Answer :

Given:


f(x) = x3+6x2+15x-12.


f’(x) = 3x2+12x+15


f’(x) = 3x2+12x+12+3


f’(x) = 3(x2+4x+4)+3


f’(x) = 3(x+2)2+3


As square is a positive number


f’(x) will be always positive for every real number


Hence f’(x) >0 for all x ϵ R


f(x) is strictly increasing.

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