(a) To express \(f(x) = 2x^2 - 16x + 23\) in the form \(2(x + a)^2 + b\), complete the square:

Start with \(2(x^2 - 8x) + 23\).

Complete the square for \(x^2 - 8x\):

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

Thus, \(2(x^2 - 8x) = 2((x-4)^2 - 16) = 2(x-4)^2 - 32\).

So, \(f(x) = 2(x-4)^2 - 32 + 23 = 2(x-4)^2 - 9\).

(b) The vertex form \(2(x-4)^2 - 9\) shows the minimum value is \(-9\) when \(x = 4\). Since \(x < 3\), the range is \(y > -7\).

(c) To find \(f^{-1}(x)\), start from \(y = 2(x-4)^2 - 9\).

\((x-4)^2 = \frac{y+9}{2}\).

\(x-4 = \pm \sqrt{\frac{y+9}{2}}\).

Since \(x < 3\), choose the negative root: \(x = 4 - \sqrt{\frac{y+9}{2}}\).

Thus, \(f^{-1}(x) = 4 - \sqrt{\frac{x+9}{2}}\).

(a) Start with \(y = \frac{x^2 - 4}{x^2 + 4}\). Clear the denominator: \((x^2 + 4)y = x^2 - 4\). Expand and rearrange: \(x^2 y + 4y = x^2 - 4\). Isolate \(x^2\): \(x^2 (1 - y) = 4y + 4\). Solve for \(x^2\): \(x^2 = \frac{4y + 4}{1 - y}\). Solve for \(x\): \(x = \frac{\sqrt{4y + 4}}{\sqrt{1 - y}}\). Thus, \(f^{-1}(x) = \frac{4x + 4}{1 - x}\).

(b) Start with \(1 - \frac{8}{x^2 + 4}\). Use a common denominator: \(\frac{x^2 + 4 - 8}{x^2 + 4} = \frac{x^2 - 4}{x^2 + 4}\). Therefore, the range of \(f\) is \(0 < f(x) < 1\).

(c) The composite function \(ff\) cannot be formed because the range of \(f\) does not include the whole of the domain of \(f\) (or any of it).

(a) To express \(-3x^2 + 12x + 2\) in the form \(-3(x-a)^2 + b\), complete the square:

\(-3(x^2 - 4x) + 2\)

\(-3((x-2)^2 - 4) + 2\)

\(-3(x-2)^2 + 12 + 2\)

\(-3(x-2)^2 + 14\)

(b) The vertex form \(-3(x-2)^2 + 14\) shows the maximum value is 14 when \(x = 2\). Thus, the largest possible value of \(k\) is 2.

(c) Given \(k = -1\), substitute into \(-3(x-2)^2 + 14\) to find the range:

\(-3((-1)-2)^2 + 14 = -3(9) + 14 = -27 + 14 = -13\)

Thus, the range is \(y \leq -13\).

(d) To find \(f^{-1}(x)\), start from \(y = -3(x-2)^2 + 14\):

\((x-2)^2 = \frac{14-y}{3}\)

\(x-2 = \pm \sqrt{\frac{14-y}{3}}\)

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

\(x = 2 - \sqrt{\frac{14-y}{3}}\)

Thus, \(f^{-1}(x) = 2 - \sqrt{\frac{14-x}{3}}\).

(a) Start with \(y = \frac{2x}{3x-1}\). Rearrange to find \(x\) in terms of \(y\):

\(y(3x-1) = 2x\)

\(3xy - y = 2x\)

\(3xy - 2x = y\)

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

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

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

(b) Start with \(\frac{2}{3} + \frac{2}{3(3x-1)}\):

\(\frac{2}{3} + \frac{2}{3(3x-1)} = \frac{2(3x-1) + 2}{3(3x-1)}\)

\(= \frac{6x}{3(3x-1)}\)

\(= \frac{2x}{3x-1}\)

(c) The range of \(f\) is \(f(x) > \frac{2}{3}\).

(i) To express \(2x^2 - 12x + 7\) in the form \(2(x + a)^2 + b\), complete the square:

\(2x^2 - 12x + 7 = 2(x^2 - 6x) + 7\)

\(= 2((x - 3)^2 - 9) + 7\)

\(= 2(x - 3)^2 - 18 + 7\)

\(= 2(x - 3)^2 - 11\)

Thus, \(a = -3\) and \(b = -11\).

(ii) The range of \(f\) is \(y > -11\) because the vertex of the parabola is at its minimum point.

(iii) For \(g\) to have an inverse, it must be one-to-one. The function is one-to-one for \(x \leq 3\), so the largest value of \(k\) is \(k < 3\).

(iv) To find \(g^{-1}(x)\), start with:

\(y = 2(x - 3)^2 - 11\)

\(y + 11 = 2(x - 3)^2\)

\(\frac{y + 11}{2} = (x - 3)^2\)

\(x - 3 = \pm \sqrt{\frac{y + 11}{2}}\)

Since \(x \leq 3\), choose the negative root:

\(x = 3 - \sqrt{\frac{y + 11}{2}}\)

Thus, \(g^{-1}(x) = 3 - \sqrt{\frac{x + 11}{2}}\).