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In the Landau theory approximation, choose the global minimum of the Landau free energy as a function of the magnetization. Since ${\mathcal{A}}_6>0$, the polynomial is a coercive function and a global minimum exists. The candidates are stationary points satisfyingCheck their values of the free-energy density; a stationary point or a metastable local minimum alone need not be the equilibrium state. Positive curvature ${\mathcal{A}}_2+3{\mathcal{A}}_4{m}^2+5{\mathcal{A}}_6{m}^4$ ensures local stability; when it vanishes, higher terms must be checked. The selected global minimum is the equilibrium magnetization within this approximation.

At zero magnetic field, equal opposite minima describe distinct phases exhibiting spontaneous symmetry breaking. A spontaneous magnetization is obtained by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A symmetric finite-volume average can instead have zero expectation even when the minima are nonzero. Thus one must specify the phase when interpreting the minimizer as the equilibrium expectation.

Let $\rho =a/\sqrt{3}$ be the distance from the origin to each fixed charge, and denote their position vectors by ${\mathbf{R}}_j$. For a small planar displacement $\mathbf{r}$, the Taylor expansioncan be summed using the threefold symmetry relationsThe electric potential due to the fixed charges is consequentlyThe moving charge has electrostatic potential energy of point charges $U=q\Phi$. Because the quadratic coefficient is positive, the origin is a strict local minimum of $U$, and the equilibrium is stable.

At $p$, the surface $\tilde{S}$ cannot cross the boundary $S=\partial \mathcal{R}$, because it lies inside $\mathcal{R}$. Consequently the two surfaces are tangent at $p$. Apply a proper Euclidean motion of Euclidean three-space so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward unit normal is $N=e_3$. Locally writewith the region $\mathcal{R}$ on the side $z\ge f(x,y)$. ThenThus $g-f$ has a local minimum at zero and its Hessian matrix is positive semidefinite. The fundamental forms of a graph surface at a horizontal tangent plane giveTaking the trace of the positive-semidefinite difference proves the mean-curvature comparison at tangential contact:

There is no analogous conclusion for $K$. In the same local coordinates takeThen $g\ge f$, so the second graph lies on the inward side of the first, while the Gaussian curvature of a graph surface givesHence the mean-curvature inequality holds butUsing smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the counterexample under the global hypotheses.

$\nabla U=(3x^2-3+y^2,2y(x+1))$, so the critical points are $(\pm 1,0)$. The Hessian matrix at $(1,0)$ is $\mathrm{diag}(6,4)$, making it the local minimum; $(-1,0)$ is degenerate and not a minimum.

Fix $d$ strictly between $f^{\prime }(a)$ and $f^{\prime }(b)$, and set $F(x)=f(x)-dx$. The function $F$ is continuous on $[a,b]$: interior differentiability gives interior continuity, and the finite one-sided derivatives give continuity at the endpoints. The extreme value theorem therefore gives a point $c$ where $F$ attains its minimum.

Since $F_{+}^{\prime }(a)=f^{\prime }(a)-d<0$, for sufficiently small $h>0$ we have $F(a+h)<F(a)$, so $a$ cannot be a minimizing point. Since $F_{-}^{\prime }(b)=f^{\prime }(b)-d>0$, the quotient $(F(b-h)-F(b))/(-h)$ is positive for sufficiently small $h>0$, giving $F(b-h)<F(b)$. Thus $b$ cannot be a minimizing point either, and $c\in (a,b)$.

Fermat's theorem on stationary points states that a differentiable function at an interior local minimum or local maximum has derivative zero there. Applying it to $F$ givesThis proves the relevant case of Darboux's theorem; the argument uses continuity of $f$, with no assumption of continuity of $f^{\prime }$.

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 $g=(x-1{)}^2+y^2-29$, the gradient $\nabla g=2(x-1,y)$ never vanishes on the circle. The Lagrange multiplier equations areIf $y=0$, the two possibilities are $(1\pm \sqrt{29},0)$. If $y\ne 0$, then $\lambda =1+x$, and the first equation gives $y^2=2x^2-2x-3$. Combining this with $y^2=28+2x-x^2$ yieldsPutBoth $h_{-}$ and $h_{+}$ are positive, giving the other four points $(r_{-},\pm h_{-})$ and $(r_{+},\pm h_{+})$. To classify all six, restrict $f$ to the circle by eliminating $y^2$:Its derivatives are $H^{\prime }=-3x^2+4x+31$ and $H^{\prime \prime }=4-6x$. In the interior, $H^{\prime \prime }(r_{-})=2s>0$ and $H^{\prime \prime }(r_{+})=-2s<0$. At either endpoint $H^{\prime }<0$; along the circle, $x$ has a strict local minimum at the left endpoint and a strict local maximum at the right endpoint. Consequently the six constrained extrema areThe circle is a compact set, so the extreme value theorem guarantees that its global minimum and global maximum occur among these six points. Their values give the corrected sharp boundThe printed lower bound $-20$ is false. A particularly short counterexample isThe sole local minimum in part (a), $(-1/2,0)$, has value $-1/4$, larger than values on this circle; it is therefore not a global minimum. In fact $f(-2,y)=2-y^2\to -\infty$, so $f$ has no global minimum on the plane.

Write $h=x-x_0$, $k=y-y_0$, and let $a=f_{xx}$, $b=f_{xy}$ and $c=f_{yy}$ be evaluated at the critical point. Assuming $f$ is twice continuously differentiable nearby, the Taylor expansion isThe linear terms vanish because the gradient is zero. For $a>0$, completing the square givesConsequently the requested constant is the Hessian determinantThe displayed identity in the question is an identity for the quadratic part; for a general $f$, its right-hand side needs the remainder $o(h^2+k^2)$.

If $D>0$, the quadratic form is positive for every nonzero $(h,k)$. On the unit circle it has a positive minimum, so it dominates the remainder sufficiently near the critical point. Hence the point is a strict local minimum.

If $D<0$, along $k=0$ the quadratic term is $a{h}^2/2>0$, whereas along $h=-bk/a$ it is $D{k}^2/(2a)<0$. Along each line the remainder is of smaller order. Thus values both above and below $f(x_0,y_0)$ occur arbitrarily nearby, giving a saddle point.

The critical points satisfySubstitute $x=-y/2$ to obtain $y(3y-1/2)=0$. Thus the only critical points are $(0,0)$ and $(-1/12,1/6)$.

The Hessian matrix and Hessian determinant areAt $(0,0)$, $D=-1$, so this is a saddle point. At $(-1/12,1/6)$, $D=1$ and $f_{xx}=2>0$, so this is a strict local minimum. ThereforeThe value at the local minimum is $-1/432$; there is no local maximum.

Put $C=q^2/(4\pi {ϵ}_0)>0$. The spring exerts the force $k(\ell -x)$ by Hooke's law, while Coulomb's law gives the attractive force $-C/x^2$. Newton's second law therefore gives the equation of motionFor $x>0$, the derivative is $f^{\prime }(x)=-k+2C/x^3$. Thus $f$ increases up to $x_{\ast }=(2C/k{)}^{1/3}$ and decreases thereafter. Also $f(x)\to -\infty$ at both ends of the positive half-axis. Its unique maximum isThere are therefore two equilibrium points precisely whenWriting them $x_{-}<x_{\ast }<x_{+}$, the linearization $m\ddot{\eta }=f^{\prime }(x_{\pm })\eta$ has exponentially growing solutions at $x_{-}$ and oscillatory solutions at $x_{+}$. Equivalently, the potential energy has a strict local maximum at $x_{-}$ and a strict local minimum at $x_{+}$. Hence the smaller equilibrium point is an unstable equilibrium; the larger is a stable equilibrium, with no damping to make it attracting.

At equality $\ell ={\ell }_1$ there is one degenerate equilibrium point $x_{\ast }$; it is unstable, since a displacement to its left gives a leftward force. For $\ell <{\ell }_1$ there are no equilibrium points.

Expanding and taking partial derivatives gives the gradient and Hessian matrixAt $p_0=(-1,0{)}^T$, these are $\nabla f(p_0)=(-6,2{)}^T$ andThus the Newton step solvesThe second equation gives $s_y=1-2s_x$; substituting into the first gives $22s_x=10$. HenceAt $p_{\ast }=(-1/2,-1/2{)}^T$, however,The eigenvalues of this Hessian matrix are $-1$ in direction $(1,1)$ and $7$ in direction $(1,-1)$. It is invertible but indefinite. Thus the Newton step is zero, andThis illustrates that stationary points are fixed points of Newton's method. Here the fixed point is a saddle point of a scalar function, not a local minimum. Indeed, direct substitution along the two eigenvector directions givesFor sufficiently small nonzero $t$, one direction decreases $f$ and the other increases it. The global minimum is instead zero: the sum of squares vanishes at $(0,0)$ and $(-1,-1)$. These are the only real zeros, since $x=-y^2$ and $y=-x^2$ imply $y(y^3+1)=0$. Thus a zero gradient is an insufficient stopping certificate for minimization problem without classifying the stationary point.

For an unconstrained differentiable function of two real variables, an interior local minimum must solve the simultaneous stationary point equationsApply Newton's method in optimization to this gradient equation. The Jacobian matrix of the gradient is the Hessian matrixAt $p_n=(x_n,y_n{)}^T$, solve the linear system $H(p_n)s_n=-\nabla f(p_n)$, then set $p_{n+1}=p_n+s_n$. When the Hessian matrix is invertible, this isThis follows by setting the first order Taylor expansion $\nabla f(p_n+s_n)\simeq \nabla f(p_n)+H(p_n)s_n$ equal to zero. Solving the linear system avoids explicitly computing a matrix inverse. A stationary point need not be a local minimum: for example, a positive-definite matrix $H(p)$ suffices to classify a stationary point as a strict local minimum, whereas an indefinite Hessian matrix identifies a saddle point of a scalar function.

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.

For a twice continuously differentiable function $F$, its gradient and Hessian matrix areA critical point solves $\nabla F=0$. The Jacobian matrix of this vector equation is $H_F$, so the multivariable Newton iteration isEquivalently ${\mathbf{x}}_{k+1}={\mathbf{x}}_k-H_F({\mathbf{x}}_k{)}^{-1}\nabla F({\mathbf{x}}_k)$ when the Hessian matrix is invertible. Solving the linear system is the direct computational form. A critical point found in this way need not be a local minimum; its type requires a separate check.

For observations ${\mathbf{x}}_i\in {\mathbb{R}}^p$, K-means clustering seeks nonempty clusters in cluster analysis $C_1,\dots ,C_K$ and centroids ${\mathbf{m}}_k$ minimizingAt fixed centroids, assigning each point to a nearest centroid minimizes its squared Euclidean distance. At fixed assignments, the centroid sum-of-squares identity shows that the sample mean uniquely minimizes each group's contribution. These alternating steps cannot increase $W$.

For finite termination of K-means clustering, more than a lower bound on $W$ is needed. Keep an assignment unchanged when nearest centroids tie, and stop when assignments no longer change. Every change then strictly decreases $W$. After updating the centroids, $W$ is determined by the partition of a set of $n$ indices; there are only finitely many choices for that partition of a set, so no strictly improving partition of a set can recur. Empty clusters in cluster analysis require a convention, such as moving a point from a nonsingleton cluster in cluster analysis to become its own centroid, which cannot increase $W$. With such a consistent convention the algorithm terminates, although it may reach only a local minimum.

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

Near a nondegenerate stationary point, the quadratic part of the Taylor expansion determines the local contour plot. For a strict local maximum, the model $f=f_0-X^2-Y^2$ gives nested closed level sets, with values decreasing as one moves outward. For a strict local minimum, the model $f=f_0+X^2+Y^2$ gives the same nested shape with values increasing outward. Unequal nonzero quadratic coefficients stretch the circles into ellipses after a rotation to the Hessian matrix's eigenvectors.

For a nondegenerate saddle point of a scalar function, the model $f=f_0+X^2-Y^2$ gives hyperbolas. Levels above $f_0$ open along the positive-curvature direction; levels below $f_0$ open along the negative-curvature direction. The critical level set is the two crossing lines $Y=\pm X$. Degenerate stationary points can have different local geometry, so these are the standard nondegenerate sketches.

The quotient rule simplifies the derivative toIts zeros are $-5$ and $-1$, both in the domain. Substitution gives the two stationary pointsThe first is a local maximum and the second a local minimum, since the numerator of $y^{\prime }$ changes sign from positive to negative at $-5$ and from negative to positive at $-1$. The denominator of the original rational function vanishes atThe numerator is nonzero at both values, so these discontinuities are poles, not removable holes.

At the origin the Hessian matrix isThus the second-derivative test gives a saddle point. Its opposite signs can also be checked without a test formula: $g(x,0)=x^2{e}^{-x^2/2}>0$ for $x\ne 0$, whereas $g(0,y)=-y^2{e}^{-y^2/2}<0$ for $y\ne 0$. Both types of values occur arbitrarily close to the zero value at the origin, excluding either a local maximum or a local minimum.

For $x>0$, differentiate the two explicit branches:The plus branch is strictly increasing and strictly convex for $x>0$. Its only zero is $x=0$, also its one-sided global minimum; it has no interior extremum or inflection point.

The minus branch rises initially, then has $y_{-}^{\prime }=0$ at $\sqrt{x}=4/5$, changing from positive to negative. Its second derivative changes sign at $\sqrt{x}=8/15$. Its zeros follow from $y_{-}=x^2(1-\sqrt{x})$. ConsequentlyThe origin is a one-sided local minimum of both graph branches, because their domain ends there; it is not an interior critical point. The minus branch has no global minimum since it tends to $-\infty$ as $x\to \infty$. Its maximum and inflection heights, useful for the sketch, are $256/3125$ and $28672/759375$, respectively.