The intermediate value theorem

Summary

This lesson proves the first great theorem on continuous functions on an interval: a continuous function cannot pass from one value to another without taking all the intermediate values. Bolzano’s theorem, which guarantees a zero when the function changes sign, is proved first, with the completeness axiom; from it follow the intermediate value theorem, the fact that the image of an interval is an interval, and the fact that the inverse of a continuous and strictly monotonic function is continuous, a result that the derivative of the inverse needs. The theorem does not say where the zero is, but the bisection method locates it with any desired precision. The lesson applies it to the existence of roots of equations, to polynomials of odd degree and to fixed points.

Learning objectives

Prove Bolzano’s theorem with the completeness axiom and deduce the intermediate value theorem from it.
Prove that the image of an interval under a continuous function is an interval, and deduce the continuity of the inverse of a continuous and strictly monotonic function.
Use the theorem to prove the existence of solutions of equations, and locate them with the bisection method.

Bolzano’s theorem

The graph of a continuous function on an interval is drawn, in the intuitive picture, without lifting the pencil; if it starts below the axis and ends above it, it has to cross it. That intuition needs a proof, because it depends on the real numbers: the function $x^2 - 2$ is negative at $0$ and positive at $2$ , but if only the rational numbers were considered it would never vanish, because $\sqrt{2}$ is not rational. The proof therefore uses the completeness axiom, by way of the supremum.

TheoremBolzano’s theorem

Let $f$ be continuous on $\icc{a}{b}$ , with $f(a) < 0 < f(b)$ . Then there exists $c$ in $\ioo{a}{b}$ such that $f(c) = 0$ . The same holds if $f(a) > 0 > f(b)$ . One takes the supremum of the points where the function is negative; preservation of sign, from the lesson on limit and continuity, prevents the function from being negative or positive there. The figure shows the set of those points and its supremum, and then the two impossible cases.

Proof

$S = \left\{ x \in \icc{a}{b} : f(x) < 0 \right\}, \qquad c = \resaltar{\sup S}$
Let $S$ be the set of the $x$ in $\icc{a}{b}$ at which $f(x) < 0$ . It is not empty, because it contains $a$ , and $b$ is an upper bound; by the completeness axiom it has a supremum $c$ , with $a \leq c \leq b$ .
$f(c) < 0 \rightarrow c + \frac{\delta}{2} \in S, \qquad c + \frac{\delta}{2} > c$
If $f(c) < 0$ , then $c \neq b$ , because $f(b) > 0$ . By preservation of sign, $f$ is negative on $\ico{c}{c + \delta}$ for some $\delta > 0$ , which can be taken with $c + \delta \leq b$ ; then $S$ contains points greater than $c$ , and $c$ would not be an upper bound.
$f(c) > 0 \rightarrow \sup S \leq c - \delta < c, \qquad \resaltar{f(c) = 0}$
If $f(c) > 0$ , then $c \neq a$ . By preservation of sign, $f$ is positive on $\ioc{c - \delta}{c}$ for some $\delta > 0$ , which can be taken with $c - \delta \geq a$ , so that no point of that interval is in $S$ ; since there are none greater than $c$ either, the number $c - \delta$ would be an upper bound smaller than the supremum. Therefore $f(c) = 0$ , and $c$ is neither $a$ nor $b$ . The second case reduces to the first with $-f$ .