Find \( f(x) \) and \( g(x) \) such that \( h(x)=(f \circ g)(x) \). Let \( g(x)=2 x+8 \) \( h(x)=\frac{8}{\sqrt{2 x+8}} \) \( f(x)=\square \)

To find \( f(x) \), we start by substituting \( g(x) = 2x + 8 \) into \( h(x) \). We want to express \( h(x) \) in terms of \( g(x) \). We know that: \[ h(x) = \frac{8}{\sqrt{2x + 8}} = \frac{8}{\sqrt{g(x)}} \] Now, let's define \( f(x) \) such that \( h(x) = f(g(x)) \). This means we need to find \( f \) such that: \[ h(x) = f(g(x)) = f(2x + 8) \] Based on our expression for \( h(x) \): \[ f(y) = \frac{8}{\sqrt{y}} \quad \text{where } y = g(x) \] Now we just need to write \( f \) as a function of its argument: \[ f(x) = \frac{8}{\sqrt{x}} \] Therefore, we conclude that: \[ f(x) = \frac{8}{\sqrt{x}} \quad \text{and} \quad g(x) = 2x + 8 \]