Answer :

It is given that function f(x) = –(x + 1)3 (x – 3)3


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


f’(x) = 3(x + 1)2 (x – 3)2[x – 3 + x + 1]


f’(x) = 6(x + 1)2 (x – 3)2(x-1)


If f’(x) = 0, then we get,


x = -1,3 and 1


So, the points x = -1, x =1 and x = 3 divides the real line into four disjoint intervals,



So, in interval


f’(x) = 6(x + 1)2 (x – 3)2(x-1) < 0


Therefore, the given function (f) is strictly decreasing in intervals .


So, in interval


f’(x) = 6(x + 1)2 (x – 3)2(x-1) > 0


Therefore, the given function (f) is strictly increasing in intervals.


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