Day 12: Extreme Values, Rolle's & the Mean Value Theorem

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

or

f'(c)

does not exist.
Every local extremum of

f

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

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Discuss the behavior of

f(x) = x^3

and

f(x) = |x|

at

x=0

.

How to think about it

Key idea: Both have a critical point at $x=0$ — check whether $f'$ actually changes sign there for each.

Full solution

1. 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.

2. 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.

3. 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.

Worked Example 02Hard

Find all critical points of

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

.

How to think about it

Key idea: Use the product rule, then factor out the common $x^{-2/5}$ and $(4-x)$ terms before setting the numerator to zero.

Full solution

1. Step 1: Apply the Product Rule

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

.

2. 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}}

.

3. Step 3: Find Where the Numerator is Zero

4-x=0 \implies x=4

.

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

.

4. Step 4: Find Where f' is Undefined

The denominator

5x^{2/5}=0

at

x=0

, and

f(0)=0

is defined, so

x=0

is also a critical point.

5. Step 5: State All Critical Points

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

.

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 — Your Turn

Try each one before opening the hint or the solution.

Practice 01Medium

Find the critical points of

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

.

Need a hint?

Differentiate and factor out the GCF first.

Show solution

1. Step 1: Differentiate

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

.

2. Step 2: Factor

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

.

3. Step 3: State the Critical Points

x=0, 1, 3

.

Practice 02Easy

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 solution

1. Step 1: Differentiate

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

.

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.

3. 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

in

(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→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Find the absolute extrema of

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

on

[-1,4]

.

How to think about it

Key idea: You already found the critical points ( $x=0,1,3$ ) — now just evaluate $f$ at those and at both endpoints.

Full solution

1. Step 1: List Candidates

Critical points

x=0,1,3

, plus endpoints

x=-1,4

.

2. Step 2: Evaluate f at Each

f(-1)=37

,

f(0)=0

,

f(1)=5

,

f(3)=-27

,

f(4)=32

.

3. Step 3: Compare

Absolute maximum is

37

at

x=-1

. Absolute minimum is

-27

at

x=3

.

Worked Example 02Medium

Find the absolute extrema of

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

on

[-2,3]

.

How to think about it

Key idea: $f'(x)$ is undefined at $x=0$ — that's a critical point too, even though $f'$ is never actually zero.

Full solution

1. 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.

2. 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

.

3. Step 3: Compare

Absolute maximum is

\sqrt[3]{9}

at

x=3

. Absolute minimum is

0

at

x=0

.

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 — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

Find the absolute extrema of

f(x) = x - 2\sin x

on

[0, 2\pi]

.

Need a hint?

Set

f'(x)=0

and solve for

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

within

[0,2\pi]

.

Show solution

1. 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}

.

2. 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

.

3. Step 3: Compare

Absolute maximum is

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

at

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

. Absolute minimum is

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

at

x=\frac{\pi}{3}

.

Practice 02Medium

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 solution

1. 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.

2. Step 2: Confirm It Actually Fails

As

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.

3. 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.

Rolle's Theorem

If a smooth curve starts and ends at exactly the same height, common sense says it has to level off somewhere in between — Rolle's Theorem makes that intuition rigorous, and it's the key building block behind the Mean Value Theorem.

Core Theorem

If

f

is continuous on

[a,b]

, differentiable on

(a,b)

, and

f(a)=f(b)

, then there exists at least one

c \in (a,b)

such that

f'(c) = 0

.

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Verify that Rolle's Theorem applies to

f(x) = x^3-x

on

[0,1]

, then find the guaranteed value of

c

.

How to think about it

Key idea: Check $f(0)$ and $f(1)$ are equal first, then solve $f'(x)=0$ and keep only the solution inside $(0,1)$ .

Full solution

1. Step 1: Check the Hypotheses

f

is a polynomial (continuous and differentiable everywhere).

f(0)=0

and

f(1)=1-1=0

— equal, so Rolle's Theorem applies.

2. Step 2: Solve f'(x) = 0

f'(x) = 3x^2-1 = 0 \implies x = \pm\frac{1}{\sqrt{3}}

.

3. Step 3: Keep Only the Solution in (0,1)

Only

c = \frac{1}{\sqrt{3}} \approx 0.577

lies in

(0,1)

.

Worked Example 02Hard

Show that

x^3+3x-1=0

has exactly one real root.

How to think about it

Key idea: Use the IVT to show a root exists, then use Rolle's Theorem (by contradiction) to show it can't have a second one.

Full solution

1. Step 1: Show a Root Exists (IVT)

Let

f(x)=x^3+3x-1

.

f(0)=-1<0

and

f(1)=3>0

, so by the IVT, there is a root in

(0,1)

.

2. Step 2: Suppose There Were Two Roots

Suppose

f(a)=f(b)=0

for some

a<b

. Since

f

is a polynomial, Rolle's Theorem would then guarantee some

c \in (a,b)

with

f'(c)=0

.

3. Step 3: Show This Is Impossible

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

for every real

x

— it can NEVER equal zero. This contradicts Step 2.

4. Step 4: Conclude

Since assuming a second root leads to a contradiction,

f(x)=x^3+3x-1=0

has exactly one real root.

Step-by-Step SOP

1
Verify the Three Hypotheses
Check continuity on $[a,b]$ , differentiability on $(a,b)$ , and that $f(a)=f(b)$ .
2
Solve f'(x) = 0
Find all solutions, then discard any outside the open interval $(a,b)$ .
3
For Uniqueness Proofs, Argue by Contradiction
If two roots of $f$ existed, Rolle's would force $f'$ to vanish somewhere between them — if you can show $f'$ never vanishes, that rules out a second root entirely.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

The polynomial

p(x)=(x-1)(x-2)(x-3)

has 3 distinct real roots. Explain why

p'(x)=0

must have at least 2 distinct real roots.

Need a hint?

Apply Rolle's Theorem to each consecutive pair of roots.

Show solution

1. Step 1: Apply Rolle's to the First Pair

Since

p(1)=p(2)=0

, Rolle's Theorem guarantees some

c_1 \in (1,2)

with

p'(c_1)=0

.

2. Step 2: Apply Rolle's to the Second Pair

Since

p(2)=p(3)=0

, Rolle's Theorem guarantees some

c_2 \in (2,3)

with

p'(c_2)=0

.

3. Step 3: Conclude

Since

c_1 \in (1,2)

and

c_2 \in (2,3)

are different intervals,

c_1 \neq c_2

— so

p'(x)=0

has at least 2 distinct real roots, exactly one between each consecutive pair of

p

's roots.

Practice 02Medium

Does Rolle's Theorem apply to

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

on

[-1,1]

?

Need a hint?

Check every hypothesis carefully, including differentiability on the ENTIRE open interval.

Show solution

1. Step 1: Check the Endpoint Values

f(-1) = (-1)^{2/3} = 1

and

f(1) = 1^{2/3} = 1

— equal.

2. Step 2: Check Differentiability

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

is undefined at

x=0

, which lies inside

(-1,1)

— so

f

is NOT differentiable on the entire open interval.

3. Step 3: Conclude

Rolle's Theorem does NOT apply here, since a key hypothesis fails. (Consistent with this:

f'(x)

is never actually

0

anywhere, so no such

c

exists.)

Common Pitfalls

⚠
All Three Hypotheses Must HoldContinuity on $[a,b]$ , differentiability on $(a,b)$ , AND $f(a)=f(b)$ are all required — missing any one means the theorem simply doesn't apply (see Example 4).
⚠
Rolle's Theorem Only Guarantees ExistenceIt tells you a $c$ exists, but not how many, or an easy way to find it beyond solving $f'(x)=0$ directly.

The Mean Value Theorem

Rolle's Theorem is really just the special case where the average rate of change happens to be zero — the Mean Value Theorem generalizes it to say the same thing must happen for ANY average rate of change.

Core Theorem

If

f

is continuous on

[a,b]

and differentiable on

(a,b)

, then there exists at least one

c \in (a,b)

such that

f'(c) = \frac{f(b)-f(a)}{b-a}

.

Watch the TikTok ExplanationMean Value Theorem (MVT)→

Worked Examples

Read these first — the full solution is shown, with the reasoning behind each move.

Worked Example 01Medium

Verify the Mean Value Theorem for

f(x) = x^3-x

on

[0,2]

, and find the guaranteed value of

c

.

How to think about it

Key idea: Compute the average rate of change first, then set $f'(x)$ equal to it and solve.

Full solution

1. Step 1: Compute the Average Rate of Change

f(0)=0

,

f(2)=8-2=6

. Average rate

= \frac{6-0}{2-0}=3

.

2. Step 2: Set f'(x) Equal to the Average Rate

f'(x) = 3x^2-1 = 3 \implies x^2 = \frac{4}{3} \implies x = \pm\frac{2}{\sqrt{3}}

.

3. Step 3: Keep Only the Solution in (0,2)

c = \frac{2}{\sqrt{3}} \approx 1.155

.

Worked Example 02Medium

The speed limit on a highway is 90 km/h. The distance between two toll booths is 300 km. A car passes the first booth at 8:00 AM and the second at 11:00 AM. Show the car must have exceeded the speed limit at some point.

How to think about it

Key idea: Compute the average speed over the 3-hour trip, then apply the MVT.

Full solution

1. Step 1: Compute the Average Speed

Average speed

= \frac{300\text{ km}}{3\text{ hr}} = 100

km/h.

2. Step 2: Apply the MVT

Assuming the position function is differentiable, the MVT guarantees some instant

c

where the car's instantaneous speed

= 100

km/h.

3. Step 3: Conclude

Since

100 > 90

, the car must have exceeded the speed limit at that instant.

Step-by-Step SOP

1
Verify Continuity and Differentiability
Confirm $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ .
2
Compute the Average Rate of Change
Calculate $\frac{f(b)-f(a)}{b-a}$ .
3
Set f'(x) Equal to It and Solve
Solve $f'(x) = \frac{f(b)-f(a)}{b-a}$ , keeping only solutions inside $(a,b)$ . For inequality proofs, bound $f'(c)$ instead of solving exactly.

Practice — Your Turn

Try each one before opening the hint or the solution.

Practice 01Hard

If

f(0)=-3

and

f'(x) \le 5

for all

x

, what is the largest possible value of

f(2)

?

Need a hint?

Apply the MVT on

[0,2]

and use the given bound on

f'

.

Show solution

1. Step 1: Apply the MVT on [0,2]

There exists

c \in (0,2)

with

f'(c) = \frac{f(2)-f(0)}{2-0} = \frac{f(2)+3}{2}

.

2. Step 2: Apply the Given Bound

Since

f'(c) \le 5

:

\frac{f(2)+3}{2} \le 5 \implies f(2)+3 \le 10 \implies f(2) \le 7

.

3. Step 3: Conclude

The largest possible value of

f(2)

is

7

.

Practice 02Hard

Prove that

|\tan x - \tan y| \ge |x-y|

for all

x,y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)

.

Need a hint?

Apply the MVT to

\tan

between

x

and

y

, then bound

\sec^2(c)

from below.

Show solution

1. Step 1: Apply the MVT to tan Between x and y

There exists

c

between

x

and

y

such that

\tan x - \tan y = \sec^2(c)(x-y)

.

2. Step 2: Bound sec²(c)

\sec^2(c) = 1+\tan^2(c) \ge 1

for every real

c

.

3. Step 3: Take Absolute Values

|\tan x - \tan y| = \sec^2(c)\,|x-y| \ge 1 \cdot |x-y| = |x-y|

.

Common Pitfalls

⚠
MVT Requires Differentiability on the Open Interval OnlyUnlike continuity (needed on the closed $[a,b]$ ), differentiability is only required on the open interval $(a,b)$ — the function can fail to be differentiable exactly at the endpoints.
⚠
The MVT Justification Has a Standard TemplateOn the AP exam, write: 'Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , by the MVT there exists $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ .' Skipping the continuity/differentiability justification costs points.

Worked Examples

1. Step 1: Analyze x³

2. Step 2: Analyze |x|

3. Step 3: Conclude

1. Step 1: Apply the Product Rule

2. Step 2: Factor Out x⁻²/⁵ and (4-x)

3. Step 3: Find Where the Numerator is Zero

4. Step 4: Find Where f' is Undefined

5. Step 5: State All Critical Points

Differentiate and Factor

Find Zeros of f'

Find Points Where f' Doesn't Exist

Practice — Your Turn

1. Step 1: Differentiate

2. Step 2: Factor

3. Step 3: State the Critical Points

1. Step 1: Differentiate

2. Step 2: Check the Sign

3. Step 3: Conclude

Worked Examples

1. Step 1: List Candidates

2. Step 2: Evaluate f at Each

3. Step 3: Compare

1. Step 1: Find Critical Points

2. Step 2: Evaluate at Candidates

3. Step 3: Compare

Confirm Continuity on a Closed Interval

List All Candidates

Evaluate and Compare

Practice — Your Turn

1. Step 1: Find Critical Points

2. Step 2: Evaluate at All Candidates

3. Step 3: Compare

1. Step 1: Check the Hypotheses

2. Step 2: Confirm It Actually Fails

3. Step 3: Conclude

Worked Examples

1. Step 1: Check the Hypotheses

2. Step 2: Solve f'(x) = 0

3. Step 3: Keep Only the Solution in (0,1)

1. Step 1: Show a Root Exists (IVT)

2. Step 2: Suppose There Were Two Roots

3. Step 3: Show This Is Impossible

4. Step 4: Conclude

Verify the Three Hypotheses

Solve f'(x) = 0

For Uniqueness Proofs, Argue by Contradiction

Practice — Your Turn

1. Step 1: Apply Rolle's to the First Pair

2. Step 2: Apply Rolle's to the Second Pair

3. Step 3: Conclude

1. Step 1: Check the Endpoint Values

2. Step 2: Check Differentiability

3. Step 3: Conclude

Worked Examples

1. Step 1: Compute the Average Rate of Change

2. Step 2: Set f'(x) Equal to the Average Rate

3. Step 3: Keep Only the Solution in (0,2)

1. Step 1: Compute the Average Speed

2. Step 2: Apply the MVT

3. Step 3: Conclude

Verify Continuity and Differentiability

Compute the Average Rate of Change

Set f'(x) Equal to It and Solve

Practice — Your Turn

1. Step 1: Apply the MVT on [0,2]

2. Step 2: Apply the Given Bound

3. Step 3: Conclude

1. Step 1: Apply the MVT to tan Between x and y

2. Step 2: Bound sec²(c)

3. Step 3: Take Absolute Values

Gary Chang