# How do you find the critical numbers of x^4-2x^3-3x^2-5?

Jun 4, 2018

#### Explanation:

Let

$f \left(x\right) = {x}^{4} - 2 {x}^{3} - 3 {x}^{2} - 5$

The first derivative is

$f ' \left(x\right) = 4 {x}^{3} - 6 {x}^{2} - 6 x$

$= 2 x \left(2 {x}^{2} - 3 x - 3\right)$

The critical points are when $f ' \left(x\right) = 0$

$2 x \left(2 {x}^{2} - 3 x - 3\right) = 0$

$\implies$, $\left\{\begin{matrix}x = 0 \\ 2 {x}^{2} - 3 x - 3 = 0\end{matrix}\right.$

$\implies$, $\left\{\begin{matrix}x = 0 \\ x = \frac{3 + \sqrt{33}}{4} = 2.186 \\ x = \frac{3 - \sqrt{33}}{4} = - 0.686\end{matrix}\right.$

The critical points are $\left(0 , - 5\right)$ , $\left(2.186 , - 17.393\right)$, and $\left(- 0.686 , - 5.545\right)$

graph{x^4-2x^3-3x^2-5 [-24.14, 27.18, -19.11, 6.56]}