# How do you find the critical points for g(x)=2x^3-3x^2-12x+5?

Nov 22, 2016

Find the first derivative and set the equation to equal zero.

#### Explanation:

$g ' \left(x\right) = 6 {x}^{2} - 6 x - 12$
$= 6 \left({x}^{2} - x - 2\right) = 6 \left(x - 2\right) \left(x + 1\right)$

$6 \left(x - 2\right) \left(x + 1\right) = 0$

Critical Values:
$x = 2 , x = - 1$