For the real-valued functions \( f(x)=x^{2}-1 \) and \( g(x)=\sqrt{x-6} \), find the composition \( f \circ g \) and specify its domain using interval notation.

To find the composition \( f \circ g \), we need to evaluate \( f(g(x)) \). Given the functions: \[ f(x) = x^2 - 1 \] \[ g(x) = \sqrt{x - 6} \] We start by substituting \( g(x) \) into \( f(x) \): \[ f(g(x)) = f(\sqrt{x - 6}) = (\sqrt{x - 6})^2 - 1 \] Calculating this gives: \[ f(g(x)) = x - 6 - 1 = x - 7 \] Thus, the composition \( f \circ g \) is: \[ f \circ g = x - 7 \] Next, we need to determine the domain of \( f \circ g \). The domain of the composition \( f \circ g \) is determined by the domain of \( g(x) \) and the values of \( g(x) \) that are valid inputs for \( f(x) \). 1. **Finding the domain of \( g(x) \)**: The function \( g(x) = \sqrt{x - 6} \) is defined when the expression under the square root is non-negative: \[ x - 6 \geq 0 \implies x \geq 6 \] Therefore, the domain of \( g(x) \) is \( [6, \infty) \). 2. **Finding the domain of \( f(g(x)) \)**: Since \( g(x) \) outputs values starting from \( g(6) = \sqrt{6 - 6} = 0 \) and increasing, we need to check if there are any restrictions from \( f(x) \). The function \( f(x) = x^2 - 1 \) is defined for all real numbers, so there are no additional restrictions from \( f \). Thus, the domain of \( f \circ g \) is the same as the domain of \( g(x) \): \[ \text{Domain of } f \circ g = [6, \infty) \] In conclusion, the composition \( f \circ g \) is: \[ f \circ g = x - 7 \] and its domain is: \[ [6, \infty) \]