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Unformatted text preview: b1 could be negative). Thus, we have a = 5 b1 is such that f ( a ) = b , so that the function is onto. 2) ind the least integer n such that f ( x ) is O ( x n ) for each of these functions. a) f ( x ) = 3 x 5 + (log x ) 4 Well, log x < x , so (log x ) 4 < x 4 , so certainly (log x ) 4 is O ( x 4 ). Obviously, 3 x 5 is O ( x 5 ), so when we add the functions to get f ( x ), we get f ( x ) is O (max( x 5 , x 4 )), which is O ( x 5 ). Thus, the answer is n = 5. b) f ( x ) = ( x 4 + x 2 + 1) / ( x 4 + 1) The top is O ( x 4 ), while the bottom is O ( 1 x 4 ). Thus, multiplying them, you get O ( x 4 x 4 ) = O (1). Thus, n = 0. 1...
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 Fall '08
 Miller
 Math

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