Given: \( g(x)=\sqrt{x-4} \) and \( h(x)=2 x-8 \). What are the restrictions on the domain of \( g \circ h \) ? \( x \geq \square \) DONE:

To find the restrictions on the domain of \( g \circ h \), we first need to determine the input for \( g \) based on the output of \( h \). The function \( h(x) = 2x - 8 \) needs to be within the valid domain of \( g(x) = \sqrt{x - 4} \). For \( g(x) \) to be defined, we require \( x - 4 \geq 0 \), which simplifies to \( x \geq 4 \). Now, substituting \( h(x) \) into this inequality, we have \( h(x) \geq 4 \): \[ 2x - 8 \geq 4 \] Solving for \( x \), we get: \[ 2x \geq 12 \implies x \geq 6. \] Thus, for \( g \circ h \) to be defined, we need \( x \geq 6 \). So, the restriction on the domain of \( g \circ h \) is \( x \geq 6 \).