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Continuity in Advanced Calculus

Extreme Value Theorem

The extreme value theorem guarantees that a continuous function on a closed interval is not only bounded but also attains both a maximum and a minimum value.

Statement of the Theorem

Extreme Value Theorem

On the closed interval [a,b][a, b], a continuous function f(x)f(x) must attain a maximum and a minimum value; that is, there exist x1,x2∈[a,b]x_1, x_2 \in [a, b] such that:

  • f(x1)=max⁡x∈[a,b]f(x)f(x_1) = \max_{x \in [a, b]} f(x)
  • f(x2)=min⁡x∈[a,b]f(x)f(x_2) = \min_{x \in [a, b]} f(x)

Geometric Meaning

The geometric meaning of the extreme value theorem is that the graph of a continuous function on a closed interval has a highest point and a lowest point.

Proof Strategy

  1. Use boundedness: the function is bounded, so it has a supremum and an infimum
  2. Construct a sequence: build sequences converging to the supremum and to the infimum
  3. Use continuity: use the continuity of the function at the limit point
  4. Prove that the extrema exist: prove that the supremum and the infimum are exactly the maximum and the minimum value

Applied Examples

Example 1: Polynomial Function

Problem: find the maximum and minimum values of the function f(x)=x2−2x+1f(x) = x^2 - 2x + 1 on [0,3][0, 3].

Solution:

  • The function is continuous on [0,3][0, 3]
  • By the extreme value theorem, the function attains a maximum and a minimum value
  • f(x)=(x−1)2f(x) = (x - 1)^2
  • The minimum value occurs at x=1x = 1, where f(1)=0f(1) = 0
  • The maximum value occurs at x=3x = 3, where f(3)=4f(3) = 4

Example 2: Trigonometric Function

Problem: find the maximum and minimum values of the function f(x)=sin⁡xf(x) = \sin x on [0,2π][0, 2\pi].

Solution:

  • The function is continuous on [0,2π][0, 2\pi]
  • Maximum value: f(π2)=1f(\frac{\pi}{2}) = 1
  • Minimum value: f(3π2)=−1f(\frac{3\pi}{2}) = -1

Exercises

Exercise 1

Let f(x)f(x) be continuous on [a,b][a, b]. Show whether f(x)f(x) is bound to attain its maximum and minimum values.

continuity on a closed intervalextreme value theoremintermediate value theorem

Approach: Use the extreme value theorem.

Steps:

  1. By the extreme value theorem, a function continuous on a closed interval must attain a maximum and a minimum value
  2. Therefore f(x)f(x) must attain a maximum and a minimum value

Answer: It must attain a maximum and a minimum value.

Exercise 2

Find the maximum and minimum values of the function f(x)=x3−3xf(x) = x^3 - 3x on [−2,2][-2, 2].

continuity on a closed intervalextreme value theoremintermediate value theorem

Approach: Use the extreme value theorem and differentiation.

Steps:

  1. The function is continuous on [−2,2][-2, 2], so it must have extrema
  2. f′(x)=3x2−3=3(x2−1)=0f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 0 gives x=±1x = \pm 1
  3. Compute the function values at the critical points:
    • f(−2)=−2f(-2) = -2
    • f(−1)=2f(-1) = 2
    • f(1)=−2f(1) = -2
    • f(2)=2f(2) = 2
  4. The maximum value is 2 and the minimum value is -2

Answer: The maximum value is 2 and the minimum value is -2.

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
π\piGreek letterPi (pi)The ratio of a circle’s circumference to its diameter, used to denote angles (such as π2\frac{\pi}{2})
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Boundedness Theorem
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Intermediate Value Theorem

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