Critical Numbers

MATH161

Reference: Stewart §3.1 • Chapter: 3 • Section: 1

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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 $f$).

Practice Problems

Level 1 Finding Critical Numbers of a Polynomial

Find all critical numbers of $f(x) = x^3 - 6x^2 + 9x + 1$.

Thought Process

For polynomials, $f'(x)$ exists everywhere, so critical numbers only come from $f'(x) = 0$.

Compute $f'(x)$
Set $f'(x) = 0$ and solve
Factor if possible

Show Answer

Step 1: Find $f'(x)$: $$f'(x) = 3x^2 - 12x + 9$$

Step 2: Set $f'(x) = 0$: $$3x^2 - 12x + 9 = 0$$ $$3(x^2 - 4x + 3) = 0$$ $$3(x - 1)(x - 3) = 0$$

Step 3: Solve: $$x = 1 \quad \text{or} \quad x = 3$$

Critical numbers: $x = 1$ and $x = 3$

(Since $f$ is a polynomial, $f'(x)$ exists everywhere, so there are no critical numbers from $f'$ being undefined.)

Level 2 Critical Numbers with Fractional Exponents

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

Thought Process

This function involves a fractional exponent $x^{2/3}$, which means:

The function is defined for all real $x$ (cube roots work for negative numbers)
The derivative may not exist at $x = 0$ due to the fractional exponent

Use the product rule: if $f(x) = g(x) \cdot h(x)$, then $f'(x) = g'(x)h(x) + g(x)h'(x)$.

Alternatively, expand first: $f(x) = x^{5/3} - 5x^{2/3}$, then differentiate term by term.

Show Answer

Method 1: Expand first

$$f(x) = x^{2/3}(x - 5) = x^{5/3} - 5x^{2/3}$$

Find $f'(x)$: $$f'(x) = \frac{5}{3}x^{2/3} - 5 \cdot \frac{2}{3}x^{-1/3} = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3}$$

Simplify by factoring out $\frac{5}{3}x^{-1/3}$: $$f'(x) = \frac{5}{3}x^{-1/3}\left(x - 2\right) = \frac{5(x-2)}{3x^{1/3}}$$

Find critical numbers:

$f'(x) = 0$: This occurs when $x - 2 = 0$, i.e., $x = 2$

$f'(x)$ undefined: The denominator $3x^{1/3} = 0$ when $x = 0$. Since $f(0) = 0$ is defined, $x = 0$ is a critical number.

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

Level 3 Critical Numbers of a Rational Function

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

Thought Process

First note the domain: $f$ is undefined at $x = 1$, so $x = 1$ cannot be a critical number.

Use the quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$

Then find where the derivative is zero (numerator = 0) and where it's undefined (but $f$ is defined).

Show Answer

Domain: $x \neq 1$ (all real numbers except 1)

Find $f'(x)$ using the quotient rule:

Let $u = x^2$ and $v = x - 1$. Then $u' = 2x$ and $v' = 1$.

$$f'(x) = \frac{u'v - uv'}{v^2} = \frac{2x(x-1) - x^2(1)}{(x-1)^2}$$

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

Find critical numbers:

$f'(x) = 0$: Numerator $= 0$ when $x(x-2) = 0$, i.e., $x = 0$ or $x = 2$

$f'(x)$ undefined: Denominator $= 0$ when $x = 1$. But $f(1)$ is also undefined, so $x = 1$ is not a critical number.

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

Level 4 Critical Numbers with 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 the chain rule: $\frac{d}{d\theta}\sin(2\theta) = 2\cos(2\theta)$
Set $f'(\theta) = 0$ and use trigonometric identities
The identity $\cos(2\theta) = 1 - 2\sin^2\theta$ or $\cos(2\theta) = 2\cos^2\theta - 1$ may help
Also useful: $\sin\theta = 0$ at $\theta = 0, \pi, 2\pi$

Show Answer

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

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

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

Factor: $$(2\sin\theta - 1)(\sin\theta + 1) = 0$$

Solve:

Case 1: $2\sin\theta - 1 = 0 \Rightarrow \sin\theta = \frac{1}{2}$

In $[0, 2\pi]$: $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$

Case 2: $\sin\theta + 1 = 0 \Rightarrow \sin\theta = -1$

In $[0, 2\pi]$: $\theta = \frac{3\pi}{2}$

Check: $f'(\theta)$ exists everywhere (trig functions are differentiable everywhere).

Critical numbers in $[0, 2\pi]$: $\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{3\pi}{2}$

Level 5 Proving Fermat's Theorem

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.

Thought Process

At a local maximum at $c$:

$f(c) \geq f(x)$ for all $x$ near $c$
So $f(c + h) \leq f(c)$ for small $\vert h\vert $
This means $f(c + h) - f(c) \leq 0$

The difference quotient is $\frac{f(c+h) - f(c)}{h}$.

When $h > 0$: dividing by positive, so quotient $\leq 0$
When $h < 0$: dividing by negative, so quotient $\geq 0$

If the limit exists (i.e., $f'(c)$ exists), it must equal both the left and right limits.

Show Answer

Proof:

Let $f$ have a local maximum at $c$, and assume $f'(c)$ exists.

Step 1: By definition of local maximum, there exists $\delta > 0$ such that $f(c) \geq f(x)$ for all $x$ with $\vert x - c\vert < \delta$.

Step 2: For $h$ with $0 < \vert h\vert < \delta$, we have $f(c + h) \leq f(c)$, so: $$f(c + h) - f(c) \leq 0$$

Step 3: Consider the right-hand limit. For $h > 0$: $$\frac{f(c + h) - f(c)}{h} \leq \frac{0}{h} = 0$$

Taking the limit as $h \to 0^+$: $$\lim_{h \to 0^+} \frac{f(c + h) - f(c)}{h} \leq 0$$

Since $f'(c)$ exists, this right-hand limit equals $f'(c)$: $$f'(c) \leq 0 \quad \text{...(1)}$$

Step 4: Consider the left-hand limit. For $h < 0$: $$\frac{f(c + h) - f(c)}{h} \geq \frac{0}{h} = 0$$ (inequality reverses because we divide by negative $h$)

Taking the limit as $h \to 0^-$: $$\lim_{h \to 0^-} \frac{f(c + h) - f(c)}{h} \geq 0$$

Since $f'(c)$ exists, this left-hand limit also equals $f'(c)$: $$f'(c) \geq 0 \quad \text{...(2)}$$

Step 5: From (1) and (2): $$f'(c) \leq 0 \quad \text{and} \quad f'(c) \geq 0$$

The only number satisfying both is: $$f'(c) = 0 \quad \blacksquare$$

Mastery Checklist

[ ] I can state Fermat's Theorem and explain its geometric meaning
[ ] I can explain why the converse of Fermat's Theorem is false
[ ] I can define "critical number" and identify both types ($f'(c) = 0$ and $f'(c)$ DNE)
[ ] I can find critical numbers of polynomials, rational functions, and functions with fractional exponents
[ ] I understand that critical numbers are necessary but not sufficient for local extrema

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