The interval on which the function f (x) = 2x³ + 9x² + 12x – 1 is decreasing is

The correct option is [-2, -1].

To find the interval on which the function $f\left(x\right)=2x^3+9x^2+12x-1$ is decreasing, we need to determine where its first derivative, $f^{\text{'}}\left(x\right)$, is less than or equal to zero (i.e., $f^{\text{'}}\left(x\right)\le 0$).

Step 1: Find the first derivative of the function.
Using the power rule of differentiation, we differentiate each term of the function $f\left(x\right)$ with respect to $x$:
$f^{\text{'}}\left(x\right)=\frac{d}{dx}(2x^3+9x^2+12x-1)$
$f^{\text{'}}\left(x\right)=6x^2+18x+12$

Step 2: Set up the inequality for a decreasing function.
A function is decreasing on an interval where its derivative is non-positive:
$f^{\text{'}}\left(x\right)\le 0$
Substituting the expression for the derivative:
$6x^2+18x+12\le 0$

Step 3: Solve the quadratic inequality.
First, we can simplify the inequality by dividing the entire expression by 6:
$x^2+3x+2\le 0$
Next, we factor the quadratic polynomial. We search for two numbers that multiply to 2 and add to 3. These numbers are 1 and 2:
$(x+1)(x+2)\le 0$

Step 4: Determine the interval.
The product of two terms is less than or equal to zero when one factor is non-negative and the other is non-positive. The critical points where the expression equals zero are $x=-2$ and $x=-1$.
We can test the sign of the product $(x+1)(x+2)$ in the intervals defined by these critical points:
1. For $x<-2$, both factors are negative, so the product is positive.
2. For $-2\le x\le -1$, the factor $(x+2)$ is non-negative and $(x+1)$ is non-positive, so the product is less than or equal to zero.
3. For $x>-1$, both factors are positive, so the product is positive.

Therefore, the inequality holds true for the closed interval:
$x\in [-2,-1]$

Thus, the function is decreasing on the interval [-2, -1].