Question

Find the solution sets of the given inequalities. $$ |2 x-1|>2 $$

Short answer

The solution set is \(x < -\frac{1}{2}\) or \(x > \frac{3}{2}\).

Step-by-step solution

Understand the Absolute Value Inequality

The inequality given is \(|2x - 1| > 2\). This means we have to consider two separate cases because the absolute value expression, \(|x|\), can be interpreted as the distance of \(x\) from zero on the number line.

Set Up the Two Inequalities

For an absolute value inequality of the form \(|A| > B\), the inequality splits into two cases: \(A > B\) or \(A < -B\). In our case, this results in the two inequalities: \(2x - 1 > 2\) and \(2x - 1 < -2\).

Solve the First Inequality

Starting with \(2x - 1 > 2\), add 1 to both sides to obtain: \(2x > 3\). Then, divide both sides by 2 to solve for \(x\): \(x > \frac{3}{2}\).

Solve the Second Inequality

Now consider \(2x - 1 < -2\). Add 1 to both sides to get: \(2x < -1\). Next, divide by 2 to solve for \(x\): \(x < -\frac{1}{2}\).

Combine the Solution Sets

The solutions \(x > \frac{3}{2}\) and \(x < -\frac{1}{2}\) indicate that the solution consists of two separate intervals on the number line, not a continuous range. The combined solution set is therefore the union: \(x < -\frac{1}{2}\) or \(x > \frac{3}{2}\).

Key concepts

Absolute Value Inequality

An absolute value inequality involves an expression within absolute value symbols and an inequality sign. The expression we are working with is \(|2x - 1| > 2\).

Absolute value \( |x| \) represents the distance of any number \(x\) from 0 on the number line. Because it's a distance, it's always non-negative.

When solving absolute value inequalities like \( |A| > B \), it means that \(|A|\) is greater than \(B\) on the number line.

It translates to two separate conditions: \(A > B\) or \(A < -B\). These conditions explore both positive and negative possibilities that could satisfy the inequality.

This kind of splitting into two separate linear inequalities is a vital step in handling absolute value inequalities efficiently.

Solution Sets

A solution set consists of all values that satisfy a given inequality or equation. In our case, it's the values of \(x\) that satisfy the inequality \(|2x - 1| > 2\).

Starting with each condition from the absolute value inequality:

• For \(2x - 1 > 2\), solve for \(x\) to get \(x > \frac{3}{2}\).
• For \(2x - 1 < -2\), solve to find \(x < -\frac{1}{2}\).

These represent the solution set of values \(x\) that make each inequality true.

Since the inequality is in the form of "greater than," the solutions form two separate intervals where the expression is greater than 2. Each interval represents a distinct part of the number line where the condition holds true.

Interval Notation

Interval notation succinctly expresses the solution set of an inequality.

For the solved values of \(x\) in our problem, interval notation helps in showing the segments of the number line where these values lie.

The two inequalities give us:

• For \(x < -\frac{1}{2}\), the interval is \((-\infty, -\frac{1}{2})\).
• For \(x > \frac{3}{2}\), the interval is \((\frac{3}{2}, \infty)\).

Combining these using the union symbol, the complete solution in interval notation is: \((-\infty, -\frac{1}{2}) \cup (\frac{3}{2}, \infty)\).

This notation is compact and widely used because it clearly indicates ranges of values and includes all possible numbers within those intervals.

Inequalities in Calculus

Inequalities are a foundational concept in calculus, essential for defining ranges and behaviors of functions.

Understanding inequalities helps in determining:

• Where a function increases or decreases.
• The constraints within which a function operates.
• Regions where solutions begin or stop.

For instance, when dealing with critical points or understanding limits, inequalities are necessary to approximate intervals where functions maintain specific properties.

In particular, handling absolute value inequalities allows us to define and solve problems involving distance and deviations, often encountered in real-world applications. By mastering these concepts, one builds a foundation for more advanced topics such as limits, derivatives, and the area under a curve.