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Mathematics LibreTexts

3.2: Intermediate Value Theorem

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    23592
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    Theorem \(\PageIndex{1}\) Intermediate value theorem

    Let \(f: [a, b] \to \mathbb{R}\) be a continuous function. Assume \(f(a)\) and \(f(b)\) have opposite signs, then \(f(t_0) = 0\) for some \(t_0 \in [a,b]\).

    截屏2021-02-02 上午11.00.07.png

    The intermediate value theorem is assumed to be known; it should be covered in any calculus course. We will use only the following corollary:

    Corollary \(\PageIndex{1}\)

    Assume that for any \(t \in [0, 1]\) we have three points in the plane \(O_t\), \(A_t\), and \(B_t\), such that 

    1. Each function \(t \mapsto O_t\), \(t \mapsto A_t\), and \(t \mapsto B_t\) is continuous.
    2. For any \(t \in [0, 1]\), the points \(O_t, A_t\), and \(B_t\) do not lie on one line. Then \(\angle A_0  O_0 B_0\) and \(\angle A_1 O_1 B_1\) have the same sign.
    Proof

    Consider the function \(f(t) = \measuredangle A_tO_tB_t\).

    Since the points \(O_t, A_t\), and \(B_t\) do not line on the one line, Theorem 2.4.1 implies that \(f(t) = \measuredangle A_tO_tB_t \ne 0\) nor \(\pi\) for any \(t \in [0, 1]\).

    Therefore, by Axiom IIIc and Exercise 1.9.2, \(f\) is a continuous function. By the intermediate value theorem, \(f(0)\) and \(f(1)\) have the same sign; hence the result follows.