# xm 1 - Version 140 EXAM 1 Fouli (58320) This print-out...

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Version 140 – EXAM 1 – Fouli – (58320) 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. no values o± x 2. none o± the other answers 3. x = 5 4. x = 5 5. x = 5 , 5 correct 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 = 5 and x 0 = 5. But at x 0 = 5, f ( 5) = 0 n = lim x →− 5 f ( x ) = 4 , while at x 0 = 5, lim x 5 f ( x ) = 4 n = lim x 5+ f ( x ) = 0 . Consequently, on ( 7 , 7) the ±unction f is discontinuous only at x = 5 , 5 . 002 10.0 points A ±unction f is defned by f ( x ) = 6 x, x ≤ − 3, x 2 , 3 < x < 3, 6 + x, x 3. Determine where f is continuous, expressing your answer in interval notation. 1. ( −∞ , 3) ( 3 , 3) (3 , ) 2. ( −∞ , 3) ( 3 , ) 3. ( −∞ , 3) (3 , ) 4. ( −∞ , 3) (3 , ) 5. ( −∞ , ) correct 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 = 3. 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 = 3 lim x 3 f ( x ) = lim x 3 x 2 = (3) 2 = 9 , lim x 3+ f ( x ) = lim x 3+ 6 + x = 9 .

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Version 140 – EXAM 1 – Fouli – (58320) 2 Thus, the function is also continuous at the point x = 3. 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 ) = 1 for some c in ( 3 , 1) 2. 1 f ( x ) 7 for all x in ( 3 , 1) 3. f ( c ) = 0 for some c in ( 3 , 0) 4. f ( c ) = 1 for some c in ( 3 , 1) correct 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 09/09/2009 for the course M 408L taught by Professor Gilbert during the Spring '09 term at University of Texas-Tyler.

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xm 1 - Version 140 EXAM 1 Fouli (58320) This print-out...

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