Intermediate values and bounded behaviour

A small sign calculation asks a bigger existence question

For f(x)=x³−2, f(1)=−1 and f(2)=6. Does some input between 1 and 2 give output zero? A continuous change of input cannot pass from below zero to above zero while skipping zero. The intermediate value theorem makes that intuition precise.

Evaluate the midpoint: f(3/2)=27/8−2=11/8>0. A root is therefore retained between 1 and 3/2. Next f(5/4)=125/64−2=−3/64<0, so retain [5/4,3/2]. These exact sign calculations locate a root without claiming that a tested rational midpoint is the exact root.

The function is defined and continuous at every point of [1,2], including the endpoints relative to that interval. A graph or five sample values cannot check this global condition. Here the polynomial formula supplies it; the continuity definition is recalled next.

Use the published preparation links for domains, squares and interval inequalities. Read the target and extreme-value examples first if desired, then return to the nested-interval and subsequence proofs. Their needed completeness and sequence ideas are defined locally, with no later calculus lesson assumed.

Continuity controls nearby outputs, within the domain

For f:D→ℝ and a∈D, continuity at a means that for every ε>0 there is δ>0 such that every x∈D with |x−a|<δ satisfies |f(x)−f(a)|<ε. Input a itself is allowed and automatically satisfies the output inequality. At an endpoint of a restricted interval, only inputs from that interval are tested.

On [u,v] with u<v, being continuous means meeting that condition at every a∈[u,v]: right-side inputs at u, left-side inputs at v and both nearby sides in the interior. A missing interior point cannot be filled by calling a function continuous only on the remaining domain.

Continuity gives a local bound. Set ε=1. For the resulting δ, |f(x)|≤|f(a)|+|f(x)−f(a)|<|f(a)|+1 for every permitted x with |x−a|<δ. This bound covers one neighbourhood, not automatically the entire domain.

For 1/x on (0,1], continuity holds at every permitted input, and each such input has a local bound. There is nevertheless no common finite upper bound for the whole domain: f(1/n)=n for each positive integer n. A bounded input domain need not give bounded outputs when an endpoint is excluded.

The polynomial rules used here are continuous by direct error bounds. If |x−a|<1, then |x²−a²|≤(2|a|+1)|x−a| and |x³−a³|≤(3a²+3|a|+1)|x−a|, by factoring each difference. Choosing δ smaller than 1 and ε divided by the positive coefficient proves continuity. Adding constants or finitely many such bounds gives the displayed polynomial combinations.

For 1/x at a≠0, require |x−a|<|a|/2 so |x|>|a|/2. Then |1/x−1/a|≤2|x−a|/a². A δ no larger than min(|a|/2,εa²/2) proves continuity on its real domain. These are local statements and do not permit a continuity claim at the excluded zero.

The intermediate value theorem reaches each between-output target

Let a<b be real, and let f:[a,b]→ℝ be continuous on the entire interval. If min(f(a),f(b))≤y≤max(f(a),f(b)), then some c∈[a,b] satisfies f(c)=y. If y lies strictly between the endpoint outputs, c lies in (a,b), because neither endpoint has that output.

Equality is allowed in the inclusive version: if y=f(a), take c=a; if y=f(b), take c=b. This does not promise an interior input for an endpoint target. If the endpoint outputs are equal, the inclusive endpoint comparison guarantees only that common output, although other values may occur inside.

The theorem gives at least one input, not exactly one. On [−2,2], x² reaches target 1 at both −1 and 1 although both endpoint outputs are 4. Failure of the endpoint-between test does not prove a target is absent. Conversely, existence of a target input does not establish continuity.

To solve f(x)=g(x), apply the zero-target case to h(x)=f(x)−g(x), after verifying that both rules are continuous throughout the chosen interval. Opposite strict signs of h at the endpoints give at least one interior intersection. The argument needs no derivative and no closed-form root.

A jump or a missing input can skip the target

Define s(x)=−1 for x<0 and s(x)=1 for x≥0 on [−1,1]. Its endpoint signs differ, yet zero is never an output. The failed hypothesis is continuity at 0: outputs on the left remain −1 while s(0)=1.

The rule r(x)=1/x is continuous at each input of [−1,1]∖{0} and has endpoint outputs −1 and 1, yet r(x)≠0 everywhere in that domain. The domain does not contain the interval connecting the endpoints. Do not join its branches across the missing zero to invoke the theorem.

Continuity is sufficient for the between-value property, not something the theorem proves from a graph or a sign test. A discontinuous function can still happen to reach a chosen target. Each failed hypothesis means the stated guarantee cannot be used; it does not force failure of every conclusion.

Completeness supplies a point inside every shrinking bracket

We use a defining completeness property of the real numbers: every nonempty set of real numbers bounded above has a least upper bound, called its supremum. An upper bound is at least every element; the least one need not itself belong to the set. For example, the numbers x with 0≤x<1 have supremum 1 but no largest element.

Suppose nonempty closed intervals Iₙ=[aₙ,bₙ] are nested, meaning Iₙ₊₁⊆Iₙ, and their widths tend to zero. Define c=sup{aₙ:n≥0}. The left endpoints are bounded above by b₀. For each fixed n, every left endpoint is at most bₙ: earlier ones are at most aₙ, and later ones belong to Iₙ. Hence aₙ≤c≤bₙ for every n.

If another point d belongs to every Iₙ, then |c−d|≤bₙ−aₙ for every n. Widths becoming arbitrarily small force c=d. Also |aₙ−c| and |bₙ−c| are at most the width, so both endpoints converge to c. Here convergence means that every positive error eventually bounds all remaining differences.

For halving intervals, the width after n halvings is (b₀−a₀)/2ⁿ. These widths tend to zero: 2ⁿ≥n+1, and positive integers can exceed any fixed real number. This uses the Archimedean property of ℝ, stated here as that absence of a finite upper bound for the positive integers.

The real-number assumption matters. For rational inputs alone, x²−2 on [1,2]∩ℚ is continuous relative to that domain and has opposite endpoint signs, but has no rational zero. If a reduced positive fraction p/q had square 2, p²=2q² would make p even, then q even, contradicting reduction. ℚ lacks the needed real completeness and full real interval.

Bisection preserves signs and proves a root exists

For a strict target between f(a₀) and f(b₀), subtract the target and, if necessary, reverse the sign so the continuous rule g has g(a₀)<0<g(b₀). At midpoint m=(aₙ+bₙ)/2, stop if g(m)=0. Otherwise retain [m,bₙ] when g(m)<0 and [aₙ,m] when g(m)>0. The new endpoints still have opposite signs.

If no exact midpoint root is found, the nested closed brackets halve indefinitely and have a unique common real point c by the previous argument. Their endpoints converge to c. Continuity makes g(aₙ) and g(bₙ) converge to g(c): given an output tolerance, eventually each endpoint is inside the continuity input tolerance.

The negative endpoint values imply g(c)≤0, and the positive endpoint values imply g(c)≥0. For instance, if g(c)>0, continuity with tolerance g(c)/2 would make all sufficiently near negative-side endpoint outputs positive, a contradiction. The negative case is analogous. Thus g(c)=0, proving the theorem rather than merely displaying a long table.

After n unresolved halvings, at least one root remains in a bracket of width w=(b₀−a₀)/2ⁿ. The bracket midpoint approximates a root with input error at most w/2. This bounds distance in x, not automatically the output error |g(m)|. A slope bound or another estimate would be needed to translate one into the other.

Finite exact bisection is meaningful only with valid signs and continuity over the connecting interval. A sign change caused by a jump or pole can narrow around a non-root. Computed decimal signs can also be uncertain through rounding; this explorer uses exact rational arithmetic for its finite steps. It is not a general numerical solver.

A bounding level can exist without being reached

A real function is bounded on D if some finite B≥0 satisfies |f(x)|≤B for every x∈D. An upper bound alone controls outputs from above; a lower bound controls them from below. A minimum is an attained output f(p) with p∈D and f(p)≤f(x) for every x∈D. A maximum reverses that inequality.

The greatest lower bound of a nonempty set bounded below is its infimum; the least upper bound is its supremum. These are levels, while p is an input where a level may be attained. Completeness gives finite infimum and supremum for a nonempty bounded output set, but does not by itself put them into that set.

For q(x)=x(2−x)=1−(x−1)² on [0,2], the squared term proves q≤1, with equality at x=1. Also x≥0 and 2−x≥0 prove q≥0, with equality at x=0 and x=2. Minimum 0 and maximum 1 are attained; the maximum input is 1, distinct from the maximum output 1 even though their numbers coincide here.

On (0,2), q(x)>0, so it has no minimum: outputs approach 0 as x approaches an excluded endpoint. It still attains maximum 1 at x=1. On [0,1), q has minimum 0 but no maximum; the supremum 1 is approached at the excluded input 1. Thus boundedness and attainment are separate questions.

A closed bounded interval keeps a subsequence limit inside

A sequence x₁,x₂,… is an ordered infinite list of real inputs. A subsequence selects indices n₁<n₂<⋯. Every sequence in a closed bounded interval [a,b] has a subsequence converging to a point c∈[a,b]. We prove the fact needed here instead of assuming a later sequences lesson.

For a<b, at least one closed half of [a,b] contains infinitely many terms of the sequence; if both contained only finitely many, their union could not contain all infinitely many indices. Keep such a half and choose a term there with an index larger than the previous selected index. Repeat, halving each retained interval. A term on their common midpoint can belong to both halves without invalidating this argument.

The nested-interval result gives a common c. Each selected term is in its corresponding shrinking interval, so its distance from c is at most that interval's width and tends to zero. It stays in [a,b], including possible endpoint limits. If a=b, the sequence is constant and the conclusion is immediate.

For continuous f and any inputs xₙ→c within its domain, f(xₙ)→f(c): given ε>0, use continuity's δ and then convergence's eventual |xₙ−c|<δ. This works even when infinitely many terms equal c. The two ingredients are where the inputs go and how continuity transfers that to outputs.

Continuity on a closed bounded interval gives a global bound

Let f:[a,b]→ℝ be continuous, with real a≤b. It is bounded. For a=b there is only f(a). For a<b, suppose instead there were no finite bound. Then for each positive integer n choose xₙ∈[a,b] with |f(xₙ)|>n.

A subsequence xₙₖ converges to c∈[a,b] by the previous proof. Continuity gives f(xₙₖ)→f(c), so eventually |f(xₙₖ)|<|f(c)|+1. But |f(xₙₖ)|>nₖ≥k grows beyond that fixed number, a contradiction. Hence a finite global bound exists.

Bounded inputs made the subsequence argument available. Closedness kept its limit in the domain, where continuity could be used. Continuity turned convergence of inputs into a finite output limit. Omitting any one of these checks can break the conclusion.

Compare the identity f(x)=x on ℝ, which has an unbounded domain and unbounded outputs; 1/x on (0,1], with an excluded endpoint and unbounded outputs; and f(0)=0,f(x)=1/x for 0<x≤1 on [0,1], which is not continuous at 0 and remains unbounded. These counterexamples remove different hypotheses.

The extreme value theorem adds attainment

A continuous real function on a nonempty closed bounded interval [a,b] attains both a minimum and a maximum. The theorem promises inputs p,q∈[a,b] with f(p)≤f(x)≤f(q) for every x in the interval. It does not locate them, force them into the interior or make them unique.

By the boundedness proof, the nonempty output set has a finite supremum S. For every positive integer n there is an input xₙ with S−1/n<f(xₙ)≤S; otherwise S−1/n would be a smaller upper bound. Select a subsequence converging to q∈[a,b]. Its outputs tend both to S, by their shrinking error bounds, and to f(q), by continuity. Therefore f(q)=S.

Apply the same maximum argument to the continuous rule −f. Its attained maximum gives an attained minimum of f. For a singleton interval, the single output is both extremes. This completes the theorem without differentiation, critical-point rules or assuming that a finite sample contains the extreme inputs.

The identity on [0,1] attains both extremes at endpoints. The bounded rule f(x)=x for 0≤x<1 with f(1)=0 on [0,1] has minimum 0 but no maximum: its supremum is 1 and no output equals 1. This second rule is not continuous at its right endpoint. Boundedness on a closed interval alone is insufficient.

Together with intermediate values, the theorem says that the full image of a continuous function on [a,b] is exactly [m,M], where m and M are its attained extreme outputs. Every between-output value is reached on the subinterval joining an extreme input to the other, ordered from left to right. This does not say the extremes equal f(a) and f(b).

Keep the guarantee, the calculation and the approximation separate

For a target question, write the complete connecting interval, establish continuity there and compare the target with its endpoint outputs. A strict between-output target gives an interior input; an endpoint target can be reached at that endpoint. Add a separate argument if uniqueness is required.

For a bound or extreme-value question, name the exact domain, distinguish upper and lower bounds from attainment and check whether the interval is closed and bounded. Use algebra for the displayed examples; a drawing that leaves its window gives neither a proof of unboundedness nor a largest value.

The explorer lets you predict target guarantees, compare endpoint inclusion and refine an exact root bracket. Return to the proofs by identifying the sign invariant for intermediate values and the convergent input subsequence for boundedness and attained extrema. Each uses real completeness; a finite table illustrates rather than replaces that step.

Worked examples

Try it yourself

1. For continuous f on [2,5] with f(2)=7 and f(5)=−1, what does target 3 guarantee?

   Show the answer and reasoning

   Some c∈(2,5) has f(c)=3.

   The target lies strictly between −1 and 7. The order of endpoint outputs does not have to be increasing.

2. With the same hypotheses, must target 7 occur at an interior input?

   Show the answer and reasoning

   No. It is already reached at input 2, but an interior occurrence is not guaranteed.

   For example f(x)=7−(8/3)(x−2) on [2,5] reaches 7 only at 2.

3. If a continuous rule has equal endpoint outputs, are all its interior outputs equal too?

   Show the answer and reasoning

   No.

   The rule x² on [−1,1] has equal endpoint outputs 1 but gives 0 at its centre. The endpoint comparison guarantees only the common endpoint output.

4. Does the step rule −1 for x<0 and 1 for x≥0 on [−1,1] have a zero?

   Show the answer and reasoning

   No.

   Its only outputs are −1 and 1. It violates continuity at zero, so the opposite endpoint signs cannot invoke the intermediate value theorem.

5. Why can 1/x not use its signs at −1 and 1 to guarantee output zero?

   Show the answer and reasoning

   Its real domain excludes the interior input 0, so it is not a continuous function on [−1,1].

   The equation 1/x=0 is impossible wherever defined. Continuity on two separate pieces does not provide the connecting interval hypothesis.

6. For x² on [−2,2], is target 1 absent because both endpoint outputs are 4?

   Show the answer and reasoning

   No. Inputs −1 and 1 both give 1.

   The endpoint-between condition is a sufficient test. Failure of that test does not establish nonexistence.

7. Starting with a bracket of width 1, what width and midpoint input-error bound follow after three unresolved halvings?

   Show the answer and reasoning

   Width 1/8; midpoint input error at most 1/16.

   Each step divides the width by 2. A root retained in the closed bracket is at most half its width from its midpoint.

8. Does a bracket midpoint's input error≤1/100 by itself give output error≤1/100?

   Show the answer and reasoning

   No.

   For g(x)=1000(x−c), a midpoint at distance 1/100 from its root has output magnitude 10. A separate output-change estimate is needed.

9. Find the infimum and minimum of f(x)=x on (0,1).

   Show the answer and reasoning

   Infimum 0; no minimum.

   Every permitted output is positive and approaches 0 along x=1/n with n≥2. Zero is the greatest lower bound but is not an attained output.

10. Must a maximum promised by the extreme value theorem occur away from the endpoints?

    Show the answer and reasoning

    No.

    The continuous identity on [0,1] has its maximum 1 at the right endpoint and minimum 0 at the left endpoint.

11. Is continuity on the bounded interval (0,1] enough to guarantee bounded outputs?

    Show the answer and reasoning

    No.

    The continuous rule 1/x there gives n at input 1/n. The excluded endpoint 0 makes the interval not closed.

12. Define f(x)=x for 0≤x<1 and f(1)=0. Does closed bounded domain [0,1] force a maximum?

    Show the answer and reasoning

    No. Its supremum is 1 and no maximum is attained.

    The rule is not continuous at 1. For every output below 1, another input closer to 1 gives a greater output, while f(1)=0.

13. What are the minimum and maximum of any real function on the singleton interval [3,3]?

    Show the answer and reasoning

    Both equal its sole output f(3).

    Every input equals 3, so both defining extreme-value inequalities are equalities. The nonempty interval convention includes this case.

14. A continuous function on [a,b] attains minimum −2 and maximum 5. Can it skip output 0?

    Show the answer and reasoning

    No. Its full image is [−2,5].

    Apply the intermediate value theorem on the subinterval between an input attaining −2 and an input attaining 5. Every intermediate output is reached.

15. Does the extreme value theorem force exactly one maximum input?

    Show the answer and reasoning

    No.

    A constant function on [0,1] attains its maximum and minimum at every input. The theorem promises existence, not unique locations.

16. Why does continuity's ε=1 local bound not alone prove a global bound on any domain?

    Show the answer and reasoning

    Its neighbourhood and bound can depend on the point.

    The identity on ℝ is locally bounded near each real input but unbounded over the entire domain. The closed bounded interval proof supplies a separate global argument.

What to carry forward

• A continuous function on the entire connecting real interval reaches every target between its endpoint outputs.
• Strictly between targets give an interior input; endpoint targets may be reached only at endpoints.
• Existence, uniqueness and approximation are different conclusions. Bisection's midpoint certificate bounds input error.
• Real completeness makes shrinking nested brackets contain a common point and lets a bounded input sequence have a convergent subsequence.
• A least bound can be approached without being attained. Minimum and maximum require a permitted input with that output.
• Continuity on a nonempty closed bounded interval gives both boundedness and attained extremes, possibly at multiple inputs or endpoints.
• Domain exclusions, jumps and finite drawing windows must remain visible; a finite table cannot replace the full theorem proof.