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Consider the original worked example $f(x)=x^3+x-1$ on $[0,1]$. First, existence: a polynomial function is a continuous function, and $f(0)=-1<0<1=f(1)$. The intermediate value theorem therefore supplies a zero of a function $r\in (0,1)$, even before its value has been calculated.

Second, uniqueness is a separate argument. The derivative is $f^{\prime }(x)=3x^2+1>0$. The derivative criterion for monotonicity, obtained from the mean value theorem, makes $f$ a strictly increasing function. Two distinct zeros of a function would contradict that strict increase. Thus this root bracket contains exactly one zero of a function; indeed the same derivative argument proves uniqueness on the entire real line.

Third, localization uses the existing bisection method. Set $[a_0,b_0]=[0,1]$; after $k$ completed halvings, retain $[a_k,b_k]$ with opposite endpoint signs. Each update tests the old midpoint $m_{k-1}=(a_{k-1}+b_{k-1})/2$. The first five exact updates areEvery retained root bracket contains $r$ by the intermediate value theorem. After ten halvings, exact arithmetic givesHence $0.681640625<r<0.6826171875$. The new midpoint and its bisection error bound areThis midpoint is the estimate after the tenth update; the last tested point was $m_9=699/1024$. In general the retained width is $2^{-k}$, so the bisection error bound is $2^{-(k+1)}$. Nine halvings already suffice for an error at most $10^{-3}$, since $2^{-10}=0.0009765625<10^{-3}$. This guarantee comes from the width of the root bracket, independently of any displayed decimal approximation to the unknown zero of a function. Endpoint signs must be established reliably if evaluations use floating-point arithmetic.

Let $f$ be differentiable on an open interval containing $c$. If the finite limit of a function $L={\lim }_{x\to c}{f}^{\prime }(x)$ exists, then $f^{\prime }(c)=L$.

Choose $h_n>0$ tending to zero with $[c,c+h_n]$ in the interval. The mean value theorem gives $x_n\in (c,c+h_n)$ such thatThe right-hand side tends to $f^{\prime }(c)$ by differentiability. Since $x_n\to c$ with $x_n\ne c$, the left-hand side tends to $L$. Uniqueness of the limit of a sequence gives equality. The same reasoning excludes an infinite limit of a function for $f^{\prime }$ at $c$.

Thus derivatives may be discontinuous, but a removable discontinuity cannot be repaired by changing the value of a derivative at a single point of differentiability.

Let $U$ be an open subset of the real line, and let $F,G$ be antiderivatives of the same function on $U$. Each connected component $I_j$ is an open interval. On that component, $(G-F{)}^{\prime }=0$, so the mean value theorem gives $G-F=C_j$. The constants on different components are independent: there is no subinterval inside $U$ joining them. Conversely, adding any chosen constant on each component preserves the derivative everywhere on $U$.

For example, the domain of $g(x)=1/x$ is $U=\mathbb{R}\setminus \{0\}$. Every antiderivative on this domain has the formwith arbitrary real $C_{-},C_{+}$. The logarithm of the absolute value differentiates to $1/x$ on each side. The function with $C_{-}=2$ and $C_{+}=-1$ is therefore a valid antiderivative, although it cannot be written as $\log |x|+C$ with one global constant.

An initial condition at $x=1$ fixes $C_{+}$ and leaves $C_{-}$ free. Fixing both constants requires one reference value in each connected component, or additional conditions that couple the components. The locally constant function $F(x)-\log |x|$ captures this remaining freedom.

The sequence is bounded and $|a_{n+1}-a_n|\le \sqrt{n+1}-\sqrt{n}\to 0$ by the mean value theorem. Nevertheless, along the integer parts of $(\pi /2+2\pi j{)}^2$ its terms tend to one, and along the integer parts of $(3\pi /2+2\pi j{)}^2$ they tend to minus one. The square root rounding error is $O(1/j)$. Thus a bounded sequence with vanishing successive increments need not be a convergent sequence.

For $a<b$ and real-valued functions $f,g$ that are continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, there exists $\xi \in (a,b)$ withApply Rolle's theorem to the linear combination $h(x)=[f(b)-f(a)]g(x)-[g(b)-g(a)]f(x)$, which has equal endpoint values. The displayed product identity permits zero derivatives. For the ratio form, additionally assume $g^{\prime }(x)\ne 0$ for every $x\in (a,b)$. Then Rolle's theorem ensures $g(b)\ne g(a)$, so both denominators below are nonzero:Neither derivative needs to be continuous, and $f^{\prime }$ may vanish. Choosing $g(x)=x$ recovers the mean value theorem; the ratio form also gives the zero-over-zero endpoint case of L'Hôpital's rule. ^[ref]

A differentiable function on an interval whose derivative is identically zero is a constant function. Indeed, for any two arguments $x<y$, the mean value theorem on $[x,y]$ gives $f(y)-f(x)=f^{\prime }(c)(y-x)=0$. For an interval with endpoints included, continuity at those endpoints is required. A disconnected domain may have different constants on different components.

On a connected open interval, if $F^{\prime }=g$, every antiderivative of $g$ has the form $F+C$, where $C$ is a constant function. Indeed, two solutions satisfy $(G-F{)}^{\prime }=0$: the zero derivative on a connected open set principle makes their difference constant. In one dimension, the mean value theorem proves this on any closed subinterval. ^[ref]

An initial condition or boundary condition can determine $C$. If a particular antiderivative is $F_0$ and $F(x_0)=y_0$, thenFor $g(x)=x^2$, $F_0(x)=x^3/3$ and $F(1)=2$, this gives $C=5/3$. Without that reference value, choosing $C=0$ simply selects one member of the family.

The freedom has a physical meaning in a simple kinematic model. A velocity $v(t)=at$, with constant acceleration $a$, determines position only as $x(t)=a{t}^2/2+C$. The initial condition $x(0)=x_0$ selects $C=x_0$; changing the starting position does not change the prescribed velocity.

For the differential equation $T^{\prime }=-B/r^2$ on $r>0$, the antiderivative is $T=B/r+C$. The boundary condition $T(R_s)=T_s$ gives $C=T_s-B/R_s$. Discarding the constant of integration therefore needs a specified condition or an approximation that quantifies its effect. On a domain with several connected components, the freedom is larger: antiderivatives on disconnected domains require a separate constant on each component.

A contraction constant for a function $T$ on a metric space is a single $q<1$ that bounds the ratio $d(Tx,Ty)/d(x,y)$ for every distinct pair. Pointwise strict inequalities $d(Tx,Ty)<d(x,y)$ do not imply such a common constant. For a differentiable function on an interval, the mean value theorem shows that $|T^{\prime }|\le q<1$ supplies a contraction constant.

Use radians and take $T(x)=\cos x$ on the closed interval $[0,1]$. This is a complete metric space with distance $|x-y|$, and $T([0,1])=[\cos 1,1]\subset [0,1]$. The mean value theorem and $|T^{\prime }(x)|=\sin x\le \sin 1$ give a contraction mapping with $L=\sin 1\approx 0.841471<1$. Both the interval and its preservation matter; a small derivative by itself would not ensure that all subsequent approximations stay in the chosen space.

Starting at $x_0=0$ gives $x_1=1$, $x_2\approx 0.540302$, $x_3\approx 0.857553$, and $x_4\approx 0.654290$. The cosine is decreasing here, so successive iterates alternate around the unique fixed point $x_{\ast }\approx 0.739085133215161$. The contraction iteration error bound needs no prior knowledge of this numerical value. At step 40, $x_{40}\approx 0.739085078689123$ and the latest-step bound is below $7.20\times 10^{-7}$, certifying error below one millionth for exact iteration. The initial-step bound requires 91 iterations for the same tolerance. The observed step-40 error is only about $5.45\times 10^{-8}$; the guarantees allow less favorable behavior than this example actually exhibits.

This is the beauty of the contraction mapping theorem: a uniform local improvement controls the entire infinite process. Completeness is essential. On the incomplete interval $(0,1)$, the contraction mapping $T(x)=x/2$ keeps every iterate inside but has no fixed point there; its missing limit is zero. Banach's 1922 paper developed abstract methods for different spaces of functions with applications to integral equations, a motivation still visible whenever repeated approximation is used to construct a function.^[ref]

Let $I$ be a real interval, and let $f:I\to \mathbb{R}$ be differentiable in its interior and continuous at any endpoints belonging to $I$. If the derivative satisfies $f^{\prime }(x)\ge 0$ throughout the interior, then $f$ is a nondecreasing function on $I$. For $x<y$ in $I$, the mean value theorem supplies $c\in (x,y)$ withChanging $\ge 0$ to $\le 0$, $>0$, or $<0$ gives a nonincreasing function, strictly increasing function, or strictly decreasing function, respectively. The endpoint continuity assumption lets the mean value theorem apply even when $x$ or $y$ is an endpoint. ^[ref].

Conversely, if a nondecreasing function on a real interval is differentiable at an interior point $x$, then $f^{\prime }(x)\ge 0$: the difference quotient $(f(x+h)-f(x))/h$ is nonnegative for either sign of $h$. A differentiable nonincreasing function has derivative $f^{\prime }(x)\le 0$. Strict monotonicity still only forces these weak signs: the strictly increasing function $x^3$ has derivative zero at zero. The sign of the derivative helps locate intervals of monotonicity when sketching a graph. ^[ref].

The real interval hypothesis matters because a connected subset of the real line contains every point between its inputs. On a disconnected domain of a function, positive derivatives need not give a nondecreasing function. For example, takeHere $f^{\prime }(x)=1$ everywhere in $D$, but $-3/2<3/2$ while $f(-3/2)=-3/2>-17/2=f(3/2)$. Each component has a strictly increasing function, yet their outputs are ordered the wrong way across the gap. This function is neither a nondecreasing function nor a nonincreasing function on all of $D$; the mean value theorem cannot cross the missing interval.

If all partial derivatives exist near a point and are continuous there, successive coordinate increments and the one-variable mean value theorem express $f(a+h)-f(a)$ as ${\sum }_j{h}_j{\partial }_{j}f(a+{\eta }_j(h))$ with ${\eta }_j(h)\to 0$. Subtracting the displayed linear map leaves $o(\Vert h\Vert )$, proving Fréchet differentiability. In two dimensions the absolute remainder is at most $ϵ(|h_1|+|h_2|)\le \sqrt{2}ϵ\Vert h\Vert$ for small enough increments.

Suppose $f$ and its partial derivative ${\partial }_{t}f$ are jointly continuous on a compact rectangle $J\times [a,b]$. Then $F(t)={\int }_a^{b}f(t,x)dx$ is continuously differentiable in the interior of $J$, and the displayed formula holds, with one-sided derivatives at parameter endpoints where the corresponding partial derivative is given.

To prove it, the extreme value theorem bounds $|{\partial }_{t}f|$ by $M_1$. For each fixed $x$, the mean value theorem bounds the difference quotient $[f(t+h,x)-f(t,x)]/h$ by $M_1$. It converges pointwise to ${\partial }_{t}f(t,x)$, and the constant $M_1$ is integrable over $[a,b]$. Dominated convergence gives the derivative formula. Apply continuity of parameter integrals on compact rectangles to ${\partial }_{t}f$ to obtain continuity of $F^{\prime }$. The proof works locally on unbounded parameter intervals as well.

Let $t_0$ be interior to a parameter interval. Suppose $h(t,\cdot )$ is a measurable function for each $t$, $h(t_0,\cdot )$ is a Lebesgue integrable function, and $h(\cdot ,x)$ is differentiable near $t_0$ for $x$ almost everywhere. If one integrable nonnegative function $D$ satisfies $|{\partial }_{t}h(t,x)|\le D(x)$ throughout that neighbourhood off one fixed null set, then $F(t)={\int }_{X}h(t,x)d\mu (x)$ is defined there and satisfies the displayed derivative formula at $t_0$.

The mean value theorem gives $|h(t,x)-h(t_0,x)|\le |t-t_0|D(x)$, proving local integrability. It also bounds the difference quotient by $D$. That quotient converges almost everywhere to ${\partial }_{t}h(t_0,x)$, so the dominated convergence theorem permits passing the limit through the Lebesgue integral.

If $f$ is continuous at an interval endpoint, differentiable in its interior, and has a finite one-sided derivative there, then its secant slope and tangent slope are arbitrarily close at some interior point. Apply the mean value theorem on a short endpoint interval; both the secant at its far endpoint and the secant at the resulting intermediate point approach the same one-sided derivative. No continuity of the derivative is required.

Suppose $f$ is continuously differentiable, $s$ is a descent direction, and $f(x+\alpha s)$ is bounded below for $\alpha \ge 0$. A positive step size satisfying the weak Wolfe conditions exists. To prove this, put $\varphi (\alpha )=f(x+\alpha s)$, $d={\varphi }^{\prime }(0)<0$, and $h(\alpha )=\varphi (\alpha )-\varphi (0)-{\beta }_1\alpha d$. The Armijo condition proof shows $h<0$ just to the right of zero. Boundedness below gives $h(\alpha )\to +\infty$. Consequently there is a first positive zero $A$, with $h\le 0$ on $[0,A]$. By the mean value theorem, some $t\in (0,A)$ satisfiesAt $t$, $h(t)\le 0$ gives the Armijo condition, and the displayed derivative gives the weak Wolfe conditions.

For every real $t$, the mean value theorem bounds $|e^{tx}-1|\le |t|x{e}^{|t|}$ on $0<x<1$. Hence the apparent singularity at zero cancels, and the remaining bound $|t|e^{|t|}/\sqrt{1-x}$ is Lebesgue integrable. On a bounded parameter interval $|t|\le M$, every positive-order parameter derivative of the integrand is bounded in absolute value by $e^M/\sqrt{1-x}$. Repeated dominated differentiation under the integral sign therefore proves that $F$ is a smooth function, withIn particular, $F(0)=0$, $F^{\prime }(0)=2$, and $F^{\prime \prime }(t)>0$. Thus $F$ is strictly convex, lies above its tangent $2t$ at zero, and satisfies $F^{\prime }(t)>2$ for $t>0$ and $F^{\prime }(t)<2$ for $t<0$. The inequalities follow directly as well by comparing $e^{tx}$ to one. The substitution $x=1-s^2$ removes the square-root endpoint singularity for numerical integration:

The fundamental theorem of calculus says that accumulation and local change undo one another, under appropriate regularity conditions. If $f$ is a continuous function on $[a,b]$ andthen $A$ is differentiable for $a<x<b$ and $A^{\prime }(x)=f(x)$. Conversely, if $G$ is an antiderivative of $f$ on this interval, continuous at its endpoints, thenThe first statement turns an integral into a function with a known derivative; the second turns an entire accumulation into two endpoint values.

For $f(t)=1+2t$, accumulation from zero gives $A(x)=x+x^2$. At $x=2$, its derivative is $A^{\prime }(2)=5$, exactly the current value $f(2)$. The accumulated value there is $A(2)=6$, and by $x=3$ it has reached $A(3)=12$. The accumulated quantity and its rate are different functions, linked by differentiation.

This connection makes the theorem useful far beyond geometric area. It recovers displacement from velocity, gives a probability from a probability density function, and converts many differential equations into integral equations. The striking economy is that information spread across a whole interval can sometimes be recovered from an antiderivative at its two boundaries.

A short explanation of the first part is thatis the local average of $f$. Continuity makes this average approach $f(x)$. For the second part, $G-A$ has zero derivative and is constant by the mean value theorem. These are the ideas behind the proof; the endpoint and regularity conditions make them valid.

An early geometric form of this connection appears in Isaac Barrow's Lectiones geometricae (1670), Lecture X, Proposition 11. The original volume is catalogued by CESIMA, and Michael Nauenberg's historical research reproduces and examines the geometric argument.^[ref][ref]

For a stationary autoregressive process of order one with unit-variance normal distribution marginals and fixed $0<\rho <1$, the Gaussian kernel covariance correction at lag $j$ has the displayed form. The difference of two observations has variance $2(1-{\rho }^j)$, and their characteristic functions reduce the Fourier integral to the difference of two Gaussian integrals. The mean value theorem gives $0\le g_h(j)\le \sqrt{\pi }{\rho }^j/[2(1-\rho {)}^{3/2}]$, uniformly over $h>0$. Thus the integrated variance differs from its independent-sample counterpart by $O(n^{-1})$. The marginal bias is unchanged, so this dependence does not alter the leading optimal bandwidth or the $n^{-4/5}$ mean integrated squared error for a second-order density kernel.

If a real function on an interval obeys $|f(x)-f(y)|\le C|x-y{|}^{\alpha }$ for finite $C$ and $\alpha >1$, its difference quotients tend to zero everywhere. The mean value theorem makes it constant. Equivalently, divide a segment into $N$ pieces to obtain $|f(x)-f(y)|\le C|x-y{|}^{\alpha }{N}^{1-\alpha }\to 0$.

For $dX=\mu (t)Xdt+\sigma (t)dB$ with deterministic coefficients, $X(t,x)=a(t)[x+{\int }_0^{t}a(s{)}^{-1}\sigma (s)d{B}_s]$, where $a(t)=\exp ({\int }_0^t\mu )$. The random part does not depend on the initial value. For $f\in C_b^1$, the mean value theorem bounds the initial-value difference quotient by $a(t)\Vert f^{\prime }{\Vert }_{\infty }$, and dominated convergence theorem gives the displayed derivative. The transition law is normal with mean $a(t)x$ and variance ${\int }_0^t\exp (2{\int }_s^t\mu )\sigma (s{)}^{2}ds$.