Problem 21 - College Algebra with Modeling and Visualization

Question

Exercises \(11-30:\) Use \(f(x)\) and \(g(x)\) to find a formula for each expression. Identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d) \((f / g)(x)\) $$ f(x)=\frac{1}{2 x-4}, \quad g(x)=\frac{x}{2 x-4} $$

Step-by-Step Solution

Answer

(a) \(\frac{1+x}{2x-4}\), \(x \neq 2\); (b) \(\frac{1-x}{2x-4}\), \(x \neq 2\); (c) \(\frac{x}{(2x-4)^2}\), \(x \neq 2\); (d) \(\frac{1}{x}\), \(x \neq 0, 2\).

Step 1: Understand Function Operations

The expressions given are basic operations on functions. To solve these, we'll use the rules for addition, subtraction, multiplication, and division of functions.1. **Addition**: \((f+g)(x) = f(x) + g(x)\)2. **Subtraction**: \((f-g)(x) = f(x) - g(x)\)3. **Multiplication**: \((f \cdot g)(x) = f(x) \cdot g(x)\)4. **Division**: \((f/g)(x) = \frac{f(x)}{g(x)}\) provided \(g(x) eq 0\).

Step 2: Addition of Functions

Compute \((f+g)(x)\) by adding the functions:\[ (f+g)(x) = \frac{1}{2x-4} + \frac{x}{2x-4} \]Since they have the same denominator, combine the numerators:\[ (f+g)(x) = \frac{1 + x}{2x-4} \]

Step 3: Identify Domain for Addition

The domain of \((f+g)(x)\) is all real numbers except where the denominator is zero. Solve:\[ 2x-4 = 0 \rightarrow x = 2 \]Therefore, the domain is all real numbers \(x eq 2\).

Step 4: Subtraction of Functions

Compute \((f-g)(x)\) by subtracting the functions:\[ (f-g)(x) = \frac{1}{2x-4} - \frac{x}{2x-4} \]Combine the numerators:\[ (f-g)(x) = \frac{1 - x}{2x-4} \]

Step 5: Identify Domain for Subtraction

The domain of \((f-g)(x)\) is similar to that of addition. Solve:\[ 2x-4 = 0 \rightarrow x = 2 \]The domain is all real numbers \(x eq 2\).

Step 6: Multiplication of Functions

Compute \((f \cdot g)(x)\) by multiplying the functions:\[ (f \cdot g)(x) = \frac{1}{2x-4} \times \frac{x}{2x-4} = \frac{x}{(2x-4)^2} \]

Step 7: Identify Domain for Multiplication

The domain is all real numbers except where \((2x-4)^2 = 0\):\[ 2x-4 = 0 \rightarrow x = 2 \]Thus, the domain is all real numbers \(x eq 2\).

Step 8: Division of Functions

Compute \((f/g)(x)\) by dividing the functions:\[ (f/g)(x) = \frac{\frac{1}{2x-4}}{\frac{x}{2x-4}} = \frac{1}{x} \]Ensure \(g(x) eq 0\), i.e., \(x eq 0\) and \(x eq 2\) (for division).

Step 9: Identify Domain for Division

The domain for \((f/g)(x)\) is where both the denominator of \(f\) and \(g\) are non-zero:- From \(g(x)\), \(x eq 2\).- From the division \(\frac{1}{x}\), \(x eq 0\).The domain is all real numbers \(x eq 2\) and \(x eq 0\).

Key Concepts

Domain of Functions

The domain of a function is the set of all possible inputs for which the function is defined. Understanding the domain is crucial because it tells us the values that can be used without causing any mathematical issues such as division by zero or the square root of a negative number.
To find the domain, we need to identify where the function might encounter such issues and exclude those values. For example:

• If a function has a denominator, like \( \frac{1}{2x-4} \), set the denominator not equal to zero: \( 2x-4 eq 0 \), leading to \( x eq 2 \).
• In functions involving division of two functions, ensure the denominator function is non-zero too.

In our exercises, the domain for addition, subtraction, and multiplication was determined by where the common denominator \( 2x-4 \) does not equal zero, which is when \( x = 2 \). For division, since \( f/g(x) = \frac{1}{x} \), we avoid both \( x = 0 \) and \( x = 2 \). Thus, the domain for these operations excludes certain critical points to ensure the function remains valid.

Addition and Subtraction of Functions

Addition and subtraction of functions involve combining their outputs. This means taking two functions, say \( f(x) \) and \( g(x) \), and either adding or subtracting their values for all \( x \) in the domain.
For addition, the formula is \((f+g)(x) = f(x) + g(x)\). If both functions share the same denominator as in our example, \( f(x) = \frac{1}{2x-4} \) and \( g(x) = \frac{x}{2x-4} \), you can simply add the numerators: \( (f+g)(x) = \frac{1 + x}{2x-4} \).
Subtraction works similarly, using \( (f-g)(x) = f(x) - g(x) \), resulting in \( (f-g)(x) = \frac{1 - x}{2x-4} \).

• Ensure that the domain is valid for both functions before adding or subtracting.
• Look for common denominators to simplify the expressions effectively.

Both operations result in the same domain here, distinctly avoiding \( x = 2 \), since it would invalidate the denominator in either operation.

Multiplication and Division of Functions

Multiplication and division of functions combine functions in a deeper manner. When you multiply functions, you multiply their outputs; for division, you divide one output by another.
Consider multiplication: \((f \cdot g)(x) = f(x) \cdot g(x)\). Using our functions, \( f(x) = \frac{1}{2x-4} \) and \( g(x) = \frac{x}{2x-4} \), you multiply them to get \( \frac{x}{(2x-4)^2} \).
Attention is required for multiplication:

• The domain will exclude any \( x \) that makes any term undefined. Here, \( x = 2 \) is skipped because it leads to division by zero.

Division, explained as \((f/g)(x) = \frac{f(x)}{g(x)}\), calculates as \( \frac{1}{x} \) from our given functions. Here's what to ensure:

• The domain must exclude values that make the denominator zero in either the original functions or the resulting fraction.
• This ensures \( x eq 2 \) (from functions' shared denominator) and \( x eq 0 \) (from \( \frac{1}{x} \)).

Thus, multiplication and division require careful attention to exclusions in the domain calculations due to possible undefined behaviors.