Problem 29 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 21-32: Find the domain of \(f\). $$ f(x)=\frac{\sqrt{4 x-3}}{x^{2}-4} $$

Step-by-Step Solution

Answer

The domain of \(f(x)\) is \([\frac{3}{4}, 2) \cup (2, \infty)\)."

Step 1: Identify conditions for the numerator

The numerator of the function is \(\sqrt{4x-3}\). A square root is defined when the expression inside the square root is greater than or equal to zero. So, set up the inequality: \(4x-3 \geq 0\). Solving this inequality gives: \[4x \geq 3\] \[x \geq \frac{3}{4}\]. This means that \(x\) must be greater than or equal to \(\frac{3}{4}\) for the square root to be defined.

Step 2: Identify conditions for the denominator

The denominator of the function is \(x^2 - 4\). The function is undefined when the denominator is zero. Setting \(x^2 - 4 = 0\) gives the solutions \(x^2 = 4\), resulting in \(x = 2\) or \(x = -2\). Therefore, \(x\) cannot be 2 or -2.

Step 3: Combine conditions for the domain

To find the domain of \(f(x)\), combine the conditions identified in Step 1 and Step 2. The function is defined where \(x \geq \frac{3}{4}\) and \(x\) is not equal to 2. Thus, the domain in interval notation is \([\frac{3}{4}, 2) \cup (2, \infty)\).

Key Concepts

Radical Expressions

Radical expressions involve the square root symbol, typically represented by \(\sqrt{}\). They are commonly used to simplify complex mathematical expressions. The notation \(\sqrt{a}\) means you are looking for the number that, when multiplied by itself, gives \(a\). In terms of domain, a radical expression with a square root must have a non-negative value inside the square root. This means the expression within the radical cannot be negative since the square root of a negative number is not defined in the set of real numbers.

For instance, if you have \(\sqrt{4x-3}\), you need \(4x-3\) to be at least 0, so the expression remains valid. This requires setting up the inequality \(4x-3 \geq 0\) and solving for \(x\). By simplifying \(4x \geq 3\), we determine \(x \geq \frac{3}{4}\). Thus, the smallest value of \(x\) that keeps \(\sqrt{4x-3}\) real is \(\frac{3}{4}\).

Rational Functions

A rational function is essentially a fraction, where both the numerator and the denominator are polynomials. It is expressed in the form \(\frac{P(x)}{Q(x)}\), where \(Q(x)\) is not equal to zero. Rational functions can become undefined if the denominator equals zero, which is a key consideration when determining their domain.

In the function \(f(x) = \frac{\sqrt{4x-3}}{x^2-4}\), the denominator \(x^2-4\) must not be zero. Solving \(x^2-4=0\) leads to \(x=2\) and \(x=-2\). Hence, these values are not permissible, making the function undefined at those points. The important point here is to identify and exclude these points to ensure the function operates smoothly across its domain. By focusing on these zero points of the denominator, you can note these exclusions when forming the domain.

Interval Notation

Interval notation is a mathematical shorthand used to describe sets of numbers, often used to specify the domain of a function. It helps indicate where a function is defined or fulfills specific conditions.

When writing intervals, square brackets \([\text{ ]}\) indicate that an endpoint is included, and parentheses \((\text{ )}\) show that it's not included. For the function \(f(x) = \frac{\sqrt{4x-3}}{x^2-4}\), we previously discovered the conditions \(x \geq \frac{3}{4}\) and that \(x\) cannot be \(2\). Combining these results gives the interval notation \([\frac{3}{4}, 2) \cup (2, \infty)\).

This indicates that \(x\) can be any number from \(\frac{3}{4}\) to \(2\) (excluding \(2\)), or any number greater than \(2\). The union symbol \(\cup\) demonstrates that the domain consists of two separate pieces, reflecting the points where the expression inside the denominator is not zero and the square root remains real.