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AP Calculus AB · Question of the Day

AP Calculus AB Question of the Day

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Friday, November 6, 2026

If f′′(x)<0f''(x)<0 on (−3,1)( -3,1) and f′(x)f'(x) changes from negative to positive at x=−1x=-1, what happens to ff at x=−1x=-1?

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If f′′(x)<0f''(x)<0 on (−3,1)( -3,1) and f′(x)f'(x) changes from negative to positive at x=−1x=-1, what happens to ff at x=−1x=-1?

  1. ff has a local maximum at x=−1x=-1.
  2. ff has a local minimum at x=−1x=-1. (correct answer)
  3. ff has an inflection point at x=−1x=-1.
  4. ff has neither a local extremum nor an inflection point at x=−1x=-1.
  5. ff must be increasing and concave up at x=−1x=-1.

Explanation: This problem uses multi-representation reasoning to analyze critical points and concavity. Since f'(x) changes from negative to positive at x = -1, and assuming f' is continuous, we have f'(-1) = 0 with f increasing after this point. This sign change in f' from negative to positive indicates a local minimum at x = -1. The condition f''(x) < 0 on (-3,1) tells us f is concave down at x = -1, which doesn't prevent the minimum. Choice A incorrectly suggests a maximum, which would require f' to change from positive to negative. When f' changes from negative to positive at a critical point, there's always a local minimum regardless of concavity.