1. **Problem Statement:** We are given the execution time of an algorithm as a function of input size $n$: $$T(n) = n^3 - 6n^2 + 12n + 20.$$ We need to find the local minima and maxima, interpret the minimum point, and analyze performance based on concavity. 2. **Finding local minima and maxima:** To find local extrema, we first compute the derivative $$T'(n) = \frac{d}{dn}(n^3 - 6n^2 + 12n + 20) = 3n^2 - 12n + 12.$$ Local extrema occur where $$T'(n) = 0,$$ so solve: $$3n^2 - 12n + 12 = 0.$$ Divide both sides by 3: $$n^2 - 4n + 4 = 0.$$ This factors as: $$(n - 2)^2 = 0,$$ so the only critical point is at $$n = 2.$$ 3. **Determining the nature of the critical point:** Compute the second derivative: $$T''(n) = \frac{d}{dn}(3n^2 - 12n + 12) = 6n - 12.$$ Evaluate at $$n=2$$: $$T''(2) = 6(2) - 12 = 12 - 12 = 0.$$ Since the second derivative test is inconclusive (equals zero), we check the third derivative: $$T'''(n) = \frac{d}{dn}(6n - 12) = 6,$$ which is positive and nonzero. This indicates a point of inflection at $$n=2$$ rather than a local minimum or maximum. 4. **Interpretation of minimum point:** Since there is no local minimum or maximum, the function has an inflection point at $$n=2$$. This means the execution time does not have a traditional local minimum or maximum but changes concavity there. In terms of optimization, this point marks a change in the rate of increase of execution time but not an optimal input size minimizing execution time. 5. **Concavity and performance:** The concavity is given by the sign of $$T''(n) = 6n - 12.$$ For $$n < 2,$$ $$T''(n) < 0,$$ so the function is concave down, indicating the execution time increases at a decreasing rate. For $$n > 2,$$ $$T''(n) > 0,$$ so the function is concave up, indicating execution time increases at an increasing rate. Thus, performance worsens as $$n$$ increases beyond 2 because execution time grows faster. **Final answers:** - No local minima or maxima; only an inflection point at $$n=2$$. - The inflection point indicates a change in growth rate, not an optimal execution time. - Performance worsens for $$n > 2$$ as execution time accelerates.