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Chapter 3: Problem 54

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. $$ \frac{(x+3)(x-2)}{x+1} \leq 0 $$

Short Answer

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

Step by step solution

01

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).
02

Test the Intervals Around the Critical Points

We divide the number line into four intervals based on the critical points: \((-鈭�, -3)\), \((-3, -1)\), \((-1, 2)\), and \((2, 鈭�)\). 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.
03

Determine the Solution Intervals

From the second step, we find that the intervals that satisfy the inequality are \((-鈭�, -3)\) 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.
04

Express the Solution in Interval Notation

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

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