#### The interval on which the function $f (x) = 2x^3 + 9x^2 + 12x - 1$ is decreasing is: A. [-1, ∞) B. [-2, -1] C. (-∞, -2] D. [-1, 1]

Given $f (x) = 2x^3 + 9x^2 + 12x - 1$

Apply first derivative and get

\begin{aligned} &f(x)=\frac{d\left(2 x^{3}+9 x^{2}+12 x-1\right)}{d x}\\ &\text { Apply the sum rule of the differentiation and } 0 \text { is the derivative of the constant, so }\\ &\Rightarrow f(x)=\frac{d\left(2 x^{3}\right)}{d x}+\frac{d\left(9 x^{2}\right)}{d x}+\frac{d(12 x)}{d x}-\frac{d(1)}{d x} \end{aligned}

Apply power rule and get

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

Split the middle term and get

$\\ { \Rightarrow f'(x)=6(x\textsuperscript{2}+2x+x+2)}\\ { \Rightarrow f'(x)=6(x(x+2)+1(x+2))}\\ { \Rightarrow f'(x)=6((x+2) (x+1))}\\$

Now f'(x)=0 gives

x=-1, -2

Three intervals are made when these points divide the real number line

$(i) \text{In the interval } (-\infty, -2), f'(x) > 0 \\ \therefore \text{ } f(x) \text{ is increasing in } (-\infty, -2) (ii) \text{In the interval } [-2, -1], f'(x) \leq 0 \\ \therefore \text{ } f(x) \text{ is decreasing in } [-2, -1] (iii) \text{In the interval } (-1, \infty), f'(x) > 0 \\ \therefore \text{ } f(x) \text{ is increasing in } (-1, \infty)$

So, the interval on which the function decreases is [-2, -1].

So, the correct answer is option B.