Absolute vs Local Extrema

9 min read

Reference: Stewart §4.1

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

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:

absolute maximum value of $f$ on $D$ if $f(c) \geq f(x)$ for all $x$ in $D$
absolute minimum value of $f$ on $D$ if $f(c) \leq f(x)$ for all $x$ in $D$

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, comparing only to nearby points. A local minimum is the smallest value nearby.

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

local maximum value if $f(c) \geq f(x)$ when $x$ is near $c$
local minimum value if $f(c) \leq f(x)$ when $x$ is near $c$

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

Common misconception

every local maximum is larger than every local minimum.

This is the concept-image-conflicts-definition error. “Local” means only locally -- in a neighborhood of the point. Two local extrema on the same function have no required relationship to each other. For $f(x) = \sin x$ on $[0, 4\pi]$: the local maxima are at $x = \pi/2$ and $x = 5\pi/2$, both with value $1$; the local minima are at $x = 3\pi/2$ and $x = 7\pi/2$, both with value $-1$. In this case, every local max IS larger than every local min -- but that is a property of sine, not a property of local extrema in general. A function can have a local maximum at height $2$ and a local minimum at height $10$ if the graph dips and rises in a complicated way.

Common misconception

the absolute maximum is the highest point on the graph, wherever it occurs.

This is the input-output-confusion error applied to extrema. The absolute maximum of $f$ on a domain $D$ is the largest OUTPUT value the function takes on $D$. If the domain is $[-1, 2]$ for $f(x) = x^2$, the absolute maximum is $f(2) = 4$ (the output), occurring at the input $x = 2$. Students sometimes report “the absolute maximum is $x = 2$” -- but $x = 2$ is the input (the location); $4$ is the maximum value (the output). The question “find the absolute maximum” asks for the output value; “find where it occurs” asks for the input.

Practice Problems

Level 1 Terminology Recognition

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

Where does $g$ achieve its largest value on $[0, 5]$?
Is this value an absolute maximum, local maximum, both, or neither?
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:

The absolute maximum value and where it occurs
The absolute minimum value and where it occurs
All local maximum values
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.

$[0, 3]$
$(0, 3)$
$[0, \infty)$

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

Is $f$ continuous on $[-2, 3]$?
Find all local and absolute extrema.
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

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?
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.
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

I can state the precise definitions of absolute and local extrema
I can identify absolute and local extrema from a graph
I understand why endpoints cannot be local extrema
I can determine which extrema exist on open vs closed vs unbounded intervals
I can analyze piecewise functions for extrema

Mental Model

The Mountain Range Perspective:

Think of zooming in and out on a topographical map:

Zoomed in (local view): A hilltop is the highest point you can see, a local maximum
Zoomed out (global view): The tallest peak in the entire range is the absolute maximum

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:

Function representations provide the foundation for reading function values
Domain and range determines where we look for extrema

Looking ahead:

The Extreme Value Theorem tells us when absolute extrema are guaranteed to exist
Fermat’s Theorem and Critical Numbers tells us where to look for local extrema

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Last updated: 2026-01-22