Introduction
A function can have a local maximum or minimum where its graph changes direction. At a differentiable interior point where such an extremum occurs, the tangent is horizontal, so the derivative is zero. This identifies a point to investigate, but it does not by itself say whether the point is a maximum, a minimum, or neither.
The Second Derivative Test uses the function’s curvature at a stationary point to make that classification. Its basic rule is concise: a positive second derivative indicates a local minimum, while a negative second derivative indicates a local maximum. The conditions matter: the point must be stationary, and the second derivative must be nonzero.
Local Extrema and Stationary Points
A point \(c\) in the domain of \(f\) is a local minimum point if there is an open interval around \(c\) such that \(f(x)\geq f(c)\) for every \(x\) in that interval and in the domain. It is a strict local minimum point if the inequality is strict whenever \(x\ne c\) in that interval. A local maximum is defined by reversing the inequality; it is strict when \(f(x)<f(c)\) for nearby \(x\ne c\).
A point \(c\) is a stationary point of a differentiable function if \(f'(c)=0\). The Second Derivative Test classifies stationary points; it is not a procedure for classifying every possible local extremum. In particular, its stated rule requires the second derivative at the stationary point to be nonzero.
- If \(f''(c)>0\), \(f\) has a strict local minimum at \(c\).
- If \(f''(c)<0\), \(f\) has a strict local maximum at \(c\).
Continuity of \(f''\) at \(c\) is a convenient sufficient condition in this version of the test. It ensures that the sign found at \(c\) persists in a sufficiently small interval around \(c\). The key conclusions concern local behavior near \(c\), not the largest or smallest value on the function’s entire domain.
How to Apply the Test
Apply the test only after checking that the point is stationary. Then evaluate the second derivative at that point, rather than deciding from the second-derivative formula alone. If the value is positive or negative, the corresponding conclusion follows from the theorem.
Verify that \(f'(c)=0\).
Calculate \(f''(c)\), making the substitution for \(c\) explicit.
If \(f''(c)>0\), conclude strict local minimum; if \(f''(c)<0\), conclude strict local maximum.
Give the extremum point \(c\), and, when useful, its graph coordinate \((c,f(c))\).
Worked Examples
Worked Example: A Quadratic with One Stationary Point
Classify the stationary point of \(f(x)=x^2-6x+11\).
Differentiate to find the stationary point:
Set the derivative equal to zero and solve:
Differentiate again and evaluate at \(x=3\):
The Second Derivative Test gives a strict local minimum at \(x=3\). Its function value is
Thus the minimum point is \((3,2)\). To check the classification directly, write \(x=3+h\). Then
For every nonzero \(h\), \(h^2>0\), so \(f(3+h)>2=f(3)\). This confirms the strict local minimum.
Worked Example: Classifying Two Stationary Points
Find and classify the stationary points of \(f(x)=x^3-3x\).
First differentiate and solve \(f'(x)=0\):
Thus the stationary points occur at \(x=1\) and \(x=-1\). The second derivative is
At \(x=-1\), \(f''(-1)=-6<0\), so the test gives a strict local maximum. The function value is
At \(x=1\), \(f''(1)=6>0\), so the test gives a strict local minimum. Its value is
For a direct check at both points, let \(x=-1+h\) and \(x=1+h\), respectively. Expansion gives
When \(0<|h|<1\), \(h-3<0\) and \(3+h>0\). Since \(h^2>0\), the first difference is negative and the second is positive. Nearby values are therefore below \(f(-1)\) and above \(f(1)\), confirming the maximum at \((-1,2)\) and minimum at \((1,-2)\).
Worked Example: A Quartic with Three Stationary Points
Classify all stationary points of \(f(x)=x^4-4x^2\).
Differentiate and factor:
Solving \(f'(x)=0\) gives \(x=0\) and \(x=\pm\sqrt{2}\). The second derivative is
At \(x=0\), \(f''(0)=-8<0\), so there is a strict local maximum. The value is \(f(0)=0\). At \(x=\sqrt{2}\) and \(x=-\sqrt{2}\),
Both points are strict local minima. Their function values are the same:
These classifications can also be checked algebraically. At \(x=0\), \(f(x)-f(0)=x^2(x^2-4)<0\) whenever \(0<|x|<2\), confirming a strict local maximum. At either minimum point,
This is positive for \(x\) sufficiently close to either \(\sqrt{2}\) or \(-\sqrt{2}\), except at the point itself. Thus both are strict local minima. The results are \((0,0)\) as a local maximum and \((-\sqrt{2},-4)\), \((\sqrt{2},-4)\) as local minima.
Why the Sign Gives the Classification
The connection between the test and the graph is concavity. When the second derivative is positive on an interval, the function is strictly convex there; when it is negative, the function is strictly concave. The tangent-line position for concave functions gives a useful justification for the test.
For example, suppose \(f'(c)=0\), \(f''(c)>0\), and the hypotheses in the theorem hold. By continuity of \(f''\), it remains positive in some interval around \(c\). The second-derivative concavity criterion therefore makes \(f\) strictly convex on a sufficiently small interval. For a strictly convex function, the graph lies strictly above its tangent line at \(c\), except at the point of tangency. That tangent line is horizontal because \(f'(c)=0\), so its height is \(f(c)\). Consequently, \(f(x)>f(c)\) for nearby \(x\ne c\): \(c\) is a strict local minimum.
If \(f''(c)<0\), continuity instead gives a nearby interval of strict concavity. The graph lies strictly below its horizontal tangent line there, giving \(f(x)<f(c)\) for nearby \(x\ne c\). Hence \(c\) is a strict local maximum. This explains why both ingredients matter: stationarity makes the tangent horizontal, and the sign of curvature determines which side of that tangent contains the graph.
A Common Misuse
A positive second derivative by itself does not license the conclusion “local minimum.” The test’s first condition is \(f'(c)=0\). For instance, consider \(g(x)=x^2+2x\) at \(x=0\). Although \(g''(0)=2>0\),
So \(x=0\) is not a stationary point, and the Second Derivative Test does not apply there. In fact, \(g(0)=0\), while \(g(-0.1)=(-0.1)^2+2(-0.1)=0.01-0.2=-0.19<0\). Thus values just to the left of \(0\) are below \(g(0)\), which rules out a local minimum at \(0\). The correct practice is to check stationarity before interpreting \(f''(c)\).
Also distinguish a local classification from a global one. A strict local minimum is lower than values sufficiently close to it; this does not establish that it is lower than every value on the whole domain. Global extrema require a broader comparison.
Summary
At a stationary point \(c\), the Second Derivative Test uses the value \(f''(c)\) to classify local behavior:
| Conditions at \(c\) | Conclusion |
|---|---|
| \(f'(c)=0\) and \(f''(c)>0\) | Strict local minimum |
| \(f'(c)=0\) and \(f''(c)<0\) | Strict local maximum |
Always find the stationary point, evaluate the second derivative there, and state the conclusion as local. The test applies when the second derivative is nonzero and its required regularity conditions hold.
Check Your Understanding
Answer these questions using the test and its conditions.
- What two derivative facts must hold at \(c\) for the stated Second Derivative Test to classify a point?
- If \(f'(c)=0\) and \(f''(c)=5\), what kind of local extremum occurs at \(c\)?
- If \(f'(c)=0\) and \(f''(c)=-3\), what kind of local extremum occurs at \(c\)?
- For \(p(x)=x^3-6x^2+9x\), find its stationary points and use the test to classify each one.
- Why does \(g''(c)>0\) alone not justify concluding that \(g\) has a local minimum at \(c\)?
