To determine the domain of [tex]\((f \circ g)(x)\)[/tex], we need to understand the domains of the individual functions [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex], and then find the restrictions they impose when combined.
1. Domain of [tex]\(g(x) = \sqrt{x+5}\)[/tex]:
- The function [tex]\(g(x)\)[/tex] involves a square root, which is defined only for non-negative values inside the root.
- Therefore, we require:
[tex]\[
x + 5 \geq 0
\][/tex]
- Solving this inequality:
[tex]\[
x \geq -5
\][/tex]
- Thus, the domain of [tex]\(g(x)\)[/tex] is [tex]\([-5, \infty)\)[/tex].
2. Domain of [tex]\(f(x) = \frac{1}{x-3}\)[/tex]:
- The function [tex]\(f(x)\)[/tex] is defined for all [tex]\(x\)[/tex] except where the denominator is zero.
- Therefore, we require:
[tex]\[
x - 3 \neq 0 \quad \Rightarrow \quad x \neq 3
\][/tex]
- So, the domain of [tex]\(f(x)\)[/tex] is [tex]\(\mathbb{R} \setminus \{ 3 \}\)[/tex].
3. Domain of [tex]\((f \circ g)(x)\)[/tex]:
- The composition [tex]\( (f \circ g)(x) \)[/tex] implies that [tex]\( g(x) \)[/tex] must produce values that are within the domain of [tex]\( f \)[/tex].
- Thus, we need:
[tex]\[
g(x) \neq 3
\][/tex]
- We find when [tex]\(g(x) = 3\)[/tex]:
[tex]\[
\sqrt{x + 5} = 3
\][/tex]
- Squaring both sides gives:
[tex]\[
x + 5 = 9
\][/tex]
- Solving for [tex]\(x\)[/tex]:
[tex]\[
x = 4
\][/tex]
- Therefore, [tex]\(x \neq 4\)[/tex] to avoid [tex]\(g(x) = 3\)[/tex].
Combining these restrictions:
- [tex]\(x \geq -5\)[/tex] (from the domain of [tex]\(g(x)\)[/tex])
- [tex]\(x \neq 4\)[/tex] (to ensure [tex]\(g(x) \neq 3\)[/tex])
The domain of [tex]\((f \circ g)(x)\)[/tex] is:
[tex]\[
[-5, 4) \cup (4, \infty)
\][/tex]
So, the correct answer is:
[tex]\[
\boxed{[-5, 4) \cup (4, \infty)}
\][/tex]
