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Absolute vs Local Extrema

MATH161
Reference: Stewart §4.1  •  Chapter: 3  •  Section: 1

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Absolute vs Local Extrema

Why Do We Need Two Types of "Highest" and "Lowest"?

Every optimization problem—from minimizing manufacturing costs to maximizing profit—comes down to finding where a function reaches its highest or lowest values. But "highest" can mean different things depending on your perspective.

Imagine hiking through a mountain range. You might reach the top of a hill and feel like you're at a high point—you're higher than everything around you. But looking across the range, you can see a much taller mountain in the distance. The hilltop is a local maximum (highest nearby), while the distant peak is the absolute maximum (highest overall).

This distinction matters enormously in applications. A company might find a pricing strategy that's "locally optimal"—better than small adjustments—but miss a completely different strategy that's globally better.

Prerequisite Map

Prerequisites
Function RepresentationsDomain & Range
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This skill
Absolute vs Local Extrema
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Unlocks
Extreme Value TheoremFermat's Theorem & Critical Numbers

Quick Reference

Property Value
Section Stewart §4.1
Course MATH161
Difficulty Beginner
Time ~15 minutes

Key Concepts

Absolute (Global) Extrema

An absolute maximum is the largest value a function attains on its entire domain. An absolute minimum is the smallest value.

Definition: Let $c$ be a number in the domain $D$ of a function $f$. Then $f(c)$ is the:

The absolute maximum and minimum values are called extreme values of $f$.

Local (Relative) Extrema

A local maximum is the largest value in some neighborhood around a point—you compare only to nearby points. A local minimum is the smallest value nearby.

Definition: The number $f(c)$ is a:

Here "near $c$" means on some open interval containing $c$.

Visual Comparison

       f(x)
        │      ★ absolute max (at endpoint)
        │     /
        │    /   ○ local max
        │   /   /\
        │  /   /  \     another endpoint
        │ /   /    \   /
        │/   /      \ /
        └────○───────●─────────── x
          local     absolute
          min       min

The Critical Distinction

Type Comparison Set Can Occur at Endpoints?
Absolute extremum Entire domain Yes
Local extremum Some open interval around the point No (requires open neighborhood)

Key insight: An absolute extremum can also be a local extremum—but not if it occurs at an endpoint. Endpoints don't have open intervals around them (within the domain), so they can't be local extrema.

Standard Examples

Function Domain Absolute Max Absolute Min Local Extrema
$f(x) = x^2$ $\mathbb{R}$ None (unbounded) $f(0) = 0$ Local min at 0
$f(x) = x^3$ $\mathbb{R}$ None None None
$f(x) = \cos x$ $\mathbb{R}$ $1$ at $x = 2n\pi$ $-1$ at $x = (2n+1)\pi$ Infinitely many
$f(x) = x^2$ $[-1, 2]$ $f(2) = 4$ $f(0) = 0$ Local min at 0 only

Practice Problems

Level 1 Terminology Recognition

For the function $g(x) = -x^2 + 4x$ on the interval $[0, 5]$:

  1. Where does $g$ achieve its largest value on $[0, 5]$?
  2. Is this value an absolute maximum, local maximum, both, or neither?
  3. Where does $g$ achieve its smallest value on $[0, 5]$?
Thought Process

The function $g(x) = -x^2 + 4x$ is a downward-opening parabola. To find the vertex:

  • Complete the square: $g(x) = -(x^2 - 4x) = -(x^2 - 4x + 4) + 4 = -(x-2)^2 + 4$
  • Vertex at $(2, 4)$

Now check the three candidates for extrema: the vertex (if in the interval) and the endpoints.

  • $g(0) = 0$
  • $g(2) = 4$ (vertex)
  • $g(5) = -25 + 20 = -5$
Show Answer

(a) The largest value is $g(2) = 4$, occurring at $x = 2$.

(b) This is both an absolute maximum and a local maximum.

  • It's an absolute maximum because $g(2) = 4$ is larger than $g(x)$ for every $x$ in $[0, 5]$.
  • It's a local maximum because $x = 2$ is an interior point, and $g(2)$ is the largest value in any small open interval around 2.

(c) The smallest value is $g(5) = -5$, occurring at $x = 5$.

This is an absolute minimum but NOT a local minimum (since $x = 5$ is an endpoint).

Level 2 Identifying Extrema from Data

A continuous function $h$ on $[-2, 4]$ has the following values at key points:

  • $h(-2) = 3$
  • $h(0) = 7$ (local maximum)
  • $h(1) = 2$ (local minimum)
  • $h(3) = 5$ (local maximum)
  • $h(4) = 1$

Find:

  1. The absolute maximum value and where it occurs
  2. The absolute minimum value and where it occurs
  3. All local maximum values
  4. All local minimum values
Thought Process

To find absolute extrema, compare ALL candidate values: endpoints and local extrema.

Candidates:

  • $h(-2) = 3$ (left endpoint)
  • $h(0) = 7$ (local max)
  • $h(1) = 2$ (local min)
  • $h(3) = 5$ (local max)
  • $h(4) = 1$ (right endpoint)

The largest among these is the absolute max; the smallest is the absolute min.

Show Answer

(a) Absolute maximum: $h(0) = 7$ at $x = 0$

(b) Absolute minimum: $h(4) = 1$ at $x = 4$

(c) Local maximum values: $h(0) = 7$ and $h(3) = 5$

(d) Local minimum values: $h(1) = 2$

Note: The absolute minimum at $x = 4$ is not a local minimum because 4 is an endpoint. The local minimum at $x = 1$ is not the absolute minimum because $h(4) = 1 < h(1) = 2$.

Level 3 Existence of Extrema on Different Domains

Consider $f(x) = \dfrac{x}{x+1}$ on each domain. Determine whether an absolute maximum and absolute minimum exist, and find them if they do.

  1. $[0, 3]$
  2. $(0, 3)$
  3. $[0, \infty)$
Thought Process

First, understand the function's behavior:

  • $f(x) = \frac{x}{x+1} = 1 - \frac{1}{x+1}$
  • As $x$ increases, $f(x)$ increases toward 1
  • $f(0) = 0$
  • $f(x) \to 1$ as $x \to \infty$

The function is increasing on $[0, \infty)$, so:

  • On a closed interval, min at left endpoint, max at right endpoint
  • On open intervals, must be careful about whether limits are attained
Show Answer

(a) On $[0, 3]$:

  • $f(0) = 0$ → absolute minimum
  • $f(3) = \frac{3}{4} = 0.75$ → absolute maximum

Both exist because the interval is closed and $f$ is continuous.

(b) On $(0, 3)$:

  • As $x \to 0^+$: $f(x) \to 0$ (not attained)
  • As $x \to 3^-$: $f(x) \to 0.75$ (not attained)
  • No absolute maximum, no absolute minimum

The function approaches but never reaches 0 and 0.75 on this open interval.

(c) On $[0, \infty)$:

  • $f(0) = 0$ → absolute minimum
  • As $x \to \infty$: $f(x) \to 1$ (never attained)
  • No absolute maximum

The minimum exists (at the included left endpoint), but the function grows arbitrarily close to 1 without reaching it.

Level 4 Extrema of Piecewise Functions

Let $f(x) = \begin{cases} 4 - x^2 & \text{if } -2 \leq x < 1 \\ x + 1 & \text{if } 1 \leq x \leq 3 \end{cases}$

  1. Is $f$ continuous on $[-2, 3]$?
  2. Find all local and absolute extrema.
  3. Sketch the function to verify your answer.
Thought Process

Check continuity at the boundary $x = 1$:

  • From left: $\lim_{x \to 1^-} (4 - x^2) = 4 - 1 = 3$
  • From right: $\lim_{x \to 1^+} (x + 1) = 1 + 1 = 2$
  • At $x = 1$: $f(1) = 1 + 1 = 2$

Since the left-hand limit doesn't match the function value, there's a discontinuity.

For extrema, analyze each piece:

  • $4 - x^2$ on $[-2, 1)$: downward parabola, vertex at $x = 0$
  • $x + 1$ on $[1, 3]$: increasing line
Show Answer

(a) No, $f$ is not continuous at $x = 1$.

  • $\lim_{x \to 1^-} f(x) = 3$
  • $f(1) = 2$
  • Jump discontinuity at $x = 1$

(b) Evaluate at critical points and endpoints:

On $[-2, 1)$:

  • $f(-2) = 4 - 4 = 0$ (endpoint)
  • $f(0) = 4$ (vertex of parabola, interior point)
  • As $x \to 1^-$: $f(x) \to 3$

On $[1, 3]$:

  • $f(1) = 2$
  • $f(3) = 4$ (endpoint)

Extrema:

  • Absolute maximum: $f(0) = 4$ and $f(3) = 4$ (tie)
  • Absolute minimum: $f(-2) = 0$
  • Local maximum: $f(0) = 4$ (at vertex, interior point)
  • Local minimum: None in the interior

Note: $x = 0$ gives both an absolute and local maximum. The endpoint $x = -2$ gives the absolute minimum but is not a local minimum.

(c) The sketch shows a downward parabola from $(-2, 0)$ reaching up to $(0, 4)$ then down toward $(1, 3)$ (open circle), jumping to $(1, 2)$, then a line going up to $(3, 4)$.

Level 5 Existence and Uniqueness of Extrema
  1. Give an example of a continuous function on $[0, 2]$ that has exactly one absolute maximum (at an interior point) and no local minimum. Is this possible?
  2. Prove or disprove: If $f$ is continuous on $[a, b]$ and has its absolute maximum at an interior point $c \in (a, b)$, then $c$ must be a local maximum.
  3. Can a function have infinitely many local maxima but only one absolute maximum? If yes, give an example. If no, explain why not.
Thought Process

(a) For "no local minimum," we need a function where every interior point has nearby points with smaller values. Consider increasing functions or functions that are flat.

(b) If $f(c) \geq f(x)$ for all $x$ in $[a, b]$, then certainly $f(c) \geq f(x)$ for all $x$ in any smaller interval around $c$.

(c) Think about periodic functions like $\cos x$ that have infinitely many local maxima, all with the same value 1.

Show Answer

(a) This is impossible for a continuous function on a closed interval.

By the Extreme Value Theorem, a continuous function on $[a, b]$ must have both an absolute maximum and an absolute minimum. The absolute minimum must occur somewhere—either at an endpoint or at an interior local minimum.

If the absolute max is at an interior point and there's "no local minimum," then the absolute minimum must be at an endpoint. But then we have both extrema, just the min at an endpoint.

(b) True. This is a direct consequence of definitions.

Proof: Suppose $f$ is continuous on $[a, b]$ and achieves its absolute maximum at $c \in (a, b)$. Then $f(c) \geq f(x)$ for all $x \in [a, b]$.

Since $c$ is an interior point, there exists $\delta > 0$ such that $(c - \delta, c + \delta) \subset [a, b]$. For all $x$ in this open interval, we have $f(c) \geq f(x)$ (since these $x$ are in $[a, b]$).

By definition, $f(c)$ is a local maximum. $\square$

(c) Yes, this is possible.

Example: $f(x) = \cos x$ on $[0, 4\pi]$

  • Local maxima at $x = 0, 2\pi, 4\pi$ with value $f = 1$
  • Actually, only $x = 2\pi$ is a local max (the endpoints aren't)

Better example: $f(x) = \cos x$ on $\mathbb{R}$

  • Infinitely many local maxima at $x = 2n\pi$ for integer $n$
  • All have the value 1, which is also the absolute maximum

So yes: there's one absolute maximum value (namely 1), achieved at infinitely many points, each of which is also a local maximum.

CCI-Style Conceptual Questions

Level 2 Conceptual: Endpoint Extrema

True or False (justify your answer):

"If a continuous function $f$ on $[1, 5]$ has its absolute maximum at $x = 5$, then $f$ must have a local maximum somewhere in the interval."

Thought Process

The question asks whether having an absolute max at an endpoint guarantees a local max exists somewhere.

Consider: If the function is strictly increasing on $[1, 5]$, does it have any local maxima?

  • Absolute max at $x = 5$ (endpoint, not local max)
  • No interior peaks
  • No local maxima at all
Show Answer

False.

Counterexample: $f(x) = x$ on $[1, 5]$.

  • Absolute maximum: $f(5) = 5$ at $x = 5$ (endpoint)
  • Absolute minimum: $f(1) = 1$ at $x = 1$ (endpoint)
  • Local extrema: None

The function is strictly increasing, so no interior point can be a local maximum or minimum. The absolute max at the endpoint is not a local max.

Mastery Checklist

Mental Model

The Mountain Range Perspective:

Think of zooming in and out on a topographical map:

A hilltop in one valley might not be the highest point overall. Similarly, a local maximum of $f$ might not be the absolute maximum—there could be a higher value somewhere else (including at an endpoint).

The key question is always: "Highest compared to what?"

Connections

Looking back:

Looking ahead:

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Last updated: 2026-01-22