Problem 30 - Thomas Calculus

Question

Solve the inequalities in Exercises \(19-34,\) expressing the solution sets as intervals or unions of intervals. Also, show each solution set on the real line. $$ |s+3| \geq \frac{1}{2} $$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -\frac{7}{2}] \cup [-\frac{5}{2}, \infty)\).

Step 1: Understanding Absolute Value Inequality

The inequality \(|s+3| \geq \frac{1}{2}\) states that the distance of \(s+3\) from 0 is at least \(\frac{1}{2}\). This means \(s+3\) can be greater than or equal to \(\frac{1}{2}\) or less than or equal to \(-\frac{1}{2}\).

Step 2: Break Down into Two Inequalities

The inequality \(|s+3| \geq \frac{1}{2}\) can be split into two separate inequalities: 1. \(s+3 \geq \frac{1}{2}\)2. \(s+3 \leq -\frac{1}{2}\)

Step 3: Solve the First Inequality

For \(s+3 \geq \frac{1}{2}\), subtract 3 from both sides: \[ s \geq \frac{1}{2} - 3 \]Simplifying this, we get:\[ s \geq -\frac{5}{2} \]

Step 4: Solve the Second Inequality

For \(s+3 \leq -\frac{1}{2}\), subtract 3 from both sides: \[ s \leq -\frac{1}{2} - 3 \]Simplifying this, we get:\[ s \leq -\frac{7}{2} \]

Step 5: Combine the Solution Sets

The solution to the inequality is the union of the two inequalities obtained: \[ s \leq -\frac{7}{2} \] or \[ s \geq -\frac{5}{2} \].The solution set can be written in interval notation as:\((-\infty, -\frac{7}{2}] \cup [-\frac{5}{2}, \infty)\).

Step 6: Represent the Solution on the Number Line

On the number line, shade the region to the left of \(-\frac{7}{2}\) and to the right of \(-\frac{5}{2}\), including both points because they are part of the solution (closed intervals).

Key Concepts

Absolute Value Inequalities

Absolute value inequalities can seem daunting, but they're just a way of expressing distances on a number line. An inequality like \(|s+3| \geq \frac{1}{2}\) implies that the distance of \(s+3\) from zero is at least \(\frac{1}{2}\). This translates into two possible scenarios.

• The expression \(s+3\) is greater than or equal to \(\frac{1}{2}\). This means that \(s+3\) is positive and at least \(\frac{1}{2}\) units away from zero in the positive direction.
• The expression \(s+3\) is less than or equal to \(-\frac{1}{2}\). Here, \(s+3\) is negative, but is still at least \(\frac{1}{2}\) units away from zero, just in the opposite direction.

Breaking down the absolute value inequality involves converting it into two separate inequalities, making it straightforward to handle.Understanding these conditions helps us see both possibilities where our solution could lie. By solving both parts separately, you capture all numbers that work for the original inequality. Everything coming from these two inequalities is valid for solving \(|s+3| \geq \frac{1}{2}\).

Interval Notation

Interval notation is a concise way to express ranges of numbers that satisfy an inequality. When your solution involves many numbers, such as those found using absolute value inequalities, this notation becomes quite useful. For the inequality \(|s+3| \geq \frac{1}{2}\), after you solve the two inequalities, you have:

• \(s \geq -\frac{5}{2}\)
• \(s \leq -\frac{7}{2}\)

These are combined to form: \((-\infty, -\frac{7}{2}] \cup [-\frac{5}{2}, \infty)\).- This interval notation lets you read these sets easily: - \((-\infty, -\frac{7}{2}]\) suggests all numbers less than or equal to \(-\frac{7}{2}\). - \([-\frac{5}{2}, \infty)\) indicates numbers greater than or equal to \(-\frac{5}{2}\).- The union symbol \(\cup\) simply means you're taking all numbers from both intervals.Interval notation provides a clear picture of solution sets without drawing a number line just yet, simplifying the communication of ranges in mathematical problems.

Real Number Line

The real number line is a visual tool that helps us understand and represent the solution of inequalities. When you solve an inequality like \(|s+3| \geq \frac{1}{2}\), the solutions are spread out over this imaginary line stretching infinitely in both directions.On this line, each number corresponds to a unique point. Here's how you can use it practically:

• Determine the critical points. From our inequalities, these points are \(-\frac{7}{2}\) and \(-\frac{5}{2}\).
• Draw a line, marking these numbers as points of interest.
• Shade the areas that belong to your solution set. For this case:
  • Shade to the left of \(-\frac{7}{2}\) and include the point, as indicated by the closed bracket in the interval \((-\infty, -\frac{7}{2}]\).
  • Shade to the right of \(-\frac{5}{2}\) and include that point as well, matching the closed bracket in the interval \([-\frac{5}{2}, \infty)\).

The shaded sections on the number line visually show all solutions to the inequality. This illustration aids in making abstract numbers more tangible, helping you ensure no numbers have been omitted or misunderstood.