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Concavity & the Second Derivative Test

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Concavity is the sign of the second derivative

Concavity describes how a graph bends. Where f″ > 0 the graph is concave up (opening upward, like a cup holding water); where f″ < 0 it is concave down (opening downward). Equivalently, concave up means the slope f′ is increasing, concave down means the slope is decreasing. Concavity is a statement about the second derivative, independent of whether the function itself is rising or falling.

Inflection points

An inflection point is where concavity changes — from up to down or down to up. It requires f″ to change sign, which usually happens where f″ = 0 (or is undefined), but f″ = 0 alone is not enough: the sign must actually flip. At an inflection point the graph switches which way it curves, and the tangent line crosses the curve.

The Second Derivative Test

At a critical point where f′(c) = 0, the Second Derivative Test classifies it quickly: if f″(c) > 0 the curve is concave up there, so c is a local minimum; if f″(c) < 0 it is concave down, so c is a local maximum. If f″(c) = 0 the test is inconclusive and you must fall back on the First Derivative Test.

Concavity and the Second Derivative Test
f″ > 0 ⇒ concave up (local min if f′ = 0); f″ < 0 ⇒ concave down (local max if f′ = 0)
Inflection point where f″ changes sign. If f″(c) = 0 at a critical point, the test is inconclusive.
Worked example

For f(x) = x³ − 3x, find the inflection point and classify the critical point at x = 1 with the Second Derivative Test.

  1. 1.First derivative: f′(x) = 3x² − 3, zero at x = ±1 (critical points).
  2. 2.Second derivative: f″(x) = 6x.
  3. 3.Concavity changes where f″ = 0: at x = 0, f″ goes from negative (x < 0) to positive (x > 0), so (0, 0) is an inflection point.
  4. 4.Apply the test at x = 1: f″(1) = 6 > 0, concave up, so x = 1 is a local minimum.
Answer: Inflection point at (0, 0); x = 1 is a local minimum because f″(1) = 6 > 0.
Watch out

f″ = 0 does not by itself make an inflection point — the concavity must actually change sign. For y = x⁴, f″(0) = 0, yet the graph stays concave up on both sides, so x = 0 is not an inflection point.

Checkpoint

If f″(x) > 0 on an interval, the graph of f is:

Checkpoint

At a critical point c, f′(c) = 0 and f″(c) = −5. What does the Second Derivative Test conclude?

On the exam

When the Second Derivative Test returns f″(c) = 0, do not guess — state that the test is inconclusive and switch to the First Derivative Test, examining the sign change of f′ around c to classify the point.

Answer the 2 checkpoints as you read.

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