Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the solution sets of the given inequalities. $$ |x-2| \geq 5 $$

Step-by-Step Solution

Answer

The solution set is \(x \leq -3\) or \(x \geq 7\).

Step 1: Understand the Inequality

The inequality given is \(|x-2| \geq 5\). This means the absolute value of \(x-2\) is greater than or equal to 5. The solution to the inequality involves finding all \(x\) values that satisfy this condition.

Step 2: Rewrite the Absolute Value Inequality

The property of absolute values shows that \(|A| \geq B\) translates into two separate inequalities: \(A \geq B\) or \(A \leq -B\). Applying this to \(|x-2| \geq 5\), we create two inequalities: \(x-2 \geq 5\) or \(x-2 \leq -5\).

Step 3: Solve the First Inequality

Consider the inequality \(x-2 \geq 5\):1. Add 2 to both sides: \[x-2 + 2 \geq 5 + 2\] \[x \geq 7\]This gives us one part of the solution set.

Step 4: Solve the Second Inequality

Consider the inequality \(x-2 \leq -5\):1. Add 2 to both sides: \[x-2 + 2 \leq -5 + 2\] \[x \leq -3\]This provides the second part of the solution set.

Step 5: Combine the Solution Sets

The solutions for the inequality \(|x-2| \geq 5\) are combined into two intervals based on the two separate inequalities:- \(x \geq 7\)- \(x \leq -3\)Therefore, the solution set is \(x \leq -3\) or \(x \geq 7\).

Key Concepts

Solution Sets

A solution set is a collection of values that satisfies a given inequality or equation. Basically, it contains all the possible answers that make our inequality true.
For the inequality \(|x-2| \geq 5\), the solution set includes all the values of \(x\) that fulfill the absolute value inequality.
When an inequality is present, particularly with absolute values, we typically end up with more than one possibility, which is why solution sets are often expressed as intervals.

• In our example, we found that the solution set is \(x \leq -3\) or \(x \geq 7\). These two intervals show every \(x\) value that makes the inequality true.
• This means: "Any number less than or equal to \(-3\), and any number greater than or equal to \(7\), is a solution."

These intervals clearly indicate the groups of numbers where our inequality holds true. This way, you can easily pick any number from these intervals and verify it will work in the original inequality.

Inequality Solving

Inequality solving is a crucial part of working with mathematical statements that include comparison symbols like \(>\), \(<\), \(\leq\), and \(\geq\). When solving inequalities, we aim to find the values that satisfy the given statement.
The process involves similar steps to solving equations, but with special rules depending on the inequality symbol.

• Just like equations, you can add, subtract, multiply, and divide both sides of an inequality by the same number. However, remember that multiplying or dividing by a negative number reverses the inequality sign.
• For instance, if we solve \(x-2 \geq 5\) by adding \(2\) to both sides, we find \(x \geq 7\).

When you have an inequality involving an absolute value, you'll end up with two separate inequalities to solve. This results in a solution set showing all the possible values for the variable in question.

Absolute Values

Absolute values are interesting, as they always provide the distance a number is from zero on a number line, regardless of its direction. This means an absolute value is always zero or positive.
When an inequality involves an absolute value, it translates into two separate scenarios to consider. This is due to the property of absolute values handling both positive and negative cases of a value.

• The expression \(|x-2| \geq 5\) indicates that \((x-2)\) must be at least 5 units away from zero, either to the right (positive) or to the left (negative).
• As a result, you write two inequalities: \(x-2 \geq 5\) and \(x-2 \leq -5\).

This process ensures that you capture the full range of values \(x\) can take to satisfy the original inequality. Understanding this property helps in solving more complex absolute value inequalities effectively.