Exam 1 Solution - Version 024 EXAM 1 Radin (57410) 1 This...

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Unformatted text preview: Version 024 EXAM 1 Radin (57410) 1 This print-out should have 17 questions. Multiple-choice 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 anti-derivatives of f ? 1. 2. correct 3. 4. 5. 6. Explanation: If F 1 and F 2 are anti-derivatives of f then F 1 ( x )- F 2 ( x ) = constant independently of x ; this means that for any two anti-derivatives 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 anti-derivative of f will be the graph of an anti-derivative of f , nor can Version 024 EXAM 1 Radin (57410) 2 a horizontal and vertical translation be the graph of an anti-derivative. 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 anti-derivatives of f , therefore, we have to look more carefully at the actual graphs. But calculus ensures that (i) an anti-derivative of f will have a local extremum at the x-intercepts of f . This eliminates two more figures since they contains graphs whose local extrema occur at points other than the x-intercepts of f . (ii) An anti-derivative of f is increasing on interval where the graph of f lies above the x-axis, and decreasing where the graph of f lies below the x-axis. Consequently, of the two remaining figures only consists entirely of graphs of anti-derivatives 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) = 2 . 1. f (1) = 3 2. f (1) = 2 3. f (1) = 1 4. f (1) = 4 5. f (1) = 5 correct 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) = 2. For f (- 1) = 2 = f ( x ) = 3 x 2- 1 . But then f ( x ) = x 3- x + D, the arbitrary constant D being given by f (- 1) = 5 = 0 + D = 5 . Thus, f ( x ) = x 3- x + 5 . Consequently, f (1) = 5 . 003 10.0 points Estimate the area, A , under the graph of f ( x ) = 5 x on [1 , 5] by dividing [1 , 5] into four equal subintervals and using right endpoints. 1. A 25 4 2. A 37 6 3. A 77 12 correct 4. A 13 2 Version 024 EXAM 1 Radin (57410) 3 5. A 19 3 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 5 2 + 5 3 + 5 4 + 1 = 77 12 ....
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Exam 1 Solution - Version 024 EXAM 1 Radin (57410) 1 This...

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