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For a monic degree-$n$ polynomial $g$ over a commutative ring $R$, let $A=R[X]/(g)$ and let $\lambda$ take the coefficient of $X^{n-1}$ in the reduced representative. The pairing $(a,b)\mapsto \lambda (ab)$ is unimodular on $1,X,\dots ,X^{n-1}$: reversing one basis order gives a triangular matrix with diagonal ones. When $A$ embeds in a separable field extension, Lagrange interpolation identifies $\lambda (a)$ with $\mathrm{Tr}(a/g^{\prime }(x))$. This yields the derivative formula for the inverse different.

Commutative algebra studies commutative rings, ideals, modules, and the spectrum of a ring. Its basic question is which familiar rules of integer and polynomial arithmetic survive in a new setting. The distinction between units, zero divisors, and integral domains makes that question concrete.

Use $D=-i{\partial }_x$. For a nonzero one-variable polynomial $P$ and $v\in {\mathcal{E}}^{\prime }(\mathbb{R})$, a compactly supported solution to $P(D)u=v$ exists exactly when $v$ annihilates every smooth solution of the transposed equation $P(-D)\phi =0$. For simple roots ${\alpha }_j$, these test solutions span the exponentials $e^{-i{\alpha }_{j}x}$, so the condition is $\hat{v}({\alpha }_j)=0$. Dividing the entire transform by $P$ and applying polynomial division preservation of exponential type and the Paley–Wiener–Schwartz theorem constructs the compactly supported solution. It is unique because an entire function killed by a nonzero polynomial must vanish identically. For a root of multiplicity $r$, the conditions become ${\hat{v}}^{(k)}(\alpha )=0$ for $0\le k<r$, corresponding to $x^k{e}^{-i\alpha x}$.

Complex analysis studies functions of a complex number, especially holomorphic functions. Its distinctive feature is that a single complex derivative controls variation in every direction of the complex plane. The map $f(z)=z^2$ gives a first picture: straight coordinate lines become curved, yet small perpendicular arrows remain perpendicular wherever $f^{\prime }(z)\ne 0$.

The bottom panels show displacements rather than absolute positions. Multiplication by $f^{\prime }(1+i)=2+2i$ predicts the image of each small arrow; the exact image differs by $h^2$. The origin is exceptional because $f^{\prime }(0)=0$. This distinction between local conformal maps and a critical point is central: complex differentiability imposes strong geometric structure, without making every holomorphic function globally one-to-one.

The remarkable next step is that holomorphic functions are determined inside a circle by their values on its boundary. Precisely, let $R>0$, let $D=\{z:|z-a|<R\}$, and suppose $f$ is a holomorphic function on an open set containing the closed disc $\overline{D}$. Orient its boundary counterclockwise. The Cauchy integral formula states that, for every $w\in D$,The contour integral combines boundary values with a kernel depending on the requested interior point. It is a reconstruction formula, not merely a trick for evaluating an integral. The hypothesis about the closed disc ensures that all boundary values and the contour integral are well defined. ^[ref]

At the centre, substituting $z=a+R{e}^{i\theta }$ cancels the denominator against $dz=iR{e}^{i\theta }d\theta$:Thus the centre value is the angular average of the boundary values. For the polynomial $f(z)=z^2$, $a=1+i$ and $R=1/2$, expand the boundary data asBoth oscillating terms have zero average, leaving $f(a)=2i$. Away from the centre the weights change: on the unit circle,With $f(z)=z^2$ and $w=1/2$, expanding the denominator as a geometric series leaves only its second-power term after integration, giving $f(1/2)=1/4$. This also explains how the Cauchy integral formula recovers an entire Taylor series from boundary data.

When a holomorphic function has isolated singularities, a different boundary calculation detects their residues. Suppose $\Gamma$ is a counterclockwise simple closed contour and $f$ is holomorphic on a neighbourhood of it and its interior except at finitely many interior isolated singularities $a_j$. The residue theorem givesA residue is the coefficient of $(z-a_j{)}^{-1}$ in the Laurent series. A short proof idea is to deform the contour to small circles around the isolated singularities using the Cauchy integral theorem. Integrating the Laurent series on each circle kills every power except $-1$. ^[ref]

Here is how a complex pole determines a real improper integral. Set $g(z)=1/(1+z^2)$ and use the contour consisting of the segment from $-R$ to $R$, followed by the upper semicircle $A_R$, with $R>1$. Only the simple pole $i$ lies inside, andThe arc must actually disappear. On $A_R$, the reverse triangle inequality gives $|1+z^2|\ge R^2-1$, and its length is $\pi R$, soThe real integrand is bounded by $x^{-2}$ for $|x|\ge 1$, so both tails converge separately. Taking the limit therefore proves the ordinary improper integralFor example, at $R=10$ the real segment contributes $2\arctan (10)\approx 2.942255$, and the arc contributes $0.199337$, below the bound $10\pi /99\approx 0.317333$. The usefulness of the residue theorem is that the same strategy works for many rational functions whose real antiderivatives are cumbersome: find the enclosed poles, compute their residues, and control the connecting arc. ^[ref]

Historically, Cauchy's 1825 memoir on integrals with imaginary limits develops complex contour integration and its use in transforming and evaluating definite integrals. The opening sections define complex path integrals, discuss changing the path, and examine what happens near singular points. These are recognizable ancestors of the modern link between holomorphic functions, contours and residues. ^[ref]

For a complex polynomial $p(z)=a{z}^2+bz+c$, assume $|a|>|c|$ and put $D=|a{|}^2-|c{|}^2$ and $E=\overline{a}b-c\overline{b}$. Both roots of a polynomial lie in the closed unit disk if and only if $|E|\le D$. Strict inequality puts both roots in the open disk. Under the strict product assumption $|c/a|<1$, any unit-modulus root is automatically simple.

To prove the strict version, define $p^{\#}(z)=z^2\overline{p(1/\overline{z})}$ and $q=\overline{a}p-c{p}^{\#}=z(Dz+E)$. On the unit circle $|p^{\#}|=|p|$, so the Rouche theorem shows that a stable $p$ and $q$ have the same two interior zeros. Conversely $Dp=aq+c{q}^{\#}$, and the same Rouche theorem implies that two interior zeros of $q$ give two interior zeros of $p$. The nonzero root of $q$ is $-E/D$, proving the criterion. The closed version follows by continuity, replacing $b$ by $(1-\epsilon )b$ in the sufficiency direction. In the necessity direction with a boundary root, replace $p(z)$ by $p(z/r)$ for $r<1$ close to one and let $r\to 1$. The product of the two roots of a polynomial excludes two unit-modulus roots and excludes a repeated unit-modulus root.

Apply the midpoint quadrature rule on each of $n$ equal panels of a partition of an interval, with $h=(b-a)/n$:Each quadrature node is a panel midpoint and each quadrature weight is $h$. For $f\in C^2[a,b]$, put $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$. Summing the panel quadrature errors givesfor some $\xi \in (a,b)$. Each panel contributes at most $K{h}^3/24$; there are $n=(b-a)/h$ panels, explaining the composite midpoint rule error of order $O(h^2)$ in big O notation. On a nonuniform partition of an interval with panel widths $h_j$ and largest width $H$, the corresponding bound is $K{\sum }_j{h}_j^3/24\le K(b-a)H^2/24$.

For $f(x)=x^2$ on $[0,1]$, summing the sampled polynomial exactly yieldsDoubling $n$ divides the absolute error by exactly four. For a general $C^2$ function, halving $h$ divides the displayed error bound by four, but the actual absolute error need not follow that ratio: cancellation can make it smaller or zero. Continuous functions still give convergence as the largest panel width tends to zero, by uniform continuity; without the bounded second derivative, the second-order rate is not guaranteed. Noise sensitivity of a quadrature rule explains a separate limitation when samples are inexact.

For $x$ in the derived algebra of a complex matrix Lie algebra $L$, define $y$ by multiplication by $\overline{\lambda }$ on each generalized eigenspace $V_{\lambda }$ of $x$. Polynomial interpolation and adjoint compatibility of additive Jordan decomposition make $\mathrm{ad}y$ a polynomial in $\mathrm{ad}x$ with zero constant term, so $[y,L]\subseteq [L,L]$. Trace orthogonality of $[L,L]$ and $L$ forces the displayed sum to vanish. Thus $x$ is a nilpotent endomorphism, and the Engel theorem implies solvability. The auxiliary $y$ and the Jordan components are not required to lie in $L$.

A trigonometric polynomial is real on the unit circle precisely when its finite coefficients have the displayed symmetry. To prove necessity, conjugate its values on the circle, use $\overline{z}=z^{-1}$, and compare coefficients. Multiplying a coefficient difference by $z^d$ gives an ordinary polynomial vanishing at infinitely many points, so it is zero. Symmetry also gives $p(z^{-1})=\overline{p(\overline{z})}$ for every nonzero complex $z$.

A continuation method follows a solution branch of a parameter-dependent equation from a known solution, using nearby solutions to guide the next solve. This is useful when one wants to understand how solutions change, rather than obtain just one isolated answer. A continuation predictor-corrector step follows a tangent vector and then uses a Newton method correction on a transverse section. The same geometry supports both pseudo-arclength continuation through a continuation fold and transversal surface intersection tracing.

Consider the original scalar polynomial exampleIts solution branch bends back twice in $\lambda$. Increasing $\lambda$ alone eventually stops following the same nearby solutions; following the curve in the $(x,\lambda )$ plane can continue around both bends. The continuation folds areFor $|\lambda |<2/(3\sqrt{3})$ there are three distinct real zeros of a function at that fixed $\lambda$, all lying on this one connected curve. A continuation method keeps their geometric relationship visible.

Every continuation method needs a seed, supplied by an independent solve, an explicitly known solution, or another search. Following one solution branch does not guarantee finding every connected component of the solution set. A large step can miss detail or make the correction converge to a distant solution. At branch intersections or a loss of rank of the full Jacobian matrix, the regular-curve argument needs separate analysis. The seed and component limitations also matter in transversal surface intersection tracing.

For a topological space $Y$ and topological spaces $X_i$, a function $f:Y\to {\prod }_i{X}_i$ is continuous into the product topology exactly when every coordinate function $f_i={\pi }_i\circ f$ is continuous.^[ref]

Proof outline: one direction uses function composition with the continuous projection maps. For the other, the preimage of a basic cylinder set isa finite set intersection of open sets in $Y$. This gives continuity on every basic test. It explains, for example, why $t\mapsto (t,t^2,t^3,\dots )$ from the real line to ${\mathbb{R}}^{\mathbb{N}}$ is continuous in the product topology: each coordinate is a polynomial function. The general coordinate criterion fails for the box topology, where infinitely many preimages can be required at once. For example, $g(t)=(t,t,\dots )$ has continuous coordinate functions, but the preimage of the box ${\prod }_{j\ge 1}(-1/j,1/j)$ is $\{0\}$, which is not an open set in the real line.

On a compact interval, positive quadrature rule weights whose sum is the interval length define uniformly bounded functionals on continuous functions. If the rules eventually integrate every fixed polynomial exactly, the Weierstrass approximation theorem proves convergence for every continuous function.

Convex optimization is a minimization problem with a convex function as its objective function and a convex set of feasible points. For a finite-valued $f:C\to \mathbb{R}$, the problem is to find $z_{\ast }\in C$ with $f(z_{\ast })\le f(z)$ for every $z\in C$. The displayed infimum describes its value even when no minimizing point exists. The two convexity requirements concern different objects: the allowed decisions and the scores assigned to them.

A convex set contains the whole line segment between any two of its points:A convex function lies below the interpolation of its endpoint values:Thus mixing two allowed decisions preserves feasibility, and the mixed score is no worse than the same mixture of endpoint scores. The objective function may have flat regions or corners. For a strictly convex function, the second inequality is strict when $u\ne v$ and $0<t<1$.

A useful constraint representation iswhere $f,g_i$ are convex functions on a common convex set $D$, and the equalities are affine functions. Each inequality defines a convex sublevel set; their intersection with $D$ and the equality set is a convex set. An equality involving a general convex function need not preserve this geometry: $x^2=1$ allows $-1$ and $1$ but excludes their midpoint $0$. See Boyd and Vandenberghe, Convex Optimization (2004), §4.2.1 for the standard problem form.

Why local information can certify a global answer. Every local minimum of a convex optimization problem is a global minimum, including minima on the boundary. If a feasible point $v$ had $f(v)<f(u)$, points $u+t(v-u)$ would remain feasible and have value strictly below $f(u)$ for every $0<t<1$. Taking $t$ arbitrarily small contradicts a local minimum at $u$. This argument explains the special power of convexity: a better distant decision would reveal an improving direction immediately nearby.

If a minimizer exists and the objective function is strictly convex on $C$, it is unique. Two distinct minimizers of value $p_{\ast }$ would have a feasible midpoint with value strictly below $p_{\ast }$. Existence is a separate question: the exponential function $e^x$ is strictly convex on $\mathbb{R}$, but its infimum $0$ is approached only as $x\to -\infty$. Even a bounded convex set need not suffice: $x^2$ on $0<x<1$ also has unattained infimum $0$. A continuous function on a nonempty compact set attains its minimum by the extreme value theorem.

Nor does strict convexity imply differentiability. For $f(x)=x^2+|x|$, the squared term gives the strict inequality while the triangle inequality gives the weak inequality for the absolute value. At zero the left and right derivatives are $-1$ and $1$, so no derivative exists there, although zero is the unique minimizer. Such objectives require subgradients or proximal operators rather than an ordinary gradient at every point.

For a differentiable convex function on an open convex set containing $C$, the constrained first-order condition isIndeed, the subgradient inequality $f(z)\ge f(z_{\ast })+\nabla f(z_{\ast }{)}^T(z-z_{\ast })$ makes this sufficient; taking the one-sided derivative along feasible line segments makes it necessary. At an interior minimizer, both directions are feasible and the gradient must vanish. At a boundary minimizer it can be nonzero: it only needs to make every feasible direction nondecreasing. See Boyd and Vandenberghe, §§3.1.3 and 4.2.3.

A fully computed nearest-point problem. Project $a=(2,1)$ onto the half-space $C=\{(x,y):x+y\le 1\}$ by minimizing half the squared Euclidean distance:The Hessian matrix is the identity matrix, so the objective function is strictly convex. The unconstrained minimizer $(2,1)$ violates the inequality. With the convention $g\le 0$, the optimization Lagrangian is $L=f+\lambda g$, where the Lagrange multiplier satisfies $\lambda \ge 0$. Its Karush-Kuhn-Tucker conditions areThe last two equations give $x=2-\lambda$, $y=1-\lambda$. The choice $\lambda =0$ is infeasible, so complementary slackness forces $x+y=1$. Consequently $3-2\lambda =1$, giving $\lambda =1$, $(x_{\ast },y_{\ast })=(1,0)$ and $f(x_{\ast },y_{\ast })=1$. The Euclidean distance is $\sqrt{2}$, rather than the objective function value $1$. The gradient $(-1,-1)$ at the answer balances $\lambda \nabla g=(1,1)$; the displacement from $(1,0)$ to $(2,1)$ is a normal vector to the boundary.

There is an exact certificate independent of numerical search. Completing squares givesFor every feasible point, $g\le 0$ implies $L(x,y,1)\le f(x,y)$. Thus every feasible objective is at least $1$, and $(1,0)$ attains this bound. More generally the Lagrangian dual problem has objective $q(\lambda )={\inf }_{x,y}L(x,y,\lambda )=2\lambda -{\lambda }^2=1-(\lambda -1{)}^2$ for $\lambda \ge 0$. Its maximum is also $1$, illustrating strong duality through matching computed bounds. This makes the example a model for Euclidean projection onto a convex set and the projection step in projected gradient descent.

What the multiplier conditions require. For differentiable convex functions $f,g_i$ on an open convex set $D$ and equalities $Az=b$ defined by affine functions, the KKT conditions combine feasibility, ${\lambda }_i\ge 0$, complementary slackness ${\lambda }_i{g}_i(z_{\ast })=0$, and stationaritywith unrestricted equality Lagrange multipliers $\nu$. They suffice to certify a global minimum without an additional regularity assumption: the optimization Lagrangian is convex, stationarity minimizes it, and feasibility with complementary slackness gives the same certificate as above. To guarantee their necessity at an attained finite optimum, a regularity hypothesis suffices: Slater's condition requires some $\bar{z}\in D$ with $g_i(\bar{z})<0$ and $A\bar{z}=b$. The projection example has the strictly feasible point $(0,0)$. See Boyd and Vandenberghe, §§5.2.3 and 5.5.3. This hypothesis guarantees suitable multipliers, not existence of a primal minimizer.

Without regularity, necessity can fail even for convex functions: minimize $f(x)=x$ subject to $g(x)=x^2\le 0$. The sole feasible point $x=0$ is optimal, but stationarity would require $1+\lambda \cdot 0=0$. No Lagrange multiplier works. There is no strictly feasible point, and the gradient of the constraint vanishes at the answer.

The nonconvex contrast. The polynomial $h(x)=(x^2-1{)}^2+0.2x$ has derivative $h^{\prime }(x)=4x^3-4x+0.2$ and second derivative $h^{\prime \prime }(x)=12x^2-4$. Numerically solving the cubic gives stationary points $-1.024120300$, $0.050125947$, and $0.973994353$. Their objective values are respectively $-0.202440434$, $1.005006281$, and $0.197434153$, with second derivatives $8.585869$, $-3.969849$, and $7.383980$. Hence both outer points are local minima, and the middle point is a local maximum. Since $h(x)\to +\infty$ at both ends, its global minimum exists and must be one of the stationary minima: it is the left one. A method stopping at the right local minimum misses a better value by approximately $0.399874587$.

The practical appeal of convex optimization is the ability to check an answer globally using geometry and matching Lagrangian duality bounds. The projection certificate certifies every feasible alternative at once; the tilted quartic shows why a zero gradient alone cannot do that for general nonlinear optimization.

Historically, an early landmark is Kuhn and Tucker's ^[ref]. Its §3 explicitly states a regularity hypothesis for inequality-constrained Lagrange multipliers; its discussion of saddle points explains how concave functions support global conclusions for maximization. This documents an early formulation of the multiplier approach; it does not establish sole discovery of today's KKT conditions or of convex optimization as a whole.

A monic polynomial over a complete discrete valuation ring factors with prescribed coprime monic residue factors. Given monic lifts $g_r,h_r$ correct modulo ${\pi }^r$, the degree-bounded map $(u,v)\mapsto u{\bar{h}}_r+v{\bar{g}}_r$ is a bijection over the residue field, by coprimality and equality of dimensions. Solve it for the error $(f-g_r{h}_r)/{\pi }^r$, and replace the lifts by $g_r+{\pi }^{r}u,h_r+{\pi }^{r}v$. Their coefficients converge by completeness and the limiting product is $f$. The completeness assumption cannot be replaced just by localization.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension of an absolute value to every finite extension. Factor a monic $g\in {\mathcal{O}}_K[X]$ into monic irreducible factors over $K$. All factors have integral coefficients because their roots have absolute value at most one. By pure-power reduction of a monic irreducible polynomial, each factor reduces to a power of a single irreducible polynomial. If $\overline{g}={\overline{g}}_1{\overline{g}}_2$ is a coprime monic factorization, assign each whole irreducible factor of $g$, with its multiplicity, to the unique side containing its residue factor. The resulting monic $g_1,g_2\in {\mathcal{O}}_K[X]$ satisfy $g=g_1{g}_2$ and reduce to the prescribed factors. Completeness of $K$ is not required for this proof.

Two polynomials over a field are coprime if their only common factors are nonzero constants. Equivalently, their greatest common divisor is a unit in the polynomial ring. In one variable this is equivalent to a Bezout identity $Af+Bg=1$ with polynomial coefficients. Independent members of a pencil spanned by coprime polynomials are themselves coprime.

Every countable set $B$ of real numbers is contained in a countable set closed under taking all real roots of a polynomial with coefficients in that set. Let $\varphi (A)$ consist of these roots of a polynomial, excluding the zero polynomial. There are countably many finite coefficient lists over a countable set, and each nonzero polynomial has finitely many roots of a polynomial; hence $\varphi (A)$ is countable. Define $B_0=B$ and $B_{n+1}=B_n\cup \varphi (B_n)$. The countable union of countable sets theorem shows that $X={\bigcup }_n{B}_n$ is countable. Every finite coefficient list in $X$ belongs to one stage $B_N$, so its real roots of a polynomial belong to $B_{N+1}\subseteq X$. Retaining $B_n$ at the next stage matters: $\varphi (A)$ need not contain $A$.

A credible interval is an interval $C(d)$ containing a prescribed amount $1-\alpha$ of posterior probability for an unknown scalar statistical parameter $\Theta$. For a continuous posterior distribution with posterior density $p(\theta \mid d)$, its endpoints satisfyThe conditioning includes the chosen statistical model and prior distribution. For a discrete posterior distribution, exact equality may be impossible because of its probability masses; an interval can instead be required to contain at least the requested posterior probability. An interval compresses the full posterior distribution into a range whose uncertainty can be communicated in the parameter's own units. See MIT 18.05, Class 16: probability intervals.

For the eight-head, two-tail example in Bayesian statistics, the posterior density is $2860{\theta }^9(1-\theta {)}^3$ on $[0,1]$. Its cumulative distribution function, obtained by integration of this polynomial, isSolving $F(a)=0.025$ and $F(b)=0.975$ by the bisection method givesThese are numerical approximations to the exact quantiles defined by those equations; the endpoints were computed from the cumulative distribution function, rather than read off a plot. The resulting equal-tailed credible interval leaves 2.5% posterior probability on each side. Its shaded area is 95% of the total area under the posterior density.

Credible intervals are not unique. The one-sided credible interval $[0,F^{-1}(0.95)]\approx [0,0.887334219]$ also has 95% posterior probability, but leaves all 5% outside it on the right. A highest density region instead selects points above a density threshold; it need not be an equal-tailed credible interval, and when the posterior density has separated peaks it need not be one interval at all. Thus a reported credible interval should specify its construction.

A 95% credible interval makes a statement about the statistical parameter conditional on these observed data and the Bayesian model. A 95% confidence interval instead refers to the coverage probability of a procedure under repeated sampling at a fixed parameter value. These are different conditioning statements: 95% posterior probability does not by itself imply 95% coverage probability for every fixed parameter. The confidence interval article develops that sampling interpretation.

When $R^2=4L/C$, the homogeneous series RLC circuit has one repeated characteristic root, $-R/(2L)$. Its current is $e^{-Rt/(2L)}(A+Bt)$, with constants determined by the initial data.

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.