Find the open intervals where $f(z)=z^{3}+6 z^{2}+z+1$ is increasing and Where it is decreasing.
Enter your answer in interval notation. For example, if the function is increasing when $-10< z< 2$ as well as when $4< z< 7$, then you would enter $(-10,2) \cup(4,7)$.
\[
f(z)=z^{3}+6 z^{2}+z+1
\]
Answer
Determine increasing/decreasing intervals: \(\boxed{(-\infty, -2 - \frac{\sqrt{33}}{3}) \cup (-2 + \frac{\sqrt{33}}{3}, \infty)}\) for increasing, \(\boxed{(-2 - \frac{\sqrt{33}}{3}, -2 + \frac{\sqrt{33}}{3})}\) for decreasing
Step 1 ：Find the first derivative: \(f'(z) = 3z^2 + 12z + 1\)
Step 2 ：Find the critical points: \(z = -2 - \frac{\sqrt{33}}{3}, -2 + \frac{\sqrt{33}}{3}\)
Step 3 ：Analyze the intervals: \((-\infty, -2 - \frac{\sqrt{33}}{3})\), \((-2 - \frac{\sqrt{33}}{3}, -2 + \frac{\sqrt{33}}{3})\), \((-2 + \frac{\sqrt{33}}{3}, \infty)\)
Step 4 ：Determine increasing/decreasing intervals: \(\boxed{(-\infty, -2 - \frac{\sqrt{33}}{3}) \cup (-2 + \frac{\sqrt{33}}{3}, \infty)}\) for increasing, \(\boxed{(-2 - \frac{\sqrt{33}}{3}, -2 + \frac{\sqrt{33}}{3})}\) for decreasing
