To find \(f'(x)\), we start with the function \(f(x) = \frac{6}{2x+3}\).

Using the quotient rule or by rewriting as \(f(x) = 6(2x+3)^{-1}\), differentiate:

\(f'(x) = 6 \times (-1)(2x+3)^{-2} \times 2\).

This simplifies to \(f'(x) = -6(2x+3)^{-2} \times 2\).

The expression \(f'(x)\) is always negative because \((2x+3)^{-2}\) is positive for \(x \geq 0\), and the negative sign makes \(f'(x)\) negative.

Since \(f'(x) < 0\) for all \(x \geq 0\), the function \(f\) is decreasing.