The function \(g(x) = 2x + 3\) is a linear function with a slope of 2 and a y-intercept of 3. To find the inverse function \(g^{-1}(x)\), we swap \(x\) and \(y\) and solve for \(y\):

\(x = 2y + 3\)

\(x - 3 = 2y\)

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

Thus, the inverse function is \(g^{-1}(x) = \frac{1}{2}(x - 3)\).

In the diagram, the graph of \(y = g(x)\) is a straight line with a slope of 2 and a y-intercept of 3. The graph of \(y = g^{-1}(x)\) is a straight line with a slope of \(\frac{1}{2}\) and a y-intercept of \(-\frac{3}{2}\).

The line \(y = x\) is also drawn to show the symmetry between \(y = g(x)\) and \(y = g^{-1}(x)\), as the inverse function is a reflection of the original function across the line \(y = x\).

The function \(f(x) = 2x - 5\) is a linear function with a slope of 2 and a y-intercept of -5. To find the inverse, solve for \(x\) in terms of \(y\):

\(y = 2x - 5\)

\(y + 5 = 2x\)

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

Thus, the inverse function is \(f^{-1}(x) = \frac{x + 5}{2}\).

Sketch the graph of \(y = 2x - 5\), which is a straight line with a slope of 2 and y-intercept at -5.

Sketch the graph of \(y = \frac{x + 5}{2}\), which is a straight line with a slope of \(\frac{1}{2}\) and y-intercept at \(\frac{5}{2}\).

The graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) should be symmetric about the line \(y = x\).

The function \(f(x) = 3x + 2\) is a linear function with a slope of 3 and a y-intercept of 2. This means the graph of \(y = f(x)\) is a straight line that crosses the y-axis at (0, 2) and has a positive slope.

The inverse function \(f^{-1}(x)\) can be found by swapping \(x\) and \(y\) and solving for \(y\):

\(x = 3y + 2\)

\(x - 2 = 3y\)

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

The graph of \(y = f^{-1}(x)\) is a line with a slope of \(\frac{1}{3}\) and an x-intercept of 2/3. This line is the reflection of \(y = f(x)\) in the line \(y = x\).

Thus, the graph of \(y = f(x)\) has a slope greater than 1 and a positive y-intercept, while the graph of \(y = f^{-1}(x)\) has a slope less than 1 and a positive x-intercept. The two graphs are reflections of each other across the line \(y = x\).

(a) To express \(f(x) = x^2 - 4x + 3\) in the form \((x-a)^2 + b\), complete the square:

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

(b) Since \(f\) is a one-one function, the smallest possible value of \(c\) is the vertex of the parabola, which is \(x = 2\).

(c) Given \(c = 5\), the function \(f(x) = (x-2)^2 - 1\) is one-one for \(x > 5\). To find \(f^{-1}(x)\), set \(y = (x-2)^2 - 1\):

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

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

Since \(x > 5\), choose the positive root: \(x = 2 + \sqrt{y+1}\).

Thus, \(f^{-1}(x) = 2 + \sqrt{x+1}\) for \(x > 8\).

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

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

The range of \(gf(x)\) is \(0 < gf(x) < \frac{1}{9}\) since \((x-2)^2 > 0\) and the minimum value of \((x-2)^2\) is 9 when \(x = 5\).

(i) Substitute \(x = -2\) and \(x = -3\) into the function:

\(2a + 4b = 8\)

\(2a^2 + 3a + 4b = 14\)

Substitute \(4b = 8 - 2a\) into the second equation:

\(2a^2 + 3a + (8 - 2a) = 14\)

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

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

Thus, \(a = -2\) or \(a = \frac{3}{2}\).

For \(a = -2\), \(2(-2) + 4b = 8\) gives \(b = 3\).

For \(a = \frac{3}{2}\), \(2(\frac{3}{2}) + 4b = 8\) gives \(b = \frac{5}{4}\).

(ii) For \(a = 1\) and \(b = -1\), the function is \(f(x) = x^2 - x - 3\).

Complete the square: \(y = \left(x - \frac{1}{2}\right)^2 - \frac{13}{4}\).

Rearrange to find \(x\):

\(x - \frac{1}{2} = \pm \sqrt{y + \frac{13}{4}}\)

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

Choose the positive root for the inverse: \(f^{-1}(x) = \frac{1}{2} \sqrt{x + \frac{13}{4}}\).

The domain of \(f^{-1}\) is \(x \geq -\frac{13}{4}\).