2. (16 pts.) Consider \( f(x)=\frac{x+3}{x-4} \) and \( g(x)=\frac{5}{x-6} \). Find the rules of the new functions below, simplifif the results, if possible, and describe their domains. (a) \( (f-g)(x)= \) (b) \( \left(\frac{g}{f}\right)(x)= \)

To solve the problem, we will find the new functions \( (f-g)(x) \) and \( \left(\frac{g}{f}\right)(x) \) step by step, simplifying the results and describing their domains. ### Given Functions - \( f(x) = \frac{x+3}{x-4} \) - \( g(x) = \frac{5}{x-6} \) ### (a) Finding \( (f-g)(x) \) The function \( (f-g)(x) \) is defined as: \[ (f-g)(x) = f(x) - g(x) = \frac{x+3}{x-4} - \frac{5}{x-6} \] To subtract these fractions, we need a common denominator, which is \( (x-4)(x-6) \). Thus, we rewrite the fractions: \[ (f-g)(x) = \frac{(x+3)(x-6) - 5(x-4)}{(x-4)(x-6)} \] Now, we will expand the numerator: 1. Expand \( (x+3)(x-6) \): \[ (x+3)(x-6) = x^2 - 6x + 3x - 18 = x^2 - 3x - 18 \] 2. Expand \( 5(x-4) \): \[ 5(x-4) = 5x - 20 \] Now, substitute these back into the numerator: \[ (f-g)(x) = \frac{x^2 - 3x - 18 - (5x - 20)}{(x-4)(x-6)} = \frac{x^2 - 3x - 18 - 5x + 20}{(x-4)(x-6)} \] Combine like terms in the numerator: \[ = \frac{x^2 - 8x + 2}{(x-4)(x-6)} \] ### Domain of \( (f-g)(x) \) The domain of \( (f-g)(x) \) is determined by the values that make the denominator zero. The denominator is zero when: 1. \( x - 4 = 0 \) → \( x = 4 \) 2. \( x - 6 = 0 \) → \( x = 6 \) Thus, the domain of \( (f-g)(x) \) is all real numbers except \( x = 4 \) and \( x = 6 \). ### (b) Finding \( \left(\frac{g}{f}\right)(x) \) The function \( \left(\frac{g}{f}\right)(x) \) is defined as: \[ \left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} = \frac{\frac{5}{x-6}}{\frac{x+3}{x-4}} = \frac{5}{x-6} \cdot \frac{x-4}{x+3} \] This simplifies to: \[ \left(\frac{g}{f}\right)(x) = \frac{5(x-4)}{(x-6)(x+3)} \] ### Domain of \( \left(\frac{g}{f}\right)(x) \) The domain of \( \left(\frac{g}{f}\right)(x) \) is determined by the values that make the denominators zero. The denominators are zero when: 1. \( x - 6 = 0 \) → \( x = 6 \) 2. \( x + 3 = 0 \) → \( x = -3 \) Thus, the domain of \( \left(\frac{g}{f}\right)(x) \) is all real numbers except \( x = 6 \) and \( x = -3 \). ### Summary of Results (a) \[ (f-g)(x) = \frac{x^2 - 8x + 2}{(x-4)(x-6)}, \quad \text{Domain: } x \in \mathbb{R}, x \neq 4, 6 \] (b) \[ \left(\frac{g}{f}\right)(x) = \frac{5(x-4)}{(x-6)(x+3)}, \quad \text{Domain: } x \in \mathbb{R}, x \neq 6, -3 \]