderivatives of f . keywords: antiderivative, graphical, graph, geometric interpretation 002 10.0 points Find f (1) when f ( x ) = 6 x and f ( 1) = 5 , f ( 1) = 1 . 1. f (1) = 3 correct 2. f (1) = 4 3. f (1) = 2 4. f (1) = 1 5. f (1) = 5 Explanation: When f ( x ) = 6 x the most general anti derivative of f is f ( x ) = 3 x 2 + C where the arbitrary constant C is determined by the condition f ( 1) = 1. For f ( 1) = 1 = f ( x ) = 3 x 2 2 . But then f ( x ) = x 3 2 x + D, the arbitrary constant D being given by f ( 1) = 5 = 1 + D = 5 . Thus, f ( x ) = x 3 2 x + 4 . Consequently, f (1) = 3 . 003 10.0 points Estimate the area, A , under the graph of f ( x ) = 3 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. 1. A 73 20 2. A 19 5 3. A 15 4 4. A 77 20 correct Version 111 EXAM 1 Radin (57410) 3 5. A 37 10 Explanation: With four equal subintervals and right end points as sample points, A braceleftBig f (2) + f (3) + f (4) + f (5) bracerightBig 1 since x i = x i = i + 1. Consequently, A 3 2 + 1 + 3 4 + 3 5 = 77 20 ....
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This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin

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