Problem 25 - Calculus with Concepts in Calculus

Question

Solve the inequality. $$ x(x-2 / 3)(x+1 / 3)<0 $$

Step-by-Step Solution

Answer

The solution is \((-\infty, -\frac{1}{3}) \cup (0, \frac{2}{3})\).

Step 1: Find the Critical Points

To solve the inequality \(x(x-\frac{2}{3})(x+\frac{1}{3})<0\), we first need to find the values of \(x\) for which the expression equals zero. Set each factor equal to zero: 1. \(x = 0\) 2. \(x - \frac{2}{3} = 0\) gives \(x = \frac{2}{3}\) 3. \(x + \frac{1}{3} = 0\) gives \(x = -\frac{1}{3}\) Therefore, the critical points are \(-\frac{1}{3}, 0, \frac{2}{3}\).

Step 2: Determine the Intervals

The critical points divide the number line into four intervals: 1. \((-\infty, -\frac{1}{3})\) 2. \((-\frac{1}{3}, 0)\) 3. \((0, \frac{2}{3})\) 4. \((\frac{2}{3}, \infty)\)

Step 3: Test the Intervals

Pick a test point from each interval and evaluate the expression \(x(x-\frac{2}{3})(x+\frac{1}{3})\):1. For \((-\infty, -\frac{1}{3})\), use \(x = -1\): \(-1((-1)-\frac{2}{3})(-1+\frac{1}{3}) = (-1)(-\frac{5}{3})(-\frac{2}{3}) < 0\).2. For \((-\frac{1}{3}, 0)\), use \(x = -\frac{1}{6}\): \((-\frac{1}{6})((-\frac{1}{6})-\frac{2}{3})((-\frac{1}{6})+\frac{1}{3}) = (-)(-)(+) > 0\).3. For \((0, \frac{2}{3})\), use \(x = \frac{1}{3}\): \(\frac{1}{3}(\frac{1}{3}-\frac{2}{3})(\frac{1}{3}+\frac{1}{3}) = (+)(-)(+) < 0\).4. For \((\frac{2}{3}, \infty)\), use \(x = 1\): \(1(1-\frac{2}{3})(1+\frac{1}{3}) = (+)(+)(+) > 0\).

Step 4: Interpret the Results

The expression is negative in the intervals \((-\infty, -\frac{1}{3})\) and \((0, \frac{2}{3})\). Since we are solving for when \(x(x-\frac{2}{3})(x+\frac{1}{3}) < 0\), the solution is the union of these intervals.

Key Concepts

Critical Points

Critical points are values of a variable where a mathematical expression reaches crucial thresholds, such as zero or undefined values. To solve a polynomial inequality like \(x(x-\frac{2}{3})(x+\frac{1}{3})<0\), it's important to first determine these points. This involves setting each factor of the polynomial to zero and solving for \(x\). By doing this, we can understand where the expression changes direction or sign.

In our case, the critical points are found by considering each part of the expression separately:

• For \(x = 0\).
• For \(x - \frac{2}{3} = 0\), solve to get \(x = \frac{2}{3}\).
• For \(x + \frac{1}{3} = 0\), solve to get \(x = -\frac{1}{3}\).

These critical points \(-\frac{1}{3}, 0, \frac{2}{3}\) help define the intervals which will be crucial for further analysis.

Interval Testing

Interval testing helps determine where a polynomial expression is positive or negative by dividing the number line based on the critical points. Each segment between critical points is an 'interval.' In our example, the critical points \(-\frac{1}{3}, 0, \frac{2}{3}\) split the number line into four intervals:

• \((-\infty, -\frac{1}{3})\)
• \((-\frac{1}{3}, 0)\)
• \((0, \frac{2}{3})\)
• \((\frac{2}{3}, \infty)\)

By selecting a test point from each interval and substituting it back into the inequality, we can check the sign of the expression. This helps in identifying intervals where the original inequality \(x(x-\frac{2}{3})(x+\frac{1}{3}) < 0\) holds true.

Mathematical Expressions

A mathematical expression is a combination of numbers, variables, operators, and sometimes constants, which together represent a particular value or set of values. In this inequality problem, the expression is \(x(x-\frac{2}{3})(x+\frac{1}{3})\). Understanding how to manipulate these expressions is essential when solving inequalities.

When you simplify or analyze expressions, it often involves:

• Factoring the expression into simpler parts.
• Identifying common factors.
• Using rules of arithmetic to simplify complex expressions.

Grasping this concept equips you to better handle variables and functions within larger equations and inequalities. It's a crucial step before applying further solving techniques.

Polynomial Inequalities

Polynomial inequalities are statements that involve polynomial expressions and inequality signs like <, >, ≤, or ≥. In this case, we are dealing with a cubic polynomial inequality \(x(x-\frac{2}{3})(x+\frac{1}{3})<0\). Solving these types of inequalities requires understanding the polynomial's factors, its zeros or critical points, and determining over which intervals the polynomial is negative or positive.

Steps to solve:

• Identify the critical points by setting each factor to zero.
• Divide the number line into intervals based on these points.
• Determine the sign of the expression within each interval (using interval testing).
• Select and interpret the intervals where the inequality holds true, i.e., where the polynomial is negative in this case.

By mastering these steps, you can handle various polynomial inequalities effectively, leading to a clear understanding of how solutions form depending on the sign of the polynomial across different intervals.