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The Hessian collects second partial derivatives and can classify critical points of a function of two variables under suitable conditions.
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Formula
Each Hessian entry has units of function value divided by the units of its two differentiation variables.
| Symbol | Meaning | Unit |
|---|---|---|
| Matrix of second partial derivatives of f | unit of f / (unit of xi · unit of xj) per entry | |
| fxx, fyy | Curvature in the x and y directions | unit of f / unit of x2 or f / unit of y2 |
| Mixed second derivative | unit of f / | |
| Determinant of the 2×2 Hessian | product of diagonal-entry units minus cross-term unit |
H_f(x,y)=\begin{pmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{pmatrix}, \quad D=f_{xx}f_{yy}-f_{xy}^2At a critical point where ∇f=0, D=fxx·fyy−fxy2 gives the two-variable second-derivative test: D>0 and fxx>0 indicate a local minimum; D>0 and fxx<0 indicate a local maximum; D<0 indicates a saddle point. If D=0, the test is inconclusive. The Hessian is symmetric when the mixed second partial derivatives are continuous. This is a local test: it does not prove a global extremum and does not handle constraints by itself.
The 2×2 test applies at a critical point of a twice continuously differentiable f; D=0 gives no classification.
The quadratic Taylor approximation near a critical point is governed by the Hessian.
Compute the Hessian determinant
For continuous mixed partials, fxy=fyx.
Classify the origin for f=x2+2xy+3y2.
∇f=0 at the origin.
H=[[2,2],[2,6]], D=8>0 and fxx=2>0: local minimum.
Also f=(x+y)2+2y2≥0.
Applying the test before checking ∇f=0.
Classify only stationary points; inspect boundaries and constraints separately.
Formula cluster
The gradient finds candidates; the Hessian evaluates local curvature.
For f=x2+2xy+3y2, the critical point is (0,0) and H=[[2,2],[2,6]]. Thus D=2·6−22=8>0 and fxx=2>0, so the point is a local minimum. Check: f=(x+y)2+2y2≥0.
Classify critical points in optimization, model fitting, and multivariable calculus as local minima, maxima, or saddle points.
Here is how to work through a typical Hessian matrix and second-derivative test (Hf = [[fxx, fxy], [fyx, fyy]]; D = det(Hf)) task step by step:
Task
Classify the origin for f=x2+2xy+3y2.
Solution path
∇f=0 at the origin.
H=[[2,2],[2,6]], D=8>0 and fxx=2>0: local minimum.
Also f=(x+y)2+2y2≥0.
First check that the gradient vanishes. For two variables, compute D=fxx·fyy−fxy2. If D is positive and fxx is positive, the point is a local minimum; if D is positive and fxx is negative, it is a local maximum. A negative D identifies a saddle point because the function curves differently in different directions. A positive D alone therefore cannot distinguish a minimum from a maximum. The test describes behavior only near the critical point, not global extrema. When D<0, the curvature eigenvalues have opposite signs, which is the geometric reason for a saddle.
When D=0, the quadratic approximation is degenerate in at least one direction. Second derivatives then do not determine the sign of the function change in every direction. A minimum, maximum or saddle point may all be possible. For example, f=x4+y4 has a minimum at the origin, while f=x2−y4 has a saddle there; both can have a zero Hessian determinant. Higher derivatives, a direct inequality or another local analysis is then needed. The right conclusion is not “there is no extremum,” but “this test does not decide.”
The second-derivative test classifies curvature around a stationary point. If ∇f is not zero, there is already a first-order direction of change, so quadratic curvature does not determine whether the point is an extremum. First solve fx=0 and fy=0, then evaluate the Hessian only at those candidates. Even after that, the analysis is not complete: boundaries, restricted domains and constraints require separate methods. To establish a global minimum, compare the function over the entire feasible region as well. Boundary points and stationary points under constraints require separate checks.
For a symmetric Hessian, the eigenvalues are the principal curvatures, and their product is the determinant D. If D is negative, the eigenvalues have opposite signs, giving both upward- and downward-curving directions: the point is a saddle. If D is positive, they share a sign; in the two-variable test, fxx determines whether both are positive or both negative. For f=x2+2xy+3y2 at the origin, H=[[2,2],[2,6]], D=8, and the eigenvalues are 4±2√2, both positive. Together with ∇f=0, this gives a local minimum. It does not replace boundary or constraint checks.
No. The Hessian examines curvature only near the stationary point. A function can have a local minimum there and take smaller values elsewhere. For a global minimum on a closed bounded domain, compare all interior critical points with boundary values; constrained problems may require Lagrange multipliers. On unbounded domains, also inspect behavior as variables grow large. A positive-definite Hessian at the point shows strict local upward curvature but gives no control over distant regions. State the result of the test explicitly as local.
Quanta exam set
Curated exam set for "Hessian matrix and second-derivative test":
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Which formula describes Hessian matrix and second-derivative test?
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How do you rearrange Hessian matrix and second-derivative test for Compute the Hessian determinant?
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Which common mistake happens with Hessian matrix and second-derivative test?
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