Problem 27 - Calculus with Concepts in Calculus

Question

Solve the inequality. $$ \frac{(2 x-1)^{2}}{(x+1)(x+3)}>0 $$

Step-by-Step Solution

Answer

The solution is \(x \in (-3, -1) \cup (-1, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\).

Step 1: Identify Sign-Changing Points

First, we identify the points where the inequality's expression changes its sign. These points come from setting the numerator and the denominator equal to zero. For the numerator, \((2x-1)^2 = 0\), resulting in \(x = \frac{1}{2}\). For the denominator, \((x+1)(x+3) = 0\), we find \(x = -1\) and \(x = -3\). Thus, the sign-changing points are \(x = \frac{1}{2}, -1, -3\).

Step 2: Create Test Intervals

The sign-changing points divide the real number line into the intervals: \((-\infty, -3)\), \((-3, -1)\), \((-1, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\). We'll test each interval to determine where the expression is positive.

Step 3: Test Each Interval

Select a test point from each interval and evaluate the expression's sign:- For \((-\infty, -3)\), choose \(x = -4\): \(\frac{(2(-4)-1)^2}{(-4+1)(-4+3)} = \frac{49}{-3} < 0\)- For \((-3, -1)\), choose \(x = -2\): \(\frac{(2(-2)-1)^2}{(-2+1)(-2+3)} = \frac{9}{1} > 0\)- For \((-1, \frac{1}{2})\), choose \(x = 0\): \(\frac{(2(0)-1)^2}{(0+1)(0+3)} = \frac{1}{3} > 0\)- For \((\frac{1}{2}, \infty)\), choose \(x = 1\): \(\frac{(2(1)-1)^2}{(1+1)(1+3)} = \frac{1}{8} > 0\).Thus, the expression is positive for the intervals \((-3, -1)\), \((-1, \frac{1}{2})\), and \((\frac{1}{2}, \infty)\).

Step 4: Consider Excluded Points

Exclude the points where the denominator becomes zero, which are \(x = -1\) and \(x = -3\), since division by zero is undefined. The roots of the numerator, \(x = \frac{1}{2}\), do not change the sign since they are squared and do not affect positivity or negativity of expression.

Step 5: Write the Solution

Taking into account the valid intervals and excluded points, the solution set for the original inequality is:\(x \in (-3, -1) \cup (-1, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\).

Key Concepts

Sign-Changing Points

When solving inequalities involving rational expressions, finding the **sign-changing points** is a crucial first step. These are the points where the expression changes its sign from positive to negative or vice-versa.

To find these points, we set both the numerator and the denominator of the rational expression equal to zero. For instance, for \(\frac{(2x-1)^2}{(x+1)(x+3)}\), we solve:

• \((2x-1)^2 = 0\) which gives \(x = \frac{1}{2}\).
• \((x+1) = 0\) or \((x+3) = 0\) giving \(x = -1\) and \(x = -3\).

These solutions mark where the rational expression might change its sign. These points help divide the number line into different intervals for further testing.

Rational Expressions

Rational expressions are similar to fractions but involve polynomials in the numerator and the denominator. In our exercise, the rational expression is \(\frac{(2x-1)^2}{(x+1)(x+3)}\).

This particular rational expression has a quadratic numerator and a linear denominator. Understanding the behavior of each part is essential.

The numerator, \((2x-1)^2\), which is always non-negative (since it's squared) achieves zero at \(x = \frac{1}{2}\). Meanwhile, the denominator \((x+1)(x+3)\) becomes zero at \(x = -1\) and \(x = -3\).

The solution to the inequality depends heavily on where the numerator is positive or zero, and where the denominator changes sign, affecting the overall sign of the expression.

Test Intervals

After identifying the sign-changing points, the next step involves creating and analyzing **test intervals**. This means dividing the real number line into intervals based on the sign-changing points.

From the previous steps, our intervals are:

• \((\infty, -3)\)
• \((-3, -1)\)
• \((-1, \frac{1}{2})\)
• \((\frac{1}{2}, \infty)\)

In each of these intervals, we pick a test point and substitute it back into the inequality to check the sign of the rational expression. This analysis determines in which intervals the original inequality holds true. For example, if substitution into the expression yields a positive result, that interval satisfies our inequality.

Solution Sets

The **solution set** of an inequality is the set of values that satisfy the inequality condition. Based on our test intervals for \(\frac{(2x-1)^2}{(x+1)(x+3)} > 0\), we identified the intervals where the expression remains positive.

These intervals were:

• \((-3, -1)\)
• \((-1, \frac{1}{2})\)
• \((\frac{1}{2}, \infty)\)

Importantly, we exclude \(x = -1\) and \(x = -3\) from the solution set because the expression is undefined at these points due to division by zero. By considering these aspects, we determine that the solution to the inequality, expressed in interval notation, is:

• \(x \in (-3, -1) \cup (-1, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\).