To find \(f'(x)\), we differentiate \(f(x) = 2 - \frac{3}{4x-p}\).

Using the chain rule, let \(u = 4x - p\), then \(\frac{du}{dx} = 4\).

The derivative of \(\frac{3}{u}\) with respect to \(u\) is \(-\frac{3}{u^2}\).

Thus, \(\frac{d}{dx} \left( \frac{3}{4x-p} \right) = -3 \cdot \frac{1}{(4x-p)^2} \cdot 4 = \frac{-12}{(4x-p)^2}\).

Therefore, \(f'(x) = 0 - \left( \frac{-12}{(4x-p)^2} \right) = \frac{12}{(4x-p)^2}\).

Since \(\frac{12}{(4x-p)^2} > 0\) for \(x > \frac{p}{4}\), \(f\) is an increasing function.

To determine when the function \(f(x) = x^3 - x^2 - 8x + 5\) is increasing, we need to find where its derivative is positive.

First, differentiate \(f(x)\):

\(f'(x) = 3x^2 - 2x - 8\).

Set \(f'(x) > 0\) to find where the function is increasing:

\(3x^2 - 2x - 8 > 0\).

Solving the inequality \(3x^2 - 2x - 8 = 0\) gives the critical points. Using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -2\), \(c = -8\).

\(x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 3 \times (-8)}}{2 \times 3}\)

\(x = \frac{2 \pm \sqrt{4 + 96}}{6}\)

\(x = \frac{2 \pm \sqrt{100}}{6}\)

\(x = \frac{2 \pm 10}{6}\)

\(x = 2\) or \(x = -\frac{4}{3}\).

The function is increasing in the interval \(x < -\frac{4}{3}\) and \(x > 2\).

Since \(x < a\), the largest possible value of \(a\) is \(-\frac{4}{3}\).

To determine when the function \(f(x) = x^3 - 3x^2 - 9x + 2\) is increasing, we need to find when its derivative \(f'(x)\) is non-negative.

First, find the derivative:

\(f'(x) = 3x^2 - 6x - 9\).

Set \(f'(x) \geq 0\) to find the critical points:

\(3x^2 - 6x - 9 = 0\).

Factor the quadratic equation:

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

\(3(x - 3)(x + 1) = 0\).

The critical points are \(x = 3\) and \(x = -1\).

Since \(f(x)\) is increasing for \(x > n\), we need \(f'(x) \geq 0\) for \(x > n\).

Test intervals around the critical points:

- For \(x > 3\), \(f'(x) > 0\).

- For \(-1 < x < 3\), \(f'(x) < 0\).

- For \(x < -1\), \(f'(x) > 0\).

Thus, \(f(x)\) is increasing for \(x > 3\).

Therefore, the least possible value of \(n\) is 3.

(i) To express \(3x^2 - 6x + 2\) in the form \(a(x+b)^2 + c\), we complete the square:

Start with \(3(x^2 - 2x) + 2\).

Complete the square for \(x^2 - 2x\):

\(x^2 - 2x = (x-1)^2 - 1\).

Substitute back: \(3((x-1)^2 - 1) + 2 = 3(x-1)^2 - 3 + 2 = 3(x-1)^2 - 1\).

Thus, \(a = 3\), \(b = -1\), \(c = -1\).

(ii) Differentiate \(f(x) = x^3 - 3x^2 + 7x - 8\):

\(f'(x) = 3x^2 - 6x + 7\).

Complete the square for \(3x^2 - 6x + 7\):

\(3(x^2 - 2x) + 7 = 3((x-1)^2 - 1) + 7 = 3(x-1)^2 + 4\).

Since \(3(x-1)^2 + 4 > 0\) for all \(x\), \(f'(x) > 0\) for all \(x\).

Therefore, \(f\) is an increasing function.

(i) To find \(f'(x)\), differentiate \(f(x) = \frac{1}{x+1} + \frac{1}{(x+1)^2}\).

The derivative of \(\frac{1}{x+1}\) is \(-(x+1)^{-2}\).

The derivative of \(\frac{1}{(x+1)^2}\) is \(-2(x+1)^{-3}\).

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

(ii) Since \(f'(x) < 0\) for \(x > -1\), the function \(f\) is decreasing.

(iii) To find the stationary point of \(g(x) = \frac{1}{x+1} + \frac{1}{(x+1)^2}\), set \(\frac{dy}{dx} = 0\).

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

Multiply by \((x+1)^3\) to get \(-(x+1) - 2 = 0\) or \(-x^2 - 4x - 3 = 0\).

Solving \(-x^2 - 4x - 3 = 0\) gives \(x = -3\).

Substitute \(x = -3\) into \(g(x)\) to find \(y = -1/4\).

Thus, the coordinates of the stationary point are \((-3, -1/4)\).

(i) To express \(9x^2 - 12x + 5\) in the form \((ax + b)^2 + c\), we complete the square:

Start with \(9x^2 - 12x\).

Factor out the 9: \(9(x^2 - \frac{4}{3}x)\).

Complete the square inside the bracket: \(x^2 - \frac{4}{3}x = (x - \frac{2}{3})^2 - (\frac{2}{3})^2\).

Thus, \(9(x^2 - \frac{4}{3}x) = 9((x - \frac{2}{3})^2 - \frac{4}{9}) = 9(x - \frac{2}{3})^2 - 4\).

So, \(9x^2 - 12x + 5 = 9(x - \frac{2}{3})^2 - 4 + 5 = 9(x - \frac{2}{3})^2 + 1\).

Therefore, \((3x - 2)^2 + 1\).

(ii) To determine if \(3x^3 - 6x^2 + 5x - 12\) is increasing, decreasing, or neither, find the derivative:

\(f'(x) = 9x^2 - 12x + 5\).

From part (i), \(f'(x) = (3x - 2)^2 + 1\).

Since \((3x - 2)^2\) is always non-negative, \((3x - 2)^2 + 1 > 0\) for all \(x\).

Therefore, \(f'(x) > 0\) for all \(x\), indicating that the function is always increasing.

(i) To find stationary points, we differentiate \(y = x^3 + ax^2 + bx\) to get \(\frac{dy}{dx} = 3x^2 + 2ax + b\). For no stationary points, the discriminant of this quadratic must be less than zero: \(b^2 - 4ac < 0\). Here, \(a = 2a\) and \(c = b\), so \(b^2 - 4(3)(b) < 0\) simplifies to \(a^2 < 3b\).

(ii) With \(a = -6\) and \(b = 9\), the equation becomes \(y = x^3 - 6x^2 + 9x\). Differentiating gives \(\frac{dy}{dx} = 3x^2 - 12x + 9\). Setting \(\frac{dy}{dx} < 0\) gives \(3x^2 - 12x + 9 < 0\). Solving \(3(x^2 - 4x + 3) < 0\) gives roots at \(x = 1\) and \(x = 3\). Thus, \(y\) is decreasing for \(1 < x < 3\).

(i) To find \(f'(x)\), use the quotient rule: if \(f(x) = \frac{u}{v}\), then \(f'(x) = \frac{u'v - uv'}{v^2}\).

Here, \(u = 5\) and \(v = 1 - 3x\).

\(u' = 0\) and \(v' = -3\).

So, \(f'(x) = \frac{0 imes (1 - 3x) - 5 imes (-3)}{(1 - 3x)^2} = \frac{15}{(1 - 3x)^2}\).

Since the derivative is negative, \(f'(x) = \frac{-15}{(1 - 3x)^2}\).

(ii) Since \(15 > 0\) and \((1 - 3x)^2 > 0\) for \(x \geq 1\), \(f'(x) > 0\).

Therefore, \(f\) is an increasing function.

To determine if \(f(x) = (2x - 5)^3 + x\) is increasing, we need to find its derivative \(f'(x)\).

Using the chain rule, the derivative of \((2x - 5)^3\) is \(3(2x - 5)^2 \times 2\).

Thus, \(f'(x) = 3(2x - 5)^2 \times 2 + 1 = 6(2x - 5)^2 + 1\).

We can also express this as \(f'(x) = 24\left(\frac{x - 5}{2}\right)^2 + 1\).

Since \((2x - 5)^2 \geq 0\) for all \(x\), it follows that \(6(2x - 5)^2 + 1 > 0\) for all \(x\).

Therefore, \(f'(x) > 0\) for all \(x \in \mathbb{R}\), which means \(f\) is an increasing function.

To determine if \(f(x) = \frac{1}{x^3} - x^3\) is a decreasing function, we need to find its derivative \(f'(x)\) and check if it is negative for \(x > 0\).

First, differentiate \(f(x)\):

\(f'(x) = \frac{d}{dx} \left( \frac{1}{x^3} \right) - \frac{d}{dx} (x^3)\).

Using the power rule, \(\frac{d}{dx} \left( x^n \right) = nx^{n-1}\), we find:

\(\frac{d}{dx} \left( \frac{1}{x^3} \right) = \frac{d}{dx} (x^{-3}) = -3x^{-4}\).

\(\frac{d}{dx} (x^3) = 3x^2\).

Thus, \(f'(x) = -3x^{-4} - 3x^2\).

Simplifying, \(f'(x) = -\frac{3}{x^4} - 3x^2\).

For \(x > 0\), both terms \(-\frac{3}{x^4}\) and \(-3x^2\) are negative, so \(f'(x) < 0\).

Therefore, \(f(x)\) is a decreasing function for \(x > 0\).

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.

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 \(\frac{5}{2}\).

To determine whether the function \(f(x) = \frac{1}{3x+2} + x^2\) is increasing or decreasing, we need to find its derivative \(f'(x)\).

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

\(f'(x) = \left[ -\left(3x+2\right)^{-2} \right] \times [3] + [2x]\)

Simplifying, we have:

\(f'(x) = -\frac{3}{(3x+2)^2} + 2x\)

For \(x < -1\), the term \(-\frac{3}{(3x+2)^2}\) is negative and dominates over \(2x\), making \(f'(x) < 0\).

Since \(f'(x) < 0\) for \(x < -1\), the function \(f\) is decreasing.

To find the least possible value of \(n\), we need \(f'(x) \geq 0\) for \(x > n\) because the function is increasing.

First, solve \(f'(x) = x^2 - 6x + 8 = 0\).

Factor the quadratic: \((x - 2)(x - 4) = 0\).

The roots are \(x = 2\) and \(x = 4\).

The quadratic \(x^2 - 6x + 8\) is positive for \(x < 2\) and \(x > 4\).

Since the function is increasing for \(x > n\), the least integer \(n\) such that \(f'(x) \geq 0\) is \(n = 4\).

First, find the critical points by setting \(f'(x) = 0\):

\(x^3 + 2x^2 - 4x + 7 = 0\).

Differentiate \(f'(x)\) to find \(f''(x)\):

\(f''(x) = 3x^2 + 4x - 4\).

Find the roots of \(f''(x)\) to determine intervals of increase/decrease:

\(3x^2 + 4x - 4 = 0\).

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find:

\(x = \frac{-4 \pm \sqrt{16 + 48}}{6} = \frac{-4 \pm 8}{6}\).

This gives \(x = \frac{2}{3}\) and \(x = -2\).

Test intervals around these points:

For \(-2 < x < \frac{2}{3}\), \(f'(x) < 0\).

For \(x > \frac{2}{3}\), \(f'(x) > 0\).

Since \(f'(x)\) changes sign, \(f\) is neither increasing nor decreasing.

(i) Differentiate the curve: \(\frac{dy}{dx} = 3x^2 - 18x + 24\).

Since the tangent is parallel to the line \(y = 2 - 3x\), the slope of the tangent is \(-3\).

Set \(3x^2 - 18x + 24 = -3\).

Solve for \(x\):

\(3x^2 - 18x + 24 + 3 = 0\)

\(3x^2 - 18x + 27 = 0\)

\(x^2 - 6x + 9 = 0\)

\((x - 3)^2 = 0\)

\(x = 3\)

Substitute \(x = 3\) into the original equation to find \(y\):

\(y = 3^3 - 9(3)^2 + 24(3) - 12 = 27 - 81 + 72 - 12 = 6\)

The point \(A\) is \((3, 6)\).

The equation of the tangent is \(y - 6 = -3(x - 3)\).

Simplify to get \(y = -3x + 15\).

(ii) Differentiate \(f(x)\): \(f'(x) = 3x^2 - 18x + 24\).

Set \(f'(x) > 0\) for \(f(x)\) to be increasing:

\(3(x^2 - 6x + 8) > 0\)

Factorize: \(3(x-2)(x-4) > 0\)

The critical points are \(x = 2\) and \(x = 4\).

For \(f(x)\) to be increasing, \(x > 4\).

Thus, the smallest value of \(k\) is 4.

To find when the function \(f(x) = (2x - 1)^{\frac{3}{2}} - 6x\) is decreasing, we need to find when its derivative \(f'(x)\) is less than or equal to zero.

First, find the derivative:

\(f'(x) = \left( \frac{3}{2} \right)(2x - 1)^{\frac{1}{2}} \times [2] - [6]\)

Simplifying, we get:

\(f'(x) = 3(2x - 1)^{\frac{1}{2}} - 6\)

Set \(f'(x) \leq 0\):

\(3(2x - 1)^{\frac{1}{2}} - 6 \leq 0\)

\(3(2x - 1)^{\frac{1}{2}} \leq 6\)

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

Square both sides:

\(2x - 1 \leq 4\)

\(2x \leq 5\)

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

Thus, the largest value of \(k\) for which \(f\) is a decreasing function is \(\frac{5}{2}\).