Zero Point Theorem

The zero point theorem is a special case of the intermediate value theorem: it shows that a continuous function whose endpoint values have opposite signs must have a zero.

Statement of the Theorem

Geometric Meaning

The geometric meaning of the zero point theorem is that if a continuous function takes opposite signs at the endpoints of an interval, its graph must cross the $x$-axis.

Proof Strategy

1. Use the intermediate value theorem: the zero point theorem is a special case of the intermediate value theorem ($C = 0$)
2. Give a direct proof: construct a nested sequence of intervals and use continuity
3. Use the extreme value theorem: prove the existence of the zero via the extreme value theorem

Applied Examples

Example 1: Polynomial Equation

Problem: prove that the equation $x^2 - 2 = 0$ has a solution in $(1, 2)$.

Solution:

• Let $f(x) = x^2 - 2$
• $f(1) = -1 < 0$, $f(2) = 2 > 0$
• The function is continuous on $[1, 2]$
• $f(1) \cdot f(2) = (-1) \times 2 = -2 < 0$
• By the zero point theorem, there exists $\xi \in (1, 2)$ with $f(\xi) = 0$

Example 2: Transcendental Equation

Problem: prove that the equation $\cos x = x$ has a solution in $(0, 1)$.

Solution:

• Let $f(x) = \cos x - x$
• $f(0) = \cos 0 - 0 = 1 > 0$
• $f(1) = \cos 1 - 1 \approx 0.54 - 1 = -0.46 < 0$
• The function is continuous on $[0, 1]$
• $f(0) \cdot f(1) < 0$
• By the zero point theorem, there exists $\xi \in (0, 1)$ with $f(\xi) = 0$

Points to Note

1. The Opposite-Sign Condition at the Endpoints

The zero point theorem requires $f(a) \cdot f(b) < 0$, that is, the function values at the endpoints must have opposite signs.

Counterexample:

• $f(x) = x^2$ is continuous on $[0, 1]$
• $f(0) = 0$, $f(1) = 1 > 0$
• Although the function does have a zero on $[0, 1]$, the condition of the zero point theorem is not satisfied

2. Continuity on a Closed Interval

The function must be continuous on the closed interval.

3. The Zero Need Not Be Unique

The zero point theorem only guarantees that at least one zero exists; there may be several.

Exercises

Exercise 1

Let $f(x)$ be continuous on $[0, 2]$, with $f(0) = -1, f(2) = 3$. Prove that $f(x)$, in $(0, 2)$, has at least one point $x_0$ with $f(x_0) = 0$.

Exercise 2

Prove that the equation $x^3 - x - 1 = 0$ has a solution in $(1, 2)$.

Exercise 3

Let $f(x)$ be continuous on $[1, 3]$, with $f(1) = 2, f(3) = -1$. Prove that $f(x)$, in $(1, 3)$, has at least one point $x_0$ with $f(x_0) = 0$.

Summary

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