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