Problem 60 - A Graphical Approach to College Algebra

Question

Solve each rational inequality by hand. Do not use a calculator. $$\frac{3 x-3}{4-2 x} \geq 0$$

Step-by-Step Solution

Answer

The solution is \(x \in [1,2)\).

Step 1: Identify the Critical Points

To solve \( \frac{3x-3}{4-2x} \geq 0 \), first identify the critical points where the expression is undefined or equal to zero. This occurs when the numerator or the denominator is zero. Set the numerator \(3x - 3 = 0\) and solve for \(x\). \[ 3x - 3 = 0 \] \[ 3x = 3 \] \[ x = 1 \] Next, set the denominator \(4 - 2x = 0\) and solve for \(x\). \[ 4 - 2x = 0 \] \[ 2x = 4 \] \[ x = 2 \] The critical points are \(x = 1\) and \(x = 2\).

Step 2: Analyze Intervals

Use the critical points to divide the number line into intervals: \((-\infty, 1)\), \((1, 2)\), and \((2, \infty)\). Determine the sign of the expression in each interval.

Step 3: Test Intervals

Pick test points in each interval to determine the sign of \( \frac{3x-3}{4-2x} \).- Interval \((-\infty, 1)\): Test point \(x = 0\). Substitute into the inequality: \( \frac{3(0)-3}{4-2(0)} = \frac{-3}{4} \lt 0 \).- Interval \((1, 2)\): Test point \(x = 1.5\). Substitute into the inequality: \( \frac{3(1.5)-3}{4-2(1.5)} = \frac{1.5}{1} = 1.5 \gt 0 \).- Interval \((2, \infty)\): Test point \(x = 3\). Substitute into the inequality: \( \frac{3(3)-3}{4-2(3)} = \frac{6}{-2} = -3 \lt 0 \).

Step 4: Consider Endpoint Behavior

Examine the endpoints. At \(x = 1\), the expression equals zero since \( \frac{3(1)-3}{4-2(1)} = 0 \). At \(x = 2\), the denominator becomes zero, making the expression undefined. Only include \(x = 1\) in the solution as it satisfies the 'greater than or equal to' condition.

Step 5: Write Solution Set

Combine the results from previous steps, the solution is the interval where the inequality is non-negative, including the endpoint where it equals zero.The solution interval is \( x \in [1, 2) \).

Key Concepts

Critical Points

Critical points are vital in solving rational inequalities because they tell us where the expression could change in behavior. For a rational expression like \( \frac{3x-3}{4-2x} \), critical points occur in two situations: when the numerator is zero or when the denominator is zero. To find these points, set the numerator equal to zero and solve for \(x\).

• For \(3x - 3 = 0\), solving gives \(x = 1\).

Next, set the denominator equal to zero to find where the expression is undefined:

• For \(4 - 2x = 0\), solving gives \(x = 2\).

These critical points, \(x = 1\) and \(x = 2\), help segment the number line into intervals that are key in determining where the expression changes sign.

Interval Testing

Once we have the critical points, interval testing lets us figure out the sign of the expression on those intervals. The number line is divided into intervals based on these critical points. In this example, the intervals are \((-\infty, 1)\), \((1, 2)\), and \((2, \infty)\). To conduct interval testing, we select a test point from each interval. By substituting these test points into the original inequality, we determine whether the resulting value is positive or negative.

• Test point in \((-\infty, 1)\): Set \(x = 0\) results in \(\frac{-3}{4} < 0\).
• Test point in \((1, 2)\): Set \(x = 1.5\) results in \(1.5 > 0\).
• Test point in \((2, \infty)\): Set \(x = 3\) results in \(-3 < 0\).

This testing establishes where the expression meets the condition \(\frac{3x-3}{4-2x} \geq 0\).

Numerator and Denominator

Understanding the roles of the numerator and denominator is essential in solving rational inequalities. Together, they determine where the expression is zero, undefined, positive, or negative. Knowing this can help in identifying solution intervals.

• The numerator is \(3x - 3\). It's zero when \(x = 1\), affecting the entire expression at that point.
• The denominator is \(4 - 2x\). It becomes zero when \(x = 2\), making the expression undefined.

Recognizing these interactions allows you to determine the behavior of the rational function on different segments of the number line.

Solution Intervals

The final step in solving rational inequalities is determining the solution intervals. This involves combining the information gathered from critical points, interval testing, and the nature of the rational function. Since the inequality is \(\geq 0\), we're interested in where the expression is non-negative (positive or zero). From interval testing, we found:

• In the interval \((1, 2)\), the expression is positive.
• Since the value at \(x=1\) makes the expression zero and is included due to the question's condition \(\geq\), \([1, 2)\) is part of the solution set.

Areas where the test points resulted in negative values are not included in the solution. Therefore, the final solution interval is \([1, 2)\), where \(x\) fulfills the inequality.