Rolle’s & Lagrange’s Mean Value Theorem

Q: What is Rolle's theorem?

If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then f'(c) = 0 for at least one c strictly between a and b. Geometrically, a smooth curve that starts and ends at the same height has a horizontal tangent somewhere in between.

Q: What is Lagrange's mean value theorem?

If f is continuous on [a, b] and differentiable on (a, b), then f'(c) = (f(b) - f(a))/(b - a) for some c in (a, b). The tangent at c is parallel to the chord joining the end points; the instantaneous rate equals the average rate.

Q: What is the difference between Rolle's theorem and the mean value theorem?

Rolle's theorem needs equal end values and gives a horizontal tangent, f'(c) = 0. The mean value theorem drops the equal-value condition and gives a tangent parallel to the chord. Rolle's theorem is the special case of the mean value theorem with a horizontal chord.

Q: What happens if a condition of Rolle's theorem fails?

The conclusion is no longer guaranteed. For example f(x) = |x| on [-1, 1] is not differentiable at 0 and has no point with zero derivative. A function that breaks a condition may still have such a point by chance, but the theorem does not promise it.

Q: How is Rolle's theorem used to count roots?

Between two roots of f there is a root of f'. So if f' has k real roots, f has at most k + 1; if f' never vanishes, f has at most one real root, as for x⁷ + x³ + λ.

Q: How do you prove an inequality using the mean value theorem?

Write the difference f(b) - f(a) as (b - a)f'(c), then bound f'(c) using the behaviour of f' on (a, b). If f' is monotonic, the bounds are f'(a) and f'(b), which gives inequalities like those for tan ⁻¹ and ln.

Q: How are Rolle's theorem and the mean value theorem asked in JEE Main?

JEE Main usually asks for the value of c in Rolle's theorem or the mean value theorem for a polynomial, exponential or logarithmic function, or for an unknown coefficient that makes Rolle's theorem hold. Remember that for a quadratic the mean value point is the midpoint.

Q: What mean value theorem problems appear in JEE Advanced?

JEE Advanced combines the theorems with given values of f to conclude something about f' or f'', uses Rolle on a cleverly built auxiliary function such as e^-xf(x), counts real roots, and proves inequalities, sometimes through Cauchy's form of the theorem.

MathsApplication Of DerivativesFor JEE aspirants

Rolle's theorem says that a smooth curve that starts and ends at the same height must have a horizontal tangent somewhere in between. Lagrange's mean value theorem tilts the picture: somewhere the tangent is parallel to the chord, so the instantaneous rate equals the average rate. Rolle's and Lagrange's mean value theorem are the bridge from the derivative to the function itself, and JEE Main and JEE Advanced use them to find points $c$ , count roots and prove inequalities.

On this page1Rolle's theorem2Conditions3Roots4Lagrange MVT5Proof and

θ

-form6Consequences7Examples

Key Formulas - Quick Reference

★ Must learnRolle: $f$ continuous on $[a, b]$ , derivable on $(a, b)$ , $f (a) = f (b)$ $\Rightarrow$ $f^{'} (c) = 0$ for at least one $c \in (a, b)$
Between two zeros of $f$ there is at least one zero of $f^{'}$ ; if $f^{'}$ never vanishes, $f$ has at most one real zero
★ Must learnLagrange (LMVT): $f$ continuous on $[a, b]$ , derivable on $(a, b)$ $\Rightarrow$ $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ for some $c \in (a, b)$
★ Must learn $θ$ -form: $f (a + h) = f (a) + h f^{'} (a + θ h)$ , $0 < θ < 1$
Quadratic $f$ : the LMVT point is the midpoint $c = \frac{a + b}{2}$ ( $θ = \frac{1}{2}$ )
★ Must learn $f^{'} (x) = 0$ on $(a, b)$ $\Rightarrow$ $f$ constant; $f^{'} (x) > 0$ on $(a, b)$ $\Rightarrow$ $f$ strictly increasing on $[a, b]$
Cauchy: $\frac{f ( b ) - f ( a )}{g ( b ) - g ( a )} = \frac{f ^{'} ( c )}{g ^{'} ( c )}$ for some $c \in (a, b)$ (when $g^{'} \neq = 0$ )

1. Rolle's Theorem

Rolle's theorem. If a function $f$ defined on $[a, b]$ is

continuous on the closed interval $[a, b]$ ,
derivable on the open interval $(a, b)$ , and
$f (a) = f (b)$ ,

then there exists at least one real number $c$ with $a < c < b$ such that $f^{'} (c) = 0$ .

1.1 Geometrical meaning

Draw $y = f (x)$ from $A (a, f (a))$ to $B (b, f (b))$ . The chord $A B$ is horizontal because the end heights are equal. Rolle's theorem says the curve has at least one point $C (c, f (c))$ where the tangent is horizontal, i.e. parallel to the $x$ -axis and to $A B$ .

Figure 1: Rolle's theorem.

f (x) = - \frac{x ^{2}}{4} + \frac{5 x}{4} + 1

[1, 4]

has

f (1) = f (4) = 2

. The chord

A B

is horizontal, and at

c = 2.5

the tangent is horizontal too:

f^{'} (2.5) = 0

There may be more than one such point. The theorem guarantees existence, not uniqueness.

Figure 2: Rolle's theorem promises at least one

c

. For

f (x) = 1.5 + sin x

[0.4, 3 π - 0.4]

(equal end values) there are three:

c = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}

, each with a horizontal tangent.

1.2 Why each condition is needed

If any one hypothesis is removed, the conclusion can fail. Note that the conditions are sufficient, not necessary: a function may break a condition and still have $f^{'} (c) = 0$ somewhere; the theorem simply no longer promises it.

Figure 3: Drop any one condition and the conclusion can fail. Left:

f (x) = x

[0, 1)

with

f (1) = 0

is not continuous at

1

. Middle:

f (x) = 1 - ∣ x ∣

[- 1, 1]

has a corner at

0

. Right:

f (x) = x

[0, 1]

has

f (0) \neq = f (1)

. None has a point with

f^{'} (c) = 0

Condition dropped	Counterexample on the interval	What goes wrong
continuity on $[a, b]$	$f (x) = x$ on $[0, 1)$ , $f (1) = 0$	jump at the end; $f^{'} = 1$ everywhere inside
derivability on $(a, b)$	$f (x) = 1 - ∣ x ∣$ on $[- 1, 1]$	corner at $0$ ; $f^{'} = \pm 1$ , never $0$
$f (a) = f (b)$	$f (x) = x$ on $[0, 1]$	$f^{'} = 1$ , never $0$

1.3 Algebraic meaning: roots of $f$ and $f^{'}$

If $a$ and $b$ are two zeros of $f$ (so $f (a) = f (b) = 0$ ) and $f$ is derivable, then $f^{'}$ has at least one zero between $a$ and $b$ . In words: between two roots of $f (x) = 0$ lies a root of $f^{'} (x) = 0$ .

Figure 4:

f (x) = (x + 1) (x - 1) (x - 2.5)

has roots

- 1, 1, 2.5

. Its derivative

f^{'} (x) = 3 x^{2} - 5 x - 1

(dashed) vanishes at

c_{1} \approx - 0.18 \in (- 1, 1)

and

c_{2} \approx 1.85 \in (1, 2.5)

: a root of

f^{'}

between every two consecutive roots of

f

Turning this around gives a root-counting tool. If $f^{'} (x) = 0$ has $k$ real roots, then $f (x) = 0$ has at most $k + 1$ real roots. In particular, if $f^{'}$ never vanishes, $f$ has at most one real root.

Figure 5:

P (x) = x^{7} + x^{3} + λ

for three values of

λ

. Each graph crosses the

x

-axis exactly once. If there were two roots, Rolle would force

P^{'} (c) = 7 c^{6} + 3 c^{2} = 0

for some

c

between them, i.e.

c = 0

; but then

P

would have to vanish on both sides of

0

while being increasing, which is impossible.

Exam Trick

Root in $(0, 1)$ ? Integrate first. To show that $P (x) = 0$ has a root between $a$ and $b$ , find $F$ with $F^{'} = P$ and check $F (a) = F (b)$ ; Rolle then gives $P (c) = 0$ . For $a x^{2} + b x + c$ with $2 a + 3 b + 6 c = 0$ , take $F (x) = \frac{a x ^{3}}{3} + \frac{b x ^{2}}{2} + c x$ : $F (0) = 0$ and $F (1) = \frac{2 a + 3 b + 6 c}{6} = 0$ .

Figure 6: With

2 a + 3 b + 6 c = 0

(here

a = 3

b = 0

c = - 1

F (x) = \frac{a x ^{3}}{3} + \frac{b x ^{2}}{2} + c x = x^{3} - x

has

F (0) = F (1) = 0

. Rolle gives

F^{'} (x) = a x^{2} + b x + c = 0

somewhere in

(0, 1)

; here at

x = \frac{1}{3} \approx 0.577

Quick Recall: tap to check

State the three conditions of Rolle's theorem.

Continuous on

[a, b]

, derivable on

(a, b)

f (a) = f (b)

Does Rolle's theorem apply to

f (x) = ∣ x ∣

[- 1, 1]

No:

f

is not derivable at

0

, and indeed

f^{'}

is never

0

f^{'}

has no real zero. How many real zeros can

f

have?

At most one.

Key idea

Equal end heights + smooth curve

\Rightarrow

a horizontal tangent inside. Between two roots of

f

lies a root of

f^{'}

2. Lagrange's Mean Value Theorem

Lagrange's mean value theorem (LMVT). If a function $f$ defined on $[a, b]$ is continuous on $[a, b]$ and derivable on $(a, b)$ , then there exists at least one $c$ with $a < c < b$ such that

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

2.1 Geometrical meaning

The right side is the slope of the chord joining $A (a, f (a))$ and $B (b, f (b))$ . So between $A$ and $B$ there is at least one point $C$ where the tangent is parallel to the chord $A B$ . Rolle's theorem is the special case where the chord is horizontal.

Figure 7: Lagrange's mean value theorem for

f (x) = x

[1, 4]

. The chord

A B

has slope

\frac{2 - 1}{4 - 1} = \frac{1}{3}

;

f^{'} (c) = \frac{1}{2 c} = \frac{1}{3}

gives

c = \frac{9}{4}

, where the tangent is parallel to

A B

Physical meaning. If $s (t)$ is the distance travelled, $\frac{s ( b ) - s ( a )}{b - a}$ is the average speed. LMVT says the speedometer shows exactly the average speed at some instant: a car that covers $120 km$ in $2 h$ must be doing exactly $60 km/h$ at least once.

Rolle's theorem

Needs $f (a) = f (b)$ . Conclusion: $f^{'} (c) = 0$ (tangent horizontal).

Lagrange's MVT

No condition on end values. Conclusion: $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ (tangent parallel to chord).

2.2 Proof

Let $g (x) = f (x) + λ x$ on $[a, b]$ , with the constant $λ$ chosen so that $g (a) = g (b)$ : $f (a) + λa = f (b) + λb$ gives $λ = - \frac{f ( b ) - f ( a )}{b - a}$ .
$g$ is a sum of a continuous, derivable function and a linear one, so $g$ is continuous on $[a, b]$ , derivable on $(a, b)$ , and $g (a) = g (b)$ .
By Rolle's theorem there is $c \in (a, b)$ with $g^{'} (c) = 0$ . But $g^{'} (x) = f^{'} (x) + λ$ , so $0 = f^{'} (c) + λ$ .
Hence $f^{'} (c) = - λ = \frac{f ( b ) - f ( a )}{b - a}$ .

Equivalently, $g$ measures the vertical gap between the curve and the chord (up to a constant); the gap is zero at both ends, so it has a turning point in between, and there the curve runs parallel to the chord.

Figure 8: The idea of the proof. Subtract the chord:

g (x) = f (x) - [f (a) + \frac{f ( b ) - f ( a )}{b - a} (x - a)]

is zero at

a

and

b

. By Rolle,

g^{'} (c) = 0

, i.e.

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

. The point

c

is where the vertical gap between curve and chord is largest (here

c = 2.25

for

x

[1, 4]

2.3 Alternative ( $θ$ ) form

Replace $b$ by $a + h$ . Any number between $a$ and $a + h$ can be written as $a + θ h$ with $0 < θ < 1$ , so

\frac{f ( a + h ) - f ( a )}{h} = f^{'} (a + θ h) or f (a + h) = f (a) + h f^{'} (a + θ h), 0 < θ < 1.

The value of $θ$ depends on $f$ , $a$ and $h$ . For $f (x) = x^{3}$ , $a = 1$ , $h = 1$ : $7 = 3 (1 + θ)^{2}$ gives $θ \approx 0.53$ .

Figure 9: For a quadratic the mean value point is always the midpoint:

f (b) - f (a) = (b - a) [p (a + b) + q]

and

f^{'} (c) = 2 p c + q

give

c = \frac{a + b}{2}

, so

θ = \frac{1}{2}

f (a + h) = f (a) + h f^{'} (a + θ h)

Exam Trick

Quadratics: no calculation needed. For any quadratic the LMVT point is the midpoint $c = \frac{a + b}{2}$ , and the Rolle point (when $f (a) = f (b)$ ) is the vertex $x = - \frac{q}{2 p}$ . Example: $f (x) = - x^{2} + 4 x - 5$ on $[- 1, 1]$ gives $c = 0$ at once.

Key idea

LMVT: average rate over

[a, b]

= instantaneous rate at some inside point

c

; Rolle is the case of a horizontal chord.

3. Consequences and Applications

3.1 Constant and increasing functions

Apply LMVT on $[x_{1}, x_{2}] \subset [a, b]$ : $f (x_{2}) - f (x_{1}) = f^{'} (c) (x_{2} - x_{1})$ for some $c$ between them. The sign of $f^{'}$ therefore controls how $f$ changes:

If on $(a, b)$ ...	then on $[a, b]$ ...	Reason (LMVT)
$f^{'} (x) = 0$	$f$ is constant	$f (x_{2}) - f (x_{1}) = 0 \cdot (x_{2} - x_{1})$
$f^{'} (x) = g^{'} (x)$	$f (x) - g (x)$ is constant	apply the first row to $f - g$
$f^{'} (x) > 0$	$f$ is strictly increasing	$x_{1} < x_{2} \Rightarrow f (x_{2}) - f (x_{1}) > 0$
$f^{'} (x) < 0$	$f$ is strictly decreasing	$x_{1} < x_{2} \Rightarrow f (x_{2}) - f (x_{1}) < 0$

The third row is the derivative test used throughout the next concept, Monotonicity.

3.2 Proving inequalities

LMVT turns a difference of values into a derivative: $f (b) - f (a) = (b - a) f^{'} (c)$ . If we can bound $f^{'}$ on $(a, b)$ , we bound the difference. When $f^{'}$ is monotonic, the bounds are the end values $f^{'} (a)$ and $f^{'} (b)$ .

Figure 10: Because

f^{'} (x) = \frac{1}{1 + x ^{2}}

decreases, the chord slope

\frac{tan ^{- 1} b - tan ^{- 1} a}{b - a} = f^{'} (c)

lies between

f^{'} (b)

and

f^{'} (a)

. For

a = 0.5

b = 2

0.2 < 0.488 < 0.8

, which is the inequality

\frac{b - a}{1 + b ^{2}} < tan^{- 1} b - tan^{- 1} a < \frac{b - a}{1 + a ^{2}}

JEE Advanced

Cauchy's mean value theorem. If $f$ and $g$ are continuous on $[a, b]$ , derivable on $(a, b)$ , and $g^{'} (x) \neq = 0$ on $(a, b)$ , then for some $c \in (a, b)$

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

Proof: apply Rolle to $h (x) = f (x) [g (b) - g (a)] - g (x) [f (b) - f (a)]$ , which has $h (a) = h (b)$ . LMVT is the case $g (x) = x$ . Example: $f = x^{2}$ , $g = x^{3}$ on $[1, 2]$ : $\frac{3}{7} = \frac{2 c}{3 c ^{2}}$ gives $c = \frac{14}{9} \in (1, 2)$ . Cauchy's theorem is the idea behind L'Hospital's rule.

Exam Trick

Two applications of LMVT give $f^{''}$ . Given three values $f (x_{1}), f (x_{2}), f (x_{3})$ , LMVT on $[x_{1}, x_{2}]$ and $[x_{2}, x_{3}]$ gives two values of $f^{'}$ ; LMVT on $f^{'}$ between those points gives information about $f^{''}$ . Compare with a known function (like $x^{3}$ ) by working with $g = f - x^{3}$ .

Quick Recall: tap to check

Write LMVT in the

θ

-form.

f (a + h) = f (a) + h f^{'} (a + θ h)

with

0 < θ < 1

For

f (x) = 3 x^{2} - 7 x + 2

[1, 5]

, where is the LMVT point?

At the midpoint,

c = 3

(quadratic).

f^{'} (x) = g^{'} (x)

on an interval, what can you say about

f

and

g

They differ by a constant:

f (x) = g (x) + C

Why is

∣ sin a - sin b ∣ \leq ∣ a - b ∣

By LMVT,

sin a - sin b = (a - b) cos c

and

∣ cos c ∣ \leq 1

Key idea

Bound

f^{'}

and LMVT bounds

f (b) - f (a)

; the sign of

f^{'}

decides whether

f

is constant, increasing or decreasing.

Figure 11: Choosing the theorem. A required point with

f^{'} (c) = 0

calls for Rolle; differences

f (b) - f (a)

and inequalities call for LMVT; a root of a polynomial in an interval often comes from Rolle applied to its antiderivative.

Figure 12: Mind map of Rolle's theorem and Lagrange's mean value theorem.

4. Solved Examples

Solved Example 1

Verify Rolle's theorem for

f (x) = (x - a)^{n} (x - b)^{m}

[a, b]

, where

m

and

n

are positive integers.

Solution:

$f$ is a polynomial, so it is continuous on $[a, b]$ and derivable on $(a, b)$ ; also $f (a) = f (b) = 0$ .
$f^{'} (x) = n (x - a)^{n - 1} (x - b)^{m} + m (x - a)^{n} (x - b)^{m - 1} = (x - a)^{n - 1} (x - b)^{m - 1} [(m + n) x - (nb + ma)]$ .
$f^{'} (c) = 0$ inside the interval gives $c = \frac{nb + ma}{m + n}$ .

Answer: $c = \frac{ma + nb}{m + n}$ , which divides $[a, b]$ internally in the ratio $n : m$ , so it lies in $(a, b)$ and the theorem is verified.

Solved Example 2

2 a + 3 b + 6 c = 0

, prove that the equation

a x^{2} + b x + c = 0

has at least one real root between

0

and

1

Solution:

Let $F (x) = \frac{a x ^{3}}{3} + \frac{b x ^{2}}{2} + c x$ , a polynomial (continuous and derivable everywhere).
$F (0) = 0$ and $F (1) = \frac{a}{3} + \frac{b}{2} + c = \frac{2 a + 3 b + 6 c}{6} = 0$ .
By Rolle's theorem $F^{'} (x) = a x^{2} + b x + c = 0$ for some $x \in (0, 1)$ .

Answer: proved (Figure 6 shows the case $a = 3$ , $b = 0$ , $c = - 1$ ).

Solved Example 3

Verify Lagrange's mean value theorem for

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

[- 1, 1]

Solution:

$f$ is a polynomial: continuous on $[- 1, 1]$ , derivable on $(- 1, 1)$ .
$f (1) = - 2$ , $f (- 1) = - 10$ , so $\frac{f ( 1 ) - f ( - 1 )}{1 - ( - 1 )} = \frac{8}{2} = 4$ .
$f^{'} (c) = - 2 c + 4 = 4$ gives $c = 0$ .

Answer: $c = 0 \in (- 1, 1)$ (the midpoint, as for every quadratic).

Solved Example 4

Using Lagrange's mean value theorem, prove that if

b > a > 0

, then

\frac{b - a}{1 + b ^{2}} < tan^{- 1} b - tan^{- 1} a < \frac{b - a}{1 + a ^{2}}

Solution:

Let $f (x) = tan^{- 1} x$ on $[a, b]$ . By LMVT, $f^{'} (c) = \frac{tan ^{- 1} b - tan ^{- 1} a}{b - a}$ for some $a < c < b$ , where $f^{'} (x) = \frac{1}{1 + x ^{2}}$ .
$f^{'}$ is decreasing on $(0, \infty)$ , so $a < c < b$ gives $f^{'} (b) < f^{'} (c) < f^{'} (a)$ .
$\frac{1}{1 + b ^{2}} < \frac{tan ^{- 1} b - tan ^{- 1} a}{b - a} < \frac{1}{1 + a ^{2}}$ ; multiply by $b - a > 0$ .

Answer: proved (Figure 10).

Solved Example 5

Let

f : R \to R

be twice differentiable with

f (2) = 8

f (4) > 64

and

f (7) = 343

. Show that there exists

c \in (2, 7)

such that

f^{''} (c) < 6 c

Solution:

Let $g (x) = f (x) - x^{3}$ . Then $g (2) = 0$ , $g (4) > 0$ , $g (7) = 0$ .
LMVT on $[2, 4]$ : $g^{'} (c_{1}) = \frac{g ( 4 ) - g ( 2 )}{2} > 0$ for some $2 < c_{1} < 4$ .
LMVT on $[4, 7]$ : $g^{'} (c_{2}) = \frac{g ( 7 ) - g ( 4 )}{3} < 0$ for some $4 < c_{2} < 7$ .
LMVT on $g^{'}$ over $[c_{1}, c_{2}]$ : $g^{''} (c) = \frac{g ^{'} ( c _{2} ) - g ^{'} ( c _{1} )}{c _{2} - c _{1}} < 0$ for some $c \in (c_{1}, c_{2}) \subset (2, 7)$ .
$g^{''} (c) = f^{''} (c) - 6 c < 0$ .

Answer: $f^{''} (c) < 6 c$ for some $c \in (2, 7)$ .

Solved Example 6

Verify Rolle's theorem for

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

[- 2, 2]

Solution:

$f$ is a polynomial, so continuous on $[- 2, 2]$ and derivable on $(- 2, 2)$ .
$f (- 2) = 6 = f (2)$ .
$f^{'} (c) = 2 c = 0$ gives $c = 0$ .

Answer: $c = 0 \in (- 2, 2)$ ; the theorem is verified.

Solved Example 7

Can Rolle's theorem be applied to

f (x) = ∣ x ∣

[- 1, 1]

? Is there a point where

f^{'} (c) = 0

Solution:

$f$ is continuous on $[- 1, 1]$ and $f (- 1) = f (1) = 1$ .
But $f$ is not derivable at $x = 0 \in (- 1, 1)$ (left derivative $- 1$ , right derivative $1$ ), so the theorem does not apply.
Indeed $f^{'} (x) = - 1$ for $x < 0$ and $1$ for $x > 0$ : never $0$ .

Answer: Rolle's theorem is not applicable, and no such $c$ exists; the derivability condition cannot be dropped.

Solved Example 8

Verify Lagrange's mean value theorem for

f (x) = x^{3}

[1, 3]

Solution:

$f$ is a polynomial, so the conditions hold.
$\frac{f ( 3 ) - f ( 1 )}{3 - 1} = \frac{27 - 1}{2} = 13$ .
$f^{'} (c) = 3 c^{2} = 13$ gives $c = \pm \frac{13}{3} = \pm 2.08$ . Only $c = 2.08$ lies in $(1, 3)$ .

Answer: $c = \frac{13}{3} \approx 2.08$ .

Solved Example 9

f (x) = x^{3} - 6 x^{2} + a x + b

satisfies Rolle's theorem on

[1, 3]

with

c = 2 + \frac{1}{3}

, then

a

equals
(A)

6

(B)

11

(C)

- 6

(D)

1

Solution:

Answer: (B). $f (1) = f (3)$ : $a + b - 5 = 3 a + b - 27$ , so $a = 11$ . Check: $f^{'} (x) = 3 x^{2} - 12 x + 11 = 0$ gives $x = 2 \pm \frac{1}{3}$ , matching $c$ . ( $b$ can be any number.)

Solved Example 10

The value of

c

in Lagrange's mean value theorem for

f (x) = x (x - 1) (x - 2)

[0, \frac{1}{2}]

is
(A)

1 - \frac{21}{6}

(B)

1 + \frac{21}{6}

(C)

\frac{1}{2}

(D)

\frac{1}{4}

Solution:

Answer: (A). $f (x) = x^{3} - 3 x^{2} + 2 x$ , $f (0) = 0$ , $f (\frac{1}{2}) = \frac{3}{8}$ , chord slope $= \frac{3}{4}$ . $3 c^{2} - 6 c + 2 = \frac{3}{4}$ gives $12 c^{2} - 24 c + 5 = 0$ , $c = 1 \pm \frac{21}{6}$ . Only $1 - \frac{21}{6} \approx 0.236$ lies in $(0, \frac{1}{2})$ .

Solved Example 11

Using the mean value theorem, prove that

∣ sin a - sin b ∣ \leq ∣ a - b ∣

for all real

a

b

Solution:

If $a = b$ both sides are $0$ . Otherwise apply LMVT to $f (x) = sin x$ on the interval between $a$ and $b$ .
$sin a - sin b = (a - b) cos c$ for some $c$ between $a$ and $b$ .
Take moduli: $∣ sin a - sin b ∣ = ∣ a - b ∣ ∣ cos c ∣ \leq ∣ a - b ∣$ , since $∣ cos c ∣ \leq 1$ .

Answer: proved. (In particular $∣ sin x ∣ \leq ∣ x ∣$ , taking $b = 0$ .)

Practice Questions

If $f (x)$ satisfies the conditions of Rolle's theorem, show that between two consecutive zeros of $f^{'} (x)$ there lies at most one zero of $f (x)$ .Answer: if $f$ had two zeros there, Rolle would give a zero of $f^{'}$ between them, contradicting consecutiveness
Show that for every real $λ$ the polynomial $P (x) = x^{7} + x^{3} + λ$ has exactly one real root.Answer: odd degree gives at least one root; $P^{'} (x) = 7 x^{6} + 3 x^{2}$ vanishes only at $x = 0$ , so $P$ is strictly increasing and has at most one root (Figure 5)
If $f$ satisfies the conditions of LMVT on $[a, b]$ and $f^{'} (x) = 0$ for all $x \in (a, b)$ , show that $f$ is constant on $[a, b]$ .Answer: $f (x) - f (a) = (x - a) f^{'} (c) = 0$ for every $x$
Using LMVT, prove that if two functions have equal derivatives at all points of $(a, b)$ , then they differ by a constant.Answer: apply the previous result to $f - g$
If $f$ is continuous on $[a, b]$ , derivable on $(a, b)$ and $f^{'} (x) > 0$ on $(a, b)$ , show that $f$ is strictly increasing on $[a, b]$ .Answer: $f (x_{2}) - f (x_{1}) = (x_{2} - x_{1}) f^{'} (c) > 0$ for $x_{1} < x_{2}$
Verify Rolle's theorem for $f (x) = x^{2} - 5 x + 4$ on $[1, 4]$ .Answer: $f (1) = f (4) = 0$ ; $c = \frac{5}{2}$
Find $c$ of Lagrange's mean value theorem for $f (x) = ln x$ on $[1, e]$ .Answer: $\frac{1}{c} = \frac{1}{e - 1}$ , so $c = e - 1 \approx 1.72$

Common Mistakes to Avoid

Watch out

Applying Rolle's theorem without checking $f (a) = f (b)$ ; if the end values differ, use LMVT instead.
Reporting a value of $c$ outside $(a, b)$ . Solve $f^{'} (c) = \dots$ , then keep only roots strictly inside the interval.
Asking for derivability on the closed interval: the theorems need continuity on $[a, b]$ but derivability only on $(a, b)$ .
Missing a corner or a jump: $∣ x - k ∣$ , $[x]$ and piecewise functions often fail a hypothesis inside the interval.
Reading the theorems as 'exactly one $c$ '. They guarantee at least one; there may be several.
Assuming the converse: $f^{'} (c) = 0$ somewhere does not mean $f (a) = f (b)$ , and a function failing a condition may still have such a point.
Reversing an inequality chain: if $f^{'}$ is decreasing and $a < c < b$ , then $f^{'} (b) < f^{'} (c) < f^{'} (a)$ , not the other way.
Treating $θ$ as fixed: in $f (a + h) = f (a) + h f^{'} (a + θ h)$ , $θ$ depends on $f$ , $a$ and $h$ , and $0 < θ < 1$ strictly.

Frequently Asked Questions

What is Rolle's theorem?

If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ and $f (a) = f (b)$ , then $f^{'} (c) = 0$ for at least one $c$ strictly between $a$ and $b$ . Geometrically, a smooth curve that starts and ends at the same height has a horizontal tangent somewhere in between.

What is Lagrange's mean value theorem?

If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a}$ for some $c$ in $(a, b)$ . The tangent at $c$ is parallel to the chord joining the end points; the instantaneous rate equals the average rate.

What is the difference between Rolle's theorem and the mean value theorem?

Rolle's theorem needs equal end values and gives a horizontal tangent, $f^{'} (c) = 0$ . The mean value theorem drops the equal-value condition and gives a tangent parallel to the chord. Rolle's theorem is the special case of the mean value theorem with a horizontal chord.

What happens if a condition of Rolle's theorem fails?

The conclusion is no longer guaranteed. For example $f (x) = ∣ x ∣$ on $[- 1, 1]$ is not differentiable at $0$ and has no point with zero derivative. A function that breaks a condition may still have such a point by chance, but the theorem does not promise it.

How is Rolle's theorem used to count roots?

Between two roots of $f$ there is a root of $f^{'}$ . So if $f^{'}$ has $k$ real roots, $f$ has at most $k + 1$ ; if $f^{'}$ never vanishes, $f$ has at most one real root, as for $x^{7} + x^{3} + λ$ .

How do you prove an inequality using the mean value theorem?

Write the difference $f (b) - f (a)$ as $(b - a) f^{'} (c)$ , then bound $f^{'} (c)$ using the behaviour of $f^{'}$ on $(a, b)$ . If $f^{'}$ is monotonic, the bounds are $f^{'} (a)$ and $f^{'} (b)$ , which gives inequalities like those for $tan^{- 1}$ and $ln$ .

How are Rolle's theorem and the mean value theorem asked in JEE Main?

JEE Main usually asks for the value of $c$ in Rolle's theorem or the mean value theorem for a polynomial, exponential or logarithmic function, or for an unknown coefficient that makes Rolle's theorem hold. Remember that for a quadratic the mean value point is the midpoint.

What mean value theorem problems appear in JEE Advanced?

JEE Advanced combines the theorems with given values of $f$ to conclude something about $f^{'}$ or $f^{''}$ , uses Rolle on a cleverly built auxiliary function such as $e^{- x} f (x)$ , counts real roots, and proves inequalities, sometimes through Cauchy's form of the theorem.

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1. Rolle's Theorem

1.1 Geometrical meaning

1.2 Why each condition is needed

1.3 Algebraic meaning: roots of f and f′

2. Lagrange's Mean Value Theorem

2.1 Geometrical meaning

2.2 Proof

2.3 Alternative (θ) form

3. Consequences and Applications

3.1 Constant and increasing functions

3.2 Proving inequalities

4. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What is Rolle's theorem?

What is Lagrange's mean value theorem?

What is the difference between Rolle's theorem and the mean value theorem?

What happens if a condition of Rolle's theorem fails?

How is Rolle's theorem used to count roots?

How do you prove an inequality using the mean value theorem?

How are Rolle's theorem and the mean value theorem asked in JEE Main?

What mean value theorem problems appear in JEE Advanced?

Ready to master Application Of Derivatives?

1.3 Algebraic meaning: roots of $f$ and $f^{'}$

2.3 Alternative ( $θ$ ) form