Exam #1 - Version 127 EXAM 1 Fouli (58395) This print-out...

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Version 127 – EXAM 1 – Fouli – (58395) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. 001 10.0 points Below is the graph o± a ±unction f . 2 4 6 2 4 6 2 4 2 4 Use this graph to determine all o± the values x on ( 7 , 7) at which f is discontinuous. 1. none o± the other answers 2. x = 2 3. x = 2 , 2 correct 4. x = 2 5. no values o± x Explanation: Since f ( x ) is defned everywhere on ( 7 , 7), the ±unction f will be discontinuous at a point x 0 in ( 7 , 7) i± and only i± lim x x 0 f ( x ) n = f ( x 0 ) or i± lim x x 0 f ( x ) n = lim x x 0 + f ( x ) . As the graph shows, the only possible candi- dates ±or x 0 are x 0 = 2 and x 0 = 2. But at x 0 = 2, f ( 2) = 4 n = lim x →− 2 f ( x ) = 0 , while at x 0 = 2, lim x 2 f ( x ) = 0 n = lim x 2+ f ( x ) = 4 . Consequently, on ( 7 , 7) the ±unction f is discontinuous only at x = 2 , 2 . 002 10.0 points A ±unction f is defned by f ( x ) = 6 x, x ≤ − 3, x 2 , 3 < x < 4, 12 + x, x 4. Determine where f is continuous, expressing your answer in interval notation. 1. ( −∞ , ) correct 2. ( −∞ , 3) (4 , ) 3. ( −∞ , 4) (4 , ) 4. ( −∞ , 3) ( 3 , ) 5. ( −∞ , 3) ( 3 , 4) (4 , ) Explanation: The ±unction is piecewise continuous, so we have to check the le±t and right hand limits at the points where the defnition o± f changes, i.e. , at x = 3 and x = 4. Now at x = 3 lim x →− 3 f ( x ) = lim x →− 3 6 x = 9 , lim x →− 3+ f ( x ) = lim x →− 3+ x 2 = 9 , hence, the ±unction is continuous at the point x = 3. On the other hand, at x = 4 lim x 4 f ( x ) = lim x 4 x 2 = (4) 2 = 16 , lim x 4+ f ( x ) = lim x 4+ 12 + x = 16 .
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Version 127 – EXAM 1 – Fouli – (58395) 2 Thus, the function is also continuous at the point x = 4. 003 10.0 points Let f be a continuous function on [ 3 , 1] such that f ( 3) = 1 , f (1) = 7 . Which of the following is a consequence of the Intermediate Value Theorem without further restrictions on f ? 1. f ( c ) = 0 for some c in ( 3 , 0) 2. f ( c ) = 1 for some c in ( 3 , 1) 3. f ( c ) = 1 for some c in ( 3 , 1) correct 4. 1 f ( x ) 7 for all x in ( 3 , 1) 5. f (0) = 0 Explanation: Because f is continuous on [ 3 , 1] the In- termediate Value Theorem ensures that for each M, 1 < M < 7 , there exists at least one choice of c in ( 3 , 1) for which f ( c ) = M . In particular,
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This note was uploaded on 07/10/2011 for the course KIN 321M taught by Professor Jensen during the Spring '11 term at University of Texas at Austin.

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Exam #1 - Version 127 EXAM 1 Fouli (58395) This print-out...

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