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 $c = (20)^{1/3} \approx 2.714...$.