# To solve a rational inequality we can extend the

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Chapter 4 / Exercise 118
Mathematics for Machine Technology
Peterson/Smith
Expert Verified
To solve a rational inequality, we can extend the graphing and algebraic techniques we used to solve polynomial inequalities. Interval Test Point Outcome for the Test Point Resulting Sign Satisfies Inequality: < 0 ? I: (0, 1.75) 0 16 0 2 + 132 (0 182 yes II: (1.75, 6.5) 3 16 (3 2 + 132 (3 182 + no III: (6.5, 8.25) 7 16 (7 2 + 132 (7 182 yes
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Chapter 4 / Exercise 118
Mathematics for Machine Technology
Peterson/Smith
Expert Verified
80 Let us start with solving by graphing. Example 7 The graph of d9 = @ ( 0 @ ( 0 is given. -6 -4 -2 2 4 6 8 10 12 14 -6 -4 -2 2 4 6 8 10 x y x = 4 Use the graph to solve the following. a. JH ( à H ( à 0 b. JH ( à H ( à < 0 Solution a. The given function is not defined at 9 = 4, a it has a zero at 9 = 2 . From the graph, we can see that the function will be on or above the x -axis when 9 ≤ 2 or 9 > 4 . The solution in interval notation is given by (− ∞, 2< ∪ (4 , ∞ . b. The graph of the function lies below the 9 -axis for 2 < 9 < 4 , or the interval (2 , 4 . Example 8 Solve the rational inequality by graphing: H ( ä H J ( ã > 0. Solution We graph d9 = @ ( @ + (  . The function has vertical asymptotes at 9 = ± 3 , and a zero at 9 = 1. -5 -4 -3 -2 -1 1 2 3 4 5 -6 -4 -2 2 4 6 x y x = -3 x = 3 The graph shows that the given function will be greater than zero for −3 < 9 < 1 or 9 > 3 . The solution in interval notation is (−3 , 1 ∪ (3 , ∞ .
81 Let us now solve rational inequalities algebraically. Solving a Rational Inequality Algebraically (1) Write the given inequality in the formd9< 0, d9> 0, d90, or d90, as appropriate. If there is more than one rational expression on the left-hand side of the inequality, combine them into a single fraction. (2) Determine any value(s) where the numerator and the denominator are zero. These will be the critical values that will determine the intervals on the number line. (3) Plot the solutions from (2) on a real number line. For the numerator, use a solid dot if the inequality contains or ; if the inequality is < or >, use an open circle. Since we need to exclude any values that would make the denominator equal to zero on any rational expression in the inequality, we always use an open circle for critical values related to the denominator. (4) Select one test point within each interval and see whether it satisfies the inequality. (5) The solution will consist of the interval(s) that will satisfy the original inequality. Example 9 Solve the rational inequality algebraically: H ( ä H J ( ã > 0 . Solution We now follow the steps for solving a rational equation algebraically. (1) The given function already satisfies this step. (2) We will find the critical values by setting the numerator and denominator equal to zero, and solving for 9 . Numerator: 9 − 1 = 0 9 = 1 Denominator: (9 + 3(9 − 3 = 0 9 = −3 or 9 = 3 (3) These critical values divide the real number line into four intervals. I II III IV 0 3 -3 1 (− ∞,  3 ) (−3 , 1 (1 , 3 (3 , ∞ (4) We will construct the sign chart using test points 4, 0, 2, and 4.
82 (5) As we can see from the sign chart, the solution for @ (  @ + (  > 0 is given by  3 , 1 ∪ (3 , ∞ , which is the same solution we found solving the problem graphically in Example 8. Example 10 Solve each the rational inequality algebraically. Confirm your answers graphically. a. @@ ( <@ ( b. @ ( @ & @ & @( 0