Absolute Extrema on Closed and Open Intervals

Absolute Extrema and the Extreme Value Theorem

By the end of this lesson, you will be able to find absolute maximum and minimum values of a function on a closed interval and apply the Extreme Value Theorem.

Local vs. Absolute Extrema

Local (relative) extrema: A function has a local maximum at c if f(c) ≥ f(x) for all x near c. A local minimum if f(c) ≤ f(x) for all x near c. These are "hills" and "valleys" in the graph.

Absolute (global) extrema: A function has an absolute maximum at c if f(c) ≥ f(x) for ALL x in the domain. An absolute minimum if f(c) ≤ f(x) for all x in the domain.

Absolute extrema are the overall highest and lowest points over the entire domain or interval.

The Extreme Value Theorem (EVT)

Statement: If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum on [a, b].

Two requirements:

1. Continuous function
2. Closed (bounded) interval [a, b]

Why both matter:

• f(x) = 1/x on (0, 1): no maximum (approaches ∞ near 0) — interval is open
• f(x) = x on [0, ∞): no maximum — interval is unbounded
• f(x) = |x|/x on [-1, 1]: discontinuous at 0; no minimum

Finding Absolute Extrema on [a, b]: The Closed Interval Method

Key theorem: The absolute extrema occur at either:

• Critical points inside (a, b) where f'(x) = 0 or f'(x) undefined, OR
• Endpoints x = a or x = b

The Closed Interval Method:

1. Find all critical points of f in (a, b)
2. Evaluate f at each critical point AND at both endpoints
3. The largest value is the absolute maximum; the smallest is the absolute minimum

Example 1

Find the absolute extrema of f(x) = 2x³ - 3x² - 12x + 1 on [-2, 3].

Step 1: Find critical points. f'(x) = 6x² - 6x - 12 = 6(x² - x - 2) = 6(x - 2)(x + 1) = 0 Critical points: x = 2 and x = -1 (both in (-2, 3) ✓)

Step 2: Evaluate f at critical points and endpoints.