rational inequalities[tex]0 \leqslant \times \frac{x - 2}{x + 1} [/tex]
Answer (1)
Answer:To solve the rational inequality \[0 \leq \frac{x + 1}{x - 2},\]we can follow these steps:### Step 1: Identify the Critical PointsSet the numerator and denominator to zero to find the critical points.1. **Numerator**: \(x + 1 = 0 \Rightarrow x = -1\)2. **Denominator**: \(x - 2 = 0 \Rightarrow x = 2\)These points will help us determine the intervals to test.### Step 2: Test the IntervalsThe critical points divide the number line into intervals. We will test the sign of the expression in each interval:1. **Interval 1**: \( (-\infty, -1) \)2. **Interval 2**: \( (-1, 2) \)3. **Interval 3**: \( (2, \infty) \)### Step 3: Choose Test Points- **For Interval 1**: Choose \(x = -2\) \[ \frac{-2 + 1}{-2 - 2} = \frac{-1}{-4} = \frac{1}{4} \quad (\text{positive}) \]- **For Interval 2**: Choose \(x = 0\) \[ \frac{0 + 1}{0 - 2} = \frac{1}{-2} = -\frac{1}{2} \quad (\text{negative}) \]- **For Interval 3**: Choose \(x = 3\) \[ \frac{3 + 1}{3 - 2} = \frac{4}{1} = 4 \quad (\text{positive}) \]### Step 4: Determine the Sign of the Expression- **Interval 1**: Positive- **Interval 2**: Negative- **Interval 3**: Positive### Step 5: Include Critical PointsWe also need to check the critical points:- At \(x = -1\): \(\frac{-1 + 1}{-1 - 2} = \frac{0}{-3} = 0\) (included)- At \(x = 2\): The expression is undefined (not included).### Step 6: Write the Solution SetThe solution to the inequality is where the expression is non-negative:\[(-\infty, -1] \cup (2, \infty)\]### Final AnswerThus, the solution to the inequality \(0 \leq \frac{x + 1}{x - 2}\) is:\[\boxed{(-\infty, -1] \cup (2, \infty)}\]
