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Unformatted text preview: Page 2 Math 171: Exam 1 So f 1 ( x ) = x + 5 2 x 1 3 . (b) [3 points] What is the domain of f ? Solution: f ( x ) is defined as long as 2 x 1 / 3 1 6 = 0. But if 2 x 1 / 3 1 = 0, then x 1 / 3 = 1 / 2 and so x = ( x 1 / 3 ) 3 = (1 / 2) 3 = 1 / 8. Thus the domain of f is { x : x 6 = 1 / 8 } , or in interval notation, (∞ , 1 / 8) ∪ (1 / 8 , ∞ ). (c) [3 points] What is the range of f ? Solution: The range of f is the same as the domain of f 1 . But f 1 ( x ) is defined as long as 2 x 1 6 = 0. Thus the domain of f is { x : x 6 = 1 / 2 } , or in interval notation, (∞ , 1 / 2) ∪ (1 / 2 , ∞ ). 6. Let p ( x ) = x 4 + 2 x 3 11 x 10 and q ( x ) = x 2 + 3 x + 5. (a) [8 points] Divide p by q . Solution: Performing the division: x 2 x 2 x 2 + 3 x + 5 ) x 4 + 2 x 3 11 x 10 x 4 3 x 3 5 x 2 x 3 5 x 2 11 x x 3 + 3 x 2 + 5 x 2 x 2 6 x 10 2 x 2 + 6 x + 10 So x 4 + 2 x 3 11 x 10 x 2 + 3 x + 5 = x 2 x 2. (b) [4 points] Factor p completely. Solution: Note that x 2 +3 x +5 has no real roots because 3 2 4 · 5 = 11 < 0. So x 2 + 3 x + 5 has no linear factors, meaning that we cannot factor it further. On the other hand, x 2 x 2 = ( x 2)( x + 1). Therefore, p ( x ) = ( x 2 +3 x +5)( x 2)( x +1), and it cannot be factored further. (c) [1 point] What are the zeros of p ? Page 3 Math 171: Exam 1 Solution: From part (b), the only (real) numbers that make p ( x ) zero are x = 2 , 1. 7. [10 points] Solve  x 3  = 1 4 x . ( Hint : You may want to break it into two cases.) Solution: We can break the problem into two cases. Case 1: If x 3 ≥ 0, then  x 3  = x 3, so the equation becomes x 3 = 1 4 x ....
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 Fall '07
 GOMEZ,JONES
 Calculus, Algebra, Trigonometry, Topology, Slope, Quadratic equation

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