Problem 69 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The solution is \((-\infty, 0) \cup [\frac{1}{2}, 2] \cup (2, \infty)\).

Step 1: Express as a Single Fraction

First, we rewrite the inequality to have a common denominator. The terms in the inequality are \(2\), \(\frac{-5}{x}\), and \(\frac{2}{x^{2}}\). The common denominator is \(x^2\), so we can rewrite each term: * \(2 = \frac{2x^2}{x^2}\) * \(-\frac{5}{x} = \frac{-5x}{x^2}\) * \(\frac{2}{x^2}\) is already in terms of \(x^2\). Thus, the inequality becomes: \[ \frac{2x^2 - 5x + 2}{x^2} \geq 0 \]

Step 2: Determine the Critical Points

Set the numerator of the rational expression equal to zero: \[ 2x^2 - 5x + 2 = 0 \] This is a quadratic equation. Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -5\), and \(c = 2\), we find the roots:\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{4} \] \[ x = \frac{5 \pm \sqrt{9}}{4} \] \[ x = \frac{5 \pm 3}{4} \] which gives us \(x = 2\) and \(x = \frac{1}{2}\). Also, the denominator \(x^2\) is zero when \(x = 0\). These values are critical points: \(x = \frac{1}{2}, 0, 2\).

Step 3: Test the Intervals

The critical points divide the number line into intervals: \((-\infty, 0)\), \((0, \frac{1}{2})\), \((\frac{1}{2}, 2)\), and \((2, \infty)\). We need to test points from each interval in the inequality to determine where it holds.- **Interval \((-\infty, 0)\):** Choose \(x = -1\) \[ \frac{2(-1)^2 - 5(-1) + 2}{(-1)^2} = \frac{2 + 5 + 2}{1} = 9 \geq 0 \]- **Interval \((0, \frac{1}{2})\):** Choose \(x = \frac{1}{4}\) \[ \frac{2(\frac{1}{4})^2 - 5(\frac{1}{4}) + 2}{(\frac{1}{4})^2} = \frac{\frac{2}{16} - \frac{5}{4} + 2}{\frac{1}{16}} < 0 \]- **Interval \((\frac{1}{2}, 2)\):** Choose \(x = 1\) \[ \frac{2(1)^2 - 5(1) + 2}{(1)^2} = \frac{2 - 5 + 2}{1} = -1 < 0 \]- **Interval \((2, \infty)\):** Choose \(x = 3\) \[ \frac{2(3)^2 - 5(3) + 2}{(3)^2} = \frac{18 - 15 + 2}{9} > 0 \] Having tested the intervals, the inequality is satisfied in \((-\infty, 0)\) and \((2, \infty)\).

Step 4: Combine Results and Write Solution

The rational inequality holds for the intervals determined in the previous step. At each critical point, check if equality holds:- At \(x = \frac{1}{2}\) and \(x = 2\), we substitute back into the factor \((2x^2 - 5x + 2 = 0)\), so these could be included based on the original inequality \(\geq\). However, \(x = 0\) makes the expression undefined, so we exclude 0.Thus, the solution to the inequality is:\[ (-\infty, 0) \cup [\frac{1}{2}, 2] \cup (2, \infty) \]

Key Concepts

Critical Points

When dealing with rational inequalities, critical points are essential for breaking down and analyzing the behavior of the inequality across different intervals on the number line. These critical points are typically where the expression might change sign, which includes the roots of the numerator and the points where the denominator is zero.

In our example inequality \( \frac{2x^2 - 5x + 2}{x^2} \geq 0 \), the critical points arise from both the numerator and the denominator.

To find the critical points:

• Numerator: Set the numerator of the expression to zero, which gives us \( 2x^2 - 5x + 2 = 0 \). Solving this using the quadratic formula yields roots, which are our critical points.
• Denominator: Identify where the denominator equals zero, \( x^2 = 0 \), which in this case is at \( x=0 \).

These points, \( x = \frac{1}{2}, 0, \) and \( 2 \), are critical because they are the values where the function changes its state of positivity or negativity.

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation of the form \( ax^2 + bx + c = 0 \). This is crucial for solving rational inequalities as it helps in determining the critical points where the expression crosses zero by solving the numerator.

The formula is expressed as follows:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \) is the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \), and \( c \) is the constant term.
For our specific problem:

• \( a = 2 \)
• \( b = -5 \)
• \( c = 2 \)

Substituting these into the formula gives us the roots \( x = 2 \) and \( x = \frac{1}{2} \). These are critical as they indicate points at which the entire rational function can potentially change its sign, crucial for solving the inequality \( \frac{2x^2 - 5x + 2}{x^2} \geq 0 \).

Testing Intervals

Once critical points have been identified, the next step is to test the intervals they create on the number line. Testing intervals is about checking the sign of the rational expression between critical points, which tells us where the inequality holds true.

This process consists of the following steps:

• Identify Intervals: Critical points divide the number line into segments. For our inequality, these intervals are \((-\infty, 0)\), \((0, \frac{1}{2})\), \((\frac{1}{2}, 2)\), and \((2, \infty)\).
• Test Each Interval: Select a test point from each interval and substitute it into the inequality. This helps determine if the function is positive or negative in that section.

Let's look at some specific interval tests:

• For the interval \((-\infty, 0)\), using \( x = -1 \), the expression evaluates to positive.
• For \((0, \frac{1}{2})\), using \( x = \frac{1}{4} \), the expression is negative.
• For \((\frac{1}{2}, 2)\), using \( x = 1 \), it's negative, and for \((2, \infty)\), using \( x = 3 \), it's positive.

The inequality \( \frac{2x^2 - 5x + 2}{x^2} \geq 0 \) holds true in the intervals \((-\infty, 0)\), \([\frac{1}{2}, 2]\), and \((2, \infty)\), which we identify through careful interval testing.