Problem 64 - College Algebra with Modeling and Visualization

Question

Solve the rational inequality. $$ \frac{2 x}{(x-2)^{2}}>0 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Rational Inequality

The inequality given is \( \frac{2x}{(x-2)^2} > 0 \). We need to find where this expression is greater than zero, considering the numerator and the denominator separately.

Step 2: Identify Critical Points

The expression will be zero or undefined where the numerator or denominator are zero. The numerator, \(2x\), is zero at \(x=0\). The denominator, \((x-2)^2\), is zero at \(x=2\). These points are critical points that we will use to analyze the sign of the expression.

Step 3: Determine Intervals

We will divide the number line into intervals using these critical points: \((-\infty, 0)\), \((0, 2)\), and \((2, \infty)\).

Step 4: Test a Value in Each Interval

Select a test point from each interval and determine the sign of the expression:- For \((-\infty, 0)\), choose \(x = -1\): \( \frac{2 \cdot (-1)}{((-1)-2)^2} = \frac{-2}{9} < 0\)- For \((0, 2)\), choose \(x = 1\): \( \frac{2 \cdot 1}{(1-2)^2} = \frac{2}{1} > 0\)- For \((2, \infty)\), choose \(x = 3\): \( \frac{2 \cdot 3}{(3-2)^2} = \frac{6}{1} > 0\)

Step 5: Determine the Solution Set

From our tests, the expression is positive in the intervals \((0, 2)\) and \((2, \infty)\). Since the inequality is strict (> 0), \(x=0\) and \(x=2\) are not included in the solution set.

Step 6: Conclusion: Write the Solution

The solution to the inequality is: \((0, 2) \cup (2, \infty)\).

Key Concepts

Critical Points

Critical points in a rational inequality are the values where the numerator or denominator equals zero. These points are essential because they indicate locations on the number line where the expression can change its sign, be undefined, or be zero.

• For the numerator, the critical point is where it equals zero. In the rational inequality \( \frac{2x}{(x-2)^2} > 0 \), the numerator \( 2x \) equals zero when \( x = 0 \).
• For the denominator, the critical point is where it equals zero because it makes the expression undefined. Here, \((x-2)^2\) equals zero when \( x = 2 \).

These critical points help us break down the number line into segments, allowing us to test each interval individually and determine where the inequality holds true. Remember, division by zero is undefined, which is why we also pay attention to these points even though they don't make the whole expression equal zero.

Intervals

Once you've determined the critical points, you can establish intervals on the number line. Each interval is bordered by the critical points, including infinity on either side when applicable.

• Starting from \(-\infty \) to the first critical point \(x=0\), we have the interval \((-\infty, 0)\).
• The next portion extends from \(0\) to the next critical point \(x=2\), giving us the interval \((0, 2)\).
• Finally, we have \((2, \infty)\) starting from \(x=2\) and extending outward.

Checking the intervals helps in verifying where the inequality holds. Testing a number within each interval tells us if the expression is positive or negative within that part. In this problem, the test points will help you identify which intervals are part of the solution.

Numerator and Denominator Analysis

A detailed analysis of the numerator and the denominator separately helps understand how the rational expression behaves. This is crucial because it determines the sign of the entire expression across different intervals.

• **Numerator Analysis**: In the expression \( \frac{2x}{(x-2)^2} \), the numerator \(2x\) is straightforward. It's positive when \(x > 0\) and negative when \(x < 0\). Thus, it contributes directly to the sign of the expression.
• **Denominator Analysis**: Since the denominator \((x-2)^2\) is squared, it's always non-negative and only zero at \(x = 2\), making the expression undefined there. The squared form means it is always positive except at \(x = 2\).

Analyzing both components together allows us to understand where and why the expression becomes positive or negative. By incorporating both numerator and denominator behaviors, you can systematically choose test points within each identified interval and conclude the validity of the inequality in those intervals.