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For a smooth function $f:U\to \mathbb{R}$ on an open set $U\subseteq {\mathbb{R}}^n$, a critical point is an input $x_{\ast }$ where the gradient vanishes: $\nabla f(x_{\ast })=0$. Equivalently, every first partial derivative is zero. Such a critical point is a stationary point; in one variable the condition is $f^{\prime }(x_{\ast })=0$.

There is a broader convention in elementary one-variable calculus: a "critical number" is an interior input where the derivative is zero or does not exist, provided the function itself is defined there. Thus a corner can be a critical number without being a stationary point. Here the smooth meaning of critical point is used unless the broader convention is explicitly stated. ^[ref]

For the polynomial $f(x)=x^4-2x^2$ on the real line,The stationary points at $x=\pm 1$ are strict local minima and attain the global minimum; the stationary point at $x=0$ is a strict local maximum. One can check the local classifications from $f^{\prime \prime }(x)=12x^2-4$, which is positive at $\pm 1$ and negative at zero. The identity $f(x)=(x^2-1{)}^2-1\ge -1$ proves the global minimum directly. Since $f(x)\to +\infty$ as $|x|\to \infty$, there is no global maximum on $\mathbb{R}$.

A zero derivative alone does not guarantee an extremum. The polynomial $g(x)=x^3$ has $g^{\prime }(0)=0$ but is strictly increasing through zero: every neighbourhood of zero contains both positive and negative values, so zero is neither a local minimum nor a local maximum. Conversely, the absolute value function $h(x)=|x|$ has a strict local minimum and a unique global minimum at zero, but its left and right derivatives are $-1$ and $+1$. It is not differentiable there, so zero is a critical number under the broader convention and is not a stationary point.

The usefulness of critical points is that they reduce the search for smooth interior extrema to equations. Every local minimum or local maximum of a differentiable function at an interior point has zero gradient; this condition is necessary, and $x^3$ shows that it is not sufficient. The striking feature is that all the first directional derivatives must vanish together, even though the surrounding values can form a valley, a peak, or a saddle point of a scalar function. ^[ref]

Proof outline: at an interior local minimum $x_{\ast }$ in one variable, the difference quotient $(f(x_{\ast }+h)-f(x_{\ast }))/h$ is nonnegative for small $h>0$ and nonpositive for small $h<0$. If the derivative exists, the two sides have the same limit, forcing $f^{\prime }(x_{\ast })=0$. Reverse the inequalities for a local maximum. In several variables, apply this argument along each coordinate line to obtain zero partial derivatives and hence zero gradient. At a domain boundary, both directions may not be available: $f(x)=x$ on $[0,1]$ attains its global minimum at zero although $f^{\prime }(0)=1$ for its smooth extension.

The origin is a saddle point of a scalar function for every real parameter. For $\alpha >0$ four further critical points have $x^2=y^2=1/(2\alpha )$. Equal-sign coordinates give global maxima and opposite-sign coordinates give global minima, with values $\pm 1/(2e\alpha )$. The Hessian matrix there is $-4\alpha xy{e}^{-1}I$. Their distance from the origin is ${\alpha }^{-1/2}$, so as $\alpha \downarrow 0$ they escape every compact region. At nonpositive $\alpha$ the origin is the only critical point. This illustrates why the number of critical points on an unbounded domain can change without a finite degenerate critical point appearing.

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.

Let $a_i,c_i$ be nonzero real constants with $a_i{c}_i>0$. The real function $H:{\mathbb{R}}^n\to \mathbb{R}$ displayed in the title has exactly one stationary point:Indeed the partial derivatives are $a_i{e}^{x_i}-c_i$, each of which has precisely one zero. The Hessian matrix at that point is $\mathrm{diag}(c_1,\dots ,c_n)$.

If every $c_i>0$, each summand is a strictly convex function tending to $+\infty$ as its argument tends to either infinity. Consequently $H$ has a unique global minimum. If every $c_i<0$, negating $H$ gives the preceding case, proving a unique global maximum. If both signs occur, restricting to two coordinate directions of opposite signs gives a saddle point of a scalar function. No degenerate Hessian matrix is possible under the nonzero assumptions.

Let $\varphi$ be a twice continuously differentiable local diffeomorphism, and suppose $\nabla f(\varphi (u_{\ast }))=0$. The chain rule gives the displayed transformation of the Hessian matrix: terms containing second partial derivatives of $\varphi$ are multiplied by first partial derivatives of $f$ and vanish at the critical point. By Sylvester's law of inertia, the numbers of positive, negative and zero eigenvalues are unchanged. Independently of nondegeneracy, the local diffeomorphism preserves the critical point's status as a local minimum, local maximum or saddle point of a scalar function, because it maps neighbourhoods to neighbourhoods.

Suppose a function near a critical point has Taylor expansion $f(p+h)=f(p)+P_m(h)+o(\Vert h{\Vert }^m)$, where $P_m$ is its first nonzero homogeneous polynomial term. If $P_m$ takes both positive and negative values on directions, then $p$ is a saddle point of a scalar function. Indeed, on rays $h=rv$ the leading term is $r^m{P}_m(v)$; for each chosen direction its sign dominates the remainder for small positive $r$. This works even when the Hessian matrix is zero. For example, $P_4(\xi ,\eta )=-{\xi }^3(\xi +\eta )$ is negative on $(1,0)$ and positive on $(1,-2)$, hence detects a saddle despite having degree four.

For $f(x,y)=g(x)+h(y)$, a degenerate Hessian matrix can be supplemented by the first nonzero term of each one-variable displacement. Two negative leading even powers give a strict local maximum, two positive ones a strict local minimum, and opposite signs give a saddle. Looking along the coordinate directions and bounding the separated terms avoids an inconclusive quadratic test.

For an explicit contrast, considerBoth have zero gradient and the same zero Hessian matrix at the origin:Nevertheless, $f_{+}$ is positive at every nonzero point, so the origin is a strict global minimum. For $f_{-}$, the restrictions $f_{-}(t,0)=t^4>0$ and $f_{-}(0,t)=-t^4<0$ for $t\ne 0$ show that the origin is a saddle point of a scalar function. Identical quadratic information can conceal different geometry when higher-order terms are decisive.

The Laplacian detects whether a value lies below or above nearby values, to second order. For a scalar field whose partial derivatives through order four are continuous functions, take the arithmetic mean of four equally spaced neighbors:Applying Taylor's theorem in each coordinate cancels the first- and third-order terms between opposite neighbors. The two pairs leaveThis is an asymptotic expansion, rather than an equality for arbitrary $h$. Rearranging gives the familiar central finite difference approximationThis finite difference replaces partial derivatives by five sampled values. The arithmetic mean lies above the center for a sufficiently small $h$ if the Laplacian is strictly positive there, and below if it is strictly negative. A zero Laplacian cancels this second-order discrepancy without requiring equality of individual values.

At $(0,0)$, the bowl $x^2+y^2$ has center value zero and four neighbor values $h^2$, so the discrepancy is exactly $h^2$. The saddle $x^2-y^2$ has neighbor values $h^2,h^2,-h^2,-h^2$, whose arithmetic mean is exactly zero. For the Gaussian function $e^{-x^2-y^2}$ the discrepancy is $e^{-h^2}-1=-h^2+O(h^4)$, agreeing with $\Delta f(0,0)=-4$.

A harmonic function need not equal this four-neighbor arithmetic mean for a finite $h$. The polynomial $f(x,y)=x^4-6x^2{y}^2+y^4$ has $\Delta f=0$ everywhere, but $A_{h}f(0,0)=h^4$. Higher-order terms remain even when the second-order terms cancel.

At a local minimum, each coordinate second derivative is at least zero, hence the Laplacian is at least zero. At a local maximum the inequality reverses. The sum alone does not characterize an extremum: cancellation at a saddle point of a scalar function can also produce zero. The local-average interpretation explains why the heat equation smooths differences, and helps motivate the heat equation maximum principle.

A critical point of a real function with continuous second partial derivatives is nondegenerate when its Hessian matrix is an invertible matrix there, equivalently when none of its eigenvalues is zero. A positive-definite matrix gives a strict local minimum, a negative definite Hessian gives a strict local maximum, and opposite-sign eigenvalues give a saddle point of a scalar function. A nondegenerate critical point need not be an extremum: $x^2-y^2$ has Hessian matrix $\mathrm{diag}(2,-2)$, which is an invertible matrix.

Such critical points are the local building blocks of Morse theory.

For $F(x,y)=(y-c{x}^2{)}^2+\alpha x^2-\beta x^4$ with $\beta >0$, its stationary points are the origin and, when $\alpha >0$, $x=\pm \sqrt{\alpha /(2\beta )}$, $y=c{x}^2$. The origin is a strict local minimum for $\alpha >0$ and a saddle point of a scalar function for $\alpha <0$. At $\alpha =0$ its Hessian matrix is singular, but the vertical line gives positive values and the parabola $y=c{x}^2$ negative values, so it remains a degenerate saddle. The other stationary points have Hessian determinant $-8\alpha$ and are saddles.

At a critical point $x_0$ of a twice continuously differentiable real function, the Taylor expansion isThe Hessian matrix is real symmetric, so an orthogonal eigenbasis diagonalizes its quadratic form. If every eigenvalue is positive, that form is bounded below by a positive multiple of $|h{|}^2$, which dominates the remainder and gives a strict local minimum. If every eigenvalue is negative, the same argument gives a strict local maximum. If both signs occur, displacement along the corresponding eigenvectors produces values above and below $f(x_0)$, so the point is a saddle point of a scalar function. These exhaust a nonsingular Hessian matrix.

A singular Hessian matrix does not always make the classification impossible: nonzero eigenvalues of both signs still give a saddle. If the Hessian matrix is positive or negative semidefinite, higher-order terms in its null directions must be examined. For example, $x^2+y^4$ and $x^2-y^4$ have the same Hessian matrix $\mathrm{diag}(2,0)$ at zero, but the first has a strict minimum and the second a saddle. The function $x^2$ has a non-strict minimum along a whole line.

For the given polynomial,A stationary point has $y=3x^2/2$ and then $x(2\alpha -x^2)=0$. Hence the origin is always critical, and two additional points exist exactly for $\alpha >0$:The Hessian matrix isAt the origin it is $\mathrm{diag}(2\alpha ,2)$, giving a strict local minimum for $\alpha >0$ and a saddle for $\alpha <0$. At either extra critical point it has determinant $64\alpha -36(2\alpha )=-8\alpha <0$, so both extra points are saddles.

At $\alpha =0$, complete the square:Along $x=0$ it is positive away from zero, whereas along $y=3x^2/2$ it is negative away from zero. Thus the origin remains a degenerate saddle despite its nonnegative Hessian matrix. This is the parabolic quartic critical-point classification. The complete result isFor $\alpha >0$ the local minimum is not global, since along the same parabola $f=\alpha x^2-x^4/4\to -\infty$.

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.

All invariants in this part refer to the selected magnetic field line. Its squared poloidal Alfvén number isAt the Alfvén surface crossing it equals unity, so ${\rho }_a={\mu }_0{k}^2$ and $M_{Ap}^2=1/y$. Subtract the two azimuthal integrals in part (a) to findFinite azimuthal fields and velocity at $y=1$ require the Alfvén-surface regularity condition for an axisymmetric wind $\ell =\omega R_a^2$. HenceHere $v_{\varphi }$ denotes the azimuthal velocity relative to field-line rotation. For gravity of a point mass at $z=0$, $\Phi =-GM/R=-GM/(R_{a}x)$.

Insert these quantities in the corotating energy integral with $w=0$, and divide by the gravitational energy scale $GM/R_a$. This gives the cold radial magnetohydrodynamic wind integral,withBoth are dimensionless. The apparent $0/0$ at $x=y=1$ must be evaluated on the smooth solution branch, not assigned an arbitrary value.

A constant-energy curve obeys $f_x+f_{y}dy/dx=0$. At a nondegenerate magnetosonic point, the vanishing coefficient of the mass density derivative requires simultaneous vanishing of its numerator:The curve must admit a real crossing slope. Expanding about such a point gives $f_{yy}{m}^2+2f_{xy}m+f_{xx}=0$, $m=dy/dx$, so a regular saddle point of a scalar function crossing has $f_{xy}^2-f_{xx}{f}_{yy}>0$. These are the form of the slow and fast regularity conditions in a warm wind, with the enthalpy contribution retained in $f$.

In the strictly cold approximation used to obtain the displayed $f$, the slow speed is zero. There is a cold-limit degeneracy of a slow magnetosonic point: a finite-speed interior slow crossing need not survive. Indeed,so this cold equation has no ordinary sub-Alfvénic slow critical point. A small retained enthalpy is needed to locate that point, which can approach the launching boundary as the sound speed tends to zero. The fast critical point still obeys the simultaneous conditions above wherever the reduced expression is nonsingular.

For this quartic potential with fourfold symmetry, the first derivatives factor:The critical points are the origin, the four axis points $(\pm 1,0),(0,\pm 1)$, and four points with both coordinates nonzero. At the latter, $x^2+b{y}^2=y^2+b{x}^2=1$. Subtracting and using $b\ne 1$ gives $x^2=y^2=1/(1+b)$. Thus the complete list iswhere the two signs in the final group are chosen separately.

The Hessian isAt the origin it is $2I$, so there is a strict local minimum with value zero. At an axis point its eigenvalues are $-4$ and $2(1-b)$, giving a saddle point of a scalar function if $b<1$ and a strict local maximum if $b>1$. All four axis values are $1/2$.

At a diagonal point the Hessian has eigenvaluesHence these four points are strict local maxima for $b<1$ and saddle points of a scalar function for $b>1$. Their value is $1/(1+b)$. None of these critical points is degenerate when $b\ne 1$.

To describe the level curves precisely, use polar coordinates and writeFor a level $f=c$, the possible radii satisfyretaining only real, nonnegative values. This formula, together with reflection in each axis and interchange of the coordinates, determines the level curves. The radial maximum in a direction is $1/(2q)$, attained at $r^2=1/q$. It also proves that the local maxima identified above are global: the largest directional value is $1/(1+b)$ on the diagonals if $b<1$, and $1/2$ on the axes if $b>1$.

For negative $c$ only the plus sign gives a positive radius, producing one outer closed level curve. At $c=0$ there is the isolated origin and the outer curve $r^2=2/q$. For $0<c<c_s$, where $c_s$ is the saddle value, both radial branches exist in every direction, giving an inner loop around the origin and an outer loop. At $c=c_s$ they meet at the four saddle points of a scalar function. Above the saddle value, four separate loops surround the four maxima; they shrink to those points at the maximum value. There are no level curves above it.

For $0<b<1$, the saddle value is $c_s=1/2$ and the four high-level loops lie along the diagonals. For $b>1$, $c_s=1/(1+b)$ and the four high-level loops lie along the coordinate axes. The sketches show representative values in both regimes; the highlighted level curve is the saddle level.

Finally, at $b=1$ the expression depends only on radius:Consequently $\boxed{\max f=1/2\text{, attained exactly on }{x}^2+y^2=1.}$ The maximum set is the whole unit circle, not merely four isolated points.

In the $(x,y)$-plane take $z=0$; holding another fixed $z$ merely shifts all contour labels by $z^2$. The level sets satisfyFor $a=1$, the origin is a local minimum and $(0,\pm 1)$ are saddle points of a scalar function at contour value $1$. For $0<c<1$, a closed contour around the origin is accompanied by unbounded outer components. At $c=1$ the contour satisfies $x=\pm (y^2-1)$ and crosses at the two saddle points of a scalar function. For $c>1$, two unbounded components remain; negative contours have only outer components.

For $a=-1$, the origin is a saddle point of a scalar function, and the zero contour is $x=\pm y\sqrt{y^2+2}$, with tangent lines $x=\pm \sqrt{2}y$. Positive contours form two components extending on opposite sides of the $y$-axis; negative contours form upper and lower components. These level sets are illustrated below.

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.

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.