exam1_2009_sol

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

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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.

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