(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 \(g(x) \geq -9\).