SECTION 4.7 Polynomial and Rational Inequalities 261 power), the sign of f changes on either side of 0, so for < < x 0 1, f is negative. Similarly, f is positive for > x 1, since the multiplicity of the zero 1 is odd. Now Work PROBLEM 21 2 Solve Rational Inequalities Graphically and Algebraically We approach solving a rational inequality in the same way we solved polynomial inequalities. We begin by writing the inequality in one of the forms ( ) ( ) ( ) ( ) < > ≤ ≥ f x f x f x f x 0 0 0 0 We then locate the real zeros of the numerator and the denominator of f . We 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—that is, ( ) > f x 0—or below the x-axis—that is, ( ) < f x 0. This enables us to identify the solution of the inequality. Now Work PROBLEM 15 Solving a Rational Inequality Using Its Graph Solve − − ≥ x x 1 4 0 2 by graphing ( ) = − − f x x x 1 4 2 . EXAMPLE 3 By Hand Graphical Solution Graph ( ) = − − f x x x 1 4 2 and determine the intervals of x such that the graph is above or on the x-axis. Do you see why? These values of x result in ( ) R x being positive or zero. We graphed ( ) = − − R x x x 1 4 2 in Example 1 from Section 4.6 (p. 249).We reproduce the graph in Figure 60. Graphing Utility Solution Graph ( ) = − − f x x x 1 4 2 . See Figure 61 using Desmos. The x-intercept of the graph of f is 1. The graph of f is above the x-axis (and, therefore, f is positive) for − < < x 2 1 or > x 2. Since the inequality is not strict, include 1 in the solution set. Therefore, the solution set is { } − < ≤ > x x x 2 1 or 2 or, using interval notation, ( ] ( ) − ∪ ∞ 2, 1 2, . Do you see why we do not include 2? From the graph, we can see that ( ) ≥ R x 0 for − < ≤ x 2 1 or > x 2. The solution set is { } − < ≤ > x x x 2 1 or 2 or, using interval notation, ( ] ( ) − ∪ ∞ 2, 1 2, . Figure 60 ( ) = − − R x x x 1 4 2 x 3 23 y 3 23 (1, 0) 0, x 5 22 x 5 2 1 – 4 ( 3 – 2( ) (23, 20.8) 2 (3, 0.4) y 5 0 ) , 2 – 7 Figure 61 ( ) = − − f x x x 1 4 2 To solve a rational inequality algebraically, we follow the same approach that we used to solve a polynomial inequality algebraically. However, we must also identify the real zeros of the denominator of the rational function because the sign of a rational function may change on either side of a vertical asymptote. Convince yourself of this by looking at Figure 60. Notice that the function values are negative for <− x 2 and are positive for >− x 2 (but less than 1).
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