Find the intervals in which the function $f(x) = -2x^{3} - 9x^{2} - 12x + 1$ is strictly increasing or strictly decreasing.

(N/A) Given function: $f(x) = -2x^{3} - 9x^{2} - 12x + 1$
First,find the derivative $f'(x)$:
$f'(x) = \frac{d}{dx}(-2x^{3} - 9x^{2} - 12x + 1) = -6x^{2} - 18x - 12$
Factorize $f'(x)$:
$f'(x) = -6(x^{2} + 3x + 2) = -6(x + 1)(x + 2)$
To find the critical points,set $f'(x) = 0$:
$-6(x + 1)(x + 2) = 0 \Rightarrow x = -1, x = -2$
The points $x = -2$ and $x = -1$ divide the real line into three intervals: $(-\infty, -2)$,$(-2, -1)$,and $(-1, \infty)$.
$1$. In the interval $(-\infty, -2)$,let $x = -3$:
$f'(-3) = -6(-3 + 1)(-3 + 2) = -6(-2)(-1) = -12 < 0$.
Thus,$f(x)$ is strictly decreasing in $(-\infty, -2)$.
$2$. In the interval $(-2, -1)$,let $x = -1.5$:
$f'(-1.5) = -6(-1.5 + 1)(-1.5 + 2) = -6(-0.5)(0.5) = 1.5 > 0$.
Thus,$f(x)$ is strictly increasing in $(-2, -1)$.
$3$. In the interval $(-1, \infty)$,let $x = 0$:
$f'(0) = -6(0 + 1)(0 + 2) = -12 < 0$.
Thus,$f(x)$ is strictly decreasing in $(-1, \infty)$.
Conclusion:
$f(x)$ is strictly increasing in $(-2, -1)$ and strictly decreasing in $(-\infty, -2) \cup (-1, \infty)$.