Welcome
To Presentation
Application of
Derivatives
Session Objectives
 Increasing and Decreasing Functions
 Use of Derivative
 Maximum and Minimum
 Extreme and Critical points
 Theorem 1 and 2
 Greatest and Least Values
I n c r e a s in g
f u n c ti o n
a x 1 bx 2
X
Y
f ( x 1 )
f ( x 2 )
O
Increasing Function
( ) ( ) ( ) ( )1 2 1 2 1 2If x < x in a, b ƒ x < ƒ x for all x , x a, b⇒ ∈
Decreasing Function
D e c r e a s in g
fu n c tio n
a x 1 bx 2
X
Y
f (x 1 )
f (x 2 )
O
( ) ( ) ( ) ( )1 2 1 2 1 2If x < x in a, b ƒ x > ƒ x for all x , x a, b⇒ ∈
Use of Derivative
( )(i) If ƒ x >0 for all x (a, b) f(x) is increasing on (a,b).′ ∈ ⇔
( )(ii) If ƒ x <0 for all x (a, b) f(x) is decresing on (a,b).′ ∈ ⇔
Use of Derivative (Con.)
[ ]As tanθ>0 for 0 <θ<90°
( )
dy
ƒ x 0
dx
′⇒ = > for all x in (a, b).
Y = f(x)
T
X
Y
O T' a b
Figure 1
P
θ
Use of Derivative (Con.)
θ
Figure 2 T'
X
Y
T a
bP
O
[ ]As tanθ <0 for 90°<θ<180°
( )
dy
ƒ x 0
dx
′⇒ = < for all x in (a, b).
For the function f(x) = 2x3
– 8x2
+ 10x +
5,
find the intervals where
(a)f(x) is increasing
(b) f(x) is decreasing
Example-1
Solution
3 2ƒ(x)=2x - 8x +10x +5
2
ƒ (x)= 6x -16x +10′∴
2
=2(3x - 8x +5)
=2(3x - 5)(x -1)
ƒ (x)= 0 2(3x - 5)(x -1)= 0′ ⇒
5
x = ,1
3
∴
Solution Cont.
5
For 1< x < , ƒ (x) is negative
3
′
5
For x > ,ƒ (x) is positive
3
′
For x < 1, is positive.ƒ (x)=3(3x - 5)(x -1)′
ƒ(x)∴ is increasing for x < 1 and
5
x >
3
and it decreases for 5
1< x <
3
Maximum and Minimum
Maximum and Minimum
The point a is called the point of maximum of the function f(x).
In the figure, y = f(x) has maximum values at Q and S.
( ) ( )ƒ a > ƒ a+δIf and ( ) ( )ƒ a > ƒ a-δ for all small values of δ.
The point b is called the point of minimum of the function f(x).
In the figure, y = f(x) has minimum values at R and T.
( ) ( )ƒ b < ƒ b+δIf and ( ) ( )ƒ b < ƒ b-δ for all small values of δ.
Let ( )y = ƒ x be a function
Extreme Points
The points of maximum or minimum of a function
are called extreme points.
At these points, ( ) ( )ƒ x = 0, if ƒ x exists.′ ′
X
Y
O
( i)
P
fincreasing
fdecreasing
X
Y
O
( ii)
Q
fincreasing
fdecreasing
( )At P and Q ƒ x does not exit.′
Critical Points
The points at which or at which
does not exist are called critical points.
( )ƒ x = 0′ ( )ƒ x′
A point of extremum must be one of the critical
points, however, there may exist a critical point,
which is not a point of extremum.
Theorem - 1
Let the function be continuous in some
interval containing x0 .
( )y = ƒ x
( )ƒ x >0′ ( )ƒ x <0′(i) If when x < x0 and When
x > x0 then f(x) has maximum value at x = x0
( )ƒ x <0′ ( )ƒ x >0′(ii) If when x < x0 and When
x > x0 ,then f(x) has minimum value at x = x0
Theorem - 2
If x0 be a point in the interval in which y = f(x) is
defined and if ( ) ( )0 0ƒ x =0 and ƒ x 0′ ′′ ≠
( ) ( ) ( )if0 0i ƒ x is a maximum ƒ x <0′′
( ) ( ) ( )if0 0ii ƒ x is a minimum ƒ x >0′′
Thank you

Application of derivatives

  • 1.
  • 2.
  • 3.
    Session Objectives  Increasingand Decreasing Functions  Use of Derivative  Maximum and Minimum  Extreme and Critical points  Theorem 1 and 2  Greatest and Least Values
  • 4.
    I n cr e a s in g f u n c ti o n a x 1 bx 2 X Y f ( x 1 ) f ( x 2 ) O Increasing Function ( ) ( ) ( ) ( )1 2 1 2 1 2If x < x in a, b ƒ x < ƒ x for all x , x a, b⇒ ∈
  • 5.
    Decreasing Function D ec r e a s in g fu n c tio n a x 1 bx 2 X Y f (x 1 ) f (x 2 ) O ( ) ( ) ( ) ( )1 2 1 2 1 2If x < x in a, b ƒ x > ƒ x for all x , x a, b⇒ ∈
  • 6.
    Use of Derivative ()(i) If ƒ x >0 for all x (a, b) f(x) is increasing on (a,b).′ ∈ ⇔ ( )(ii) If ƒ x <0 for all x (a, b) f(x) is decresing on (a,b).′ ∈ ⇔
  • 7.
    Use of Derivative(Con.) [ ]As tanθ>0 for 0 <θ<90° ( ) dy ƒ x 0 dx ′⇒ = > for all x in (a, b). Y = f(x) T X Y O T' a b Figure 1 P θ
  • 8.
    Use of Derivative(Con.) θ Figure 2 T' X Y T a bP O [ ]As tanθ <0 for 90°<θ<180° ( ) dy ƒ x 0 dx ′⇒ = < for all x in (a, b).
  • 9.
    For the functionf(x) = 2x3 – 8x2 + 10x + 5, find the intervals where (a)f(x) is increasing (b) f(x) is decreasing Example-1
  • 10.
    Solution 3 2ƒ(x)=2x -8x +10x +5 2 ƒ (x)= 6x -16x +10′∴ 2 =2(3x - 8x +5) =2(3x - 5)(x -1) ƒ (x)= 0 2(3x - 5)(x -1)= 0′ ⇒ 5 x = ,1 3 ∴
  • 11.
    Solution Cont. 5 For 1<x < , ƒ (x) is negative 3 ′ 5 For x > ,ƒ (x) is positive 3 ′ For x < 1, is positive.ƒ (x)=3(3x - 5)(x -1)′ ƒ(x)∴ is increasing for x < 1 and 5 x > 3 and it decreases for 5 1< x < 3
  • 12.
  • 13.
    Maximum and Minimum Thepoint a is called the point of maximum of the function f(x). In the figure, y = f(x) has maximum values at Q and S. ( ) ( )ƒ a > ƒ a+δIf and ( ) ( )ƒ a > ƒ a-δ for all small values of δ. The point b is called the point of minimum of the function f(x). In the figure, y = f(x) has minimum values at R and T. ( ) ( )ƒ b < ƒ b+δIf and ( ) ( )ƒ b < ƒ b-δ for all small values of δ. Let ( )y = ƒ x be a function
  • 14.
    Extreme Points The pointsof maximum or minimum of a function are called extreme points. At these points, ( ) ( )ƒ x = 0, if ƒ x exists.′ ′ X Y O ( i) P fincreasing fdecreasing X Y O ( ii) Q fincreasing fdecreasing ( )At P and Q ƒ x does not exit.′
  • 15.
    Critical Points The pointsat which or at which does not exist are called critical points. ( )ƒ x = 0′ ( )ƒ x′ A point of extremum must be one of the critical points, however, there may exist a critical point, which is not a point of extremum.
  • 16.
    Theorem - 1 Letthe function be continuous in some interval containing x0 . ( )y = ƒ x ( )ƒ x >0′ ( )ƒ x <0′(i) If when x < x0 and When x > x0 then f(x) has maximum value at x = x0 ( )ƒ x <0′ ( )ƒ x >0′(ii) If when x < x0 and When x > x0 ,then f(x) has minimum value at x = x0
  • 17.
    Theorem - 2 Ifx0 be a point in the interval in which y = f(x) is defined and if ( ) ( )0 0ƒ x =0 and ƒ x 0′ ′′ ≠ ( ) ( ) ( )if0 0i ƒ x is a maximum ƒ x <0′′ ( ) ( ) ( )if0 0ii ƒ x is a minimum ƒ x >0′′
  • 18.