Find $$ \mathrm{f}+\mathrm{g}, \mathrm{f}-\mathrm{g}, \mathrm{fg} $$ and $$ \frac{\mathrm{l}}{\mathrm{g}} $$. Determine the domain for each function. $$ f(x)=\frac{3 x+8}{x^{2}-81}, g(x)=\frac{2 x-1}{x^{2}-81} $$ A. ine comani oi $$ \mathrm{r}+\mathrm{g} $$ is $$ (-\infty,-y) \cup(-y, y) \cup(y, \infty) $$. (Type your answer in interval notation.) B. The domain of $$ f+g $$ is $$ \{\quad\} $$. (Use a comma to separate answers as needed.) $$ \square $$ C. The domain of $$ f+g $$ is $$ \varnothing $$. $$ (f-g)(x)=\frac{1}{x-9}(\text { Simplify your answer.) } $$ What is the domain of $$ f-g $$ ? A. The domain of $$ f-g $$ is $$ (-\infty,-9) \cup(-9,9) \cup(9, \infty) $$. B. The domain of $$ f-g $$ is $$ \} $$. (Use a comma to separate answers as needed.) C. The domain of f fg is $$ \varnothing $$. $$ (\mathrm{fg})(\mathrm{x})= $$ $$ \square $$ $$ 3 x+8 $$ (Simplify your answer.)

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UpStudy AI · Accepted Answer

To solve the problem, we will first find the functions $$ f+g $$, $$ f-g $$, $$ fg $$, and $$ \frac{f}{g} $$. Then, we will determine the domain for each function.

Given:
$$
f(x) = \frac{3x + 8}{x^2 - 81}, \quad g(x) = \frac{2x - 1}{x^2 - 81}
$$

### Step 1: Find $$ f + g $$

$$
f + g = \frac{3x + 8}{x^2 - 81} + \frac{2x - 1}{x^2 - 81} = \frac{(3x + 8) + (2x - 1)}{x^2 - 81} = \frac{5x + 7}{x^2 - 81}
$$

### Step 2: Find $$ f - g $$

$$
f - g = \frac{3x + 8}{x^2 - 81} - \frac{2x - 1}{x^2 - 81} = \frac{(3x + 8) - (2x - 1)}{x^2 - 81} = \frac{x + 9}{x^2 - 81}
$$

### Step 3: Find $$ fg $$

$$
fg = \left( \frac{3x + 8}{x^2 - 81} \right) \left( \frac{2x - 1}{x^2 - 81} \right) = \frac{(3x + 8)(2x - 1)}{(x^2 - 81)^2}
$$

Now, we will expand the numerator:
$$
(3x + 8)(2x - 1) = 6x^2 - 3x + 16x - 8 = 6x^2 + 13x - 8
$$
Thus,
$$
fg = \frac{6x^2 + 13x - 8}{(x^2 - 81)^2}
$$

### Step 4: Find $$ \frac{f}{g} $$

$$
\frac{f}{g} = \frac{\frac{3x + 8}{x^2 - 81}}{\frac{2x - 1}{x^2 - 81}} = \frac{3x + 8}{2x - 1}
$$

### Step 5: Determine the domain for each function

The domain of $$ f(x) $$ and $$ g(x) $$ is determined by the denominator $$ x^2 - 81 \neq 0 $$, which gives:
$$
x^2 - 81 = 0 \implies x = \pm 9
$$
Thus, the domain for $$ f $$ and $$ g $$ is:
$$
(-\infty, -9) \cup (-9, 9) \cup (9, \infty)
$$

#### Domain of $$ f + g $$

The domain of $$ f + g $$ is the same as $$ f $$ and $$ g $$:
$$
(-\infty, -9) \cup (-9, 9) \cup (9, \infty)
$$

#### Domain of $$ f - g $$

The domain of $$ f - g $$ is also the same:
$$
(-\infty, -9) \cup (-9, 9) \cup (9, \infty)
$$

#### Domain of $$ fg $$

The domain of $$ fg $$ is the same:
$$
(-\infty, -9) \cup (-9, 9) \cup (9, \infty)
$$

#### Domain of $$ \frac{f}{g} $$

The domain of $$ \frac{f}{g} $$ is determined by both $$ g \neq 0 $$ and $$ x^2 - 81 \neq 0 $$:
$$
2x - 1 \neq 0 \implies x \neq \frac{1}{2}
$$
Thus, the domain is:
$$
(-\infty, -9) \cup (-9, \frac{1}{2}) \cup (\frac{1}{2}, 9) \cup (9, \infty)
$$

### Final Answers

A. The domain of $$ f + g $$ is $$ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) $$.

B. The domain of $$ f + g $$ is $$ (-\infty, -9), (-9, 9), (9, \infty) $$.

C. The domain of $$ f - g $$ is $$ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) $$.

D. The simplified form of $$ fg $$ is $$ \frac{6x^2 + 13x - 8}{(x^2 - 81)^2} $$.

E. The domain of $$ \frac{f}{g} $$ is $$ (-\infty, -9) \cup (-9, \frac{1}{2}) \cup (\frac{1}{2}, 9) \cup (9, \infty) $$.
- **$$ f + g $$:** Domain is $$ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) $$. - **$$ f - g $$:** Domain is $$ (-\infty, -9) \cup (-9, 9) \cup (9, \infty) $$. - **$$ fg $$:** Simplified form is $$ \frac{6x^2 + 13x - 8}{(x^2 - 81)^2} $$. - **$$ \frac{f}{g} $$:** Domain is $$ (-\infty, -9) \cup (-9, \frac{1}{2}) \cup (\frac{1}{2}, 9) \cup (9, \infty) $$.

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