Question

determine the open intervals on which the function is increasing, decreasing, or constant. (enter your answers using interval notation. if an answer does not exist, enter dne.)
$f(x)=x^{3}-9x^{2}+1$

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=x^{3}-9x^{2}+1\) is \(f^{\prime}(x) = 3x^{2}-18x=3x(x - 6)\)

Step2: Find the critical points

Set \(f^{\prime}(x)=0\), then \(3x(x - 6)=0\). Solving for \(x\), we get \(x = 0\) and \(x=6\)

Step3: Test the intervals

• For the interval \((-\infty,0)\), let's choose \(x=-1\). Then \(f^{\prime}(-1)=3\times(-1)\times(-1 - 6)=21>0\), so the function is increasing on \((-\infty,0)\)
• For the interval \((0,6)\), let's choose \(x = 1\). Then \(f^{\prime}(1)=3\times1\times(1 - 6)=-15<0\), so the function is decreasing on \((0,6)\)
• For the interval \((6,\infty)\), let's choose \(x = 7\). Then \(f^{\prime}(7)=3\times7\times(7 - 6)=21>0\), so the function is increasing on \((6,\infty)\)

Answer:

increasing: \((-\infty,0)\cup(6,\infty)\)
decreasing: \((0,6)\)
constant: NONE