\(f'(t) = \frac{d}{dt}(-2t^3 + 3t^2 + 5)\) \(= -6t^2 + 6t\)
\(-6t^2 + 6t = 0\)
Factor the equation:
\(6t(t - 1) = 0\)
This gives two solutions: \( t = 0 \) and \( t = 1 \).
\(f''(t) = \frac{d}{dt}(-6t^2 + 6t)\) \(= -12t + 6\)
Therefore, the correct answer is option \(1\).