We need to find the domain of the composite function $(f \circ g)(x) = f(g(x))$ given that:

$f(x) = \sqrt{x+1} \quad \text{and} \quad g(x) = \frac{2}{x - 3}$

Step 1: Identify the domain of $g(x)$

The function $g(x) = \frac{2}{x - 3}$ is undefined when the denominator is zero. Therefore, $g(x)$ is undefined at $x = 3$. Thus, the domain of $g(x)$ is: $x \in (-\infty, 3) \cup (3, \infty)$

Step 2: Identify the domain of $f(x)$

For $f(x) = \sqrt{x+1}$, the expression inside the square root must be non-negative: $x + 1 \geq 0 \Rightarrow x \geq -1$ Therefore, the domain of $f(x)$ is: $x \in [-1, \infty)$

Step 3: Find the domain of $(f \circ g)(x)$

The composite function $(f \circ g)(x) = f(g(x))$ is defined only when:

1. $g(x)$ is defined.
2. $g(x)$ produces values within the domain of $f(x)$.

Since the domain of $f(x)$ is $x \geq -1$, we need: $g(x) \geq -1$ Substitute $g(x) = \frac{2}{x - 3}$ into this inequality: $\frac{2}{x - 3} \geq -1$ Solving this inequality will give us the valid $x$-values that make $f(g(x))$ defined.

Solving $\frac{2}{x - 3} \geq -1$

1. Multiply both sides by $x - 3$ (considering sign changes):
   • If $x - 3 > 0$ (i.e., $x > 3$), we get $2 \geq -1(x - 3)$.
   • If $x - 3 < 0$