Problem 66 - A Graphical Approach to Precalculus with Limits

Question

Solve each rational inequality by hand. $$\frac{5-x}{x^{2}-x-2}<0$$

Step-by-Step Solution

Answer

The solution to the inequality is \((-1, 2) \cup (5, \infty)\).

Step 1: Set the Inequality to Zero

The inequality is \( \frac{5-x}{x^2-x-2} < 0 \). Begin by setting the numerator equal to zero: \( 5 - x = 0 \). Solving this gives \( x = 5 \).

Step 2: Factor the Denominator

Factor the quadratic expression in the denominator: \( x^2 - x - 2 = (x - 2)(x + 1) \). Now, the inequality is \( \frac{5-x}{(x-2)(x+1)} < 0 \).

Step 3: Identify Critical Points

Find the critical points by setting the numerator and the factors of the denominator to zero: \( x = 5, x = 2, \text{ and } x = -1 \). These points will help in testing intervals.

Step 4: Test Intervals

The critical points divide the number line into intervals: \((-\infty, -1), (-1, 2), (2, 5), (5, \infty)\). Test a point from each interval to determine the sign of the inequality. - Choose \( x = -2 \) for \((-\infty, -1)\): evaluates to a positive value.- Choose \( x = 0 \) for \((-1, 2)\): evaluates to a negative value.- Choose \( x = 3 \) for \((2, 5)\): evaluates to a positive value.- Choose \( x = 6 \) for \((5, \infty)\): evaluates to a negative value.

Step 5: Write the Solution Set

From these tests, the intervals where the inequality \( \frac{5-x}{(x-2)(x+1)} < 0 \) holds are \((-1, 2) \cup (5, \infty)\).

Key Concepts

Factoring Quadratics

When dealing with rational inequalities like \( \frac{5-x}{x^2-x-2}<0 \), factoring quadratics in the denominator is crucial. Factoring simplifies the expression and helps to identify the critical points needed for solving the inequality.

The expression \( x^2 - x - 2 \) is a quadratic polynomial and can be tricky if you're not familiar with factoring techniques. To factor this, you need two numbers whose product is the constant term (-2) and whose sum is the linear coefficient (-1). Here, these numbers are -2 and +1.

So, we rewrite the quadratic as \((x-2)(x+1)\). This shows the roots of the quadratic, which are critical to finding where the expression may change sign. Factoring is a beneficial skill as it breaks down complex expressions into simpler parts, making them easier to manipulate and solve.

Critical Points

Once you factor the quadratic, finding the critical points is the next step. These points are where the numerator or the denominator equals zero, as they are essential for understanding the behavior of the inequality.

For \(\frac{5-x}{(x-2)(x+1)} < 0\), set the numerator and the factors of the denominator to zero:

• For the numerator: \(5-x = 0\), solve to find \(x = 5\).
• For the denominator: \((x-2)(x+1) = 0\), solve to find \(x = 2\) and \(x = -1\).

These points \(x = 5, 2, \text{and } -1\) divide the real number line into different intervals. Remember, the critical points help us evaluate the inequality's sign over different intervals.

Testing Intervals

After identifying critical points, testing intervals is the method used to determine the sign of the inequality across these sections. Critical points divide the number line into intervals, and you need to test points from each interval to see where the inequality holds true.

In our example, these intervals are: \((-fty, -1), (-1, 2), (2, 5), (5, fty)\). For each interval, pick a simple representative point and substitute it into the inequality \( \frac{5-x}{(x-2)(x+1)} \). Here’s the breakdown of the interval testing:

• Test \(x = -2\) for \((-fty, -1)\): It results in a positive value.
• Test \(x = 0\) for \((-1, 2)\): It results in a negative value.
• Test \(x = 3\) for \((2, 5)\): It results in a positive value.
• Test \(x = 6\) for \((5, fty)\): It results in a negative value.

This process tells you which intervals satisfy the inequality condition by revealing the sign of the expression there.

Solution Sets

Finally, compiling a solution set is the objective after interval testing. The solution set is where the inequality is satisfied, meaning it holds true. This involves bringing together all intervals with negative values since the inequality is \( \frac{5-x}{(x-2)(x+1)}<0 \).

From our interval testing, the expression is negative in the intervals \((-1, 2) \text{ and } (5, fty)\). This means the solution set of the inequality is \((-1, 2) \cup (5, fty)\).

Solution sets tell you all possible values of \(x\) that satisfy the original inequality. Recall, any boundary points where the denominator equals zero (like at \(x = 2\) and \(x = -1\)) must be excluded from the solution set as they cause the denominator to be zero.