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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 \( x - 3 < 0 \)

If f(x) = √(x+1), g(x) = 2/(x-3), then find the domain of (f ∘ g)(x).

Learn how to find the domain of the composite function (f ∘ g)(x) where f(x) = √(x+1) and g(x) = 2/(x-3). This problem covers key algebraic concepts such as domain restrictions for square root and rational functions. Ideal for students in Grades 10-12.

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