1. The problem is to analyze the function $$f(x) = x^3 - 9$$. 2. This is a cubic function where the general form is $$f(x) = ax^3 + bx^2 + cx + d$$. Here, $$a=1$$, $$b=0$$, $$c=0$$, and $$d=-9$$. 3. To find the roots (where $$f(x)=0$$), solve: $$x^3 - 9 = 0$$ $$x^3 = 9$$ $$x = \sqrt[3]{9}$$ 4. To find critical points (where the slope is zero), compute the derivative: $$f'(x) = 3x^2$$ Set $$f'(x) = 0$$: $$3x^2 = 0$$ $$x = 0$$ 5. Evaluate $$f(x)$$ at $$x=0$$: $$f(0) = 0^3 - 9 = -9$$ This is a critical point, specifically a point of inflection since the second derivative changes sign. 6. The second derivative is: $$f''(x) = 6x$$ At $$x=0$$, $$f''(0) = 0$$, confirming the inflection point. 7. Summary: - Root at $$x = \sqrt[3]{9}$$ - Inflection point at $$(0, -9)$$ Final answer: The function $$f(x) = x^3 - 9$$ has one real root at $$x = \sqrt[3]{9}$$ and an inflection point at $$(0, -9)$$.