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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 $I$ 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.

Let $h:\mathbb{R}\to \mathbb{R}$ be continuous, strictly increasing, and satisfy Lusin condition N. Thendefines a measure on the Lebesgue measurable sets, and $\nu \ll \lambda$. Moreover,for $a\le b$.

Interval bisection solves a target equation $f(x)=t$ by repeatedly halving a closed interval known to contain a solution. It is the bisection method applied to $g(x)=f(x)-t$, so the target becomes a zero of a function. This reduction from an equation to root finding is described in Demanet's notes.^[ref] Here the emphasis is on a continuous strictly increasing function: the target identifies one input, making interval bisection a way to evaluate its inverse function without an explicit inverse formula.

Let $f:[a_0,b_0]\to \mathbb{R}$ be a continuous function, with $a_0<b_0$ and $f(a_0)\le t\le f(b_0)$. The intermediate value theorem guarantees existence; a strictly increasing function also guarantees uniqueness. Check endpoint equality first. Otherwise, for the current root bracket $[a_k,b_k]$, evaluate its midpoint $m_k=(a_k+b_k)/2$. If $f(m_k)<t$, retain $[m_k,b_k]$; if $f(m_k)>t$, retain $[a_k,m_k]$; if $f(m_k)=t$, return that exact solution. These comparisons preserve the target bracket. General bisection method needs continuity and opposite signs of $g$ at the endpoints, without requiring monotonicity or a unique zero of a function; see limitations of bisection.

An original dimensionless polynomial example isHere $f(0)=0<10<30=f(3)$ and the derivative $f^{\prime }(x)=3x^2+1>0$, so there is exactly one solution. In fact $x_{\ast }=2$; knowing this lets us check the computed absolute error independently of the guarantee. The first four halvings give the following exact rational numbers:

The beauty of interval bisection is that the enclosure itself supplies the certificate: one comparison removes half the possible inputs, regardless of the steepness of the function graph. In exact arithmetic, after $k$ completed halvings without exact termination, the width and the bisection error bound for the new midpoint areThe reported midpoint after update four is $m_4=63/32$, rather than the last tested point $m_3=33/16$. Its certified absolute error is at most $3/32$, and its actual absolute error is $1/32$. For a requested position tolerance $10^{-3}$, the bisection stopping criterion requires eleven halvings: ten give $3/2048>10^{-3}$, while eleven giveThe actual absolute error is $1/4096$. The new midpoint need not be evaluated to obtain this certificate.

The signed output residual is $r=f(m)-t$; the position approximation error is $m-x_{\ast }$. They answer different questions: how close is the output to the target, and how close is the input to the solution? Driscoll and Braun explain why a small residual needs information about the derivative to imply small position error.^[ref] For this particular polynomial, direct factorization givesThus at $m_4=63/32$ the residual is $-13121/32768\approx -0.400421$, despite position absolute error $1/32=0.03125$. If a differentiable function satisfies $0<\ell \le f^{\prime }(x)\le L$ throughout the bracket, the mean value theorem gives the rigorous conversionOur example permits $\ell =1$ and $L=28$ on $[0,3]$. Strict increase alone supplies no positive lower derivative bound: $f(x)=x^3$ is strictly increasing but $f^{\prime }(0)=0$. A small residual alone is therefore not a universal position certificate; even the target equation $10^{-6}(x-2)=0$ has residual $10^{-6}$ at $x=3$, a full unit from its solution.

The guarantees require continuity and correct evaluations, not just narrowing endpoints. For the strictly increasing function $h(x)=x$ when $x<0$ and $h(x)=x+1$ when $x\ge 0$, the target $1/2$ lies between $h(-1)$ and $h(1)$ but is never attained: a jump discontinuity defeats the existence argument. With floating-point arithmetic or an approximate evaluation, an incorrect comparison can discard the true solution. A certified interval enclosure of $f(m)$ wholly below or above $t$ certifies the corresponding decision; an enclosure containing $t$ is inconclusive, rather than proof of equality. If rounding makes the midpoint equal an endpoint, report the precision limit; see bisection stopping criterion for the rounded position bound. Both figures are exact mathematical examples, not measured data.

The log-likelihood is the natural logarithm of the likelihood function:For positive finite likelihood, the natural logarithm is a strictly increasing function, so the maximizing parameter values are unchanged. Set $\ell =-\infty$ where $L=0$; the same ordering then includes zero-likelihood boundaries whenever a positive finite maximum exists. For independent and identically distributed random variables, a product becomes a sum,This makes derivatives simpler and avoids forming products of tiny numbers in floating-point arithmetic. These equal-maximizer and additive formulas underlie the worked calculations in MIT 18.05, Reading 10b, section 3.1.

For the ordered sample of eight successes and two failures in likelihood function, on $0<p<1$,The score equation ${\ell }^{\prime }(p)=0$ gives $8(1-p)=2p$, hence $\hat{p}=8/10=0.8$. The negative second derivative proves strict concavity, so this is the unique interior maximum. Both counts are positive, so $L(0)=L(1)=0$, while $L(p)>0$ inside: the same point is the unique global maximum on $[0,1]$. Counting only successes adds $\log 45$ to the log-likelihood, leaving both derivatives and the MLE unchanged. For all-success or all-failure samples, the boundary analysis in maximum-likelihood estimator applies instead of an interior score equation.

The two panels below compare $k=2,5,8$ successes out of ten Bernoulli trials. Their peaks are at $\hat{p}=k/10$. The left panel uses relative height; the right panel uses the log-likelihood difference from each peak. These are the same comparison on two scales:A difference of $-2$ means relative height $e^{-2}$, and a difference of $-10$ means $e^{-10}$; the log-likelihood scale makes small relative heights easier to see. Neither vertical axis is a parameter probability density function. The zero likelihood endpoints have log-likelihood $-\infty$ and are outside the finite vertical range in the right panel.

Multiplying a likelihood function by a positive parameter-independent constant adds a constant to its log-likelihood. Subtracting its peak removes that constant. This is why the ordered-data and count-data curves coincide in this figure after rescaling. When differentiable, the gradient of the log-likelihood is the score function; the negative Hessian matrix is the observed information. A solution of a score equation still needs checking against other stationary points and boundaries: maximizing the log-likelihood does not automatically make it concave.

A real-valued function $f:D\to \mathbb{R}$, with domain of a function $D\subseteq \mathbb{R}$ in the usual order of the real numbers, is a monotonic function if it is either a nondecreasing function throughout $D$ or a nonincreasing function throughout $D$. The same direction must work for every pair $x<y$ in $D$. Being a strictly increasing function on one part and a strictly decreasing function on another does not make the whole function monotone: for example, $x^2$ on the real line changes direction at zero. A constant function is both a nondecreasing function and a nonincreasing function.

The word "increasing" is ambiguous across textbooks: some allow equal outputs and some require unequal outputs. Use nondecreasing function for the weak inequality and strictly increasing function for the strict inequality; use nonincreasing function and strictly decreasing function in the other direction. Neither continuity nor differentiability is part of the definition of a monotonic function.

Strict monotonicity gives injectivity, so an equation $f(x)=b$ has at most one solution and an inverse function exists on the image of a function. Weak monotonicity permits repeated outputs and jumps. In probability theory, a cumulative distribution function is a nondecreasing function; its quantile function uses a generalized inverse to handle those flat parts and jumps. Monotonicity alone does not guarantee continuity, even for a strictly increasing function.

A real-valued function $f:D\to \mathbb{R}$ is a nondecreasing function if every pair $x<y$ in its domain of a function satisfies $f(x)\le f(y)$. Flat intervals are allowed, unlike for a strictly increasing function. For example,is a continuous nondecreasing function, but it repeats its output on each flat interval.

A nondecreasing function can also be discontinuous. The step function $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge 0$ has a jump discontinuity at zero without ever decreasing. It is the cumulative distribution function of a random variable that equals zero with probability one; quantile functions handle such jumps through a generalized inverse. Nondecreasing allocation functions permit critical-value payments in single-parameter mechanisms.

First, equality of two $U_0$ values forces equality of the corresponding $V_0$ values: neither strict comparison holds for $U_0$, and the same is then true for $V_0$ in both directions. Thus $V_0=g\circ U_0$ for a well-defined strictly increasing function $g$ on $J=U_0(\mathcal{P})$. Since $\mathcal{P}$ is a convex set and $U_0,V_0$ are affine functions, $J$ is a real interval andIf $J$ contains $u<v$, let $a=[g(v)-g(u)]/(v-u)>0$ and $b=g(u)-au$. The mixture identity gives $g(w)=aw+b$ for $w\in [u,v]$. For $w>v$, express $v$ as a convex combination of $u$ and $w$ and solve the same identity for $g(w)$; for $w<u$, express $u$ as a mixture of $w$ and $v$. Hence the formula holds on all of $J$, not merely between the chosen anchors. We obtain positive affine uniqueness of affine preference representations:If $U_0$ is constant, the same comparisons force $V_0$ to be constant; choose $a=1$ and the appropriate $b$. If the probability-measure set is empty, the assertion is vacuous. These degenerate cases do not require distinct anchors.

Fix $a<x<b$. Injectivity excludes $f(x)=c$ or $d$. If $f(x)<c$, the intermediate value theorem on $[x,b]$ gives a point other than $a$ with value $c$, contradicting injective function. If $f(x)>d$, the same theorem on $[a,x]$ gives a point other than $b$ with value $d$. ThereforeTo prove strict increase, take $x<y$. If $x=a$, the displayed endpoint ordering already gives $f(x)<f(y)$. Otherwise $f(a)<f(y)$, and the same interior-point argument on the subinterval $[a,y]$ yields $f(a)<f(x)<f(y)$. Hence $f$ is a strictly increasing function. This is the interval order theorem for continuous injections; the assumed endpoint order rules out strict decrease.

If $f$ is a strictly increasing function and $x\ne y$, one of $x<y$ or $y<x$ holds. The corresponding strict inequality of function values shows that $f$ is injective. Conversely, if $f(x)<f(y)$, then $x=y$ is impossible and $x>y$ would imply $f(x)>f(y)$. ThusTogether with strict increase, this says that such a function preserves and reflects the ordering of points.

PutBecause $dR/ds=-n\cdot v$, differentiation givesA massive charge has the subluminal speed $|v|<c$, so $F^{\prime }(s)>0$. Hence $F$ is strictly increasing, and $F(s)=t$ has at most one root. This is the uniqueness of retarded time for a subluminal source.

The equation requires $\sinh z=\pm i\cosh a$. Write $z=x+iy$, with real $x,y$. Comparing real parts and imaginary parts using part (b)(i) givesIf $x=0$, the second equality would require $|\sin y|=\cosh a>1$, impossible because $a>0$. Thus $\sinh x\ne 0$, forcing $\cos y=0$ andNow $|\sin y|=1$, so $\cosh x=\cosh a$. The hyperbolic cosine is even and is a strictly increasing function on $[0,\infty )$, hence $x=\pm a$. Conversely, substituting these values gives $\sinh (x+iy)=i(-1{)}^k\cosh a$, so every value found does satisfy the original equation. Therefore the complete set is

For fixed $F>0$, the left side $L(v)=F^2{v}^2$ is a strictly increasing function on $v>1$, with limit $F^2$ at $1$ and unbounded growth. The right side $R(v)=1+3/[2(v-1)]+1/[2(v-1{)}^2]$ is a strictly decreasing function, sinceIt tends to infinity as $v\downarrow 1$ and to one as $v\to \infty$. Their continuity and the intermediate value theorem give a crossing, and strict opposite monotonicity makes it unique. There is exactly one admissible bore speed $V>U$ for every $F>0$.

The illustrated case is $F=1$; changing $F>0$ moves the increasing curve without changing the existence or uniqueness argument. The vertical asymptote at $v=1$ expresses the growing depth ratio $H/h_0=v/(v-1)$.

If $n$ is positive and odd, $n^2$ is odd. The function $x\mapsto x^{n^2}$ is a strictly increasing function on the real numbers, so it is injective. Every real $y$ has the real odd root $\mathrm{sgn}(y)|y{|}^{1/n^2}$, so the function is also surjective.

If $n$ is positive and even, the exponent is positive and even. Then $f(1)=f(-1)$, while $f(x)\ge 0$ for all real $x$. These facts disprove injectivity and surjectivity, respectively. ThereforeIf the convention $0\in \mathbb{N}$ is used, $n=0$ gives exponent zero. Interpreting $x^0$ as the constant polynomial $1$ makes the function neither injective nor surjective. If $0^0$ is left undefined, that case does not define the stated function on all the real numbers. Neither convention adds a qualifying value of $n$.

A wall moving at constant speed $U>0$ into stationary water of depth $h_0$ produces a hydraulic bore of speed $V>U$ in the uniform-state shallow water equations model. Behind the bore the fluid velocity equals $U$ and its depth $H$ obeys $H/h_0=V/(V-U)$. Put $F=U/\sqrt{g{h}_0}$ and $v=V/U$. The moving Rankine-Hugoniot conditions give $F^{2}v=[(H/h_0{)}^2-1]/2$, hence the displayed equation for $v>1$. Its left side is a strictly increasing function from $F^2$ to infinity, while its right side is a strictly decreasing function from infinity to one. The intermediate value theorem therefore gives exactly one admissible $V$ for every $F>0$.

A real-valued function $f:D\to \mathbb{R}$ is a strictly decreasing function if every pair $x<y$ in its domain of a function satisfies $f(x)>f(y)$. It is a nonincreasing function and is injective, with a strictly decreasing inverse function on its image of a function. Negating a strictly increasing function gives a strictly decreasing function: $f(x)=-x^3$ is an example on the real line, even though its derivative vanishes at zero. Equal outputs are forbidden, unlike for a general nonincreasing function.

A real-valued function $f:D\to \mathbb{R}$ is a strictly increasing function if every pair $x<y$ in its domain of a function satisfies $f(x)<f(y)$. It is therefore injective, and each output has at most one preimage. Its inverse function on its image of a function is also strictly increasing. Every strictly increasing function is a nondecreasing function; the converse fails when outputs repeat.

The polynomial $f(x)=x^3$ is a strictly increasing function on the real line, although its derivative $f^{\prime }(x)=3x^2$ is zero at zero. Indeed, for $x<y$,because $x^2+xy+y^2=(x+y/2{)}^2+3y^2/4$ can vanish only when $x=y=0$. Thus a positive derivative everywhere is sufficient on a real interval, but is not necessary for a strictly increasing function.

Strict monotonicity does not ensure continuity: $g(x)=x$ for $x<0$ and $g(x)=x+1$ for $x\ge 0$ is a strictly increasing function with a jump discontinuity. Its image of a function omits $[0,1)$, so the inverse function exists only on the actual image of a function, rather than on all the real numbers.