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

A critical point (where

f' = 0

or is undefined) is only a *candidate* for an extremum.

x^3

has

f'(0)=0

but no extremum, because

f'

never changes sign there — the sign change is what matters.

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

On a closed interval, the absolute max and min live in a short finite list: the critical numbers inside, plus the two endpoints. Evaluate

f

at each and compare — no sign charts needed.

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

Rolle's is the special case of the MVT where the two endpoints sit at the same height, so the slope it guarantees is exactly zero. Confirm all three hypotheses first — continuous, differentiable, and

f(a) = f(b)

— then solving

f'(c) = 0

is ordinary algebra.

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

The MVT says the instantaneous rate at some interior point must equal the average rate across the whole interval — a tangent parallel to the secant. Compute the secant slope

\frac{f(b) - f(a)}{b - a}

first, then set

f'(c)

equal to 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