For \( f(x)=\sqrt{x} \) and \( g(x)=6 x+7 \), find the following composite functions and state the domain of each. \( \begin{array}{lll}\text { (a) } f \circ g & \text { (b) } g \circ f & \text { (c) } f \circ f\end{array} \quad \) (d) \( g \circ g \) \( \begin{array}{ll}\text { (a) }(f \circ g)(x)=\sqrt{6 x+7} \\ \text { (Simplify your answer.) } \\ \text { The domain of } f \circ g \text { is } x \geq-\frac{7}{6} & \text { (Type your answer in interval notation. Use integers or fractions for any numbers in the expression. Simplify } \\ \text { your answer.) }\end{array} \)

To find the composite functions and their domains, we will follow these steps for each part: ### (a) \( f \circ g \) 1. **Definition of Composite Function**: \[ (f \circ g)(x) = f(g(x)) \] Given \( g(x) = 6x + 7 \), we substitute this into \( f(x) \): \[ (f \circ g)(x) = f(6x + 7) = \sqrt{6x + 7} \] 2. **Domain of \( f \circ g \)**: The function \( f(x) = \sqrt{x} \) is defined for \( x \geq 0 \). Therefore, we need: \[ 6x + 7 \geq 0 \] Solving for \( x \): \[ 6x \geq -7 \implies x \geq -\frac{7}{6} \] Thus, the domain of \( f \circ g \) is: \[ \text{Domain of } f \circ g: \left[-\frac{7}{6}, \infty\right) \] ### (b) \( g \circ f \) 1. **Definition of Composite Function**: \[ (g \circ f)(x) = g(f(x)) \] Given \( f(x) = \sqrt{x} \), we substitute this into \( g(x) \): \[ (g \circ f)(x) = g(\sqrt{x}) = 6\sqrt{x} + 7 \] 2. **Domain of \( g \circ f \)**: The function \( f(x) = \sqrt{x} \) is defined for \( x \geq 0 \). Therefore, the domain of \( g \circ f \) is: \[ \text{Domain of } g \circ f: [0, \infty) \] ### (c) \( f \circ f \) 1. **Definition of Composite Function**: \[ (f \circ f)(x) = f(f(x)) \] Substituting \( f(x) = \sqrt{x} \): \[ (f \circ f)(x) = f(\sqrt{x}) = \sqrt{\sqrt{x}} = x^{1/4} \] 2. **Domain of \( f \circ f \)**: The function \( f(x) = \sqrt{x} \) is defined for \( x \geq 0 \). Therefore, the domain of \( f \circ f \) is: \[ \text{Domain of } f \circ f: [0, \infty) \] ### (d) \( g \circ g \) 1. **Definition of Composite Function**: \[ (g \circ g)(x) = g(g(x)) \] Substituting \( g(x) = 6x + 7 \): \[ (g \circ g)(x) = g(6x + 7) = 6(6x + 7) + 7 = 36x + 42 + 7 = 36x + 49 \] 2. **Domain of \( g \circ g \)**: The function \( g(x) = 6x + 7 \) is defined for all real numbers. Therefore, the domain of \( g \circ g \) is: \[ \text{Domain of } g \circ g: (-\infty, \infty) \] ### Summary of Results - (a) \( (f \circ g)(x) = \sqrt{6x + 7} \) with domain \( \left[-\frac{7}{6}, \infty\right) \) - (b) \( (g \circ f)(x) = 6\sqrt{x} + 7 \) with domain \( [0, \infty) \) - (c) \( (f \circ f)(x) = x^{1/4} \) with domain \( [0, \infty) \) - (d) \( (g \circ g)(x) = 36x + 49 \) with domain \( (-\infty, \infty) \)