Intermediate Value Theorem

The intermediate value theorem states that a continuous function can attain every intermediate value between two of its values.

Statement of the Theorem

Geometric Meaning

The geometric meaning of the intermediate value theorem is that the graph of a continuous function on a closed interval “connects” the function values at the two endpoints.

Proof Strategy

1. Construct an auxiliary function: let $g(x) = f(x) - C$
2. Apply the zero point theorem: $g(a) \cdot g(b) < 0$
3. Use the zero point theorem: there exists $\xi \in (a, b)$ with $g(\xi) = 0$
4. Draw the conclusion: $f(\xi) = C$

Applied Examples

Example 1: Solving an Equation

Problem: prove that the equation $x^3 - 3x + 1 = 0$ has a solution in $(0, 1)$.

Solution:

• Let $f(x) = x^3 - 3x + 1$
• $f(0) = 1 > 0$, $f(1) = -1 < 0$
• The function is continuous on $[0, 1]$
• 0 lies between $f(0)$ and $f(1)$
• By the intermediate value theorem, there exists $\xi \in (0, 1)$ with $f(\xi) = 0$

Example 2: Trigonometric Function

Problem: prove that the function $f(x) = \sin x$ on $[0, \pi]$ can attain the value $\frac{1}{2}$.

Solution:

• $f(0) = 0$, $f(\frac{\pi}{2}) = 1$
• The function is continuous on $[0, \frac{\pi}{2}]$
• $\frac{1}{2}$ lies between $f(0) = 0$ and $f(\frac{\pi}{2}) = 1$
• By the intermediate value theorem, there exists $\xi \in (0, \frac{\pi}{2})$ with $f(\xi) = \frac{1}{2}$

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Exercises

Exercise 1

Let $f(x)$ be continuous on $[0, 2]$, with $f(0) = 1, f(2) = 5$. Prove that there exists $\xi \in (0, 2)$ with $f(\xi) = 3$.

Exercise 2

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

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Summary

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