Problem 65 - A Graphical Approach to College Algebra

Question

Solve each rational inequality by hand. Do not use a calculator. $$\frac{2 x-5}{x^{2}-1} \geq 0$$

Step-by-Step Solution

Answer

The solution is \((-1, 1)\cup \left[\frac{5}{2}, \infty\right)\).

Step 1: Finding Critical Points

First, identify the critical points where the expression could be zero or undefined. This occurs when the numerator or the denominator is zero. For the numerator, solve: \(2x - 5 = 0\). Hence, \(x = \frac{5}{2}\). For the denominator, \(x^2 - 1 = 0\). Factor as \((x - 1)(x + 1) = 0\), giving \(x = 1\) and \(x = -1\). Thus, the critical points are \(-1, 1,\) and \(\frac{5}{2}\).

Step 2: Test Intervals

Divide the number line into intervals based on the critical points: \((-\infty, -1), (-1, 1), (1, \frac{5}{2}), \left(\frac{5}{2}, \infty\right)\). For each interval, choose a test point to determine the sign of the expression in that interval.

Step 3: Testing Interval \((-\infty, -1)\)

Choose \(x = -2\): \(\frac{2(-2) - 5}{(-2)^2 - 1} = \frac{-4 - 5}{4 - 1} = \frac{-9}{3} = -3\). Since the result is negative, \(\frac{2x - 5}{x^2 - 1} < 0\) for \((-\infty, -1)\).

Step 4: Testing Interval \((-1, 1)\)

Choose \(x = 0\): \(\frac{2(0) - 5}{0^2 - 1} = \frac{-5}{-1} = 5\). Since the result is positive, \(\frac{2x - 5}{x^2 - 1} > 0\) for \((-1, 1)\).

Step 5: Testing Interval \((1, \frac{5}{2})\)

Choose \(x = 2\): \(\frac{2(2) - 5}{2^2 - 1} = \frac{4 - 5}{4 - 1} = \frac{-1}{3}\). Since the result is negative, \(\frac{2x - 5}{x^2 - 1} < 0\) for \((1, \frac{5}{2})\).

Step 6: Testing Interval \(\left(\frac{5}{2}, \infty\right)\)

Choose \(x = 3\): \(\frac{2(3) - 5}{3^2 - 1} = \frac{6 - 5}{9 - 1} = \frac{1}{8}\). Since the result is positive, \(\frac{2x - 5}{x^2 - 1} > 0\) for \(\left(\frac{5}{2}, \infty\right)\).

Step 7: Assembling the Solution Set

Based on the signs from the intervals, the inequality \(\frac{2x - 5}{x^2 - 1} \geq 0\) holds for the intervals where it is positive or zero. This is \((-1, 1)\) and \(\left(\frac{5}{2}, \infty\right)\), including the endpoint \(x = \frac{5}{2}\) since the expression is zero at this point.

Key Concepts

Critical Points

Rational inequalities often involve finding the values that make the numerator zero or the denominator zero. These special values are known as critical points. The numerator of our inequality is determined by solving the equation where the denominator does not come into play, meaning we set it to zero:

• For the numerator in this case, we solve: \(2x - 5 = 0\). This gives \(x = \frac{5}{2}\).
• The denominator is \(x^2 - 1 = (x-1)(x+1) = 0\). So, setting these factors to zero provides solutions \(x = 1\) and \(x = -1\).

Ultimately, we have three critical points: \(-1, 1,\) and \(\frac{5}{2}\). Identifying critical points is crucial because they help us define the intervals we need for further analysis. These points highlight where the rational expression could change its sign or become undefined, setting the stage for evaluating the behavior of the expression across the real line.

Sign Analysis

Sign analysis is the process of determining whether the rational inequality is positive or negative over certain intervals of real numbers. We analyze these signs between and beyond our critical points.The goal is to determine in which intervals the expression \(\frac{2x - 5}{x^2 - 1}\) is positive (or zero) since the inequality stipulates \(\geq 0\). It involves selecting a test point in each interval created by the critical points and substituting these values back into the inequality to evaluate its sign:

• For instance, choosing \(x = -2\) in the interval \((-\infty, -1)\) results in a negative sign, while \(x = 0\) in the interval \((-1, 1)\) turns the expression positive.
• Continuing similarly with intervals \((1, \frac{5}{2})\) using \(x = 2\) and \((\frac{5}{2}, \infty)\) with \(x = 3\) gives negative and positive signs, respectively.

This step clarifies where the inequality holds true and where it does not. Sign analysis lays down the groundwork necessary to systematically assemble a solution set for the inequality.

Interval Testing

Interval testing links closely with sign analysis. It involves choosing test points from each defined interval from our critical points and substituting those into our inequality. Evaluating these test points helps in confirming whether specific intervals satisfy the conditions of our inequality.Determine the sign (whether greater than or equal to zero) by substituting:

• From \((-\infty, -1)\) with a test point like \(x = -2\), you calculate \(\frac{-9}{3} = -3\), showing it's negative.
• Choose \(x = 0\) in \((-1, 1)\) to get \(5\), indicating a positive value.
• For \((1, \frac{5}{2})\), use \(x = 2\) resulting in \(\frac{-1}{3}\), another negative.
• Finally, test \(x = 3\) from \((\frac{5}{2}, \infty)\) to obtain \(\frac{1}{8}\), which is positive.

This method validates our intervals for solution assembly, clarifying the sequence before deciding the valid intervals that make our inequality true.

Solution Set Assembly

Once we have tested the intervals and know which ones satisfy the inequality, it's time to assemble the solution set. We’re looking for where the expression is non-negative.Based on previous tests, the expression is positive or zero in:

• The interval \((-1, 1)\), where we found a positive result.
• The interval \((\frac{5}{2}, \infty)\), another region of positivity. Remember, \(x = \frac{5}{2}\) sets the numerator zero, indicating a boundary point satisfying \(\geq 0\).

So the solution set can be expressed as a union of these intervals: \((-1, 1) \cup [\frac{5}{2}, \infty)\)Assembling this solution set concludes our approach to see where the original inequality holds true. Each part of our process—critical points, sign analysis, interval testing—leads us to a comprehensive result about in which intervals the inequality is fulfilled.