The Intermediate Value Theorem

In this lesson you’ll learn the intermediate value theorem, the supremum proof, why it fails over $\mathbb{Q}$, and the fixed-point and root-finding consequences that follow from it.

Intermediate value theorem. If $f$ is continuous on $[a,b]$ and $y$ lies between $f(a)$ and $f(b)$, then there exists $c \in [a,b]$ with $f(c) = y$.

The special case $y = 0$ is the one used most: a continuous function that changes sign has a root.

Two hypotheses, and both are doing work.

• Continuity. A jump can step over the value entirely, as the right panel shows.
• A closed interval. On an open interval the crossing point may be exactly the endpoint you excluded.

Take $f(a) < 0 < f(b)$ and let

$$S = \{x \in [a,b] : f(x) < 0\}$$

$S$ is non-empty ($a \in S$) and bounded above by $b$. By completeness, $c = \sup S$ exists. That is where the axiom gets spent, and it is the only non-trivial step.

Now show $f(c) = 0$ by ruling out both alternatives.

If $f(c) < 0$: by continuity there is a neighbourhood of $c$ where $f$ stays negative. (Take $\varepsilon = |f(c)|$ in the definition.) So points slightly to the right of $c$ are in $S$, contradicting $c$ being an upper bound. Note $c \ne b$ here, since $f(b) > 0$.

If $f(c) > 0$: similarly $f$ stays positive on a neighbourhood of $c$, so some interval $(c - \delta, c]$ contains no points of $S$. Then $c - \delta$ is a smaller upper bound, contradicting leastness.

Hence $f(c) = 0$. ∎

Notice the shape: the supremum is the candidate, and continuity kills both ways it could be wrong. Sup for existence, continuity for correctness. That pairing recurs in the next lesson too.

Let $f(x) = x^2 - 2$ on $[0,2] \cap \mathbb{Q}$. Then $f(0) = -2 < 0$ and $f(2) = 2 > 0$, and $f$ is continuous as a function on the rationals.

But there is no rational $c$ with $c^2 = 2$. The theorem fails.

The proof breaks at exactly one place: $\sup S$ does not exist in $\mathbb{Q}$. This is the cleanest demonstration that IVT is a completeness theorem wearing geometric clothing. The picture of a curve crossing a line looks like it should be about geometry; it is about the number line having no holes.

It does not say the root is unique. $f(x) = \sin x$ on $[0, 4\pi]$ has many.

It does not say a value not between $f(a)$ and $f(b)$ is skipped. Values outside that range may also be attained; the theorem is one-directional.

The converse is false. A function can attain every intermediate value and still be discontinuous. $f(x) = \sin\frac1x$ for $x \ne 0$ with $f(0)=0$ has the intermediate value property on any interval containing 0 and is discontinuous there. So “hits every value in between” is strictly weaker than continuity.

Root existence and bisection. If $f(a)$ and $f(b)$ have opposite signs, a root lies between them, and halving repeatedly finds it. That is the bisection algorithm, and IVT is its correctness proof.

Fixed points. If $g: [0,1]\to[0,1]$ is continuous, it has a fixed point $g(c) = c$. Apply IVT to $f(x) = g(x) - x$: $f(0) = g(0) \ge 0$ and $f(1) = g(1)-1 \le 0$, so $f$ has a zero. This is the one-dimensional case of Brouwer’s fixed point theorem.

Odd-degree polynomials have a real root. As $x \to \pm\infty$ an odd-degree polynomial takes both signs, so IVT applies on a large enough interval. Even-degree polynomials need not: $x^2+1$ has none.

The pancake and ham sandwich style results. Any continuous function on a circle takes the same value at some pair of antipodal points, which follows from IVT applied to $f(x) - f(-x)$.

Example 1: Show $x^3 + x - 1 = 0$ has a real root.

Solution. Let $f(x) = x^3+x-1$, a polynomial and so continuous everywhere.

$$f(0) = -1 < 0, \qquad f(1) = 1 > 0$$

By IVT there is $c \in (0,1)$ with $f(c) = 0$. ∎

To narrow it: $f(0.5) = 0.125+0.5-1 = -0.375 < 0$, so the root is in $(0.5, 1)$. Then $f(0.75) = 0.4219+0.75-1 = 0.1719 > 0$, so it is in $(0.5, 0.75)$. That is bisection, and each step halves the interval.

Example 2: Show every positive real has a square root.

Solution. Fix $y > 0$ and let $f(x) = x^2$ on $[0, y+1]$, continuous.

$$f(0) = 0 \le y, \qquad f(y+1) = (y+1)^2 = y^2+2y+1 > y$$

By IVT there is $c$ with $c^2 = y$. ∎

The existence of $\sqrt y$ is a theorem, not a definition. It requires completeness, which is why the rationals have no $\sqrt2$.

Example 3: A continuous self-map has a fixed point.

Let $g:[0,1]\to[0,1]$ be continuous. Show $g(c) = c$ for some $c$.

Solution. Let $f(x) = g(x) - x$, continuous as a difference of continuous functions.

Since $g(0) \in [0,1]$, $f(0) = g(0) \ge 0$. Since $g(1) \in [0,1]$, $f(1) = g(1) - 1 \le 0$.

If either is zero we are done. Otherwise $f(0) > 0 > f(1)$, and IVT gives $c$ with $f(c) = 0$, that is $g(c) = c$. ∎

Interpretation: any continuous way of stretching and folding $[0,1]$ into itself leaves at least one point where it started.

Example 4: Where the closed interval matters.

Does $f(x) = \frac1x$ on $(0,1)$ attain the value 5? What about IVT?

Solution. It does: $f(0.2) = 5$. But note IVT does not apply as usually stated, because $(0,1)$ is not closed and $f$ has no values at the endpoints to compare.

A better illustration of the failure: on $(0,1)$ consider $f(x) = x$. It takes values arbitrarily close to 0 and 1 without attaining either, so the value 0 is “between” the limits at the endpoints and is never attained. Closedness is what makes the endpoints available.

Example 5: An odd-degree polynomial.

Show $p(x) = x^5 - 3x^2 + 1$ has a real root.

Solution. $p(0) = 1 > 0$ and $p(-1) = -1 - 3 + 1 = -3 < 0$.

By IVT there is a root in $(-1, 0)$. ∎

Generally, for odd degree the leading term dominates at large $|x|$ and has opposite signs at $\pm\infty$, so IVT always applies. Every odd-degree real polynomial has a real root, while $x^2+1$ shows even degree gives no such guarantee.

Example 6: Antipodal points.

Show that at any moment there are two diametrically opposite points on the equator with the same temperature.

Solution. Let $T(\theta)$ be the temperature at angle $\theta$, assumed continuous, and define

$$f(\theta) = T(\theta) - T(\theta + \pi)$$

Then $f(0) = T(0) - T(\pi)$ and $f(\pi) = T(\pi) - T(2\pi) = T(\pi) - T(0) = -f(0)$.

So $f(0)$ and $f(\pi)$ have opposite signs (or both are 0). By IVT there is $\theta_0$ with $f(\theta_0) = 0$, meaning

$$T(\theta_0) = T(\theta_0 + \pi)$$

∎

The trick of forming $f(x) - f(x+\text{shift})$ and noting it changes sign is worth remembering; it powers a whole family of these results.

Example 7: The converse fails.

Give a discontinuous function with the intermediate value property.

Solution. Let $f(x) = \sin\frac1x$ for $x \ne 0$ and $f(0) = 0$.

On any interval containing 0, $f$ oscillates through every value in $[-1,1]$ infinitely often, so it attains every intermediate value.

Yet $f$ is discontinuous at 0, since $\lim_{x\to0}\sin\frac1x$ does not exist.

The intermediate value property is strictly weaker than continuity. IVT is a one-way implication.

Bisection and numerical solvers. Every root-finding routine that brackets a root relies on IVT for correctness. It is slower than Newton’s method but it cannot fail once a sign change is found, which is why robust solvers fall back to it.

Control systems. Showing a control input exists that achieves a target output is an IVT argument: the response is continuous and straddles the target, so some input hits it exactly. That is the formal basis of a calibration sweep.

Thermostats and equilibrium. Any continuous process that is too cold at one setting and too hot at another has a setting that is exactly right. Trivial-sounding, and it is the argument used to prove equilibrium existence in far less obvious settings.

Economics. Existence of a market-clearing price is IVT in one dimension and Brouwer’s fixed point theorem in many. Excess demand is positive at low prices and negative at high ones, so it vanishes somewhere.

Game theory. Nash’s existence theorem is a fixed-point argument, generalizing Example 3 to higher dimensions. The one-dimensional case here is the whole idea in miniature.

Engineering tolerance. If a design parameter gives too little clearance at one extreme and too much at the other, some intermediate value is exact. Bisection on a physical prototype is IVT applied by hand.