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Unformatted text preview: Fall 2007 ARE211 Midterm Exam  Answer key Problem 1 : Continuity . (14 points) In the first two parts of this problem, determine whether or not the specified functions are contin uous. If they are, prove it. If they are not, identify the subset X in the domain of the function such that for every x X , f is continuous at x . a) f ( x ) = braceleftbigg x if x is rational 0 if x is irrational Ans: A function f : X R k is continuous at a point x X if epsilon1 > , there exists > such that  x x  < implies  f ( x ) f ( x )  < epsilon1 . The function f is continuous if it is continuous at x , for every x X . At x = 0 : Fix epsilon1 > and set = epsilon1 . When  x  < , then  f ( x ) f (0)  = braceleftbigg  x  < epsilon1 if x rational   < epsilon1 if x irrational Therefore the function is continuous at x = 0 . At x negationslash = 0 , where x is rational: in this case, f ( x ) = x so that  f ( x ) f ( x )  = braceleftbigg  x x  if x rational x if x irrational Set epsilon1 = 0 . 5 x and choose > arbitrarily. Regardless of the size of , we can choose an irrational x such that  x x  < ; we then have  f ( x ) f ( x )  = x > epsilon1 . Hence f is not continuous at x . At x negationslash = 0 , where x is irrational: In this case, f ( x ) = 0 and  f ( x ) f ( x )  = braceleftbigg  x  if x rational if x irrational Once again, set epsilon1 = 0 . 5 x , and pick a rational x (0 . 5 x ,x ) ,  f ( x ) f ( x )  = x > . 5 x = epsilon1 . Hence f is not continuous at x . Conclude that f is continuous only at x = 0 b) f ( x ) = 1 if x = 0 1 /q if x = p/q is rational if x is irrational Ans: At x = 0 : In this case f ( x ) = 1 . Pick epsilon1 = 0 . 5 and choose > arbitrarily. There exists an irrational x such that  x  < , and for this x  f ( x ) f ( x )  = 1 > epsilon1 . Hence f is not continuous at x . 2 At x = p/q rational: In this case, f ( x ) = 1 /q . Pick epsilon1 = 1 / 2 q , choose > arbitrarily and pick an irrational xinB ( x , ) Since  f ( x ) f ( x )  = 1 /q > epsilon1 , f is not continuous at x . At x irrational: In this case, f ( x ) = 0 . Fix epsilon1 > and define q ( epsilon1 ) = min { q N : 1 /q < epsilon1 } . Next, define p ( epsilon1 ) = min { p N : p/q ( epsilon1 ) > x } . Now define R = { x = p/q : p p ( epsilon1 ); q q ( epsilon1 ) } . Since R is a finite set which does not contain x (since, by assumption, x is irrational), we can pick > sufficiently small such that B ( x , ) R is empty. Finally, note that if x = p/q ( B ( x , ) R C ) , then q > q ( epsilon1 ) . (This it true because if q q ( epsilon1 ) and p/q / R then necessarily p > p ( epsilon1 ) , and hence p/q p/q ( epsilon1 ) > p ( epsilon1 ) /q ( epsilon1 ) > x + )....
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at University of California, Berkeley.
 Fall '07
 Simon

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