The Second Derivative Test

At a critical point where \(f'(c) = 0\), one number settles the question: the sign of \(f''(c)\). Positive means the curve cups upward around a local minimum; negative means it caps over a local maximum; zero means the test stays silent.

By the end you'll be able to classify a critical point by computing \(f''(c)\) and reading its sign — and know to fall back to the First Derivative Test when \(f''(c) = 0\) leaves the verdict open.

Predict: for \(f(x) = x^3 - 3x\) we have \(f''(x) = 6x\), so \(f''(1) = 6\). From that sign alone — is \(x = 1\) the top of a cap or the bottom of a cup? Toggle to x = 1 below and check the verdict.

Both critical points of \(f(x) = x^3 - 3x\) have a flat tangent — the first derivative can't tell them apart. The colored arc overlays the local concavity that \(f''(c)\) measures: a green cup when \(f''(c) > 0\), a red cap when \(f''(c) < 0\). The third button switches to \(g(x) = x^4\), where \(f'' = 0\) at the critical point and the test goes silent. Hover or focus any dot — or hover the curve — for exact values.

Critical point under test:

f″(−1) = −6 < 0 ⇒ local maximum at x = −1 — concave down: a cap.

The curve, the flat tangent at the critical point, and the concavity arc — the arc's color is the verdict
the function flat tangent (f′ = 0) cup: f″(c) > 0 ⇒ min cap: f″(c) < 0 ⇒ max

The Second Derivative Test classifies a critical point where \(f'(c) = 0\) by the sign of \(f''(c)\): positive gives a local minimum, negative a local maximum, and zero is inconclusive.

Formal

Suppose \(f'(c) = 0\) and \(f''\) exists at \(c\). Then: \(f''(c) > 0\) ⇒ \(f\) has a local minimum at \(c\); \(f''(c) < 0\) ⇒ \(f\) has a local maximum at \(c\); \(f''(c) = 0\) ⇒ the test is inconclusive. The logic comes straight from the concavity lesson: a positive second derivative means the curve is concave up near \(c\), so a point with a horizontal tangent must sit at the bottom of the cup. Negative \(f''(c)\) means concave down — the point sits on top of the cap. Two cautions: the test only applies where \(f'(c) = 0\) (never at points where \(f'\) is merely undefined, and never at non-critical points), and \(f''(c) = 0\) is not a verdict of "neither" — it means the test has nothing to say, so you switch to the First Derivative Test's sign chart.

The method, in three steps

  1. Solve \(f'(c) = 0\) to find the critical points.
  2. Compute \(f''\) and evaluate it at each critical point.
  3. Read off the verdict from the sign: \(+\) ⇒ min, \(-\) ⇒ max, \(0\) ⇒ switch to the First Derivative Test.

When it applies, this is usually faster than building a full sign chart of \(f'\): you evaluate one number per critical point instead of testing intervals on both sides.

Worked example

\(f(x) = x^3 - 3x\). Step 1: \(f'(x) = 3x^2 - 3 = 0\) gives \(x = \pm 1\). Step 2: \(f''(x) = 6x\), so \(f''(1) = 6\) and \(f''(-1) = -6\). Step 3: \(f''(1) = 6 > 0\) ⇒ local minimum at \(x = 1\) (value \(f(1) = -2\)); \(f''(-1) = -6 < 0\) ⇒ local maximum at \(x = -1\) (value \(f(-1) = 2\)).

Your turn

Same drill, new function: \(g(x) = x^3 - 27x\). Step 1: \(g'(x) = 3x^2 - 27 = 0\) gives \(x = \pm 3\). Step 2: \(g''(x) = 6x\), so \(g''(3) = 18 > 0\) ⇒ local minimum at \(x = 3\). Now finish it: \(g''(-3) = \) ____, so \(x = -3\) is a ____.

Reveal the answer

\(g''(-3) = 6(-3) = -18 < 0\), so the curve is concave down there and \(x = -3\) is a local maximum (value \(g(-3) = -27 + 81 = 54\)). Same sign map as the demo above: negative \(g''\) is the red cap.

More info — why the sign of f″ decides, and what to do when it can't

Near a critical point the tangent is flat, so the quadratic term takes over: \(f(x) \approx f(c) + \tfrac{1}{2}f''(c)(x-c)^2\). That approximating parabola is exactly the colored arc in the demo — it opens upward when \(f''(c) > 0\) (so nearby values are larger than \(f(c)\): a minimum) and downward when \(f''(c) < 0\). When \(f''(c) = 0\) the parabola degenerates to a flat line and carries no information — higher-order terms decide, which is why \(x^4\) (a minimum), \(-x^4\) (a maximum), and \(x^3\) (neither) all look identical to this test at \(0\). The rescue is the previous lesson's First Derivative Test: check the sign of \(f'\) on each side of \(c\) — it always gives an answer when \(f\) is differentiable nearby. The OpenStax chapter in Dive deeper below walks the same comparison.

Check your understanding

Question 1 of 4

Let \(h(x) = 2x^3 - 6x^2\). Then \(h'(x) = 6x^2 - 12x\) vanishes at \(x = 0\) and \(x = 2\), and \(h''(x) = 12x - 12\). What does the Second Derivative Test say about \(x = 0\)?

Question 2 of 4

At a critical point \(c\) you compute \(f'(c) = 0\) and \(f''(c) = 0\). What can you conclude from the Second Derivative Test alone?

Question 3 of 4

For some function \(g\) you find \(g''(5) = 7 > 0\), but also \(g'(5) = 4\). What does the Second Derivative Test tell you about \(x = 5\)?

Question 4 of 4

\(g(x) = x^4\) has \(g'(x) = 4x^3\) and \(g''(x) = 12x^2\), so \(g'(0) = 0\) and \(g''(0) = 0\). Which test settles what happens at \(x = 0\), and what is the verdict?

Recap

  • The test applies only at critical points with \(f'(c) = 0\) — never at points where \(f'\) is undefined or nonzero.
  • \(f''(c) > 0\) ⇒ concave up (cup) ⇒ local minimum at \(c\).
  • \(f''(c) < 0\) ⇒ concave down (cap) ⇒ local maximum at \(c\).
  • \(f''(c) = 0\) ⇒ inconclusive — not "neither"; fall back to the First Derivative Test's sign chart (e.g. \(x^4\) still has a local min at 0).
  • Method: solve \(f' = 0\), evaluate \(f''\) at each critical point, read the sign.

Dive deeper

Sources

  • The Second Derivative Test