Problem 33 - Calculus: One and Several Variables

Question

Solve the inequality and express the solution set as an interval or as the union of intervals. $$|2 x+1| < \frac{1}{4}$$.

Step-by-Step Solution

Answer

The solution to the inequality is \((-\frac{5}{8}, -\frac{3}{8})\)

Step 1 - Resolve the inequality into two separate cases

Absolute value denotes the distance of a number from zero, therefore, it can either be a positive or a negative number. We can resolve \(|2x + 1| < \frac{1}{4}\) into two separate cases, namely, \(2x + 1 < \frac{1}{4}\) and \(2x + 1 > -\frac{1}{4}\)

Step 2 - Solve for x in both cases

Solving the inequality \(2x + 1 < \frac{1}{4}\), we subtract 1 from both sides to get \(2x < \frac{1}{4} - 1 = -\frac{3}{4}\). Therefore x < -\frac{3}{8}. Similarly, for the inequality \(2x + 1 > -\frac{1}{4}\), we subtract 1 from both sides to get \(2x > -\frac{1}{4} - 1 = -\frac{5}{4}\). Therefore x > -\frac{5}{8}.

Step 3 - Express as an interval

Values of x that satisfy both inequalities are the solution set. Expressing this in interval notation gives \((-\frac{5}{8}, -\frac{3}{8})\) which indicates that x is greater than -\frac{5}{8} and less than -\frac{3}{8}

Key Concepts

Interval Notation

Interval notation is a way of expressing the set of all solutions to an inequality using parentheses and brackets to specify the range of numbers. Parentheses, like \(a, b\), indicate that the endpoints are not included. Meanwhile, brackets, like [a, b], show that the endpoints are included.
When we solved the inequality \(|2x + 1| < \frac{1}{4}\), we got the solution set \((-\frac{5}{8}, -\frac{3}{8})\). This interval notation tells us that the value of \(x\) lies between \(-\frac{5}{8}\) and \(-\frac{3}{8}\) exclusively, meaning the endpoints are not part of the solution.
Understanding how to express answer sets in interval notation is crucial when solving inequalities because it concisely communicates which values satisfy the inequality.

Solving Inequalities

Solving inequalities is similar to solving equations but with some key differences, especially regarding how inequalities change direction. When you multiply or divide both sides of an inequality by a negative number, you must flip the inequality sign. To solve \(|2x + 1| < \frac{1}{4}\), we need to interpret the absolute value condition first.
This results in two separate inequalities: \(2x + 1 < \frac{1}{4}\) and \(2x + 1 > -\frac{1}{4}\). Each inequality gives a range of values for \(x\). Solving each involves traditional methods:

• Subtract any constants from both sides of the inequality to isolate terms involving \(x\).
• If necessary, divide or multiply to solve for \(x\).

For example, from \(2x + 1 < \frac{1}{4}\), we'd subtract 1 to get \(2x < -\frac{3}{4}\), leading to \(x < -\frac{3}{8}\) after dividing by 2. This systematic approach helps us find ranges of \(x\) that satisfy the original inequality.

Absolute Value

The absolute value of a number is its distance from zero on a number line, regardless of direction. It's always a non-negative value. The absolute value has crucial applications in inequalities, especially when combined with symbols like <, >, \(\leq\), or \(\geq\). For the inequality \(|2x + 1| < \frac{1}{4}\), the absolute value notation indicates we're looking at how far the expression \(2x + 1\) is from zero.
Breaking it down, \(|a| < b\) implies two conditions: \(a < b\) and \(a > -b\). In our problem, this set the stage for creating the inequalities \(2x + 1 < \frac{1}{4}\) and \(2x + 1 > -\frac{1}{4}\).
Comprehending absolute value helps one manage the two-output nature of expressions like \(|2x + 1|\), ensuring both potential negative and positive differences from zero are considered when solving inequalities.