For \( f(x)=x^{2}+5 \) and \( g(x)=\sqrt{x-4} \), find the following composite functions and state the domain of each. \( \begin{array}{llll}\text { (a) } f \circ g & \text { (b) } g \circ f & \text { (c) } f \circ f & \text { (d) } g \circ g\end{array} \) (a) \( (f \circ g)(x)=\square \) (Simplify your answer.)

To find the composite functions and their domains, let's begin with each part one at a time. **(a)** For \( (f \circ g)(x) = f(g(x)) \): First, we need to compute \( g(x) = \sqrt{x-4} \). Now plug that into \( f(x) \): \[ f(g(x)) = f(\sqrt{x-4}) = (\sqrt{x-4})^2 + 5 = x - 4 + 5 = x + 1. \] The domain of \( g(x) \) is \( x \geq 4 \) (since we cannot take the square root of a negative number), so the domain of \( f \circ g \) is also \( x \geq 4 \). So, \( (f \circ g)(x) = x + 1 \) with domain \( x \geq 4 \). **(b)** For \( (g \circ f)(x) = g(f(x)) \): We start with \( f(x) = x^2 + 5 \). Plug this into \( g(x) \): \[ g(f(x)) = g(x^2 + 5) = \sqrt{(x^2 + 5) - 4} = \sqrt{x^2 + 1}. \] Since there are no restrictions on the values we can plug into \( f(x) \) (as it is defined for all \( x \)), the output \( x^2 + 5 \) will always be at least 5, ensuring that \( g \) is well-defined. Thus, the domain of \( g \circ f \) is all real numbers \( x \). Hence, \( (g \circ f)(x) = \sqrt{x^2 + 1} \) with domain \( (-\infty, \infty) \). **(c)** For \( (f \circ f)(x) = f(f(x)) \): Here, we compute: \[ f(f(x)) = f(x^2 + 5) = (x^2 + 5)^2 + 5. \] Upon expanding, we have: \[ (x^2 + 5)^2 = x^4 + 10x^2 + 25. \] So, \[ f(f(x)) = x^4 + 10x^2 + 25 + 5 = x^4 + 10x^2 + 30. \] The function \( f(x) \) is defined for all real \( x \), so the domain remains all real numbers. Thus, \( (f \circ f)(x) = x^4 + 10x^2 + 30 \) with domain \( (-\infty, \infty) \). **(d)** For \( (g \circ g)(x) = g(g(x)) \): Start with \( g(x) = \sqrt{x-4} \) and find: \[ g(g(x)) = g(\sqrt{x-4}) = \sqrt{\sqrt{x-4} - 4}. \] To evaluate this, note that for \( x-4 \geq 0 \), i.e., \( x \geq 4 \), we also need \( \sqrt{x-4} - 4 \geq 0 \): \[ \sqrt{x-4} \geq 4 \implies x - 4 \geq 16 \implies x \geq 20. \] Thus, the domain of \( g \circ g \) is \( x \geq 20 \). Therefore, \( (g \circ g)(x) = \sqrt{\sqrt{x-4} - 4} \) with domain \( x \geq 20 \).