Question

multiple choice 2 points
which of the following statements is true of ( f(x)=-x^{3}-6 x^{2}-9 x-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 \(f(x)\)

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 \(f^\prime(x)\)

Set \(f^\prime(x)>0\) (for increasing intervals) and \(f^\prime(x)<0\) (for decreasing intervals).

• For \(f^\prime(x)>0\): \(-3(x + 1)(x + 3)>0\). Divide both sides by \(-3\) (inequality sign flips), we get \((x + 1)(x + 3)<0\). The solution of \((x + 1)(x + 3)<0\) is \(x\in(-3,-1)\).
• For \(f^\prime(x)<0\): \(-3(x + 1)(x + 3)<0\). Divide both sides by \(-3\) (inequality sign flips), we get \((x + 1)(x + 3)>0\). The solutions are \(x\in(-\infty,-3)\cup(-1,\infty)\).

Answer:

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