Mathematics
Session
Applications of Derivatives - 3
Session Objectives

Rolle’s Theorem

Geometrical Meaning

Lagrange’s Mean Value Theorem

Geometrical Meaning

Approximation of Differentials

Class Exercise
Rolle’s Theorem
Let f(x) be a real function defined in the closed
interval [a, b] such that
(i) f(x) is continuous in the closed interval [a, b]
(ii) f(x) is differentiable in the open interval (a, b).
(iii) f(a) = f(b)
Then, there is a point c in the open interval
(a, b), such that ƒ  c  = 0
Geometrical Meaning
Y
Y
f(a)
f(a) = f(b)
A
f(b)
B
II
f(b)
B
f(a)
f(a) = f(b)
I
O
x=a
x=b
Figure (1)
X
IV III
O
x=a
Figure (2)
x=b
ƒ  c  = 0
There will be at least one point with in [a, b] where
tangent of the curve will be parallel to x-axis.
X
Example - 1
Verify Rolle’s theorem for the function
f(x) = x2 – 8x + 12 on the interval [2, 6].
Solution :
We have f(x) = x2 – 8x + 12
(1) Given function f(x) is polynomial function.
\ f(x) is continuous on [2, 6]
(2) f'(x) = 2x – 8 exists in (2, 6)
\ f(x) is differentiable in (2, 6)
Solution
(3) f(2) = 22 – 8 x 2 + 12 = 0 and f(6) = 62 – 8 x 6 + 12 = 0
\ f(2) = f(6)
\ All the conditions of Rolle’s theorem is satisfied.
\ Three exists some c  2, 6 such that f'(c) = 0
 2c - 8 = 0  c = 4  2, 6
Hence, Rolle’s theorem is verified.
Example - 2
Using Rolle’s theorem, find the points on the curve
y = x(x - 4), x  [0, 4], where the tangent is parallel to x-axis.
Solution: (1) Being a polynomial function, the given function is
continuous on [0, 4].
2
y = 2(x - 2) exits in 0, 4.
\ Function is differentiable in (0, 4).
3
y x = 0 = 0 and y x = 4 = 0
Con.
\
All conditions of Rolle’s theorem are satisfied.
So,  x  0, 4 such that
ƒ (x) = 0  2(x - 2) = 0  x = 2
\ y = 2(2 - 4)  y = - 4
Required point is (2, –4)
Lagrange’s Mean Value Theorem
Let f(x) be a function defined on [a, b] such that
(i) it is continuous on [a, b].
(ii) it is differentiable on (a, b).
Then,there exists a real number c  (a, b)
such that ƒ  c  =
ƒ b  - ƒ  a
b-a
Geometrical Meaning
B(b, f(b))
Y
(a, f(a))
A
O
D

C  (c, f(c))
F
E
X
Geometrical Meaning
From the triangle AFB,
tan =
ƒ b  - ƒ  a
BF
 tan =
AF
b-a
By Lagrange’s Mean Value theorem,
ƒ  c  =
ƒ  b  - ƒ  a
b-a
 tan = ƒ  c 
 Slope of the chord AB = Slope of the tangent at  c, ƒ  c  
Example - 3
Verify Lagranges Mean Value theorem for
the function f(x) = x2 – 3x + 2 on [-1, 2].
Solution :
(1) The function f(x) being a polynomial
function is continuous in [-1, 2].
(2) f'(x) = 2x – 3 exists in (-1, 2)
\
f(x) is differentiable in (-1, 2)
Solution
So, there exists at least one c  -1, 2 such that
f'  c  =
f 2 - f -1
2 - -1
 2c - 3 = 0-6
2+1
 2c - 3 = -2  c =
1
 -1, 2 
2
Hence, Lagrange's mean value theorem is verified.
Example - 4
Using Lagrange’s mean value theorem, find the point on the curve
y = x3 - 3x,
and (2, 1).
where tangent is parallel to the chord joining (1, –2)
Solution: (1) The function being a polynomial function is continuous
on [1, 2].
2 
y  = 3x2 - 3 exists in 1, 2 
\ Function is differentiable in (1, 2).
So,  x  1, 2
such that tangent is parallel to chord joining
(1, –2) and (2, 1)
Contd.
\ ƒ (x) =


ƒ(2) - ƒ(1)
2 - (-2)
 3 x2 - 1 =
2-1
1
 x2 - 1 =
4
7
7
 x2 =  x = ±
3
3
3
3
 7 2

7
\ y =   - 3 × ± 

3 
3

2 7
\y =+
3 3

\ The points are 


7
2 7  7 2 7
, ,
 , .
3
3 3  
3 3 3 
Approximation of Differentials
As by the definition of
dy
dx
y dy
=
dx
δx 0 x
lim
Hence, for small increment in x, change in y will be
 dy 
y = 
 × x
dx


 dy 
Hence, y + y = y + 
 .x
dx


Example - 5
Using differentials, find the approximate
value of 37.
Solution :
Let y = x
Taking x = 36, Δx = 1  x + Δx = 37
Δy = x + Δx - x
 Δy = 37 - 6
 37 = 6+Δy
Contd.
Δy =
dy
Δx
dx
 Δy =

1
2 x
×1

dy
1 
y
=
x

=


dx
2
x


1
1

 0.08
2  6 12
\ 37  6  y
= 6 + 0.08 = 6.08
Example - 6
Using differentials, find the approximate
value of 29 
1
3
Solution :
Let y = 3 x
Taking x = 27, Δx = 2  x + Δx = 29
Δy = 3 x + Δx - 3 x
 Δy = 3 29 - 3
 3 29 = 3+ Δy
Contd.
2
Δy= dy Δx= 1 x 3 ×2
3
dx
2
2
= 27 3
3
=
2
2
=
= 0.074
3×9 27
\ 3 29 =3+0.074 =3.074


1

3  dy = 1 
y
=
x

2
dx

3x 3 
Solution
So, there exists at least one c  3, 5 where tangent is
parallel to chord joining (3, 0) and (5, 4).
\ ƒ (c) =
ƒ(5) - ƒ(3)
5-3
4-0
2
 c = 4  3, 5
 2(c - 3) =
At x = 4
y  (4  3)2  1
\ Required point is (4, 1)
Class Exercise - 5
Using differentials, find the approximate
value of
82 
1
4 up to 3 places of decimals.
Solution :
Let y = 4 x
Taking x = 81, Δx =1  x + Δx = 82
Δy = 4 x + Δx - 4 x
 Δy = 4 82 - 3
 4 82 =3+ Δy
Solution
3
dy
1
Δy =
Δx = x 4 ×1
dx
4


1

dy
1 
4
y
=
x

=


3
dx


4

4x 
3
1
= 81 4
4
=
=
1
4×27
25
0.926
=
= 0.00926
100×27
100
\ 4 82 = 3 + 0.00926 = 3.00926  3.009
Thank you