Problem 72 - Precalculus Student Solutions Manual 5th

Question

Solve each rational inequality. Write each solution set in interval notation. $$\frac{6-x}{x+2}>1$$

Step-by-Step Solution

Answer

The solution set in interval notation is \((-2, 2)\).

Step 1: Bring Inequality to One Side

First, subtract 1 from both sides of the inequality to set it to 0: \[\frac{6-x}{x+2} - 1 > 0\]

Step 2: Combine the Fractions

Combine the terms to have a single rational expression:\[\frac{6-x-(x+2)}{x+2} > 0\]

Step 3: Simplify the Numerator

Simplify the expression in the numerator:\[\frac{6-x-x-2}{x+2} > 0\]This simplifies to:\[\frac{4-2x}{x+2} > 0\]

Step 4: Determine Critical Points

Set the numerator and denominator equal to zero to find critical points:\[4-2x = 0 \rightarrow x = 2\]\[x+2 = 0 \rightarrow x = -2\]

Step 5: Test Intervals Around Critical Points

Test the intervals created by the critical points (-∞, -2), (-2, 2), and (2, ∞) using test points in the inequality \[\frac{4-2x}{x+2}>0\].

Step 6: Evaluate and Determine the Sign in Each Interval

Choose test points:- For interval \((-∞, -2)\), choose x = -3:\[\frac{4-2(-3)}{(-3)+2} = \frac{4+6}{-1} = -10 \; (negative)\]- For interval \((-2, 2)\), choose x = 0:\[\frac{4-2(0)}{0+2} = \frac{4}{2} = 2 \; (positive)\]- For interval \((2, ∞)\), choose x = 3:\[\frac{4-2(3)}{3+2} = \frac{4-6}{5} = -\frac{2}{5} \; (negative)\]

Step 7: Combine Interval Analysis

The inequality is greater than zero in the interval (-2, 2). Since the rational expression is undefined at x = -2 and equals zero at x = 2, the solution in interval notation is:\((-2, 2)\).

Key Concepts

Interval Notation

Interval notation is a way of writing the set of all numbers between two endpoints. It provides a clean and simple way to describe segments on the number line. For example, the interval (-2, 2) means all numbers greater than -2 and less than 2.

There are two kinds of intervals:

• Open intervals, which do not include their endpoints, and are represented with parentheses, like (-2, 2).
• Closed intervals, which include their endpoints, and use square brackets, such as [-2, 2].

Conjunctions with a mix of both open and closed interval are also possible, like (-2, 2] or [-2, 2).

When solving inequalities and expressing their solutions in interval notation, it is important to carefully interpret the behavior at the critical points to determine the correct type of interval to use.

Critical Points

Critical points in a rational inequality are values of the variable that make the numerator or the denominator equal to zero. These points are essential because they partition the number line into separate intervals where the inequality may change its truth value.

In the given problem, to identify the critical points, we set both the numerator and the denominator to zero:

• For the numerator: \( 4 - 2x = 0 \rightarrow x = 2 \)
• For the denominator: \( x + 2 = 0 \rightarrow x = -2 \)

These critical points, x = 2 and x = -2, help us break down the problem into simpler parts. They signal where the expression might switch from positive to negative or vice versa.

Test Intervals

Test intervals are created by the critical points. They help us understand where the rational expression is positive or negative.

For the inequality problem at hand, the critical points -2 and 2 divide the number line into three intervals:

• \( (-∞, -2) \)
• \( (-2, 2) \)
• \( (2, ∞) \)

We select a test point from each interval and plug it into the simplified inequality \( \frac{4 - 2x}{x + 2} > 0 \). Testing tells us whether each interval makes the inequality true or false.

Here are the test values we used:

• For \( (-∞, -2) \), using x = -3 showed a negative result.
• For \( (-2, 2) \), using x = 0 showed a positive result.
• For \( (2, ∞) \), using x = 3 showed a negative result.

Rational Expressions

A rational expression is a fraction with polynomials in both its numerator and denominator. These expressions can have undefined values when the denominator equals zero.

To solve rational inequalities, such as \( \frac{6 - x}{x + 2} > 1 \), we start by manipulating the inequality to bring all terms to one side, resulting in a single rational expression like \( \frac{4 - 2x}{x + 2} > 0 \).

We combine fractions and simplify polynomials using basic algebra:

• First, combine terms to create one fraction like \( \frac{6-x - (x+2)}{x+2} \).
• Simplify the expression in the numerator \( 4 - 2x \).

By understanding and manipulating these expressions, we can solve inequalities step by step and verify results through test intervals, ensuring all critical points and intervals are considered.