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 .
UGC NET Paper 1 Mathematical Reasoning & Aptitude.pdf
Rolles theorem
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
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.
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
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.
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)
