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Unformatted text preview: R . Part 2: SCRAP WORK: y = f ( x ) = 2 x +  x  1 = ± 3 x1 : x ≥ x1 : x < Solving for x (in terms of y ) we get, x = ± y +1 3 : x ≥ y + 1 : x < For x = y +1 3 ≥ 0, this solution will only work for y ≥ 1. Similarly for x = y + 1 < 0, this solution will only work for y <1. ACTUAL WRITTEN SOLUTION: Given y ∈ S = R , if y ≥ 1, choose x = y +1 3 ∈ A = R and if y <1, choose x = y + 1 ∈ A = R . (Notice we had to change the format just a bit to 1 account for cases). For y ≥ 1, we have x ≥ 0 and then we have f ( x ) = f ( y +1 3 ) = 2( y +1 3 ) + y +1 31 = y . For y <1, we have x < 0 and then we have f ( x ) = f ( y + 1) = 2( y + 1)( y + 1) + 1 = y . 2...
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 Spring '07
 GHEORGHICIUC
 Math, Want

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