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INECUACIONES RESOLVER CON EL METODO DE PUNTOS CRITICOS

The given inequality is:

\[
\frac{(x - 2)(x + 2)}{(x + 3)(x - 3)} \leq 0
\]

### Step 1: Identify the critical points

The critical points occur where the numerator or the denominator equals zero:

1. **Numerator**: \((x - 2)(x + 2) = 0\)  
   Critical points: \(x = 2\) and \(x = -2\)

2. **Denominator**: \((x + 3)(x - 3) = 0\)  
   Critical points: \(x = -3\) and \(x = 3\)

### Step 2: Determine intervals

These critical points divide the real line into intervals. We test each interval to see where the inequality is satisfied. The intervals are:

1. \( (-\infty, -3) \)
2. \( (-3, -2) \)
3. \( (-2, 2) \)
4. \( (2, 3) \)
5. \( (3, \infty) \)

### Step 3: Test each interval

- **Interval \((- \infty, -3)\)**: Choose \(x = -4\)
  \[
  \frac{(-4 - 2)(-4 + 2)}{(-4 + 3)(-4 - 3)} = \frac{(-6)(-2)}{(-1)(-7)} = \frac{12}{7} > 0
  \]
  Inequality is not satisfied.

- **Interval \((-3, -2)\)**: Choose \(x = -2.5\)
  \[
  \frac{(-2.5 - 2)(-2.5 + 2)}{(-2.5 + 3)(-2.5 - 3)} = \frac{(-4.5)(-0.5)}{(0.5)(-5.5)} = \frac{2.25}{-2.75} < 0
  \]
  Inequality is satisfied.

- **Interval \((-2, 2)\)**: Choose \(x = 0\)
  \[
  \frac{(0 - 2)(0 + 2)}{(0 + 3)(0 - 3)} = \frac{(-2)(2)}{(3)(-3)} = \frac{-4}{-9} > 0
  \]
  Inequality is not satisfied.

- **Interval \((2, 3)\)**: Choose \(x = 2.5\)
  \[
  \frac{(2.5 - 2)(2.5 + 2)}{(2.5 + 3)(2.5 - 3)} = \frac{(0.5)(4.5)}{(5.5)(-0.5)} = \frac{2.25}{-2.75} < 0
  \]
  Inequality is satisfied.

- **Interval \((3, \infty)\)**: Choose \(x = 4\)
  \[
  \frac{(4 - 2)(4 + 2)}{(4 + 3)(4 - 3)} = \frac{(2)(6)}{(7)(1)} = \frac{12}{7} > 0
  \]
  Inequality is not satisfied.

### Step 4: Include critical points where the expression equals zero

The inequality is non-strict (≤), so we include points where the numerator equals zero but exclude points where the denominator equals zero.

- \(x = 2\) and \(x = -2\) (included)
- \(x = -3\) and \(x = 3\) (excluded)

### Final solution:

\[
x \in (-3, -2] \cup [2, 3)
\]

This matches the option \( \boxed{X \in (-3, -2] \cup [2, 3)} \).

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Would you like further details or have any questions? 

Here are 8 related questions for more practice:

1. How would the solution change if the inequality were strict (i.e., < instead of ≤)?
2. What happens to the solution if the numerator had a higher degree polynomial?
3. Can you find the solution graphically by sketching the function?
4. How would adding a constant to the inequality affect the solution set?
5. How do you handle inequalities with multiple fractions?
6. What is the impact of squaring both sides of an inequality?
7. How does solving a quadratic inequality differ from solving a linear one?
8. Can you generalize the solution method for any rational inequality?

**Tip:** Always remember to check the points where the denominator is zero—these must be excluded from the solution set!

Learn how to solve inequalities using the critical points method. This detailed solution covers identifying critical points, dividing intervals, and testing inequalities. Suitable for advanced high school and college students.

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