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At the origin the gradient vanishes and the Hessian matrix is $\mathrm{diag}(0,2)$, which is positive semidefinite. Nevertheless, $F(t,0)=t^3$ takes both signs arbitrarily close to the origin. Thus the origin is a saddle point of a scalar function. This example shows why a zero eigenvalue prevents the ordinary second-derivative test from establishing a local minimum.

A critical point whose Hessian matrix is a singular matrix. The second-derivative test does not completely classify such a point. Higher-order terms can produce a local minimum, a local maximum, or a saddle point of a scalar function.

Define $f(0,0)=0$ and, elsewhere,All first and second partial derivatives exist at the origin: $f_x=f_y=f_{xy}=f_{yx}=0$ and $f_{xx}=f_{yy}=2$. Thus its Hessian matrix there is the positive-definite matrix $2I$. Nevertheless $f(t,0)=t^2>0$ and $f(t,t)=-t^2<0$ for $t\ne 0$, so the origin is a saddle point of a scalar function. A valid second-derivative test needs a second-order Taylor expansion with a remainder $o(x^2+y^2)$; continuous second partial derivatives in a neighborhood are a sufficient hypothesis. Existence of those derivatives at the point alone does not supply this hypothesis.

The determinant of the Hessian matrix. For a twice continuously differentiable function of two variables, it is $f_{xx}{f}_{yy}-f_{xy}^2$. A negative value at a critical point implies a saddle point; a positive value combines with the sign of $f_{xx}$ in the second-derivative test. A zero value is inconclusive.

The Hessian matrix describes how the slope of a function changes in different directions. For a real-valued function $f$ with continuous second partial derivatives on an open subset of ${\mathbb{R}}^n$, it isThe symmetry of second derivatives makes $H_f$ a symmetric matrix under these hypotheses. It is also the Jacobian matrix of the gradient: near $x$, a small displacement $h$ changes $\nabla f$ by $H_f(x)h+o(\Vert h\Vert )$. Thus the gradient measures the current slope, while the Hessian measures its local variation. ^[ref]

For example, take the quadratic formIts gradient and Hessian matrix areThe only critical point is the origin. Completing the square gives $q=(x+y{)}^2+2y^2$, so the origin is a strict global minimum. The eigenvalues $4-2\sqrt{2}$ and $4+2\sqrt{2}$ are both positive, recovering the same conclusion from the positive definiteness of the Hessian. Its eigenvectors identify the rotated directions in which the quadratic form has its smallest and largest bending.

For a fixed vector $v$, restriction to a straight line gives the directional interpretationThis is the quadratic form associated with the Hessian matrix. For a unit eigenvector $v$, it equals the corresponding eigenvalue. Opposite signs therefore mean that some directions bend upwards and others downwards. The second-derivative test turns this information into a classification when the gradient vanishes.

The same object is useful for computation. Newton's method in optimization chooses a displacement $h$ by solving $H_f(x)h=-\nabla f(x)$ when the Hessian matrix is an invertible matrix; this makes the gradient of the local quadratic model vanish. For the displayed $q$, the equation gives $h=-(x,y{)}^T$, so one exact step reaches the minimum from any starting point. For a general function, a Hessian with opposite-sign eigenvalues or a large condition number need not give a useful minimizing step, and the local minimum need not be a global minimum.

The name honors Otto Hesse. ^[ref] An older geometric use is the Hessian criterion for flexes of a plane cubic, connecting a determinant of second partial derivatives with inflection points of an algebraic curve. Hesse's 1844 paper Über die Wendepuncte der Curven dritter Ordnung is an early source on that geometric problem. ^[ref]

Multivariable calculus studies derivatives and integrals of functions with several independent variables. A temperature field $T(x,y,z)$, for example, has partial derivatives describing coordinate-wise changes, while its gradient gives the direction and rate of steepest increase. The Fréchet derivative combines directional changes into one linear map; its coordinate representation is the Jacobian matrix.

For $f(x,y)=x^2+y^2$, the gradient is $(2x,2y)$ and the directional derivative along a unit vector $v$ is $\nabla f\cdot v$. The Hessian matrix describes second-order variation and, under the hypotheses of the second-derivative test, distinguishes a local minimum from a saddle point. Constrained variation leads to Lagrange multipliers and the implicit function theorem.

Multiple integrals accumulate quantities over regions, as when integrating a mass density to obtain total mass. A change of variables formula uses the absolute Jacobian determinant to account for stretching of volume. For vector fields, the fundamental theorem for line integrals, Stokes theorem and divergence theorem relate interior derivatives to boundary measurements. They extend the endpoint difference of the ordinary fundamental theorem of calculus to circulation and flux.

At a smooth interior stationary point, define $D=t_{11}{t}_{22}-t_{12}^2$. The second-derivative test gives a minimum for $D>0,t_{11}>0$, a maximum for $D>0,t_{11}<0$, and a saddle point for $D<0$; $D=0$ is inconclusive.

For the stated time function, putPositive physical speeds make all three coefficients positive. The positive Hessian of a layered travel-time functional here isThus the stationary path is a strict minimum, not merely a candidate of unspecified type. In fact the same positivity holds for all finite crossing positions, so the time function is strictly convex and the minimum is unique.

For a differentiable function at an interior stationary point, both first partial derivatives vanish. If the second derivatives are continuous, form the symmetric Hessian matrixThe second-derivative test gives a strict local minimum if $D>0$ and $f_{xx}>0$, a strict local maximum if $D>0$ and $f_{xx}<0$, and a saddle point if $D<0$. If $D=0$, higher-order terms must be examined. For example, $x^4+y^4$, $-x^4-y^4$ and $x^4-y^4$ all have zero Hessian matrix at the origin but respectively a minimum, maximum and saddle. These are local interior criteria; boundary extrema require separate treatment.

For a twice continuously differentiable function $F$, first check the stationary-point condition $F_x=F_y=0$. Write its Hessian matrix as $H=\left(\begin{array}{cc}A& B\\ B& C\end{array}\right)$, with $A=F_{xx}$, $B=F_{xy}$ and $C=F_{yy}$. The second-derivative test uses $D=AC-B^2$: if $D>0$ and $A>0$, the point is a strict local minimum; if $D>0$ and $A<0$, it is a strict local maximum; if $D<0$, it is a saddle point. If $D=0$, the quadratic test is inconclusive and higher-order terms or a direct argument are needed.

For $F=e^{y^2-x^2}$, the first derivatives vanish at the origin andThus $\boxed{(0,0)\text{ is a saddle point, with }F(0,0)=1}$. Directly, $F(x,0)=e^{-x^2}<1$ for nearby nonzero $x$, while $F(0,y)=e^{y^2}>1$ for nearby nonzero $y$. The presence of both increasing and decreasing directions confirms the Hessian matrix classification.

The gradient vanishes when $3x^2-2=3y^2-2=0$. Put $s=\sqrt{2/3}$. The Hessian matrix is $\mathrm{diag}(6x,6y)$, so the second-derivative test gives:

The positive-definite and negative definite cases give small closed level curves around the minimum and maximum. At each saddle, the zero contour crosses itself. In fact,so the entire zero level curve is the union of the line $y=-x$ and the ellipse $x^2-xy+y^2=2$. This factorization fixes the topology of a rough contour sketch. Neither local extremum is global, since the cubic is unbounded above and below.

At a stationary point, the linear terms vanish, so $\Delta f\simeq \frac{1}{2}(A{h}^2+2Bhk+C{k}^2)$, with $A=f_{xx}$, $B=f_{xy}$, $C=f_{yy}$. Let $D=AC-B^2$ be the Hessian determinant. For $A\ne 0$, completing the square givesThe second-derivative test therefore yieldsWhen $D<0$ the quadratic form is indefinite, including the case $A=0$, $B\ne 0$. When $D=0$, the quadratic approximation is degenerate and the test is inconclusive: higher-order terms may produce a minimum, maximum or saddle. For nondegenerate cases the quadratic term dominates the Taylor remainder sufficiently close to the point.

An interior stationary point has $f_x=f_y=0$. Assuming twice continuous differentiability nearby, the two-dimensional Taylor series about it iswhere $A=f_{xx}$, $B=f_{xy}$ and $C=f_{yy}$. Let $D=AC-B^2$. The second-derivative test gives a strict local minimum if $D>0,A>0$, a strict local maximum if $D>0,A<0$, and a saddle point if $D<0$.

For example, if $A\ne 0$, the quadratic form is $A(h_x+B{h}_y/A{)}^2+(D/A)h_y^2$, which proves these sign tests. If $D=0$, the quadratic part is inconclusive: $x^4+y^4$ has a minimum while $x^4-y^4$ has a saddle at the origin, although both have zero Hessian matrix there. Higher-order terms or direct inequalities must then be examined.

Writing $r=\sqrt{x^2+y^2}$, the absolute lower bound is $\cos r\ge -1$, attained exactly whenEvery point on these circles is a global and local minimum, but not a strict minimum: neighbouring points on the same circle have exactly the same value. At such a point the radial Hessian eigenvalue is one and the tangential eigenvalue is zero, illustrating why a degenerate second-derivative test does not exclude a minimum.

There are no other local minima: between successive multiples of $\pi$, moving radially changes the value monotonically. The positive even-radius circles are maxima. At the origin the apparently singular radial expression is smooth, since $\cos \sqrt{x^2+y^2}=1-(x^2+y^2)/2+O(r^4)$, and the origin is a strict local maximum.

At a critical point $p$ of a twice continuously differentiable real function, $\nabla f(p)=0$ and the Taylor expansion isThe real symmetric Hessian matrix has an orthonormal eigenbasis, so writing $h={\sum }_i{a}_i{u}_i$ gives $h^{T}Hh={\sum }_i{\lambda }_i{a}_i^2$. If every eigenvalue is positive, this quadratic form is at least ${\lambda }_{\min }|h{|}^2$, and the remainder cannot change its sign for sufficiently small nonzero $h$. Thus $p$ is a strict local minimum. If every eigenvalue is negative, the analogous argument gives a strict local maximum. If positive and negative eigenvalues both occur, displacement along the corresponding eigenvectors gives values above and below $f(p)$, so $p$ is a saddle point of a scalar function.

The second-derivative test can be inconclusive when the Hessian matrix is singular and semidefinite: higher-order terms along directions in the kernel of a linear map matter. For example $x^2+y^4$ and $x^2-y^4$ have the same Hessian $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle point of a scalar function. A zero eigenvalue does not prevent a saddle conclusion if other eigenvalues already have both signs.

For the given function write $r^2=x^2+y^2$. Its first derivatives areThe origin is a critical point for every real $\alpha$, withIts eigenvalues are $1,-1$, so it is always a saddle point of a scalar function. If one coordinate at a critical point is zero, the other must also be zero. If both are nonzero, the derivative equations force $x^2=y^2=1/(2\alpha )$, possible only for $\alpha >0$. Hence for $\alpha \le 0$ the origin is the only critical point.

For $\alpha >0$ there are four additional critical points,At such a point,Thus the two equal-sign points are strict maxima, and the two opposite-sign points are strict minima. They are global: $2|xy|\le r^2$ implieswhere the last maximum occurs at $r^2=1/\alpha$ and equality also requires $x^2=y^2$.

As $\alpha \downarrow 0$, the four nonzero critical points have distance ${\alpha }^{-1/2}$ from the origin and escape to infinity, rather than merging with the origin. Their values diverge as $\pm 1/(2e\alpha )$. As $\alpha \uparrow 0$ from below there are no extra critical points, while the origin stays a saddle throughout. This is critical points escaping to infinity in a Gaussian-weighted bilinear function.

Suppose $u$ were positive somewhere. Since it tends to zero at infinity and to $-\infty$ at its vortex zero, it would attain a positive interior maximum away from the origin. At such a maximum the second-derivative test gives $\Delta u\le 0$, whereas the smooth Taubes equation givesThis contradiction with the maximum principle for subharmonic functions provesConsequently the Higgs magnitude of a vortex satisfies $|\varphi |=e^{u/2}\le 1$ everywhere.

The gradient of the given polynomial isIts second component vanishes when $y=0$ or $x=-1$. The first component then gives exactly three critical points:The Hessian matrix and its determinant areAt $(-1/2,0)$ the Hessian matrix is $\mathrm{diag}(2,1)$, whose positive definiteness proves a strict local minimum by the second-derivative test. At $(-1,\pm 1)$ its determinant is $-4$, so both points are saddle points of a scalar function. Their values and classifications areThere are no local maxima.

For a twice continuously differentiable function, an interior local extremum requires $f_x=f_y=0$. At such a critical point, put $D=f_{xx}{f}_{yy}-f_{xy}^2$. The second-derivative test gives a strict local minimum if $D>0$ and $f_{xx}>0$, a strict local maximum if $D>0$ and $f_{xx}<0$, and a saddle point if $D<0$. When $D=0$, the test is inconclusive. More generally the Hessian matrix is necessarily positive semidefinite at a local minimum and negative semidefinite at a local maximum, but these non-strict signs alone are not sufficient.

For this polynomial,Subtracting the two stationarity equations gives $y(2x-y)=0$. If $y=0$, then $x=0$; if $y=2x$, the remaining equation gives $x=0$ or $x=3/2$. Thus the only critical points are $(0,0)$ and $(3/2,3)$.

At $(3/2,3)$, $\det H=-18<0$, so it is a saddle point. At $(0,0)$ the determinant is zero, but along $y=-x$ the value is $-x^3$, which takes both signs arbitrarily near zero. ThereforeIn particular a vanishing Hessian determinant does not exclude a saddle point.

Write $A=a^2>0$ and $E=e^{-(x^2+y^2)/A}>0$. The partial derivatives factor asA critical point has both partial derivatives zero. If $x,y\ne 0$, this would require $x^2-2y^2=A$ and $x^2-2y^2=-2A$, which is impossible. On the axes the remaining conditions giveAt either critical point with $x\ne 0$, the Hessian matrix isIt is negative definite, so the second-derivative test shows that both $(\pm |a|,0)$ are strict local maxima, of value $a^2/e$. They also attain the global maximum: $f\le x^2{e}^{-x^2/A}\le A/e$. For completeness, the Hessian matrix at $(0,\pm |a|)$ is $\mathrm{diag}(6/e,8/e)$, giving strict local minima, while at the origin it is $\mathrm{diag}(2,-4)$, giving a saddle point.

At an interior local extremum of a differentiable function, $f^{\prime }(x)=0$. The second-derivative test classifies a stationary point: a positive second derivative gives a strict local minimum, and a negative second derivative gives a strict local maximum. A zero second derivative is inconclusive: $x^4$, $-x^4$, and $x^3$ at zero illustrate a minimum, a maximum, and neither.

HereThus $f^{\prime \prime }(1/3)=-8$ and $f^{\prime \prime }(3)=8$. Substitution into $f$ gives

Set both first partial derivatives equal to zero:Thus $x=0,\pm 1/\sqrt{2}$ and $y=0,\pm \sqrt{2}$, independently, giving exactly nine stationary points. The Hessian matrix is diagonal everywhere:Its diagonal entries are its eigenvalues, so the second-derivative test immediately classifies the points:

In particular,All nine stationary points are nondegenerate, so no higher-order test is needed. Neither type of local extremum is global: $f(x,0)=x^4-x^2\to +\infty$ as $|x|\to \infty$, whereas $f(0,y)=-y^4+4y^2\to -\infty$ as $|y|\to \infty$.

To organize the contour plot, complete the squares:The level set through all four outer saddle points of a scalar function factors exactly:It is therefore the union of the circle $x^2+y^2=5/2$ and the hyperbola $y^2-x^2=3/2$, which cross at those four saddle points of a scalar function. Near the origin, $f=-x^2+4y^2+O((x^2+y^2{)}^2)$, so the zero contours have tangents $y=\pm x/2$. These two exact structural features, together with the nested contours around the four local extrema, determine the requested sketch. They are an example of a separable quartic landscape with nine stationary points.