1. **Problem Statement:** Find the maxima, minima, and inflection points of the function $$f(x) = (x-12)^3$$. 2. **Formula and Rules:** - To find maxima and minima, we find the first derivative $$f'(x)$$ and solve $$f'(x) = 0$$. - To classify these points, we use the second derivative $$f''(x)$$: - If $$f''(x) > 0$$, the point is a local minimum. - If $$f''(x) < 0$$, the point is a local maximum. - If $$f''(x) = 0$$, the test is inconclusive; check for inflection points. - Inflection points occur where $$f''(x) = 0$$ and the concavity changes. 3. **First Derivative:** $$f(x) = (x-12)^3$$ Using the power rule: $$f'(x) = 3(x-12)^2$$ 4. **Find Critical Points:** Set $$f'(x) = 0$$: $$3(x-12)^2 = 0$$ $$\Rightarrow (x-12)^2 = 0$$ $$\Rightarrow x = 12$$ 5. **Second Derivative:** $$f''(x) = \frac{d}{dx} [3(x-12)^2] = 6(x-12)$$ 6. **Classify Critical Point:** Evaluate $$f''(12)$$: $$f''(12) = 6(12-12) = 0$$ Since $$f''(12) = 0$$, the second derivative test is inconclusive. 7. **Check for Inflection Point:** Since $$f''(x) = 6(x-12)$$ changes sign at $$x=12$$ (from negative to positive or vice versa), there is an inflection point at $$x=12$$. 8. **Summary:** - The function has no local maxima or minima. - There is an inflection point at $$x=12$$. **Final answer:** - No maxima or minima. - Inflection point at $$x=12$$.