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Unformatted text preview: Version 085 – EXAM 3 – Fouli – (58320) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the most general antiderivative, F , of the function f ( x ) = 12 x 2 − 16 x + 6 . 1. F ( x ) = 4 x 3 − 8 x 2 + 6 x 2. F ( x ) = 4 x 3 − 8 x 2 + 6 x + C correct 3. F ( x ) = 4 x 3 + 8 x 2 + 6 x + C 4. F ( x ) = 4 x 3 + 8 x 2 + 6 x 5. F ( x ) = 12 x 3 − 16 x 2 + 6 x + C Explanation: Since d dx x r = rx r − 1 , the most general antiderivative of f is the function F ( x ) = 12 parenleftbigg x 3 3 parenrightbigg − 16 parenleftbigg x 2 2 parenrightbigg + 6 x + C with C an arbitrary constant. Consequently, F ( x ) = 4 x 3 − 8 x 2 + 6 x + C . 002 10.0 points If the graph of f is which one of the following contains only graphs of antiderivatives of f ? 1. cor rect 2. 3. 4. Version 085 – EXAM 3 – Fouli – (58320) 2 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. In general, no horizontal translation of the graph of an antiderivative can be the graph of an antiderivative, nor can a hori zontal 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 /* If you use any of these, fix the comment symbols. 003 10.0 points Find the value of f (0) when f ′ ( t ) = cos 2 t , f parenleftBig π 4 parenrightBig = 3 . 1. f (0) = 5 2 correct 2. f (0) = 3 3. f (0) = 4 4. f (0) = 7 2 5. f (0) = 2 Explanation: Since d dx sin mt = m cos mt , for all m negationslash = 0, we see that f ( t ) = 1 2 sin 2 t + C Version 085 – EXAM 3 – Fouli – (58320) 3 where the arbitrary constant C is determined by the condition f ( π/ 4) = 3. But sin2 t vextendsingle vextendsingle vextendsingle t = π/ 4 = sin π 2 = 1 ....
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This note was uploaded on 09/09/2009 for the course M 408L taught by Professor Gilbert during the Spring '09 term at University of TexasTyler.
 Spring '09
 GILBERT

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