Find $$ f+g, f-g $$, $$ f g $$, and $$ \frac{f}{g} $$. Determine the domain for each function $$ (f+g)(x)=\sqrt{x} ; g(x)=x-9 $$ What is the domain of $$ f+g $$ ? A. The domain of $$ f+g $$ is $$ \} $$. (Use a comma to separate answers as needed) C. The domain of $$ f+g $$ is $$ \{0, \infty) $$. (Type your answer in interval notation ) $$ f+g $$ is $$ \varnothing $$ ( $$ f-g)(x)=\sqrt{x}-x+9 $$ (Simplify your answer) What is the domain of $$ f-g $$ ? A. The domain of $$ f-g $$ is $$ \square $$. (Type your answer in interval notation.) B. The domain of $$ f-g $$ is $$ \} $$ (Use a comma to separate answers as needed) C. The domain of $$ f-g $$ is $$ \varnothing $$.

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To solve the problem, we need to find the functions $$ f $$ and $$ g $$ based on the given information and then compute $$ f+g $$, $$ f-g $$, $$ fg $$, and $$ \frac{f}{g} $$. We will also determine the domain for each function.

### Step 1: Identify $$ f $$ and $$ g $$

We know:
- $$ (f+g)(x) = \sqrt{x} $$
- $$ g(x) = x - 9 $$

From the equation $$ (f+g)(x) = f(x) + g(x) = \sqrt{x} $$, we can express $$ f(x) $$ as:
$$
f(x) = \sqrt{x} - g(x) = \sqrt{x} - (x - 9) = \sqrt{x} - x + 9
$$

### Step 2: Calculate $$ f+g $$

We already have:
$$
f+g = \sqrt{x}
$$

### Step 3: Calculate $$ f-g $$

Now, we can find $$ f-g $$:
$$
f-g = f(x) - g(x) = (\sqrt{x} - x + 9) - (x - 9) = \sqrt{x} - x + 9 - x + 9 = \sqrt{x} - 2x + 18
$$

### Step 4: Calculate $$ fg $$

Next, we calculate $$ fg $$:
$$
fg = f(x) \cdot g(x) = (\sqrt{x} - x + 9)(x - 9)
$$

### Step 5: Calculate $$ \frac{f}{g} $$

Now, we calculate $$ \frac{f}{g} $$:
$$
\frac{f}{g} = \frac{\sqrt{x} - x + 9}{x - 9}
$$

### Step 6: Determine the domains

1. **Domain of $$ f+g $$**:
   - Since $$ f+g = \sqrt{x} $$, the domain is $$ x \geq 0 $$.
   - In interval notation, the domain is $$ [0, \infty) $$.

2. **Domain of $$ f-g $$**:
   - The expression $$ f-g = \sqrt{x} - 2x + 18 $$ requires $$ \sqrt{x} $$ to be defined, so $$ x \geq 0 $$.
   - Additionally, we need to ensure that $$ f-g $$ does not lead to any undefined expressions. Since $$ \sqrt{x} $$ is defined for $$ x \geq 0 $$, the domain remains $$ [0, \infty) $$.

3. **Domain of $$ fg $$**:
   - The product $$ fg $$ is defined as long as both $$ f $$ and $$ g $$ are defined. Since $$ g(x) = x - 9 $$ is defined for all $$ x $$, we only need to consider $$ f(x) $$ which requires $$ x \geq 0 $$.
   - Thus, the domain is $$ [0, \infty) $$.

4. **Domain of $$ \frac{f}{g} $$**:
   - The function $$ \frac{f}{g} $$ is defined as long as $$ g(x) \neq 0 $$. This means $$ x - 9 \neq 0 $$ or $$ x \neq 9 $$.
   - Therefore, the domain is $$ [0, 9) \cup (9, \infty) $$.

### Final Answers

- The domain of $$ f+g $$ is $$ [0, \infty) $$.
- The domain of $$ f-g $$ is $$ [0, \infty) $$.
- The domain of $$ fg $$ is $$ [0, \infty) $$.
- The domain of $$ \frac{f}{g} $$ is $$ [0, 9) \cup (9, \infty) $$.

### Summary of Domains
- **Domain of $$ f+g $$**: $$ [0, \infty) $$
- **Domain of $$ f-g $$**: $$ [0, \infty) $$
- **Domain of $$ fg $$**: $$ [0, \infty) $$
- **Domain of $$ \frac{f}{g} $$**: $$ [0, 9) \cup (9, \infty) $$
- **Domain of $$ f+g $$**: $$ [0, \infty) $$ - **Domain of $$ f-g $$**: $$ [0, \infty) $$ - **Domain of $$ fg $$**: $$ [0, \infty) $$ - **Domain of $$ \frac{f}{g} $$**: $$ [0, 9) \cup (9, \infty) $$

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