4_Cal_Solution of Calculus_6e

4_Cal_Solution of Calculus_6e - f (x) = 0 if 0 x < 1, 3 f...

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f ( x )=0 if 0 5 x< 1 3 , f ( x )=1 if 1 3 5 2 3 , f (2) = 2 if 2 3 5 1; moreover, f ( x )= 3i f 1 5 2 3 , f ( x 2i f 2 3 5 1 3 , f ( x 1i f 1 3 5 0 . In general, if m is any integer, then f ( x )=3 m if m 5 x<m + 1 3 , f ( x m +1 if m + 1 3 5 + 2 3 , f ( x m +2 if m + 2 3 5 +1 . Because every integer is equal to 3 m or to 3 m + 1 or to 3 m + 2 for some integer m , we see that the range of f includes the set Z of all integers. Because f can assume no values other than integers, we can conclude that the range of f is exactly Z . C01S01.019: Given f ( x )=( 1) [ x ] , we Frst note that the values of the exponent [[ x ]] consist of all the integers and no other numbers. So all that matters about the exponent is whether it is an even integer or an odd integer, for if even then f ( x ) = 1 and if odd then f ( x 1. No other values of f ( x ) are possible, so the range of f is the set consisting of the two numbers 1 and 1. C01S01.020: If 0 <x 5 1, then f ( x ) = 34. If 1 5 2 then f ( x ) = 34 + 21 = 55. If 2 5 3 then f ( x )=34+2 · 21 = 76. We continue in this way and conclude with the observation that if 11
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This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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