The interval on which \(f(x)=2x^3+9x^2+12x-1\) is decreasing is

\((-\infty,-2]\) \([1,\infty)\) [-2,-1] none

[-2,-1]

\(f^{'}(x)=6x^2+18x+12=3(x^2+3x+2) = 3(x^2 + 2x+x+2) = 3(x+2)(x+1)<0\) $\Rightarrow -2<x<-1 $