The function in question is $\frac{f}{g}(x) = \frac{x+7}{x+3}$, and the task is to determine the domain of this function.

Step-by-step explanation:

1. Identify restrictions on the domain: The domain of $\frac{f}{g}(x)$ is all real numbers except where the denominator equals zero. The denominator is $x + 3$, so we set it equal to zero: $x + 3 = 0 \implies x = -3.$ Therefore, $x = -3$ is excluded from the domain.

2. Examine the given domain interval: The problem asks for the domain in the interval $(-3, \infty)$. However, the domain of $\frac{f}{g}(x)$ will also be restricted at any other points where the denominator makes the function undefined.

3. Check for discontinuities in the numerator: The numerator $x+7$ is a linear function and does not impose any additional restrictions on the domain, as it is defined for all real numbers.

Conclusion:

The only restriction on the domain is $x = -3$. Hence, the correct domain of $\frac{f}{g}(x)$ is: $(-\infty, -3) \cup (-3, \infty).$

Verification:

• Choice A corresponds to the domain $(-7, -3) \cup (-3, \infty)$, which seems to contain a mistake because the lower bound should extend to $-\infty$, not $-7$.
• Choice B represents the empty set, which is incorrect.
• Choice C also represents an empty set, which is incorrect.

So, the correct domain should actually be $(-\infty, -3) \cup (-3, \infty)$, but this choice is not given in the options.

Would you like further details or clarifications?

Here are 5 related questions:

1. What are the steps to find the domain of a rational function in general?
2. How does the numerator of a fraction affect the domain of the function?
3. Why does division by zero make a function undefined?
4. Can the function $\frac{f}{g}(x)$ have a domain that is smaller than the domain of either $f(x)$ or $g(x)$ alone?
5. How would the domain change if the denominator were $x^2 + 3x$ instead of $x + 3$?

Tip: Always check both the numerator and denominator for restrictions when finding the domain of a rational function.