Problem 8 - College Algebra

Question

Determine the domain for each function in interval notation. Given \(f(x)=\frac{1}{x-4}\) and \(g(x)=\frac{1}{6-x},\) find \(f+g, f-g, f g,\) and \(\frac{f}{g}\)

Step-by-Step Solution

Answer

The domain for all expressions is \((-\infty, 4) \cup (4, 6) \cup (6, \infty)\).

Step 1: Identify Domain Restrictions for f(x)

The function is given as \( f(x) = \frac{1}{x-4} \). The denominator cannot be zero, so set it to zero and solve for \( x \), which gives \( x = 4 \). Therefore, \( f(x) \) is undefined at \( x = 4 \). The domain of \( f(x) \) is all real numbers except 4, written in interval notation as \((-\infty, 4) \cup (4, \infty)\).

Step 2: Identify Domain Restrictions for g(x)

The function is \( g(x) = \frac{1}{6-x} \). Similarly, the denominator cannot be zero, so solve \( 6-x=0 \) to find \( x = 6 \). Thus, \( g(x) \) is undefined at \( x=6 \). The domain of \( g(x) \) is all real numbers except 6, written in interval notation as \((-\infty, 6) \cup (6, \infty)\).

Step 3: Determine Domain for f + g and f - g

To find the domain for \( f+g \) and \( f-g \), both functions must be defined. The restrictions from both \( f(x) \) (\( x eq 4 \)) and \( g(x) \) (\( x eq 6 \)) apply. Thus, the domain is their intersection: \((-\infty, 4) \cup (4, 6) \cup (6, \infty)\).

Step 4: Determine Domain for fg

The domain for the product \( fg \) is also limited by the points where either function is undefined. Therefore, the domain is the same as the domain for \( f+g \) and \( f-g \): \((-\infty, 4) \cup (4, 6) \cup (6, \infty)\).

Step 5: Determine Domain for \(\frac{f}{g}\)

For \( \frac{f}{g} \), in addition to where \( f(x) \) and \( g(x) \) are defined, \( g(x) \) must not equal zero to avoid zero in the denominator of the fraction. Since \( g(x) = \frac{1}{6-x} \) is only zero when the denominator is zero (which is already checked), no additional points are removed from the domain. Hence, the domain is \((-\infty, 4) \cup (4, 6)\cup (6, \infty)\).

Key Concepts

Interval Notation

When discussing domains of functions, interval notation is a crucial tool for representing sets of real numbers. It helps express the range of values a function can accept. In the case of the functions \(f(x) = \frac{1}{x-4}\) and \(g(x) = \frac{1}{6-x}\), certain values make the functions undefined, such as when the denominator equals zero. For \(f(x)\), this occurs at \(x = 4\), and for \(g(x)\), it's at \(x = 6\).
Interval notation allows us to exclude these points from the domain by using parentheses, which indicate non-inclusive boundaries.

• The domain of \(f(x)\), avoiding 4, is \((-\infty, 4) \cup (4, \infty)\).
• Conversely, the domain of \(g(x)\), excluding 6, is \((-\infty, 6) \cup (6, \infty)\).

This notation conveys clear information about which values are permissible for the input of a function and highlights exclusion points where the function becomes undefined.

Function Operations

Function operations involve combining two functions using basic arithmetic: addition, subtraction, multiplication, and division. When working with operations on \(f(x)\) and \(g(x)\), determining the resulting domain is essential. Each operation adheres to rules to ensure validity:

• For \(f+g\) and \(f-g\), the domain is the intersection of the domains of \(f(x)\) and \(g(x)\) because both functions must be valid for the result to be defined.
• For \(fg\), the process is similar; the domain is where both functions overlap in terms of validity.
• When dividing, as in \(\frac{f}{g}\), an additional consideration is required: the denominator \(g(x)\) cannot be zero.

Overall, the key is ensuring each operation's resulting domain accounts for initial restrictions, maintaining integrity across each mathematical step.

Undefined Points

Identifying undefined points in a function is a critical aspect of finding its domain. These points occur where the function's denominator becomes zero, leading to an undefined expression.
For example, \(f(x) = \frac{1}{x-4}\) is undefined at \(x = 4\) since it results in division by zero. Similarly, \(g(x) = \frac{1}{6-x}\) is undefined at \(x = 6\). These points show critical areas of concern:

• At these undefined points, you typically exclude them from the domain using interval notation.
• Each undefined point is crucial in that it marks boundaries where the function's behavior changes significantly.
• For function operations, these undefined points are considered collectively to determine where the combined functions remain defined.

Remember, addressing undefined points ensures clarity in defining domains and helps avoid errors in calculations or misunderstandings in the behavior of functions.