Differential Calculus

• Understand Rolle’s Mean Theorem.

• Understand Lagrange mean Value Theorem and Cauchy’s

mean Value Theorem

• Distinguish between three Mean Value theorem

• Understand geometrical interpretation of Mean Value

Theorem

• Apply Mean Value theorem for solving Examples

➢ Continuity of function
➢Differentiability of function
➢Derivative of standard Functions
 Statement :Let f(x) be a function that satisfies the following three conditions:
❖ f(x) is continuous on the closed interval [a, b].
❖. f(x) is differentiable on the open interval (a, b).
❖ f (a) = f (b)
Then there is at least one value ‘c’ of x in (a, b)
such that f ′(c) = 0.
 Y

▪This theorem states that if the graph of a f ′(c) = 0

differentiable function intersects the x-axis at

two places, a and b there must be at least one
place where the tangent line is horizontal. a c b
f (a) = f (b) X

▪There exist a point 𝑐 ∈ (𝑎, 𝑏) where tangent to

the graph of function is horizontal
Ex1: Verify Rolle’s Theorem for f x = x 2 − 5x + 4 on the given
interval [1, 4]. then find all numbers c that satisfy the conclusion of
Rolle’s Theorem.

Sol: f(x) is polynomial function

i) f(x) is continuous on [1,4]
ii) f(x) differentiable on (1,4)
iii) f(1)=0 and f(4)=0
All three conditions of rolle’s theorem are Satisfied.
Therefore the derivative of f x must vanish for
at least one value of 𝑥𝜖(1, 4)
𝑓 ′ x = 2x − 5 vanish for some 𝑥 = 𝜖(1, 4
Now we apply rolle’s theorem,
for some cϵ 1,4
f′ c = 0
f ′ x = 2x − 5
f ′ c = 0 ⇒ 2c − 5 = 0
5
⇒c=
2

c = 2.5 which is lies in interval (1, 4)

Rolle’s theorem verified
Ex:2 Verify rolle’s theorem for f x = x x + 3 e−1Τ2x in −3, 0 then
find all numbers c that satisfy the conclusion of Rolle’s Theorem.
Sol: given f x = x x + 3 e−1Τ2x
i) f x is continuous on [−3,0].
ii)f x differentiable on (−3, 0)
iii) f −3 = f 0
All three conditions of rolle’s theorem are Satisfied
therefore the derivative of f x must vanish for at least one value of xϵ(−3, 0)
Now we apply rolle’s theorem
f ′ c = 0 for c ∈ −3 , 0
f x = x x + 3 e−1Τ2x

1 −1Τ2x
f′ x = 2x + 3 e−1Τ2x + (x 2 + 3x)(− e )
2
 1
f ′ c = 0 ⟹ 2c + 3 e−1Τ2c + (c 2 + 3c)(− e−1Τ2c )=0
2

−1
⇒ 2c + 3 + c 2 + 3c =0
2

c2 −3c
2c + 3 + − + =0
2 2

c2 c
− + +3=0
2 2

−c 2 + c + 6 = 0
c2 − c − 6 = 0
⇒ c = 3 or c = −2
∴ c = −2 ∈ −3, 0
 Ex:3 Verify rolle’s theorem f x = tanx in 0, π then find all numbers c
that satisfy the conclusion of Rolle’s Theorem.

So given f x = tanx
π
i) f x = tanx is discontinuous at x = ϵ 0, π
2

therefore the Rolle’s theorem is not applicable here

Ex:4 verify rolle’s theorem f x = x on interval [-1,1]
then find all numbers c that satisfy the conclusion of Rolle’s Theorem.

Sol: given f x = x
i) f x = x is continuous on −1, 1
ii) f x = x is not differential at x = 0
The Rolle’s theorem fails here
because f(x) is not differentiable over the whole (−1,1).
 Multiple choice Questions`
Ex. 1:Rolle’s Theorem tells about the
a) Existence of point c where derivative of a function becomes zero
b) Existence of point c where derivative of a function is positive
c) Existence of point c where derivative of a function is negative
d) Existence of point c where derivative of a function is either positive or negative

Ex.2 :Rolle’s theorem is applicable to the

a) Functions differentiable in closed interval [a, b] and continuous in open
interval (a, b) only and having same value at point ‘a’ and ‘b’
b) Functions continuous in closed interval [a, b] only and having same value
at point ‘a’ and ‘b’
c) Functions continuous in closed interval [a, b] and differentiable in
open interval (a, b) only and having same value at point ‘a’ and ‘b’
d) Monotonically Increasing functions
 Multiple choice Questions`

Ex.3: Value of c where f(x) = Sin(x) satisfied condition of rolle’s theorem on [0, 𝜋]
a) 𝜋 b) 𝜋 Τ 2 c) 𝜋 Τ 6 d) 𝜋 Τ 4

Ex 4: In which interval Rolle’s Theorem is applicable for function f(x) = x 2

a) interval [0, 1]
b) interval [−2, 1]
c) interval [−1, 1]
d) interval [2, 3]

Ex 5:For 𝑦 = −𝑥 2 + 2𝑥 there exists a value of c in the interval (-2, 2)

such that f ′ c = 0
a) True
b) False
 Multiple choice Questions`
Ex.6: For f(x) = sin(x), How many points exist such that f'(c) = 0
in the interval [0, 2π].

a) 1 b) 3 c) 2 d) 4

Ex 7: Find value of c(a point in a curve where slope of tangent to curve is zero)
where f(x) = tan(x) in interval [0, π].
a) 𝜋 Τ2
b) Rolle’s Theorem is not applied, because function is not
continuous in interval [0, π].
c) Rolle’s Theorem is not applied, because function is not
differential in interval [0, π].
d) Function is both continuous and differentiable but
Rolle’s theorem is not applicable as f(0) ≠ f(π)
 Multiple choice Questions`

Ex 8: f(x) = Sin(2x), is continuous over interval [0, π] and

differentiable over interval (0, 𝜋) and 𝑐 ∈ (0, 𝜋) is given by
a) 𝜋 b) 𝜋 Τ 2 c) 𝜋 Τ 4 d) 𝜋 Τ 8
Ex.1 Ex.2 Ex.3 Ex.4

a) c) b) c)

Ex.5 Ex.6 Ex.7 Ex.8

a) c) b) c)

19
Statement:
Let f x be a function that satisfies the following two conditions:
❖ f is continuous on the closed interval [a, b].
❖ f is differentiable on the open interval (a, b).
Then there exist at least one point c in (a, b) such that
f b −f(a)
f′ c =
b−a
If a curve 𝐴𝐵 has tangent at each of its points then there exits at least one
point ‘c’ on the curve, the tangent at which is parallel to the chord 𝐴𝐵
y Tangent parallel to chord
Slope of tangent:

f  (c) B

Slope of chord
f (b ) − f ( a )
b−a
A

x
0 a c b
➢ If we assume that f(t) represents the position of a body moving along
𝑓 𝑏 −𝑓(𝑎)
a line, depending on the time t, then the ratio of is the
𝑏−𝑎

average velocity of the body in the period of time b−a.

➢ Since f′(t) is the instantaneous velocity, this theorem means that there
exists a moment of time c, in which the instantaneous speed is equal
to the average speed.
Ex:1 Verify Lagrange’s mean value theorem for the function
𝑓 𝑥 = 𝑥 2 − 3𝑥 +5 on [1,4]. If the theorem holds, find a point c
satisfying the conditions of the theorem.

Sol: 𝑓(𝑥) is polynomial function

∴ 𝑓(𝑥) is continuous on 1,4 and
𝑓 𝑥 is differential on (1,4)
Apply Lagrange’s mean value theorem
𝑓 ′ (x)=2𝑥 − 3
There exists c ∈ (1,4) such that
𝑓 4 −𝑓(1)
𝑓′ 𝑐 =
4−1
 9−3
2c − 3 =
3

2𝑐 − 3 = 2
2𝑐 = 5
5
𝑐=
2

5
Point 𝑐 = lies in the interval 1,4 .
2

Lagrange’s mean value theorem Verified.

 2
Ex.2: Apply Lagrange’s Mean value theorem for 𝑓 𝑥 = 𝑥 Τ3 on
interval [-1,2] .If the theorem holds, find a point c satisfying the
conditions of the theorem.
2Τ
Sol: Given 𝑓(𝑥) = 𝑥 3 which is polynomial function
∴ 𝑓(𝑥) is continuous on −1,2
2 −1Τ
𝑓′ 𝑥 = 𝑥 3
3

2 1
𝑓′ 𝑥 =
3 𝑥 1ൗ3

𝑓(𝑥) is not differentiable at 𝑥 = 0

Lagrange’s mean value theorem is not applicable.
Ex.3: Apply Lagrange’s Mean value theorem for 𝑓 𝑥 = 𝑙𝑜𝑔𝑥 on interval
[1,e]. If the theorem holds, find a point c satisfying the conditions of
the theorem.
Sol: 𝑓(𝑥) is continuous on 1, 𝑒
𝑓(𝑥) is differentiable on 1, 𝑒
Now we apply LMVT,
1
𝑓′ 𝑋 =
𝑥

There exists c ∈ (1, 𝑒) such that

𝑓 𝑒 −𝑓(1)
𝑓′ 𝑐 =
𝑒−1

1 𝑙𝑜𝑔𝑒−𝑙𝑜𝑔1
=
𝑐 𝑒−1

1 1−0
=
𝑐 𝑒−1
1 1
=
𝑐 𝑒−1

⇒ 𝑐 = 𝑒 − 1𝑐 = 𝑐𝑒 −
= 1𝑒 −
∈ (1,
1 ∈𝑒)(1, 𝑒)
Ex.4: Apply Lagrange’s Mean value theorem for 𝑓 𝑥 = 𝑒 𝑥 on interval
[0,1]. If the theorem holds, find a point c satisfying the conditions
of the theorem.
Sol: 𝑓 𝑥 = 𝑒 𝑥
𝑓(𝑥) is continuous on 0,1
𝑓(𝑥) is differentiable on 0,1
Now we apply LMVT,
𝑓′ X = 𝑒𝑥
There exists c ∈ (0,1) such that
𝑓 1 −𝑓(0)
𝑓′ 𝑐 =
1−0

𝑒 1 −𝑒 0
𝑒𝑐 =
1

𝑒𝑐 = 𝑒 − 1
Taking log on the both side
⇒ c = log(𝑒 − 1)
 𝑏−𝑎 𝑏−𝑎
Ex.5: Prove that < 𝑡𝑎𝑛−1 𝑏 − 𝑡𝑎𝑛−1 𝑎 < where 0 < 𝑎 < 𝑏 < 1.
1+𝑏2 1+𝑎2

𝜋 3 4 𝜋 1
hence, show that + < 𝑡𝑎𝑛−1 < + .
4 25 3 4 6

Sol : Let 𝑓 𝑥 = 𝑡𝑎𝑛−1 𝑥 in 𝑎, 𝑏 , 0 < 𝑎 < 𝑏 < 1

𝑓(𝑥) is continuous on 𝑎, 𝑏
1
As 𝑓 ′ 𝑥 = , 𝑖𝑛 (𝑎, 𝑏)
1+𝑥 2

𝑓(𝑥) is differentiable in 𝑎, 𝑏
By Lagrange’s Mean Value Theorem, there exists 𝑐 ∈ (𝑎, 𝑏) such that
𝑓 𝑏 −𝑓(𝑎)
𝑓′ 𝑐 =
𝑏−𝑎

We have 0 < 𝑎 < 𝑏 < 1 and 𝑎 < 𝑐 < 𝑏

1 + 𝑎2 < 1 + 𝑐 2 < 1 + 𝑏2
1 1 1
> >
1+𝑎2 1+𝑐 2 1+𝑏2

1 𝑡𝑎𝑛−1 𝑏−𝑡𝑎𝑛−1 𝑎 1
< <
1+𝑏2 𝑏−𝑎 1+𝑎2
 𝑏−𝑎 𝑏−𝑎
Hence < 𝑡𝑎𝑛−1 𝑏 − 𝑡𝑎𝑛−1 𝑎 <
1+𝑏2 1+𝑎2
4
Let 𝑎 = 1, 𝑏 =
3

4 4
−1 4 −1
3
4 2
< 𝑡𝑎𝑛−1 − 𝑡𝑎𝑛−1 1 < 3

1+ 3 1+ 1 2
3

1 1
4 𝜋
3
16 < 𝑡𝑎𝑛−1 − < 3
1+ 9 3 4 1+1

3 4 𝜋 1
< 𝑡𝑎𝑛−1 − <
25 3 4 6

Hence
𝜋 3 4 𝜋 1
+ < 𝑡𝑎𝑛−1 < +
4 25 3 4 6
 𝑎−𝑏 𝑎−𝑏
Ex.6: Prove that > 𝑐𝑜𝑠 −1 𝑏 − 𝑐𝑜𝑠 −1 𝑎 > where 0 < 𝑎 < 𝑏 < 1.
1−𝑎2 1−𝑏2

𝜋 1 3 𝜋 1
hence, show that + > 𝑐𝑜𝑠 −1 > − .
3 5 3 5 3 8
Sol : Let 𝑓 𝑥 = 𝑐𝑜𝑠 −1 𝑥 in 𝑎, 𝑏 , 0 < 𝑎 < 𝑏 < 1
1
𝑓′ 𝑥 = − , 𝑒𝑥𝑖𝑠𝑡𝑠 𝑖𝑛 (𝑎, 𝑏)
1−𝑥 2

By Lagrange’s Mean Value Theorem, 𝑐 ∈ (𝑎, 𝑏) such that

𝑓 𝑏 −𝑓(𝑎) 1 𝑐𝑜𝑠 −1 𝑏−𝑐𝑜𝑠 −1 𝑎
𝑓′ 𝑐 = ⇒ − =
𝑏−𝑎 1−𝑥 2 𝑏−𝑎

We have 𝑎 < 𝑐 < 𝑏

1 1 1
< <
1−𝑎2 1−𝑐 2 1−𝑏2

1 −1 −1
− > >
1−𝑎2 1−𝑐 2 1−𝑏2

−1 𝑐𝑜𝑠 −1 𝑏−𝑐𝑜𝑠 −1 𝑎 −1
> >
1−𝑎2 𝑏−𝑎 1−𝑏2

𝑎−𝑏 𝑎−𝑏
> 𝑐𝑜𝑠 −1 𝑏 − 𝑐𝑜𝑠 −1 𝑎 >
1−𝑎2 1−𝑏2
 1 3
Put 𝑎 = , 𝑏 =
2 5

1 3 1 3
− 3 1 −
2 5
> 𝑐𝑜𝑠 −1 − 𝑐𝑜𝑠 −1 > 2 5
1 5 2 9
1−4 1−
25

1 1
− 3 𝜋 −
10
> 𝑐𝑜𝑠 −1 − > 10
3 5 3 16
4 25

−1 3 𝜋 −1
> 𝑐𝑜𝑠 −1 − >
5 3 5 3 8

Hence
𝜋 1 3 𝜋 1
− > 𝑐𝑜𝑠 −1 > −
3 5 3 5 3 8
 Multiple choice Questions`
Ex.1: Lagrange’s Mean Value Theorem tells about the
a) Existence of point c in a curve where slope of a tangent to curve is
equal to the slope of line joining two points in which curve is continuous
and differentiable
b) Existence of point c in a curve where slope of a tangent to curve is equal to
zero
c) Existence of point c in a curve where curve meets y axis
d) Existence of point c in a curve where curve meets x axis
Ex.2: If f(a) is equals to f(b) in Lagrange’s Mean Value Theorem, then it becomes
a) Lebniz Theorem
b) Rolle’s Theorem
c) Taylor Series of a function
d) Leibnit’z Theorem
 Multiple choice Questions`
Ex.3: Mean Value theorem is applicable to the
a) Functions differentiable in closed interval [a, b] and continuous in open
interval (a, b)
b) Functions continuous in closed interval [a, b] only and having same value
at point ‘a’ and ‘b’
c) Functions continuous in closed interval [a, b] and differentiable in open
interval (a, b)
d) Functions differentiable in open interval (a, b) only and having same value
at point ‘a’ and ‘b’
Ex.4: Find the point c in the curve 𝑓 𝑥 = 𝑥 3 + 𝑥 2 + 𝑥 + 1 in the interval
[0, 1] where slope of a tangent to a curve is equals to the slope of a line
joining (0,1)
a) 0.64 b) 0.54 c) 0.44 d) 0.34
 Multiple choice Questions`
Ex.5: Find point c between [0,2] where, the function f(x) = x2 − x
satisfies condition of Lagrange's mean value theorem
a) 1.5 b) 0.5 c) 1 d) -2.5

Ex.6: Find point c between [0,1] where, the function f(x) = ex satisfies
condition of Lagrange's mean value theorem
a) log(𝑒 + 1) b) log(𝑒 − 1) c) 𝑒 d) 𝑒 − 1

Ex.7: Lagranges mean Value theorem is applicable to function

𝑓(𝑥) = |x| in [−1,1]
a) True
b) False
 Multiple choice Questions`
𝑏−𝑎 𝑏−𝑎
Ex.8: Using Identity 1+𝑏 2 < 𝑡𝑎𝑛 −1
𝑏 − 𝑡𝑎𝑛 −1
𝑎 < 1+𝑎2
where 0 < 𝑎 < 𝑏 < 1.
𝜋 3 4 𝜋 1
find value of a and b such that + 25 < 𝑡𝑎𝑛−1 3 < 4 + 6.
4

4 4
a) a = ,b=1 b) a = 1 , b =
3 3
3 4
c) a = 1 , b = 4 d) a = 1.5 , b = 3

𝑏−𝑎 𝑏−𝑎
Ex.9: Using Identity < 𝑠𝑖𝑛−1 𝑏 − 𝑠𝑖𝑛−1 𝑎 < find value of
1−𝑎2 1−𝑏2
𝜋 1 1 𝜋 1
a and b such that − < 𝑠𝑖𝑛−1 < − .
6 2 3 4 6 15

1
a) a = 1 , b = 4
1
b) a =
2
, b = 14
1 1 1
c) a = 3 , b = 4 d) a =
4
,b=1
 Ex.1 Ex.2 Ex.3 Ex.4

a) b) c) b)

Ex.5 Ex.6 Ex.7 Ex.8 Ex.9

c) b) b) b) b)

39
Statement: If f(x) and g(x) are any two functions of x such that they are ,
➢ Continuous on closed interval a, b
➢ Differential on open interval a, b
➢ g ′ x ≠ 0 ∀ x ∈ a, b
Then there exists at least one point x = c in (a, b) such that,

f′ c f b −f(a)
=
g′ c g(b)−g(a)
Ex.1: Verify Cauchy’s mean value theorem for 𝑓 𝑥 = 𝑒 𝑥 and 𝑔 𝑥 = 𝑒 −𝑥 in
𝑎, 𝑏 .

Sol: Given 𝑓 𝑥 = 𝑒 𝑥 and 𝑔 𝑥 = 𝑒 −𝑥

𝑓 𝑥 and 𝑔 𝑥 both are continuous on 𝑎, 𝑏 and differential on 𝑎, 𝑏 .
And 𝑔′ 𝑥 ≠ 0 ∀ 𝑥 ∈ 𝑎, 𝑏
Here 𝑔 𝑥 = 𝑒 −𝑥 𝑔′ 𝑥 = −𝑒 −𝑥
𝑓 𝑥 = 𝑒𝑥 𝑓′ 𝑥 = 𝑒𝑥
By Cauchy’s Mean Value Theorem , we have
𝑓′ 𝑐 𝑓 𝑏 −𝑓(𝑎)
=
𝑔′ 𝑐 𝑔(𝑏)−𝑔(𝑎)

𝑒𝑐 𝑒 𝑏 −𝑒 𝑎
=
−𝑒 −𝑐 𝑒 −𝑏 −𝑒 −𝑎
 2𝑐 𝑒 𝑏 −𝑒 𝑎
−𝑒 = 1 1
𝑏 −𝑒𝑎 𝑒

2𝑐 𝑒 𝑏 −𝑒 𝑎
−𝑒 = 𝑒𝑎 −𝑒𝑏
𝑒𝑏 𝑒𝑎
𝑒 2𝑐 = 𝑒 𝑎+𝑏

2𝑐 = 𝑎 + 𝑏
𝑎+𝑏
𝑐= which lies in (a,b)
2

Cauchy’s mean Value Theorem Verified.

 Ex.2: Verify Cauchy’s mean value theorem for 𝑓 𝑥 = 𝑥 3 and
𝑞 𝑥 = 𝑥 4 in 0,2 .
Sol: Given 𝑓 𝑥 = 𝑥 3 and 𝑞 𝑥 = 𝑥 4 are polynomial function
𝑓 𝑥 and 𝑔 𝑥 both are continuous on 0,2 and differentiable on
0,2 .
𝑓 ′ 𝑥 = 3𝑥 2 and𝑔′ 𝑥 = 4𝑥 3
And 𝑔′ 𝑥 ≠ 0; ∀ 𝑥 ∈ 0,2
By CMVT, we have
𝑓′ 𝑐 𝑓 𝑏 −𝑓(𝑏) 𝑓 2 −𝑓(0)
= =
𝑔′ 𝑐 𝑔(𝑏)−𝑔(𝑎) 𝑔(2)−𝑔(0)

3𝑐 2 8−0 1
= =
4𝑐 3 16−0 2

31 1
=
4𝑐 2

3
⇒ 𝑐 = ∈ 0,2
2

Hence CMVT is verified.

Ex.3: Verify Cauchy’s mean value theorem for 𝑓 𝑥 = 𝑠𝑖𝑛𝑥 and
𝜋
𝑔 𝑥 = 𝑐𝑜𝑠𝑥 in − , 0 .
2
Sol: Given 𝑓 𝑥 = 𝑠𝑖𝑛𝑥 and 𝑔 𝑥 = 𝑐𝑜𝑠𝑥
𝜋
𝑓 𝑥 and 𝑔 𝑥 both are continuous on − , 0
2
𝜋
differentiable on − , 0 .
2
𝑓 ′ 𝑥 = 𝑐𝑜𝑠𝑥 and 𝑔′ 𝑥 = 𝑠𝑖𝑛𝑥
𝜋
And 𝑔′ 𝑥 ≠ 0; ∀ 𝑥 ∈ − , 0
2

By CMVT, we have
𝑓′ 𝑐 𝑓 𝑏 −𝑓(𝑏)
=
𝑔′ 𝑐 𝑔(𝑏)−𝑔(𝑎)

𝜋
𝑐𝑜𝑠𝑐 𝑓 0 −𝑓(− 2 ) 0+1
= 𝜋 = =1
−𝑠𝑖𝑛𝑐 𝑔(0)−𝑔(− 2 ) 1−0
−𝜋 𝜋
𝑡𝑎𝑛𝑐 = −1 ⇒ 𝐶 = ∈ − ,0
4 2

Hence CMVT is verified.