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:
- By EVT, absolute extrema exist.
- If an absolute extremum is at an interior point $c$, it’s also a local extremum at $c$.
- By Fermat’s Theorem (extended), $c$ must be a critical number.
- 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
the absolute maximum always occurs at a critical number.
This is the concept-image-conflicts-definition error. On a closed interval $[a, b]$, the absolute maximum occurs at either a critical number in the open interior $(a, b)$ OR at one of the endpoints $a$ or $b$. The endpoints are not critical numbers (they are not interior points), but they are still candidates. For $f(x) = x$ on $[0, 3]$: $f'(x) = 1 \neq 0$ everywhere, so there are no critical numbers at all. The absolute maximum is $f(3) = 3$, occurring at the right endpoint. Forgetting to check endpoints is the single most common error in applying the Closed Interval Method.
the candidate with the largest $f'$ value is the maximum.
This is the height-vs-slope error in optimization. The Closed Interval Method compares function VALUES $f(c)$ at each candidate, not derivative values $f'(c)$. At a critical number, $f'(c) = 0$ or $f'(c)$ is undefined -- the derivative gives no information about which candidate is largest. The comparison is entirely among the output values $f(c_1), f(c_2), \ldots, f(a), f(b)$. Whichever output is largest is the absolute maximum; whichever is smallest is the absolute minimum. The derivative has already done its job (identifying the candidates); only the function values determine the winner.
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
Find the absolute maximum and minimum values of $f(x) = x^2 - 4x + 5$ on $[0, 3]$.
Find the absolute maximum and minimum values of $f(x) = 2x^3 - 9x^2 + 12x - 3$ on $[0, 3]$.
Find the absolute maximum and minimum values of $f(x) = x^{2/3}(x - 4)$ on $[-1, 8]$.
Find the absolute maximum and minimum values of $f(x) = x - 2\sin x$ on $[0, 2\pi]$.
A rectangular box with no top is to be made from a square piece of cardboard by cutting equal squares from each corner and folding up the sides. If the cardboard is 12 inches on each side, what size square should be cut from each corner to maximize the volume of the box?
Set up the problem, identify the constraint, apply the Closed Interval Method, and verify your answer makes physical sense.
Mastery Checklist
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:
- Round up all contestants (find critical numbers, note endpoints)
- Score each one (evaluate $f$ at each point)
- 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.
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
| Previous | Up | Next |
|---|---|---|
| Critical Numbers | Section Index | Mean Value Theorem |
Last updated: 2026-01-22