JEE Mathematics/ Lakshmikanta Satapathy/ Application of Derivative QA part 5/ Question on Absolute maximum and Minimum values of a function in a closed interval solved with the related concepts

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
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