Copyright Cengage Learning. All rights reserved Inequalities 2. Equations and Inequalities 2 An inequality is a statement that two quantities or expressions are not equal. It may be the case that one quantity is less than ( ), or greater than or equal to ( ) another quantity. Consider the inequality 2x + 3 > 11, where x is a variable. 3 Inequalities As illustrated in the following table, certain numbers yield true statements when substituted for x, and others yield false statements. If a true statement is obtained when a number b is substituted for x, then b is a solution of the inequality. Thus, x = 5 is a solution of 2x + 3 > 11 since 13 > 11 is true, but x = 3 is not a solution since 9 > 11 is false. 4 Inequalities To solve an inequality means to find all solutions. Two inequalities are equivalent if they have exactly the same solutions. Most inequalities have an infinite number of solutions. To illustrate, the solutions of the inequality 2 < x < 5 consist of every real number x between 2 and 5. We call this set of numbers an open interval and denote it by (2, 5). 5 Inequalities The graph of the open interval (2, 5) is the set of all points on a coordinate line that lie betweenbut do not includethe points corresponding to x = 2 and x = 5. The graph is represented by shading an appropriate part of the axis, as shown in Figure 1. Figure 1 6 Inequalities We refer to this process as sketching the graph of the interval. The numbers 2 and 5 are called the endpoints of the interval (2, 5). The parentheses in the notation (2, 5) and in Figure 1 are used to indicate that the endpoints of the interval are not included. If we wish to include an endpoint, we use a bracket instead of a parenthesis. For example, the solutions of the inequality 2 x 5 are denoted by [2, 5] and are referred to as a closed interval. 7 Inequalities The graph of [2, 5] is sketched in Figure 2, where brackets indicate that endpoints are included. We shall also consider half-open intervals [a, b) and (a, b] and infinite intervals, as described in the next chart. The symbol (read infinity) used for infinite intervals is merely a notational device and does not represent a real number. Figure 2 8 Inequalities Intervals 9 Inequalities Methods for solving inequalities in x are similar to those used for solving equations. In particular, we often use properties of inequalities to replace a given inequality with a list of equivalent inequalities, ending with an inequality from which solutions are easily obtained. 10 Inequalities The properties in the following chart can be proved for real numbers a, b, c, and d. Properties of Inequalities 11 Inequalities It is important to remember that multiplying or dividing both sides of an inequality by a negative real number reverses the inequality sign (see property 4). Properties similar to those above are true for other inequalities and for and . Thus, if a > b, then a + c > b + c; if a b and c < 0, then ac bc; and so on. 12 Example 3 Solving an inequality Solve the inequality 6 < 2x 4 < 2. Solution: A real number x is a solution of the given inequality if and only if it is a solution of both of the inequalities 6 < 2x 4 and 2x 4 < 2. This first inequality is solved as follows: 6 < 2x 4 6 + 4 < (2x 4) + 4 given add 4 13 Example 3 Solution 2 < 2x 1 < x x > 1 contd simplify divide by 2 simplify equivalent inequality 14 Example 3 Solution The second inequality is then solved: 2x 4 < 2 2x < 6 x < 3 contd given add 4 divide by 2 15 Example 3 Solution Thus, x is a solution of the given inequality if and only if both x > 1 and x < 3; that is, 1 < x < 3. Hence, the solutions are all numbers in the open interval (1, 3) sketched in Figure 5. contd Figure 5 16 Example 3 Solution An alternative (and shorter) method is to solve both inequalities simultaneouslythat is, solve the continued inequality: 6 < 2x 4 < 2 6 + 4 < 2x < 2 < 2x < 6 1 < x < 3 contd given add 4 simplify divide by 2 17 Example 4 Solving a continued inequality Solve the continued inequality Solution: A number x is a solution of the given inequality if and only if and We can either work with each inequality separately or solve both inequalities simultaneously, as follows (keep in mind that our goal is to isolate x): given 18 Example 4 Solution 10 4 3x < 2 10 4 3x < 2 4 14 3x < 2 contd multiply by 2 subtract 4 simplify divide by 3; reverse the inequality signs simplify 19 Example 4 Solution Thus, the solutions of the inequality are all numbers in the half-open interval sketched in Figure 6. contd equivalent inequality Figure 6 20 Example 5 Solving a rational inequality Solve the inequality Solution: Since the numerator is positive, the fraction is positive if and only if the denominator, x 2, is also positive. Thus, x 2 > 0 or, equivalently, x > 2, and the solutions are all numbers in the infinite interval (2, ) sketched in Figure 7. Figure 7 21 Inequalities It follows that the solutions of an inequality such as | x | < 3 consist of the coordinates of all points whose distance from O is less than 3. This is the open interval (3, 3) sketched in Figure 10. Figure 10 22 Inequalities Thus, | x | < 3 is equivalent to 3 < x < 3. Similarly, for | x | > 3, the distance between O and a point with coordinate x is greater than 3; that is, | x | > 3 is equivalent to x 3. 23 Inequalities The graph of the solutions to | x | > 3 is sketched in Figure 11. We often use the union symbol and write (, 3) (3, ) to denote all real numbers that are in either (, 3) or (3, ). Figure 11 24 Inequalities The notation (, 2) (2, ) represents the set of all real numbers except 2. The intersection symbol is used to denote the elements that are common to two sets. For example, (, 3) (3, ) = (3, 3), since the intersection of (, 3) and (3, ) consists of all real numbers x such that both x 3. 25 Inequalities The preceding discussion may be generalized to obtain the following properties of absolute values. In the next example we use property 1 with a = x 3 and b = 0.5. 26 Example 7 Solving an inequality containing an absolute value Solve the inequality | x 3 | < 0.5. Solution: | x 3 | < 0.5. 0.5 < x 3 < 0.5 < (x 3) + 3 < < x < 3.5 given property 1 isolate x by adding 3 simplify 27 Example 7 Solution Thus, the solutions are the real numbers in the open interval (2.5, 3.5). The graph is sketched in Figure 12. contd Figure 12 28 Inequalities In the next example we use property 2 with a = 2x + 3 and b = 9. 29 Example 8 Solving an inequality containing an absolute value Solve the inequality | 2x + 3 | > 9. Solution: | 2x + 3 | > 9 2x x 6 x 3 given property 2 subtract 3 divide by 2 30 Example 8 Solution Consequently, the solutions of the inequality | 2x + 3 | > 9 consist of the numbers in (, 6) (3, ). The graph is sketched in Figure 13. contd Figure 13 31 Inequalities The trichotomy law states that for any real numbers a and b exactly one of the following is true: a > b, a < b, or a = b Thus, after solving | 2x + 3 | > 9 in Example 8, we readily obtain the solutions for | 2x + 3 | < 9 and | 2x + 3 | = 9 namely, (6, 3) and {6, 3}, respectively. Note that the union of these three sets of solutions is necessarily the set of real numbers. 32 Inequalities When using the notation a 3 (in Example 8) as 3 < x < 6. Another misuse of inequality notation is to write a b, since when several inequality symbols are used in one expression, they must point in the same direction. 33 Copyright Cengage Learning. All rights reserved. 2.7 More on Inequalities 34 More on Inequalities To solve an inequality involving polynomials of degree greater than 1, we shall express each polynomial as a product of linear factors ax + b and/or irreducible quadratic factors ax 2 + bx + c. If any such factor is not zero in an interval, then it is either positive throughout the interval or negative throughout the interval. 35 More on Inequalities Hence, if we choose any k in the interval and if the factor is positive (or negative) for x = k, then it is positive (or negative) throughout the interval. The value of the factor at x = k is called a test value of the factor at the test number k. This concept is exhibited in the following example. 36 Example 1 Solving a quadratic inequality Solve the inequality 2x 2 x < 3. Solution: To use test values, it is essential to have 0 on one side of the inequality sign. Thus, we proceed as follows: 2x 2 x < 3 2x 2 x 3 < 0 (x + 1)(2x 3) < 0 The factors x + 1 and 2x 3 are zero at 1 and respectively. given make one side 0 factor 37 Example 1 Solution The corresponding points on a coordinate line (see Figure 1) determine the nonintersecting intervals (, 1), (1, ) and We may find the signs of x + 1 and 2x 3 in each interval by using a test value taken from each interval. contd Figure 1 38 Example 1 Solution To illustrate, if we choose k = 10 in (, 1), the values of both x + 1 and 2x 3 are negative, and hence they are negative throughout (, 1). A similar procedure for the remaining two intervals gives us the following sign chart, where the term resulting sign in the last row refers to the sign obtained by applying laws of signs to the product of the factors. contd 39 Example 1 Solution Note that the resulting sign is positive or negative according to whether the number of negative signs of factors is even or odd, respectively. Sometimes it is convenient to represent the signs of x + 1 and 2x 3 by using a coordinate line and a sign diagram, of the type illustrated in Figure 2. contd Figure 2 40 Example 1 Solution The vertical lines indicate where the factors are zero, and signs of factors are shown above the coordinate line. The resulting signs are shown in red. The solutions of (x + 1)(2x 3) < 0 are the values of x for which the product of the factors is negativethat is, where the resulting sign is negative. This corresponds to the open interval (1, ). contd 41 More on Inequalities The following warning shows this incorrect extension applied to the inequality in Example 1. 42 Example 3 Using a sign diagram to solve an inequality Solve the inequality Solution: Since 0 is already on the right side of the inequality and the left side is factored, we may proceed directly to the sign diagram in Figure 4, where the vertical lines indicate the zeros (2, 1, and 3) of the factors. Figure 4 43 Example 3 Solution The frame around the 1 indicates that 1 makes a factor in the denominator of the original inequality equal to 0. Since the quadratic factor x is always positive, it has no effect on the sign of the quotient and hence may be omitted from the diagram. The various signs of the factors can be found using test values. Alternatively, we need only remember that as x increases, the sign of a linear factor ax + b changes from negative to positive if the coefficient a of x is positive, and the sign changes from positive to negative if a is negative. contd