Problem 62 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify critical points

The inequality is \( \frac{x(x-3)}{x+2} \geq 0 \). To find the critical points, set the numerator \( x(x-3) \) and the denominator \( x+2 \) to zero. The numerator is zero when \( x=0 \) or \( x=3 \). The denominator is zero when \( x=-2 \). These critical points divide the number line into sections.

Step 2: Determine intervals

The critical points \( x=-2 \), \( x=0 \), and \( x=3 \) divide the number line into four intervals: \(( -\infty, -2 )\), \((-2, 0)\), \((0,3)\), and \((3, \infty )\).

Step 3: Test each interval

Choose test points from each interval: \( x = -3 \) for \(( -\infty, -2 )\), \( x = -1 \) for \((-2, 0)\), \( x = 1 \) for \((0, 3)\), and \( x = 4 \) for \((3, \infty)\). Substitute each test point into the expression \( \frac{x(x-3)}{x+2} \): - For \( x = -3 \), \( \frac{-3(-3-3)}{-3+2} = \frac{18}{-1} = -18 \).- For \( x = -1 \), \( \frac{-1(-1-3)}{-1+2} = \frac{4}{1} = 4 \).- For \( x = 1 \), \( \frac{1(1-3)}{1+2} = \frac{-2}{3} \).- For \( x = 4 \), \( \frac{4(4-3)}{4+2} = \frac{4}{6} = \frac{2}{3} \).

Step 4: Determine where inequality holds

The inequality \( \frac{x(x-3)}{x+2} \geq 0 \) holds for intervals where the test point gives a non-negative result. From the test points:- \((-2, 0)\) produces a positive value.- \((3, \infty)\) produces a positive value.Thus the solution intervals are \((-2, 0) \cup (3, \infty) \).

Step 5: Check boundary points

Examine the critical points where the expression is zero or undefined. - At \( x=0 \), \( \frac{0(0-3)}{0+2} = 0 \), so \( x=0 \) is included in the solution. - At \( x=3 \), \( \frac{3(3-3)}{3+2} = 0 \), so \( x=3 \) is included in the solution. - At \( x=-2 \), the expression is undefined, so \( x=-2 \) is not included. Thus, the solution is \([-2, 0) \cup [3, \infty)\).

Key Concepts

Critical Points

In the context of a rational inequality, critical points are key values that impact the number line intervals and solution of the inequality. Critical points emerge from the values that make either the numerator or the denominator of the rational expression equal to zero. This is because those values will either make the entire expression zero or undefined. Notably, they divide the entire number line into distinct intervals which we later test to determine where the inequality holds. Here, in the inequality \( \frac{x(x-3)}{x+2} \geq 0 \), we find the critical points by setting the numerator \( x(x-3) \) to zero, finding solutions \( x=0 \) and \( x=3 \), and the denominator \( x+2 \) to zero, where \( x=-2 \). These critical points are \(-2\), \(0\), and \(3\). They're crucial to understanding how to solve the inequality effectively.

Number Line Intervals

Once we have identified the critical points, the next step is to work with the number line intervals. We plot these critical points on a number line to divide it into separate regions or intervals. Each interval encompasses all real numbers between two successive critical points. For example, with critical points \(-2\), \(0\), and \(3\), the number line is split into four intervals:

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

These intervals represent all possible sections of real number values that we need to test to identify where the original rational inequality holds true. This splitting of the number line helps us manage the checking of different possibilities without missing any sections.

Boundary Points

Boundary points are special types of critical points responsible for defining the edges of our number line intervals. In a rational inequality, boundary points include those where the expression is either zero or undefined. Knowing which boundary points contribute to the solution set is essential, as some of them might be included, while others are excluded due to making the expression undefined. For this inequality \( \frac{x(x-3)}{x+2} \geq 0 \), we closely examine each boundary point:

• At \( x=0 \), the expression equals zero, so this point is included.
• At \( x=3 \), the expression also equals zero, so this point is included too.
• At \( x=-2 \), the expression is undefined, so this point is not included in the solution.

Identifying which boundary points are part of the solution is important, because it influences the intervals we consider in our final answer.

Test Points

Test points are specific values chosen from within each number line interval. They are necessary for determining whether the rational inequality's given condition holds true in a particular interval. For each interval formed between critical points, we choose at least one test point, plug it into the inequality, and evaluate whether the condition is satisfied (i.e., whether the expression is non-negative). For the inequality \( \frac{x(x-3)}{x+2} \geq 0 \), consider:

• For the interval \(( -\infty, -2 )\), use test point \( x = -3 \), which returns a negative result.
• For \((-2, 0)\), use \( x = -1 \), giving a positive output that satisfies the inequality.
• For \((0, 3)\), use \( x = 1 \), resulting in a negative outcome.
• For \((3, \infty)\), use \( x = 4 \), where the result is positive.

These test points validate which intervals are included in the solution. In this context, only intervals \((-2, 0)\) and \((3, \infty)\) satisfy the inequality, as their test points produce non-negative results. Hence, these are included in the solution set.