Problem 87 - Precalculus Student Solutions Manual 5th

Question

Solve each rational inequality. Write each solution set in interval notation.4 $$\frac{2 x-3}{x^{2}+1} \geq 0$4

Step-by-Step Solution

Answer

The solution set is \(\left[ \frac{3}{2}, \infty \right)\).

Step 1: Denominator Analysis

Examine the denominator to determine if it affects the inequality's domain. The denominator is given by the expression \(x^2 + 1\). Since \(x^2 + 1 > 0\) for all real numbers, the inequality applies to all real numbers. Thus, no values need to be excluded from the domain.

Step 2: Set Numerator Greater Than or Equal to Zero

Solve the inequality with respect to the numerator: \(2x - 3 \geq 0\). Solve for \(x\):\[2x - 3 \geq 0\]

Step 3: Solving the Inequality

Add 3 to both sides and then divide by 2:\[2x \geq 3 \ x \geq \frac{3}{2}\]

Step 4: Combine Results

The solution set is all values of \(x\) such that \(x \geq \frac{3}{2}\). In interval notation, this is written as: \[\left[ \frac{3}{2}, \infty \right)\]

Key Concepts

Denominator Analysis

Understanding the denominator in a rational inequality is crucial. The 'denominator analysis' helps determine the domains where the inequality is valid. For the given problem, the denominator is expressed as \(x^2 + 1\).
It is important to analyze whether this expression can ever be zero because zero in the denominator would make the fraction undefined. However, for any real number \(x\), \(x^2 + 1 > 0\) always holds true.
This tells us that no values of \(x\) need to be excluded from our domain because \(x^2 + 1\) is never zero and is always positive. The inequality can be considered over all real numbers.

Solving Inequalities

Now, let's dive into the process of 'solving inequalities.' For the inequality \(\frac{2x-3}{x^2+1} \geq 0\), since the denominator is always positive, we only need to focus on the numerator.
We set the numerator greater than or equal to zero: \(2x - 3 \geq 0\).
From here, solving for \(x\) involves basic algebraic steps:

• Add 3 to both sides resulting in: \(2x \geq 3\)
• Divide both sides by 2 to isolate \(x\): \(x \geq \frac{3}{2}\)

Thus, the values of \(x\) that make the numerator nonnegative are all \(x\) such that \(x \geq \frac{3}{2}\).

Interval Notation

Having found the solution set, it's essential to present it in 'interval notation' for clarity and precision. The result from our inequality solving gives us \(x \geq \frac{3}{2}\).
In interval notation, we write this as \[\left[ \frac{3}{2}, \infty \right)\].
Here's a breakdown:

• \(\left[ \frac{3}{2}, \infty \right)\) signifies that our solution starts at \(\frac{3}{2}\) and includes \(\frac{3}{2}\) as indicated by the bracket \([\; ]\)
• The interval extends towards infinity, represented by \(\infty\), and since infinity is not a number we can reach, it is denoted with a parenthesis \(( \; )\)

By using interval notation, we precisely communicate the range of \(x\) values that satisfy the rational inequality.