Start with the function \(y = h(x) = 4x^2 - 12x + 13\).

Complete the square for the quadratic expression:

\(4x^2 - 12x + 13 = 4(x^2 - 3x) + 13\).

\(= 4((x - \frac{3}{2})^2 - \frac{9}{4}) + 13\).

\(= 4(x - \frac{3}{2})^2 - 9 + 13\).

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

Set \(y = 4(x - \frac{3}{2})^2 + 4\).

Rearrange to solve for \(x\):

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

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

\(x - \frac{3}{2} = \pm \sqrt{\frac{y - 4}{4}}\).

Since \(x < 0\), choose the negative root:

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

Thus, \(h^{-1}(x) = \frac{3}{2} - \frac{\sqrt{x - 4}}{2}\).

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

\(7 - 2(x^2 + 6x) = 7 - 2((x + 3)^2 - 9) = 7 - 2(x + 3)^2 + 18 = 25 - 2(x + 3)^2\).

Thus, \(a = 25\) and \(b = 3\).

(ii) The stationary point occurs where the derivative is zero. The derivative of \(f(x) = 7 - 2x^2 - 12x\) is \(f'(x) = -4x - 12\).

Setting \(f'(x) = 0\), we get \(-4x - 12 = 0\) which gives \(x = -3\).

Substitute \(x = -3\) into \(f(x)\):

\(f(-3) = 7 - 2(-3)^2 - 12(-3) = 7 - 18 + 36 = 25\).

Thus, the stationary point is \((-3, 25)\).

(iii) For \(g\) to have an inverse, it must be one-to-one. The function \(g(x) = 7 - 2x^2 - 12x\) is a downward-opening parabola, so it is one-to-one for \(x \geq -3\).

Thus, the smallest value of \(k\) is \(-3\).

(iv) To find \(g^{-1}(x)\), start with \(y = 7 - 2x^2 - 12x\) and solve for \(x\):

Rearrange to \(2x^2 + 12x + y - 7 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 12\), \(c = y - 7\):

\(x = \frac{-12 \pm \sqrt{12^2 - 8(y - 7)}}{4} = \frac{-12 \pm \sqrt{144 - 8y + 56}}{4} = \frac{-12 \pm \sqrt{200 - 8y}}{4}\).

Since \(x \geq -3\), choose the positive root:

\(x = \frac{-12 + \sqrt{200 - 8y}}{4}\).

Thus, \(g^{-1}(x) = \sqrt{\frac{25 - x}{2}} - 3\).

(iii) To express \(6x - x^2 - 5\) in the form \(a - (x - b)^2\), we complete the square:

\(6x - x^2 - 5 = -(x^2 - 6x) - 5\)

\(= -(x^2 - 6x + 9 - 9) - 5\)

\(= -(x - 3)^2 + 9 - 5\)

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

Thus, \(a = 4\) and \(b = 3\).

(iv) The function \(g\) has an inverse when it is one-to-one. The vertex of the parabola \(y = 6x - x^2 - 5\) is at \(x = 3\). Therefore, the smallest value of \(k\) for which \(g\) is one-to-one is \(k = 3\).

(v) For \(k = 3\), the function \(g(x) = 4 - (x - 3)^2\). To find the inverse, set \(y = 4 - (x - 3)^2\) and solve for \(x\):

\(y = 4 - (x - 3)^2\)

\((x - 3)^2 = 4 - y\)

\(x - 3 = \pm \sqrt{4 - y}\)

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

\(x = \sqrt{4 - y} + 3\)

Thus, \(g^{-1}(x) = \sqrt{4 - x} + 3\).

(ii) To express \(g(x) = 2x^2 - 6x + 5\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 - 6x + 5 = 2(x^2 - 3x) + 5\)

\(= 2((x - \frac{3}{2})^2 - \frac{9}{4}) + 5\)

\(= 2(x - \frac{3}{2})^2 - \frac{9}{2} + 5\)

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

Thus, \(a = 2, b = -\frac{3}{2}, c = \frac{1}{2}\).

(iii) The vertex form \(2(x - \frac{3}{2})^2 + \frac{1}{2}\) shows the minimum value is \(\frac{1}{2}\) when \(x = \frac{3}{2}\). The maximum value is \(13\) when \(x = 4\). Therefore, the range is \(\frac{1}{2} \leq g(x) \leq 13\).

(iv) For \(h\) to have an inverse, it must be one-to-one. This occurs when \(x \geq \frac{3}{2}\), so the smallest \(k = \frac{3}{2}\).

(v) For \(k = \frac{3}{2}\), express \(h(x) = 2(x - \frac{3}{2})^2 + \frac{1}{2}\). To find \(h^{-1}(x)\), solve for \(x\):

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

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

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

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

Since \(x \geq \frac{3}{2}\), choose the positive root:

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

Thus, \(h^{-1}(x) = \frac{3}{2} + \sqrt{\frac{x}{2} - \frac{1}{4}}\).

To find the smallest possible value of \(k\), we need the function \(h(x) = x^2 + 4x\) to be one-to-one, which occurs when it is either strictly increasing or decreasing. The vertex of the parabola \(x^2 + 4x\) is at \(x = -\frac{b}{2a} = -\frac{4}{2} = -2\). Thus, the smallest possible value of \(k\) is \(-2\) so that the function is increasing for \(x \geq -2\).

To find \(h^{-1}(x)\), start by setting \(y = x^2 + 4x\).

Complete the square: \(y = (x+2)^2 - 4\).

Rearrange to solve for \(x\):

\(y + 4 = (x+2)^2\)

\(\sqrt{y + 4} = x + 2\)

\(x = \sqrt{y + 4} - 2\)

Thus, the inverse function is \(h^{-1}(x) = \sqrt{x+4} - 2\).

(ii) To express \(f(x) = 2x^2 - 8x + 14\) in the form \(a(x+b)^2 + c\), complete the square:

\(2x^2 - 8x + 14 = 2(x^2 - 4x) + 14\)

\(= 2((x-2)^2 - 4) + 14\)

\(= 2(x-2)^2 - 8 + 14\)

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

Thus, \(a = 2, b = -2, c = 6\).

(iii) The vertex form \(2(x-2)^2 + 6\) shows the minimum value is 6, so the range is \(f \geq 6\).

(iv) For \(g(x) = 2x^2 - 8x + 14\) to have an inverse, it must be one-to-one. The function is one-to-one for \(x \geq 2\) (the vertex of the parabola).

(v) To find \(g^{-1}(x)\), solve \(y = 2x^2 - 8x + 14\) for \(x\):

\(y = 2(x-2)^2 + 6\)

\(y - 6 = 2(x-2)^2\)

\(\frac{y-6}{2} = (x-2)^2\)

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

Since \(x \geq 2\), choose the positive root:

\(x = \sqrt{\frac{y-6}{2}} + 2\)

Thus, \(g^{-1}(x) = \sqrt{\frac{x-6}{2}} + 2\).

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

First, factor out the 2 from the quadratic terms: \(2(x^2 - 6x) + 7\).

Complete the square inside the parentheses: \(x^2 - 6x = (x-3)^2 - 9\).

Thus, \(2(x^2 - 6x) = 2((x-3)^2 - 9) = 2(x-3)^2 - 18\).

Adding the constant term: \(2(x-3)^2 - 18 + 7 = 2(x-3)^2 - 11\).

So, \(f(x) = 2(x-3)^2 - 11\).

(ii) The range of \(f\) is determined by the vertex form \(2(x-3)^2 - 11\). The minimum value occurs at \(x = 3\), giving \(f(3) = -11\). Therefore, the range is \(f \geq -11\).

(iii) To find the set of values of \(x\) for which \(f(x) < 21\), solve:

\(2x^2 - 12x + 7 < 21\).

Rearrange: \(2x^2 - 12x + 7 - 21 < 0\) or \(2x^2 - 12x - 14 < 0\).

Complete the square: \(2(x-3)^2 - 11 < 21\).

Rearrange: \(2(x-3)^2 < 32\).

Divide by 2: \((x-3)^2 < 16\).

Take the square root: \(-4 < x-3 < 4\).

Thus, \(-1 < x < 7\).

(i) To express \(f(x) = 2x^2 - 12x + 13\) in the form \(a(x+b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 6x) + 13\)

\(= 2((x-3)^2 - 9) + 13\)

\(= 2(x-3)^2 - 18 + 13\)

\(= 2(x-3)^2 - 5\)

Thus, \(a = 2\), \(b = -3\), \(c = -5\).

(ii) The line of symmetry is at \(x = 3\). Since \(0 \leq x \leq A\), \(A = 6\) for symmetry about \(x = 3\).

(iii) The vertex form \(2(x-3)^2 - 5\) shows the minimum value is \(-5\) at \(x = 3\). The maximum value is \(f(0) = 13\). Thus, the range is \(-5 \leq y \leq 13\).

(iv) The function \(g\) is defined for \(x \geq 4\), which is a one-to-one function because it is strictly increasing (as \(4 > 3\)).

(v) To find \(g^{-1}(x)\), start with \(y = 2x^2 - 12x + 13\) and solve for \(x\):

\(y = 2(x-3)^2 - 5\)

\(y + 5 = 2(x-3)^2\)

\(\frac{y+5}{2} = (x-3)^2\)

\(x-3 = \pm \sqrt{\frac{y+5}{2}}\)

Since \(x \geq 4\), take the positive root:

\(x = \sqrt{\frac{y+5}{2}} + 3\)

Thus, \(g^{-1}(x) = \sqrt{\frac{x+5}{2}} + 3\).

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

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

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

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

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

(iv) To find \(h^{-1}(x)\), start with \(y = 6x - x^2\).

Rearrange to express \(x\) in terms of \(y\):

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

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

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

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

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

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

(ii) To express \(f(x) = x^2 - 3x\) in the form \((x-a)^2 - b\), complete the square:

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

Thus, \(a = \frac{3}{2}\) and \(b = \frac{9}{4}\).

(iii) The minimum value of \(f(x) = (x - \frac{3}{2})^2 - \frac{9}{4}\) is \(-\frac{9}{4}\), so the range is \(f(x) \geq -\frac{9}{4}\).

(iv) The function \(f(x) = x^2 - 3x\) is not one-to-one because it is a quadratic function, which is symmetric about its vertex. Therefore, \(f\) does not have an inverse.

(v) Solve \(g(x) = x - 3\sqrt{x} = 10\).

Let \(y = \sqrt{x}\), then \(x = y^2\).

Substitute into the equation: \(y^2 - 3y = 10\).

Rearrange to form a quadratic: \(y^2 - 3y - 10 = 0\).

Factorize: \((y - 5)(y + 2) = 0\).

Thus, \(y = 5\) or \(y = -2\).

Since \(y = \sqrt{x} \geq 0\), \(y = 5\).

Therefore, \(x = y^2 = 25\).

(i) To find the set of values of \(x\) for which \(f(x) > 15\), solve the inequality:

\(x^2 - 2x > 15\)

Rearrange to form a quadratic equation:

\(x^2 - 2x - 15 = 0\)

Factorize the quadratic:

\((x - 5)(x + 3) = 0\)

The solutions are \(x = 5\) and \(x = -3\).

Test intervals around the roots to find where \(f(x) > 15\):

For \(x < -3\), choose \(x = -4\):

\(f(-4) = 16 + 8 = 24 > 15\)

For \(x > 5\), choose \(x = 6\):

\(f(6) = 36 - 12 = 24 > 15\)

Thus, \(x < -3\) and \(x > 5\).

(ii) To find the range of \(f\), complete the square:

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

The minimum value of \((x-1)^2\) is 0, occurring at \(x = 1\).

Thus, the minimum value of \(f(x)\) is \(-1\).

Therefore, the range of \(f\) is \(y \geq -1\).

Since \(f\) is a quadratic function and not one-to-one, it does not have an inverse.

(a) The function \(f(x) = -3x^2 + 2\) is a downward-opening parabola. The vertex occurs at \(x = 0\), but since \(x \leq -1\), the maximum value of \(f(x)\) is at \(x = -1\). Calculating \(f(-1) = -3(-1)^2 + 2 = -1\). Therefore, the range of \(f\) is \(y \leq -1\).

(b) To find the inverse, start with \(y = -3x^2 + 2\). Rearrange to solve for \(x\):

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

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

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

Since \(x \leq -1\), we take the negative root:

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

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

(i) To express \(2x^2 + 8x - 10\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 + 8x - 10 = 2(x^2 + 4x) - 10\)

\(= 2((x+2)^2 - 4) - 10\)

\(= 2(x+2)^2 - 8 - 10\)

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

Thus, \(a = 2\), \(b = 2\), \(c = -18\).

(ii) The least value of \(y\) occurs at the vertex of the parabola, which is at \(x = -2\). Substituting \(x = -2\) into the equation:

\(y = 2(-2)^2 + 8(-2) - 10 = 8 - 16 - 10 = -18\)

Thus, the least value of \(y\) is \(-18\) when \(x = -2\).

(iii) To find the set of values of \(x\) for which \(y \geq 14\), solve:

\(2x^2 + 8x - 10 \geq 14\)

\(2x^2 + 8x - 24 \geq 0\)

\(x^2 + 4x - 12 \geq 0\)

\((x+6)(x-2) \geq 0\)

The solution is \(x \geq 2\) or \(x \leq -6\).

(iv) For \(f\) to be one-one, the domain must be restricted to where the function is either increasing or decreasing. The vertex is at \(x = -2\), so the smallest \(k\) is \(-2\).

(v) To find \(f^{-1}(x)\), solve \(y = 2x^2 + 8x - 10\) for \(x\):

\(y = 2(x+2)^2 - 18\)

\(y + 18 = 2(x+2)^2\)

\(\frac{y+18}{2} = (x+2)^2\)

\(x+2 = \pm \sqrt{\frac{y+18}{2}}\)

Since \(x \geq -2\), choose the positive root:

\(x = \sqrt{\frac{y+18}{2}} - 2\)

Thus, \(f^{-1}(x) = \sqrt{\frac{x+18}{2}} - 2\).

(a) To express \(f(x) = -2x^2 - 8x - 13\) in the form \(-2(x + a)^2 + b\), complete the square:

\(f(x) = -2(x^2 + 4x) - 13\)

\(= -2((x + 2)^2 - 4) - 13\)

\(= -2(x + 2)^2 + 8 - 13\)

\(= -2(x + 2)^2 - 5\)

Thus, \(a = 2\) and \(b = -5\).

(b) The vertex form \(-2(x + 2)^2 - 5\) shows the maximum value is \(-5\) when \(x = -2\). Since \(x < -3\), \(f(x) < -7\).

(c) To find \(f^{-1}(x)\), start with \(y = -2(x + 2)^2 - 5\).

Rearrange to solve for \(x\):

\(y + 5 = -2(x + 2)^2\)

\((x + 2)^2 = -\frac{y + 5}{2}\)

\(x + 2 = \pm \sqrt{-\frac{y + 5}{2}}\)

\(x = -2 \pm \sqrt{-\frac{y + 5}{2}}\)

Thus, \(f^{-1}(x) = -2 - \sqrt{\frac{x+5}{2}}\) or \(f^{-1}(x) = -2 + \sqrt{\frac{x+5}{2}}\).

To find \(f^{-1}(x)\), start by setting \(y = f(x) = 2 - \frac{3}{4x - p}\).

Rearrange to solve for \(x\):

\(y - 2 = -\frac{3}{4x - p}\)

\((y - 2)(4x - p) = -3\)

\(4xy - 8x = py - 2p - 3\)

\(4x(y - 2) = p(y - 2) - 3\)

\(4x = \frac{p(y - 2) - 3}{y - 2}\)

\(x = \frac{p(y - 2) - 3}{4(y - 2)}\)

\(x = \frac{p}{4} - \frac{3}{4(y - 2)}\)

Thus, \(f^{-1}(x) = \frac{p}{4} - \frac{3}{4x - 8}\).

For \(f^{-1}(x) \equiv f(x)\), equate the expressions:

\(2 - \frac{3}{4x - p} = \frac{p}{4} - \frac{3}{4x - 8}\)

Equating the coefficients gives \(p = 8\).

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