Chapter 1: Problem 25
Solve the inequality. Express the answer using interval notation. $$ |x+1| \geq 1 $$
Short Answer
Expert verified
The solution is \((-\infty, -2] \cup [0, \infty)\).
Step by step solution
01
Understand Absolute Value Inequality
An absolute value inequality of the form \(|x+a| \geq b\) can be rewritten as two separate inequalities: \(x+a \geq b\) or \(x+a \leq -b\). Our task is to solve for \(x\) in these two cases.
02
Rewrite Using Two Cases
Given the inequality \(|x+1| \geq 1\), rewrite it as two separate cases: 1. \(x+1 \geq 1\)2. \(x+1 \leq -1\)
03
Solve First Inequality
Solve \(x+1 \geq 1\): Subtract 1 from both sides to isolate \(x\).\[x + 1 - 1 \geq 1 - 1\]\[x \geq 0\]
04
Solve Second Inequality
Solve \(x+1 \leq -1\): Subtract 1 from both sides to isolate \(x\).\[x + 1 - 1 \leq -1 - 1\]\[x \leq -2\]
05
Combine Results Using Union
The solutions from both inequalities are combined using a union because the absolute value inequality uses "greater than or equal to." Thus, the solution is the set of all \(x\) such that \(x \geq 0\) or \(x \leq -2\).
06
Express Solution in Interval Notation
Express the combined solution in interval notation. From the solutions of the two inequalities:- \(x \geq 0\) corresponds to the interval \([0, \infty)\).- \(x \leq -2\) corresponds to the interval \((-\infty, -2]\).Combine them to form the final answer: \[(-\infty, -2] \cup [0, \infty)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interval Notation
Interval notation is a shorthand way of expressing a range of numbers along the number line. It's a critical skill in mathematics and helps in communicating solutions to inequalities efficiently.
In interval notation:
When using interval notation to express the solution of our inequality \( |x+1| \geq 1 \), we see that the solution consists of two parts:
In interval notation:
- Brackets [ ] are used when the end number is included in the set, known as a closed interval.
- Parentheses ( ) are used when the end number is not included, known as an open interval.
When using interval notation to express the solution of our inequality \( |x+1| \geq 1 \), we see that the solution consists of two parts:
- One interval for \( x \geq 0 \), written as \( [0, \infty) \)
- Another interval for \( x \leq -2 \), written as \( (-\infty, -2] \)
Inequality Solution
Solving inequalities involves finding the range of values that satisfy a given condition. When we deal with absolute value inequalities like \( |x+1| \geq 1 \), we're looking for values of \( x \) that make this expression true.
The absolute value expression \( |x+1| \) can represent two scenarios:
The absolute value expression \( |x+1| \) can represent two scenarios:
- When the expression inside the absolute value is non-negative, i.e., \( x+1 \geq 0 \).
- When the expression is negative, i.e., \( x+1 < 0 \).
- Case 1: \( x+1 \geq 1 \), which simplifies to \( x \geq 0 \).
- Case 2: \( x+1 \leq -1 \), which simplifies to \( x \leq -2 \).
Algebraic Expression
An algebraic expression is a combination of numbers, symbols, and operators (like +, -, \, *, etc.) that represent a specific value.
The expression within our inequality, \( x+1 \), is an example of a simple algebraic expression. It represents the values that \( x \) can take when modified by the addition of 1.
Algebraic expressions are fundamental in solving inequalities because they:
To solve, we split the absolute value expression into two scenarios:
The expression within our inequality, \( x+1 \), is an example of a simple algebraic expression. It represents the values that \( x \) can take when modified by the addition of 1.
Algebraic expressions are fundamental in solving inequalities because they:
- Allow us to manipulate and simplify equations and inequalities.
- Help translate word problems into mathematical statements.
- Provide a method to evaluate different scenarios through substitution.
To solve, we split the absolute value expression into two scenarios:
- Firstly, considering if \( x+1 \) is greater than or equal to the right side, leading to \( x \geq 0 \).
- Secondly, considering if \( -(x+1) \) is less than or equal to the right side, leading to \( x \leq -2 \).