Find \( f+g, f-g, f g \) and \( \frac{f}{g} \). Determine the domain for each function \( f(x)=3 x-7, g(x)=x-1 \)

Let's find the functions \( f + g \), \( f - g \), \( fg \), and \( \frac{f}{g} \) based on the given functions: \[ f(x) = 3x - 7 \quad \text{and} \quad g(x) = x - 1 \] After finding each operation, we'll determine the domain for each resulting function. ### 1. \( f + g \) \[ f + g = (3x - 7) + (x - 1) = 4x - 8 \] **Domain:** Both \( f(x) \) and \( g(x) \) are linear functions defined for all real numbers. Therefore, \( f + g \) is also defined for all real numbers. \[ \text{Domain of } f + g: \quad \mathbb{R} \] ### 2. \( f - g \) \[ f - g = (3x - 7) - (x - 1) = 2x - 6 \] **Domain:** Similar to \( f + g \), both \( f(x) \) and \( g(x) \) are defined for all real numbers, so \( f - g \) is also defined for all real numbers. \[ \text{Domain of } f - g: \quad \mathbb{R} \] ### 3. \( f \cdot g \) \[ fg = (3x - 7)(x - 1) = 3x^2 - 3x - 7x + 7 = 3x^2 - 10x + 7 \] **Domain:** Both \( f(x) \) and \( g(x) \) are defined for all real numbers, so their product \( fg \) is also defined for all real numbers. \[ \text{Domain of } fg: \quad \mathbb{R} \] ### 4. \( \frac{f}{g} \) \[ \frac{f}{g} = \frac{3x - 7}{x - 1} \] **Domain:** For the division \( \frac{f}{g} \) to be defined, the denominator \( g(x) = x - 1 \) must not be zero. \[ x - 1 \neq 0 \quad \Rightarrow \quad x \neq 1 \] \[ \text{Domain of } \frac{f}{g}: \quad \mathbb{R} \setminus \{1\} \] ### Summary - **\( f + g = 4x - 8 \)** - **Domain:** All real numbers (\( \mathbb{R} \)) - **\( f - g = 2x - 6 \)** - **Domain:** All real numbers (\( \mathbb{R} \)) - **\( fg = 3x^2 - 10x + 7 \)** - **Domain:** All real numbers (\( \mathbb{R} \)) - **\( \frac{f}{g} = \frac{3x - 7}{x - 1} \)** - **Domain:** All real numbers except \( x = 1 \) (\( \mathbb{R} \setminus \{1\} \))