Intermediate Value Theorem

The intermediate value theorem states that for each value between the least upper bound and greatest lower bound of the image of a continuous function there is at least one point in its domain that the function maps to that value.

There are three ways of stating the intermediate value theorem:

1. Version I: If $f$ is a real-valued continuous function on the interval $[a, b]$, and $u$ is a number between $f(a)$ and $f(b)$, then there is a $c \in [a, b]$ such that $f(c) = u$.
2. Version 2: Suppose that $f : [a, b] \to R$ is continuous and that u is a real number satisfying $f(a) < u < f(b)$ or $f(a) > u > f(b)$. Then for some $c \in [a, b]$, $f(c) = u$.
3. Version 3: Suppose that $I$ is an interval $[a, b]$ in the real numbers $\mathbb{R}$ and that $f : I \to R$ is a continuous function. Then the image set $f(I)$ is also an interval, and either it contains $[f(a), f(b)]$, or it contains $[f(b), f(a)]$; that is, $f(I) \supseteq [f(a), f(b)]$, or $f(I) \supseteq [f(b), f(a)]$.

This captures an intuitive property of continuous functions: given $f$ continuous on $[1, 2]$, if $f(1) = 3$ and $f(2) = 5$, then $f$ must take the value $4$ somewhere between $1$and $2$. It represents the idea that the graph of a continuous function on a closed interval can be drawn without lifting your pencil from the paper.

The theorem depends on (and is actually equivalent to) the completeness of the real numbers. It is false for the rational numbers $\mathbb{Q}$. For example, the function $f(x) = x^2 − 2$ for $x \in \mathbb{Q}$ satisfies $f(0) = −2$ and $f(2) = 2$. However there is no rational number $x$ such that $f(x) =0$, because $\sqrt 2$ is irrational.

The intermediate value theorem can be used to show that a polynomial has a solution. For example, $x^2-1$. The IVT shows that this has at least one point where $x=0$ because at $x=0$ it is negative and at $x=2$ it is positive. Therefore, since it is continuous, there must be at least one point where $x$ is $0$.