(a) The function \(f(x) = 1 + \frac{3}{x-2}\) has a vertical asymptote at \(x = 2\) and a horizontal asymptote at \(y = 1\). As \(x \to \infty\), \(f(x) \to 1\). Therefore, the range of \(f\) is \(y > 1\).

(b) To find \(f^{-1}(x)\), start with \(y = 1 + \frac{3}{x-2}\).

Rearrange to find \(x\):

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

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

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

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

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

The domain of \(f^{-1}\) is \(x > 1\) because the original range of \(f\) was \(y > 1\).

To find the inverse function \(f^{-1}(x)\), we start with the equation:

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

Rearrange to make \(x\) the subject:

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

Square both sides:

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

Multiply both sides by 2:

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

Expand \((y - 1)^2\):

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

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

Rearrange to solve for \(x\):

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

\(x = 2y^2 - 4y - 1\)

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

For the domain of \(f^{-1}\), since \(f(x) = \sqrt{\frac{x+3}{2}} + 1\) is defined for \(x \geq -3\), the range of \(f(x)\) is \(y \geq 1\). Therefore, the domain of \(f^{-1}\) is \(x \geq 1\).

(i) To express \(f(x) = 4x^2 - 24x + 11\) in the form \(a(x-b)^2 + c\), complete the square:

\(f(x) = 4(x^2 - 6x) + 11\)

\(= 4((x-3)^2 - 9) + 11\)

\(= 4(x-3)^2 - 36 + 11\)

\(= 4(x-3)^2 - 25\)

The vertex form is \(4(x-3)^2 - 25\), so the vertex is \((3, -25)\).

(ii) The range of \(g(x) = 4x^2 - 24x + 11\) for \(x \leq 1\) is determined by the vertex, which is the minimum point. Thus, the range is \(g(x) \geq -9\).

(iii) To find \(g^{-1}(x)\), start with \(y = 4(x-3)^2 - 25\).

\((x-3)^2 = \frac{1}{4}(y + 25)\)

\(x-3 = \pm \frac{1}{2}\sqrt{y + 25}\)

Since \(x \leq 1\), choose the negative root:

\(x = 3 - \frac{1}{2}\sqrt{y + 25}\)

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

The domain of \(g^{-1}\) is \(x \geq -9\) because the range of \(g\) is \(g(x) \geq -9\).

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

Rearrange to make \(x\) the subject: \(2x + 5 = \frac{3}{y}\).

Multiply both sides by \(y\): \(y(2x + 5) = 3\).

Expand: \(2xy + 5y = 3\).

Rearrange to isolate \(x\): \(2xy = 3 - 5y\).

Divide by \(2y\): \(x = \frac{3 - 5y}{2y}\).

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

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

To find \(f^{-1}(x)\), solve for \(x\) in terms of \(y\):

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

Add 5 to both sides: \(y + 5 = (3x + 2)^3\)

Take the cube root: \(\sqrt[3]{y + 5} = 3x + 2\)

Subtract 2: \(\sqrt[3]{y + 5} - 2 = 3x\)

Divide by 3: \(x = \frac{\sqrt[3]{y + 5} - 2}{3}\)

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

The domain of \(f^{-1}\) is the range of \(f\), which is \(x \geq 3\).

(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 \(y \leq -a\).

(a) The function \(f(x) = 2 - \frac{5}{x+2}\) has a horizontal asymptote at \(y = 2\). As \(x \to \infty\), \(f(x) \to 2\). Therefore, the range of \(f\) is \(y < 2\).

(b) To find the inverse, start with \(y = 2 - \frac{5}{x+2}\).

Rearrange to solve for \(x\):

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

\(yx + 2y = 2x + 4 - 5\)

\(yx + 2y = 2x - 1\)

\(2y + 1 = 2x - yx\)

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

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

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

The domain of \(f^{-1}\) is determined by the range of \(f\), which is \(x < 2\).

First, complete the square for \(g(x) = x^2 - 6x + 7\).

Rewrite \(x^2 - 6x + 7\) as \((x-3)^2 - 9 + 7\).

This simplifies to \((x-3)^2 - 2\).

Set \(y = (x-3)^2 - 2\) and solve for \(x\).

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

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

Since \(x > 4\), choose the positive root: \(x = 3 + \sqrt{y + 2}\).

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

The domain of \(g^{-1}\) is \(x > -1\) because \(x + 2 > 0\).

(i) To express \(9x^2 - 6x + 6\) in the form \((ax + b)^2 + c\), complete the square:

\(9x^2 - 6x + 6 = 9(x^2 - \frac{2}{3}x) + 6\)

\(= 9((x - \frac{1}{3})^2 - \frac{1}{9}) + 6\)

\(= 9(x - \frac{1}{3})^2 - 1 + 6\)

\(= 9(x - \frac{1}{3})^2 + 5\)

Thus, \(a = 3\), \(b = -1\), \(c = 5\).

(ii) The function \(f(x) = 9x^2 - 6x + 6\) is a quadratic function that opens upwards. For \(f\) to be one-one, it must be restricted to the domain where it is either increasing or decreasing. The vertex of the parabola is at \(x = \frac{1}{3}\). Therefore, the smallest value of \(p\) is \(\frac{1}{3}\).

(iii) For \(p = \frac{1}{3}\), the function is one-one for \(x \geq \frac{1}{3}\). To find \(f^{-1}(x)\), start with \(y = (3x-1)^2 + 5\):

\(y - 5 = (3x-1)^2\)

\(\sqrt{y-5} = \pm (3x-1)\)

Since \(x \geq \frac{1}{3}\), we take the positive root:

\(3x - 1 = \sqrt{y-5}\)

\(3x = \sqrt{y-5} + 1\)

\(x = \frac{1}{3}\sqrt{y-5} + \frac{1}{3}\)

Thus, \(f^{-1}(x) = \frac{1}{3}\sqrt{x-5} + \frac{1}{3}\) with domain \(x \geq 5\).

(iv) The equation \(f(x) = q\) has no solution when \(q < 5\) because the minimum value of \(f(x)\) is 5, which occurs at the vertex \(x = \frac{1}{3}\).

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

\(-x^2 + 6x - 5 = -1(x^2 - 6x) - 5\)

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

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

Thus, \(-1(x^2 - 6x) = -1((x - 3)^2 - 9) = -1(x - 3)^2 + 9\)

Therefore, \(-x^2 + 6x - 5 = -1(x - 3)^2 + 9 - 5 = -1(x - 3)^2 + 4\)

(ii) The function \(f(x) = -x^2 + 6x - 5\) is a downward-opening parabola. It is one-one when restricted to the domain where \(x\) is greater than or equal to the vertex. The vertex is at \(x = 3\), so the smallest \(m\) is 3.

(iii) For \(m = 5\), we find the inverse:

Start with \(y = -x^2 + 6x - 5\)

Rearrange to solve for \(x\):

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

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

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

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

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

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

The domain of \(f^{-1}\) is \(x \leq 0\) because the range of \(f\) for \(x \geq 5\) is \(-\infty \leq y \leq 0\).

Let \(y = \frac{15}{2x+3}\).

Rearrange to solve for \(x\):

\(2x + 3 = \frac{15}{y}\)

\(2x = \frac{15}{y} - 3\)

\(x = \frac{15 - 3y}{2y}\)

Thus, \(f^{-1}(x) = \frac{15 - 3x}{2x}\).

For the range of \(f^{-1}\), since \(0 \leq x \leq 6\) for \(f(x)\), the range of \(f^{-1}\) is \(0 \leq f^{-1}(x) \leq 6\).

For the domain of \(f^{-1}\), solve \(0 \leq \frac{15}{2x+3} \leq 6\) for \(x\):

\(1 \leq x \leq 5\).

Let \(y = \frac{5}{1 - 3x}\).

Rearrange to make \(x\) the subject:

\(3x = 1 - \frac{5}{y}\)

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

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

Thus, \(f^{-1}(x) = \frac{x - 5}{3x}\) or \(\frac{1}{3} - \frac{5}{3x}\).

The range of \(f\) is \(x \geq 1\), so the domain of \(f^{-1}\) is \(-2.5 \leq x < 0\).

The domain of \(f\) is \(x \geq 1\), so the range of \(f^{-1}\) is \(x \geq 1\).

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

\(2x^2 - 12x + 13 = 2(x^2 - 6x) + 13\)

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

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

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

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

(ii) For \(f(x)\) to be one-one, the derivative \(f'(x) = 4x - 12\) must be non-negative for \(x \geq k\). Solving \(4x - 12 \geq 0\) gives \(x \geq 3\). Therefore, the smallest possible value of \(k\) is 3.

(iii) With \(k = 7\), the minimum value of \(f(x)\) is \(f(7)\):

\(f(7) = 2(7)^2 - 12(7) + 13 = 98 - 84 + 13 = 27\)

Thus, the range of \(f\) is \(y \geq 27\).

(iv) To find \(f^{-1}(x)\), start with \(y = 2(x-3)^2 - 5\) and solve for \(x\):

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

\((x-3)^2 = \frac{1}{2}(y + 5)\)

\(x-3 = \pm \sqrt{\frac{1}{2}(y + 5)}\)

Since \(x \geq 7\), choose the positive root:

\(x = 3 + \sqrt{\frac{1}{2}(y + 5)}\)

Thus, \(f^{-1}(x) = 3 + \sqrt{\frac{1}{2}(x + 5)}\) for \(x \geq 27\).