Problem 61 - A Graphical Approach to Precalculus with Limits

Question

Solve each rational inequality by hand. $$\frac{(x+1)(x-2)}{(x+3)}<0$$

Step-by-Step Solution

Answer

The solution is \((-1, 2)\).

Step 1: Find the Critical Points

The critical points occur where the numerator is zero or where the denominator is undefined. For \((x+1)(x-2) = 0\), the solutions are \(x = -1\) and \(x = 2\). The denominator \(x+3 = 0\) when \(x = -3\). Therefore, the critical points are \(x = -3, -1, 2\).

Step 2: Identify Intervals

The critical points divide the number line into intervals: \((-\infty, -3)\), \((-3, -1)\), \((-1, 2)\), and \((2, \infty)\). We will test a value from each interval to determine the sign of the expression in that interval.

Step 3: Test Intervals

Choose test points: \(x = -4\) for \((-\infty, -3)\), \(x = -2\) for \((-3, -1)\), \(x = 0\) for \((-1, 2)\), and \(x = 3\) for \((2, \infty)\). Calculate: - \(x = -4: \frac{((-4+1)(-4-2))}{(-4+3)} = \frac{(3)(-6)}{-1} = 18\) (Positive)- \(x = -2: \frac{((-2+1)(-2-2))}{(-2+3)} = \frac{(-1)(-4)}{1} = 4\) (Positive)- \(x = 0: \frac{((0+1)(0-2))}{(0+3)} = \frac{(1)(-2)}{3} = \frac{-2}{3}\) (Negative)- \(x = 3: \frac{((3+1)(3-2))}{(3+3)} = \frac{(4)(1)}{6} = \frac{2}{3}\) (Positive)

Step 4: Determine Solution Set

From the tests in Step 3, we identified the intervals where the expression is negative: \((-1, 2)\). Since the inequality is strictly less than zero, the intervals where the expression is zero (i.e., the points \(x = -1\) and \(x = 2\)) are not included.

Key Concepts

Critical Points

When solving rational inequalities like \( \frac{(x+1)(x-2)}{(x+3)} < 0 \), the first step is to find the critical points. These are the values of \( x \) where the expression could change its sign.

• Critical points occur either where the numerator of the fraction is zero or where the denominator is zero (making the fraction undefined).
• In this case, the numerator \((x+1)(x-2) = 0\) gives us \(x = -1\) and \(x = 2\).
• The denominator \(x + 3 = 0\) gives \(x = -3\).

So, the critical points are: \( x = -3, -1, \) and \( 2 \). These points divide the entire number line into distinct sections. They serve as boundaries that we can test to see where the inequality might hold true.

Test Intervals

Once the critical points are identified, they divide the number line into intervals. Each interval is defined between consecutive critical points. We call these 'test intervals.'

• For this problem, the intervals are: \((-\infty, -3)\), \((-3, -1)\), \((-1, 2)\), and \((2, \infty)\).
• Within each test interval, we pick any test point (a value that falls within the interval) and determine the sign of the inequality's expression at that point.

These tests help us understand whether the entire interval satisfies the inequality by seeing if the expression yields a positive or negative value at the test point.

Solution Intervals

After testing each interval, the goal is to determine the solution set for the original inequality. This is where the expression satisfies the inequality's sign condition (\(< 0\) in this example).

Interval Analysis:

• For \((-\infty, -3)\) and \((-3, -1)\), the test shows positive results. Thus, they do not satisfy the inequality \(\frac{(x+1)(x-2)}{(x+3)} < 0\).
• The interval \((-1, 2)\) yields a negative value, so it satisfies the inequality.
• The interval \((2, \infty)\) results in a positive value, thus not included in the solution.

Hence, the solution interval is \((-1, 2)\), excluding \(x = -1\) and \(x = 2\) since the inequality is strictly less than zero and not equal to zero.