173 1.7 Inequalities Step 3 Choose a test value in each interval. Interval Test Value Is 2x2 +5x−12 #0 True or False? A: 1-∞, -42 -5 21-522 + 51-52 - 12 Ú ? 0 13 Ú 0 True B: A -4, 3 2B 0 21022 + 5102 - 12 Ú ? 0 -12 Ú 0 False C: A3 2, ∞B 2 21222 + 5122 - 12 Ú ? 0 6 Ú 0 True 2 The values in Intervals A and C make the inequality true, so the solution set is a disjoint interval—that is, the union of the two intervals—written 1-∞, -44 ´c 3 2 , ∞b. The graph of the solution set is shown in Figure 15. S Now Try Exercise 41. –4 0 3 2 Figure 15 NOTE Inequalities that use the symbols 6and 7are strict inequalities, while … and Ú are used in nonstrict inequalities. The solutions of the equation in Example 5 were not included in the solution set because the inequality was a strict inequality. In Example 6, the solutions of the equation were included in the solution set because of the nonstrict inequality. EXAMPLE 7 Finding Projectile Height If a projectile is launched from ground level with an initial velocity of 96 ft per sec, its height s in feet t seconds after launching is given by the following equation. s = -16t2 + 96t When will the projectile be greater than 80 ft above ground level? SOLUTION -16t2 + 96t 780 Set s greater than 80. -16t2 + 96t - 80 70 Subtract 80. t2 - 6t + 5 60 Divide by -16. Reverse the direction of the inequality symbol. Now solve the corresponding equation. t2 - 6t + 5 = 0 1t - 121t - 52 = 0 Factor. t - 1 = 0 or t - 5 = 0 Zero-factor property t = 1 or t = 5 Solve each equation. Use these values to determine the intervals 1-∞, 12, 11, 52, and 15, ∞2. We are solving a strict inequality, so solutions of the equation t2 - 6t + 5 = 0 are not included. Choose a test value from each interval to see whether it satisfies the inequality t2 - 6t + 5 60. See Figure 16 on the next page.
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