Theorem

The intermediate value theorem

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Statement

Let \( a, b \in \mathbb{R} \) with \( a \lt b \), and let \( f : [a,b] \to \mathbb{R} \) be continuous at every point of \( [a,b] \) (two-sidedly at interior points, one-sidedly at \( a \) and \( b \)). Let \( y \in \mathbb{R} \) lie strictly between the endpoint values, that is \( f(a) \lt y \lt f(b) \) or \( f(b) \lt y \lt f(a) \). Then there exists \( c \in (a,b) \) with \( f(c) = y \). Equivalently, in the form used in practice: if \( f \) is continuous on \( [a,b] \) and \( f(a)f(b) \lt 0 \), then \( f \) vanishes somewhere in the open interval \( (a,b) \).

Why it matters

The intermediate value theorem (IVT) is the first theorem of the degree that manufactures a solution out of nothing but qualitative data. No formula, no algorithm, no smoothness: two numerical evaluations and the word “continuous” are enough to certify that an equation has a root. Everything downstream that asserts existence rather than exhibiting an answer — \( n \)-th roots, real roots of odd-degree polynomials, fixed points of self-maps of an interval, equilibria of continuous dynamics, the operating point of a nonlinear circuit or a nonlinear equation of state — ultimately reduces to this one move.

It is also the theorem that exposes what the real numbers are for. Every hypothesis except continuity is a statement about the domain and the field: the domain must be an interval (connected), and the field must have no gaps. Over \( \mathbb{Q} \) the theorem is false, and the failure is not an accident of the counterexample but structural — for an ordered field, “every continuous function with a sign change has a zero” is equivalent to Dedekind completeness. IVT is therefore the point where analysis stops being algebra and starts depending on the topology of the line, and it is the one-dimensional shadow of the general fact that continuous images of connected sets are connected.

Hypotheses

\( f \) is continuous at every point of \( [a,b] \).One bad point suffices to destroy the conclusion. On \( [-1,1] \) put \( f(x) = -1 \) for \( x \lt 0 \) and \( f(x) = 1 \) for \( x \ge 0 \). Then \( f(-1) = -1 \lt \tfrac12 \lt 1 = f(1) \), yet the range is the two-point set \( \{-1,1\} \) and the value \( \tfrac12 \) is never attained. Continuity almost everywhere is not enough; the theorem is not robust to a single jump.

The domain is an interval (a connected set).On \( D = [-1,0) \cup (0,1] \) the function \( f(x) = 1/x \) is continuous at every point of \( D \) and satisfies \( f(-1) = -1 \lt 0 \lt 1 = f(1) \), but \( 1/x = 0 \) has no solution. The sign change is real; the path between the two sample points is not. This is the trap behind “\( \tan \) changes sign across \( \pi/2 \), so it has a root there”.

\( y \) lies strictly between \( f(a) \) and \( f(b) \).Strictness is what buys the interior conclusion \( c \in (a,b) \). For \( f(x) = x^2 \) on \( [0,1] \) and \( y = 0 = f(0) \), the only preimage is the endpoint \( c = 0 \). With \( y \) equal to an endpoint value the weaker statement “some \( c \in [a,b] \)” is true but vacuous.

The codomain is the complete ordered field \( \mathbb{R} \).Invisible in the statement, load-bearing in the proof. Over \( \mathbb{Q} \), take \( f(x) = x^2 - 2 \) on \( [1,2] \cap \mathbb{Q} \). It is continuous in the verbatim \( \varepsilon\text{-}\delta \) sense, and \( f(1) = -1 \lt 0 \lt 2 = f(2) \), yet \( f(c) = 0 \) forces \( c = \sqrt{2} \notin \mathbb{Q} \). In the proof below, the nested intervals still shrink but their common point is missing from the field, so Step 4 has nothing to converge to.

The codomain is ordered at all.“Between” is an order-theoretic word. For \( f : [0,2\pi] \to \mathbb{R}^2 \), \( f(\theta) = (\cos\theta, \sin\theta) \), there is no sense in which \( (0,0) \) lies “between” \( f(0) \) and \( f(2\pi) \), and indeed the origin is never attained. Vector-valued statements need Brouwer or Poincaré–Miranda, not IVT.

Proof

We prove the case \( f(a) \lt y \lt f(b) \); the mirrored case is reduced to it in Step 8. The argument is the bisection proof: it uses only the order axioms, the monotone convergence theorem for real sequences (equivalently, the completeness axiom), and the sequential characterisation of continuity. It is chosen over the supremum proof because it is quantitative — Step 9 falls out for free and is the numerical method used in both worked examples below.

Set \( g := f - y \). Then \( g : [a,b] \to \mathbb{R} \) is continuous and \[ g(a) = f(a) - y \lt 0 \lt f(b) - y = g(b). \] It suffices to produce \( c \in (a,b) \) with \( g(c) = 0 \).

A continuous function minus a constant is continuous (algebra of continuous functions), and \( g(c) = 0 \iff f(c) = y \). This reduction to Bolzano’s form costs nothing and removes \( y \) from every later line. A

Define \( a_0 = a \), \( b_0 = b \), and recursively, given \( a_n \lt b_n \) in \( [a,b] \) with \( g(a_n) \lt 0 \lt g(b_n) \), put \( m_n = \dfrac{a_n + b_n}{2} \) and \[ (a_{n+1}, b_{n+1}) \;=\; \begin{cases} (m_n,\, b_n) , & g(m_n) \lt 0, \\ (a_n,\, m_n) , & g(m_n) \gt 0. \end{cases} \] If \( g(m_n) = 0 \) for some \( n \), take \( c = m_n \) and go to Step 7; so assume the recursion never terminates.

Definition by recursion on \( \mathbb{N} \), with an explicit rule — no axiom of choice is used, unlike a proof that picks arbitrary points. Trichotomy of the order on \( \mathbb{R} \) makes the three cases exhaustive and exclusive, and each surviving case reproduces the invariant \( g(a_{n+1}) \lt 0 \lt g(b_{n+1}) \). Induction then gives the invariant for every \( n \). B

For every \( n \), \[ a \le a_n \le a_{n+1} \lt b_{n+1} \le b_n \le b, \qquad b_n - a_n = \frac{b-a}{2^{n}} . \]

Both cases of Step 2 keep the new interval inside the old one and halve its length, so the length claim is an immediate induction; nesting gives the chain of inequalities. Thus \( (a_n) \) is increasing and bounded above by \( b \), and \( (b_n) \) is decreasing and bounded below by \( a \). A

There is a single point \( c \) with \[ a_n \to c, \qquad b_n \to c, \qquad c \in [a,b], \qquad a_n \le c \le b_n \ \ \text{for all } n. \]

Monotone convergence theorem (a consequence of the least-upper-bound axiom): the increasing bounded sequence \( (a_n) \) converges to \( c := \sup_n a_n \). Since \( b_n = a_n + (b-a)2^{-n} \) and \( (b-a)2^{-n} \to 0 \), the algebra of limits gives \( b_n \to c \) as well. The order limit theorem applied to \( a \le a_n \le b \) puts \( c \in [a,b] \); \( a_n \le c \) because \( c \) is an upper bound of \( \{a_n\} \), and \( c \le b_n \) because \( b_n \) is an upper bound of \( \{a_k\} \) by the nesting in Step 3. This is the only place completeness is used, and it is exactly where the argument dies over \( \mathbb{Q} \). C

\[ \lim_{n\to\infty} g(a_n) \;=\; g(c) \;=\; \lim_{n\to\infty} g(b_n). \]

Sequential characterisation of continuity at \( c \in [a,b] \): \( x_n \to c \) with \( x_n \in [a,b] \) implies \( g(x_n) \to g(c) \). Applied twice, to \( (a_n) \) and to \( (b_n) \), both of which converge to \( c \) by Step 4. At an endpoint the one-sided continuity assumed in the Statement is precisely what makes this legal. B

\[ g(a_n) \lt 0 \ \forall n \implies g(c) \le 0, \qquad g(b_n) \gt 0 \ \forall n \implies g(c) \ge 0, \qquad \text{hence } g(c) = 0 . \]

Order limit theorem: weak inequalities pass to limits (strict ones do not, which is why \( \lt \) becomes \( \le \) here — and why the argument needs both sequences, not one). Antisymmetry of \( \le \) then forces \( g(c) = 0 \). B

\( c \in (a,b) \), and \( f(c) = y \).

If \( c = a \) then \( 0 = g(c) = g(a) \lt 0 \), a contradiction; if \( c = b \) then \( 0 = g(c) = g(b) \gt 0 \), likewise. Hence \( c \) is interior. Undoing Step 1, \( f(c) = g(c) + y = y \). The early-exit branch of Step 2 lands here with \( c = m_n \in (a_n, b_n) \subseteq (a,b) \). A

Mirrored case. If \( f(b) \lt y \lt f(a) \), apply Steps 1–7 to \( \tilde f := -f \) and \( \tilde y := -y \): \[ \tilde f(a) = -f(a) \lt -y \lt -f(b) = \tilde f(b), \] so some \( c \in (a,b) \) has \( -f(c) = -y \), i.e. \( f(c) = y \). \(\blacksquare\)

\( -f \) is continuous, and negation reverses the order on \( \mathbb{R} \), so “strictly between” is preserved. No new analysis is needed. A

Quantitative form (the bisection error bound). With \( m_n \) as in Step 2, \[ \lvert m_n - c \rvert \;\le\; \frac{b_n - a_n}{2} \;=\; \frac{b-a}{2^{\,n+1}} . \]

Step 4 gives \( c \in [a_n, b_n] \), and \( m_n \) is the midpoint of that interval, so the distance from \( m_n \) to any point of it is at most half its length; Step 3 supplies the length. This bound is a priori: it is known before any evaluation of \( f \), needs no derivative, and holds for every continuous \( f \) — the reason bisection is the fail-safe of root-finding. To reach an absolute accuracy \( \tau \gt 0 \) it suffices to take \( n \ge \log_2\!\bigl((b-a)/\tau\bigr) - 1 \). C

Result

\[ f \in C\bigl([a,b],\mathbb{R}\bigr), \quad y \ \text{strictly between} \ f(a) \ \text{and} \ f(b) \quad \implies \quad \exists\, c \in (a,b) : f(c) = y \]

Reading. A continuous function cannot step over a value. If the graph starts below a horizontal line and ends above it, it meets the line somewhere strictly inside — but the theorem says nothing about where, and nothing about how many times.

Scope. Real-valued continuous functions on any interval of \( \mathbb{R} \) (for two arbitrary points of the domain, apply the theorem to the closed subinterval they span). It needs completeness of \( \mathbb{R} \) and connectedness of the domain, and it is purely an existence statement: no formula, no uniqueness, no location. The constructive content it does carry is the bisection bound \( \lvert m_n - c\rvert \le (b-a)2^{-(n+1)} \) of Step 9.

Corollaries & converses

Bolzano’s theorem. \( f \) continuous on \( [a,b] \) with \( f(a)f(b) \lt 0 \) has a zero in \( (a,b) \). (Take \( y = 0 \).) This is the form quoted in every numerical method.
Existence of \( n \)-th roots. For \( \alpha \gt 0 \) and \( n \in \mathbb{N} \), \( x \mapsto x^n \) is continuous with \( 0^n = 0 \lt \alpha \lt (1+\alpha)^n \), so \( \alpha \) has a positive \( n \)-th root; strict monotonicity makes it unique. The existence of \( \sqrt{2} \) is a corollary of IVT, not an axiom.
Odd-degree real polynomials have a real root. A monic \( p \) of odd degree satisfies \( p(-R) \lt 0 \lt p(R) \) for \( R \) large; apply Bolzano. Even-degree polynomials need not have one (\( x^2+1 \)), so parity, not degree, is what IVT can see.
One-dimensional Brouwer fixed point. Every continuous \( f : [a,b] \to [a,b] \) has a fixed point: apply Bolzano to \( g(x) = f(x) - x \), which satisfies \( g(a) \ge 0 \ge g(b) \) because \( f \) maps into \( [a,b] \). (If either inequality is an equality that endpoint is already fixed; otherwise the sign change is strict.)
Continuous images of intervals are intervals. \( f \) continuous on an interval \( I \) makes \( f(I) \) an interval. Combined with the extreme value theorem, \( f\bigl([a,b]\bigr) = [\min f, \max f] \): EVT supplies the endpoints, IVT fills the inside.
Continuous injections are strictly monotone. A continuous injective \( f : [a,b] \to \mathbb{R} \) is strictly monotone, hence has a continuous inverse. This is the route to continuity of \( \sqrt[n]{\;\cdot\;} \), \( \log \), \( \arcsin \).
Converse is false. The intermediate-value property does not imply continuity. \( f(x) = \sin(1/x) \) for \( x \ne 0 \), \( f(0) = 0 \) has the property on every interval yet is discontinuous at \( 0 \); Conway’s base-13 function attains every real value on every interval and is continuous nowhere. Darboux’s theorem (1875) says every derivative has the intermediate-value property, and derivatives can be discontinuous — so “Darboux” is strictly weaker than “continuous”.

Fails without

Continuity dropped at a single point: \( f : [-1,1] \to \mathbb{R} \) with \( f(x) = -1 \) for \( x \lt 0 \) and \( f(x) = 1 \) for \( x \ge 0 \). Then \( f(-1) \lt \tfrac12 \lt f(1) \) but \( f^{-1}\bigl(\{\tfrac12\}\bigr) = \emptyset \). In the proof, Step 5 fails: \( a_n \to 0 \) and \( b_n \to 0 \) but \( g(a_n) \to -1 - \tfrac12 \) while \( g(b_n) \to 1 - \tfrac12 \), so the two limits in Step 5 are different and Step 6 concludes nothing.
Completeness of the field dropped: \( f(x) = x^2 - 2 \) on \( [1,2] \cap \mathbb{Q} \), with values in \( \mathbb{Q} \). Continuity and the sign change \( f(1) = -1 \lt 0 \lt 2 = f(2) \) hold verbatim, but there is no rational zero. The bisection of Step 2 runs forever and produces the rational left endpoints \( 1,\ 1.25,\ 1.375,\ 1.40625, \dots \) whose limit \( \sqrt 2 \) is simply not in the field: Step 4 has no \( c \) to hand to Step 5.
Connectedness of the domain dropped: \( f(x) = 1/x \) on \( [-1,0) \cup (0,1] \) is continuous on its whole domain and changes sign, yet never vanishes. Bisection breaks at the first midpoint, \( m_0 = 0 \), which is not in the domain — the recursion of Step 2 is not even well defined.
Order on the codomain dropped: \( f(\theta) = (\cos\theta, \sin\theta) \) on \( [0, 2\pi] \) is continuous with \( f(0) = f(2\pi) = (1,0) \), and the origin is not attained. “Between” presupposes a linear order; in \( \mathbb{R}^2 \) the correct replacement is a degree or winding-number argument, and the conclusion changes character entirely (Brouwer, Poincaré–Miranda).

Common errors

“\( f \) takes all intermediate values, therefore \( f \) is continuous.” The converse is false and badly so — see Conway’s base-13 function, which is nowhere continuous. IVT is a one-way implication.
“IVT gives the root.” It gives at least one. \( \sin x = \tfrac12 \) has infinitely many solutions on \( [0, 100\pi] \). Uniqueness always requires a separate input, normally strict monotonicity or a contraction estimate.
“\( f \) changes sign across the singularity, so it has a root.” Applying IVT to \( 1/x \) on \( [-1,1] \) or to \( \tan x \) on \( [0,\pi] \). The function is not defined, let alone continuous, on the whole closed interval; the hypothesis simply fails.
“By IVT, \( c = \dots \)” followed by a formula. IVT locates nothing. What the proof yields is Step 9, an interval containing \( c \); digits come from running bisection or another algorithm, never from the theorem statement.
Confusing IVT with the mean value theorem. IVT is about values of \( f \) and needs only continuity; MVT is about values of \( f' \) and needs differentiability. “Intermediate” refers to the vertical axis, not to slopes.
Checking the endpoints and forgetting the interior. The hypothesis is continuity at every point of \( [a,b] \). A function beautifully continuous except at one interior point satisfies the endpoint test and still fails the conclusion.
Using it on a set that merely looks like an interval. \( [0,1] \cap \mathbb{Q} \), a Cantor set, or a union of two intervals all admit sign-changing continuous functions with no zero. Connectedness, not “lots of points”, is the operative property.

Discussion

Historically, IVT is where analysis stopped trusting pictures. Through the eighteenth century the statement was treated as geometrically self-evident: a curve that starts below a line and finishes above it obviously meets it. Bernard Bolzano’s 1817 pamphlet, whose title advertises a “purely analytic proof”, is the founding document of rigorous analysis precisely because it insists that the claim needs a proof, that the proof needs an \( \varepsilon\text{-}\delta \) definition of continuity, and that it must rest on a completeness principle for the reals. Cauchy’s Cours d’analyse of 1821 gives an interval-subdivision argument of the same shape as the one reproduced above, set out in its Note III as a procedure for solving equations numerically as well as a proof that a solution exists.

The modern reading is topological and much shorter. The interval \( [a,b] \) is connected; continuous maps send connected sets to connected sets; the connected subsets of \( \mathbb{R} \) are exactly the intervals; therefore \( f([a,b]) \) is an interval containing \( f(a) \) and \( f(b) \), hence containing everything between them. Every ingredient of the bisection proof reappears in the proof that \( [a,b] \) is connected, so nothing has been gained logically — but the factorisation identifies the two independent hypotheses cleanly (connected domain, continuous map) and shows immediately what generalises: IVT holds for a continuous real-valued function on any connected topological space, path-connected or not. What does not generalise is the vector-valued statement; in \( \mathbb{R}^n \) the correct successors are Brouwer’s fixed-point theorem and the Poincaré–Miranda theorem, both of which need degree theory rather than order.

Algorithmically the theorem is unusually generous. Step 9 gives a convergence rate — one bit of accuracy per function evaluation, error \( (b-a)2^{-(n+1)} \) — that is independent of \( f \). Newton’s method is quadratically faster when it converges, but it needs a derivative, a good starting point, and a non-degenerate root, and it can diverge; bisection needs only a bracket and continuity, and can never fail. Production root-finders (Brent’s method and its descendants) are hybrids that attempt fast interpolation and fall back on bisection whenever the fast step would leave the bracket — the fallback is exactly the guarantee proved above. Worked Example 1 runs the method on a real equation of state and the a priori bound of Step 9 is compared with the error actually achieved.

Foundationally IVT is a delicate instrument, and its exact strength is well mapped. Over an ordered field \( F \), the statement “every continuous \( f : [a,b] \to F \) with \( f(a)f(b) \lt 0 \) has a zero” is equivalent to Dedekind completeness of \( F \), so IVT can serve as an axiom characterising \( \mathbb{R} \) among ordered fields. Constructively, in Bishop’s sense, the exact theorem is not provable: the trichotomy used in Step 2 to decide the sign of \( g(m_n) \) is not an algorithm on real inputs, and a function that hovers at the target value defeats any purported procedure. What survives is the approximate IVT — for every \( \varepsilon \gt 0 \) there is \( x \) with \( \lvert f(x) - y\rvert \lt \varepsilon \) — together with the exact theorem for functions that are locally non-constant, which covers every function met in practice. In reverse mathematics IVT for continuous functions presented with a modulus of continuity is provable outright in the weak base system \( \mathsf{RCA}_0 \) — strictly weaker than the extreme value theorem, which is equivalent to weak König’s lemma \( \mathsf{WKL}_0 \), and weaker again than the Bolzano–Weierstrass theorem, which is equivalent to arithmetical comprehension \( \mathsf{ACA}_0 \). So although all three are “consequences of completeness”, IVT is the cheapest of them, which is the formal counterpart of the fact that bisection is implementable and the supremum in EVT is not.

Common misconceptions. IVT does not say the value is attained once; it does not say where; it does not require differentiability, monotonicity or boundedness; and its converse fails so badly that a nowhere-continuous function can satisfy the intermediate-value property in the strongest possible sense. Nor is it “obvious”: it is false over \( \mathbb{Q} \), false on disconnected domains, and false for maps into \( \mathbb{R}^2 \), so every proof must at some point cash in the completeness of \( \mathbb{R} \) and the connectedness of \( [a,b] \).

Worked examples

Example 1 (molar volume from a van der Waals equation of state). Carbon dioxide is modelled by the van der Waals equation \[ p(V) \;=\; \frac{RT}{V-b} \;-\; \frac{a}{V^{2}}, \] with the standard tabulated constants for \( \mathrm{CO_2} \), \( a = 0.3640\ \mathrm{Pa\,m^{6}\,mol^{-2}} \) and \( b = 4.267\times10^{-5}\ \mathrm{m^{3}\,mol^{-1}} \), and the molar gas constant \( R = 8.3145\ \mathrm{J\,mol^{-1}\,K^{-1}} \). At \( T = 350\ \mathrm{K} \) (above the critical temperature \( 304\ \mathrm{K} \), so the isotherm is strictly decreasing) show that the equation \( p(V) = 8.000\ \mathrm{MPa} \) has a solution with \( V \) between \( 2.000\times10^{-4} \) and \( 8.000\times10^{-4}\ \mathrm{m^{3}\,mol^{-1}} \), and locate it by five bisections.

Put \( \Phi(V) := p(V) - p_\star \) with \( p_\star = 8.000\times10^{6}\ \mathrm{Pa} \), on the interval \( [V_1, V_2] = [2.000\times10^{-4},\, 8.000\times10^{-4}]\ \mathrm{m^{3}\,mol^{-1}} \). Then \( \Phi \) is continuous on \( [V_1,V_2] \).

On this interval \( V - b \ge 2.000\times10^{-4} - 4.267\times10^{-5} = 1.5733\times10^{-4}\ \mathrm{m^{3}\,mol^{-1}} \gt 0 \) and \( V \ne 0 \), so both denominators are bounded away from zero; \( \Phi \) is a quotient-and-difference of continuous functions with non-vanishing denominators. Checking this before quoting IVT is the whole discipline: the theorem is worthless across a singularity. A

Clearing the denominators, which are both strictly positive on \( [V_1,V_2] \), the sign of \( \Phi(V) \) is the sign of \[ \Psi(V) \;=\; RT\,V^{2} \;-\; \bigl(p_\star V^{2} + a\bigr)(V-b). \] Numerically, with \( RT = 8.3145 \times 350 = 2910.1\ \mathrm{J\,mol^{-1}} \), \[ p(V_1) = 9.397\ \mathrm{MPa}, \qquad p(V_2) = 3.274\ \mathrm{MPa}, \] so \( \Phi(V_1) = +1.397\ \mathrm{MPa} \gt 0 \gt -4.726\ \mathrm{MPa} = \Phi(V_2) \).

Rearranged before substitution so the sign test is a polynomial comparison and no rounding can flip it. Both evaluations use only the two tabulated constants and \( R \); every term of \( p \) carries units of \( \mathrm{Pa} \), since \( \mathrm{J\,mol^{-1}}/\mathrm{m^{3}\,mol^{-1}} = \mathrm{Pa} \) and \( \mathrm{Pa\,m^{6}\,mol^{-2}}/\mathrm{m^{6}\,mol^{-2}} = \mathrm{Pa} \). A

By the intermediate value theorem there exists \( V_\star \in (V_1, V_2) \) with \( \Phi(V_\star) = 0 \), i.e. \( p(V_\star) = 8.000\ \mathrm{MPa} \).

Hypotheses discharged in Steps 1–2: continuity on the closed interval, and \( 0 \) strictly between the endpoint values. Existence is established before a single digit of \( V_\star \) is computed — that ordering is the point of the theorem. B

Bisection (Step 2 of the proof), lengths in \( 10^{-4}\ \mathrm{m^{3}\,mol^{-1}} \): \[ \begin{array}{lll} m_0 = 5.0000 : & p = 4.907\ \mathrm{MPa} \lt p_\star & \Rightarrow\ [2.0000,\,5.0000] \\ m_1 = 3.5000 : & p = 6.497\ \mathrm{MPa} \lt p_\star & \Rightarrow\ [2.0000,\,3.5000] \\ m_2 = 2.7500 : & p = 7.712\ \mathrm{MPa} \lt p_\star & \Rightarrow\ [2.0000,\,2.7500] \\ m_3 = 2.3750 : & p = 8.483\ \mathrm{MPa} \gt p_\star & \Rightarrow\ [2.3750,\,2.7500] \\ m_4 = 2.5625 : & p = 8.082\ \mathrm{MPa} \gt p_\star & \Rightarrow\ [2.5625,\,2.7500] \end{array} \]

At each stage the half retaining the sign change is kept, exactly as in the proof; \( p \) decreasing means \( p(m) \gt p_\star \) puts the root to the right of \( m \). Five evaluations of \( p \), no derivative. B

\[ V_\star \approx m_5 = \frac{2.5625 + 2.7500}{2}\times 10^{-4} = 2.656\times10^{-4}\ \mathrm{m^{3}\,mol^{-1}}, \qquad \lvert m_5 - V_\star\rvert \le \frac{V_2 - V_1}{2^{6}} = 9.375\times10^{-6}\ \mathrm{m^{3}\,mol^{-1}}. \]

Step 9 of the proof with \( n = 5 \). The true root, from continued bisection, is \( V_\star = 2.6027\times10^{-4}\ \mathrm{m^{3}\,mol^{-1}} \); the actual error \( 5.4\times10^{-6} \) does indeed sit under the guaranteed bound, as it must. A

\[ V_\star \;=\; \bigl(2.656 \pm 0.094\bigr)\times10^{-4}\ \mathrm{m^{3}\,mol^{-1}} \quad\text{after 5 bisections}; \qquad V_\star = 2.6027\times10^{-4}\ \mathrm{m^{3}\,mol^{-1}} \ \text{exactly to 5 s.f.} \]

Reading. Two evaluations of the equation of state prove that a physical state at \( 350\ \mathrm{K} \) and \( 8.000\ \mathrm{MPa} \) exists in the model; five more pin it to \( \pm 9.4\times10^{-6}\ \mathrm{m^{3}\,mol^{-1}} \), about \( \pm 3.5\% \). For comparison the ideal-gas value is \( RT/p_\star = 3.638\times10^{-4}\ \mathrm{m^{3}\,mol^{-1}} \), so at this pressure attraction and excluded volume together compress \( \mathrm{CO_2} \) to \( 72\% \) of the ideal molar volume — a correction no linearisation about the ideal gas would have found.

Scope. Existence needs only continuity on a bracket with a sign change. Uniqueness here comes from outside IVT: above the critical temperature \( p \) is strictly decreasing in \( V \), so the root is unique. Below \( T_c \) the van der Waals loop makes \( p \) non-monotone and the same pressure can be met at three molar volumes — IVT would still certify existence and would still say nothing about how many.

Example 2 (the half-mass point of a non-uniform rod). A rod of length \( L = 1.20\ \mathrm{m} \) has linear mass density \( \lambda(x) = \lambda_0\bigl(1 + 2x/L\bigr) \) with \( \lambda_0 = 0.500\ \mathrm{kg\,m^{-1}} \). Prove that there is a point at which a cut divides the rod into two pieces of equal mass, and find it.

\[ M \;=\; \int_0^{L} \lambda(x)\,\mathrm{d}x \;=\; \lambda_0\left[x + \frac{x^{2}}{L}\right]_0^{L} \;=\; 2\lambda_0 L \;=\; 2(0.500)(1.20) = 1.20\ \mathrm{kg}. \]

Direct integration of an affine density; symbols first, numbers last. The target is \( M/2 = \lambda_0 L = 0.600\ \mathrm{kg} \). A

Define the accumulated-mass function \[ m(t) \;=\; \int_0^{t} \lambda(x)\,\mathrm{d}x \;=\; \lambda_0\left(t + \frac{t^{2}}{L}\right), \qquad t \in [0,L], \] which is continuous on \( [0,L] \).

Continuity does not need the closed form: \( 0 \lt \lambda(x) \le \lambda_0(1+2) = 1.50\ \mathrm{kg\,m^{-1}} \) on \( [0,L] \) gives \( \lvert m(t) - m(s)\rvert \le 1.50\,\lvert t-s\rvert \), so \( m \) is Lipschitz, hence continuous. That argument survives for any bounded density, integrable or not in closed form — which is why the method generalises. B

\[ m(0) = 0 \;\lt\; \tfrac{M}{2} = 0.600\ \mathrm{kg} \;\lt\; 1.20\ \mathrm{kg} = m(L), \] so by the intermediate value theorem there exists \( t_\star \in (0,L) \) with \( m(t_\star) = M/2 \).

IVT with \( y = M/2 \) strictly between the endpoint values. The cut at \( t_\star \) leaves \( m(t_\star) = M/2 \) to the left and \( M - m(t_\star) = M/2 \) to the right. B

Solve symbolically: \( \lambda_0\bigl(t_\star + t_\star^{2}/L\bigr) = \lambda_0 L \) gives \[ t_\star^{2} + L\,t_\star - L^{2} = 0 \quad\Longrightarrow\quad t_\star = \frac{\sqrt5 - 1}{2}\,L \;=\; \frac{L}{\varphi}, \qquad \varphi = \frac{1+\sqrt5}{2}. \]

The quadratic formula, with the negative root discarded because \( t_\star \in (0,L) \). Note \( \lambda_0 \) cancels: the half-mass point depends on the density profile, not its scale. The golden ratio appears because this particular profile triples the density across the rod, \( \lambda(L) = 3\lambda(0) \), which is exactly the ratio that reduces the half-mass condition to \( t_\star^{2} + L t_\star - L^{2} = 0 \). B

\[ t_\star = 0.618034 \times 1.20\ \mathrm{m} = 0.7416\ \mathrm{m}, \qquad m(t_\star) = 0.500\left(0.7416 + \frac{0.7416^{2}}{1.20}\right) = 0.600\ \mathrm{kg}. \ \checkmark \]

Substitution and an arithmetic check against the target of Step 1. Uniqueness: \( \lambda \gt 0 \) makes \( m \) strictly increasing, so \( t_\star \) is the only such point — again supplied by monotonicity, not by IVT. A

\[ t_\star \;=\; \frac{\sqrt5 - 1}{2}\,L \;=\; 0.7416\ \mathrm{m} \quad\text{measured from the light end, with } M = 1.20\ \mathrm{kg} \]

Reading. The cut sits at \( 61.8\% \) of the length, biased towards the heavy end, because the far half of the rod is denser. IVT is what turns “the accumulated mass grows continuously from \( 0 \) to \( M \)” into “every intermediate mass is realised by some cut”.

Scope. The existence argument uses only \( 0 \lt \lambda \le \lambda_{\max} \lt \infty \) and needs no formula for \( m \); the closed form in Step 4 is a bonus of this particular profile. The same argument gives a cut realising any prescribed fraction of \( M \), and it is the one-dimensional case of the ham-sandwich family of bisection results. It fails for a rod with a gap: on a disconnected body \( m \) jumps, and an intermediate mass can be skipped.

Problems

Show that \( x^{3} - 2x - 5 = 0 \) has a root in \( (2,3) \), then perform four bisections and state both the guaranteed error bound and the error actually achieved (the root is \( 2.0945515 \) to 8 s.f.).
Solution

Let \( f(x) = x^{3} - 2x - 5 \), a polynomial, hence continuous on \( [2,3] \). Then \( f(2) = 8 - 4 - 5 = -1 \lt 0 \) and \( f(3) = 27 - 6 - 5 = 16 \gt 0 \), so \( 0 \) lies strictly between \( f(2) \) and \( f(3) \) and IVT gives \( c \in (2,3) \) with \( f(c) = 0 \).

Bisection: \( m_0 = 2.5 \), \( f = 15.625 - 5 - 5 = +5.625 \Rightarrow [2, 2.5] \). \( m_1 = 2.25 \), \( f = 11.390625 - 4.5 - 5 = +1.890625 \Rightarrow [2, 2.25] \). \( m_2 = 2.125 \), \( f = 9.595703125 - 4.25 - 5 = +0.345703125 \Rightarrow [2, 2.125] \). \( m_3 = 2.0625 \), \( f = 8.773681640625 - 4.125 - 5 = -0.351318359375 \Rightarrow [2.0625,\ 2.125] \).

Take \( m_4 = 2.09375 \). The guaranteed bound (Step 9 with \( n = 4 \), \( b - a = 1 \)) is \( \lvert m_4 - c\rvert \le 2^{-5} = 0.03125 \). The actual error is \( \lvert 2.09375 - 2.0945515\rvert = 8.0\times10^{-4} \), comfortably inside it. Note the bound is not tight — it cannot be, since it is independent of \( f \).
Prove that \( \mathrm{e}^{-x} = x^{2} \) has a solution in \( (0,1) \) and locate it to within \( 0.05 \). Is the solution unique in \( \mathbb{R} \)?
Solution

Let \( g(x) = \mathrm{e}^{-x} - x^{2} \), continuous on \( [0,1] \) as a difference of continuous functions. \( g(0) = 1 \gt 0 \) and \( g(1) = \mathrm{e}^{-1} - 1 = -0.632121 \lt 0 \), so IVT gives a zero in \( (0,1) \).

Bisection: \( g(0.5) = 0.606531 - 0.25 = +0.356531 \Rightarrow [0.5, 1] \); \( g(0.75) = 0.472367 - 0.5625 = -0.090133 \Rightarrow [0.5, 0.75] \); \( g(0.625) = 0.535261 - 0.390625 = +0.144636 \Rightarrow [0.625, 0.75] \); \( g(0.6875) = 0.502832 - 0.472656 = +0.030175 \Rightarrow [0.6875, 0.75] \). The bracket has length \( 0.0625 \), so its midpoint \( 0.71875 \) is within \( 0.03125 \lt 0.05 \) of the root. (The root is \( 0.703467 \); the actual error is \( 0.0153 \).)

Uniqueness on \( \mathbb{R} \): for \( x \le 0 \), \( \mathrm{e}^{-x} \ge 1 \) while \( x^{2} \lt 1 \) on \( (-1,0] \) and, for \( x \le -1 \), \( \mathrm{e}^{-x} \ge \mathrm{e}^{\lvert x\rvert} \gt x^{2} \) since \( \mathrm{e}^{t} \gt t^{2} \) for all \( t \ge 1 \); so \( g \gt 0 \) there and no root exists. On \( [0,\infty) \), \( \mathrm{e}^{-x} \) is strictly decreasing and \( x^{2} \) strictly increasing, so \( g \) is strictly decreasing and has at most one zero. Hence the solution is unique. As always, IVT supplied existence and monotonicity supplied uniqueness.
Let \( f : [0,3] \to \mathbb{R} \) be continuous with \( f(0) = f(3) \). Prove that there is \( x \in [0,2] \) with \( f(x) = f(x+1) \). Then find such an \( x \) explicitly for \( f(x) = \sin(2\pi x/3) \).
Solution

Put \( h(x) = f(x+1) - f(x) \) on \( [0,2] \); \( h \) is continuous (difference of \( f \) with a shift of \( f \), both continuous). Telescoping, \[ h(0) + h(1) + h(2) = \bigl(f(1)-f(0)\bigr) + \bigl(f(2)-f(1)\bigr) + \bigl(f(3)-f(2)\bigr) = f(3) - f(0) = 0. \] Three reals summing to \( 0 \) cannot all be strictly positive, nor all strictly negative. So either some \( h(k) = 0 \), in which case \( x = k \) works; or two of them have strictly opposite signs, say \( h(j) \lt 0 \lt h(k) \), and IVT applied to \( h \) on the closed interval with endpoints \( j \) and \( k \) (a subinterval of \( [0,2] \)) produces \( x \) strictly between them with \( h(x) = 0 \), i.e. \( f(x) = f(x+1) \).

For \( f(x) = \sin(2\pi x/3) \): \( f(0) = 0 = \sin 2\pi = f(3) \), so the hypothesis holds. Using \( \sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \) with \( A = 2\pi(x+1)/3 \), \( B = 2\pi x/3 \), \[ h(x) = 2\sin\frac{\pi}{3}\,\cos\frac{\pi(2x+1)}{3} = \sqrt3\,\cos\frac{\pi(2x+1)}{3}. \] This vanishes when \( \pi(2x+1)/3 = \pi/2 \), i.e. \( x = \tfrac14 \), and again at \( x = \tfrac74 \). Check: \( f(0.25) = \sin(\pi/6) = 0.5 \) and \( f(1.25) = \sin(5\pi/6) = 0.5 \). Both roots lie in \( [0,2] \), illustrating that the conclusion asserts existence, not uniqueness.
Let \( f \) be continuous on \( [a,b] \) and let \( x_1,\dots,x_n \in [a,b] \). Prove that \( f \) attains the average of the sampled values: there is \( c \in [a,b] \) with \( f(c) = \frac1n\sum_{k=1}^{n} f(x_k) \). Verify with \( f(x) = x^{2} \) on \( [0,3] \) and the samples \( 0, 1, 3 \).
Solution

Write \( A = \frac1n\sum_{k} f(x_k) \). The finite set \( \{f(x_1),\dots,f(x_n)\} \) has a least element \( f(x_j) =: \mu \) and a greatest \( f(x_i) =: \nu \) (a finite non-empty set of reals attains both — no analysis needed). Since every term of the average lies in \( [\mu,\nu] \), \[ \mu = \frac1n\sum_k \mu \;\le\; A \;\le\; \frac1n\sum_k \nu = \nu . \] If \( A = \mu \) take \( c = x_j \); if \( A = \nu \) take \( c = x_i \). Otherwise \( \mu \lt A \lt \nu \), and \( A \) lies strictly between the values of \( f \) at the two points \( x_j, x_i \in [a,b] \). Apply IVT on the closed interval with endpoints \( x_j \) and \( x_i \), which is contained in \( [a,b] \) and on which \( f \) is continuous: it yields \( c \) strictly between \( x_j \) and \( x_i \) with \( f(c) = A \). \(\blacksquare\)

Verification: \( f(0) = 0 \), \( f(1) = 1 \), \( f(3) = 9 \), so \( A = (0+1+9)/3 = 10/3 = 3.3333 \). Here \( \mu = 0 \) at \( x_j = 0 \) and \( \nu = 9 \) at \( x_i = 3 \), and \( 0 \lt 10/3 \lt 9 \). Solving \( c^{2} = 10/3 \) with \( c \gt 0 \) gives \( c = \sqrt{10/3} = 1.82574 \in (0,3) \), as predicted. Note the same argument with weights \( w_k \ge 0 \), \( \sum w_k = 1 \), gives any convex combination, and the limiting integral version \( f(c) = \frac{1}{b-a}\int_a^b f \) follows by combining IVT with the extreme value theorem.
(Universal chord theorem.) Let \( f : [0,1] \to \mathbb{R} \) be continuous with \( f(0) = f(1) \), and let \( n \in \mathbb{N} \). Prove there is \( x \in \bigl[0, 1 - \tfrac1n\bigr] \) with \( f\bigl(x + \tfrac1n\bigr) = f(x) \). Then show the analogous statement is false for chord length \( \lambda = \tfrac23 \), using \( f(x) = \sin^{2}\bigl(\tfrac{3\pi x}{2}\bigr) - x \).
Solution

Reciprocal lengths. Put \( h(x) = f\bigl(x+\tfrac1n\bigr) - f(x) \) on \( \bigl[0, 1-\tfrac1n\bigr] \), continuous as before. Telescoping over the \( n \) grid points \( x = k/n \), \( k = 0,\dots,n-1 \), \[ \sum_{k=0}^{n-1} h\!\left(\frac{k}{n}\right) = \sum_{k=0}^{n-1}\left[ f\!\left(\frac{k+1}{n}\right) - f\!\left(\frac{k}{n}\right)\right] = f(1) - f(0) = 0 . \] If any \( h(k/n) = 0 \) we are done. Otherwise the \( n \) numbers are non-zero and sum to zero, so at least one is positive and at least one negative, say \( h(j/n) \lt 0 \lt h(k/n) \). Both \( j/n \) and \( k/n \) lie in \( \bigl[0,1-\tfrac1n\bigr] \), \( h \) is continuous on the closed interval between them, and \( 0 \) is strictly between the two values; IVT delivers \( x \) there with \( h(x) = 0 \). \(\blacksquare\)

Concrete instance, \( n = 3 \), \( f(x) = x(1-x) \) (so \( f(0) = f(1) = 0 \)): \( h(x) = \bigl(x+\tfrac13\bigr)\bigl(\tfrac23 - x\bigr) - x(1-x) = \tfrac29 - \tfrac23 x \), which vanishes at \( x = \tfrac13 \). Check \( f(1/3) = \tfrac29 = f(2/3) \). \(\checkmark\)

Failure for \( \lambda = \tfrac23 \). Take \( f(x) = \sin^{2}\bigl(\tfrac{3\pi x}{2}\bigr) - x \) on \( [0,1] \), which is continuous. Endpoints: \( f(0) = 0 \) and \( f(1) = \sin^{2}\bigl(\tfrac{3\pi}{2}\bigr) - 1 = 1 - 1 = 0 \), so \( f(0) = f(1) \) as required. For any \( x \in \bigl[0,\tfrac13\bigr] \), since \( \tfrac{3\pi}{2}\bigl(x + \tfrac23\bigr) = \tfrac{3\pi x}{2} + \pi \) and \( \sin^{2}(\theta + \pi) = \sin^{2}\theta \), \[ f\!\left(x+\tfrac23\right) - f(x) = \sin^{2}\!\left(\tfrac{3\pi x}{2} + \pi\right) - \sin^{2}\!\left(\tfrac{3\pi x}{2}\right) - \tfrac23 = -\tfrac23 \;\ne\; 0 . \] So no chord of length \( \tfrac23 \) is horizontal, and the telescoping argument above genuinely needs \( \lambda = 1/n \): with \( \lambda \) not the reciprocal of an integer the sum of the increments over a partition never closes up. The general statement — horizontal chords of length \( \lambda \) are forced exactly for \( \lambda \in \{1, \tfrac12, \tfrac13, \dots\} \), with counterexamples of the above shape otherwise — is the universal chord theorem.