Problem 63 - A Graphical Approach to College Algebra

Question

Solve each rational inequality by hand. Do not use a calculator. $$\frac{(x+1)^{2}}{x-2} \leq 0$$

Step-by-Step Solution

Answer

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

Step 1: Identify Critical Points

To solve the inequality \( \frac{(x+1)^2}{x-2} \leq 0 \), we need to find the values of \( x \) that make the numerator zero and the denominator undefined. The numerator \((x+1)^2 = 0\) when \(x = -1\). The denominator \(x-2=0\) when \(x = 2\). So, the critical points are \(x = -1\) and \(x = 2\).

Step 2: Determine the Sign of Each Interval

Divide the real number line into intervals using the critical points \(-∞, -1, 2, ∞\). Test a point from each interval to determine the sign of \( \frac{(x+1)^{2}}{x-2} \). - For \(x < -1\), pick \(x = -2\): \( \frac{((-2)+1)^2}{-2-2} = \frac{1}{-4} < 0 \). - For \(-1 < x < 2\), pick \(x = 0\): \( \frac{(0+1)^2}{0-2} = \frac{1}{-2} < 0 \). - For \(x > 2\), pick \(x = 3\): \( \frac{(3+1)^2}{3-2} = \frac{16}{1} > 0 \).

Step 3: Analyze the Inequality's Conditions

Since the inequality is \( \frac{(x+1)^{2}}{x-2} \leq 0 \), we need the expression to be less than or equal to zero. We observed that the expression is negative for intervals \((-\infty, -1)\) and \((-1, 2)\). Check the behavior at critical points. - At \(x = -1\), \(\frac{(x+1)^{2}}{x-2} = 0\) (since the numerator is zero), and it satisfies the inequality. - At \(x = 2\), the expression is undefined (denominator is zero).

Step 4: Write the Solution Interval

Combine the intervals where the expression satisfies the inequality: \((-fty, -1] \cup (-1, 2)\). This includes \(x = -1\) where the expression equals zero, but excludes \(x = 2\) where it is undefined.

Key Concepts

Critical Points

Critical points are important values of the variable that influence the behavior of a rational expression. For rational inequalities, these points occur where the numerator is zero or the denominator is undefined. Identifying these points is the first step in solving a rational inequality.

• Numerator: In the expression \((x+1)^2\), the numerator becomes zero when \(x+1=0\), leading to \(x = -1\).
• Denominator: For \(x-2\), the denominator is undefined when \(x=2\).

The critical points for this inequality are \(x = -1\) and \(x = 2\). These points divide the number line into distinct intervals.

Sign Analysis

After identifying critical points, we divide the number line into intervals and select test points to determine the sign of the rational expression in each interval. This allows us to know where the expression is positive or negative, which helps in solving the inequality.

• Interval \((-\infty, -1)\): Choose a point like \(x = -2\). The expression \(\frac{(x+1)^{2}}{x-2}\) is negative because \(\frac{1}{-4} < 0\).
• Interval \((-1, 2)\): Choose a point like \(x = 0\). Here, the expression \(\frac{1}{-2} < 0\) is also negative.
• Interval \((2, \infty)\): Choose a point like \(x = 3\). The expression becomes positive as \(\frac{16}{1} > 0\).

This analysis reveals that the expression changes its sign across these intervals.

Solution Intervals

By examining sign changes and critical points, we find intervals where the inequality \(\frac{(x+1)^{2}}{x-2} \leq 0\) holds true. This involves combining intervals where the expression is negative and properly handling critical points.

• Interval \((-\infty, -1]\): Includes \(x = -1\) because the expression equals zero, satisfying the \(\leq\) condition.
• Interval \((-1, 2)\): The expression is negative, matching the condition \(\leq 0\).

At \(x = 2\), the expression is undefined, so it is excluded from the solution. Thus, the correct solution intervals are \((-\infty, -1] \cup (-1, 2)\). This illustrates how critical and solution intervals guide us in solving rational inequalities effectively.