1. **State the problem:** We need to find the intervals where the function $f(x) = x^4 - 4x^3$ is decreasing. 2. **Recall the rule:** A function is decreasing where its derivative is negative, i.e., $f'(x) < 0$. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x^4 - 4x^3) = 4x^3 - 12x^2$$ 4. **Factor the derivative:** $$f'(x) = 4x^2(x - 3)$$ 5. **Find critical points by setting $f'(x) = 0$:** $$4x^2(x - 3) = 0 \implies x = 0 \text{ or } x = 3$$ 6. **Determine the sign of $f'(x)$ on intervals defined by critical points:** - For $x < 0$, choose $x = -1$: $f'(-1) = 4(-1)^2(-1 - 3) = 4(1)(-4) = -16 < 0$ (decreasing) - For $0 < x < 3$, choose $x = 1$: $f'(1) = 4(1)^2(1 - 3) = 4(1)(-2) = -8 < 0$ (decreasing) - For $x > 3$, choose $x = 4$: $f'(4) = 4(4)^2(4 - 3) = 4(16)(1) = 64 > 0$ (increasing) 7. **Conclusion:** The function $f(x)$ is decreasing on the intervals $$(-\infty, 0) \cup (0, 3)$$. Note that at $x=0$, the derivative is zero but the sign does not change (since $x^2$ is always positive), so the function is decreasing on both sides of 0. **Final answer:** $$f(x) \text{ is decreasing on } (-\infty, 3)$$