# How do you find the critical points of b'(x)=x^3+3x^2-4x-12?

Jun 10, 2017

$x = - 3 \text{ and } x = \pm 2$

#### Explanation:

$\text{find the critical points by equating " b'(x)" to zero}$

$\Rightarrow {x}^{3} + 3 {x}^{2} - 4 x - 12 = 0$

$\Rightarrow {x}^{2} \left(x + 3\right) - 4 \left(x + 3\right) = 0$

$\Rightarrow \left(x + 3\right) \left({x}^{2} - 4\right) = 0$

$x + 3 = 0 \Rightarrow x = - 3 \leftarrow \text{ is a critical point}$

$\left(x - 2\right) \left(x + 2\right) = 0 \Rightarrow x = \pm 2 \leftarrow \text{ are critical points}$