Day 13

Mean Value Theorem And Rolles Theorem

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

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

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 Exercises

Example 01Medium

Verify that Rolle's Theorem applies to

f(x) = x^3-x

[0,1]

, then find the guaranteed value of

c

NEED A HINT?

Check

f(0)

and

f(1)

are equal first, then solve

f'(x)=0

and keep only the solution inside

(0,1)

SHOW DETAILED EXPLANATION

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.

Step 2: Solve f'(x) = 0

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

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

Only

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

lies in

(0,1)

Example 02Hard

Show that

x^3+3x-1=0

has exactly one real root.

NEED A HINT?

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

SHOW DETAILED EXPLANATION

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)

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

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.

Step 4: Conclude

Since assuming a second root leads to a contradiction,

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

has exactly one real root.

Example 03Hard

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 DETAILED EXPLANATION

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

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

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.

Example 04Medium

Does Rolle's Theorem apply to

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

[-1,1]

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Step 1: Check the Endpoint Values

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

and

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

— equal.

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.

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

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)→

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 Exercises

Example 01Medium

Verify the Mean Value Theorem for

f(x) = x^3-x

[0,2]

, and find the guaranteed value of

c

NEED A HINT?

Compute the average rate of change first, then set

f'(x)

equal to it and solve.

SHOW DETAILED EXPLANATION

Step 1: Compute the Average Rate of Change

f(0)=0

f(2)=8-2=6

. Average rate

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

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}}

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

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

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.

NEED A HINT?

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

SHOW DETAILED EXPLANATION

Step 1: Compute the Average Speed

Average speed

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

km/h.

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.

Step 3: Conclude

Since

100 > 90

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

Example 03Hard

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 DETAILED EXPLANATION

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}

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

Step 3: Conclude

The largest possible value of

f(2)

7

Example 04Hard

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 DETAILED EXPLANATION

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)

Step 2: Bound sec²(c)

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

for every real

c

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.

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.

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Verify the Three Hypotheses

Solve f'(x) = 0

For Uniqueness Proofs, Argue by Contradiction

Practice Exercises

Step 1: Check the Hypotheses

Step 2: Solve f'(x) = 0

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

Step 1: Show a Root Exists (IVT)

Step 2: Suppose There Were Two Roots

Step 3: Show This Is Impossible

Step 4: Conclude

Step 1: Apply Rolle's to the First Pair

Step 2: Apply Rolle's to the Second Pair

Step 3: Conclude

Step 1: Check the Endpoint Values

Step 2: Check Differentiability

Step 3: Conclude

Verify Continuity and Differentiability

Compute the Average Rate of Change

Set f'(x) Equal to It and Solve

Practice Exercises

Step 1: Compute the Average Rate of Change

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

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

Step 1: Compute the Average Speed

Step 2: Apply the MVT

Step 3: Conclude

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

Step 2: Apply the Given Bound

Step 3: Conclude

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

Step 2: Bound sec²(c)

Step 3: Take Absolute Values

Gary Chang