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Magazine Chapter 6: Problem 79
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
Expert verified
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
01
- Understanding the Functions
Given\[ f(x) = 3x + 8 \] and \[ g(x) = x - 5 \]. These will be combined in various ways to find \((f+g)(x), (f-g)(x), (f \times g)(x)\text{ and }\frac{f}{g}(x)\).
02
- 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: \[ (f+g)(x) = 3x + x + 8 - 5 = 4x + 3 \].
03
- 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: \((-\infty, \infty)\).
04
- 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: \[ (f-g)(x) = 3x - x + 8 + 5 = 2x + 13 \].
05
- 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 \((-\infty, \infty)\).
06
- 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: \[(f \times g)(x) = 3x^2 - 15x + 8x - 40 = 3x^2 - 7x - 40 \].
07
- Finding the Domain of \((f \times g)(x)\)
By the same reasoning, the domain of \((f \times g)(x)\) is all real numbers: \((-\infty, \infty)\).
08
- 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: \[ \left( \frac{f}{g} \right)(x) = \frac{3x+8}{x-5} \].
09
- 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 \( (-\infty, 5) \,\cup\, (5, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Function addition
In mathematics, adding two functions means creating a new function by pointwise addition of the two original functions. Consider the functions given:
- \( f(x) = 3x + 8 \)
- \( g(x) = x - 5 \).
Function subtraction
Subtracting two functions results in creating a new function by pointwise subtraction. Using
- \( f(x) = 3x + 8 \)
- \( g(x) = x - 5 \)
Function multiplication
Multiplying two functions results in a new function derived by pointwise multiplication. Using:
- \( f(x) = 3x + 8 \)
- \( g(x) = x - 5 \)
Function division
Dividing two functions involves creating a new function by pointwise division. Using:
- \( f(x) = 3x + 8 \)
- \( g(x) = x - 5 \)
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.
