helpful slide

1. Welcome
To Presentation

2. Application of
Derivatives

3. Session Objectives
 Increasing and Decreasing Functions
 Use of Derivative
 Maximum and Minimum
 Extreme and Critical points
 Theorem 1 and 2
 Greatest and Least Values

4. 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⇒ ∈

5. 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⇒ ∈

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 function f(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. Maximum and Minimum

13. 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

14. 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.′

15. 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.

16. 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

17. 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′′

18. Thank you