For \( f(x)=x^{2}+5 \) and \( g(x)=\sqrt{x-1} \), find the following composite functions and state the domain of each. \( \begin{array}{lll}\text { (a) } f \circ g & \text { (b) } g \circ f & \text { (c) } f \circ f \\ \text { (d) } g \circ g \\ \text { (a) }(f \circ g)(x)=x+4 \\ \text { (Simplify your answer.) } \\ \text { The domain of } f \circ g \text { is } x \geq 1 .\end{array} \) (Type your answer in interval notation. Use integers or fractions for any numbers in the expression. Simplify your answer.)

To find the composite functions and their domains, we will follow these steps for each composite function: 1. **Calculate the composite function.** 2. **Determine the domain of the composite function.** Let's start with each part: ### (a) \( f \circ g \) The composite function \( (f \circ g)(x) \) is defined as \( f(g(x)) \). 1. **Calculate \( f(g(x)) \)**: \[ g(x) = \sqrt{x-1} \] Now substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = f(\sqrt{x-1}) = (\sqrt{x-1})^2 + 5 = x - 1 + 5 = x + 4 \] 2. **Domain of \( f \circ g \)**: The domain of \( g(x) \) is \( x \geq 1 \) (since the square root function is defined for \( x-1 \geq 0 \)). Therefore, the domain of \( f \circ g \) is: \[ \text{Domain of } f \circ g: [1, \infty) \] ### (b) \( g \circ f \) The composite function \( (g \circ f)(x) \) is defined as \( g(f(x)) \). 1. **Calculate \( g(f(x)) \)**: \[ f(x) = x^2 + 5 \] Now substitute \( f(x) \) into \( g(x) \): \[ g(f(x)) = g(x^2 + 5) = \sqrt{(x^2 + 5) - 1} = \sqrt{x^2 + 4} \] 2. **Domain of \( g \circ f \)**: The function \( g(x) \) requires \( x - 1 \geq 0 \), which translates to \( x^2 + 5 - 1 \geq 0 \). Since \( x^2 + 4 \) is always non-negative for all real \( x \), the domain of \( g \circ f \) is: \[ \text{Domain of } g \circ f: (-\infty, \infty) \] ### (c) \( f \circ f \) The composite function \( (f \circ f)(x) \) is defined as \( f(f(x)) \). 1. **Calculate \( f(f(x)) \)**: \[ f(f(x)) = f(x^2 + 5) = (x^2 + 5)^2 + 5 \] 2. **Domain of \( f \circ f \)**: Since \( f(x) \) is a polynomial, it is defined for all real numbers. Therefore, the domain of \( f \circ f \) is: \[ \text{Domain of } f \circ f: (-\infty, \infty) \] ### (d) \( g \circ g \) The composite function \( (g \circ g)(x) \) is defined as \( g(g(x)) \). 1. **Calculate \( g(g(x)) \)**: \[ g(g(x)) = g(\sqrt{x-1}) = \sqrt{\sqrt{x-1} - 1} \] 2. **Domain of \( g \circ g \)**: For \( g(x) \) to be defined, we need \( x \geq 1 \). Now, we need to ensure that \( \sqrt{x-1} - 1 \geq 0 \): \[ \sqrt{x-1} \geq 1 \implies x - 1 \geq 1 \implies x \geq 2 \] Therefore, the domain of \( g \circ g \) is: \[ \text{Domain of } g \circ g: [2, \infty) \] ### Summary of Results - (a) \( (f \circ g)(x) = x + 4 \) with domain \( [1, \infty) \) - (b) \( (g \circ f)(x) = \sqrt{x^2 + 4} \) with domain \( (-\infty, \infty) \) - (c) \( (f \circ f)(x) = (x^2 + 5)^2 + 5 \) with domain \( (-\infty, \infty) \) - (d) \( (g \circ g)(x) = \sqrt{\sqrt{x-1} - 1} \) with domain \( [2, \infty) \)