Question

x - 4 ; the domain of (f ∘ g)(x) is all real numbers except 3
\\(\frac{3}{x + 9}\\) ; the domain of (f ∘ g)(x) is all real numbers except 4
\\(\frac{3}{x - 2}\\) ; the domain of (f ∘ g)(x) is all real numbers except 2
x + 2 ; the domain of (f ∘ g)(x) is all real numbers except 9

Explanation:

To solve this problem, we need to analyze the composition of functions \((f \circ g)(x)\) and determine its domain. Let's assume we have functions \(f\) and \(g\) such that \((f \circ g)(x) = f(g(x))\). The domain of \((f \circ g)(x)\) is the set of all \(x\) for which \(g(x)\) is defined and \(f(g(x))\) is defined.

Let's analyze each option:

Option 1: \(\frac{3}{x + 9}\); the domain of \((f \circ g)(x)\) is all real numbers except \(4\)

• For the function \(\frac{3}{x + 9}\), the denominator \(x + 9\) cannot be zero. So, \(x + 9

eq 0 \implies x
eq -9\). But the domain is stated as all real numbers except \(4\), which is inconsistent. So, this option is incorrect.

Option 2: \(\frac{3}{x - 2}\); the domain of \((f \circ g)(x)\) is all real numbers except \(2\)

• For the function \(\frac{3}{x - 2}\), the denominator \(x - 2\) cannot be zero. So, \(x - 2

eq 0 \implies x
eq 2\). The domain is stated as all real numbers except \(2\), which is consistent. Let's check if this makes sense. If \(g(x)\) is such that when we compose \(f\) and \(g\), we get \(\frac{3}{x - 2}\), then the domain of \((f \circ g)(x)\) would exclude \(x = 2\) because the denominator becomes zero at \(x = 2\). This is consistent.

Option 3: \(x + 2\); the domain of \((f \circ g)(x)\) is all real numbers except \(9\)

• The function \(x + 2\) is a polynomial, and its domain is all real numbers. But the domain is stated as all real numbers except \(9\), which is inconsistent. So, this option is incorrect.

Option 4: \(x - 4\); the domain of \((f \circ g)(x)\) is all real numbers except \(3\)

• The function \(x - 4\) is a polynomial, and its domain is all real numbers. But the domain is stated as all real numbers except \(3\), which is inconsistent. So, this option is incorrect.

Step 1: Analyze the function form

The function \(\frac{3}{x - 2}\) is a rational function. For a rational function \(\frac{a}{b}\), the denominator \(b\) cannot be zero. Here, the denominator is \(x - 2\), so \(x - 2
eq 0 \implies x
eq 2\).

Step 2: Check the domain statement

The domain is stated as all real numbers except \(2\), which matches the restriction from the denominator \(x - 2
eq 0\). Other options have inconsistencies between the function form and the stated domain (e.g., polynomial functions with restricted domains or rational functions with incorrect domain restrictions). Thus, the correct option is \(\frac{3}{x - 2}\) with domain all real numbers except \(2\).

Answer:

\(\boldsymbol{\frac{3}{x - 2}}\); the domain of \((f \circ g)(x)\) is all real numbers except \(2\)