How To Find The Solution Set Of An Inequality Graph

Solving Inequalities

Contents: This page corresponds to § 2.five (p. 216) of the text.

Linear Inequalities

An inequality is a comparison of expressions past 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 nosotros use <= and >= for these relations.

Example 1. 10 + 3 <= 10

A solution for an inequality in ten is a number such that when we substitute that number for x we have a truthful statement. And then, 4 is a solution for example 1, while eight is non. The solution ready of an inequality is the set up of all solutions. Typically an inequality has infinitely many solutions and the solution set is hands described using interval notation.

The solution fix of example i is the set of all x <= 7. In interval annotation this prepare is (-inf, 7], where nosotros use inf to correspond infinity.

A linear inequality is one such that if we replaced the inequality with the equals relation, then nosotros would have a linear equation. Solving linear inequalities is very much like solving linear equations, with 1 important departure.

When you multiply or split both sides of an inequality past a negative number, the direction of the inequality is reversed.

You can see this using an inequality with no variables.

Example ii.

3 < seven. This is TRUE.

(three)(-2) < (7)(-2). This is Fake, because -vi is to the right of -14 on the number line. Hence, -vi > -fourteen.

(three)(-2) > (seven)(-2). This is TRUE. And then, when nosotros multiply the original inequality by -2, we must reverse the direction to obtain another truthful argument.

Annotation: In general nosotros may not multiply or divide both sides of an inequality by an expression with a variable, because some values of the variable may brand the expression positive and some may make information technology negative.

Example 3.

vii - 2x < iii.

-2x < -4.

x > 2.

Note: When we divided both sides of the inequality past -2 nosotros changed the direction of the inequality.

Wait at the graphs of the functions on either side of the inequality.

To satisfy the inequality, seven - 2x needs to exist less than 3. So nosotros are looking for numbers ten such that the point on the graph of y = vii - 2x is below the point on the graph of y = 3. This is truthful for 10 > 2. In interval notation the solution set is (ii, inf).

There is another mode to employ a graphing utility to solve this inequality. In the Java Grapher the expression (7-ii*x)L3 has the value ane for numbers x that satisfy the inequality, and the value 0 for other numbers 10. The picture below shows the graph of (7-two*ten)L3 as drawn past the Grapher.

Exercise i:

Solve the inequality 4 - x > 1 + 3x. Answer

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Combinations of Inequalities

Case 4.

Find all numbers x such that -3 < 5 - 2x and 5 - 2x < 9.

-three < 5 - 2x
-8 < -2x

4 > ten

(-inf, iv)

AND

5 - 2x < nine
-2x < 4

x > -ii

(-2, inf)

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 2 solution sets, which is the prepare (-2, 4) in interval notation.

The problem higher up is ordinarily written as a double inequality.

-3 < 5 - 2x < 9 stands for -3 < 5 - 2x and 5 - 2x < ix.

Note: When nosotros solved the two inequalities separately, the steps in the 2 problems were the same. Therefore, the double inequality notation may be used to solve the inequalities simultaneously.

-iii < 5 - 2x < ix.

-8 < -2x < 4.

4 > x > -two.

In terms of graphs, this problem corresponds to finding the values of x such that the corresponding indicate on the graph of y = 5 - 2x is between the graphs of y = -iii and y = 9.

Example five.

Find all numbers x such that x + 1 < 0 or x + 1 > 3.

In Example 4 to 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 union of solution sets instead of an intersection.

Do not use the double inequality notation in this situation.

10 + ane < 0
x < -1

(-inf, -1)
OR 10 + 1 > 3
x > ii

(two, inf)

The solution set is the union of the two intervals (-inf, -ane) and (2, inf).

Exercise 2:

(a) 1 < three + 5x < 7 Answer

(b) ii - x < 1, or 2 - x > 5 Reply

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Inequalities Involving Absolute Values

Inequalities involving absolute values can exist rewritten equally combinations of inequalities.

Let a be a positive number.

|ten| < 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, call back about a number line. The absolute value of a number is the altitude the number is from 0 on the number line. Then the inequality |ten| < a is satisfied by numbers whose distance from 0 is less than a. This is the fix of numbers between -a and a.

The inequality |x| > a is satisfied by numbers whose altitude from 0 is larger than a. This means numbers that are either larger than a, or less than -a.

Instance 6.

| three + 2x | <= 7.

-seven <= 3 + 2x <= 7.

-10 <= 2x <= 4.

-five <= ten <= ii.

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 = | iii + 2x | is either beneath or equal to the signal on the graph of y = 7.

Example vii.

| 5 - 2x | > 3.

five - 2x < -3 or five - 2x > three.

-2x < -8 or -2x > -two.

x > 4 or x < 1.

x is in (four, 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.

Exercise 3 :

Solve the following inequalities. Utilise a graphing utility to check your answers.

(a) | three + ten | < 4.

(b) | 2 - x | > three.

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Polynomial Inequalities

Example eight.

x² - x - 6 < 0.

The first step is to find the zeros of the polynomial x² - x - vi.

tenⁱⁱ - x - 6 = 0.

(10 + 2)(x - 3) = 0.

x = -2, or ten = three.

-two and 3 are called the disquisitional numbers of the inequality.

Note: -2 and 3 are not in the solution ready 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 10-axis. The disquisitional numbers -two and 3 are the places where the graph intersects the x-axis.

The critical numbers divide the 10-axis into three intervals chosen exam intervals for the inequality.

Test intervals: (-inf, -2), (-two, iii), (3, inf).

We are going to use the fact that polynomial functions are continuous. This means that their graphs do non have whatsoever breaks or jumps.

Since we have found all the x-intercepts of the graph of ten² - ten - vi, throughout each examination interval the graph must exist either above the x-axis or below it. This is where nosotros need to know that the graph does not have whatsoever breaks. This means that we may choose any number nosotros like in a test interval and evaluate the polynomial at that number to see if the graph is in a higher place or below the ten-axis throughout that examination interval.

(-inf, -two): -5 is in the interval. (-5)² - (-5) - six = 24 > 0, and so the graph of y = xⁱⁱ - x - 6 is higher up the ten-centrality on the unabridged interval (-inf, -2).

(-ii, 3): 0 is in the interval. 0ⁱⁱ - 0 - 6 = -6 < 0, so the graph of y = x² - x - six is below the x-axis on the entire interval.

(3, inf): iv is in the interval. 4² - four - six = 6 > 0, so the graph of y = ten² - x - half-dozen is to a higher place the x-axis on the unabridged interval.

Since nosotros are looking for regions where the graph is beneath the axis, the solution prepare is -2 < x < 3, or (-2, 3).

Mutual Fault

Nosotros volition use the problem in Example 8 to illustrate a common mistake.

tenⁱⁱ - x - vi < 0.

(ten + 2)(x - iii) < 0 OK to this point.

x + 2 < 0 or x - three < 0 WRONG!

When a production of ii numbers is equal to 0, then at to the lowest degree one of the numbers must exist 0. Nonetheless, a production of two negative numbers is not negative, so this approach is not useful for solving inequalities.

Case ix.

1.2 xⁱⁱⁱ + 3.07 x² - 10 - 3.71 > 0.

This problem is much more hard than the inequality in the previous example! It is non piece of cake to factor, and then nosotros will not be able to find the exact values of the disquisitional numbers. Nosotros will use a graphing utility to judge the critical numbers. The graph of the polynomial is shown beneath.

y = 1.two x³ + 3.07 10² - x - 3.71

The critical numbers are approximately -2.35, -1.25, and 1.05. In this trouble we looking for regions where the graph is to a higher place the axis.

Solution Fix: (-2.35, -i.25) spousal relationship (1.05, inf).

Exercise 4 :

Solve the inequality 10ⁱⁱ + 3x - 4 > 0. Use a graphing utility to check your solution.

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Rational Inequalities

A rational expression is ane of the form polynomial divided past polynomial. In general, graphs of rational functions do accept breaks. They are non divers at the zeros of the denominator. These are the merely places where there are breaks, so we can employ the aforementioned technique to solve rational inequalities that we use for polynomial inequalities.

Example 10.

The disquisitional 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 disquisitional numbers are -3, five, and 1.

The three disquisitional numbers split the number line into iv test intervals.

(-inf, -three): -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, one): 0 is in the interval. The value of the function at 0 is 5, which is positive. The graph of the role is above the x-centrality throughout the interval.

(one, v): ii is in the interval. The value at 2 is -5. The graph of the function is below the x-axis.

(5, inf): half-dozen is in the interval. The value at six is 9/xv. The graph of the function is above the x-axis.

We are looking for regions where the graph is in a higher place the 10-axis, so the solution set up is (-3, 1) union (v, inf).

Note: A graphing utility tin can exist used to see which side of the 10-axis the graph is on over the diverse exam intervals. In some cases y'all must solve algebraically to notice the exact values of the disquisitional numbers, but once this is done, a grapher provides a fast way to finish the problem.

Graph of y = (x + iii)(10 - v)/3(x - ane)

At that place 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

xⁱⁱ - 3x - 11 < x + 10, we would subtract 10 and 10 from both sides to obtain

ten² - 4x - 21 < 0.

2. Do not multiply both sides of an inequality by an expression with a variable.

For example, given the problem , practice not multiply both sides by x. The correct fashion to handle this problem is as follows:

Now nosotros see that the disquisitional numbers are 0 (from denominator), 1, and -i.

Exercise 5 :

(a) Stop solving x² - 3x - 11 < 10 + 10, and check your solution with a graphing utility.

(b) Finish solving , and check your solution with a graphing utility.

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