1. **State the problem:** We are given the function $$f(x) = -x^4 - 12x^3 - 36x^2$$ and need to find the intervals where $$f$$ is increasing. 2. **Recall the rule:** A function is increasing where its derivative $$f'(x)$$ is positive. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(-x^4 - 12x^3 - 36x^2) = -4x^3 - 36x^2 - 72x$$ 4. **Factor the derivative:** $$f'(x) = -4x^3 - 36x^2 - 72x = -4x(x^2 + 9x + 18)$$ 5. **Factor the quadratic:** $$x^2 + 9x + 18 = (x + 3)(x + 6)$$ 6. **Rewrite derivative fully factored:** $$f'(x) = -4x(x + 3)(x + 6)$$ 7. **Find critical points by setting $$f'(x) = 0$$:** $$-4x(x + 3)(x + 6) = 0 \implies x = 0, -3, -6$$ 8. **Determine sign of $$f'(x)$$ on intervals defined by critical points:** - For $$x < -6$$, pick $$x = -7$$: $$f'(-7) = -4(-7)(-7+3)(-7+6) = -4(-7)(-4)(-1) = -4 \times -7 \times -4 \times -1 = -112 < 0$$ - For $$-6 < x < -3$$, pick $$x = -5$$: $$f'(-5) = -4(-5)(-5+3)(-5+6) = -4(-5)(-2)(1) = -4 \times -5 \times -2 \times 1 = 40 > 0$$ - For $$-3 < x < 0$$, pick $$x = -1$$: $$f'(-1) = -4(-1)(-1+3)(-1+6) = -4(-1)(2)(5) = -4 \times -1 \times 2 \times 5 = 40 > 0$$ - For $$x > 0$$, pick $$x = 1$$: $$f'(1) = -4(1)(1+3)(1+6) = -4(1)(4)(7) = -112 < 0$$ 9. **Conclusion:** The function $$f$$ is increasing where $$f'(x) > 0$$, which is on the intervals $$(-6, -3)$$ and $$(-3, 0)$$. **Final answer:** $$f$$ is increasing on $$(-6, 0)$$ (since the union of $$(-6, -3)$$ and $$(-3, 0)$$ is $$(-6, 0)$$).