# How do you find the critical points for f'(x)=11+30x+18x^2+2x^3?

Jul 27, 2015

The critical numbers for $11 + 30 x + 18 {x}^{2} + 2 {x}^{3}$ are:
$- 5$ and $- 1$.

#### Explanation:

I'm not sure why you called the function $f ' \left(x\right)$. Since we need to take the derivative to find the critical numbers, I'll refer to the function as $g \left(x\right)$.

The critical numbers for a function $g$ are the numbers in the domain of $g$ at which the derivative is either $0$ or does not exist. Some people use "critical point" to mean the same thing, others use it to mean a point on the graph (so it has 2 coordinates).

$g \left(x\right) = 11 + 30 x + 18 {x}^{2} + 2 {x}^{3}$

$g ' \left(x\right) = 30 + 36 x + 6 {x}^{2}$

This is never undefined, so we need only find the zeros:

$6 {x}^{2} + 36 x + 30 = 0$

$6 \left({x}^{2} + 6 x + 5\right) = 0$

$6 \left(x + 5\right) \left(x + 1\right) = 0$

$x = - 5$ or $x = - 1$

The critical numbers for the function: $11 + 30 x + 18 {x}^{2} + 2 {x}^{3}$ are $- 5$ and $- 1$.

If you wish to find the $y$ values, you can do so.

At $x = - 5$, we get:

$11 + 30 \left(- 5\right) + 18 {\left(- 5\right)}^{2} + 2 {\left(- 5\right)}^{3}$
(I prefer to do arithmetic with smaller numbers. Using $30 = 6 \cdot 5$, we can get some multiples of $25$. So I'll regroup and use the distributive property.)

$11 - 6 \left(25\right) + 18 \left(25\right) - 10 \left(25\right) = 11 + 18 \left(25\right) - 16 \left(25\right)$

$= 11 + \left[18 - 16\right] \left(25\right) = 11 + 2 \left(25\right)$

$= 11 + 50 = 61$

At $x = - 1$, we get:

$11 - 30 + 18 - 2 = 29 - 32 = - 3$

So, if you use "critical points" to mean points on the graph (rather than points in the domain), then they will be:

$\left(- 5 , 61\right)$ and $\left(- 1 , - 3\right)$