Case 1 if x 3 then x 3 x 3 and x 2 x 2 so the

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Case 1: If x 3, then | x - 3 | = x - 3 and | x + 2 | = x + 2, so the equation becomes x + x + 2 + x - 3 = 7. Solving this gives us x = 8 / 3. But 8 / 3 6≥ 3, so this solution is extraneous. Page 3
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Math 171: Final Exam Case 2: If - 2 x 3, then | x - 3 | = - x + 3 but | x + 2 | = x + 2, so the equation becomes x + x + 2 - x + 3 = 7. Solving this gives us x = 2. Since - 2 2 3, this solution is valid. Case 3: If x ≤ - 2, then | x - 3 | = - x + 3 and | x + 2 | = - x - 2, so the equation becomes x - x - 2 - x + 3 = 7. Solving this gives us x = - 6. Since - 6 ≤ - 2, this solution is also valid. So the only solutions are x = 2 , - 6. 8. A young alien on a distant planet throws a ball into the air. It reaches its maximum height of 30ft after 4 seconds. Assume that the ball is at 2ft when it is released at time t = 0. (a) [6 points] Find a formula for the height of the ball as a function of time. ( Hint : The graph of your function will be a parabola.) Solution: We know that the formula for the height h at time t will be a quadratic, so let h ( t ) = at 2 + bt + c . The problem gives us enough informa- tion to solve for a , b and c . First, we know that h (0) = 2, so a · 0 2 + b · 0+ c = 2. Therefore, c = 2. Second, we are told that h (4) = 30, so a · 4 2 + b · 4 + 2 = 30. Simplifying, 4 a + b = 7. Finally, we are told that h has its maximum value at t = 4, meaning that the parabola y = h ( t ) = at 2 + bt + 2 has its vertex there. Completing the square, h ( t ) = a t 2 + b a t + 2 = a " t + b 2 a 2 - b 2 a 2 # + 2 . Therefore, the vertex is at t = - b 2 a , so 4 = - b 2 a . Simplifying, b = - 8 a . (Note that we could have gotten this equation a little easier by either (1) using the fact that h 0 (4) = 0 because h has a local maximum at t = 4 or (2) that by symmetry, after 8 seconds the ball will be at height 2ft again, so h (8) = 2.) We have found that 4 a + b = 7 and b = - 8 a . Solving for a and b , we have a = - 7 / 4 and b = 14. Thus h ( t ) = - 7 4 t 2 + 14 t + 2 .
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