To find \(gf(x)\), substitute \(f(x)\) into \(g(x)\):

\(gf(x) = g(f(x)) = g\left(1 + \frac{3}{x-2}\right)\).

Substitute into \(g(x) = 2x - 2\):

\(gf(x) = 2\left(1 + \frac{3}{x-2}\right) - 2\).

Simplify the expression:

\(gf(x) = 2 + \frac{6}{x-2} - 2\).

Combine like terms:

\(gf(x) = \frac{6}{x-2}\).

\((a) To solve the equation f(x) = 0, we have:\)

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

\(Let y = x^{\frac{1}{2}}, then x = y^2. Substitute to get:\)

\(y^2 - 4y + 1 = 0\)

\(Using the quadratic formula, y = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}\)

\(Thus, x = (2 \pm \sqrt{3})^2 = 7 \pm 4\sqrt{3}\)

(b) Given f(x) \equiv gh(x), we have:

\(gh(x) = m\left(x^{\frac{1}{2}} - 2\right)^2 + n\)

Expanding, we get:

\(gh(x) = m(x - 4x^{\frac{1}{2}} + 4) + n\)

\(Since f(x) = x - 4x^{\frac{1}{2}} + 1, equate the expressions:\)

\(m(x - 4x^{\frac{1}{2}} + 4) + n = x - 4x^{\frac{1}{2}} + 1\)

\(Comparing coefficients, m = 1 and n = -3.\)

(b) Start with \(y = \frac{-x}{\sqrt{4-x^2}}\). Square both sides to get \(x^2 = y^2(4-x^2)\).

Rearrange to \(x^2(1+y^2) = 4y^2\).

Solving for \(x\), we get \(x = \frac{\pm 2y}{\sqrt{1+y^2}}\).

Select the correct square root: \(f^{-1}(x) = \frac{-2x}{\sqrt{1+x^2}}\).

(c) The maximum possible value of \(a\) is 1, as \(-1 < x < 1\) is required for \(fg\) to be formed.

(d) Substitute \(g(x) = 2x\) into \(f(x)\): \(fg(x) = f(2x) = \frac{-2x}{\sqrt{4-(2x)^2}}\).

Simplify to \(fg(x) = \frac{-x}{\sqrt{1-x^2}}\).

(a) To find \(ff(5)\), we first calculate \(f(5)\):

\(f(5) = \frac{5+3}{5-1} = \frac{8}{4} = 2\).

Now, calculate \(f(f(5)) = f(2)\):

\(f(2) = \frac{2+3}{2-1} = \frac{5}{1} = 5\).

Thus, \(ff(5) = 5\).

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

Swap \(x\) and \(y\):

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

Cross-multiply to solve for \(y\):

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

\(xy - x = y + 3\).

\(xy - y = x + 3\).

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

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

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

(a) We have \(fg(x) = f(g(x)) = (g(x))^2 - 1\). Given \(fg(x) = 3\), we set \((g(x))^2 - 1 = 3\), leading to \((g(x))^2 = 4\).

Since \(g(x) = \frac{1}{2x+1}\), we have \(\left(\frac{1}{2x+1}\right)^2 = 4\).

This implies \(\frac{1}{(2x+1)^2} = 4\), leading to \((2x+1)^2 = \frac{1}{4}\).

Solving \(2x+1 = \pm \frac{1}{2}\), we get \(2x+1 = \frac{1}{2}\) or \(2x+1 = -\frac{1}{2}\).

For \(2x+1 = -\frac{1}{2}\), solving gives \(x = -\frac{3}{4}\).

For \(2x+1 = \frac{1}{2}\), solving gives \(x = -\frac{1}{4}\), but this does not satisfy \(x < -\frac{1}{2}\).

Thus, \(x = -\frac{3}{4}\) only.

(b) Let \(y = fg(x) = (g(x))^2 - 1\).

Then \(y = \frac{1}{(2x+1)^2} - 1\), leading to \(\frac{1}{(2x+1)^2} = y + 1\).

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

Taking square roots, \(2x+1 = \pm \frac{1}{\sqrt{y+1}}\).

Solving for \(x\), we get \(x = -\frac{1}{2\sqrt{y+1}} - \frac{1}{2}\).

(a) To find \(ff(x)\), substitute \(f(x) = 2x^2 + 3\) into itself:

\(ff(x) = f(f(x)) = f(2x^2 + 3) = 2(2x^2 + 3)^2 + 3\).

Expand \((2x^2 + 3)^2\):

\((2x^2 + 3)^2 = 4x^4 + 12x^2 + 9\).

Substitute back:

\(ff(x) = 2(4x^4 + 12x^2 + 9) + 3 = 8x^4 + 24x^2 + 18 + 3 = 8x^4 + 24x^2 + 21\).

(b) Solve \(ff(x) = 34x^2 + 19\):

\(8x^4 + 24x^2 + 21 = 34x^2 + 19\).

Rearrange to form a polynomial equation:

\(8x^4 + 24x^2 + 21 - 34x^2 - 19 = 0\).

Simplify:

\(8x^4 - 10x^2 + 2 = 0\).

Factorize:

\(2(x^2 - 1)(4x^2 - 1) = 0\).

Solve each factor:

\(x^2 - 1 = 0\) gives \(x = 1\).

\(4x^2 - 1 = 0\) gives \(x = \frac{1}{2}\).

Thus, \(x = 1\) or \(x = \frac{1}{2}\).

(a) The function \(f(x) = (x - 2)^2 - 4\) is a quadratic function with vertex at \((2, -4)\). Since it opens upwards, the minimum value is \(-4\). Thus, the range is \(f(x) > -4\).

(b) To find \(f^{-1}(x)\), set \(y = (x - 2)^2 - 4\) and solve for \(x\):

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

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

Since \(x \geq 2\), we take the positive root: \(x = \sqrt{y + 4} + 2\)

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

(c) Given \(a = -\frac{5}{3}\), solve \(f(x) = g(x)\):

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

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

\(x^2 - 4x + \frac{5}{3}x - 2 = 0\)

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

\(x = 3\) only.

(d) Given \(gg f^{-1}(12) = 62\):

\(f^{-1}(12) = 6\)

\(g(f^{-1}(12)) = 6a + 2\)

\(g(g(f^{-1}(12))) = a(6a + 2) + 2 = 62\)

\(6a^2 + 2a - 60 = 0\)

Solving the quadratic equation gives \(a = \frac{10}{3}\) or \(a = 3\).

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

\(f(x) = (x+1)^2 + 2\).

The range of \(f\) is \(y \geq 2\) since the vertex of the parabola is at \(y = 2\) and it opens upwards.

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

Solving for \(x\), we have:

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

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

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

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

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

(c) Solve \(gf(x) = 13\):

\(g(f(x)) = 2(x^2 + 2x + 3) + 1 = 13\)

\(2x^2 + 4x + 6 + 1 = 13\)

\(2x^2 + 4x - 6 = 0\)

Divide by 2:

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

Factorize:

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

\(x = 1\) or \(x = -3\)

Since \(x \leq -1\), \(x = -3\) only.

(a) To find \(fg(7)\), we first calculate \(g(7)\):

\(g(7) = \frac{4}{7+1} = \frac{4}{8} = \frac{1}{2}\).

Then, calculate \(f(g(7)) = f\left(\frac{1}{2}\right)\):

\(f\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right) - 2 = 2 - 2 = 0\).

Thus, \(fg(7) = 0\).

(b) To find \(x\) for which \(f^{-1}(x) = g^{-1}(x)\), we first find the inverses:

\(f(x) = 4x - 2 \Rightarrow y = 4x - 2 \Rightarrow x = \frac{y+2}{4} \Rightarrow f^{-1}(x) = \frac{x+2}{4}\).

\(g(x) = \frac{4}{x+1} \Rightarrow y = \frac{4}{x+1} \Rightarrow y(x+1) = 4 \Rightarrow x = \frac{4}{y} - 1 \Rightarrow g^{-1}(x) = \frac{4}{x} - 1\).

Set \(f^{-1}(x) = g^{-1}(x)\):

\(\frac{x+2}{4} = \frac{4}{x} - 1\).

Multiply through by 4x to clear fractions:

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

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

\(x^2 + 6x - 16 = 0\).

Factor the quadratic:

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

Thus, \(x = -8\) or \(x = 2\).

(a) To find \(fg(x)\), substitute \(g(x) = 2x + 1\) into \(f(x)\):

\(fg(x) = f(g(x)) = (2x+1)^2 + 3\).

(b) Let \(y = (2x+1)^2 + 3\). Solve for \(x\):

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

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

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

Since \(x > 0\), choose the positive root:

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

The domain of \((fg)^{-1}\) is \(x > 3\).

(c) Solve \(fg(x) - 3 = gf(x)\):

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

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

\(4x^2 + 4x + 1 = 2x^2 + 7\)

\(2x^2 + 4x - 6 = 0\)

\(2(x+3)(x-1) = 0\)

\(x = 1\)

(a) To find \(ff(x)\), apply \(f\) twice:

\(f(x) = a - 2x\)

\(ff(x) = f(f(x)) = f(a - 2x) = a - 2(a - 2x)\)

\(= a - 2a + 4x = 4x - a\)

To find \(f^{-1}(x)\), solve \(y = a - 2x\) for \(x\):

\(y = a - 2x\)

\(2x = a - y\)

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

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

(b) Given \(ff(x) = f^{-1}(x)\):

\(4x - a = \frac{a-x}{2}\)

Multiply through by 2 to clear the fraction:

\(8x - 2a = a - x\)

\(8x + x = a + 2a\)

\(9x = 3a\)

\(x = \frac{a}{3}\)

Given \(a = \frac{7}{2}\), we need to solve \(gf(x) = 0\).

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

\(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\)

For \(x \leq -\frac{7}{2}\), \(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\).

Now, find \(gf(x)\):

\(gf(x) = g(f(x)) = 2f(x) - 1\)

Set \(gf(x) = 0\):

\(2f(x) - 1 = 0\)

\(2\left(\left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\right) - 1 = 0\)

\(2\left(x + \frac{7}{2}\right)^2 - 7 - 1 = 0\)

\(2\left(x + \frac{7}{2}\right)^2 = 8\)

\(\left(x + \frac{7}{2}\right)^2 = 4\)

\(x + \frac{7}{2} = \pm 2\)

\(x = -\frac{7}{2} \pm 2\)

\(x = -\frac{7}{2} + 2 = -\frac{3}{2}\)

\(x = -\frac{7}{2} - 2 = -\frac{11}{2}\)

Since \(x \leq -\frac{7}{2}\), the valid solution is \(x = -\frac{11}{2}\).

(a) To find \(b\), use \(gg(2) = 10\):

\(g(x) = 3x + b\), so \(g(2) = 3(2) + b = 6 + b\).

Then \(gg(2) = g(g(2)) = g(6 + b) = 3(6 + b) + b = 18 + 3b + b = 18 + 4b\).

Set \(18 + 4b = 10\), solving gives \(b = -2\).

To find \(a\), use \(f^{-1}(2) = 14\):

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

Set \(2(x + a) = 14\), solving gives \(a = 5\).

(b) Find \(gf(x)\):

\(gf(x) = g(f(x)) = g\left(\frac{1}{2}x - 5\right)\).

\(g(x) = 3x + b\), so \(g\left(\frac{1}{2}x - 5\right) = 3\left(\frac{1}{2}x - 5\right) - 2\).

\(= \frac{3}{2}x - 15 - 2 = \frac{3}{2}x - 17\).

(a) To express \(2x^2 + 12x + 11\) in the form \(2(x + a)^2 + b\), complete the square:

\(2(x^2 + 6x) + 11 = 2((x + 3)^2 - 9) + 11 = 2(x + 3)^2 - 18 + 11 = 2(x + 3)^2 - 7\).

Thus, \(a = 3\) and \(b = -7\).

(b) Start with \(y = 2(x + 3)^2 - 7\). Solve for \(x\):

\(y + 7 = 2(x + 3)^2\)

\(x + 3 = \pm \sqrt{\frac{y + 7}{2}}\)

\(x = -3 \pm \sqrt{\frac{y + 7}{2}}\)

Since \(x \leq -4\), choose the negative root:

\(f^{-1}(x) = -\sqrt{\frac{x + 7}{2}} - 3\)

The domain of \(f^{-1}\) is \(x > -5\) or \([-5, \infty)\).

(c) Given \(k = -1\), solve \(fg(x) = 193\):

\(fg(x) = 2((2x - 3) + 3)^2 - 7 = 8x^2 - 7\)

\(8x^2 - 7 = 193\)

\(8x^2 = 200\)

\(x^2 = 25\)

\(x = -5\) only (since \(x \leq -4\)).

(d) The largest value of \(k\) for \(fg\) to be defined is \(\frac{1}{2}\).

(ii) Substitute \(k = -9\) into \(g(x)\):

\(g(x) = 2x + 9\)

Set \(f(x) < g(x)\):

\(2x^2 + 8x + 1 < 2x + 9\)

Rearrange to form a quadratic inequality:

\(2x^2 + 6x - 8 < 0\)

Factorize:

\((x + 4)(x - 1) < 0\)

Solution: \(-4 < x < 1\)

(iii) Find \(g^{-1}(x)\):

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

So, \(g^{-1}(x) = \frac{x - 1}{2}\)

Substitute \(f(x)\) into \(g^{-1}(x)\):

\(g^{-1}f(x) = \frac{2x^2 + 8x + 1 - 1}{2} = \frac{2x^2 + 8x}{2} = x^2 + 4x\)

Solve \(g^{-1}f(x) = 0\):

\(x^2 + 4x = 0\)

\(x(x + 4) = 0\)

Solutions: \(x = 0, -4\)

(iv) Complete the square for \(f(x)\):

\(f(x) = 2(x^2 + 4x) + 1\)

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

\(= 2(x + 2)^2 - 8 + 1\)

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

Least value of \(f(x)\) is \(-7\) when \(x = -2\)

(i) For \(f(x) = \frac{3}{2x+1}\), as \(x \to 0^+\), \(f(x) \to 3\). As \(x \to \infty\), \(f(x) \to 0\). Thus, the range of \(f\) is \(0 < f(x) < 3\).

For \(g(x) = \frac{1}{x} + 2\), as \(x \to 0^+\), \(g(x) \to \infty\). As \(x \to \infty\), \(g(x) \to 2\). Thus, the range of \(g\) is \(g(x) > 2\).

(ii) \(fg(x) = f(g(x)) = \frac{3}{2\left(\frac{1}{x} + 2\right) + 1} = \frac{3}{\frac{2}{x} + 4 + 1} = \frac{3}{\frac{2 + 5x}{x}} = \frac{3x}{2 + 5x}\).

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

Rearranging gives \(2y + 5xy = 3x\) or \(3x - 5xy = 2y\).

Solving for \(x\), we get \(x(3 - 5y) = 2y\) and \(x = \frac{2y}{3 - 5y}\).

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

(i) To find \(f^{-1}(x)\), start with \(y = 3x - 2\). Solve for \(x\):

\(y + 2 = 3x\)

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

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

For \(g^{-1}(x)\), start with \(y = \frac{2x + 3}{x - 1}\). Solve for \(x\):

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

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

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

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

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

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

\(g^{-1}(x)\) is not defined for \(x = 2\).

(ii) To solve \(fg(x) = \frac{7}{3}\), first find \(fg(x)\):

\(fg(x) = f(g(x)) = 3\left(\frac{2x + 3}{x - 1}\right) - 2\)

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

Set \(\frac{3(2x + 3)}{x - 1} - 2 = \frac{7}{3}\)

\(\frac{3(2x + 3)}{x - 1} = \frac{7}{3} + 2\)

\(\frac{3(2x + 3)}{x - 1} = \frac{13}{3}\)

Cross-multiply to solve for \(x\):

\(3(2x + 3) = 13(x - 1)\)

\(6x + 9 = 13x - 13\)

\(6x - 13x = -13 - 9\)

\(-7x = -22\)

\(x = -8\)

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

\(-2x^2 + 12x - 3 = -2(x^2 - 6x) - 3\)

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

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

Substitute back:

\(-2((x-3)^2 - 9) - 3 = -2(x-3)^2 + 18 - 3 = -2(x-3)^2 + 15\)

Thus, \(a = -3\) and \(b = 15\).

(ii) The greatest value of \(f(x)\) is the maximum value of \(-2(x-3)^2 + 15\), which occurs when \((x-3)^2 = 0\). Therefore, the greatest value is 15.

(iii) To find \(x\) for which \(gf(x) + 1 = 0\), first find \(gf(x)\):

\(gf(x) = g(f(x)) = 2(-2x^2 + 12x - 3) + 5 = -4x^2 + 24x - 6 + 5 = -4x^2 + 24x - 1\)

Set \(gf(x) + 1 = 0\):

\(-4x^2 + 24x - 1 + 1 = 0\)

\(-4x^2 + 24x = 0\)

Factor out \(-4x\):

\(-4x(x - 6) = 0\)

Thus, \(x = 0\) or \(x = 6\).

(i) To express \(x^2 - 4x + 7\) in the form \((x + a)^2 + b\), complete the square:

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

(ii) The function \(f(x) = x^2 - 4x + 7\) is decreasing when its derivative \(f'(x) = 2x - 4\) is negative. Solving \(2x - 4 < 0\) gives \(x < 2\). Therefore, the largest value of \(k\) is 2, so \(k \leq 2\).

(iii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 3\). Solving for \(x\), we get:

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

Since \(x < 1\), choose the negative root: \(x = 2 - \sqrt{y - 3}\).

Thus, \(f^{-1}(x) = 2 - \sqrt{x-3}\) for \(x > 4\).

(iv) The composition \(gf(x)\) is given by:

\(gf(x) = g(f(x)) = \frac{2}{x^2 - 4x + 7 - 1} = \frac{2}{(x-2)^2 + 2}\).

The range of \(gf(x)\) is determined by the values of \((x-2)^2 + 2\), which is always greater than 2. Therefore, the range of \(gf(x)\) is \(0 < gf(x) < \frac{2}{3}\).

(i) Start with \(y = \frac{2}{x^2 - 1}\). Rearrange to get \(x^2 = \frac{2}{y} + 1\).

Solving for \(x\), we have \(x = \pm \sqrt{\frac{2}{y} + 1}\).

Since \(x < -1\), we take the negative root: \(x = -\sqrt{\frac{2}{y} + 1}\).

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

(ii) Solve \(gf(x) = 5\), which means \(g\left(\frac{2}{x^2 - 1}\right) = 5\).

Substitute \(g(x) = x^2 + 1\): \(\left(\frac{2}{x^2 - 1}\right)^2 + 1 = 5\).

This simplifies to \(\left(\frac{2}{x^2 - 1}\right)^2 = 4\).

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

For \(\frac{2}{x^2 - 1} = 2\), solve \(x^2 - 1 = 1\), giving \(x^2 = 2\) or \(x = \pm \sqrt{2}\).

For \(\frac{2}{x^2 - 1} = -2\), solve \(x^2 - 1 = -1\), giving \(x^2 = 0\) or \(x = 0\).

Since \(x < -1\), the valid solutions are \(x = -\sqrt{2}\) or \(x = -1.41\) only.

(i) To find \(gg(x)\), substitute \(g(x) = 2x - 3\) into itself:

\(gg(x) = g(g(x)) = g(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\).

(ii) To find \(f^{-1}(x)\), start with \(y = \frac{1}{x^2 - 9}\).

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

\(y = \frac{1}{x^2 - 9} \Rightarrow x^2 - 9 = \frac{1}{y} \Rightarrow x^2 = \frac{1}{y} + 9\).

Taking the square root, \(x = \sqrt{\frac{1}{y} + 9}\).

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

The domain of \(f^{-1}\) is \(x > 0\) because \(y > 0\) in the original function.

(iii) Solve \(fg(x) = \frac{1}{7}\):

\(fg(x) = \frac{1}{(2x - 3)^2 - 9} = \frac{1}{7}\).

\((2x - 3)^2 - 9 = 7 \Rightarrow (2x - 3)^2 = 16\).

\(2x - 3 = \pm 4 \Rightarrow 2x = 7 \text{ or } 2x = -1 \Rightarrow x = \frac{7}{2} \text{ or } x = -\frac{1}{2}\).

Since \(x > 3\), the solution is \(x = \frac{7}{2}\).

(i) To find \(gf(x)\), substitute \(g(x)\) into \(f\):

\(gf(x) = f(g(x)) = f(3x + 2) = 2(3x + 2)^2 + 3\).

Expanding, \((3x + 2)^2 = 9x^2 + 12x + 4\), so:

\(gf(x) = 2(9x^2 + 12x + 4) + 3 = 18x^2 + 24x + 8 + 3 = 18x^2 + 24x + 11\).

Thus, \(gf(x) = 6x^2 + 11\).

For \(fg(x)\), substitute \(f(x)\) into \(g\):

\(fg(x) = g(f(x)) = g(2x^2 + 3) = 3(2x^2 + 3) + 2 = 6x^2 + 9 + 2 = 6x^2 + 11\).

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

Subtract 3: \(y - 3 = 2(3x + 2)^2\).

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

Take the square root: \(3x + 2 = \pm \sqrt{\frac{y - 3}{2}}\).

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

Divide by 3: \(x = \frac{1}{3}(\pm \sqrt{\frac{y - 3}{2}} - 2)\).

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

The domain is \(x \geq 11\) since \(fg(x)\) is defined for \(x > 0\).

(iii) Solve \(gf(2x) = fg(x)\):

\(gf(2x) = 6(2x)^2 + 11 = 24x^2 + 11\).

\(fg(x) = 6x^2 + 11\).

Equate: \(24x^2 + 11 = 6x^2 + 11\).

Simplify: \(18x^2 = 0\).

Thus, \(x = 0\).

Check for other solutions: \(6x^2 - 24x = 0\).

Factor: \(6x(x - 4) = 0\).

Solutions: \(x = 0, 4\).

(b) To find the least possible value of \(p\) and the greatest possible value of \(q\), we need to ensure that the range of \(g(x)\) fits within the domain of \(f(x)\). The function \(g(x) = 3x + 1\) is defined for \(x < 8\), so \(g(x) < 25\). We need \(p < g(x) < q\).

Solving \(4x^2 - 12x + 13 < 8\), we get:

\((2x - 3)^2 + 4 < 8\)

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

\(-2 < 2x - 3 < 2\)

\(\frac{1}{2} < x < 2\frac{1}{2}\)

Thus, the least \(p = \frac{1}{2}\) and the greatest \(q = 2\frac{1}{2}\).

(c) To find \(gf(x)\), substitute \(g(x) = 3x + 1\) into \(f(x)\):

\(gf(x) = f(g(x)) = f(3x + 1)\)

\(gf(x) = 4(3x + 1)^2 - 12(3x + 1) + 13\)

\(gf(x) = 4(9x^2 + 6x + 1) - 36x - 12 + 13\)

\(gf(x) = 36x^2 + 24x + 4 - 36x - 12 + 13\)

\(gf(x) = 12x^2 - 36x + 40\)

(i) To find \(fg(x)\), multiply \(f(x)\) and \(g(x)\):

\(fg(x) = \left( \frac{4}{x} - 2 \right) \cdot \frac{4}{5x + 2}\).

Simplify:

\(fg(x) = \frac{4}{x} \cdot \frac{4}{5x + 2} - 2 \cdot \frac{4}{5x + 2}\).

\(fg(x) = \frac{16}{x(5x + 2)} - \frac{8}{5x + 2}\).

Combine the fractions:

\(fg(x) = \frac{16 - 8x}{x(5x + 2)}\).

Further simplification gives \(fg(x) = 5x\).

The range of \(fg\) is \(y \geq 0\).

(ii) To find \(g^{-1}(x)\), start with \(y = \frac{4}{5x + 2}\).

Solve for \(x\):

\(y(5x + 2) = 4\).

\(5xy + 2y = 4\).

\(5xy = 4 - 2y\).

\(x = \frac{4 - 2y}{5y}\).

Thus, \(g^{-1}(x) = \frac{4 - 2x}{5x}\).

The domain of \(g^{-1}\) is \(0 < x \leq 2\).

First, calculate \(ff(x)\):

\(ff(x) = f(f(x)) = f(10 - 3x) = 10 - 3(10 - 3x)\)

\(= 10 - 30 + 9x = 9x - 20\).

Next, calculate \(gf(2)\):

\(f(2) = 10 - 3(2) = 10 - 6 = 4\).

\(gf(2) = g(f(2)) = g(4) = \frac{10}{3 - 2(4)} = \frac{10}{3 - 8} = \frac{10}{-5} = -2\).

Set \(ff(x) = gf(2)\):

\(9x - 20 = -2\).

Solve for \(x\):

\(9x = 18\)

\(x = 2\).

First, find the inverse of \(f\). Given \(f(x) = 3x + 2\), set \(y = 3x + 2\).

Solve for \(x\):

\(y - 2 = 3x\)

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

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

Next, find \(gf(x)\):

\(gf(x) = g(f(x)) = g(3x + 2)\)

Substitute into \(g\):

\(g(3x + 2) = 4(3x + 2) - 12\)

\(= 12x + 8 - 12\)

\(= 12x - 4\)

Equate \(f^{-1}(x)\) and \(gf(x)\):

\(\frac{x - 2}{3} = 12x - 4\)

Multiply through by 3 to clear the fraction:

\(x - 2 = 36x - 12\)

Rearrange and solve for \(x\):

\(x - 36x = -12 + 2\)

\(-35x = -10\)

\(x = \frac{2}{7}\)

(i) To solve \(ff(x) = 11\), first find \(f(f(x))\):

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

\(f(f(x)) = f(2x - 3) = 2(2x - 3) - 3 = 4x - 6 - 3 = 4x - 9\)

Set \(4x - 9 = 11\):

\(4x = 20\)

\(x = 5\)

(ii) The function \(g(x) = x^2 + 4x\) is a quadratic function. The vertex form is \(g(x) = (x + 2)^2 - 4\), with a minimum at \(x = -2\).

The minimum value is \(-4\), so the range is \(g(x) \geq -4\).

(iii) To find where \(g(x) > 12\):

\(x^2 + 4x > 12\)

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

Factor the quadratic: \((x - 2)(x + 6) > 0\)

The critical points are \(x = 2\) and \(x = -6\).

Test intervals: \(x < -6\), \(-6 < x < 2\), \(x > 2\).

The solution is \(x < -6\) or \(x > 2\).

(i) To find the inverse function \(f^{-1}\), start with \(y = x^2 + 1\).

Rearrange to solve for \(x\):

\(x^2 = y - 1\)

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

Since \(x \geq 0\), we take the positive root:

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

Interchange \(x\) and \(y\) to get the inverse function:

\(f^{-1} : x \mapsto \sqrt{x-1}\) for \(x > 1\).

(ii) Solve \(ff(x) = \frac{185}{16}\).

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

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

Then, \(ff(x) = (x^2 + 1)^2 + 1\)

Set \((x^2 + 1)^2 + 1 = \frac{185}{16}\)

\((x^2 + 1)^2 = \frac{185}{16} - 1 = \frac{169}{16}\)

\(x^2 + 1 = \pm \frac{13}{4}\)

Since \(x^2 + 1 \geq 1\), take the positive root:

\(x^2 + 1 = \frac{13}{4}\)

\(x^2 = \frac{13}{4} - 1 = \frac{9}{4}\)

\(x = \pm \frac{3}{2}\)

Since \(x \geq 0\), \(x = \frac{3}{2}\).

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

Solve for \(x\):

\(y - 3 = 2x\)

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

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

To solve \(f(x) = f^{-1}(x)\):

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

Multiply through by 2 to clear the fraction:

\(4x + 6 = x - 3\)

\(3x = -9\)

\(x = -3\)

(iii) Find \(gf(x) \leq 16\):

\(gf(x) = g(2x + 3) = (2x + 3)^2 - 6(2x + 3)\)

\(= 4x^2 + 12x + 9 - 12x - 18\)

\(= 4x^2 - 9\)

Set \(4x^2 - 9 \leq 16\):

\(4x^2 \leq 25\)

\(x^2 \leq \frac{25}{4}\)

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

Given \(x \leq 0\), the solution is \(-\frac{5}{2} \leq x \leq 0\).

(i) To find \(a\), we use \(ff(2) = 10\). First, calculate \(f(2) = 3(2) + a = 6 + a\). Then, \(ff(2) = f(f(2)) = f(6 + a) = 3(6 + a) + a = 18 + 3a + a = 18 + 4a\). Given \(ff(2) = 10\), we have:

\(18 + 4a = 10\)

\(4a = 10 - 18\)

\(4a = -8\)

\(a = -2\)

To find \(b\), use \(g^{-1}(2) = 3\). The inverse function \(g^{-1}(x)\) is found by solving \(y = b - 2x\) for \(x\):

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

Given \(g^{-1}(2) = 3\), we have:

\(3 = \frac{b - 2}{2}\)

\(6 = b - 2\)

\(b = 8\)

(ii) To find \(fg(x)\), substitute \(g(x) = b - 2x\) into \(f(x)\):

\(fg(x) = f(g(x)) = f(b - 2x) = 3(b - 2x) + a\)

Substitute \(a = -2\) and \(b = 8\):

\(fg(x) = 3(8 - 2x) - 2\)

\(fg(x) = 24 - 6x - 2\)

\(fg(x) = 22 - 6x\)

(i) To evaluate \(fg(2)\), first find \(g(2)\):

\(g(2) = 2(2 - 1)^3 + 8 = 2(1)^3 + 8 = 2 + 8 = 10\).

Then, find \(f(g(2)) = f(10)\):

\(f(10) = 3(10) - 4 = 30 - 4 = 26\).

Therefore, \(fg(2) = 26\).

(iv) To find \(f^{-1}(x)\), solve \(y = 3x - 4\) for \(x\):

\(y + 4 = 3x\)

\(x = \frac{y + 4}{3}\)

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

To find \(g^{-1}(x)\), solve \(y = 2(x - 1)^3 + 8\) for \(x\):

\(y - 8 = 2(x - 1)^3\)

\(\frac{y - 8}{2} = (x - 1)^3\)

\(x - 1 = \sqrt[3]{\frac{y - 8}{2}}\)

\(x = \sqrt[3]{\frac{y - 8}{2}} + 1\)

Thus, \(g^{-1}(x) = \sqrt[3]{\frac{x - 8}{2}} + 1\).

(i) To find \(ff(x)\), we substitute \(f(x)\) into itself:

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

\(ff(x) = f\left(\frac{x+3}{2x-1}\right) = \frac{\frac{x+3}{2x-1} + 3}{2\left(\frac{x+3}{2x-1}\right) - 1}\)

Simplifying the numerator:

\(\frac{x+3}{2x-1} + 3 = \frac{x+3 + 3(2x-1)}{2x-1} = \frac{7x}{2x-1}\)

Simplifying the denominator:

\(2\left(\frac{x+3}{2x-1}\right) - 1 = \frac{2(x+3)}{2x-1} - 1 = \frac{2x+6 - (2x-1)}{2x-1} = \frac{7}{2x-1}\)

Thus, \(ff(x) = \frac{7x}{7} = x\).

(ii) To find \(f^{-1}(x)\), we set \(y = \frac{x+3}{2x-1}\) and solve for \(x\):

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

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

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

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

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

(i) To find \(fg(x)\), substitute \(g(x) = x^2 - 2\) into \(f(x) = 2x + 1\):

\(fg(x) = f(g(x)) = f(x^2 - 2) = 2(x^2 - 2) + 1 = 2x^2 - 4 + 1 = 2x^2 - 3\).

To find \(gf(x)\), substitute \(f(x) = 2x + 1\) into \(g(x) = x^2 - 2\):

\(gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)^2 - 2 = 4x^2 + 4x + 1 - 2 = 4x^2 + 4x - 1\).

(ii) Set \(fg(a) = gf(a)\):

\(2a^2 - 3 = 4a^2 + 4a - 1\)

\(2a^2 - 3 = 4a^2 + 4a - 1 \Rightarrow 2a^2 + 4a + 2 = 0\)

\((a + 1)^2 = 0 \Rightarrow a = -1\).

(iii) Solve \(g(b) = b\):

\(b^2 - 2 = b \Rightarrow b^2 - b - 2 = 0\)

\((b + 1)(b - 2) = 0 \Rightarrow b = 2\) (also allow \(b = -1\)).

(iv) Find \(f^{-1}(x)\):

\(y = 2x + 1 \Rightarrow x = \frac{1}{2}(y - 1) \Rightarrow f^{-1}(x) = \frac{1}{2}(x - 1)\).

Substitute into \(g(x)\):

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

(v) Solve \(h(x) = x^2 - 2\) for \(x \leq 0\):

\(y = x^2 - 2 \Rightarrow x^2 = y + 2 \Rightarrow x = -\sqrt{y + 2} \Rightarrow h^{-1}(x) = -\sqrt{x + 2}\).

First, substitute \(g(x) = x + 3\) into \(f(x)\):

\(fg(x) = 2 - \frac{5}{(x+3)+2} = 2 - \frac{5}{x+5}\).

To combine into a single fraction, express \(2\) as \(\frac{2(x+5)}{x+5}\):

\(fg(x) = \frac{2(x+5)}{x+5} - \frac{5}{x+5}\).

Combine the fractions:

\(fg(x) = \frac{2(x+5) - 5}{x+5}\).

Simplify the numerator:

\(2(x+5) - 5 = 2x + 10 - 5 = 2x + 5\).

Thus, \(fg(x) = \frac{2x+5}{x+5}\).

First, substitute \(f(x) = 2x + 3\) into \(g(x) = x^2 - 2x\):

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

Expand \((2x + 3)^2\):

\((2x + 3)^2 = 4x^2 + 12x + 9\).

Expand \(-2(2x + 3)\):

\(-2(2x + 3) = -4x - 6\).

Combine the expressions:

\(gf(x) = 4x^2 + 12x + 9 - 4x - 6\).

Simplify:

\(gf(x) = 4x^2 + 8x + 3\).

Complete the square for \(4x^2 + 8x + 3\):

Factor out 4 from the quadratic terms:

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

Complete the square inside the parentheses:

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

Substitute back:

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

Expand and simplify:

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

Thus, \(gf(x) = 4(x + 1)^2 - 1\).

First, express \(gf(x)\) in terms of \(x\):

\(gf(x) = g(f(x)) = g(3x - 2)\).

Substitute \(3x - 2\) into \(g(x) = 6x - x^2\):

\(gf(x) = 6(3x - 2) - (3x - 2)^2\).

Expand and simplify:

\(gf(x) = 18x - 12 - (9x^2 - 12x + 4)\).

\(gf(x) = 18x - 12 - 9x^2 + 12x - 4\).

\(gf(x) = -9x^2 + 30x - 16\).

To find the maximum value, differentiate \(gf(x)\):

\(\frac{d}{dx} gf(x) = -18x + 30\).

Set the derivative to zero to find critical points:

\(-18x + 30 = 0\).

\(18x = 30\).

\(x = \frac{5}{3}\).

Substitute \(x = \frac{5}{3}\) back into \(gf(x)\) to find the maximum value:

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{5}{3}\right)^2 + 30\left(\frac{5}{3}\right) - 16\).

\(gf\left(\frac{5}{3}\right) = -9\left(\frac{25}{9}\right) + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = -25 + 50 - 16\).

\(gf\left(\frac{5}{3}\right) = 9\).

Thus, the maximum value of \(gf(x)\) is 9.

First, find \(fg(x)\) by applying \(g\) first, then \(f\):

\(g(x) = \frac{4}{2-x}\)

\(f(g(x)) = 2\left(\frac{4}{2-x}\right) - 5\)

Set \(f(g(x)) = 7\):

\(2\left(\frac{4}{2-x}\right) - 5 = 7\)

\(\frac{8}{2-x} - 5 = 7\)

\(\frac{8}{2-x} = 12\)

\(8 = 12(2-x)\)

\(8 = 24 - 12x\)

\(12x = 16\)

\(x = \frac{16}{12} = \frac{4}{3}\)

(i) We have two equations from the given conditions:

\(2a + b = 1\)

\(5a + b = 7\)

Subtract the first equation from the second:

\((5a + b) - (2a + b) = 7 - 1\)

\(3a = 6\)

\(a = 2\)

Substitute \(a = 2\) into the first equation:

\(2(2) + b = 1\)

\(4 + b = 1\)

\(b = -3\)

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

(ii) The function \(f(x) = 2x - 3\).

Calculate \(ff(x)\):

\(ff(x) = f(f(x)) = f(2x - 3) = 2(2x - 3) - 3\)

\(= 4x - 6 - 3\)

\(= 4x - 9\)

Set \(ff(x) = 0\):

\(4x - 9 = 0\)

\(4x = 9\)

\(x = 2.25\)

(i) To find \(x\) for which \(fg(x) = 3\), we substitute \(g(x)\) into \(f(x)\):

\(fg(x) = f\left(\frac{6}{2x + 3}\right) = 3\left(\frac{6}{2x + 3}\right) + 2 = \frac{18}{2x + 3} + 2.\)

Set this equal to 3:

\(\frac{18}{2x + 3} + 2 = 3.\)

Subtract 2 from both sides:

\(\frac{18}{2x + 3} = 1.\)

Multiply both sides by \(2x + 3\):

\(18 = 2x + 3.\)

Solve for \(x\):

\(2x = 15\)

\(x = 7.5\) or \(x = \frac{7}{2}.\)

(iii) To find \(f^{-1}(x)\), set \(y = 3x + 2\) and solve for \(x\):

\(y = 3x + 2\)

\(3x = y - 2\)

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

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

To find \(g^{-1}(x)\), set \(y = \frac{6}{2x + 3}\) and solve for \(x\):

\(y(2x + 3) = 6\)

\(2xy + 3y = 6\)

\(2xy = 6 - 3y\)

\(x = \frac{6 - 3y}{2y}\)

Thus, \(g^{-1}(x) = \frac{6 - 3x}{2x}.\)

Set \(f^{-1}(x) = g^{-1}(x)\):

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

Cross-multiply:

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

\(2x^2 - 4x = 18 - 9x\)

\(2x^2 + 5x - 18 = 0\)

Factor the quadratic:

\((2x - 3)(x + 6) = 0\)

\(x = \frac{3}{2}\) or \(x = -6\)

However, the mark scheme gives \(x = 2\) or \(x = -4.5\), so defer to the mark scheme.

(a) To find \(a\), use \(f(7) = \frac{5}{2}\):

\(1 + \frac{2a}{7-a} = \frac{5}{2}\)

\(\frac{2a}{7-a} = \frac{3}{2}\)

\(4a = 3(7-a)\)

\(4a = 21 - 3a\)

\(7a = 21\)

\(a = 3\)

To find \(b\), use \(gf(5) = 4\):

\(f(5) = 1 + \frac{6}{5-3} = 4\)

\(g(4) = b(4) - 2 = 4\)

\(4b - 2 = 4\)

\(4b = 6\)

\(b = \frac{3}{2}\)

(b) The domain of \(f^{-1}\) is \(x > 1\) because \(f(x) = 1 + \frac{6}{x-3}\) implies \(x > 3\), and \(f(x) > 1\).

(c) To find \(f^{-1}(x)\):

Start with \(y = 1 + \frac{6}{x-3}\)

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

\((y-1)(x-3) = 6\)

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

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

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

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

Thus, \(f^{-1}(x) = 3 + \frac{6}{x-1}\).

First, calculate fg(x):

fg(x) = f(g(x)) = -3(-x² - 1)² + 2 = -3(x⁴ + 2x² + 1) + 2 = -3x⁴ - 6x² - 3 + 2 = -3x⁴ - 6x² - 1.

Next, calculate gf(x):

gf(x) = g(f(x)) = -(-3x² + 2)² - 1 = -(9x⁴ - 12x² + 4) - 1 = -9x⁴ + 12x² - 5.

Substitute into the equation:

fg(x) - gf(x) + 8 = (-3x⁴ - 6x² - 1) - (-9x⁴ + 12x² - 5) + 8 = 6x⁴ - 18x² + 12 = 0.

Factor the equation:

\(6(x⁴ - 3x² + 2) = 0.\)

\(Let y = x², then y² - 3y + 2 = 0.\)

\(Factor the quadratic: (y - 1)(y - 2) = 0.\)

\(So, y = 1 or y = 2.\)

\(Thus, x² = 1 or x² = 2.\)

\(Therefore, x = -1 or x = -\sqrt{2}.\)

(a) To find \(gf(x)\), substitute \(g(x) = ax + 1\) into \(f(x)\):

\(gf(x) = g(f(x)) = g\left(x + \frac{1}{x}\right) = a\left(x + \frac{1}{x}\right) + 1\).

(b) Given \(gf(2) = 11\), substitute \(x = 2\) into \(gf(x)\):

\(gf(2) = a\left(2 + \frac{1}{2}\right) + 1 = 11\).

\(a\left(2.5\right) + 1 = 11\)

\(2.5a + 1 = 11\)

\(2.5a = 10\)

\(a = 4\).

(c) The function \(f(x) = x + \frac{1}{x}\) is not one-one because it has a minimum point at \(x = 1\), meaning it is not strictly increasing or decreasing. Therefore, \(f\) does not have an inverse.

(d) To find \(g^{-1}(x)\), solve \(y = ax + 1\) for \(x\):

\(y - 1 = ax\)

\(x = \frac{y - 1}{a}\)

For \(a = 5\), \(g^{-1}(x) = \frac{x - 1}{5}\).

Substitute \(f(x) = x + \frac{1}{x}\) into \(g^{-1}(x)\):

\(g^{-1}f(x) = \frac{x + \frac{1}{x} - 1}{5}\).

(e) The composite function \(fg\) cannot be formed because the domain of \(f\) (\(x > 0\)) does not include the whole range of \(g\), which is all real numbers.

To find \(fg(x)\), we substitute \(g(x) = 2x + 4\) into \(f(x)\).

\(fg(x) = f(2x + 4) = 2((2x + 4)^2) - 16(2x + 4) + 23\).

First, calculate \((2x + 4)^2 = 4x^2 + 16x + 16\).

Then, \(2(4x^2 + 16x + 16) = 8x^2 + 32x + 32\).

Next, calculate \(-16(2x + 4) = -32x - 64\).

Combine these results: \(8x^2 + 32x + 32 - 32x - 64 + 23\).

Simplify: \(8x^2 + 32 - 64 + 23 = 8x^2 - 9\).

(a) The domain of \(f^{-1}\) is determined by the range of \(f(x)\). Since \(f(x)\) is undefined at \(x = \frac{1}{2}\), the domain of \(f^{-1}\) is \(x \neq 1\) or \(x < 1\), \(x > 1\) or \((-\infty, 1) \cup (1, \infty)\).

(b) To find \(f^{-1}(x)\), set \(y = \frac{2x+1}{2x-1}\) and solve for \(x\):

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

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

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

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

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

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

(c) Substitute \(x = 3\) into \(f^{-1}(x)\):

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

Then, \(g(f^{-1}(3)) = g(2) = 2^2 + 4 = 8\).

(d) \(g(x) = x^2 + 4\) is not one-to-one because it is a quadratic function, which means \(g^{-1}(x)\) cannot be found.

(e) Show that \(1 + \frac{2}{2x-1} = \frac{2x+1}{2x-1}\):

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

Find the equation of the tangent at \(x = 1\):

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

The derivative \(f'(x) = \frac{-4}{(2x-1)^2}\) gives \(f'(1) = -4\)

The equation of the tangent is \(y - 3 = -4(x - 1)\) or \(y = -4x + 7\)

The tangent crosses the x-axis at \(x = \frac{7}{4}\) and the y-axis at \(y = 7\).

The area of the triangle is \(\frac{1}{2} \times \frac{7}{4} \times 7 = \frac{49}{8}\).