Absolute and Local Extreme Values
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.1: “Related Rates” is nearby; the extrema content is at Section 4.3: “Maxima and Minima” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-3-maxima-and-minima |
| Textbook used in class | Stewart, Calculus, Section 3.1: “Maximum and Minimum Values” |
Opening Scenario
Imagine recording the elevation along a hiking trail from the trailhead to the summit and back. At the very top of the mountain the elevation is higher than at any other point on the entire route. That peak is the absolute maximum. But partway up there is a small knoll: higher than everything within a short distance on either side, yet clearly not the highest point of the whole hike. That knoll is a local maximum.
After the summit, the trail dips into a saddle before climbing again to a second, lower summit. The bottom of the saddle is lower than everything nearby, so it is a local minimum, even though the trailhead is at a lower elevation still.
Before you compute anything, sketch the rough shape in your mind: one dominant peak, a smaller knoll, a saddle, and a flat start. You already know where the extrema are -- the definitions will now give you precise language for what you already see.
Quick Reference
| Term | Informal meaning | Formal meaning |
|---|---|---|
| Absolute max of $f$ on $D$ | highest output on the whole domain | $f(c) \geq f(x)$ for all $x \in D$ |
| Absolute min of $f$ on $D$ | lowest output on the whole domain | $f(c) \leq f(x)$ for all $x \in D$ |
| Local max at $c$ | highest output in some neighborhood | $f(c) \geq f(x)$ for $x$ near $c$ |
| Local min at $c$ | lowest output in some neighborhood | $f(c) \leq f(x)$ for $x$ near $c$ |
| Critical number | input where derivative is zero or undefined | $f'(c) = 0$ or $f'(c)$ does not exist |
Key Concepts
1. Absolute Versus Local Extrema
The words absolute and local answer two different questions.
Absolute asks: is this the highest (or lowest) output over the entire domain?
Local asks: is this the highest (or lowest) output compared to everything nearby?
Every absolute extremum that occurs at an interior point of the domain is also a local extremum. The reverse is false: a local extremum need not be the largest or smallest value overall.
A function can have many local extrema and at most one absolute maximum value (though that value might be achieved at several inputs). A function can also have no absolute extrema at all, if the domain is open or the function is not continuous.
“the local max is the highest nearby, so it must matter globally.” A local maximum is only a claim about a neighborhood. The function can be larger anywhere outside that neighborhood. On the hiking trail, the knoll is the high point within a short stretch; the summit dwarfs it over the whole trail.
2. The Extreme Value Theorem
The Extreme Value Theorem tells you when you are guaranteed absolute extrema exist.
Extreme Value Theorem. If $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ attains an absolute maximum value and an absolute minimum value somewhere on $[a, b]$.
The two hypotheses matter. Drop either one and the guarantee fails.
- If the interval is open -- say $(0, 1)$ -- then $f(x) = x$ has no maximum on $(0, 1)$: the outputs approach $1$ but never reach it.
- If $f$ is not continuous -- say it jumps upward at one point -- then the function can “miss” the supremum.
When both hypotheses hold, the theorem says extrema exist; it does not say where they are. That is the job of the Closed Interval Method.
3. Fermat’s Theorem and Critical Numbers
Where can an absolute extremum live? The Extreme Value Theorem says somewhere on $[a, b]$. There are only two possibilities for an interior point $c$ (that is, $a < c < b$):
- $f'(c) = 0$ (the tangent line is horizontal), or
- $f'(c)$ does not exist (there is a corner, cusp, or vertical tangent).
This is Fermat’s Theorem: if $f$ has a local extremum at an interior point $c$ and $f$ is differentiable at $c$, then $f'(c) = 0$.
A number $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ does not exist is called a critical number. Every local extremum at an interior point must be a critical number. But the converse fails:
“if $f'(c) = 0$, then $f$ has a local extremum at $c$.” This is false. The function $f(x) = x^3$ has $f'(0) = 3 \cdot 0^2 = 0$, yet $f$ is increasing on both sides of $0$. The graph passes through the origin with a horizontal tangent but does not reverse direction. A critical number is a candidate for a local extremum, not a guarantee.
4. The Closed Interval Method
Combining everything: on a closed interval $[a, b]$ with $f$ continuous, the absolute extrema must occur either at a critical number in the interior or at an endpoint. This gives a complete procedure.
Closed Interval Method. To find the absolute maximum and minimum of a continuous function $f$ on $[a, b]$:
- Find all critical numbers of $f$ in the open interval $(a, b)$: solve $f'(c) = 0$ and find where $f'$ does not exist.
- Evaluate $f$ at each critical number and at both endpoints: $f(a)$ and $f(b)$.
- Compare all the values. The largest is the absolute maximum; the smallest is the absolute minimum.
No further analysis is needed. Comparing a short list of values is enough.
5. Worked Example: Polynomial on a Closed Interval
Example 1. Find the absolute maximum and minimum values of $f(x) = x^3 - 3x + 1$ on $[-2, 2]$.
Goal. Apply the Closed Interval Method. (This parallels Stewart 3.1, Example 5.)
Prediction. The function is a cubic with a positive leading coefficient, so it falls to the left and rises to the right. There should be a local max somewhere to the left and a local min somewhere to the right. The absolute max might be at the right endpoint or at the local max; the absolute min could be at the left endpoint or the local min.
Step 1: Find critical numbers.
$f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$
Setting $f'(x) = 0$: $x = 1$ or $x = -1$. Both are in $(-2, 2)$.
The derivative is a polynomial, so it exists everywhere. No additional critical numbers from non-differentiability.
Step 2: Evaluate $f$ at the critical numbers and endpoints.
$f(-2) = (-2)^3 - 3(-2) + 1 = -8 + 6 + 1 = -1$
$f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$
$f(1) = 1^3 - 3(1) + 1 = 1 - 3 + 1 = -1$
$f(2) = 2^3 - 3(2) + 1 = 8 - 6 + 1 = 3$
Step 3: Compare.
| Input | Value |
|---|---|
| $x = -2$ (endpoint) | $f = -1$ |
| $x = -1$ (critical) | $f = 3$ |
| $x = 1$ (critical) | $f = -1$ |
| $x = 2$ (endpoint) | $f = 3$ |
Boxed answer: Absolute maximum value is $3$, achieved at $x = -1$ and $x = 2$. Absolute minimum value is $-1$, achieved at $x = -2$ and $x = 1$.
Recap. Notice that two different inputs share the absolute maximum and two share the absolute minimum. The theorem guarantees the extreme values are attained; it says nothing about uniqueness. The prediction was roughly right: there is a local max at $x = -1$ (which ties for absolute max with the right endpoint) and a local min at $x = 1$ (which ties for absolute min with the left endpoint).
6. Worked Example: A Function with a Non-Differentiable Point
Example 2. Find the absolute maximum and minimum of $f(x) = |x|$ on $[-2, 3]$.
Goal. Handle a critical number from non-differentiability.
Step 1: Find critical numbers.
For $x > 0$, $f(x) = x$ so $f'(x) = 1$. For $x < 0$, $f(x) = -x$ so $f'(x) = -1$. At $x = 0$, the derivative does not exist (left derivative is $-1$, right derivative is $1$). So $x = 0$ is a critical number; it is a point where $f'$ is undefined.
There are no points where $f'(x) = 0$.
Step 2: Evaluate.
$f(-2) = 2$, $f(0) = 0$, $f(3) = 3$.
Step 3: Compare.
Boxed answer: Absolute maximum value is $3$ at $x = 3$. Absolute minimum value is $0$ at $x = 0$.
Recap. The minimum is at a non-differentiable point. If you only checked where $f'(x) = 0$, you would miss the minimum entirely. Always include points where the derivative does not exist in your candidate list.
7. When the Theorem Does Not Apply
The Extreme Value Theorem requires both a closed interval and continuity. Here are two reminders of what can go wrong without those conditions.
(a) $f(x) = x$ on $(0, 1)$ (open interval). The outputs get arbitrarily close to $1$ and $0$ but never reach them. No absolute maximum, no absolute minimum.
(b) $f(x) = 1/x$ on $[1, \infty)$ (not bounded). The function decreases toward $0$ but never reaches $0$. No absolute minimum.
These are not hypothetical edge cases; they appear in physics and economics whenever a quantity approaches but never reaches a bound.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Concluding $f'(c)=0$ means a local extremum | $f(x)=x^3$, $f'(0)=0$, no extremum at $0$ | $f'(c)=0$ makes $c$ a candidate, not a guarantee; check the sign of $f'$ on either side |
| Forgetting endpoints | finding only critical numbers and comparing them | Evaluate $f$ at $a$ and $b$ too; the absolute extremum is often at an endpoint |
| Forgetting non-differentiable points | missing $x=0$ for $f(x) = \lvert x\rvert$ | Critical numbers include where $f'$ does not exist |
| Confusing local and absolute | claiming the highest local max is the absolute max | The absolute max is the highest value overall; check all candidates |
| Applying the method to an open interval | declaring the extremum from the list without checking the theorem applies | Confirm the interval is closed and bounded and $f$ is continuous first |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find all critical numbers of $f(x) = 3x^2 - 12x + 5$.
Show answer
$f'(x) = 6x - 12$. Setting $f'(x) = 0$: $6x - 12 = 0$, so $x = 2$.
The derivative exists everywhere (it is a polynomial).
Boxed answer: The only critical number is $x = 2$.
Problem 2. Find the absolute maximum and minimum of $f(x) = 3x^2 - 12x + 5$ on $[0, 4]$.
Show answer
From Problem 1, the only critical number in $(0, 4)$ is $x = 2$.
Evaluate at the critical number and both endpoints: $f(0) = 5$, $f(2) = 3(4) - 24 + 5 = 12 - 24 + 5 = -7$, $f(4) = 3(16) - 48 + 5 = 48 - 48 + 5 = 5$.
Boxed answer: Absolute minimum $-7$ at $x = 2$; absolute maximum $5$ at $x = 0$ and $x = 4$.
Problem 3. Does $f(x) = x^2$ on $(-1, 1)$ attain an absolute minimum? Explain.
Show answer
Yes. The value $f(0) = 0$ is attained (since $0 \in (-1, 1)$) and $f(x) \geq 0 = f(0)$ for all $x$.
The Extreme Value Theorem does not apply here (the interval is open, not closed), but the function still attains a minimum. The theorem gives a guarantee when its hypotheses hold; it does not say the extremum fails to exist otherwise.
Boxed answer: Yes, the absolute minimum value is $0$, attained at $x = 0$.
Level 2 -- Multiple Steps
Problem 4. Find the absolute maximum and minimum of $f(x) = x^3 - 3x^2$ on $[-1, 4]$.
Show answer
$f'(x) = 3x^2 - 6x = 3x(x - 2)$. Setting $f'(x) = 0$: $x = 0$ or $x = 2$. Both lie in $(-1, 4)$.
Evaluate: $f(-1) = -1 - 3 = -4$
$f(0) = 0$
$f(2) = 8 - 12 = -4$
$f(4) = 64 - 48 = 16$
Boxed answer: Absolute maximum $16$ at $x = 4$; absolute minimum $-4$ at $x = -1$ and $x = 2$.
Problem 5. Find all critical numbers of $g(x) = x^{2/3}$ on $[-1, 1]$, and state whether the Extreme Value Theorem applies.
Show answer
$g'(x) = \dfrac{2}{3} x^{-1/3} = \dfrac{2}{3x^{1/3}}$.
Setting $g'(x) = 0$: the numerator $2$ is never zero, so $g'(x) \neq 0$ for any $x$.
But $g'(0)$ does not exist (the denominator is zero). So $x = 0$ is a critical number from non-differentiability.
The EVT applies: $g$ is continuous on $[-1, 1]$ (the apparent issue at $0$ is a non-differentiable point, not a discontinuity -- $g(0) = 0$ is perfectly defined), and $[-1, 1]$ is closed and bounded.
$g(-1) = 1$, $g(0) = 0$, $g(1) = 1$.
Boxed answer: One critical number: $x = 0$. Absolute maximum is $1$ at $x = -1$ and $x = 1$; absolute minimum is $0$ at $x = 0$.
Level 3 -- Deeper Problems
Problem 6. Suppose $f$ is differentiable on all of $\mathbb{R}$ and $f(0) = 3$. If $f$ has exactly two critical numbers, at $x = -1$ and $x = 4$, and you evaluate $f$ at many inputs, must you check any inputs outside $\{-1, 4\}$ to locate the absolute max and min? What does the answer depend on?
Show answer
Yes, the answer depends on the interval or domain of interest.
If the question asks for the absolute extremum over a closed interval like $[-3, 5]$, then you must also check the endpoints $x = -3$ and $x = 5$. The endpoints might give the largest or smallest value.
If the question asks about the entire real line $(-\infty, \infty)$, then only the critical numbers $x = -1$ and $x = 4$ can be absolute extrema at interior points. However, the function might have no absolute extremum at all: if $f(x) \to +\infty$ as $x \to \infty$, then no finite input gives the maximum.
Boxed answer: Over a closed interval, always check endpoints too. Over an unbounded domain, the absolute extremum may not exist; the Extreme Value Theorem does not apply.
Problem 7. The function $f(x) = \sin x$ is continuous on all of $\mathbb{R}$. Identify all absolute maximum and minimum values of $f$ on $[0, 2\pi]$, and verify each one is achieved.
Show answer
$f'(x) = \cos x$. Setting $f'(x) = 0$: $\cos x = 0$ on $[0, 2\pi]$ gives $x = \pi/2$ and $x = 3\pi/2$.
Evaluate: $f(0) = 0$, $f(\pi/2) = 1$, $f(3\pi/2) = -1$, $f(2\pi) = 0$.
Boxed answer: Absolute maximum value $1$ at $x = \pi/2$; absolute minimum value $-1$ at $x = 3\pi/2$. Both values are attained at interior critical points, not endpoints.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Picture the graph of a continuous function on a closed interval as a single connected curve with both ends anchored to walls. The Extreme Value Theorem says the curve must have a highest point and a lowest point somewhere -- it cannot escape to infinity because the walls trap it.
To find those extreme points, trace the curve with a pencil. The pencil moves up and down. Every time the pencil is at a peak or a valley (a local extremum), the tangent is horizontal -- the derivative is zero -- or the curve has a sharp corner where no tangent exists. These are exactly the critical numbers. Combine them with the two anchored endpoints and you have a short list. The highest value on that list is the absolute maximum; the lowest is the absolute minimum.
The key picture is: endpoints anchor the curve; critical numbers are where the curve turns around; the winner is the highest or lowest output from that list.
Connections
Within Chapter 3
- Rolle’s Theorem and the Mean Value Theorem (Section 3.2): Both use the Extreme Value Theorem in their proofs. Rolle’s Theorem says that if $f(a) = f(b)$, there is a critical number in $(a, b)$. It relies on the function attaining its absolute max or min in the interior.
- The First and Second Derivative Tests (Section 3.3): These give more information about a critical number -- whether it is a local max, local min, or neither. The Closed Interval Method does not need them for the absolute extremum on $[a, b]$; they matter when you want to classify interior behavior.
- Curve sketching (Section 3.5): Absolute and local extrema are the landmarks you plot first when sketching a function by hand.
Toward MATH162
- Optimization (Section 3.7): The Closed Interval Method is the engine behind every applied optimization problem. You set up the function, find its domain (often a closed interval), and apply the method. The setup changes; the method does not.
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