Determine the intervals on which function $f (x) = x^3 + 6x^2 + 9x + 2$ is increasing. Determine the intervals on which the function with equation $y = x^3 + 6x^2 + 9x + 2$ is increasing. What is the solution for this question? Help would be greatly appreciated. 
 A: A differentiable function is increasing when its derivative is positive. 
Computing the derivative:
$3x^2+12x+9$
We know that the function is increasing when this is positive. Can you find values of x for which:
$3x^2+12x+9>0$?
A: $$f(x)=x^3+6x^2+9x+2\implies f'(x)=3x^2+12x+9=$$
$$=3(x^2+4x+3)=3(x+3)(x+1)$$
So now you have that
$$f'(x)\ge0\iff (x+3)(x+1)\ge0$$
Observe the above is a quadratic inequality, so if you identify its roots you can solve the inequality, and the problem, easily.
A: When $$y = x^3 + 6x^2 + 9x + 2,$$ the intervals on which the function increases are given by the solutions to $y' > 0$. 
In order to compute $y'$, we'll take the power rule on all of the terms. This provides $$y' = 3x^2 + 12x + 9.$$
To find the solutions to the inequality, $y' > 0$, we'll plug in $y'$ and observe.
This gives $$3x^2 + 12x + 9 > 0 \implies x^2+4x+3 > 0.$$
The solutions can be found from the quadratic formula.
$$x = \frac{-4 \pm \sqrt{16-12}}{2} \implies x = -2 \pm 1 \implies x=-1, \space x=-3$$
To find the total interval on which the derivative is greater than $0$, we need to check the point in-between the two critical points (at $x=-2$).
$$3x^2+12x+9 = -3$$
Thus, it is decreasing at the center of the two, so the interval must be outside of the interval $(-3, -1)$.
The intervals are at $(-\infty, -3)$ and $(-1, \infty)$
