Three Hard Theorems

This page will include information and problems from "Calculus : Chapter 7 : Three Hard Theorems" By Micheal Spivak

Theorem 7.1

If $f$ is continuous on $[a, b]$ and $f(a) < 0 < f(b)$, then there is some $x$ in $[a, b]$ such that $f(x) = 0$

Theorem 7.2

If $f$ is continuous on $[a, b]$, then $f$ is bounded above on $[a, b]$, that is, there is some number $N$ such that $f(x) \le N$ for all $x$ in $[a, b]$

Theorem 7.3

If $f$ is continuous on $[a, b]$, then there is some number $y$ in $[a, b]$ such that $f(y) \ge f(x)$ for all $x$ in $[a, b]$

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The following theorems are trivial consequences of theorem 7.1, 7.2 and 7.3

Theorem 7.4

If $f$ is continuous on $[a, b]$ and $f(a) < c < f(b)$, then there is some $x$ in $[a, b]$ such that $f(x) = c$

Proof:

Let $g = f - c$. Then $g$ is continuous, and $g(a) < 0 < g(b)$. By Theorem 7.1, there is some $x$ in $[a, b]$ such that $g(x) = 0$. But this means that $f(x) = c. \quad \Box$

Theorem 7.5

If $f$ is continuous on $[a, b]$ and $f(a) > c > f(b)$, then there is some $x$ in $[a, b]$ such that $f(x) = c$

Proof:

The function $-f$ is continuous on $[a, b]$ and $-f(a) < -c < -f(b)$. By Theorem 7.4 there is some $x$ in $[a, b]$ such that $-f(x) = -c$, which means that $f(x) = c .\quad\Box$.

Theorem 7.6

If $f$ is continuous on $[a, b]$, then $f$ is bounded below on $[a, b]$, that is, there is some number $N$ such that $f(x) \ge N$ for all $x$ in $[a, b]$.

Proof:

The function $-f$ is continuous on $[a, b]$, so by Theorem 7.2 there is a number $M$ such that $-f(x) \le M$ for all $x$ in $[a, b]$. But this means that $f(x) \ge -M$ for all $x$ in $[a, b]$, so we can let $N = -M. \quad \Box$

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Problems:

Question Number	Difficulty	Completed
1	()	No
2	()	No
3	()	No
4	()	No
5	()	No
6	()	No
7	()	No
8	()	No
9	()	No
10	()	Yes
11	()	Yes
12	()	No
13	()	No
14	()	No
15	(*)	No
16	(*)	No
17	(*)	No
18	(*)	No
19	(*)	No
20	()	No
21	(*)	No