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Let's break down and solve this problem step by step.

### Given Functions:
\[
f(x) = \frac{7x + 2}{x^2 - 36}, \quad g(x) = \frac{6x - 4}{x^2 - 36}
\]
We are asked to find the following:
1. \( f(x) + g(x) \)
2. \( f(x) - g(x) \)
3. \( f(x) \cdot g(x) \)
4. \( \frac{f(x)}{g(x)} \)

We also need to determine the domain for each function.

### Domain
The domain of a rational function is the set of all real numbers except those that make the denominator equal to zero. Both \( f(x) \) and \( g(x) \) have the denominator \( x^2 - 36 \), which can be factored as:
\[
x^2 - 36 = (x - 6)(x + 6)
\]
Thus, \( x = 6 \) and \( x = -6 \) are the values that make the denominator zero. Therefore, the domain of both \( f(x) \) and \( g(x) \) is:
\[
\text{Domain of } f(x), g(x): x \in \mathbb{R}, x \neq \pm 6
\]

### 1. \( f(x) + g(x) \)

Since \( f(x) \) and \( g(x) \) have the same denominator, we can directly add the numerators:
\[
f(x) + g(x) = \frac{7x + 2}{x^2 - 36} + \frac{6x - 4}{x^2 - 36} = \frac{(7x + 2) + (6x - 4)}{x^2 - 36} = \frac{13x - 2}{x^2 - 36}
\]
The domain of \( f(x) + g(x) \) is the same as that of \( f(x) \) and \( g(x) \): \( x \neq \pm 6 \).

### 2. \( f(x) - g(x) \)

Similarly, subtracting the numerators:
\[
f(x) - g(x) = \frac{7x + 2}{x^2 - 36} - \frac{6x - 4}{x^2 - 36} = \frac{(7x + 2) - (6x - 4)}{x^2 - 36} = \frac{x + 6}{x^2 - 36}
\]
The domain of \( f(x) - g(x) \) is also \( x \neq \pm 6 \).

### 3. \( f(x) \cdot g(x) \)

Multiplying the two functions:
\[
f(x) \cdot g(x) = \frac{7x + 2}{x^2 - 36} \cdot \frac{6x - 4}{x^2 - 36} = \frac{(7x + 2)(6x - 4)}{(x^2 - 36)^2}
\]
Now, expanding the numerator:
\[
(7x + 2)(6x - 4) = 42x^2 - 28x + 12x - 8 = 42x^2 - 16x - 8
\]
Thus:
\[
f(x) \cdot g(x) = \frac{42x^2 - 16x - 8}{(x^2 - 36)^2}
\]
The domain remains \( x \neq \pm 6 \).

### 4. \( \frac{f(x)}{g(x)} \)

Dividing \( f(x) \) by \( g(x) \):
\[
\frac{f(x)}{g(x)} = \frac{\frac{7x + 2}{x^2 - 36}}{\frac{6x - 4}{x^2 - 36}} = \frac{7x + 2}{6x - 4}
\]
Here, we must exclude any values that make the denominator zero, i.e., \( 6x - 4 = 0 \), or \( x = \frac{2}{3} \). Therefore, the domain is:
\[
\text{Domain of } \frac{f(x)}{g(x)}: x \in \mathbb{R}, x \neq \pm 6, x \neq \frac{2}{3}
\]

### Summary:
1. \( f(x) + g(x) = \frac{13x - 2}{x^2 - 36} \), Domain: \( x \neq \pm 6 \)
2. \( f(x) - g(x) = \frac{x + 6}{x^2 - 36} \), Domain: \( x \neq \pm 6 \)
3. \( f(x) \cdot g(x) = \frac{42x^2 - 16x - 8}{(x^2 - 36)^2} \), Domain: \( x \neq \pm 6 \)
4. \( \frac{f(x)}{g(x)} = \frac{7x + 2}{6x - 4} \), Domain: \( x \neq \pm 6, x \neq \frac{2}{3} \)

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Would you like more details or have any questions?

Here are some related questions for further understanding:
1. How would the domain change if the denominators were different?
2. How would the functions behave at the boundaries of their domain (near \( x = 6 \) and \( x = -6 \))?
3. What would happen if the numerator of one function was a quadratic equation?
4. Can you simplify complex rational expressions like \( f(x) \cdot g(x) \) further?
5. How would adding more terms to the numerator affect the overall function?

**Tip**: Always simplify the function as much as possible before analyzing its domain to avoid mistakes with undefined values.

Find f + g, f - g, fg, and f/g. Determine the domain for each function.
 f(x) = (7x + 2) / (x^2 - 36), g(x) = (6x - 4) / (x^2 - 36)

Learn how to add, subtract, multiply, and divide rational functions. This example uses f(x) = (7x + 2) / (x^2 - 36) and g(x) = (6x - 4) / (x^2 - 36), and includes a detailed explanation of domain restrictions due to the denominator.

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