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.

(i) Differentiate the function: \(\frac{dy}{dx} = 3x^2 - 6x - 9\).

Set \(\frac{dy}{dx} = 0\) to find critical points: \(3x^2 - 6x - 9 = 0\).

Factorize: \(3(x^2 - 2x - 3) = 0\).

Solve: \(x^2 - 2x - 3 = 0\) gives \((x - 3)(x + 1) = 0\).

Thus, \(x = 3\) or \(x = -1\).

Since the minimum point is on the \(x\)-axis, \(y = 0\) at \(x = 3\).

Substitute \(x = 3\) into the original equation: \(3^3 - 3(3)^2 - 9(3) + k = 0\).

Simplify: \(27 - 27 - 27 + k = 0\).

Thus, \(k = 27\).

(ii) Substitute \(x = -1\) into the original equation to find the maximum point:

\(y = (-1)^3 - 3(-1)^2 - 9(-1) + 27\).

Simplify: \(y = -1 - 3 + 9 + 27 = 32\).

Thus, the maximum point is \((-1, 32)\).

(iii) The function is decreasing where \(\frac{dy}{dx} < 0\).

From \(3x^2 - 6x - 9 = 0\), the roots are \(x = 3\) and \(x = -1\).

Test intervals: \(-1 < x < 3\) is where \(\frac{dy}{dx} < 0\).

Thus, the function is decreasing for \(-1 < x < 3\).

\(To find f'(x), we differentiate f(x) = (2x − 3)³ − 8.\)

\(Using the chain rule, let u = 2x − 3, then f(x) = u³ − 8.\)

The derivative of u³ is 3u², and the derivative of u with respect to x is 2.

\(Thus, f'(x) = 3(2x − 3)² × 2 = 6(2x − 3)².\)

Since (2x − 3)² is always positive for all x, f'(x) is positive for all x in the interval [2, 4].

Therefore, f is an increasing function.

First, find the derivative \(f'(x)\) to determine the behavior of the function:

\(f'(x) = 5x^4 - 30x^2 + 50\).

To analyze whether \(f(x)\) is increasing or decreasing, we need to check the sign of \(f'(x)\).

Consider \(f'(x) = 5(x^2 - 3)^2 + 5\).

Since \((x^2 - 3)^2 \geq 0\) for all \(x\), it follows that \(5(x^2 - 3)^2 + 5 > 0\) for all \(x\).

Thus, \(f'(x) > 0\) for all \(x\), indicating that \(f(x)\) is an increasing function.

To find the maximum possible value of \(a\), we need to ensure that the function \(f(x) = \frac{1}{3}(2x - 1)^{\frac{3}{2}} - 2x\) is decreasing for \(\frac{1}{2} < x < a\).

The derivative \(f'(x)\) is given by:

\(f'(x) = \left( (2x-1)^{\frac{1}{2}} \right) \times \left( \frac{1}{3} \times 2x \times \frac{3}{2} \right) \times (-2)\)

\(= (2x-1)^{\frac{1}{2}} \times (-2)\)

For \(f(x)\) to be decreasing, \(f'(x) \leq 0\).

\((2x-1)^{\frac{1}{2}} - 2 \leq 0\)

\(2x - 1 \leq 4\)

\(2x - 1 < 4\)

Solving \(2x - 1 < 4\), we get:

\(2x < 5\)

\(x < \frac{5}{2}\)

Thus, the maximum possible value of \(a\) is \(\frac{5}{2}\).