For the given functions, find \( (f \circ g)(x) \) and \( (g \circ f)(x) \) and the domain of each. \( f(x)=8 x+5, g(x)=\sqrt{x} \)

To find the compositions \( (f \circ g)(x) \) and \( (g \circ f)(x) \), we will substitute one function into the other. 1. **Finding \( (f \circ g)(x) \)**: \[ (f \circ g)(x) = f(g(x)) = f(\sqrt{x}) \] Now, substitute \( \sqrt{x} \) into \( f(x) = 8x + 5 \): \[ f(\sqrt{x}) = 8(\sqrt{x}) + 5 = 8\sqrt{x} + 5 \] **Domain of \( (f \circ g)(x) \)**: The function \( g(x) = \sqrt{x} \) is defined for \( x \geq 0 \). Since \( f(x) \) is defined for all real numbers, the domain of \( (f \circ g)(x) \) is determined by \( g(x) \): \[ \text{Domain of } (f \circ g)(x) = [0, \infty) \] 2. **Finding \( (g \circ f)(x) \)**: \[ (g \circ f)(x) = g(f(x)) = g(8x + 5) \] Now, substitute \( 8x + 5 \) into \( g(x) = \sqrt{x} \): \[ g(8x + 5) = \sqrt{8x + 5} \] **Domain of \( (g \circ f)(x) \)**: The function \( g(x) = \sqrt{x} \) is defined for \( x \geq 0 \). Therefore, we need \( 8x + 5 \geq 0 \): \[ 8x + 5 \geq 0 \implies 8x \geq -5 \implies x \geq -\frac{5}{8} \] Thus, the domain of \( (g \circ f)(x) \) is: \[ \text{Domain of } (g \circ f)(x) = \left[-\frac{5}{8}, \infty\right) \] In summary: - \( (f \circ g)(x) = 8\sqrt{x} + 5 \) with domain \( [0, \infty) \) - \( (g \circ f)(x) = \sqrt{8x + 5} \) with domain \( \left[-\frac{5}{8}, \infty\right) \)