Problem 62 - A Graphical Approach to Precalculus with Limits

Question

Solve each rational inequality by hand. $$\frac{x(x-3)}{x+2} \geq 0$$

Step-by-Step Solution

Answer

The solution set is \((-\infty, -2) \cup [0, 3] \cup (3, \infty)\).

Step 1: Identify Critical Points

Determine when the numerator and the denominator of the inequality \( \frac{x(x-3)}{x+2} \geq 0 \) are equal to zero.\\(x(x-3) = 0\) gives critical points \(x = 0\) and \(x = 3\).\\(x+2 = 0\) gives critical point \(x = -2\).\These are the points where the sign of the inequality might change.

Step 2: Set Up a Number Line

Draw a number line and mark the critical points: \(x = -2\), \(x = 0\), and \(x = 3\). These points divide the number line into intervals. We have four intervals to test: \((-\infty, -2)\), \((-2, 0)\), \((0, 3)\), and \((3, \infty)\).

Step 3: Test Each Interval

Select test points from each interval: choose \(x = -3\) for \((-\infty, -2)\), \(x = -1\) for \((-2, 0)\), \(x = 1\) for \((0, 3)\), and \(x = 4\) for \((3, \infty)\). Substitute these values into the expression \( \frac{x(x-3)}{x+2}\) to determine if the inequality is positive or negative in each interval.

Step 4: Determine Sign in Each Interval

- For \((-\infty, -2)\), with \(x = -3\), the expression is \(\frac{(-3)((-3)-3)}{-3+2} = \frac{(-3)(-6)}{-1} = 18 > 0\).\- For \((-2, 0)\), with \(x = -1\), the expression is \(\frac{(-1)((-1)-3)}{-1+2} = \frac{(-1)(-4)}{1} = 4 > 0\).\- For \((0, 3)\), with \(x = 1\), the expression is \(\frac{1(1-3)}{1+2} = \frac{1(-2)}{3} = -\frac{2}{3} < 0\).\- For \((3, \infty)\), with \(x = 4\), the expression is \(\frac{4(4-3)}{4+2} = \frac{4(1)}{6} = \frac{2}{3} > 0\).

Step 5: Include Boundary Points

Examine the critical points to determine if they should be included in the solution:\- \(x = -2\) makes the denominator zero, resulting in an undefined expression. Exclude it from the solution.\- For \(x = 0\) and \(x = 3\), the inequality holds as zero is an acceptable value because the inequality is \(\geq\). Include these in the solution.

Step 6: Final Answer: Write the Solution Set

Based on testing the intervals and the boundary points, the solution set is the union of the intervals where the inequality holds: \[(-\infty, -2) \cup [-2, 0] \cup [3, \infty)\] Note that \(-2\) is not included because it results in an undefined expression.

Key Concepts

Critical Points

Critical points are essential when solving rational inequalities as they help identify potential changes in the inequality's sign. In the exercise, we focus on the numerator and denominator of the inequality \( \frac{x(x-3)}{x+2} \geq 0 \). We set both the numerator and denominator to zero, leading to potential critical points.

For the numerator, \( x(x-3) = 0 \), solving gives critical points at \( x = 0 \) and \( x = 3 \). For the denominator, \( x+2 = 0 \), we find the critical point \( x = -2 \).

These points are critical because they indicate where the expression might change from positive to negative or vice versa. Identifying these points allows us to divide the number line and test the inequality in different intervals.

Number Line Analysis

After finding the critical points, the next step is to perform a number line analysis. This involves drawing a number line and marking all critical points: \( x = -2 \), \( x = 0 \), and \( x = 3 \). These marks divide the number line into distinct sections, or intervals.

Each section created by these critical points needs to be tested for the sign of the inequality. In this problem, the intervals are

• \(( -\infty, -2 )\)
• \((-2, 0)\)
• \((0, 3)\)
• \((3, \infty)\)

This comprehensive setup allows us to test values within each interval and understand how the inequality behaves along the entire number line.

Interval Testing

Interval testing is a fundamental technique where we select test points from each interval to determine the sign of the rational expression \( \frac{x(x-3)}{x+2} \) within those intervals.

For our critical points:

• In the interval \(( -\infty, -2 )\), a test point like \( x = -3 \) is chosen, and the inequality evaluates to a positive value (\(18 > 0\)).
• In \((-2, 0)\), using \( x = -1 \) results in another positive value (\(4 > 0\)).
• In \((0, 3)\), selecting \( x = 1 \) yields a negative value (\(-\frac{2}{3} < 0\)).
• Finally, in \((3, \infty)\), \( x = 4 \) produces a positive value (\(\frac{2}{3} > 0\)).

This testing helps us conclude that the inequality is positive in multiple intervals, which is crucial for formulating the overall solution.

Boundary Points

Boundary points refer to the critical points themselves and their role in the inequality. It's important to analyze each point to decide whether they are included in the solution.

First, observe \( x = -2 \), which makes the denominator zero, causing the expression to be undefined. This point should naturally be excluded from the solution.

Now, consider the remaining critical points \( x = 0 \) and \( x = 3 \). In our inequality \( \frac{x(x-3)}{x+2} \geq 0 \), these points satisfy the condition \( \geq \), meaning that at these points, the expression equals zero, which is acceptable.

Therefore, \( x = 0 \) and \( x = 3 \) are included in the solution as part of the intervals that satisfy the inequality. Understanding the role of each boundary point helps us finalize the solution set.