Question

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

Open in App
Solution

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.

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Extrema
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program