Problem 64 - A Graphical Approach to College Algebra

Question

Solve each rational inequality by hand. Do not use a calculator. $$\frac{(x-2)^{2}}{2 x}>0$$

Step-by-Step Solution

Answer

The solution is \( (0, 2) \cup (2, \infty) \).

Step 1: Understand the Inequality

We need to solve the inequality \( \frac{(x-2)^{2}}{2x} > 0 \). This inequality is true when the fraction is positive, meaning the numerator and the denominator must have the same sign.

Step 2: Analyze the Numerator

The numerator is \((x-2)^2\), which is always non-negative since it is a square. The value of \((x-2)^2\) is zero when \(x = 2\) and positive otherwise.

Step 3: Analyze the Denominator

The denominator is \(2x\). It is positive when \(x > 0\) and negative when \(x < 0\). The denominator is zero when \(x = 0\), which makes the expression undefined at this point.

Step 4: Determine Critical Points

The critical points are where the numerator is zero \((x = 2)\) and where the expression is undefined \((x = 0)\). These critical points help to define test intervals for the inequality.

Step 5: Test Intervals

Use the critical points to divide the number line into intervals: \((-\infty, 0)\), \((0, 2)\), and \((2, \infty)\). Test each interval:- For \(x = -1\) in \((-\infty, 0)\): \(\frac{((-1)-2)^2}{2(-1)} = \frac{9}{-2} < 0\)- For \(x = 1\) in \((0, 2)\): \(\frac{(1-2)^2}{2(1)} = \frac{1}{2} > 0\)- For \(x = 3\) in \((2, \infty)\): \(\frac{(3-2)^2}{2(3)} = \frac{1}{6} > 0\)

Step 6: Solution Interpretation

From the test intervals, the inequality holds true for intervals \((0, 2)\) and \((2, \infty)\). Remember that \(x = 2\) makes the numerator zero, and \(x = 0\) makes the denominator zero, so these points are excluded.

Key Concepts

Numerator and Denominator Analysis

Understanding the individual components of a fraction in a rational inequality is crucial. Here, the fraction is \( \frac{(x-2)^{2}}{2x} \). The numerator \((x-2)^2\) is a perfect square, which means:

• It is non-negative, because any real number squared is zero or positive.
• It equals zero when \(x = 2\).
• It is positive for all other values of \(x\).

The denominator is \(2x\). Its behavior is more variable:

• Positive when \(x > 0\).
• Negative when \(x < 0\).
• Zero at \(x = 0\), making the whole expression undefined at this point.

By comprehensively analyzing both the numerator and denominator, we prepare ourselves to solve the inequality rationally, focusing on where both parts have the same sign to satisfy the inequality \(> 0\).

Critical Points

Critical points play an essential role in solving rational inequalities. These points arise where:

• The numerator equals zero.
• The denominator equals zero, making the expression undefined.

In this exercise:

• Numerator: \((x-2)^2 = 0\) at \(x = 2\).
• Denominator: \(2x = 0\) at \(x = 0\).

These critical points \((x = 2\) and \(x = 0\)) help segment the number line into intervals. They mark transitions where the sign of the rational expression may change, guiding us in understanding which portions of the number line satisfy the inequality.

Test Intervals

After identifying critical points, \((x = 0\) and \(x = 2\)), the next step is to check intervals created by these points:

• \((-\infty, 0)\)
• \((0, 2)\)
• \((2, \infty)\)

Each interval represents a range of \(x\) values, potentially satisfying our inequality. To find out, we select a test point from each interval:

• In \((-\infty, 0)\), test \(x = -1\). We find the result negative (\(< 0\)).
• In \((0, 2)\), test \(x = 1\). The result is positive (\(> 0\)).
• In \((2, \infty)\), test \(x = 3\). This too gives a positive result (\(> 0\)).

Thus, the inequality holds during \((0, 2)\) and \((2, \infty)\) intervals—whenever the fraction yields a positive value.

Graphical Solutions

Visualizing the rational inequality \(\frac{(x-2)^{2}}{2x}>0\) on a graph can provide intuitive insights:

• The graph shows \(x = 0\) and \(x = 2\) as critical points.
• At \(x = 0\), the graph will exhibit a vertical asymptote due to the fraction's undefined nature.
• The graph crosses the \(x\)-axis at the critical point \(x = 2\), where the numerator is zero.

These features help confirm the analytical solution:

• On intervals \((0, 2)\) and \((2, \infty)\), the graph remains above the \(x\)-axis, aligning with the positive inequality requirement.
• It dips below the \(x\)-axis in the \((-\infty, 0)\) interval.

Through graphics, you can visually verify if the mathematical solutions make sense, offering clarity on the interval sign behavior.