Find \( f+g, f-g, f g \) and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=5 x+2, g(x)=x-6 \) \( (f+g)(x)=\square \) (Simplify your answer.)

Divide the polynomials by following steps: - step0: Using Long Division to Divide Polynomials: \(\frac{\left(5x+2\right)}{\left(x-6\right)}\) - step1: Calculate: \(5+\frac{32}{x-6}\) Calculate or simplify the expression \( (5*x + 2) + (x - 6) \). Simplify the expression by following steps: - step0: Solution: \(\left(5x+2\right)+\left(x-6\right)\) - step1: Remove the parentheses: \(5x+2+x-6\) - step2: Add the terms: \(6x-4\) Calculate or simplify the expression \( (5*x + 2) - (x - 6) \). Simplify the expression by following steps: - step0: Solution: \(\left(5x+2\right)-\left(x-6\right)\) - step1: Remove the parentheses: \(5x+2-\left(x-6\right)\) - step2: Remove the parentheses: \(5x+2-x+6\) - step3: Subtract the terms: \(4x+8\) Calculate or simplify the expression \( (5*x + 2) * (x - 6) \). Simplify the expression by following steps: - step0: Solution: \(\left(5x+2\right)\left(x-6\right)\) - step1: Apply the distributive property: \(5x\times x-5x\times 6+2x-2\times 6\) - step2: Multiply the terms: \(5x^{2}-30x+2x-12\) - step3: Add the terms: \(5x^{2}-28x-12\) Let's calculate \( f+g \), \( f-g \), \( fg \), and \( \frac{f}{g} \) step by step. 1. **Finding \( f+g \)**: \[ (f+g)(x) = f(x) + g(x) = (5x + 2) + (x - 6) = 6x - 4 \] 2. **Finding \( f-g \)**: \[ (f-g)(x) = f(x) - g(x) = (5x + 2) - (x - 6) = 4x + 8 \] 3. **Finding \( fg \)**: \[ (fg)(x) = f(x) \cdot g(x) = (5x + 2)(x - 6) = 5x^2 - 30x + 2x - 12 = 5x^2 - 28x - 12 \] 4. **Finding \( \frac{f}{g} \)**: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{5x + 2}{x - 6} = 5 + \frac{32}{x - 6} \] Now, let's determine the domain for each function: - **Domain of \( f+g \)**: Since both \( f(x) \) and \( g(x) \) are polynomials, their sum \( 6x - 4 \) is also a polynomial. The domain is all real numbers: \( \mathbb{R} \). - **Domain of \( f-g \)**: Similarly, \( 4x + 8 \) is a polynomial, so the domain is all real numbers: \( \mathbb{R} \). - **Domain of \( fg \)**: The product \( 5x^2 - 28x - 12 \) is a polynomial, thus the domain is all real numbers: \( \mathbb{R} \). - **Domain of \( \frac{f}{g} \)**: The function \( \frac{5x + 2}{x - 6} \) is defined for all \( x \) except where the denominator is zero. Thus, the domain is \( x \neq 6 \) or in interval notation: \( (-\infty, 6) \cup (6, \infty) \). In summary: - \( (f+g)(x) = 6x - 4 \) - \( (f-g)(x) = 4x + 8 \) - \( (fg)(x) = 5x^2 - 28x - 12 \) - \( \left(\frac{f}{g}\right)(x) = 5 + \frac{32}{x - 6} \) **Domains**: - \( f+g \): \( \mathbb{R} \) - \( f-g \): \( \mathbb{R} \) - \( fg \): \( \mathbb{R} \) - \( \frac{f}{g} \): \( (-\infty, 6) \cup (6, \infty) \)