MHT-CET Maths · Applications of Derivative

Rolle's Theorem and the Mean Value Theorem

If a function is smooth on an interval with equal endpoint values, its graph must level off somewhere (Rolle); more generally, some tangent must be parallel to the chord joining the endpoints (Lagrange). The whole subtopic is checking the three hypotheses, then solving f'(c) for the point c.

All Applications of Derivative notes MHT-CET Maths strategy

Why this matters

One of the most reliably tested subtopics in MHT-CET calculus: 18 PYQs sit here (3 HARD, 12 MODERATE, 3 EASY). The paper recycles a small set of moves — verify Rolle and find c, count how many c work, apply LMVT and solve for c, or use 'Rolle holds' to back out unknown coefficients a and b. The recurring traps are always the same: rejecting a root of f'(c)=0 that falls outside the open interval, forgetting that MVT needs differentiability (not just continuity), and confusing Rolle's f(a)=f(b) requirement with LMVT's chord slope.

Concept 1 of 5

Rolle's Theorem — the Three Hypotheses and the Conclusion

Intuition

If a smooth curve starts and ends at the same height over [a, b], it cannot rise without coming back down — somewhere in between its tangent must be flat. That flat spot is a point c where f'(c) = 0. Rolle's theorem promises at least one such c, provided three hypotheses hold.

Definition

Rolle's theorem. If $f$ satisfies all three hypotheses:

$f$ is continuous on the closed interval $[a, b]$ ,
$f$ is differentiable on the open interval $(a, b)$ ,
$f(a) = f(b)$ (equal endpoint values),

then there exists at least one $c \in (a, b)$ with $f'(c) = 0$ .

In practice: check the hypotheses (for a polynomial, continuity and differentiability are automatic — only $f(a)=f(b)$ needs verifying), then solve $f'(x)=0$ and keep the root that lies inside $(a,b)$ .

Rolle's theorem

f \text{ cont. on }[a,b],\ \text{diff. on }(a,b),\ f(a)=f(b)\ \Rightarrow\ \exists\, c\in(a,b):\ f'(c)=0

cpoint inside (a, b) where the tangent is horizontal
f(a)=f(b)the equal-endpoint hypothesis unique to Rolle

Worked example

Verify Rolle's theorem for

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

[1, 3]

and find

c

$f$ is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$ .
Endpoint values: $f(1) = 1 - 4 + 3 = 0$ and $f(3) = 9 - 12 + 3 = 0$ , so $f(1) = f(3)$ . All three hypotheses hold.
Solve $f'(x) = 2x - 4 = 0 \Rightarrow x = 2$ .
$c = 2 \in (1,3)$ , so Rolle's theorem is verified with $c = 2$ .

Answer:

c = 2

Practice this conceptself-check · 4 quick reps

Try it yourself

For

f(x) = x^2 - 6x + 8

[2, 4]

, verify the hypotheses of Rolle's theorem and find

c

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
State the third (interval-endpoint) hypothesis of Rolle's theorem.
2.
What does Rolle guarantee about $f'$ ?
3.
For a polynomial on $[a,b]$ , which hypothesis actually needs checking?
4.
Find $c$ for $f(x)=x^2-2x$ on $[0,2]$ .

From the bank · past-year question

Example 1Applications of DerivativeMODERATE

The value of

c

for which Rolle's theorem for the function

f(x)=x^{3}-3x^{2}+2x

in the interval

[0,2]

are

[Q142 · 3rd May Shift 2 · 2023]

All THREE hypotheses are required, not just f(a) = f(b)

Rolle needs continuity on

[a,b]

AND differentiability on

(a,b)

AND

f(a)=f(b)

. A function like

f(x)=|x|

[-1,1]

has

f(-1)=f(1)=1

but is NOT differentiable at

0

, so Rolle does not apply — there is no

c

with

f'(c)=0

. Never skip the differentiability check.

Rolle gives f'(c) = 0, not the chord slope

Rolle's conclusion is specifically

f'(c)=0

(a horizontal tangent), because the endpoints are equal. If the endpoints differ, use the Mean Value Theorem instead — its conclusion is

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

, which reduces to

0

only when

f(a)=f(b)

Concept 2 of 5

Finding c and Rejecting Roots Outside the Interval

Intuition

Once the hypotheses hold, finding

c

is just solving

f'(x)=0

. But

f'

often has more than one root — and Rolle only promises a

c

that lies STRICTLY between

a

and

b

. So compute every root, then discard any that fall outside the open interval

(a,b)

Definition

To find the Rolle point $c$ :

Solve $f'(x) = 0$ to get all candidate values.
**Keep only the candidates that lie inside the open interval $(a,b)$ **; reject the rest.

For products/exponential-times-polynomial forms, differentiate carefully (product rule + chain rule) — the exponential factor $e^{kx}$ is never zero, so it only scales $f'$ ; the roots come entirely from the polynomial factor. Set that polynomial factor to zero and filter by the interval.

Rolle point from f'(x) = 0

f'(c) = 0,\quad c \in (a,b)\ \ \text{(reject any root outside the interval)}

Worked example

Rolle's theorem holds for

f(x) = (x-1)(x-4)

[1,4]

. Find

c

, rejecting any root outside the interval.

Expand: $f(x) = x^2 - 5x + 4$ , so $f'(x) = 2x - 5$ .
Solve $f'(x) = 0$ : $x = \tfrac{5}{2} = 2.5$ .
Check the interval: $2.5 \in (1,4)$ ✓ — accept it.

Answer:

c = \dfrac{5}{2}

Practice this conceptself-check · 4 quick reps

Try it yourself

f(x) = x(x+3)e^{-x/2}

satisfies Rolle's theorem on

[-3, 0]

, find

c

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$f'(x) = (x-1)(x-5)$ on $[0,3]$ . Which root is the Rolle point?
2.
Rolle point must lie in which interval — $[a,b]$ or $(a,b)$ ?
3.
For $f'(x) = e^{x}(x^2 - 9)$ on $[-4, 0]$ , find $c$ .
4.
If both roots of $f'$ lie in $(a,b)$ , how many valid $c$ ?

From the bank · past-year question

Example 2Applications of DerivativeMODERATE

Value of c satisfying the conditions and conclusions of Rolle's theorem for the function

f(x) = x\sqrt{x+6},\; x \in [-6,0]

[Q109 · 11th May Shift 1 · 2024]

Reject any root of f'(c) = 0 that lies OUTSIDE (a, b)

Solving

f'(x)=0

can give roots that Rolle never promised. For

f(x)=x(x+3)e^{-x/2}

[-3,0]

f'

vanishes at

x=3

and

x=-2

, but only

-2

lies in

(-3,0)

. Picking

x=3

is the single most common mistake — always filter candidates by the interval.

The exponential factor is never zero

In a form like

f'(x)=e^{-x/2}\,(\text{polynomial})

, the factor

e^{-x/2}>0

always. So

f'(c)=0

forces the POLYNOMIAL factor to be zero — never set the exponential to zero. It only affects sign/scaling, not the location of the Rolle point.

Drill 1 more on finding c and rejecting roots outside the interval

Concept 3 of 5

Counting the Number of Valid c

Intuition

Some questions do not ask for the value of

c

but for HOW MANY values of

c

satisfy the conclusion. For a trigonometric function like

\sin(k\pi x)

f'(x)=0

has many solutions — so you count all the roots of

f'

that fall inside the open interval.

Definition

To count the Rolle/MVT points:

Write $f'(x)=0$ and solve the resulting trigonometric (or polynomial) equation over the open interval.
Count every solution that lies strictly inside $(a,b)$ .

For $f(x)=\sin(k\pi x)$ , $f'(x)=k\pi\cos(k\pi x)$ ; $f'=0$ where $\cos(k\pi x)=0$ , i.e. $k\pi x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \dots$ . List these, then keep the ones inside the interval and count them.

Number of Rolle points

\#\{\,c\in(a,b): f'(c)=0\,\} = \text{number of roots of } f'(x)=0 \text{ inside } (a,b)

Worked example

How many values of

c

satisfy Rolle's theorem for

f(x) = \sin \pi x

[0, 2]

$f(0) = \sin 0 = 0$ , $f(2) = \sin 2\pi = 0$ : endpoints equal, hypotheses hold.
$f'(x) = \pi\cos \pi x = 0 \Rightarrow \cos \pi x = 0 \Rightarrow \pi x = \tfrac{\pi}{2}, \tfrac{3\pi}{2}$ .
So $x = \tfrac12, \tfrac32$ , both in $(0,2)$ .

Answer:Two values of

c

x = \tfrac12

and

x = \tfrac32

Practice this conceptself-check · 4 quick reps

Try it yourself

How many values of

c

satisfy the conclusion of Rolle's theorem for

f(x) = \sin 2\pi x

[-1, 1]

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$f'(x)=\cos \pi x = 0$ on $(0,1)$ : how many roots?
2.
How many $c$ for $f(x)=\sin \pi x$ on $[0,3]$ ?
3.
To count Rolle points, count roots of $f'$ that lie where?
4.
$f(x)=\sin 2\pi x$ on $[0,1]$ : how many $c$ ?

From the bank · past-year question

Example 3Applications of DerivativeMODERATE

The number of values of

C

that satisfy the conclusion of Rolle's theorem in case of following function

f(x) = \sin 2\pi x,\, x \in [-1,1]

[Q104 · 15th May Shift 2 · 2023]

Count roots INSIDE the open interval only — mind the endpoints

When counting

c

, include only roots strictly between

a

and

b

. For

\sin 2\pi x

[-1,1]

the roots of

f'

are

\pm\tfrac14, \pm\tfrac34

— four values, all interior. A root landing exactly on an endpoint would NOT count. Sketch or list them; don't guess the count.

Solve the full trig equation — don't stop at the first solution

\cos(k\pi x)=0

has infinitely many solutions; over a finite interval several survive. Writing

k\pi x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \dots

systematically and filtering by the interval avoids under-counting (a common way to pick '02' when the answer is '04').

Concept 4 of 5

Lagrange's Mean Value Theorem — Statement and Finding c

Intuition

Rolle needs equal endpoints; Lagrange removes that restriction. Over any smooth arc from

(a, f(a))

(b, f(b))

, there is a point where the tangent is parallel to the CHORD joining the ends. The chord's slope is

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

, so LMVT says

f'(c)

equals that slope for some

c

inside.

Definition

Lagrange's Mean Value Theorem (LMVT). If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then there exists $c \in (a,b)$ with

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

To find

c

: compute the chord slope

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

, set

f'(x)

equal to it, solve, and keep the root inside

(a,b)

. Rolle is the special case

f(a)=f(b)

, where the chord slope is

0

Lagrange's Mean Value Theorem

f'(c) = \dfrac{f(b) - f(a)}{b - a},\qquad c \in (a,b)

\frac{f(b)-f(a)}{b-a}slope of the chord joining the endpoints
f'(c)slope of the tangent at the guaranteed point c

Worked example

Find the value of

c

of LMVT for

f(x) = x^2

[1, 3]

Chord slope: $\dfrac{f(3) - f(1)}{3 - 1} = \dfrac{9 - 1}{2} = 4$ .
$f'(x) = 2x$ ; set equal to the chord slope: $2c = 4 \Rightarrow c = 2$ .
$c = 2 \in (1,3)$ ✓.

Answer:

c = 2

Practice this conceptself-check · 4 quick reps

Try it yourself

Find

c

of LMVT for

f(x) = x^2 - 3x

[1, 4]

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Chord slope of $f(x)=x^2$ on $[0,4]$ ?
2.
LMVT for $f(x)=\log x$ on $[1,e]$ : find $c$ .
3.
Rolle is the special case of LMVT when the chord slope is?
4.
If $f(1)=1$ and $f'(x)\le 5$ on $[1,5]$ , max of $f(5)$ ?

From the bank · past-year question

Example 4Applications of DerivativeMODERATE

The value of

c

of Lagrange's mean value theorem for

f(x)=\sqrt{25-x^2}

[1,5]

[Q126 · 9th May Shift 1 · 2024]

MVT needs DIFFERENTIABILITY, not just continuity

Both hypotheses must hold: continuous on

[a,b]

AND differentiable on

(a,b)

. A function continuous but with a corner (non-differentiable) inside the interval can fail to have a valid

c

. Continuity alone is not enough — always confirm differentiability before applying LMVT.

Chord slope uses f(b) − f(a), not f'(a) or f'(b)

The right-hand side of LMVT is the average slope

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

— a difference of FUNCTION values over the interval width, not a derivative at an endpoint. A frequent slip is to compute

f'(a)

f'(b)

instead of the chord slope.

Bounding trick: f(b) − f(a) ≤ (b − a)·max f'

When

f'(x) \le M

everywhere, LMVT gives

f(b) - f(a) = f'(c)(b-a) \le M(b-a)

. So if

f(1)=1

and

f'\le 5

[1,5]

, then

f(5) \le 1 + 5\cdot 4 = 21

. Use the theorem as an INEQUALITY to bound a value.

Drill 8 more on lagrange's mean value theorem — statement and finding c

Concept 5 of 5

Solving Unknown Parameters and the Tangent-Parallel-to-Chord View

Intuition

A whole family of questions says 'Rolle (or MVT) holds for

f

at a GIVEN point

c

' and asks you to find unknown coefficients

a, b

. Each piece of the theorem is an equation:

f(a)=f(b)

gives one, and

f'(c)=0

(Rolle) or

f'(c)=\text{chord slope}

(MVT) gives another — two equations, two unknowns. Geometrically, MVT is just 'find the point where the tangent is parallel to the chord.'

Definition

Back-solving parameters. Given that the theorem holds with a specified $c$ :

Write the endpoint equation: Rolle ⇒ $f(a)=f(b)$ ; MVT ⇒ (usually) also $f(a)=f(b)$ or the given chord.
Write the interior equation: Rolle ⇒ $f'(c)=0$ ; MVT ⇒ $f'(c)=\frac{f(b)-f(a)}{b-a}$ .
Solve the resulting simultaneous equations for the unknowns.

Tangent parallel to chord (geometric MVT). 'Find the point(s) where the tangent is parallel to the chord $AB$ ' means: compute the chord slope through $A, B$ , set $f'(x)$ equal to it, and solve for $x$ — the same computation as finding the LMVT point.

Two equations from 'the theorem holds at c'

\underbrace{f(a)=f(b)}_{\text{endpoint}} \quad \text{and} \quad \underbrace{f'(c)=0}_{\text{Rolle at given }c}

Worked example

Rolle's theorem holds for

f(x) = x^3 + ax^2 + bx

[0, 3]

at the point

c = 1

. Find

a

and

b

Endpoint equation $f(0) = f(3)$ : $0 = 27 + 9a + 3b \Rightarrow 3a + b = -9$ .
Interior equation $f'(c) = 0$ : $f'(x) = 3x^2 + 2ax + b$ , so $f'(1) = 3 + 2a + b = 0 \Rightarrow 2a + b = -3$ .
Solve the pair $3a + b = -9,\ 2a + b = -3$ : subtract to get $a = -6$ , then $b = -3 - 2(-6) = 9$ .

Answer:

a = -6,\ b = 9

Practice this conceptself-check · 4 quick reps

Try it yourself

A(1,-3), B(4,3)

lie on the curve

y = x - \dfrac{4}{x}

. Find the points on the curve where the tangent is parallel to the chord

AB

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Rolle for $x^3 - 6x^2 + ax + b$ on $[1,3]$ needs $f(1)=f(3)$ . Find $a$ .
2.
'Tangent parallel to chord' means set $f'(x)$ equal to?
3.
Two unknowns $a,b$ need how many equations?
4.
Chord slope through $(1,-3),(4,3)$ ?

From the bank · past-year question

Example 5Applications of DerivativeMODERATE

If Rolle's theorem holds for the function

x^{3}+ax^{2}+bx,1 \leq x\leq 2

at the point

\frac{4}{3}

, then the values of

a

and

b

are respectively

[Q110 · 25 April Shift I · 2025]

You need BOTH equations — endpoint and interior

Using only

f(a)=f(b)

fixes just one relation between

a

and

b

; some options may share that value. You must ALSO use

f'(c)=0

(or the chord condition) at the given

c

to pin down both. Solving one equation and matching a partial answer is how students land on the wrong option.

Watch coefficient order — 'a and b respectively'

Options like

(11, -6)

(-6, 11)

differ only by which value is

a

and which is

b

. Track exactly which unknown you solved for; re-substitute into the original

f

to confirm the pair actually satisfies

f(a)=f(b)

and

f'(c)=0

before selecting.

Drill 4 more on solving unknown parameters and the tangent-parallel-to-chord view

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Rolle's Theorem — the Three Hypotheses and the Conclusion
Rolle's theorem
$f \text{ cont. on }[a,b],\ \text{diff. on }(a,b),\ f(a)=f(b)\ \Rightarrow\ \exists\, c\in(a,b):\ f'(c)=0$
Finding c and Rejecting Roots Outside the Interval
Rolle point from f'(x) = 0
$f'(c) = 0,\quad c \in (a,b)\ \ \text{(reject any root outside the interval)}$
Counting the Number of Valid c
Number of Rolle points
$\#\{\,c\in(a,b): f'(c)=0\,\} = \text{number of roots of } f'(x)=0 \text{ inside } (a,b)$
Lagrange's Mean Value Theorem — Statement and Finding c
Lagrange's Mean Value Theorem
$f'(c) = \dfrac{f(b) - f(a)}{b - a},\qquad c \in (a,b)$
Solving Unknown Parameters and the Tangent-Parallel-to-Chord View
Two equations from 'the theorem holds at c'
$\underbrace{f(a)=f(b)}_{\text{endpoint}} \quad \text{and} \quad \underbrace{f'(c)=0}_{\text{Rolle at given }c}$

Watch out for (11)

All THREE hypotheses are required, not just f(a) = f(b)
→ Rolle's Theorem — the Three Hypotheses and the Conclusion
Rolle gives f'(c) = 0, not the chord slope
→ Rolle's Theorem — the Three Hypotheses and the Conclusion
Reject any root of f'(c) = 0 that lies OUTSIDE (a, b)
→ Finding c and Rejecting Roots Outside the Interval
The exponential factor is never zero
→ Finding c and Rejecting Roots Outside the Interval
Count roots INSIDE the open interval only — mind the endpoints
→ Counting the Number of Valid c
Solve the full trig equation — don't stop at the first solution
→ Counting the Number of Valid c
MVT needs DIFFERENTIABILITY, not just continuity
→ Lagrange's Mean Value Theorem — Statement and Finding c
Chord slope uses f(b) − f(a), not f'(a) or f'(b)
→ Lagrange's Mean Value Theorem — Statement and Finding c
Bounding trick: f(b) − f(a) ≤ (b − a)·max f'
→ Lagrange's Mean Value Theorem — Statement and Finding c
You need BOTH equations — endpoint and interior
→ Solving Unknown Parameters and the Tangent-Parallel-to-Chord View
Watch coefficient order — 'a and b respectively'
→ Solving Unknown Parameters and the Tangent-Parallel-to-Chord View

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Applications of DerivativeMODERATE

If the function

f(x) =x(x+ 3)e^{-\frac{x}{2}}

satisfies all the conditions of Rolle's theorem in

\lbrack - 3,0\rbrack

, then c is

[Q132 · 21 April Shift I · 2025]

Example 2Applications of DerivativeEASY

Let f be a function which is continuous and differentiable for all

x

. If

f(1) = 1

and

f^{'}(x) \leq 5

for all

x

\lbrack 1,5\rbrack

, then the maximum value of f (5) is

[Q107 · 22 April Shift II · 2025]

Example 3Applications of DerivativeHARD

If Rolle's theorem holds for the function

f(x) = x^3 + bx^2 + ax + 5

[1,3]

with

c = 2 + \dfrac{1}{\sqrt{3}}

, then the values of

a

and

b

respectively are

[Q138 · 2nd May Shift 1 · 2023]

Example 4Applications of DerivativeMODERATE

The value of C for which the Mean Value Theorem holds for

f(x) = \log_e x

[1,3]

[Q117 · 2nd May Shift 2 · 2023]

Example 5Applications of DerivativeMODERATE

A(1,-3), B(4,3) are two points on the curve

y=x-\frac{4}{x}

. The points on the curve, the tangents at which are parallel to the chord AB, are

[Q123 · 13th May Shift 2 · 2024]

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