Question

given the function ( f(x)=-x^{3}-18x^{2}-81x ), determine all intervals on which ( f ) is increasing.

Explanation:

Step1: Find the derivative of the function

Use the power rule \( (x^n)^\prime=nx^{n - 1}\).
For \(y = f(x)=-x^{3}-18x^{2}-81x\), the derivative \(f^\prime(x)=-3x^{2}-36x - 81\).
Factor out \(-3\): \(f^\prime(x)=-3(x^{2}+12x + 27)\).
Factor the quadratic expression: \(x^{2}+12x + 27=(x + 3)(x+9)\), so \(f^\prime(x)=-3(x + 3)(x + 9)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\).
\(-3(x + 3)(x + 9)=0\).
Since \(-3
eq0\), then \(x+3 = 0\) or \(x + 9=0\).
The critical points are \(x=-9\) and \(x=-3\).

Step3: Test the intervals

We have three intervals to test: \((-\infty,-9)\), \((-9,-3)\) and \((-3,\infty)\).

• For the interval \((-\infty,-9)\), let \(x=-10\). Then \(f^\prime(-10)=-3(-10 + 3)(-10 + 9)=-3\times(-7)\times(-1)=-21<0\).
• For the interval \((-9,-3)\), let \(x=-6\). Then \(f^\prime(-6)=-3(-6 + 3)(-6 + 9)=-3\times(-3)\times3 = 27>0\).
• For the interval \((-3,\infty)\), let \(x=0\). Then \(f^\prime(0)=-3(0 + 3)(0 + 9)=-81<0\).

Answer:

The function \(f(x)\) is increasing on the interval \((-9,-3)\).