The problem asks for the domain and range of the function $f - g$, where:

$f(x) = \sqrt{x + 5}, \quad g(x) = \sqrt{x - 5}$

Step 1: Determine the domain of $f(x)$ and $g(x)$

1. Domain of $f(x) = \sqrt{x + 5}$: $x + 5 \geq 0 \implies x \geq -5$ So, the domain of $f(x)$ is $x \geq -5$.

2. Domain of $g(x) = \sqrt{x - 5}$: $x - 5 \geq 0 \implies x \geq 5$ So, the domain of $g(x)$ is $x \geq 5$.

Since $f - g$ exists only where both $f(x)$ and $g(x)$ are defined, the domain of $f - g$ is the intersection of the two domains: $\text{Domain of } f - g: x \geq 5$

Step 2: Determine the range of $f - g$

• For $f(x) = \sqrt{x + 5}$:

  • Minimum value: $f(5) = \sqrt{10}$
  • As $x$ increases, $f(x)$ increases without bound, so the range of $f(x)$ is $[\sqrt{10}, \infty)$.

• For $g(x) = \sqrt{x - 5}$:

  • Minimum value: $g(5) = 0$
  • As $x$ increases, $g(x)$ increases without bound, so the range of $g(x)$ is $[0, \infty)$.

• Range of $f - g$:

  • Minimum value: $f(5) - g(5) = \sqrt{10} - 0 = \sqrt{10}$
  • As $x$ increases, $f(x) - g(x)$ increases without bound.

Thus, the range of $f - g$ is $[\sqrt{10}, \infty)$.

Final Answer

• Domain: $x \geq 5$
• Range: $y \geq \sqrt{10}$

This corresponds to option A in the provided choices.

Would you like more details on any part of this explanation, or do you have any other questions?

Here are some related questions to expand your understanding:

1. How do you find the domain and range of a function that involves square roots?
2. What is the difference between the union and intersection of domains?
3. How does the range of a function change when you subtract two functions?
4. How would the domain and range change if $g(x)$ was a linear function instead of a square root function?
5. What are the implications if $x$ in the square root functions is replaced with a quadratic expression?

Tip: When dealing with square roots, always remember that the expression inside the root must be non-negative to ensure real number outputs.