3.2: Intermediate Value Theorem

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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

Each function $t \mapsto O_t$, $t \mapsto A_t$, and $t \mapsto B_t$ is continuous.
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.