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Magazine Chapter 8: Problem 35
Rational Inequalities Solve. $$\frac{x+1}{x-3} \geq 0$$
Short Answer
Expert verified
Solution: \( (-\infty, -1] \cup (3, \infty) \)
Step by step solution
01
Identify Critical Points
Set the numerator and the denominator equal to zero to find the critical points. Solve for when the fraction is zero and when the denominator is undefined. Numerator: \( x + 1 = 0 \) \( x = -1 \) Denominator: \( x - 3 = 0 \) \( x = 3 \)
02
Create Number Line Intervals
Divide the number line into intervals based on the critical points found in Step 1. The intervals are: \( (-\infty, -1) \), \( (-1, 3) \), and \( (3, \infty) \)
03
Test Each Interval
Choose a test point in each interval and substitute it into the inequality \( \frac{x+1}{x-3} \geq 0 \) to check the sign. For \( (-\infty, -1) \): Choose \( x = -2 \), \( \frac{-2 + 1}{-2 - 3} = \frac{-1}{-5} = \frac{1}{5} > 0 \) (True) For \( (-1, 3) \): Choose \( x = 0 \), \( \frac{0 + 1}{0 - 3} = \frac{1}{-3} < 0 \) (False) For \( (3, \infty) \): Choose \( x = 4 \), \( \frac{4 + 1}{4 - 3} = \frac{5}{1} = 5 > 0 \) (True)
04
Consider Critical Points
Examine the critical points. At \( x = -1 \): \( \frac{-1+1}{-1-3} = \frac{0}{-4} = 0 \) (satisfies \( \geq 0\) ) At \( x = 3 \): The denominator makes the expression undefined, hence do not include it in the solution.
05
Combine the Results
Compile the intervals where the inequality holds true and include the valid critical point: Solution: \( (-\infty, -1] \cup (3, \infty) \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
Critical points help determine where the rational inequality changes its behavior. To find these points, you need to set both the numerator and denominator of the inequality to zero individually. For the given inequality \( \frac{x+1}{x-3} \geq 0 \), we start with setting the numerator and denominator equal to zero.
- Numerator: \( x + 1 = 0 \)鈫抃( x = -1 \)
- Denominator: \( x - 3 = 0 \)鈫抃( x = 3 \)
Number Line Intervals
Once you've determined the critical points, you can create segments on a number line. These segments, called intervals, help you analyze the behavior of the inequality across different ranges of \( x \). For our inequality with critical points -1 and 3, the number line is divided into:
- \( (-\infty, -1) \)
- \( (-1, 3) \)
- \( (3, \infty) \)
Test Intervals
To determine which intervals satisfy the inequality, we select a test point from each interval and substitute it into the inequality \( \frac{x+1}{x-3} \geq 0 \). We need to check the sign resulting from these substitutions:
- For the interval \( (-\infty, -1) \) select \( x = -2 \): \( \frac{-2 + 1}{-2 - 3} = \frac{-1}{-5} = \frac{1}{5} > 0 \). This interval works.
- For the interval \( (-1, 3) \) select \( x = 0 \): \( \frac{0 + 1}{0 - 3} = \frac{1}{-3} < 0 \). This interval does not work.
- For the interval \( (3, \infty) \) select \( x = 4 \): \( \frac{4 + 1}{4 - 3} = \frac{5}{1} = 5 > 0 \). This interval works.
Inequality Solutions
Finally, we interpret the results from our interval tests and consider the critical points.
- At \( x = -1 \), the fraction becomes \( \frac{-1+1}{-1-3} = \frac{0}{-4} = 0 \). Since \( 0 \geq 0 \), point \( x = -1 \) is included in our solution.
- At \( x = 3 \), the denominator becomes zero, making the fraction undefined. Thus, \( x = 3 \) is not included.
- The interval \( (-\infty, -1] \)
- The interval \( (3, \infty) \)
