1. **Problem statement:** We are given the function $f(x) = x^3 + 3x^2 + kx - 1$ and need to find the range of values for $k$ such that $f$ is one-to-one. 2. **Key concept:** A function is one-to-one if it is strictly monotonic (either strictly increasing or strictly decreasing) over its entire domain. 3. **Approach:** To check monotonicity, analyze the derivative $f'(x)$: $$f'(x) = 3x^2 + 6x + k$$ 4. **Monotonicity condition:** For $f$ to be one-to-one, $f'(x)$ must not change sign, i.e., it should be always positive or always negative. 5. **Analyze $f'(x)$:** The quadratic $3x^2 + 6x + k$ has a discriminant: $$\Delta = 6^2 - 4 \cdot 3 \cdot k = 36 - 12k$$ 6. **No real roots condition:** For $f'(x)$ to not change sign, the quadratic must have no real roots, so: $$\Delta < 0 \implies 36 - 12k < 0$$ 7. **Solve inequality:** $$36 < 12k \implies k > 3$$ 8. **Check leading coefficient:** Since the coefficient of $x^2$ in $f'(x)$ is positive (3), $f'(x)$ opens upward, so if no real roots, $f'(x) > 0$ for all $x$. 9. **Conclusion:** The function $f$ is strictly increasing and one-to-one if and only if: $$k > 3$$ 10. **Answer choice:** Among the options, the closest is **C) k \geq 3**. Since strict inequality is required, $k > 3$ is the exact condition, but $k \geq 3$ is the best match. **Final answer:** $\boxed{k \geq 3}$