AP Calculus BC · Question of the Day

AP Calculus BC Question of the Day

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Thursday, September 17, 2026

A function $f'$ has a local maximum at $x=1$ and crosses the $x$ -axis there from positive to negative; which could be $f$ near $x=1$ ?

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A function $f'$ has a local maximum at $x=1$ and crosses the $x$ -axis there from positive to negative; which could be $f$ near $x=1$ ?

$f$ has a local minimum at $x=1$ and is concave up there.
$f$ has a local maximum at $x=1$ and is concave down there. (correct answer)
$f$ has an inflection point at $x=1$ and is increasing on both sides.
$f$ has a local maximum at $x=1$ and is concave up there.
$f$ has a local minimum at $x=1$ and is concave down there.

Explanation: This problem tests understanding of how features of f' translate to features of f. Since f' has a local maximum at x = 1, the second derivative f''(1) < 0, which means f is concave down at x = 1. Since f' crosses the x-axis from positive to negative at x = 1, this means f changes from increasing to decreasing, giving f a local maximum at x = 1. Therefore, f has a local maximum at x = 1 and is concave down there. Choice D might seem plausible as it correctly identifies the local maximum, but it incorrectly states f is concave up when f''(1) < 0 tells us f must be concave down. When analyzing critical points, remember that f' = 0 gives potential extrema, the sign change of f' determines the type of extremum, and the sign of f'' determines concavity.