Question

Given \(f(x)=3 x+8\) and \(g(x)=x-5,\) find $$ (f+g)(x),(f-g)(x),(f \cdot g)(x), \text { and }\left(\frac{f}{g}\right)(x) $$ State the domain of each.

Short answer

The results are: \((f+g)(x) = 4x + 3, \; (f-g)(x) = 2x + 13, \; (f \times g)(x) = 3x^2 - 7x - 40,\; \left( \frac{f}{g} \right)(x) = \frac{3x+8}{x-5}\). The domains are: \((f+g)(x): \; (-\infty, \infty), \; (f-g)(x): \; (-\infty, \infty), \; (f \times g)(x):\; (-\infty, \infty), \; \left( \frac{f}{g} \right)(x): \; (-\infty, 5) \cup (5, \infty) \).

Step-by-step solution

- Understanding the Functions

Given\[ f(x) = 3x + 8 \] and \[ g(x) = x - 5 \]. These will be combined in various ways to find \((f+g)(x), (f-g)(x), (f \times g)(x)\text{ and }\frac{f}{g}(x)\).

- Finding \((f+g)(x)\)

Add the functions together: \( (f+g)(x) = f(x) + g(x) \). Substitute the expressions: \[ (f+g)(x) = (3x + 8) + (x - 5) \]. Combine like terms: \[ (f+g)(x) = 3x + x + 8 - 5 = 4x + 3 \].

- Finding the Domain of \((f+g)(x)\)

Both \( f(x) \) and \( g(x) \) are linear functions defined for all real numbers. Therefore, the domain of \((f+g)(x)\) is also all real numbers: \((-\infty, \infty)\).

- Finding \((f-g)(x)\)

Subtract\( g(x)\) from \( f(x)\): \[ (f-g)(x) = f(x) - g(x) \]. Substitute the expressions: \[ (f-g)(x) = (3x + 8) - (x - 5) \]. Combine like terms: \[ (f-g)(x) = 3x - x + 8 + 5 = 2x + 13 \].

- Finding the Domain of \((f-g)(x)\)

Like \((f+g)(x)\), both functions are defined for all real numbers, so the domain of \((f-g)(x)\) is \((-\infty, \infty)\).

- Finding \((f \times g)(x)\)

Multiply the functions: \[ (f \times g)(x) = f(x) \times g(x) \]. Substitute the expressions: \[(f \times g)(x) = (3x + 8)(x - 5) \]. Expand the equation: \[(f \times g)(x) = 3x^2 - 15x + 8x - 40 = 3x^2 - 7x - 40 \].

- Finding the Domain of \((f \times g)(x)\)

By the same reasoning, the domain of \((f \times g)(x)\) is all real numbers: \((-\infty, \infty)\).

- Finding \(\left( \frac{f}{g} \right)(x)\)

Divide \( f(x) \) by \( g(x) \): \[ \left( \frac{f}{g} \right)(x) = \frac{f(x)}{g(x)} \]. Substitute the expressions: \[ \left( \frac{f}{g} \right)(x) = \frac{3x+8}{x-5} \].

- Finding the Domain of \(\left( \frac{f}{g} \right)(x)\)

The domain of this function is all real numbers except where the denominator is zero. Set the denominator equal to zero and solve: \(x - 5 = 0\), so \(x = 5\). Therefore, the domain is \( (-\infty, 5) \,\cup\, (5, \infty)\).

Key concepts

Function addition

In mathematics, adding two functions means creating a new function by pointwise addition of the two original functions. Consider the functions given:

• \( f(x) = 3x + 8 \)
• \( g(x) = x - 5 \).

To find the sum \( (f+g)(x) \) you simply combine them as follows: \[ (f+g)(x) = f(x) + g(x) = (3x + 8) + (x - 5) = 4x + 3 \]. This new function represents the addition result for any value of \( x \). The important part here is understanding that addition combines like terms from both functions. For any linear functions like these, the domain remains all real numbers, written as \( (-fty, fty) \). This is because there are no restrictions or undefined values within the linear function itself.

Function subtraction

Subtracting two functions results in creating a new function by pointwise subtraction. Using

• \( f(x) = 3x + 8 \)
• \( g(x) = x - 5 \)

, we subtract \( g(x) \) from \( f(x) \): \[ (f-g)(x) = f(x) - g(x) = (3x + 8) - (x - 5) = 2x + 13 \]. Combining the like terms, we derive the function for subtraction. Understanding subtraction is critical because it is only about changing the operation of combining terms. Here, as with addition, the domain remains all real numbers \( (-fty, fty) \). Linear functions do not have restrictions that would limit their domain.

Function multiplication

Multiplying two functions results in a new function derived by pointwise multiplication. Using:

• \( f(x) = 3x + 8 \)
• \( g(x) = x - 5 \)

, we get \[ (f \times g)(x) = f(x) \times g(x) = (3x + 8)(x - 5) \]. To find the product, you must distribute each term and combine like terms:\[ (f \times g)(x) = 3x^2 - 15x + 8x - 40 = 3x^2 - 7x - 40 \]. This results in a quadratic function. Quadratic functions also have a domain of all real numbers, expressed as \( (-\fty, fty) \), because they are continuous and unbounded in both directions.

Function division

Dividing two functions involves creating a new function by pointwise division. Using:

• \( f(x) = 3x + 8 \)
• \( g(x) = x - 5 \)

, the division is calculated as follows: \[ \frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{3x+8}{x-5} \]. Unlike addition, subtraction, and multiplication, the division of functions can introduce restrictions. Specifically, division by zero is undefined. Thus, the domain excludes any values of \( x \) that make the denominator zero. Solve for when the denominator is zero: \[ x-5=0 \rightarrow x=5 \]. Therefore, \( x = 5 \) is not included in the domain of \( \frac{f}{g}(x) \). The domain becomes all real numbers except \( x = 5 \), represented as \( (-fty, 5) \bigcup (5, fty) \).

Domain

The domain of a function represents all possible values of \( x \) for which the function is defined. In function operations:

• For addition and subtraction of linear functions, the domain typically remains all real numbers, \( (-fty, fty) \). There are no restrictions, as these operations do not affect the continuity or existence of linear functions.
• For multiplication of functions, even with quadratic results, the domain typically remains all real numbers as quadratics extend infinitely in both directions.
• For division of functions, the primary concern arises with avoiding division by zero. This requires identifying values that create undefined results, adjusting the domain accordingly.

Understanding the domain is crucial for comprehending where a function and its operations are valid. Always verify potential restrictions and adjust the domain appropriately to ensure proper evaluation of function operations.