Problem 36 - Precalculus : Building Concepts and Connections

Question

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

Step-by-Step Solution

Answer

The solution to the inequality is \((-\infty, -1) \cup (4, \infty)\).

Step 1: Deal with it as an equation

If we treat the inequality as an equation for now, it becomes \(\frac{2}{x+1}=\frac{1}{x-2}\). Then, the cross-multiplication method is used to clear the fractions, resulting in \(2(x-2)=1(x+1)\). Solving for \(x\) gives \(x=4\).

Step 2: Identify vertical asymptotes

Rational functions are undefined where the denominator is equal to zero. Therefore, finding the roots of the denominators, \(x+1=0\) and \(x-2=0\), gives \(x=-1\) and \(x=2\) respectively. These values are the vertical asymptotes for the rational function and they divide the domain into four intervals: \((-∞, -1)\), \((-1, 2)\), \((2, 4)\) and \((4, ∞)\).

Step 3: Choose test points and apply the original inequality

Choose a test point from each of the four intervals. For instance, choose -2, 0, 3, and 5. Substitute these values into the original inequality, \(\frac{2}{x+1}>\frac{1}{x-2}\), to determine which intervals satisfy the inequality. Doing this will reveal that the inequality is satisfied for the intervals \((-∞, -1)\) and \((4, ∞)\).

Step 4: Write the solution

The solution to the inequality is \((-\infty, -1) \cup (4, \infty)\). Because inequalities do not include the points where the function is undefined (vertical asymptotes), the solution does not include -1 and 2. The inequality also does not include 4 because the inequality is 'greater than' rather than 'greater than or equal to'. So, 4 is not part of the solution.

Key Concepts

Cross-Multiplication

When faced with a rational inequality like \(\frac{2}{x+1} > \frac{1}{x-2}\), cross-multiplication is a strategic move to clear the fractions. This technique involves multiplying each side of the inequality by the denominators of each fraction. The principle behind cross-multiplication is to create a common denominator, allowing us to compare the numerators directly. For our example, we multiply the left fraction by \(x-2\) and the right fraction by \(x+1\), resulting in the equation \(2(x-2) > 1(x+1)\).

By simplifying, we work out the inequality in terms of \(x\) without fractions, which is easier to solve. It is essential to remember that cross-multiplication can change the direction of the inequality if one or both of the denominators are negative. That's because multiplying by a negative number reverses the inequality sign. Therefore, we must always check the sign of the denominators before finalizing the inequality's direction.

Vertical Asymptotes

When dealing with rational functions, vertical asymptotes mark the x-values at which the function approaches infinity. They occur where the denominator equals zero, since division by zero is undefined. For the inequality \(\frac{2}{x+1} > \frac{1}{x-2}\), the denominators set to zero, \(x+1=0\) and \(x-2=0\), give us the vertical asymptotes \(x=-1\) and \(x=2\).

Understanding Vertical Asymptotes

Vertical asymptotes are important because they divide the number line into different intervals, each of which must be considered separately when solving rational inequalities. To solve such inequalities, we must remember that the values at the vertical asymptotes are not included in the solution set, as the function is not defined at these points.

Test Points

Choosing test points is crucial when determining which intervals satisfy a rational inequality. After establishing vertical asymptotes, we select a value (test point) within each interval to see if it fulfills the original inequality. It's like sampling a few fruits from a tree to guess the taste of all of them – a correct pick can tell us much about the rest.

Applying Test Points

For our example, test points were chosen from within each interval created by the vertical asymptotes and the solution of the related equation: -2, 0, 3, and 5, representing the intervals \((-∞, -1)\), \((-1, 2)\), \((2, 4)\), and \((4, ∞)\) respectively. Substituting these points into the inequality helps us to determine which intervals satisfy the conditions set by the inequality.

Intervals of Inequality

Once test points are used to determine where the inequality holds true, we can define the intervals of inequality. These intervals are the regions on the number line where the inequality is satisfied. For the given problem, the inequality is satisfied in the intervals \((-∞, -1)\) and \((4, ∞)\).

Interpreting the Intervals

The solution to the inequality, namely \((-\text{∞}, -1) \text{U} (4, +\text{∞})\), tells us that all values of \(x\) within these intervals make the original inequality true. However, the number 4 is not included because the inequality is strictly 'greater than' and not 'greater than or equal to'. Thus, we represent the intervals with parentheses to show that the end points are not part of the solution.