Proving Images - R Part 2 SCRAP WORK y = f x = 2 x | x |-1...

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Determining the Image of a function: Given f : A B we want to determine the image of f . We claim that it is the set S . Part 1: Image ( f ) S . Here you are trying to show f ( x ) S for all x A . This is usually the easier proof. For example, if S = B , then this is trivial, the image is always contained in the target. Or if S is an interval, say ( a,b ), then this is the same as showing a < f ( x ) < b . Part 2: S Image ( f ). This is the part I’ve been getting a lot of questions about. Here you want to let y S and show y is in the image. To show y is in the image, you need to find an x A with f ( x ) = y . Here is a model for how your answer can be formatted: Given y S = , choose x = A . Then we have f ( x ) =. ....(do math here) . ... = y . This is NOT the only way to format your answer, but it reduces the need for double arrows, etc. Example: Let f : R R , f ( x ) = 2 x + | x | - 1. Claim: the image is
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Unformatted text preview: R . Part 2: SCRAP WORK: y = f ( x ) = 2 x + | x | -1 = ± 3 x-1 : x ≥ x-1 : 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 3-1 = 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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Proving Images - R Part 2 SCRAP WORK y = f x = 2 x | x |-1...

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