# EXAM1_pdf - Version 035 EXAM 1 Fouli (58395) 1 This...

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Unformatted text preview: Version 035 EXAM 1 Fouli (58395) 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 . 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. x = 2 2. x = 2 , 2 correct 3. x = 2 4. no values of x 5. none of the other answers Explanation: Since f ( x ) is defined everywhere on ( 7 , 7), the function f will be discontinuous at a point x in ( 7 , 7) if and only if lim x x f ( x ) negationslash = f ( x ) or if lim x x f ( x ) negationslash = lim x x + f ( x ) . As the graph shows, the only possible candi- dates for x are x = 2 and x = 2. But at x = 2, f ( 2) = 4 negationslash = lim x 2 f ( x ) = 0 , while at x = 2, lim x 2 f ( x ) = 0 negationslash = lim x 2+ f ( x ) = 4 . Consequently, on ( 7 , 7) the function f is discontinuous only at x = 2 , 2 . 002 10.0 points A function f is defined by f ( x ) = 2 x, x 2, x 2 , 2 < x < 2, 2 + x, x 2. Determine where f is continuous, expressing your answer in interval notation. 1. ( , ) correct 2. ( , 2) (2 , ) 3. ( , 2) ( 2 , 2) (2 , ) 4. ( , 2) (2 , ) 5. ( , 2) ( 2 , ) 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 = 2. 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 = 2 lim x 2 f ( x ) = lim x 2 x 2 = (2) 2 = 4 , lim x 2+ f ( x ) = lim x 2+ 2 + x = 4 . Version 035 EXAM 1 Fouli (58395) 2 Thus, the function is also continuous at the point x = 2. 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. 1 f ( x ) 7 for all x in ( 3 , 1) 2. f ( c ) = 0 for some c in ( 3 , 0) 3. f ( c ) = 1 for some c in ( 3 , 1) correct 4. f (0) = 0 5. f ( c ) = 1 for some c 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 ....
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## This note was uploaded on 04/13/2010 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Fall '08 term at University of Texas at Austin.

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EXAM1_pdf - Version 035 EXAM 1 Fouli (58395) 1 This...

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