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, we perform the following calculations: (a) \( f \circ g(x) = f(g(x)) = f(\sqrt{x-1}) = (\sqrt{x-1})^2 + 5 = x - 1 + 5 = x + 4 \). The domain of \( f \circ g \) is \( x \geq 1 \) because \( g(x) \) must be defined, which requires \( x - 1 \geq 0 \). (b) \( g \circ f(x) = g(f(x)) = g(x^2 + 5) = \sqrt{(x^2 + 5) - 1} = \sqrt{x^2 + 4} \). The domain of \( g \circ f \) is all real numbers \( \mathbb{R} \) since \( x^2 + 4 \) is always non-negative. (c) \( f \circ f(x) = f(f(x)) = f(x^2 + 5) = (x^2 + 5)^2 + 5 = x^4 + 10x^2 + 25 + 5 = x^4 + 10x^2 + 30 \). The domain of \( f \circ f \) is also all real numbers \( \mathbb{R} \). (d) \( g \circ g(x) = g(g(x)) = g(\sqrt{x-1}) = \sqrt{\sqrt{x-1} - 1} \). The domain of \( g \circ g \) is \( x \geq 2 \) because \( \sqrt{x-1} \) must be at least 1 for \( g \) to be defined. To summarize: - \( (f \circ g)(x) = x + 4 \) with domain \( [1, \infty) \) - \( (g \circ f)(x) = \sqrt{x^2 + 4} \) with domain \( (-\infty, \infty) \) - \( (f \circ f)(x) = x^4 + 10x^2 + 30 \) with domain \( (-\infty, \infty) \) - \( (g \circ g)(x) = \sqrt{\sqrt{x-1} - 1} \) with domain \( [2, \infty) \)