Question

question
given the function $f(x)=-x^{4}+12x^{3}-36x^{2}$, determine all intervals on which $f$ is decreasing.
answer attempt 1 out of 2

Explanation:

Step1: Find the second - derivative

First, we have \(f^{\prime}(x)=-x^{4}+12x^{3}-36x^{2}\).
Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the second - derivative \(f^{\prime\prime}(x)\) is:
\(f^{\prime\prime}(x)=-4x^{3}+36x^{2}-72x=-4x(x^{2}-9x + 18)\).
Factor \(x^{2}-9x + 18=(x - 3)(x - 6)\), so \(f^{\prime\prime}(x)=-4x(x - 3)(x - 6)\).

Step2: Find the critical points of \(f^{\prime\prime}(x)\)

Set \(f^{\prime\prime}(x)=0\), then \(-4x(x - 3)(x - 6)=0\).
The critical points are \(x = 0\), \(x=3\), and \(x = 6\).

Step3: Test the intervals

We consider the intervals \((-\infty,0)\), \((0,3)\), \((3,6)\), and \((6,\infty)\).

• For \(x\in(-\infty,0)\), let \(x=-1\). Then \(f^{\prime\prime}(-1)=-4\times(-1)\times(-1 - 3)\times(-1 - 6)=-4\times(-1)\times(-4)\times(-7)=112>0\).
• For \(x\in(0,3)\), let \(x = 1\). Then \(f^{\prime\prime}(1)=-4\times1\times(1 - 3)\times(1 - 6)=-4\times1\times(-2)\times(-5)=-40<0\).
• For \(x\in(3,6)\), let \(x = 4\). Then \(f^{\prime\prime}(4)=-4\times4\times(4 - 3)\times(4 - 6)=-4\times4\times1\times(-2)=32>0\).
• For \(x\in(6,\infty)\), let \(x = 7\). Then \(f^{\prime\prime}(7)=-4\times7\times(7 - 3)\times(7 - 6)=-4\times7\times4\times1=-112<0\).

Since \(f^{\prime}(x)\) is decreasing when \(f^{\prime\prime}(x)<0\).

Answer:

The function \(f^{\prime}(x)\) is decreasing on the intervals \((0,3)\) and \((6,\infty)\).