Fermat's Theorem and Critical Numbers

8 min read

Reference: Stewart §4.1

Where Should We Look for Extrema?

The Extreme Value Theorem tells us that a continuous function on $[a, b]$ has a maximum and minimum; but it doesn’t say where to find them. Do we need to check every point? That’s infinitely many!

Fortunately, calculus gives us a shortcut. Fermat’s Theorem narrows down the search: if you’re looking for local extrema, you only need to check points where the derivative is zero or doesn’t exist. These special points are called critical numbers.

Think of a critical number as a “suspect” in the search for extrema. Not every suspect is guilty (has an extremum), but every guilty point (local extremum) is among the suspects.

Prerequisite Map

Quick Reference

Property	Value
Section	Stewart §4.1
Course	MATH161
Difficulty	Intermediate
Time	~20 minutes

Key Concepts

Fermat’s Theorem

$$\boxed{\text{If } f \text{ has a local extremum at } c \text{ and } f'(c) \text{ exists, then } f'(c) = 0.}$$

Geometric interpretation: At a local maximum or minimum where the function is smooth, the tangent line is horizontal.

    Local Maximum               Local Minimum
         ___                        |
       /     \   slope = 0      ___|___   slope = 0
      /       \                /       \
     /         \              /         \

The Proof (for local maximum)

Suppose $f$ has a local maximum at $c$ and $f'(c)$ exists.

From the right: For small positive $h$, $f(c + h) \leq f(c)$, so: $$\frac{f(c+h) - f(c)}{h} \leq 0$$

Taking $\lim_{h \to 0^+}$ gives $f'(c) \leq 0$.

From the left: For small negative $h$, $f(c + h) \leq f(c)$, but now we divide by negative $h$: $$\frac{f(c+h) - f(c)}{h} \geq 0$$

Taking $\lim_{h \to 0^-}$ gives $f'(c) \geq 0$.

Combining: $f'(c) \leq 0$ and $f'(c) \geq 0$ implies $f'(c) = 0$. $\square$

Caution: The Converse is False!

Fermat’s Theorem says: local extremum → derivative zero (if it exists)

It does NOT say: derivative zero → local extremum

Counterexample: $f(x) = x^3$ has $f'(0) = 0$, but $x = 0$ is not a local extremum.

         f(x) = x³
            /
           /
    ------●------  ← horizontal tangent at (0,0)
         /           but NOT an extremum
        /

Critical Numbers

Definition: A critical number of $f$ is a number $c$ in the domain of $f$ such that either:

$f'(c) = 0$, or
$f'(c)$ does not exist

Why Include “Derivative DNE”?

Fermat’s Theorem has a hypothesis: “if $f'(c)$ exists.” What about points where the derivative doesn’t exist?

Example: $f(x) = \vert x\vert $ has a local minimum at $x = 0$, but $f'(0)$ doesn’t exist.

      |x|
     /\
    /  \
   /    \  ← sharp corner at x = 0
  /      \    (derivative DNE)

So we must also check points where the derivative doesn’t exist; hence the definition of critical number includes both cases.

Rephrasing Fermat’s Theorem

Corollary: If $f$ has a local extremum at $c$, then $c$ is a critical number of $f$.

This tells us: all local extrema occur at critical numbers. So to find all candidates for local extrema, find all critical numbers.

Types of Critical Numbers

Type	Condition	Example	Could Be Extremum?
Stationary point	$f'(c) = 0$	$f(x) = x^2$ at $c = 0$	Yes
Singular point	$f'(c)$ DNE	$f(x) = \\|x\\|$ at $c = 0$	Yes

Common misconception

Fermat’s theorem says every critical number is a local extremum.

This is the concept-image-conflicts-definition error. Fermat’s theorem runs in ONE direction only: if $f$ has a local extremum at $c$, then $c$ is a critical number ($f'(c) = 0$ or $f'(c)$ undefined). The CONVERSE is false: a critical number need not be an extremum. For $f(x) = x^3$: $f'(0) = 0$, so $x = 0$ is a critical number, but the graph passes through the origin with a horizontal tangent and keeps increasing -- there is no local maximum or minimum there. Fermat’s theorem is a filter for candidates, not a guarantee.

Common misconception

if $f'(c) = 0$, the graph of $f$ must equal zero at $c$ too.

This is the height-vs-slope error. $f'(c) = 0$ means the SLOPE of the tangent is zero at $x = c$ -- the graph is momentarily flat. It says nothing about the HEIGHT $f(c)$. For $f(x) = x^2 + 7$: $f'(0) = 0$, but $f(0) = 7$, not zero. The graph has a horizontal tangent at height $7$. The zero of the derivative locates a potential extremum on the x-axis scale, but the function value there can be anything.

Finding Critical Numbers: Procedure

Find $f'(x)$
Find where $f'(x) = 0$ (solve the equation)
Find where $f'(x)$ doesn’t exist (but $f(x)$ does)
List all such $x$-values: these are the critical numbers

Practice Problems

Level 1 Finding Critical Numbers (Polynomial)

Find all critical numbers of $f(x) = x^3 - 12x + 5$.

Thought Process

For a polynomial:

The derivative exists everywhere, so we only need to find where $f'(x) = 0$
Differentiate: $f'(x) = 3x^2 - 12$
Set equal to zero and solve

Show Answer

Find the derivative: $$f'(x) = 3x^2 - 12$$

Set $f'(x) = 0$: $$3x^2 - 12 = 0$$ $$x^2 = 4$$ $$x = \pm 2$$

Since $f$ is a polynomial, $f'(x)$ exists for all $x$. No additional critical numbers from DNE.

Critical numbers: $x = -2$ and $x = 2$

Level 2 Finding Critical Numbers (Rational Function)

Find all critical numbers of $g(x) = \dfrac{x^2}{x - 1}$.

Thought Process

For a rational function:

First note the domain: $x \neq 1$
Use quotient rule to find $g'(x)$
Find where $g'(x) = 0$ (numerator = 0, denominator ≠ 0)
Check where $g'(x)$ might not exist (but $g(x)$ does)

Important: $x = 1$ is NOT a critical number because it’s not in the domain of $g$.

Show Answer

Domain: $x \neq 1$

Find the derivative using the quotient rule: $$g'(x) = \frac{(2x)(x-1) - (x^2)(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$$

Find where $g'(x) = 0$: $$\frac{x(x-2)}{(x-1)^2} = 0$$

Numerator equals zero when $x = 0$ or $x = 2$. Both are in the domain of $g$.

Check where $g'(x)$ DNE: $g'(x)$ is undefined when $x = 1$, but $x = 1$ is not in the domain of $g$, so it’s not a critical number.

Critical numbers: $x = 0$ and $x = 2$

Level 3 Critical Numbers with Fractional Exponents

Find all critical numbers of $h(x) = x^{2/3}(x - 5)$.

Thought Process

The function involves $x^{2/3}$, which means:

Domain is all real numbers (even negative, since $2/3$ has odd denominator in lowest terms)
Use the product rule: $h(x) = x^{2/3} \cdot (x-5)$
Watch for where the derivative doesn’t exist, likely at $x = 0$ due to $x^{-1/3}$

Show Answer

Domain: All real numbers (since cube roots are defined for all reals)

Find the derivative using the product rule: $$h'(x) = \frac{2}{3}x^{-1/3}(x-5) + x^{2/3}(1)$$

$$= \frac{2(x-5)}{3x^{1/3}} + x^{2/3}$$

Get a common denominator $3x^{1/3}$: $$= \frac{2(x-5)}{3x^{1/3}} + \frac{3x}{3x^{1/3}} = \frac{2(x-5) + 3x}{3x^{1/3}} = \frac{2x - 10 + 3x}{3x^{1/3}} = \frac{5x - 10}{3x^{1/3}}$$

$$= \frac{5(x-2)}{3x^{1/3}}$$

Where is $h'(x) = 0$? Numerator = 0: $5(x-2) = 0 \Rightarrow x = 2$

Where does $h'(x)$ not exist? Denominator = 0: $3x^{1/3} = 0 \Rightarrow x = 0$

Check: $x = 0$ is in the domain of $h$ (since $h(0) = 0$), so $x = 0$ is a critical number.

Critical numbers: $x = 0$ and $x = 2$

Level 4 Critical Numbers of Trigonometric Functions

Find all critical numbers of $f(\theta) = 2\cos\theta + \sin(2\theta)$ on the interval $[0, 2\pi]$.

Thought Process

Differentiate using chain rule for $\sin(2\theta)$
Set derivative equal to zero
Use trig identities to solve: $\sin(2\theta) = 2\sin\theta\cos\theta$
Factor and find all solutions in $[0, 2\pi]$

Show Answer

Find the derivative: $$f'(\theta) = -2\sin\theta + 2\cos(2\theta)$$

Set $f'(\theta) = 0$: $$-2\sin\theta + 2\cos(2\theta) = 0$$ $$\cos(2\theta) = \sin\theta$$

Use the identity $\cos(2\theta) = 1 - 2\sin^2\theta$: $$1 - 2\sin^2\theta = \sin\theta$$ $$2\sin^2\theta + \sin\theta - 1 = 0$$

Let $u = \sin\theta$: $$2u^2 + u - 1 = 0$$ $$(2u - 1)(u + 1) = 0$$ $$u = \frac{1}{2} \quad \text{or} \quad u = -1$$

Solve for $\theta$ in $[0, 2\pi]$:

If $\sin\theta = \frac{1}{2}$: $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$

If $\sin\theta = -1$: $\theta = \frac{3\pi}{2}$

Since trig functions are differentiable everywhere, no additional critical numbers from DNE.

Critical numbers: $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$

Level 5 Proof: Cubic Functions

Let $f(x) = ax^3 + bx^2 + cx + d$ where $a \neq 0$.

How many critical numbers can $f$ have? (Consider all cases.)
Prove that if $f$ has two distinct critical numbers, it has exactly one local maximum and one local minimum.
Prove that if $f$ has exactly one critical number, it has no local extrema.

Thought Process

(a) $f'(x) = 3ax^2 + 2bx + c$ is a quadratic. A quadratic can have 0, 1, or 2 real roots depending on its discriminant.

(b) If two critical numbers exist, the cubic changes from increasing to decreasing (or vice versa) at each. Use the sign of $f'$ between critical numbers.

(c) If only one critical number exists, it’s a point of inflection for the cubic, not an extremum.

Show Answer

(a) Number of critical numbers:

$$f'(x) = 3ax^2 + 2bx + c$$

Critical numbers occur where $f'(x) = 0$. This quadratic has discriminant: $$\Delta = (2b)^2 - 4(3a)(c) = 4b^2 - 12ac$$

If $\Delta > 0$: Two distinct critical numbers
If $\Delta = 0$: One critical number (repeated root)
If $\Delta < 0$: No critical numbers (no real roots)

(b) Proof: Two critical numbers implies one local max and one local min:

Let $c_1 < c_2$ be the two critical numbers where $f'(c_1) = f'(c_2) = 0$.

Since $f'(x) = 3a(x - c_1)(x - c_2)$ and the leading coefficient is $3a$:

Case 1: $a > 0$

For $x < c_1$: $(x-c_1) < 0$ and $(x-c_2) < 0$, so $f'(x) > 0$ (increasing)
For $c_1 < x < c_2$: $(x-c_1) > 0$ and $(x-c_2) < 0$, so $f'(x) < 0$ (decreasing)
For $x > c_2$: $(x-c_1) > 0$ and $(x-c_2) > 0$, so $f'(x) > 0$ (increasing)

Pattern: ↗ at $c_1$ ↘ at $c_2$ ↗

Therefore: $c_1$ is a local maximum, $c_2$ is a local minimum.

Case 2: $a < 0$ Signs reverse: $c_1$ is a local minimum, $c_2$ is a local maximum.

Either way, exactly one local max and one local min. $\square$

(c) Proof: One critical number implies no local extrema:

If $\Delta = 0$, then $f'(x) = 3a(x - c)^2$ for some $c$.

Note that $(x - c)^2 \geq 0$ for all $x$, with equality only at $x = c$.

Case 1: $a > 0$ $f'(x) = 3a(x-c)^2 \geq 0$ for all $x$, so $f$ is non-decreasing everywhere. At $x = c$, $f'(c) = 0$ but $f' \geq 0$ on both sides. $f$ is increasing on $(-\infty, c)$ and $(c, \infty)$. So $c$ is not a local extremum (the function doesn’t change direction).

Case 2: $a < 0$ $f'(x) \leq 0$ for all $x$, so $f$ is non-increasing everywhere. Similarly, $c$ is not a local extremum.

The single critical number is a point of horizontal inflection, not an extremum. $\square$

CCI-Style Conceptual Questions

Level 2 Conceptual: Interpreting the Theorem

True or False (justify each):

If $f'(c) = 0$, then $f$ has a local extremum at $c$.
If $f$ has a local maximum at $c$, then $f'(c) = 0$.
If $f'(c)$ does not exist, then $c$ is a critical number of $f$.
Every continuous function on $[a, b]$ has at least one critical number in $(a, b)$.

Thought Process

(a) This is the converse of Fermat’s Theorem. Check if it’s true. (b) This is almost Fermat’s Theorem, but missing a hypothesis. (c) Check the definition of critical number. What else is required? (d) Think of functions like $f(x) = x$ on $[0, 1]$.

Show Answer

(a) False.

Counterexample: $f(x) = x^3$ has $f'(0) = 0$, but $x = 0$ is not a local extremum.

(b) False.

Missing hypothesis: this is only true if $f'(c)$ exists.

Counterexample: $f(x) = \vert x\vert $ has a local minimum at $x = 0$, but $f'(0)$ does not exist (so it’s not equal to 0).

(c) False.

A critical number must be in the domain of $f$. If $f'(c)$ DNE but $c$ is not in the domain of $f$, then $c$ is not a critical number.

Example: $f(x) = 1/x$ has $f'(0)$ undefined, but $x = 0$ is not a critical number because $0$ is not in the domain of $f$.

(d) False.

Counterexample: $f(x) = x$ on $[0, 1]$ is continuous, and $f'(x) = 1 \neq 0$ everywhere. There are no critical numbers in $(0, 1)$.

(Note: This function still has absolute max and min at the endpoints.)

Mastery Checklist

I can state Fermat’s Theorem and explain its geometric meaning
I understand why the converse of Fermat’s Theorem is false
I can define “critical number” precisely (both conditions)
I can find critical numbers by setting $f'(x) = 0$ and checking where $f'(x)$ DNE
I know that $c$ must be in the domain of $f$ to be a critical number
I can find critical numbers for polynomials, rational functions, and functions with fractional exponents

Mental Model

The Suspects Analogy:

Think of critical numbers as “suspects” in the search for local extrema:

Fermat’s Theorem says: “Every local extremum is at a critical number” (i.e., “the criminal is among the suspects”)
But not every suspect is guilty; $f'(c) = 0$ doesn’t guarantee an extremum at $c$
We also can’t forget suspects with “no alibi” (where $f'$ doesn’t exist)

To find extrema: first round up all the suspects (critical numbers), then investigate each one to determine if it’s actually an extremum.

Connections

Looking back:

Derivative definition is used in the proof of Fermat’s Theorem
Absolute vs Local Extrema defines what we’re trying to find

Looking ahead:

The Closed Interval Method uses critical numbers to find absolute extrema
The First Derivative Test determines if a critical number is a max, min, or neither
The Second Derivative Test provides another way to classify critical numbers

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Extreme Value Theorem	Section Index	Closed Interval Method

Last updated: 2026-01-22