Solving this gives us x = 4 5 but 4 5 3 6≥ 0 so

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Unformatted text preview: Solving this gives us x = 4 / 5. But 4 / 5- 3 6≥ 0, so this solution is extraneous. Case 2: If x- 3 ≤ 0, then | x- 3 | =- x +3, so the equation becomes- x +3 = 1- 4 x . Solving this gives us x =- 2 / 3. Since- 2 / 3- 3 ≤ 0, this solution is valid. So the only solution is x =- 2 / 3. 8. Suppose f has domain [- 1 , 1] and range [- 1 , 1]. Define a new function g by: g ( x ) = 2 f (3 x + 4) + 5 . (a) [5 points] What is the domain of g ? Solution: Note that g has the same domain as f (3 x + 4). To get from the graph of f to the graph of f (3 x +4), we first shift the graph left by 4 and then compress it by a factor of 3. So the domain of g is [(- 1- 4) / 3 , (1- 4) / 3] = [- 5 / 3 ,- 1]. (b) [5 points] What is the range of g ? Solution: Note that g has the same range as 2 f ( x )+5. To get from the graph of f to the graph of 2 f ( x ) + 5, we stretch the graph vertically by a factor of 2 and then shift it up by 5. So the range of g is [2(- 1) + 5 , 2 · 1 + 5] = [3 , 7]. 9. [10 points] Solve 3 x- 2 ≤ x . Write the solution in interval notation. ( Hint : You may want to break it into two cases.) Solution: We want to multiply both sides by x- 2, but the direction of the inequality changes if x- 2 is negative, so we have to consider two cases. Case 1: If x > 2, then x- 2 > 0. So the inequality becomes 3 ≤ x ( x- 2), or x 2- 2 x- 3 ≥ 0. Solving x 2- 2 x- 3 = 0 gives x = 3 ,- 1. So x 2- 2 x- 3 ≥ 0 when x is in (-∞ ,- 1] ∪ [3 , ∞ ). But we have assumed that x > 2, so only [3 , ∞ ) is valid in this case. Page 4 Math 171: Exam 1 Case 2: If x < 2, then x- 2 < 0. So the inequality becomes 3 ≥ x ( x- 2), or x 2- 2 x- 3 ≤ 0. Solving x 2- 2 x- 3 = 0 gives x = 3 ,- 1. So x 2- 2 x- 3 ≤ 0 when x is in [- 1 , 3]. But we have assumes that x < 2, so only [- 1 , 2) is valid in this case. Putting the two cases together and writing the answer in interval notation: [- 1 , 2) ∪ [3 , ∞ ). Page 5...
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Solving this gives us x = 4 5 But 4 5 3 6≥ 0 so this...

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