Next: 10. The Derivative Up: 9. Properties of Continuous Previous: 9.1 Extreme Values   Index

# 9.2 Intermediate Value Theorem

9.11   Theorem (Intermediate Value Theorem.) Let with , and let be a continuous function. Suppose . Then there is some point with .

Proof: We will construct a binary search sequence with such that

 (9.12)

Let

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

Let be the number such that . Then and (cf theorem 7.87), so by continuity of , and . Since for all , it follows by the inequality theorem that , and since , we have . Hence, .

9.13   Exercise (Intermediate value theorem.) A Let with and let be a continuous function with . Let be a number in the interval . Show that there is some with . (Use theorem 9.11. Do not reprove it.)

9.14   Notation ( is between and .) Let . I say is between and if either or .

9.15   Corollary (Intermediate value theorem.) Let with . Let be a continuous function with . If is any number between and , then there is some such that . In particular, if and have opposite signs, there is a number with .

Proof: By exercise 9.13A, the result holds when . If , let . Then is continuous on and , so by exercise 9.13A there is a with , so so .

9.16   Example. Let be real numbers with , and let

We will show that there is a number such that . Suppose, in order to get a contradiction, that no such number exists, and let

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

It follows that for some , so and have opposite signs for some , and is continuous on , so by the intermediate value theorem, for some , contradicting the assumption that 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)
, has no maximum.
b)
, is continuous, has no maximum.
c)
, is continuous, has no maximum or minimum.
d)
, is bounded and continuous, has no maximum.

9.18   Exercise. Let . Prove that the equation has at least three solutions in . s

9.19   Exercise. A Let be a continuous function from to such that
a)
For all , .
b)
.
Prove that .

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.

Next: 10. The Derivative Up: 9. Properties of Continuous Previous: 9.1 Extreme Values   Index