Problem 63 - College Algebra with Modeling and Visualization

Question

Solve the rational inequality. $$ \frac{(x+1)^{2}}{x-2} \leq 0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify Critical Points

To solve the inequality \( \frac{(x+1)^{2}}{x-2} \leq 0 \), first identify the critical points where the expression equals zero or is undefined. The expression is zero when the numerator \((x+1)^2 = 0\) and undefined when the denominator \(x-2 = 0\).- \((x+1)^2 = 0\) gives \(x=-1\).- \(x-2=0\) gives \(x=2\). So, the critical points are \(x = -1\) and \(x = 2\).

Step 2: Create a Sign Chart

With the critical points \(x = -1\) and \(x = 2\), divide the number line into intervals: \((-\infty, -1)\), \((-1, 2)\), and \((2, \infty)\). Evaluate the expression \( \frac{(x+1)^{2}}{x-2} \) on each interval to determine the sign. - For \((-\infty, -1)\): Choose \(x=-2\). \(\frac{((-2)+1)^2}{-2-2} = \frac{1}{-4} < 0\).- For \((-1, 2)\): Choose \(x=0\). \(\frac{(0+1)^2}{0-2} = \frac{1}{-2} < 0\).- For \((2, \infty)\): Choose \(x=3\). \(\frac{(3+1)^2}{3-2} = \frac{16}{1} > 0\).

Step 3: Determine Where the Inequality Holds

The inequality \(\frac{(x+1)^2}{x-2} \leq 0\) holds where the expression is less than or equal to zero. - The expression is negative on \((-\infty, -1)\) and \((-1, 2)\).- Check the endpoints: At \(x=-1\), the expression is zero (equal), included in the solution. At \(x=2\), it is undefined (excluded).Therefore, the inequality holds on \((-\infty, -1]\) and \((-1, 2)\).

Step 4: Write the Solution

Based on the sign chart and the conditions of the inequality, the solution is the union of the intervals found in the previous step, which are \((-\infty, -1]\) and \((-1, 2)\).

Key Concepts

Critical Points

Critical points are essential when solving rational inequalities because they indicate where the expression may change its behavior. In the inequality \( \frac{(x+1)^{2}}{x-2} \leq 0 \), critical points occur where the expression equals zero or is undefined.

The expression is zero when the numerator is zero, i.e., \((x+1)^2 = 0\). Solving this, we find \(x = -1\). On the other hand, the expression is undefined when the denominator is zero, i.e., \(x - 2 = 0\). Solving this gives \(x = 2\).

These points, \(x = -1\) and \(x = 2\), split the number line into distinct intervals, which can then be analyzed for signs of positivity or negativity. Given these critical points, you now have nodes on the number line to deeply explore how the inequality behaves in each segment.

Sign Chart

To solve the inequality, creating a sign chart helps to evaluate the behavior of the rational expression in different intervals. The sign chart involves splitting the number line based on the critical points identified.

For \( \frac{(x+1)^{2}}{x-2} \leq 0 \), the critical points divide the number line into three intervals: \((-fty, -1)\), \((-1, 2)\), and \((2, \infty)\).

Understanding how the inequality behaves in these intervals allows you to determine where it holds:

• In \((-fty, -1)\), pick \(x=-2\). The expression \( \frac{((-2)+1)^2}{-2-2} \) evaluates to \(\frac{1}{-4}<0\), indicating negativity.
• In \((-1, 2)\), pick \(x=0\). The expression \( \frac{(0+1)^2}{0-2} \) results in \(\frac{1}{-2}<0\) as well, indicating negativity.
• In \((2, \infty)\), try \(x=3\). The value \( \frac{(3+1)^2}{3-2} = 16 > 0 \) indicates positivity.

By analyzing the sign in each interval, you can observe where the inequality \( \leq 0 \) holds.

Inequality Solutions

Once the sign chart is complete, the final step is to determine where the inequality solutions exist. You’re looking for intervals where the expression is less than or equal to zero. These represent where the inequality \( \frac{(x+1)^2}{x-2} \leq 0 \) holds true.

In our analysis:

• Between \((-fty, -1)\), the expression is negative, meeting the inequality condition.
• Between \((-1, 2)\), the expression remains negative, again satisfying the inequality.
• The expression turns positive in \((2, \infty)\), hence, it does not satisfy the inequality there.

It’s critical to re-check endpoints to ensure they meet any equality conditions. The solution includes:

• The point \(x = -1\), where the numerator equals zero, thus equal to zero is included.
• The point \(x = 2\), where the expression is undefined, hence excluded.

Hence, the solution intervals are \((-fty, -1]\) and \((-1, 2)\). By combining these, you get the final solution for the rational inequality.