What Is the Solution to the Inequality ?
Solving Inequalities |
Contents: This folio corresponds to § 2.5 (p. 216) of the text.
Suggested Issues from Text:
p. 225 #xi, 12, thirteen, xiv, 16, 28, 33, 35, 38, 41, 53, 56, 62, 63, 68, 69
Linear Inequalities
Combinations of Inequalities
Inequalities Involving Absolute Values
Polynomial Inequalities
Rational Inequalities
Linear Inequalities
An inequality is a comparison of expressions by either "less than" (<), "less than or equal to" (<=), "greater than" (>), or "greater than or equal to" (>=). Notation 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 <= ten
A solution for an inequality in 10 is a number such that when we substitute that number for 10 we have a true statement. So, 4 is a solution for example 1, while viii is non. The solution gear up of an inequality is the gear up of all solutions. Typically an inequality has infinitely many solutions and the solution set is hands described using interval notation.
The solution set of example 1 is the gear up of all x <= 7. In interval note this prepare is (-inf, 7], where we utilise inf to stand for infinity.
A linear inequality is one such that if nosotros replaced the inequality with the equals relation, then we would take a linear equation. Solving linear inequalities is very much like solving linear equations, with i important difference.
When you multiply or split up both sides of an inequality by a negative number, the direction of the inequality is reversed.
You can encounter this using an inequality with no variables.
Instance 2.
3 < 7. This is True.
(3)(-2) < (7)(-2). This is FALSE, because -6 is to the right of -fourteen on the number line. Hence, -6 > -fourteen.
(iii)(-2) > (vii)(-ii). This is TRUE. So, when we multiply the original inequality past -2, nosotros must opposite the management to obtain another true statement.
Annotation: In general we may not multiply or divide both sides of an inequality by an expression with a variable, considering some values of the variable may make the expression positive and some may make information technology negative.
Case three.
seven - 2x < 3.
-2x < -four.
x > 2.
Note: When we divided both sides of the inequality by -ii we changed the direction of the inequality.
Look at the graphs of the functions on either side of the inequality.
To satisfy the inequality, vii - 2x needs to exist less than iii. So we are looking for numbers x such that the signal on the graph of y = 7 - 2x is below the point on the graph of y = 3. This is true for x > two. 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 (vii-two*x)L3 has the value 1 for numbers 10 that satisfy the inequality, and the value 0 for other numbers 10. The picture below shows the graph of (7-2*10)L3 equally fatigued past the Grapher.
Exercise ane:
Solve the inequality 4 - x > i + 3x. Answer
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Combinations of Inequalities
Example 4.
Observe all numbers ten such that -3 < v - 2x and 5 - 2x < 9.
-8 < -2x
iv > x
(-inf, iv)
AND
-2x < 4
x > -2
(-2, inf)
In order to satisfy both inequalities, a number must be in both solution sets. And so the numbers that satisfy both inequalities are the values in the intersection of the 2 solution sets, which is the set (-two, four) in interval notation.
The problem higher up is usually written as a double inequality.
-iii < 5 - 2x < 9 stands for -3 < 5 - 2x and 5 - 2x < 9.
Note: When nosotros solved the two inequalities separately, the steps in the 2 problems were the same. Therefore, the double inequality notation may exist used to solve the inequalities simultaneously.
-3 < v - 2x < 9.
-8 < -2x < 4.
4 > x > -2.
In terms of graphs, this problem corresponds to finding the values of x such that the respective point on the graph of y = 5 - 2x is between the graphs of y = -iii and y = 9.
Example five.
Discover all numbers 10 such that x + 1 < 0 or ten + one > three.
In Instance 4 in a higher place 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 matrimony of solution sets instead of an intersection.
Exercise not employ the double inequality notation in this situation.
x + one < 0 10 < -1
(-inf, -1)
OR x + 1 > 3 x > ii
(ii, inf)
The solution set is the union of the two intervals (-inf, -i) and (ii, inf).
Exercise ii:
(a) 1 < three + 5x < 7 Answer
(b) 2 - x < 1, or 2 - x > 5 Answer
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Inequalities Involving Accented Values
Inequalities involving absolute values can be rewritten as combinations of inequalities.
Let a be a positive number.
|10| < a if and simply if -a < x < a.
|x| > a if and just if x < -a or x > a.
To brand sense of these statements, remember about a number line. The accented value of a number is the distance the number is from 0 on the number line. Then the inequality |x| < a is satisfied by numbers whose distance from 0 is less than a. This is the set up of numbers between -a and a.
The inequality |ten| > 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.
Example six.
| three + 2x | <= seven.
-seven <= 3 + 2x <= 7.
-10 <= 2x <= 4.
-five <= x <= 2.
10 is in [-5, two].
In terms of graphs, nosotros are looking for x values such that the corresponding point on the graph of y = | 3 + 2x | is either beneath or equal to the point on the graph of y = 7.
Example seven.
| 5 - 2x | > 3.
5 - 2x < -3 or 5 - 2x > 3.
-2x < -viii or -2x > -2.
ten > 4 or x < 1.
ten is in (4, inf) union (-inf, 1).
This solution fix corresponds to the region where the graph of y = | 5 - 2x | is higher up the graph of y = three.
Practice 3 :
Solve the following inequalities. Employ a graphing utility to check your answers.
(a) | 3 + ten | < 4.
(b) | 2 - x | > iii.
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Polynomial Inequalities
Example 8.
x2 - ten - 6 < 0.
The first step is to find the zeros of the polynomial 10ii - x - vi.
x2 - x - 6 = 0.
(ten + 2)(10 - 3) = 0.
10 = -2, or x = 3.
-ii and three are called the critical numbers of the inequality.
Notation: -2 and 3 are not in the solution set of the inequality. We are looking for values of ten 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-centrality. The critical numbers -2 and 3 are the places where the graph intersects the x-centrality.
The critical numbers divide the ten-centrality into three intervals chosen exam intervals for the inequality.
Test intervals: (-inf, -2), (-ii, 3), (3, inf).
We are going to use the fact that polynomial functions are continuous. This means that their graphs do not accept whatever breaks or jumps.
Since we have found all the ten-intercepts of the graph of ten2 - x - half dozen, throughout each examination interval the graph must be either above the x-axis or beneath it. This is where we need to know that the graph does not accept any breaks. This means that nosotros may cull 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 examination interval.
(-inf, -2): -5 is in the interval. (-five)ii - (-5) - 6 = 24 > 0, so the graph of y = x2 - x - 6 is above the x-axis on the unabridged interval (-inf, -2).
(-two, iii): 0 is in the interval. 02 - 0 - vi = -6 < 0, then the graph of y = tenii - x - 6 is below the x-axis on the entire interval.
(three, inf): 4 is in the interval. fourtwo - 4 - 6 = 6 > 0, so the graph of y = x2 - x - vi is above the ten-axis on the entire interval.
Since nosotros are looking for regions where the graph is beneath the axis, the solution set is -ii < ten < 3, or (-2, 3).
Common Error
Nosotros will use the problem in Example 8 to illustrate a common mistake.
x2 - x - vi < 0.
(10 + ii)(x - 3) < 0 OK to this point.
10 + two < 0 or ten - 3 < 0 WRONG!
When a product of two numbers is equal to 0, and then at least one of the numbers must be 0. Nevertheless, a product of 2 negative numbers is not negative, so this arroyo is not useful for solving inequalities.
Case ix.
1.two xthree + iii.07 x2 - 10 - three.71 > 0.
This problem is much more difficult than the inequality in the previous example! Information technology is not easy to cistron, and so we will not be able to discover the verbal values of the critical numbers. We volition use a graphing utility to approximate the critical numbers. The graph of the polynomial is shown below.
y = i.two x3 + 3.07 xii - 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: (-ii.35, -ane.25) union (ane.05, inf).
Do 4 :
Solve the inequality x2 + 3x - iv > 0. Employ a graphing utility to check your solution.
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Rational Inequalities
A rational expression is one of the form polynomial divided by polynomial. In full general, graphs of rational functions do take breaks. They are not defined at the zeros of the denominator. These are the only places where there are breaks, then nosotros tin utilise the same technique to solve rational inequalities that we use for polynomial inequalities.
Instance ten.
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 case, we see that the critical numbers are -3, v, and one.
The iii critical numbers divide the number line into four test intervals.
(-inf, -3): -4 is in the interval, and the rational office evaluated at -4 is -9/15. Since the value is negative, the graph of the rational office is below the x-axis throughout the interval.
(-3, i): 0 is in the interval. The value of the role at 0 is v, which is positive. The graph of the function is above the x-axis throughout the interval.
(1, five): ii is in the interval. The value at 2 is -five. The graph of the function is below the ten-centrality.
(5, inf): six is in the interval. The value at 6 is 9/15. The graph of the function is to a higher place the x-axis.
We are looking for regions where the graph is to a higher place the x-centrality, so the solution ready is (-3, 1) wedlock (v, inf).
Notation: 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 detect the exact values of the disquisitional numbers, but in one case this is done, a grapher provides a fast fashion to finish the trouble.
Graph of y = (ten + 3)(x - 5)/3(10 - 1)
There are two of import points to keep in mind when working with inequalities:
1. We demand to compare an expression to 0. So, if we outset with the trouble
xii - 3x - 11 < x + 10, we would subtract ten and x from both sides to obtain
ten2 - 4x - 21 < 0.
ii. Do not multiply both sides of an inequality by an expression with a variable.
For example, given the problem
, practice non multiply both sides by x. The correct way to handle this problem is as follows:
Now nosotros see that the disquisitional numbers are 0 (from denominator), i, and -ane.
Exercise 5 :
(a) Finish solving xii - 3x - eleven < x + ten, and check your solution with a graphing utility.
(b) Finish solving
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Source: http://dl.uncw.edu/digilib/Mathematics/Algebra/mat111hb/IZS/inequalities/inequalities.html
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