\begin{tabular}{l} Find the domain for \( (g \circ f)(x) \). \\ \( \qquad f(x)=\sqrt{x-10} \) and \( g(x)=x^{2}+4 \) \\ \hline\end{tabular}

To find the domain of \( (g \circ f)(x) \), we first determine the domain of \( f(x) = \sqrt{x - 10} \). The expression under the square root must be non-negative, so: \[ x - 10 \geq 0 \] This gives us: \[ x \geq 10 \] Thus, the domain of \( f \) is \( [10, \infty) \). Next, we find \( g(f(x)) \) where \( g(x) = x^2 + 4 \). Since \( f(x) \) is only defined for \( x \geq 10 \), we need to check the output of \( f(x) \) for this domain. Calculating \( f(10) \): \[ f(10) = \sqrt{10 - 10} = 0 \] As \( x \) increases from 10, \( f(x) \) will produce values greater than or equal to 0. Since the function \( g(x) \) is defined for all real numbers, we now investigate \( g(f(x)) \): \[ g(f(x)) = g(\sqrt{x - 10}) = (\sqrt{x - 10})^2 + 4 = x - 10 + 4 = x - 6 \] The output will naturally restrict to the range of \( f \), but since \( g(x) \) is defined everywhere, the overall domain of \( (g \circ f)(x) \) is still governed by the domain of \( f \). Thus, the domain of \( (g \circ f)(x) \) is: \[ [10, \infty) \]