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Intermediate Value Theorem





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>