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Absolute and Local Extrema

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Reference: Stewart §3.1

Why Do We Care About Maximum and Minimum Values?

Every optimization problem (from minimizing manufacturing costs to maximizing profit, from finding the fastest route to designing the strongest beam) comes down to finding where a function reaches its highest or lowest values. Before we can solve these problems, we need precise language to describe what kind of high or low point we’re looking at.

Consider hiking in the mountains. The summit of a local hill might be the highest point in your immediate area, but it’s not necessarily the highest point in the entire mountain range. Similarly, a function can have peaks and valleys at different scales, and distinguishing between them is essential for understanding the function’s behavior.

Prerequisite Map

PrerequisitesDomain and Range
This skillAbsolute and Local Extrema
Leads toCritical NumbersThe Extreme Value Theorem

Quick Reference

Property Value
Section Stewart §3.1
Course MATH161
Difficulty Beginner
Time ~15 minutes

Key Concepts

Absolute (Global) Extrema

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

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

Together, 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. A local minimum is the smallest value nearby.

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

“Near $c$” means on some open interval containing $c$.

Visual Comparison

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

Key Distinction

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

Important: An absolute extremum can also be a local extremum, but not always. If the absolute max or min occurs at an endpoint, it is NOT a local extremum because endpoints don’t have open intervals around them within the domain.

Examples from Standard Functions

Function Domain Absolute Max Absolute Min Local Extrema
$f(x) = x^2$ $\mathbb{R}$ None $f(0) = 0$ 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
$f(x) = x^2$ on $[-1, 2]$ $[-1, 2]$ $f(2) = 4$ $f(0) = 0$ Local min at 0
Common misconception

an absolute maximum must also be a local maximum.

This is the concept-image-conflicts-definition error. A local maximum requires an open interval around the point where the function is largest. Endpoints of a closed domain cannot be local maxima because there is no open interval around them that stays within the domain. For $f(x) = x^2$ on $[-1, 2]$: the absolute maximum is $f(2) = 4$, occurring at the right endpoint. But $x = 2$ is NOT a local maximum because there is no open interval around $x = 2$ within $[-1, 2]$ where $f(2)$ is the largest value -- points just to the left of $2$ give smaller values, but there are no points to the right. Endpoints can be absolute extrema without being local extrema.

Common misconception

the maximum is the x-value where the function is largest.

This is the input-output-confusion error. The maximum VALUE of a function is the largest OUTPUT. The x-value where it occurs is the input. For $f(x) = -x^2 + 4$ on $[-2, 2]$: the maximum value is $f(0) = 4$ (the output); it occurs at $x = 0$ (the input). Saying “the maximum is $x = 0$” confuses the location with the value. A complete answer names both: “the maximum value is $4$, occurring at $x = 0$.”

Practice Problems

Level 1 Identifying Extrema from a Description

For the function $f(x) = x^2$ with domain $[-2, 3]$:

  1. What is the absolute maximum value and where does it occur?
  2. What is the absolute minimum value and where does it occur?
  3. Are there any local extrema? If so, identify them.
Thought Process

The parabola $y = x^2$ opens upward with vertex at the origin. On the closed interval $[-2, 3]$:

  • The lowest point of the parabola is at $x = 0$
  • Check endpoint values: $f(-2) = 4$ and $f(3) = 9$
  • The highest point must be at one of the endpoints since the parabola opens upward

For local extrema, remember that endpoints cannot be local extrema (they require an open interval around them).

Show Answer

(a) The absolute maximum is $f(3) = 9$, occurring at $x = 3$.

(b) The absolute minimum is $f(0) = 0$, occurring at $x = 0$.

(c) Yes, there is a local minimum at $x = 0$ with value $f(0) = 0$. The vertex of the parabola is the lowest point in any neighborhood around it.

Note: The absolute maximum at $x = 3$ is NOT a local maximum because $x = 3$ is an endpoint.

Level 2 Reading Extrema from a Graph

A continuous function $f$ is defined on $[0, 6]$ with the following values:

  • $f(0) = 2$
  • $f(1) = 5$ (local maximum)
  • $f(2) = 3$
  • $f(4) = 1$ (local minimum)
  • $f(6) = 4$

Identify:

  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

For absolute extrema, compare ALL values: endpoints AND local extrema.

List of values to compare:

  • $f(0) = 2$ (endpoint)
  • $f(1) = 5$ (local max)
  • $f(4) = 1$ (local min)
  • $f(6) = 4$ (endpoint)

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

Show Answer

(a) Absolute maximum: $f(1) = 5$ at $x = 1$

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

(c) Local maximum values: $f(1) = 5$

(d) Local minimum values: $f(4) = 1$

Note that in this case, both the absolute max and absolute min happen to coincide with the local extrema. This isn’t always the case. Sometimes absolute extrema occur at endpoints.

Level 3 Extrema with Open vs Closed Intervals

Consider $f(x) = \dfrac{1}{x}$ on each of the following domains. For each, determine whether absolute maximum and minimum values exist, and if so, find them.

  1. $[1, 4]$
  2. $(1, 4)$
  3. $[1, \infty)$
Thought Process

The function $f(x) = 1/x$ is decreasing on $(0, \infty)$.

For each domain:

  • Closed intervals: check endpoints and any interior critical points
  • Open intervals: the function approaches but may not attain certain values
  • Unbounded domains: the function may have no maximum or minimum

Key insight: On a closed interval, a continuous function always has absolute extrema. On an open interval, it might not.

Show Answer

(a) On $[1, 4]$:

  • $f(1) = 1$ (absolute maximum)
  • $f(4) = 0.25$ (absolute minimum)
  • Both exist because the interval is closed and $f$ is continuous.

(b) On $(1, 4)$:

  • As $x \to 1^+$, $f(x) \to 1$, but $f$ never equals 1
  • As $x \to 4^-$, $f(x) \to 0.25$, but $f$ never equals 0.25
  • No absolute maximum, no absolute minimum (the supremum and infimum exist but are not attained)

(c) On $[1, \infty)$:

  • $f(1) = 1$ (absolute maximum)
  • As $x \to \infty$, $f(x) \to 0$, but never equals 0
  • No absolute minimum (the infimum is 0 but not attained)
Level 4 Piecewise Function Extrema

Let $f(x) = \begin{cases} x^2 & \text{if } -2 \leq x \leq 0 \\ 3 - 2x & \text{if } 0 < x \leq 2 \end{cases}$

  1. Is $f$ continuous on $[-2, 2]$?
  2. Find all local and absolute extrema of $f$ on $[-2, 2]$.
Thought Process

First check continuity at the boundary $x = 0$:

  • From the left: $\lim_{x \to 0^-} x^2 = 0$
  • From the right: $\lim_{x \to 0^+} (3 - 2x) = 3$
  • At $x = 0$: $f(0) = 0^2 = 0$

There’s a jump discontinuity at $x = 0$.

For extrema, analyze each piece:

  • $x^2$ on $[-2, 0]$: parabola opening up, min at $x = 0$, check endpoint $x = -2$
  • $3 - 2x$ on $(0, 2]$: decreasing line, check boundary behavior and endpoint
Show Answer

(a) No, $f$ is not continuous at $x = 0$.

  • $f(0) = 0$
  • $\lim_{x \to 0^+} f(x) = 3$
  • Since $f(0) \neq \lim_{x \to 0^+} f(x)$, there’s a jump discontinuity.

(b) Finding extrema:

On $[-2, 0]$: The parabola $x^2$ has:

  • $f(-2) = 4$
  • $f(0) = 0$ (local and absolute minimum on this piece)

On $(0, 2]$: The line $3 - 2x$ has:

  • As $x \to 0^+$: $f(x) \to 3$ (not attained)
  • $f(2) = -1$

Comparing all values: $f(-2) = 4$, $f(0) = 0$, $f(2) = -1$

  • Absolute maximum: $f(-2) = 4$ at $x = -2$
  • Absolute minimum: $f(2) = -1$ at $x = 2$
  • Local minimum: $f(0) = 0$ at $x = 0$ (smallest in a neighborhood on the left)

Note: Despite the discontinuity, absolute extrema still exist because $f$ attains both its supremum and infimum.

Level 5 Constructing Functions with Specified Extrema
  1. Sketch a continuous function on $[0, 4]$ that has an absolute maximum at $x = 2$ but no local maximum at $x = 2$. Is this possible? Explain.
  2. Give an example of a function on $[0, 2]$ that has a local maximum at $x = 1$ but no absolute maximum on the entire interval. What must be true about this function?
  3. Prove that if $f$ is continuous on a closed interval $[a, b]$ and has a local maximum at an interior point $c$, then $f(c)$ is not necessarily the absolute maximum. Construct a specific counterexample.
Thought Process

(a) Think about what “local maximum” requires: $f(c) \geq f(x)$ for $x$ near $c$. If $x = 2$ is an interior point and gives an absolute max, is it automatically a local max?

(b) For a function to have a local max but no absolute max, the function must take arbitrarily large values somewhere. On a closed interval with a continuous function, this is impossible (EVT). So the function must be...?

(c) Construct a function where the local max value is smaller than a value at an endpoint.

Show Answer

(a) This is not possible for a continuous function.

If $f$ is continuous on $[0, 4]$ and has its absolute maximum at an interior point $x = 2$, then by definition $f(2) \geq f(x)$ for all $x \in [0, 4]$. In particular, this holds for all $x$ near 2, so $f(2)$ is also a local maximum.

Key insight: Every interior absolute maximum of a continuous function is automatically a local maximum.

(b) The function must be discontinuous.

Example: $f(x) = \begin{cases} x & \text{if } x \neq 1 \\ 2 & \text{if } x = 1 \end{cases}$

Then $f(1) = 2$ is a local maximum (it’s the largest value near $x = 1$), but as $x \to 2^-$, $f(x) \to 2$, and the function never attains a value larger than 2 on $[0, 2]$. Wait, actually $f(2) = 2$ as well.

Better example: $f(x) = \begin{cases} 2 & \text{if } x = 1 \\ x & \text{if } x \neq 1 \end{cases}$ on $[0, 3)$

Here $f(1) = 2$ is a local max, but the function has no absolute maximum because $f(x) \to 3$ as $x \to 3^-$.

(c) Counterexample: Let $f(x) = -x^2 + 2x$ on $[-1, 3]$.

  • $f'(x) = -2x + 2 = 0$ gives $x = 1$
  • $f(1) = -1 + 2 = 1$ (local maximum at interior point)
  • $f(-1) = -1 - 2 = -3$
  • $f(3) = -9 + 6 = -3$

Actually, in this case $f(1) = 1$ IS the absolute maximum. Try again.

Better counterexample: $f(x) = \sin x$ on $[0, 5\pi/2]$

  • Local maximum at $x = \pi/2$ with $f(\pi/2) = 1$
  • Local maximum at $x = 5\pi/2$ with $f(5\pi/2) = 1$
  • Both are absolute maxima (tied)

Actual counterexample: $f(x) = x^3 - 3x$ on $[-1, 3]$

  • $f'(x) = 3x^2 - 3 = 0$ gives $x = \pm 1$
  • $f(-1) = -1 + 3 = 2$ (local maximum)
  • $f(1) = 1 - 3 = -2$ (local minimum)
  • $f(3) = 27 - 9 = 18$ (endpoint)

Here, $f(-1) = 2$ is a local maximum, but the absolute maximum is $f(3) = 18$ at the right endpoint.

Mastery Checklist

Mental Model

The Mountain Range Analogy:

Think of a function’s graph as a hiking trail through mountains:

Just as a hilltop might not be the highest peak in the range, a local maximum might not be the absolute maximum. And just as you can’t stand on the “highest point” of an infinite plain (there isn’t one), some functions on unbounded domains have no absolute extrema.

Connections

Looking back:

Looking ahead:

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Last updated: 2026-01-22