Problem 68 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Critical Points

Determine where the expression is equal to zero or undefined. The numerator is zero when \(x(x-3)=0\), giving critical points at \(x=0\) and \(x=3\). The denominator is zero when \(x+2=0\), giving a critical point at \(x=-2\).

Step 2: Analyze Intervals

Use the critical points \(-2\), \(0\), and \(3\) to divide the number line into intervals: \((-\infty, -2)\), \((-2, 0)\), \((0, 3)\), and \((3, \infty)\).

Step 3: Test Each Interval

Select a test point from each interval to determine the sign of the expression \(\frac{x(x-3)}{x+2}\). For \((-\infty, -2)\), choose \(x=-3\). For \((-2, 0)\), choose \(x=-1\). For \((0, 3)\), choose \(x=1\). For \((3, \infty)\), choose \(x=4\).

Step 4: Evaluate Sign of Expression

Calculate the sign for each test point. \(x=-3\) gives negative, \(x=-1\) gives positive, \(x=1\) gives negative, \(x=4\) gives positive values. Record these signs for each interval.

Step 5: Determine Solution Set

The inequality is \(\geq 0\), which means we include intervals where the expression is positive or zero. The solution intervals are \([-2, 0) \cup (3, \infty)\). Since\(x=-2\) makes the expression undefined, and \(x=0\) and \(x=3\) make it zero, include 0 but not -2 or 3.

Key Concepts

Critical Points

To solve a rational inequality effectively, identifying critical points is essential. Critical points occur where the rational expression is zero or undefined. To identify these points, we need to find the values of the variable in the numerator and denominator separately. In this case, the critical points of the expression \(\frac{x(x-3)}{x+2}\) are derived as follows:

• Numerator Zero: Set \(x(x-3) = 0\). Solving this equation gives us \(x = 0\) and \(x = 3\), where the expression equals zero.
• Denominator Zero: Set \(x+2 = 0\). Solving this provides \(x = -2\), where the expression becomes undefined.

The critical points found are \(-2\), \(0\), and \(3\). These points divide the real number line into separate intervals that must be considered individually.

Number Line Intervals

Once we have the critical points, the next step is to divide the number line into intervals. Each interval is determined by two adjacent critical points. This separation helps simplify the analysis of the expression's sign within each segment. The intervals based on the critical points \(-2\), \(0\), and \(3\) become:

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

In each interval, the rational expression can behave differently, either being positive, negative, or zero, which is key to solving the inequality.

Test Points

In evaluating the expression's behavior within each interval, selecting test points becomes critical. A test point is just a value chosen within the interval that will help determine the sign of the expression there. It's important because it gives insight into how the expression behaves across that segment. For the intervals chosen:

• For \((-\infty, -2)\), we use \(x = -3\).
• For \((-2, 0)\), we use \(x = -1\).
• For \((0, 3)\), we use \(x = 1\).
• For \((3, \infty)\), we use \(x = 4\).

By substituting these points into the expression \(\frac{x(x-3)}{x+2}\), we determine whether the outcome is positive or negative for each of the intervals:

Solution Set

Identifying the solution set is the final step in solving the rational inequality. For the inequality \(\frac{x(x-3)}{x+2} \geq 0\), we're looking for where the expression is greater than or equal to zero:

• The test point \(-3\) gave a negative value, indicating \((-\infty, -2)\) is not part of the solution.
• The test point \(-1\) resulted in a positive value, meaning \((-2, 0)\) is part of the solution set.
• The test point \(1\) gave a negative value, excluding the interval \((0, 3)\).
• The test point \(4\) yielded a positive value, so \((3, \infty)\) is included in the solution.

Critical points where the expression equals zero, such as \(x = 0\), are included in the solution, but points like \(x = -2\), where the expression is undefined, are excluded. Thus, the solution set is \([-2, 0) \cup (3, \infty)\). This captures all intervals where the expression aligns with \(\geq 0\).