Unformatted text preview: MAC1105: Quiz #13 Solutions
July 20, 2010 In the topright corner of a clean sheet of paper, write your name, UFID, and section number. Please use a pen with blue or black ink. When you are nished, FOLD your paper in half lengthwise and write your name on the back. Find the solution sets for the following inequalities:
1. x3  3x2 4x First, we obtain:x3  3x2  4x 0 x(x  4)(x + 1) 0. So we have four intervals to test: (, 1), (1, 0), (0, 4), (4, ). Setting up the sign chart and lling it in, we have: (x + 1) x (x  4) Total
2. 3x  5 + 2 7 (, 1)     (1, 0) +   + (0, 4) + +   (4, ) + + + + So our solution set is the positive parts, i.e., [1, 0] [4, ). 3x  5 5 3x 10 x 10 3 (3x  5) 5 3x  5 5 3x 0 x 0 Since both must be true at the same time, we get 0 x 10 , or 3 [0, 10 ] in interval notation. 3 x1 3. 1 x+1 3x  5 5 x1 x1 x+1 2 10  0 0 x+1 x+1 x+1 x+1
Since the numerator is negative, the inequality will be true when the denominator is also negative (negative divided by negative is positive). This occurs whenever x < 1, or (, 1) in interval notation. Notice that we do not include 1 even though the inequality is nonstrict this is because x = 1 would make the denominator 0, which is not allowed. 1 ...
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 Summer '10
 Picklesimer
 Algebra, Division, Divisor, x1 x1

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