Question

given the function ( f(x)=-x^{4}+16x^{3} ), determine all intervals on which ( f^{prime} ) is decreasing

Explanation:

Step1: Find the first - derivative of \(f(x)\)

Using the power rule \((x^n)^\prime=nx^{n - 1}\), if \(f(x)=-x^{4}+16x^{3}\), then \(f^\prime(x)=-4x^{3}+48x^{2}\).

Step2: Find the second - derivative of \(f(x)\)

Differentiate \(f^\prime(x)\) with respect to \(x\). Using the power rule again, \(f^{\prime\prime}(x)=-12x^{2}+96x=-12x(x - 8)\).

Step3: Find where \(f^{\prime\prime}(x)<0\)

Set \(f^{\prime\prime}(x)<0\), so \(-12x(x - 8)<0\).
First, find the roots of the equation \(-12x(x - 8)=0\). The roots are \(x = 0\) and \(x = 8\).
We can test the intervals \((-\infty,0)\), \((0,8)\), and \((8,\infty)\) using test points.

• For the interval \((-\infty,0)\), let \(x=-1\). Then \(f^{\prime\prime}(-1)=-12\times(-1)\times(-1 - 8)=-108<0\).
• For the interval \((0,8)\), let \(x = 1\). Then \(f^{\prime\prime}(1)=-12\times1\times(1 - 8)=84>0\).
• For the interval \((8,\infty)\), let \(x = 9\). Then \(f^{\prime\prime}(9)=-12\times9\times(9 - 8)=-108<0\).

Answer:

The function \(f^\prime(x)\) is decreasing on the intervals \((-\infty,0)\cup(8,\infty)\).