Problem 43 - Intermediate Algebra

Question

Solve each inequality. $$ \frac{2 x-1}{x} \geq 0 $$

Step-by-Step Solution

Answer

The solution is \(( -\infty, 0 )\cup [\frac{1}{2}, \infty)\).

Step 1: Finding Critical Points

To find the critical points, identify where the numerator and the denominator of the rational inequality equal zero. For the numerator, set \(2x - 1 = 0\), leading to \(x = \frac{1}{2}\). The denominator \(x = 0\) is another critical point since division by zero is undefined.

Step 2: Divide Number Line into Intervals

Using the critical points \(x = \frac{1}{2}\) and \(x = 0\), divide the number line into intervals \((-\infty, 0)\), \((0, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\). These intervals will help in testing the inequality in each interval.

Step 3: Test Each Interval

Choose a test point from each interval and substitute it into the inequality. For \((-\infty, 0)\), choose \(x = -1\): \(\frac{2(-1) - 1}{-1} = \frac{-3}{-1} = 3\), which is greater than 0. For \((0, \frac{1}{2})\), choose \(x = \frac{1}{4}\): \(\frac{2(\frac{1}{4}) - 1}{\frac{1}{4}} = \frac{0.5 - 1}{0.25} = \frac{-0.5}{0.25} = -2\), which is less than 0. For \((\frac{1}{2}, \infty)\), choose \(x = 1\): \(\frac{2(1) - 1}{1} = \frac{1}{1} = 1\), which is greater than 0.

Step 4: Consider the Critical Points

Check if the critical points satisfy the inequality. For \(x = \frac{1}{2}\), substitute into inequality: \(\frac{2(\frac{1}{2}) - 1}{\frac{1}{2}} = \frac{0}{\frac{1}{2}} = 0\), which meets the \(\geq 0\) condition. However, \(x = 0\) is not part of the solution as it makes the denominator zero and is undefined.

Step 5: Write the Solution Set

Based on the tests, the solution set includes the intervals where the inequality holds true. Thus, the solution is \(( -\infty, 0 )\cup [\frac{1}{2}, \infty)\), where \([\frac{1}{2}\) includes the boundary point that satisfies the inequality.

Key Concepts

Critical Points

Critical points are vital in solving rational inequalities. They are the values of the variable that make the numerator or the denominator equal to zero. To find them, first set the numerator equal to zero: - For the inequality \( \frac{2x-1}{x} \geq 0 \), the numerator is \( 2x-1 \). Solving \( 2x-1 = 0 \) gives \( x = \frac{1}{2} \) as one critical point. Next, consider the denominator. Since division by zero is undefined, find where the denominator equals zero:- In our example, the denominator \( x = 0 \) is another critical point. These critical points are used to divide the number line into intervals for testing. Remember:

• The critical points where the numerator equals zero are potential solutions because they could satisfy the inequality.
• However, where the denominator equals zero, the point is not a solution since division by zero is not possible.

Interval Testing

Interval testing is a technique used to determine in which intervals of critical points the inequality holds true. Once you identify your critical points, you divide your number line into distinct intervals. For example:- Given the critical points \( x = 0 \) and \( x = \frac{1}{2} \), divide the number line into three intervals: \(( -\infty , 0 )\), \((0, \frac{1}{2})\), and \(( \frac{1}{2}, \infty )\). For each interval, choose a number (a test point) and substitute it into the initial inequality:

• In \(( -\infty , 0 )\), using \( x = -1 \) results in a positive value (3), satisfying \( \frac{2x-1}{x} \geq 0 \).
• In \((0, \frac{1}{2})\), using \( x = \frac{1}{4} \) results in a negative value (-2), which does not satisfy the inequality.
• In \(( \frac{1}{2}, \infty )\), using \( x = 1 \) results in a positive value (1), satisfying the inequality.

These outcomes tell us where the inequality holds and where it doesn't, transforming a potentially complex problem into manageable pieces. Conducting tests is crucial because it determines the valid intervals for the solution.

Solution Set

After identifying where the inequality holds through interval testing and considering the critical points, you can define the solution set. This set consists of all the intervals that satisfy the inequality, along with any boundary points that meet the inequality condition exactly. From the lessons of interval testing:

• The interval \(( -\infty , 0 ) \) yielded a positive test result, meaning it satisfies the inequality.
• The interval \(( \frac{1}{2}, \infty ) \) also satisfied the inequality.

The critical point \( x = \frac{1}{2} \) was tested separately. Since it resulted in zero, which is allowed by the \( \geq 0 \) condition, it is included in the solution.- Consequently, the solution set includes \([\frac{1}{2}, \infty)\) because this incorporates \(\frac{1}{2}\) and all values greater than it.- The interval \(( -\infty , 0 ) \) does not include \( 0 \) because it is an undefined point, but it is still part of the solution set in terms of valid values nearer to zero. By adopting this structured approach, determining the solution set of a rational inequality becomes both clear and straightforward.