260 CHAPTER 4 Polynomial and Rational Functions The results of Example 1 lead to the following approach to solving polynomial inequalities algebraically. Suppose that the polynomial inequality is in one of the forms ( ) ( ) ( ) ( ) < > ≤ ≥ f x f x f x f x 0 0 0 0 Locate the real zeros of f . Use these zeros to divide the real number line into intervals because on each interval, the graph of f is either above the x-axis [ ] ( ) > f x 0 or below the x-axis [ ] ( ) < f x 0 .This enables us to identify the solution of the inequality. How to Solve a Polynomial Inequality Algebraically Solve the inequality > x x 4 algebraically, and graph the solution set. EXAMPLE 2 Figure 59 24 22 26 0 2 4 6 x Step-by-Step Solution Step 1 Write the inequality so that a rational 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 0 4 4 Subtract x from both sides of the inequality. This inequality is equivalent to the one we are solving. Step 2 Determine the real zeros (x-intercepts of the graph) of f . Find the real zeros of ( ) = − f x x x 4 by solving − = x x 0 4 . ( ) ( ) ( ) − = − = − + + = = − = + + = = = x x x x x x x x x x x x x x 0 1 0 1 1 0 0 or 1 0 or 1 0 0 or 1 4 3 2 2 The equation + + = x x 1 0 2 has no real solutions. Do you see why? Step 3 Use the real zeros found in Step 2 to divide the real number line into intervals. Use the real zeros to separate the real number line into three intervals: ( ) ( ) ( ) −∞ ∞ , 0 0, 1 1, 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 found in Step 3 and evaluate ( ) = − f x x x 4 at each number to determine whether the value of f is positive or negative. See Table 19. Conclude that ( ) > f x 0 for all numbers x for which < x 0 or > x 1. The solution set of the inequality > x x 4 is { } < > x x x 0 or 1 or, using interval notation, ( ) ( ) −∞ ∪ ∞ , 0 1, . Figure 59 shows the graph of the solution set. 1 0 x Interval ( ) −∞, 0 ( ) 0, 1 ( )∞1, Number chosen −1 1 2 2 Value of f ( ) − = f 1 2 ( ) = − f 1 2 7 16 ( ) = f 2 14 Conclusion Positive Negative Positive Table 19 NOTE If the inequality is not strict (that is, if it is ≤ or ≥), include the solutions of ( ) = f x 0 in the solution set. j Factor out x. Factor the difference of two cubes. Use the Zero-Product Property. The Role of Multiplicity in Solving Polynomial Inequalities In Example 2, we used the number −1 and found that f is positive for all < x 0. Because the “cut point” of 0 is a zero of odd multiplicity (x is a factor to the first
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