# How do you find the critical numbers for x^4-2x^3-3x^2-5 to determine the maximum and minimum?

Apr 20, 2016

Please see the explanation section below,

#### Explanation:

$f \left(x\right) = {x}^{4} - 2 {x}^{3} - 3 {x}^{2} _ 5$ is a polynomial, its domain is $\left(- \infty , \infty\right)$

$f ' \left(x\right) = 4 {x}^{3} - 6 {x}^{2} - 6 x$ is never undefined and is $0$ at the solutions to

$4 {x}^{3} - 6 {x}^{2} - 6 x = 0$

Factor out the common factor of $2 x$ to get

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

This leads to two lower degree polynomial equations:

$2 x = 0$ $\text{ }$ or $\text{ }$ $2 {x}^{2} - 3 x - 3 = 0$

The first has solution $0$ and the solutions to the second may be found using the quadratic formula

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

We get $x = \frac{3 \pm \sqrt{33}}{4}$.

$f ' \left(x\right) = 0$ at

$0$, $\text{ }$ $\frac{3 + \sqrt{33}}{4}$,$\text{ }$ and $\text{ }$ $\frac{3 - \sqrt{33}}{4}$

All of these solutions are in the domain of $f$, so they are all critical numbers for $f$.