Critical Points and the Hessian

First-order silence does not decide the shape

At an interior local extremum of a differentiable function, the gradient is zero. Such stationary points are candidates, not automatic maxima or minima. Critical points can also include places where derivatives fail to exist; the smooth tests below do not apply there without further analysis.

For a function with continuous second partials near a stationary point, form the Hessian $H$ and $D=f_{xx}f_{yy}-f_{xy}^2$ . If $D>0$ and $f_{xx}>0$ , there is a strict local minimum. If $D>0$ and $f_{xx}<0$ , there is a strict local maximum. If $D<0$ , the point is a saddle. When $D=0$ , the test is inconclusive.

Visual guide

VISUAL GUIDEClosed contours surround a strict minimum

Level above the minimum 0.36
Level above the minimum 1.44
Level above the minimum 3.24

Contours of x² + 2xy + 3y² − 4x are centered at (3, −1). The positive-definite Hessian makes every small displacement increase the function. The tilted ellipses reflect the mixed xy term.

Worked example: a tilted bowl

For $f=x^2+2xy+3y^2-4x$ , stationarity gives $2x+2y-4=0$ and $2x+6y=0$ . Solving yields $(x,y)=(3,-1)$ . The Hessian is $\begin{pmatrix}2&2\\2&6\end{pmatrix}$ , with determinant $8>0$ and positive first diagonal entry. The point is a strict local minimum.

It is also the global minimum here because the quadratic Hessian is positive definite everywhere. That global conclusion uses the full quadratic structure, not just the local second-derivative test.

Worked example: same zero Hessian, different outcomes

At zero, both $f=x^4+y^4$ and $g=x^4-y^4$ have zero gradient and zero Hessian. The test gives $D=0$ for both. Directly, $f\ge0$ with equality only at zero, so it has a strict minimum. For $g$ , the $x$ -axis gives positive values and the $y$ -axis negative ones, so zero is a saddle.

In higher dimensions, positive definite Hessian implies a strict local minimum, negative definite a maximum, and an indefinite Hessian a saddle. Semidefinite cases require higher-order analysis. The Linear Algebra unit on quadratic forms explains why eigenvalue signs determine these local shapes.

Practice

Classify the origin for $x^2-y^2$ .
Classify the origin for $-x^2-2y^2$ .
Does a zero gradient guarantee an extremum?

Show worked solutions

$D=-4<0$ , so it is a saddle.
The Hessian is negative definite, giving a strict local maximum.
No. The first example has zero gradient but increases in one direction and decreases in another.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.