3_Cal_Solution of Calculus_6e

3_Cal_Solution of Calculus_6e - f ( x ) = x | x | = x x = 1...

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f ( a + h ) f ( a )=[1 2( a + h )] [1 2 a ]=1 2 a 2 h 1+2 a = 2 h. C01S01.013: If f ( x )= x 2 , then f ( a + h ) f ( a )=( a + h ) 2 a 2 = a 2 +2 ah + h 2 a 2 =2 ah + h 2 = h · (2 a + h ) . C01S01.014: If f ( x )= x 2 +2 x , then f ( a + h ) f ( a )=[( a + h ) 2 +2( a + h )] [ a 2 +2 a ] = a 2 +2 ah + h 2 +2 a +2 h a 2 2 a =2 ah + h 2 +2 h = h · (2 a + h +2) . C01S01.015: If f ( x )= 1 x , then f ( a + h ) f ( a )= 1 a + h 1 a = a a ( a + h ) a + h a ( a + h ) = a ( a + h ) a ( a + h ) = h a ( a + h ) . C01S01.016: If f ( x )= 2 x +1 , then f ( a + h ) f ( a )= 2 a + h +1 2 a +1 = 2( a +1) ( a + h + 1)( a +1) 2( a + h +1) ( a + h + 1)( a +1) = 2 a +2 ( a + h + 1)( a +1) 2 a +2 h +2 ( a + h + 1)( a +1) = (2 a +2) (2 a +2 h +2) ( a + h + 1)( a +1) = 2 a +2 2 a 2 h 2 ( a + h + 1)( a +1) = 2 h ( a + h + 1)( a +1) . C01S01.017: If x> 0 then
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Unformatted text preview: f ( x ) = x | x | = x x = 1 . If x < 0 then f ( x ) = x | x | = x − x = − 1 . We are given f (0) = 0, so the range of f is {− 1 , , 1 } . That is, the range of f is the set consisting of the three real numbers − 1, 0, and 1. C01S01.018: Given f ( x ) = [[3 x ]], we see that 3...
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