(i) To express \(f(x) = 2x^2 - 8x + 11\) in the form \(a(x + b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 4x) + 11\)

\(= 2((x - 2)^2 - 4) + 11\)

\(= 2(x - 2)^2 - 8 + 11\)

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

(ii) The range of \(f\) is \(\geq 3\) because the vertex of the parabola is at its minimum point \((2, 3)\).

(iii) \(f\) does not have an inverse because it is not one-to-one. It is a quadratic function, which means it is symmetric and has two x-values for each y-value.

(iv) The largest value of \(A\) for which \(g\) has an inverse is \(A = 2\), as this makes the function one-to-one on the interval \(x \leq 2\).

(v) To find \(g^{-1}(x)\), start with \(y = 2(x - 2)^2 + 3\) and solve for \(x\):

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

\(\frac{y - 3}{2} = (x - 2)^2\)

\(x - 2 = \pm \sqrt{\frac{y - 3}{2}}\)

Since \(x \leq 2\), choose the negative root:

\(x = 2 - \sqrt{\frac{y - 3}{2}}\)

Thus, \(g^{-1}(x) = 2 - \sqrt{\frac{x - 3}{2}}\).

The range of \(g^{-1}\) is \(\leq 2\) because the inverse function reflects the domain of \(g\).

\(Let y = (2x − 3)³ − 8.\)

To find the inverse, solve for x in terms of y:

\((2x − 3)³ = y + 8\)

\(2x − 3 = \sqrt[3]{y + 8}\)

\(2x = \sqrt[3]{y + 8} + 3\)

\(x = \frac{\sqrt[3]{y + 8} + 3}{2}\)

\(Thus, f⁻¹(x) = \frac{\sqrt[3]{x + 8} + 3}{2}.\)

The range of f is the domain of f⁻¹. Calculate the range of f for 2 ≤ x ≤ 4:

\(For x = 2, f(2) = (2(2) − 3)³ − 8 = −7.\)

\(For x = 4, f(4) = (2(4) − 3)³ − 8 = 117.\)

Therefore, the domain of f⁻¹ is −7 ≤ x ≤ 117.

(iii) To express \(x^2 - 6x\) in the form \((x-p)^2 - q\), we complete the square:

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

Thus, \(p = 3\) and \(q = 9\).

(iv) To find \(h^{-1}(x)\), start with \(y = (x-3)^2 - 9\).

Solve for \(x\):

\(y + 9 = (x-3)^2\)

\(x - 3 = \pm \sqrt{y + 9}\)

Since \(x \geq 3\), we take the positive root:

\(x = \sqrt{y + 9} + 3\)

Thus, \(y = h^{-1}(x) = \sqrt{x + 9} + 3\).

The domain of \(h^{-1}\) is \(x \geq -9\) because the expression under the square root must be non-negative.

(i) To express \(8x - x^2\) in the form \(a - (x + b)^2\), we complete the square:

\(8x - x^2 = -(x^2 - 8x) = -(x^2 - 8x + 16 - 16) = -(x - 4)^2 + 16\).

Thus, \(a = 16\) and \(b = -4\).

(ii) The stationary point occurs where \(\frac{dy}{dx} = 0\). Differentiating, \(\frac{dy}{dx} = 8 - 2x\).

Setting \(8 - 2x = 0\) gives \(x = 4\).

Substituting \(x = 4\) into \(y = 8x - x^2\), we get \(y = 16\).

Thus, the stationary point is \((4, 16)\).

(iii) Solve \(8x - x^2 \geq -20\):

\(x^2 - 8x - 20 \leq 0\).

Factorizing gives \((x - 10)(x + 2) \leq 0\).

The solution is \(-2 \leq x \leq 10\).

(iv) From the function \(g(x) = 8x - x^2\) for \(x \geq 4\), the maximum value is at \(x = 4\), giving \(y = 16\).

Thus, the domain of \(g^{-1}\) is \(x \leq 16\) and the range is \(x \geq 4\).

(v) To find \(g^{-1}(x)\), solve \(y = 8x - x^2\) for \(x\):

\(x^2 - 8x + y = 0\).

Using the quadratic formula, \(x = \frac{8 \pm \sqrt{64 - 4y}}{2}\).

Since \(x \geq 4\), we take the positive root: \(x = 4 + \sqrt{16 - y}\).

Thus, \(g^{-1}(x) = 4 + \sqrt{16 - x}\).

(a) To find the inverse, start with \(y = (x + a)^2 - a\).

Rearrange to solve for \(x\):

\(y + a = (x + a)^2\)

\(x + a = \pm \sqrt{y + a}\)

Since \(x \leq -a\), choose the negative root:

\(x = -\sqrt{y + a} - a\)

Thus, \(f^{-1}(x) = -\sqrt{x + a} - a\).

(b)(i) The domain of \(f^{-1}\) is the range of \(f\), which is \(x > -a\).

(b)(ii) The range of \(f^{-1}\) is the domain of \(f\), which is \(y \leq -a\).