Problem 74 - College Algebra with Modeling and Visualization

Question

Solve the rational inequality. $$ \frac{1}{x-1}+\frac{1}{x+1}>\frac{3}{4} $$

Step-by-Step Solution

Answer

The solution is \(-1 < x < -\frac{3}{4} \) and \(-\frac{3}{4} < x < 1\).

Step 1: Combine the Fractions on the Left-Hand Side

To combine the fractions, find a common denominator for the two fractions on the left-hand side. This will be \((x-1)(x+1)\). Thus, the expression becomes:\[\frac{(x+1) + (x-1)}{(x-1)(x+1)} = \frac{2x}{x^2-1}\].

Step 2: Bring All Terms to One Side

Subtract \(\frac{3}{4}\) from both sides to set the inequality to zero:\[\frac{2x}{x^2-1} - \frac{3}{4} > 0\].

Step 3: Create a Single Fraction

Express the left-hand side as a single fraction by finding a common denominator (the product of the two denominators):\[\frac{8x - 3x^2 + 3}{4(x^2-1)} > 0\].

Step 4: Analyze Critical Points and Intervals

Identify where the expression is undefined (denominator zero) or zero (numerator zero). The critical points are: solutions of \(x^2 = 1\) (i.e., \(x = 1\), \(x = -1\)), as well as solutions of \(8x - 3x^2 + 3 = 0\). Solving the quadratic gives roots \(-1\) and \(-\frac{3}{4}\).

Step 5: Test Intervals Between Critical Points

Evaluate the inequality signs in intervals set by critical points \(-\infty, -1, -\frac{3}{4}, 1\), and \(\infty\). Use test points from each interval to determine where the expression is positive. The signs change at \(x = -\frac{3}{4}\) and \(x = 1\).

Step 6: Determine the Solution Set

From the interval test, the expression is positive in \(-1 < x < -\frac{3}{4}\) and \(-\frac{3}{4} < x < 1\). These intervals avoid division by zero and satisfy the inequality.

Key Concepts

Fraction Operations

Fraction operations are crucial when dealing with rational inequalities because they enable us to manipulate expressions and find common denominators. In the given inequality \( \frac{1}{x-1} + \frac{1}{x+1} > \frac{3}{4} \), we first need to combine the fractions on the left-hand side. To do this, we find a common denominator, which is \((x-1)(x+1)\).
This allows us to rewrite the expression as one fraction:

• Combine numerators: \((x+1) + (x-1) = 2x\).
• Denominator: \((x-1)(x+1)\) simplifies to \(x^2 - 1\).

Now, the inequality becomes \( \frac{2x}{x^2-1} > \frac{3}{4} \). By subtracting \(\frac{3}{4}\) from both sides, we can work with a single, simplified fraction form.

Quadratic Equations

Quadratic equations play a vital role in finding critical points in rational inequalities. After formulating the single fraction \(\frac{8x - 3x^2 + 3}{4(x^2-1)} > 0\), we need solutions of its numerator to find where the expression equals zero. These solutions are obtained by solving the quadratic equation \(8x - 3x^2 + 3 = 0\).
The standard method of solving quadratic equations involves factoring or using the quadratic formula. Here, the equation factors easily, showing roots at \(x = -1\) and \(x = -\frac{3}{4}\). These roots signify crucial points where the expression potentially changes its sign as we determine the solution set for the inequality.

Critical Points

Critical points in rational inequalities are the values of \(x\) where the expression switches from positive to negative or vice versa, or where the denominator equals zero, leading to undefined expressions. For the expression \( \frac{8x - 3x^2 + 3}{4(x^2-1)} \), critical points must be identified.
The denominator \(x^2 - 1 = 0\) gives critical points \(x = 1\) and \(x = -1\). These points do not make the inequality true since they cause division by zero. The numerator \(8x - 3x^2 + 3 = 0\), creating zeros at \(x = -1\) and \(x = -\frac{3}{4}\), reveals switching points.

• The segments between these critical values need to be tested for positivity or negativity of the expression.

Test Intervals

Test intervals are used to analyze which segments between critical points satisfy the inequality. Between each pair of consecutive critical points \(-\infty, -1, -\frac{3}{4}, 1, \infty\), we choose test points and substitute them into the inequality \(\frac{8x - 3x^2 + 3}{4(x^2-1)} > 0\) to check for positivity.
Select test points like \(x = -2, -0.9, 0, 0.5\). Evaluate each:

• Substitute into simplified fraction.
• Determine if the result is greater than or less than zero.

Findings reveal the expression stays positive over the intervals \(-1 < x < -\frac{3}{4}\) and \(-\frac{3}{4} < x < 1\). These are the intervals where the inequality holds true, ensuring no division by zero, and leading us to identify the correct solution set.