# How do you find the Critical Points for x^3 + 3x^2-24x?

Sep 11, 2015

Critical points are 2, -4

#### Explanation:

critical point 'c' of a function is where either f'(c)=0 or f'(c) does not exist.

To find critical point f'(x)= $3 {x}^{2} + 6 x - 24 = 0$

That gives ${x}^{2} + 2 x - 8 = 0$

$\left(x + 4\right) \left(x - 2\right) = 0$

Critical points are 2, -4