In mathematical analysis, the intermediate value theorem states that for each value between the least upper bound and greatest lower bound of the image of a continuous function there is a corresponding point in its domain that the function maps to that value.
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Intermediate value theorem
 Version I. The intermediate value theorem states the following: If f is a realvalued continuous function on the interval [a, b], and u is a number between f(a) and f(b), then there is a c ∈ [a, b] such that f(c) = u.
 Version II. Suppose that I is an interval [a, b] in the real numbers R and that f : I → R is a continuous function. Then the image set f(I) is also an interval, and either it contains [f(a), f(b)], or it contains [f(b), f(a)]; that is,
It is frequently stated in the following equivalent form: Suppose that f : [a, b] → R is continuous and that u is a real number satisfying f(a) < u < f(b) or f(a) > u > f(b). Then for some c ∈ [a, b], f(c) = u.
This captures an intuitive property of continuous functions: given f continuous on [1, 2], if f(1) = 3 and f(2) = 5 then f must take the value 4 somewhere between 1 and 2. It represents the idea that the graph of a continuous function on a closed interval can only be drawn without lifting your pencil from the paper.
The theorem depends on (and is actually equivalent to) the completeness of the real numbers. It is false for the rational numbers Q. For example, the function f(x) = x^{2} − 2 for x ∈ Q satisfies f(0) = −2 and f(2) = 2. However there is no rational number x such that f(x) = 0, because √2 is irrational.
Proof
We shall prove the first case f(a) < u < f(b); the second is similar.
Let S be the set of all x in [a, b] such that f(x) ≤ u. Then S is nonempty since a is an element of S, and S is bounded above by b. Hence, by the completeness property of the real numbers, the supremum c = sup S exists. That is, c is the lowest number that is greater than or equal to every member of S. We claim that f(c) = u.
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