Intermediate value theorem

The intermediate value theorem says that a real-valued continuous function cannot skip a value between two of its outputs. Let $a<b$ and let $f:[a,b]\to \mathbb{R}$ be a continuous function on the entire closed interval $[a,b]$. ThenThe word "between" includes both endpoint values: choose $c=a$ when $y=f(a)$, or $c=b$ when $y=f(b)$. If $y$ is strictly between them, every such $c$ lies in the open interval $(a,b)$. Continuity on $[a,b]$ means continuity at every interior point, together with the endpoint one-sided limits ${\lim }_{x\to a^{+}}f(x)=f(a)$ and ${\lim }_{x\to b^{-}}f(x)=f(b)$. No derivative is required.^[ref]

For finding a zero of a function, the useful special case isStrictly opposite endpoint signs certify a root bracket, because zero is strictly between $f(a)$ and $f(b)$. If an endpoint value is already zero, that endpoint supplies a zero of a function directly; an interior zero of a function need not exist. For example, $f(x)=x$ on $[0,1]$ has its only zero of a function at the left endpoint. The intermediate value theorem guarantees existence, but neither uniqueness nor a necessary sign-change test. Its numerical usefulness is that the bisection method can turn existence into successively tighter root brackets.

Three explicit functions separate the hypotheses from the conclusion. These are original mathematical illustrations, not measured data.

The polynomial function $h(x)=x^3-x=x(x-1)(x+1)$ is a continuous function on $[-3/2,3/2]$, with $h(-3/2)=-15/8$ and $h(3/2)=15/8$. The intermediate value theorem guarantees at least one zero of a function in the interior, while the polynomial factorization displays three: $-1,0,1$. Thus strict opposite signs do not establish uniqueness.

The everywhere-defined functionhas $j(-1)=-1$ and $j(1)=1$, yet no zero of a function. Its jump discontinuity at zero is exactly where the continuity hypothesis fails. A drawn vertical connector would be misleading: the function graph contains neither the intermediate heights nor a zero of a function at that jump. Even continuity only on the interior is insufficient: setting $q(x)=-1$ for $0\le x<1$ and $q(1)=1$ gives opposite endpoint signs on $[0,1]$ with no zero of a function, because the left one-sided limit at $1$ is not $q(1)$.

The polynomial function $t(x)=x^2$ on $[-1,1]$ is a continuous function, and $t(-1)=t(1)=1$, yet $t(0)=0$. Its root of a polynomial at zero has multiplicity of a root two. The function graph touches its horizontal tangent line there and stays positive on both sides: existence of a zero of a function does not require a sign change. More generally, near a real root of a polynomial $r$, a polynomial factorization $p(x)=(x-r{)}^{2m}u(x)$ with $m\ge 1$ and $u(r)\ne 0$ has the same sign on both sides sufficiently close to $r$, since the continuous function $u$ retains its sign. The intermediate value theorem does not rule out values outside the range between endpoint outputs.^[ref]

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.

For a short proof outline, endpoint values give the non-strict cases immediately. If $y$ is strictly between $f(a)$ and $f(b)$, apply the bisection method to the continuous function $g(x)=f(x)-y$. A midpoint with $g=0$ finishes the proof. Otherwise retain nested intervals $[a_k,b_k]$ with strictly opposite signs and widths $(b-a)2^{-k}$. The nested interval theorem, using completeness of the real numbers, gives a common point $c$ to which both endpoint sequences converge. After reversing the sign of $g$ if necessary, $g(a_k)<0<g(b_k)$. Continuity makes both endpoint value sequences converge to $g(c)$, forcing $g(c)\le 0$ and $g(c)\ge 0$, hence $f(c)=y$. The original endpoint values differ from $y$, so $c\in (a,b)$. This proof explains why a jump can defeat the construction: shrinking a candidate closed interval cannot supply the missing continuity.

A continuous function on any real interval has image a real interval. Given two image values and an intermediate value, choose preimages and apply the intermediate value theorem on the closed segment joining them, which stays inside the domain. Neither compactness nor boundedness of the original interval is needed. In particular $f(x)=\sin (1/x)/x$ on $(0,1]$ is a continuous surjection onto $\mathbb{R}$: its values are unbounded above and below, so its interval image contains every real number.

A continuous injective real-valued function on a real interval is strictly monotone. A value outside the interval between two endpoint values would, by the intermediate value theorem, create a repeated value. Applying this observation to subintervals gives consistent strict ordering. If the left endpoint value is smaller than the right endpoint value, the function is strictly increasing; otherwise it is strictly decreasing.

If $f:[0,1]\to \mathbb{R}$ is continuous and $f(0)=f(1)$, then for every positive integer $n$ there is $\alpha \in [0,1-1/n]$ with $f(\alpha +1/n)=f(\alpha )$. For $n>1$, the continuous difference $h_n(x)=f(x+1/n)-f(x)$ satisfies ${\sum }_{j=0}^{n-1}{h}_n(j/n)=0$. Either one sampled difference is zero or two have opposite signs; the intermediate value theorem then supplies a zero between them. For $n=1$, the endpoint chord suffices.

If a continuous real-valued function on a real or complex vector space has opposite signs at two points, restricting it to the line segment between them and applying the intermediate value theorem gives a zero.

If a continuous $g:[0,na]\to \mathbb{R}$ satisfies $g(na)-g(0)=na$, then the incrementssatisfy ${\sum }_{k=0}^{n-1}h(ka)=0$. Continuity forces $h$ to vanish somewhere on $[0,(n-1)a]$.

Every continuous injective function from a real interval to the real line is strictly monotone. A continuous bijection $\mathbb{R}\to \mathbb{R}$ therefore has a continuous inverse.