Answer :

We have, f (x) = x3 – 3x2 + 3x – 100

=> f’(x) = 3x2 -6x + 3


= 3(x2 -2x + 1)


= 3(x-1)2


For any x ϵ R, (x -1)2 > 0


Thus, f’(x) is always positive in R.


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


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