Day 12

Extreme Value Theorem And Critical Points

Critical Points & Local Extrema The Extreme Value Theorem & Absolute Extrema

Critical Points & Local Extrema

Local maxima and minima can only happen at very specific x-values — but having a flat tangent there is no guarantee you've actually found one. This is where the search for extrema always begins.

Core Theorem

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 critical point is a point

c

in the domain of

f

where

f'(c) = 0

f'(c)

does not exist.
Every local extremum of

f

occurs at either a critical point or an endpoint of the domain.

Step-by-Step SOP

1
Differentiate and Factor
Find $f'(x)$ and factor it as much as possible to expose its zeros clearly.
2
Find Zeros of f'
Solve $f'(x)=0$ for all real solutions in the domain of $f$ .
3
Find Points Where f' Doesn't Exist
Check for any domain points where $f'$ is undefined (but $f$ itself is defined) — corners, cusps, vertical tangents.

Practice Exercises

Example 01Medium

Discuss the behavior of

f(x) = x^3

and

f(x) = |x|

x=0

NEED A HINT?

Both have a critical point at

x=0

— check whether

f'

actually changes sign there for each.

SHOW DETAILED EXPLANATION

Step 1: Analyze x³

f'(x) = 3x^2 \ge 0

everywhere, so

x=0

IS a critical point (

f'(0)=0

), but

f'

never changes sign around it —

f(x)=x^3

is increasing on both sides.

x=0

is NOT a local extremum, just a flat inflection point.

Step 2: Analyze |x|

f'(x) = -1

for

x<0

and

f'(x)=1

for

x>0

—

f'

is undefined at

x=0

(a corner), making

x=0

a critical point. Since

f

decreases then increases,

x=0

IS a local (and absolute) minimum.

Step 3: Conclude

Both functions have a critical point at

x=0

, but only one is actually an extremum — critical points are only candidates, not guarantees.

Example 02Hard

Find all critical points of

f(x) = x^{3/5}(4-x)^2

NEED A HINT?

Use the product rule, then factor out the common

x^{-2/5}

and

(4-x)

terms before setting the numerator to zero.

SHOW DETAILED EXPLANATION

Step 1: Apply the Product Rule

f'(x) = \frac{3}{5}x^{-2/5}(4-x)^2 + x^{3/5}\cdot(-2)(4-x)

Step 2: Factor Out x⁻²/⁵ and (4-x)

f'(x) = x^{-2/5}(4-x)\left[\frac{3}{5}(4-x) - 2x\right] = x^{-2/5}(4-x)\cdot\frac{12-13x}{5} = \frac{(4-x)(12-13x)}{5x^{2/5}}

Step 3: Find Where the Numerator is Zero

4-x=0 \implies x=4

12-13x=0 \implies x=\frac{12}{13}

Step 4: Find Where f' is Undefined

The denominator

5x^{2/5}=0

x=0

, and

f(0)=0

is defined, so

x=0

is also a critical point.

Step 5: State All Critical Points

x=0, \frac{12}{13}, 4

Example 03Medium

Find the critical points of

f(x) = 3x^4-16x^3+18x^2

NEED A HINT?

Differentiate and factor out the GCF first.

SHOW DETAILED EXPLANATION

Step 1: Differentiate

f'(x) = 12x^3-48x^2+36x

Step 2: Factor

f'(x) = 12x(x^2-4x+3) = 12x(x-1)(x-3)

Step 3: State the Critical Points

x=0, 1, 3

Example 04Easy

Show that

g(x) = x^5+2x-3

has no critical points at all.

NEED A HINT?

Differentiate and check the sign of

g'(x)

for every real

x

SHOW DETAILED EXPLANATION

Step 1: Differentiate

g'(x) = 5x^4+2

Step 2: Check the Sign

Since

x^4 \ge 0

for all real

x

g'(x) = 5x^4+2 \ge 2 > 0

always — it is never zero and always defined.

Step 3: Conclude

g

has no critical points, and (since

g'>0

everywhere)

g

is strictly increasing on all of

\mathbb{R}

Common Pitfalls

⚠
A Critical Point Is Only a CandidateFinding $f'(c)=0$ or $f'(c)$ undefined does NOT guarantee a local extremum at $c$ — see $f(x)=x^3$ at $x=0$ . You still need the First or Second Derivative Test to confirm it.
⚠
Don't Forget Points Where f' Is UndefinedCritical points aren't just zeros of $f'$ — any point in the domain where the derivative fails to exist (like a corner or vertical tangent) also counts, as long as $f$ itself is still defined there.

The Extreme Value Theorem & Absolute Extrema

A continuous function on a closed interval is guaranteed to actually reach its highest and lowest values somewhere — and finding them is just a matter of checking a short, finite list of candidates.

Core Theorem

Extreme Value Theorem (Weierstrass): If

f

is continuous on

[a,b]

, then

f

is bounded on

[a,b]

and attains both an absolute maximum and an absolute minimum on

[a,b]

.
Closed Interval Method:
(1) Find all critical points of

f

(a,b)

.
(2) Evaluate

f

at those critical points and at the endpoints

a, b

.
(3) The largest value is the absolute max; the smallest is the absolute min.

Watch the TikTok ExplanationExtreme Value Theorem→

Step-by-Step SOP

1
Confirm Continuity on a Closed Interval
Verify $f$ is continuous on $[a,b]$ before applying the closed interval method.
2
List All Candidates
Find every critical point in $(a,b)$ , then add both endpoints $a$ and $b$ to the list.
3
Evaluate and Compare
Plug every candidate into $f$ and simply compare the resulting values — the largest is the absolute max, the smallest is the absolute min.

Practice Exercises

Example 01Medium

Find the absolute extrema of

f(x) = 3x^4-16x^3+18x^2

[-1,4]

NEED A HINT?

You already found the critical points (

x=0,1,3

) — now just evaluate

f

at those and at both endpoints.

SHOW DETAILED EXPLANATION

Step 1: List Candidates

Critical points

x=0,1,3

, plus endpoints

x=-1,4

Step 2: Evaluate f at Each

f(-1)=37

f(0)=0

f(1)=5

f(3)=-27

f(4)=32

Step 3: Compare

Absolute maximum is

37

x=-1

. Absolute minimum is

-27

x=3

Example 02Medium

Find the absolute extrema of

f(x) = x^{2/3}

[-2,3]

NEED A HINT?

f'(x)

is undefined at

x=0

— that's a critical point too, even though

f'

is never actually zero.

SHOW DETAILED EXPLANATION

Step 1: Find Critical Points

f'(x) = \frac{2}{3}x^{-1/3}

, which is never

0

but is undefined at

x=0

. So

x=0

is the only critical point.

Step 2: Evaluate at Candidates

f(-2) = (-2)^{2/3} = \sqrt[3]{4} \approx 1.587

f(0)=0

f(3) = 3^{2/3} = \sqrt[3]{9} \approx 2.080

Step 3: Compare

Absolute maximum is

\sqrt[3]{9}

x=3

. Absolute minimum is

0

x=0

Example 03Hard

Find the absolute extrema of

f(x) = x - 2\sin x

[0, 2\pi]

NEED A HINT?

Set

f'(x)=0

and solve for

\cos x = \frac{1}{2}

within

[0,2\pi]

SHOW DETAILED EXPLANATION

Step 1: Find Critical Points

f'(x) = 1-2\cos x = 0 \implies \cos x = \frac{1}{2} \implies x = \frac{\pi}{3}, \frac{5\pi}{3}

Step 2: Evaluate at All Candidates

f(0)=0

f(\pi/3) = \pi/3 - \sqrt{3} \approx -0.685

f(5\pi/3) = 5\pi/3+\sqrt{3} \approx 6.968

f(2\pi) = 2\pi \approx 6.283

Step 3: Compare

Absolute maximum is

\frac{5\pi}{3}+\sqrt{3}

x=\frac{5\pi}{3}

. Absolute minimum is

\frac{\pi}{3}-\sqrt{3}

x=\frac{\pi}{3}

Example 04Medium

Why does

f(x) = \frac{1}{x}

on the OPEN interval

(0,1)

have no absolute maximum or minimum?

NEED A HINT?

Check every hypothesis of the Extreme Value Theorem carefully.

SHOW DETAILED EXPLANATION

Step 1: Check the Hypotheses

f

is continuous on

(0,1)

, but

(0,1)

is an OPEN interval, not closed — the Extreme Value Theorem's hypothesis is not satisfied.

Step 2: Confirm It Actually Fails

x \to 0^+

f(x) \to \infty

(unbounded, no maximum). As

x \to 1^-

f(x) \to 1

, but that value is never attained on the open interval, so there's no minimum either.

Step 3: Conclude

Without a closed, bounded interval, the EVT gives no guarantee — and here, that guarantee genuinely fails.

Common Pitfalls

⚠
The Interval Must Be Closed AND BoundedThe EVT requires $[a,b]$ specifically — drop either the closed endpoints or boundedness, and a continuous function might not attain an absolute max or min at all (see Example 4).
⚠
Don't Skip the EndpointsEven if a critical point gives a locally extreme value, an ENDPOINT could still be the true absolute max or min — always include both endpoints in the comparison.

Gary Chang

Calculus Educator

5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

Discuss with Gary on Reddit

Differentiate and Factor

Find Zeros of f'

Find Points Where f' Doesn't Exist

Practice Exercises

Step 1: Analyze x³

Step 2: Analyze |x|

Step 3: Conclude

Step 1: Apply the Product Rule

Step 2: Factor Out x⁻²/⁵ and (4-x)

Step 3: Find Where the Numerator is Zero

Step 4: Find Where f' is Undefined

Step 5: State All Critical Points

Step 1: Differentiate

Step 2: Factor

Step 3: State the Critical Points

Step 1: Differentiate

Step 2: Check the Sign

Step 3: Conclude

Confirm Continuity on a Closed Interval

List All Candidates

Evaluate and Compare

Practice Exercises

Step 1: List Candidates

Step 2: Evaluate f at Each

Step 3: Compare

Step 1: Find Critical Points

Step 2: Evaluate at Candidates

Step 3: Compare

Step 1: Find Critical Points

Step 2: Evaluate at All Candidates

Step 3: Compare

Step 1: Check the Hypotheses

Step 2: Confirm It Actually Fails

Step 3: Conclude

Gary Chang