Problem 54 - Algebra and Trigonometry

Question

Solve each rational inequality in Exercises \(43-60\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{(x+3)(x-2)}{x+1} \leq 0 $$

Step-by-Step Solution

Answer

The solution to the rational inequality is \((-\infty, -3] \cup (-1, 2]\)

Step 1: Find the Critical Points

Solve \((x+3)(x-2) = 0\) and \(x+1 = 0\) to find the critical points. This will result in three critical points: \(x = -3, 2, -1\) (where \(x = -1\) makes the denominator zero and thus, is undefined).

Step 2: Test the Intervals Around the Critical Points

We divide the number line into four intervals based on the critical points: \((-∞, -3)\), \((-3, -1)\), \((-1, 2)\), and \((2, ∞)\). We then choose a test point from each interval (e.g., \(x=-4, -2, 0, 3\)) and substitute it back into the inequality. If it makes the inequality true, then all numbers within that interval are part of the solution.

Step 3: Determine the Solution Intervals

From the second step, we find that the intervals that satisfy the inequality are \((-∞, -3)\) and \((-1, 2)\). However, because the inequality is \(\leq 0\), it includes the points where the expression equals zero. Therefore, the solution set also includes \(x = -3\) and \(x = 2\). But \(x=-1\) is not included because it makes the denominator zero.

Step 4: Express the Solution in Interval Notation

The solution set in interval notation is: \((-\infty, -3] \cup (-1, 2]\)