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Unformatted text preview: Version 095 Exam 3 Zheng (57255) 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 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 2. x y correct 3. x y 4. x y 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 < 0. Consequently, only x y consists entirely of graphs of antiderivatives of f . keywords: antiderivative, linear function, graph, 002 10.0 points Find the value of f (0) when f ( t ) = 2(9 t 4) and f (1) = 3 , f (1) = 2 . 1. f (0) = 1 Version 095 Exam 3 Zheng (57255) 2 2. f (0) = 3 3. f (0) = 0 4. f (0) = 1 correct 5. f (0) = 2 Explanation: The most general antiderivative of f has the form f ( t ) = 9 t 2 8 t + C where C is an arbitrary constant. But if f (1) = 3, then f (1) = 9 8 + C = 3 , i.e., C = 2 . From this it follows that f ( t ) = 9 t 2 8 t + 2 , and the most general antiderivative of the latter is f ( t ) = 3 t 3 4 t 2 + 2 t + D , where D is an arbitrary constant. But if f (1) = 2, then f (1) = 3 4 + 2 + D = 2 , i.e., D = 1 . Consequently, f ( t ) = 3 t 3 4 t 2 + 2 t + 1 . At x = 0, therefore, f (0) = 1 . 003 10.0 points Find the inverse of f ( x ) = 1 + 6 x 9 5 x . 1. f 1 ( x ) = 9 x 1 6 x + 5 2. f 1 ( x ) = 9 x + 1 5 x + 6 3. f 1 ( x ) = 9 x + 1 6 x + 5 4. f 1 ( x ) = 6 x 1 5 x + 9 5. f 1 ( x ) = 9 x 1 5 x + 6 correct Explanation: To determine the inverse of f we first solve for x in y = 1 + 6 x 9 5 x . In this case x = 9 y 1 5 y + 6 . The inverse f 1 is now obtained by inter changing x, y . Thus y = f 1 ( x ) = 9 x 1 5 x + 6 . 004 10.0 points Find the inverse, f 1 , of f ( x ) = 2 e x . 1. f 1 ( x ) = ln(2 x ) , x < 2 2. f 1 ( x ) = ln parenleftBig 1 2 x parenrightBig 3. f 1 ( x ) = ln(2 x ) 4. f 1 ( x ) = ln parenleftBig 1 x 2 parenrightBig , x > 2 5. f 1 ( x ) = ln parenleftBig 1 2 x parenrightBig , x < 2 correct 6. f 1 ( x ) = ln( x 2) 7. f 1 ( x ) = ln parenleftBig 1 x 2 parenrightBig 8. f 1 ( x ) = ln( x 2) , x > 2 Version 095 Exam 3 Zheng (57255) 3 Explanation: As e x is defined and decreasing for all x , the inverse f 1 exists because f is increasing; in addition, since e x > 0, we see that domain( f ) = ( , ) , range( f ) = ( , 2) , so by properties of inverse functions, domain( f 1 ) = ( , 2) , range( f 1 ) = ( , ) ....
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This note was uploaded on 08/24/2011 for the course M 408k taught by Professor Schultz during the Fall '08 term at University of Texas at Austin.
 Fall '08
 schultz

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