Problem 13 - Algebra for College Students

Question

Solve each inequality and graph its solution set on a number line. $$\frac{x+1}{x-2}>0$$

Step-by-Step Solution

Answer

The solution is \((-\infty, -1) \cup (2, \infty)\).

Step 1: Understanding the Expression

The inequality \( \frac{x+1}{x-2}>0 \) involves a rational function. The expression is positive when both the numerator \(x+1\) and the denominator \(x-2\) are either both positive or both negative. Our task is to find these conditions.

Step 2: Critical Points Identification

To identify where the expression changes sign, find the critical points by setting the numerator and denominator equal to zero: \( x+1 = 0 \) and \( x-2 = 0 \). Solving these gives us the critical points \( x = -1 \) and \( x = 2 \). These points divide the number line into intervals.

Step 3: Sign Analysis on Intervals

We test intervals around the critical points \(x = -1\) and \(x = 2\). The intervals are \( (-\infty, -1) \), \( (-1, 2) \), and \( (2, \infty) \). Choose a test point in each interval to determine the sign of \( \frac{x+1}{x-2} \) within that interval.For \(x = -2\) in \((-\infty, -1)\): \( \frac{-2+1}{-2-2} = \frac{-1}{-4} > 0\). For \(x = 0\) in \((-1, 2)\): \( \frac{0+1}{0-2} = \frac{1}{-2} < 0\).For \(x = 3\) in \((2, \infty)\): \( \frac{3+1}{3-2} = \frac{4}{1} > 0\). These results suggest that the expression is positive in intervals \((-\infty, -1)\) and \((2, \infty)\).

Step 4: Constructing the Solution Set

Since we seek the intervals where the expression is greater than zero and not equal to zero, we don't include the critical points themselves. Thus, the solution set is \((-\infty, -1) \cup (2, \infty)\).

Step 5: Graphing the Solution Set

To graph \((-\infty, -1) \cup (2, \infty)\) on a number line, draw open circles at \(x = -1\) and \(x = 2\) to indicate these points are not included. Shading the region to the left of \(x = -1\) and also the region to the right of \(x = 2\) shows the solution set.

Key Concepts

Critical Points

Understanding critical points is crucial when dealing with rational inequalities, like the one given in our exercise: \( \frac{x+1}{x-2} > 0 \). These critical points emerge from identifying the values of \( x \) where either the numerator or the denominator equals zero.

In this case, the numerator \( x+1 \) results in the critical point \( x = -1 \), and the denominator \( x-2 \) results in \( x = 2 \).

• The expressions that we care about are \( x+1 = 0 \) and \( x-2 = 0 \).
• By solving these, we find \( x = -1 \) and \( x = 2 \) respectively.

These points are crucial because they split the number line into separate intervals where the expression could change its sign.

Sign Analysis

Sign analysis is the technique used to determine whether a function is positive or negative within a given interval. After identifying critical points—here \( x = -1 \) and \( x = 2 \)—we divide the number line into intervals and check the sign of the expression in each.

For our inequality, the intervals are:

• \((-\infty, -1)\)
• \((-1, 2)\)
• \((2, \infty)\)

Select test points within each interval, plug these into the rational expression \( \frac{x+1}{x-2} \), and determine its sign.

- For \( x = -2 \) in \((-\infty, -1)\), calculate \( \frac{-1}{-4} > 0 \).- For \( x = 0 \) in \((-1, 2)\), calculate \( \frac{1}{-2} < 0 \).- For \( x = 3 \) in \((2, \infty)\), the result is \( \frac{4}{1} > 0 \).These test values show us where the inequality is satisfied.

Number Line Graphing

Graphing the solution of a rational inequality like \( \frac{x+1}{x-2} > 0 \) involves marking intervals on the number line based on sign analysis results.In our case, the expression is positive in the intervals \((-\infty, -1)\) and \((2, \infty)\).

To represent this on the number line:

• Draw open circles at \( x = -1 \) and \( x = 2 \), indicating these points are not part of the solution set.
• Shade the regions to the left of \( x = -1 \) and to the right of \( x = 2 \).

This visual approach gives a clear and immediate understanding of the solution set, making it easy to see where the inequality holds true.

Intervals

Intervals form the backbone of solving rational inequalities. Once critical points have been determined, they split the number line into distinct segments where different properties of the inequality may hold.

For the inequality \( \frac{x+1}{x-2} > 0 \), the intervals of interest are \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\).

• \((-\infty, -1):\) This interval is where both the numerator and denominator yield a positive value.
• \((-1, 2):\) In this interval, the expression takes a negative value.
• \((2, \infty):\) Here, the expression is positive again.

Including the appropriate intervals in the solution without the endpoints, since the inequality is strict \( (>0) \), defines the set of solutions to the inequality.