Intermediate Value Theorem Article Index for
Intermediate 		Website Links For
Intermediate 		 

Information About

Intermediate Value Theorem
  CATEGORIES ABOUT INTERMEDIATE VALUE THEOREM
   
continuous mappings
   
mathematical theorems
   
articles containing proofs
   
APPAREL
BABY
BEAUTY
BOOKS
CAR TOYS
CELL PHONES
DVD'S
ELECTRONICS
GOURMET FOOD
GROCERIES
HEALTH & PERSONAL
HOME & GARDEN
JEWELRY
MUSIC
MUSIC INSTRUMENTS
OFFICE PRODUCTS
SOFTWARE
SPORTING GOODS
TOOLS & HARDWARE
TOYS
VIDEO GAMES
SHOPPING HOME
MORE SHOPPING...




INTERMEDIATE VALUE THEOREM


The intermediate value theorem states the following: Suppose that \displaystyle I is an Interval \displaystyle in the Real Number s \mathbb{R} and that f\colon I ightarrow \mathbb{R} is a Continuous Function . Then the image set \displaystyle f(I) is also an interval, and either it contains \displaystyle [f(a),f(b) , or it contains \displaystyle [f(b),f(a)]. I.e.
  • \displaystyle f(I) \supseteq {Link without Title} ,

    • or

  • \displaystyle f(I) \supseteq {Link without Title} .


It is frequently stated in the following equivalent form: Suppose that f\colon {Link without Title} ightarrow \mathbb{R} is continuous and that \displaystyle u is a real number satisfying \displaystyle f(a) < u < f (b)\, or \displaystyle f(a) > u > f (b). Then for some \displaystyle c \in (a,b), \,\displaystyle f(c) = u.

This captures an intuitive property of continuous functions: given \displaystyle f continuous on \displaystyle {Link without Title} , if \displaystyle f(1) = 3 and \displaystyle f(2) = 5 then \displaystyle f must be equal to 4 somewhere between 1 and 2. It represents the idea that the graph of a continuous function can be drawn without lifting your pencil from the paper.

The theorem depends on the completeness of the Real Number s. It is false for the Rational Number s \mathbb{Q}. For example, the function \displaystyle f(x) = x^{2} - 2, \, x \in\mathbb{Q} satisfies f(0) = -2,\, f(2) = 2. However there is no rational number \displaystyle x such that \displaystyle f(x) = 0.


Proof

We shall prove the first case \displaystyle f(a) < u < f (b); the second is similar.

Let \displaystyle ext{S} = \{ x \in {Link without Title} : f(x) \leq u \}. Then \displaystyle S is non-empty (since \displaystyle a \in \displaystyle S) and bounded above by \displaystyle b. Hence by the Completeness property of the real numbers, the Supremum c = \sup ext{S} exists. We claim that \displaystyle f(c) = u.

  Suppose Next That <math>\displaystyle F(c) < U</math> Again, By Continuity, There Is A <math>\displaystyle \delta > 0</math> Such That <math>\displaystyle f(x) - F(c) < U - F(c)</math> Whenever <math>\displaystyle x - C < \delta</math> Then <math>\displaystyle F(x) < F(c) + (u - F(c)) u</math> for <math>\displaystyle x</math> in <math>\displaystyle(c - \delta, c + \delta)</math> and there are numbers <math>\displaystyle x</math> greater than <math>\displaystyle c</math> for which <math>\displaystyle f(x) < u</math>, again a contradiction to the definition of <math>\displaystyle c</math>