To determine where the function \(f\) is increasing, we need to find where \(f'(x) > 0\).

Given \(f'(x) = 3x^2 + 2x - 5\), we set up the inequality:

\(3x^2 + 2x - 5 > 0\).

Factor the quadratic expression:

\((3x + 5)(x - 1) > 0\).

The critical points are found by setting each factor to zero:

\(3x + 5 = 0 \Rightarrow x = -\frac{5}{3}\)

\(x - 1 = 0 \Rightarrow x = 1\)

These critical points divide the number line into intervals. Test each interval to determine where the inequality holds:

• For \(x < -\frac{5}{3}\), choose \(x = -2\): \((3(-2) + 5)((-2) - 1) = (-1)(-3) = 3 > 0\).
• For \(-\frac{5}{3} < x < 1\), choose \(x = 0\): \((3(0) + 5)(0 - 1) = (5)(-1) = -5 < 0\).
• For \(x > 1\), choose \(x = 2\): \((3(2) + 5)(2 - 1) = (11)(1) = 11 > 0\).

Thus, \(f(x)\) is increasing for \(x \leq -\frac{5}{3}\) and \(x > 1\).

(a) Differentiate \(y = k(3x-k)^{-1} + 3x\) to find \(\frac{dy}{dx} = -\frac{k}{(3x-k)^2} + 3\). Set \(\frac{dy}{dx} = 0\) to find stationary points:

\(-\frac{k}{(3x-k)^2} + 3 = 0\)

\(\Rightarrow 3(3x-k)^2 = k\)

\(\Rightarrow 3x-k = \pm \sqrt{k}\)

\(\Rightarrow x = \frac{k \pm \sqrt{k}}{3}\)

(b) Substitute \(x = a\) when \(k = 4\) into \(f(x) = 4(3x-4)^{-1} + 3x\):

\(a = \frac{4 \pm \sqrt{4}}{3} \Rightarrow a = 2\)

Find the second derivative \(f''(x) = 72(3x-4)^{-3}\), which is positive at \(x = 2\), indicating a minimum.

(c) Substitute \(k = -1\) into \(g(x) = -(3x+1)^{-1} + 3x\) to find \(g'(x) = \frac{3}{(3x+1)^2} + 3\), which is always positive, indicating \(g\) is an increasing function.

(i) Differentiate \(y = \frac{1}{6}(2x - 3)^3 - 4x\) with respect to \(x\).

Using the chain rule, \(\frac{dy}{dx} = \frac{1}{6} \times 3 \times (2x - 3)^2 \times 2 - 4\).

(ii) To find the y-intercept, set \(x = 0\):

\(y = \frac{1}{6}(2 \times 0 - 3)^3 - 4 \times 0 = -\frac{27}{6} = -\frac{9}{2}\).

The slope of the tangent is \(\frac{dy}{dx}\) at \(x = 0\):

\(\frac{dy}{dx} = \frac{1}{6} \times 3 \times (-3)^2 \times 2 - 4 = 5\).

The equation of the tangent is \(y + \frac{9}{2} = 5x\).

Rearranging gives \(2y + 9 = 10x\).

(iii) The function is increasing when \(\frac{dy}{dx} > 0\).

\(\frac{1}{6} \times 3 \times (2x - 3)^2 \times 2 - 4 > 0\).

\((2x - 3)^2 > 4\).

Solving \((2x - 3)^2 - 4 = 0\) gives \(x = \frac{5}{2}\) or \(x = \frac{1}{2}\).

The set of values for which the function is increasing is \(x > \frac{5}{2}, x < \frac{1}{2}\).

To find \(f'(x)\), we use the quotient rule for differentiation. The function \(f(x) = \frac{3}{2x+5}\) can be rewritten as \(3(2x+5)^{-1}\).

The derivative of \((2x+5)^{-1}\) is \(-1(2x+5)^{-2} \times 2\) using the chain rule.

Thus, \(f'(x) = 3 \times -1(2x+5)^{-2} \times 2 = -3(2x+5)^{-2} \times 2\).

Since \(f'(x)\) is negative for all \(x \neq -2.5\), the function \(f\) is decreasing.

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.