Q. 64

# Mark (√) against

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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