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) $$ \left(\frac{g}{f}\right)(x)= $$

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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) 
$$
\left(\frac{g}{f}\right)(x) = \frac{5(x-4)}{(x-6)(x+3)}, \quad \text{Domain: } x \in \mathbb{R}, x \neq 6, -3
$$
(a) $$ (f-g)(x) = \frac{x^2 - 8x + 2}{(x-4)(x-6)} $$, Domain: All real numbers except $$ x = 4 $$ and $$ x = 6 $$. (b) $$ \left(\frac{g}{f}\right)(x) = \frac{5(x-4)}{(x-6)(x+3)} $$, Domain: All real numbers except $$ x = 6 $$ and $$ x = -3 $$.

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) \( \left(\frac{g}{f}\right)(x)= \)

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