262 CHAPTER 4 Polynomial and Rational Functions How to Solve a Rational Inequality Algebraically Solve the inequality ( ) + + + ≤ x x x 3 13 9 2 3 2 2 algebraically, and graph the solution set. EXAMPLE 4 Figure 62 24 22 26 0 2 4 6 x NOTE If the inequality is not strict (≤ or ≥), include the solutions of ( ) = f x 0 in the solution set. j Step-by-Step Solution Step 1 Write the inequality so that a polynomial function f is on the left side and zero is on the right side. Rearrange the inequality so that 0 is on the right side. ( ) ( ) + + + ≤ + + + + − ≤ + + + + − ⋅ + + + + ≤ + + − − − + + ≤ − + ≤ x x x x x x x x x x x x x x x x x x x x x x x 3 13 9 2 3 3 13 9 4 4 3 0 3 13 9 4 4 3 4 4 4 4 0 3 13 9 3 12 12 4 4 0 3 2 0 2 2 2 2 2 2 2 2 2 2 2 2 Subtract 3 from both sides of the inequality; Expand ( ) +x 2 2. Multiply 3 by + + + + x x x x 4 4 4 4 2 2 . Write as a single quotient. Combine like terms. Step 2 Determine the real zeros (x-intercepts of the graph) of f and the real numbers for which f is undefined. The real zero of ( ) ( ) = − + f x x x 3 2 2 is 3. Also, f is undefined for = − x 2. Use the real zero and the undefined value to divide the real number line into three intervals: ( ) ( ) ( ) −∞ − − ∞ , 2 2, 3 3, Step 3 Use the real zeros and undefined values found in Step 2 to divide the real number line into intervals. Step 4 Select a number in each interval, evaluate f at the number, and determine whether the value of f is positive or negative. If the value of f is positive, all values of f in the interval are positive. If the value of f is negative, all values of f in the interval are negative. Select a test number in each interval from Step 3, and evaluate f at each number to determine whether the value of f is positive or negative. See Table 20. We conclude that ( ) ≤ f x 0 for all numbers for which <− x 2 or − < ≤ x 2 3. Notice that we do not include −2 in the solution because −2 is not in the domain of f . The solution set of the inequality ( ) + + + ≤ x x x 3 13 9 2 3 2 2 is { } <− − < ≤ x x x 2 or 2 3 or, using interval notation, ( ) ( ] −∞ − ∪ − , 2 2, 3 . Figure 62 shows the graph of the solution set. 3 −2 x Interval ( ) −∞ −, 2 ( ) −2, 3 ( )∞ 3, Number chosen −3 0 4 Value of f ( ) − = − f 3 6 ( ) = − f 0 3 4 ( ) = f 4 1 36 Conclusion Negative Negative Positive Table 20 The Role of Multiplicity in Solving Rational Inequalities In Example 4, we used the number −3 and found that ( ) f x is negative for all <− x 2. Because the “cut point” of −2 is a zero of even multiplicity, we know the sign of ( ) f x does not change on either side of −2, so for − < < x 2 3, ( ) f x is negative. Because the “cut point” of 3 is a zero of odd multiplicity, the sign of ( ) f x changes on either side of 3, so for > x 3, ( ) f x is positive.Therefore, the solution set of ( ) + + + ≤ x x x 3 13 9 2 3 2 2 is { } <− − < ≤ x x x 2 or 2 3 or, using interval notation, ( ) ( ] −∞ − ∪ − , 2 2, 3 . Now Work PROBLEMS 35 AND 41
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