Solution: Suppose, for contradiction, that there exist two distinct real numbers \( c_1 \) and \( c_2 \) (with \( c_1 < c_2 \)) such that \( f(c_1) = f(c_2) = 0 \). Since \( f \) is differentiable on \( \mathbb{R} \), it is continuous on \( [c_1, c_2] \) and differentiable on \( (c_1, c_2) \). By the Mean Value Theorem, there exists \( \xi \in (c_1, c_2) \) such that: \[ f'(\xi) = \frac{f(c_2) - f(c_1)}{c_2 - c_1} = \frac{0 - 0}{c_2 - c_1} = 0. \] However, this contradicts the assumption that \( f'(x) \geq 3 \) for all \( x \in \mathbb{R} \). Therefore, there can be at most one real number \( c \) such that \( f(c) = 0 \).