# exam 3 - Version 084 Exam 3 Radin (56520) This print-out...

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Version 084 – Exam 3 – Radin – (56520) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – fnd 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. 3. correct 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

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Version 084 – Exam 3 – Radin – (56520) 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 Fg- 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 Fgures 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 Fgures 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 Fgures only consists entirely of graphs of anti-derivatives of f . keywords: antiderivative, graphical, graph, geometric interpretation 002 10.0 points ±ind the value of f (0) when f ( t ) = cos 2 t, f p π 4 P = 1 . 1. f (0) = 3 2 correct 2. f (0) = 2 3. f (0) = 1 4. f (0) = 1 2 5. f (0) = 0 Explanation: Since d dx sin mt = m cos mt, for all m n = 0, we see that f ( t ) = 1 2 sin 2 t + C where the arbitrary constant C is determined by the condition f ( π/ 4) = 1. But sin 2 t v v v t = π/ 4 = sin π 2 = 1 . Thus f p π 4 P = 1 2 + C = 1 , and so f ( t ) = 1 2 sin 2 t 3 2 . Consequently, f (0) = 3 2 . 003 10.0 points ±ind f (1) when f ′′ ( x ) = 3 x and f ( 1) = 4 , f ( 1) = 3 . 1. f (1) = 8 correct 2. f (1) = 10 3. f (1) = 6 4. f (1) = 9
Version 084 – Exam 3 – Radin – (56520) 3 5. f (1) = 7 Explanation: When f ′′ ( x ) = 3 x the most general anti- derivative of f ′′ is f ( x ) = 3 2 x 2 + C where the arbitrary constant C is determined by the condition f ( 1) = 3. For f ( 1) = 3 = f ( x ) = 3 2 x 2 + 3 2 . But then f ( x ) = 1 2 x 3 + 3 2 x + D, the arbitrary constant D being given by f ( 1) = 4 = ⇒ − 2 + D = 4 .

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## This note was uploaded on 06/05/2010 for the course MATH 57420 taught by Professor Radin during the Fall '09 term at University of Texas at Austin.

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exam 3 - Version 084 Exam 3 Radin (56520) This print-out...

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