www.mhhe.com/messersmith Step 4: Choose a test number in each interval. Step 5: The solution set of 1 3a a 2 0 aand therefore 7 a 2 3b will contain the numbers in the intervals where 1 3a a 2 is negative. These are the fi rst and last intervals. 2 13 1 3a a 2 Negative Positive Negative Step 6: Determine whether the endpoints of the intervals, 2 and 1 3 , are included in the solution set. The endpoint 1 3 is included because it does not make the denominator equal 0. But 2 is not included because it makes the denominator equal 0. The graph of the solution set of 7 a 2 3 is 1 5 2 1 1 2 5 The solution set is (q, 2) ´ c 1 3 , qb. Although an inequality symbol may be or , an endpoint cannot be included in the solution set if it makes the denominator equal 0. YOU TRY 6 Solve 3 z 4 2. Graph the solution set, and write the solution in interval notation. ANSWERS TO YOU TRY EXERCISES 1) a) x y 5 5 5 5 y x2 x 5 b) 5, 1 c) (q, 5) ´ (1, q) 2) 5 2 1 1 2 5 4, 1 3) a) b) (q, q) 4) 5 2 1 1 2 5 3, 1 ´ 1, q) 5) 2 1 1 2 5 (q, 6) 6) 5 2 5 21 1 2 5 a4, 5 2 d Interval a 2 2 a 1 3 a 1 3 Test number a 3 a 0 a 1 Evaluate 1 3a a 2 1 3(3) 3 2 10 1 10 1 3(0) 0 2 1 2 1 3(1) 1 2 2 3 Sign Negative Positive Negative 688 CHAPTER 10 Quadratic Equations and Functions
messersmith_power_intermediate_algebra_1e_ch4_7_10
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