Second Derivative Test for Stationary Points

What Are Stationary Points?

A stationary point on a curve y = f(x) is a point where the gradient is zero: f′(a) = 0. At such a point, the curve momentarily has a horizontal tangent — it is neither increasing nor decreasing at that instant. Stationary points may be local maxima, local minima, or stationary points of inflection (where the curve flattens but does not change direction).

Previously, we classified stationary points using the first derivative sign test: check the sign of f′ just to the left and just to the right of x = a, and look at how the sign changes. While reliable, this requires substituting two test values and interpreting the pattern. The second derivative test offers a more streamlined approach in many cases.

The Second Derivative Test: Logic and Reasoning

Recall that f″(a) measures the rate of change of the gradient at x = a.

Suppose x = a is a stationary point (f′(a) = 0) and f″(a) > 0. This means the gradient is increasing at x = a. Just to the left of a, the gradient was even lower (more negative), and just to the right it will be positive. So the gradient goes from negative to zero to positive — exactly the pattern for a local minimum. The curve is concave up at a (like a bowl), confirming it is a trough.

Conversely, if f″(a) < 0, the gradient is decreasing at x = a. Just left of a it was positive, just right it becomes negative — the pattern for a local maximum. The curve is concave down at a (like an arch), confirming it is a peak.

This is why the memory aid works: concave up (∪) means a local min; concave down (∩) means a local max.

The Inconclusive Case: When f″(a) = 0

The second derivative test fails when f″(a) = 0. In this situation, we cannot determine the nature of the stationary point from f″ alone — the point could be a local minimum, a local maximum, or a stationary inflection point.

When this occurs, revert to the first derivative sign test: evaluate the sign of f′(x) at a point slightly to the left of a and at a point slightly to the right of a.

• If f′ goes from negative to positive (− then +): local minimum
• If f′ goes from positive to negative (+ then −): local maximum
• If f′ does not change sign (+ then + or − then −): stationary inflection point

A classic example is f(x) = x⁴: f′(0) = 0, f″(0) = 0 (inconclusive). The first derivative test shows f′(−1) = −4 < 0 and f′(1) = 4 > 0, giving a sign change from − to +, so (0, 0) is a local minimum.

Another example is f(x) = x³: f′(0) = 0, f″(0) = 0 (inconclusive). f′(−1) = 3 > 0 and f′(1) = 3 > 0 — no sign change — so (0, 0) is a stationary inflection point.

Full Procedure for Classifying Stationary Points

Step 1: Differentiate f(x) to get f′(x). Solve f′(x) = 0 to find the x-coordinates of all stationary points.

Step 2: Find f″(x). Evaluate f″ at each stationary point x = a.

Step 3: Apply the second derivative test: f″(a) > 0 ⇒ min; f″(a) < 0 ⇒ max; f″(a) = 0 ⇒ use first derivative sign test.

Step 4: Calculate y = f(a) for each stationary point to get the full coordinates.

Step 5: State each stationary point with its coordinates and type (local max, local min, or stationary inflection point).

Combining First and Second Derivatives for Curve Sketching

A complete curve sketch uses information from both f′ and f″. From f′: find stationary points, intervals of increase/decrease. From f″: classify stationary points using the second derivative test, find inflection points, determine intervals of concavity. Together, these give you the shape of the curve in rich detail — the location and type of every turning point, the direction of every bend, and the exact coordinates of every key feature.