Application of derivatives 2 maxima and minima

Session
Applications of Derivatives - 2

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

Increasing Function
I n c r e a s i n g
f u n c t i o n
a x 1 x 2 b
X
Y
f ( x 2 )
f ( x 1 )
O

Increasing Function
A function is said to be a strictly
increasing function of x on (a, b).
If x1 < x2 in ( a, b) Þ ƒ ( x1 ) < ƒ ( x2 ) for all x1, x2 Î ( a, b)
‘Strictly increasing’ is also referred to as
‘Monotonically increasing’.

Decreasing Function
D e c r e a s i n g
f u n c t i o n
a x 1 x 2 b
X
Y
f ( x 2 )
f ( x 1 )
O

Decreasing Function
A function ƒ(x) is said to be a strictly
decreasing function of x on (a, b).
If x1 < x2 in ( a, b) Þ ƒ ( x1 ) > ƒ ( x2 ) for all x1, x2 Î ( a, b)
‘Strictly decreasing’ is also referred to
as ‘Monotonically decreasing’.

Use of Derivative
Let f(x) be a differentiable real function
defined on an open interval (a, b).
(i) If ƒ¢( x) >0 for all x Î(a, b)Û f(x) is increasing on (a,b).
(ii) If ƒ¢( x) <0 for all x Î(a, b)Û f(x) is decresing on (a,b).

Use of Derivative (Con.)
Y = f(x) T
X
Y
P
O T' a b
Figure 1
q
Slope of tangent at any point in (a, b) > 0
[ As tanθ>0 for 0<θ<90°]
ƒ ( x) dy 0
Þ ¢ = > for all x in (a, b).
dx

Use of Derivative (Con.)
q
Figure 2 T'
X
Y
T a
P b
O
Slope of tangent at any point in (a, b) < 0
[ As tanθ<0 for 90°<θ<180°]
ƒ ( x) dy 0
Þ ¢ = < for all x in (a, b).
dx

Example-1
For the function f(x) = 2x3 – 8x2 + 10x + 5,
find the intervals where
(a)f(x) is increasing
(b) f(x) is decreasing

Solution
We have
ƒ(x)=2x3 - 8x2 +10x +5
ƒ¢(x)=6x2 -16x+10
=2(3x2 - 8x+5)
=2(3x - 5) (x -1)
ƒ¢(x)=0Þ2(3x - 5) (x -1)=0
5
x= ,1
3

Solution Cont.
For x < 1, is ƒ¢(x)=3(3x - 5) (x -1) positive.
5
For 1<x< , ƒ (x) is negative
3
¢
5
For x> ,ƒ (x) is positive
3
¢
ƒ(x) is increasing for x < 1 and
5
x>
3
and it decreases for 5
1<x<
3

Example-2
Find the intervals in which the function
in increases or decreases.
ƒ(x)= x + cosx
[0, 2p]
Solution: We have ƒ(x)= x + cosx
ƒ¢(x)=1 - sinx
As sinx is £ 1 for all x Î[0, 2p]
And sinx =1 for x =
p
2
¢ p
ƒ ( x) >0 for all x except x =
2
Þ ¢ p .
ƒ ( x) is increasing for all x except x =
2

Maximum and Minimum
Let y= ƒ( x) be a function
Ifƒ( a) > ƒ( a+δ ) and ƒ( a) > ƒ( a-δ ) for all small values of δ.
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.
If ƒ(b) < ƒ(b+δ ) and ƒ(b) <ƒ(b-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.

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
f i n c r e a s i n g
f d e c r e a s i n g
X
Y
O
( ii)
Q
f i n c r e a s i n g
f d e c r e a s i n g
At P and Q ƒ¢(x) does not exit.

Critical Points
ƒ¢( x) =0 ƒ¢( x)
The points at which or at which
does not exist are called critical points.
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
y= ƒ( x)
Let the function be continuous in some
interval containing x0 .
(i) If ƒ ¢ ( x ) > 0 when x < x ƒ¢( x) <0 0 and When
x > x0 then f(x) has maximum value at x = x0
(ii) If ƒ ¢ ( x ) < 0 when x < x ƒ¢( x) >0 0 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 ƒ¢( x0 ) =0 and ƒ¢¢( x0 ) ¹ 0
(i) ƒ( x0 ) is amaximumif ƒ¢¢( x0 ) <0
(ii) ƒ( x0 ) is aminimum if ƒ¢¢( x0 ) >0

Greatest and Least Values
The greatest or least value of a continuous function
f(x) in an interval [a, b] is attained either at the
critical points or at the end points of the interval.
So, obtain the values of f(x) at these points and
compare them to determine the greatest and the
least value in the interval [a, b].

Example-3
Find all the points of maxima and minima
and the corresponding maximum and
minimum values of the function:
f ( x ) 3 45 = - x 4 - 8x 3 - x 2
+105
4 2
(CBSE 1993)

Solution
We have
( ) 3 4 3 45 2
f x = - x - 8x - x +105
4 2
f' ( x) = -3x3 - 24x2 - 45x
Þ f' ( x) = -3x (x2 +8x+15)
For maximum or minimum f’(x) = 0
-3x (x2 +8x+15) = 0
Þ -3x ( x+3) ( x+5) = 0
Þ x = 0, - 3, - 5

Solution Cont.
f'' ( x) = -9x2 - 48x - 45
At x = 0, f'' (0) = -45 < 0
f(x) is maximum at x = 0
The maximum value at x = 0 is f(0) = 105
( ) ( )2 ( ) At x = -3, f'' -3 = -9 -3 - 48 -3 - 45 =18 > 0
f(x) is minimum at x = -3
The minimum value at x = -3 is
( ) 3 ( )4 ( )3 45 ( )2 231
f -3 = - -3 - 8 -3 - -3 +105 =
4 2 4

Solution Cont.
( ) ( )2 ( ) f'' -5 = -9 -5 - 48 -5 - 45 = -30 < 0
At x = -5,
f(x) is maximum at x = -5
The maximum value at x = -5 is
( ) 3 ( )4 ( )3 45 ( )2 295
f -5 = - -5 - 8 -5 - -5 +105 =
4 2 4

Example-4
Show that the total surface area of a cuboid with a square base
and given volume is minimum, when it is a cube.
Solution: Let the cuboid has a square base of edge x and height y.
The volume of cuboid, V = x2y
The surface area of cuboid, S =2( x×x+x×y+x×y)
=2x2 + 4xy
2
V
2
=2x + 4x.
x

Con.
2 2V
S=2 é x +
ù êë x
úû
dS
For minimum surface area, =0
dx
é 2V
ù
Þ ê ú
2 2x - =0
2
ë x
û
Þ x3 - V =0 Þ x =3 V
2
2 3
é ù
ê ú
ë û
d S 4V
=2 2+
dx x

Con.
é 2V
ù
ê ú
ë 3
û
= 4 1+
x
æ ö é ù çç ¸¸ ê ú è ø ë û
x =3 V
2
2
d S 2V
= 4 1+ = 4×3=12
dx V
2
3
d V
As >0 at x = V
2
dx
At x=3 V , surface area is minimum.

Con.
x = 3 V Þ V = x3
Þ x2y = x3 Þ y = x
Þ Cuboid is a cube.

Application of derivatives 2 maxima and minima

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