# Lecture 12: Maximum and Minimum Values

## Calculus I — Differential Calculus

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## Learning Objectives

By the end of this lecture, students will be able to:

1. Define absolute (global) and local (relative) extrema
2. State and apply the Extreme Value Theorem
3. Find critical numbers of a function
4. Use the Closed Interval Method to find absolute extrema on a closed interval
5. Apply Fermat's Theorem to understand why critical numbers matter

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## Lecture Content

### I. Definitions of Extrema

An **absolute (global) maximum** occurs at c if f(c) >= f(x) for all x in the domain of f, meaning f(c) is the largest value f attains anywhere. An **absolute (global) minimum** occurs at c if f(c) <= f(x) for all x in the domain. A **local (relative) maximum** at c means f(c) >= f(x) for all x in some open interval containing c, so f(c) is the largest value in a neighborhood of c. A **local (relative) minimum** is defined analogously. The term **extremum** (plural: extrema) refers to either a maximum or minimum. Every absolute extremum is also a local extremum, but the reverse is not true.

### II. The Extreme Value Theorem

The **Extreme Value Theorem** states that if f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum on [a, b]. This theorem guarantees **existence** but does not specify where the extrema occur.

Both hypotheses are essential. If f is not continuous, the conclusion may fail -- for instance, a function with a removed point at its would-be peak has no actual maximum. If the interval is not closed, the conclusion may also fail -- the function f(x) = x on the open interval (0, 1) approaches 1 but never attains it, so it has no maximum.

<image>Three panels illustrating the Extreme Value Theorem. Panel A: A continuous function on [a, b] with the absolute maximum and minimum clearly marked with dots, demonstrating the theorem. Panel B: A discontinuous function on [a, b] where the maximum "should" occur at a removed point (open dot) — the function has no actual maximum. Panel C: A continuous function on the open interval (a, b) that approaches but never reaches its supremum at the endpoints — no absolute maximum exists. Title: "Why continuity on a closed interval is needed."</image>

### III. Fermat's Theorem

**Fermat's Theorem** states that if f has a local extremum at c and f'(c) exists, then f'(c) = 0. In other words, local extrema can only occur where f'(c) = 0 (a horizontal tangent) or where f'(c) does not exist. However, the converse is false: f'(c) = 0 does **not** guarantee a local extremum. The function f(x) = x^3 has f'(0) = 0, but x = 0 is neither a maximum nor a minimum -- it is an inflection point.

### IV. Critical Numbers

A **critical number** (or critical point) of f is a number c in the domain of f where f'(c) = 0 or f'(c) does not exist. By Fermat's Theorem, all local extrema must occur at critical numbers. However, not every critical number yields an extremum; critical numbers are candidates that must be tested further.

To find critical numbers, compute f'(x), set f'(x) = 0 and solve, and then find where f'(x) is undefined while f(x) itself is still defined.

### V. The Closed Interval Method

To find the **absolute maximum and minimum** of a continuous function f on a closed interval [a, b], follow these four steps. First, find all critical numbers of f in the open interval (a, b). Second, evaluate f at each critical number. Third, evaluate f at the endpoints f(a) and f(b). Fourth, compare all these values: the largest is the absolute maximum and the smallest is the absolute minimum. This method is guaranteed to work by the Extreme Value Theorem.

### VI. Worked Examples

Consider f(x) = x^3 - 3x + 1 on [-2, 3]. The derivative is f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1), giving critical numbers x = 1 and x = -1, both in (-2, 3). Evaluating: f(-2) = -1, f(-1) = 3, f(1) = -1, and f(3) = 19. The absolute maximum is f(3) = 19, and the absolute minimum is f(-2) = f(1) = -1.

For f(x) = x^{2/3} on [-1, 8], the derivative is f'(x) = (2/3)x^{-1/3}, which is undefined at x = 0 (but f(0) = 0 is defined), making x = 0 a critical number. Since f'(x) is never zero (the numerator is always 2), x = 0 is the only critical number. Evaluating: f(-1) = 1, f(0) = 0, and f(8) = 4. The absolute maximum is f(8) = 4, and the absolute minimum is f(0) = 0.

<image>The graph of f(x) = x^3 - 3x + 1 on [-2, 3]. The curve is plotted with the critical points at x = -1 (local max, f = 3) and x = 1 (local min, f = -1) marked with dots. The endpoints (-2, -1) and (3, 19) are also marked. The absolute maximum (3, 19) at the right endpoint and the absolute minimum values at (-2, -1) and (1, -1) are highlighted with labels. Title: "Closed Interval Method: f(x) = x^3 - 3x + 1 on [-2, 3]."</image>

### VII. Functions on Open Intervals or Unbounded Domains

The Closed Interval Method does not apply to open or unbounded domains. On such domains, absolute extrema may or may not exist, and if they do, they must occur at critical numbers since there are no endpoints to check. The behavior as x approaches the endpoints of the domain or as x approaches plus or minus infinity must be analyzed.

For example, f(x) = x + 4/x for x > 0 has f'(x) = 1 - 4/x^2 = 0, giving x = 2. Evaluating gives f(2) = 4, and as x approaches 0 from the right or x approaches infinity, f(x) approaches infinity. Therefore f(2) = 4 is an absolute minimum on (0, infinity).