question_answer 101)

Find the points at which f(x) = (x – 2)4 (x + 1)3 has (1) Local maxima (II) Local minima (III) Point of inflexion.  

Answer:

f(x) = (x ? 2)4 (x + 1)3             = (x ? 2)3 (x + 1)2 [3x ? 6 + 4x + 4]       = (x ? 2)3 (x + 1)2 (7x ? 2)             f?(x) = 0       Now f?(x) = (x ? 2)3 [(x + 1)2 × 7 +                   (7x ? 2) × 2 (x + 1) ]       + (x + 1)2 (7x ? 2) × 3 (x ? 2)2       f?(x)|x = 2 = 0 + 0 = 0        order derivative test fails.       Hence we will use 1st order derivative test.       At x =2                      When  x isslightly less than 2, then f?(x) < 0        When x is slightly greater than 2, then f?(x) > 0       Therefore f(x) has local minima at x = 2       At x = ? 1       When x is slightly < ?1 then f?(x) > 0       When x is slightly > ?1, then f?(x) > 0       Therefore f(x) has neither maxima nor minima at x = ?1       is the point of in flexion.       At       When x is slightly <  then f?(x) > 0       When x is slightly >  then f?(x) < 0       Therefore f(x) has local maxima at .