Question

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

Answer 1

The given function is,

f( x )= ( x−2 ) 4 ( x+1 ) 3

First derivative of f( x )is,

f ′ ( x )= d( ( x−2 ) 4 ( x+1 ) 3 ) dx

Use product rule to differentiate the function.

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

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

f ′ ( x )=0 ( x−2 ) 3 ( x+1 ) 2 [ 7x−2 ]=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.