Contents: This page corresponds to § 2.5 (p. 216) of the text.
Suggested Problems from Text:
p. 225 #11, 12, 13, 14, 16, 28, 33, 35, 38, 41, 53, 56, 62, 63, 68, 69
Combinations of Inequalities
Inequalities Involving Absolute Values
An inequality is a comparison of expressions by either "less than" (<), "less than or equal to" (<=), "greater than" (>), or "greater than or equal to" (>=). Note that Html does not support the standard symbols for "less than or equal to" and "greater than or equal to", so we use <= and >= for these relations.
Example 1. x + 3 <= 10
A solution for an inequality in x is a number such that when we substitute that number for x we have a true statement. So, 4 is a solution for example 1, while 8 is not. The solution set of an inequality is the set of all solutions. Typically an inequality has infinitely many solutions and the solution set is easily described using interval notation.
The solution set of example 1 is the set of all x <= 7. In interval notation this set is (-inf, 7], where we use inf to stand for infinity.
A linear inequality is one such that if we replaced the inequality with the equals relation, then we would have a linear equation. Solving linear inequalities is very much like solving linear equations, with one important difference.
When you multiply or divide both sides of an inequality by a negative number, the direction of the inequality is reversed.
You can see this using an inequality with no variables.
3 < 7. This is TRUE.
(3)(-2) < (7)(-2). This is FALSE, because -6 is to the right of -14 on the number line. Hence, -6 > -14.
(3)(-2) > (7)(-2). This is TRUE. So, when we multiply the original inequality by -2, we must reverse the direction to obtain another true statement.
Note: In general we may not multiply or divide both sides of an inequality by an expression with a variable, because some values of the variable may make the expression positive and some may make it negative.
7 - 2x < 3.
-2x < -4.
x > 2.
Note: When we divided both sides of the inequality by -2 we changed the direction of the inequality.
Look at the graphs of the functions on either side of the inequality.
To satisfy the inequality, 7 - 2x needs to be less than 3. So we are looking for numbers x such that the point on the graph of y = 7 - 2x is below the point on the graph of y = 3. This is true for x > 2. In interval notation the solution set is (2, inf).
There is another way to use a graphing utility to solve this inequality. In the Java Grapher the expression (7-2*x)L3 has the value 1 for numbers x that satisfy the inequality, and the value 0 for other numbers x. The picture below shows the graph of (7-2*x)L3 as drawn by the Grapher.
Solve the inequality 4 - x > 1 + 3x. Answer
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Find all numbers x such that -3 < 5 - 2x and 5 - 2x < 9.
-3 < 5 - 2x
-8 < -2x
4 > x
5 - 2x < 9
-2x < 4
x > -2
In order to satisfy both inequalities, a number must be in both solution sets. So the numbers that satisfy both inequalities are the values in the intersection of the two solution sets, which is the set (-2, 4) in interval notation.
The problem above is usually written as a double inequality.
-3 < 5 - 2x < 9 stands for -3 < 5 - 2x and 5 - 2x < 9.
Note: When we solved the two inequalities separately, the steps in the two problems were the same. Therefore, the double inequality notation may be used to solve the inequalities simultaneously.
-3 < 5 - 2x < 9.
-8 < -2x < 4.
4 > x > -2.
In terms of graphs, this problem corresponds to finding the values of x such that the corresponding point on the graph of y = 5 - 2x is between the graphs of y = -3 and y = 9.
Find all numbers x such that x + 1 < 0 or x + 1 > 3.
In Example 4 above we were looking for numbers that satisfied both inequalities. Here we want to find the numbers that satisfy either of the inequalities. This corresponds to a union of solution sets instead of an intersection.
Do not use the double inequality notation in this situation.
x + 1 < 0
x < -1
OR x + 1 > 3
x > 2
The solution set is the union of the two intervals (-inf, -1) and (2, inf).
(a) 1 < 3 + 5x < 7 Answer
(b) 2 - x < 1, or 2 - x > 5 Answer
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Inequalities involving absolute values can be rewritten as combinations of inequalities.
Let a be a positive number.
|x| < a if and only if -a < x < a.
|x| > a if and only if x < -a or x > a.
To make sense of these statements, think about a number line. The absolute value of a number is the distance the number is from 0 on the number line. So the inequality |x| < a is satisfied by numbers whose distance from 0 is less than a. This is the set of numbers between -a and a.
The inequality |x| > a is satisfied by numbers whose distance from 0 is larger than a. This means numbers that are either larger than a, or less than -a.
| 3 + 2x | <= 7.
-7 <= 3 + 2x <= 7.
-10 <= 2x <= 4.
-5 <= x <= 2.
x is in [-5, 2].
In terms of graphs, we are looking for x values such that the corresponding point on the graph of y = | 3 + 2x | is either below or equal to the point on the graph of y = 7.
| 5 - 2x | > 3.
5 - 2x < -3 or 5 - 2x > 3.
-2x < -8 or -2x > -2.
x > 4 or x < 1.
x is in (4, inf) union (-inf, 1).
This solution set corresponds to the region where the graph of y = | 5 - 2x | is above the graph of y = 3.
Solve the following inequalities. Use a graphing utility to check your answers.
(a) | 3 + x | < 4.
(b) | 2 - x | > 3.
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x2 - x - 6 < 0.
The first step is to find the zeros of the polynomial x2 - x - 6.
x2 - x - 6 = 0.
(x + 2)(x - 3) = 0.
x = -2, or x = 3.
-2 and 3 are called the critical numbers of the inequality.
Note: -2 and 3 are not in the solution set of the inequality. We are looking for values of x where the polynomial is negative. The solution set of the inequality corresponds to the region where the graph of the polynomial is below the x-axis. The critical numbers -2 and 3 are the places where the graph intersects the x-axis.
The critical numbers divide the x-axis into three intervals called test intervals for the inequality.
Test intervals: (-inf, -2), (-2, 3), (3, inf).
We are going to use the fact that polynomial functions are continuous. This means that their graphs do not have any breaks or jumps.
Since we have found all the x-intercepts of the graph of x2 - x - 6, throughout each test interval the graph must be either above the x-axis or below it. This is where we need to know that the graph does not have any breaks. This means that we may choose any number we like in a test interval and evaluate the polynomial at that number to see if the graph is above or below the x-axis throughout that test interval.
(-inf, -2): -5 is in the interval. (-5)2 - (-5) - 6 = 24 > 0, so the graph of y = x2 - x - 6 is above the x-axis on the entire interval (-inf, -2).
(-2, 3): 0 is in the interval. 02 - 0 - 6 = -6 < 0, so the graph of y = x2 - x - 6 is below the x-axis on the entire interval.
(3, inf): 4 is in the interval. 42 - 4 - 6 = 6 > 0, so the graph of y = x2 - x - 6 is above the x-axis on the entire interval.
Since we are looking for regions where the graph is below the axis, the solution set is -2 < x < 3, or (-2, 3).
We will use the problem in Example 8 to illustrate a common mistake.
x2 - x - 6 < 0.
(x + 2)(x - 3) < 0 OK to this point.
x + 2 < 0 or x - 3 < 0 WRONG!
When a product of two numbers is equal to 0, then at least one of the numbers must be 0. However, a product of two negative numbers is not negative, so this approach is not useful for solving inequalities.
1.2 x3 + 3.07 x2 - x - 3.71 > 0.
This problem is much more difficult than the inequality in the previous example! It is not easy to factor, so we will not be able to find the exact values of the critical numbers. We will use a graphing utility to approximate the critical numbers. The graph of the polynomial is shown below.
y = 1.2 x3 + 3.07 x2 - x - 3.71
The critical numbers are approximately -2.35, -1.25, and 1.05. In this problem we looking for regions where the graph is above the axis.
Solution Set: (-2.35, -1.25) union (1.05, inf).
Solve the inequality x2 + 3x - 4 > 0. Use a graphing utility to check your solution.
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A rational expression is one of the form polynomial divided by polynomial. In general, graphs of rational functions do have breaks. They are not defined at the zeros of the denominator. These are the only places where there are breaks, so we can use the same technique to solve rational inequalities that we use for polynomial inequalities.
The critical numbers for a rational inequality are all the zeros of the numerator and the denominator. Since the numerator and denominator are already factored in this example, we see that the critical numbers are -3, 5, and 1.
The three critical numbers divide the number line into four test intervals.
(-inf, -3): -4 is in the interval, and the rational function evaluated at -4 is -9/15. Since the value is negative, the graph of the rational function is below the x-axis throughout the interval.
(-3, 1): 0 is in the interval. The value of the function at 0 is 5, which is positive. The graph of the function is above the x-axis throughout the interval.
(1, 5): 2 is in the interval. The value at 2 is -5. The graph of the function is below the x-axis.
(5, inf): 6 is in the interval. The value at 6 is 9/15. The graph of the function is above the x-axis.
We are looking for regions where the graph is above the x-axis, so the solution set is (-3, 1) union (5, inf).
Note: A graphing utility can be used to see which side of the x-axis the graph is on over the various test intervals. In some cases you must solve algebraically to find the exact values of the critical numbers, but once this is done, a grapher provides a fast way to finish the problem.
Graph of y = (x + 3)(x - 5)/3(x - 1)
There are two important points to keep in mind when working with inequalities:
1. We need to compare an expression to 0. So, if we start with the problem
x2 - 3x - 11 < x + 10, we would subtract x and 10 from both sides to obtain
x2 - 4x - 21 < 0.
2. Do not multiply both sides of an inequality by an expression with a variable.
For example, given the problem , do not multiply both sides by x. The correct way to handle this problem is as follows:
Now we see that the critical numbers are 0 (from denominator), 1, and -1.
(a) Finish solving x2 - 3x - 11 < x + 10, and check your solution with a graphing utility.
(b) Finish solving , and check your solution with a graphing utility.
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