For \( f(x)=x^{2}+5 \) and \( g(x)=\sqrt{x-4} \), find the following composite functions and state the domain of each. \( \begin{array}{llll}\text { (a) } f \circ g & \text { (b) } g \circ f & \text { (c) } f \circ f & \text { (d) } g \circ g \\ \text { (a) }(f \circ g)(x)=x+1 & \text { (Simplify your answer.) } \\ \text { Select the correct choice below and fill in any answer boxes within your choice. } \\ \text { A. The domain of } f \circ g \text { is }\{x \llbracket\} \\ \text { (Type an inequality. Simplify your answer. Use integers or fractions for any numbers in the expression.) } \\ \text { B. The domain of } f \circ g \text { is all real numbers }\end{array} \)

To find the composite functions \( f \circ g \), \( g \circ f \), \( f \circ f \), and \( g \circ g \), we will follow these steps: 1. **Calculate each composite function.** 2. **Determine the domain of each composite function.** ### Given Functions: - \( f(x) = x^2 + 5 \) - \( g(x) = \sqrt{x - 4} \) ### (a) Find \( f \circ g \) The composite function \( f \circ g \) is defined as: \[ (f \circ g)(x) = f(g(x)) = f(\sqrt{x - 4}) \] Substituting \( g(x) \) into \( f(x) \): \[ f(\sqrt{x - 4}) = (\sqrt{x - 4})^2 + 5 = x - 4 + 5 = x + 1 \] Thus, \[ (f \circ g)(x) = x + 1 \] **Domain of \( f \circ g \)**: The domain of \( g(x) = \sqrt{x - 4} \) is \( x - 4 \geq 0 \) or \( x \geq 4 \). Therefore, the domain of \( f \circ g \) is: \[ \text{Domain of } f \circ g: \{ x \mid x \geq 4 \} \] ### (b) Find \( g \circ f \) The composite function \( g \circ f \) is defined as: \[ (g \circ f)(x) = g(f(x)) = g(x^2 + 5) \] Substituting \( f(x) \) into \( g(x) \): \[ g(x^2 + 5) = \sqrt{(x^2 + 5) - 4} = \sqrt{x^2 + 1} \] Thus, \[ (g \circ f)(x) = \sqrt{x^2 + 1} \] **Domain of \( g \circ f \)**: Since \( g(x) \) is defined for all \( x \) such that \( x - 4 \geq 0 \), we need \( x^2 + 5 \geq 4 \): \[ x^2 + 1 \geq 0 \quad \text{(always true for all real numbers)} \] Thus, the domain of \( g \circ f \) is: \[ \text{Domain of } g \circ f: \text{all real numbers} \] ### (c) Find \( f \circ f \) The composite function \( f \circ f \) is defined as: \[ (f \circ f)(x) = f(f(x)) = f(x^2 + 5) \] Substituting \( f(x) \) into itself: \[ f(x^2 + 5) = (x^2 + 5)^2 + 5 \] Expanding this: \[ = x^4 + 10x^2 + 25 + 5 = x^4 + 10x^2 + 30 \] Thus, \[ (f \circ f)(x) = x^4 + 10x^2 + 30 \] **Domain of \( f \circ f \)**: Since \( f(x) \) is a polynomial, it is defined for all real numbers. Thus, the domain of \( f \circ f \) is: \[ \text{Domain of } f \circ f: \text{all real numbers} \] ### (d) Find \( g \circ g \) The composite function \( g \circ g \) is defined as: \[ (g \circ g)(x) = g(g(x)) = g(\sqrt{x - 4}) \] Substituting \( g(x) \) into itself: \[ g(\sqrt{x - 4}) = \sqrt{\sqrt{x - 4} - 4} \] This expression is valid only if \( \sqrt{x - 4} \geq 4 \): \[ x - 4 \geq 16 \quad \Rightarrow \quad x \geq 20 \] Thus, \[ (g \circ g)(x) = \sqrt{\sqrt{x - 4} - 4} \] **Domain of \( g \circ g \)**: The domain of \( g \circ g \) is: \[ \text{Domain of } g \circ g: \{ x \mid x \geq 20 \} \] ### Summary of Results: - \( (f \circ g)(x) = x + 1 \) with domain \( \{ x \mid x \geq 4 \} \) - \( (g \circ f)(x) = \sqrt{x^2 + 1} \) with domain all real numbers - \( (f \circ f)(x) = x^4 + 10x^2 + 30 \) with domain all real numbers - \( (g \circ g)(x) = \sqrt{\sqrt{x - 4} - 4} \) with domain \( \{ x \mid x \geq 20 \} \)