(i) To find the stationary points, set \(\frac{dy}{dx} = 0\):

\(-x^2 + 5x - 4 = 0\)

Solving the quadratic equation:

\(x^2 - 5x + 4 = 0\)

\((x - 1)(x - 4) = 0\)

Thus, \(x = 1\) and \(x = 4\).

(ii) Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -2x + 5\)

Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 1\):

\(\frac{d^2y}{dx^2} = -2(1) + 5 = 3\)

Since \(3 > 0\), \(x = 1\) is a minimum.

Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 4\):

\(\frac{d^2y}{dx^2} = -2(4) + 5 = -3\)

Since \(-3 < 0\), \(x = 4\) is a maximum.