Problem 74 - College Algebra

Question

Find the domain of each function. $$ f(x)=\frac{\sqrt{x-3}}{x-6} $$

Step-by-Step Solution

Answer

The domain of the function \(f(x)=\frac{\sqrt{x-3}}{x-6}\) consists of all real numbers \(x\) that are greater than or equal to 3 and not equal to 6.

Step 1: Address the Square Root

The first condition to consider is the value under the square root. In this case, it is \(x-3\). Since the square root of a negative number is not a real number, \(x-3\) must be greater than or equal to 0. Solve \(x-3 \geq 0\) for \(x\), which gives \(x \geq 3\). So, for the square root to be a real number, \(x\) must be greater than or equal to 3.

Step 2: Address the Denominator

The second condition to consider is the denominator. In this case, it is \(x-6\). A fraction with a denominator equal to zero is undefined. Therefore, \(x-6\) cannot be 0. Solve \(x-6 \neq 0\) for \(x\), which gives \(x \neq 6\). So, for the fraction to be defined, \(x\) cannot be equal to 6.

Step 3: Combine Both Conditions

Combine both conditions to find the domain of the function. The solutions to both conditions are \(x \geq 3\) and \(x \neq 6\). Hence, the domain of \(f(x)\) are all real numbers \(x\) such that \(x \geq 3\) and \(x \neq 6\).

Key Concepts

Square Root Functions

Square root functions are equations that involve the square root of a variable, such as \( \sqrt{x} \). These functions are important because they require special attention when determining their domain. The domain of a square root function consists of all the values that can be substituted for the variable under the square root without resulting in a negative value.

For example, if you have a square root function like \( \sqrt{x-3} \), the expression inside the square root, \( x-3 \), must be greater than or equal to zero for the result to be a real number. This means that:

• We solve for \( x \) by setting the expression \( x-3 \) greater than or equal to zero, like this: \( x-3 \geq 0 \).
• Solving this inequality gives \( x \geq 3 \).

The domain for this square root function is thus all real numbers \( x \) such that \( x \geq 3 \). Ensuring the expression under the square root is non-negative is crucial for defining the domain properly.

Rational Functions

A rational function is expressed as the ratio of two polynomials, where one polynomial is in the numerator, and another in the denominator, such as \( \frac{a(x)}{b(x)} \).

One vital characteristic of rational functions is that they are undefined when the denominator equals zero, since division by zero is not possible.

• For instance, in the function \( \frac{\sqrt{x-3}}{x-6} \), we must ensure that the denominator \( x-6 \) does not equal zero.
• We set up the inequality \( x-6 eq 0 \) and solve for \( x \), resulting in \( x eq 6 \).

This solution means the function does not exist for \( x = 6 \), and therefore this value must be excluded from the domain. Understanding this helps in establishing the domain of rational functions properly by identifying excluded values where the denominator equals zero.

Real Numbers

Real numbers are a broad category of numbers that include both rational and irrational numbers. They cover every possible number on the number line without any gaps: positive, negative, zero, fractions, and decimals are all part of real numbers.

When discussing the domain of a function, specifically in contexts like square root and rational functions, we're usually talking about a subset of the real numbers. The goal is to specify the values within this set that make the function well-defined and produce real outputs.

• For square root functions, all numbers under the root must be non-negative to ensure the result remains real.
• For rational functions, any number that makes the denominator zero must be excluded from the domain.

In summary, the domain for the function \( f(x)=\frac{\sqrt{x-3}}{x-6} \) in real numbers requires us to start from all real numbers and exclude any that fall outside these rules. Therefore, the domain consists of real numbers \( x \) such that \( x \geq 3 \) and \( x eq 6 \). This means combining the constraints from both square root and rational function requirements.