Step 1: Apply Rolle’s Theorem.
We know that \( f(0) = 4 \) and \( f(1) = -2 \), so the function changes sign over the interval \( [0, 1] \). By Rolle’s Theorem, since \( f(x) \) is continuous and differentiable on the closed interval \( [0, 1] \), and \( f(0) = f(1) \), there must exist some \( c_1 \in (0, 1) \) such that \( f'(c_1) = 0 \).
Step 2: Conclusion.
From Rolle's Theorem and the function's values, we conclude that \( f'(x) = 0 \) for some \( x \in [0, 3] \), so option (C) is true.
Final Answer:
\[
\boxed{f'(x_2) = 0 \text{ for some } x_2 \in [0, 3].}
\]