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# Exam1_3 - Version 140 EXAM 1 Radin(58305 This print-out...

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Version 140 – EXAM 1 – Radin – (58305) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Below is the graph of a function f . -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 2 4 6 2 4 6 2 4 2 4 Use this graph to determine all of the values of x on ( 7 , 7) at which f is discontinuous. 1. none of the other answers 2. x = 3 3. no values of x 4. x = 3 , 3 correct 5. x = 3 Explanation: Since f ( x ) is defined everywhere on ( 7 , 7), the function f will be discontinuous at a point x 0 in ( 7 , 7) if and only if lim x x 0 f ( x ) negationslash = f ( x 0 ) or if lim x x 0 f ( x ) negationslash = lim x x 0 + f ( x ) . As the graph shows, the only possible candi- dates for x 0 are x 0 = 3 and x 0 = 3. But at x 0 = 3, f ( 3) = 0 negationslash = lim x → − 3 f ( x ) = 4 , while at x 0 = 3, lim x 3 f ( x ) = 4 negationslash = lim x 3+ f ( x ) = 0 . Consequently, on ( 7 , 7) the function f is discontinuous only at x = 3 , 3 . 002 10.0 points A function f is defined by f ( x ) = 2 x, x ≤ − 2, x 2 , 2 < x < 4, 12 + x, x 4. Determine where f is continuous, expressing your answer in interval notation. 1. ( −∞ , 2) (4 , ) 2. ( −∞ , 2) ( 2 , 4) (4 , ) 3. ( −∞ , ) correct 4. ( −∞ , 2) ( 2 , ) 5. ( −∞ , 4) (4 , ) Explanation: The function is piecewise continuous, so we have to check the left and right hand limits at the points where the definition of f changes, i.e. , at x = 2 and x = 4. Now at x = 2 lim x →− 2 f ( x ) = lim x →− 2 2 x = 4 , lim x →− 2+ f ( x ) = lim x →− 2+ x 2 = 4 , hence, the function is continuous at the point x = 2. 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 140 – EXAM 1 – Radin – (58305) 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 ) = 1 for some c in ( 3 , 1) 2. f (0) = 0 3. f ( c ) = 1 for some c in ( 3 , 1) correct 4. f ( c ) = 0 for some c in ( 3 , 0) 5. 1 f ( x ) 7 for all x in ( 3 , 1) 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, f ( c ) = 1 for some c in ( 3 , 1) .
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Exam1_3 - Version 140 EXAM 1 Radin(58305 This print-out...

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