Critical Numbers
Where Should We Look for Extrema?
The Extreme Value Theorem tells us that extrema exist for continuous functions on closed intervals, but it doesn’t tell us where to find them. This is where critical numbers come in.
Here’s the key insight: at a local maximum or minimum that occurs at an interior point, something special happens to the derivative. Either the tangent line is horizontal ($f'(c) = 0$), or the tangent line doesn’t exist ($f'(c)$ is undefined). These special points (called critical numbers) are exactly where we need to look for local extrema.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Section | Stewart §3.1 |
| Course | MATH161 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Key Concepts
Fermat’s Theorem
$$\boxed{\text{If } f \text{ has a local max or min at } c, \text{ and } f'(c) \text{ exists, then } f'(c) = 0.}$$
Geometric interpretation: At a local extremum where the function is differentiable, the tangent line is horizontal.
At a local maximum: At a local minimum:
___ \ /
/ \ \ /
/ \ \_/
/ \
tangent is horizontal tangent is horizontal
Why Fermat’s Theorem is True (Intuition)
At a local maximum $c$:
- Points slightly to the left of $c$ have smaller function values, so the function is increasing as we approach $c$ from the left: $f'(c) \geq 0$
- Points slightly to the right of $c$ also have smaller function values, so the function is decreasing as we leave $c$ to the right: $f'(c) \leq 0$
- Both conditions together force $f'(c) = 0$
Important Warnings About Fermat’s Theorem
Warning 1: The converse is FALSE
$f'(c) = 0$ does NOT imply $f$ has an extremum at $c$.
Counterexample: $f(x) = x^3$
- $f'(x) = 3x^2$, so $f'(0) = 0$
- But $f$ has no local extremum at $x = 0$ (it’s an inflection point)
y = x³
/
/
─────●───── ← horizontal tangent but no extremum
/
/
Warning 2: Extrema can occur where $f'$ doesn’t exist
Counterexample: $f(x) = \vert x\vert $
- Local (and absolute) minimum at $x = 0$
- But $f'(0)$ does not exist (corner point)
y = |x|
\ /
\ /
V ← minimum but f'(0) DNE
Definition of Critical Number
Because of these warnings, we need a broader search:
$$\boxed{\text{A \textbf{critical number} of } f \text{ is a number } c \text{ in the domain of } f \text{ where } f'(c) = 0 \text{ or } f'(c) \text{ does not exist.}}$$
Types of Critical Numbers
| Type | Condition | Graph Appearance | Example |
|---|---|---|---|
| Horizontal tangent | $f'(c) = 0$ | Smooth peak/valley/inflection | $f(x) = x^2$ at $x = 0$ |
| Corner | $f'(c)$ DNE (one-sided limits differ) | Sharp point | $f(x) = \|x\|$ at $x = 0$ |
| Cusp | $f'(c)$ DNE (vertical tangent) | Sharp point with vertical tangent | $f(x) = x^{2/3}$ at $x = 0$ |
| Vertical tangent | $f'(c)$ DNE ($f'(c) = \pm\infty$) | Smooth but steep | $f(x) = x^{1/3}$ at $x = 0$ |
The Key Takeaway
$$\boxed{\text{If } f \text{ has a local extremum at } c, \text{ then } c \text{ is a critical number of } f.}$$
This is the reformulation of Fermat’s Theorem: all local extrema occur at critical numbers.
The converse is false: not every critical number gives an extremum.
Finding Critical Numbers: Algorithm
Step 1: Find $f'(x)$.
Step 2: Find where $f'(x) = 0$. These are critical numbers.
Step 3: Find where $f'(x)$ does not exist but $f(x)$ does exist. These are also critical numbers.
Important: A point where $f$ is undefined is NOT a critical number (it must be in the domain of $f$).
a critical number is where $f$ equals zero, not where $f'$ equals zero.
This is the height-vs-slope error. A zero of $f$ (where $f(c) = 0$) is an x-intercept of the graph -- a height of zero. A critical number is where $f'(c) = 0$ (or $f'(c)$ is undefined) -- a slope of zero, meaning the tangent line is horizontal. For $f(x) = x^2 - 4$: the zeros of $f$ are $x = \pm 2$ (where the graph crosses the x-axis). The critical number is $x = 0$ (where $f'(x) = 2x = 0$, the vertex). The zeros and the critical number are different points. Setting $f = 0$ finds x-intercepts; setting $f' = 0$ finds potential extrema.
every critical number is a local extremum.
This is the concept-image-conflicts-definition error. The mental picture of “critical number” often carries the image of a peak or valley, but the definition only requires $f'(c) = 0$ or $f'(c)$ undefined. For $f(x) = x^3$: $f'(x) = 3x^2 = 0$ at $x = 0$, so $x = 0$ is a critical number. But the graph of $x^3$ passes through the origin without turning around -- it is increasing on both sides. The tangent at $x = 0$ is horizontal but the function has neither a maximum nor minimum there. A critical number is a CANDIDATE for an extremum; whether it is one depends on how the sign of $f'$ behaves around it.
Practice Problems
Find all critical numbers of $f(x) = x^3 - 6x^2 + 9x + 1$.
Find all critical numbers of $f(x) = x^{2/3}(x - 5)$.
Find all critical numbers of $f(x) = \dfrac{x^2}{x - 1}$.
Find all critical numbers of $f(\theta) = 2\cos\theta + \sin(2\theta)$ on the interval $[0, 2\pi]$.
Prove Fermat’s Theorem: If $f$ has a local maximum at $c$ and $f'(c)$ exists, then $f'(c) = 0$.
Hint: Use the definition of derivative as a limit and consider the signs of the difference quotient for $h > 0$ and $h < 0$ separately.
Mastery Checklist
Mental Model
The Detective’s Suspect List:
Finding extrema is like solving a mystery: you know the extreme value exists (EVT), but you need to find where.
Critical numbers are your suspects. Every local extremum is at a critical number (no innocent critical numbers at crime scenes), but not every critical number is guilty of being an extremum.
- $f'(c) = 0$: The function “paused” here: horizontal tangent. Suspicious, but might be innocent (inflection point).
- $f'(c)$ DNE: Something “broke” here: a corner, cusp, or vertical tangent. Also suspicious.
Your job: round up all the suspects (find all critical numbers), then test them to see who’s really responsible for the extrema.
Connections
Looking back:
- Absolute and local extrema defines what we’re looking for
- Derivative computation provides the tools to find $f'$
Looking ahead:
- Closed Interval Method uses critical numbers to find absolute extrema
- First Derivative Test classifies critical numbers as max, min, or neither
- Second Derivative Test provides another classification method
| Previous | Up | Next |
|---|---|---|
| Extreme Value Theorem | Section Index | Closed Interval Method |
Last updated: 2026-01-22