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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