HOME David Terr Ph.D. Math, UC Berkeley << 2.7. Rational Functions and Their Graphs   2.8. Polynomial and Rational Inequalities In Section P.11 we showed how to solve linear inequalities. In this section we will show how to solve polynomial and rational inequalities. Polynomial inequalities have one of the four forms p(x) > 0, p(x) ≥ 0, p(x) ≤ 0, or p(x) < 0, where p(x) is a polynomial. These inequalities are solved by first solving the equation p(x) = 0 for each of the zeros of p. These are the endpoints of the solution to the corresponding inequality. Once we have found them, it is simply a matter of determining between which pairs of endpoints the solutions belong. We start with a simple example.   Example 1: Solve the polynomial inequality x2 < 4. Solution: Taking the square root of both sides, we see that |x| < 2, or equivalently, -2 < x < 2. Thus the solution set is the interval (-2, 2), whose graph is shown below.   Example 2: Solve the polynomial inequality x2 + x ≥ 2. Solution: First we solve the corresponding polynomial equation x2 + x = 2, or equivalently, p(x) = x2 + x - 2 = 0. By trial and error, we see that p(x) has the factorization (x + 2)(x - 1), implying that its zeros are -2 and 1, which are the endpoints of the solution set. We must now test every interval with these endpoints and containing these endpoints, since the inequality is not strict. The first such interval is (-∞,-2]. To see whether this interval is part of the solution set, we test one of its points. We try x = -3. The left side of the intequality becomes (-3)2 - 3 = 6, which is clearly greater than or equal to -2, so the inequality holds for this value of x. Thus we see that the interval (-∞,-2] is part of the solution set. Next we test the interval [-2,1]. Plugging in x=0, we test whether the inequality x2 + x ≥ 2 holds. Clearly it does not, since the left side becomes zero when we substitute 0 for x, and 0 is not greater than or equal to 2. Thus we see that the interval [-2,1] is not part of the solution set. Finally we test the interval [1,∞]. We try the value x = 2. Plugging into the left side of the inequality, we find it becomes 22 +2 = 6 ≥ 2, whence we see that this interval is part of the solution set. Since there are no more endpoints, we have found the entire solution set, namely (-∞,-2] ∪ [-1,∞). Its graph is shown below.   Example 3: Solve the polynomial inequality x2 - x ≤ 1. Solution: First we solve the corresponding polynomial equation x2 - x = 1, or equivalently, p(x) = x2 - x - 1 = 0. By applying the quadratic formula P.10.7, we see that the solutions to this equation are x = (1 ± √5)/2, which are approximately -0.618 and 1.618. (Incidently, the number (1 + √5)/2 is an important mathematical constant known as the golden mean and denoted by the greek letter φ.) As in the previous example, there are three intervals to test. First we test the interval (-∞,-0.618] by testing x = -1. Plugging this into the left side of the inequality, we obtain 2, which is not less than or equal to -1, whence the solution set does not contain the interval (-∞ ,-0.618]. The next interval to test is the interval [-0.618, 1.618]. We test the point x=0. Substituing into the left side of the inequality, we obtain 0, which is clearly less than or equal to 1. Thus the solution set comtains the interval [-0.618, 1.618]. The last interval to test is the interval [1.618, ∞). We test the value x=2. Plugging into the left side of the inequality, we obtain 2, which is not less than or equal to -1. Thus, the solution set does not comtain this interval. Putting everything together, we see that the solution set consists of the sole interval [(1 - √5)/2, (1 + √5)/2] ≈ [-0.618, 1.618]. A graph of this solution set is shown below.   Example 4: Solve the polynomial inequality x3 - 6x2 + 11x > 6. Solution: The corresponding polynomial equation is p(x) = x3 - 6x2 + 11x - 6 = 0. By synthetic division (see Section 2.6), we find the solutions of this equation are x = 1, x = 2, and x = 3. This time there are four interval to test. First we test the interval (-∞,1). The easiest point in this interval to text is x=0. Substituting x=0 into the inequality, we get 0 on the left side, which is clearly not greater than 6. Thus, the interval (-∞,1) does not belong to the solution set. Next we test the interval (1, 2). Plugging in x = 3/2 to the left side, we obtain 51/8 = 6 3/8, which is greater than 6. Thus, the solution set comtains the interval (1, 2). Next we test the interval (2, 3). Plugging in x = 5/2, we find the left side of the inequality becomes 45/8 = 5 5/8, which is less than 6, whence the inequality does not hold for this point and thus the solution set does not contain the interval (2,3). The last interval we must test is (3, ∞). Plugging in x = 4 to the left side of the inequality, we obtain 12, which is greater than 6. Thus the solution set contains the interval (3, ∞). Putting everything together, we see that the solution set is equal to (1, 2) ∪(3, ∞). A graph of this solution set is shown below.   In every example we have looked at thus far, the intervals we have tested have alternated between belonging and not belonging to the solution set. This is usually but not always the case. Below is an exception to this rule.   Example 5: Solve the polynomial inequality x3 - 3x < 2. Solution: The corresponding polynomial equation is p(x) = x3 - 3x - 2 = 0. By means of synthetic division, we find that p(x) factors as (x + 1)2 (x - 2). Thus, the endpoint intervals are x = -1 and x = 2. The first interval we must test is (-∞, -1). Plugging in x=-2, we obtain -2 on the left side of the inequality, which is less than 2, whence the solution set contains the interval (-∞, -1). The next interval to test is (-1 ,2). Plugging in x=0, the left side the inequality becomes 0, which is once again less than 2. Thus the solution set also contains the interval (-1, 2). Note that it does not contain -1, however, since substituting -1 into the left side of the inequality yields 2, which is not less than 2. The final interval to test is (2, ∞). Plugging in x=3, we see that the left side of the inequality becomes 18, which is not less than 2. Thus we see that the interval (2, ∞) is not part of the solution set. Putting everything together, we see that the solution set is (-∞, -1) ∪(-1, 2), whose graph is shown below.   Solving rational inequalities is somewhat more difficult, but the strategy is the same. This time, one must keep track of both the zeros and the singularities of the rational function and proceed in the same way as before.   Example 6: Solve the rational inequality (x2 - 4) / (3x2 + 8x - 3) ≥ 0. Solution: In Example 2 of the previous section we graphed the rational function f(x) = (x2 - 4) / (3x2 + 8x - 3), so we can see at a glance for which values of x we have f(x) > 0. However, it is better to suppose we do not have access to this graph. (We can check our result at the end by comparing against the graph.) Our strategy is now to first find all zeros and singularities of f(x). Since the numerator factors as (x + 2)(x - 2) and the denominator factos as (x + 3)(3x - 1), we see that the zeros of f are -2 and 2 and the singularities are -3 and 1/3. These are the endpoints of the intervals in question. The first interval we test is (-∞,-3]. Since f(-10) = 96 / 217 > 0, we see that this interval belongs to the solution set. The second interval in question is [-3,-2]. We have f(-5/2) = -17/4 < 0, so this interval does not belong to the solution set. The third interval we need to test is [-2,1/3]. Since f(0) = 4/3 > 0, we see that this interval belongs to the solution set. The fourth interval is [1/3,2]. We have f(1) = -3/8 < 0, so this interval does not belong to the solution set. The last interval is [2,∞). We have f(10) = 96 / 377 > 0, so this interval belongs to the solution set. Putting everything together, we find the solution set is (-∞, -3] ∪[-2, 1/3] ∪[2, ∞), whose graph is shown below.   << 2.7. Rational Functions and Their Graphs Copyright © 2007-2009 - MathAmazement.com. All Rights Reserved.