quiz4bANS - b-1 could be negative). Thus, we have a = 5 b-1...

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Math 240, Quiz 4 Name: Circle One: T 12:05 T 2:25 R 12:05 R 2:25 Instructions: Answer all questions fully, showing work where necessary. 1)Determine whether the following function from R to R are one-to-one and/or onto. f ( x ) = x 5 + 1 To see if it is one-to-one, we need to see that f ( a ) = f ( b ) implies a = b . Suppose f ( a ) = f ( b ). Then a 5 + 1 = b 5 + 1, or a 5 = b 5 . Since 5 is odd, we can take the Ffth root of both sides (note: we couldn’t do this if the power were even). Thus a = b , and the function is indeed one-to-one. To see if it is onto, we need to see if any b R , there exists an a R such that f ( a ) = b . So we want a 5 + 1 = b , or a 5 = b - 1. Since 5 is odd, we can take the Ffth root without any problems (again, if the pwoer was even, then we would be in trouble, because
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Unformatted text preview: b-1 could be negative). Thus, we have a = 5 b-1 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 func-tions. 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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