Given the function f(x)=-x^3-6x^2-9x determine all intervals on which f' is decreasing.

To determine the intervals on which $f'$ is decreasing, we first need to find the derivative of $f(x)$, which is $f'(x)$. Then, we can find the critical points by setting $f'(x) = 0$ and solving for $x$. Finally, we can use these critical points to determine the intervals on which $f'$ is decreasing.<br /><br />Step 1: Find the derivative of $f(x)$.<br />$f'(x) = -3x^2 - 12x - 9$<br /><br />Step 2: Find the critical points by setting $f'(x) = 0$ and solving for $x$.<br />$-3x^2 - 12x - 9 = 0$<br />$-3(x^2 + 4x + 3) = 0$<br />$-3(x + 3)(x + 1) = 0$<br />$x = -3$ or $x = -1$<br /><br />Step 3: Determine the intervals on which $f'$ is decreasing.<br />We can use the critical points to divide the number line into three intervals: $(-\infty, -3)$, $(-3, -1)$, and $(-1, \infty)$. We can then test a point from each interval in $f'(x)$ to determine whether $f'$ is increasing or decreasing in that interval.<br /><br />For $x \in (-\infty, -3)$, let's choose $x = -4$:<br />$f'(-4) = -3(-4)^2 - 12(-4) - 9 = -48 + 48 - 9 = -9 < 0$<br /><br />For $x \in (-3, -1)$, let's choose $x = -2$:<br />$f'(-2) = -3(-2)^2 - 12(-2) - 9 = -12 + 24 - 9 = 3 > 0$<br /><br />For $x \in (-1, \infty)$, let's choose $x = 0$:<br />$f'(0) = -3(0)^2 - 12(0) - 9 = -9 < 0$<br /><br />Therefore, $f'$ is decreasing on the intervals $(-\infty, -3)$ and $(-1, \infty)$.