1. **State the problem:** We have a particle moving along a horizontal line with position function $$s(t) = t^3 - 9t^2 + 15t + 4$$ for $$t \geq 0$$. We want to find when the particle is moving to the left. 2. **Recall the rule:** The particle moves to the left when its velocity is negative. Velocity $$v(t)$$ is the derivative of position $$s(t)$$ with respect to time $$t$$. 3. **Find the velocity function:** $$v(t) = \frac{ds}{dt} = 3t^2 - 18t + 15$$ 4. **Set velocity less than zero to find when moving left:** $$3t^2 - 18t + 15 < 0$$ 5. **Simplify inequality by dividing both sides by 3:** $$\cancel{3}t^2 - \cancel{3}6t + \cancel{3}5 < 0 \Rightarrow t^2 - 6t + 5 < 0$$ 6. **Factor quadratic:** $$t^2 - 6t + 5 = (t - 1)(t - 5)$$ 7. **Analyze inequality:** The quadratic opens upward (positive leading coefficient), so $$t^2 - 6t + 5 < 0$$ between the roots: $$1 < t < 5$$ 8. **Consider domain:** Since $$t \geq 0$$, the particle moves left for $$t$$ in the interval $$1 < t < 5$$. **Final answer:** The particle moves to the left when $$t$$ is between 1 and 5, i.e., $$1 < t < 5$$.