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The Closed Interval Method

The Complete Algorithm for Finding Absolute Extrema

We now have all the pieces to systematically find absolute maximum and minimum values:

• The Extreme Value Theorem guarantees they exist (for continuous functions on closed intervals)
• Critical numbers tell us where interior extrema can occur
• Endpoints are the only other places to check

The Closed Interval Method combines these into a foolproof 3-step algorithm.

Prerequisite Map

Quick Reference

Property	Value
Section	Stewart §3.1
Course	MATH161
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

The Closed Interval Method

To find the absolute maximum and absolute minimum values of a continuous function $f$ on a closed interval $[a, b]$:

$$\boxed{ \begin{aligned} &\textbf{Step 1: } \text{Find all critical numbers of } f \text{ in } (a, b) \\ &\textbf{Step 2: } \text{Evaluate } f \text{ at critical numbers and endpoints} \\ &\textbf{Step 3: } \text{Compare: largest = abs max, smallest = abs min} \end{aligned} }$$

Why This Works

Theorem: If $f$ is continuous on $[a, b]$, then $f$ attains its absolute maximum and minimum at either:

• A critical number in $(a, b)$, OR
• An endpoint ($a$ or $b$)

Proof idea:

1. By EVT, absolute extrema exist.
2. If an absolute extremum is at an interior point $c$, it's also a local extremum at $c$.
3. By Fermat's Theorem (extended), $c$ must be a critical number.
4. Otherwise, the extremum is at an endpoint.

So we only need to check critical numbers and endpoints—that's our complete list of candidates.

The Algorithm Visualized

                    f(x)
                     │
    endpoint         │      ★ critical number
    f(a) = ?    ───▶ │     (f'(c) = 0)
                     │    /\
                     │   /  \
                     │  /    \
                     │ /      \_____ endpoint
                     │/              f(b) = ?
                     └────────────────────────── x
                     a    c                    b

Candidates: {f(a), f(c), f(b)}
Compare all values → identify max and min

Common Mistakes to Avoid

Mistake	Why It's Wrong	Correct Approach
Forgetting to check endpoints	Absolute extrema can occur at endpoints	Always evaluate $f(a)$ and $f(b)$
Including critical numbers outside $[a,b]$	Only interior critical numbers matter	Only use critical numbers in $(a, b)$
Stopping after finding $f'(x) = 0$	Must compare values, not just find critical points	Evaluate and compare all candidates
Claiming the method works for open intervals	EVT requires closed intervals	Only applies to $[a, b]$, not $(a, b)$

Worked Example

Find the absolute maximum and minimum of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$.

Step 1: Find critical numbers in $(-2, 2)$. $$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$$ $$f'(x) = 0 \Rightarrow x = 1 \text{ or } x = -1$$

Both are in $(-2, 2)$. ✓

Step 2: Evaluate $f$ at critical numbers and endpoints.

$x$	Type	$f(x) = x^3 - 3x + 1$
$-2$	endpoint	$(-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1$
$-1$	critical	$(-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$
$1$	critical	$(1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1$
$2$	endpoint	$(2)^3 - 3(2) + 1 = 8 - 6 + 1 = 3$

Step 3: Compare.

• Largest value: $3$ (occurs at $x = -1$ and $x = 2$)
• Smallest value: $-1$ (occurs at $x = -2$ and $x = 1$)

Answer:

• Absolute maximum: $3$, attained at $x = -1$ and $x = 2$
• Absolute minimum: $-1$, attained at $x = -2$ and $x = 1$

Practice Problems

Mastery Checklist

• [ ] I can state the three steps of the Closed Interval Method
• [ ] I can explain why the method works (connecting EVT and critical numbers)
• [ ] I can find critical numbers of polynomials, rational functions, and functions with fractional exponents
• [ ] I can correctly evaluate functions at critical numbers and endpoints
• [ ] I can identify absolute max/min from a list of candidate values
• [ ] I can apply the method to word problems by first setting up the function and domain

Mental Model

The Talent Show Analogy:

Finding the absolute max/min on $[a, b]$ is like judging a talent show with a limited contestant pool.

The contestants:

• Endpoints ($a$ and $b$): They're automatically in the competition—everyone who shows up gets judged.
• Critical numbers: These are the only interior points "talented" enough to possibly win (horizontal tangent or special behavior).

The judging:

1. Round up all contestants (find critical numbers, note endpoints)
2. Score each one (evaluate $f$ at each point)
3. Compare scores (the highest wins "max," the lowest wins "min")

You don't need to check every interior point—only the critical numbers have a chance at winning.

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Connections

Looking back:

• Extreme Value Theorem guarantees the winner exists
• Critical numbers identifies the only possible interior winners

Looking ahead:

• Optimization problems applies this method to real-world scenarios
• The First Derivative Test and Second Derivative Test classify critical numbers as local max, local min, or neither

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Previous	Up	Next
Critical Numbers	Section Index	Mean Value Theorem

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Last updated: 2026-01-22