Topic 5.7 Using the Second Derivative Test to Determine Extrema: classify critical points using the sign of the second derivative.

What this topic is asking

The College Board (Topic 5.7) gives the Second Derivative Test: at a critical point where $f'(c) = 0$, the sign of $f''(c)$ classifies it. Concave down means a maximum; concave up means a minimum. You must apply it and know when it is inconclusive.

The test

A worked classification

When the test is inconclusive

The Second Derivative Test fails exactly when $f''(c) = 0$. For $f(x) = x^4$, $f'(0) = 0$ and $f''(0) = 0$, so the test says nothing; but the First Derivative Test (or just inspection) shows a relative minimum. For $f(x) = x^3$, again $f''(0) = 0$, and there the critical point is neither a max nor a min. Because $f''(c) = 0$ can accompany any of the three outcomes, the test simply cannot decide, and you must fall back on the sign change of $f'$. Knowing this limit is itself examined.

Choosing between the two tests

The Second Derivative Test is often faster when $f''$ is easy to compute and you only need a single evaluation per critical point, rather than testing the sign of $f'$ on intervals. But the First Derivative Test is more general: it works at critical points where $f'$ is undefined (cusps), and it never returns "inconclusive". A practical rule is to use the Second Derivative Test when $f''$ is cheap and nonzero at the critical points, and the First Derivative Test otherwise. On free-response questions either test earns full credit, provided the justification cites the relevant derivative sign at the critical point. State $f''(c)$'s sign (or the sign change of $f'$) explicitly.

The geometric picture

The Second Derivative Test is easiest to remember through the shape of the graph at the critical point. A critical point sits at the bottom of a valley when the curve around it bends upward, which is exactly concave up, $f''(c) > 0$, and that bottom is a relative minimum. The same point sits at the top of a hill when the curve bends downward, concave down, $f''(c) < 0$, and that top is a relative maximum. Because the second derivative reports concavity, evaluating it at a horizontal-tangent critical point tells you immediately which way the curve cups, and therefore whether you are at a low point or a high point. Holding this valley-and-hill image alongside the algebra keeps the sign-to-conclusion direction correct: concave up cups upward to a minimum, concave down cups downward to a maximum. When $f''(c)$ is exactly zero the curve is momentarily flat in its bending too, which is why no valley or hill is guaranteed and the test cannot decide.