(i) To find the stationary point, set the derivative equal to zero:

\(f'(x) = 2x - 6 = 0\)

Solving for \(x\), we get:

\(2x = 6\)

\(x = 3\)

Thus, the value of \(x\) for which \(f(x)\) has a stationary value is 3.

(ii) To find \(f(x)\), integrate \(f'(x)\):

\(f(x) = \int (2x - 6) \, dx\)

\(f(x) = x^2 - 6x + c\)

Given \(f(x) \geq -4\), substitute \(x = 3\) to find \(c\):

\(f(3) = 3^2 - 6(3) + c = -4\)

\(9 - 18 + c = -4\)

\(c = 5\)

Thus, \(f(x) = x^2 - 6x + 5\).