The document discusses finding the maximum and minimum values of the function f(x) = 4x^3 - 8x^2 + 12x - 48 over the interval [0, 3]. It details the process of determining critical points by differentiating the function and applying tests for local maxima and minima, concluding with the calculations for absolute maximum and minimum values. The critical point is identified at x = 2, leading to the final results of absolute maximum 25 and absolute minimum -3.

1. Physics Helpline
L K Satapathy
Maxima and Minima 1

2. Physics Helpline
L K Satapathy
Application of Derivative 5
Maxima and Minima
D
B
A C
E
F
maxf
minf
1c 2ca b
Maximum at B 1( )x c Minimum at E 2( )x c
Tangents at B and E are parallel to x-axis
 f (x) = 0 at B and E [ critical points ]
f changes from +ve  –ve across B  f  is –ve at B (max at B)
f changes from –ve  +ve across E  f  is +ve at E (min at E)
 is a Local maximum And is a Local minimum
In [ a , b ] , the maximum of = absolute maximum1 2( ) , ( ) , ( ) , ( )f a f c f c f b
and the minimum of = absolute minimum
1x c 2x c
1 2( ) , ( ) , ( ) , ( )f a f c f c f b
In the figure , the value of f(x) is

3. Physics Helpline
L K Satapathy
Application of Derivative 5
Maxima and Minima
Question : Find the maximum and minimum values of the function
in the interval [ 0 , 3 ] .4 3 2
( ) 3 8 12 48 25f x x x x x    
Answer :
4 3 2
( ) 3 8 12 48 25f x x x x x    
2
12( 2)( 2)x x  
2
( ) 36 48 24 . . . (1)f x x x   
The given function
Differentiating , we get 3 2 3 2
( ) 12 24 24 48 12( 2 2 4)f x x x x x x x        
Again differentiating , we get
We need tot find the absolute maximum and minimum values of f (x) in [0 , 3]

4. Physics Helpline
L K Satapathy
Application of Derivative 5
Maxima and Minima
2 2
2 0 2 ( )x x no soln     
(1) (2) 36 4 48 2 24Now f      
To get the critical points , we solve the equation f (x) = 0
2
( ) 12( 2)( 2) 0f x x x    
2 0 2or x x   
144 96 24 72   
We observe that is a local minimum(2) 0 2f x   

5. Physics Helpline
L K Satapathy
Application of Derivative 5
Maxima and Minima
We have One critical point at x = 2 and the closed interval is [ 0 , 3 ]
 We need to find f (0) , f (2) and f (3)
4 3 2
( ) 3 8 12 48 25f x x x x x    
(0) 25f 
(2) 3 16 8 8 12 4 48 2 25 48 64 48 96 25 39f                
(3) 3 81 8 27 12 9 48 3 25 243 216 108 144 25 16f               
max 25 [ ]f Ans
min 3 [ ]9f Ans 
 Absolute maximum
And absolute minimum

6. Physics Helpline
L K Satapathy
For More details:
www.physics-helpline.com
Subscribe our channel:
youtube.com/physics-helpline
Follow us on Facebook and Twitter:
facebook.com/physics-helpline
twitter.com/physics-helpline