https://file.aiquickdraw.com/mathbot/file/17265929147177iw5d5dg.jpg

We are given the rational inequality:

\[
\frac{1}{x^2 + 2x - 3} \leq 0
\]

Let's solve this step by step.

### 1. Factor the denominator:
The denominator is a quadratic expression, \(x^2 + 2x - 3\). We can factor this as follows:

\[
x^2 + 2x - 3 = (x + 3)(x - 1)
\]

So, the inequality becomes:

\[
\frac{1}{(x + 3)(x - 1)} \leq 0
\]

### 2. Analyze the critical points:
The critical points come from the denominator, which will determine where the expression is undefined or equal to zero. These occur when \(x = -3\) and \(x = 1\), since:

\[
(x + 3) = 0 \quad \text{when} \quad x = -3
\]
\[
(x - 1) = 0 \quad \text{when} \quad x = 1
\]

At these points:
- The expression is undefined at \(x = -3\), since division by zero is not allowed.
- The expression is zero at \(x = 1\).

### 3. Test intervals:
Now we test the sign of the expression \(\frac{1}{(x + 3)(x - 1)}\) in the intervals determined by the critical points: 
- \( (-\infty, -3) \)
- \( (-3, 1) \)
- \( (1, \infty) \)

For each interval, we check the sign of the expression:

- **For \(x \in (-\infty, -3)\):** Pick \(x = -4\):
  \[
  \frac{1}{(-4 + 3)(-4 - 1)} = \frac{1}{(-1)(-5)} = \frac{1}{5} > 0
  \]

- **For \(x \in (-3, 1)\):** Pick \(x = 0\):
  \[
  \frac{1}{(0 + 3)(0 - 1)} = \frac{1}{(3)(-1)} = \frac{1}{-3} < 0
  \]

- **For \(x \in (1, \infty)\):** Pick \(x = 2\):
  \[
  \frac{1}{(2 + 3)(2 - 1)} = \frac{1}{(5)(1)} = \frac{1}{5} > 0
  \]

### 4. Determine where the inequality is satisfied:
We are looking for where the expression is less than or equal to 0, so:
- The expression is negative in the interval \((-3, 1)\).
- At \(x = 1\), the expression is exactly 0, so \(x = 1\) is included.
- \(x = -3\) is excluded because the expression is undefined there.

### 5. Solution in interval notation:
The solution is:

\[
(-3, 1]
\]

Thus, the correct answer is the last option: **\([-3, 1]\)**.

---

Would you like more details on any step?

Here are 5 related questions you might explore:
1. How do you solve rational inequalities in general?
2. What is the difference between a strict inequality and a non-strict inequality?
3. Why do we exclude points where the denominator is zero?
4. How do you test intervals when solving inequalities?
5. How would the solution change if the inequality were strict, \( < \) instead of \( \leq \)?

**Tip:** Always remember to factor the denominator or numerator when solving rational inequalities—it helps you find critical points quickly.

Choose the correct solution to the rational inequality, 1/(x^2 + 2x - 3) ≤ 0, in interval notation.

Learn how to solve the rational inequality 1/(x^2 + 2x - 3) ≤ 0 by factoring the quadratic expression and testing intervals. Understand critical points, sign testing, and interval notation. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation