Problem 14 - Algebra 2

Question

Find \((f+g)(x),(f-g)(x),(f \cdot g)(x),\) and \(\left(\frac{f}{g}\right)\) for each \(f(x)\) and \(g(x)\) $$ \begin{array}{l}{f(x)=x+9} \\ {g(x)=x-9}\end{array} $$

Step-by-Step Solution

Answer

The solutions are: \((f+g)(x) = 2x\), \((f-g)(x) = 18\), \((f \cdot g)(x) = x^2 - 81\), \(\left(\frac{f}{g}\right)(x) = \frac{x+9}{x-9}\) with \(x \neq 9\).

Step 1: Understanding the Functions

We are given two functions: \( f(x) = x + 9 \) and \( g(x) = x - 9 \). We need to find their sum, difference, product, and quotient.

Step 2: Calculating the Sum \((f+g)(x)\)

To find \((f+g)(x)\), we add \(f(x)\) and \(g(x)\): \[ (f+g)(x) = (x+9) + (x-9) = x + 9 + x - 9 = 2x. \]

Step 3: Calculating the Difference \((f-g)(x)\)

To find \((f-g)(x)\), we subtract \(g(x)\) from \(f(x)\): \[ (f-g)(x) = (x+9) - (x-9) = x + 9 - x + 9 = 18. \]

Step 4: Calculating the Product \((f \, \cdot \, g)(x)\)

To find \((f \, \cdot \, g)(x)\), we multiply \(f(x)\) by \(g(x)\): \[ (f \, \cdot \, g)(x) = (x+9)(x-9) = x^2 - 9x + 9x - 81 = x^2 - 81. \]

Step 5: Calculating the Quotient \(\left(\frac{f}{g}\right)(x)\)

To find \(\left(\frac{f}{g}\right)(x)\), we divide \(f(x)\) by \(g(x)\): \[ \left(\frac{f}{g}\right)(x) = \frac{x+9}{x-9}. \] Note that this expression is undefined for \(x=9\), as it makes the denominator zero.

Key Concepts

Function Addition

Function addition involves combining two functions by adding their corresponding outputs. Given the functions \( f(x) = x + 9 \) and \( g(x) = x - 9 \), we simply add these together. This means you sum up the expressions:

• \( f(x) = x + 9 \)
• \( g(x) = x - 9 \)

Thus, the operation becomes:\[ (f+g)(x) = (x + 9) + (x - 9) \]Now, distribute and combine like terms:\[ (f+g)(x) = x + 9 + x - 9 = 2x \]The result \( 2x \) represents the sum of the functions, effectively doubling the variable \( x \) in the expression.

Function Subtraction

Function subtraction is about finding the difference between two functions. Here, you take the outputs of one function and subtract the other. With our functions \( f(x) = x + 9 \) and \( g(x) = x - 9 \), perform the following subtraction:

• \( f(x) = x + 9 \)
• \( g(x) = x - 9 \)

We compute:\[ (f-g)(x) = (x + 9) - (x - 9) \]When distributing and combining like terms, it simplifies to:\[ (f-g)(x) = x + 9 - x + 9 = 18 \]The constant \( 18 \) highlights how subtracting these functions removes the variable terms and leaves a fixed number represented as the result.

Function Multiplication

In function multiplication, you multiply the outputs of two functions. This can also be recognized as the product of the functions. Starting from \( f(x) = x + 9 \) and \( g(x) = x - 9 \), the process is as follows:

• Multiply: \((x + 9)(x - 9)\)

Apply the distributive property (also known as FOIL for binomials):\[ (f \cdot g)(x) = x^2 - 9x + 9x - 81 \]Combine like terms to simplify:\[ (f \cdot g)(x) = x^2 - 81 \]The final result is a quadratic expression where the middle terms cancel out.

Function Division

Function division involves dividing one function by another, similar to handling fractions with numerical values. For our functions, \( f(x) = x + 9 \) divided by \( g(x) = x - 9 \), it looks like this:

• \( \frac{x+9}{x-9} \)

It's essential to observe where the denominator becomes zero, as division by zero is undefined. This happens when \( x = 9 \). Thus, the quotient function:\[ \left(\frac{f}{g}\right)(x) = \frac{x+9}{x-9} \]exists for all \( x eq 9 \). Remember, checking for undefined values is crucial during division because it influences the domain of the resulting function.