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Question

Find the points at which the function f given by has (i) local maxima (ii) local minima (ii) point of inflexion

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Solution

The given function is,

f( x )= ( x2 ) 4 ( x+1 ) 3

First derivative of f( x )is,

f ( x )= d( ( x2 ) 4 ( x+1 ) 3 ) dx

Use product rule to differentiate the function.

f ( x )=4 ( x2 ) 3 ( x+1 ) 3 +3 ( x+1 ) 2 ( x2 ) 4 = ( x2 ) 3 ( x+1 ) 2 [ 4( x+1 )+3( x2 ) ] = ( x2 ) 3 ( x+1 ) 2 [ 7x2 ]

By substituting the value of f ( x ) zero, we get

f ( x )=0 ( x2 ) 3 ( x+1 ) 2 [ 7x2 ]=0

Solve for roots of x , we get,

x=2,1, 2 7

At x=1,

Sign of f ( x ) does not change for its nearest value so, this is point of inflexion.

At x= 2 7 ,

Sign of f ( x ) changes from positive to negative value for its nearest points so, it is maxima.

At x=2,

Sign of f ( x ) changes from negative to positive value for its nearest points so, it is minima.

Thus, point of local maxima is x= 2 7 , point of local minima is x=2 and point of inflexion is x=1.


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