1. **Problem Statement:** We are given the function $$F(x) = x^3 - 6x^2 + 9x + 1$$ and need to analyze its behavior by finding derivatives, critical points, inflection points, intervals of increase/decrease, and concavity. 2. **First Derivative and Critical Points:** The first derivative $$F'(x)$$ helps us find critical points where the slope is zero or undefined. $$F'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 1) = 3x^2 - 12x + 9$$ Set $$F'(x) = 0$$ to find critical points: $$3x^2 - 12x + 9 = 0$$ Divide both sides by 3: $$x^2 - 4x + 3 = 0$$ Factor: $$(x - 3)(x - 1) = 0$$ So, critical points are at $$x = 1$$ and $$x = 3$$. 3. **Second Derivative and Inflection Points:** The second derivative $$F''(x)$$ tells us about concavity and inflection points. $$F''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12$$ Set $$F''(x) = 0$$ to find inflection points: $$6x - 12 = 0 \implies x = 2$$ 4. **Intervals of Increase and Decrease:** - For $$x < 1$$, pick $$x=0$$: $$F'(0) = 3(0)^2 - 12(0) + 9 = 9 > 0$$, so increasing. - For $$1 < x < 3$$, pick $$x=2$$: $$F'(2) = 3(4) - 24 + 9 = 12 - 24 + 9 = -3 < 0$$, so decreasing. - For $$x > 3$$, pick $$x=4$$: $$F'(4) = 3(16) - 48 + 9 = 48 - 48 + 9 = 9 > 0$$, so increasing. 5. **Intervals of Concavity:** - For $$x < 2$$, pick $$x=0$$: $$F''(0) = 6(0) - 12 = -12 < 0$$, so concave down. - For $$x > 2$$, pick $$x=3$$: $$F''(3) = 18 - 12 = 6 > 0$$, so concave up. 6. **Summary:** - Critical points at $$x=1$$ and $$x=3$$. - Inflection point at $$x=2$$. - Increasing on $$(-\infty, 1) \cup (3, \infty)$$. - Decreasing on $$(1, 3)$$. - Concave down on $$(-\infty, 2)$$. - Concave up on $$(2, \infty)$$. 7. **Final function for graphing:** $$F(x) = x^3 - 6x^2 + 9x + 1$$