Given f (x) = 2x3 + 9x2 + 12x – 1
Applying the first derivative we ge

Applying the sum rule of differentiation and also the derivative of the constant is 0, so we get

Applying the power rule we get
⇒ f’(x)=6x2+18x+12-0
⇒ f’(x)=6(x2+3x+2)
By splitting the middle term, we get
⇒ f’(x)=6(x2+2x+x+2)
⇒ f’(x)=6(x(x+2)+1(x+2))
⇒ f’(x)=6((x+2) (x+1))
Now f’(x)=0 gives us
x=-1, -2
These points divide the real number line into three intervals
(-∞, -2), [-2,-1] and (-1,∞)
(i) in the interval (-∞, -2), f’(x)>0
∴ f(x) is increasing in (-∞,-2)
(ii) in the interval [-2,-1], f’(x)≤0
∴ f(x) is decreasing in [-2, -1]
(iii) in the interval (-1, ∞), f’(x)>0
∴ f(x) is increasing in (-1, ∞)
Hence the interval on which the function f (x) = 2x3 + 9x2 + 12x – 1 is decreasing is [-2, -1].
So the correct option is option B.