Mathematics

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

Increasingfunction

a x1 bx2

X

Y

f(x1)

f(x2)

O

Increasing Function

A function is said to be a strictly increasing function of x on (a, b).

1 2 1 2 1 2I f x < x in a, b ƒ x < ƒ x for all x , x a, b

‘Strictly increasing’ is also referred to as ‘Monotonically increasing’.

Decreasing Function

Decreasingfunction

a x1 bx2X

Y

f(x1)

f(x 2)

O

Decreasing Function

A function ƒ(x) is said to be a strictly decreasing function of x on (a, b).

1 2 1 2 1 2I f x < x in a, b ƒ x > ƒ x for all x , x 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) I f ƒ x >0 for all x (a, b) f(x) is increasing on (a,b).

(ii) I f ƒ x <0 for all x (a, b) f(x) is decresing on (a,b).

Use of Derivative (Con.)

Y = f(x) T

X

Y

O T' a bFigure 1

P

Slope of tangent at any point in (a, b) > 0

As tanθ >0 for 0<θ <90°

dyƒ x 0

dx for all x in (a, b).

Use of Derivative (Con.)

Figure 2 T'X

YT a

bP

O

Slope of tangent at any point in (a, b) < 0

As tanθ <0 for 90° <θ <180°

dyƒ x 0

dx for all x in (a, b).

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

3 2ƒ (x) =2x - 8x +10x +5

2ƒ (x) =6x - 16x +10

2=2(3x - 8x +5)

=2(3x - 5) (x - 1)

ƒ (x) =0 2(3x - 5) (x - 1) =0

5x = , 1

3

Solution Cont.

5For 1<x < , ƒ (x) is negative

3

5For 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 and5

x >3

and it decreases for 51<x <

3

Example-2

Find the intervals in which the function in increases or decreases.

ƒ (x) = x + cosx[0, 2 ]

Solution: We have ƒ (x) = x + cosx

ƒ (x) =1 - sinx

As sinx is 1 for all x 0, 2

And sinx =1 for x =2

ƒ x > 0 for all x except x =2

ƒ x is increasing for all x except x =2

.

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

X

Y

O(ii)

Q

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

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:

4 3 23 45f x = - x - 8x - x +105

4 2(CBSE 1993)

Solution

4 3 23 45f x = - x - 8x - x +105

4 2

3 2f' x = -3x - 24x - 45x

We have

2f' x = -3x x +8x +15

For maximum or minimum f’(x) = 0

2-3x x +8x +15 = 0

-3x x +3 x +5 = 0

x = 0, - 3, - 5

Solution Cont.

2f'' x = -9x - 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

f(x) is minimum at x = -3

The minimum value at x = -3 is

2f'' -3 = -9 -3 - 48 -3 - 45 =18 > 0At x = -3,

4 3 23 45 231f -3 = - -3 - 8 -3 - -3 +105 =

4 2 4

Solution Cont.

The maximum value at x = -5 is

2f'' -5 = -9 -5 - 48 -5 - 45 = -30 < 0

f(x) is maximum at x = -5

At x = -5,

4 3 23 45 295f -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.

2The volume of cuboid, V = x y

The surface area of cuboid, S = 2 x×x +x×y +x×y

2= 2x + 4xy

22

V= 2x + 4x.

x

Con.

2 2VS = 2 x +

x

dSFor minimum surface area, = 0

dx

2

2V2 2x - = 0

x

3 3x - V = 0 x = V

2

2 3

d S 4V= 2 2 +

dx x

Con.

3

2V= 4 1 +

x

3x = V

2

2

d S 2V= 4 1 + = 4×3 =12

Vdx

23

2

d VAs > 0 at x = V

dx

3x = V , surface area is minimum. At

Con.

33x = V V = x

2 3x y = x y = x

Cuboid is a cube.

Thank you