Problem 58
Question
The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the interval(s) on a number line. $$\left|x+\frac{\pi}{2}\right|>1$$
Step-by-Step Solution
Verified Answer
The solution is \((-\infty, -1-\frac{\pi}{2})\cup(1-\frac{\pi}{2}, \infty)\).
1Step 1: Understand the Absolute Value Inequality
We are given the inequality \( \left| x + \frac{\pi}{2} \right| > 1 \). The absolute value \( |A| > B \) can be rewritten as two separate inequalities: \( A > B \) or \( A < -B \).
2Step 2: Set Up the Two Inequalities
From \( \left| x + \frac{\pi}{2} \right| > 1 \), we get two inequalities: \( x + \frac{\pi}{2} > 1 \) and \( x + \frac{\pi}{2} < -1 \).
3Step 3: Solve the First Inequality
For \( x + \frac{\pi}{2} > 1 \), subtract \( \frac{\pi}{2} \) from both sides: \( x > 1 - \frac{\pi}{2} \).
4Step 4: Solve the Second Inequality
For \( x + \frac{\pi}{2} < -1 \), subtract \( \frac{\pi}{2} \) from both sides: \( x < -1 - \frac{\pi}{2} \).
5Step 5: Combine the Solutions
The solutions for the inequality are the values where \( x < -1 - \frac{\pi}{2} \) or \( x > 1 - \frac{\pi}{2} \). These give us two intervals on the number line.
6Step 6: Write the Solution in Interval Notation
The solution set in interval notation is \( (-\infty, -1 - \frac{\pi}{2}) \cup (1 - \frac{\pi}{2}, \infty) \).
7Step 7: Graph the Solution on the Number Line
Draw a number line, mark and label the points \( -1 - \frac{\pi}{2} \) and \( 1 - \frac{\pi}{2} \). Shade the regions to the left of \( -1 - \frac{\pi}{2} \) and to the right of \( 1 - \frac{\pi}{2} \). Use open circles to indicate that these points are not included in the solution.
Key Concepts
Interval NotationNumber Line RepresentationReal NumbersInequality Solutions
Interval Notation
When dealing with absolute value inequalities, expressing solutions in interval notation is crucial to effectively communicate the range of values that satisfy the inequality. Interval notation provides a concise way of showing all the numbers between a specified range, using parentheses and brackets to show which values are included or excluded.
In our original problem, after solving for the inequalities derived from the absolute value inequality \( \left| x + \frac{\pi}{2} \right| > 1 \), we end up with two separate intervals: \( (-\infty, -1 - \frac{\pi}{2}) \) and \( (1 - \frac{\pi}{2}, \infty) \).
The union of these two intervals \( (-\infty, -1 - \frac{\pi}{2}) \cup (1 - \frac{\pi}{2}, \infty) \) is used in the solution to represent all possible solutions. The parentheses indicate that the boundary values, such as \( -1 - \frac{\pi}{2} \) and \( 1 - \frac{\pi}{2} \), are not included in the solution set.
In our original problem, after solving for the inequalities derived from the absolute value inequality \( \left| x + \frac{\pi}{2} \right| > 1 \), we end up with two separate intervals: \( (-\infty, -1 - \frac{\pi}{2}) \) and \( (1 - \frac{\pi}{2}, \infty) \).
The union of these two intervals \( (-\infty, -1 - \frac{\pi}{2}) \cup (1 - \frac{\pi}{2}, \infty) \) is used in the solution to represent all possible solutions. The parentheses indicate that the boundary values, such as \( -1 - \frac{\pi}{2} \) and \( 1 - \frac{\pi}{2} \), are not included in the solution set.
Number Line Representation
A number line representation of inequality solutions visually shows the interval of values that satisfy the inequality. It is a simple, yet powerful tool to aid in understanding and analyzing mathematical concepts.
For a given inequality, like \( \left| x + \frac{\pi}{2} \right| > 1 \), after obtaining the solutions \( x < -1 - \frac{\pi}{2} \) and \( x > 1 - \frac{\pi}{2} \), we can plot these regions on a number line to gain a clear visual interpretation.
To do this, mark the critical points \( -1 - \frac{\pi}{2} \) and \( 1 - \frac{\pi}{2} \) on the line. Use open circles at these points to show they are not included in the solution. Then, shade the regions to the left of \( -1 - \frac{\pi}{2} \) and to the right of \( 1 - \frac{\pi}{2} \). This shading represents that all numbers in these intervals are included, reinforcing the concept of solutions lying outside of these critical points.
For a given inequality, like \( \left| x + \frac{\pi}{2} \right| > 1 \), after obtaining the solutions \( x < -1 - \frac{\pi}{2} \) and \( x > 1 - \frac{\pi}{2} \), we can plot these regions on a number line to gain a clear visual interpretation.
To do this, mark the critical points \( -1 - \frac{\pi}{2} \) and \( 1 - \frac{\pi}{2} \) on the line. Use open circles at these points to show they are not included in the solution. Then, shade the regions to the left of \( -1 - \frac{\pi}{2} \) and to the right of \( 1 - \frac{\pi}{2} \). This shading represents that all numbers in these intervals are included, reinforcing the concept of solutions lying outside of these critical points.
Real Numbers
Real numbers include all the numbers on the number line without any gaps. This set of numbers includes rational numbers, such as integers and fractions, and irrational numbers such as pi (\( \pi \)) and square roots of non-perfect squares.
In the given inequality \( \left| x + \frac{\pi}{2} \right| > 1 \), the solutions are expressed as intervals on the number line indicating the set of all real numbers that make the inequality true. Thus, it's important to recognize that interval solutions encapsulate a continuous set of real numbers, rather than discrete points.
The use of real numbers also underscores the continuous nature of mathematical solutions, bridging the gap between integer and rational numbers, and including values such as \( -\frac{\pi}{2} \) as part of the comprehensive set. In our solution, we venture into both the negative and positive direction infinitely, showcasing the vast nature of real numbers beyond the confined finite integers.
In the given inequality \( \left| x + \frac{\pi}{2} \right| > 1 \), the solutions are expressed as intervals on the number line indicating the set of all real numbers that make the inequality true. Thus, it's important to recognize that interval solutions encapsulate a continuous set of real numbers, rather than discrete points.
The use of real numbers also underscores the continuous nature of mathematical solutions, bridging the gap between integer and rational numbers, and including values such as \( -\frac{\pi}{2} \) as part of the comprehensive set. In our solution, we venture into both the negative and positive direction infinitely, showcasing the vast nature of real numbers beyond the confined finite integers.
Inequality Solutions
Solving absolute value inequalities requires transforming the original inequality into one or more simpler inequalities. Understanding the underlying principles helps solve this type of inequality more efficiently.
In the exercise \( \left| x + \frac{\pi}{2} \right| > 1 \), the absolute value is split into two separate inequalities: \( x + \frac{\pi}{2} > 1 \) and \( x + \frac{\pi}{2} < -1 \). This effectively divides the inequality into scenarios that can be more easily managed. Each solution path exposes different possibilities, and solving both provides a full range of solutions.
After solving these inequalities separately and combining their solutions, it demonstrates the techniques of managing absolute values that encounter different real-number solutions. Linear inequality solutions offer critical insights for graphing and understanding where real-world application constraints are effective.
In the exercise \( \left| x + \frac{\pi}{2} \right| > 1 \), the absolute value is split into two separate inequalities: \( x + \frac{\pi}{2} > 1 \) and \( x + \frac{\pi}{2} < -1 \). This effectively divides the inequality into scenarios that can be more easily managed. Each solution path exposes different possibilities, and solving both provides a full range of solutions.
After solving these inequalities separately and combining their solutions, it demonstrates the techniques of managing absolute values that encounter different real-number solutions. Linear inequality solutions offer critical insights for graphing and understanding where real-world application constraints are effective.
Other exercises in this chapter
Problem 57
Solve each equation for \(x\) in terms of the other letters. $$(a x+b)^{2}-(b x+a)^{2}=0, \text { where } a \neq \pm b$$
View solution Problem 57
Give an example of irrational numbers a and b such that the indicated expression is (a) rational and (b) irrational. $$a b$$
View solution Problem 58
(a) Verify that the point (3,7) is on the circle $$x^{2}+y^{2}-2 x-6 y-10=0$$ (b) Find the equation of the line tangent to this circle at the point \((3,7) .\)
View solution Problem 58
Solve each equation for \(x\) in terms of the other letters. $$(x-p)^{2}+(x-q)^{2}=p^{2}+q^{2}$$
View solution