To find \(f'(x)\), we differentiate \(f(x) = (3x + 2)^3 - 5\).

Using the chain rule, let \(u = 3x + 2\), then \(f(x) = u^3 - 5\).

The derivative \(\frac{du}{dx} = 3\).

Differentiate \(u^3\) with respect to \(u\): \(\frac{d}{du}(u^3) = 3u^2\).

Thus, \(\frac{d}{dx}(u^3) = 3u^2 \cdot \frac{du}{dx} = 3u^2 \cdot 3 = 9u^2\).

Substitute back \(u = 3x + 2\): \(f'(x) = 9(3x + 2)^2\).

Expanding \(9(3x + 2)^2\) gives \(81x^2 + 108x + 36\).

Since \((3x + 2)^2\) is always positive for \(x \geq 0\), \(f'(x) > 0\).

Therefore, f is an increasing function.