Answer :

Given:


f(x) = (x+1)3.(x-3)3


f’(x) = 3(x+1)2(x-3)3 + 3(x-3)3 (x+1)3


Put f’(x) = 0


3(x+1)2(x-3)3 = -3(x-3)2(x+1)3


x-3 = -(x+1)


2x = 2


x =1


When x>1 the function is increasing.


x<1 function is decreasing.


So, f(x) is increasing in (1, ∞).

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