Problem 55 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$\frac{x^{2}(3-x)}{x+2} \leq 0$$

Step-by-Step Solution

Answer

The solution is \((-2, 0] \cup \{ 3 \}\).

Step 1: Analyze the Inequality

The inequality given is \( \frac{x^{2}(3-x)}{x+2} \leq 0 \). Notice that the expression is a fraction, so we need to analyze both the numerator \( x^{2}(3-x) \) and the denominator \( x+2 \). The inequality holds when the fraction is either zero or negative.

Step 2: Find Critical Points

To find when the fraction is zero, we set the numerator equal to zero: \( x^{2}(3-x) = 0 \). Solving this, \( x^{2} = 0 \) gives \( x = 0 \), and \( 3-x = 0 \) gives \( x = 3 \). The denominator \( x+2 = 0 \) provides another critical point, \( x = -2 \), which is undefined (makes the expression inapplicable). Hence, critical points are \( x = -2, 0, 3 \).

Step 3: Determine Sign for Each Interval

Use the critical points to divide the number line into intervals: \( (-\infty, -2), (-2, 0), (0, 3), (3, \infty) \). Choose test points from each interval to determine the sign of the fraction in that interval. Evaluate the expression for test points in each interval:1. \( x = -3 \) in \( (-\infty, -2) \): The fraction is positive.2. \( x = -1 \) in \( (-2, 0) \): The fraction is negative.3. \( x = 1 \) in \( (0, 3) \): The fraction is positive.4. \( x = 4 \) in \( (3, \infty) \): The fraction is negative.

Step 4: Determine Solution Including Critical Points

The inequality \( \frac{x^{2}(3-x)}{x+2} \leq 0 \) is satisfied wherever the fraction is zero or negative. From Step 3, it's negative in the interval \( (-2, 0) \). It is zero at \( x = 0 \) and \( x = 3 \). Also, note that \( x = -2 \) is not included in the solution because it makes the denominator zero. So, combine this to get the solution intervals.

Step 5: Express Solution in Interval Notation

Based on the analysis, the solution to the inequality includes where the fraction is negative or zero: the intervals are \( (-2, 0] \) and \{ 3 \}. Therefore, the final solution in interval notation is \((-2, 0] \cup \{ 3 \}\).

Key Concepts

Critical Points

When dealing with inequalities involving rational expressions, determining critical points is a crucial step. Critical points are values of the variable where the expression can potentially change sign or become undefined. For the inequality given, \( \frac{x^{2}(3-x)}{x+2} \leq 0 \), critical points arise from:

• The numerator \( x^{2}(3-x) \): Solving \( x^{2}(3-x) = 0 \) gives critical points \( x = 0 \) and \( x = 3 \), as it turns the entire expression into zero.
• The denominator \( x+2 \): Solving \( x+2 = 0 \) gives \( x = -2 \), where the expression becomes undefined, indicating a point of discontinuity.

Finding these critical points helps divide the number line into intervals for testing the sign of the inequality. This division is key to understanding where the expression satisfies the inequality.

Interval Notation

Once the critical points are identified, the next task is to express the solution using interval notation. Interval notation provides a clear and concise way to show where the inequality holds true on the number line.The inequality \( \frac{x^{2}(3-x)}{x+2} \leq 0 \) divides the number line based on the critical points into the following intervals:

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

We then test each interval with a point to determine where the inequality is true:- Solution occurs in intervals where the analytical test yields non-positive results (the fraction is negative or zero).Finally, the solution is expressed in interval notation: \((-2, 0] \cup \{3\}\), marking open and closed intervals appropriately.

Rational Expressions

A rational expression is a fraction where the numerator and/or the denominator are polynomials. The inequality \( \frac{x^{2}(3-x)}{x+2} \leq 0 \) is a classic example.Understanding rational expressions involves analyzing both the numerator and the denominator:

• Numerator: In our expression, it is \( x^2(3-x) \), which is a product of polynomials. It affects where the entire expression equals zero, giving potential solutions.
• Denominator: \( x + 2 \), its roots (where it becomes zero) help us locate discontinuities since division by zero is undefined.

When solving the inequality, we consider where the rational expression is less than or equal to zero by testing values around critical points. Employing these critical points reveals intervals which make the entire rational expression non-positive, ultimately leading us to formulating the solution in interval notation.