9.2 Intermediate Value Theorem

9.11   Theorem (Intermediate Value Theorem.) Let $a,b\in\mbox{{\bf R}}$ with $a<b$, and let $f\colon [a,b]\to\mbox{{\bf R}}$ be a continuous function. Suppose $f(a)<0<f(b)$. Then there is some point $c\in(a,b)$ with $f(c)=0$.

Proof: We will construct a binary search sequence $[a_n,b_n]$ with $[a_0,b_0]=[a,b]$ such that

$$f(a_n)\leq 0\leq f(b_n)\mbox{ for all } n.$$ (9.12)

Let

$$\begin{eqnarray*}[a_0,b_0]&=&[a,b] \cr [a_{n+1},b_{n+1}]&=&\cases{ \left[a_n,{{a... ...er 2},b_n\right] &if $f\left( {{a_n+b_n}\over 2}\right) <0$.\cr} \end{eqnarray*}$$

This is a binary search sequence satisfying condition (9.12).

Let $c$ be the number such that $\{[a_n,b_n]\} \to c$. Then $\{a_n\} \to c$ and $\{b_n\} \to c$ (cf theorem 7.87), so by continuity of $f$, $\{f(a_n)\}\to f(c)$ and $\{f(b_n)\}\to f(c)$. Since $f(b_n)\geq 0$ for all $n$, it follows by the inequality theorem that $f(c)=\lim\{f(b_n)\}\geq 0$, and since $f(a_n)\leq 0$, we have $f(c)=\lim\{f(a_n)\}$$\leq 0$. Hence, $f(c)=0$. $\mid\!\mid\!\mid$

9.13   Exercise (Intermediate value theorem.) A Let $a,b\in\mbox{{\bf R}}$ with $a<b$ and let $f:[a,b]\to\mbox{{\bf R}}$ be a continuous function with $f(a)<f(b)$. Let $y$ be a number in the interval $\left(f(a),f(b)\right)$. Show that there is some $c\in(a,b)$ with $f(c)=y$. (Use theorem 9.11. Do not reprove it.)

9.14   Notation ($x$ is between $a$ and $b$.) Let $a,b,x\in\mbox{{\bf R}}$. I say $x$ is between $a$ and $b$ if either $a<x<b$ or $b<x<a$.

9.15   Corollary (Intermediate value theorem.) Let $a,b\in\mbox{{\bf R}}$ with $a<b$. Let $f\colon [a,b]\to\mbox{{\bf R}}$ be a continuous function with $f(a)\neq f(b)$. If $y$ is any number between $f(a)$ and $f(b)$, then there is some $c\in(a,b)$ such that $f(c)=y$. In particular, if $f(a)$ and $f(b)$ have opposite signs, there is a number $c\in(a,b)$ with $f(c)=0$.

Proof: By exercise 9.13A, the result holds when $f(a)<f(b)$. If $f(a)>f(b)$, let $g=-f$. Then $g$ is continuous on $[a,b]$ and $g(a)<g(b)$, so by exercise 9.13A there is a $c\in(a,b)$ with $g(c)=0$, so $-f(c)=0$ so $f(c)=0$. $\mid\!\mid\!\mid$

9.16   Example. Let $A,B,C,D$ be real numbers with $A\neq 0$, and let

$$f(x)=Ax^3+Bx^2+Cx+D.$$

We will show that there is a number $c\in\mbox{{\bf R}}$ such that $f(c)=0$. Suppose, in order to get a contradiction, that no such number $c$ exists, and let

$$g(x)={{f(-x)}\over {f(x)}}={{-Ax^3+Bx^2-Cx+D}\over {Ax^3+Bx^2+C+D}}\mbox{ for all } x\in\mbox{{\bf R}}.$$

(I use the fact that $f(x)$ has no zeros here.) Then

$$\begin{eqnarray*} \lim\{g(n)\}_{n\geq 1}&=&\lim \left\{ {{-A+{B\over n}-{C\over ... ...{n^3}}}\right\}_{n\geq 1} \\ &=&{{-A+0+0+0}\over {A+0+0+0}}=-1. \end{eqnarray*}$$

It follows that $g(n)<0$ for some $n$, so $f(-n)$ and $f(n)$ have opposite signs for some $n$, and $g$ is continuous on $[-n,n]$, so by the intermediate value theorem, $g(c)=0$ for some $c\in(-n,n)$, contradicting the assumption that $g$ is never zero.

9.17   Exercise. A Give examples of the requested functions, or explain why no such function exists. Describe your functions by formulas if you can, but pictures of graphs will do if a formula seems too complicated.

a)

$f\colon[0,1]\to\mbox{{\bf R}}$, $f$ has no maximum.

b)

$g\colon[0,\infty)\to\mbox{{\bf R}}$, $g$ is continuous, $g$ has no maximum.

c)

$k\colon[0,\infty)\to\mbox{{\bf R}}$, $k$ is continuous, $k$ has no maximum or minimum.

d)

$l\colon(0,1)\to\mbox{{\bf R}}$, $l$ is bounded and continuous, $l$ has no maximum.

9.18   Exercise. Let $f(x)=x^3-3x+1$. Prove that the equation $f(x)=0$ has at least three solutions in $\mbox{{\bf R}}$. s

9.19   Exercise. A Let $F$ be a continuous function from $\mbox{{\bf R}}$ to $\mbox{{\bf R}}$ such that

a)

For all $x\in\mbox{{\bf R}}$, $\Big((F(x) = 0) \hspace{1ex}\Longleftrightarrow\hspace{1ex}(x^2=1)\Big)$.

b)

$F(2)>0$.

Prove that $F(4)>0$.

9.20   Note. The intermediate value theorem was proved independently by Bernhard Bolzano in 1817 [42], and Augustin Cauchy in 1821[23, pp 167-168]. The proof we have given is almost identical with Cauchy's proof.

The extreme value theorem was proved by Karl Weierstrass circa 1861.