Second Derivative Test

Classifying Critical Points with Second-Order Information

At a critical point, the gradient vanishes, so the first-order approximation does not distinguish a local minimum from a local maximum or a saddle point. The Hessian supplies the next-order information: its quadratic form describes the leading change in the function along small displacements. The Second-Order Necessary Condition for a Local Extremum, introduced earlier in this course, says that the Hessian quadratic form must be nonnegative at a local minimum and nonpositive at a local maximum. The Second Derivative Test strengthens this observation when the quadratic form has a strict sign in every nonzero direction.

In this tutorial, assume that \(f:U\to\mathbb{R}\), where \(U\subseteq\mathbb{R}^n\) is open, has continuous second partial derivatives near a point \(a\in U\). This regularity ensures that the Hessian is symmetric and that the second-order Taylor formula applies. The test concerns critical points, meaning points where \(\nabla f(a)=0\).

Definiteness of the Hessian

The Hessian acts on a direction \(v\) through the scalar \(v^T H_f(a)v\). Its sign in every direction, rather than the sign of any one entry, determines whether the second-order change consistently points upward, downward, or in different directions.

The word “definite” matters: positive definiteness excludes zero values in every nonzero direction, while positive semidefiniteness allows them. The same distinction applies to negative definite and negative semidefinite matrices. The strict classification in the test below relies on definiteness.

Proof. By the Taylor Formula with Peano Remainder from earlier in the course, and because \(\nabla f(a)=0\), for \(h\) sufficiently small we have

First suppose \(H_f(a)\) is positive definite. On the unit sphere, \(u\mapsto u^TH_f(a)u\) is continuous and positive. There is a constant \(c>0\) such that \(u^TH_f(a)u\ge c\) for every unit vector \(u\). To see why the lower bound is strictly positive, if no such \(c\) existed, there would be unit vectors \(u_k\) with \(u_k^TH_f(a)u_k\to0\). The unit sphere is compact by the Heine–Borel Theorem, so a subsequence would converge to a unit vector \(u\). Continuity would give \(u^TH_f(a)u=0\), contradicting positive definiteness.

For any nonzero \(h\), apply this bound to \(u=h/\|h\|_2\). It follows that \(h^TH_f(a)h\ge c\|h\|_2^2\). Because \(r(h)=o(\|h\|_2^2)\), there is a radius \(\delta>0\) such that \(0<\|h\|_2<\delta\) implies \(|r(h)|<(c/4)\|h\|_2^2\). Therefore

Thus \(f(a+h)>f(a)\) for all sufficiently small nonzero \(h\), proving that \(a\) is a strict local minimum. If \(H_f(a)\) is negative definite, apply the positive-definite argument to \(-f\), whose Hessian is \(-H_f(a)\). This proves that \(a\) is a strict local maximum of \(f\).

Finally, suppose \(H_f(a)\) is indefinite. Choose nonzero directions \(u\) and \(v\) with \(u^TH_f(a)u>0\) and \(v^TH_f(a)v<0\). Along the first line, Taylor’s formula gives

for all sufficiently small nonzero \(t\). Along the second line the corresponding expression is negative for all sufficiently small nonzero \(t\). Every neighborhood of \(a\) therefore contains points where \(f\) is above \(f(a)\) and points where it is below \(f(a)\). Hence \(a\) is neither a local minimum nor a local maximum. \(\square\)

Worked Examples in Two Variables

For \(f(x,y)\), write the symmetric Hessian at a critical point as \(H=\begin{pmatrix}A&B\\B&C\end{pmatrix}\), and let \(D=AC-B^2\). When \(A\ne0\), its quadratic form can be written as

This identity gives a useful two-variable version of the test. If \(D>0\), then \(A\ne0\), and both terms have the sign of \(A\), so the Hessian is positive definite when \(A>0\) and negative definite when \(A<0\). If \(D<0\) and \(A\ne0\), the two coefficients \(A\) and \(D/A\) have opposite signs, so the quadratic form takes both signs. If \(A=0\), then \(D=-B^2<0\) forces \(B\ne0\), and the form \(2Bxy+Cy^2\) takes both signs by setting \(y=1\) and taking \(x\) sufficiently large with either sign. Thus \(D<0\) gives an indefinite Hessian. When \(D=0\), this determinant test does not establish definiteness.

When the Test Is Inconclusive

A zero determinant in two variables does not mean that the point is not an extremum. It means that this second-order test has not decided the question. A direction in which the Hessian quadratic form is zero may leave the behavior to higher-order terms. The following functions have the same Hessian at the origin, yet behave differently there.

At a local minimum, the Second-Order Necessary Condition for a Local Extremum requires \(v^TH_f(a)v\ge0\) for every \(v\); at a local maximum, it requires \(v^TH_f(a)v\le0\) for every \(v\). Consequently, an indefinite Hessian rules out both kinds of local extremum, while a semidefinite Hessian is compatible with an extremum but does not guarantee one. Do not confuse these necessary conditions with the sufficient conclusions of the Second Derivative Test.

For a practical classification, first verify that the point is critical and that the regularity assumptions hold. Then evaluate the Hessian quadratic form, or in two variables use its determinant and the sign of \(f_{xx}\). A definite Hessian gives a strict extremum; an indefinite Hessian gives neither type of extremum; a degenerate, semidefinite case calls for further analysis, such as examining higher-order terms or comparing function values directly.