To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\).
We have:
\[ f(x) = \frac{2}{x + 3} \]
\[ g(x) = \frac{1}{2x} \]
Now, we find \(f(g(x))\):
\[ f(g(x)) = f\left(\frac{1}{2x}\right) = \frac{2}{\frac{1}{2x} + 3} \]
Next, we simplify the denominator:
\[ \frac{1}{2x} + 3 = \frac{1}{2x} + \frac{3 \cdot 2x}{2x} = \frac{1 + 6x}{2x} \]
Now substituting this back into our expression for \(f(g(x))\):
\[ f(g(x)) = \frac{2}{\frac{1 + 6x}{2x}} = 2 \cdot \frac{2x}{1 + 6x} = \frac{4x}{1 + 6x} \]
Therefore, we have:
\[ (f \circ g)(x) = \frac{4x}{1 + 6x} \]
The correct response is:
\((f \circ g)(x) = \frac{4x}{1 + 6x}\)