1. **State the problem:** We are given the derivative of a function $f'(x) = -x^3 - 6x^2 - 9x$ and need to find the intervals where the original function $f$ is decreasing. 2. **Recall the rule:** A function $f$ is decreasing where its derivative $f'(x)$ is less than zero, i.e., $f'(x) < 0$. 3. **Analyze the derivative:** Given $$f'(x) = -x^3 - 6x^2 - 9x,$$ we can factor it to find critical points: 4. **Factor the derivative:** $$f'(x) = -x^3 - 6x^2 - 9x = -x(x^2 + 6x + 9).$$ Note that $x^2 + 6x + 9 = (x+3)^2$, so $$f'(x) = -x(x+3)^2.$$ 5. **Find critical points:** Set $f'(x) = 0$: $$-x(x+3)^2 = 0 \\ \Rightarrow x=0 \text{ or } x=-3.$$ 6. **Determine sign intervals:** The critical points divide the real line into intervals: - $(-\infty, -3)$ - $(-3, 0)$ - $(0, \infty)$ 7. **Test each interval:** - For $x < -3$, pick $x = -4$: $$f'(-4) = -(-4)((-4)+3)^2 = 4 \times 1^2 = 4 > 0,$$ so $f'(x) > 0$ here, function is increasing. - For $-3 < x < 0$, pick $x = -1$: $$f'(-1) = -(-1)((-1)+3)^2 = 1 \times 2^2 = 4 > 0,$$ so $f'(x) > 0$ here, function is increasing. - For $x > 0$, pick $x = 1$: $$f'(1) = -(1)(1+3)^2 = -1 \times 4^2 = -16 < 0,$$ so $f'(x) < 0$ here, function is decreasing. 8. **Conclusion:** The function $f$ is decreasing on the interval $$\boxed{(0, \infty)}.$$