This ppt covers following topics of Unit - 2 of B.Sc. 2 Mathematics Rolle's Theorem , Lagrange's mean value theorem , Mean value theorem & its example .

1. B.SC. - II (MATHEMATICS)
ADVANCED CALCULUS
ROLLE’S THEOREM
By Preeti Shrivastava

2. SYNOPSIS
 Rolle’s Theorem
 Rolle’s theorem’s Example
 Lagrange’s Mean Value Theorem (first mean value
theorem)
 Example of Lagrange’s mean value theorem
 Cauchy’s Mean Value Theorem(second mean value
theorem)
 Example of Cauchy’s mean value theorem

3. ROLLE’S
THEOREM

4. Statement:-
If f(x) is a function of the variable x such that :-
 f(x) is continuous in the closed interval [a,b].
 f(x) is differentiable for every point in the open
interval (a,b).
 f(a) = f(b), then there is at least one point c such that
f’(c) =0.

5. Proof:-
 We need to know Close Interval and Open Interval:
 Open Interval:- An open interval is an interval that
does not include its two mid- point .The open interval
{x:a<x<b} is denoted by (a,b).
 Close Interval:- A closed interval is an interval that
include all its limit point. A finite number a and b
then the interval {x:a<x<b} is denoted by [a,b].
 Case 1:- If f(x) is constant then,
f(x) = k (constant)
f(a) = k
f(b) = k

6.  so, f(a) = f(b) f(a) f(b)
since, f(x) =k a b
therefore f’(x) =0
and f’(c) =0
 Case 2:- If function f(x) will first increase and then decreases.
Then at some point C consequently f(x) will be maximum.
Then f(c-h) < f (c) and f(c+h) < f(c)
f(c-h) – f(c) < 0 and f(c+h) - f(c) < 0
Taking limit on both sides:



x
h
cfhcf

 )()(
0 0
)()(


h
cfhcf
and

7. Case 3:- - If function f(x) w ill first decrease and then
increases. Then at some point C consequently f(x)
w ill be minimum.
Then
 – > 0

Taking limits on both sides:-




8. Verify Rolle’s theorem in the interval
(2,4] for the
function 𝒇 𝒙 = 𝒙 𝟐
− 𝟔𝒙 + 𝟖
Solution:-
Given that ,
𝑓 𝑥 = 𝑥2
− 6𝑥 + 8
[a,b] =[2,4]
a =2, b=4.
By Rolle’s theorem
∵ 𝑓 𝑥 = 𝑥2
− 6𝑥 + 8
∴ 𝑓′
𝑥 = 2𝑥 − 6
𝑓’(𝑐) = 2𝑐 − 6

9. ∵ 𝑓’(𝑐) = 0
∴ 2𝑐 − 6 = 0
2𝑐 = 6
𝐶 = 3 ∈ [2,4]
Here a<c<b
Hence Rolle ’s Theorem is verified.

10. LAGRANGE’S
MEAN VALUE
THEOREM

11. Statement:-
If f(x) is a function of the variable x and :-
(1) f(x) is continuous in the closed interval [a,b].
(2) f(x) differentiable in the open interval (a,b), then there
exists at least one point c in the open interval (a,b) such
that
Proof:-
If f(x) is a function of the variable x.
Where A is constant to be determined such that
∵ f(x) is continuous in the closed interval [a,b).
∴ ∅(x)is also continuous in the closed interval [a,b].
Again, ∵ f(x) differentiable in the open interval (a,b).
∴ ∅(x) is also differentiable in the open interval (a,b).

12. Hence by the Rolle’s theorem,
Then there exist at least one point c such that ∅’ c = 0
…. III
From equation (I)
∅(x) = f(x) + A.x
∅(a) = f(a) + A.a ….. IV
∅(b) = f(b) + A.b …… V
From equation (II)
∅(𝑎) = ∅(𝑏)
𝑓(𝑎) + 𝐴. 𝑎 = 𝑓(𝑏) + 𝐴. 𝑥
𝐴. 𝑎 – 𝐴. 𝑏 = 𝑓(𝑏) – 𝑓(𝑎)
−𝐴(𝑏 − 𝑎) = 𝑓(𝑏) – 𝑓(𝑎)
−𝐴 =
𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
… . . (𝑉𝐼)
Again, from equation (I)
∅(𝑥) = 𝑓(𝑥) + 𝐴𝑥
∅’(𝑥) = 𝑓’(𝑥) + 𝐴

13. …… VII
Example:-
Verify Lagrange’s mean value theorem for the function
in the interval [2,4].
Solution:-
It is given that

14. 𝑓’(𝑥) =
𝑥
𝑥2−4
… … (𝐼𝐼)
∵ 𝑓(𝑥) = 𝑥2 − 4
∴ 𝑓(𝑎) = 𝑓(2) = 22 − 4 = 0
𝐹(𝑏) = 𝑓(4) = 42 − 4 = 12
By Lagrange’s mean value theorem,
𝑓′
𝑐 =
𝑓 𝑏 − 𝑓(𝑎)
𝑏−𝑎
𝑐
𝑐2−4
=
12−0
4−2
𝑐
𝑐2−4
=
12
2

15.  Squaring on both side
bCa
C
C
C
CC
CC
C
C









462
6
6
488
41248
48124
4
12
4
2
2
22
22
2
2

Hence Lagrange’s Theorem is Verified.

16. CAUCHY’ MEAN
VALUE THEOREM

17. Statement:-
If two function f(x) and g(x) are:-
Continuous in the Closed Interval [a,b].
Differentiable in the Open Interval [a,b].
Then there is at least one point c such that:-
where a<c<b
 Proof:-
Now consider a function F(x) defined by:-
F(x)= f(x)+ A(x) ……….(I)
Where A is a constant to be determined
)()(
)()'(
ab
aFbF



 )('
)('
c
cF

18. such that :
F(a) = F(b) ……….(II)
Since f(x) are continuous in closed interval
[a,b].
So, F(x) is also continuous in closed interval
[a,b].
 Again,
f(x) are differentiable in Open interval
(a,b).
So, F(x) is also differentiable in the open
interval (a,b)
Then all condition of Rolle’s Theorem will be
satisfied .

19.  Hence by the Rolle’s Theorem there at least
one Point C in
 open interval (a,b) such that F’(c) =0
……(III)
from Equation (1):-
 F(x) = f(x) + A ∅(x)
 F(a) = f(a)+ A ∅(a) ……..(IV)
 F(b) = f(b)+ A ∅(b) ………(V)
 Now from Equation (II), we get:-
F(a) =F(b)

20. 𝑓 𝑎 + 𝐴 ∅ 𝑎 = 𝑓 𝑏 + 𝐴 ∅ 𝑏
𝐴 ∅ 𝑎 − 𝐴 ∅ 𝑏 = 𝑓 𝑏 − 𝑓(𝑎)
−𝐴 ∅ 𝑏 − ∅ 𝑎 = 𝑓 𝑏 − 𝑓(𝑎)
− 𝐴 =
𝑓 𝑏 − 𝑓(𝑎)
∅ 𝑏 − ∅(𝑎)
……….(VI)
Again from Equation (I) we get :
𝐹 𝑥 = 𝑓 𝑥 + 𝐴 ∅ 𝑥
𝐹′
𝑥 = 𝑓′
𝑥 + 𝐴 ∅ 𝑥
𝐹′
𝑐 = 𝑓′
𝑐 + 𝐴 ∅(𝑐)

21. Now From Equation (III)
Ans..
Hence Proved

22. EXAMPLE:-
VERIFY CAUCHY’S MEAN VALUE THEOREM FOR FUNCTION
AND IN INTERVAL [1,2]
It is given that,
𝑓 𝑥 = 𝑥2
+ 1
∅ 𝑥 = 2𝑥3
𝑎, 𝑏 = 1,2
Since a=1 , b=2
By the Cauchy’s Mean Value Theorem,
𝑓′ (𝑐)
∅′ (𝑐)
=
𝑓 𝑏 −𝑓(𝑎)
∅ 𝑏 −∅(𝑎)
……(I)
𝑓 𝑥 = 𝑥2
+ 1
𝑓 𝑎 = 𝑎2
+ 1
𝑓 𝑎 = 1 + 1

23. And 𝑓 𝑏 = 22
+ 1
𝑓 𝑏 = 4 + 1 = 5
∅ 𝑥 = 2𝑥3
∅ 𝑎 = 2𝑎3
∅ 𝑎 = 2 × 13
∅ 𝑎 = 2
And ∅ 𝑏 = 2 + 23
= 16
Since 𝑓 𝑥 = 𝑥2
+ 1
There fore 𝑓′
𝑥 = 2𝑥
So 𝑓′
𝑐 = 2𝑐
And ∅ 𝑥 = 2𝑥3
∅′
𝑥 = 6𝑥2
∅′
𝑐 = 6𝑐2

24. Hence from Equation (I) :
Since [a<c<b]
Clearly lies in (1,2)
Hence Cauchy’s Mean Value Theorem is
verified.

25. THANK YOU