Let's go through each part of the problem step-by-step.
### Part (a): Finding the Domains
#### Domain of \( f \)
The function \( f \) is defined as \( f(x) = 8x + 3 \). This is a linear function, which is defined for all real numbers. The domain of \( f \) is:
[tex]\[ (-\infty, \infty) \][/tex]
#### Domain of \( g \)
The function \( g \) is defined as \( g(x) = 1 - 6x \). This is also a linear function, which is defined for all real numbers. The domain of \( g \) is:
[tex]\[ (-\infty, \infty) \][/tex]
#### Domain of \( f + g \)
The function \((f + g)(x)\) is the sum of two linear functions, which is also a linear function and thus defined for all real numbers. The domain of \( f + g \) is:
[tex]\[ (-\infty, \infty) \][/tex]
#### Domain of \( f - g \)
The function \((f - g)(x)\) is the difference of two linear functions, which is also a linear function and thus defined for all real numbers. The domain of \( f - g \) is:
[tex]\[ (-\infty, \infty) \][/tex]
#### Domain of \( fg \)
The function \((fg)(x)\) is the product of two linear functions, which results in a quadratic function. A quadratic function is defined for all real numbers. The domain of \( fg \) is:
[tex]\[ (-\infty, \infty) \][/tex]
#### Domain of \( ff \)
The function \((ff)(x)\) means the function \( f \) composed with itself, \( f(f(x)) \). Since \( f \) is defined for all real numbers and its output is also a real number, the domain of \( ff \) is:
[tex]\[ (-\infty, \infty) \][/tex]
#### Domain of \(\frac{f}{g}\)
The function \(\left(\frac{f}{g}\right)(x)\) is the quotient of \( f \) and \( g \). This quotient is defined for all \( x \) except where \( g(x) = 0 \).
[tex]\[ g(x) = 1 - 6x = 0 \implies x = \frac{1}{6} \][/tex]
So, the domain of \(\frac{f}{g}\) is:
[tex]\[ (-\infty, \frac{1}{6}) \cup (\frac{1}{6}, \infty) \][/tex]
#### Domain of \(\frac{g}{f}\)
The function \(\left(\frac{g}{f}\right)(x)\) is the quotient of \( g \) and \( f \). This quotient is defined for all \( x \) except where \( f(x) = 0 \).
[tex]\[ f(x) = 8x + 3 = 0 \implies x = -\frac{3}{8} \][/tex]
So, the domain of \(\frac{g}{f}\) is:
[tex]\[ (-\infty, -\frac{3}{8}) \cup (-\frac{3}{8}, \infty) \][/tex]
### Part (b): Finding the Functions
Now, let's find the expressions for each function.
#### \((f + g)(x)\)
[tex]\[ (f + g)(x) = f(x) + g(x) = (8x + 3) + (1 - 6x) = 2x + 4 \][/tex]
#### \((f - g)(x)\)
[tex]\[ (f - g)(x) = f(x) - g(x) = (8x + 3) - (1 - 6x) = 8x + 3 - 1 + 6x = 14x + 2 \][/tex]
#### \((fg)(x)\)
[tex]\[ (fg)(x) = f(x) \cdot g(x) = (8x + 3)(1 - 6x) = 8x - 48x^2 + 3 - 18x = -48x^2 - 10x + 3 \][/tex]
#### \((ff)(x)\)
[tex]\[ (ff)(x) = f(f(x)) \][/tex]
[tex]\[ f(x) = 8x + 3 \][/tex]
[tex]\[ f(f(x)) = f(8x + 3) = 8(8x + 3) + 3 = 64x + 24 + 3 = 64x + 27 \][/tex]
#### \(\left(\frac{f}{g}\right)(x)\)
[tex]\[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{8x + 3}{1 - 6x} \][/tex]
#### \(\left(\frac{g}{f}\right)(x)\)
[tex]\[ \left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} = \frac{1 - 6x}{8x + 3} \][/tex]
In summary:
### Part (a): Domains
- Domain of \( f \): \((- \infty, \infty)\)
- Domain of \( g \): \((- \infty, \infty)\)
- Domain of \( f + g \): \((- \infty, \infty)\)
- Domain of \( f - g \): \((- \infty, \infty)\)
- Domain of \( fg \): \((- \infty, \infty)\)
- Domain of \( ff \): \((- \infty, \infty)\)
- Domain of \(\frac{f}{g}\): \((- \infty, \frac{1}{6}) \cup (\frac{1}{6}, \infty)\)
- Domain of \(\frac{g}{f}\): \((- \infty, -\frac{3}{8}) \cup (-\frac{3}{8}, \infty)\)
### Part (b): Functions
- \((f + g)(x) = 2x + 4\)
- \((f - g)(x) = 14x + 2\)
- \((fg)(x) = -48x^2 - 10x + 3\)
- \((ff)(x) = 64x + 27\)
- \(\left(\frac{f}{g}\right)(x) = \frac{8x + 3}{1 - 6x}\)
- \(\left(\frac{g}{f}\right)(x) = \frac{1 - 6x}{8x + 3}\)
I hope this detailed solution helps you understand how to find the domains and the resulting functions!
