1. **Problem statement:** We have the function $$f(x) = x^6 (x - 6)^3$$ defined for $$x \in [-10, 11]$$. We need to find the intervals where $$f$$ is increasing, the region where $$f$$ is positive, and where $$f$$ achieves its minimum. 2. **Find the derivative:** To find where $$f$$ is increasing, we compute $$f'(x)$$ and find where $$f'(x) > 0$$. Using the product rule: $$f'(x) = \frac{d}{dx}[x^6] \cdot (x-6)^3 + x^6 \cdot \frac{d}{dx}[(x-6)^3]$$ Calculate derivatives: $$\frac{d}{dx}[x^6] = 6x^5$$ $$\frac{d}{dx}[(x-6)^3] = 3(x-6)^2$$ So, $$f'(x) = 6x^5 (x-6)^3 + x^6 \cdot 3(x-6)^2 = 3x^5 (x-6)^2 (2(x-6) + x)$$ Simplify inside the parentheses: $$2(x-6) + x = 2x - 12 + x = 3x - 12$$ Therefore, $$f'(x) = 3x^5 (x-6)^2 (3x - 12) = 3x^5 (x-6)^2 3(x - 4) = 9x^5 (x-6)^2 (x-4)$$ 3. **Find critical points:** Set $$f'(x) = 0$$: $$9x^5 (x-6)^2 (x-4) = 0$$ This gives critical points at $$x=0$$, $$x=6$$, and $$x=4$$. 4. **Determine sign of $$f'(x)$$ to find increasing intervals:** - The factors $$x^5$$ and $$(x-6)^2$$ are always non-negative except $$x^5$$ changes sign at 0. - $$(x-6)^2 \geq 0$$ always. - The sign of $$f'(x)$$ depends on $$x^5$$ and $$(x-4)$$. Test intervals: - For $$x < 0$$: $$x^5 < 0$$, $$(x-4) < 0$$, product of signs: negative * negative = positive, so $$f'(x) > 0$$. - For $$0 < x < 4$$: $$x^5 > 0$$, $$(x-4) < 0$$, product positive * negative = negative, so $$f'(x) < 0$$. - For $$4 < x < 6$$: $$x^5 > 0$$, $$(x-4) > 0$$, product positive * positive = positive, so $$f'(x) > 0$$. - For $$x > 6$$: $$x^5 > 0$$, $$(x-4) > 0$$, product positive * positive = positive, so $$f'(x) > 0$$. 5. **Intervals where $$f$$ is increasing:** - $$(-10, 0)$$ - $$(4, 11)$$ (since domain ends at 11 and $$f'(x) > 0$$ for $$x > 6$$) 6. **Find where $$f(x) > 0$$:** $$f(x) = x^6 (x-6)^3$$. - $$x^6 \geq 0$$ always. - Sign depends on $$(x-6)^3$$. For $$x < 6$$, $$(x-6)^3 < 0$$, so $$f(x) < 0$$. For $$x > 6$$, $$(x-6)^3 > 0$$, so $$f(x) > 0$$. At $$x=6$$, $$f(6) = 6^6 \cdot 0 = 0$$. Therefore, $$f(x) > 0$$ on $$(6, 11]$$. 7. **Find minimum:** Check critical points and endpoints: - At $$x=0$$, $$f(0) = 0^6 (0-6)^3 = 0$$. - At $$x=4$$, $$f(4) = 4^6 (4-6)^3 = 4^6 (-2)^3 = 4^6 (-8) < 0$$ (negative large value). - At $$x=6$$, $$f(6) = 0$$. - At $$x=-10$$ and $$x=11$$, values are large but positive or negative depending on sign. Since $$f(4)$$ is negative and less than values at other critical points, the minimum is at $$x=4$$. **Final answers:** - Increasing intervals: $$(-10, 0)$$ and $$(4, 11)$$ - Region where $$f$$ is positive: $$(6, 11)$$ - Minimum at $$x=4$$