Problem 41 - Calculus with Concepts in Calculus

Question

Solve the inequality. $$ \frac{x+1}{x-1} \leq \frac{1}{2} $$

Step-by-Step Solution

Answer

The solution is \([-3, 1)\).

Step 1: Convert the Inequality to a Single Fraction

Begin by rewriting the given inequality: \( \frac{x+1}{x-1} \leq \frac{1}{2} \). We can convert this to a single fraction by bringing all terms to one side of the inequality: \( \frac{x+1}{x-1} - \frac{1}{2} \leq 0 \).

Step 2: Find a Common Denominator

The common denominator for the expressions on the left-hand side is \( 2(x-1) \). Rewrite each fraction with this common denominator:\[\frac{2(x+1)}{2(x-1)} - \frac{x-1}{2(x-1)} \leq 0.\]

Step 3: Combine the Fractions

Combine the fractions into one expression:\[\frac{2(x+1) - (x-1)}{2(x-1)} \leq 0.\]Simplify the numerator: \[2(x+1) - (x-1) = 2x + 2 - x + 1 = x + 3.\]

Step 4: Simplify the Expression

The inequality is now:\[\frac{x + 3}{2(x-1)} \leq 0.\]

Step 5: Find Critical Points

Identify the points where the fraction is zero or undefined. Set the numerator to zero: \( x + 3 = 0 \), yielding \( x = -3 \). The denominator is zero at \( x = 1 \). These are critical points: \( x = -3 \) and \( x = 1 \).

Step 6: Test Intervals Around Critical Points

Choose test points in the intervals determined by the critical points: \((-\infty, -3)\), \((-3, 1)\), and \((1, \infty)\).- Test point \(x = -4\) in \((-\infty, -3)\): \( \frac{-1}{-10} > 0\)- Test point \(x = 0\) in \((-3, 1)\): \( \frac{3}{-2} < 0\)- Test point \(x = 2\) in \((1, \infty)\): \( \frac{5}{2} > 0\)

Step 7: Determine Solution Interval

The expression is less than or equal to zero in the interval \((-3, 1)\). Since we have \( \leq \), include \(x = -3\) but not \(x = 1\) in the solution set: \([-3, 1)\).

Key Concepts

Critical Points

In the context of inequalities involving rational expressions, critical points are values of the variable that make the expression either zero or undefined.
These points are crucial because they typically divide the number line into intervals, where the inequality will behave differently in terms of being positive or negative.

• First, to find the zero point, you set the numerator of the inequality to zero. For instance, solving for when the numerator equals zero gives us a critical point.
• Second, find where the expression is undefined by setting the denominator equal to zero; this helps identify points where the variable cannot take a certain value as it would make the expression undefined.

For the given inequality, the critical points come out to be two values: when the numerator ( \(x + 3 \ = 0\)) results in \(x = -3\), and when the denominator ( \(x - 1 \ = 0\)) results in \(x = 1\). These are crucial for partitioning the expression into testable intervals.

Numerator and Denominator

Understanding the numerator and denominator in rational inequalities is essential for correctly interpreting the expression's behavior.
The numerator represents the part of the fraction that can make the whole expression zero, while the denominator might cause the expression to be undefined due to division by zero.

• The numerator in our inequality is a function of \(x\), specifically \(x + 3\). Setting the numerator to zero, i.e., solving \(x + 3 = 0\), determines when the entire fraction equals zero.
• Conversely, the denominator is \(2(x-1)\). Setting the denominator to zero, i.e., \(x - 1 = 0\), helps us identify where the expression is undefined, hence the asymptotic behavior.

Through the interaction of these two components, you gauge where the inequality shifts between positive and negative values on the number line.

Test Intervals

Test intervals are segments on the number line divided by critical points, over which the inequality's sign can change.
These intervals help us determine where our inequality is true or false. By picking test points—values of \(x\) within each interval—we can decide if the expression satisfies the inequality in that specific range.

• In our situation, the critical points \(x = -3\) and \(x = 1\) divide the line into three main intervals: \((- \infty, -3)\), \((-3, 1)\), and \((1, \infty)\).
• You test each interval with values (test points) like \(x = -4\), \(0\), and \(2\), respectively, to check the sign of the entire expression.

Knowing where the expression is positive or negative lets you identify which intervals solve the inequality \( \leq 0\), helping you pinpoint the solution accurately.

Common Denominator

Dealing with expressions that include more than one fraction often involves finding a common denominator. This step allows you to combine and simplify fractions easily.

• To align the fractions on the same baseline, you determine the least common multiple of their individual denominators.
• In our inequality, the fractions \(\frac{x+1}{x-1} \) and \(\frac{1}{2} \) were combined under a common denominator of \(2(x-1)\).

This unified form transforms the overall fraction, letting us simplify it effectively and concentrate on simpler components, which greatly aids in solving the inequality accurately.