Question

which of the following statements is true of $f(x)=-x^{3}-6x^{2}-9x - 2$?
$f$ is increasing on $(-\infty,-3)$
$f$ is increasing on $(-3,-1)$
$f$ is decreasing for all real values
$f$ is increasing on $(-2,\infty)$

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=-x^{3}-6x^{2}-9x - 2\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), the derivative \(f^\prime(x)=-3x^{2}-12x - 9\).
Factor out \(-3\): \(f^\prime(x)=-3(x^{2}+4x + 3)=-3(x + 1)(x + 3)\).

Step2: Determine the sign of the derivative

Set \(f^\prime(x)=0\), then \(-3(x + 1)(x + 3)=0\), so \(x=-3\) or \(x=-1\).
We consider the intervals \((-\infty,-3)\), \((-3,-1)\) and \((-1,\infty)\).
Take a test - point in \((-\infty,-3)\), say \(x=-4\). Then \(f^\prime(-4)=-3(-4 + 1)(-4 + 3)=-3\times(-3)\times(-1)=-9<0\).
Take a test - point in \((-3,-1)\), say \(x=-2\). Then \(f^\prime(-2)=-3(-2 + 1)(-2 + 3)=-3\times(-1)\times1 = 3>0\).
Take a test - point in \((-1,\infty)\), say \(x=0\). Then \(f^\prime(0)=-3(0 + 1)(0 + 3)=-9<0\).

Answer:

C. \(f\) is increasing on \((-3,-1)\)