Problem 28 - Algebra and Trigonometry

Question

Solve the inequality. Express the answer using interval notation. $$ |x+1| \geq 3 $$

Step-by-Step Solution

Answer

The solution is (-\infty, -4] \cup [2, \infty).

Step 1: Understand Absolute Value Inequality

The inequality \(|x+1| \geq 3\) means that the expression inside the absolute value, \(x+1\), can be either greater than or equal to 3 or less than or equal to -3. We need to solve both these cases separately.

Step 2: Solve for \(x+1 \geq 3\)

Firstly, solve the inequality \(x + 1 \geq 3\). Subtract 1 from both sides to isolate \(x\).\[\begin{align*}x + 1 &\geq 3 \x &\geq 2 \end{align*}\]

Step 3: Solve for \(x+1 \leq -3\)

Next, solve the inequality \(x + 1 \leq -3\) by subtracting 1 from both sides.\[\begin{align*}x + 1 &\leq -3 \x &\leq -4 \end{align*}\]

Step 4: Combine Solutions and Express in Interval Notation

Combine the solutions from Steps 2 and 3. The solution for \(|x+1| \geq 3\) is the union of the intervals from Step 2, \(x \geq 2\), and Step 3, \(x \leq -4\). In interval notation, this is \[(-\infty, -4] \cup [2, \infty)\]

Key Concepts

Interval Notation

Interval notation is a way of representing a set of numbers on the number line. It is especially useful in conveying the solutions to inequalities. When we solve an inequality, the answer is often a range of numbers rather than a single number, and interval notation helps convey these ranges concisely.

An interval can be "open" or "closed" at either end.

• **Open intervals** do not include their endpoints and are denoted with parentheses, for instance, \((a, b)\), meaning the values between \(a\) and \(b\), but not \(a\) or \(b\) themselves.
• **Closed intervals** include their endpoints and are noted with brackets, like \([a, b]\), which includes all the values from \(a\) to \(b\), including \(a\) and \(b\).
• Mixed intervals can have one endpoint included and the other excluded, such as \((a, b]\) or \([a, b)\).

When dealing with infinity, we always use parentheses, for example \((-fty, x)\).

By solving inequalities like the one in the exercise, we find solutions that we can write using interval notation. Here, the inequality with absolute value gives us two separate intervals, \((-fty, -4]\) and \([2, fty)\), meaning all numbers less than or equal to \(-4\) and all numbers greater than or equal to \(2\).

Solving Inequalities

Solving inequalities involves finding which values make the inequality true. Basically, inequalities compare the sizes of two expressions—whether one is less than, greater than, or not equal to the other.

The inequality \(|x+1| \geq 3\) implies considering two cases, as it involves an absolute value, which measures the distance of a number from zero. The expression inside the absolute value, \(x+1\), can lead us to two possible inequalities:

• The positive case: \(x+1 \geq 3\), meaning the expression inside is greater than or equal to 3.
• The negative case: \(x+1 \leq -3\), means it's less than or equal to \(-3\).

Each case is solved separately, just like you would solve any other linear inequality. For \(x+1 \geq 3\), subtracting 1 from each side results in \(x \geq 2\). For \(x+1 \leq -3\), you again subtract 1, yielding \(x \leq -4\). These steps help gather the solutions, which can be written neatly in interval notation.

Union of Intervals

The union of intervals is used when combining multiple solution sets into one comprehensive answer. In mathematics, the union symbol \(\cup\) is used to denote the combination of two sets where either condition can satisfy the total solution.

In the problem given, the inequality \(|x+1| \geq 3\) resulted in two conditions:

• One condition is \(x \geq 2\).
• The other is \(x \leq -4\).

Since both conditions must be included to satisfy the inequality, the results are combined using a union.

Therefore, the solution can be expressed as the union of two intervals: \((-\infty, -4]\) and \([2, \infty)\), illustrated as \((-\infty, -4] \cup [2, \infty)\). It encapsulates all numbers \(x\) wherein the expression \(|x+1| \geq 3\) holds true, covering all numbers within \(x \leq -4\) together with those where \(x \geq 2\). This understanding of union is crucial in handling solutions where multiple conditions might be correct simultaneously.