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Magazine Chapter 4: Problem 64
Solve each rational inequality by hand. Do not use a calculator. $$\frac{(x-2)^{2}}{2 x}>0$$
Short Answer
Expert verified
The solution is \( (0, 2) \cup (2, \infty) \).
Step by step solution
01
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.
02
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.
03
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.
04
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.
05
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 \((2, \infty)\): \(\frac{(3-2)^2}{2(3)} = \frac{1}{6} > 0\)
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \(x\).
- Positive when \(x > 0\).
- Negative when \(x < 0\).
- Zero at \(x = 0\), making the whole expression undefined at this point.
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.
- Numerator: \((x-2)^2 = 0\) at \(x = 2\).
- Denominator: \(2x = 0\) at \(x = 0\).
Test Intervals
After identifying critical points, \((x = 0\) and \(x = 2\)), the next step is to check intervals created by these points:
- \((-\infty, 0)\)
- \((0, 2)\)
- \((2, \infty)\)
- 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 (\(> 0\)).
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.
- 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.
