Bolzano's Intermediate Value Theorem

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

# Bolzano's Intermediate Value Theorem

Recall from The Location of Roots Theorem that if $f : I \to \mathbb{R}$ is a continuous function from the closed and bounded interval $I = [a, b]$ into the real numbers, then if $f(a) < 0 < f(b)$ or $f(a) > 0 > f(b)$, then there exists at least one root on $I$.

The following theorem known simply as The Intermediate Value Theorem or Bolzano's Intermediate Value Theorem is a stronger generalization of The Location of Roots Theorem.

 Theorem 1 (Bolzano's Intermediate Value): Let $I$ be an interval and let $f : I \to \mathbb{R}$ be a continuous function. If $a, b \in I$ and if $k \in \mathbb{R}$ is such that $f(a) < k < f(b)$, then there exists a $c \in I$, where $c$ is between $a$ and $b$ such that $f(c) = k$.
• Proof: Let $I$ be an interval and let $f : I \to \mathbb{R}$ be a continuous function.
• First consider the case where $a < b$, and define $g(x) = f(x) - k$. Then we have that $g(a) = f(a) - k < 0$, and $g(b) = f(b) - k > 0$. Therefore $g(a) < 0 < g(b)$. By The Location of Roots Theorem, there exists a $c$ such that $a < c < b$ where $0 = g(c)$. But $g(c) = f(c) - k$ and so $0 = f(c) - k$, in other words, $k = f(c)$.
• Now consider the case where $a > b$, and define $g(x) = k - f(x)$. Therefore $g(b) = k - f(b) < 0$ and $g(a) = k - f(a) > 0$. Thus $g(b) < 0 < g(a)$, and once again by The Location of Roots Theorem there exists a $c$ such that $b < c < a$ where $0 = g(c)$. But $g(c) = k - f(c)$, and so $0 = k - f(c)$, in other words $f(c) = k$. $\blacksquare$
 Corollary 1: Let $I$ be an interval and let $f : I \to \mathbb{R}$ be a continuous function. If $k \in \mathbb{R}$ satisfies $\inf f(I) ≤ k ≤ \sup f(I)$, then there exists a $c \in I$ such that $f(c) = k$.
• Proof: Suppose that $\inf f(I) ≤ K ≤ \sup f(I)$. From The Maximum-Minimum Theorem, we have that $\exists x_*, x^* \in I$ such that $\inf f(I) = f(x_*) ≤ k ≤ f(x^*) = \sup f(I)$. Therefore by Bolzano's Intermediate Value Theorem, there exists a $c \in I$ such that $f(c) = k$. $\blacksquare$

## Example 1

Let $f : [1, 4] \to \mathbb{R}$ be defined by $f(x) = x^3$. Show that there exists a point $c \in I$ such that $f(c) = 20$.

First notice that $f$ is a continuous function, and $I = [1, 4]$ is an interval. Notice also that $f(1) = 1^3 = 1$ and $f(4) = 4^3 = 64$. Now $1 < 20 < 64$, and so by Bolzano's Intermediate Value Theorem, there exists a $c$ such that $1 < c < 4$ such that $f(c) = 20$. In fact, we obtain that $c = (20)^{1/3} \approx 2.714...$.